PKj„ìJÕ$:!}[}[;A TEXTBOOK OF ELECTRICAL TECHNOLOGY VOL III/chapter47.ipynb{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# CHAPTER 47 : ELECTRIC HEATING\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 47.1 , PAGE NO :- 1841"
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Length of wire = 16.07 m.\n",
"Diameter of wire = 2.72 mm.\n"
]
}
],
"source": [
"'''A resistance oven employing nichrome wire is to be operated from 220 V single-phase supply and is to be rated at 16 kW.\n",
"If the temperature of the element is to be limited to 1,170Â°C and average temperature of the charge is 500Â°C, find the\n",
"diameter and length of the element wire.\n",
"Radiating efficiency = 0.57, Emmissivity=0.9, Specific resistance of nichrome=(109eâ€“8)ohm-m.'''\n",
"\n",
"\n",
"P = 16000.0 #W (output power)\n",
"V = 220.0 #V (applied voltage)\n",
"rho = 109.0e-8 #ohm-m (resistivity)\n",
"e = 0.9 # (Emmisivity)\n",
"K = 0.57 # (Radiating efficiency)\n",
"T1 = 1170.0 + 273.0 #K (Temp of hot body)\n",
"T2 = 500.0 + 273.0 #K (Temp of cold body)\n",
"\n",
"#Now , l/d^2 = pi*V^2/4*rho*P = a .Therefore a is\n",
"a = 3.14*(V**2)/(4*rho*P) # (a = l/d^2) ------ 1\n",
" \n",
"#Using Stefan's law of radiation \n",
"H = 5.72*e*K*((T1/100)**4-(T2/100)**4) #W/m^2\n",
"\n",
"#Total heat dissipated = electrical power input\n",
"# (pi*d)*l*H = P . Therefore ,let b = l*d. So,\n",
"b = P/(H*3.14)\n",
"b2 = b**2 #(b2 = l^2*d^2)------------------------ 2\n",
"\n",
"#Multiplying 1 and 2.\n",
"l3 = a*b2 # ( = l^3)\n",
"l = l3**(1/3.0) #m (length)\n",
"d = b/l*1000 #mm (diameter)\n",
"\n",
"print \"Length of wire =\",round(l,2),\"m.\"\n",
"print \"Diameter of wire =\",round(d,2),\"mm.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 47.2 , PAGE NO :- 1841"
]
},
{
"cell_type": "code",
"execution_count": 18,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Width of strip = 7.4 mm.\n"
]
}
],
"source": [
"'''A 30-kW, 3-phase, 400-V resistance oven is to employ nickel-chrome strip 0.254 mm thick for the three star-connected heating\n",
"elements.If the wire temperature is to be 1,100Â°C and that of the charge to be 700Â°C, estimate a suitable width for the strip.\n",
"Assume emissivity = 0.9 and radiating efficiency to be 0.5 and resistivity of the strip material is 101.6e-8 ohm- m.What would be\n",
"the temperature of the wire if the charge were cold ?'''\n",
"\n",
"import math as m\n",
"\n",
"P = 30.0*1000 *(1/3.0) #W (Power/phase)\n",
"V = 400.0/m.sqrt(3) #V (Phase voltage)\n",
"rho = 101.6e-8 #ohm-m (resistivity)\n",
"e = 0.9 # (emmisivity)\n",
"K = 0.5 # (radiating efficiency)\n",
"t = 0.254e-3 #m (thickness of strip)\n",
"T1 = 1100.0 + 273 #K (Temp. of hot wire)\n",
"T2 = 700.0 + 273 #K (Temp. of charge) \n",
"R = V*V/P #ohm (Resistance => P = V^2/R)\n",
"\n",
"# R = rho*l/(w*t) l/w = R*t/rho = a.Therefore a is\n",
"a = R*t/rho #(=l/w)------------------------------------ 1\n",
"\n",
"#Using stefan's law\n",
"H = 5.72*e*K*((T1/100)**4-(T2/100)**4) #W/m^2\n",
"\n",
"#Surface area of strip = 2*w*l .\n",
"#Total Heat dissipated = electrical power => wl*2H = P . Let b = wl\n",
"b = P/(2*H) #(=wl)----------------------------------------- 2\n",
"\n",
"#Dividing 1 by 2\n",
"w2 = b/a #(=w*w)\n",
"w = m.sqrt(w2)*1000 #mm (width)\n",
"print \"Width of strip =\",round(w,2),\"mm.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 47.3 , PAGE NO :- 1842"
]
},
{
"cell_type": "code",
"execution_count": 19,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Loading in kW = 13.51 kW.\n",
"Efficency of Tank = 87.41 %\n"
]
}
],
"source": [
"'''A cubic water tank has surface area of 6.0 m^2 and is filled to 90% capacity six times daily. The water is heated from 20Â°C \n",
"to 65Â°C.The losses per square metre of tank surface per 1Â°C temperature difference are 6.3 W. Find the loading in kW and the\n",
"efficiency of the tank.Assume specific heat of water = 4,200 J/kg/Â°C and one kWh = 3.6 MJ.'''\n",
"\n",
"import math as m\n",
"sa = 6.0 #m^2 (surface area)\n",
"T2 = 65.0 #*C (final temp)\n",
"T1 = 20.0 #*C (initial temp)\n",
"loss = 6.3 #W/*C/m^2 (loss per square metre per 1*C)\n",
"s = 4200.0 #J/Kg/*C (Specific heat)\n",
"\n",
"#Now, sa = 6*l^2\n",
"l = m.sqrt(sa/6) #m\n",
"\n",
"#Volume = l^3\n",
"V = l**3 #m^3\n",
"\n",
"#Volume of water to be heated daily is\n",
"V2 = 6*V*0.9 #m^3\n",
"\n",
"#Since 1m^3 = 1000 kg => mass of water to be heated is\n",
"mass = V2*1000.0 #kg\n",
"\n",
"#Heat required to raise temp =\n",
"H = mass*s*(T2-T1) #MJ\n",
"H = H/(3.6*10e+5) #kWh\n",
"\n",
"#Daily loss from surface\n",
"L = 6*loss*(T2-T1)*(24.0/1000) #kWh\n",
"\n",
"#Total energy required\n",
"Tot = L + H #kWh \n",
"#(i) Loading in KW is\n",
"load = Tot/24 #kW\n",
"\n",
"#(ii)Efficiency of tank is\n",
"eff = (H/Tot)*100.0 # (% efficency)\n",
"print \"Loading in kW = \",round(load,2),\"kW.\"\n",
"print \"Efficency of Tank =\",round(eff,2),\"%\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 47.4 , PAGE NO :- 1844"
]
},
{
"cell_type": "code",
"execution_count": 20,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Power factor = 0.9487\n",
"Power drawn from supply = 900.0 kW.\n",
"Time required for melting steel = 40.0 minutes. and 46.0 seconds.\n"
]
}
],
"source": [
"'''A 4-phase electric arc furnace has the following data :\n",
"Current drawn = 5000 A ; Arc voltage = 50 V\n",
"Resistance of transformer referred to secondary = 0.002 ohm\n",
"Resistance of transformer referred to secondary = 0.004 ohm\n",
"(i) Calculate the power factor and kW drawn from the supply.\n",
"(ii) If the overall efficiency of the furnace is 65%, find the time required to melt 2 tonnes of steel if\n",
"latent heat of steel = 8.89 kcal/kg, specific heat of steel = 0.12, melting point of steel = 1370Â°C and\n",
"initial temperature of steel = 20Â°C.'''\n",
"\n",
"import math as m\n",
"\n",
"I = 5000.0 #A (current drawn)\n",
"V = 50.0 #V (Arc Voltage) \n",
"Rs = 0.002 #ohm (transformer resistance on secondary)\n",
"Xs = 0.004 #ohm (transformer reactance on secondary)\n",
"T2 = 1370.0 #*C (final temp)\n",
"T1 = 20.0 #*C (initial temp)\n",
"\n",
"# Voltage drop due to resistance =\n",
"Vr = I*Rs #V\n",
"\n",
"# Voltage drop due to reactance =\n",
"Vx = I*Xs #V\n",
"\n",
"#Total Voltage is (Using vector sum)\n",
"V_tot = m.sqrt((V+Vr)**2 + Vx**2) #V\n",
"\n",
"#(i) Supply power factor is\n",
"pf = (V+Vr)/V_tot\n",
"\n",
"#Total Power drawn => P = 3*VI*(power factor)\n",
"P = 3*V_tot*I*pf/1000 #kW\n",
"\n",
"print \"Power factor =\",round(pf,4)\n",
"print \"Power drawn from supply =\",round(P,2),\"kW.\"\n",
"#Energy required to melt 2 tonnes of steel\n",
"m = 2000.0 #kg\n",
"s = 0.12 # (specific heat of steel)\n",
"L = 8.89 #kcal/kg (latent heat of steel)\n",
"\n",
"enrgy = m*s*(T2-T1) + m*L #kcal\n",
"enrgy = enrgy/860.0 #kWh\n",
"\n",
"#Utilised power\n",
"P = 0.65*P #kW\n",
"\n",
"#Time required for melting steel\n",
"Time = enrgy/P #hr\n",
"Time = Time*60 #min\n",
"Sec = (round(Time,2) - round(Time,-1) )*60 #sec\n",
"\n",
"print \"Time required for melting steel =\",round(Time,-1),\"minutes. and \",round(Sec),\"seconds.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 47.5 , PAGE NO :- 1845"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"#Total KVA taken from supply line = 2145.63 KVA .\n"
]
}
],
"source": [
"'''If a 3-phase arc furnace is to melt 10 tonne steel in 2 hours, estimate the average input to the furnace if overall\n",
"efficiency is 50%. If the current input is 9,000 A with the above kW input and the resistance and reactance of furnace leads\n",
"(including transformer) are 0.003 ohm and 0.005 ohm respectively, estimate the arc voltage and total kVA taken from the supply\n",
"Specific heat of steel = 444 J /kg/Â°C ,Latent heat of fusion of steel = 37.25 kJ/kg , Melting point of steel = 1,370 Â°C.'''\n",
"\n",
"from sympy import Symbol,solve,Eq,sqrt\n",
"import math as m\n",
"\n",
"mass = 10000.0 #kg (mass in kg)\n",
"t = 2.0 #hr (time taken to melt)\n",
"eff = 50.0 #% (overall efficiency)\n",
"I = 9000.0 #A (current input)\n",
"Rs = 0.003 #ohm (secondary resistance)\n",
"Xs = 0.005 #ohm (secondary reactance)\n",
"s = 444.0 #J/kg/*C (specific heat of steel)\n",
"L = 37250 #J/kg (latent heat of fusion)\n",
"T2 = 1370.0 #*C (final temp)\n",
"T1 = 20.0 #*C (initial temp)\n",
"\n",
"#Energy required to melt 10 tonnes of steel\n",
"\n",
"enrgy = mass*s*(T2-T1) + mass*L #J\n",
"enrgy = enrgy/(1000*3600) #kWh\n",
"\n",
"#Avg output power = energy/time\n",
"P = enrgy/t #kW\n",
"#Avg input power =\n",
"Pin = P/eff*100\n",
"\n",
"#Voltage drop due to resistance\n",
"Vr = I*Rs #V\n",
"#Voltage drop due to reactance\n",
"Vx = I*Xs #V\n",
"\n",
"#Now,Let Va is arc drop voltage\n",
"Va = Symbol('Va') #V\n",
"Vt = sqrt((Va+Vr)**2 + Vx**2) \n",
"pf = (Va+Vr)/Vt \n",
"#Total power input = 3*(Vt*It*pf)\n",
"\n",
"eq = Eq(Pin*1000,3*Vt*I*pf)\n",
"Va = solve(eq) #V\n",
"\n",
"Va1 = Va[0]\n",
"Vt = sqrt((Va1+Vr)**2 + Vx**2)\n",
"\n",
"#Total KVA taken from supply line =\n",
"power = 3*Vt*I/1000\n",
"\n",
"print \"#Total KVA taken from supply line =\",round(power,2),\"KVA .\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 47.6 , PAGE NO :- 1850"
]
},
{
"cell_type": "code",
"execution_count": 27,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Efficiency of induction furnace = 59.35 %.\n"
]
}
],
"source": [
"'''Determine the efficiency of a high-frequency induction furnace which takes 10 minutes to melt 2 kg of a aluminium initially \n",
"at a temperature of 20Â°C. The power drawn by the furnace is 5 kW, specific heat of aluminium = 0.212, melting point of \n",
"aluminium = 660Â° C and latent heat of fusion of aluminium. = 77 kcal/kg.'''\n",
"\n",
"m = 2.0 #kg (mass of alluminium)\n",
"L = 77.0 #kcal/kg (Latent heat of fusion)\n",
"T2 = 660.0 #*C (final temp)\n",
"T1 = 20.0 #*C (initial temp)\n",
"s = 0.212 # (specific heat of alluminium)\n",
"Pin = 5.0 #kW (input power)\n",
"#Heat required to melt alluminium\n",
"H1 = m*L #kcal\n",
"#Heat required to raise the temperature\n",
"H2 = m*s*(T2 - T1) #kcal\n",
"#Total heat\n",
"heat_tot = H1 + H2 #kcal\n",
"#Heat required per hour\n",
"enrgy = heat_tot/(10.0/60) #kcal\n",
"#Power delivered to alluminium\n",
"Power = enrgy/860 #kW\n",
"\n",
"eff = Power/Pin*100 #(% efficiency)\n",
"\n",
"print \"Efficiency of induction furnace = \",round(eff,2),\"%.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 47.7 , PAGE NO :- 1850"
]
},
{
"cell_type": "code",
"execution_count": 21,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Power absorbed = 577.0 kW.\n",
"power factor = 0.83\n"
]
}
],
"source": [
"'''A low-frequency induction furnace has a secondary voltage of 20V and takes 600 kW at 0.6 p.f. when the hearth is full. If the\n",
"secondary voltage is kept constant, determine the power absorbed and the p.f. when the hearth is half-full. Assume that the\n",
"resistance of the secondary circuit is doubled but the reactance remains the same.'''\n",
"\n",
"import math as m\n",
"\n",
"V = 20.0 #V (secondary voltage)\n",
"P = 600*1000 #W (Input Power)\n",
"pf = 0.6 # (power factor)\n",
"\n",
"#Inital secondary current using P = VI*pf\n",
"I = P/(V*pf) #A (secondary current)\n",
"\n",
"#Now, pf = cosQ , .'. sinQ = sqrt(1-pf^2)\n",
"Vr = V*pf #V (Voltage across resistance)\n",
"Vx = V*m.sqrt(1-pf**2) #V (Voltage across reactance)\n",
"\n",
"#As, Vr = I*R and Vx = I*X\n",
"\n",
"R = Vr/I #ohm\n",
"X = Vx/I #ohm\n",
"\n",
"#When hearth is half-full\n",
"R2 = 2*R\n",
"X2 = X\n",
"pf = R2/m.sqrt(R2**2 + X2**2) #(new power factor)\n",
"\n",
"Vr = V*pf #V (Voltage across resistance)\n",
"#As, Vr = I*R\n",
"I = Vr/R2 #A\n",
"\n",
"power = V*I*pf/1000 #kW\n",
"\n",
"print \"Power absorbed =\",round(power),\"kW.\"\n",
"print \"power factor =\",round(pf,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 47.8 , PAGE NO :- 1851"
]
},
{
"cell_type": "code",
"execution_count": 22,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total furnace input = 119.37 kWh .\n"
]
}
],
"source": [
"'''Estimate the energy required to melt 0.5 tonne of brass in a single-phase induction furnace. If the melt is to be carried out\n",
"in 0.5 hour, what must be the average power input to the furnace?\n",
"Specific heat of brass = 0.094\n",
"Latent heat of fusion of brass = 39 kilocal/kg\n",
"Melting point of brass = 920Â°C\n",
"Furnace efficiency = 60.2%\n",
"The temperature of the cold charge may be taken as 20Â°C.'''\n",
"\n",
"m = 0.5*1000 #kg (mass of brass)\n",
"s = 0.094 # (specific heat)\n",
"T2 = 920.0 #*C (final temp)\n",
"T1 = 20.0 #*C (initial temp)\n",
"L = 39.0 #kcal/kg(Latent heat of fusion)\n",
"eff = 60.2 # (% efficiency)\n",
"\n",
"#Total amount of heat req to melt 0.5 kg brass\n",
"H = m*L + m*s*(T2-T1) #kcal\n",
"H = H/860 #kWh\n",
"H_tot = H/(eff)*100 #kWh (input energy required)\n",
"\n",
"print \"Total furnace input =\",round(H_tot,2),\"kWh .\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 47.9 , PAGE NO :- 1851"
]
},
{
"cell_type": "code",
"execution_count": 23,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"height for maximum heat = 0.75 *H.\n"
]
}
],
"source": [
"'''A low-frequency induction furnace whose secondary voltage is maintained constant at 10 V, takes 400 kW at 0.6 p.f. when the\n",
"hearth is full.Assuming the resistance of the secondary circuit to vary inversely as the height of the charge and reactance to \n",
"remain constant,find the height upto which the hearth should be filled to obtain maximum heat.'''\n",
"\n",
"\n",
"import math as m\n",
"V2 = 10.0 #V (secondary voltage)\n",
"P = 400.0*1000 #kW (power)\n",
"pf = 0.6 # (power factor)\n",
"\n",
"\n",
"#Secondary current is (Using P = VI*cosQ)\n",
"I = P/(V2*pf) #A\n",
"\n",
"#Impedance of secondary circuit is\n",
"Z = V2/I #ohm\n",
"#Now, R = Z*cosQ X = Z*sinQ\n",
"R = Z*pf #ohm (resistance)\n",
"X = Z*m.sqrt(1-pf**2) #ohm (reactance)\n",
"\n",
"#Let height of charge be 'x' times of the full hearth h = x*H\n",
"#Resistance varies inersely as height .Therefore,\n",
"# R' = R/x\n",
"\n",
"#Now ,for max heat resistance should be equal to reactance.Therefore,\n",
"x = R/X\n",
"\n",
"print \"height for maximum heat = \",round(x,2),\"*H.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 47.10 , PAGE NO :- 1853"
]
},
{
"cell_type": "code",
"execution_count": 24,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Voltage = 798.04 V.\n"
]
}
],
"source": [
"'''A slab of insulating material 150 cm^2 in area and 1 cm thick is to be heated by dielectric heating. The power required is \n",
"400 W at 30 MHz.Material has relative permittivity of 5 and p.f. of 0.05. Determine the necessary voltage. Absolute\n",
"permittivity = 8.854e - 12 F/m.'''\n",
"\n",
"import math as m\n",
"\n",
"P = 400.0 #W (power)\n",
"f = 30.0e+6 #Hz (frequency)\n",
"A = 150.0e-4 #m^2 (area)\n",
"d = 1.0e-2 #m (thickness)\n",
"er = 5.0 # (relative permitivity)\n",
"e0 = 8.89e-12#F/m (absolute permitivity)\n",
"pf = 0.05 # (power factor) \n",
"#Capacitance\n",
"C = (A/d)*(er*e0) #F\n",
"\n",
"#Now, P = (2*pi*f)*(C*V^2)*pf .Therefore V is\n",
"V = m.sqrt(P/(2*3.14*f*C*pf)) #V\n",
"print \"Voltage =\",round(V,2),\"V.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 47.11 , PAGE NO :- 1853"
]
},
{
"cell_type": "code",
"execution_count": 25,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Voltage = 846.45 V.\n",
"Current through material = 9.45 A.\n",
"Frequency = 58.49 MHz.\n"
]
}
],
"source": [
"'''An insulating material 2 cm thick and 200 cm^2 in area is to be heated by dielectric heating. The material has relative \n",
"permitivity of 5 and power factor of 0.05.Power required is 400 W and frequency of 40 MHz is to be used. Determine the necessary\n",
"voltage and the current that will flow through the material.If the voltage were to be limited to 700 V, what will be the \n",
"frequency to get the same loss? '''\n",
"\n",
"import math as m\n",
"\n",
"d = 2.0e-2 #m (Thickness)\n",
"A = 200e-4 #m^2 (Area)\n",
"er = 5 # (relative permitivity)\n",
"pf = 0.05 # (power factor)\n",
"f = 40.0e+6 #Hz (frequency)\n",
"P = 400.0 #W (power)\n",
"e0 = 8.89e-12 # (absolute permitivity)\n",
"\n",
"C = (A/d)*(e0*er) #F (Capacitance)\n",
"\n",
"#Now, P = 2*pi*f*C*V^2*pf\n",
"V = m.sqrt(P/(2*3.14*f*C*pf)) #V\n",
"\n",
"#Also, P = VI*cosQ .Therefore,current through material\n",
"I = P/(V*pf) #A\n",
"\n",
"#Heat produced is propotional to V^2*f. (V2/V1)^2 = (f1/f2)\n",
"f2 = f*(V/700)**2 #Hz\n",
"f2 = f2/(10e+5) #MHz \n",
"\n",
"print \"Voltage = \",round(V,2),\"V.\"\n",
"print \"Current through material =\",round(I,2),\"A.\"\n",
"print \"Frequency = \",round(f2,2),\"MHz.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 47.12 , PAGE NO :- 1853"
]
},
{
"cell_type": "code",
"execution_count": 26,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Power input = 750.0 W .\n"
]
}
],
"source": [
"'''A plywood board of 0.5*0.25*0.02 metre is to be heated from 25 to 125Â°C in 10 minutes by dielectric heating employing a\n",
"frequency of 30 MHz. Determine the power required in this heating process. Assume specific heat of wood 1500/J/kg/Â°C; \n",
"weight of wood 600 kg/m3 and efficiency of process 50%.'''\n",
"\n",
"\n",
"Vol = 0.5*0.25*0.02 #m^3 (Volume of plywood to be heated)\n",
"f = 30.0e+6 #Hz (frequency)\n",
"t = 10.0 #minutes (time)\n",
"T2 = 125.0 #*C (final temperature)\n",
"T1 = 25.0 #*C (initial temperature)\n",
"s = 1500.0 #J/kg/*C (specific heat of wood)\n",
"den = 600.0 #kg/m^3 (weight of wood)\n",
"eff = 50.0 #% (efficiency of process)\n",
"\n",
"wt = den*Vol #kg (weight of plywood)\n",
"#Heat required to raise the temp is\n",
"H = wt*s*(T2-T1) #J\n",
"H = H/3600 #Wh\n",
"\n",
"#As P = H/t .Therfore power required for heating\n",
"P = H/(10.0/60) #W\n",
"\n",
"#As efficiency is 50%\n",
"Pin = P/eff*100 #W\n",
"\n",
"print \"Power input =\",round(Pin,2),\"W .\""
]
},
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"cell_type": "code",
"execution_count": null,
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PKj„ìJÇÓñíí;A TEXTBOOK OF ELECTRICAL TECHNOLOGY VOL III/chapter40.ipynb{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# CHAPTER 40: D.C TRANSMISSION AND DISTRIBUTION\n",
"\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.1 ,Page No :- 1574"
]
},
{
"cell_type": "code",
"execution_count": 20,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"percentage saving in copper is = 50.0 %.\n"
]
}
],
"source": [
"#A DC 2-wire feeder supplies a constant load with a sending-end voltage of 220V.Calculate the saving in copper\n",
"#if this voltage is doubled with power transmitted remaining the same.\n",
"##################################################################################################################\n",
"\n",
"\n",
"\n",
"#Given\n",
"V1 = 220.0\n",
"V2 = 440.0\n",
"##Let us assume the wire has##\n",
"#length -> length \n",
"#area -> area\n",
"#current density -> cd\n",
"#power -> P\n",
"P = 10000.0 #assumption\n",
"length = 1000.0 #assumption \n",
"cd = 5.0 #assumption\n",
"#The values are assumed as these terms cancel out while calculating percentage.\n",
"I1 = P/V1\n",
"area = I1/cd\n",
"#Vol of Cu required for 220V ->vol1\n",
"vol1 = 2*area*length\n",
"\n",
"\n",
"I2 = P/V2\n",
"area = I2/cd\n",
"#Vol of Cu required for 440V ->vol2\n",
"vol2 = 2*area*length\n",
"\n",
"#percentage saving of copper is\n",
"per_cent = ((vol1-vol2)/vol1)*100\n",
"print 'percentage saving in copper is ',per_cent,'%.'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.2 ,Page No :- 1577"
]
},
{
"cell_type": "code",
"execution_count": 19,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Maximum voltage drop from one end is = 12.0 V.\n",
"Maximum voltage drop from both end is = 3.0 V.\n"
]
}
],
"source": [
"#A uniform 2-wire d.c distributor 200 metres long is loaded with 2 amperes/metre.Resistance of\n",
"#single wire is 0.3 ohm/kilometre.Calculate the maximum voltage drop if the distributor is fed\n",
"#(a)from one end (b)from both ends with equal voltages.\n",
"#################################################################################################\n",
"\n",
"#Given\n",
"length = 200.0 #metres\n",
"#current per unit length is\n",
"cur = 2.0 #amp/metre\n",
"#resistance per unit length is\n",
"res = 0.3/1000 #ohm/metre\n",
"\n",
"#total resistance is\n",
"R = res*length #ohm\n",
"#total current is\n",
"I = cur*length #amp\n",
"#Total drop over a distributor fed from one end is given by\n",
"drop1 = (1/2.0)*I*R #volts\n",
"#Total drop over a distributor fed from both ends is given by\n",
"drop2 = (1/8.0)*I*R\n",
"print 'Maximum voltage drop from one end is = ',drop1,'V.'\n",
"print 'Maximum voltage drop from both end is = ',drop2,'V.'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.3 ,Page No :- 1577"
]
},
{
"cell_type": "code",
"execution_count": 21,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Cross sectional area of distributor = 116.412 cm^2\n"
]
}
],
"source": [
"#A 2-wire d.c distributor AB is 300 metres long.It is fed at point A.The various loads and\n",
"#their positions are given below.\n",
"# At point distance from A in metres concentrated load in A\n",
"# C 40 30\n",
"# D 100 40 \n",
"# E 150 100\n",
"# F 250 50\n",
"#If the maximum permissible voltage drop is not to exceed 10V,find the cross-sectional\n",
"#area of the distributor.Take resistivity = 1.78*10^(-8) ohm-m.\n",
"###########################################################################################\n",
"\n",
"\n",
"#Given\n",
"resistivity = 1.78e-8 #ohm-m\n",
"drop_max = 10.0 #V\n",
"#loads and their positions\n",
"I1 = 30.0 #A\n",
"l1 = 40.0 #m\n",
"I2 = 40.0 #A\n",
"l2 = 100.0 #m\n",
"I3 = 100.0 #A\n",
"l3 = 150.0 #m\n",
"I4 = 50 #A\n",
"l4 = 250 #m\n",
"#We know that R = resistivity*length/Area\n",
"#Also max drop = I1*R1 + I2*R2 + I3*R3 + I4*R4 , using this\n",
"area = 2*(I1*l1 + I2*l2 + I3*l3 + I4*l4)*resistivity/drop_max #m^2\n",
"area = area*1000000 #cm^2 \n",
"print 'Cross sectional area of distributor =',area,'cm^2'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.4 ,Page No :- 1578"
]
},
{
"cell_type": "code",
"execution_count": 23,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Hence drop at minimum potential where load is 70 A is = 48.4 V.\n"
]
}
],
"source": [
"#A 2-wire d.c distributor F1F2 1000 metres long is loaded as under:\n",
"#Distance from F1(in metres): 100 250 500 600 700 800 850 920\n",
"#Load in amperes: 20 80 50 70 40 30 10 15\n",
"#The feeding points F1 and F2 are maintained at the same potential.Find which point will have the\n",
"#minimum potential and what will be the drop at this point?Take the cross-section of the conductors\n",
"#as 0.35 cm^2 and specific resistance of copper as 1.764*10^(-6) ohm-cm\n",
"#####################################################################################################\n",
"\n",
"#Given\n",
"import numpy as np\n",
"resistivity = 1.764e-8 #ohm-m\n",
"area = 0.35e-4 #m^2 \n",
"#loads and their positions taking from F1\n",
"I1 = 20 #A\n",
"l1 = 100 #m\n",
"I2 = 80 #A\n",
"l2 = 150 #m\n",
"I3 = 50 #A\n",
"l3 = 250 #m\n",
"I4 = 70 #A\n",
"l4 = 100 #m\n",
"I5 = 40 #A\n",
"l5 = 100 #m\n",
"I6 = 30 #A\n",
"l6 = 50 #m\n",
"I7 = 10 #A\n",
"l7 = 70 #m\n",
"I8 = 15 #A\n",
"l8 = 80 #m \n",
"\n",
"#sum of loads from F1\n",
"load1 = I1*l1 + I2*(l1+l2) + I3*(l1+l2+l3) #A-m\n",
"load2 = I8*l8 + I7*(l7+l8) + I6*(l6+l7+l8) + I5*(l5+l6+l7+l8) #A-m\n",
"\n",
"#guessing the point of minimum potential\n",
"if load1>load2:\n",
" load2_new = load2 + I4*(l4+l5+l6+l7+l8)\n",
" if load2_new>load1:\n",
" pivot = I4\n",
"\n",
"#solving 2 equations simultaneously\n",
"# x + y = 70(pivot) & 47000(load1) + 600(l1+l2+l3)x = 20,700(load2) + 400(l5+l6+l7+l8)y)\n",
"line1 = l1+l2+l3+l4 #m\n",
"line2 = l4+l5+l6+l7+l8 #m \n",
"\n",
"a = [[1,1],[line1,-line2]]\n",
"b = [pivot,load2-load1]\n",
"soln = np.linalg.solve(a,b) #soln is array with its elements[x,y]\n",
"#drop at minimum potential per conductor (in A-m)\n",
"drop_m = load1 + soln[0]*line1 #A-m\n",
"\n",
"#resistance per metre = resistivity/Area\n",
"res = resistivity/area #ohm/m\n",
"\n",
"#Hence, drop in voltage per conductor is\n",
"drop = drop_m*res #V \n",
"#drop due to both is\n",
"drop = drop*2 #V\n",
"\n",
"print 'Hence drop at minimum potential where load is',pivot,'A is =',round(drop,2),'V.'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.5 ,Page No :- 1579"
]
},
{
"cell_type": "code",
"execution_count": 33,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The current entering at A is = 88.6 A \n",
"The current entering at B is = 211.4 A.\n"
]
}
],
"source": [
"#The resistance of a cable is 0.1ohm per 1000 metre for both conductors.It is loaded as shown in Fig.40.14(a).\n",
"#Find the current supplied at A and at B.If both the ends are supplied at 200 V\n",
"##############################################################################################################\n",
"\n",
"#Given\n",
"#resistance per metre\n",
"res = 0.1/1000 #ohm/m\n",
"#loads and their positions taking from A\n",
"I1 = 50.0 #A\n",
"l1 = 500.0 #m\n",
"I2 = 100.0 #A\n",
"l2 = 700.0 #m\n",
"I3 = 150.0 #A\n",
"l3 = 300.0 #m\n",
"l4 = 250.0 #m \n",
"\n",
"#Assuming I flows from A to B\n",
"# equation is res*[500*i + 700(i-50) + 300(i-150) + 250(i-300)] = 0\n",
"current_i = (I1*l2+(I1+I2)*l3 + (I1+I2+I3)*l4)/(l1+l2+l3+l4)\n",
"current_AC = current_i\n",
"current_CD = current_i-I1\n",
"current_DE = current_CD-I2\n",
"current_EB = current_DE-I3\n",
"if current_EB<0:\n",
" current_EB = -current_EB;\n",
"print 'The current entering at A is = ',round(current_i,1),'A '\n",
"print 'The current entering at B is = ',round(current_EB,1),'A.' "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.6 ,Page No :- 1580"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Current supplied at A is = 88.6 A.\n",
"Current supplied at B is = -211.4 A.\n",
"Current in AC is = 88.6 A.\n",
"Current in CD is = 38.6 A.\n",
"Current in DE is = -61.4 A.\n",
"Current in EB is = -211.4 A.\n",
"Drop over AC is = 4.4 V.\n",
"Drop over CD is = 2.7 V.\n",
"Drop over DE is = -1.8 V.\n",
"Voltage at C is = 195.6 V.\n",
"Voltage at D is = 192.9 V.\n",
"Voltage at E is = 194.7 V.\n"
]
}
],
"source": [
"#The resistance of two conductors of a 2-conductor distributor shown in Fig.39.15 is 0.1ohm per 1000m\n",
"#for both conductors.Find (a)the current supplied at A(b)the current supplied at B\n",
"#(c)the current in each section (d)the voltages at C,D and E.Both A and B are maintained at 200V.\n",
"######################################################################################################\n",
"\n",
"#Given\n",
"#resistance per metre\n",
"res = 0.1/1000 #ohm/m\n",
"#loads and their positions taking from A\n",
"I1 = 50.0 #A\n",
"l1 = 500.0 #m\n",
"I2 = 100.0 #A\n",
"l2 = 700.0 #m\n",
"I3 = 150.0 #A\n",
"l3 = 300.0 #m\n",
"l4 = 250.0 #m \n",
"\n",
"#Assuming I flows from A to B\n",
"# equation is res*[500*i + 700(i-50) + 300(i-150) + 250(i-300)] = 0\n",
"current_i = (I1*l2+(I1+I2)*l3 + (I1+I2+I3)*l4)/(l1+l2+l3+l4)\n",
"current_AC = current_i\n",
"current_CD = current_i-I1\n",
"current_DE = current_CD-I2\n",
"current_EB = current_DE-I3\n",
"print \"Current supplied at A is = \",round(current_i,1),\"A.\"\n",
"print \"Current supplied at B is = \",round(current_EB,1),\"A.\"\n",
"print \"Current in AC is = \",round(current_i,1),\"A.\"\n",
"print \"Current in CD is = \",round(current_CD,1),\"A.\"\n",
"print \"Current in DE is = \",round(current_DE,1),\"A.\"\n",
"print \"Current in EB is = \",round(current_EB,1),\"A.\"\n",
"#Drop in volts is (resistance/metre)*length*current\n",
"drop_AC = res*l1*current_AC #V\n",
"drop_CD = res*l2*current_CD #V \n",
"drop_DE = res*l3*current_DE #V\n",
"print \"Drop over AC is = \",round(drop_AC,1),\"V.\"\n",
"print \"Drop over CD is = \",round(drop_CD,1),\"V.\"\n",
"print \"Drop over DE is = \",round(drop_DE,1),\"V.\"\n",
"\n",
"#Voltages at C,D,E are\n",
"volt_C = 200-drop_AC #V\n",
"volt_D = volt_C-drop_CD #V\n",
"volt_E = volt_D-drop_DE #V\n",
"print 'Voltage at C is = ',round(volt_C,1),'V.'\n",
"print 'Voltage at D is =',round(volt_D,1),'V.'\n",
"print 'Voltage at E is = ',round(volt_E,1),'V.'\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.7 ,Page No :- 1581"
]
},
{
"cell_type": "code",
"execution_count": 31,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Therefore point of minimum potential is D and minimum potential is = 246.0 V.\n"
]
}
],
"source": [
"#A 200 m long distributor is fed from both ends A and B at the same voltage of 250V.The\n",
"#concentrated loads of 50,40,30 and 25 A are coming on the distributor at distances of 50,75,\n",
"#100 and 150 m respectively from end A.Determine the minimum potential and locate its positions.\n",
"#Also,determine the current in each section of the distributor.It is given that the resistance\n",
"#of the distributor is 0.08ohm per 100 metres for go and return.\n",
"##################################################################################################\n",
"\n",
"\n",
"#Given\n",
"#resistance per metre\n",
"res = 0.08/100 #ohm/m\n",
"V_A = 250.0 #V\n",
"V_B = 250.0 #V\n",
"#load currents and their positions\n",
"I_C = 50.0 #A\n",
"I_D = 40.0 #A\n",
"I_E = 30.0 #A\n",
"I_F = 25.0 #A\n",
"l_AC = 50.0 #m\n",
"l_CD = 75.0 - l_AC #m\n",
"l_DE = 100.0 - l_CD - l_AC #m\n",
"l_EF = 150.0 - l_DE - l_CD - l_AC #m\n",
"l_FB = 200.0-150.0\n",
"#Assuming I flows from A to B\n",
"# equation is res*[50*i + 25(i-50) + 25(i-90) + 50(i-120)+50(i-145)] = 0\n",
"current_i = (l_CD*I_C + l_DE*(I_C+I_D)+l_EF*(I_C+I_D+I_E) + l_FB*(I_C+I_D+I_E+I_F))/200.0\n",
"current_AC = current_i\n",
"current_CD = current_i-I_C\n",
"current_DE = current_CD-I_D\n",
"current_EF = current_DE-I_E\n",
"current_FB = current_EF-I_F\n",
"#As from figure in the book , point D is at minimum potential\n",
"if (current_CD>0) & (current_DE<0):\n",
" point = \"D\"\n",
" \n",
"#drop in volts = resistance/metre*sum(length*current) \n",
"drop_d = res*(l_AC*current_AC + l_CD*current_CD)\n",
"min_pot = V_A-drop_d\n",
"print \"Therefore point of minimum potential is\",point,\"and minimum potential is = \",round(min_pot,1),\"V.\" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.8 ,Page No :- 1582"
]
},
{
"cell_type": "code",
"execution_count": 35,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Voltage at point C is = 250.13 V.\n",
"Voltage at point D is = 247.73 V.\n"
]
}
],
"source": [
"#Each conductor of a 2-core distributor,500 metres long has a cross-sectional area\n",
"#of 2 cm^2.The feeding point A is supplied at 255V and the feeding point B at\n",
"#250V and load currents of 120A and 160A are taken at points C and D which are\n",
"#150 metres and 350 metres respectively from the feeding point A.Calculate the\n",
"#voltage at each load.Specific Resistance of copper is 1.7*10^(-6) ohm-cm\n",
"##################################################################################\n",
"\n",
"#Given\n",
"area = 2e-4 #m^2\n",
"resistivity = 1.7e-8 #ohm-m\n",
"#load currents and their positions\n",
"i_c = 120.0 #A\n",
"i_d = 160.0 #A\n",
"l_ac = 150.0 #m\n",
"l_cd = 200.0 #m\n",
"l_db = 150.0 #m\n",
"V_a = 255.0 #V\n",
"V_b = 250.0 #V\n",
"#Resistance = resistivity*length/Area\n",
"#It is a 2 core distributor.Therefore,resistance per metre is\n",
"res = 2*resistivity/area #ohm/m\n",
"#drop over whole distributor is equal to\n",
"drop = V_a - V_b #V\n",
"#Therefore equation of total drop can be written as\n",
"# resistivity*(150i+200(i-120)+150(i-280))=5\n",
"current_i = (drop/res + l_cd*i_c + l_db*(i_c+i_d))/(l_ac+l_cd+l_db) #A\n",
"current_ac = current_i #A\n",
"current_cd = current_ac-i_c #A\n",
"current_db = current_cd-i_d #A\n",
"\n",
"#Voltage at C = 255-drop over AC\n",
"volt_c = V_a-res*l_ac*current_ac #V\n",
"#Voltage at D = 250-drop over DB \n",
"volt_d = V_b -res*l_db*abs(current_db) #V\n",
"print \"Voltage at point C is = \",round(volt_c,2),\"V.\"\n",
"print \"Voltage at point D is = \",round(volt_d,2),\"V.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.9 ,Page No :- 1583"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Volatge at point Q is = 225.25 V.\n",
"Voltage at point B is = 236.56 V.\n"
]
}
],
"source": [
"#A 2-wire distributor 500 metres long is fed at P at 250V and loads of 40A,20A,60A,30A are tapped off\n",
"#off from points A,B,C and D which are at distances of 100 metres,150 metres,300 metres and 400 metres\n",
"#from P respectively.The distributor is also uniformly loaded at the rate of 0.1A/m.If the resistance of\n",
"#the distributor per metre(go and return) is 0.0005 ohm,calculate the voltage at(i)pointQ and(ii)point B.\n",
"###########################################################################################################\n",
"\n",
"#Given\n",
"V_P = 250.0 #V\n",
"resistance = 0.0005 #ohm/m\n",
"\n",
"#loads and their positions\n",
"i_a = 40.0 #A\n",
"i_b = 20.0 #A\n",
"i_c = 60.0 #A\n",
"i_d = 30.0 #A\n",
"l_pa = 100.0 #m\n",
"l_ab = 150.0-100.0 #m\n",
"l_bc = 300.0-150.0 #m\n",
"l_cd = 400.0-300.0 #m\n",
"#uniform dstributed load\n",
"cur_uni = 0.1 #A/m\n",
"\n",
"\n",
"#considering drop due to concentrated loading\n",
"drop_pa = (i_a+i_b+i_c+i_d)*l_pa*resistance #V\n",
"drop_ab = (i_b+i_c+i_d)*l_ab*resistance #V \n",
"drop_bc = (i_c+i_d)*l_bc*resistance #V\n",
"drop_cd = i_d*l_cd*resistance #V\n",
"tot_drop = drop_pa + drop_ab + drop_bc + drop_cd #V\n",
"\n",
"#considering drop due to uniform loading(drop = irl^2/2) l = 500m\n",
"drop_uni = cur_uni*resistance*(500.0*500.0)/2 #V\n",
"\n",
"V_Q = V_P - (tot_drop + drop_uni) #V\n",
"#for point B\n",
"#drop due to concentrated loading\n",
"drop_b = drop_pa + drop_ab #V\n",
"#drop due to uniform loading (drop = ir(lx-x^2/2)) l=500m x=150m\n",
"drop_ub = cur_uni*resistance*(500*(l_pa+l_ab)-(l_pa+l_ab)*(l_pa+l_ab)/2) #V\n",
"\n",
"V_B = V_P - (drop_b + drop_ub) #V\n",
"\n",
"print \"Volatge at point Q is = \",round(V_Q,2),\"V.\"\n",
"print \"Voltage at point B is = \",round(V_B,2),\"V.\" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.10 ,Page No :- 1583"
]
},
{
"cell_type": "code",
"execution_count": 38,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Current in section AC is = 53.75 A.\n",
"Current in section CD is = 33.75 A.\n",
"Current in section DE is = -6.25 A.\n",
"Current in section EF is = -31.25 A.\n",
"Current in section FB is = -61.25 A.\n",
"Minimum voltage is at point D and minimum voltage is = 233.18 V.\n"
]
}
],
"source": [
"#A distributor AB is fed from both ends.At feeding point A,the voltage is maintained at 236V and at B at 237V.\n",
"#The total length of the distributor is 200 metres and loads are tapped off as under:\n",
"#(i) 20A at 50 metres from A (ii) 40A at 75 metres from A. (iii)25A at 100 metres from A (iv)30A at 150 metres from A\n",
"#The reistance per kilometre of one conductor is 0.4ohm.Calculate the currents in the various sections of the distributor,\n",
"#the minimum voltage and the point at which it occurs.\n",
"###########################################################################################################################\n",
"\n",
"\n",
"#Given\n",
"#resistance per metre\n",
"res = 2*0.4/1000 #ohm/m\n",
"V_a = 236.0 #V\n",
"V_b = 237.0 #V\n",
"#loads and their positions\n",
"i_c = 20.0 #A\n",
"i_d = 40.0 #A\n",
"i_e = 25.0 #A\n",
"i_f = 30.0 #A\n",
"l_ac = 50.0 #m\n",
"l_cd = 25.0 #m\n",
"l_de = 25.0 #m\n",
"l_ef = 50.0 #m\n",
"l_fb = 50.0 #m\n",
"#Voltage drop equation res*(50i + 25(i-20)+25(i-60) + 50(i-85) + 50(i-115)=-1)\n",
"current_i = ((V_a-V_b)/res + l_cd*(i_c)+l_de*(i_c+i_d)+l_ef*(i_c+i_d+i_e)+l_fb*(i_c+i_d+i_e+i_f))/200.0\n",
"current_ac = current_i\n",
"current_cd = current_ac-i_c\n",
"current_de = current_cd-i_d\n",
"current_ef = current_de-i_e\n",
"current_fb= current_ef-i_f\n",
"if current_cd>0:\n",
" if current_de<0:\n",
" point = \"D\"\n",
"#Minimum potential is at D as shown in figure\n",
"drop = res*(current_ac*l_ac + current_cd*l_cd)\n",
"V_d = V_a-drop\n",
"print \"Current in section AC is = \",round(current_ac,2),\"A.\"\n",
"print \"Current in section CD is = \",round(current_cd,2),\"A.\"\n",
"print \"Current in section DE is = \",round(current_de,2),\"A.\"\n",
"print \"Current in section EF is = \",round(current_ef,2),\"A.\"\n",
"print \"Current in section FB is = \",round(current_fb,2),\"A.\"\n",
"print \"Minimum voltage is at point\",point,\"and minimum voltage is = \",round(V_d,2),\"V.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.11 ,Page No :- 1584"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Current supplied by feeder at point A is 46.29 A and that by point B is 109.71 A.\n",
"Voltage at point B is = 240.55 V.\n",
"Voltage at point C is = 239.63 V.\n",
"Voltage at point D is = 239.42 V.\n",
"Voltage at point E is = 239.38 V.\n"
]
}
],
"source": [
"#A distributor cable AB is fed at its ends A and B.Loads of 12,24,72 and 48 A are taken from the cable at\n",
"#points C,D,E and F.The resistances of sections AC,CD,DE,EF and FB of the cable are 8,6,4,10 and 5 milliohm\n",
"#respecively(for the go and return conductors together). The potential difference at point A is 240V,the p.d\n",
"#at the load F is also to be 240V.Calculate the voltages at the feeding point B,the current supplied by each\n",
"#feeder and the p.d.s at the loads C,D and E.\n",
"##############################################################################################################\n",
"\n",
"#Given\n",
"V_a = 240.0 #V \n",
"V_f = 240.0 #V\n",
"#loads and their resistances.\n",
"i_c = 12.0 #A\n",
"i_d = 24.0 #A\n",
"i_e = 72.0 #A\n",
"i_f = 48.0 #A\n",
"\n",
"r_ac = 8e-3 #ohm\n",
"r_cd = 6e-3 #ohm\n",
"r_de = 4e-3 #ohm\n",
"r_ef = 10e-3 #ohm\n",
"r_fb = 5e-3 #ohm\n",
"\n",
"#Voltage drop accross AF is zero.Therefore equation 8i +6(i-12) + 4(i-36)+10(i-108)*10^(-3)\n",
"current_i = (r_cd*i_c + r_de*(i_c+i_d) + r_ef*(i_c+i_d+i_e))/(28.0e-3) #A\n",
"#currents in different sections\n",
"current_ac = current_i #A\n",
"current_cd= current_ac-i_c #A\n",
"current_de = current_cd-i_d #A\n",
"current_ef = current_de-i_e #A \n",
"current_fb = current_ef-i_f #A\n",
"#voltage at different points are\n",
"V_b = V_f - current_fb*r_fb #V\n",
"V_c = V_a - current_ac*r_ac #V\n",
"V_d = V_c - current_cd*r_cd #V\n",
"V_e = V_d - current_de*r_de #V \n",
"\n",
"print \"Current supplied by feeder at point A is\",round(current_ac,2),\"A and that by point B is\",round(abs(current_fb),2),\"A.\"\n",
"print \"Voltage at point B is = \",round(V_b,2),\"V.\"\n",
"print \"Voltage at point C is = \",round(V_c,2),\"V.\"\n",
"print \"Voltage at point D is = \",round(V_d,2),\"V.\"\n",
"print \"Voltage at point E is = \",round(V_e,2),\"V.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.12 ,Page No :- 1585"
]
},
{
"cell_type": "code",
"execution_count": 42,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The current supplied at P is = 143.75 A.\n",
"The current supplied at Q is = 116.25 A.\n",
"Power dissipated in distributor is = 847.34 W.\n"
]
}
],
"source": [
"#A two-wire d.c sdistributor PQ,800 metre long is loaded as under:\n",
"#Distance from P(metres): 100 250 500 600 700\n",
"#Loads in amperes: 20 80 50 70 40\n",
"#The feeding point at P is maintained at 248V and that at Q at 245V.The total resistance of\n",
"#the distributor(lead and return) is 0.1 ohm.Find (a)the current supplied at P and Q and\n",
"#(b)the power dissipated in the distributor.\n",
"##################################################################################################\n",
"\n",
"#Given\n",
"V_p = 248.0 #V\n",
"V_q = 245.0 #V\n",
"res = 0.1/800 #ohm/m \n",
"#loads and their positions\n",
"i1 = 20.0 #A\n",
"i2 = 80.0 #A\n",
"i3 = 50.0 #A\n",
"i4 = 70.0 #A\n",
"i5 = 40.0 #A\n",
"l1 = 100.0 #m\n",
"l2 = 250.0-100.0 #m\n",
"l3 = 500.0 -250.0 #m\n",
"l4 = 600.0-500.0 #m\n",
"l5 = 700.0-600.0 #m\n",
"l6 = 800.0-700.0 #m\n",
"#drop accross the distributor :- 1/8000(100i + 150(i-20) + 250(i-100)+ 100(i-150)+100(i-220)+100(i-260) )=3\n",
"current_i = ((V_p-V_q)/res + l2*i1+l3*(i1+i2)+l4*(i1+i2+i3)+l5*(i1+i2+i3+i4)+l6*(i1+i2+i3+i4+i5))/800.0\n",
"current_p = current_i #A\n",
"current_2 = current_p-i1 #A\n",
"current_3 = current_2-i2 #A\n",
"current_4 = current_3-i3 #A\n",
"current_5 = current_4-i4 #A\n",
"current_q = current_5-i5 #A\n",
"#Power loss = sum(I^2R)\n",
"loss = res*(current_p*current_p*l1 + current_2*current_2*l2 + current_3*current_3*l3 + current_4*current_4*l4 + current_5*current_5*l5 + current_q*current_q*l6)\n",
"print \"The current supplied at P is = \",round(current_p,2),\"A.\"\n",
"print \"The current supplied at Q is = \",round(abs(current_q),2),\"A.\"\n",
"print \"Power dissipated in distributor is =\",round(loss,2),\"W.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.13 ,Page No :- 1586"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The point of minimum potential is D and minimum potential is = 231.76 V.\n",
"Current fed at the end A is = 366.0 A.\n",
"Current fed at the end B is = 454.0 A.\n"
]
}
],
"source": [
"#The two conductors of a d.c distributor cable 1000m long have a total resistance of 0.1 ohm.\n",
"#The ends A and B are fed at 240V.The cable is uniformly loaded at 0.5 A per metre length\n",
"#and has concentrated loads of 120A,60A,100A and 40A at points distant 200,400,700 and 900m.\n",
"#respectively from the end A.Calculate (i)the point of minimum potential on the distributor\n",
"#(ii)the value of minimum potential and (iii) currents fed at the ends A and B.\n",
"###############################################################################################\n",
"\n",
"#Given\n",
"V_a = 240.0 #V\n",
"V_b = 240.0 #V\n",
"res = 0.1/1000 #ohm/m\n",
"#concentrated loads and their positions\n",
"i_c = 120.0 #A\n",
"i_d = 60.0 #A\n",
"i_e = 100.0 #A\n",
"i_f = 40.0 #A\n",
"l_ac = 200.0 #m\n",
"l_cd = 400.0-200.0 #m\n",
"l_de = 700.0-400.0 #m\n",
"l_ef = 900.0-700.0 #m\n",
"l_fb = 1000.0-900.0 #m\n",
"#Uniform loading\n",
"cur_uni = 0.5 #A/m\n",
"#Equation for drop from A to B -> (1/10000)*(200i + 200(i-120)+ 300(i-180)+200(i-280)+100(i-320))=0\n",
"current_i = (l_cd*i_c + l_de*(i_c+i_d)+l_ef*(i_c+i_d+i_e)+l_fb*(i_c+i_d+i_e+i_f))/1000\n",
"\n",
"#point of minimum potential\n",
"current_ac = current_i #A\n",
"current_cd = current_ac-i_c #A\n",
"current_de = current_cd-i_d #A\n",
"current_ef = current_de-i_e #A\n",
"current_fb = current_ef-i_f #A\n",
"\n",
"if current_cd>0:\n",
" if current_de<0:\n",
" point = \"D\"\n",
"#As from figure it is inferred that point of minimum potential is D.\n",
"#Therefore,uniform load from point A to D(supplied from A)\n",
"cur_uni_A = cur_uni*(l_ac + l_cd) #A\n",
"cur_A = cur_uni_A + current_ac #A\n",
"#Therefore,uniform load from point B to D(supplied from B)\n",
"cur_uni_B = cur_uni*(l_de + l_ef + l_fb) #A\n",
"cur_B = cur_uni_B + abs(current_fb) #A\n",
"\n",
"#drop at D due to concentrated load(from A)\n",
"drop_con = res*(current_ac*l_ac + current_cd*l_cd)\n",
"#drop at D due to uniform load(from A)[formula-> irl^2/2]\n",
"drop_uni = cur_uni*res*(l_ac+l_cd)*(l_ac+l_cd)/2\n",
"#total drop is\n",
"drop_tot = drop_con + drop_uni\n",
"\n",
"#potential at D is\n",
"V_d = V_a - drop_tot\n",
"print \"The point of minimum potential is\",point,\"and minimum potential is = \",round(V_d,2),\"V.\"\n",
"print \"Current fed at the end A is = \",round(cur_A,2),\"A.\"\n",
"print \"Current fed at the end B is = \",round(cur_B,2),\"A.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.14 ,Page No :- 1587"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Voltage V is = 8.62 V.\n",
"Cross-sectional Area A is = 2.78 cm^2.\n",
"Cross-sectional Area A1 is = 0.26 cm^2.\n",
"Cross-sectional Area A2 is = 2.24 cm^2.\n"
]
}
],
"source": [
"#It is proposed to lay out a d.c distribution system comprising three sections-the first section consists\n",
"#of a cable from the sub-station to a point distant 800 metres from which two cables are taken,one 350 metres\n",
"#long supplying a load of 22kW and the other 1.5 kilometre long and supplying a load of 44kW.Calculate the\n",
"#cross-sectional area of each cable so that the total weight of copper required shall be minimum if the maximum\n",
"#drop of voltage along the cable is not to exceed 5% of the normal voltage of 440V at the consumer's premises.\n",
"#Take specific resistance of copper at working temperature equal to 2*10e-7 ohm-cm.\n",
"###################################################################################################################\n",
"\n",
"#Given\n",
"resistivity = 2*10e-7 #ohm-cm\n",
"dist = 800.0*100 #cm\n",
"#Current taken from 350m section\n",
"cur_1 = 22000.0/440\n",
"#Current taken from 1.5km section\n",
"cur_2 = 44000.0/440\n",
"#Therefore,current in first section\n",
"cur = cur_1 + cur_2\n",
"#Let us assume\n",
"#V->voltage drop accross first section\n",
"#R->resistance of the first section\n",
"#A->cross-sectional area of te first section\n",
"\n",
"from sympy import Eq, var, solve\n",
"var('V') \n",
"#Now , R = V/I\n",
"R = V/cur\n",
"# A = resistivity*l/R -> A = resistivity*l*I/V \n",
"A = resistivity*dist/R\n",
"#Max allowable drop\n",
"max_drop = (5.0/100)*440.0\n",
"#Voltage drop along other sections\n",
"vol_drop = max_drop - V\n",
"#Cross-sectional area of 350 m A = resistivity*l/R \n",
"A1 = resistivity*350.0*100*cur_1/(vol_drop)\n",
"#Cross-sectional area of 1.5km A = resistivity*l/R \n",
"A2 = resistivity*1500.0*100*cur_2/(vol_drop)\n",
"\n",
"\n",
"#Now,Total weight is propotional to total volume \n",
"W = 800.0*A + 350.0*A1+1500.0*A2\n",
"Diff = W.diff(V)\n",
"eq = Eq(Diff,0)\n",
"\n",
"V = solve(eq)\n",
"#We get 2 values of V of which Negative is not possible.Therefore,\n",
"V = float(V[1])\n",
"A = resistivity*dist*cur/V\n",
"vol_drop = max_drop - V\n",
"A1 = resistivity*350.0*100*cur_1/vol_drop\n",
"A2 = resistivity*1500.0*100*cur_2/vol_drop\n",
"print \"Voltage V is = \",round(V,2),\"V.\"\n",
"print \"Cross-sectional Area A is = \",round(A,2),\"cm^2.\"\n",
"print \"Cross-sectional Area A1 is = \",round(A1,2),\"cm^2.\"\n",
"print \"Cross-sectional Area A2 is = \",round(A2,2),\"cm^2.\"\n",
"\n",
"\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.15 ,Page No :- 1588"
]
},
{
"cell_type": "code",
"execution_count": 48,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The point of minimum potential is at 261.74 m from A.\n",
"The minimum potential is = 247.34 V.\n"
]
}
],
"source": [
"#A d.c two-wire distributor AB is 450m long and is fed at both ends at 250 volts.It is loaded as follows:20A at 60m from A,\n",
"#40A at 100m from A and a uniform loading of 1.5A/m from 200 to 450m from A.The resistance of each conductor is\n",
"#0.05ohm/km.Find the point of minimum potential and its potential.\n",
"####################################################################################################################\n",
"\n",
"#Given\n",
"V_a = 250.0 #V\n",
"V_b = 250.0 #V\n",
"res = 0.05/1000 #ohm/m\n",
"cur_uni = 1.5 #A/m (uniform loading)\n",
"#loads and positions\n",
"i_c = 20.0 #A\n",
"i_d = 40.0 #A\n",
"l_ac = 60.0 #m\n",
"l_cd = 40.0 #m\n",
"l_de = 100.0 #m\n",
"l_eb = 250.0 #m\n",
"\n",
"#Let us assume that point of minimum potential is D and let i be current in section CD.\n",
"#Therefore,current from B is (40-i).If r is resistance then\n",
"#(20+i)*60r + i*40r = (40-i)*350r + 1.5*r*250^2/2 [drop over AD = drop over BD as V_a = V_b]\n",
"\n",
"cur_i = (i_d*(l_de+l_eb)*res + cur_uni*res*l_eb*l_eb/2 - i_c*l_ac*res)/((l_ac+l_cd+l_de+l_eb)*res) #A\n",
"\n",
"#cur_i > 40 i.e 40-i is negative,it means D is not point of minimum potential.Let F be point of minimum potential(between DB)\n",
"#current in section DF is\n",
"cur_df = cur_i-i_d #A\n",
"\n",
"#distance EF\n",
"dist_ef = cur_df/cur_uni #m\n",
"\n",
"#distance of F from A is\n",
"dist = l_ac + l_cd + l_de + dist_ef #m\n",
"\n",
"#total drop over AF is [(20+i)*60r + i*40r+ (i-40)*161.7r - 1.5*r*61.7^2/2\n",
"drop_af = 2*res*((i_c+cur_i)*l_ac + cur_i*l_cd + cur_df*(l_de+dist_ef)-cur_uni*dist_ef*dist_ef/2) #V\n",
"#potential at F\n",
"V_f = V_a - drop_af #V\n",
"print \"The point of minimum potential is at\",round(dist,2),\"m from A.\"\n",
"print \"The minimum potential is = \",round(V_f,2),\"V.\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.16 ,Page No :- 1588"
]
},
{
"cell_type": "code",
"execution_count": 21,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Current fed at A is = 225.0 A.\n",
"Current fed at B is = 475.0 A.\n",
"Point of minimum potential from B is = 475.0 metres.\n",
"Voltage at minimum potential is = 230.72 V.\n"
]
}
],
"source": [
"#A two-wire d.c distributor AB,1000 metres long,is supplied from both ends,240V at A and 242V at B.There is a\n",
"#concentrated load of 200A at a distance of 400 metre from A and a uniformly distrubuted load of 1.0A/m between\n",
"#the mid-point and end B.Determine (i)the currents fed at A and B(ii)the point of minimum potential and\n",
"#(iii)voltage at this point.Take cable resistance as 0.005 ohm per 100 metre each core.\n",
"#####################################################################################################################\n",
"\n",
"#Given\n",
"#resistance per 100 metres\n",
"res = 2*0.005/100 #ohm/m\n",
"cur_uni = 1.0 #A/m\n",
"cur_con = 200.0 #A\n",
"len_uni = 500.0\n",
"#Let us assume that Ib current flows from point B.\n",
"#Considering a element dx in BD(500 metres) at a distance of X units(100 m each)\n",
"#voltage drop over dx = (1-100*x)*res*dx\n",
"#voltage drop over BD by integrating is = 0.05*Ib - 12.5\n",
"#voltage drop over DC = (Ib-500)*0.01\n",
"#voltage drop over CA = (Ib-700)*0.01*4\n",
"#total drop over AB = \n",
"tot_drop = 242.0-240.0\n",
"#summation of drops from AC + CD + DB\n",
"from sympy import Eq, var, solve\n",
"var('Ib') \n",
"sum = (Ib-500)*0.01 +(Ib-700)*0.01*4 + 0.05*Ib - 12.5\n",
"\n",
"eq = Eq(sum,tot_drop)\n",
"\n",
"Ib = solve(eq)\n",
"Ib = float(Ib[0])\n",
"#Total current\n",
"cur_tot = len_uni*cur_uni + cur_con\n",
"Ia = cur_tot - Ib #A\n",
"#Current in distributed load\n",
"cur_dis = Ia-cur_con #A\n",
"#point of minimum potential from D is\n",
"distD = cur_dis/cur_uni\n",
"#Therefore distance from B is\n",
"distB = len_uni-distD\n",
"#Therefore voltage drop is\n",
"from scipy.integrate import quad\n",
"\n",
"def integrand(x):\n",
" return (Ib-100*x)*res*100\n",
"\n",
"ans, err = quad(integrand, 0, (distB/100))\n",
"#Therefore potential of M is\n",
"pot_M = 242.0-ans #V\n",
"print \"Current fed at A is = \",Ia,\"A.\"\n",
"print \"Current fed at B is = \",Ib,\"A.\"\n",
"print \"Point of minimum potential from B is = \",distB,\"metres.\"\n",
"print \"Voltage at minimum potential is = \",round(pot_M,2),\"V.\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.17 ,Page No :- 1590"
]
},
{
"cell_type": "code",
"execution_count": 51,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Voltage at B is = 236.9 V.\n",
"Voltage at C is = 235.98 V.\n",
"Voltage at D is = 237.45 V.\n"
]
}
],
"source": [
"#A 400-metre ring distributor has loads as shown in Fig. 40.29(a) where distances are in metres.The resistance\n",
"#of each conductor is 0.2 ohm per 1000 metres and the loads tapped off at points B,C,D are as shown.If the\n",
"#distributor is fed at A,find voltages at B,C and D.\n",
"#################################################################################################################\n",
"\n",
"#Given\n",
"\n",
"res = 0.2/1000 #ohm/m\n",
"V_a = 240.0 #V\n",
"#loads and positions\n",
"i_b = 100.0 #A\n",
"i_c = 70.0 #A\n",
"i_d = 50.0 #A\n",
"l_ab = 60.0 #m\n",
"l_bc = 80.0 #m\n",
"l_cd = 90.0 #m\n",
"l_da = 70.0 #m\n",
"\n",
"#total drop ->70i + 90(i-50)+80(i-120)+60(i-220)=0\n",
"cur_i = (l_cd*i_d + l_bc*(i_d+i_c) + l_ab*(i_d+i_c+i_b))/(l_ab+l_bc+l_cd+l_da)\n",
"#drops in different sections\n",
"drop_da = 2*cur_i*l_da*res\n",
"drop_cd = 2*(cur_i-i_d)*l_cd*res\n",
"drop_bc = 2*abs(cur_i-i_d-i_c)*l_bc*res\n",
"drop_ab = 2*abs(cur_i-i_d-i_c-i_b)*l_ab*res\n",
"\n",
"#voltages at different points\n",
"V_d = V_a - drop_da\n",
"V_c = V_d - drop_cd\n",
"V_b = V_a - drop_ab\n",
"print \"Voltage at B is = \",round(V_b,2),\"V.\"\n",
"print \"Voltage at C is = \",round(V_c,2),\"V.\"\n",
"print \"Voltage at D is = \",round(V_d,2),\"V.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.18 ,Page No :- 1591"
]
},
{
"cell_type": "code",
"execution_count": 53,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Voltage at B is = 394.2 V.\n",
"Voltage at C is = 393.42 V.\n",
"Current in section BC is = 43.33 A.\n"
]
}
],
"source": [
"#In a direct current ring main,a voltage of 400V is maintained at A.At B,500 metres away from A,a load of 150A is taken\n",
"#and at C,300 metres from B,a load of 200A is taken.The distance between A and C is 700 metres.The resistance of each\n",
"#conductor of the mains is 0.03ohm per 1000 metres.Find the voltage at B and C and also find the current in the section BC.\n",
"##############################################################################################################################\n",
"\n",
"#Given\n",
"V_a = 400.0 #V\n",
"res = 0.03/1000 #ohm/m\n",
"#loads and positions\n",
"i_b = 150.0 #A\n",
"i_c = 200.0 #A\n",
"l_ab = 500.0 #m\n",
"l_bc = 300.0 #m\n",
"l_ca = 700.0 #m\n",
"\n",
"#total drop-> 500i + 300(i-150) + 700(i-350) = 0\n",
"cur_i = (l_bc*i_b + l_ca*(i_b+i_c))/(l_ab+l_bc+l_ca)\n",
"#current in different sections\n",
"cur_ab = cur_i\n",
"cur_bc = cur_i-i_b\n",
"cur_ca = abs(cur_bc-i_c)\n",
"#drops in different sections\n",
"drop_ab = 2*cur_ab*l_ab*res\n",
"drop_bc = 2*cur_bc*l_bc*res\n",
"#voltages in different sections\n",
"V_b = V_a-drop_ab\n",
"V_c = V_b-drop_bc\n",
"print \"Voltage at B is = \",round(V_b,2),\"V.\"\n",
"print \"Voltage at C is = \",round(V_c,2),\"V.\"\n",
"print \"Current in section BC is = \",round(cur_bc,2),\"A.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.19 ,Page No :- 1591"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Current in AB,BC,CD,DE,EA is 29.04 A, 19.04 A, 0.96 A, 30.96 A, 40.96 A respectively.\n",
"\n",
"Voltage at B,C,D,E is 217.1 V, 216.14 V, 216.15 V, 216.93 V respectively\n",
"\n",
"Current in AB,BC,DE,CE,EA is 27.72 A, 17.72 A, 32.28 A, 9.76 A, 42.28 A respectively.\n",
"\n",
"Voltage at B,C,D,E is 217.23 V, 216.34 V, 216.02 V, 216.83 V respectively\n"
]
}
],
"source": [
"#A d.c ring main ABCDE is fed at point A from a 220-V supply and the resistances(including both lead and return)\n",
"#of the various sections are as follows(in ohms):AB=0.1;BC=0.05;CD=0.01;DE=0.025 and EA=0.075.The main supplies\n",
"#loads of 10A at B; 20A at C; 30A at D and 10A at E.Find the magnitude and direction of the current flowing in each\n",
"#section and the voltage at each load point.\n",
"#If the points C and E are further linked together by a conductor of 0.05 ohm resistance and the output currents\n",
"#from the mains remain unchanged,find the new distribution of the current and voltage in the network.\n",
"#####################################################################################################################\n",
"\n",
"#Given\n",
"\n",
"V_a = 220.0 #V\n",
"#resistances of different sections\n",
"r_ab = 0.1 #ohm\n",
"r_bc = 0.05 #ohm\n",
"r_cd = 0.01 #ohm\n",
"r_de = 0.025 #ohm\n",
"r_ea = 0.075 #ohm\n",
"#loads\n",
"i_b = 10.0 #A\n",
"i_c = 20.0 #A\n",
"i_d = 30.0 #A\n",
"i_e = 10.0 #A\n",
"#total drop -> 0.1i + 0.05(i-10) + 0.01(i-30) + 0.025(i-60) + 0.075(i-70)=0\n",
"cur_i = (r_bc*i_b + r_cd*(i_b+i_c) + r_de*(i_b+i_c+i_d) + r_ea*(i_b+i_c+i_d+i_e))/(r_ab+r_bc+r_cd+r_de+r_ea)\n",
"#current in different sections\n",
"cur_ab = cur_i\n",
"cur_bc = cur_ab-i_b\n",
"cur_cd = cur_bc-i_c\n",
"cur_de = cur_cd-i_d\n",
"cur_ea = cur_de-i_e\n",
"\n",
"#drops in different sections\n",
"drop_ab = cur_ab*r_ab\n",
"drop_bc = cur_bc*r_bc\n",
"drop_de = abs(cur_de)*r_de\n",
"drop_ea = abs(cur_ea)*r_ea\n",
"#voltages at different points\n",
"V_b = V_a - drop_ab\n",
"V_c = V_b - drop_bc\n",
"V_e = V_a - drop_ea\n",
"V_d = V_e - drop_de\n",
"print \"Current in AB,BC,CD,DE,EA is\",round(cur_ab,2),\"A,\",round(cur_bc,2),\"A,\",round(abs(cur_cd),2),\"A,\",round(abs(cur_de),2),\"A,\",round(abs(cur_ea),2),\"A respectively.\" \n",
"print \"\"\n",
"print \"Voltage at B,C,D,E is\",round(V_b,2),\"V,\",round(V_c,2),\"V,\",round(V_d,2),\"V,\",round(V_e,2),\"V respectively\"\n",
"print \"\"\n",
"#part-2\n",
"#Potential difference between end points of interconnector(CE)\n",
"V_ce = V_e-V_c\n",
"#Resistance between CE ,as shown in figure\n",
"r1 = r_ab+r_bc+r_ea\n",
"r2 = r_de + r_cd\n",
"res_ce = r1*r2/(r1+r2)+ 0.05\n",
"\n",
"#Current in interconnector [I = V/R Ohm's Law]\n",
"cur_ce = V_ce/res_ce\n",
"#Current goes from E to C as E is at higher potential.\n",
"\n",
"#The current in other sections will also change.\n",
"#let us assume i1 along ED, voltage round the closed mesh EDC is zero.\n",
"#total drop -> -0.025*i1-0.01*(i1-30)+0.05*9.75 = 0\n",
"\n",
"cur_i1 = (0.05*cur_ce + r_cd*i_d)/(r_cd+r_de)\n",
"\n",
"current_ea = i_e+cur_i1+cur_ce\n",
"current_ab = (i_b+i_c+i_d+i_e)-current_ea\n",
"current_bc = current_ab-i_b\n",
"current_de = current_ea-i_e\n",
"#new drops\n",
"drop_ab = current_ab*r_ab\n",
"drop_bc = current_bc*r_bc\n",
"drop_ea = current_ea*r_ea\n",
"drop_de = current_de*r_de\n",
"\n",
"#new potentials\n",
"V_b = V_a - drop_ab\n",
"V_c = V_b - drop_bc\n",
"V_e = V_a - drop_ea\n",
"V_d = V_e - drop_de\n",
"\n",
"print \"Current in AB,BC,DE,CE,EA is\",round(current_ab,2),\"A,\",round(current_bc,2),\"A,\",round(current_de,2),\"A,\",round(cur_ce,2),\"A,\",round(current_ea,2),\"A respectively.\"\n",
"print \"\"\n",
"print \"Voltage at B,C,D,E is\",round(V_b,2),\"V,\",round(V_c,2),\"V,\",round(V_d,2),\"V,\",round(V_e,2),\"V respectively\" \n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.20 ,Page No :- 1594"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Voltage across 3 ohm load is = 244.9 V.\n",
"Voltage across 4 ohm load is = 247.9 V.\n"
]
}
],
"source": [
"#In a 3-wire distribution system,the supply voltage is 250V on each side.The load on one side is a 3 ohm\n",
"#resistance and on the other, a 4 ohm resistance.The resistance of each of the 3 conductors is 0.05 ohm.\n",
"#Find the load voltages.\n",
"#########################################################################################################\n",
"\n",
"import numpy as np\n",
"#Given\n",
"#Resistances\n",
"res_1 = 3.0 #ohm\n",
"res_2 = 4.0 #ohm\n",
"res_con = 0.05 #ohm\n",
"V_sup = 250.0 #V\n",
"\n",
"#Let the assumed directions of unknown currents be as shown in figure.\n",
"#KVL for ABCD\n",
"# (3+0.05)x + 0.05(x-y) = 250 -------------- eqn 1\n",
"a = res_1 + 2*res_con\n",
"b = -(res_con)\n",
"#KVL for DCEFD\n",
"# 0.05(y-x) + (4+0.05)y = 250 -------------- eqn 2\n",
"c = res_2+ 2*res_con \n",
"#Solving eqn 1 and eqn2\n",
"m = [[a,b],[b,c]]\n",
"n = [V_sup,V_sup]\n",
"soln = np.linalg.solve(m,n) #soln is array with its elements[x,y]\n",
"#Calculating the load voltages\n",
"#V1 = 250-0.05*x-0.05(x-y)\n",
"vol1 = V_sup - res_con*soln[0]-res_con*(soln[0]-soln[1]) #V\n",
"#V2 = 250 + 0.05(x-y)- 0.05y\n",
"vol2 = V_sup + res_con*(soln[0]-soln[1]) - res_con*soln[1] #V\n",
"print \"Voltage across 3 ohm load is = \",round(vol1,1),\"V.\"\n",
"print \"Voltage across 4 ohm load is = \",round(vol2,1),\"V.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.21 ,Page No :- 1594"
]
},
{
"cell_type": "code",
"execution_count": 57,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Potential Difference across AB is = 248.62 V.\n",
"Potential Difference across QK is = 247.83 V.\n",
"Potential Difference across CD is = 248.4 V.\n",
"Potential Difference across FE is = 247.65 V.\n"
]
}
],
"source": [
"#A 3-wire d.c distributor PQ,250 metres long,is supplied at end P at 500/250V and is loaded as under:\n",
"#Positive side: 20A 150 metres from P ; 30A 250 metres from P.\n",
"#Negative side: 24A 100 metres from P ; 36A 220 metres from P.\n",
"#The resistance of each outer wire is 0.02 ohm per 100 metres and the cross-section of the middle wire\n",
"#is one-half of the outer.Find the voltage across each load point.\n",
"##########################################################################################################\n",
"\n",
"#Given\n",
"V_PN = 250.0 #V\n",
"V_NR = 250.0 #V\n",
"res_out = 0.02/100 #ohm/m\n",
"res_mid = 2*res_out #ohm/m (Area of middle wire is half.As, R = rho*l/A .Therefore,Resistance doubles)\n",
"\n",
"#Given Currents\n",
"i_ab = 20.0 #A\n",
"i_qk = 30.0 #A\n",
"i_cd = 24.0 #A\n",
"i_fe = 36.0 #A\n",
"\n",
"#Currents in different sections\n",
"i_pa = i_ab+i_qk #A\n",
"i_aq = i_qk #A\n",
"i_fk = i_qk #A\n",
"i_bf = i_fe-i_qk #A\n",
"i_bc = i_ab-i_bf #A\n",
"i_cn = i_cd-i_bc #A\n",
"i_de = i_fe #A\n",
"i_dr = i_cd+i_fe #A\n",
"\n",
"\n",
"#lengths of different sections\n",
"l_pa = 150.0 #m\n",
"l_aq = 100.0 #m\n",
"l_kf = 250.0-220.0 #m\n",
"l_bc = 150.0-100.0 #m\n",
"l_bf = 220.0-150.0 #m\n",
"l_cn = 100.0 #m\n",
"l_de = 220.0-100.0 #m\n",
"l_dr = 100.0 #m\n",
"\n",
"#Resistances of different sections\n",
"r_pa = l_pa*res_out #ohm\n",
"r_aq = l_aq*res_out #ohm\n",
"r_kf = l_kf*res_mid #ohm\n",
"r_bc = l_bc*res_mid #ohm\n",
"r_bf = l_bf*res_mid #ohm\n",
"r_cn = l_cn*res_mid #ohm\n",
"r_de = l_de*res_out #ohm\n",
"r_dr = l_dr*res_out #ohm\n",
"\n",
"#Drop across different sections\n",
"drop_pa = r_pa*i_pa #V\n",
"drop_aq = r_aq*i_aq #V\n",
"drop_kf = r_kf*i_fk #V\n",
"drop_bc = r_bc*i_bc #V\n",
"drop_bf = r_bf*i_bf #V\n",
"drop_cn = r_cn*i_cn #V\n",
"drop_de = r_de*i_de #V\n",
"drop_dr = r_dr*i_dr #V\n",
"\n",
"#Voltages across different sections\n",
"vol_ab = V_PN - drop_pa - drop_bc + drop_cn #V\n",
"vol_qk = vol_ab - drop_aq - drop_kf + drop_bf #V\n",
"vol_cd = V_NR - drop_cn - drop_dr #V \n",
"vol_fe = vol_cd + drop_bc - drop_bf - drop_de #V\n",
"\n",
"print \"Potential Difference across AB is = \",round(vol_ab,2),\"V.\"\n",
"print \"Potential Difference across QK is = \",round(vol_qk,2),\"V.\"\n",
"print \"Potential Difference across CD is = \",round(vol_cd,2),\"V.\"\n",
"print \"Potential Difference across FE is = \",round(vol_fe,2),\"V.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.22 ,Page No :- 1597"
]
},
{
"cell_type": "code",
"execution_count": 58,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total load on main generator is = 155.0 kW.\n",
"Load on Balancer 1 is = 22.5 kW.\n",
"Load on Balancer 2 is = 27.5 kW.\n"
]
}
],
"source": [
"#A d.c 3-wire system with 500-V between outers has lighting load of 100kW on the positive and 50kW on the\n",
"#negative side.If,at this loading,the balancer machines have each a loss of 2.5kW,Calculate the kW loading\n",
"#of each balancer machine and the total load on the system.\n",
"###########################################################################################################\n",
"\n",
"#Given\n",
"V_out = 500.0 #V\n",
"load_p = 100.0 #kW (positive side)\n",
"load_n = 50.0 #KW (negative side)\n",
"load_b = 2.5 #kW (balancer machine)\n",
"#total load on main generator\n",
"load_tot = load_p + load_n + 2*load_b #kW\n",
"#Output current of main generator\n",
"cur_out = load_tot*1000/V_out #W/V->A\n",
"#load current on positive side\n",
"cur_p = load_p*1000/(V_out/2) #A\n",
"#load current on negative side\n",
"cur_n = load_n*1000/(V_out/2) #A\n",
"#Current through neutral(Out of balance)\n",
"cur_o = cur_p-cur_n #A\n",
"\n",
"#Currents of balancer\n",
"cur_b1 = cur_p-cur_out #A\n",
"cur_b2 = cur_o - cur_b1 #A\n",
"\n",
"#Load on balancer\n",
"load_b1 = (V_out/2)*cur_b1/1000 #kW\n",
"load_b2 = (V_out/2)*cur_b2/1000 #kW\n",
"\n",
"print \"Total load on main generator is = \",round(load_tot,2),\"kW.\"\n",
"print \"Load on Balancer 1 is = \",round(load_b1,2),\"kW.\"\n",
"print \"Load on Balancer 2 is = \",round(load_b2,2),\"kW.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.23 ,Page No :- 1598"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total load on main generator is = 1216.0 kW.\n",
"Current through Balancer 1 is = 168.0 A.\n",
"Current through Balancer 2 is = 232.0 A.\n"
]
}
],
"source": [
"#In a 500/250-V d.c 3-wire system,there is a current of 2000A on the +ve side, 1600A on the negative side\n",
"#and a load of 300 kW across the outers.The loss in each balancer set is 8 kW.Calculate the current in each\n",
"#armature of the balancer set and total load on the main generator.\n",
"#############################################################################################################\n",
"\n",
"#Given\n",
"V_out = 500.0 #V\n",
"cur_p = 2000.0 #A (current on positive side)\n",
"cur_n = 1600.0 #A (current on negative side)\n",
"load_ext = 300.0 #kW (across outers)\n",
"load_b = 8.0 #kW (loss in balancer set)\n",
"#loading on positive side\n",
"load_p = (cur_p*(V_out/2))/1000 #kW\n",
"#loading on negative side\n",
"load_n = (cur_n*(V_out/2))/1000 #kW\n",
"#Total loading on main generator\n",
"load_tot = load_p + load_n + 2*load_b + load_ext #kW\n",
"\n",
"#current on main generator -> I = W/V\n",
"cur_tot = load_tot*1000/V_out #A\n",
"\n",
"#current through neutral(out of balance)\n",
"cur_o = cur_p-cur_n #A\n",
"\n",
"#current through external resistance\n",
"cur_ext = load_ext*1000/V_out #A\n",
"\n",
"#current through balancer sets\n",
"cur_b1 = (cur_p+cur_ext)-cur_tot #A\n",
"cur_b2 = cur_o - cur_b1 #A\n",
"\n",
"print \"Total load on main generator is = \",round(load_tot,2),\"kW.\"\n",
"print \"Current through Balancer 1 is = \",round(cur_b1,2),\"A.\"\n",
"print \"Current through Balancer 2 is = \",round(cur_b2,2),\"A.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.24 ,Page No :- 1598"
]
},
{
"cell_type": "code",
"execution_count": 60,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Current supplied by generator is = 7000.0 A.\n",
"Current in positive side is = 6000.0 A.\n",
"Current in negative side is = 8000.0 A.\n",
"Current in neutral is = 2000.0 A.\n",
"Current through armature 1 is = 1000.0 A.\n",
"Current through armature 2 is = 1000.0 A.\n"
]
}
],
"source": [
"#On a 3-wire d.c distribution system with 500V between outers,there is a load of 1500kW on the positive\n",
"#side and 2000 kW on the negative side.Calculate the current in the neutral and in each of the balancer\n",
"#armatures and the total current supplied by the generator.Neglect losses.\n",
"##########################################################################################################\n",
"\n",
"#Given\n",
"V_out = 500.0 #V\n",
"load_p = 1500.0 #kW (load on positive side)\n",
"load_n = 2000.0 #kW (load on negative side)\n",
"#total loading on main generator\n",
"load_tot = load_p + load_n #kW\n",
"#current supplied by generator\n",
"cur_tot = load_tot*1000/V_out #A\n",
"#current on positive side\n",
"cur_p = load_p*1000/(V_out/2) #A\n",
"#current on negative side\n",
"cur_n = load_n*1000/(V_out/2) #A\n",
"#current in neutral(out of balance)\n",
"cur_o = abs(cur_p-cur_n) #A\n",
"#current through armatures\n",
"cur_b1 = cur_tot-cur_p #A\n",
"cur_b2 = cur_o-cur_b1 #A\n",
"\n",
"print \"Current supplied by generator is = \",cur_tot,\"A.\"\n",
"print \"Current in positive side is = \",cur_p,\"A.\"\n",
"print \"Current in negative side is = \",cur_n,\"A.\"\n",
"print \"Current in neutral is = \",cur_o,\"A.\"\n",
"print \"Current through armature 1 is = \",cur_b1,\"A.\"\n",
"print \"Current through armature 2 is = \",cur_b2,\"A.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.25 ,Page No :- 1599"
]
},
{
"cell_type": "code",
"execution_count": 61,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Current in balancer set 1 is = 22.0 A.\n",
"Current in balancer set 2 is = 28.0 A.\n",
"Output of main generator is = 119.5 kW.\n"
]
}
],
"source": [
"#A 125/250 V,3-wire distributor has an out-of-balance current of 50 A and larger load of 500 A.The balancer\n",
"#set has a loss of 375 W in each machine.Calculate the current in each of the balancer machines and output\n",
"#of main generator.\n",
"############################################################################################################\n",
"\n",
"#Given\n",
"V_out = 250.0 #V\n",
"#Currents\n",
"cur_p = 500.0 #A\n",
"cur_o = 50.0 #A\n",
"cur_n = cur_p - cur_o #A\n",
"#larger Load\n",
"load_p = cur_p*(V_out/2)/1000 #kW\n",
"#smaller Load\n",
"load_n = cur_n*(V_out/2)/1000 #kW\n",
"#Balancer loss\n",
"loss_b = 2*375.0/1000 #kW\n",
"#total load on generator\n",
"load_tot = load_p + load_n + loss_b\n",
"#current from main generator -> VI = W\n",
"cur_tot = load_tot*1000/V_out #A\n",
"\n",
"#Current in balancer sets\n",
"cur_b1 = cur_p - cur_tot #A\n",
"cur_b2 = cur_o - cur_b1 #A\n",
"print \"Current in balancer set 1 is = \",cur_b1,\"A.\"\n",
"print \"Current in balancer set 2 is = \",cur_b2,\"A.\"\n",
"print \"Output of main generator is = \",load_tot,\"kW.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.26 ,Page No :- 1599"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total load on main generator is = 1210.0 kW.\n",
"Load on Balancer set 1 is = 20.0 kW.\n",
"Load on balancer set 2 is = 30.0 kW.\n"
]
}
],
"source": [
"#The load on d.c 3-wire system with 500 V between outers consists of lighting current of 1500 A on the\n",
"#positive side and 1300 A on the negative side while motors connected across the outers absorb 500kW.\n",
"#Assuming that at this loading,the balancer machines have each a loss of 5kW,calculate the load on the\n",
"#main generator and on each of the balancer machines.\n",
"##########################################################################################################\n",
"\n",
"#Given\n",
"cur_p = 1500.0 #A\n",
"cur_n = 1300.0 #A\n",
"V_out = 500.0 #V\n",
"load_ext = 500.0 #kW\n",
"loss_b = 2*5.0 #kW\n",
"\n",
"#current through external load\n",
"cur_ext = load_ext*1000/V_out #A\n",
"#larger load\n",
"load_p = cur_p*(V_out/2)/1000 #kW\n",
"#smaller load\n",
"load_n = cur_n*(V_out/2)/1000 #kW\n",
"#total load on generator\n",
"load_tot = load_p + load_n + loss_b + load_ext #kW\n",
"#current from generator -> VI = W\n",
"cur_tot = load_tot*1000/V_out #A\n",
"#current through neutral(out of balance)\n",
"cur_o = cur_p-cur_n #A\n",
"#current through balancer sets\n",
"cur_b1 = (cur_p+cur_ext)-cur_tot #A\n",
"cur_b2 = cur_o-cur_b1 #A\n",
"#load of balancer sets\n",
"load_b1 = cur_b1*(V_out/2)/1000 #kW\n",
"load_b2 = cur_b2*(V_out/2)/1000 #kW\n",
"\n",
"print \"Total load on main generator is = \",load_tot,\"kW.\"\n",
"print \"Load on Balancer set 1 is = \",load_b1,\"kW.\"\n",
"print \"Load on balancer set 2 is = \",load_b2,\"kW.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.27 ,Page No :- 1599"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Voltage across Balancer 1 is = 230.0 A.\n",
"Voltage across Balancer 2 is = 250.0 A.\n",
"Load current on main generator is = 1110.0 A.\n"
]
}
],
"source": [
"#A d.c 3-wire system with 480 V across outers supplies 1200 A on the positive and 1000 A on the negative side.\n",
"#The balancer machines have each an armature resistances of 0.1W and take 10 A on no-load.Find\n",
"#(a)the voltage across each balancer and\n",
"#(b)the total load on the main generator and the current loading of each balancer machine.\n",
"#The balancer field windings are in series across the outers\n",
"################################################################################################################\n",
"\n",
"#Given\n",
"V_out = 480.0 #V\n",
"#currents\n",
"cur_p = 1200.0 #A\n",
"cur_n = 1000.0 #A\n",
"cur_o = cur_p - cur_n #A (out of balance)\n",
"#armature resistance \n",
"res_arm = 0.1 #ohm\n",
"#no-load current\n",
"cur_nold = 10.0 #A\n",
"\n",
"#Let us assume current Im flows through mtoring machine,then (200-Im) flows through generating machine.\n",
"#Let Vg and Vm be potential difference of 2 machines.\n",
"\n",
"#Total losses in sets = no-load losses + Cu losses in two machines\n",
"#loss_set = V_out*cur_nold + 0.1*Im^2+ 0.1*(200-Im)^2\n",
"#Vm*Im = Vg*Ig + loss_set\n",
"#Now, Vm = Eb+Im*Ra Vg = Eb-Ig*Ra\n",
"Eb = V_out/2-res_arm*cur_nold\n",
"\n",
"#Therefore, Vm = 239 + Im*0.1 and Vg = 239 - (200-Im)*0.1\n",
"#Hence,Equation is \n",
"#(239+0.1*Im)*Im = [239 - (200-Im)*0.1]*(200-Im) + loss_set\n",
"#Simplified -> 239Im = 239*(200-Im)+4800\n",
"\n",
"#Solving this equation\n",
"from sympy import Eq, var, solve\n",
"var('Im') \n",
"eq = Eq(Eb*(2*Im-cur_o),V_out*cur_nold)\n",
"Im = solve(eq)\n",
"Im = int(Im[0])\n",
"Ig = cur_o-Im\n",
"#Voltage across balancers\n",
"\n",
"Vm = Eb + Im*res_arm #V\n",
"Vg = Eb - Ig*res_arm #V \n",
"\n",
"#Load on main generator\n",
"cur_load = cur_p - Ig #A\n",
"print \"Voltage across Balancer 1 is = \",round(Vg,2),\"A.\"\n",
"print \"Voltage across Balancer 2 is = \",round(Vm,2),\"A.\"\n",
"print \"Load current on main generator is = \",round(cur_load,2),\"A.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.28 ,Page No :- 1600"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Voltage on positive side is = 283.0 V.\n",
"Voltage on negative side is = 177.0 V.\n"
]
}
],
"source": [
"#A d.c 3-wire system with 460V between outers supplies 250kW on the positive and 400kW on the negative side,\n",
"#the voltages being balanced.Calculate the voltage on the positive and negative side,the voltages being balanced.\n",
"#Calculate the voltage on the positive and negative sides repectively,if the neutral wire becomes disconnected\n",
"#from balancer set.\n",
"#################################################################################################################\n",
"\n",
"#Given\n",
"V_mid = 230.0 #V\n",
"V_out = 460.0 #V\n",
"#loads\n",
"load_p = 250.0 #kW\n",
"load_n = 400.0 #kW\n",
"#resistance on positive side -> (V^2/R) = W\n",
"res_p = (V_mid*V_mid)/(load_p*1000) #ohm\n",
"\n",
"#resistance on negative side -> (V^2/R) = W\n",
"res_n = (V_mid*V_mid)/(load_n*1000) #ohm\n",
"\n",
"#Voltages after disconnecting balancer set\n",
"vol_p = (res_p/(res_p+res_n))*V_out #V\n",
"vol_n = V_out - vol_p #V\n",
"\n",
"print \"Voltage on positive side is = \",round(vol_p),\"V.\"\n",
"print \"Voltage on negative side is = \",round(vol_n),\"V.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 40.29 ,Page No :- 1601"
]
},
{
"cell_type": "code",
"execution_count": 66,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Terminal potential difference of the booster is = 180.0 V.\n",
"Output of booster is = 21.6 kW.\n"
]
}
],
"source": [
"#A 2-wire system has the voltage at the supply end maintained at 500.The line is 3 km long.If the full-load\n",
"#current is 120 A,what must be the booster voltage and output in order that the far end voltage may also be 500 V.\n",
"#Take the resistance of the cable at the working temperature as 0.5ohm/kilometre.\n",
"####################################################################################################################\n",
"\n",
"#Total resistance of line\n",
"res_tot = 0.5*3 #ohm\n",
"#Full load current\n",
"cur_full = 120.0 #A\n",
"\n",
"#drop in the line-> V=IR\n",
"drop = res_tot*cur_full #V\n",
"\n",
"#Output of booster ->VI = W\n",
"output = drop*cur_full/1000 #kW\n",
"\n",
"print \"Terminal potential difference of the booster is = \",drop,\"V.\"\n",
"print \"Output of booster is = \",round(output,2),\"kW.\""
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": []
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 2",
"language": "python",
"name": "python2"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 2
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython2",
"version": "2.7.11"
}
},
"nbformat": 4,
"nbformat_minor": 0
}
PKj„ìJxÕÏ¹°¹°;A TEXTBOOK OF ELECTRICAL TECHNOLOGY VOL III/chapter41.ipynb{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# CHAPTER 41 : A.C. Transmission and Distribution"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.1 , PAGE NO :- 1613"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Cu for 3-phase system = 0.29 * Cu for dc system.\n"
]
}
],
"source": [
"'''A 3-phase, 4-wire system is used for lighting. Compare the amount of copper required with that needed for a\n",
"2-wire D.C. system with same line voltage. Assume the same losses and balanced load. The neutral is one half\n",
"the cross-section of one of the respective outers.'''\n",
"\n",
"from sympy import Symbol\n",
"\n",
"#(a) Two-wire DC\n",
"#We know that, I = P/V .Therefore\n",
"I = Symbol('I')\n",
"#Also let the resistance be R1\n",
"R1 = Symbol('R1')\n",
"#power loss is\n",
"loss1 = 2*(I*I)*R1 \n",
"\n",
"#(b) Three-phase,4-wire\n",
"\n",
"#We know that, I2 = P/3*V .Therefore I2 = I/3\n",
"#Also let the resistance be R2\n",
"R2 = Symbol('R2')\n",
"#power loss is\n",
"loss2 = 3*(I/3*I/3)*R2\n",
"\n",
"#loss1/loss2 = 2*I^2*R1/(I^2*R2*1/3) .Let ratio of resistances is R1/R2 = r1_r2\n",
"#As loss1 = loss2\n",
"r1_r2 = 1.0/6\n",
"\n",
"#Let the ratio of areas of conductors be a1_a2 As R o< 1/A\n",
"a2_a1 = 1/r1_r2\n",
"\n",
"#Cu loss of 3-phase/Cu loss of dc system = 3.5*A2*l/2*A1*l\n",
"ratio = (3.5/2)/a2_a1\n",
"\n",
"print \"Cu for 3-phase system = \",round(ratio,2),\" * Cu for dc system.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.2 , PAGE NO :- 1613"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Wt of copper for 3-conductors = 9349.58 kg.\n"
]
}
],
"source": [
"'''Estimate the weight of copper required to supply a load of 100 MW at upf by a 3-phase, 380-kV system over a distance\n",
"of 100 km. The neutral point is earthed. The resistance of the conductor is 0.045 ohm/cm^2/km. The weight of copper\n",
"is 0.01 kg/cm^3. The efficiency of transmission can be assumed to be 90 percent.'''\n",
"\n",
"#Power loss in the line\n",
"loss1 = (1 - 0.9) * 100.0 #MW\n",
"#Line current Il = P/vl*cosQ\n",
"Il = 100 * 1.0e+6/(1.732*380*(1.0e+3)*1) #A\n",
"#Since I^2R loss in 3-conductors is loss1, loss per conductor is\n",
"loss_c = loss1*1.0e+6/3 #W\n",
"\n",
"#Resistance per conductor Using loss = I^2*R\n",
"R_c = loss_c/(Il*Il) #ohm\n",
"#Resistance per conductor per km\n",
"R_km = R_c/100 #ohm\n",
"#Conductor cross-section\n",
"Vol = 0.045/R_km #m^3\n",
"#Volume of copper per meter run\n",
"Vol = Vol*100 #cm^3\n",
"#Weight of copper for 3-conductor for 100 km length\n",
"wt = 3 * (Vol * 0.01) * 100 * 1000 #kg\n",
"\n",
"print \"Wt of copper for 3-conductors = \",round(wt,2),\"kg.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.3 , PAGE NO :- 1614"
]
},
{
"cell_type": "code",
"execution_count": 38,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Additional power transmitted = 80.5 %.\n"
]
}
],
"source": [
"'''A d.c. 2-wire distribution system is converted into a.c. 3-phase, 3-wire system by adding a third conductor of\n",
"the same size as the two existing conductors. If voltage between conductors and percentage power loss remain the same, calculate\n",
"the percentage additional balanced load which can now be carried by the conductors at 0.95 p.f.'''\n",
"\n",
"from sympy import Symbol\n",
"\n",
"#(a)DC 2-wire system\n",
"\n",
"#Let us assume\n",
"V = 1.0\n",
"R = 1.0\n",
"I1 = Symbol('I1')\n",
"#Power transmitted\n",
"P1 = V*I1\n",
"#Power loss\n",
"loss1 = 2*(I1**2)*R \n",
"#%power loss = power loss/Power transmitted\n",
"ploss1 = loss1/P1\n",
"\n",
"#(b)3-phase, 3-wire system\n",
"\n",
"I2 = Symbol('I2')\n",
"#Power transmitted P = 1.732*VI*cosQ\n",
"P2 = 1.732*V*I2*(0.95)\n",
"#Power loss\n",
"loss2 = 3*(I2**2)*R\n",
"#%power loss = power loss/Power transmitted\n",
"ploss2 = loss2/P2\n",
"\n",
"#As ploss1 == ploss2\n",
"I2 = 2*(0.95)*I1/1.732\n",
"P2 = 1.732*V*I2*(0.95)\n",
"#Add. power transmitted\n",
"ptrans = (P2 - P1)/P1 *100 #%\n",
"\n",
"print \"Additional power transmitted = \",round(ptrans,2),\"%.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.4 , PAGE NO :- 1614"
]
},
{
"cell_type": "code",
"execution_count": 41,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Additional power transmitted = 30.0 MW.\n"
]
}
],
"source": [
"'''A 2-phase, 3-wire a.c. system has a middle conductor of same cross-sectional area as the outer and supplies a load\n",
"of 20 MW. The system is converted into 3-phase, 4-wire system by running a neutral wire. Calculate the new power which\n",
"can be supplied if voltage across consumer terminal and percentage line losses remain the same. Assume balanced load.'''\n",
"\n",
"from sympy import Symbol\n",
"\n",
"#(a)2-phase 3-wire system\n",
"\n",
"#Let us assume\n",
"V = 1.0\n",
"R = 1.0\n",
"cosQ = 1.0\n",
"I1 = Symbol('I1')\n",
"#Power transmitted\n",
"P1 = 2*V*I1*cosQ\n",
"#Power loss\n",
"loss1 = 2*(I1**2)*R \n",
"#%power loss = power loss/Power transmitted\n",
"ploss1 = loss1/P1\n",
"\n",
"#(b)3-phase, 4-wire system\n",
"\n",
"I2 = Symbol('I2')\n",
"#Power transmitted P = 1.732*VI*cosQ\n",
"P2 = 3*V*I2*cosQ\n",
"#Power loss\n",
"loss2 = 3*(I2**2)*R\n",
"#%power loss = power loss/Power transmitted\n",
"ploss2 = loss2/P2\n",
"\n",
"#As ploss1 == ploss2\n",
"I2 = I1\n",
"P2 = 3*V*I2*cosQ\n",
"#New Power that can be supplied P1/P2 = 20/x\n",
"pnew = 20*P2/P1 #MW\n",
"\n",
"print \"Additional power transmitted = \",round(pnew,2),\"MW.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.5 , PAGE NO :- 1617"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"L = 1.02 uH/m.\n",
"L = 10.92 uH/m.\n"
]
}
],
"source": [
"'''What is the inductance per loop metre of two parallel conductors of a single phase system if each has a diameter of 1\n",
"cm and their axes are 5 cm apart when conductors have a relative permeability of (a) unity and (b) 100. The relative\n",
"permeability of the surrounding medium is unity in both cases. End effects may be neglected and the current may be assumed\n",
"uniformly distributed over cross-section of the wires.'''\n",
"\n",
"import math as m\n",
"# (a)\n",
"# u = u0\n",
"u0 = 4*3.14*1.0e-7 #H/m\n",
"ui = 1.0\n",
"L = u0/3.14*(m.log(5.0/0.5) + ui/4)*1.0e+6 #uH/m\n",
"print \"L = \",round(L,2),\"uH/m.\"\n",
"# (b)\n",
"# u = u0\n",
"u0 = 4*3.14*1.0e-7 #H/m\n",
"ui = 100.0\n",
"L = u0/3.14*(m.log(5.0/0.5) + ui/4)*1.0e+6 #uH/m\n",
"print \"L = \",round(L,2),\"uH/m.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.6 , PAGE NO :- 1617"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"loop impedance is = 19.83 ohm.\n"
]
}
],
"source": [
"'''A 20-km single-phase transmission line having 0.823 cm diameter has two line conductors separated by 1.5 metre.\n",
"The conductor has a resistance of 0.311 ohm per kilometre. Find the loop impedance of this line at 50 Hz.'''\n",
"\n",
"import math as m\n",
"\n",
"#Given\n",
"length = 20.0*1000 #m\n",
"u = 4*3.14*1.0e-7 #H\n",
"ui = 1.0\n",
"D = 1.5 #m\n",
"r = 0.823/2*1.0e-2 #m\n",
"#inductance is\n",
"L = length*(u/3.14*(m.log(D/r) + ui/4 )) #H\n",
"#reactance is\n",
"X = 2*3.14*50.0*L #ohm\n",
"#loop resistance\n",
"R = 2*length*0.311/1000 #ohm\n",
"#impedance is\n",
"Z = m.sqrt(X*X + R*R) #ohm\n",
"\n",
"print \"loop impedance is = \",round(Z,2),\"ohm.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.7 , PAGE NO :- 1618"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Capacitance (excluding ground effect) = 0.00928 uF/km\n",
"Capacitance (including ground effect) = 0.00932 uF/km\n"
]
}
],
"source": [
"'''The conductors in a single-phase transmission line are 6 m above ground. Each conductor has a diameter of 1.5 cm\n",
"and the two conductors are spaced 3 m apart. Calculate the capacitance per km of the line\n",
"(i) excluding ground effect and (ii) including the ground effect.'''\n",
"\n",
"import math as m\n",
"#Given\n",
"D = 3.0 #m (distance between conductors)\n",
"r = 1.5/2*1e-2 #m (radius)\n",
"h = 6.0 #m (height from ground)\n",
"eps = 8.85e-12 #epsilon (constant)\n",
"\n",
"#(i)Capacitance per km excluding ground effect\n",
"Cn = 2*3.14*eps/(m.log(D/r))*1.0e+9 #uF/km\n",
"print \"Capacitance (excluding ground effect) = \",round(Cn,5),\"uF/km\"\n",
"\n",
"#(ii)Capacitance including ground effect\n",
"Cn = 2*3.14*eps/(m.log(D/(r*m.sqrt(1 + D*D/(4*h*h)))))*1.0e+9 #uF/km\n",
"print \"Capacitance (including ground effect) = \",round(Cn,5),\"uF/km\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.8 , PAGE NO :- 1620"
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Sending-end voltage = 3859.08 V.\n"
]
}
],
"source": [
"'''A single-phase line has an impedance of 5ang(60) and supplies a load of 120 A,3,300 V at 0.8 p.f. lagging.\n",
"Calculate the sending-end voltage and draw a vector diagram.'''\n",
"\n",
"import cmath as cm\n",
"import math as m\n",
"#Given\n",
"Er = cm.rect(3300.0,0) #V (Voltage) \n",
"Z = cm.rect(5.0,3.14/3) #ohm (Impedance)\n",
"pf = 0.8 #power factor\n",
"theta = m.acos(pf) #Q (power factor angle)\n",
"\n",
"I = cm.rect(120.0,-theta) #A (current)\n",
"\n",
"#Voltage drop\n",
"V = (I)*(Z) #V\n",
"\n",
"#Sending-end voltage is\n",
"Es = Er + V #V\n",
"\n",
"Es = m.sqrt(Es.real**2 + Es.imag**2)#V\n",
"\n",
"print \"Sending-end voltage = \",round(Es,2),\"V.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPE 41.9 , PAGE NO :- 1620"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Sending end voltage is = 33.71 V.\n",
"Power factor = 0.796 lag\n"
]
}
],
"source": [
"'''An overhead, single-phase transmission line delivers 1100 kW at 33 kV at 0.8 p.f. lagging. The total resistance of the line is\n",
"10ohm and total inductive reactance is 15ohm . Determine\n",
"(i) sending-end voltage (ii) sending-end p.f. and (iii) transmission efficiency.'''\n",
"\n",
"import math as m\n",
"import cmath as cm\n",
"#Given\n",
"P = 1100.0 #kW (Power delivered)\n",
"V = 33.0 #kV (Voltage)\n",
"pf = 0.8 # (Power factor)\n",
"R = 10.0 #ohm (Resistance)\n",
"X = 15.0 #ohm (Reactance) \n",
"#Full-load line current is\n",
"I = P/(V*pf) #A\n",
"theta = m.acos(pf) #Q (power factor angle)\n",
"I = cm.rect(I,-theta) #ohm (impedance)\n",
"#Line-loss\n",
"loss = I*I*R/1000 #kW\n",
"\n",
"\n",
"#(iii)Transmission efficiency\n",
"eff = (P/(P+loss))*100 #%\n",
"#Line voltage drop is IZ\n",
"Z = R + 1j * X\n",
"\n",
"\n",
"#Sending end voltage is\n",
"Es = V + I*Z/1000 #V\n",
"Es1 = m.sqrt(Es.real**2 + Es.imag**2) #V\n",
"print \"Sending end voltage is = \",round(Es1,2),\"V.\"\n",
"\n",
"\n",
"#Sending end pf angle is\n",
"theta2 = theta + cm.phase(Es)\n",
"pf2 = m.cos(theta2) #power factor\n",
"\n",
"print \"Power factor = \",round(pf2,3),\"lag\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.10 , PAGE NO :- 1621"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Max. length in km is = 13.59 km.\n"
]
}
],
"source": [
"'''What is the maximum length in km for a 1-phase transmission line having copper conductors of 0.775 cm^2 cross-section\n",
"over which 200 kW at unity power factor and at 3300 V can be delivered ? The efficiency of transmission is 90 per cent.\n",
"Take specific resistance as (1.725 * 10â€“8) ohm-m.'''\n",
"\n",
"#Given\n",
"A = 0.775e-4 #m^2 (Area of copper conductor)\n",
"P = 200.0 #kW (Power)\n",
"V = 3300.0 #V (Voltage)\n",
"pf = 1.0 # (Power factor)\n",
"rho = 1.725e-8 #ohm-m (Specific Resistance)\n",
"\n",
"#Sending-end power is\n",
"Es = P/0.9 #kW\n",
"#Line losses\n",
"loss = Es - P #kW\n",
"#Line current\n",
"I = P/(V*pf)*1000 #A\n",
"\n",
"\n",
"#If R is resistance of consuctor then 2*I^2*R = loss\n",
"R = loss/(2*I*I)*1000 #ohm\n",
"\n",
"#Now, using R = rho*l/A . The length is\n",
"l = R*A/rho #m \n",
"l = l/1000 #km\n",
"print \"Max. length in km is = \",round(l,2),\"km.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.11 , PAGE NO :- 1621"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"load current = 209.19 A\n",
"Voltage at sending end is = 2419.76 V.\n",
"Sending power = 526.03 kW.\n",
"Sending end reactive power = 701.38 kVAR.\n",
"Sending end volt ampere kVA = 876.72 kVA.\n"
]
}
],
"source": [
"'''An industrial load consisting of a group of induction motors which aggregate 500 kW at 0.6 power factor lagging is\n",
"supplied by a distribution feeder having an equivalent impedance of (0.15 + j0.6) ohm. The voltage at the load end of\n",
"the feeder is 2300 volts.\n",
"(a) Determine the load current.\n",
"(b) Find the power, reactive power and voltampere supplied to the sending end of the feeder.\n",
"(c) Find the voltage at the sending end of the feeder.'''\n",
"\n",
"import math as m\n",
"import cmath as cm\n",
"#Given\n",
"P = 500.0 #kW (Power)\n",
"V = 2300.0 #V (Voltage)\n",
"pf = 0.6 # (Power factor)\n",
"#(a)line current\n",
"I = P*1000/(V*pf*1.732) #A (Current)\n",
"theta = m.acos(pf) # (Power factor angle) \n",
"I1 = cm.rect(I,-theta) #A\n",
"Z = 0.15 + 1j*0.6 #ohm (Impedance)\n",
"\n",
"#Voltage drop is\n",
"drop = I1*Z #V\n",
"#Sending end voltage is\n",
"Es = V + drop #V\n",
"Es = abs(Es)\n",
"\n",
"#Sending end pf angle is\n",
"theta2 = theta + cm.phase(Es)\n",
"pf2 = m.cos(theta2) #power factor\n",
"pf21 = m.sin(theta2) #sinQ component\n",
"\n",
"#Sending power = root(3)*Vl*Il*cosQ\n",
"Ps = 1.732*Es*I*pf2/1000 #kW\n",
"\n",
"#Sending end reactive power = root(3)*Vl*Il*sinQ\n",
"Prs = 1.732*Es*I*pf21/1000 #kVAR\n",
"\n",
"#Sending end volt ampere kVA = root(3)*Vl*Il\n",
"Pvs = 1.732*Es*I/1000 #kVA\n",
"\n",
"print \"load current = \",round(I,2),\"A\"\n",
"print \"Voltage at sending end is = \",round(Es,2),\"V.\"\n",
"print \"Sending power = \",round(Ps,2),\"kW.\"\n",
"print \"Sending end reactive power = \",round(Prs,2),\"kVAR.\"\n",
"print \"Sending end volt ampere kVA = \",round(Pvs,2),\"kVA.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.12 , PAGE NO :- 1622"
]
},
{
"cell_type": "code",
"execution_count": 15,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Resistance R = 3.24 ohm.\n",
"Reactance X = 6.11 ohm.\n"
]
}
],
"source": [
"'''A 33-kV, 3-phase generating station is to supply 10 MW load at 31 kV and 0.9 power factor lagging over a 3-phase\n",
"transmission line 3 km long. For the efficiency of the line to be 96% , what must be the resistance and reactance of the line?'''\n",
"\n",
"import math as m\n",
"#Given\n",
"#Power Output\n",
"pout = 10.0 #MW (Power output)\n",
"eff = 0.96 # (Efficiency)\n",
"pin = pout/eff #MW (Power input)\n",
"\n",
"#Total loss\n",
"loss = pin - pout #MW\n",
"\n",
"#Full-load current I = P/V*pf*root(3)\n",
"I = pout*1e+6/(31.0e+3*0.9*1.732) #A\n",
"\n",
"#If R is resistance per phase,then 3*I*I*R = loss\n",
"R = loss*1e+6/(3*I*I) #ohm\n",
"\n",
"#Now, Vs per phase is\n",
"Vs = 33/1.732 #kV\n",
"#Vr per phase is\n",
"Vr = 31/1.732 #kV\n",
"#Using Vs = Vr + I(RcosQ + XsinQ )\n",
"X = ((Vs - Vr)/I*1000 - R*0.9)/m.sqrt(1 - 0.9*0.9)\n",
"\n",
"print \"Resistance R = \",round(R,2),\"ohm.\"\n",
"print \"Reactance X = \",round(X,2),\"ohm.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 41.13 , PAGE NO :- 1622"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Voltage at recieving end = 59.91 V.\n",
"Angle between voltages = 3.12 degrees.\n",
"Transmission efficiency is = 92.59 %.\n"
]
}
],
"source": [
"'''A balanced Y-connected load of (300 + j100) ohm is supplied by a 3-phase line 40 km long with an impedance of\n",
"(0.6 + j0.7) ohm per km (line-to-neutral). Find the voltage at the receiving end when the voltage at the sending\n",
"end is 66 kV. What is the phase angle between these voltages? Also, find the transmission efficiency of the line.'''\n",
"\n",
"import math as m\n",
"from sympy import Symbol,solve,Eq\n",
"#Resistance for 40 km conductor length\n",
"R = 40 * 0.6 #ohm\n",
"#Reactance for 40 km conductor length\n",
"X = 40 * 0.7 #ohm \n",
"#Total resistance/phase\n",
"R1 = R + 300 #ohm\n",
"#Total reactance/phase\n",
"X1 = X + 100 #ohm\n",
"#Total impedance/phase\n",
"Z = m.sqrt(R1**2 + X1**2) #ohm\n",
"#Line current\n",
"Il = 66000.0/1.732/Z #A \n",
"\n",
"#Now,\n",
"theta = m.atan(100.0/300.0)\n",
"cosQ = m.cos(theta)\n",
"sinQ = m.sin(theta)\n",
"#Voltage drop in conductor resistance\n",
"dropR = Il*R #V\n",
"#Voltage drop in conductor reactance\n",
"dropX = Il*X #V\n",
"\n",
"#Let us assume recieving end voltage as Vr\n",
"Vr = Symbol('Vr')\n",
"#Sending-end voltage is\n",
"Vs1 = 66000.0/1.732 #V\n",
"Vs2 = (Vr + dropR*cosQ + dropX*sinQ)**2 + (dropX*cosQ - dropR*sinQ)**2 #V\n",
"eq = Eq(Vs1*Vs1,Vs2)\n",
"Vr = solve(eq)\n",
"Vr1 = Vr[1] #V\n",
"\n",
"#Line-voltage across load\n",
"Vrl = Vr1*1.732/1000 #kV\n",
"print \"Voltage at recieving end = \",round(Vrl,2),\"V.\"\n",
"\n",
"#Angle between voltages\n",
"a_b = (dropX*cosQ - dropR*sinQ)/(Vr1 + dropR*cosQ + dropX*sinQ)\n",
"theta2 = m.atan(a_b)*180/3.14 #angle\n",
"print \"Angle between voltages = \",round(theta2,2),\"degrees.\"\n",
"#Transmission Efficiency\n",
"eff = (300.0/R1)*100\n",
"print \"Transmission efficiency is = \",round(eff,2),\"%.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.14 , PAGE NO :- 1623"
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Sending end voltage is = 12.12 V.\n",
"Power factor = 0.76\n",
"Transmission Efficiency = 93.93 %\n",
"Voltage Regulation = 10.2 %\n"
]
}
],
"source": [
"'''Define â€˜regulationâ€™ and â€˜efficiencyâ€™ of a short transmission line.A 3-phase, 50-Hz, transmission line having resistance of \n",
"5ohm per phase and inductance of 30 mH per phase supplies a load of 1000 kW at 0.8 lagging and 11 kV at the receiving end. Find.\n",
"(a) sending end voltage and power factor (b) transmission efficiency (c) regulation.'''\n",
"\n",
"import math as m\n",
"import cmath as cm\n",
"\n",
"#Recieving-end Voltage\\phase\n",
"Vr = 11000.0/1.732 #V\n",
"#Line current\n",
"Il = 1000.0e+3/(1.732*11000.0*0.8) #A\n",
"#Inductive reactance\n",
"Xl = 2*3.14*50*30.0e-3 #ohm\n",
"R = 5.0 #ohm\n",
"#Impedance\n",
"Z = R + 1j*Xl #ohm\n",
"#drop per conductor\n",
"theta = m.atan(0.8)\n",
"Il1 = cm.rect(Il,-theta)\n",
"drop = Il1*(Z) #ohm\n",
"\n",
"#(a)Sending end voltage\n",
"Vs = Vr + drop #V\n",
"Vs1 = abs(Vs)*1.732/1000 #kV\n",
"#For power factor\n",
"theta2 = theta + cm.phase(Vs)\n",
"#Power factor\n",
"pf = m.cos(theta2)\n",
"\n",
"#(b)\n",
"#Power loss\n",
"loss = 3*Il*Il*R/1000 #kW\n",
"#Input Power\n",
"pin = 1000.0 + loss #kW\n",
"#Transmission efficiency\n",
"eff = 1000.0/pin *100 #% \n",
"# (c)% Voltage regulation\n",
"reg = (Vs1 - 11.0)/11.0*100 #%\n",
"\n",
"print \"Sending end voltage is = \",round(Vs1,2),\"V.\"\n",
"print \"Power factor = \",round(pf,2)\n",
"print \"Transmission Efficiency = \",round(eff,2),\"%\"\n",
"print \"Voltage Regulation = \",round(reg,2),\"%\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.15 , PAGE NO :- 1624"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Active power = 111397.0 kW.\n",
"Reactive power = 53952.0 kVAR.\n"
]
}
],
"source": [
"'''A short 3-Ï† line with an impedance of (6 + j8) ohm per line has sending and receiving end line voltages of 120 and\n",
"110 kV respectively for some receiving-end load at a p.f. of 0.9. Find the active power and the reactive power at the\n",
"receiving end.'''\n",
"\n",
"import math as m\n",
"\n",
"#Given\n",
"R = 6.0 #ohm\n",
"X = 8.0 #ohm\n",
"cosQ = 0.9\n",
"sinQ = m.sqrt(1 - 0.9*0.9)\n",
"#Sending end-voltage\n",
"Vs = 120.0/1.732*1000 #V\n",
"#Recieving end-voltage\n",
"Vr = 110.0/1.732*1000 #V\n",
"#Now Vs = Vr + IRcosQ + IXsinQ.Therefore,line current is\n",
"I = (Vs - Vr)/(R*cosQ + X*sinQ) #A\n",
"\n",
"#Active Power at recieving end\n",
"act_pwr = 1.732*110.0*I*cosQ #kW\n",
"#Reactive Power at recieving end\n",
"rct_pwr = 1.732*110.0*I*sinQ #kVAR\n",
"\n",
"print \"Active power = \",round(act_pwr),\"kW.\"\n",
"print \"Reactive power = \",round(rct_pwr),\"kVAR.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.16 , PAGE NO :- 1624"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"-------pf = 0.707------- \n",
"% regulation = 15.18\n",
"Efficiency = 93.8\n",
"-------pf = 0.9--------- \n",
"% regulation = 9.36\n",
"Efficiency = 96.08\n"
]
}
],
"source": [
"'''A 3-phase, 20 km line delivers a load of 10 MW at 11 kV having a lagging p.f. of 0.707 at the receiving end.\n",
"The line has a resistance of 0.02 ohm/km phase and an inductive reactance of 0.07 ohm/km/phase. Calculate the regulation\n",
"and efficiency of the line. If, now, the receiving end p.f. is raised to 0.9 by using static capacitors, calculate the\n",
"new value of regulation and efficiency.'''\n",
"\n",
"import cmath as cm\n",
"import math as m\n",
"\n",
"print \"-------pf = 0.707------- \"\n",
"#(i) When pf = 0.707 (lag)\n",
"pf = 0.707\n",
"#line current \n",
"Il = 10.0e+6/(1.732*11000*pf) #A\n",
"#Vr per phase\n",
"Vr = 11000.0/1.732 #V\n",
"#Total resistance/phase for 20km\n",
"R = 20*0.02 #W\n",
"#Total reactance/phase for 20km\n",
"X = 20*0.07 #W\n",
"\n",
"#Total impedance/phase\n",
"Z = R + 1j*X #ohm\n",
"#If Vr is taken as reference vector,then drop per phase is\n",
"theta = m.acos(pf)\n",
"Il1 = cm.rect(Il,-theta) #A\n",
"#drop/phase\n",
"drop = Il1*Z #V\n",
"#Sending end voltage\n",
"Vs = Vr + drop #V\n",
"Vs1 = abs(Vs) #V\n",
"#% regulation\n",
"reg = (Vs1 - Vr)/Vr*100\n",
"print \"% regulation = \",round(reg,2)\n",
"\n",
"#Total line loss\n",
"loss = 3*Il*Il*R/1.0e+6 #MW\n",
"#Total output\n",
"tot_op = 10.0 + loss #MW\n",
"#Efficiency\n",
"eff = 10.0/tot_op*100\n",
"print \"Efficiency = \",round(eff,2)\n",
"#-----------------------------------------------------------------------------------------------------------------------#\n",
"print \"-------pf = 0.9--------- \"\n",
"#(ii) When pf = 0.9 (lag)\n",
"pf = 0.9\n",
"#line current \n",
"Il = 10.0e+6/(1.732*11000*pf) #A\n",
"\n",
"#If Vr is taken as reference vector,then drop per phase is\n",
"theta = m.acos(pf)\n",
"Il1 = cm.rect(Il,-theta) #A\n",
"#drop/phase\n",
"drop = Il1*Z #V\n",
"#Sending end voltage\n",
"Vs = Vr + drop #V\n",
"Vs1 = abs(Vs) #V\n",
"#% regulation\n",
"reg = (Vs1 - Vr)/Vr*100\n",
"print \"% regulation = \",round(reg,2)\n",
"\n",
"#Total line loss\n",
"loss = 3*Il*Il*R/1.0e+6 #MW\n",
"#Total output\n",
"tot_op = 10.0 + loss #MW\n",
"#Efficiency\n",
"eff = 10.0/tot_op*100\n",
"print \"Efficiency = \",round(eff,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.17 , PAGE NO :- 1625"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Sending end-voltage is = 11.86 kV.\n",
"Total loss = 68.57 kW.\n",
"Reduction in loss = 24.69 kW.\n"
]
}
],
"source": [
"'''A load of 1,000 kW at 0.8 p.f. lagging is received at the end of a 3-phase line 10 km long. The resistance and inductance\n",
"of each conductor per km are 0.531 W and 1.76 mH respectively. The voltage at the receiving end is 11 kV at 50 Hz.\n",
"Find the sending-end voltage and the power loss in the line. What would be the reduction in the line loss if the p.f.\n",
"of the load were improved to unity?'''\n",
"\n",
"import cmath as cm\n",
"import math as m\n",
"\n",
"#Line current\n",
"Il = 1000.0 * 1000/(1.732 * 11 * 1000 * 0.8) #A\n",
"#Voltage/phase\n",
"V = 11000/1.732 #V\n",
"X = 2*3.14*50* 1.76e-3 *10 #ohm\n",
"R= 0.531 * 10 #ohm\n",
"Z = R + 1j*X\n",
"#Voltage drop/phase\n",
"theta = m.acos(0.8)\n",
"Il1 = cm.rect(Il,-theta)\n",
"drop = Il1*Z #V\n",
"#Sending end voltage is\n",
"Vs = V + drop #V\n",
"#line-to-line sending-end voltage\n",
"Vs1= abs(Vs)*1.732/1000 #kV\n",
"#Total loss \n",
"loss = 3*Il*Il*R/1000 #kW\n",
"\n",
"#Line current for unity p.f.\n",
"Il2 = 1000/(11*1.732) #A\n",
"#New losses\n",
"new_loss = 3*Il2*Il2*R/1000 #kW\n",
"\n",
"#Reduction in loss\n",
"red_loss = loss - new_loss #kW\n",
"\n",
"print \"Sending end-voltage is = \",round(Vs1,2),\"kV.\"\n",
"print \"Total loss = \",round(loss,2),\"kW.\"\n",
"print \"Reduction in loss = \",round(red_loss,2),\"kW.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.18 , PAGE NO :- 1625"
]
},
{
"cell_type": "code",
"execution_count": 18,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Length of line = 69.55 km.\n"
]
}
],
"source": [
"'''Estimate the distance over which a load of 15,000 kW at 0.85 p.f. can be delivered by a 3-phase transmission line\n",
"having conductors of steel-cored aluminium each of resistance 0.905 W per kilometre. The voltage at the receiving end\n",
"is to be 132 kV and the loss in transmission is to be 7.5% of the load.'''\n",
"\n",
"#Line current\n",
"Il = 15000/(132 * 1.732 * 0.85) #A\n",
"#Total loss\n",
"loss = (7.5/100)*15000 #kW\n",
"\n",
"#If R is the resistance of one conductor, then 3*I^2*R = loss\n",
"R = loss*1000/(3*Il*Il) #ohm \n",
"\n",
"#Length of the line\n",
"length = R/0.905 #km.\n",
"print \"Length of line = \",round(length,2),\"km.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 41.19 , PAGE NO :- 1625"
]
},
{
"cell_type": "code",
"execution_count": 19,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Power factor = 0.797\n",
"Efficiency of transmission line is = 97.15\n"
]
}
],
"source": [
"'''A 3-phase line has a resistance of 5.31 ohm and inductance of 0.0176 H. Power is transmitted at 33 kV, 50-Hz from one end\n",
"and the load at the receiving end is 3,600 kW at 0.8 p.f. lagging. Find the line current, receiving-end voltage,\n",
"sending-end p.f. and efficiency of transmisson.'''\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"import math as m\n",
"import cmath as cm\n",
"\n",
"#Let us assume that Vr is receiving end voltage\n",
"Vr = Symbol('Vr')\n",
"#Power delivered/phase = Vr*I*cosQ.Therefore I is\n",
"I = (3600.0/3)*1000/(0.8*Vr)\n",
"#Sending end voltage/phase =\n",
"Vs = 33000.0/1.732 #V\n",
"R = 5.31 #ohm\n",
"X = 2*3.14*0.0176*50 #ohm\n",
"\n",
"#Now,\n",
"cosQ = 0.8\n",
"sinQ = m.sqrt(1 - 0.8*0.8)\n",
"\n",
"#As we know, Vs = Vr + IRcosQ + IXsinQ\n",
"Vs2 = Vr + I*R*cosQ + I*X*sinQ #V\n",
"eq = Eq(Vs,Vs2)\n",
"Vr = solve(eq)\n",
"Vr1 = Vr[1] #V\n",
"#Line voltage at receiving end\n",
"Vrl = Vr1*1.732/1000 #kV\n",
"I = (3600.0/3)*1000/(0.8*Vr1) #A\n",
"\n",
"Vs = Vr1 + I*(cosQ - 1j*sinQ)*(R + 1j*X) #V\n",
"#Power factor\n",
"theta = m.acos(0.8) + cm.phase(Vs)\n",
"pf2 = m.cos(theta)\n",
"print \"Power factor = \",round(pf2,3)\n",
"#Power lost in line is\n",
"loss = 3*I*I*R/1000 #kW\n",
"#Power at sending end is\n",
"tot_pwr = 3600.0 + loss #kW\n",
"#Eficiency of transmission is\n",
"eff = 3600.0/tot_pwr*100\n",
"print \"Efficiency of transmission line is = \",round(eff,2) "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.20 , PAGE NO :- 1626"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total Maximum power = 18.0 MW.\n",
"Total kVAR supplied = 28801.69 kW.\n"
]
}
],
"source": [
"'''A 3-phase short transmission line has resistance and reactance per phase of 15 ohm and 20 ohm respectively. If the sending-end\n",
"voltage is 33 kV and the regulation of the line is not to exceed 10%, find the maximum power in kW which can be transmitted\n",
"over the line. Find also the kVAR supplied by the line when delivering the maximum power.'''\n",
"\n",
"import math as m\n",
"\n",
"#Given\n",
"Vs = 33000.0/1.732 #V (Sending end voltage)\n",
"Vr = Vs/(1 + 10.0/100) #V (Receiving end voltage)\n",
"Z = m.sqrt(15**2 + 20**2) #ohm (impedance)\n",
"R = 15.0 #ohm (resistance)\n",
"X = 20.0 #ohm (reactance) \n",
"#Maximum Power transmitted is given by\n",
"Pmax = (Vr/Z)**2*(Z*(Vs/Vr) - R) #watts/phase\n",
"\n",
"#Total max. power\n",
"ptot = Pmax*3/1e+6 #MW\n",
"\n",
"#kVAR supplied per phase is given by\n",
"pkvar = (Vr/Z)**2*X/1e+3 #kW\n",
"#Total kVAR supplied\n",
"kvartot = 3*pkvar #kW\n",
"\n",
"print \"Total Maximum power = \",round(ptot,2),\"MW.\"\n",
"print \"Total kVAR supplied = \",round(kvartot,2),\"kW.\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.21 , PAGE NO :- 1627"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Sending end voltage = 67.69 kV.\n",
"Efficiency = 98.8 %\n",
"Max. value of Q for 3-phases are = 346840.0 kVA.\n"
]
}
],
"source": [
"'''A 3-phase, 50-Hz generating station supplies a load of 9,900 kW at 0.866 p.f. (lag) through a short overhead transmission line.\n",
"Determine the sending-end voltage if the receiving-end voltage is 66 kV and also the efficiency of transmission.\n",
"The resistance per km is 4ohm and inductance 40 mH. What is the maximum power in kVA that can be transmitted through\n",
"the line if both the sending and receiving-end voltages are kept at 66 kV and resistance of the line is negligible.?'''\n",
"\n",
"import math as m\n",
"#Resistance\n",
"R = 4.0 #ohm\n",
"#Reactance\n",
"X = 40.0e-3*(2*3.14*50)#ohm\n",
"#Impedance\n",
"Z = m.sqrt(R*R + X*X) #ohm\n",
"#Line current\n",
"I = 9900.0/(1.732*66*0.866) #A\n",
"#Receiving end voltage\n",
"Vr = 66000.0/1.732 #V\n",
"#Now,\n",
"cosQ = 0.866\n",
"sinQ = m.sqrt(1 - 0.866**2)\n",
"#Sending end voltage is\n",
"Vs = Vr + I*R*cosQ + I*X*sinQ #V\n",
"#Line value of sending end voltage\n",
"Vs1 = Vs*1.732/1000 #kV\n",
"print \"Sending end voltage = \",round(Vs1,2),\"kV.\"\n",
"#Total line loss\n",
"loss = 3*I*I*R/1000 #kW\n",
"#Efficiency is\n",
"eff = 9900.0/(9900.0 + loss)*100#%\n",
"print \"Efficiency = \",round(eff,2),\"%\"\n",
"\n",
"#Max. value of Q for 3-phases are (As Vs = Vr R is negligible)\n",
"Z = X #ohm \n",
"max_value = (3*Vr*Vr)/(Z*Z)*X*1e-3 #kVA\n",
"print \"Max. value of Q for 3-phases are = \",round(max_value,-1),\"kVA.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.22 , PAGE NO :- 1627"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Power factor at sending end = 0.79\n",
"Transmission efficiency = 94.73 %\n"
]
}
],
"source": [
"'''3-phase load of 2000 kVA,0.8 p.f. is supplied at 6.6 kV, 50-Hz by means of a 33 kV transmission line 20 km long and a 5 : 1\n",
"transformer. The resistance per km of each conductor is 0.4 ohm and reactance 0.5 ohm. The resistance and reactance of the\n",
"transformer primary are 7.5 ohm and 13.2 ohm, whilst the resistance of the secondary is 0.35 ohm and reactance 0.65 ohm.\n",
"Find the voltage necessary at the sending end of transformission line when 6.6 kV is maintained at the load-end and find\n",
"the sending-end power factor. Determine also the efficiency of transmission.'''\n",
"\n",
"import math as m\n",
"#Impedance of high voltage line\n",
"Zh = 8.0 + 1j*10.0 #ohm\n",
"#Impedance of transformer(primary side)\n",
"Zt = 7.5 + 1j*13.2 #ohm\n",
"#Total impedance on high tension side\n",
"Ztot_h = Zh + Zt #ohm\n",
"#Impedance as referred to secondary side\n",
"Zsec = Ztot_h/(5**2) #ohm\n",
"\n",
"#Total impedance on high tension side\n",
"Ztot_l = Zsec + (0.35 + 1j*0.65) #ohm\n",
"\n",
"#Now,kVA load per phase\n",
"load = 2000.0/3 #kVA \n",
"#Receiving-end voltage per phase\n",
"Vr = 6.6/1.732 #kV\n",
"#current in line is\n",
"I = load/Vr #A\n",
"#Now,\n",
"cosQ = 0.8\n",
"sinQ = m.sqrt(1 - 0.8*0.8)\n",
"#Drop per conductor\n",
"drop = I*(Zsec.real*cosQ + Zsec.imag*sinQ) #V\n",
"#Sending end voltage is\n",
"Vs = Vr + drop/1000 #kV\n",
"#Sending end voltage referred from high voltage side\n",
"Vs = Vs*5 #kV\n",
"#Line sending end voltage\n",
"Vsl = Vs*1.732 #kV\n",
"\n",
"#If theta is phase angle at sending end then\n",
"tantheta = (sinQ + I*Zsec.imag/(Vr*1000))/(cosQ + I*Zsec.real/(Vr*1000))\n",
"theta = m.atan(tantheta)\n",
"pf = m.cos(theta)\n",
"print \"Power factor at sending end = \",round(pf,2)\n",
"#power loss/phase\n",
"loss = (I*I)*0.97/1000 #kW\n",
"#power at the receiving end/phase\n",
"power = 2000.0*cosQ/3 #kW\n",
"#Transmission efficiency\n",
"eff = power/(power + loss)*100\n",
"print \"Transmission efficiency = \",round(eff,2),\"%\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.23 , PAGE NO :- 1629"
]
},
{
"cell_type": "code",
"execution_count": 20,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Sending end current is 240.3 A.\n",
"Sending end voltage is 79598.0 V.\n",
"Voltage regulation = 20.6 %.\n"
]
}
],
"source": [
"'''A (medium) single-phase transmission line 50 km long has the following constants :\n",
"\n",
"resistance/km = 0.5 ohm ; reactance/km = 1.6 ohm\n",
"susceptance/km = 28 * 10âˆ’6 S ; receiving-end line voltage = 66,000 V\n",
"Assuming that total capacitance of the line is located at receiving end alone, determine the\n",
"sending-end voltage, the sending-end current and regulation. The line is delivering 15,000 kW at\n",
"0.8 p.f. lagging. Draw a vector diagram to illustrate your answer.'''\n",
"\n",
"\n",
"import math as m\n",
"import cmath as cm\n",
"\n",
"#Load current at receiving end Using I = P/VcosQ\n",
"Ir = 15.0e+6/(66e+3*0.8) #A\n",
"\n",
"#Total resistance is\n",
"R = 0.5*50.0 #ohm\n",
"#Total reactance is\n",
"X = 1.6*50.0 #ohm\n",
"#Susceptance\n",
"B = 28e-6*50.0 #Siemens\n",
"#Capacitive admittance\n",
"Y = B #Siemens\n",
"#Sending end current Is is vector sum of load current Ir and capacitive current Ic\n",
"Er = 66000.0\n",
"Ic = 1j*Er*Y #A\n",
"theta = m.acos(0.8)\n",
"Irl = cm.rect(Ir,-theta) #A\n",
"\n",
"#Sending end current is\n",
"Is = Irl + Ic #A\n",
"Z = R + 1j*X #ohm\n",
"print \"Sending end current is\",round(abs(Is),2),\"A.\"\n",
"#line drop\n",
"drop = Is*Z #V\n",
"#Sending end voltage is\n",
"Es = Er + drop #V\n",
"print \"Sending end voltage is\",round(abs(Es)),\"V.\"\n",
"#Voltage regulation is\n",
"reg = (abs(Es)-abs(Er))/abs(Er)*100 #%\n",
"print \"Voltage regulation = \",round(reg,2),\"%.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.24 , PAGE NO :- 1631"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Line value of sending end voltage = 155.43 kV.\n",
"Power factor = 0.774 lagging.\n"
]
}
],
"source": [
"'''A 3-phase, 50-Hz overhead transmission line 100 km long with 132 kV between lines at the receiving end has the\n",
"following constants :\n",
"\n",
"resistance/km/phase = 0.15 ohm inductance/km/phase = 1.20 mH\n",
"capacitance/km/phase = 0.01 mF\n",
"\n",
"Determine, using an approximate method of allowing for capacitance, the voltage, current and\n",
"p.f. at the sending end when the load at the receiving end is 72 MW at 0.8 p.f. lagging. Draw vector\n",
"diagram for the circuit assumed.'''\n",
"\n",
"import math as m\n",
"import cmath as cm\n",
"#For a 100-km length of line\n",
"#Resistance of line is\n",
"R = 0.15*100 #ohm\n",
"Xl = (2*3.14*50)*(1.2e-3)*100 #ohm\n",
"Xc = 1/(2*3.14*50*0.1e-3)*100 #ohm\n",
"\n",
"#Using nominal T-method\n",
"Vr = 132/1.732 #kV\n",
"#Load current\n",
"Ir = (72e+6)/(1.732*132e+3*0.8) #A\n",
"#Load current is\n",
"theta = m.acos(0.8)\n",
"Irl = cm.rect(Ir,-theta) #A\n",
"#Impedance Zbc is\n",
"Zbc = R/2 + 1j*Xl/2 #ohm\n",
"#Drop/phase over BC is\n",
"drop = Irl*Zbc #V\n",
"#Now, voltage V1 is\n",
"V1 = Vr*1000 + drop #V\n",
"\n",
"#From fig. Ic is\n",
"Ic = V1/(-1j*Xc) #A\n",
"\n",
"#Sending end current is\n",
"Is = Ic + Irl #A\n",
"\n",
"#Impedance Zab is\n",
"Zab = R/2 + 1j*Xl/2 #ohm\n",
"#Drop/phase over AB is\n",
"drop2 = Is*Zab #V\n",
"#Sending end voltage is\n",
"Vs = V1 + drop2 #V\n",
"#Line value of sending end voltage is\n",
"Vsl = 1.732*abs(Vs)/1000 #kV\n",
"print \"Line value of sending end voltage =\",round(Vsl,2),\"kV.\"\n",
"#Phase angle between Vs and Is is\n",
"angle = abs(cm.phase(Vs)) + abs(cm.phase(Is))\n",
"pf = m.cos(angle) #(lag)\n",
"print \"Power factor = \",round(pf,3),\"lagging.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.25 , PAGE NO :- 1632"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Line value of sending end voltage = 117.15 kV.\n",
" sending end current = 110.11 A.\n",
"Efficiency = 95.95 %.\n"
]
}
],
"source": [
"'''A 3-phase, 50-Hz transmission line, 100 km long delivers 20 MW at 0.9 p.f. lagging and at 110 kV. The resistance and\n",
"reactance of the line per phase per km are 0.2 ohm and 0.4 ohm respectively while the capacitive admittance is 2.5 * 10eâˆ’6 S per\n",
"km. Calculate (a) the voltage and current at the sending end and (b) the efficiency of transmission. Use the nominal T-method.'''\n",
"\n",
"import math as m\n",
"import cmath as cm\n",
"\n",
"#Resistance for 100 km is\n",
"R = 0.2*100 #ohm\n",
"#Reactance for 100 km is\n",
"X = 0.4*100 #ohm\n",
"#Capacitive admittance for 100 km is\n",
"Y = 2.5e-6*100 #Siemens\n",
"#Receiving end voltage Er is\n",
"Er = 110.0/1.732 #kV\n",
"Ir = 20.0e+6/(1.732*110e+3*0.9) #A\n",
"#Now\n",
"cosQ = 0.9\n",
"sinQ = m.sqrt(1 - 0.9*0.9)\n",
"theta = m.acos(0.9) \n",
"Irl = cm.rect(Ir,-theta) #A\n",
"\n",
"#Impedance is\n",
"Zbc = R/2 + 1j*X/2 #ohm\n",
"#Voltage drop between point B and C is\n",
"dropbc = Irl*Zbc #V\n",
"V1 = Er*1000 + dropbc #V\n",
"\n",
"#Current through capacitor \n",
"Ic = V1*1j*Y #A\n",
"\n",
"#Sending end current is\n",
"Is = Ic + Irl #A\n",
"Zab = Zbc\n",
"#Voltage drop between point A and B is\n",
"dropab = Irl*Zab #V\n",
"Vs = V1 + dropab #V\n",
" \n",
"#Line value of sending end voltage is\n",
"Vsl = 1.73*abs(Vs)/1000 #kV\n",
"print \"Line value of sending end voltage =\",round(Vsl,2),\"kV.\"\n",
"print \" sending end current =\",round(abs(Is),2),\"A.\"\n",
" \n",
"#Phase angle between Vs and Is is\n",
"angle = abs(cm.phase(Vs)) + abs(cm.phase(Is))\n",
"pf = m.cos(angle) #(lag)\n",
"\n",
"#Input power\n",
"pin = 1.73*abs(Vsl)*abs(Is)*pf/1e+3 #MW\n",
"\n",
"#Efficiency\n",
"eff = 20.0/pin*100\n",
"print \"Efficiency = \",round(eff,2),\"%.\" \n",
" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.26 , PAGE NO :- 1635"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Line value of sending-end voltage 174.83 kV.\n",
"power factor = 0.75\n"
]
}
],
"source": [
"'''A 3-phase transmission line,100 km long has following constants:\n",
"resistance per km per phase = 0.28 ohm ; inductive reactance per km per phase = 0.63 ohm .\n",
"Capacitive susceptance per km per phase = 4 * 10e-6 siemens.\n",
"If the load at the receiving end is 75 MVA at 0.8 p.f. lagging with 132 kV between lines calculate sending-end voltage,\n",
"current and p.f. Use nominal-pi-method.'''\n",
"\n",
"import math as m\n",
"import cmath as cm\n",
"\n",
"#For 100 km length line\n",
"#Resistance/phase\n",
"R = 0.28*100 #ohm\n",
"#Inductive Reactance/phase\n",
"Xl = 0.63*100 #ohm\n",
"#Capacitive Susceptance/phase\n",
"Y = 4.0e-6*100 #S\n",
"#Capacitive Susceptance at each end\n",
"Y = 1j*Y/2 #S\n",
"#Receiving end voltage Vr is\n",
"Vr = 132e+3/1.732 #V\n",
"#Receiving end current Ir is\n",
"Ir = 75.0e+6/(1.732*132e+3*0.8) #A\n",
"theta = m.acos(0.8)\n",
"Irl = cm.rect(Ir,-theta) #A\n",
"#Current through capacitance is\n",
"Ic = Vr*Y #A\n",
"#Now\n",
"Il = Ic + Irl #A\n",
"\n",
"#Drop per conductor is\n",
"Zl = R +1j*Xl #ohm\n",
"drop = Il*Zl #V\n",
"\n",
"#sending end voltage\n",
"Vs = Vr + drop\n",
"#Line value of sending-end voltage\n",
"Vsl = abs(Vs)*1.732/1000 #kV\n",
"print \"Line value of sending-end voltage\",round(Vsl,2),\"kV.\"\n",
"#\n",
"Ic2 = Vs*Y #A\n",
"Is = Ic2 + Il\n",
"\n",
"#Angle between VS and IS\n",
"angle = abs(cm.phase(Vs)) + abs(cm.phase(Is)) \n",
"pf = m.cos(angle)\n",
"#cos41.4 = 0.75 \n",
"print \"power factor = \",round(pf,2) "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.27 , PAGE NO :- 1637"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Regulation = 15.18 %\n",
"Transmission efficiency = 95.02 %.\n"
]
}
],
"source": [
"'''A 100-km long, three-phase, 50-Hz transmission line has resistance/phase/km = 0.1 ohm ; reactance/phase/km = 0.5 ohm ; \n",
"susceptance/phase/km = 10 * 10âˆ’6 siemens.If the line supplies a load of 20 MW at 0.9 p.f. lagging at 66 kV at the receiving end,\n",
"calculate by nominal â€˜pâ€™ method, the regulation and efficiency of the line.Neglect leakage.'''\n",
"\n",
"import math as m\n",
"import cmath as cm\n",
"#For a 100 km line\n",
"#resistance/phase\n",
"R = 0.1 * 100 #ohm\n",
"#inductive reactance/phase\n",
"Xl = 0.5 * 100 #ohm\n",
"#Capacitive susceptance/phase\n",
"Yc = 10.0 * 1.0e-6 * 100 #siemens\n",
"#Susceptance at each end\n",
"Yc = 1j*Yc/2 #siemens\n",
"#Voltage at receiving end is\n",
"Vr = 66.0e+3/1.732 #V\n",
"Ir = 20.0e+6/(1.732*66e+3*0.9) #A\n",
"theta = m.acos(0.9)\n",
"Irl = cm.rect(Ir,-theta) #A\n",
"Ic1 = Vr*Yc #A\n",
"Il = Irl + Ic1 #A\n",
"\n",
"#drop/conductor\n",
"Zl = R + 1j*Xl #ohm\n",
"drop = Il*Zl #V\n",
"#Sending end voltage is\n",
"Vs = Vr + drop #V\n",
"#Line value of sending end voltage\n",
"Vsl = abs(Vs)*1.732/1000 #kV\n",
"Ic2 = Vs*Yc #A\n",
"#Sending end current is\n",
"Is = Ic2 + Il #A \n",
"\n",
"#Phase angle between Vs and Is is\n",
"angle = abs(cm.phase(Vs)) + abs(cm.phase(Is))\n",
"pf = m.cos(angle) #(lag)\n",
"\n",
"\n",
"#(i) Regulation is\n",
"reg = (abs(Vsl) - 66.0)/66.0 * 100 \n",
"print \"Regulation = \",round(reg,2),\"%\"\n",
"#(ii)Efficiency\n",
"#Input power is\n",
"pin = 1.732*Vsl*abs(Is)*pf*1000.0/1.0e+6 #MW\n",
"\n",
"eff = 20.0/pin*100 #%\n",
"\n",
"print \"Transmission efficiency = \",round(eff,2),\"%.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.28 , PAGE NO :- 1638"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Line value of sending end voltage = 122.24 kV.\n",
"Sending end current = 195.65 A.\n",
"Transmission efficiency = 95.36 %.\n",
"Receiving end voltage = 63148.47 V.\n",
"Receiving end current = 19.0 A.\n"
]
}
],
"source": [
"'''(a) A 50-Hz, 3-phase, 100-km long line delivers a load of 40 MVA at 110 kV and 0.7 p.f. lag. The line constants\n",
"(line to neutral) are :\n",
"resistance of 11 ohms, inductive reactance of 38 ohms and capacitive susceptance of 3 * 10âˆ’4 siemens. Find the sending-end\n",
"voltage, current,power factor and efficiency of power transmission.\n",
"(b) draw the vector diagram.\n",
"(c) If the sending-end voltage is held constant and load is removed, calculate the receiving-end voltage and current.'''\n",
"\n",
"\n",
"import math as m\n",
"import cmath as cm\n",
"\n",
"#resistance/phase\n",
"R = 11.0 #ohm\n",
"#inductive reactance/phase\n",
"Xl = 38.0 #ohm\n",
"#Capacitive susceptance/phase\n",
"Yc = 3.0e-4 #siemens\n",
"#Susceptance at each end\n",
"Yc = 1j*Yc/2 #siemens\n",
"#Voltage at receiving end is\n",
"Vr = 110.0e+3/1.732 #V\n",
"Ir = 40.0e+6/(1.732*110e+3) #A\n",
"theta = m.acos(0.7)\n",
"Irl = cm.rect(Ir,-theta) #A\n",
"Ic1 = Vr*Yc #A\n",
"Il = Irl + Ic1 #A\n",
"\n",
"#drop/conductor\n",
"Zl = R + 1j*Xl #ohm\n",
"drop = Il*Zl #V\n",
"#Sending end voltage is\n",
"Vs = Vr + drop #V\n",
"#Line value of sending end voltage\n",
"Vsl = abs(Vs)*1.732/1000 #kV\n",
"Ic2 = Vs*Yc #A\n",
"#Sending end current is\n",
"Is = Ic2 + Il #A \n",
"print \"Line value of sending end voltage = \",round(Vsl,2),\"kV.\"\n",
"print \"Sending end current = \",round(abs(Is),2),\"A.\"\n",
"#Phase angle between Vs and Is is\n",
"angle = abs(cm.phase(Vs)) + abs(cm.phase(Is))\n",
"pf = m.cos(angle) #(lag)\n",
"\n",
"#Efficiency\n",
"#Input power is\n",
"pin = 1.732*Vsl*abs(Is)*pf*1000.0/1.0e+6 #MW\n",
"\n",
"eff = (40.0*0.7)/pin*100 #%\n",
"\n",
"print \"Transmission efficiency = \",round(eff,2),\"%.\"\n",
"\n",
"\n",
"#(c)Under no-load condition, current in the conductor is Ic1\n",
"#Drop/phase =\n",
"drop = Ic1*Zl #V\n",
"#Sending end voltage is\n",
"Vs = Vr + drop #V\n",
"Ic2 = Vs*Yc #A\n",
"Is = Ic1 + Ic2 #A\n",
"print \"Receiving end voltage = \",round(abs(Vs),2),\"V.\"\n",
"print \"Receiving end current = \",round(abs(Is),2),\"A.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.29 , PAGE NO :- 1640"
]
},
{
"cell_type": "code",
"execution_count": 21,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Sending end voltage is = 59558.07 V.\n",
"Sending end current is = 282.23 A.\n",
"Sending end power is = 48.99 MW.\n",
"Transmission efficiency = 81.65 %\n"
]
}
],
"source": [
"'''Find the following for a single-circuit transmission line delivering a load of 50 MVA at 110 kV and p.f. 0.8 lagging :\n",
"(i) sending-end voltage, (ii) sending-end current, (iii) sending-end power, (iv) efficiency of\n",
"transmission. (Given A = D = 0.98 âˆ 3Âº , B = 110 âˆ 75Âº ohm, C = 0.0005 âˆ 80Âº ohm).'''\n",
"\n",
"import math as m\n",
"import cmath as cm\n",
"\n",
"#Receiving end voltage Vr is\n",
"Vr = 110.0/1.732 #kV\n",
"Ir = 50.0e+6/(1.732*110e+3) #A\n",
"theta = m.acos(0.8)\n",
"Irl = cm.rect(Ir,theta) #A\n",
"\n",
"#Now,\n",
"A = D = cm.rect(0.98,3*3.14/180)\n",
"B = cm.rect(110.0,75*3.14/180)\n",
"C = cm.rect(0.0005,80*3.14/180)\n",
"\n",
"#(i)Sending end voltage is\n",
"Vs = A*Vr*1000 + B*Irl #V\n",
"print \"Sending end voltage is = \",round(abs(Vs),2),\"V.\"\n",
"#(ii)Sending end current is\n",
"Is = C*Vr*1000 + D*Irl #A\n",
"print \"Sending end current is = \",round(abs(Is),2),\"A.\"\n",
"#Angle between Vs and Is is\n",
"angle = cm.phase(Is) - cm.phase(Vs)\n",
"pf = m.cos(angle)\n",
"\n",
"#(iii)Sending end power is\n",
"s_pwr = 3*abs(Vs)*abs(Is)*pf/1.0e+6 #MW\n",
"print \"Sending end power is = \",round(s_pwr,2),\"MW.\"\n",
"#Receiving end power is\n",
"r_pwr = 50.0*0.8 #MW\n",
"#(iv)Transmission efficiency\n",
"eff = r_pwr/s_pwr*100 #%\n",
"print \"Transmission efficiency = \",round(eff,2),\"%\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.30 , PAGE NO :- 1640"
]
},
{
"cell_type": "code",
"execution_count": 22,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"A = (0.9325+0.016875j)\n",
"B = (22.5+90j)\n",
"C = (-1.265625e-05+0.001449375j)\n",
"D = (0.9325+0.016875j)\n",
"Voltage regulation = 34.77 %.\n"
]
}
],
"source": [
"'''A 150 km, 3-Ï†, 110-V, 50-Hz transmission line transmits a load of 40,000 kW at 0.8 p.f. lag at receiving end.\n",
"resistance/km/phase = 0.15 ohm, reactance/km/phase = 0.6 ohm ; susceptance/km/phase = 10eâˆ’5 S\n",
"(a) determine the A, B, C and D constants of the line . (b) find regulation of the line.'''\n",
"\n",
"import math as m\n",
"import cmath as cm\n",
"\n",
"#For a 150km length line\n",
"R = 0.15*150.0 #ohm\n",
"X = 0.6*150.0 #ohm\n",
"Y = 1j*1.0e-5*150.0 #S\n",
"Z = R + 1j*X #ohm\n",
"\n",
"#(a)\n",
"A = D = 1 + Y*Z/2\n",
"B = Z\n",
"C = Y*(1 + Y*Z/4)\n",
"print \"A = \" ,A\n",
"print \"B = \" ,B\n",
"print \"C = \" ,C\n",
"print \"D = \" ,D\n",
"#(b)\n",
"Vr = 110/1.732*1000 #V\n",
"Ir = 40.0e+6/(1.732*110*0.8*1000)#A\n",
"theta = m.acos(0.8)\n",
"Irl = cm.rect(Ir,-theta) #A\n",
"#Sending end voltage is\n",
"Vs = A*Vr + B*Irl #V\n",
"\n",
"#Now, Vs = A*Vro => Vro = Vs/A\n",
"Vro = Vs/A #V\n",
"\n",
"#Voltage regulation\n",
"reg = (abs(Vro) - abs(Vr))/abs(Vr)*100\n",
"print \"Voltage regulation = \",round(reg,2),\"%.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.31 , PAGE NO :- 1641"
]
},
{
"cell_type": "code",
"execution_count": 24,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Regulation = 29.97 %\n",
"Charging current Ic = (-7.1+127.64j) A.\n"
]
}
],
"source": [
"'''A 132-kV, 50-Hz, 3-phase transmission line delivers a load of 50 MW at 0.8 p.f. lagging at receiving-end.\n",
"The generalised constants of the transmission line are A = D = 0.95 âˆ 1.4Âº ; B = 96 âˆ 7.8Âº ; C = 0.0015 âˆ 90Âº\n",
"Find the regulation of the line and the charging current. Use nominal T-method.'''\n",
"\n",
"import math as m\n",
"import cmath as cm\n",
"\n",
"#Given\n",
"A = D = cm.rect(0.95,1.4*3.14/180)\n",
"B = cm.rect(96.0,78*3.14/180)\n",
"C = cm.rect(0.0015,90*3.14/180)\n",
"\n",
"#Receiving end voltage\n",
"Ir = 50e+6/(1.732*132e+3*0.8) #A\n",
"theta = m.acos(0.8)\n",
"Irl = cm.rect(Ir,-theta) #A\n",
"Vr = 132e+3/1.732 #V\n",
"#Sending end voltage\n",
"Vs = A*Vr + B*Irl #V\n",
"#Sending end current\n",
"Is = C*Vr + D*Irl #A\n",
"Vro = abs(Vs)/abs(A)\n",
"\n",
"#% Regulation\n",
"reg = (abs(Vro) - abs(Vr))/abs(Vr)*100 #%\n",
"\n",
"print \"Regulation = \",round(reg,2),\"%\"\n",
"#Charging current\n",
"Ic = Is - Irl #A\n",
"print \"Charging current Ic = \",complex(round(Ic.real,2),round(Ic.imag,2)),\"A.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 41.32 , PAGE NO :- 1641"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"A = A1*A2 + B1*C2\n",
"C = A2*C1 + C2*D1\n",
"B = A1*B2 + B1*D2\n",
"D = B2*C1 + D1*D2\n"
]
}
],
"source": [
"'''A 3-phase transmission line consists of two lines 1 and 2 connected in series,line 1 being at the sending end and\n",
"2 at the receiving end. The respective auxiliary constants of the two lines are :\n",
"A1, B1, C1, D1 and A2, B2, C2, D2. Find the A, B, C, D constants of the whole line which is equivalent to two\n",
"series-connected lines.'''\n",
"\n",
"from sympy import Symbol\n",
"A1 = Symbol('A1')\n",
"B1 = Symbol('B1')\n",
"C1 = Symbol('C1')\n",
"D1 = Symbol('D1')\n",
"\n",
"A2 = Symbol('A2')\n",
"B2 = Symbol('B2')\n",
"C2 = Symbol('C2')\n",
"D2 = Symbol('D2')\n",
"\n",
"Vr = Symbol('Vr')\n",
"Ir = Symbol('Ir')\n",
"\n",
"#--------For A &C---------------#\n",
"Ir = 0\n",
"Vr = 1\n",
"#For line no. 2\n",
"V = A2*Vr + B2*Ir\n",
"I = C2*Vr + D2*Ir \n",
"\n",
"#For line no. 1\n",
"Vs = A1*V + B1*I\n",
"Is = C1*V + D1*I\n",
"\n",
"#For A\n",
"print \"A = \",Vs\n",
"print \"C = \",Is\n",
"\n",
"#--------For B & D---------------#\n",
"Ir = 1\n",
"Vr = 0\n",
"#For line no. 2\n",
"V = A2*Vr + B2*Ir\n",
"I = C2*Vr + D2*Ir \n",
"\n",
"#For line no. 1\n",
"Vs = A1*V + B1*I\n",
"Is = C1*V + D1*I\n",
"\n",
"#For A\n",
"print \"B = \",Vs\n",
"print \"D = \",Is\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.33 , PAGE NO :- 1644"
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Critical voltage = 136.01 kV.\n"
]
}
],
"source": [
"'''Find the disruptive critical voltage for a transmission line having :\n",
"conductor spacing = 1 m ; conductor (stranded) radius = 1 cm\n",
"barometric pressure = 76 cm of Hg ; temperature = 40ÂºC\n",
"Air break-down potential gradient (at 76 cm of Hg and at 25ÂºC) = 21.1 kV (r.m.s.)/cm.'''\n",
"\n",
"import math as m\n",
"\n",
"#Given\n",
"g0 = 21.1 #kV/cm (air breakdown potential gradient)\n",
"m0 = 0.85 #(assumed)\n",
"d = 3.92*76/(273 + 40)\n",
"r = 1.0 #cm\n",
"D = 100.0 #cm\n",
"\n",
"#Vc = 2.3*m0*g0*d*r*log(D/r)\n",
"Vc = 2.3*m0*g0*d*r*m.log10(D/r) #kV\n",
"#Line voltage\n",
"Vc = 1.732*Vc\n",
"print \"Critical voltage = \",round(Vc,2),\"kV.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.34 , PAGE NO :- 1644"
]
},
{
"cell_type": "code",
"execution_count": 23,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Critical voltage = 152.4 kV.\n",
"visual corona voltage = 157.67 kV.\n"
]
}
],
"source": [
"'''Find the disruptive critical and visual corona voltage of a grid-line operating at 132 kV.\n",
"conductor dia = 1.9 cm ; conductor spacing = 3.81 m\n",
"temperature = 44ÂºC ; barometric pressure = 73.7 cm\n",
"conductor surface factor : fine weather = 0.8 ; rough weather = 0.66.'''\n",
"\n",
"\n",
"import math as m\n",
"\n",
"#Given\n",
"\n",
"m0 = 0.8 #\n",
"d = 3.92*73.7/(273 + 44)\n",
"r = 1.9 #cm\n",
"D = 381.0 #cm\n",
"\n",
"#Vc = 2.3*m0*g0*d*r*log(D/r)\n",
"Vc = 48.8*m0*d*r/2*m.log10(2*D/r) #kV\n",
"#Line voltage\n",
"Vc = 1.732*Vc\n",
"print \"Critical voltage = \",round(Vc,2),\"kV.\"\n",
"\n",
"#(b)\n",
"#Vv = 2.3*g0*mv*d*r*(1 + 0.3/root(d*r))*log10(D/r)\n",
"mv = 0.66\n",
"Vv = 48.8*mv*d*r*(1 + 0.3/m.sqrt(d*r))*m.log10(D/r)\n",
"print \"visual corona voltage = \",round(Vv,2),\"kV.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 41.35 , PAGE NO :- 1644"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Critical disruptive voltage = 53.34 kV/km.\n",
"Corona loss = 121.49 kW.\n"
]
}
],
"source": [
"'''A certain 3-phase equilateral transmission line has a total corona loss of 53 kW at 106 kV and a loss of 98 kW at 110.9 kV.\n",
"What is the disruptive critical voltage between lines? What is the corona loss at 113 kV?'''\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"\n",
"#total corona loss P o< (V- Vc)^2\n",
"# P1/P2 = (V1 - Vc)^2/(V2 - Vc)^2\n",
"\n",
"V1 = 106/1.732 #kV\n",
"V2 = 110.9/1.732 #kV\n",
"#Let Vc be distruptive critical voltage\n",
"Vc = Symbol('Vc')\n",
"#ratio P1/P2 =\n",
"P1_P2a = 53.0/98.0\n",
"#Ratio (V1 - Vc)^2/(V2 - Vc)^2\n",
"P1_P2b = (V1 - Vc)**2/(V2 - Vc)**2\n",
"#Equating the ratios\n",
"eq = Eq(P1_P2a,P1_P2b)\n",
"Vc = solve(eq)\n",
"Vc1 = Vc[0] #kV/km\n",
"\n",
"#Now, W/98 = (Vb - Vc)^2/(V2 - Vc)^2\n",
"W = 98*(113/1.732 - Vc1)**2/(V2 - Vc1)**2 #kW\n",
"print \"Critical disruptive voltage = \",round(Vc1,2),\"kV/km.\"\n",
"print \"Corona loss = \",round(W,2),\"kW.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.36 , PAGE NO :- 1645"
]
},
{
"cell_type": "code",
"execution_count": 25,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"corona power loss per km = 1.76 kW/km/phase.\n"
]
}
],
"source": [
"'''A 3-phase, 50-Hz, 220-kV transmission line consists of conductors of 1.2 cm radius spaced 2 metres at the corners of an\n",
"equilateral triangle. Calculate the corona power loss per km of the line at a temperature of 20ÂºC and barometric pressure of \n",
"72.2 cm.Take the surface factors of the conductor as 0.96.'''\n",
"\n",
"import math as m\n",
"\n",
"#As we know P = 241*(f+25)/d*root(r/D)*(V-Vc)^2\n",
"#Here,\n",
"d = 3.92*(72.2)/(273 + 20)\n",
"#Given\n",
"m0 = 0.96 # (surface factors) \n",
"D = 200.0 #cm (distance btwn condr.)\n",
"r = 1.2 #cm (radius of condr.)\n",
"#Critical Voltage\n",
"Vc = 48.8*m0*d*r*m.log10(D/r) #kV/phase\n",
"V = 220.0/1.732 #kV/phase\n",
"P = 241*(50+25)/d*m.sqrt(r/D)*(V-Vc)**2*1.0e-5 #kW/km/phase\n",
"#Total loss for 3-phase\n",
"loss = 3*P #kW/km/phase\n",
"print \"corona power loss per km = \",round(loss,2),\"kW/km/phase.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.37 , PAGE NO :- 1648"
]
},
{
"cell_type": "code",
"execution_count": 26,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Sheath diameter = 8.29 cm.\n"
]
}
],
"source": [
"'''A single-core lead-covered cable is to be designed for 66-kV to earth. Its conductor radius is 1.0 cm and its three insulating\n",
"materials A, B, C have permittivities of 5.4 and 3 respectively with corresponding maximum safe working stress of 38 kV per cm\n",
"(r.m.s. value), 26-kV per cm and 20-kV per cm respectively. Find the minimum diameter of the lead sheath.'''\n",
"\n",
"import math as m\n",
"from sympy import solve,Symbol,Eq\n",
"\n",
"#Given\n",
"gA = 38.0 #kV/cm (working stress on A)\n",
"gB = 26.0 #kV/cm (working stress on B)\n",
"gC = 20.0 #kV/cm (working stress on C)\n",
"ea = 5.0 # (rel. permittivity of A)\n",
"eb = 4.0 # (rel. permittivity of B)\n",
"ec = 3.0 # (rel. permittivity of C)\n",
"ra = 1.0 #cm (radius A)\n",
"#Working stress O< 1/(rel. permittivty)*(condr. radius)\n",
"#gA/gB = rb*eb/ra*ea\n",
"#Radius of B is\n",
"rb = (gA/gB)*(ra*ea/eb) #cm \n",
"#gA/gC = rc*ec/ra*ea\n",
"#Radius of C is\n",
"rc = (gA/gC)*(ra*ea/ec) #cm\n",
"\n",
"#Now , V = g*r*2.3*log10(r1/r)\n",
"\n",
"V1 = gA*ra*2.3*m.log10(rb/ra) #kV\n",
"V2 = gB*rb*2.3*m.log10(rc/rb) #kV\n",
"V = 66.0 #kV\n",
"V3 = V - V1 - V2 #V\n",
"\n",
"#Let the radius of sheath be rs\n",
"#V3 = gC*rc*2.3*m.log10(rs/rc) #kV . Therefore, rs is\n",
"\n",
"rs = rc * 10**(V3/(gC*rc*2.3)) #cm\n",
"print \"Sheath diameter = \",round(2*rs,2),\"cm.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.38 , PAGE NO :- 1649"
]
},
{
"cell_type": "code",
"execution_count": 27,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Charging current = 4.31 A.\n"
]
}
],
"source": [
"'''The capacitances per kilometer of a 3-phase cable are 0.63 Î¼F between the three cores bunched and the sheath and 0.37 Î¼F\n",
"between one core and the other two connected to sheath. Calculate the charging current taken by eight kilometres of this cable\n",
"when connected to a 3-phase, 50-Hz, 6,600-V supply.'''\n",
"\n",
"#According to question 0.63 = 3*Cs\n",
"Cs = 0.63/3 #uF/km\n",
"#Also 0.37 = 2*C1 + Cs\n",
"C1 = (0.37 - Cs)/2 #uF/km\n",
"#For 8 km\n",
"Cs = Cs*8 #uF\n",
"C1 = C1*8 #uF\n",
"\n",
"Cn = Cs + 3*C1 #uF\n",
"Vp = 6600.0/1.732 #V\n",
"w = 2*3.14*50.0 #rad/s\n",
"#Charging current is\n",
"Ic = Vp*(w*Cn*1e-6) #A\n",
"print \"Charging current = \",round(Ic,2),\"A.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.39 , PAGE NO :- 1649"
]
},
{
"cell_type": "code",
"execution_count": 23,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Charging for 15 km = 48.0 A.\n"
]
}
],
"source": [
"'''A 3-core, 3-phase belted cable tested for capacitance between a pair of cores on single phase with the third core earthed,\n",
"gave a capacitance of 0.4 mF per km. Calculate the charging current for 1.5 km length of this cable when connected to 22 kV, \n",
"3-phase,50-Hz supply.'''\n",
"\n",
"\n",
"Cl = 0.4 #uF (Capacitance)\n",
"Vl = 22000.0 #V (line voltage)\n",
"w = 2*3.14*50 #rad/s\n",
"#Charging current\n",
"Ic = (2.0/1.732)*Vl*Cl*w*1.0e-6 #A/km\n",
"#Charging current for 15 km\n",
"Ic = 15*Ic #A\n",
"print \"Charging for 15 km = \",round(Ic),\"A.\"\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.40 , PAGE NO :- 1649"
]
},
{
"cell_type": "code",
"execution_count": 24,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Capacitance (i) = 0.33 uF.\n",
"Capacitance (ii) = 0.13 uF.\n",
"capacitance between 2 shorted conductors and the other = 0.49 uF.\n"
]
}
],
"source": [
"'''A 3-core, 3-phase metal-sheathed cable has (i) capacitance of 1 Î¼F between shorted conductors and sheath and (ii) capacitance\n",
"between two conductors shorted with the sheath and the third conductor 0.6 Î¼F. Find the capacitance\n",
"(a) between any two conductors (b) between any two shorted conductors and the third conductor.'''\n",
"\n",
"#(a)\n",
"#(i)As we know 3*Cs = 1.0\n",
"Cs = 1.0/3 #uF\n",
"#(ii) 2*C1 + Cs = 0.6 #uF\n",
"C1 = (0.6 - Cs)/2 #uF\n",
"print \"Capacitance (i) = \",round(Cs,2),\"uF.\"\n",
"print \"Capacitance (ii) = \",round(C1,2),\"uF.\"\n",
"#(b)\n",
"#capacitance between 2 shorted conductors and the other is given by\n",
"Cn = 2*C1 + (2.0/3)*Cs #uF\n",
"print \"capacitance between 2 shorted conductors and the other = \",round(Cn,2),\"uF.\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.41 , PAGE NO :- 1650"
]
},
{
"cell_type": "code",
"execution_count": 28,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Voltage drop from A to B = (7.47+2.78j) V.\n"
]
}
],
"source": [
"'''A 2-wire a.c. feeder 1 km long supplies a load of 100 A at 0.8 p.f. lag 200 volts\n",
"at its far end and a load of 60 A at 0.9 p.f. lag at its mid-point. The resistance and reactance per km\n",
"(lead and return) are 0.06 ohm and 0.08 ohm respectively. Calculate the voltage drop along the\n",
"distributor from sending end to mid-point and from mid-point to far end.'''\n",
"\n",
"import math as m\n",
"import cmath as cm\n",
"\n",
"#Voltage at C\n",
"Vc = 200.0 #V\n",
"theta = m.acos(0.8)\n",
"Ic = 100.0 #A\n",
"Icl = cm.rect(Ic,-theta) #A\n",
"#Loop impedance of feeder BC is\n",
"Z = (0.06 + 1j*0.08)/2 #ohm\n",
"#Voltage drop in BC \n",
"drop = Icl*Z #V\n",
"#Voltage at B is\n",
"Vb = Vc + drop #V\n",
"#Current at B is\n",
"Ib = 60.0 #A\n",
"theta = m.acos(0.9)\n",
"Ibl = cm.rect(Ib,-theta) #A\n",
"#Current in feeder AB is\n",
"Iab = Ibl + Icl #A\n",
"#Drop in AB is\n",
"dropab = Iab*Z #V\n",
"print \"Voltage drop from A to B = \",complex(round(dropab.real,2),round(dropab.imag,2)),\"V.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.42 , PAGE NO :- 1651"
]
},
{
"cell_type": "code",
"execution_count": 26,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"drop = 5.0 V.\n"
]
}
],
"source": [
"'''A single-phase a.c. distributor 500 m long has a total impedance of (0.02 + j 0.04) ohm and is fed from one end at\n",
"250V. It is loaded as under :\n",
"(i) 50 A at unity power factor 200 m from feeding point.\n",
"(ii) 100 A at 0.8 p.f. lagging 300 m from feeding point.\n",
"(iii) 50 A at 0.6 p.f. lagging at the far end.\n",
"Calculate the total voltage drop and voltage at the far end.'''\n",
"\n",
"#Using 3rd method\n",
"import math as m\n",
"import cmath as cm\n",
"\n",
"#The center of gravity(C.G) of load is at the following distance from feeding end\n",
"\n",
"cg = (50.0*200 + 100.0*300 + 50.0*500)/(50 + 100 + 50) #m\n",
"\n",
"#Value of resistance upto CG\n",
"R = cg*0.02/500 #ohm\n",
"#Value of reactance upto CG\n",
"X = cg*0.04/500 #ohm\n",
"#Average pf\n",
"pf = (50*1 + 100.0*0.8 + 50.0*0.6)/200\n",
"cosQ = pf\n",
"sinQ = m.sqrt(1 - pf*pf)\n",
"#Drop is\n",
"Itot = 50.0 + 100.0 + 50.0 #A\n",
"drop = 200.0*(R*cosQ + X*sinQ) #V\n",
"print \"drop = \",round(drop),\"V.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.43 , PAGE NO :- 1652"
]
},
{
"cell_type": "code",
"execution_count": 29,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Supply voltage = 345.0 V.\n",
"Supply Voltage lead B by = 4.59 degrees.\n"
]
}
],
"source": [
"'''A single-phase distributor, one km long has resistance and reactance per conductor of 0.2 ohm and 0.3 ohm respectively. At the\n",
"far end, the voltage VB = 240 V and the current is 100 A at a power factor of 0.8 lag. At the mid-point A of the distributor \n",
"current of 100 A is tapped at a power factor of 0.6 lag with reference to the voltage VA at the mid-point. Calculate the supply\n",
"voltage VS for the distributor and the phase angle between VS and VB.'''\n",
"\n",
"import math as m\n",
"import cmath as cm\n",
"\n",
"#Given\n",
"Vb = 240.0 #V (Voltage at B)\n",
"Ib = 100.0 #A (current)\n",
"theta = m.acos(0.8)\n",
"Ibl = cm.rect(Ib,-theta) #A\n",
"X = 0.2 + 1j*0.3 #ohm\n",
"#Drop over AB\n",
"dropab = Ibl*(X) #V\n",
"\n",
"#Voltage at A\n",
"Va = Vb + dropab #V\n",
"\n",
"#Phase angle between A and B\n",
"ang = cm.phase(Va)\n",
"theta2 = m.acos(0.6)\n",
"theta2 = theta2 - ang\n",
"Ia = 100.0 #A\n",
"Ial = cm.rect(Ia,-theta2) #A\n",
"\n",
"#Total I\n",
"I = Ial + Ibl #A\n",
"#Drop in section\n",
"dropsa = I*X #V\n",
"\n",
"#Voltage at S\n",
"Vs = Va + dropsa #V\n",
"print \"Supply voltage = \",round(abs(Vs)),\"V.\"\n",
"print \"Supply Voltage lead B by = \",round(cm.phase(Vs)*180/3.14,2),\"degrees.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.44 , PAGE NO :- 1653"
]
},
{
"cell_type": "code",
"execution_count": 27,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ia = (25-24j) A.\n"
]
}
],
"source": [
"'''A 1-phase ring distributor ABC is fed at A. The loads at B and C are 20 A at 0.8 p.f. lagging and 15 A at 0.6 p.f.\n",
"lagging respectively, both expressed with reference to voltage at A. The total impedances of the sections AB, BC and\n",
"CA are (1 + j1), (1 + j2) and (1 + j3) ohm respectively. Find the total current fed at A and the current in each section.'''\n",
"\n",
"import math as m\n",
"import cmath as cm\n",
"\n",
"\n",
"#Current in AB is\n",
"theta = m.acos(0.8) \n",
"Iab = cm.rect(20,-theta)\n",
"\n",
"#Current in AC is\n",
"theta = m.acos(0.6) \n",
"Iac = cm.rect(15,-theta)\n",
"\n",
"#Impedances of dif. section\n",
"Zab = 1 + 1j #ohm\n",
"Zbc = 1 + 1j*2 #ohm\n",
"Zca = 1 + 1j*3 #ohm\n",
"\n",
"#Drop over AB is\n",
"dropab = Iab*Zab #V\n",
"\n",
"#Drop over AC is\n",
"dropac = Iac*Zca #V\n",
"\n",
"#Potential diff. B and C is\n",
"pd_bc = dropac - dropab #V\n",
"\n",
"#Equivalent Thevenin's thm impedance across bc\n",
"Ztot = Zab + Zca + Zbc #ohm\n",
"\n",
"#Current in BC\n",
"Ibc = pd_bc/Ztot #A\n",
"\n",
"#Total current in AB is\n",
"Iab2 = Iab + Ibc #A\n",
"\n",
"#Total current in BC is\n",
"Ibc2 = Iac - Ibc #A\n",
"\n",
"#Total current fed at point A\n",
"Ia = Iab2 + Ibc2 #A\n",
"\n",
"print \"Ia = \",Ia,\"A.\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.45 , PAGE NO :- 1654"
]
},
{
"cell_type": "code",
"execution_count": 28,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"I1 = (19.5-16.5j) A.\n",
"I2 = (14.5-19.5j) A.\n",
"I3 = (3.5-4.5j) A.\n",
"Vb = (347.5-22.5j) V.\n",
"Vc = (327-24j) V.\n"
]
}
],
"source": [
"'''A 2-wire ring distributor ABC is supplied at A at 400 V. Point loads of 20 A at a p.f. of 0.8 lagging and 30 A at a p.f. 0.6\n",
"lagging are tapped off at B and C respectively. Both the power factors refer to the voltage at A. The respective go-and-return \n",
"impedances of sections AB,BC and CA are (1 + j2) ohm, (2 + j3) ohm and (1 + j3) ohm. Calculate the current flowing through each\n",
"section and the potentials at B and C. Use Superposition theorem.'''\n",
"\n",
"import math as m\n",
"import cmath as cm\n",
"\n",
"\n",
"#Current in AB is\n",
"theta = m.acos(0.8) \n",
"Iab = cm.rect(20,-theta)\n",
"\n",
"#Current in AC is\n",
"theta = m.acos(0.6) \n",
"Iac = cm.rect(30,-theta)\n",
"\n",
"#Impedances of dif. section\n",
"Zab = 1 + 1j*2 #ohm\n",
"Zbc = 2 + 1j*3 #ohm\n",
"Zca = 1 + 1j*3 #ohm\n",
"\n",
"#Let the currents in AB and AC is Ia1 and Ia2\n",
"Ia1 = Iab*(Zbc + Zca)/(Zbc + Zca + Zab) #A\n",
"\n",
"\n",
"Ia2 = Iab - Ia1 #A\n",
"\n",
"#Let the currents in AB and AC is Ia11 and Ia22\n",
"Ia11 = Iac*(Zca)/(Zbc + Zca + Zab) #A\n",
"Ia22 = Iac - Ia11 #A\n",
"\n",
"#As per superposition thm,\n",
"I1 = Ia1 + Ia11 #A\n",
"I2 = Ia2 + Ia22 #A\n",
"I3 = Ia11 - Ia2 #A\n",
"\n",
"#Potential Of B is\n",
"Vb = 400.0 - I1*Zab #V\n",
"#Potential of C is \n",
"Vc = 400.0 - I2*Zca #V\n",
"\n",
"print \"I1 = \",I1,\"A.\"\n",
"print \"I2 = \",I2,\"A.\"\n",
"print \"I3 = \",I3,\"A.\"\n",
"print \"Vb = \",Vb,\"V.\"\n",
"print \"Vc = \",Vc,\"V.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.46 , PAGE NO :- 1655"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Iab = (139.76-42.81j) A\n",
"Ibc = (99.76-12.81j) A\n",
"Icd = (-20.24-12.81j) A\n",
"Ida = (-80.86+22.2j) A\n"
]
}
],
"source": [
"'''A 3-phase ring main ABCD, fed from end A, supplies balanced loads of 50 A at 0.8 p.f. lagging at B, 120 A at u.p.f. at C and\n",
"70 A at 0.866 p.f. lagging at D, the load currents being referred to the voltage at point A.The impedance per phase of the various\n",
"line sections are :\n",
"section AB = (1 + j0.6)ohm ; section BC = (1.2 + j0.9) ohm\n",
"section CD = (0.8 + j0.5)ohm ; section DA = (3 + j2) ohm\n",
"Determine the currents in the various sections.'''\n",
"\n",
"from sympy import Symbol\n",
"import math as m\n",
"import cmath as cm\n",
"import numpy as np\n",
"\n",
"#Let the current in section AB = (x + jy)A\n",
"x = Symbol('x')\n",
"y = Symbol('y')\n",
"Iab = x + 1j*y #A\n",
"#Current in BC\n",
"theta = m.acos(0.8)\n",
"Icl = cm.rect(50.0,-theta) #A\n",
"Ibc = Iab - Icl #A\n",
"#Current in CD\n",
"Icd = Ibc - 120 #A\n",
"#Current in DA\n",
"theta = m.acos(0.866)\n",
"Idl = cm.rect(70.0,-theta) #A\n",
"Ida = Icd - Idl #A\n",
"#Applying kirchoff's law\n",
"V = (1 + 1j*0.6)*Iab + (1.2 + 1j*0.9)*Ibc + (0.8 + 1j*0.5)*Icd + (3 + 1j*2)*Ida #V\n",
"#Equating V to 0 we get following two equations\n",
"# 6*x - 4*y + 1009.8 = 0 - 1 \n",
"# 4*x + 6*y - 302.2 = 0 - 2\n",
"\n",
"\n",
"a = np.array([[6,-4], [4,6]])\n",
"b = np.array([1009.8,302.2])\n",
"vec = np.linalg.solve(a, b)\n",
"x = vec[0]\n",
"y = vec[1]\n",
"\n",
"\n",
"Iab = x + 1j*y #A\n",
"#Current in BC\n",
"Ibc = Iab - Icl #A\n",
"#Current in CD\n",
"Icd = Ibc - 120 #A\n",
"#Current in DA\n",
"Ida = Icd - Idl #A\n",
"\n",
"print \"Iab = \",complex(round(Iab.real,2),round(Iab.imag,2)),\"A\"\n",
"print \"Ibc = \",complex(round(Ibc.real,2),round(Ibc.imag,2)),\"A\"\n",
"print \"Icd = \",complex(round(Icd.real,2),round(Icd.imag,2)),\"A\"\n",
"print \"Ida = \",complex(round(Ida.real,2),round(Ida.imag,2)),\"A\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.47 , PAGE NO :- 1656"
]
},
{
"cell_type": "code",
"execution_count": 29,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Power transmitted by 1st line = 5283.01 kW.\n",
"Power transmitted by 2nd line = 6716.99 kW.\n",
"Total power(check) = 12000.0 kW.\n"
]
}
],
"source": [
"'''A total load of 12,000 kW at a power factor of 0.8 lagging is transmitted to a substation by two overhead\n",
"three-phase lines connected in parallel. One line has a conductor resistance of 2 ohm per conductor and reactance\n",
"(line to neutral) of 1.5 ohm, the corresponding values for the other line being 1.5 and 1.2 ohm respectively. Calculate\n",
"the power transmitted by each overhead line.'''\n",
"\n",
"import math as m\n",
"import cmath as cm\n",
"\n",
"#Let us assume a line voltage of 1000 kV for convinience.\n",
"Z1 = 2 + 1j*1.5 #ohm \n",
"Z2 = 1.5 + 1j*1.2 #ohm\n",
"#Total load current\n",
"Il = 12000/(1.732*1000*0.8) #A\n",
"theta = m.acos(0.8)\n",
"Il1 = cm.rect(Il,-theta) #A\n",
"\n",
"#Now\n",
"I1 = Il1*Z2/(Z1 + Z2) #A\n",
"\n",
"#Power transmitted by 1st line\n",
"P1 = 1.732*I1.real*1000 #kW\n",
"\n",
"#Now\n",
"I2 = Il1*Z1/(Z1 + Z2) #A\n",
"\n",
"#Power transmitted by 1st line\n",
"P2 = 1.732*I2.real*1000 #kW\n",
"\n",
"#Total power\n",
"power = P1 + P2 #kW\n",
"print \"Power transmitted by 1st line = \",round(P1,2),\"kW.\"\n",
"print \"Power transmitted by 2nd line = \",round(P2,2),\"kW.\"\n",
"print \"Total power(check) = \",round(power,2),\"kW.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 41.48 , PAGE NO :- 1658"
]
},
{
"cell_type": "code",
"execution_count": 30,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"V1 = 7.54 kV\n",
"V2 = 8.29 kV\n",
"V3 = 9.87 kV\n",
"V4 = 12.45 kV\n",
"string efficiency = 76.55 %\n"
]
}
],
"source": [
"'''For a string insulator with four discs, the capacitance of the disc is ten times the capacitance between the pin and earth.\n",
"Calculate the voltage across each disc when used on a 66-kV line. Also, calculate the string efficiency.'''\n",
"\n",
"import math as m\n",
"from sympy import Symbol,solve,Eq\n",
"#Let C be self capacitance and kC be capacitance b/w each link and earth,\n",
"#Given\n",
"k = 10.0\n",
"#Let voltage across 1st disc be\n",
"V1 = Symbol('V1')\n",
"V2 = (1+k)/k*V1 #V\n",
"V3 = (1 + 3/k + 1/k**2)*V1 #V\n",
"V4 = (1 + 6/k + 5/k**2 + 1/k**3)*V1 #V\n",
"#Voltage V is\n",
"Va = V1 + V2 + V3 + V4 #kV\n",
"Vb = 66/1.73 #kV\n",
"eq = Eq(Va,Vb)\n",
"V1 = solve(eq)\n",
"V1l = V1[0] #kV\n",
"V2 = (1+k)/k*V1l #V\n",
"V3 = (1 + 3/k + 1/k**2)*V1l #V\n",
"V4 = (1 + 6/k + 5/k**2 + 1/k**3)*V1l #V\n",
"print \"V1 = \",round(V1l,2),\"kV\"\n",
"print \"V2 = \",round(V2,2),\"kV\"\n",
"print \"V3 = \",round(V3,2),\"kV\"\n",
"print \"V4 = \",round(V4,2),\"kV\"\n",
"#string efficiency\n",
"eff = 66.0/(4*1.732*V4)*100 #%\n",
"print \"string efficiency = \",round(eff,2),\"%\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.49 , PAGE NO :- 1659"
]
},
{
"cell_type": "code",
"execution_count": 31,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"V1 = 5.51 kV\n",
"V2 = 6.12 kV\n",
"V3 = 7.42 kV\n",
"string efficiency = 85.63 %\n"
]
}
],
"source": [
"'''Explain what is meant by string efficiency and how it can be improved. Each line of a 3-phase, 33-kV system is suspended by a\n",
"string of 3 identical insulator discs. The capacitance of each disc is 9 times the capacitance to ground. Find voltage distribution\n",
"across each insulator and the string efficiency. Suggest a method for improving the string efficiency.'''\n",
"\n",
"import math as m\n",
"from sympy import Symbol,solve,Eq\n",
"#Let C be self capacitance and kC be mutual capacitance b/w each\n",
"#Given\n",
"k = 9.0\n",
"#Let voltage across 1st disc be\n",
"V1 = Symbol('V1')\n",
"V2 = (1+k)/k*V1 #V\n",
"V3 = (1 + 3/k + 1/k**2)*V1 #V\n",
"\n",
"#Voltage V is\n",
"Va = V1 + V2 + V3 #kV\n",
"Vb = 33/1.732 #kV \n",
"eq = Eq(Va,Vb)\n",
"V1 = solve(eq)\n",
"V1l = V1[0] #kV\n",
"V2 = (1+k)/k*V1l #V\n",
"V3 = (1 + 3/k + 1/k**2)*V1l #V\n",
"\n",
"print \"V1 = \",round(V1l,2),\"kV\"\n",
"print \"V2 = \",round(V2,2),\"kV\"\n",
"print \"V3 = \",round(V3,2),\"kV\"\n",
"\n",
"#string efficiency\n",
"eff = 33.0/(3*1.732*V3)*100 #%\n",
"print \"string efficiency = \",round(eff,2),\"%\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.50 , PAGE NO :- 1660"
]
},
{
"cell_type": "code",
"execution_count": 30,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Bus-bar voltage of station A = 33.79 kV\n",
"phase angle = 1.75 degrees\n"
]
}
],
"source": [
"'''The bus-bar voltages of two stations A and B are 33 kV and are in phase. If station A sends 8.5 MW power at u.p.f.\n",
"to station B through an interconnector having an impedance of (3 + j4) ohm , determine the bus-bar voltage of station A\n",
"and the phase angle shift between the busbar voltages.'''\n",
"\n",
"import cmath as cm\n",
"\n",
"#Voltage of station B is\n",
"Vb = 33000.0/1.732 #V/phase\n",
"\n",
"#Power transferred\n",
"pin = 8.5 #MW\n",
"\n",
"#Current in interconnector\n",
"I = pin*1e+6/(1.732*33000)\n",
"\n",
"#Voltage Va is\n",
"Va = Vb + I*(3+1j*4)\n",
"#Bus-bar voltage of station A is\n",
"Va1 = 1.732*abs(Va)/1000 #kV\n",
"\n",
"print \"Bus-bar voltage of station A = \",round(Va1,2),\"kV\"\n",
"print \"phase angle = \",round(cm.phase(Va)*180/3.14,2),\"degrees\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.51 , PAGE NO :- 1666"
]
},
{
"cell_type": "code",
"execution_count": 32,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Clearance between conductor and water at mid-way = 62.0 m.\n"
]
}
],
"source": [
"'''A transmission line conductor at a river crossing is supported from two towers at heights of 70 m above water level.\n",
"The horizontal distance between towers is 300 m. If the tension in conductor is 1,500 kg, find the clearance at a point\n",
"midway between the towers. The size of conductor is 0.9 cm2. Density of conductor material is 8.9 g/cm3 and suspension\n",
"length of the string is 2 metres.'''\n",
"\n",
"#Given\n",
"l = 300.0/2 #m (distance of mid pt from one tower)\n",
"T = 1500.0 #kg-wt (tension)\n",
"\n",
"#density = mass/volume. Therefore Weight (W) = rho*l*A\n",
"W = (0.9e-4)*(8.9e+3) #kg-wt\n",
"\n",
"#Sag in the line\n",
"sag = (W*l**2)/(2*T) #m\n",
"\n",
"\n",
"#Clearance between conductor and water at mid-way\n",
"cler = 70.0 - sag - 2.0 #m\n",
"print \"Clearance between conductor and water at mid-way = \",round(cler),\"m.\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.52 , PAGE NO :- 1666"
]
},
{
"cell_type": "code",
"execution_count": 32,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total weight per metre run of line = 2.46 kg-wt/m.\n"
]
}
],
"source": [
"'''The effective diameter of a line is 1.96 cm and it weighs 90 kg per 100 metre length. What would be the additional loading due\n",
"to ice of radial thickness 1.25 cm and a horizontal wind pressure of 30 kg/m^2 of projected area? Also, find the total weight \n",
"per metre run of the line.Density of ice is 920 kg/m^3.'''\n",
"\n",
"import math as m\n",
"\n",
"#Given\n",
"W = 90.0/100 #kg/m (weight of condr)\n",
"#Weight due to ice is Wi = 3.14*rho*R*(D+R)\n",
"Wi = 3.14*920.0*0.0125*(0.0196 + 0.0125) #kg/m (Weight of ice)\n",
"#Weight due to wind is Ww = P*(D + 2*R)\n",
"Ww = 30.0*(0.0196 + 2*0.0125) #kg/m\n",
"\n",
"#Total weight is given by\n",
"\n",
"Wt = m.sqrt((W+Wi)**2 + Ww**2) #kg-wt/m\n",
"\n",
"print \"Total weight per metre run of line = \",round(Wt,2),\"kg-wt/m.\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.53 , PAGE NO :- 1666"
]
},
{
"cell_type": "code",
"execution_count": 33,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"sag = 3.27 m.\n"
]
}
],
"source": [
"'''A transmission line has a span of 150 metres between supports, the supports being at the same level. The conductor has a\n",
"cross-sectional area of 2 cm^2. The ultimate strength is 5,000 kg/cm^2. The specific gravity of the material is 8.9. If the wind\n",
"pressure is 1.5 kg/m length of the conductor, calculate the sag at the centre of the conductor if factor of safety is 5.'''\n",
"\n",
"\n",
"import math as m\n",
"\n",
"#Safety factor = Ultimate stress/Working stress\n",
"sf = 5.0 #Safety factor\n",
"wstress = 5000.0/sf #kg/cm^2\n",
"#Tension\n",
"A = 2.0 #cm^2 (cross-sectional area)\n",
"T = wstress*A #kg-wt\n",
"#Volume of 1m of conductor\n",
"V = A*100.0 #cm^3\n",
"#Wt of 1m of material Using density = mass/volume\n",
"W = 8.9*200.0/1000.0 #kg-wt\n",
"#W is 1.78 and not 1.98\n",
"\n",
"#Total weight per metre\n",
"Wt = m.sqrt(W**2 + 1.5**2)\n",
"\n",
"#Sag = W*l^2/2T\n",
"l = 150.0/2 #m (length from pole to centre)\n",
"sag = Wt*l**2/(2*T) #m\n",
"\n",
"\n",
"print \"sag = \",round(sag,2),\"m.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.54 , PAGE NO :- 1667"
]
},
{
"cell_type": "code",
"execution_count": 34,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Verticle Sag = 5.38 m.\n"
]
}
],
"source": [
"'''A transmission line has a span of 200 metres between level supports. The conductor has a cross-sectional area of 1.29 cm^2,\n",
"weighs 1,170 kg/km and has a breaking stress of 4,218 kg/cm2. Calculate the sag for a factor of safety of 5 allowing a wind pressure\n",
"of 122 kg per m^2 of projected area. What is the vertical sag?'''\n",
"\n",
"import math as m\n",
"\n",
"#Safety factor = ultimate stress/working stress \n",
"wstress = 4218.0/5 #kg/cm2 (working stress)\n",
"#Working tension\n",
"A = 1.29 #cm^2 (cross-sectional area) \n",
"T = wstress*A #kg-wt\n",
"W = 1170.0/1000 #kg-wt/m\n",
"#Let us now find diameter of the conductor from the equation\n",
"# A = 3.14 * d^2/4\n",
"d = m.sqrt(4*A/3.14)/100 #m \n",
"#Projected area of the conductor per metre length\n",
"proArea = d*1 #m^2\n",
"#Weight of wind exerted\n",
"Ww = 122*proArea #kg-wt/m\n",
"\n",
"#Total weight\n",
"Wt = m.sqrt(W**2 + Ww**2) #kg-wt/m\n",
"\n",
"#Slant sag\n",
"l = 200.0/2 #m\n",
"sag = Wt*l**2/(2*T) #m\n",
"\n",
"#Now, tan Î¸ = Ww/W\n",
"tanQ = Ww/W\n",
"#vertical sag\n",
"vsag = sag*m.cos(m.atan(tanQ)) #m\n",
"print \"Verticle Sag = \",round(vsag,2),\"m.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.55 , PAGE NO :- 1667"
]
},
{
"cell_type": "code",
"execution_count": 35,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Max. sag produced = 3.93 m.\n",
"Verticle sag = 2.36 m.\n"
]
}
],
"source": [
"'''A transmission line has a span of 214 metres. The line conductor has a cross-section of 3.225 cm^2 and has an ultimate \n",
"breaking strengh of 2,540 kg/cm^2. Assuming that the line is covered with ice and provides a combined copper and ice load of\n",
"1.125 kg/m while the wind pressure is 1.5 kg/m run (i) calculate the maximum sag produced. Take a factor of safety of 3\n",
"(ii) also determine the vertical sag.'''\n",
"\n",
"import math as m\n",
"\n",
"#(i) Maximum sag\n",
"#Here,\n",
"W = 1.125 #kg-wt/m\n",
"Ww = 1.5 #kg-wt/m\n",
"Wt = m.sqrt(1.125**2 + 1.5**2) #kg-wt/m\n",
"#Safety factor = ultimate stress/working stress\n",
"wstress = 2540.0/3 #kg-wt/cm2\n",
"#Permissible tension\n",
"A = 3.225 #cm^2 (cross-sectional area) \n",
"T = wstress*(3.225) #kg-wt\n",
"#Sag is\n",
"l = 214.0/2 #m \n",
"sag = Wt*l**2/(2*T) #m\n",
"\n",
"#(ii)Verticle Sag\n",
"#Now,\n",
"tanQ = Ww/W\n",
"cosQ = m.cos(m.atan(tanQ))\n",
"#vertical sag\n",
"vsag = sag*cosQ #m\n",
"print \"Max. sag produced = \",round(sag,2),\"m.\"\n",
"print \"Verticle sag = \",round(vsag,2),\"m.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.56 , PAGE NO :- 1658"
]
},
{
"cell_type": "code",
"execution_count": 33,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Minimum clearance = 23.02 m.\n",
"Hence, clearance point = 30.7 m.\n"
]
}
],
"source": [
"'''Two towers of height 30 and 90 m respectively support a transmission line conductor at water crossing. The horizontal\n",
"distance between the towers is 500 m. If the tension in the conductor is 1,600 kg, find the minimum clearance of the\n",
"conductor and the clearance of the conductor mid-way between the supports. Weight of the conductor is 1.5 kg/m. Bases of\n",
"the towers can be considered to be at the water level.'''\n",
"\n",
"\n",
"#Given\n",
"l = 500.0/2 #m (mid-way of towers)\n",
"h = 90.0 - 30.0 #m (height)\n",
"T = 1600.0 #kg-wt (tension)\n",
"W = 1.5 #kg-wt/m (weight)\n",
"\n",
"#Point of maximum sag is\n",
"x1 = l - h*T/(2*W*l)\n",
"\n",
"#Sag is\n",
"sag = W*x1**2/(2*T) #m\n",
"\n",
"#Minimum clearance is\n",
"cler = 30.0 - sag #m\n",
"print \"Minimum clearance = \",round(cler,2),\"m.\"\n",
"#Taking this point as reference , sag of mid-pt is\n",
"sag2 = W*(l-x1)**2/(2*T) #m\n",
"\n",
"#Mid-point Clearance\n",
"cler2 = cler + sag2\n",
"print \"Hence, clearance point = \",round(cler2,2),\"m.\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 41.57 , PAGE NO :- 1658"
]
},
{
"cell_type": "code",
"execution_count": 34,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Clearance of conductor from water surface = 63.89 m.\n"
]
}
],
"source": [
"'''An overhead transmission line at a river crossing is supported from two towers at heights of 50 m and 100 m above the\n",
"water level, the horizontal distance between the towers being 400 m. If the maximum allowable tension is 1,800 kg and the\n",
"conductor weighs 1 kg/m, find the clearance between the conductor and water at a point mid-way between the towers.'''\n",
"\n",
"#Given\n",
"h = 100.0 - 50.0 #m (height)\n",
"l = 400.0/2 #m (mid-way length)\n",
"T = 1800.0 #kg-wt (tension)\n",
"W = 1.0 #kg-wt/m (Weigth)\n",
"#Point of max sag\n",
"x1 = l - h*T/(2*W*l) #m\n",
"x2 = l + h*T/(2*W*l) #m\n",
"\n",
"#Distance of mid-pt from the above point is\n",
"d1 = (x2 - x1)/2 #m\n",
"\n",
"#Therefore height of P above max sag point is\n",
"h1 = W*d1**2/(2*T) #m\n",
"h2 = W*x2**2/(2*T) #m\n",
"\n",
"#Therefore point P from top is\n",
"Psag = h2 - h1 #m\n",
"\n",
"#Clearance above ground is\n",
"cler = 100.0 - Psag #m\n",
"\n",
"print \"Clearance of conductor from water surface = \",round(cler,2),\"m.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 41.58 , PAGE NO :- 1659"
]
},
{
"cell_type": "code",
"execution_count": 18,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Horizontal Tension = 2373.42 kg-wt/m.\n"
]
}
],
"source": [
"'''A conductor is strung across a river, being supported at the two ends at heights of 20 m and 16 m respectively, from the bed\n",
"of the river. The distance between the supports is 375 m and the weight of the conductor = 1.2 kg/m.If the clearance of the\n",
"conductor from the river bed be 9 m, find the horizontal tension in the conductor. Assume a parabolic configuration and that\n",
"there is no wind or ice loading.'''\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"\n",
"#Given\n",
"l = 375.0/2 #m (mid-pt of towers)\n",
"h = 20.0 - 16.0 #m (height)\n",
"W = 1.2 #kg-wt/m\n",
"\n",
"#Let T be the tension in conductor\n",
"T = Symbol('T')\n",
"x1 = l - h*T/(2*W*l) #m\n",
"\n",
"#Minimum clearance\n",
"cler = 9.0 #m \n",
"sag = 16.0 - cler #m\n",
"\n",
"#Now ,sag is also\n",
"sag2 = (W*x1*x1)/(2*T) #m\n",
"\n",
"#Equating the two equations\n",
"eq = Eq(sag,sag2)\n",
"T = solve(eq)\n",
"T1 = T[0] #kg-wt/m\n",
"print \"Horizontal Tension = \",round(T1,2),\"kg-wt/m.\""
]
}
],
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"kernelspec": {
"display_name": "Python 2",
"language": "python",
"name": "python2"
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"language_info": {
"codemirror_mode": {
"name": "ipython",
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"file_extension": ".py",
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PKj„ìJlÞÿ\Å¤Å¤;A TEXTBOOK OF ELECTRICAL TECHNOLOGY VOL III/chapter43.ipynb{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# CHAPTER 43: ELECTRIC TRACTION"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.1 , PAGE NO :- 1716"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Max speed of train is = 39.2 km/h.\n"
]
}
],
"source": [
"#A suburban train runs with an average speed of 36 km/h between two stations 2 km apart.Values of acceleration\n",
"#and retardation are 1.8 km/h/s and 3.6 km/h/s.Compute the maximum speed of the train assuming trapezoidal\n",
"#speed/time curve\n",
"################################################################################################################\n",
"import math\n",
"\n",
"#Given\n",
"\n",
"Va = 36.0*5/18.0 #m/s (average speed)\n",
"alpha = 1.8*5/18.0 #m/s^2 (acc.)\n",
"beta = 3.6*5/18.0 #m/s^2 (ret.)\n",
"D = 2000.0 #m (distance )\n",
"t = D/Va #s (time)\n",
"K = (alpha + beta)/(2*alpha*beta) #(constant)\n",
"Vm = (t-math.sqrt(t*t-4*K*D))/(2*K) #m/s (max speed)\n",
"Vm = Vm*18/5.0 #km/h\n",
"print \"Max speed of train is = \",round(Vm,2),\"km/h.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.2 , PAGE NO :- 1716"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The acceleration is = 1.7 km/h/s.\n"
]
}
],
"source": [
"#A train is required to run between two stations 1.5 km apart at a schedule speed of 36 km/h,the duration of stops being\n",
"#25 seconds.The braking retardation is 3 km/h/s.Assuming a trapezoidal speed/time curve,calculate the acceleration if the\n",
"#ratio of maximum speed to average speed is to be 1.25\n",
"##########################################################################################################################\n",
"\n",
"#Given\n",
"D = 1500.0 #m (distance)\n",
"Vsch = 36.0*5/18.0 #m/s (scheduled speed)\n",
"beta = 3.0*5/18.0 #km/h/s (ret.)\n",
"\n",
"time = D/Vsch #s (scheduled time)\n",
"t_stop = 25.0 #s (stopping time)\n",
"t = time-t_stop #s (actual time of run)\n",
"Va = D/t #m/s (average speed)\n",
"Vm = 1.25*Va #m/s (max speed)\n",
"\n",
"#The value of K is\n",
"K = (D/(Vm*Vm))*(Vm/Va-1)\n",
"#We know that K = 1/2*(1/alpha + 1/beta).Therefore,value of alpha is\n",
"alpha = beta/(2*K*beta-1) #m/s^2\n",
"alpha = alpha*18/5.0 #km/h/s\n",
"print \"The acceleration is =\",round(alpha,1),\"km/h/s.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.3 , PAGE NO :- 1716"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Schedule speed of train is 38.7 Km/h.\n"
]
}
],
"source": [
"#Find the scheduled speed of an electric train for a run of 1.5km if the ratio of its maximum to average speed is\n",
"#1.25.It has a braking retardation of 3.6 km/h/s,acceleration of 1.8km/h/s and stop time of 21 seconds.Assume\n",
"#trapezoidal speed/time curve.\n",
"##################################################################################################################\n",
"import math\n",
"\n",
"#Given\n",
"alpha = 1.8*5/18.0 #m/s^2 (acceleration)\n",
"beta = 3.6*5/18.0 #m/s^2 (retardation)\n",
"D = 1500.0 #m (distance)\n",
"ratio = 1.25 #(Vm/Va) \n",
"t_stop = 21 #s (stopping time) \n",
"K = 0.5*(1/alpha + 1/beta)\n",
"\n",
"#Also constant K = (D/Vm^2)*(Vm/Va-1)\n",
"#Therefore Vm^2 = (D/K)*((Vm/Va)-1).\n",
"Vm = math.sqrt(D/K*(ratio-1)) #m/s (Max. speed)\n",
"Va = Vm/ratio #m/s (Avg. speed) \n",
"t = D/Va #s (travel time)\n",
"\n",
"tsch = t + t_stop #s (schedule time)\n",
"Vsch = D/tsch #m/s (scheduled speed)\n",
"Vsch = Vsch*18/5.0 #km/h\n",
"print \"Schedule speed of train is\",round(Vsch,1),\"Km/h.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.4 , PAGE NO :- 1717"
]
},
{
"cell_type": "code",
"execution_count": 19,
"metadata": {
"collapsed": false,
"scrolled": true
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Duration of acceleration = 26.7 s.\n",
"Duration of coasting = 119.3 s.\n",
"Duration of braking = 14.0 s.\n",
"Total duration = 160.0 s.\n"
]
}
],
"source": [
"#A train runs between two stations 1.6km apart at an average speed of 36 km/h.If the maximum speed is to be limited\n",
"#to 72km/h,acceleration to 2.7km/h/s,coasting retardation to 0.18 km/h/s and braking retardation to 3.2 km/h/s,compute\n",
"#the duration of acceleration,coasting and braking periods.Assume a simplified speed/time curve.\n",
"########################################################################################################################\n",
"\n",
"from sympy import Eq,Symbol, solve\n",
"#Given\n",
"D = 1600 #m (distance)\n",
"Va = 36.0*5/18.0 #m/s (avg speed)\n",
"V1 = 72.0*5/18.0 #m/s (max speed)\n",
"alpha = 2.7*5/18.0 #m/s^2 (acceleration)\n",
"beta = 3.6*5/18.0 #m/s^2 (braking retardation)\n",
"beta_c = 0.18*5/18.0 #m/s^2 (coasting retardation)\n",
"#Duration of acceleration\n",
"t1 = V1/alpha #s\n",
"\n",
"#Actual time of run\n",
"t = D/Va #s\n",
"#Let us assume V2 be the speed at start of braking.\n",
"\n",
"V2 = Symbol('V2')\n",
"t3 = V2/beta #(braking period)\n",
"t2 = (V1-V2)/beta_c #(coasting period)\n",
"\n",
"eq = Eq(t1+t2+t3,t)\n",
"V2 = solve(eq)\n",
"Ve2 = V2[0]\n",
"\n",
"#Now we know the value of V2 that is Ve2.\n",
"t3 = Ve2/beta #(braking period)\n",
"t2 = (V1-Ve2)/beta_c #(coasting period)\n",
"print \"Duration of acceleration = \",round(t1,1),\"s.\"\n",
"print \"Duration of coasting = \",round(t2,1),\"s.\"\n",
"print \"Duration of braking = \",round(t3,1),\"s.\"\n",
"print \"Total duration = \",round(t,0),\"s.\"\n",
"\n",
"\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.5 , PAGE NO :- 1724"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The limiting value of train speed = 82.0 km/h.\n"
]
}
],
"source": [
"#The peripheral speed of a railway traction motor cannot be allowed to exceed 44m/s.If gear ratio is 18/75,motor armature\n",
"#diameter 42cm and wheel diameter 91cm,calculate the limiting value of the train speed.\n",
"###########################################################################################################################\n",
"\n",
"#Given\n",
"gratio = 18/75.0 #(gear ratio)\n",
"Vmot = 44.0 #m/s (speed of traction motor)\n",
"wdia = 0.91 #m (wheel diameter)\n",
"mdia = 0.42 #m (motor armature diameter)\n",
"\n",
"#Maximum number of revolutions by armature (in 1s)\n",
"mrev = Vmot/(3.14*mdia) #rps\n",
"#Maximum number of revolutions by driving wheel(in 1s)\n",
"wrev = mrev*gratio #rps\n",
"#Maximum distance travelled by driving wheel (in 1s)\n",
"dist = wrev*(3.14*wdia) #m\n",
"#Therefore,limiting speed is\n",
"vel = dist*18/5.0 #km/h\n",
"print \"The limiting value of train speed = \",round(vel),\"km/h.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.6 , PAGE NO :- 1724"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Torque developed by each motor = 7181.0 N-m.\n"
]
}
],
"source": [
"#A 250-tonne motor coach driven by four motors takes 20 seconds to attain a speed of 42 km/h,starting from rest on an ascending\n",
"#gradient of 1 in 80.The gear ratio is 3.5,gear efficiency 92%,wheel diameter 92cm train resistance 40N/t and rotational inertia\n",
"#10 percent of the dead weight.Find the torque developed by each motor.\n",
"##################################################################################################################################\n",
"\n",
"#Given\n",
"M = 250.0 #tonne (mass of motor)\n",
"Me = 1.1*250.0 #tonne (mass of rotating motor)\n",
"Vm = 42.0 #km/h (speed)\n",
"t1 = 20.0 #s (time)\n",
"G = 1/80.0*100 # (% gradient)\n",
"r = 40.0 #N/tonne(train reistance)\n",
"D = 0.92 #m (wheel diameter)\n",
"gratio = 3.5 # (gear ratio)\n",
"geff = 0.92 # (gear efficiency)\n",
"\n",
"a = Vm/t1 # (acceleration) \n",
"#Now,tractive force is given by\n",
"Ft = 277.8*Me*a + 98*M*G + M*r #N\n",
"#Now Ft = 2*gratio*geff*T/D.Therefore torque 'T' is\n",
"T = Ft*D/(2*gratio*geff) #N-m\n",
"#There are motors.so,torque by each motor is\n",
"torque = T/4 #N-m\n",
"print \"Torque developed by each motor = \",round(torque),\"N-m.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.7 , PAGE NO :- 1724"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Time taken to achieve speed of 80km/h is = 44.3 s.\n",
"Current drawn per motor is 488.0 A.\n"
]
}
],
"source": [
"#A 250-tonne motor coach having 4 motors,each developing a torque of 8000 N-m during acceleration,starts from rest.If\n",
"#up-gradient is 30 in 1000,gear ratio 3.5,gear transmission efficiency 90%,wheel diameter 90cm,train resistance 50N/t,\n",
"#rotational inertia effect 10%,compute the time taken by the coach to attain a speed of 80km/h.\n",
"#If supply voltage is 3000V and motor efficiency 85%,calculate the current taken during the acceleration period.\n",
"#########################################################################################################################\n",
"\n",
"#Given\n",
"M = 250.0 #tonne (mass of motor)\n",
"Me = 1.1*250 #tonne (mass of rotating motor)\n",
"r = 50.0 #N/t (train resistance)\n",
"G = 30.0/1000*100 # (% gradient)\n",
"torque = 8000 #N-m (torque of each motor)\n",
"D = 0.90 #m (wheel diameter)\n",
"gratio = 3.5 # (gear ratio)\n",
"geff = 0.90 # (gear efficiency)\n",
"Vm = 80 #km/h (speed)\n",
"meff = 0.85 # (Motor efficiency)\n",
"V = 3000.0 # (Voltage)\n",
"\n",
"T = 4*torque # (total torque)\n",
"#Now,we Know that\n",
"Ft = 2*gratio*geff*T/D #N\n",
"#Also,we know that Ft = 277.8*Me*a + 98MG + Mr\n",
"a = (Ft-98*M*G-M*r)/(277.8*Me) #km/h/s(acceleration)\n",
"#Time taken to attain the speed\n",
"t1 = Vm/a #s\n",
"#Power taken by motor is given by\n",
"Power = Ft*(5.0/18*Vm)/meff #W\n",
"#Total current drawn is (using P = VI)\n",
"I = Power/V #A\n",
"#Current drawn/motor is\n",
"cur = I/4 #A\n",
"\n",
"print \"Time taken to achieve speed of 80km/h is =\",round(t1,1),\"s.\"\n",
"print \"Current drawn per motor is\",round(cur,1),\"A.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 43.8 , PAGE NO :- 1725"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Weight of locomotive is = 142.0 tonne.\n",
"Number of axles required are = 7.0\n"
]
}
],
"source": [
"#A goods train weighing 500 tonne is to be hauled by a locomotive up an ascending gradient of 2% with an\n",
"#acceleation of 1 km/h/s.If the coefficient of adhesion is 0.25,train resistance 40N/t and effect of rotational\n",
"#inertia 10%,find the weight of locomotive and number of axles if load is not to increase beyond 21 tonne/axle.\n",
"###############################################################################################################\n",
"\n",
"from sympy import Eq,Symbol, solve\n",
"#Given\n",
"mass = 500.0 #tonne (mass of goods train)\n",
"a = 1 #km/h/s (acceleration)\n",
"coef_ad = 0.25 #(coefficient of adhesion)\n",
"r = 40.0 #N/t (train resistance)\n",
"G = 2.0 # (% gradient)\n",
"w_axle = 21.0 #tonne/axle (weight per axle)\n",
"#Let Ml be mass of locomotive\n",
"Ml = Symbol('Ml')\n",
"#Total mass is given by\n",
"M = mass + Ml\n",
"#Now,tractive force is given by\n",
"Ft1 = M*(277.8*1.1*a + 98*G + r) #N\n",
"#Also,Maximum tractive force is given by\n",
"Ft2 = 1000*coef_ad*Ml*9.8\n",
"#As both Ft1 and Ft2 are same, therefore equating\n",
"eq = Eq(Ft1,Ft2)\n",
"Ml = solve(eq)\n",
"#Mass of locomotive 'M_l' is\n",
"M_l = Ml[0]\n",
"#No of axles required are\n",
"no_axle = round(M_l)/w_axle\n",
"print \"Weight of locomotive is =\",round(M_l),\"tonne.\"\n",
"print \"Number of axles required are = \",round(no_axle)\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.9 , PAGE NO :- 1725"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The trailing weight that can now be hauled up = 1100.0 tonne.\n"
]
}
],
"source": [
"#An electric locomotive weighing 100 tonne can just accelerate a train of 500 tonne(trailing weight) with an acceleration of \n",
"#1 km/h/s on an up-gradient of 0.1%.Train resistance is 45N/t and rotational inertia is 10%.If this locomotive is helped by another\n",
"#locomotive of 120 tonne,find:\n",
"#(i)the trailing weight that can now be hauled up the same gradient under the same conditions.\n",
"#(ii)the maximum gradient,if the trailing hauled load remains unchanged.\n",
"#Assume adhesive weight expressed as percentage of total dead weight as 0.8 for both locomotives.'''\n",
"#############################################################################################################################\n",
"\n",
"#Given\n",
"Ml1 = 100.0 #tonne (locomotive 1)\n",
"Ml2 = Ml1 + 120.0 #tonne (locomotive 1&2)\n",
"mass = 500.0 #tonne (trailing part)\n",
"r = 45.0 #N/t (train resistance)\n",
"G = 0.1 # (% gradient)\n",
"a = 1.0 #km/h/s (acceleration)\n",
"\n",
"#Total mass of train and locomotive\n",
"M = Ml1 + mass #tonne\n",
"Me = 1.1*M #tonne \n",
"#Traction Force is\n",
"Ft = 277.8*Me*a + 98*M*G + M*r #N\n",
"#Let coefficient_of_adhesion be 'ua'.Then the maximium tractive effort by locomotive 1 is Ft = 1000*ua*Ml1*9.8\n",
"ua = Ft/(1000*9.8*Ml1)\n",
"\n",
"#Maximum tractive effort by locomotive 1&2 is\n",
"Ft2 = 1000*ua*Ml2*9.8 #N\n",
"#Also,Ft2 = M2*(277.8*1.1*a + G + r).Therefore,\n",
"M2 = Ft2/(277.8*1.1*a + 98*G + r) #tonne\n",
"#(i)The trailing weight that can be hauled up is\n",
"wtrail = M2-Ml2 #tonne\n",
"#(ii)Now, in this case the total weight of train & locomotive is\n",
"M3 = mass + Ml2\n",
"#Tractive force Ft2 = M3*(277.8*1.1*a + 98G2 + r).Therefore,for max gradient 'G2'\n",
"G2 = (Ft2/M3-277.8*1.1*a-r)/98\n",
"\n",
"print \"The trailing weight that can now be hauled up =\",round(wtrail,2),\"tonne.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.10 , PAGE NO :- 1726"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Max speed = 44.32 km/h.\n",
"Specific energy output = 30.18 Wh/t-km.\n"
]
}
],
"source": [
"#The average distance between stops on a level section of a railway is 1.25km.Motor-coach train weighing 200 tonne has a\n",
"#schedule speed of 30 km/h,the duration of stops being 30 seconds.The acceleration is 1.9 km/h/s and the braking retardation\n",
"#is 3.2 km/h/s.Train resistance to traction is 45 N/t.Allowance for rotational inertia is 10%.Calculate the specific energy\n",
"#output in Wh/t-km.Assume a trapezoidal speed/time curve.\n",
"##############################################################################################################################\n",
"import math as m\n",
"#Given\n",
"alpha = 1.9*5/18.0 #km/h/s (acceleration)\n",
"beta = 3.2*5/18.0 #km/h/s (retardation)\n",
"D = 1.25*1000.0 #m (distance)\n",
"tstop = 30.0 #s (stop time)\n",
"Vsch = 30*5/18.0 #m/s (schedule speed) \n",
"r = 45.0 #N/t (train resistance) \n",
"K = round((alpha + beta)/(2*alpha*beta),1) #(constant K)\n",
"\n",
"tsch = D/Vsch #s (schedule time)\n",
"t = tsch-tstop #s (running time)\n",
"#Now,max speed is given by\n",
"Vm = (t - m.sqrt(t*t-4*K*D))/(2*K) #m/s\n",
"#Braking distance is given by(using newton's III equation of motion)\n",
"dist = Vm*Vm/(2*beta) #m\n",
"#Therefore non-braking distance\n",
"D2 = (D - dist)/1000 #km\n",
"Vm = Vm*18/5.0 #km/h\n",
"D = D/1000 #km\n",
"#Specific energy output is\n",
"spengy = 0.01072*(Vm*Vm/D)*1.1 + 0.2778*r*(D2/D)\n",
"\n",
"print \"Max speed = \",round(Vm,2),\"km/h.\"\n",
"print \"Specific energy output = \",round(spengy,2),\" Wh/t-km.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.11 , PAGE NO :- 1726"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The value of specific energy consumption is 21.6 Wh/t-km.\n"
]
}
],
"source": [
"#A 300-tonne EMU is started with a uniform acceleration and reaches a speed of 40 Km/h in 24 seconds on a level track.\n",
"#Assuming trapezoidal speed/time curve,find specific energy consumption if rotational inertia is 8%,retardation is 3 km/h/s,\n",
"#distance between stops is 3 km,motor efficiency is 0.9 and train resistance is 49N/tonne.\n",
"#############################################################################################################################\n",
"\n",
"#Given\n",
"M = 300.0 #tonne (mass of train)\n",
"Mratio = 1.08 # (Me/M)\n",
"Vm = 40.0 #km/h (speed)\n",
"beta = 3.0 #km/h/s (retardation)\n",
"r = 49.0 #N/t (train resistance)\n",
"meff = 0.9 # (motor efficiency)\n",
"D = 3.0 #km (distance)\n",
"\n",
"#Braking time is(using Newton's Ist eqn)\n",
"t3 = Vm/beta\n",
"#Distance travelled in braking time(using Newton's IInd eqn)\n",
"dist = 0.5*beta*t3*t3/3600 #km\n",
"#Non braking distance D2 is\n",
"D2 = D - dist #km\n",
"#Specific energy consumption is given by,\n",
"spengy = 0.01072*(Vm*Vm/(meff*D))*Mratio + 0.2778*r/meff*(D2/D) #Wh/t-km\n",
"print \"The value of specific energy consumption is\",round(spengy,1),\"Wh/t-km.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.12 , PAGE NO :- 1726"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"acceleration = 2.0 km/h/s.\n",
"coasting retardation = 0.2 km/h/s.\n",
"The schedule speed = 30.3 km/h.\n",
"The new schedule speed = 31.6 km/h.\n"
]
}
],
"source": [
"#An electric train accelerates uniformly from rest to a speed of 50km/h in 25 seconds.It then coasts for 70 seconds\n",
"#against a constant resistance of 60N/t and is then braked to rest with uniform retardation of 3.0km/h/s in 12 seconds.\n",
"#Compute :- (i)uniform acceleration (ii)coasting retardation\n",
"#(iii)schedule speed if station stops are of 20-second duration\n",
"#Allow 10% for rotational inertia.How will the scedule speed be affected if duration of stops is reduced to 15 seconds,\n",
"#other factors remaining the same?\n",
"##########################################################################################################################\n",
"\n",
"#Given\n",
"V1 = 50.0 #km/h (Max speed)\n",
"t1 = 25.0 #s (accelerating time)\n",
"t2 = 70.0 #s (coasting time)\n",
"r = 60.0 #N/t (train resistance)\n",
"beta = 3.0 #km/h/s (retardation)\n",
"t3 = 12.0 #s (braking time)\n",
"\n",
"#(i) Uniform acceleration is (using Newton's 1st eqn.)\n",
"alpha = V1/t1 #km/h/s\n",
"print \"acceleration = \",round(alpha,1),\"km/h/s.\"\n",
"#Now in braking period (using Newton's 1st eqn.)-> 0 = V2 - beta*t3\n",
"V2 = beta*t3 #km/h\n",
"#(ii)Now in coasting period (using Newton's 1st eqn.)-> V2 = V1 - beta_c*t2\n",
"beta_c = (V1-V2)/t2 #km/h/s\n",
"print \"coasting retardation = \",round(beta_c,1),\"km/h/s.\"\n",
"#distance travelled in accelerating period (using Newton's 2nd eqn.)\n",
"dis1 = 0.5*alpha*t1*t1/3600 #km\n",
"#distance travelled in coasting period (using Newton's 3rd eqn.)\n",
"dis2 = -(V2*V2 - V1*V1)/(2*beta_c*3600) #km\n",
"#distance travelled in braking period (using Newton's 2nd eqn.)\n",
"dis3 = 0.5*beta*t3*t3/3600 #km\n",
"#Total distance\n",
"D = dis1 + dis2 + dis3 #km\n",
"#Total time\n",
"T = t1 + t2 + t3 + 20.0 #s\n",
"#Scheduled Speed Vsch is\n",
"Vsch = D/T*3600 #km/h\n",
"print \"The schedule speed = \",round(Vsch,1),\"km/h.\"\n",
"#New Total time\n",
"T = t1 + t2 + t3 + 15.0\n",
"#New scheduled speed is\n",
"Vsch = D/T*3600 #km/h\n",
"print \"The new schedule speed = \",round(Vsch,1),\"km/h.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.13 , PAGE NO :- 1727"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The specific energy consumption is = 63.1 Wh/t-km.\n"
]
}
],
"source": [
"#A 350-tonne electric train runs up an ascending gradient of 1% with the following speed/time curves:\n",
"# 1.uniform acceleration of 1.6km/h/s for 25 seconds 2.constant speed for 50 seconds\n",
"# 3.coasting for 30 seconds 4.braking at 2.56 km/h/s to rest.\n",
"#Compute the specific energy consumption if train resistance is 50N/t,effect of rotational inertia 10%,overall efficiency of\n",
"#transmission gear and motor,75%.\n",
"#############################################################################################################################\n",
"\n",
"#Given\n",
"M = 350.0 #tonne (weight of train)\n",
"Me = 1.1*M #tonne (effect of rot. inertia)\n",
"G = 1.0 # (% gradient)\n",
"alpha = 1.6 #km/h/s (acceleration)\n",
"t1 = 25.0 #s (acceleration time)\n",
"t2 = 50 #s (constant speed time)\n",
"t3 = 30 #s (coasting time)\n",
"beta = 2.56 #km/h/s (braking retardation)\n",
"r = 50.0 #N/t (train resistance)\n",
"meff = 0.75 # (motor efficiency)\n",
"\n",
"V1 = alpha*t1 #km/h (Max speed attained)\n",
"#Tractive force during coasting is\n",
"Ft = 98*M*G + M*r #N\n",
"#Also,Tractive force during coasting is given by\n",
"#Ft = 277.8*Me*beta_c . Therefore beta_c is\n",
"beta_c = Ft/(277.8*Me) #km/h/s\n",
"\n",
"#During Coasting period.(Using Newton's 1st eqn.) . \"V2\" is given by\n",
"V2 = V1 - beta_c*t3 #km/h\n",
"t4 = V2/beta #s (braking time) \n",
"\n",
"#Distance travelled during acceleration period.\n",
"dis1 = 0.5*alpha*t1*t1/3600 #km\n",
"#Distance travelled during constant speed period.\n",
"dis2 = V1*t2/3600 #km\n",
"#Distance travelled during coasting period.\n",
"dis3 = (V1+V2)/2*t3/3600 #km\n",
"#Distance travelled during braking period.\n",
"dis4 = 0.5*V2*t4/3600 #km\n",
"#Total distance between stops\n",
"D = dis1 + dis2 + dis3 + dis4 #km\n",
"#Distance travelled during acceleration and constant speed period\n",
"D2 = dis1 + dis2 #km\n",
"#Specific energy consumption is given by\n",
"spengy = (0.01072*(V1*V1/D)*(Me/M) + 27.25*G*(D2/D) + 0.2778*r*(D2/D))/meff #Wh/t-km\n",
"\n",
"print \"The specific energy consumption is =\",round(spengy,1),\"Wh/t-km.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.14 , PAGE NO :- 1728"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total number of locomotives required are = 4.0\n"
]
}
],
"source": [
"#An ore-carrying train weighing 5000 tonne is to be hauled down a gradient of 1:50 at a maximum speed of 30km/h\n",
"#and started on a level track at an acceleration of 0.29 km/h/s.How many locomotives,each weighing 75 tonne,will\n",
"#have to be employed?\n",
"#Train resistance during starting = 29.4 N,Train resistance at 30km/h = 49N/t,Coefficient of adhesion = 0.3\n",
"#Rotational inertia = 10%\n",
"################################################################################################################################\n",
"\n",
"#Given\n",
"M = 5000.0 #tonne (Mass of train)\n",
"Ml = 75.0 #tonne (Mass of locomotive) \n",
"G = 1.0/50 # (gradient)\n",
"Vm = 30.0 #km/h (Max speed)\n",
"a = 0.29 #km/h/s (acceleration) \n",
"r1 = 49.0 # (Train resistance at 30km/h)\n",
"r2 = 29.4 # (Train resistance at starting)\n",
"ua = 0.3 # (coefficient of adhesion)\n",
"\n",
"#Downward force due to gravity\n",
"F = M*G*9.8*1000 #N\n",
"#Train resistance\n",
"Fres = r1*M #N\n",
"#Braking force required is\n",
"Fbrk = F-Fres #N\n",
"#Max. braking force which one locomotive can produce.\n",
"#F = 1000*ua*M*g\n",
"Fbrk_1 = 1000*ua*Ml*9.8 #N\n",
"#Therefore, Number of locomotives required for braking\n",
"num1 = Fbrk/Fbrk_1\n",
"num1 = round(num1)+ 1\n",
"#Considering the starting case.Tractive force required is\n",
"Ft = 277.8*a*M*1.1 + M*r2\n",
"#Therefore, Number of locomotives required for starting\n",
"num2 = Ft/Fbrk_1\n",
"#Number of locomotives required\n",
"if num1>num2:\n",
" num = num1\n",
"else:\n",
" num = num2\n",
"print \"Total number of locomotives required are = \",round(num)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.15 , PAGE NO :- 1729"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Scheduled speed = 32.9 Km/h.\n",
"Specific energy consumption for up-gradient is = 69.08 Wh/t-km.\n",
"Specific energy consumption for down-gradient is = 39.58 Wh/t-km\n"
]
}
],
"source": [
"#A 200-tonne electric train runs according to the following quadrilateral speed/time curve:\n",
"# 1. uniform acceleration from rest at 2 km/h/s for 30 seconds.\n",
"# 2. coasting for 50 seconds\n",
"# 3. duration of braking:15 seconds\n",
"#If up-gradient is =1%,train resistance = 40N/t,rotational inertia effect = 10%,duration of stops = 15s and overall\n",
"#efficiency of gear and motor = 75%,find\n",
"#(i)schedule speed (ii)specific energy consumption (iii)how will the value of specific energy consumption change if\n",
"#there is a down-gradient of 1% rather than the up-gradient ?\n",
"########################################################################################################################\n",
"\n",
"#Given\n",
"M = 200.0 #tonne (Mass of train)\n",
"Me = 1.1*200.0 #tonne (rotational inertia effect on mass)\n",
"alpha = 2.0 #km/h/s (acceleration)\n",
"t1 = 30.0 #s (acceleration time)\n",
"t2 = 50.0 #s (coasting time)\n",
"t3 = 15.0 #s (braking time)\n",
"G = 1.0 # (% gradient)\n",
"r = 40.0 #N/t (train resistance)\n",
"tstop = 15.0 #s (stop time)\n",
"meff = 0.75 # (motor efficiency)\n",
"\n",
"V1 = alpha*t1 #km/h (Max speed)\n",
"#During coasting retardation force is\n",
"Fr = 98*M*G + M*r #N\n",
"#Also this retardation force in terms of coasting retardation beta_c is\n",
"#Fr = 277.8*Me*beta_c.Therefore,\n",
"beta_c = Fr/(277.8*Me) #km/h/s\n",
"#Using newton's Ist eqn in coasting period\n",
"V2 = V1-beta_c*t2 #km/h\n",
"\n",
"#Braking retardation is\n",
"beta = V2/t3\n",
"#Distance travelled during acceleration\n",
"dist1 = 0.5*alpha*t1*t1/3600 #m\n",
"#Distance travelled during coasting\n",
"dist2 = (V1+V2)/2*t2/3600 #m\n",
"#Distance travelled during braking\n",
"dist3 = 0.5*beta*t3*t3/3600 #m\n",
"#Total distance travelled\n",
"D = dist1 + dist2 + dist3 #km\n",
"#Total schedule time\n",
"T = t1 + t2 + t3 + tstop #s\n",
"#(i)Schedule speed\n",
"Vsch = D/T*3600 #km/h\n",
"print \"Scheduled speed = \",round(Vsch,1),\"Km/h.\"\n",
"#(ii)Specific energy consumption is given by\n",
"spengy = (0.01072*(V1*V1/D)*(Me/M) + 27.25*G*(dist1/D)+ 0.2778*r*(dist1/D))/meff #Wh/t-km\n",
"print \"Specific energy consumption for up-gradient is = \",round(spengy,2),\"Wh/t-km.\"\n",
"#If there is a down gradient then during coasting accelerative force is\n",
"Fa = 98*M*(-G) + M*r\n",
"#Also this accelerative force in terms of coasting acceleration alpha_c is\n",
"#Fa = 277.8*Me*alpha_c.Therefore,\n",
"alpha_c = Fa/(277.8*Me) #km/h/s\n",
"#Using newton's Ist eqn in coasting period\n",
"V2 = V1 - alpha_c*t2 #km/h\n",
"#Braking retardation is\n",
"beta = V2/t3\n",
"#Distance travelled during acceleration\n",
"dist1 = 0.5*alpha*t1*t1/3600 #m\n",
"#Distance travelled during coasting\n",
"dist2 = (V1+V2)/2*t2/3600 #m\n",
"#Distance travelled during braking\n",
"dist3 = 0.5*beta*t3*t3/3600 #m\n",
"#Total distance travelled\n",
"D = dist1 + dist2 + dist3 #km\n",
"#(iii)Specific energy consumption is given by\n",
"spengy = (0.01072*(V1*V1/D)*(Me/M) - 27.25*G*(dist1/D)+ 0.2778*r*(dist1/D))/meff #Wh/t-km\n",
"print \"Specific energy consumption for down-gradient is = \",round(spengy,2),\"Wh/t-km\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.16 , PAGE NO :- 1730"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
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"text/plain": [
""
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"#An electric train has an average speed of 45 kmph on a level track between stops 1500m apart.It is accelerated at\n",
"#1.8km/h/s and is braked at 3 km/h/s.Draw the speed-time curve for the run.\n",
"####################################################################################################################\n",
"\n",
"import math as m\n",
"import matplotlib.pyplot as plt\n",
"%matplotlib inline \n",
"\n",
"#Given\n",
"alpha = 1.8 #km/h/s (acceleration)\n",
"beta = 3.0 #km/h/s (retardation)\n",
"S = 1.5 #km (Distance of run)\n",
"Va = 45.0 #km/h (Average speed)\n",
"\n",
"T = S/Va*3600 #s (Time of run)\n",
"\n",
"#Constant K is given by\n",
"K = 0.5*(1/alpha + 1/beta)\n",
"#Max speed is given by\n",
"Vm = T/(2*K)-m.sqrt(T*T/(4*K*K) - 3600*S/K) #km/h\n",
"#Acceleration time\n",
"t1 = Vm/alpha #s\n",
"#Braking time\n",
"t3 = Vm/beta #s\n",
"#Free running time\n",
"t2 = T - (t1+t3) #s\n",
"\n",
"#SPEED TIME CURVE\n",
"\n",
"plt.plot([0,t1,t1+t2,t1+t2+t3], [0,Vm,Vm,0])\n",
"plt.ylabel('Speed(in km/h)')\n",
"plt.xlabel('Time(in seconds)')\n",
"plt.show()"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.17 , PAGE NO :- 1731"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Maximum Speed is 72.85 Km/h.\n"
]
}
],
"source": [
"#A train has schedule speed of 60 km per hour between the stops which are 9 km apart.Determine the crest speed over\n",
"#the run,assuming trapezoidal speed-time curve.The train accelerates at 3km/h/s and retards at 4.5 km/h/s.Duration\n",
"#of stops is 75 seconds.\n",
"#########################################################################################################################\n",
"import math as m\n",
"\n",
"#Given\n",
"alpha = 3.0 #km/h/s (acceleration)\n",
"beta = 4.5 #Km/h/s (retardation)\n",
"S = 9.0 #km (distance)\n",
"Vsch = 60.0 #km/h (schedule speed)\n",
"\n",
"Tsch = S/Vsch*3600 #s (schedule time)\n",
"T = Tsch - 75.0 #s (time of run)\n",
"#Constant K is\n",
"K = 0.5*(1/alpha + 1/beta)\n",
"\n",
"#Maximum speed Vm is\n",
"Vm = (T/(2*K)) - m.sqrt(T*T/(4*K*K) - 3600*S/K)\n",
"\n",
"print \"Maximum Speed is\",round(Vm,2),\"Km/h.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.18 , PAGE NO :- 1732"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Schedule speed is 60.0 km/h.\n"
]
}
],
"source": [
"#An electric train is to have acceleration and braking retardation of 1.2 km/h/s and 4.8 km/h/s respectively.If the ratio of\n",
"#maximum to average speed is 1.6 and time for stop is 35 seconds,find schedule speed for a run of 3km.Assume simplified\n",
"#trapezoidal speed-time curve.\n",
"#############################################################################################################################\n",
"import math as m\n",
"alpha = 1.2 #km/h/s (acceleration)\n",
"beta = 4.8 #km/h/s (retardation)\n",
"S = 3.0 #km (distance)\n",
"\n",
"#Constant K is\n",
"K = 0.5*(1/alpha + 1/beta)\n",
"\n",
"#The average speed is Va = S/T . Therefore Va*T = S*3600\n",
"Va_T = S*3600 \n",
"#Now Vm/Va = 1.6 .Therefore,\n",
"Vm_T = Va_T*1.6\n",
"\n",
"#Since, Vm^2*K - Vm*T + 3600*S = 0, Therefore\n",
"#Vm^2 = Vm*T - 3600*S\n",
"Vm = m.sqrt((Vm_T - 3600*S)/K) #km/h\n",
"#Average Speed is\n",
"Va = Vm/1.5 #km/h\n",
"#Actual Time of Run\n",
"T = 3600*S/Va #s\n",
"#Schedule Time\n",
"Tsch = T + 35.0 #s\n",
"#Schedule Speed\n",
"Vsch = S/Tsch*3600 #km/h\n",
"print \"Schedule speed is\",round(Vsch),\"km/h.\" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.19 , PAGE NO :- 1732"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The rate of acceleration required to operate this service = 1.85 km/h/s.\n"
]
}
],
"source": [
"#An electric train has a schedule speed of 25 kmph between stations 800m apart.The duration of stop is 20 seconds,the\n",
"#maximum speed is 20 percent higher than the average running speed and the braking retardation is 3 km/h/s.Calculate\n",
"#the rate of acceleration required to operate this service.\n",
"###################################################################################################################\n",
"\n",
"#Given\n",
"Vsch = 25.0 #km/h (Schedule speed)\n",
"S = 0.8 #km (Distance)\n",
"beta = 3.0 #km/h/s (Retardation)\n",
"\n",
"Tsch = S/Vsch*3600 #s\n",
"\n",
"#Actual time of run = Tsch - (time of stop)\n",
"T = Tsch - 20.0 #s\n",
"Va = S/T*3600 #km/h (Average speed)\n",
"Vm = 1.2*Va #km/h (Maximum speed)\n",
"\n",
"#Since, Vm^2*K -Vm*T + 3600*S = 0\n",
"#Constant K = (Vm*T - 3600*S)/(Vm^2)\n",
"K = (Vm*T - 3600*S)/(Vm*Vm)\n",
"\n",
"#Also,K = 0.5*(1/alpha + 1/beta).Therefore,\n",
"alpha = 1/(2*K - 1/beta) #km/h/s\n",
"\n",
"print \"The rate of acceleration required to operate this service =\",round(alpha,2),\"km/h/s.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.20 , PAGE NO :- 1733"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Retardation = 0.51 km/h/s.\n"
]
}
],
"source": [
"#A suburban electric train has a maximum speed of 80 kmph.The schedule speed including a station stop of 35 seconds is 50kmph.\n",
"#If the acceleration is 1.5 km/h/s,find the value of retardation when the average distance between stops is 5 km.\n",
"################################################################################################################################\n",
"\n",
"#Given\n",
"Vsch = 50.0 #km/h (schedule speed)\n",
"S = 5.0 #km (distance)\n",
"alpha = 1.5 #km/h/s (acceleration) \n",
"Vm = 80.0 #km/h (Max speed)\n",
"tstop = 30.0 #s (Time of stop)\n",
"\n",
"\n",
"Tsch = S/Vsch*3600 #s (Schedule time)\n",
"T = Tsch - tstop #s (Actual time of run)\n",
"\n",
"#Since, Vm^2*K -Vm*T + 3600*S = 0\n",
"#Constant K = (Vm*T - 3600*S)/(Vm^2)\n",
"K = (Vm*T - 3600*S)/(Vm*Vm)\n",
"\n",
"#Also,K = 0.5*(1/alpha + 1/beta).Therefore,\n",
"beta = 1/(2*K - 1/alpha) #km/h/s\n",
"\n",
"print \"Retardation =\",round(beta,2),\"km/h/s.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 43.21 , PAGE NO :- 1733"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Duration of acceleration = 32.0 s.\n",
"Duration of coasting retardation = 96.84 s.\n",
"Duration of braking retardation = 15.16 s.\n"
]
}
],
"source": [
"#A train is required to run between two stations 1.6 km apart at the average speed of 40 kmph.The run is to be made to a\n",
"#simplified quadrilateral speed-time curve.If the maximum speed is to be limited to 64 kmph,accelerating to 2 km/h/s and\n",
"#coasting and braking retardation to 0.16 km/h/s and 3.2 km/h/s respectively,determine the duration of acceleration,coasting\n",
"#and braking periods.\n",
"#################################################################################################################################\n",
"from sympy import Eq,Symbol,solve\n",
"#Given\n",
"S = 1.6 #km (Distance)\n",
"Va = 40.0 #km/h (Average speed)\n",
"Vm = 64.0 #km/h (Maximum speed)\n",
"alpha = 2.0 #km/h/s (Acceleration)\n",
"beta_c = 0.16 #km/h/s (Coasting retardation)\n",
"beta = 3.2 #km/h/s (braking retardation)\n",
"\n",
"T = S/Va*3600 #s (Actual time of run)\n",
"t1 = Vm/alpha #s (acceleration time)\n",
"#Let us assume the speed at starting of braking be V2\n",
"V2 = Symbol('V2')\n",
"t2 = (Vm-V2)/beta_c #s (coasting period)\n",
"t3 = V2/beta #s (braking period)\n",
"eq = Eq(t1+t2+t3,T)\n",
"V2 = solve(eq)\n",
"Ve2 = V2[0]\n",
"#Therfore coasting and braking periods are\n",
"t2 = (Vm-Ve2)/beta_c #s (coasting period)\n",
"t3 = Ve2/beta #s (braking period)\n",
"\n",
"print \"Duration of acceleration = \",round(t1,2),\"s.\"\n",
"print \"Duration of coasting retardation = \",round(t2,2),\"s.\"\n",
"print \"Duration of braking retardation = \",round(t3,2),\"s.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.22 , PAGE NO :- 1737"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Speed-Armature current graph\n"
]
},
{
"data": {
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"text/plain": [
""
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"#The torque-armature current characteristics of a series traction motor are given as:\n",
"#Armature Current(amp): 5 10 15 20 25 30 35 40\n",
"#Torque(N-m): 20 50 100 155 215 290 360 430\n",
"#The motor resistance is 0.3 ohm.If this motor is connected across 230V,deduce the speed armature current characteristics.\n",
"############################################################################################################################\n",
"\n",
"import matplotlib.pyplot as plt\n",
"%matplotlib inline\n",
"#Given\n",
"V = 230 #V (Supply voltage)\n",
"r = 0.3 #ohm (Motor resistance)\n",
"\n",
"#Current Data Ia\n",
"i1 = 5.0 #A\n",
"i2 = 10.0 #A\n",
"i3 = 15.0 #A\n",
"i4 = 20.0 #A\n",
"i5 = 25.0 #A\n",
"i6 = 30.0 #A\n",
"i7 = 35.0 #A\n",
"i8 = 40.0 #A\n",
"\n",
"#Torque Data \n",
"tau1 = 20.0 #N-m\n",
"tau2 = 50.0 #N-m\n",
"tau3 = 100.0 #N-m\n",
"tau4 = 155.0 #N-m\n",
"tau5 = 215.0 #N-m\n",
"tau6 = 290.0 #N-m\n",
"tau7 = 360.0 #N-m\n",
"tau8 = 430.0 #N-m\n",
"\n",
"#Back EMF Eb = V - Ia*r\n",
"e1 = V - i1*r #V\n",
"e2 = V - i2*r #V\n",
"e3 = V - i3*r #V\n",
"e4 = V - i4*r #V\n",
"e5 = V - i5*r #V\n",
"e6 = V - i6*r #V\n",
"e7 = V - i7*r #V\n",
"e8 = V - i8*r #V\n",
"\n",
"\n",
"#Speed Na = 9.55*Eb*Ia/T\n",
"N1 = 9.55*e1*i1/tau1 #rpm\n",
"N2 = 9.55*e2*i2/tau2 #rpm\n",
"N3 = 9.55*e3*i3/tau3 #rpm\n",
"N4 = 9.55*e4*i4/tau4 #rpm\n",
"N5 = 9.55*e5*i5/tau5 #rpm\n",
"N6 = 9.55*e6*i6/tau6 #rpm\n",
"N7 = 9.55*e7*i7/tau7 #rpm\n",
"N8 = 9.55*e8*i8/tau8 #rpm\n",
"\n",
"print \"Speed-Armature current graph\"\n",
"\n",
"plt.plot([i1,i2,i3,i4,i5,i6,i7,i8], [N1,N2,N3,N4,N5,N6,N7,N8])\n",
"plt.axis([0,40,0,600])\n",
"plt.ylabel('Speed(in r.p.m)')\n",
"plt.xlabel('Armature current Ia(in A)')\n",
"plt.show()"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.23 , PAGE NO :- 1737"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Speed - Torque Curve\n"
]
},
{
"data": {
"image/png": 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"text/plain": [
""
]
},
"metadata": {},
"output_type": "display_data"
}
],
"source": [
"#The following figures give the magnetization curve of d.c. series motor when working as a seperately excited generator at 600 rpm.\n",
"#Field Current(amperes) : 20 40 60 80\n",
"#E.M.F(volts): 215 381 485 550\n",
"#The total resistance of the motor is 0.8 ohm.Deduce the speed-torque curve for this motor when operating at a constant voltage of\n",
"#600 volts.\n",
"################################################################################################################################\n",
"\n",
"#Given\n",
"V = 600.0 #V (Supply voltage)\n",
"r = 0.8 #ohm (total motor resistance)\n",
"\n",
"#Field Current Data(If=Ia)\n",
"i1 = 20.0 #A\n",
"i2 = 40.0 #A\n",
"i3 = 60.0 #A\n",
"i4 = 80.0 #A\n",
"\n",
"#E.M.F data at 600 rpm(E)\n",
"e1 = 215.0 #V\n",
"e2 = 381.0 #V\n",
"e3 = 485.0 #V\n",
"e4 = 550.0 #V\n",
"\n",
"#At constant voltage of 600V. EMF are Eb = V - Ia*Rm\n",
"eb1 = V - i1*r #V\n",
"eb2 = V - i2*r #V\n",
"eb3 = V - i3*r #V\n",
"eb4 = V - i4*r #V\n",
"\n",
"#For DC series motor N1/E1 = N2/E2 . Therefore N2 = N/E*Eb\n",
"n1 = 600/e1*eb1 #rpm\n",
"n2 = 600/e2*eb2 #rpm\n",
"n3 = 600/e3*eb3 #rpm\n",
"n4 = 600/e4*eb4 #rpm\n",
"\n",
"#Now, Torque T = 9.55*Eb*Ia/N2\n",
"tau1 = 9.55*eb1*i1/n1 #N-m\n",
"tau2 = 9.55*eb2*i2/n2 #N-m\n",
"tau3 = 9.55*eb3*i3/n3 #N-m\n",
"tau4 = 9.55*eb4*i4/n4 #N-m\n",
"\n",
"print \"Speed - Torque Curve\"\n",
"\n",
"plt.plot([tau1,tau2,tau3,tau4], [n1,n2,n3,n4])\n",
"#plt.axis([0,40,0,600])\n",
"plt.ylabel('Speed(in r.p.m)')\n",
"plt.xlabel('Torque T(in N-m)')\n",
"plt.show()"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 43.24 , PAGE NO :- 1738"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Speed of both motors is = 447.87 rpm.\n",
"Voltage across motor A is = 256.35 V.\n",
"Voltage across motor B is = 243.65 V.\n"
]
}
],
"source": [
"#Two d.c. traction motors run at a speed of 900 r.p.m and 950 r.p.m respectively when each takes a current of 50A from 500V mains.\n",
"#Each motor has an effective resistance of 0.3 ohm . Calculate the speed and voltage across each machine when mechanically coupled\n",
"#and electrically connected in series and taking a current of 50A from 500V mains,the resistance of each motor being unchanged.\n",
"#################################################################################################################################\n",
"from sympy import Eq,Symbol,solve\n",
"#Let the two motors be A and B .\n",
"Na = 900.0 #rpm (speed of motor A)\n",
"Nb = 950.0 #rpm (speed of motor B)\n",
"r = 0.3 #ohm (resistance of each motor)\n",
"V = 500.0 #V (applied voltage)\n",
"I = 50.0 #A (current)\n",
"\n",
"#Back emf of motor A when taking current of 50A,\n",
"Eb_a = V - I*r #V\n",
"#Back emf of motor B when taking current of 50A,\n",
"Eb_b = V - I*r #V\n",
"\n",
"#As the machines are mechanically coupled and are connected in series. Therefore,\n",
"#speed will be same. Say 'N'\n",
"#current will be same and equal to 50A\n",
"#summation of voltage drop is 500V\n",
"\n",
"N = Symbol('N') #rpm\n",
"#Let the voltage drop across motors be Va & Vb .Therefore,for Va & Vb we will find back emf of both motors\n",
"eb_A = Eb_a*N/Na #V (using N1/N2 = E1/E2)\n",
"eb_B= Eb_b*N/Nb #V (using N1/N2 = E1/E2)\n",
"\n",
"#For Va & Vb \n",
"Va = eb_A + I*r #V (using Eb = V - Ia*r)\n",
"Vb = eb_B + I*r #V (using Eb = V - Ia*r)\n",
"\n",
"#We know that Va + Vb = 500 . Therefore,\n",
"eq = Eq(Va+Vb,V)\n",
"N = solve(eq)\n",
"Ne = N[0]\n",
"\n",
"#As we know the speed N . Therefore, Va & Vb are\n",
"eb_A = Eb_a*Ne/Na #V (using N1/N2 = E1/E2)\n",
"eb_B= Eb_b*Ne/Nb #V (using N1/N2 = E1/E2)\n",
"\n",
"Va = eb_A + I*r #V (using Eb = V - Ia*r)\n",
"Vb = eb_B + I*r #V (using Eb = V - Ia*r)\n",
"\n",
"print \"Speed of both motors is = \",round(Ne,2),\"rpm.\"\n",
"print \"Voltage across motor A is = \",round(Va,2),\"V.\"\n",
"print \"Voltage across motor B is = \",round(Vb,2),\"V.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.25 , PAGE NO :- 1739"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Current drawn by supply mains is = 138.32 A.\n"
]
}
],
"source": [
"#A tram car is equipped with two motors which are operating in parallel.Calculate the current drawn from the supply main at 500 V\n",
"#when the car is running at steady speed of 50kmph and each motor is developing a tractive effort of 2100 N.The resistance of each \n",
"#motor is 0.4 ohm.The friction,windage and other losses may be assumed as 3500 watts per motor.\n",
"#################################################################################################################################\n",
"from sympy import Eq,Symbol,solve\n",
"\n",
"V = 500.0 #V (applied voltage)\n",
"Vm = 50.0 #km/h (max. speed)\n",
"Ft = 2100 #N (tractive effort)\n",
"Rm = 0.4 #ohm (motor resistance)\n",
"mloss = 3500.0 #W (losses per motor)\n",
"\n",
"#Power output of each motor\n",
"Pout = Ft*Vm*5.0/18 #W\n",
"#Copper loss = I^2*Rm\n",
"I = Symbol('I')\n",
"cu_loss = (I*I)*Rm #W\n",
"#Input power = Output power + constant losses + Copper losses\n",
"Pin = V*I #W\n",
"eq = Eq(Pin,Pout+mloss+cu_loss)\n",
"I = solve(eq) #A\n",
"#current drawn by motor is\n",
"Ie = I[0] #A\n",
"#Therefore,current drawn by supply mains\n",
"Ie = 2*Ie #A\n",
"print \"Current drawn by supply mains is =\",round(Ie,2),\"A.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.26 , PAGE NO :- 1740"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Speed of first motor is = 217.17 rpm.\n",
"Speed of second motor is = 227.27 rpm.\n"
]
}
],
"source": [
"#A motor coach is being driven by two identical d.c. series motors.First motor is geared to driving wheel having diameter of 90cm and\n",
"#other motor to driving wheel having diameter of 86cm.The speed of the first motor is 500 rpm when connected in parallel with the other\n",
"#across 600V supply.Find the motor speeds when connected in series across the same supply.Assume armature current to remain same and\n",
"#armature voltage drop of 10% at this current.\n",
"###############################################################################################################################\n",
"from sympy import Eq,Symbol,solve\n",
"#Given\n",
"V = 600.0 #V (applied voltage)\n",
"d1= 0.90 #m (diameter 1) \n",
"d2 = 0.86 #m (diameter 2) \n",
"N1 = 500.0 #rpm (Speed of first motor)\n",
"\n",
"Eb1 = V-0.1*V #V (Back emf)\n",
"\n",
"#Let the supply voltage across motor 1 & 2 be V1 and V2 respectively\n",
"#Therefore, V1 + V2 = 600\n",
"V1 = Symbol('V1') #V\n",
"V2 = V - V1 #V\n",
"\n",
"#Now , N is directly propotional to (V - IR).Therefore, ratio (N1/N2) is\n",
"\n",
"Nratio1 = (V1 - 0.1*V)/(V2 - 0.1*V)\n",
"\n",
"#Also, we know that N1*d1 = N2*d2 . Therefore (N1/N2) is\n",
"Nratio2 = d2/d1\n",
"\n",
"#As, N1/N2 = Nratio1 = Nratio2\n",
"eq = Eq(Nratio1,Nratio2)\n",
"V1 = solve(eq)\n",
"Ve1 = V1[0] #V\n",
"Ve2 = V - Ve1 #V\n",
"#Now , for the speed of motor 1, we know that for N1 ,back emf is Eb1 and for Ne1 ,back emf is (Ve1-0.1*Ve1)\n",
"Ne1 = N1*(Ve1 - 0.1*V)/Eb1 #rpm (using N1/N2 = E1/E2)\n",
"#Nratio is Ne1/Ne2 .Therefore,\n",
"Ne2 = Ne1/Nratio2 #rpm\n",
"\n",
"print \"Speed of first motor is =\",round(Ne1,2),\"rpm.\"\n",
"print \"Speed of second motor is =\",round(Ne2,2),\"rpm.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 43.27 , PAGE NO :- 1741"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Speed of first motor is = 173.74 rpm.\n",
"Speed of second motor is = 181.82 rpm.\n"
]
}
],
"source": [
"#Two similar series type motors are used to drive a locomotive.The supply fed to their parallel connection is 650V.If the first \n",
"#motor 'A' is geared to drive wheels of radius 45 cms and other motor 'B' to 43 cms.And if the speed of first motor 'A' when\n",
"#connected in parallel to 2nd motor 'B' across the main supply lines is 400 rpm,find voltages and speeds of motors when connected\n",
"#in series.Assume Ia to be constant and armature voltage drop of 10% at this current.\n",
"################################################################################################################################\n",
"\n",
"from sympy import Eq,Symbol,solve\n",
"#Given\n",
"V = 650.0 #V (applied voltage)\n",
"r1 = 0.45 #m (radius 1) \n",
"r2 = 0.43 #m (radius 2) \n",
"Na = 400.0 #rpm (Speed of first motor)\n",
"\n",
"Eb_a = V-0.1*V #V (Back emf)\n",
"\n",
"#Let the supply voltage across motor A & B be Va and Vb respectively\n",
"#Therefore, Va + Vb = 650.0 V\n",
"Va = Symbol('Va') #V\n",
"Vb = V - Va #V\n",
"\n",
"#Now , N is directly propotional to (V - IR).Therefore, ratio (N1/N2) is\n",
"\n",
"Nratio1 = (Va - 0.1*V)/(Vb - 0.1*V)\n",
"\n",
"#Also, we know that N1*r1 = N2*r2 . Therefore (N1/N2) is\n",
"Nratio2 = r2/r1\n",
"\n",
"#As, N1/N2 = Nratio1 = Nratio2\n",
"eq = Eq(Nratio1,Nratio2)\n",
"Va = solve(eq)\n",
"Vea = Va[0] #V\n",
"Veb = V - Vea #V\n",
"\n",
"#Now , for the speed of motor A, we know that for Na ,back emf is Eb_a and for Nea ,back emf is (Vea-0.1*Vea)\n",
"Nea = Na*(Vea - 0.1*V)/Eb_a #rpm (using N1/N2 = E1/E2)\n",
"#Nratio is Ne1/Ne2 .Therefore,\n",
"Neb = Nea/Nratio2 #rpm\n",
"\n",
"print \"Speed of first motor is =\",round(Nea,2),\"rpm.\"\n",
"print \"Speed of second motor is =\",round(Neb,2),\"rpm.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 43.28 , PAGE NO :- 1744"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Time for which motors are operated in series = 8.44 s.\n",
"Energy loss during starting period = 480.01 W-h.\n"
]
}
],
"source": [
"#Two motors of a motor coach are started on series-parallel system,the current per motor being 350A during the starting period\n",
"#which is 18 sec.If the acceleration during starting period is uniform,the line voltage is 600V and resistance of each motor is\n",
"#0.1W.Find (a)the time during which the motors are operated in series. (b)the energy loss in the rheostat during starting period.\n",
"#####################################################################################################################################\n",
"\n",
"#Given\n",
"V = 600.0 #V (line voltage)\n",
"I = 350.0 #A (current)\n",
"R = 0.1 #ohm (resistance)\n",
"T = 18.0 #s (starting period)\n",
"\n",
"#time for which motors are in series\n",
"ts = 0.5*(V-2*I*R)/(V-I*R)*T #s\n",
"\n",
"#time for which motors are in parallel\n",
"tp = T - ts #s\n",
"\n",
"#back emf of 1 motor during series operation\n",
"Eb_s = V/2 - I*R #V\n",
"#total back emf\n",
"Eb_s = 2*Eb_s #V\n",
"#Emf during parallel operation\n",
"Eb_p = V - I*R #V\n",
"\n",
"#Energy lost during series connection\n",
"enrgy_s = 0.5*(Eb_s)*I*ts/3600 #W-h\n",
"#Energy lost during parallel connection\n",
"enrgy_p = 0.5*(Eb_p/2)*(2*I)*tp/3600 #W-h\n",
"\n",
"#Total energy lost\n",
"enrgy_t = enrgy_s + enrgy_p #W-h\n",
"\n",
"print \"Time for which motors are operated in series = \",round(ts,2),\"s.\"\n",
"print \"Energy loss during starting period = \",round(enrgy_t,2),\"W-h.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.29 , PAGE NO :- 1749"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Energy lost in rheostat = 441.02 W-h.\n",
"Energy lost in motor = 75.0 W-h.\n",
"Motor output = 909.77 W-h.\n",
"Efficiency at starting = 69.07 %.\n",
"Train speed at which transition from series to parallel = 11.98 km/h.\n"
]
}
],
"source": [
"#Two 750V D.C. series motors each having a resistance of 0.1 ohm are started on series-parallel system.Mean current through-out \n",
"#the starting period is 300A.Starting period is 15 sec. and train speed at the end of this period is 25 km/hr.Calculate\n",
"#(i)Rheostatic losses during series and parallel combination of motors\n",
"#(ii)Energy lost in motor\n",
"#(iii)Motor output\n",
"#(iv)Starting eff\n",
"#(v)Train speed at which transition from series to parallel must be made.\n",
"##############################################################################################################################\n",
"\n",
"V = 750.0 #V (applied voltage)\n",
"R = 0.1 #ohm (resistance)\n",
"I = 300.0 #A (current) \n",
"T = 15.0 #s (starting time)\n",
"S = 25.0 #km/h(speed after staring period)\n",
"\n",
"#Time when motors are in series\n",
"ts = 0.5*(V-2*I*R)/(V-I*R)*T #s\n",
"#Time when motors are in parallel\n",
"tp = T - ts #s\n",
"\n",
"#(i)Energy lost in rheostat\n",
"enrgy_lostr = 0.5*(2*(V/2-I*R))*I*ts + 0.5*(V-I*R)/2*2*I*tp #W-s\n",
"enrgy_lostr = enrgy_lostr/3600 #W-h\n",
"\n",
"#(ii)Total energy supplied\n",
"enrgy_supp = V*I*ts + 2*V*I*tp #W-s\n",
"enrgy_supp = enrgy_supp/3600 #W-h\n",
"\n",
"#Energy lost in armature of 2 motors\n",
"enrgy_losta = 2*(I*I*R)*T\n",
"enrgy_losta = enrgy_losta/3600 #W-h\n",
"\n",
"#Motor output is\n",
"outp = enrgy_supp - enrgy_lostr - enrgy_losta #W-h\n",
"\n",
"#Efficiency at starting\n",
"eff_s = (enrgy_supp-enrgy_lostr)/(enrgy_supp)*100\n",
"\n",
"#Acceleraton is uniform during starting period .Therefore,\n",
"speed = S/T*ts #km/h\n",
"print \"Energy lost in rheostat = \",round(enrgy_lostr,2),\"W-h.\"\n",
"print \"Energy lost in motor = \",round(enrgy_losta,2),\"W-h.\"\n",
"print \"Motor output =\",round(outp,2),\"W-h.\"\n",
"print \"Efficiency at starting =\",round(eff_s,2),\"%.\"\n",
"print \"Train speed at which transition from series to parallel = \",round(speed,2),\"km/h.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.30 , PAGE NO :- 1750"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Energy loss in series = 213.16 W-h.\n",
"Energy loss in parallel = 263.16 W-h.\n",
"The transition speed = 14.21 km/h.\n"
]
}
],
"source": [
"#Two 600V motors each having a resistance of 0.1 ohm are started on the series-parallel system,the mean current per motor throughout\n",
"#the starting period being 300A.The starting period is 20 seconds and the train speed at the end of this period is 30km per hour.\n",
"Calculate\n",
"#(i)the rheostatic losses during (a)the series and (b) the parallel combination of motors\n",
"#(ii)the train speed at which transition from series to parallel must be made.\n",
"###############################################################################################################################\n",
"\n",
"num_m = 2 # (number of motors)\n",
"V = 600.0 #V (line voltage)\n",
"I = 300.0 #A (current per motor)\n",
"Ts = 20.0 #s (starting period)\n",
"R = 0.1 #ohm (motor resistance)\n",
"Vm = 30.0 #km/h (maximum speed)\n",
"\n",
"#Back emf of each motor in series position,\n",
"Eb_s = V/2 - I*R #V\n",
"#Back emf of each motor in parallel position,\n",
"Eb_p = V - I*R #V\n",
"#time for which motors were in series\n",
"ts = Ts*Eb_s/Eb_p #s\n",
"#time for which motors are in parallel\n",
"tp = Ts - ts #s\n",
"\n",
"#(a) Voltage drop in the starting rheostat in series combination\n",
"V_s = V - 2*I*R #V\n",
"#Energy loss in series is\n",
"enrgy_s = (V_s/2)*I*(ts/3600) #W-h\n",
"#(b) Voltage drop in the starting rheostat in parallel combination\n",
"V_p = V/2 #V\n",
"#Energy loss in parallel is\n",
"enrgy_p = (V_p/2)*(2*I)*(tp/3600)\n",
"\n",
"#Acceleration is\n",
"acc = Vm/Ts #km/h/s\n",
"#Speed at the end of series period\n",
"speed = acc*ts #km/h\n",
"\n",
"print \"Energy loss in series =\",round(enrgy_s,2),\"W-h.\"\n",
"print \"Energy loss in parallel =\",round(enrgy_p,2),\"W-h.\"\n",
"print \"The transition speed = \",round(speed,2),\"km/h.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.31 , PAGE NO :- 1751"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Period of series operation = 11.36 s.\n",
"Period of parallel operation = 13.64 s.\n",
"Speed at which series connection are to be changed = 36.36 km/h.\n"
]
}
],
"source": [
"#Two d.c. series motors of a motor coach have resistance of 0.1ohm each.These motors draw a current of 500A from 600V mains during\n",
"#series-parallel starting period of 25 seconds.If the acceleration during starting period remains uniform,determine:\n",
"#(i)time during which the motors operate in (a)series (b)parallel.\n",
"#(ii)the speed at which the series connections are to be changed if the speed just after starting period is 80kmph.\n",
"####################################################################################################################################\n",
"\n",
"\n",
"num_m = 2 # (number of motors)\n",
"V = 600.0 #V (line voltage)\n",
"I = 500.0 #A (current per motor)\n",
"Ts = 25.0 #s (starting period)\n",
"R = 0.1 #ohm (motor resistance)\n",
"Vm = 80.0 #km/h (maximum speed)\n",
"\n",
"#Back emf of each motor in series position,\n",
"Eb_s = V/2 - I*R #V\n",
"#Back emf of each motor in parallel position,\n",
"Eb_p = V - I*R #V\n",
"#time for which motors were in series\n",
"ts = Ts*Eb_s/Eb_p #s\n",
"#time for which motors are in parallel\n",
"tp = Ts - ts #s\n",
"\n",
"#(ii) Speed at which series connection are to be changed = acc*ts and acc = Vm/Ts.Therefore,\n",
"speed = Vm/Ts*ts #km/h\n",
"\n",
"print \"Period of series operation = \",round(ts,2),\"s.\"\n",
"print \"Period of parallel operation = \",round(tp,2),\"s.\"\n",
"print \"Speed at which series connection are to be changed = \",round(speed,2),\"km/h.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 43.32 , PAGE NO :- 1751"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Braking torque = 930.0 N-m.\n"
]
}
],
"source": [
"#The following figures refer to the speed-current and torque-current characteristics of a 600V d.c. series traction motor.\n",
"#Current(A) : 50 100 150 200 250\n",
"#Speed(Kmph): 73.6 48 41.1 37.3 35.2\n",
"#Torque(N-m): 150 525 930 1335 1750\n",
"#Determine the braking torque at a speed of 48kmph when operating as self excited d.c. generator.Assume resistance of motor and\n",
"#braking rheostat to be 0.6 ohm and 3.0 ohm respectively.\n",
"#############################################################################################################################\n",
"\n",
"Rm = 0.6 #ohm(resistance of motor)\n",
"Rrh = 3.0 #ohm(braking rheostat)\n",
"\n",
"#As a motor\n",
"V = 600.0 #V (terminal voltage)\n",
"#From speed-current characteristics . For a speed of 48kmph,\n",
"I = 100.0 #A (current)\n",
"Eb = V - I*Rm #V (back emf)\n",
"\n",
"#As a generator\n",
"\n",
"#At instant of applying rheostatic braking ,the terminal voltage will be equal to emf developed by machine i.e\n",
"V2 = Eb #V\n",
"Rt = Rm + Rrh #ohm (total resistance)\n",
"\n",
"#Using V=IR\n",
"I2 = V2/Rt #A (current)\n",
"\n",
"if (I2==50):\n",
" tau = 150.0 #N-m\n",
"elif (I2==100):\n",
" tau = 525.0 #N-m\n",
"elif (I2==150):\n",
" tau = 930.0 #N-m\n",
"elif (I2==200):\n",
" tau = 1335.0 #N-m\n",
"elif (I2==250):\n",
" tau = 1750.0 #N-m\n",
"\n",
"print \"Braking torque = \",round(tau,2),\"N-m.\" "
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 2",
"language": "python",
"name": "python2"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 2
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython2",
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},
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}
PKj„ìJÀ9 úÏ
Ï
;A TEXTBOOK OF ELECTRICAL TECHNOLOGY VOL III/chapter46.ipynb{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# CHAPTER 46 : ELECTRONIC CONTROL OF A.C MOTORS"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 46.1 , PAGE NO :- 1827"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Ton = 1.89 ms.\n",
"magnitude of the current pulse I0 = 102.31 A.\n"
]
}
],
"source": [
"'''The wound-rotor induction motor of Fig.43.5 is rated at 30-kW,975 rpm,440-V,50 Hz.The open-circuit line voltage is 400V and the load\n",
"resistance is 0.5 ohm.If chopper frequency is 200 Hz,calculate Ton so that the motor develops a gross torque of 200 N-m at 750 rpm.Also,\n",
"calculate the magnitude of the current pulses drawn from the capacitor.'''\n",
"\n",
"import math as m\n",
"Ns = 1000.0 #rpm (Synchronus speed)\n",
"N1 = 750.0 #rpm (rotating speed)\n",
"E2 = 400.0 #V (OC line voltage)\n",
"Tg = 200.0 #N-m (gross torque)\n",
"R0 = 0.5 #ohm (load resistance)\n",
"f = 200.0 #Hz (chopper frequency)\n",
"s = (Ns-N1)/Ns # slip\n",
"Vrl = s*E2 #V (rotor line voltage)\n",
"Vdc = 135.0 #V (DC voltage of 3-phase bridge rectifier)\n",
"#Now, Tg = P^2/(2*3.14*Ns).Therefore,\n",
"P2 = Tg*2*3.14*Ns/60 #W\n",
"#Power dissipated as heat\n",
"sP2 = s*P2 #W\n",
"#Power is actually dissipated in R and is equal to rectifier output Vdc*Idc.Therefore,\n",
"Idc = sP2/Vdc #A\n",
"#The apparent resistance at the input of chopper is\n",
"Ra = Vdc/Idc #ohm\n",
"#Now, Ra = Ro/(f*Ton)^2 .Therefore,\n",
"Ton = m.sqrt(R0/(f*f*Ra))*1000 #ms\n",
"#Current in R0 can be found from relation,I0^2*R0 = sP2\n",
"I0 = m.sqrt(sP2/R0) #A\n",
"print \"Ton = \",round(Ton,2),\"ms.\"\n",
"print \"magnitude of the current pulse I0 =\",round(I0,2),\"A.\""
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": []
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 2",
"language": "python",
"name": "python2"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 2
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython2",
"version": "2.7.11"
}
},
"nbformat": 4,
"nbformat_minor": 0
}
PKj„ìJe"Z8;A TEXTBOOK OF ELECTRICAL TECHNOLOGY VOL III/chapter45.ipynb{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# CHAPTER 45 : RATING AND SERVICE CAPACITY"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.1 , PAGE NO :- 1796"
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Size of motor = 66.0 KW.\n"
]
}
],
"source": [
"'''An electric motor operates at full-load of 100KW for 10 minutes,at 3/4 full load for the next 10 minutes and at 1/2 load for\n",
"next 20 minutes,no-load for the next 20 minutes and this cycle repeats continously.Find the continous rating of the suitable \n",
"motor.'''\n",
"\n",
"import math as m\n",
"#Loads\n",
"l1 = 100.0 #kW (load 1)\n",
"l2 = 0.75*l1 #kW (load 2)\n",
"l3 = 0.5*l1 #kW (load 3)\n",
"l4 = 0.0 #kW (no-load)\n",
"\n",
"#coresponding time\n",
"t1 = 10.0 #minutes\n",
"t2 = 10.0 #minutes\n",
"t3 = 20.0 #minutes\n",
"t4 = 20.0 #minutes\n",
"\n",
"#size of motor required\n",
"size = m.sqrt((l1*l1*t1 + l2*l2*t2 + l3*l3*t3 + l4*l4*t4)/(t1+t2+t3+t4/3)) #kW\n",
"print \"Size of motor =\",round(size),\"KW.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.2 , PAGE NO :- 1797"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Size of motor = 141.0 KW.\n"
]
}
],
"source": [
"'''An electric motor has to be selected for a load which rises uniformly from zero to 200KW in 10 minutes after which it remains \n",
"constant at 200KW for the next 10 minutes,followed by a no-load period of 15 minutes before the cycle repeats itself.Estimate a \n",
"suitable size of continuosly rated motor.'''\n",
"\n",
"import math as m\n",
"#Loads\n",
"l1 = 200.0/2 #kW (load 1)\n",
"l2 = 200.0 #kW (load 2)\n",
"l3 = 0.0 #kW (no-load)\n",
"\n",
"#coresponding time\n",
"t1 = 10.0 #minutes\n",
"t2 = 10.0 #minutes\n",
"t3 = 15.0 #minutes\n",
"\n",
"\n",
"#size of motor required\n",
"size = m.sqrt((l1*l1*t1 + l2*l2*t2 + l3*l3*t3)/(t1+t2+t3/3)) #kW\n",
"print \"Size of motor =\",round(size),\"KW.\"\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.3 , PAGE NO :- 1797 "
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Size of motor = 70.0 KW.\n"
]
}
],
"source": [
"'''A certain motor has to perform the following duty cycle:\n",
"(a) 100KW for 10 minutes (No-load for 5 minutes)\n",
"(b) 50KW for 8 minutes (No-load for 4 minutes)\n",
"The duty cycle is repeated indefinitely.Draw the curve for the load cycle.Assuming that the heating is propotional to the square of\n",
"the load,determine suitable size of a continuosly-rated motor.'''\n",
"\n",
"import math as m\n",
"\n",
"#Loads\n",
"l1 = 100.0 #KW (load 1)\n",
"l3 = 50.0 #KW (load 2)\n",
"\n",
"#Time\n",
"t1 = 10.0 #minutes\n",
"t2 = 5.0 #minutes\n",
"t3 = 8.0 #minutes\n",
"t4 = 4.0 #minutes\n",
"\n",
"#Size of the motor is\n",
"size = m.sqrt((l1*l1*t1 + l3*l3*t3)/(t1+t2+t3+t4)) #KW\n",
"print \"Size of motor = \",round(size,-1),\"KW.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.4 , PAGE NO :- 1799"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Size of motor = 69.07 H.P.\n"
]
}
],
"source": [
"'''A motor has to perform the following duty cycle :-\n",
"(i) 100 H.P (10 mins) (ii) No Load (5 mins)\n",
"(iii)60 H.P (8 mins) (iv) No Load (4 mins)\n",
"which is repeated infinitely.Determine the suitable size of continuosly rated motor.'''\n",
"\n",
"import math as m\n",
"\n",
"#Loads\n",
"l1 = 100.0 #H.P (load 1)\n",
"l3 = 60.0 #H.P (load 2)\n",
"\n",
"#Time\n",
"t1 = 10.0 #minutes\n",
"t2 = 5.0 #minutes\n",
"t3 = 8.0 #minutes\n",
"t4 = 4.0 #minutes\n",
"\n",
"#Size of the motor is\n",
"size = m.sqrt((l1*l1*t1 + l3*l3*t3)/(t1+t2+t3+t4)) #H.P\n",
"print \"Size of motor = \",round(size,2),\"H.P.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.5 , PAGE nO :- 1800"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Size of motor = 48.11 H.P .\n"
]
}
],
"source": [
"'''A motor working in a coal mine has to exert power starting from zero and rising uniformly to 100 H.P in 5 min\n",
"after which it works at a constant rate of 50 H.P for 10 min.Then, a no load period of 3 min.The cycle is repeated\n",
"indefinitely,estimate suitable size of motor.'''\n",
"\n",
"import math as m\n",
"\n",
"#Load\n",
"l1 = 100.0 #H.P (load 1)\n",
"l2 = 50.0 #H.P (load 2)\n",
"l3 = 0.0 #H.P (no-load)\n",
"\n",
"#Time\n",
"t1 = 5.0 #min (time 1) \n",
"t2 = 10.0 #min (time 2)\n",
"t3 = 3.0 #min (time 3)\n",
"\n",
"#Using Simpson's one-third rule of Integration\n",
"rating = m.sqrt((1.0/3*l1*l1*t1 + l2*l2*t2)/(t1 + t2 + t3) ) #H.P\n",
"\n",
"print \"Size of motor =\",round(rating,2),\"H.P .\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.6 , PAGE NO :- 1800"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Size of motor = 263.19 H.P .\n",
"Therefore, suitable size of motor is 300.0 H.P\n"
]
}
],
"source": [
"'''A motor has following duty cycle\n",
"Load rising from 200 to 400 H.P - 4 min.\n",
"Uniform load 300 H.P - 2 min.\n",
"Regenerative braking (50 H.P to 0) - 1 min.\n",
"Idle - 1 min.\n",
"Estimate suitable H.P rating of the motor that can be used.'''\n",
"\n",
"import math as m\n",
"\n",
"#Loads\n",
"l1 = 200.0 #H.P (load 1)\n",
"l2 = 400.0 #H.P (load 2)\n",
"l3 = 300.0 #H.P (load 3)\n",
"l4 = 50.0 #H.P (load 4)\n",
"\n",
"#Time\n",
"t1 = 4.0 #min (time 1)\n",
"t2 = 2.0 #min (time 2)\n",
"t3 = 1.0 #min (time 3)\n",
"t4 = 1.0 #min (idle time)\n",
"\n",
"rating = m.sqrt((1.0/3*(l1*l1 + l1*l2 + l2*l2)*t1 + l3*l3*t2 + 1.0/3*l4*l4*t3)/(t1 + t2 + t3 + t4)) #H.P\n",
"\n",
"print \"Size of motor = \",round(rating,2),\"H.P .\"\n",
"print \"Therefore, suitable size of motor is\",round(rating,-2),\"H.P\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.7 , PAGE NO :- 1802"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Suitable motor size = 30.0 H.P .\n"
]
}
],
"source": [
"'''The load cycle of a motor for 15 min in driving some equipment is as follows :\n",
"0 - 5 min - 30 H.P\n",
"5 - 9 min - No load\n",
"9 - 12 min - 45 H.P\n",
"12 - 15 min - No load\n",
"The load cycle is repeated indefinitely.Suggest a suitable size of continuosly rated motor.'''\n",
"\n",
"import math as m\n",
"\n",
"#Loads\n",
"l1 = 30.0 #H.P\n",
"l3 = 45.0 #H.P\n",
"\n",
"#Time\n",
"t1 = 5.0 #min\n",
"t2 = 4.0 #min\n",
"t3 = 3.0 #min\n",
"t4 = 3.0 #min\n",
"\n",
"#Size of motor is\n",
"Size = m.sqrt((l1*l1*t1 + l3*l3*t3)/(t1+t2+t3+t4)) #H.P\n",
"print \"Suitable motor size =\",round(Size,-1),\"H.P .\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.8 , PAGE NO :- 1802"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Size of continously rated motor = 505.0 H.P .\n"
]
}
],
"source": [
"'''A motor driving a colliery winder has the following acceleration period :\n",
" load cycle 0-15 sec : Load rises uniformly from 0-1000 H.P .\n",
" Full speed period : 15-85 sec. Load constant at 600 H.P .\n",
" Decceleration period : 85-95 sec. Regenerative braking the H.P returned uniformly from 200 to 0 H.P.\n",
" 95 - 120 sec : Motor stationary.\n",
"Estimate the size of continuosly rated motor.'''\n",
"\n",
"import math as m\n",
"\n",
"#Loads\n",
"l1 = 1000.0 #H.P (load 1)\n",
"l2 = 600.0 #H.P (load 2)\n",
"l3 = 200.0 #H.P (load 3)\n",
"\n",
"#Time\n",
"t1 = 15.0 #s\n",
"t2 = 70.0 #s\n",
"t3 = 10.0 #s\n",
"t4 = 25.0 #s\n",
"\n",
"#Size of motor is\n",
"\n",
"size = m.sqrt((l1*l1*t1/3 + l2*l2*t2 + l3*l3*t3/3)/(t1+t2+t3+t4)) #H.P\n",
"\n",
"while(round(size)%5!=0):\n",
" size = size + 1\n",
" \n",
"print \"Size of continously rated motor = \",round(size),\"H.P .\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.9 , PAGE NO :- 1807"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" 1/2 hour rating = 75.13 KW.\n"
]
}
],
"source": [
"'''A 40KW motor when run continuosly on full load,attains a temperature of 35C , above the surrounding air.Its heating time \n",
"constant is 90 min.What would be the 1/2 hour rating of the motor for this temperature rise?Assume that the machine cools down \n",
"completely between each load period and that the losses are propotional to square of the load.'''\n",
"\n",
"\n",
"from sympy import Symbol,Eq,solve\n",
"import math as m\n",
"# Let 'P' KW be 1/2 hour rating of the motor\n",
"# theta1 - final temp rise at P KW\n",
"# theta2 - final temp rise at 40 KW\n",
"#Losses at P KW is directlt propotional to P^2\n",
"\n",
"theta2 = 35.0 # *C\n",
"tau = 1.5 #hr (time constant)\n",
"t = 0.5 #hr (motor running time)\n",
"\n",
"#Now, (theta1/theta2) = loss at P KW/loss at 40KW = (P/40)^2\n",
"P = Symbol('P')\n",
"theta1 = theta2*(P/40)*(P/40) #*C\n",
"\n",
"#Now, theta2 = theta1*(1 - exp(-t/tau))\n",
"\n",
"theta2a = theta1*(1-m.exp(-t/tau)) #*C\n",
"eq = Eq(theta2,theta2a)\n",
"P = solve(eq)\n",
"P1 = P[1] #KW\n",
"\n",
"print \"1/2 hour rating = \",round(P1,2),\"KW.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.10 , PAGE NO :- 1807"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"1 hour rating = 24.0 H.P.\n"
]
}
],
"source": [
"'''Determine the one-hour rating of a 15 H.P motor having heating time contant of 2 hours.The motor attains the temperature of\n",
"40*C on continuos run at full load.Assume that the losses are propotional to square of the load and the motor is allowed to cool\n",
"down to the ambient temperature before being loaded again.'''\n",
"\n",
"from sympy import Symbol,Eq,solve\n",
"import math as m\n",
"# Let 'P' H.P be 1 hour rating of the motor\n",
"# theta2 - final temp rise at P H.P\n",
"# theta1 - final temp rise at 15 H.P\n",
"#Losses at P H.P is directlt propotional to P^2\n",
"\n",
"theta1 = 40.0 # *C\n",
"tau = 2.0 #hr (time constant)\n",
"t = 1.0 #hr (motor running time)\n",
"\n",
"#Now, (theta2/theta1) = loss at P H.P/loss at 15 H.P = (P/15)^2\n",
"P = Symbol('P')\n",
"theta2 = theta1*(P/15)*(P/15) #*C\n",
"\n",
"#Now, theta1 = theta2*(1 - exp(-t/tau))\n",
"\n",
"theta1a = theta2*(1-m.exp(-t/tau)) #*C\n",
"eq = Eq(theta1,theta1a)\n",
"P = solve(eq)\n",
"P1 = P[1] #H.P\n",
"\n",
"print \"1 hour rating = \",round(P1),\"H.P.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.11 , PAGE NO :- 1808"
]
},
{
"cell_type": "code",
"execution_count": 15,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Temperature rise of motor = 35.6 *C .\n"
]
}
],
"source": [
"'''The heating and cooling time constants of a motor are 1 hour and 2 hours respectively.Final temperature rise attained is \n",
"100*C.This motor runs at full load for 30 minutes and then kept idle for 12 min and the cycle is repeated indefinitely.Determine \n",
"the temperature rise of motor after one cycle.'''\n",
"\n",
"import math as m\n",
"\n",
"theta2 = 100.0 #*C (Final temperature rise)\n",
"tau_h = 1.0 #hr (heating time constant)\n",
"tau_c = 2.0 #hr (cooling time constant)\n",
"t1 = 30.0/60 #hr (motor running time)\n",
"t2 = 12.0/60 #hr (motor idle time)\n",
"\n",
"#Heating cycle\n",
"theta1 = theta2*(1 - m.exp(-t1/tau_h))\n",
"\n",
"#Cooling cycle\n",
"thetac = theta1*m.exp(-t2/tau_c)\n",
"\n",
"print \"Temperature rise of motor = \",round(thetac,2),\"*C .\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.12 , PAGE NO :- 1808"
]
},
{
"cell_type": "code",
"execution_count": 68,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Maximum overload that can be carried by motor = 25.82 KW.\n"
]
}
],
"source": [
"'''Calculate the maximum overload that can be carried by a 20KW output motor,if the temperature rise is not to exceed 50*C after\n",
"one hour on overload .The temperature rise on full load,after 1 hour is 30*C and after 2 hour is 40*C . Assume losses propotional\n",
"to square of load.'''\n",
"\n",
"from sympy import solve,Symbol,Eq\n",
"import math as m\n",
"\n",
"# As theta = thetaf*(1 - exp(-t/T))\n",
"\n",
"theta1 = 30.0 #*C (temperature rise in time1)\n",
"t1 = 1 #hr (time 1)\n",
"theta2 = 40.0 #*C (temperature rise in time2)\n",
"t2 = 2 #hr (time 2)\n",
"#Now theta1/theta2 = (1-exp(-t1/T))/(1-exp(-t2/T))\n",
"#Let us assume that x = exp(-1/T).Therefore\n",
"x = Symbol('x')\n",
"ratio1 = (1 - x**t1)/(1-x**t2) #(theta1/theta2)\n",
"ratio2 = theta1/theta2\n",
"\n",
"#As ratio1 = ratio2\n",
"eq = Eq(ratio1,ratio2)\n",
"x1 = solve(eq)\n",
"x = x1[0] #variable \n",
"\n",
"#x = exp(-1/T) . Therefore,\n",
"T = -1/m.log(x)\n",
"\n",
"#Now, theta1 = thetaf1*(1 - exp(-t1/T))\n",
"\n",
"thetaf1 = theta1/(1-x**t1) #*C\n",
"\n",
"#Also theta3 = thetaf3*(1 - exp(-t3/T))\n",
"theta3 = 50.0 #*C (max temp)\n",
"t3 = 1 #hr (time 3) \n",
"thetaf3 = theta3/(1-x**t3) #*C\n",
"\n",
"#Given that temp is directly propotional to square of power output i.e thetaf1/thetaf3 = (Power1/Power3)^2\n",
"P1 = 20.0 #KW\n",
"P3 = m.sqrt(thetaf3/thetaf1)*P1 #KW\n",
"\n",
"print \"Maximum overload that can be carried by motor = \",round(P3,2),\"KW.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.13 , PAGE NO :- 1809"
]
},
{
"cell_type": "code",
"execution_count": 67,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Final steady temperature rise = 50.0 *C.\n",
"Cooling time constant = 0.93 hr\n"
]
}
],
"source": [
"'''In a transformer the temperature rise is 25*C after 1 hour and 37.5*C after 2 hours,starting from cold conditions.Calculate \n",
"its final steady temperature rise and the heating time constant.If the transformer temerature falls from the final steady state\n",
"value to 40*C in 1.5 hours when disconnected,calculate its cooling time constant.Ambient temperature is 30*C.'''\n",
"\n",
"from sympy import solve,Symbol,Eq\n",
"import math as m\n",
"\n",
"# As theta = thetaf*(1 - exp(-t/T))\n",
"\n",
"theta1 = 25.0 #*C (temperature rise in time1)\n",
"t1 = 1 #hr (time 1)\n",
"theta2 = 37.5 #*C (temperature rise in time2)\n",
"t2 = 2 #hr (time 2)\n",
"#Now theta1/theta2 = (1-exp(-t1/T))/(1-exp(-t2/T))\n",
"#Let us assume that x = exp(-1/T).Therefore\n",
"x = Symbol('x')\n",
"ratio1 = (1 - x**t1)/(1-x**t2) #(theta1/theta2)\n",
"ratio2 = theta1/theta2\n",
"\n",
"#As ratio1 = ratio2\n",
"eq = Eq(ratio1,ratio2)\n",
"x1 = solve(eq)\n",
"x = x1[0] #variable \n",
"\n",
"#x = exp(-1/T) . Therefore,\n",
"T = -1/m.log(x)\n",
"\n",
"#As theta1 = thetaf1*(1 - exp(-t1/T))\n",
"thetaf1 = theta1/(1-x**t1) #*C\n",
"print \"Final steady temperature rise =\",round(thetaf1,2),\"*C.\"\n",
"\n",
"#Cooling conditions\n",
"theta_rise = 40.0 - 30.0 #*C (temp rise above ambient conditions)\n",
"t3 = 1.5 #hr (time taken)\n",
"\n",
"#Now, theta_rise = thetaf1*exp(-t3/T)\n",
"T = -t3/m.log(theta_rise/thetaf1) #hr\n",
"print \"Cooling time constant =\",round(T,2),\"hr\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.14 , PAGE NO :- 1809"
]
},
{
"cell_type": "code",
"execution_count": 66,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Temperature of machine = 70.58 *C.\n"
]
}
],
"source": [
"'''The initial temperature of machine is 20*C.Calculate the temperature of machine after 1.2 hours,if its final steady \n",
"temperature rise is 85*C and the heating time constant is 2.4 hours.Ambient temperature is 25*C.''' \n",
"\n",
"import math as m\n",
"thetaf = 85.0 #*C (final temp. rise)\n",
"theta1 = 20.0 #*C (initial temp)\n",
"t1 = 1.2 #hr (time taken)\n",
"T = 2.4 #hr (heat time constant)\n",
"#Now, Temperature rise above coling medium is\n",
"theta = thetaf - (thetaf - theta1)*m.exp(-t1/T) #*C\n",
"\n",
"#Therefore, temp. of machine after t1 time is\n",
"temp = theta + 25.0\n",
"\n",
"print \"Temperature of machine =\",round(temp,2),\"*C.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.15 , PAGE NO :- 1809"
]
},
{
"cell_type": "code",
"execution_count": 69,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Final steady temperature rise = 45.0 *C .\n",
"Time constant = 2.47 hr.\n",
"The steady temperature rise = 30.0 *C .\n"
]
}
],
"source": [
"'''The following rises were observed in a teperature rise test on a D.C machine at full loads. :-\n",
"After 1 hour - 15*C\n",
"After 2 hours - 25*C\n",
"Find out (i) Final steady temperature rise and time constant.\n",
" (ii)The steady temperature rise after 1 hour at 50% overload,from cold.\n",
"Assume that the final temperature rise on 50% overload is 90*C.'''\n",
"\n",
"\n",
"from sympy import solve,Symbol,Eq\n",
"import math as m\n",
"\n",
"# As theta = thetaf*(1 - exp(-t/T))\n",
"\n",
"theta1 = 15.0 #*C (temperature rise in time1)\n",
"t1 = 1 #hr (time 1)\n",
"theta2 = 25.0 #*C (temperature rise in time2)\n",
"t2 = 2 #hr (time 2)\n",
"#Now theta1/theta2 = (1-exp(-t1/T))/(1-exp(-t2/T))\n",
"#Let us assume that x = exp(-1/T).Therefore\n",
"x = Symbol('x')\n",
"ratio1 = (1 - x**t1)/(1-x**t2) #(theta1/theta2)\n",
"ratio2 = theta1/theta2\n",
"\n",
"#As ratio1 = ratio2\n",
"eq = Eq(ratio1,ratio2)\n",
"x1 = solve(eq)\n",
"x = x1[0] #variable \n",
"\n",
"#x = exp(-1/T) . Therefore,\n",
"T = -1/m.log(x) #hr (time constant)\n",
"\n",
"#As theta1 = thetaf1*(1 - exp(-t/T))\n",
"thetaf1 = theta1/(1-x**t1) #*C (Final steady temp. rise)\n",
"print \"Final steady temperature rise =\",round(thetaf1,2),\"*C .\"\n",
"print \"Time constant =\",round(T,2),\"hr.\"\n",
"\n",
"#(ii) Now at 50% overload .Final temp rise is\n",
"thetaf3 = 90.0 #*C\n",
"t3 = 1 #hr (time taken)\n",
"#As , theta = thetaf*(1 - exp(-t/T))\n",
"theta3 = thetaf3*(1 - m.exp(-t3/T)) #*C\n",
"\n",
"print \"The steady temperature rise =\",round(theta3,2),\"*C .\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.16 , PAGE NO :- 1813"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Moment of Inertia = 2.947543e+06 kg-m^2\n"
]
}
],
"source": [
"'''The following data refers to a 500 H.P rolling mill,induction motor equipped with a flywheel.\n",
" No load speed -> 40 rpm\n",
" Slip at full load(torque) -> 12%\n",
" Load torque during actual rolling -> 41500 kg-m\n",
" Duration of each rolling period -> 10 sec.\n",
"Determine inertia of flywheel required in the above case to limit motor torque to twice its full load value.Neglect no-load \n",
"losses and assume that the rolling mill torque falls to zero between each rolling period.Assume motor slip propotional to\n",
"full load torque.'''\n",
"\n",
"\n",
"import math as m\n",
"N = 40.0 #rpm (No load speed)\n",
"P = 500.0*(735.5) #W (Power)\n",
"w = 2*(3.14)*N/60 #rad/sec (angular speed)\n",
"T0 = 0 #kg-m (initial torque)\n",
"Tl = 41500.0 #kg-m (Torque load) \n",
"t = 10.0 #sec (time taken)\n",
"s = 0.12 # (slip)\n",
"g = 9.81 #m/s^2 \n",
"Tfull = P/(w*(1-s)) #N-m (full load torque)\n",
"Tfull = Tfull/g #kg-m\n",
"Tm = 2*Tfull #kg-m (Max torque)\n",
"S = 2*3.14*(0.12*40)/60\n",
"#Now, S = K*Tfl\n",
"K = S/Tfull #constant\n",
"#Also, Tm = Tl - (Tl-T0)*exp(-tg/IK) .Therefore I is\n",
"I =(-t*g)/(K*m.log((Tl-Tm)/(Tl-T0))) #kg-m^2\n",
"\n",
"print \"Moment of Inertia = %e kg-m^2\" %round(I,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.17 , PAGE NO :- 1814 "
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Tm = 47.84 kg-m.\n",
"Actual Speed = 942.59 rpm.\n"
]
}
],
"source": [
"'''A 6 pole,50 Hz Induction Motor has a flywheel of 1200 kg-m^2 as moment of inertia.Load torque is 100 kg-m for 10 sec.No load\n",
"period is long enough for the flywheel,to regain its full speed.Motor has a slip of 6% at a torque of 50 kg-m.Calculate\n",
"(i)Maximum torque exerted by motor\n",
"(ii)Speed at the end of deacceleration period.'''\n",
"\n",
"import math as m\n",
"\n",
"Tl = 100.0 #kg-m (load torque)\n",
"t = 10.0 #s (time taken)\n",
"g = 9.81 #m/s^2 (gravitational acceleration)\n",
"I = 1200.0 #kg-m^2 (moment of inertia)\n",
"p = 6 # (poles)\n",
"f = 50.0 #Hz (frequency)\n",
"s = 0.06 # (slip)\n",
"Tfull = 50.0 #kg-m (full load torque)\n",
"Ns = 120*f/p\n",
"Nr = (1-s)*Ns\n",
"\n",
"#Now, S = K*T\n",
"S = 2*3.14*(Ns - Nr)/60 #rad/sec\n",
"K = S/Tfull #constant\n",
"\n",
"#As Tm = Tl*(1-exp(-t*g/I*K))\n",
"Tm = Tl*(1 - m.exp(-t*g/(I*K))) #kg-m\n",
"print \"Tm = \",round(Tm,2),\"kg-m.\"\n",
"\n",
"#(ii)Slip speed\n",
"S = K*Tm #rad/sec (slip speed)\n",
"S = S*(60/(2*3.14)) #rpm\n",
"N = Ns - S #rpm (actual speed)\n",
"print \"Actual Speed =\",round(N,2),\"rpm.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.18 , PAGE NO :- 1815"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Speed 1 = 691.75 rpm.\n",
"Speed 2 = 731.62 rpm.\n",
"Speed 1 = 683.89 rpm.\n",
"Speed 2 = 731.0 rpm.\n",
"Speed 1 = 683.62 rpm.\n",
"Speed 2 = 730.98 rpm.\n"
]
}
],
"source": [
"'''An Induction Motor equipped with a flywheel is driving a rolling mill which requires a load torque of 1900 N-m for 10 sec\n",
"followed by 250 N-m for 30 sec.This cycle being repeated indefinitely.The synchronus speed of motor is 750 rpm and it has slip of\n",
"10% when delivering 1400 N-m torque.The total Moment of Inertia of the flywheel and other rotating parts is 2100 kg-m^2.Draw the \n",
"curves showing the torque exerted by the motor and the speed for five complete cycles,assuming the initial torque is zero.'''\n",
"\n",
"import math as m\n",
"\n",
"Tl1 = 1900.0 #N-m (load torque 1)\n",
"t1 = 10.0 #s (time 1)\n",
"Tl2 = 280.0 #N-m (load torque 2)\n",
"t2 = 30.0 #s (time 2)\n",
"s = 0.1 # (slip)\n",
"Ns = 750.0 #rpm (synchronus speed)\n",
"I = 2100.0 #kg-m^2 (moment of inertia)\n",
"Tm = 1400.0 #N-m\n",
"S = Ns*s #rpm (slip speed)\n",
"S = S*(2*3.14/60) #rad/sec \n",
"\n",
"K = S/Tm #constant\n",
"T0 = 0 #N-m\n",
"#(i) During First Cycle\n",
"Tm = Tl1 -(Tl1-T0)*m.exp(-t1/(I*K)) #N-m\n",
"s2 = K*Tm*(60/(2*3.14)) #rpm\n",
"Speed1 = Ns - s2 #rpm\n",
"print \"Speed 1 = \",round(Speed1,2),\"rpm.\"\n",
"\n",
"# Now, Tm = T0 + (Tm' - T0)*exp(-t/(I*K))\n",
"Tmb = Tm #N-m\n",
"T0 = Tl2 #N-m (No Load Torque)\n",
"\n",
"Tm = T0 + (Tmb - T0)*m.exp(-t2/(I*K))\n",
"S2 = K*Tm*(60/(2*3.14)) #rpm\n",
"Speed2 = Ns - S2 #rpm\n",
"print \"Speed 2 = \",round(Speed2,2),\"rpm.\"\n",
"#################################################################\n",
"\n",
"#(ii) During Second cycle\n",
"T0 = Tm\n",
"Tm2 = Tl1 -(Tl1-T0)*m.exp(-t1/(I*K)) #N-m\n",
"s2 = K*Tm2*(60/(2*3.14)) #rpm\n",
"Speed1 = Ns - s2 #rpm\n",
"print \"Speed 1 = \",round(Speed1,2),\"rpm.\"\n",
"\n",
"# Now, Tm = T0 + (Tm' - T0)*exp(-t/(I*K))\n",
"Tm2b = Tm2 #N-m\n",
"T0 = Tl2 #N-m (No Load Torque)\n",
"\n",
"Tm = T0 + (Tm2b - T0)*m.exp(-t2/(I*K))\n",
"S2 = K*Tm*(60/(2*3.14)) #rpm\n",
"Speed2 = Ns - S2 #rpm\n",
"print \"Speed 2 = \",round(Speed2,2),\"rpm.\"\n",
"###################################################################\n",
"\n",
"#(iii) During Third cycle\n",
"T0 = Tm\n",
"Tm3 = Tl1 -(Tl1-T0)*m.exp(-t1/(I*K)) #N-m\n",
"s2 = K*Tm3*(60/(2*3.14)) #rpm\n",
"Speed1 = Ns - s2 #rpm\n",
"print \"Speed 1 = \",round(Speed1,2),\"rpm.\"\n",
"\n",
"# Now, Tm = T0 + (Tm' - T0)*exp(-t/(I*K))\n",
"Tm3b = Tm3 #N-m\n",
"T0 = Tl2 #N-m (No Load Torque)\n",
"\n",
"Tm = T0 + (Tm3b - T0)*m.exp(-t2/(I*K))\n",
"S2 = K*Tm*(60/(2*3.14)) #rpm\n",
"Speed2 = Ns - S2 #rpm\n",
"print \"Speed 2 = \",round(Speed2,2),\"rpm.\"\n",
"####################################################################"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 45.19 , PAGE NO :- 1817"
]
},
{
"cell_type": "code",
"execution_count": 33,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Moment of inertia = 2860.94 kg-m^2.\n"
]
}
],
"source": [
"'''A motor fitted with a flywheel supplies a load torque of 150 kg-m for 15 sec.During the no-load period,the flywheel regains \n",
"its original speed.The motor torque is required to be limited to 85 kg-m.Determine moment of inertia of flywheel.\n",
"The no-load speed of motor is 500 rpm and it has slip of 10% on full load.'''\n",
"\n",
"from sympy import Symbol,solve,Eq,exp\n",
"import math as m\n",
"\n",
"Tm = 85.0 #kg-m (Max torque)\n",
"Tl = 150.0 #kg-m (load torque with flywheel)\n",
"T0 = 0 #kg-m (constant load torque)\n",
"t = 15.0 #s (time)\n",
"N = 500.0 #rpm (no load speed)\n",
"s = 0.1 # (slip)\n",
"g = 9.82 #m/s^2 \n",
"# s = K*T\n",
"K = 2*(3.14)*N*s/(60*Tm) #constant\n",
"\n",
"# As Tm = Tl*(1 - exp(-t*g/(I*K)))\n",
"\n",
"I =(-t*g)/(K*m.log(1 - Tm/Tl)) #kg-m^2 (Moment of inertia)\n",
"\n",
"print \"Moment of inertia =\",round(I,2),\"kg-m^2.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.20 , PAGE NO :- 1817"
]
},
{
"cell_type": "code",
"execution_count": 35,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" Moment of inertia of flywheel = 1908.86 kg-m^2.\n",
"Time taken after removal of additional load = 8.88 s.\n"
]
}
],
"source": [
"'''A 3-phase ,50 KW,6 pole,960 rpm induction motor has a constant load torque of 300 N-m and at wide intervals additional \n",
"torque of 1500 N-m for 10 sec.Calculate\n",
"(a)The moment of inertia of the flywheel used for load equalization,if the motor torque is not to exceed twice the rated torque.\n",
"(b)Time taken after removal of additional load,before the motor torque becomes 700 N-m.'''\n",
"\n",
"from sympy import Symbol,Eq,solve\n",
"import math as m\n",
"\n",
"P = 50.0e+3 #W (output power)\n",
"Nr = 960.0 #rpm (rotational speed)\n",
"p = 6.0 # (no. of poles)\n",
"t = 10.0 #s (time)\n",
"T0 = 300.0 #N-m (constant load torque)\n",
"Tl = T0 + 1500.0 #N-m (total load torque)\n",
"f = 50.0 #Hz (frequency) \n",
"\n",
"# Power = T*w (torque*ang_speed)\n",
"T = P/(2*3.14*Nr/60) #N-m (Full-load torque)\n",
"Tm = 2*T #N-m (Max torque)\n",
"\n",
"Ns = 120*f/p #rpm (synchronus speed)\n",
"\n",
"#Slip speed\n",
"sl = Ns-Nr #rpm\n",
"\n",
"#Now, s = K*T\n",
"K = 2*3.14*sl/(60*T) #constant\n",
"\n",
"#As Tm = Tl - (Tl - T0)*exp(-t/I*K)\n",
"\n",
"I = (-t)/(K*m.log((Tl - Tm)/(Tl - T0))) #kg-m^2 (moment of inertia)\n",
"\n",
"print \"Moment of inertia of flywheel =\",round(I,2),\"kg-m^2.\"\n",
"\n",
"#(b) Tm2 = T0 + (Tm-T0)*exp(-t/I*K)\n",
"Tm2 = 700.0 #N-m (Max torque - case 2)\n",
"\n",
"t1= (-I*K)*m.log((Tm2 - T0)/(Tm - T0)) #s (time after removal of load)\n",
"print \"Time taken after removal of additional load =\",round(t1,2),\"s.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 45.21 , PAGE NO :- 1818"
]
},
{
"cell_type": "code",
"execution_count": 39,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Moment of Inertia = 551.35 kg-m^2.\n"
]
}
],
"source": [
"'''A 3-phase,8 pole,50 cps.Induction Motor equipped with a flywheel supplies a constant load torque of 100 N-m and at wide \n",
"intervals an additional load torque of 300 N-m for 6 sec.The motor runs at 735 rpm at 100 N-m torque.Find moment of inertia of \n",
"the flywheel,if the motor torque is not to exceed 250 N-m.'''\n",
"\n",
"from sympy import Symbol,Eq,solve\n",
"import math as m\n",
"\n",
"T0 = 100.0 #N-m (constant load torque)\n",
"Tl = T0 + 300.0 #N-m (Total load torque)\n",
"f = 50.0 #Hz (frequency)\n",
"P = 8.0 # (poles)\n",
"Tm = 250.0 #N-m (Max torque) \n",
"Ns = 120*f/P #rpm (Synchronus speed)\n",
"sl = Ns - 735.0 #rpm (Slip speed)\n",
"t = 6.0 #s (time)\n",
"#Now, s = K*T0\n",
"K = 2*3.14*sl/(60*T0) #constant\n",
"\n",
"#Also, Tm = Tl - (Tl-T0)*exp(-t/I*K)\n",
"\n",
"I = -t/(K*m.log((Tl - Tm)/(Tl-T0))) #kg-m^2 (moment of inertia)\n",
"\n",
"print \"Moment of Inertia =\",round(I,2),\"kg-m^2.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.22 , PAGE NO :- 1818"
]
},
{
"cell_type": "code",
"execution_count": 42,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Maximum torque developed by motor = 615.35 N-m.\n",
"Speed at the end of deacceleration period = 938.47 rpm.\n"
]
}
],
"source": [
"'''A 6 pole,50 Hz,3-phase wound rotor Induction Motor has a flywheel coupled to its shaft.The total moment of inertia is \n",
"1000kg-m^2.Load torque is 1000 N-m for 10 sec followed by a no-load period which is long enough for the motor to reach its \n",
"no-load speed.Motor has a slip of 5% at a torque of 500 N-m.Find\n",
"(a)Maximum torque developed by motor\n",
"(b)Speed at the end of deacceleration period.'''\n",
"\n",
"import math as m\n",
"\n",
"P = 6.0 # (No of poles)\n",
"I = 1000.0 #kg-m^2 (Moment of Inertia)\n",
"Tl = 1000.0 #N-m (Load torque with flywheel)\n",
"t = 10.0 #s (time)\n",
"s = 0.05 # (slip)\n",
"Tfl = 500.0 #N-m (full load Torque)\n",
"f = 50.0 #Hz (frequency)\n",
"\n",
"Ns = 120*f/P #rpm (Synchronus speed)\n",
"\n",
"#Now, s = K*Tfl\n",
"K = 2*3.14*(Ns*s)/(60*Tfl) #constant\n",
"\n",
"#K = 6.2e-3 #(considered value) \n",
"\n",
"#Also Tm = Tl*(1-exp(-t/I*K)\n",
"Tm = Tl*(1 - m.exp(-t/(I*K))) #N-m\n",
"print \"Maximum torque developed by motor = \",round(Tm,2),\"N-m.\"\n",
" \n",
"#(b) s = K*Tfl where s = 2*3.14*(Ns - N)/60\n",
"N = Ns - (60/(2*3.14))*K*Tm #rpm\n",
"print \"Speed at the end of deacceleration period =\",round(N,2),\"rpm.\" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 45.23 , PAGE NO :- 1819"
]
},
{
"cell_type": "code",
"execution_count": 53,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Moment of inertia of flywheel = 275.67 kg-m^2.\n"
]
}
],
"source": [
"'''A motor fitted with a flywheel supplies a load torque of 1000 N-m for 2 seconds.During no load period,the flywheel regains its\n",
"original speed.The motor torque is to be limited to 500 N-m.Find moment of inertia of the flywheel.No load speed of the motor is \n",
"500 rpm and its full load slip is 10%.'''\n",
"\n",
"from sympy import solve,Eq,Symbol\n",
"import math as m\n",
"N = 500.0 #rpm (No load speed)\n",
"s = 0.1 # (slip)\n",
"Tfl = 500.0 #N-m (full load torque)\n",
"Tl = 1000.0 #N-m (load torque with flywheel)\n",
"t = 2.0 #s (time) \n",
"#Now, s = K*Tfl\n",
"K = (2*3.14*(N*s))/(Tfl*60) #constant\n",
"\n",
"\n",
"#Also, Tm = Tl*(1 - exp(-t/I*K))\n",
"I =-t/(K*m.log(1 - Tfl/Tl)) #(moment of inertia)\n",
"\n",
"print \"Moment of inertia of flywheel =\",round(I,2),\"kg-m^2.\""
]
},
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"cell_type": "code",
"execution_count": null,
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"collapsed": true
},
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PKj„ìJR‹D»l€l€;A TEXTBOOK OF ELECTRICAL TECHNOLOGY VOL III/chapter44.ipynb{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# CHAPTER 44 : INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 44.1 , PAGE NO :- 1769"
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total annual cost in group drive = RS. 11000.0\n",
"Total annual cost in individual drive = RS. 11400.0\n",
"Hence,Individual drive is costlier than group drive.\n"
]
}
],
"source": [
"'''A motor costing Rs. 10,000 is used for group drive in a certain installation.How will its total annual cost compare with\n",
"the case where four individuals motors each costing Rs. 4000 were used? With group drive,the energy consumption is\n",
"50MWh whereas it is 45MWh for individual drive.The cost of electric energy is 20 paise/kWh.Assume depriciation,\n",
"maintenance and other fixed charges at 10% in the case of group drive and 15 percent in the case of individual drive.'''\n",
"\n",
"\n",
"#Group drive\n",
"cost_g = 10000.0 #Rs (Capital cost)\n",
"other_g = 0.1*cost_g #Rs (Annual depriciation,maintenance and other charges)\n",
"enrgy_g = 50.0*1000*20/100.0 #Rs (Annual cost of energy)\n",
"total_g = enrgy_g + other_g #Rs (total annual cost)\n",
"\n",
"#Individual drive\n",
"cost_i = 4*4000.0 #Rs (Capital cost)\n",
"other_i = 0.15*cost_i #Rs (Annual depriciation,maintenance and other charges)\n",
"enrgy_i = 45.0*1000*20/100.0 #Rs (Annual cost of energy)\n",
"total_i = enrgy_i + other_i #Rs (total annual cost)\n",
"print \"Total annual cost in group drive = RS.\",total_g\n",
"print \"Total annual cost in individual drive = RS.\",total_i\n",
"print \"Hence,Individual drive is costlier than group drive.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 44.2 , PAGE NO :- 1775 "
]
},
{
"cell_type": "code",
"execution_count": 25,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Additional resistance = 5.7 ohm.\n",
"Initial braking Torque = 955.41 N-m.\n",
"Torque at 360 rpm = 766.53 N-m.\n"
]
}
],
"source": [
"'''A 40-kW,440-V,d.c shunt motor is braked by plugging.Calculate(i)the value of resistance that must be placed in series with the\n",
"armature circuit to limit the initial braking current to 150A (ii)the braking torque and(iii)the torque when motor speed falls\n",
"to 360 rpm. Armature resistance Ra = 0.1 ohm,ful-load Ia=100A,full-load speed=600 rpm.'''\n",
"\n",
"V = 440.0 #V (applied voltage)\n",
"Ib = 150.0 #A (initial braking current)\n",
"Ra = 0.1 #ohm (armature resistance)\n",
"Ia = 100.0 #A (armature current)\n",
"N1 = 600.0 #rpm (full load speed)\n",
"N2 = 360.0 #rpm (decreased speed)\n",
"Eb = V - Ia*Ra #V (back emf)\n",
"#Voltage across the armature at the start of braking\n",
"V2 = V + Eb #V\n",
"#(i)Since initial braking current is limited to 150A,total armature circuit resistance required is\n",
"Rt = V2/Ib #ohm\n",
"#Therefore additional resistance R is\n",
"R = Rt - Ra #ohm\n",
"#(ii)For a shunt motor,Torque(Tb) is propotional to Ia\n",
"Tb = 40*1000/(2*3.14*600/60) #N/m\n",
"# .'. Initial braking torque/full-load torque = initial braking current/full-load current\n",
"\n",
"T_ini = Tb*(Ib/Ia) #N/m\n",
"#(iii)The decrease in Eb is directly propotional to decrease in motor speed\n",
"Eb_360 = Eb*(N2/N1) #V\n",
"Ia_360 = (V+Eb_360)/Rt #A\n",
"Tb_360 = Tb*(Ia_360/Ia) #N-m\n",
"\n",
"print \"Additional resistance = \",R,\"ohm.\"\n",
"print \"Initial braking Torque = \",round(T_ini,2),\"N-m.\"\n",
"print \"Torque at 360 rpm = \",round(Tb_360,2),\"N-m.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 44.3 , PAGE NO :- 1776"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Plugging torque = 131.92 N-m.\n"
]
}
],
"source": [
"'''A 30-kW,400-V,3-phase,4 pole,50-Hz induction motor has full-load slip of 5%.If the ratio of standstill reactance to resistance\n",
"per motor phase is 4 ,estimating the plugging torque at full speed.'''\n",
"\n",
"Power = 30.0 #kW (power consumed)\n",
"f = 50.0 #Hz (frequency)\n",
"P = 4 # (pole)\n",
"s1 = 0.05 # (slip)\n",
"Ns = 120*f/P #rpm (Synchronus speed)\n",
"Nf = Ns*(1-s1) #rpm (Full-load speed)\n",
"Tf = Power*1000/(2*3.14*Nf/60) #N-m (Full-load torque)\n",
"R1_X1 = 4.0 \n",
"\n",
"# As T is propotional to (s*R2*E2^2)/(R2^2 + s^2*X2^2) i.e (T2/T1) = (s2/s1)*(1+s1^2(X2/R2)^2/1+s2^2(X2/R2)^2)\n",
"s2 = 2-s1\n",
"Tp = (s2/s1)*(1 +(R1_X1)*(R1_X1)*s1*s1)/(1 +(R1_X1)*(R1_X1)*s2*s2)*Tf\n",
"print \"Plugging torque = \",round(Tp,2),\"N-m.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 44.4 , PAGE NO :- 1780"
]
},
{
"cell_type": "code",
"execution_count": 19,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Hence,energy returned to supply lines = 43.43 kWh.\n"
]
}
],
"source": [
"'''A 500 tonne electric trains travel down a descending gradient of 1 in 80 for 90 seconds during which period its speed is\n",
"reduced from 100 km/h to 60 km/h by regenerative braking.Compute the energy returned to the lines of kWh if tractive\n",
"resistance = 50N/t;allowance for rotational inertia = 10%;overall efficiency of the system = 75%.'''\n",
"\n",
"\n",
"G = 1/80.0*100 # (percent gradient)\n",
"M = 500.0 #tonne (electric train)\n",
"Me_M = 1.1 # (ratio of rotational mass to stationary mass)\n",
"V1 = 100.0 #km/h (initial speed)\n",
"V2 = 60.0 #km/h (decreased speed)\n",
"t = 90.0 #s (braking period)\n",
"r = 50.0 #N/t (tractive resistance)\n",
"eff = 0.75 # (efficiency)\n",
"d = (V1+V2)/2*t/3600\n",
"#Hence,energy returned to supply line is\n",
"enrgy = 0.75*(0.01072*(Me_M)*(V1*V1 - V2*V2) + d*(27.25*G - 0.2778*r)) #Wh/tonne\n",
"enrgy = enrgy*500/1000.0 #kWh\n",
"print \"Hence,energy returned to supply lines =\",round(enrgy,2),\"kWh.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 44.5 , PAGE NO :- 1780"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Power returned to line = 509.0 kW.\n"
]
}
],
"source": [
"'''A 350-t electric train has its speed reduced by regenerative braking from 60 to 40km/h over a distance of 2km along down gradient\n",
"of 1.5%.Calculate (i)electrical energy and (ii)average power returned to the line.Assume specific train resistance = 50 N/t;rotational\n",
"inertia effect = 10% ; conversion efficency of the system = 75%.'''\n",
"\n",
"\n",
"M = 350.0 #tonne (Mass of train)\n",
"eff = 0.75 # (efficiency)\n",
"V1 = 60.0 #km/h (initial speed)\n",
"V2 = 40.0 #km/h (final speed)\n",
"Me = 1.1*M # (rotational mass )\n",
"G = 1.5 # (percent gradient)\n",
"d = 2.0 #km (distance)\n",
"r = 50.0 #N/t (train resistance)\n",
"\n",
"#Energy returned to line is\n",
"enrgy = eff*(0.01072*Me/M*(V1*V1 - V2*V2) + d*(27.25*G - 0.2778*r)) #Wh/t\n",
"enrgy = enrgy*M/1000 #kWh\n",
"\n",
"#(ii)\n",
"\n",
"speed = (V1+V2)/2 #km/h (Average speed)\n",
"time = d/speed #h (time taken)\n",
"\n",
"power = enrgy/time #kW (power returned)\n",
"\n",
"print \"Power returned to line =\",round(power),\"kW.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 44.6 , PAGE NO :- 1780"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Power fed to line = 424.38 kW.\n"
]
}
],
"source": [
"'''If in Example 44.5,regenerative braking is applied in such a way that train speed on down gradient remains constant\n",
"at 60 km/h,what would be the power fed into the line?'''\n",
"\n",
"\n",
"M = 350.0 #tonne (Mass of train)\n",
"G = 1.5 # (percent gradient)\n",
"r = 50.0 #N/t (train resistance)\n",
"V = 60.0 #km/h (speed)\n",
"eff = 0.75\n",
"#In down-gradient motors act as generators .Force generated\n",
"Ft = 98*M*G - M*r #N\n",
"#Power that can be recuperated is\n",
"P = Ft*(1000.0/3600)*V #W\n",
"#Power actually sent to line\n",
"P = eff*P/1000 #kW\n",
"print \"Power fed to line = \",round(P,2),\"kW.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 44.7 , PAGE NO :- 1780 "
]
},
{
"cell_type": "code",
"execution_count": 20,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Power fed = 650.0 kW.\n"
]
}
],
"source": [
"'''A train weighing 500 tonne is going down a gradient of 20 in 1000.It is desired to maintain train speed at 40 km/h by regenerative\n",
"braking.Calculate the power fed into the line.Tractive resistance is 40 N/t and allow rotational inertia of 10% and efficiency\n",
"of conversion of 75%.'''\n",
"\n",
"M = 500.0 #tonne (Mass of train)\n",
"G = 20/1000.0*100 # (percent gradient)\n",
"r = 40.0 #N/t (Tractive resistance)\n",
"V = 40.0 #km/h (speed)\n",
"eff = 0.75 # (efficiency) \n",
"#Tractive Force when motors are driven as generators is\n",
"Ft = 98*M*G - M*r #N\n",
"#Power that can be drawn is\n",
"P = 0.2778*Ft*V #W\n",
"#Power actually fed\n",
"P = eff*P/1000 #kW\n",
"print \"Power fed = \",round(P,0),\"kW.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 44.8 , PAGE NO :- 1781"
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Additional resistance = 1.94 ohm.\n",
"Initial braking torque = 1042.27 N-m.\n",
"Braking torque at 200rpm = 719.84 N-m.\n"
]
}
],
"source": [
"'''A 250V d.c shunt motor,taking an armature current of 150 A and running at 550 r.p.m is braked by reversing the connections to the\n",
"armature and inserting additional resistance in series with it.Calculate:\n",
"(a)the value of series resistance required to limit the initial current to 240A.\n",
"(b)the initial value of braking torque.\n",
"(c)the value of braking torque when the speed has fallen to 200 r.p.m.\n",
"The armature resistance is 0.09 ohm.Neglect winding friction and iron losses.'''\n",
"\n",
"V = 250.0 #V (applied voltage)\n",
"Ia = 150.0 #A (armature current)\n",
"Ib = 240.0 #A (initial braking current)\n",
"N = 550.0 #rpm (speed)\n",
"N2 = 200.0 #rpm (decreased speed) \n",
"Ra = 0.09 #ohm \n",
"#Induced emf at full-load\n",
"Eb = V - Ia*Ra #V\n",
"#Voltage across the armature at braking\n",
"Vb = V + Eb #V\n",
"#Resistance to limit the current to 240A\n",
"Rt = Vb/240 #ohm\n",
"#Resistance to be added in the circuit\n",
"R = Rt - Ra #ohm\n",
"\n",
"#(ii)\n",
"Tf = V*Ia/(2*3.14*N/60) #N-m (Full load torque)\n",
"# Initial braking torque/full-load torque = initial braking current/full-load current\n",
"\n",
"T_ini = Tf*(Ib/Ia) #N-m (initial braking torque)\n",
"\n",
"Eb_200 = Eb*N2/N #V (Back emf at 200 rpm)\n",
"Ia_200 = (V + Eb_200)/Rt #A (Current drawn at 200 rpm)\n",
"Tb_200 = Tf*Ia_200/Ia #N-m\n",
"\n",
"print \"Additional resistance = \",round(R,2),\"ohm.\"\n",
"print \"Initial braking torque = \",round(T_ini,2),\"N-m.\"\n",
"print \"Braking torque at 200rpm = \",round(Tb_200,2),\"N-m.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 44.9 , PAGE NO :- 1781"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Starting torque Ts = 0.25 *Tfl.\n"
]
}
],
"source": [
"'''A 400V 3-phase squirrel cage induction motor has a full load slip of 4%.A stand-still impedance of 1.54 ohm and the full load current\n",
"equal to 30A.The maximum starting current which may be taken from line is 75A.What taping must be provided on an auto-transformer starter\n",
"to limit the current to this value and what would be the starting torque available in terms of full-load torque ?'''\n",
"\n",
"import math as m\n",
"from sympy import Eq,solve,Symbol\n",
"\n",
"#considering Transformer action V2/V1 = I1/I2 = X\n",
"V1 = 400.0/(m.sqrt(3)) #V (applied voltage)\n",
"I1 = 75.0 #A (max starting current)\n",
"Z = 1.54 #ohm (impedance)\n",
"X = Symbol('X')\n",
"I2 = I1/X #A\n",
"V2 = Z*I2 #V\n",
"eq = Eq(V1*I1,V2*I2)\n",
"X = solve(eq) \n",
"X1 = X[1] #ohm\n",
"\n",
"I2 = I1/X1 #A\n",
"sfl = 0.04 # (full-load slip)\n",
"Ifl = 30.0 #A (full-load current)\n",
"\n",
"#Ts/Tfl = X^2*(Is/Ifl)^2*sfl\n",
"\n",
"Ts_Tfl = (X1*X1)*(I2/Ifl)*(I2/Ifl)*sfl # (Ts/Tfl)\n",
"print \"Starting torque Ts = \",round(Ts_Tfl,2),\"*Tfl.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 44.10 , PAGE NO :- 1782"
]
},
{
"cell_type": "code",
"execution_count": 21,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Additional resistance(constant torque) = 1.02 ohm.\n",
"Additional resistance(torque propotional to speed square) = 1.77 ohm.\n"
]
}
],
"source": [
"'''A 220V,10 HP shunt motor has field and armature resistances of 122ohm and 0.3ohm respectively.Calculate the resistance to be\n",
"inserted in the armature circuit to reduce the speed to 80% assuming motor efficiency at full load to be 80%.\n",
"(a)When torque is to remain constant.\n",
"(b)When torque is propotional to square of the speed.'''\n",
"\n",
"\n",
"V = 220.0 #V (applied voltage)\n",
"Rf = 122.0 #ohm (field resistance)\n",
"Ra = 0.3 #ohm (armature resistance)\n",
"\n",
"If = V/Rf #A (field current)\n",
"\n",
"m_out = 10*746 #W (motor output)\n",
"m_in = m_out/0.8 #W (motor input)\n",
"Il = m_in/V #A (line current)\n",
"\n",
"Ia = Il - If #A (armature current)\n",
"\n",
"Eb1 = V - Ia*Ra #V (back emf)\n",
"\n",
"#As flux is constant N2/N1 = Eb2/Eb1 and N2 = N1*0.8\n",
"\n",
"Eb2 = Eb1*0.8 #V (back emf at reduced speed)\n",
"\n",
"#(a) Torque remains constant ,hence Ia remains constant .Using Eb2 = V - Ia*R\n",
"Rt = (V-Eb2)/Ia #ohm (Total resistance required)\n",
"\n",
"#Therefore, additional resistance is\n",
"R = Rt - Ra #ohm\n",
"print \"Additional resistance(constant torque) = \",round(R,2),\"ohm.\"\n",
"\n",
"#(b) As (T2/T1) = (N2/N1)^2 and (T2/T1) = (Ia2/Ia1)\n",
"\n",
"T2_T1 = (0.8)*(0.8) # (T2/T1)\n",
"\n",
"Ia2 = Ia*T2_T1 #A (Changed armature current)\n",
"\n",
"#(b) Using Eb2 = V - Ia*R\n",
"Rt = (V-Eb2)/Ia2 #ohm (Total resistance required)\n",
"\n",
"\n",
"#Therefore, additional resistance is\n",
"R = Rt - Ra #ohm\n",
"print \"Additional resistance(torque propotional to speed square) = \",round(R,2),\"ohm.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 44.11 , PAGE NO :- 1783"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Initial braking torque = 713.33 N-m.\n",
"Electric braking torque at 1/2 speed = 539.96 N-m.\n"
]
}
],
"source": [
"'''A 37.5 HP, 220V DC shunt motor with a full load speed of 535 rpm is to be braked by plugging.Estimate the value of resistance \n",
"which should be placed in series with it to limit the initial braking current to 200 A.What would be the initial value of the\n",
"electric braking torque and the value when the speed had fallen to half its full load value?Armature resistance of motor is \n",
"0.086 ohm and full load armature current is 140A.'''\n",
"\n",
"V = 220.0 #V (applied voltage)\n",
"Ia = 140.0 #A (full-load armature current)\n",
"Ra = 0.086 #ohm (armature resistance)\n",
"Ib = 200.0 #A (braking current)\n",
"P = 37.5*746 #W (Power)\n",
"N = 535.0 #rpm (Speed)\n",
"Eb = V - Ia*Ra #V (Back emf)\n",
"#Total voltage during braking\n",
"Vb = Eb + V #V\n",
"\n",
"#Total resistance required is (using R = V/I)\n",
"Rt = Vb/Ib #ohm\n",
"Rt = round(Rt,2) #ohm\n",
"#Therefore,additional resistance required is\n",
"R = Rt - Ra #ohm\n",
"#We know that P = Torque*w where w is\n",
"w = 2*3.1416*N/60 #rad/s (angular velocity)\n",
"Tfl = P/w #N-m (full-load torque)\n",
"#As torque is propotional to I\n",
"#(Initial braking torque/Initial braking current) = (Full load torque/Full load current)\n",
"T_ini = Tfl*(Ib/Ia) #N-m (Initial braking torque)\n",
"\n",
"#As speed is propotional to back emf\n",
"Eb_2 = Eb/2 #V (Back emf at 1/2 speed)\n",
"Ib_2 = (V+Eb_2)/Rt #A (initial braking current at 1/2 speed)\n",
"\n",
"T_ini2 = Tfl*(Ib_2/Ia) #N-m (Initial braking torque at 1/2 speed)\n",
"print \"Initial braking torque = \",round(T_ini,2),\"N-m.\"\n",
"print \"Electric braking torque at 1/2 speed = \",round(T_ini2,2),\"N-m.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 44.12 , PAGE NO :- 1783"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Speed when torque is constant is = 600.61 rpm.\n",
"Speed when torque is propotional to speed square = 546.14 rpm.\n"
]
}
],
"source": [
"'''A 500V series motor having armature and field resistances of 0.2 and 0.3 ohm runs at 500 rpm when taking 70A.Assumimg unsaturated\n",
"field find out its speed when field diverter of 0.648 ohm is used for following load whose torque\n",
"(a) remains constant\n",
"(b) varies as square of speed '''\n",
"\n",
"\n",
"from sympy import Eq,solve,Symbol\n",
"V = 500.0 #V (applied voltage)\n",
"Ra = 0.2 #ohm (armature resistance)\n",
"Rf = 0.3 #ohm (field resistance)\n",
"N1 = 500.0 #rpm (speed)\n",
"Ia = 70.0 #A (armature current)\n",
"Rd = 0.684 #ohm (diverter resistance)\n",
"Eb1 = V - Ia*(Ra + Rf) #V (Back emf)\n",
"\n",
"#(a)Let Ia2 be armature current when diverter is used\n",
"\n",
"Ia2 = Symbol('Ia2')\n",
"\n",
"#Now If2 (field current when diverter is used) is\n",
"If2 = Ia2*Rd/(Rf+Rd)\n",
"\n",
"#As Torque is constant Ia1*(Flux1) = Ia2*(Flux2) Also,Ia1=If1 is propotional to (flux1)\n",
"eq = Eq(Ia*Ia/If2-Ia2,0)\n",
"Ia2 = solve(eq) \n",
"Ia_2 = Ia2[1] #A (Armature current when diverter is used)\n",
"\n",
"If2 = Ia_2*Rd/(Rf+Rd) #A (field current when diverter is used)\n",
"\n",
"#Resistance of field with diverter\n",
"Rfd = Rf*Rd/(Rf+Rd) #ohm\n",
"#Total resistance\n",
"Rt = Rfd + Ra #ohm\n",
"\n",
"Eb2 = V - Ia_2*Rt #V (Back emf when diverter is used)\n",
"\n",
"#Now, (N1/N2) = (Eb1*flux1/Eb2*flux2) and flux is propotional to If\n",
"N2 = (Eb2/Eb1)*N1*(Ia/If2) #rpm\n",
"print \"Speed when torque is constant is = \",round(N2,2),\"rpm.\"\n",
"####################################################################################\n",
"\n",
"#(B)Let Ia22 be armature current when diverter is used\n",
"#Now, (N1/N2)^2 = T1/T2 As (T1/T2) = Ia1*Ia1/(Ia2*If2)\n",
"\n",
"Ia22 = Symbol('Ia22')\n",
"#Now If2 (field current when diverter is used) is\n",
"If2 = Ia22*Rd/(Rf+Rd)\n",
"\n",
"N1_N2a = Ia*Ia/(Ia22*If2) #N1_N2a -> (N1/N2)^2\n",
"#Also, N1/N2 = Eb1*flux2/(Eb2*flux1)\n",
"N1_N2b = Eb1*If2/((V - Ia22*Rt)*Ia) #N1_N2b -> (N1/N2)\n",
"\n",
"eq = Eq(N1_N2a,N1_N2b*N1_N2b)\n",
"Ia22 = solve(eq)\n",
"Ia_22 = Ia22[1] #A (Armature current when diverter is used)\n",
"\n",
"If2 = Ia_22*Rd/(Rf+Rd)\n",
"N1_N2b = Eb1*If2/((V - Ia_22*Rt)*Ia) \n",
"#Using equation of N1_N2b = N1/N2\n",
"N2 = N1/N1_N2b #rpm\n",
"\n",
"print \"Speed when torque is propotional to speed square = \",round(N2,2),\"rpm.\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 44.13 , PAGE NO :- 1784 "
]
},
{
"cell_type": "code",
"execution_count": 22,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Additonal resistance required is = 5.9 ohm.\n"
]
}
],
"source": [
"'''A 200V series motor runs at 1000 rpm and takes 20A . Armature and field resistance is 0.4 ohm.Calculate the\n",
"resistance to be inserted in series so as to reduce the speed to 800 rpm,assuming torque to vary as cube of the\n",
"speed and unsaturated field.'''\n",
"\n",
"from sympy import Eq,Symbol,solve\n",
"\n",
"V = 200.0 #V (applied voltage)\n",
"N1 = 1000.0 #rpm (speed 1)\n",
"Raf = 0.4 #ohm (armature and field resistance)\n",
"N2 = 800.0 #rpm (speed 2)\n",
"Ia1 = 20.0 #A (armature current)\n",
"\n",
"\n",
"#Given, T1/T2 = (N1/N2)^3\n",
"\n",
"T1_T2a = (N1/N2)*(N1/N2)*(N1/N2) # (Ratio T1/T2)\n",
"\n",
"#Also T1/T2 = Ia1*flux1/Ia2*flux2 and Ia is propotional to flux\n",
"#Let Ia2 be armature current when speed is 800 rpm.\n",
"\n",
"Ia2 = Symbol('Ia2')\n",
"T1_T2b = Ia1*Ia1/(Ia2*Ia2)\n",
"eq = Eq(T1_T2a,T1_T2b)\n",
"Ia2 = solve(eq)\n",
"Ia_2 = Ia2[1] #A (armature current 2)\n",
"\n",
"Eb1 = V - Ia1*Raf #V (Back emf 1)\n",
"\n",
"#As Eb1/Eb2 = N1*I1/(N2*I2).Therefore Back emf 2 is\n",
"Eb2 = Eb1*(N2/N1)*(Ia_2/Ia1) #V (Back emf 2)\n",
"\n",
"#Also Eb2 = V - Ia2*Rt .Therefore total resistance required is\n",
"Rt = (V - Eb2)/Ia_2 #ohm\n",
"#Therefore,additional resistance required is\n",
"R = Rt - Raf #ohm\n",
"print \"Additonal resistance required is = \",round(R,2),\"ohm.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 44.14 , PAGE NO :- 1785"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Speed at which torque is 75% of initial value = 242.79 rpm.\n"
]
}
],
"source": [
"'''A 220V,500 rpm DC shunt motor with an armature resistance of 0.08 ohm and full load armature current of 150A is to be braked by \n",
"plugging.Estimate the value of resistance which is to be placed in series with the armature to limit initial braking current to\n",
"200A.What would be the speed at which the electric braking torque is 75% of its initial value.'''\n",
"\n",
"from sympy import Eq,solve,Symbol\n",
"\n",
"V = 220.0 #V (applied voltage)\n",
"N1 = 500.0 #rpm (speed 1)\n",
"Ra = 0.08 #ohm (armature resistance)\n",
"Ia = 150.0 #A (armature current)\n",
"Ib = 200.0 #A (initial braking current)\n",
"\n",
"\n",
"Eb1 = V - Ia*Ra #V (Back emf) \n",
"#Voltage across armature when braking starts\n",
"Vb = V + Eb1 #V\n",
"\n",
"#Resistance in armature circuit\n",
"Rt = Vb/Ib #ohm \n",
"#Additional resistance required\n",
"R = Rt - Ra #ohm \n",
"#Since field flux is constant therefore 75% torque is produced when armature current is 75% of Ib.\n",
"#As (Eb1/Eb2) = (N1/N2)\n",
"N2 = Symbol('N2') #rpm\n",
"Eb2 = Eb1*(N2/N1) #V\n",
"\n",
"#Voltage across armature when braking starts is V1=V2 =>\n",
"V1 = (0.75*Ib)*Rt #V\n",
"V2 = V + Eb2 #V\n",
"eq = Eq(V1,V2)\n",
"N2 = solve(eq) #rpm\n",
"N_2 = N2[0] #rpm\n",
"print \"Speed at which torque is 75% of initial value =\",round(N_2,2),\"rpm.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 44.15 , PAGE NO :- 1785"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Speed of motor with the shunted armature connection = 986.07 rpm.\n",
"Series motor can't be started on no-load.\n"
]
}
],
"source": [
"'''A D.C series motor operating at 250V D.C mains and draws 25A and runs at 1200 rpm Ra = 0.1 ohm and Rs = 0.3 ohm.A resistance of\n",
"25 ohm is placed in parallel with the armature of motor.Determine:\n",
"(i)The speed of motor with the shunted armature connection,if the magnetic circuit remains unsaturated and the load torque remains\n",
"constant.\n",
"(ii)No load speed of motor.'''\n",
"\n",
"from sympy import Eq,Symbol,solve\n",
"\n",
"V = 250.0 #V (applied voltage)\n",
"Ia = 25.0 #A (armature current)\n",
"N1 = 1200.0 #rpm (speed)\n",
"Ra = 0.1 #ohm (armature resistance)\n",
"Rse = 0.3 #ohm (series field resistance)\n",
"Rd = 25.0 #ohm (diverter resistance)\n",
"\n",
"#Let I2 flow from series winding , Ia2 be new armature current and Id be diverter current\n",
"\n",
"I2 = Symbol('I2')\n",
"Vd = V - Rse*I2 #V (Voltage across diverter)\n",
"Id = Vd/Rd #A (V=IR) \n",
"Ia2 = I2 - Id #A (new armature current)\n",
"\n",
"#AS T is constant, (flux1)*Ia1 = (flux2)*Ia2\n",
"\n",
"eq = Eq(Ia*Ia,(I2)*Ia2) # As, flux1/flux2 = Ia/I2\n",
"I2 = solve(eq)\n",
"I_2 = I2[1] #A (current through series winding)\n",
"\n",
"Ia2 = Ia*Ia/I_2 #A\n",
"Eb1 = V - Ia*(Ra+Rse) #V (Back emf 1) \n",
"\n",
"Eb2 = V - I_2*Rse - Ia2*Ra #V (Back emf 2)\n",
"\n",
"#N2/N1 = Eb2*flux1/Eb1*flux2\n",
"N2 = N1*(Eb2/Eb1)*(Ia/I_2) #rpm\n",
"print \"Speed of motor with the shunted armature connection =\",round(N2,2),\"rpm.\"\n",
"print \"Series motor can't be started on no-load.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 44.16 , PAGE NO :- 1786"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Additional Resistance required in (i) is = 0.8 ohm.\n",
"Additional Resistance required in (ii) is = 1.27 ohm.\n"
]
}
],
"source": [
"'''A 4 pole,50 Hz,slip ring Induction Motor has rotor resistance and stand still reactance referred to stator of 0.2 ohm and 1 ohm\n",
"per phase respectively.At full load,it runs at 1440 rpm.Determine the value of resistance to be inserted in rotor in ohm/phase to\n",
"operate at a speed of 1200 rpm,if:\n",
"(i)Load torque remains constant (ii)Load torque varies as square of the speed\n",
"Neglect stator resistance and leakage reactance.'''\n",
"\n",
"from sympy import Eq,solve,Symbol\n",
"\n",
"p = 4.0 # poles\n",
"f = 50.0 #Hz (frequency)\n",
"R2 = 0.2 #ohm (rotor resistance)\n",
"X2 = 1.0 #ohm (stand still reactance)\n",
"N1 = 1440.0 #rpm (speed)\n",
"N2 = 1200.0 #rpm (new speed)\n",
"\n",
"Ns = 120*f/p #rpm (synchronus speed)\n",
"s1 = (Ns-N1)/Ns #rpm (slip 1)\n",
"s2 = (Ns-N2)/Ns #rpm (slip 2)\n",
"\n",
"#(i)Load torque is constant i.e (T1 = T2)\n",
"#T is propotional to (s/R2) , (T1/T2) = (s1/S2)*(R2/Rt).Therefore, new resistance required is \n",
"\n",
"Rt = (s2/s1)*R2 #ohm (total resistance)\n",
"r = Rt - R2 #ohm (additional resistance)\n",
"\n",
"print \"Additional Resistance required in (i) is =\",round(r,2),\"ohm.\"\n",
"\n",
"#(ii) Load torque varies as square of speed (i.e T1/T2 = (N1/N2)^2 )\n",
"T1_T2a = (N1/N2)*(N1/N2) #(T1/T2)\n",
"\n",
"Rt = Symbol('Rt')\n",
"T1_T2b = (s1*R2/(R2*R2 + (s1*X2)*(s1*X2)))/(s2*Rt/(Rt*Rt + (s2*X2)*(s2*X2)))\n",
"eq = Eq(T1_T2a,T1_T2b)\n",
"Rt = solve(eq)\n",
"R_t = Rt[1] #ohm (total resistance)\n",
"\n",
"r = R_t - R2 #ohm (additional resistance)\n",
"print \"Additional Resistance required in (ii) is =\",round(r,2),\"ohm.\""
]
},
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"execution_count": null,
"metadata": {
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"outputs": [],
"source": []
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PKj„ìJÂ \ùPÑPÑ;A TEXTBOOK OF ELECTRICAL TECHNOLOGY VOL III/chapter49.ipynb{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# CHAPTER 49 : ILLUMINATION"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.1 , PAGE NO :- 1899"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Illumination at point = 0.76 lm/m^2 .\n"
]
}
],
"source": [
"'''A lamp giving out 1200 lm in all directions is suspended 8 m above the working plane. Calculate the illumination at a point on\n",
"the working plane 6 m away from the foot of the lamp.'''\n",
"\n",
"import math as m\n",
"\n",
"I = 1200/(4*3.14) #Cd (luminous intensity of lamp)\n",
"h = 8.0 #m (height)\n",
"b = 6.0 #m (breadth) \n",
"length = m.sqrt(h**2 + b**2) #m\n",
"\n",
"cosQ = h/length\n",
"E = I*cosQ/length**2 #lm/m^2\n",
"\n",
"print \"Illumination at point =\",round(E,2),\"lm/m^2 .\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.2 , PAGE NO :- 1899"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Distance between A and B is = 19.08 m.\n"
]
}
],
"source": [
"'''A small light source with intensity uniform in all directions is mounted at a height of 10 metres above a horizontal surface.\n",
"Two points A and B both lie on the surface with point A directly beneath the source. How far is B from A if the illumination at B\n",
"is only 1/10 as great as at A ?'''\n",
"\n",
"from sympy import Eq,solve,Symbol\n",
"\n",
"#let the intensity of lamp be I and distance between A and B be x metres\n",
"x = Symbol('x')\n",
"\n",
"l = 10.0 #m (vertical distance)\n",
"#Illumination at point A\n",
"Ea = I/l**2 #lux\n",
"#Illumination at point B\n",
"\n",
"Eb = I/(l**2)*(l/(l**2 + x**2)**0.5)**3\n",
"\n",
"I = 10.0 #lm (assumed value as the equation does not depend on I)\n",
"#As Eb = 1/10*Ea\n",
"eq = Eq(Eb,Ea/10.0)\n",
"x = solve(eq)\n",
"\n",
"x1 = x[1]\n",
"\n",
"print \"Distance between A and B is =\",round(x1,2),\"m.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 49.3 , PAGE NO :- 1900"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Illumination due to 4 lamps = 6.16 lm/m^2 .\n"
]
}
],
"source": [
"'''A corridor is lighted by 4 lamps spaced 10 m apart and suspended at a height of 5 m above the centre line of the floor.\n",
"If each lamp gives 200 C.P. in all directions below the horizontal,find the illumination at the point on the floor mid-way\n",
"between the second and third lamps.'''\n",
"\n",
"import math as m\n",
"\n",
"I = 200.0 #C.P (luminous intensity)\n",
"h = 5.0 #m (height between lamps and ground)\n",
"l1 = 15.0 #m (horizantal distance 1)\n",
"l2 = 5.0 #m (horizantal distance 2)\n",
"\n",
"d1 = m.sqrt(h**2 + l1**2) #m (Dist btwn L1 and mid-pt)\n",
"d2 = m.sqrt(h**2 + l2**2) #m (Dist btwn L2 and mid-pt)\n",
"\n",
"#(i)Illumination due to L1\n",
"#L = (I/r^2)*cosQ\n",
"L1 = (I/d1**2)*(h/d1) #lm/m^2\n",
"\n",
"#(ii)Illumination due to L2\n",
"L2 = (I/d2**2)*(h/d2) #lm/m^2\n",
"\n",
"#Illumination at mid-pt due to 4-lamps\n",
"Lt = 2*(L1+L2) #lm/m^2\n",
"\n",
"print \"Illumination due to 4 lamps = \",round(Lt,2),\"lm/m^2 .\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.4 , PAGE NO :- 1901"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Reading of photometer = 0.066 lm/m^2.\n"
]
}
],
"source": [
"'''Two lamps A and B of 200 candela and 400 candela respectively are situated 100 m apart. The height of A above the ground \n",
"level is 10 m and that of B is 20 m. If a photometer is placed at the centre of the line joining the two lamp posts,\n",
"calculate its reading.'''\n",
"\n",
"import math as m\n",
"I1 = 200.0 #Cd (lamp 1 intensity)\n",
"I2 = 400.0 #Cd (lamp 2 intensity)\n",
"h1 = 10.0 #m (height between lamp 1 and ground)\n",
"h2 = 20.0 #m (height between lamp 2 and ground)\n",
"d1 = 50.0 #m (horizontal distance from 1)\n",
"d2 = 50.0 #m (horizontal distance from 2)\n",
"\n",
"l1 = m.sqrt(h1**2 + d1**2)\n",
"l2 = m.sqrt(h2**2 + d2**2)\n",
"\n",
"cosQ1 = h1/l1\n",
"cosQ2 = h2/l2\n",
"\n",
"#Illumination at point C = Illumination due to 1 + Illumination due to 2\n",
"I_tot = (I1/l1**2)*cosQ1 + (I2/l2**2)*cosQ2 #lm/m^2\n",
"\n",
"print \"Reading of photometer =\",round(I_tot,3),\"lm/m^2.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.5 , PAGE NO :- 1901"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Average Brightness = 6375.22 cd/m^2.\n",
"cost of running = Rs 150.0\n"
]
}
],
"source": [
"'''The average luminous output of an 80-W fluorescent lamp 1.5 metre in length and 3.5 cm diameter is 3300 lumens. Calculate its\n",
"average brightness.If the auxiliary gear associated with the lamp consumes a load equivalent to 25 percent of the lamp,\n",
"calculate the cost of running a twin unit for 2500 hours at 30 paise per kWh.'''\n",
"\n",
"\n",
"length = 1.5 #m (lamp output length)\n",
"dia = 3.5e-2 #m (lamp output diameter)\n",
"l_flux = 3300.0 #lumens (luminous flux)\n",
"P = 80.0 #W (Power output)\n",
"\n",
"#Surface area of lamp\n",
"sa = 3.14*dia*length #m^2\n",
"\n",
"#Flux emmited per unit area\n",
"fluxA = l_flux/sa #lm/m^2\n",
"#Therefore,\n",
"B = fluxA/3.14 #cd/m^2\n",
"print \"Average Brightness =\",round(B,2),\"cd/m^2.\"\n",
"#Total load of twin fitting\n",
"load = 2*(P +0.25*P) #W\n",
"time = 2500.0 #hr\n",
"enrgy = load*time/1000 #kWh (Energy consumed)\n",
"\n",
"#Total cost\n",
"cost = enrgy*0.3 #Rs\n",
"print \"cost of running = Rs\",round(cost)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.6 , PAGE NO :- 1901"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"lamp wattage = 4415.63 W.\n"
]
}
],
"source": [
"'''A small area 7.5 m in diameter is to be illuminated by a lamp suspended at a height of 4.5 m over the centre of the area.\n",
"The lamp having an efficiency of 20 lm/w is fitted with a reflector which directs the light output only over the surface to be\n",
"illuminated,giving uniform candle power over this angle. Utilisation coefficient = 0.40. Find out the wattage of the lamp.\n",
"Assume 800 lux of illumination level from the lamp.'''\n",
"\n",
"dia = 7.5 #m (diameter)\n",
"h = 4.5 #m (height)\n",
"E = 800.0 #lux (illumination) \n",
"eff = 20.0 #lm/w (lamp efficiency)\n",
"A = 3.14*(dia**2)/4\n",
"#Luminous flux reaching the surface\n",
"flux = A*E #lm\n",
"\n",
"#Total flux emmited is\n",
"f_out = flux/0.4 #lm\n",
"\n",
"#Lamp in watts\n",
"watt = f_out/eff #W\n",
"print \"lamp wattage =\",round(watt,2),\"W.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.7 , PAGE NO :- 1902"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Illumination = 5.0 lux.\n"
]
}
],
"source": [
"'''A lamp of 100 candela is placed 1 m below a plane mirror which reflects 90% of light falling on it. The lamp is hung 4 m above\n",
"ground.Find the illumination at a point on the ground 3 m away from the point vertically below the lamp.'''\n",
"\n",
"import math as m\n",
"\n",
"h1 = 4.0 #m (height of 1 from ground)\n",
"d1 = 3.0 #m (horizontal distance 1)\n",
"I1 = 100.0 #cd (intenstity)\n",
"\n",
"l1 = m.sqrt(h1**2 + d1**2) #m\n",
"cosQ1 = h1/l1\n",
"#The lamp L1 will produce the image L2 1m behind the mirror.Therefore,\n",
"\n",
"h2 = h1+1+1 #m (height of 2 from ground)\n",
"d2 = 3.0 #m (horizontal distance 2)\n",
"I2 = 0.9*I1 #cd (intensity)\n",
"\n",
"l2 = m.sqrt(h2**2 + d2**2) #m\n",
"cosQ2 = h2/l2\n",
"\n",
"#Illumination at the required point is\n",
"\n",
"E = I1/l1**2*cosQ1 + I2/l2**2*cosQ2 #lux\n",
"\n",
"print \"Illumination = \",round(E),\"lux.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.8 , PAGE NO :- 1902"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Average illumination = 924.08 lux .\n"
]
}
],
"source": [
"'''A light source having an intensity of 500 candle in all directions is fitted with a reflector so that it directs 80% of its\n",
"light along a beam having a divergence of 15Âº. What is the total light flux emitted along the beam? What will be the average\n",
"illumination produced on a surface normal to the beam direction at a distance of 10 m? '''\n",
"\n",
"import math as m\n",
"\n",
"I = 500.0 #cd (intensity)\n",
"Q = 15.0 #degrees (Beam angle)\n",
"h = 10.0 #m (height)\n",
"\n",
"#Total flux emmited is\n",
"flux = 0.8*(4*3.14*I) #lm\n",
"#radius of circle to be illuminated\n",
"r = h*m.tan(Q/2*(3.14/180)) #m\n",
"\n",
"#Area of surface to be illuminated is\n",
"A = 3.14*(r*r) #m^2\n",
"\n",
"#Avg illumination\n",
"avg = flux/A #lux\n",
"\n",
"print \"Average illumination =\",round(avg,2),\"lux .\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.9 , PAGE NO :- 1902"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"----Without reflector ----\n",
"Illumination at centre = 0.75 lm/m^2.\n",
"Illumination at edge = 0.54 lm/m^2.\n",
"----With reflector ----\n",
"Illumination at every point = 6.0 lm/m^2\n"
]
}
],
"source": [
"'''A lamp has a uniform candle power of 300 in all directions and is fitted with a reflector which directs 50% of the total\n",
"emitted light uniformly on to a flat circular disc of 20 m diameter placed 20 m vertically below the lamp. Calculate the\n",
"illumination (a) at the centre and (b)at the edge of the surface without the reflector. Repeat these two calculations with\n",
"the reflector provided.'''\n",
"\n",
"import math as m\n",
"\n",
"I = 300.0 #Cd (intensity)\n",
"h = 20.0 #m (height)\n",
"dia = 20.0 #m (diameter of luminous area)\n",
"\n",
"#(i)Without reflector\n",
"Ec = I/h**2 #lm/m^2 (illumination at centre)\n",
"\n",
"theta = m.atan((dia/2)/h)\n",
"l = m.sqrt(h**2 + (dia/2)**2) #m (distance between edge and source lamp)\n",
"\n",
"Eb = I/l**2*m.cos(theta) #lm/m^2 (illuminaton at edge)\n",
"print \"----Without reflector ----\"\n",
"print \"Illumination at centre =\",round(Ec,2),\"lm/m^2.\"\n",
"print \"Illumination at edge =\",round(Eb,2),\"lm/m^2.\"\n",
"\n",
"#(ii)With reflector\n",
"#Luminous output of lamp\n",
"lflux = I*4*3.14 #lm\n",
"\n",
"#flux directed by reflector\n",
"reflux = 0.5*lflux #lm\n",
"\n",
"#Area of disc\n",
"A = 3.14*(dia*dia)/4 #m^2\n",
"\n",
"#Illumination at every point will be same and will be equal to\n",
"Et = reflux/A #lm/m^2\n",
"\n",
"print \"----With reflector ----\"\n",
"print \"Illumination at every point =\",round(Et,2),\"lm/m^2\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.10 , PAGE NO :- 1903"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Iluminaton at P = 11.11 cd/m^2\n",
"Iluminaton at Q = 3.93 cd/m^2\n",
"Total radiations sent = 314.0 lumens.\n"
]
}
],
"source": [
"'''A light is placed 3 m above the ground and its candle power is 100 cos Î¸ in any downward direction making an angle q with the\n",
"vertical. If P and Q are two points on the grond, P being vertically under the light and the distance PQ being 3 m, calculate.\n",
"(a) the illumination of the ground at P and also at Q.\n",
"(b) the total radiations sent down by the lamp.'''\n",
"\n",
"import math as m\n",
"from scipy.integrate import quad\n",
"\n",
"r1 = 3.0 #m\n",
"r2 = m.sqrt(3**2 + 3**2) #m\n",
"#(a)\n",
"#Candela Power along LP\n",
"CP1 = 100.0*m.cos(0) #cd\n",
"#Illumination at P is\n",
"Ep = CP1/(r1**2) #cd/m^2\n",
"\n",
"#Candela Power along LQ\n",
"CP2 = 100.0*m.cos(45*3.14/180) #cd\n",
"\n",
"#Illumination at Q is\n",
"Eq = CP2/(r2**2) #cd/m^2\n",
"\n",
"print \"Iluminaton at P = \",round(Ep,2),\"cd/m^2\"\n",
"print \"Iluminaton at Q = \",round(Eq,2),\"cd/m^2\"\n",
"\n",
"#After working out , total flux = integral (100*pi*sin2Q*dQ) 0->pi/2\n",
"\n",
"def integrand(Q):\n",
" return 100*3.14*m.sin(2*Q)\n",
"\n",
"ans, err = quad(integrand, 0,3.14/2)\n",
"print \"Total radiations sent = \",round(ans),\"lumens.\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.11 , PAGE NO :- 1903"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Number of lamps = 21.0\n"
]
}
],
"source": [
"'''A drawing office containing a number of boards and having a total effective area of 70 m2 is lit by a number of 40 W\n",
"incandescent lamps giving 11 lm/W. An illumination of 80 lux is required on the drawing boards. Assuming that 60% of the\n",
"total light emitted by the lamps is available for illuminating the drawing boards, estimate the number of lamps required.'''\n",
"\n",
"A = 70.0 #m^2 (area)\n",
"watt = 40.0 #W (each bulb wattage)\n",
"eff = 11.0 #lm/W (luminous efficacy)\n",
"E = 80.0 #lux (Illumination)\n",
"\n",
"#Output per lamp is\n",
"oplamp = watt*eff #lm\n",
"\n",
"#Flux actually used per lamp is\n",
"flux = 0.6*oplamp #lm\n",
"\n",
"#Now, Total flux required is Illumination*Area\n",
"flux_tot = E*A #lm\n",
"\n",
"#Therefore number of lamps required are\n",
"N = flux_tot/flux\n",
"\n",
"print \"Number of lamps =\",round(N)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.12 , PAGE NO :- 1904"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total flux radiated = 62.74 lm.\n"
]
}
],
"source": [
"'''A perfectly diffusing surface has a luminous intensity of 10 candles at an angle of 60Âº to the normal. If the area of the \n",
"surface is 100 cm2, determine the brightness and total flux radiated.'''\n",
"\n",
"import math as m\n",
"\n",
"I = 10.0 #Cd (Intensity)\n",
"theta = 60.0 #degrees (angle to normal)\n",
"A = 100.0 #cm^2 (Area)\n",
"\n",
"proA = A*m.cos(theta*3.14/180) #cm^2\n",
"\n",
"B = I/proA*(10000) #cd/m^2 (Brightness)\n",
"B = B*3.14 #lm/m^2 (Brightness)\n",
"\n",
"flux = B*A*10e-5 #lm (Total flux radiated)\n",
"print \"Total flux radiated =\",round(flux,2),\"lm.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.13 , PAGE NO :- 1904"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Brightness 1 = 1.191083e+04 cd/m^2.\n",
"Brightness 2 = 0.06 cd/m^2.\n"
]
}
],
"source": [
"'''Calculate the brightness (or luminance) of snow under an illumination of (a) 44,000 lux and (b) 0.22 lux. Assume that snow\n",
"behaves like a perfect diffusor having a reflection factor of 85 per cent.'''\n",
"\n",
"E1 = 44000.0 #lux (illumination 1)\n",
"E2 = 0.22 #lux (illumination 2)\n",
"rf = 0.85 # (reflection factor)\n",
"\n",
"L1 = (E1*rf/3.14) #cd/m^2 (Brightness 1)\n",
"L2 = (E2*rf/3.14) #cd/m^2 (Brightness 2)\n",
"\n",
"print \"Brightness 1 = %e cd/m^2.\" %round(L1,2)\n",
"print \"Brightness 2 =\",round(L2,2),\"cd/m^2.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.14 , PAGE NO :- 1904"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"luminous intensity of globe is = 44.0 Cd.\n",
"percentage absorption = 44.61 %.\n"
]
}
],
"source": [
"'''A 21 cm diameter globe of dense opal glass encloses a lamp emitting 1000 lumens and has uniform brightness of 4e+3 lumen/m^2\n",
"when viewed in any direction. What would be the luminous intensity of the globe in any direction? Find what percentage of the\n",
"flux emitted by the lamp is absorbed by the globe.'''\n",
"\n",
"d = 21.0 #cm (diameter)\n",
"flux = 1000.0 #lumens (luminous flux) \n",
"B = 4e+3 #lm/m^2 (uniform Brightness)\n",
"\n",
"#Surface Area of the globe\n",
"sa = 3.14*(d*d)*10e-5 #m^2\n",
"#Flux emitted by globe is\n",
"fluxe = sa*B #lm\n",
"\n",
"#luminous intensity of globe is\n",
"lint = fluxe/(4*3.14) #Cd\n",
"print \"luminous intensity of globe is =\",round(lint),\"Cd.\"\n",
"#Flux absorbed by globe is\n",
"fluxab = flux - fluxe #lm\n",
"\n",
"#% absorption is\n",
"absrp = fluxab/flux*100 #% absorption\n",
"print \"percentage absorption = \",round(absrp,2),\"%.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.15 , PAGE NO :- 1904"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Luminous intensity is = 1005440.0 Cd\n",
"The beam spread is = 14.25 degrees\n"
]
}
],
"source": [
"'''A 2.5 cm diameter disc source of luminance 1000 cd/cm2 is placed at the focus of a specular parabolic reflector\n",
"normal to the axis. The focal length of the reflector is 10 cm, diameter 40 cm and reflectance 0.8. Calculate the axial\n",
"intensity and beam-spread. Also show diagrammatically what will happen if the source were moved away from the reflector\n",
"along the axis in either direction.'''\n",
"\n",
"import math as m\n",
"\n",
"dia = 0.025 #m (diameter of disc)\n",
"d = 0.4 #m (diameter of relector) \n",
"L = 1000.0e+4 #Cd/m^2 (luminance)\n",
"\n",
"#Surface area is\n",
"A = 3.142*d*d/4 #m^2 (Area)\n",
"\n",
"#Luminous intensity is\n",
"I = 0.8*A*L #Cd\n",
"print \"Luminous intensity is =\",round(I,2),\"Cd\"\n",
"\n",
"#Let us assume 'theta' as the beam-spread .Then\n",
"r = dia/2 #m (radius)\n",
"f = 0.1 #m (focal length) \n",
"theta = 2*m.degrees(m.atan((r/f)))\n",
"\n",
"print \"The beam spread is =\",round(theta,2),\"degrees\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.16 , PAGE NO :- 1905"
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Flux emmited by source is = 6942.15 lm/m^2 .\n",
"C.P of globe is = 84.0 Cd.\n"
]
}
],
"source": [
"'''A 22cm diameter globe of opal glass encloses a lamp of uniform luminous intensity 120 C.P. Thirty per cent of light emitted \n",
"by the lamp is absorbed by globe. Determine (a) luminance of globe (b) C.P. of globe in any direction.'''\n",
"\n",
"\n",
"d = 0.22 #m (diameter)\n",
"I = 120.0 #Cd (luminous intensity)\n",
"\n",
"#surface area is\n",
"sa = 3.14*d*d #m^2\n",
"\n",
"#Flux emmited by source is\n",
"flux = I*(4*3.14) #lm\n",
"#Flux emmited by globe is\n",
"reflux = 0.7*flux #lm\n",
"\n",
"#(a)Luminance of globe is\n",
"L = reflux/sa #lm/m^2\n",
"print \"Flux emmited by source is =\",round(L,2),\"lm/m^2 .\"\n",
"#(b) C.P of globe is\n",
"cp = reflux/(4*3.14) #Cd\n",
"print \"C.P of globe is = \",round(cp,2),\"Cd.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.17 , PAGE NO :- 1905"
]
},
{
"cell_type": "code",
"execution_count": 15,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Average luminance = 7722.93 lm/m^2 .\n"
]
}
],
"source": [
"'''A 0.4 m diameter diffusing sphere of opal glass (20 percent absorption) encloses an incandescent lamp with a luminous flux of\n",
"4850 lumens. Calculate the average luminance of the sphere.'''\n",
"\n",
"\n",
"d = 0.4 #m (diameter)\n",
"lflux = 4850.0 #lm (luminous flux)\n",
"reflux = 0.8*lflux #lm (flux emmited by globe)\n",
"\n",
"sa = 3.14*d*d #m^2 (surface area)\n",
"\n",
"#Brightness B = flux emmited/surface area . i.e\n",
"B = reflux/sa #lm/m^2 (brightness)\n",
"\n",
"print \"Average luminance =\",round(B,2),\"lm/m^2 .\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.18 , PAGE NO :- 1907"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Illumination produced is = 122.55 lm/m^2.\n"
]
}
],
"source": [
"'''A show case is lighted by 4 metre of architectural tubular lamps arranged in a continuous line and placed along the top of the\n",
"case.Determine the illumination produced on a horizontal surface 2 metres below the lamps in a position directly underneath the\n",
"centre of the 4 m length of the lamps on the assumption that in tubular lamps emit 1,880 lm per metre run.\n",
"Neglect the effect of any reflectors which may be used.'''\n",
"\n",
"import math as m\n",
"\n",
"L = 4.0 #m (length of source of light)\n",
"d = 2.0 #m (height)\n",
"flux = 1880.0 #lumens (flux) \n",
"#Now\n",
"theta = m.atan(L/(2*d))\n",
"\n",
"#As I = flux/(3.14*3.14*L)\n",
"I = 4*flux/(3.14*3.14*L) #cd/m\n",
"\n",
"#Illumination produced is\n",
"E = I/(2*d)*(m.sin(2*theta) + 2*theta) #lm/m^2\n",
"\n",
"print \"Illumination produced is =\",round(E,2),\"lm/m^2.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 49.19 , PAGE NO :- 1913"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Luminous flux yield of the source = 1109.2 lm.\n"
]
}
],
"source": [
"'''If an integrating sphere 0.6 m in diameter whose inner surface has a reflection coefficient of 0.8 contains a lamp producing\n",
"on the portion of the sphere, screened from direct radiation,a luminance of 1000 cd/m2, what is the luminous flux yield of\n",
"the source ?'''\n",
"\n",
"from sympy import Eq,solve,Symbol\n",
"\n",
"coef = 0.8 # (reflection coefficient)\n",
"L = 1000.0 #cd/m^2 (luminance)\n",
"d = 0.6 #m (diameter)\n",
"\n",
"Fl = Symbol('Fl')\n",
"E = coef*Fl/(3.14*d*d*(1-coef)) #lm/m^2\n",
"L1 = coef*E/3.14 #cd/m^2\n",
"#As L is equal to L1\n",
"eq = Eq(L,L1)\n",
"Fl = solve(eq)\n",
"Fl1 = Fl[0] #lumens\n",
"\n",
"print \"Luminous flux yield of the source =\",round(Fl1,2),\"lm.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 49.20 , PAGE NO :- 1919 "
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Utilization coefficient = 0.4\n"
]
}
],
"source": [
"'''A room 8 m * 12 m is lighted by 15 lamps to a fairly uniform illumination of 100 lm/m^2. Calculate the utilization coefficient\n",
"of the room given that the output of each lamp is 1600 lumens.'''\n",
"\n",
"Area = 8*12 #m^2 (area of room)\n",
"num = 15.0 # (number of lamps)\n",
"I = 1600.0 #lumens (output of each lamp)\n",
"E = 100.0 #lm/m^2 (illumination)\n",
"\n",
"#Lumens emmited by lamp\n",
"I_tot = num*I #lumens\n",
"\n",
"#Lumens recieved by working plane\n",
"I1 = Area*E #lumens\n",
"\n",
"#Utilization coefficient is\n",
"coef = I1/I_tot\n",
"\n",
"print \"Utilization coefficient = \",round(coef,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.21 , PAGE NO :- 1919"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Number of lamps required = 50.0\n"
]
}
],
"source": [
"'''The illumination in a drawing office 30 m*10 m is to have a value of 250 lux and is to be provided by a number of 300 W\n",
"filament lamps. If the coefficient of utilization is 0.4 and the depreciation factor 0.9, determine the number of lamps\n",
"required. The luminous efficiency of each lamp is 14 lm/W.'''\n",
"\n",
"A = 30*10.0 #m^2 (area)\n",
"E = 250.0 #lm/m^2 (illumination)\n",
"coef = 0.4 # (coefficient of utilization)\n",
"p = 0.9 # (depriciation factor)\n",
"eff = 14.0 #lm/W (luminous efficiency)\n",
"watt = 300.0 #W (wattage of eacch lamp)\n",
"#Now, flux = E*A/coef*p\n",
"flux = E*A/(coef*p) #lm (output in lumens)\n",
"\n",
"#Flux emmited per lamp is\n",
"Fl2 = watt*eff #lm\n",
"\n",
"#No. of lamps required are\n",
"num = flux/Fl2\n",
"\n",
"print \"Number of lamps required =\",round(num)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.22 , PAGE NO :- 1919"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Net saving in load = 600.0 W.\n",
"Increase in illumination = 42.22 %\n"
]
}
],
"source": [
"'''Find the total saving in electrical load and percentage increase in illumination if instead of using twelve 150 W tungsten-\n",
"filament lamps,we use twelve 80 W fluorescent tubes. It may be assumed that (i) there is a choke loss of 25 per cent of rated\n",
"lamp wattage (ii) average luminous efficiency throughout life for each lamp is 15 lm/W and for each tube 40 lm/W and \n",
"(iii) coefficient of utilization remains the same in both cases.'''\n",
"\n",
"#Luminous efficiency\n",
"eff1 = 15.0 #lm/W\n",
"eff2 = 40.0 #lm/W\n",
"\n",
"#Total load in filament-lamps\n",
"flamp = 12*150.0 #W\n",
"#Total load in fluoroscent tubes\n",
"tube = 12*(80 + 0.25*80) #W\n",
"#Net saving\n",
"load = flamp - tube #W\n",
"print \"Net saving in load =\",round(load,2),\"W.\"\n",
"\n",
"#Let us assume that\n",
"#E1 -> illumination with lamps\n",
"#E2 -> illumination with tubes\n",
"#Now E1/E2 = (O/P in lumens 1)/(O/P in lumens 2)\n",
"\n",
"tube2 = 12*80.0 #W\n",
"\n",
"E1_E2 = flamp*eff1/(tube2*eff2)\n",
"\n",
"#Increase in illumination is given by %increase = (E2/E1 - 1)*100\n",
"increase = (1/E1_E2 - 1)*100.0\n",
"\n",
"print \"Increase in illumination = \",round(increase,2),\"%\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.23 , PAGE NO :- 1919"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Number of lamps on each tower is = 50.0\n"
]
}
],
"source": [
"'''A football pitch 120 m * 60 m is to be illuminated for night play by similar banks of equal 1000 W lamps supported on twelve \n",
"towers which are distributed around the ground to provide approximately uniform illumination of the pitch.Assuming that 40% of\n",
"the total light emitted reaches the playing pitch and that an illumination of 1000 lm/m2 is necessary for television purposes,\n",
"calculate the number of lamps on each tower. The overall efficiency of the lamp is to be taken as 30 lm/W.'''\n",
" \n",
"Area = 120.0*60.0 #m^2 (Area of pitch)\n",
"E = 1000.0 #lm/m^2 (Illumination of pitch)\n",
"\n",
"#Flux required is\n",
"flux = Area*E #lm \n",
"\n",
"#Since only 40% reaches the ground.Total flux required is\n",
"lflux = flux/0.4 #lm\n",
"\n",
"#There are 12 tower banks . Therefore flux by each tower bank is\n",
"flux_each = lflux/12 #lm\n",
"\n",
"#Output of each 1000 W lamp is\n",
"I = 30.0*1000 #lm\n",
"\n",
"#Therefore, number of each lamps is\n",
"num = flux_each/I\n",
"\n",
"print \"Number of lamps on each tower is =\",round(num)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.24 , PAGE NO :- 1920"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"number of flouroscent tubes = 384.0\n",
"number of twin fittings = 192.0\n",
"These can be arranged in 8 rows and 24 columns with space/height ratio = 1.\n"
]
}
],
"source": [
"'''Design a suitable lighting scheme for a factory 120 m * 40 m with a height of 7 m. Illumination required is 60 lux.\n",
"State the number, location and mounting height of 40 W fluorescent tubes giving 45 lm/W. Depreciation factor = 1.2;\n",
"utilization factor = 0.5 .'''\n",
"\n",
"A = 120.0*40.0 #m^2 (Area)\n",
"h = 7.0 #m (Height)\n",
"E = 60.0 #lm/m^2 (Illumination)\n",
"watt = 40.0 #W (Wattage of bulb)\n",
"eff = 45.0 #lm/W (Luminous efficiency)\n",
"dep = 1.2 # (Depriciation factor)\n",
"uti = 0.5 # (Utilization factor)\n",
"\n",
"#Total output flux is\n",
"flux = E*A/(uti*1/dep) #lm\n",
"#Flux per tube is\n",
"flux_tube = eff*watt #lm\n",
"\n",
"#Therefore,number of flouroscent tubes required\n",
"num = flux/flux_tube\n",
"print \"number of flouroscent tubes =\",round(num)\n",
"#For twin fittings\n",
"num = num/2\n",
"print \"number of twin fittings =\",round(num)\n",
"print \"These can be arranged in 8 rows and 24 columns with space/height ratio = 1.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.25 , PAGE NO :- 1920"
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Number of 500-W lamps required = 24.0\n",
"Number of 300-W lamps required = 49.0\n"
]
}
],
"source": [
"'''A drawing hall in an engineering college is to be provided with a lighting installation. The hall is 30 m * 20 m * 8 m (high).\n",
"The mounting height is 5 m and the required level of illumination is 144 lm/m^2. Using metal filament lamps, estimate\n",
"the size and number of single lamp luminaries and also draw their spacing layout. Assume :\n",
"Utilization coefficient = 0.6; maintenance factor = 0.75; space/height ratio=1 lumens/watt for 300-W lamp = 13,\n",
"lumens/watt for 500-W lamp = 16.'''\n",
"\n",
"\n",
"A = 30.0*20 #m^2 (Area)\n",
"E = 144.0 #lm/m^2 (Illumination)\n",
"coef = 0.6 # (Utilization coefficient)\n",
"mfac = 0.75 # (maintenance factor)\n",
"\n",
"\n",
"#The flux is given by\n",
"flux = E*A/(coef*mfac) #lm\n",
"\n",
"#Lumen output for 500-W lamp\n",
"I5 = 500.0*16 #lm\n",
"\n",
"#Lumen output for 500-W lamp\n",
"I3 = 300.0*13 #lm\n",
"\n",
"#No. of 500 W lamps required is\n",
"num5 = flux/I5\n",
"print \"Number of 500-W lamps required =\",round(num5)\n",
"\n",
"#No. of 300 W lamps required is\n",
"num3 = flux/I3\n",
"print \"Number of 300-W lamps required =\",round(num3)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.26 , PAGE NO :- 1920"
]
},
{
"cell_type": "code",
"execution_count": 15,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total number of lamps = 36.0\n",
"Wattage of each lamp is = 500.0 W.\n"
]
}
],
"source": [
"'''Estimate the number and wattage of lamps which would be required to illuminate a workshop space 60 * 15 metres by means of\n",
"lamps mounted 5 metres above the working plane. The average illumination required is about 100 lux.Coefficient of\n",
"utilization=0.4 ; Luminous efficiency=16 lm/W.Assume a spacing/height ratio of unity and a candle power depreciation of 20%.'''\n",
"\n",
"A = 60.0*15.0 #m^2\n",
"E = 100.0 #lm/m^2\n",
"coef = 0.4 # (coefficient of utilization)\n",
"lum = 16.0 #lm/W (luminous efficiency)\n",
"dep = 1+0.2 # (depriciation factor)\n",
"\n",
"#Total flux is given by\n",
"flux = E*A/(coef*1/dep) #lm\n",
"\n",
"#Total wattage required is\n",
"watt = flux/lum #W\n",
"\n",
"#Now,space/height ratio is 1.\n",
"h = 5.0 #m\n",
"\n",
"#Therefore along breadth , lamps are\n",
"num_b = 15.0/h\n",
"\n",
"#Therefore along length , lamps are\n",
"num_l = 60.0/h\n",
"\n",
"#Total number of lamps are\n",
"num_tot = num_b*num_l\n",
"\n",
"#Wattage of each lamp is\n",
"watt_each = watt/num_tot\n",
"\n",
"print \"Total number of lamps =\",num_tot\n",
"print \"Wattage of each lamp is =\",round(watt_each,-2),\"W.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 49.27 , PAGE NO :- 1921"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Number of 200-W lamps required are = 68.0\n",
"Number of 300-W lamps required are = 40.0\n",
"Number of 500-W lamps required are = 22.0\n"
]
}
],
"source": [
"'''A drawing hall 40 m * 25 m * 6 high is to be illuminated with metal-filament gas-filled lamps to an average illumination of\n",
"90 lm/m^2 on a working plane 1 metre above the floor.Estimate suitable number, size and mounting height of lamps. Sketch the\n",
"spacing layout.Assume coefficient of utilization of 0.5, depreciation factor of 1.2 and spacing/height ratio of 1.2\n",
"\n",
"Size of lamps : 200 W 300 W 500 W\n",
"Luminous efficiency (in lm/W) : 16 18 20 '''\n",
"\n",
"A = 40.0*25.0 #m^2 (Area)\n",
"E = 90.0 #lm/m^2 (Illumination)\n",
"coef = 0.5 # (Coefficient of utilization)\n",
"dep = 1.2 # (Depreciation factor)\n",
"\n",
"#Total flux required is\n",
"flux = E*A/(coef*1/1.2) #lumens\n",
"\n",
"#Lumen output of each 200-W lamp is\n",
"flux_200 = 200.0*16 #lumens\n",
"\n",
"#Lumen output of each 200-W lamp is\n",
"flux_300 = 300.0*18 #lumens\n",
"\n",
"#Lumen output of each 200-W lamp is\n",
"flux_500 = 500.0*20 #lumens\n",
"\n",
"#Number of 200-W lamps required is\n",
"num_200 = flux/flux_200\n",
"#Number of 200-W lamps required is\n",
"num_300 = flux/flux_300\n",
"#Number of 200-W lamps required is\n",
"num_500 = flux/flux_500\n",
"\n",
"print \"Number of 200-W lamps required are =\",round(num_200)\n",
"print \"Number of 300-W lamps required are =\",round(num_300)\n",
"print \"Number of 500-W lamps required are =\",round(num_500)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.28 , PAGE NO :- 1922"
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total wattage required is = 2019.23 W.\n"
]
}
],
"source": [
"'''A school classroom, 7 m * 10 m * 4 m high is to be illuminated to 135 lm/m^2 on the working plane. If the coefficient of\n",
"utilization is 0.45 and the sources give 13 lumens per watt,work out the total wattage required, assuming a depreciation factor\n",
"of 0.8 .Sketch roughly the plan of the room, showing suitable positions for fittings, giving reasons for the positions chosen.'''\n",
"\n",
"A = 7.0*10.0 #m^2 (Area)\n",
"E = 135.0 #lm/m^2 (Illumination)\n",
"coef = 0.45 # (Coefficient of utilization)\n",
"dep = 0.8 # (Depreciation factor)\n",
"eff = 13.0 #lm/W (luminous efficiency) \n",
"\n",
"#Total flux is\n",
"flux = E*A/(coef*dep) #lumens\n",
"\n",
"#Therefore,total wattage required is\n",
"watt = flux/eff #W\n",
"\n",
"print \"Total wattage required is =\",round(watt,2),\"W.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.29 , PAGE NO :- 1922"
]
},
{
"cell_type": "code",
"execution_count": 18,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Number of 100-W lamps required = 29.0\n",
"Number of 200-W lamps required = 13.0\n",
"Number of 300-W lamps required = 10.0\n",
"Number of 500-W lamps required = 5.0\n",
"Number of 1000-W lamps required = 2.0\n",
"If we take the mounting height of 5 m, then 300 W lamps would be suitable.\n",
"The No.of lamps required would be 10, arranged in two rows, each row having 5 lamps thus giving space/height ratio of 6/5\n"
]
}
],
"source": [
"'''A hall 30 m long and 12 m wide is to be illuminated and the illumination required is 50 lm/m2. Calculate the number,\n",
"the wattage of each unit and the location and mounting height of the units, taking a depreciation factor of 1.3 and\n",
"utilization factor of 0.5, given that the outputs of the different types of lamp are as under :\n",
"Watts : 100 200 300 500 1000\n",
"Lumens : 1615 3650 4700 9950 21500 '''\n",
"\n",
"A = 30.0*12.0 #m^2 (Area)\n",
"E = 50.0 #lm/m^2 (Illumination)\n",
"coef = 0.5 # (coefficient of utilization)\n",
"dep = 1/1.3 # (depreciation factor)\n",
"\n",
"#Total flux required is\n",
"flux = E*A/(coef*dep) #lumens\n",
"\n",
"#For 100-W lamps are used ,Number required\n",
"num_100 = flux/1615.0\n",
"#For 200-W lamps are used ,Number required\n",
"num_200 = flux/3650.0\n",
"#For 300-W lamps are used ,Number required\n",
"num_300 = flux/4700.0\n",
"#For 500-W lamps are used ,Number required\n",
"num_500 = flux/9950.0\n",
"#For 1000-W lamps are used ,Number required\n",
"num_1000 = flux/21500.0\n",
"\n",
"print \"Number of 100-W lamps required =\",round(num_100)\n",
"print \"Number of 200-W lamps required =\",round(num_200)\n",
"print \"Number of 300-W lamps required =\",round(num_300)\n",
"print \"Number of 500-W lamps required =\",round(num_500)\n",
"print \"Number of 1000-W lamps required =\",round(num_1000)\n",
"print \"If we take the mounting height of 5 m, then 300 W lamps would be suitable.\"\n",
"print \"The No.of lamps required would be 10, arranged in two rows, each row having 5 lamps thus giving space/height ratio of 6/5\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.30 , PAGE NO :- 1924"
]
},
{
"cell_type": "code",
"execution_count": 19,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Number of lamps required are = 10.0\n"
]
}
],
"source": [
"'''It is desired to floodlight the front of a building 42 m wide and 16 m high.Projectors of 30Â° beam spread and 1000-W lamps\n",
"giving 20 lumen/watt are available. If the desired level of illumination is 75 lm/m2 and if the projectors are to be located\n",
"at ground level 17 m away,design and show a suitable scheme. Assume the following :\n",
"Coefficient of utilization = 0.4 ; Depreciation factor = 1.3 ; Waste-light factor = 1.2. '''\n",
"\n",
"A = 42.0*16.0 #m^2 (Area)\n",
"E = 75.0 #lm/m^2 (Illumination)\n",
"W = 1.2 # (Waste-light factor)\n",
"coef = 0.4 # (Coefficient of utilization)\n",
"dep = 1/1.3 # (Depreciation factor)\n",
"eff = 20.0 #lm/W (luminous efficiency)\n",
"\n",
"#Total flux is\n",
"flux = E*A*W/(coef*dep) #lm/m^2\n",
"\n",
"#Lumen output of each 1000-W lamp is\n",
"flux_each = 1000.0*eff\n",
"\n",
"#Number of lamps required are\n",
"num = flux/flux_each\n",
"\n",
"print \"Number of lamps required are =\",round(num)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.31 , PAGE NO :- 1925"
]
},
{
"cell_type": "code",
"execution_count": 20,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The number of floodlight projectors required = 24.0\n"
]
}
],
"source": [
"'''Estimate the number of 1000-W floodlight projectors required to illuminate the up per 75 m of one face of a 96 m tower of \n",
"width 13 m if approximate initial average luminance is to be 6.85 cd/m2. The projectors are mounted at ground level 51m from base\n",
"of the tower.Utilization factor is = 0.2; reflection factor of wall = 25% and efficiency of each lamp = 18 lm/W.'''\n",
"\n",
"\n",
"A = 13.0*75.0 #m^2 (Area to be flood-lighted)\n",
"B = 6.85 #cd/m^2 (Average luminance)\n",
"watt = 1000.0 #W (Wattage floodlight projectors)\n",
"coef = 0.2 # (Utilization factor)\n",
"ref = 0.25 # (Reflection factor)\n",
"\n",
"#Illumination E = pi*B/reflection factor\n",
"E = 3.14*B/ref #lm/m^2\n",
"\n",
"#Therefore, total flux required is\n",
"flux = E*A #lm\n",
"\n",
"#Flux to be emmited by lamp is\n",
"lflux = flux/coef #lm\n",
"\n",
"#Flux from each lamp is\n",
"flux_each = 18.0*watt #lm\n",
"\n",
"#The number of floodlight projectors required are\n",
"num = lflux/flux_each\n",
"\n",
"print \"The number of floodlight projectors required =\",round(num)+1"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.32 , PAGE NO :- 1928"
]
},
{
"cell_type": "code",
"execution_count": 24,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Diameter d2 = 0.4 *d1.\n",
"Length l2 = 1.26 *l1.\n"
]
}
],
"source": [
"'''If the filament of a 32 candela, 100-V lamp has a length l and diameter d,calculate the length and diameter of the filament\n",
"of a 16 candela 200-V lamp,assuming that the two lamps run at the same intrinsic brilliance.'''\n",
"\n",
"\n",
"#As l*d is directly propotional luminous intensity .\n",
"#Let a = l1*d1 & b = l2*d2 . Then, l1*d1/l2*d2 = 32/16\n",
"\n",
"a_b = 32.0/16 # (= a/b = l1*d1/l2*d2)\n",
"\n",
"#As luminous intensity is directly propotional to power output\n",
"# 32 o< 100*I1 & 16 o< 200*I2\n",
"\n",
"I1_I2 = 32*200.0/(16*100.0) #( = I1/I2)\n",
"\n",
"\n",
"#Also , I o< d^3/2 \n",
"d1_d2 = (I1_I2)**(2.0/3) #( = d1/d2)\n",
"\n",
"print \"Diameter d2 = \",round(1/d1_d2,2),\"*d1.\"\n",
"\n",
"#As , d1_d2 = d1/d2\n",
"l1_l2 = a_b/d1_d2 #( = l1/l2)\n",
"print \"Length l2 = \",round(1/l1_l2,2),\"*l1.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.33 , PAGE NO :- 1928"
]
},
{
"cell_type": "code",
"execution_count": 25,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Diameter d2 = 0.00198 cm.\n",
"Length l2 = 126.0 cm.\n"
]
}
],
"source": [
"'''An incandescent lamp has a filament of 0.005 cm diameter and one metre length. It is required to construct another lamp of\n",
"similar type to work at double the supply voltage and give half the candle power. Assuming that the new lamp operates at the same\n",
"brilliancy,determine suitable dimensions for its filament.'''\n",
"\n",
"d1 = 0.005 #cm (diameter)\n",
"l1 = 100.0 #cm (length) \n",
"#As l*d is directly propotional luminous intensity .\n",
"#Let a = l1*d1 & b = l2*d2 . Then, l1*d1/l2*d2 = 2.0/1.0\n",
"\n",
"a_b = 2.0/1 # (= a/b = l1*d1/l2*d2)\n",
"\n",
"#As luminous intensity is directly propotional to power output\n",
"# I1 o< V1*i1 & I2 o< V2*i2\n",
"\n",
"I1_I2 = (2.0/1)*(2.0/1) #( = I1/I2)\n",
"\n",
"\n",
"#Also , I o< d^3/2 \n",
"d1_d2 = (I1_I2)**(2.0/3) #( = d1/d2)\n",
"d2 = d1/d1_d2 #cm\n",
"print \"Diameter d2 = \",round(d2,5),\"cm.\"\n",
"\n",
"#As , d1_d2 = d1/d2\n",
"l1_l2 = a_b/d1_d2 #( = l1/l2)\n",
"l2 = l1/l1_l2 #cm \n",
"print \"Length l2 = \",round(l2),\"cm.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 49.34 , PAGE NO :- 1928"
]
},
{
"cell_type": "code",
"execution_count": 52,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Hence expression for candle power is C = 1.155816e-09 *V^ 4.47\n",
"Change of candle power per volt = 268.1 cd/m.\n",
"% change in candle power for increase = 19.16\n",
"% change in candle power for decrease = -16.67\n"
]
}
],
"source": [
"'''A 60 candle power, 250-V metal filament lamp has a measured candle power of 71.5 candela at 260 V and 50 candela at 240 V.\n",
"(a) Find the constant for the lamp in the expression C = aV^b where C = candle power and V = voltage.\n",
"(b) Calculate the change of candle power per volt at 250 V. Determine the percentage variation of candle power due to a voltage\n",
"variation of Ã¦ 4% from the normal value. '''\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"\n",
"#Given expression is C = a*V^b\n",
"b = Symbol('b')\n",
"# 71.5/50 = (260/240)^b\n",
"lhs = 71.5/50.0\n",
"rhs = (260.0/240)**b\n",
"eq = Eq(lhs,rhs)\n",
"b = solve(eq)\n",
"b1 = b[0] #constant\n",
"\n",
"a = 71.5/(260.0)**b1\n",
"print \"Hence expression for candle power is C = %e\" %a,\"*V^\",round(b1,2)\n",
"\n",
"#Change of candle power per volt\n",
"# dC/dV = b*a*V^b\n",
"change = b1*a*((250.0)**(b1)) #cd/V\n",
"print \"Change of candle power per volt = \",round(change,1),\"cd/m.\"\n",
"\n",
"#When voltage is increases by 4% C2/C1 = (1.04)^b\n",
"per_change = ( (1.04)**b1 - 1 ) * 100\n",
"\n",
"print \"% change in candle power for increase =\",round(per_change,2)\n",
"\n",
"#When voltage is decreases by 4% C2/C1 = (0.96)^b\n",
"per_change = ( (0.96)**b1 - 1)* 100\n",
"\n",
"print \"% change in candle power for decrease =\",round(per_change,2)\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": []
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PKj„ìJìc4Ó4Ó;A TEXTBOOK OF ELECTRICAL TECHNOLOGY VOL III/chapter50.ipynb{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# CHAPTER 50 : Tariffs and Economic Considerations"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.1 , PAGE NO :- 1946"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total annual charge on installation = Rs. 28572.0\n"
]
}
],
"source": [
"'''Find the total annual charge on an installation costing Rs. 500,000 to buy and install, the estimated life being 30 years\n",
"and negligible scrap value. Interest is 4% compounded annually.'''\n",
"\n",
"import math as m\n",
"\n",
"Q = 5e+5 #(installation cost)\n",
"r = 0.04 #(rate of interest)\n",
"\n",
"q = Q*(r/(1+r))/(m.pow(1+r,30) - 1)\n",
"\n",
"#Hence, total annual charge on installation is\n",
"charge = r*Q + q # (As P=Q)\n",
"\n",
"print \"Total annual charge on installation = Rs.\",round(charge)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.2 , PAGE NO :- 1946"
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Annual Deposit for Straight-line depriciation method = Rs. 28500.0\n",
"Annual Deposit for Sink fund method = Rs. 22515.0\n"
]
}
],
"source": [
"'''A power plant having initial cost of Rs. 2.5 lakhs has an estimated salvage value of Rs. 30,000 at the end of its\n",
"useful life of 20 years. What will be the annual deposit necessary if it is calculated by :\n",
"(i) straight-line depreciation method.\n",
"(ii) sinking-fund method with compound interest at 7%.'''\n",
"\n",
"import math as m\n",
"\n",
"P = 250000.0 #Rs (Initial investment)\n",
"Q = P - 30000.0 #Rs (Replacement cost)\n",
"r = 0.07 # (Interest rate)\n",
"n = 20.0 # (Number of years)\n",
"\n",
"#(i)straight-line depriciation method\n",
"\n",
"#Annual depriciation\n",
"q= Q/n\n",
"\n",
"deposit1 = r*P + q\n",
"\n",
"\n",
"#(ii)sinking fund method\n",
"\n",
"#Annual depriciation\n",
"q= Q*(r/(1+r))/(m.pow(1+r,n) - 1)\n",
"\n",
"deposit2 = r*P + q\n",
"\n",
"print \"Annual Deposit for Straight-line depriciation method = Rs.\",round(deposit1)\n",
"print \"Annual Deposit for Sink fund method = Rs.\",round(deposit2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.3 , PAGE NO :- 1947"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Annual Deposit for Straight-line depriciation method = Rs. 300000.0\n",
"Annual Deposit for Sink fund method = Rs. 373375.0\n"
]
}
],
"source": [
"'''A plant initially costing Rs. 5 lakhs has an estimated salvage value of Rs. 1 lakh at the end of its useful life of 20 years.\n",
"What will be its valuation half-way through its life (a) on the basis of straight-line depreciation and\n",
"(b) on the sinking-fund basis at 8% compounded annually?'''\n",
"\n",
"import math as m\n",
"\n",
"P = 500000.0 #Rs (Initial investment)\n",
"Q = P - 100000.0 #Rs (Replacement cost)\n",
"r = 0.08 # (Interest rate)\n",
"n = 20.0 # (Number of years)\n",
"\n",
"#(i)straight-line depriciation method\n",
"\n",
"#Annual depriciation in 10 years\n",
"q= Q/2\n",
"\n",
"#Value after 10 years\n",
"value1 = P - q\n",
"\n",
"\n",
"#(ii)sinking fund method\n",
"\n",
"#Annual depriciation\n",
"q= Q*(r/(1+r))/(m.pow(1+r,n) - 1)\n",
"\n",
"#Value after 10 years i.e n/2 years\n",
"value2 = P - q*((1+r)/r)*(m.pow(1+r,n/2) - 1)\n",
"\n",
"print \"Annual Deposit for Straight-line depriciation method = Rs.\",round(value1)\n",
"print \"Annual Deposit for Sink fund method = Rs.\",round(value2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.4 , PAGE NO :- 1955"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total connected load = 2600.0 W\n",
"Monthly energy consumption = 252.0 kWh\n",
"Load factor = 0.23\n"
]
}
],
"source": [
"'''A consumer has the following connected load : 10 lamps of 60 W each and two heaters of 1000 W each. His maximum demand is 1500 W.\n",
"On the average, he uses 8 lamps for 5 hours a day and each heater for 3 hours a day. Find his total load, montly energy\n",
"consumption and load factor.'''\n",
"\n",
"\n",
"#Total connected load is\n",
"con_load = 10*60.0 + 2*1000.0 #W \n",
"\n",
"#Daily energy consumption is\n",
"enrgy = 8*60.0*5 + 2*1000.0*3 #Wh \n",
"\n",
"#Monthly energy consumption is\n",
"menrgy = enrgy*30.0/1000.0 #kWh\n",
"\n",
"#Load factor = Avg load/Max load is\n",
"lf = menrgy*1000/(1500.0*24*30)\n",
"\n",
"print \"Total connected load =\" ,con_load ,\"W\"\n",
"print \"Monthly energy consumption =\",menrgy ,\"kWh\"\n",
"print \"Load factor =\",round(lf,2) \n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.5 , PAGE NO :- 1955"
]
},
{
"cell_type": "code",
"execution_count": 19,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Capacity of substation = 1656.0 kW\n"
]
}
],
"source": [
"'''The load survey of a small town gives the following categories of expected loads.\n",
" Type Load in kW % D.F. Group D.F.\n",
"1. Residential lighting 1000 60 3\n",
"2. Commercial lighting 300 75 1.5\n",
"3. Street lighting 50 100 1.0\n",
"4. Domestic power 300 50 1.5\n",
"5. Industrial power 1800 55 1.2\n",
"What should be the kVA capacity of the S/S assuming a station p.f. of 0.8 lagging ?'''\n",
"\n",
"#Using Demand factor = Max demand/Connected load \n",
"#(i) Residential lighting\n",
"maxd1 = 1000.0*(60.0/100.0) #kW (Max demand)\n",
"grpd1 = maxd1/3.0 #kW (Max demand of group)\n",
"\n",
"#(ii) Commercial lighting\n",
"maxd2 = 300.0*(75.0/100.0) #kW (Max demand)\n",
"grpd2 = maxd2/1.5 #kW (Max demand of group)\n",
"\n",
"#(iii) Street lighting\n",
"maxd3 = 50.0*(100.0/100.0) #kW (Max demand)\n",
"grpd3 = maxd3/1.0 #kW (Max demand of group)\n",
"\n",
"#(iv) Domestic Power\n",
"maxd4 = 300.0*(50.0/100.0) #kW (Max demand)\n",
"grpd4 = maxd4/1.5 #kW (Max demand of group)\n",
"\n",
"#(v) Industrial Power \n",
"maxd5 = 1800.0*(55.0/100.0) #kW (Max demand)\n",
"grpd5 = maxd5/1.2 #kW (Max demand of group)\n",
"\n",
"#Total maximum demand is\n",
"tot = grpd1 + grpd2 + grpd3 + grpd4 + grpd5 #kW\n",
"\n",
"#KVA capacity of substation is\n",
"capacity = tot/0.8 #kW\n",
"\n",
"print \"Capacity of substation = \",round(capacity),\"kW\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.6 , PAGE NO :- 1956"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Load factor = 0.46\n",
"Daily bill of consumer = Rs 17.5\n"
]
}
],
"source": [
"'''A consumer has the following load-schedule for a day :\n",
"From midnight (12 p.m.) to 6 a.m. = 200 W ;\n",
"From 6 a.m. to 12 noon = 3000 W\n",
"From 12 noon to 1 p.m. = 100 W ;\n",
"From 1 p.m. to 4 p.m. = 4000 W\n",
"From 4 p.m. to 9 p.m. = 2000 W ;\n",
"From 9 p.m. to mid-night (12 p.m.) = 1000 W\n",
"\n",
"Find the load factor.\n",
"If the tariff is 50 paisa per kW of max. demand plus 35 paisa per kWh, find the daily bill the consumer has to pay.'''\n",
"\n",
"\n",
"maxp = 4000.0 #W (Maximum power)\n",
"\n",
"#Total consumption is\n",
"enrgy = 6*200.0 + 6*3000.0 + 1*100.0 + 3*4000.0 + 5*2000.0 + 3*1000.0 #W\n",
"\n",
"#Average power\n",
"pwer = enrgy/24 #kW\n",
"\n",
"#Daily load factor = Average Power/Max Power\n",
"lf = pwer/maxp\n",
"\n",
"#Max Demand charge.\n",
"charge1 = (maxp/1000)*0.5 #Rs\n",
"\n",
"#Consumption charge\n",
"charge2 = (enrgy/1000)*0.35 #Rs\n",
"\n",
"#Total Bill\n",
"bill = charge1 + charge2 #Rs\n",
"\n",
"print \"Load factor =\",round(lf,2)\n",
"print \"Daily bill of consumer = Rs\",round(bill,1)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.7 , PAGE NO :- 1956"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Demand factor = 0.47\n",
"Average Power = 7020.0 W.\n",
"Load factor = 0.35\n"
]
}
],
"source": [
"'''A generating station has a connected load of 43,000 kW and a maximum demand of 20,000 kW, the units generated being\n",
"61,500,000 for the year. Calculate the load factor and demand factor for this case.'''\n",
"\n",
"\n",
"maxp = 20000.0 #W (Maximum Demand)\n",
"conn = 43000.0 #W (Connected load)\n",
"\n",
"#Demand factor = Max demand/Connected load\n",
"df = maxp/conn\n",
"\n",
"#Average power\n",
"pwer = 61500000/(365*24) #W\n",
"\n",
"#Load factor = Avg Power/Max Power\n",
"lf = pwer/maxp\n",
"\n",
"print \"Demand factor =\",round(df,2)\n",
"print \"Average Power =\",round(pwer,2),\"W.\"\n",
"print \"Load factor =\",round(lf,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.8 , PAGE NO :- 1956"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"load factor = 0.21\n"
]
}
],
"source": [
"'''A 100 MW power station delivers 100 MW for 2 hours, 50 MW for 6 hours and is shut down for the rest of each day.\n",
"It is also shut down for maintenance for 45 days each year.Calculate its annual load factor.'''\n",
"\n",
"#Given\n",
"days = 365 - 45 # (Operating days)\n",
"maxp = 100.0 #MW (Maximum demand)\n",
"\n",
"#Energy consumption in a year\n",
"enrgy = (100.0*2 + 50.0*6)*days #MWh\n",
"\n",
"#Load factor = Total Energy Consumption/(Max demand*24*no of days)\n",
"lf = enrgy/(maxp*24*days)\n",
"\n",
"print \"load factor =\",round(lf,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.9 , PAGE NO :- 1956"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Cost per unit = 1.6 paise\n"
]
}
],
"source": [
"'''Differentiate between fixed and running charges in the operation of a power company.\n",
"Calculate the cost per kWh delivered from the generating station whose\n",
"\n",
"(i) capital cost = Rs. 10^6 , (ii) annual cost of fuel = Rs. 10^5 , (iii) wages and taxes = Rs. 5*10^5\n",
"(iv) maximum demand laod = 10,000 kW , (v) rate of interest and depreciation = 10% , (vi) annual load factor = 50%.\n",
"Total number of hours in a year is 8,760.'''\n",
"\n",
"#Given\n",
"maxp = 10000.0 #kW (Maximum demand)\n",
"Avgload = 0.5*maxp #kW (Average load)\n",
"\n",
"#Total energy consumption\n",
"enrgy = Avgload*8760.0 #kWh\n",
"\n",
"#Total Annual Charge\n",
"charge = (1.0e+5 + 5.0e+5) + (10.0e+5)*(10.0/100) #Rs\n",
"\n",
"#Cost per unit\n",
"cost = charge/enrgy*100.0 #paise\n",
"print \"Cost per unit =\",round(cost,2),\"paise\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.10 , PAGE NO :- 1957"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Peak demand of the city sub-station = 113.1 kW.\n"
]
}
],
"source": [
"'''A new colony of 200 houses is being established, with each house having an average connected load of 2 kW. The business\n",
"centre of the colony will have a total connected load of 200 kW. Find the peak demand of the city sub-station given the\n",
"following data.\n",
" Demand factor Group D.F. Peak D.F.\n",
"Residential load 50% 3.2 1.5\n",
"Business load 60% 1.4 1.2 .'''\n",
"\n",
"\n",
"#Given\n",
"dem_res = 0.5 # (demand factor Residential load)\n",
"dem_bus = 0.6 # (demand factor Business load)\n",
"gf_res = 3.2 # (Group demand factor Residential load)\n",
"gf_bus = 1.4 # (Group demand factor Business load)\n",
"pk_res = 1.5 # (Peak demand factor Residential load)\n",
"pk_bus = 1.2 # (Peak demand factor Business load)\n",
"\n",
"#######Residential Load##########\n",
"\n",
"\n",
"#Demand factor = Max demand/Connected load\n",
"\n",
"maxp_res = dem_res*2.0 #kW (Max demand of each house)\n",
"\n",
"#Group D.F = Sum of individual max demand/Actual max demand is\n",
"maxtot_res = (200*maxp_res)/gf_res #kW (Actual max demand)\n",
"\n",
"#Peak D.F = Max demand of consumer/Max demand during Peak time is\n",
"maxpeak_res = maxtot_res/pk_res #kW (Max demand during Peak time)\n",
"\n",
"\n",
"\n",
"########Business Load############\n",
"\n",
"\n",
"#Demand factor = Max demand/Connected load\n",
"\n",
"maxp_bus = dem_bus*200.0 #kW (Max demand of Commercial Load)\n",
"\n",
"#Group D.F = Sum of individual max demand/Actual max demand is\n",
"maxtot_bus = maxp_bus/gf_bus #kW (Actual max demand)\n",
"\n",
"#Peak D.F = Max demand of consumer/Max demand during Peak time is\n",
"maxpeak_bus = maxtot_bus/pk_bus #kW (Max demand during Peak time)\n",
"\n",
"\n",
"#Total Max demand during Peak time\n",
"peaktot = maxpeak_bus + maxpeak_res #kW\n",
"\n",
"print \"Peak demand of the city sub-station =\",round(peaktot,2),\"kW.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.11 , PAGE NO :- 1957"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Maximum demand of Substation = 3636.36 kW\n",
"kVA capacity of transformer = 4545.45 kVA\n"
]
}
],
"source": [
"'''In Fig. 50.6 is shown the distribution network from main sub-station. There are four feeders connected to each load centre\n",
"sub-station. The connected loads of different feeders and their maximum demands are as follows :\n",
"Feeder No. Connected load, kW Maximum Demand, kW\n",
"1. 150 125\n",
"2. 150 125\n",
"3. 500 350\n",
"4. 750 600\n",
"If the actual demand on each load centre is 1000 kW, what is the diversity factor on the feeders?If load centres B, C and D are\n",
"similar to A and the diversity factor between different load centres is 1.1, calculate the maximum demand of the main sub-station.\n",
"What would be the kVA capacity of the transformer required at the main sub-station if the overall p.f. at the main sub-station\n",
"is 0.8 ?'''\n",
"\n",
"#Maximum Demands\n",
"fed1 = 125.0 #kW (feeder 1)\n",
"fed2 = 125.0 #kW (feeder 2)\n",
"fed3 = 350.0 #kW (feeder 3)\n",
"fed4 = 600.0 #kW (feeder 4)\n",
"df_load = 1.1 # (Diversity Factor of load centres) \n",
"\n",
"#Diversity factor of feeders is\n",
"df_fed = (fed1 + fed2 + fed3 + fed4)/1000.0 #(df = Individual Max demand/Simultaneous Max demand)\n",
"\n",
"#Simultaneous Max demand of load centres = Total Individual Max demand/D.F of load centres is \n",
"maxp = (4*1000.0)/df_load #kW\n",
"\n",
"#KVA capacity of transformer is\n",
"capacity = maxp/0.8 #kW\n",
"\n",
"print \"Maximum demand of Substation =\",round(maxp,2),\"kW\"\n",
"print \"kVA capacity of transformer =\",round(capacity,2),\"kVA\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.12 , PAGE NO :- 1958"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Units generated per year = 4.809e+07 kWh\n",
"Diversity Factor = 1.29\n"
]
}
],
"source": [
"'''If a generating station had a maximum load for the year of 18,000 kW and a load factor of 30.5% and the maximum loads on\n",
"the sub-stations were 7,500, 5,000, 3,400, 4,600 and 2,800 kW, calculate the units generated for the year and the\n",
"diversity factor.'''\n",
"\n",
"\n",
"#Given\n",
"lf = 0.305 # (load factor)\n",
"maxp = 18000.0 #kW (Maximum demand)\n",
"sum_max = 7500.0+5000+3400+4600+2800.0 #kW (Sum of individual max demand)\n",
"\n",
"\n",
"\n",
"# Average Power consumption is (Using Load factor = Avg load/Max load)\n",
"avg_pwer = lf*maxp #kW\n",
"\n",
"#Energy generated per year\n",
"enrgy = avg_pwer*8760 #kWh\n",
"\n",
"#Diversity factor = Sum of Individual Max demand/Simultaneous Max demand\n",
"df = sum_max/maxp\n",
"\n",
"print 'Units generated per year = %1.3e kWh' %enrgy\n",
"print \"Diversity Factor =\",round(df,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.13 , PAGE NO :- 1958"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Maximum demand = 20.0 MW.\n",
"Annual energy supllied = 1.05e+08 kWh\n",
"Installed capacity = 30.000000 MW . With 2 -- 10 MW generators and 2 -- 5 MW generators.\n"
]
}
],
"source": [
"'''A power station is supplying four regions of load whose peak loads are 10MW,5 MW, 8 MW and 7 MW. the diversity factor of the \n",
"load at the station is 1.5 and the average annual load factor is 60%. Calculate the maximum demand on the station and the annual\n",
"energy supplied from the station. Suggest the installed capacity and the number of units taking all aspects into account.'''\n",
"\n",
"#Given\n",
"df = 1.5 # (diversity factor)\n",
"lf = 0.6 # (load factor)\n",
"sum_max = 10.0+5+8+7.0 #MW (Sum of individual max. load)\n",
"\n",
"#Max demand is (Using Diversity factor = Sum of individual max. demand/Simultaneous Max. demand)\n",
"maxdem = sum_max/df #MW\n",
"\n",
"#Average load is (Using Load factor = Avg load/Max. load)\n",
"avg_load = lf*maxdem #MW\n",
"\n",
"#Total Energy consumption in a year is\n",
"enrgy = avg_load*8760*1000.0 #kWh\n",
"\n",
"#Considering 50% more for future use\n",
"install = maxdem*1.5 #MW (installed capacity)\n",
"\n",
"print \"Maximum demand =\",round(maxdem,2),\"MW.\"\n",
"print \"Annual energy supllied = %1.2e kWh\" %enrgy\n",
"print \"Installed capacity = %f MW . With 2 -- 10 MW generators and 2 -- 5 MW generators.\" %install"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.14 , PAGE NO :- 1959"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Generating cost per unit is 3.15 paise.\n"
]
}
],
"source": [
"'''The capital cost of 30 MW generating station is Rs. 15*10^6. The annual expenses incurred on account of fuel, taxes,\n",
"salaries and maintenance amount to Rs. 1.25*10^6. The station operates at an annual load factor of 35%. Determine the\n",
"generating cost per unit delivered, assuming rate of interest 5% and rate of depreciation 6%.'''\n",
"\n",
"#Given\n",
"lf = 0.35 # (load factor)\n",
"maxp = 30.0*1000 #kW (max power)\n",
"mntnce = 1.25e+6 #Rs (maintainance cost) \n",
"cost = 15.0e+6 #Rs (capital cost) \n",
"\n",
"#load factor = Avg load/Max load . Therefore , Avg Load is\n",
"avg_load = lf*maxp #kW\n",
"\n",
"#Units produced per year\n",
"units = avg_load*8760 #kWh\n",
"\n",
"dep = 0.11*cost #(Depriciation + Interest cost)\n",
"\n",
"#Total cost per year\n",
"tot_cost = dep + mntnce #Rs\n",
"\n",
"#Cost per unit\n",
"cost_unit = tot_cost/units*100 #Paise/kWh\n",
"\n",
"print \"Generating cost per unit is\",round(cost_unit,2),\"paise.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.15 , PAGE NO :- 1959"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Fixed charge(100% load factor) = 4.11 paise.\n",
"Fixed charge(25% load factor) = 16.44 paise.\n"
]
}
],
"source": [
"'''A generating plant has a maximum capacity of 100 kW and costs Rs.300,000.The fixed charges are 12% consisting of 5% interest,\n",
"5% depreciation and 2% taxes etc. Find the fixed charges per kWh generated if load factor is (i) 100% and (ii) 25%.'''\n",
"\n",
"#Annual fixed charges are\n",
"charge1 = 300.0e+3*(12.0/100) #Rs\n",
"\n",
"#Number of kWh generated per year with 100% load factor are\n",
"units1 = 100.0*8760*1 #kWh\n",
"\n",
"#Number of kWh generated per year with 25% load factor are\n",
"units2 = 100.0*8760*0.25 #kWh\n",
"\n",
"#Fixed charges per unit is\n",
"\n",
"cost1 = charge1/units1*100 #paise (100% load factor)\n",
"cost2 = charge1/units2*100 #paise (25% load factor)\n",
"\n",
"print \"Fixed charge(100% load factor) = \",round(cost1,2),\"paise.\"\n",
"print \"Fixed charge(25% load factor) = \",round(cost2,2),\"paise.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.16 , PAGE NO :- 1959"
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"a = 80000.0\n",
"b = 900.0\n",
"c = 0.00025\n"
]
}
],
"source": [
"'''The annual working cost of a thermal station is represented by the formula Rs. (a + b kW + c kWh) where a, b and c are constants\n",
"for that particular station, kW is the total installed capacity and kWh is the energy produced per annum.\n",
"\n",
"Determine the values of a, b and c for a 100 MW station having annual load factor of 55% and for which\n",
"(i) capital cost of buildings and equipment is Rs. 90 million\n",
"(ii) the annual cost of fuel, oil,taxation and wages and salaries of operating staff is Rs. 1,20,000\n",
"(iii) interest and depreciation on buildings and equipment are 10% p.a.\n",
"(iv) annual cost of orginasation, interest on cost of site etc. is Rs. 80,000.'''\n",
"\n",
"\n",
"#Given\n",
"lf = 0.55 # (load factor)\n",
"cap = 100.0e+3 #kW (capacity of station)\n",
"\n",
"#As a represents the fixed-cost,b semi-fixed cost and c the running cost.\n",
"\n",
"a = 80000.0 #Rs (Given)\n",
"\n",
"\n",
"#Now, b*kW minimum demand = semi-fixed cost\n",
"b = 90.0e+6/cap\n",
"\n",
"\n",
"#Total units generated per annum = (max demand in kW)*(load factor)*8760 \n",
"units = cap*lf*8760 #kWh\n",
"\n",
"# Now, c*(no. of units) = 120000.0 . Therefore,\n",
"c = 120000.0/units\n",
"\n",
"print \"a =\",round(a,2)\n",
"print \"b = \",round(b,2)\n",
"print \"c = \",round(c,5)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.17 , PAGE NO :- 1960"
]
},
{
"cell_type": "code",
"execution_count": 18,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Limiting value of water evaporated = 6.0 kg.\n",
"Coal consumption per hour on no-load is = 750.0 kg.\n"
]
}
],
"source": [
"'''In a steam generating station, the relation between the water evaporated W kg and coal consumed C kg and power in kW\n",
"generated per 8-hour shift is as follows :\n",
"W = 28,000 + 5.4 kWh; C = 6000 + 0.9 kWh\n",
"What would be the limiting value of the water evaporated per kg of coal consumed as the station output increases ?\n",
"Also, calculate the amount of coal required per hour to keep the station running at no-load.'''\n",
"\n",
"\n",
"#For an 8-hour shift,Wt of water evaporated per kg of coal consumed is\n",
"# W/C = 28000 + 5.4kWh/6000 + 0.9kWh\n",
"\n",
"#Limiting value will approach\n",
"lim = 5.4/0.9\n",
"print \"Limiting value of water evaporated =\",lim,\"kg.\"\n",
"\n",
"#Since at no load there is no generation of power\n",
"#Coal consumption per 8 hour shift is 6000.0 kg\n",
"#Coal consumption per hour on no-load is\n",
"cons = 6000.0/8\n",
"print \"Coal consumption per hour on no-load is =\",round(cons,2),\"kg.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.18 , PAGE NO :- 1960"
]
},
{
"cell_type": "code",
"execution_count": 19,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Cost per kWh(unit) delivered = 4.28 paise.\n"
]
}
],
"source": [
"'''Estimate the generating cost per kWh delivered from a generating station from the following data :\n",
"Plant capacity = 50 MW ; annual load factor = 40%; capital cost = Rs. 3.60 crores; annual cost\n",
"of wages, taxation etc. = Rs. 4 lakhs; cost of fuel, lubrication, maintenance etc. = 2.0 paise per kWh\n",
"generated, interest 5% per annum, depreciation 5% per annum of initial value.'''\n",
"\n",
"#Given\n",
"maxp = 50.0e+3 #kW (Plant capacity)\n",
"lf = 0.4 # (load factor)\n",
"\n",
"#Average power over a year is = Max. power * load factor\n",
"\n",
"avg_p = maxp*lf #kW\n",
"\n",
"#Units produced per year\n",
"units = avg_p*8760 #kWh\n",
"\n",
"#Depriciation + Interest (5 + 5 = 10% of capital cost)\n",
"charge1 = (10.0/100)*3.60e+7 #Rs\n",
"\n",
"#Annual wage and taxation\n",
"charge2 = 4.0e+5 #Rs\n",
"\n",
"#Total cost/year =\n",
"tot_charge= charge1 + charge2 #Rs\n",
"\n",
"\n",
"#Cost per unit\n",
"cost = tot_charge/units*100 #paise\n",
"\n",
"#Cost per kWh(unit) delivered = base cost + maintenance cost\n",
"tot_cost = cost + 2.0 #paise\n",
"\n",
"print \"Cost per kWh(unit) delivered =\",round(tot_cost,2),\"paise.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.19 , PAGE NO :- 1960"
]
},
{
"cell_type": "code",
"execution_count": 20,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Cost per kWh generated = 5.67 Paise.\n",
"Cost per kW is = 268.2 Rs.\n"
]
}
],
"source": [
"'''The following data relate to a 1000 kW thermal station :\n",
"Cost of Plant = Rs. 1,200 per kW Interest, insurance and taxes = 5% p.a.\n",
"Depreciation = 5% p.a. Cost of primary distribution system = Rs. 4,00,000\n",
"Interest, insurance, taxes and depreciaton = 5% p.a. Cost of coal including transportation = Rs. 40 per tonne\n",
"Operating cost = Rs. 4,00,000 p.a. Plant maintenance cost : fixed = Rs. 20,000 p.a.\n",
"variable = Rs. 30,000 p.a. Installed plant capacity = 10,000 kW\n",
"Maximum demand = 9,000 kW Annual load factor = 60%\n",
"Consumption of coal = 25,300 tonne\n",
"Find the cost of power generation per kilowatt per year, the cost per kilowatt-hour generated\n",
"and the total cost of generation per kilowatt-hour. Transmission/primary distribution is chargeable\n",
"to generation.'''\n",
"\n",
"\n",
"#Given\n",
"maxp = 9000.0 #kW (Maximum demand)\n",
"lf = 0.6 # (load factor)\n",
"#Fixed cost\n",
"costp = 400000.0 #Rs (cost of primary distribution)\n",
"costplant = 1200.0*10000.0 #Rs (Total cost of plant)\n",
"mntnce = 20000.0 #Rs (Maintenance cost)\n",
"\n",
"#Variable cost\n",
"fuel = 25300.0*40.0 #Rs (fuel cost)\n",
"var_mntnce = 30000.0 #Rs (variable maintenance cost)\n",
"op_cost = 400000.0 #Rs (operating cost)\n",
"\n",
"\n",
"#Total fixed cost is\n",
"fixed = costplant*(10.0/100) + mntnce + costp*(5.0/100) #Rs \n",
"\n",
"#Total variable cost is\n",
"var = op_cost + fuel + var_mntnce #Rs\n",
"\n",
"#Total cost is\n",
"cost_tot = fixed + var #Rs\n",
"\n",
"#Average demand (kWh) = Load factor * Max demand *(no. of hours in a year)\n",
"avg_p = lf*maxp*8760 #kWh\n",
"\n",
"#Cost per kWh generated = Total annual cost/units produced in a year\n",
"cost = cost_tot/avg_p*100.0 #paise\n",
"\n",
"#Since installed capacity is 10000.0 kW , cost per kW is\n",
"cost_kw = cost_tot/10000.0 #Rs\n",
"\n",
"print \"Cost per kWh generated = \",round(cost,2),\"Paise.\"\n",
"print \"Cost per kW is =\",round(cost_kw,2),\"Rs.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.20 , PAGE NO :- 1961"
]
},
{
"cell_type": "code",
"execution_count": 21,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Annual Bill = 13768.0 Rs.\n",
"Cost/kWh (i) = 7.81 paise .\n",
"Cost/kWh (ii) = 7.81 paise .\n",
"Cost/kWh (iii)= 9.07 paise .\n"
]
}
],
"source": [
"'''Consumer has an annual consumption of 176,400 kWh. The charge is Rs. 120 per kW of maximum demand plus 4 paisa\n",
"per kWh.\n",
"(i) Find the annual bill and the overall cost per kWh if the load factor is 36%.\n",
"(ii) What is the overall cost per kWh, if the consumption were reduced 25% with the same load factor ?\n",
"(iii) What is the overall cost per kWh, if the load factor is 27% with the same consumption as in (i) '''\n",
"\n",
"\n",
"#Given\n",
"an_con = 176400.0 #kWh (Annual consumption)\n",
"lf = 0.36 # (load factor)\n",
"\n",
"###(i)###\n",
"#Average demand = annual consumption/No of hours in a year is\n",
"avg_dem = an_con/8760 #kW\n",
"\n",
"#Maximum demand = Avg demand/Load factor\n",
"maxp = avg_dem/lf #kW\n",
"\n",
"#Total bill = charge due to maximum demand + 4 paise per kWh\n",
"bill = maxp*120.0 + 0.04*an_con #Rs\n",
"\n",
"#Cost/kWh = annual bill/annual consumption(in kWh)\n",
"cost = bill/an_con*100 #paise\n",
"\n",
"print \"Annual Bill = \",round(bill),\"Rs.\"\n",
"print \"Cost/kWh (i) = \",round(cost,2),\"paise .\"\n",
"\n",
"\n",
"\n",
"###(ii)###\n",
"an_con = 176400.0 * (0.75) #kWh (Annual consumption is 75% only)\n",
"lf = 0.36 # (load factor)\n",
"\n",
"#Average demand = annual consumption/No of hours in a year is\n",
"avg_dem = an_con/8760 #kW\n",
"\n",
"#Maximum demand = Avg demand/Load factor\n",
"maxp = avg_dem/lf #kW\n",
"\n",
"#Total bill = charge due to maximum demand + 4 paise per kWh\n",
"bill = maxp*120.0 + 0.04*an_con #Rs\n",
"\n",
"#Cost/kWh = annual bill/annual consumption(in kWh)\n",
"cost = bill/an_con*100 #paise\n",
"\n",
"print \"Cost/kWh (ii) = \",round(cost,2),\"paise .\"\n",
"\n",
"\n",
"###(iii)###\n",
"an_con = 176400.0 #kWh (Annual consumption)\n",
"lf = 0.27 # (changed load factor)\n",
"\n",
"#Average demand = annual consumption/No of hours in a year is\n",
"avg_dem = an_con/8760 #kW\n",
"\n",
"#Maximum demand = Avg demand/Load factor\n",
"maxp = avg_dem/lf #kW\n",
"\n",
"#Total bill = charge due to maximum demand + 4 paise per kWh\n",
"bill = maxp*120.0 + 0.04*an_con #Rs\n",
"\n",
"#Cost/kWh = annual bill/annual consumption(in kWh)\n",
"cost = bill/an_con*100 #paise\n",
"\n",
"print \"Cost/kWh (iii)= \",round(cost,2),\"paise .\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.21 , PAGE NO :- 1964"
]
},
{
"cell_type": "code",
"execution_count": 22,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"load factor = 0.71\n",
"capacity factor = 0.61\n",
"daily fuel requirement = 125.84 tonne.\n"
]
}
],
"source": [
"'''A power station has a load cycle as under :\n",
"\n",
"260 MW for 6 hr ; 200 MW for 8 hr ; 160 MW for 4 hr ; 100 MW for 6 hr\n",
"If the power station is equipped with 4 sets of 75 MW each, calculate the load factor and the\n",
"capacity factor from the above data. Calculate the daily fuel requirement if the calorific value of the\n",
"oil used were 10,000 kcal/kg and the average heat rate of the station were 2,860 kcal/kWh.'''\n",
"\n",
"\n",
"#Given\n",
"maxp = 260.0 #MW (Maximum demand)\n",
"heat = 2860.0 #kcal/kWh (Heat rate of station)\n",
"cal = 100000.0 #kcal/kg (calorific value)\n",
"\n",
"#Energy consumed per day\n",
"enrgy = maxp*6 + 200.0*8 + 160.0*4 + 100.0*6 #MW\n",
"\n",
"#Daily load factor = Energy consumed/Max demand*24 is\n",
"lf = enrgy/(maxp*24)\n",
"\n",
"#Installed capacity is\n",
"cap = (75.0)*4 #MW\n",
"\n",
"#Capacity factor = Energy consumed/Installed capacity*24 is\n",
"cf = enrgy/(cap*24)\n",
"\n",
"#Heat produced in station in kcal\n",
"heat_p = heat*enrgy*1000.0 #kcal\n",
"\n",
"#Daily fuel requirement is\n",
"fuel = (heat_p/cal)/1000.0 #tonne\n",
"\n",
"print \"load factor =\",round(lf,2)\n",
"print \"capacity factor =\",round(cf,2)\n",
"print \"daily fuel requirement =\",round(fuel,2),\"tonne.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.22 , PAGE NO :- 1964"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"load factor = 0.57\n",
"utilization factor = 0.73\n"
]
}
],
"source": [
"'''A generating station has two 50 MW units each running for 8,500 hours in a year and one 30 MW unit running for\n",
"1,250 hours in one year. The station output is 650 * 10^6 kWh per year. Calculate\n",
"(i) station load factor, (ii) the utilization factor.'''\n",
"\n",
"#Given\n",
"\n",
"#Output of power station(Energy used)\n",
"enrgy = 650.0e+6 #kWh\n",
"\n",
"#Maximum demand\n",
"maxp = 2*50.0 + 30.0 #MW\n",
"maxp = maxp*1000.0 #kW\n",
"\n",
"#Total energy produced in kWh is\n",
"enrgy_pro = 2*(50.0e+3)*(8500.0) + (30.0e+3)*1250.0 #kWh\n",
"\n",
"#Annual load factor = Energy consumed/Max demand*8760 is\n",
"lf = enrgy/(maxp*8760)\n",
"\n",
"#Utilization factor = Energy consumed/Energy produced\n",
"uf = enrgy/enrgy_pro\n",
"\n",
"print \"load factor = \",round(lf,2)\n",
"print \"utilization factor =\",round(uf,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.23 , PAGE NO :- 1965"
]
},
{
"cell_type": "code",
"execution_count": 24,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Installed Capacity = 50.0 MW\n",
"Plant factor = 0.48\n",
"Maximum demand = 40000.0 kW\n",
"Load factor = 0.6\n",
"Utilization factor = 0.8\n"
]
}
],
"source": [
"'''The yearly duration curve of a certain plant may be considered as a straight line from 40,000 kW to 8,000 kW.\n",
"To meet this load, three turbo-generators, two rated at 20,000 kW each and one at 10,000 kW are installed.\n",
"Determine (a) the installed capacity, (b) plant factor,(c) maximum demand, (d) load factor and (e) utilization factor.'''\n",
"\n",
"\n",
"#(a) Installed capacity is\n",
"cap = 2*20.0 + 10.0 #MW\n",
"\n",
"#(b)plant factor\n",
"\n",
"#Average demand\n",
"avg_dem = (40000.0 + 8000.0)/2 #kW\n",
"\n",
"#Plant factor = Average demand/Installed capacity\n",
"pf = avg_dem/(cap*1000.0)\n",
"\n",
"#(c) Maximum demand\n",
"maxp = 40000.0 #kW (Given)\n",
"\n",
"#(d)load factor = Avg demand/Max demand\n",
"lf = avg_dem/maxp\n",
"\n",
"#(e)Utilization factor = Max demand/Plant capacity\n",
"uf = maxp/(cap*1000.0)\n",
"\n",
"print \"Installed Capacity = \",round(cap),\"MW\"\n",
"print \"Plant factor = \",round(pf,2)\n",
"print \"Maximum demand = \",maxp,\"kW\"\n",
"print \"Load factor = \",round(lf,2)\n",
"print \"Utilization factor = \",round(uf,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.24 , PAGE NO :- 1965"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Maximum demand of Run of River = 40.0 MW.\n",
"Maximum demand of Reservoir = 40.0 MW.\n",
"Maximum demand of Steam = 80.0 MW.\n",
"Load factor(Run of river) = 1.0\n",
"Load factor(Reservoir) = 0.25\n",
"Load factor(Steam) = 0.88\n"
]
}
],
"source": [
"'''The load duration curve of a system is as shown in Fig. 50.10. The system is\n",
"supplied by three stations; a steam station, a run-of-river station and a reservoir hydro-electric\n",
"station. The ratios of number of units supplied by the three stations are as below :\n",
"Steam : Run of river : Reservoir\n",
"7 : 4 : 1\n",
"The run-of-river station is capable of generating power continuosuly and works as a peak load\n",
"station. Estimate the maximum demand on each station and also the load factor of each station.'''\n",
"\n",
"\n",
"from sympy import Symbol,Eq,solve\n",
"\n",
"#Let 100% time be 1 year = 8760 hours\n",
"\n",
"#Area under curve is energy produced in a year\n",
"enrgy = (0.5*(160-80)*8760.0 + 80*8760.0)*1000.0 #kWh\n",
"\n",
"#For the given ratio ,units supplied by each station is\n",
"steam = enrgy*(7.0/12) #kWh (Steam station)\n",
"runriv = enrgy*(4.0/12) #kWh (Run of river station)\n",
"reservoir = enrgy*(1.0/12) #kWh (Reservoir)\n",
"\n",
"#Maximum demand of run of river is\n",
"max_ror = runriv/(8760.0*1000.0) #MW (From figure)\n",
"\n",
"#For Maximum demand of Reservoir,\n",
"x = Symbol('x') #(As shown in figure)\n",
"\n",
"#No of hours reservoir will work is\n",
"y = x/80*(8760)\n",
"\n",
"#Area under the reservoir curve(Energy produced) is\n",
"area = 0.5*(y)*(x)*1000.0 #kWh\n",
"\n",
"eq = Eq(area,reservoir)\n",
"x = solve(eq)\n",
"max_res = x[1] #MW\n",
"#Maximum demand of steam station is\n",
"max_steam = 160.0 - max_ror - max_res #MW\n",
"\n",
"#Load factors of each station load factor = Energy produced/Max demand*8760\n",
"\n",
"#Run of River operates continously\n",
"lf_ror = 1.0\n",
"#Load factor Reservoir\n",
"lf_res = reservoir/(max_res*1000*8760)\n",
"#Load factor Steam\n",
"lf_steam = steam/(max_steam*1000*8760)\n",
"\n",
"\n",
"print \"Maximum demand of Run of River = \",round(max_ror,2),\"MW.\"\n",
"print \"Maximum demand of Reservoir = \",round(max_res,2),\"MW.\"\n",
"print \"Maximum demand of Steam = \",round(max_steam,2),\"MW.\"\n",
"\n",
"print \"Load factor(Run of river) = \",round(lf_ror,2)\n",
"print \"Load factor(Reservoir) = \",round(lf_res,2)\n",
"print \"Load factor(Steam) = \",round(lf_steam,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.25 , PAGE NO :- 1966"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Load factor = 0.34\n"
]
}
],
"source": [
"'''A load having a maximum value of 150 MW can be supplied by either a hydroelectric plant or a steam power plant.\n",
"The costs are as follows :\n",
"Capital cost of steam plant = Rs. 700 per kW installed\n",
"Capital cost of hydro-electric plant = Rs. 1,600 per kW installed\n",
"Operating cost of steam plant = Rs. 0.03 per kWh\n",
"Operating cost of hydro-electric plant = Rs. 0.006 per kWh\n",
"Interest on capital cost 8 per cent. Calculate the minimum load factor above which the hydroelectric plant will be\n",
"more economical.'''\n",
"\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"\n",
"#Given\n",
"maxp = 150.0e+3 #kW\n",
"\n",
"#Let x be the total number of units generated per annum\n",
"x = Symbol('x')\n",
"\n",
"## Steam plant ##\n",
"\n",
"#Capital cost is\n",
"cap_stm = 700.0*(maxp) #Rs\n",
"#Interest charges\n",
"charge1 = cap_stm*(8.0/100) #Rs \n",
"\n",
"#Fixed charge/unit\n",
"fxd_stm = charge1/x #Rs \n",
"\n",
"#Total cost per kWh for steam plant is\n",
"cost_stm = fxd_stm + 0.03 #Rs\n",
"\n",
"\n",
"## Hydro-electric plant ##\n",
"\n",
"#Capital cost is\n",
"cap_hyd = 1600.0*(maxp) #Rs\n",
"#Interest charges\n",
"charge1 = cap_hyd*(8.0/100) #Rs \n",
"\n",
"#Fixed charge/unit\n",
"fxd_hyd = charge1/x #Rs \n",
"\n",
"#Total cost per kWh for steam plant is\n",
"cost_hyd = fxd_hyd + 0.006 #Rs\n",
"\n",
"\n",
"#Let us find out x when both cost will be equal\n",
"eq = Eq(cost_hyd,cost_stm)\n",
"x = solve(eq)\n",
"x1 = x[0]\n",
"\n",
"#If units generated are more than x1 then cost of hydro-electricity will be cheaper\n",
"\n",
"#Load factor = Energy produced/Max demand*8760\n",
"\n",
"lf = x1/(maxp*8760)\n",
"\n",
"print \"Load factor =\",round(lf,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.26 , PAGE NO :- 1967"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Overall cost/kWh for (a) = 10.97 paise.\n",
"Overall cost/kWh for (b) = 10.71 paise.\n",
"Overall cost/kWh for (c) = 11.21 paise.\n",
"Overall cost/kWh for steam station with 90% load factor = 6.9 paise.\n",
"Overall cost/kWh for steam station with 90% load factor = 4.87 paise.\n"
]
}
],
"source": [
"'''A power system having maximum demand of 100 MW has a load 30% and is to be supplied by either of the following schemes :\n",
"(a) a steam station in conjunction with a hydro-electric station, the latter supplying 100 * 10^6\n",
" units per annum with a max. output of 40 MW,\n",
"(b) a steam station capable of supply the whole load,\n",
"(c) a hydro station capable of supplying the whole load,\n",
"Compare the overall cost per unit generated assuming the following data :\n",
" Steam Hydro\n",
"Capital cost / kW Rs. 1,250 Rs. 2,500\n",
"Interest and depreciation on the capital cost 12% 10%\n",
"Operating cost/kWh 5 paise 1.5 paise\n",
"Transmission cost/kWh Negligible 0.2 paise\n",
"Show how overall cost would be affected in case (ii) and (iii) above if the system load factor\n",
"were improved to 90 per cent.'''\n",
"\n",
"\n",
"#Average power consumption is\n",
"avg_pwr = 100.0*0.3*1000.0 #kW\n",
"\n",
"#Units generated in a year = Power*time(no of hours in a year)\n",
"units = avg_pwr*(8760) #kWh\n",
"\n",
"#----------------------------------------------------------------------------------------------#\n",
"#(a)Steam station in conjunction with Hydro station\n",
"\n",
"hyd_units = 100.0e+6 #kWh (units supplied by hydro station)\n",
"stm_units = units - hyd_units #kWh (units supplied by steam station)\n",
"hyd_max = 40.0e+3 #kW (max. output of Hydro station)\n",
"stm_max = 100.0e+3 - hyd_max #kW (max. output of Steam station)\n",
"\n",
"\n",
"#(i)Steam station\n",
"cap_cost = stm_max*(1250) #Rs (capital cost)\n",
"fxd_charge = (12.0/100)*cap_cost #Rs (fixed charge: interest & depreciation)\n",
"op_charge = (5.0/100)*stm_units #Rs (operating charge)\n",
"stm_cost = fxd_charge + op_charge #Rs (total steam charges)\n",
"\n",
"#(ii)Hydro station\n",
"cap_cost = hyd_max*(2500) #Rs (capital cost)\n",
"fxd_charge = (10.0/100)*cap_cost #Rs (fixed charges: interest & depreciation)\n",
"op_charge = (1.5/100)*hyd_units #Rs (operating charges)\n",
"tr_charge = (0.2/100)*hyd_units #Rs (transmission charges)\n",
"hyd_cost = fxd_charge + op_charge + tr_charge #Rs (total hydro charges)\n",
"\n",
"#Total cost\n",
"tot_cost = stm_cost + hyd_cost #Rs (total cost) \n",
"#Overal cost per kWh\n",
"cost_kwh = tot_cost/units*100 #paisa\n",
"print \"Overall cost/kWh for (a) = \",round(cost_kwh,2),\"paise.\"\n",
"\n",
"#------------------------------------------------------------------------------------------------#\n",
"#(b)Steam station alone\n",
"stm_max = 100.0e+3 #kW (max. output of Steam station)\n",
"stm_units = units #kWh (units supplied by Steam station)\n",
"\n",
"cap_cost = stm_max*(1250) #Rs (capital cost)\n",
"fxd_charge = (12.0/100)*cap_cost #Rs (fixed charge: interest & depreciation)\n",
"op_charge = (5.0/100)*stm_units #Rs (operating charge)\n",
"stm_cost = fxd_charge + op_charge #Rs (total steam charges)\n",
"stm_fxdkWh = fxd_charge/stm_units*100#paise (fixed charge per unit)\n",
"#Overal cost per kWh\n",
"cost_kwh = stm_cost/stm_units*100 #paisa\n",
"print \"Overall cost/kWh for (b) = \",round(cost_kwh,2),\"paise.\"\n",
"\n",
"#------------------------------------------------------------------------------------------------#\n",
"#(c)Hydro station alone\n",
"hyd_max = 100.0e+3 #kW (max. output of Hydro station)\n",
"hyd_units = units #kWh (units supplied by Hydro station)\n",
"\n",
"cap_cost = hyd_max*(2500) #Rs (capital cost)\n",
"fxd_charge = (10.0/100)*cap_cost #Rs (fixed charges: interest & depreciation)\n",
"op_charge = (1.5/100)*hyd_units #Rs (operating charges)\n",
"tr_charge = (0.2/100)*hyd_units #Rs (transmission charges)\n",
"hyd_cost = fxd_charge + op_charge + tr_charge #Rs (total hydro charges)\n",
"hyd_fxdkWh = fxd_charge/hyd_units*100#paise (fixed charge per unit)\n",
"#Overall cost per kWh\n",
"cost_kwh = hyd_cost/hyd_units*100 #paisa\n",
"print \"Overall cost/kWh for (c) = \",round(cost_kwh,2),\"paise.\"\n",
"\n",
"#-------------------------------------------------------------------------------------------------#\n",
"#(d)As load factor is made 3 times. No. of units generated are 3 times and therefore fixed price\n",
"#decrease to 1/3 times the previous price\n",
"\n",
"#(i)Steam station\n",
"cost_kwh = stm_fxdkWh/3 + 5.0 #paise (new total steam cost per kWh)\n",
"print \"Overall cost/kWh for steam station with 90% load factor = \",round(cost_kwh,2),\"paise.\"\n",
"#(ii)Hydro station\n",
"cost_kwh = hyd_fxdkWh/3 + 1.5 + 0.2 #paise (new total hydro cost per kWh)\n",
"print \"Overall cost/kWh for steam station with 90% load factor = \",round(cost_kwh,2),\"paise.\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.27 , PAGE NO :- 1969"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Generation cost is = Rs ( 154.25 kW + 0.0119 )kWh.\n"
]
}
],
"source": [
"'''The capital cost of a hydro-power station of 50,000 kW capacity is Rs. 1,200 per kW. The annual charge on investment\n",
"including depreciation etc. is 10%. A royality of Rs. 1 per kW per year and Rs. 0.01 per kWh generated is to be paid for\n",
"using the river water for generation of power. The maximum demand is 40,000 kW and the yearly load factor is 80%.\n",
"Salaries,maintenance charges and supplies etc. total Rs. 6,50,000. If 20% of this expense is also chargeable\n",
"as fixed charges, determine the generation cost in the form of A per kW plus B per kWh.'''\n",
"\n",
"\n",
"maxp = 40000.0 #kW (Maximum demand)\n",
"#Capital cost of station = Cost/kW *(Capacity of plant in kW)\n",
"cap_cost = 1200.0*50000.0 #Rs\n",
"\n",
"#Annual charge on investment includin depriciation\n",
"anl_charge = (10.0/100)*cap_cost #Rs\n",
"\n",
"#Total running charge\n",
"run_charge = (80.0/100)*650000.0 #Rs \n",
"\n",
"#Fixed charges\n",
"fxd_charge = (20.0/100)*650000.0 #Rs\n",
"\n",
"#Cost per Max demand kW due to fixed charge is\n",
"cost_md1 = (anl_charge + fxd_charge)/maxp #Rs\n",
"#Cost per Max demand kW due to royalty is\n",
"cost_md2 = 1.0 #Rs\n",
"#Total Cost per Max demand kW is\n",
"cost_md = cost_md1 + cost_md2 #Rs\n",
"\n",
"\n",
"#Total number of unit generated per annum = Max demand*load factor*(No of hours)\n",
"units = maxp*0.8*8760 #kWh\n",
"\n",
"#Cost per unit due to running charges\n",
"cost_unit = run_charge/units #Rs\n",
"#Royalty cost\n",
"cost_roy = 0.01 #Rs\n",
"#Total cost per unit is\n",
"cost_tot = cost_unit + cost_roy #Rs\n",
"\n",
"print \"Generation cost is = Rs (\",round(cost_md,2),\"kW + \",round(cost_tot,4),\")kWh.\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 50.28 , PAGE NO :- 1969"
]
},
{
"cell_type": "code",
"execution_count": 25,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"load factor = 0.472\n",
"generating cost = Rs. 41314200.0\n"
]
}
],
"source": [
"'''The capital costs of steam and water power stations are Rs. 1,200 and Rs. 2,100 per kW of the installed capacity.\n",
"The corresponding running costs are 5 paise and 3.2 paise per kWh respectively.The reserve capacity in the case of the\n",
"steam station is to be 25% and that for the water power station is to be 33.33% of the installed capacity.At what\n",
"load factor would the overall cost per kWh be the same in both cases ? Assume interest and depreciation charges\n",
"on the capital to be 9% for the thermal and 7.5% for the hydro-electric station. What would be the cost of generating\n",
"500 million kWh at this load factor ? '''\n",
"\n",
"from sympy import Symbol,Eq,solve\n",
"\n",
"#Let x be the maximum demand in kWh and y the load factor\n",
"x = Symbol('x')\n",
"y = Symbol('y')\n",
"#Total number of units produced\n",
"units = x*y*8760.0 #kWh\n",
"\n",
"\n",
"###### Steam station ######\n",
"\n",
"#Installed capacity\n",
"capacity = 1.25*x #kW (including reserve capacity)\n",
"#Capital cost\n",
"cap_cost = capacity*1200.0 #Rs\n",
"#Annual depreciation\n",
"anl_charge = (9.0/100)*cap_cost #Rs\n",
"#Annual running cost\n",
"run_charge = (8760*x*y)*(5.0/100) #Rs\n",
"#Total annual cost\n",
"cost_stm = run_charge + anl_charge\n",
"#Total cost/unit\n",
"stmcost_unt = cost_stm/units\n",
"\n",
"###### Hydro station ######\n",
"\n",
"#Installed capacity\n",
"capacity = 1.33*x #kW (including reserve capacity)\n",
"#Capital cost\n",
"cap_cost = capacity*2100.0 #Rs\n",
"#Annual depreciation\n",
"anl_charge = (7.5/100)*cap_cost #Rs\n",
"#Annual running cost\n",
"run_charge = (8760*x*y)*(3.2/100) #Rs\n",
"#Total annual cost\n",
"cost_hyd = run_charge + anl_charge #Rs\n",
"#Total cost/unit\n",
"hydcost_unt = cost_hyd/units #Rs\n",
"\n",
"#To solve we will assume a value of x\n",
"x = 10\n",
"eq = Eq(hydcost_unt,stmcost_unt)\n",
"y = solve(eq)\n",
"y1 = y[0]\n",
"for key in y1:\n",
" y1 = y1.get(key);\n",
"#Maximum demand is\n",
"x = 500.0e+6/(8760.0*y1) #kW\n",
"\n",
"#Calculating cost\n",
"capacity = 1.25*x\n",
"cap_cost = capacity*1200.0\n",
"anl_charge = (9.0/100)*cap_cost\n",
"run_charge = (8760*x*y1)*(5.0/100)\n",
"cost_stm = run_charge + anl_charge #Rs\n",
"print \"load factor =\",round(y1,3)\n",
"print \"generating cost = Rs.\",round(cost_stm,-1)"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 50.29 , PAGE NO :- 1970"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Hydro cost/kWh = 13.56 paise.\n",
"Steam cost/kWh = 11.85 paise.\n",
"Steam is economical for lf = 0.1\n",
"Hydro cost/kWh = 3.51 paise.\n",
"Steam cost/kWh = 6.37 paise.\n",
"Hydro is economical for lf = 0.5\n"
]
}
],
"source": [
"'''In a particular area, both steam and hydro-stations are equally possible. It has been estimated that capital cost and the running\n",
"costs of these two types will be as follows:\n",
" Capital cost/kW Running cost/kWh Interest\n",
"Hydro : Rs. 2,200 1 Paise 5%\n",
"Steam : Rs. 1,200 5 Paise 5%\n",
"If expected average load factor is only 10%, which is economical to operate : steam or hydro?\n",
"If the load factor is 50%, would there be any change in the choice ? If so, indicate with calculation.'''\n",
"\n",
"from sympy import Symbol\n",
"\n",
"#Let x be the capacity of power station in kW.\n",
"x = Symbol('x')\n",
"#------------------------------------------------------------------------#\n",
"#Case I :- Load factor = 10%\n",
"lf1 = 0.1\n",
"#Total units generated per annum\n",
"units = x*lf1*(8760) #kWh\n",
"\n",
"#(a) Hydro Station\n",
"#--------------------------------#\n",
"cap_cost = 2200.0*x #Rs (Capital cost)\n",
"fxd_charge = (5.0/100)*cap_cost #Rs (Fixed cost)\n",
"run_charge = (1.0/100)*units #Rs (Running cost -> given, 1 unit costs 1 paisa)\n",
"tot_charge = fxd_charge + run_charge #Rs (Total charge)\n",
"hyd_costkwh = tot_charge/units*100 #paise (Cost per kWh)\n",
"\n",
"#(b) Steam Station\n",
"#--------------------------------#\n",
"cap_cost = 1200.0*x #Rs (Capital cost)\n",
"fxd_charge = (5.0/100)*cap_cost #Rs (Fixed cost)\n",
"run_charge = (5.0/100)*units #Rs (Running cost -> given, 1 unit costs 5 paisa)\n",
"tot_charge = fxd_charge + run_charge #Rs (Total charge)\n",
"stm_costkwh = tot_charge/units*100 #paise (Cost per kWh)\n",
"\n",
"print \"Hydro cost/kWh =\",round(hyd_costkwh,2),\"paise.\"\n",
"print \"Steam cost/kWh =\",round(stm_costkwh,2),\"paise.\"\n",
"\n",
"if (hyd_costkwh >= stm_costkwh) :\n",
" print \"Steam is economical for lf =\",round(lf1,2)\n",
"else:\n",
" print \"Hydro is economical for lf =\",round(lf1,2)\n",
"\n",
"#-----------------------------------------------------------------------------------------------------------#\n",
"\n",
"#Case II :- Load factor = 50%\n",
"lf2 = 0.5\n",
"#Total units generated per annum\n",
"units = x*lf2*(8760) #kWh\n",
"\n",
"#(a) Hydro Station\n",
"#--------------------------------#\n",
"cap_cost = 2200.0*x #Rs (Capital cost)\n",
"fxd_charge = (5.0/100)*cap_cost #Rs (Fixed cost)\n",
"run_charge = (1.0/100)*units #Rs (Running cost -> given, 1 unit costs 1 paisa)\n",
"tot_charge = fxd_charge + run_charge #Rs (Total charge)\n",
"hyd_costkwh = tot_charge/units*100 #paise (Cost per kWh)\n",
"\n",
"#(b) Steam Station\n",
"#--------------------------------#\n",
"cap_cost = 1200.0*x #Rs (Capital cost)\n",
"fxd_charge = (5.0/100)*cap_cost #Rs (Fixed cost)\n",
"run_charge = (5.0/100)*units #Rs (Running cost -> given, 1 unit costs 5 paisa)\n",
"tot_charge = fxd_charge + run_charge #Rs (Total charge)\n",
"stm_costkwh = tot_charge/units*100 #paise (Cost per kWh)\n",
"\n",
"print \"Hydro cost/kWh =\",round(hyd_costkwh,2),\"paise.\"\n",
"print \"Steam cost/kWh =\",round(stm_costkwh,2),\"paise.\"\n",
"\n",
"if (hyd_costkwh >= stm_costkwh) :\n",
" print \"Steam is economical for lf =\",round(lf2,2)\n",
"else:\n",
" print \"Hydro is economical for lf =\",round(lf2,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.30 , PAGE NO :- 1971"
]
},
{
"cell_type": "code",
"execution_count": 28,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"a = 50000.0\n",
"b = 0.833\n",
"c = 0.000342\n"
]
}
],
"source": [
"'''The annual working cost of a thermal station can be represented by a formula Rs. (a + b kW + c kWh) where a, b and c\n",
"are constants for a particular station, kW is the total installed capactiy and kWh the energy produced per annum.\n",
"Explain the significance of the constants a, b and c and the factors on which their values depend.\n",
"Determine the values of a, b and c for a 60 MW station operating with annual load factor of 40% for which :\n",
"(i) capital cost of buildings and equipment is Rs. 5 * 10^5\n",
"(ii) the annual cost of fuel, oil, taxation and wages and salaries of operating staff is Rs.90,000\n",
"(iii) the interest and depreciation on buildings and equipment are 10% per annum\n",
"(iv) annual cost of organisation and interest on cost of site etc. is Rs. 50,000.'''\n",
"\n",
"#Constant a is fixed cost i.e annual ineterest cost\n",
"a = 50000.0\n",
"\n",
"#Constand b is semi-fixed cost such that\n",
"#b*(Max demand in kW) = annual interest on capital cost\n",
"b = 0.1*5.0e+5/(60.0e+3)\n",
"\n",
"#Constant c is running cost such that\n",
"#c*(units in kWh) = annual fuel cost etc.\n",
"\n",
"avg_pwr = 0.5*(60.0e+3) #kW (Average power)\n",
"units = avg_pwr*8760.0 #kWh (No of units produced)\n",
"c = 90000.0/units\n",
"print \"a =\",a\n",
"print \"b =\",round(b,3)\n",
"print \"c =\",round(c,6)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.31 , PAGE NO :- 1972"
]
},
{
"cell_type": "code",
"execution_count": 29,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total cost = 133.75 Rs\n",
"Cost per unit(kWh) is = 35.67 paise\n"
]
}
],
"source": [
"'''Compute the cost of electrical energy and average cost for consuming 375 kWh under ' block rate tariff ' as under :\n",
"First 50 kWh at 60 paisa per kWh ; next 50 kWh at 50 paisa per kWh; next 50 kWh at 40 paisa per kWh; next 50 kWh at 30 paisa per kWh.\n",
"Excess over 200 kWh at 25 paisa per kWh.'''\n",
"\n",
"#Energy charge for first 50 kWh is\n",
"cost1 = 0.6*50.0 #Rs\n",
"#Energy charge for next 50 kWh is\n",
"cost2 = 0.5*50.0 #Rs\n",
"#Energy charge for next 50 kWh is\n",
"cost3 = 0.4*50.0 #Rs\n",
"#Energy charge for next 50 kWh is\n",
"cost4 = 0.3*50.0 #Rs\n",
"#Energy charge for the rest (375-200) kWh is\n",
"cost5 = 0.25*(375.0 - 200.0) #Rs\n",
"#Total cost is\n",
"tot_cost = cost1 + cost2 + cost3 + cost4 + cost5 #Rs\n",
"\n",
"#Cost/kWh = Total cost/No of units\n",
"cost_unit = tot_cost/375.0 * 100 #paisa\n",
"\n",
"print \"Total cost =\",round(tot_cost,2),\"Rs\"\n",
"print \"Cost per unit(kWh) is = \",round(cost_unit,2),\"paise\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.32 , PAGE NO :- 1972"
]
},
{
"cell_type": "code",
"execution_count": 30,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Cost of generating is = 5.37 paise.\n"
]
}
],
"source": [
"'''The output of a generating station is 390 * 10^6 units per annum and installed capacity is 80,000 kW. If the annual fixed \n",
"charges are Rs. 18 per kW of isntalled plant and running charges are 5 paisa per kWh, what is the cost per unit at the generating\n",
"station ?'''\n",
"\n",
"#Given\n",
"cap = 80000.0 #kW (installed capacity)\n",
"units = 390.0e+6 #kWh (no of units produced)\n",
"\n",
"#Annual Fixed charge is\n",
"anl_charge = cap*18.0 #Rs\n",
"\n",
"#Fixed charge per kW is\n",
"fxd_charge = anl_charge/units*100 #paise\n",
"\n",
"#Running charges per kW is\n",
"run_charge = 5.0 #paise\n",
"\n",
"#Cost of generating station is\n",
"gen_cost = run_charge + fxd_charge #paise\n",
"\n",
"print \"Cost of generating is = \",round(gen_cost,2),\"paise.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.33 , PAGE NO :- 1973"
]
},
{
"cell_type": "code",
"execution_count": 31,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Cost of generation per kWh(load factor 100%) = 2.97 paise.\n",
"Cost of generation per kWh(load factor 100%) = 7.72 paise.\n"
]
}
],
"source": [
"'''A power station has an installed capacity of 20 MW. The capital cost of station\n",
"is Rs. 800 per kW. The fixed costs are 13% of the cost of investment. On full-load at 100% load factor,\n",
"the variable costs of the station per year are 1.5 times the fixed cost. Assume no reserve capacity and\n",
"variable cost to be proportional to the energy produced, find the cost of generation per kWh at load\n",
"factors of 100% and 20%. Comment on the results.'''\n",
"\n",
"#Given\n",
"capacity = 20000.0 #kW (installed capacity)\n",
"lf1 = 1.0 # (load factor 1)\n",
"lf2 = 0.2 # (load factor 2)\n",
"\n",
"#Capital cost of station is\n",
"cap_cost = capacity*(800.0) #Rs\n",
"\n",
"#(a) At 100% load factor\n",
"fxd_cost = (13.0/100)*cap_cost #Rs (fixed cost)\n",
"var_cost = 1.5*fxd_cost #Rs (variable cost)\n",
"\n",
"#Total operating cost is\n",
"tot_cost = fxd_cost + var_cost #Rs\n",
"\n",
"#Total no. of units generated = (Max.demand*8760)*(load factor)\n",
"units = capacity*8760*lf1 #kWh\n",
"#Cost of unit per kWh generated is\n",
"cost_kwh = tot_cost/units*100 #paise\n",
"\n",
"#error in the answer of book\n",
"print \"Cost of generation per kWh(load factor 100%) = \",round(cost_kwh,2),\"paise.\"\n",
"\n",
"#(a) At 20% load factor\n",
"fxd_cost = (13.0/100)*cap_cost #Rs (fixed cost)\n",
"#Since variable cost is propotional to load\n",
"var_cost = 0.2*(1.5*fxd_cost) #Rs (variable cost)\n",
"\n",
"#Total operating cost is\n",
"tot_cost = fxd_cost + var_cost #Rs\n",
"\n",
"#Total no. of units generated = (Max.demand*8760)*(load factor)\n",
"units = capacity*8760*lf2 #kWh\n",
"#Cost of unit per kWh generated is\n",
"cost_kwh = tot_cost/units*100 #paise\n",
"\n",
"print \"Cost of generation per kWh(load factor 100%) = \",round(cost_kwh,2),\"paise.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.34 , PAGE NO :- 1973"
]
},
{
"cell_type": "code",
"execution_count": 32,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Annual running charge/kWh is = 1.38 paise.\n"
]
}
],
"source": [
"'''The annual output of a generating sub-station is 525.6 * 10^6 kWh and the average load factor is 60%. If annual fixed charges are\n",
"Rs. 20 per kW installed plant and the annual running charges are 1 paisa per kWh, what would be the cost per kWh at the bus bars ?'''\n",
"\n",
"#Given\n",
"enrgy = 525.6e+6 #kWh (Energy produced)\n",
"lf = 0.6 # (Load factor) \n",
"\n",
"#Average power supplied per annum = Energy produced/(No of hours in a year)\n",
"avg_pwr = enrgy/8760 #kW\n",
"#Maximum demand = average power/load factor is\n",
"maxp = avg_pwr/lf #kW\n",
"\n",
"#Annual fixed charges\n",
"charge = 20.0*maxp #Rs\n",
"\n",
"#Fixed charge/kWh\n",
"fxd_charge = charge/enrgy*100 #paise\n",
"\n",
"#Running charge/kWh\n",
"run_charge = 1.0 #paise\n",
"\n",
"#Annual running charge/kWh is\n",
"cost = fxd_charge + run_charge #paise\n",
"print \"Annual running charge/kWh is = \",round(cost,2),\"paise.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.35 , PAGE NO :- 1974"
]
},
{
"cell_type": "code",
"execution_count": 33,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"M.D charges = Rs. 5162.0\n",
"Monthly Bill = Rs. 14163.0\n",
"Load factor = 0.61\n",
"Power factor = 0.6\n"
]
}
],
"source": [
"'''A certain factory working 24 hours a day is charged at Rs. 10 per kVA of max.demand plus 5 paisa per kV ARh. The meters record\n",
"for a month of 30 days; 135,200 kWh, 180,020 kV ARh and maximum demand 310 kW. Calculate\n",
"(i) M.D. charges, (ii) monthly bill, (iii) load factor, (iv) average power factor.'''\n",
"\n",
"import math as m\n",
"\n",
"#Given\n",
"act_nrgy = 135200.0 #kWh (Active energy consumed in a month)\n",
"re_nrgy = 180020.0 #kVARh (Reactive energy consumed in a month)\n",
"maxp = 310.0 #kW (Maximum demand) \n",
"\n",
"#Average power\n",
"avgact_pwr = act_nrgy/(24*30) #kW (Active)\n",
"avgre_pwr = re_nrgy/(24*30) #kVAR (Reactive) \n",
"#Now, tan(theta) = kVAR/kW\n",
"theta = m.atan(avgre_pwr/avgact_pwr)\n",
"\n",
"#Maximum demand in kVA is\n",
"md_kva = maxp/(m.cos(theta))\n",
"\n",
"#(i)Maximum demand charge is\n",
"md_charge = md_kva*10.0 #Rs\n",
"\n",
"#(ii)Monthly bill\n",
"#Reactive energy charges\n",
"re_charge = (5.0/100)*re_nrgy #Rs\n",
"mon_bill = re_charge + md_charge #Rs (Monthly bill)\n",
"\n",
"#(iii)Load factor = Avg Demand/Max demand is\n",
"lf = avgact_pwr/maxp\n",
"\n",
"#(iv)Average power factor\n",
"pf = m.cos(theta)\n",
"\n",
"print \"M.D charges = Rs.\",round(md_charge)\n",
"print \"Monthly Bill = Rs.\",round(mon_bill)\n",
"print \"Load factor = \",round(lf,2)\n",
"print \"Power factor = \",round(pf,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.36 , PAGE NO :- 1974"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"overall cost/unit = 6.37 Paise.\n",
"Two-part tariff = Rs 146.67 per kW + 1.77 paise per kWh.\n"
]
}
],
"source": [
"'''The cost data of a power supply company is as follows :\n",
"Station maximum demand = 50 MW ; station load factor = 60% ; Reserve capacity = 20% ; capital cost = Rs. 2,000 per kW;\n",
"interest and depreciation = 12% ; salaries (annual) = Rs. 5 Ã— 105; fuel cost (annual) = Rs. 5 Ã— 106 ; \n",
"maintenance and repairs (annual) Rs. 2 Ã— 105 ; losses indistribution = 8% ; load diversity factor = 1.7.\n",
"Calculate the average cost per unit and the two-part tariff, assuming 80 per cent of salaries and repair and maintenace cost\n",
"to be fixed.'''\n",
"\n",
"#Station capacity \n",
"stn_cap = (1 + 20.0/100)*50 #MW\n",
"#Average power\n",
"avg_pwr = stn_cap * 0.6*1000 #kW \n",
"#Capital investment\n",
"cap_inv = stn_cap*1000 * 2000 #Rs\n",
"#Interest + depreciation cost\n",
"charge1= 0.12*cap_inv #Rs\n",
"#Total cost both fixed and running\n",
"cost_tot= charge1 + 5.0e+5 + 5e+6 + 2e+5 #Rs\n",
"#No. of units generated annually \n",
"units = 8760*avg_pwr\n",
"#overall cost/unit \n",
"cost_uni = cost_tot/units*100 #Paise.\n",
"print \"overall cost/unit = \",round(cost_uni,2),\"Paise.\"\n",
"#----------------------------------------------------------------------------------------------#\n",
"#Fixed charges\n",
"#Annual interest and depreciation \n",
"charge1 = charge1\n",
"#80% of salaries \n",
"charge2 = 0.8 * 5e+5 #Rs\n",
"#80% of repair and maintenance cost \n",
"charge3 = 0.8 * 2e+5 #Rs\n",
"#Total fixed charges #Rs\n",
"charge_tot = charge1 + charge2 + charge3 #Rs \n",
"#Aggregate maximum demand of all consumers = Max. demand on generating station * diversity factor\n",
"max_dem = 60*1000*1.7 #kW\n",
"#annual cost/kW of maximum demand \n",
"cost = charge_tot/max_dem #Rs \n",
"\n",
"#-----------------------------------------------------------------------------------------------#\n",
"#Running Charges\n",
"#Cost of fuel \n",
"charge1 = 50e+5 \n",
"#20% of salaries \n",
"charge2 = 0.2 * 5e+5 #Rs\n",
"#20% of repair and maintenance cost \n",
"charge3 = 0.2 * 2e+5 #Rs\n",
"#Total running charges\n",
"charge_run = charge1 + charge2 + charge3 #Rs\n",
"#Cost per unit delivered (considering 8% losses)\n",
"cost_uni = charge_run/(0.92*units)*100 #Paise\n",
"print \"Two-part tariff = Rs\",round(cost,2),\"per kW + \",round(cost_uni,2),\"paise per kWh.\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.37 , PAGE NO :- 1975"
]
},
{
"cell_type": "code",
"execution_count": 34,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Two-part tariff = Rs. 91.18 kW + Paise. 1.32 kWh\n"
]
}
],
"source": [
"'''A certain electric supply undertaking having a maximum demand of 110 MW generates 400 * 10^6 kWh per year.The supply\n",
"undertaking supplies power to consumers having an aggregate demand of 170 MW. The annual expenses including capital charges are :\n",
"Fuel = Rs. 5 * 10^6\n",
"Fixed expenses connected with generation = Rs. 7 * 10^6\n",
"Transmission and distribution expenses = Rs. 8 * 10^6\n",
"Determine a two-part tariff for the consumers on the basis of actual cost.\n",
"Assume 90% of the fuel cost as variable charges and transmission and distrbution losses as 15% of energy generated.'''\n",
"\n",
"\n",
"#Given\n",
"maxp = 170.0e+3 #kW (Maximum demand)\n",
"units = 400.0e+6 #kWh (Number of units generated)\n",
"fxd_charge = 7.0e+6 #Rs (Fixed charges)\n",
"tr_charge = 8.0e+6 #Rs (Transmission and distribution charges)\n",
"fuel_charge = (10.0/100)*5.0e+6 #Rs\n",
"#Total cost\n",
"tot_charge = fxd_charge + tr_charge + fuel_charge #Rs\n",
"#Cost per kW of Max. demand\n",
"cost_kw = tot_charge/maxp #Rs\n",
"\n",
"#Running charges = 90% of fuel cost\n",
"run_charge = (90.0/100)*5.0e+6 #Rs\n",
"#Units consumed after losses\n",
"units_con = (1 - 15.0/100)*units #kWh\n",
"#Cost per kWh of energy\n",
"cost_kwh = run_charge/units_con*100 #Paise\n",
"\n",
"print \"Two-part tariff = Rs.\",round(cost_kw,2),\"kW + Paise.\",round(cost_kwh,2),\"kWh\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.38 , PAGE NO :- 1975"
]
},
{
"cell_type": "code",
"execution_count": 35,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total annual bill is = Rs. 58257.0\n"
]
}
],
"source": [
"'''Customer is offered power at Rs. 80 per annum per kVA of maximum demand plus 8 paisa per unit metered.\n",
"He proposes to install a motor to carry his estimated maximum demand of 300 b.h.p. (223.8 kW). The motor available has a power\n",
"factor of 0.85 at full-load. How many units will he require at 20% load factor and what willl be his annual bill ? '''\n",
"\n",
"#Given\n",
"mot_eff = 90.0/100 # (motor efficiency)\n",
"maxp = 223.8/mot_eff #kW (full-load power intake)\n",
"lf = 0.2 # (load factor)\n",
"\n",
"#Avg demand = Max demand * Load factor\n",
"avg_dem = maxp*lf #kW\n",
"#Annual consumption is\n",
"units = avg_dem*8760 #kWh\n",
"#Cost of units consumed per annum\n",
"cost1 = units*(8.0/100) #Rs\n",
"\n",
"#Maximum kVA of demand is\n",
"max_kva = maxp/0.85 #kVA\n",
"#Cost per kVA of Maximum demand is\n",
"cost2 = max_kva*(80.0) #Rs\n",
"\n",
"#Total annual bill is\n",
"tot_cost = cost1 + cost2\n",
"print \"Total annual bill is = Rs.\",round(tot_cost)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.39 , PAGE NO :- 1975"
]
},
{
"cell_type": "code",
"execution_count": 36,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Annual energy bill = Rs. 1475000.0\n",
"Annual savings = Rs. 155000.0\n"
]
}
],
"source": [
"'''How two-part tariff is modified for penalising low p.f. consumers ?\n",
"An industry consumes 4 million kWh/year with a maximum demand of 1000 kW at 0.8 p.f. What is its load factor ?\n",
"(a) Calculate the annual energy charges if tariff in force is as under :\n",
" Max. demand charge = Rs. 5 per kVA per month. Energy charges = Rs. 0.35 per kWh\n",
"(b) Also calculate reduction in this bill if the maximum demand is reduced to 900 kW at 0.9 p.f. lagging.'''\n",
"\n",
"#Given\n",
"units = 4.0e+6 #kWh (No. of units generated)\n",
"maxp = 1000.0 #kW (Maximum demand)\n",
"avg_dem = units/8760 #kW (Average demand)\n",
"\n",
"#Load factor = Avg demand/Max demand\n",
"lf = avg_dem/maxp\n",
"#Max kVA demand is\n",
"max_kVA = maxp/0.8 #kVA\n",
"#--------------------------------------------#\n",
"#(a)\n",
"#Charge per kVA is\n",
"charge = 5.0*12 #Rs\n",
"#Annual charge per kVA is\n",
"cost1 = charge*max_kVA #Rs\n",
"#Annual charge per unit is\n",
"cost2 = 0.35*units #Rs\n",
"#Total energy bill is\n",
"tot_bill = cost1 + cost2 #Rs\n",
"print \"Annual energy bill = Rs.\",tot_bill\n",
"#--------------------------------------------#\n",
"#(b)\n",
"maxp = 900.0 #kW (New Maximum demand)\n",
"#Since load factor remains same\n",
"#Average load = Max demad * load factor\n",
"avg_dem = maxp*lf #kW\n",
"#No of units generated in a year are\n",
"units = avg_dem*8760 #kWh\n",
"\n",
"#Max kVA demand is\n",
"max_kVA = maxp/0.9 #kVA\n",
"#Annual charge per kVA is\n",
"cost1 = charge*max_kVA #Rs\n",
"#Annual charge per unit is\n",
"cost2 = 0.35*units #Rs\n",
"#Total energy bill is\n",
"tot_bill2 = cost1 + cost2 #Rs\n",
"#Total savings\n",
"savings = tot_bill - tot_bill2 #Rs\n",
"print \"Annual savings = Rs.\",savings"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 50.40 , PAGE NO :- 1976"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Number of units taken per annum = 1300.0 kWh.\n"
]
}
],
"source": [
"'''A supply is offered on the basis of fixed charges of Rs. 30 per annum plus 3 paise per unit or alternatively,\n",
"at the rate of 6 paisa per unit for the first 400 units per annum and 5 paise per unit for all the additional units.\n",
"Find the number of units taken per annum for which the cost under these two tariffs becomes the same.'''\n",
"\n",
"\n",
"from sympy import Symbol,Eq,solve\n",
"#Let x kWh be the annual consumption of the consumer\n",
"x = Symbol('x')\n",
"\n",
"#Cost under tariff 1\n",
"cost1 = 30.0 + (3.0/100)*x #Rs\n",
"#Cost under tariff 2\n",
"cost2 = (6.0/100)*400.0 + (5.0/100)*(x-400.0) #Rs\n",
"\n",
"#Since charge in both cases are equal\n",
"eq = Eq(cost1,cost2)\n",
"x = solve(eq)\n",
"x1 = x[0]\n",
"print \"Number of units taken per annum = \",round(x1,2),\"kWh.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.41 , PAGE NO :- 1976"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Cost per unit at unity p.f = 7.42 paise\n",
"Cost per unit at 0.7 p.f = 8.89 paise\n"
]
}
],
"source": [
"'''If power is charged for at the rate of Rs. 75 per kVA of maximum demand and 4 paisa per unit, what is the cost per unit at\n",
"25% yearly load factor (a) for unity power factor demand and (b) for 0.7 power factor demand.'''\n",
"\n",
"#Given\n",
"lf = 0.25 # (load factor)\n",
"charge = 75.0 #Rs (Cost per kVA of max. demand)\n",
"\n",
"#(a) At unity power factor\n",
"#Max. demand charge per unit\n",
"cost1 = charge/(8760*0.25)*100 #paise\n",
"#Cost per unit\n",
"tot_cost = cost1 + 4.0 #paise\n",
"print \"Cost per unit at unity p.f = \",round(tot_cost,2),\"paise\"\n",
"\n",
"#(b) At 0.7 power factor\n",
"#Max. demand charge per unit\n",
"cost2 = charge/(8760*0.25*0.7)*100 #paise\n",
"#Cost per unit\n",
"tot_cost = cost2 + 4.0 #paise\n",
"print \"Cost per unit at 0.7 p.f = \",round(tot_cost,2),\"paise\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.42 , PAGE NO :- 1976"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total savings in annual bill = Rs. 69.44\n",
"Total cost per unit at 0.6 load factor = 11.19 paise.\n",
"Total cost per unit at 0.8 load factor = 10.89 paise.\n",
"Cost per unit is reduced with increase in lf.\n"
]
}
],
"source": [
"'''Explain different methods of tariff. A tariff in force is Rs. 50 per kVA of max. demand per year plus 10 p per kWh.\n",
"A consumer has a max. demand of 10 kW with a load factor of 60% and p.f. 0.8 lag.\n",
"(i) Calculate saving in his annual bill if he improves p.f. to 0.9 lag.\n",
"(ii) Show the effect of improving load factor to 80% with the same max. demand and p.f. 0.8 lag\n",
"on the total cost per kWh.'''\n",
"\n",
"#Given\n",
"maxp = 10.0 #kW (Maximum demand)\n",
"\n",
"#(i)\n",
"max_kva = maxp/0.8 #kVA (Maximum demand in kVA)\n",
"#Maximum demand charges\n",
"charge1 = max_kva*(50.0) #Rs\n",
"\n",
"max_kva2 = maxp/0.9 #kVA (Maximum demand in kVA at 0.9 pf)\n",
"#Maximum demand charges\n",
"charge2 = max_kva2*(50.0) #Rs\n",
"\n",
"#Since energy consumed is same,savings is due to reduction in M.D charges\n",
"savings = charge1 - charge2 #Rs\n",
"print \"Total savings in annual bill = Rs. \",round(savings,2)\n",
"#------------------------------------------------------------------------------------------------------#\n",
"#(ii)\n",
"lf = 0.6 #(load factor)\n",
"#Avg demand = Max. demand * load factor\n",
"avg_dem = maxp*lf #kW\n",
"#No. of units consumed\n",
"units = avg_dem*8760 #kWh\n",
"#M.D charge per unit consumed\n",
"md_unit = charge1/units*100.0 #Paise\n",
"#Total charges\n",
"tot_cost1 = 10.0 + md_unit #paise\n",
"\n",
"print \"Total cost per unit at 0.6 load factor = \",round(tot_cost1,2),\"paise.\"\n",
"\n",
"lf = 0.8 #(load factor)\n",
"#Avg demand = Max. demand * load factor\n",
"avg_dem = maxp*lf #kW\n",
"#No. of units consumed\n",
"units = avg_dem*8760 #kWh\n",
"#M.D charge per unit consumed\n",
"md_unit = charge1/units*100.0 #Paise\n",
"#Total charges\n",
"tot_cost2 = 10.0 + md_unit #paise\n",
"\n",
"print \"Total cost per unit at 0.8 load factor = \",round(tot_cost2,2),\"paise.\"\n",
"\n",
"if (tot_cost1 > tot_cost2):\n",
" print \"Cost per unit is reduced with increase in lf.\"\n",
"else:\n",
" print \"Cost per unit is increase with increase in lf.\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.43 , PAGE NO :- 1977"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Annual charges in (a) = Rs. 8154.0\n",
"Annual charges in (b) = Rs. 8608.0\n",
"(a) is economical.\n"
]
}
],
"source": [
"'''A consumer has a maximum demand (M.D.) of 20 kW at 0.8 p.f. lagging and an annual load factor of 60%. There are two\n",
"alternative tariffs (i) Rs. 200 per kVA of M.D. plus 3p per kWh consumed and (ii) Rs. 50 per kVA of M.D. plus 7p per kWh\n",
"consumed. Determine which of the tariffs will be economical for him.'''\n",
"\n",
"\n",
"#Given\n",
"maxp = 20.0 #kW (Maximum demand)\n",
"lf = 0.6 # (load factor)\n",
"\n",
"#Maximum demand in kVA\n",
"max_kva = maxp/0.8 #kVA\n",
"#Average power = Max. demand * load factor\n",
"avg_pwr = maxp*lf #kW\n",
"#Energy consumed in a year\n",
"units = avg_pwr*8760 #kWh\n",
"#(a)\n",
"charge1 = 200.0*max_kva #Rs (Max. demand charge)\n",
"charge2 = (3.0/100)*units #Rs (Units consumption charge)\n",
"#Total charges\n",
"tot_costa = charge1 + charge2 #Rs\n",
"print \"Annual charges in (a) = Rs.\",round(tot_costa)\n",
"\n",
"#(b)\n",
"charge1 = 50*max_kva #Rs (Max. demand charge)\n",
"charge2 = (7.0/100)*units #Rs (Units consumption charge)\n",
"#Total charges\n",
"tot_costb = charge1 + charge2 #Rs\n",
"print \"Annual charges in (b) = Rs.\",round(tot_costb)\n",
"\n",
"if (tot_costa > tot_costb):\n",
" print \"(b) is economical.\"\n",
"else:\n",
" print \"(a) is economical.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 50.44 , PAGE NO :- 1977"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Load Factor = 0.55\n"
]
}
],
"source": [
"'''Determine the load factor at which the cost of supplying a unit of electricity from a Diesel station and from a steam\n",
"station is the same if the respective annual fixed and running charges are as follows.\n",
"Station Fixed charges Running charges\n",
"Diesel Rs. 300 per kW 25 paise/kWh\n",
"Steam Rs. 1200 per kW 6.25 paise/kWh .'''\n",
"\n",
"from sympy import Symbol,Eq,solve\n",
"#(i)Diesel Station\n",
"#Suppose that the energy supplied in one year is one unit(1 kWh)\n",
"#Annual average power is\n",
"avg_pwr = 1.0/8760 #kW\n",
"\n",
"#Annual load factor(L) = Annual average power/Annual max. demand\n",
"L = Symbol('L')\n",
"maxp = avg_pwr/L #kW\n",
"fxd_charge = 300.0*(maxp)*100 #Paise\n",
"run_charge = 25.0 #Paise\n",
"#Total charge\n",
"tot_charge1 = fxd_charge + run_charge #Paise\n",
"#(ii)Steam Station\n",
"fxd_charge = 1200.0*(maxp)*100 #Paise\n",
"run_charge = 6.25 #Paise\n",
"#Total charge\n",
"tot_charge2 = fxd_charge + run_charge #Paise\n",
"\n",
"eq = Eq(tot_charge1,tot_charge2)\n",
"L = solve(eq) \n",
"L1 = L[0]\n",
"print \"Load Factor = \",round(L1,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.45 , PAGE NO :- 1978"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total charge per unit = 6.69 paise.\n",
"Total Savings = Rs. 525.0\n"
]
}
],
"source": [
"'''A factory has a maximum load of 300 kW at 0.72 p.f. with an annual consumption of 40,000 units, the tariff is\n",
"Rs. 4.50 per kVA of maximum demand plus 2 paisa/unit. Find out the average price per unit. What will be the annual\n",
"saving if the power factor be improved to units ? '''\n",
"\n",
"#Given\n",
"maxp = 300.0 #kW (Maximum demand)\n",
"pf = 0.72 # (Power factor)\n",
"units = 40000.0 #kWh (No of units consumed)\n",
"#Maximum demand in kVA\n",
"max_kva = maxp/pf #kVA\n",
"\n",
"#Max. KVA demand charge\n",
"charge1 = 4.5*max_kva #Rs\n",
"#M.D charge per unit\n",
"md_unit = charge1/units*100 #paise\n",
"\n",
"#Total charge per unit\n",
"tot_charge = md_unit + 2.0 #paise\n",
"print \"Total charge per unit = \",round(tot_charge,2),\"paise.\"\n",
"\n",
"#Maximum demand in kVA at unity power factor\n",
"max_kva = maxp #kVA\n",
"\n",
"#Max. KVA demand charge\n",
"charge2 = 4.5*max_kva #Rs\n",
"\n",
"#Total savings is\n",
"saving = charge1 - charge2 #Rs\n",
"print \"Total Savings = Rs.\",saving"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.46 , PAGE NO :- 1978"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total cost/hr for lamp 1 = 0.49 paise.\n",
"Total cost/hr for lamp 2 = 0.61 paise.\n",
"Therefore,Lamp 1 is advantageous.\n",
"Both lamp will be equally advantageous at load factor = 0.282\n"
]
}
],
"source": [
"'''There is a choice of two lamps, one costs Rs. 1.2 and takes 100 W and the other costs Rs. 5.0 and takes 30 W ;\n",
"each gives the same candle power and has the same useful life of 1000 hours. Which will prove more economical with\n",
"electrical energy at Rs. 60 per annum per kW of maximum demand plus 3 paise per unit ? At what load factor would\n",
"they be equally advantageous? '''\n",
"\n",
"from sympy import Symbol,Eq,solve\n",
"\n",
"#(i)First Lamp\n",
"#Initial cost per hour\n",
"charge11 = 120.0/1000 #Paise\n",
"#Maximum demand/hr\n",
"maxp = 0.1 #kW\n",
"#Maximum demand charge/hr\n",
"charge21 = (60.0*maxp/8760)*100 #Paise\n",
"#Energy charge/hr\n",
"charge31 = 3*maxp #Paise\n",
"\n",
"#Total cost/hr\n",
"tot_cost1 = charge11 + charge21 + charge31 #Paise\n",
"\n",
"print \"Total cost/hr for lamp 1 = \",round(tot_cost1,2),\"paise.\"\n",
"#-----------------------------------------------------------------------------#\n",
"\n",
"#(ii)Second Lamp\n",
"#Initial cost per hour\n",
"charge12 = 500.0/1000 #Paise\n",
"#Maximum demand/hr\n",
"maxp = 0.03 #kW\n",
"#Maximum demand charge/hr\n",
"charge22 = (60.0*maxp/8760)*100 #Paise\n",
"#Energy charge/hr\n",
"charge32 = 3*maxp #Paise\n",
"\n",
"#Total cost/hr\n",
"tot_cost2 = charge12 + charge22 + charge32 #Paise\n",
"print \"Total cost/hr for lamp 2 = \",round(tot_cost2,2),\"paise.\"\n",
"\n",
"if (tot_cost1 > tot_cost2):\n",
" print \"Therefore,Lamp 2 is advantageous.\"\n",
"else:\n",
" print \"Therefore,Lamp 1 is advantageous.\"\n",
"\n",
"#As Maximum demand charge will only vary with load factor and it varies inversely\n",
"#with maximum demand charge\n",
"\n",
"x = Symbol('x')\n",
"tot_cost1 = charge11 + charge21/x + charge31 #Paise\n",
"tot_cost2 = charge12 + charge22/x + charge32 #Paise\n",
"\n",
"eq = Eq(tot_cost1,tot_cost2)\n",
"x = solve(eq)\n",
"x1 = x[0]\n",
"print \"Both lamp will be equally advantageous at load factor =\",round(x1,3)"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 50.47 , PAGE NO :- 1979"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Annual Load factor = 0.437\n"
]
}
],
"source": [
"'''The following data refers to a public undertaking which supplies electric energy to its consumers at a fixed tariff of 11.37 \n",
"paise per unit.Total installed capacity = 344 MVA ; Total capital investment = Rs. 22.4 crores;\n",
"Annual recurring expenses = Rs. 9.4 crores ; Interest charge = 6% ; depreciation charge = 5%\n",
"Estimate the annual load factor at which the system should operate so that there is neither profit nor loss to the undertaking.\n",
"Assume distribution losses at 7.84% and the average system p.f. at 0.86.'''\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"#Given\n",
"capacity = 344.0e+3 #kVA (installed capacity)\n",
"cap_cost = 22.4e+7 #Rs (capital cost)\n",
"pf = 0.86 # (power factor)\n",
"\n",
"#Load factor (L) = No of kWh (units)supplied/Max. no of kWh(units) that can be supplied\n",
"L = Symbol('L')\n",
"units = (capacity*pf)*8760*L #kWh (No. of units supplied)\n",
"#Considering distribution losses,actual units supplied are\n",
"units = (1-7.84/100)*units #kWh\n",
"#Total amount for consumption of units\n",
"cost1 = units*(11.37/100) #Rs\n",
"\n",
"\n",
"\n",
"#Fixed charges\n",
"charge1 = (11.0/100)*cap_cost #Rs (interest and depreciation cost)\n",
"charge2 = 9.4e+7 #Rs\n",
"cost2 = charge1 + charge2 #Rs\n",
"#As both charges should be equal for no profit no gain\n",
"eq = Eq(cost1,cost2)\n",
"L = solve(eq)\n",
"L1 = L[0]\n",
"print \"Annual Load factor = \",round(L1,3)"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 50.48 , PAGE NO :- 1979"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Cost per unit for Steam station = 8.81 Paise.\n",
"Cost per unit for Nuclear station = 8.09 Paise.\n",
"Load factor = 0.34\n"
]
}
],
"source": [
"'''An area has a M.D. of 250 MW and a load factor of 45%. Calculate the overall cost per unit generated by\n",
"(i) steam power station with 30 per cent reserve generating capacity and\n",
"(ii) nuclear station with no reserve capacity.\n",
"Steam station : Capital cost per kW = Rs. 1000 ; interest and depreciation on capital costs =15% ;\n",
" operating cost per unit = 5 paise.\n",
"\n",
"Nucelar station : Capital cost per kW = Rs. 2000 ; interest and depreciation on capital cost =12% ;\n",
" operating cost per unit = 2 paise.\n",
"For which load factor will the overall cost in the two cases become equal ?'''\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"\n",
"#(i)Steam station\n",
"#Installed capacity of steam station with 30% reserve capacity\n",
"capacity1 = (250.0e+3)*(1 + 30.0/100) #kW\n",
"#Average power = Max. demand * load factor\n",
"avg_pwr = capacity1*0.45 #kW\n",
"#No. of units produced in a year\n",
"units = avg_pwr*8760 #kWh\n",
"#Capital investment\n",
"cap_cost = 1000.0*capacity1 #Rs\n",
"#Annual interest and depreciation\n",
"charge1 = (0.15)*cap_cost #Rs\n",
"#Fixed charge/unit\n",
"cost_unit = charge1/units*100 #Paise\n",
"#Overall cost per unit\n",
"tot_cost = cost_unit + 5.0 #Paise\n",
"print \"Cost per unit for Steam station = \",round(tot_cost,2),\"Paise.\"\n",
"#-------------------------------------------------------------------------------------------------------------#\n",
"\n",
"#(ii)Nuclear station\n",
"#Installed capacity of steam station with 30% reserve capacity\n",
"capacity2 = (250.0e+3) #kW\n",
"#Average power = Max. demand * load factor\n",
"avg_pwr = capacity2*0.45 #kW\n",
"#No. of units produced in a year\n",
"units = avg_pwr*8760 #kWh\n",
"#Capital investment\n",
"cap_cost = 2000.0*capacity2 #Rs\n",
"#Annual interest and depreciation\n",
"charge2 = (0.12)*cap_cost #Rs\n",
"#Fixed charge/unit\n",
"cost_unit = charge2/units*100 #Paise\n",
"#Overall cost per unit\n",
"tot_cost = cost_unit + 2 #Paise\n",
"print \"Cost per unit for Nuclear station = \",round(tot_cost,2),\"Paise.\"\n",
"#---------------------------------------------------------------------------------------------------------------#\n",
"\n",
"#Let L be the load factor\n",
"L = Symbol('L')\n",
"\n",
"#Cost per unit of steam station is\n",
"stm_cost = charge1/(capacity1*8760*L)*100 + 5 #Paise\n",
"\n",
"#Cost per unit of nuclear station is\n",
"nuc_cost = charge2/(capacity2*8760*L)*100 + 2 #Paise\n",
"\n",
"#As overall cost should be equal\n",
"eq = Eq(stm_cost,nuc_cost)\n",
"L = solve(eq)\n",
"L1 = L[0]\n",
"print \"Load factor = \",round(L1,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.49 , PAGE NO :- 1980"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Annual Bill = Rs. 900.0\n",
"Equivalent flat rate = 9.23 Paise.\n"
]
}
],
"source": [
"'''The maximum demand of a customer is 25 amperes at 220 volt and his total energy consumption is 9750 kWh. If the\n",
"energy is charged at the rate of 20 paise per kWh for 500 hours' use of the maximum demand plus 5 paise power unit\n",
"for all additional units, estimate his annual bill and the equivalent flat rate.'''\n",
"\n",
"#Given\n",
"maxp = 25.0*220.0/1000 #kW (Max. demand)\n",
"units = maxp*500.0 #kWh (No. of units consumed)\n",
"charge1 = units*(20.0/100) #Rs (Max. demand charge)\n",
"\n",
"#Units to be charged at lower rate\n",
"units2 = 9750.0 - units #kWh\n",
"charge2 = units2*(5.0/100) #Rs (Charge)\n",
"\n",
"#Annual Bill is\n",
"bill = charge1 + charge2 #Rs\n",
"\n",
"#Equivalent flat rate\n",
"rate = bill/9750.0*100 #Paise\n",
"print \"Annual Bill = Rs.\",bill\n",
"print \"Equivalent flat rate = \",round(rate,2),\"Paise.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.50 , PAGE NO :- 1980"
]
},
{
"cell_type": "code",
"execution_count": 20,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Amount that can be borrowed = Rs. 125000.0\n"
]
}
],
"source": [
"'''A workshop having a number of induction motors has a maximum demand of 750 kW with a power factor of 0.75 and a\n",
"load factor of 35%. If the tariff is Rs. 75 per annum per kVA of maximum demand plus 3 paise per unit, estimate what\n",
"expenditure would it pay to incur to raise the power factor of 0.9.'''\n",
"\n",
"#Given\n",
"maxp = 750.0 #kW (Maximum demand)\n",
"max_kva = maxp/0.75 #kVA (Maximum demand in kVA)\n",
"\n",
"#Max. demand charge\n",
"charge1 = max_kva*75 #Rs\n",
"\n",
"#If pf = 0.9\n",
"max_kva2 = maxp/0.9 #kVA (Maximum demand in kVA)\n",
"\n",
"#Max. demand charge\n",
"charge2 = max_kva2*75 #Rs\n",
"\n",
"#Difference of amounts in one year\n",
"cost = charge1 - charge2 #Rs\n",
"\n",
"#Assuming interest of 10% . The capital cost that can be put to increase power factor is\n",
"# cost = (10.0/100)*cap_cost\n",
"\n",
"cap_cost = (100.0/10)*cost #Rs\n",
"print \"Amount that can be borrowed = Rs.\",cap_cost"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.51 , PAGE NO :- 1981"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Public supply charges per unit = 6.3 Paise.\n",
"Private supply charges per unit = 6.69 Paise.\n"
]
}
],
"source": [
"'''The owner of a new factory is comparing a private oil-engine generating station with public supply. Calculate the average\n",
"price per unit his supply would cost him in each case, using the following data :\n",
"\n",
"Max. demand, 600 kW; load factor, 30% ; supply tariff, Rs. 70 per kW of maximum demand plus 3 paise per unit;\n",
"capital cost of plant required for public supply, Rs. 105; capital cost of plant required for private generating station,\n",
"Rs. 4 * 10^5 ; cost of fuel, Rs. 80 per tonne ; consumption of fuel oil; 0.3 kg per unit generated. Other work costs for\n",
"private plant are as follows : lubricating oil,stores and water = 0.35 paise per unit generated ; wages 1.1 paise ;\n",
"repairs and maintenance 0.3 paise per unit.'''\n",
"\n",
"\n",
"#Given\n",
"maxp = 600.0 #kW (Maximum demand)\n",
"lf = 0.3 # (load factor)\n",
"\n",
"avg_pwr = maxp * lf #kW (Average demand = Max demand * load factor)\n",
"units = avg_pwr*8760 #kWh (Energy consumption)\n",
"\n",
"#(i)Public supply\n",
"fxd_charge = 70.0*maxp #Rs (Fixed annual charges)\n",
"cap_charge = (10.0/100)*1.0e+5 #Rs (Capital annual charges)\n",
"\n",
"tot_charge = fxd_charge + cap_charge #Rs (Total annual charges)\n",
"fxdcost_unit = tot_charge/units*100 #Paise (Fixed cost per unit)\n",
"totcost_unit = fxdcost_unit + 3.0 #Paise (Total cost per unit)\n",
"print \"Public supply charges per unit = \",round(totcost_unit,2),\"Paise.\"\n",
"#--------------------------------------------------------------------------------------------------#\n",
"#(ii)Private supply\n",
"cap_charge = (10.0/100)*4.0e+5 #Rs (Capital annual charges)\n",
"fxdcost_unit = cap_charge/units*100 #Paise (Fixed cost per unit)\n",
"oilcost_unit = (80.0/1000)*(0.3)*100 #Paise (Oil cost per unit)\n",
"runcost_unit = 0.35 + 0.3 + 1.1 #Paise (Running cost per unit)\n",
"totcost_unit = runcost_unit + oilcost_unit + fxdcost_unit #Paise (Total cost per unit)\n",
"print \"Private supply charges per unit = \",round(totcost_unit,2),\"Paise.\"\n",
"#--------------------------------------------------------------------------------------------------#\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.52 , PAGE NO :- 1981"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Therefore minimum charges are Rs 4.17 per kW and 4.0 paise per kWh consumed.\n"
]
}
],
"source": [
"'''Calculate the minimum two-part tariff to be charged to the consumers of a supply undertaking from the following data :\n",
"Generating cost per kWh; 3.6 paise ; Generating cost per kW of maximum demand, Rs. 50 Total energy generated per year ;\n",
"4,380 * 10^4 kWh Load factor at the generating station, 50% Annual charges for distribution Rs. 125,000\n",
"Diversity factor for the distribution network, 1.25 Total loss between station and consumer, 10%.'''\n",
"\n",
"#Given\n",
"units = 4380.0e+4 #kWh (No of units consumed in a year)\n",
"lf = 0.5 # (Load factor)\n",
"df = 1.25 # (Diversity factor)\n",
"avg_pwr = units/8760 #kW (Average generating power)\n",
"maxp = avg_pwr/lf #kW (Max. load on generator , lf = Avg Power/Max Power)\n",
"\n",
"#Annual fixed charges are\n",
"charges = maxp*50 #Rs\n",
"#Total fixed charges are\n",
"tot_charges = charges + 125000 #Rs\n",
"#Consumer's Max demand is\n",
"max_dem = maxp*df #kW\n",
"\n",
"#Cost per kW of Max demand is\n",
"cost_kw = tot_charges/max_dem #Rs\n",
"#Monthly cost per kW of maximum demand is\n",
"cost_kw = cost_kw/12 #Rs\n",
"\n",
"\n",
"#Since there are 10% losses ,energy reached to consumers per kWh is\n",
"units = 1*(1-0.1) #kWh\n",
"#Charges per kWh generated is\n",
"charge1 = 3.6/units #Paise\n",
"print \"Therefore minimum charges are Rs\",round(cost_kw,2),\"per kW and\",round(charge1,2),\"paise per kWh consumed.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.53 , PAGE NO :- 1982"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"High-voltage is cheaper by = 618.75 Rs.\n"
]
}
],
"source": [
"'''Two systems of tariffs are available for a factory working 8 hours a day for 300 working days in a year.\n",
"(a) High-voltage supply at 5 paise per unit plus Rs. 4.50 per month per kVA of maximum demand.\n",
"(b) Low-voltage supply at Rs. 5 per month per kVA of maximum demand plus 5.5 paise per unit.\n",
" The factory has an average load of 200 kW at 0.8 power factor and a maximum demand of\n",
" 250 kW at the same p.f.\n",
"The high-voltage equipment costs Rs. 50 per kVA and losses can be taken as 4 per cent. Interest and depreciation charges\n",
"are 12 per cent. Calculate the difference in the annual cost between the two systems.'''\n",
"\n",
"#Given\n",
"maxp = 250.0 #kW (Maximum demand)\n",
"avg_dem = 200.0 #kW (Average demand)\n",
"pf = 0.8 # (power factor)\n",
"max_kva = maxp/pf #kVA (Maximum demand in kVA)\n",
"max_kva = (100.0/96)*max_kva #kVA (considering losses)\n",
"#(a)\n",
"#Annual interest on capital investment\n",
"charge1 = max_kva*50.0*0.12 #Rs\n",
"\n",
"#Annual charge due to kVA max. demand is\n",
"charge2 = max_kva*12*4.5 #Rs\n",
"\n",
"#Annual charge due to kWh consumption\n",
"charge3 = avg_dem*(100.0/96)*(5.0/100)*8*300 #Rs\n",
"\n",
"#Total charges\n",
"tot_chargesa = charge1 + charge2 + charge3 #Rs\n",
"#------------------------------------------------------------------#\n",
"#(b)\n",
"#Annual charge due to kVA max. demand is\n",
"charge1 = maxp/pf*12*5 #Rs\n",
"\n",
"#Annual charge due to kWh consumption\n",
"charge2 = avg_dem*(5.5/100)*8*300 #Rs\n",
"\n",
"#Total charges\n",
"tot_chargesb = charge1 + charge2 #Rs\n",
"\n",
"#Hence high-voltage is cheaper by\n",
"cost = tot_chargesb - tot_chargesa #Rs\n",
"\n",
"print \"High-voltage is cheaper by = \",round(cost,2),\"Rs.\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 50.54 , PAGE NO :- 1982"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Number of units consumed = 2400.0 units.\n"
]
}
],
"source": [
"'''Estimate what the consumption must be in order to justify the following maximum demand tariff in preference to the flat\n",
"rate if the maximum demand is 6 kW.On Maximum Demand Tariff. A max. demand rate of 37 paise per unit for the first 200 hr. at the\n",
"maximum demand rate plus 3 paisa for all units in excess.Flat-rate tariff, 20 paise per unit.'''\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"#Let x = number of units to be consumed (within a specific period)\n",
"x = Symbol('x')\n",
"\n",
"#Units consumed at max. demand rate\n",
"units = 6 * 200.0 #kWh\n",
"#Units in excess of the max. demand units\n",
"excess = (x - 1200) #kWh \n",
"#Cost of max. demand units is\n",
"cost = units*37 #Paise\n",
"#Cost of excess units\n",
"cost_excss = 3.0*(excess) #Paise\n",
"#Total cost on tariff\n",
"tot_cost1 = cost + cost_excss #Paise\n",
"\n",
"\n",
"#Flat rate tariff\n",
"tot_cost2 = 20*x #Paise\n",
"\n",
"eq = Eq(tot_cost1,tot_cost2)\n",
"x = solve(eq)\n",
"x1 = x[0]\n",
"print \"Number of units consumed = \",round(x1,2),\"units.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.55 , PAGE NO :- 1986"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Current density = 41.12 A/cm^2\n"
]
}
],
"source": [
"'''If the cost of an overhead line is Rs. 2000 A (where A is the cross-sectional area in cm^2) and if the interest and depreciation\n",
"charges on the line are 8%, estimate the most economical current density to use for a transmission requiring full-load current for\n",
"60% of the year.The cost of generating electric energy is 5 paise/kWh. The resistance of a conductor one kilometre long and \n",
"1 cm^2 cross-section is 0.18 ohm.'''\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"\n",
"#Given\n",
"A = 1.0 #cm^2\n",
"R = 0.18 #ohm\n",
"#Let the current through the overhead line be I\n",
"I = Symbol('I')\n",
"#Power loss in line is\n",
"loss = 2*(I*I)*R/1000 #kW\n",
"\n",
"#No. of units consumed\n",
"units = loss*(0.6)*(365*24) #kWh\n",
"\n",
"#Annual charges\n",
"charge = units*0.05 #Rs\n",
"\n",
"#Interest and depreciation charges\n",
"charge2 = (8.0/100)*2000 #Rs\n",
"\n",
"#Equating two charges\n",
"eq = Eq(charge,charge2)\n",
"I = solve(eq)\n",
"I1 = I[1] #A\n",
"#Current density\n",
"J = I1/A #A/cm^2\n",
"print \"Current density = \",round(J,2),\"A/cm^2\"\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.56 , PAGE NO :- 1986"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Most economical cross-section is = 0.446 cm^2.\n"
]
}
],
"source": [
"'''A 500-V, 2-core feeder cable 4 km long supplies a maximum current of 200 A and the demand is such that the copper\n",
"loss per annum is such as would be produced by the full-load current flowiing for six months. The resistance of the\n",
"conductor 1 km long and 1 sq cm. cross-sectional area is 0.17 ohm. The cost of cable including installation is\n",
"Rs. (120 A + 24) per metre where A is the area of cross-section in sq. cm and interest and depreciation charges are\n",
"10%. The cost of energy is 4 paise per kWh. Find the most economical cross-section.'''\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"\n",
"#Let us consider 1 km length of feeder cable\n",
"#Also, let A be the area of cross-section\n",
"A = Symbol('A')\n",
"\n",
"#Cable cost/metre\n",
"cab_cost = 120.0*A + 24 #Rs\n",
"#Cost of 1km long cable\n",
"cost1 = 120.0*A*(1000) #Rs\n",
"#Interest and depreciation per annum\n",
"charge1 = (10.0/100)*cost1 #Rs\n",
"\n",
"#Resistance of 1km long cable is\n",
"res = 0.17/A #ohm\n",
"#Cu loss in cable = 2*I^2*R\n",
"I = 200.0 #A\n",
"loss = 2*(I*I)*res/1000 #kW\n",
"#Energy loss over 6 months\n",
"units = loss*(8760.0/2) #kWh\n",
"#Cost of this energy loss is\n",
"charge2 = 0.04*units #Rs\n",
"#For most economical cross-section charge1= charge2\n",
"eq = Eq(charge1,charge2)\n",
"A = solve(eq)\n",
"A1 = A[1] #cm^2\n",
"print \"Most economical cross-section is = \",round(A1,3),\"cm^2.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.57 , PAGE NO :- 1987"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The most economical cross-sectional area is = 0.23 cm^2.\n",
"The most economical diameter is = 0.54 cm.\n"
]
}
],
"source": [
"'''A 2-core, 11-kV cable is to supply 1 MW at 0.8 p.f. lag for 3000 hours in a year. Capital cost of the cable is\n",
"Rs. (20 + 400a) per metre where a is the cross-sectional area of core in cm^2. Interest and depreciation total 10% and cost\n",
"per unit of energy is 15 paise. If the length of the cable is 1 km, calculate the most economical cross-section of the\n",
"conductor. The specific resistance of copper is 1.75 * 10^â€“6 ohm-cm.'''\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"import math as m\n",
"\n",
"#Let A be the cross-sectional area of core\n",
"A = Symbol('A') \n",
"#Cost of 1 km length of cable\n",
"cost1 = (400.0*A)*1000 #Rs\n",
"#Resistance of 1km cable length is R = rho*l/A\n",
"res = 1.75e-6*(1000*100)/A #ohm\n",
"\n",
"#Now,\n",
"V = 11.0e+3 #V (applied voltage)\n",
"P = 1.0e+6 #W (power consumed)\n",
"pf = 0.8 # (power factor) \n",
"\n",
"#P = V*I*pf.Therefore, full-load current is\n",
"I = P/(V*pf) #A\n",
"\n",
"#Power loss in cable\n",
"loss = 2*(I*I)*res #W\n",
"\n",
"#Annual cost of energy loss\n",
"charge1 = loss*(3000)*(15.0/100)*(1/1000.0) #Rs\n",
"#Interest and depreciation per annum\n",
"charge2 = (10.0/100)*cost1 #Rs\n",
"\n",
"#The most economical cross-section will be when charge1 = charge2\n",
"eq = Eq(charge1,charge2)\n",
"A = solve(eq)\n",
"A1 = A[1] #cm^2\n",
"d = m.sqrt(4*A1/3.14) #cm (diameter) \n",
"print \"The most economical cross-sectional area is = \",round(A1,2),\"cm^2.\"\n",
"print \"The most economical diameter is = \",round(d,2),\"cm.\"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.58 , PAGE NO :- 1988"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Most economical cross-section is = 0.55 cm^2.\n"
]
}
],
"source": [
"'''The cost of a two-core feeder cable including insulation is Rs. (130 A + 24) per metre and the interest and depreciation\n",
"charges 10% per annum. The cable is two km in length and the cost of energy is 4 paisa per unit. The maximum current in the\n",
"feeder is 250 amperes and the demand is such that the copper loss is equal to that which would be produced by the full current\n",
"flowing for six months. If the resistance of a conductor of 1 sq. cm cross-sectional area and one km in length be 0.18 ohm,\n",
"find the most economical section of the same.'''\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"\n",
"#Let us consider 1 km length of feeder cable\n",
"#Also, let A be the area of cross-section\n",
"A = Symbol('A')\n",
"\n",
"#Cable cost/metre\n",
"cab_cost = 130.0*A + 24 #Rs\n",
"#Cost of 1km long cable\n",
"cost1 = 130.0*A*(1000) #Rs\n",
"#Interest and depreciation per annum\n",
"charge1 = (10.0/100)*cost1 #Rs\n",
"\n",
"#Resistance of 1km long cable is\n",
"res = 0.18/A #ohm\n",
"#Cu loss in cable = 2*I^2*R\n",
"I = 250.0 #A\n",
"loss = 2*(I*I)*res/1000 #kW\n",
"#Energy loss over 6 months\n",
"units = loss*(8760.0/2) #kWh\n",
"#Cost of this energy loss is\n",
"charge2 = 0.04*units #Rs\n",
"#For most economical cross-section charge1= charge2\n",
"eq = Eq(charge1,charge2)\n",
"A = solve(eq)\n",
"A1 = A[1] #cm^2\n",
"print \"Most economical cross-section is = \",round(A1,2),\"cm^2.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.59 , PAGE NO :- 1988"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Most economical cross-sectional area is = 0.1 cm^2.\n",
"Most economical current density is = 230.94 A/cm^2.\n",
"Most economical diameter is = 0.36 cm.\n"
]
}
],
"source": [
"'''An 11-kV, 3-core cable is to supply a works with 500-kW at 0.9 p.f. lagging for 2,000 hours p.a. Capital cost of the cable per\n",
"core when laid is Rs. (10,000 + 32,00 A) per km where A is the cross-sectional area of the core in sq. cm. The resistance per km\n",
"of conductor of 1 cm^2 crosssection is 0.16 ohm.If the energy losses cost 5 paise per unit and the interest and sinking fund is\n",
"recovered by a charge of 8% p.a., calculate the most economical current density and state the conductor diameter.'''\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"import math as m\n",
"\n",
"#Let A be the cross-sectional area of conductor\n",
"A = Symbol('A')\n",
"#The annual charge on cost of conductor per km is\n",
"charge1 = 0.08*32000.0*A #Rs\n",
"#Current per conductor is I = P/V*pf\n",
"I = (500000/1.73)/(11000.0)*(0.9)\n",
"#Resistance of conductor is\n",
"R = 0.16/A #ohm\n",
"\n",
"#Losses in 3-core cable\n",
"loss = 3*(I*I)*R/1000.0 #kW\n",
"\n",
"#Annual cost of this loss\n",
"charge2 = loss*(5.0/100)*2000 #Rs\n",
"\n",
"#For the most economical cross-section (charge1 = charge2)\n",
"eq = Eq(charge1,charge2)\n",
"A = solve(eq) \n",
"A1 = A[1] #cm^2\n",
"#Current density is\n",
"J = I/A1 #A/cm^2\n",
"#Conductor diameter is Using pi*d^2/4 = A\n",
"d = m.sqrt(A1*4/3.14) #cm\n",
"print \"Most economical cross-sectional area is = \",round(A1,2),\"cm^2.\"\n",
"print \"Most economical current density is = \",round(J,2),\"A/cm^2.\"\n",
"print \"Most economical diameter is = \",round(d,2),\"cm.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.60 , PAGE NO :- 1989"
]
},
{
"cell_type": "code",
"execution_count": 20,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Most economical cross-sectional area is = 0.63 cm^2.\n"
]
}
],
"source": [
"'''Discuss limitations of the application of Kelvinâ€™s law.An industrial load is supplied by a 3-phase cable from a sub-station\n",
"at a distance of 6 km. The voltage at the load is 11 kV. The daily load cycle for six days in a week for the entire year\n",
"is as given below :\n",
"(i) 700 kW at 0.8 p.f. for 7 hours (ii) 400 kW at 0.9 p.f. for 3 hours,\n",
"(iii) 88 kW at unity p.f. for 14 hours.\n",
"Compute the most economical cross-section of conductors for a cable whose cost is Rs. (5000 A +1500) per km (including the\n",
"cost of laying etc.). The tariff for the energy consumed at the load is Rs. 150 per annum per kVA of M.D. plus 5 paise per unit.\n",
"Assume the rate of interest and depreciation as 15%. The resistance per km of the conductor is (0.173/A) ohm.'''\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"\n",
"#Let us assume cross-section of conductor to be 'A'\n",
"A = Symbol('A')\n",
"#Capital cost of the cable is\n",
"cost1 = 6*5000.0*A #Rs\n",
"#Annual cost of interest and depreciation\n",
"charge1 = (15.0/100)*cost1 #Rs\n",
"\n",
"#Resistance of conductor is\n",
"R = 6*(0.173/A) #ohm\n",
"\n",
"#Line currents due to different loads Using I = P/V*pf\n",
"\n",
"I1 = (700000.0/1.73)/(11.0e+3*0.8) #A \n",
"I2 = (400000.0/1.73)/(11.0e+3*0.9) #A\n",
"I3 = (88000.0/1.73)/(11.0e+3*1) #A\n",
"\n",
"#Corresponding energy loss per week Using loss = 3*I*I*R\n",
"loss1 = 3*(I1*I1)*R*(6*7)/1000 #kWh\n",
"loss2 = 3*(I2*I2)*R*(6*3)/1000 #kWh\n",
"loss3 = 3*(I3*I3)*R*(6*14)/1000 #kWh\n",
"\n",
"#Total weekly loss is\n",
"tot_loss = loss1 + loss2 + loss3 #kW\n",
"#Annual cost from loss \n",
"cost2 = tot_loss*(52)*(5.0/100) #Rs (Assumed 52 weeks per year)\n",
"\n",
"#Max. Voltage drop in each conductor Using V = I*R\n",
"V = I1*R #V (As I1 is Max.)\n",
"#Max kVA demand charge\n",
"cost3 = 3*V*I1*(150.0)/1000 #Rs\n",
"\n",
"#Total annual charge due to cable loss is\n",
"charge2 = cost2 + cost3 #Rs\n",
"\n",
"#For most economical size of cable charge1 = charge2\n",
"eq = Eq(charge1,charge2)\n",
"A = solve(eq)\n",
"A1 = A[1]\n",
"print \"Most economical cross-sectional area is = \",round(A1,2),\"cm^2.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.61 , PAGE NO :- 1994"
]
},
{
"cell_type": "code",
"execution_count": 21,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Capacitance of each capacitor is = 157.03 uF.\n"
]
}
],
"source": [
"'''A 3-phase, 50-Hz, 3,000-V motor develops 600 h.p. (447.6 kW), the power factor being 0.75 lagging and the efficiency\n",
"0.93. A bank of capacitors is connected in delta across the supply terminals and power factor raised to 0.95 lagging.\n",
"Each of the capacitance units is built of five similar 600-V capacitors. Determine capacitance of each capacitor.'''\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"import math as m\n",
"#Given\n",
"V = 3000.0 #V (supplied voltage)\n",
"pout = 447600.0 #W (output power)\n",
"eff = 0.93 # (efficiency)\n",
"\n",
"pin = pout/eff #W (input power)\n",
"\n",
"#As cosQ = power factor Q = cos-1(pf)\n",
"phi1 = m.acos(0.75) # (angle 1)\n",
"phi2 = m.acos(0.95) # (angle 2)\n",
"\n",
"#Now, taking tanQ of angles\n",
"tan1 = m.tan(phi1)\n",
"tan2 = m.tan(phi2)\n",
"\n",
"#Leading VAR supplied by capacitor bank is\n",
"pvar = pin*(tan1 - tan2) # VAR \n",
"#Leading VAR supplied by each capacitor bank is\n",
"pvar = pvar/3 # VAR ----------------------(i)\n",
"\n",
"#Phase current of capacitor is I = V/Xc where Xc = 1/w*C C-> Capacitance\n",
"C = Symbol('C')\n",
"w = 2*3.14*50.0 #Hz\n",
"Icp = V*w*C #A\n",
"#Reactive power is V*Icp\n",
"pvar2 = V*Icp #VAR -----------------------(ii)\n",
"#Equating (i) & (ii)\n",
"eq = Eq(pvar,pvar2)\n",
"C = solve(eq)\n",
"C1 = C[0]*1000000.0 #uF (capacitance)\n",
"\n",
"#As it is made of 5 capacitance in series\n",
"cap_each = 5*C1 #uF\n",
"print \"Capacitance of each capacitor is =\",round(cap_each,2),\"uF.\" "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.62 , PAGE NO :- 1994"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Power factor = 0.87\n"
]
}
],
"source": [
"'''A synchronous motor having a power consumption of 50 kW is connected in parallel with a load of 200 kW having a\n",
"lagging power factor of 0.8. If the combined load has a p.f. of 0.9, what is the value of leading reactive kVA\n",
"supplied by the motor and at what p.f. is it working?'''\n",
"\n",
"import math as m\n",
"\n",
"#Given\n",
"phi2 = m.acos(0.8) #(power factor angle of load)\n",
"phit = m.acos(0.9) #(combined power factor angle)\n",
"\n",
"#Now, taking tanQ of angles\n",
"tan2 = m.tan(phi2)\n",
"tant = m.tan(phit)\n",
"\n",
"#Combined power\n",
"power = 200.0 + 50.0 #kW\n",
"#Total kVAR is\n",
"kvar = power*tant #kVAR\n",
"#Load kVAR is\n",
"load_kvar = 200.0*tan2 #kVAR\n",
"\n",
"#KVAR supplied by motor is\n",
"mot_kvar = kvar - load_kvar #kVAR\n",
"\n",
"#Now, 50*tan(phi1) = mot_kvar\n",
"tan1 = mot_kvar/50.0\n",
"phi1 = m.atan(tan1)\n",
"\n",
"#Power factor = cos(phi)\n",
"pf = m.cos(phi1) #leading\n",
"print \"Power factor = \",round(pf,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.63 , PAGE NO :- 1994"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Power factor is = 0.21\n"
]
}
],
"source": [
"'''A generating station supplies power to the following, lighting load 100-kW ;an induction motor 400 h.p. (298.4 kW),\n",
"power factor 0.8, efficiency, 0.92 ; a rotary converter giving 100 A at 500 V at an efficiency of 0.94.\n",
"What must be the power factor of the rotary converter in order that the power factor of the supply station may be unity.'''\n",
"\n",
"import math as m\n",
"\n",
"#Motor Power input = Motor output/Efficiency\n",
"pin = 298.4/0.92 #kW\n",
"phi1 = m.acos(0.8) #(motor power factor angle)\n",
"tan1 = m.tan(phi1)\n",
"\n",
"#Lagging motor kVAR is\n",
"mot_kvar = pin*tan1 #kVAR\n",
"#Leading kVAR to be supplied by rotary converter is same that of motor.\n",
"rot_kvar = mot_kvar\n",
"#Input power of rotary converter is\n",
"pin_rot = (500.0*100.0)/(0.94*1000) #kW\n",
"\n",
"#For rotary converter, tanQ = kVAR/kW\n",
"tan2 = rot_kvar/pin_rot\n",
"phi2 = m.atan(tan2)\n",
"\n",
"#Power factor = cosQ\n",
"pf = m.cos(phi2) #leading\n",
"\n",
"print \"Power factor is =\",round(pf,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.64 , PAGE NO :- 1995"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Power factor at maximum savings = 0.8\n",
"Annual savings = Rs. 833.33\n"
]
}
],
"source": [
"'''A factory has an average annual demand of 50 kW and an annual load factor of 0.5. The power factor is 0.75 lagging.\n",
"The tariff is Rs. 100 per kVA maximum demand per annum plus five paise per kWh. If loss-free capacitors costing Rs. 600\n",
"per kVAR are to be utilized, find the value of the power factor at which maximum saving will result. The interest and\n",
"depreciation together amount to ten per cent. Also, determine the saving affected by improving the power factor to this value.'''\n",
"\n",
"import math as m\n",
"#Given\n",
"A = 100.0 #Rs (charge of maximum demand per kVA)\n",
"C = (10.0/100)*600 #Rs (interest and depreciation charge)\n",
"#The most economical power factor angle is given by sinQ = C/A\n",
"phi = m.asin(C/A) #angle\n",
"#Power factor = cosQ\n",
"pf = m.cos(phi)\n",
"\n",
"#Max demand = Avg demand/load factor\n",
"maxp = 50.0/0.5 #kW (Maximum demand)\n",
"\n",
"\n",
"#(i)At initial pf = 0.75 the Max load in kVA is\n",
"max_kva1 = maxp/0.75 #kVA\n",
"#Max. demand charge is\n",
"charge1 = max_kva1*A #Rs\n",
"#(ii)At economical 'pf' the Max load in kVA is\n",
"max_kva2 = maxp/pf #kVA\n",
"#Max. demand charge is\n",
"charge2 = max_kva2*A #Rs\n",
"\n",
"#Annual Savings\n",
"savings = charge1 - charge2 #Rs\n",
"print \"Power factor at maximum savings = \",round(pf,2)\n",
"print \"Annual savings = Rs.\",round(savings,2)"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 50.65 , PAGE NO :- 1995"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Therefore, Maximum cost per kVA of pf corrections is = Rs. 314.76\n"
]
}
],
"source": [
"'''For increasing the kW capacity of a plant working at 0.7 lag p.f. the necessary increase of power can be obtained by raising\n",
"the p.f. to 0.85 or by installing additional plant. What is the maximum cost per kVA of p.f. correction apparatus to make its use\n",
"more economical than additional plant at Rs. 500 kVA ?'''\n",
"\n",
"from sympy import Symbol\n",
"import math as m\n",
"\n",
"#Let kVA1 be the initial capacity of plant and kVA2 be its increased capacity\n",
"kVA1 = Symbol('kVA1')\n",
"#kVA1 * cosQ1 = kVA2 * cosQ2\n",
"kVA2 = kVA1*(0.85/0.7)\n",
"#KVA of the additional plant\n",
"kva_add = kVA2 - kVA1\n",
"#Capital cost of additional plant is\n",
"cost_add = 500.0*kva_add #Rs\n",
"\n",
"#Now, cosQ1 and cosQ2 are known and we have to find sinQ1 and sinQ2\n",
"phi1 = m.acos(0.7)\n",
"phi2 = m.acos(0.85)\n",
"\n",
"sin1 = m.sin(phi1)\n",
"sin2 = m.sin(phi2)\n",
"\n",
"#kVAR supplied by pf corrections\n",
"kvar_supp = kVA2*sin1 - kVA1*sin2 #kVAR\n",
"#Let Cost of pf corrections be x\n",
"#Now ,cost_add = cost of pf corrections . Therefore\n",
"x = cost_add/kvar_supp #Rs\n",
"\n",
"print \"Therefore, Maximum cost per kVA of pf corrections is = Rs.\",round(x,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.66 , PAGE NO :- 1996"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Annual savings = Rs. 3054.0\n"
]
}
],
"source": [
"'''A consumer taking a steady load of 160 kW at a p.f. of 0.8 lag is charged at Rs. 80 per annum per kVA of\n",
"maximum demand plus 5 paise per kWh consumed. Calculate the value to which he should improve the p.f. in order to\n",
"affect the maximum saving if the leading kVA cost Rs.100 per kVA and interest and depreciation be at 12% per annum.\n",
"Calculate also the saving.'''\n",
"\n",
"import math as m\n",
"\n",
"#Let the cost per kVAR is Rs. B and rate of interest and depreciation is P,then\n",
"#C = BP/100\n",
"A = 80.0 #Rs (cost per kVA of max. demand)\n",
"C = 100*12/100 #Rs\n",
"#sinQ = C/A\n",
"phi = m.asin(C/A) # angle\n",
"pf = m.cos(phi) # power factor\n",
"\n",
"maxp = 160.0 #kW (max. demand)\n",
"#Max. demand in kVA for pf = 0.8\n",
"max_kva1 = maxp/0.8 #kVA\n",
"#Max.demand charge is\n",
"charge1 = max_kva1*80.0 #Rs\n",
"\n",
"#----------------------------------------------------------------------------------#\n",
"#Max. demand in kVA for most economical pf \n",
"max_kva2 = maxp/pf #kVA\n",
"#Max.demand charge is\n",
"charge2 = max_kva2*80.0 #Rs\n",
"#Annual savings is\n",
"savings = charge1 - charge2 #Rs\n",
"print \"Annual savings = Rs.\",round(savings)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.67 , PAGE NO :- 1996"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"power factor = 0.987\n",
"kVA supplied by plant is = 1245.0 kVA\n"
]
}
],
"source": [
"'''A consumer takes a steady load of 1500 kW at a p.f. of 0.71 lagging and pays Rs. 50 per annum per kVA of maximum\n",
"demand. Phase advancing plant costs Rs. 80 per kVA. Determine the capacity of the phase advancing plant required for\n",
"minimum overall annual expenditure.Interest and depreciation total 10%. What will be the value of the new power factor\n",
"of the supply?'''\n",
"\n",
"import math as m\n",
"\n",
"#Let the cost per kVAR is Rs. B and rate of interest and depreciation is P,then\n",
"#C = BP/100\n",
"A = 50.0 #Rs (cost per kVA of max. demand)\n",
"C = 80*10/100 #Rs\n",
"#sinQ = C/A\n",
"phi = m.asin(C/A) # angle\n",
"pf = m.cos(phi) # power factor\n",
"\n",
"#As power factor = cosQ\n",
"phi1 = m.acos(0.71) #angle\n",
"tan1 = m.tan(phi1) #tanQ\n",
"phi2 = m.acos(pf) #angle\n",
"tan2 = m.tan(phi2) #tanQ\n",
"\n",
"#kVA supplied by plant is\n",
"kva_supp = 1500.0*(tan1 - tan2)\n",
"print \"power factor = \",round(pf,3)\n",
"print \"kVA supplied by plant is = \",round(kva_supp),\"kVA\""
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 50.68 , PAGE NO :- 1996"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total annual savings = Rs. 488.29\n"
]
}
],
"source": [
"'''A factory takes a load of 200 kW at 0.85 p.f. (lagging) for 2,500 hours per annum and buys energy on tariff of Rs. 150\n",
"per kVA plus 6 paise per kWh consumed. If the power factor is improved to 0.9 lagging by means of capacitors costing\n",
"Rs. 525 per kVA and having a power loss of 100 W per kVA, calculate the annual saving affected by their use.\n",
"Allow 8% per annum for interest and depreciation on the capacitors.'''\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"import math as m\n",
"\n",
"#Given\n",
"maxp = 200.0 #kW (factory load)\n",
"pf = 0.85 # (power factor = cosQ)\n",
"phi = m.acos(pf) #angle\n",
"tan1 = m.tan(phi) # (tanQ)\n",
"#Lagging kVAR of factory load\n",
"max_kvar = maxp*tan1 #kVAR\n",
"\n",
"#Let x be capacitor's kVAR. Therefore, total kVAR is\n",
"x = Symbol('x')\n",
"tot_kvar = max_kvar - x\n",
"#Because loss per kVA is 100W i.e 1/10kW per kVA\n",
"cap_loss = x/10.0 #kW\n",
"\n",
"#Total kW is\n",
"tot_kw = maxp + cap_loss #kW\n",
"#Overall pf is cosQ = 0.9\n",
"phi2 = m.acos(0.9)\n",
"tan2 = m.tan(phi2)\n",
"eq = Eq(tan2,tot_kvar/tot_kw)\n",
"x = solve(eq)\n",
"x1 = x[0] #kVAR\n",
"\n",
"#(i)cost per annum before improvement\n",
"#Max demand in kVA\n",
"max_kva = maxp/pf #kVA \n",
"#Units consumed per annum\n",
"units = maxp*2500.0 #kWh\n",
"\n",
"#Total annual cost\n",
"cost1 = max_kva*150.0 + units*(6.0/100) #Rs\n",
"\n",
"\n",
"#(ii)cost per annum after improvement\n",
"max_kva = maxp/0.9 #kVA\n",
"#Units consumed per annum\n",
"units = maxp*2500.0 #kWh\n",
"\n",
"#Charges due to losses in capacitor\n",
"charge1 = x1*100*2500*6/(1000*100) #Rs\n",
"\n",
"#Annual interest and depreciation cost\n",
"charge2 = x1*525*(8.0/100) #Rs \n",
"\n",
"#Total annual cost\n",
"cost2 = max_kva*150.0 + units*(6.0/100) + charge1 + charge2 #Rs\n",
"\n",
"#Total annual savings are\n",
"savings = cost1 - cost2 #Rs\n",
"\n",
"print \"Total annual savings = Rs.\",round(savings,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.69 , PAGE NO :- 1997"
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"B is cheaper than A by = Rs. 52.04\n"
]
}
],
"source": [
"'''A 30 h.p. (22.38 kW) induction motor is supplied with energy on a two-part tariff of Rs. 60 per kVA of maximum demand\n",
"per annum plus 5 paise per unit. Motor (A) has an efficiency of 89% and a power factor of 0.83.\n",
"Motor (B) with an efficiency of 90% and a p.f. of 0.91 costs Rs. 160 more. With motor (A) the p.f. would be raised to\n",
"0.91 (lagging) by installing capacitors at a cost Rs. 50 per kVA.If the service required from the motor is equivalent\n",
"to 2,280 hr. per annum at full load, compare the annual charges in the two cases. Assume interest and depreciation charges\n",
"to be 12.5% per annum for the motor and 8% per annum for the capacitors.'''\n",
"\n",
"import math as m\n",
"\n",
"#(i)For Motor A\n",
"pin = 22.38/0.89 #kW (Motor Input in kW)\n",
"pin_kva = pin/0.83 #kVA (Motor Input in kVA)\n",
"#If power factor is changed to 0.91 then\n",
"pin_kva2 = pin/0.91 #kVA (Motor Input in kVA)\n",
"#Annual cost of energy supplied to motor is\n",
"charge1 = pin_kva2*60.0 + pin*2280.0*(5.0/100) #Rs\n",
"\n",
"#Now, as we know pf = cosQ\n",
"phi1 = m.acos(0.83) #angle\n",
"phi2 = m.acos(0.91) #angle\n",
"tan1 = m.tan(phi1) #tanQ\n",
"tan2 = m.tan(phi2) #tanQ\n",
"\n",
"#kVAR necessary for this improvement is\n",
"tot_kvar = pin*(tan1 - tan2) #kVAR\n",
"#Annual charges on capacitors\n",
"charge2 = 50.0*tot_kvar*(8.0/100) #Rs\n",
"\n",
"#Total charges per annum is\n",
"tot_chargeA = charge1 + charge2 #Rs\n",
"\n",
"\n",
"#-----------------------------------------------------------------------------#\n",
"#(ii)For Motor B\n",
"pin = 22.38/0.9 #kW (Motor Input in kW)\n",
"#If power factor is changed to 0.91 then\n",
"pin_kva2 = pin/0.91 #kVA (Motor Input in kVA)\n",
"#Annual cost of energy supplied to motor is\n",
"charge1 = pin_kva2*60.0 + pin*2280.0*(5.0/100) #Rs\n",
"\n",
"#Annual charges on capacitors\n",
"charge2 = (12.5/100)*160 #Rs\n",
"\n",
"#Total charges per annum is\n",
"tot_chargeB = charge1 + charge2 #Rs\n",
"\n",
"#Hence B is cheaper than A by\n",
"cheap = tot_chargeA - tot_chargeB #Rs\n",
"\n",
"\n",
"print \"B is cheaper than A by = Rs. \",round(cheap,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.70 , PAGE NO :- 1997"
]
},
{
"cell_type": "code",
"execution_count": 15,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total annual cost for Motor A is = Rs. 15059.48\n",
"Total annual cost for Motor B is = Rs. 15430.78\n",
"Motor A would be recommended.\n"
]
}
],
"source": [
"'''The motor of a 22.5 kW condensate pump has been burnt beyond economical repairs. Two alternatives have been\n",
"proposed to replace it by\n",
"Motor A. Cost = Rs. 6000 ; Î· at full-load=90% ; at half-load = 86%.\n",
"Motor B. Cost = Rs. 4000 ; Î· at full-load=85% ; at half-load= 82%.\n",
"The life of each motor is 20 years and its salvage value is 10% of the initial cost. The rate of\n",
"interest is 5% annually. The motor operates at full-load for 25% of the time and at half-load for the\n",
"remaining period. The annual maintenance cost of motor A is Rs. 420 and that of motor B is Rs. 240.\n",
"The energy rate is 10 paise per kWh.Which motor will you recommend ?'''\n",
"\n",
"pout = 22.5 #kW (output power)\n",
"#(i)For Motor A\n",
"eff_fl = 0.9 # (efficiency at full-load)\n",
"eff_hl = 0.86 # (efficiency at half-load)\n",
"#Annual interest on capital cost\n",
"charge1 = 6000.0*(5.0/100) #Rs\n",
"#Annual depreciation charges = (original cost - salvage value)/20 years\n",
"charge2 = (6000.0 - 6000.0*(10.0/100))/20 #Rs\n",
"#Annual maintenance cost\n",
"charge3 = 420.0 #Rs\n",
"#Energy input per annum\n",
"units = pout*0.25*8760/eff_fl + (pout/2)*0.75*8760/eff_hl #kWh\n",
"#Annual energy cost is\n",
"charge4 = units*(10.0/100) #Rs\n",
"#Total annual cost is\n",
"tot_costA = charge1 + charge2 + charge3 + charge4 #Rs\n",
"\n",
"#---------------------------------------------------------------------------------------#\n",
"\n",
"#(ii)For Motor B\n",
"eff_fl = 0.85 # (efficiency at full-load)\n",
"eff_hl = 0.82 # (efficiency at half-load)\n",
"#Annual interest on capital cost\n",
"charge1 = 4000.0*(5.0/100) #Rs\n",
"#Annual depreciation charges = (original cost - salvage value)/20 years\n",
"charge2 = (4000.0 - 4000.0*(10.0/100))/20 #Rs\n",
"#Annual maintenance cost\n",
"charge3 = 240.0 #Rs\n",
"#Energy input per annum\n",
"units = pout*0.25*8760/eff_fl + (pout/2)*0.75*8760/eff_hl #kWh\n",
"#Annual energy cost is\n",
"charge4 = units*(10.0/100) #Rs\n",
"#Total annual cost is\n",
"tot_costB = charge1 + charge2 + charge3 + charge4 #Rs\n",
"\n",
"print \"Total annual cost for Motor A is = Rs.\",round(tot_costA,2)\n",
"print \"Total annual cost for Motor B is = Rs.\",round(tot_costB,2)\n",
"\n",
"if(tot_costA>tot_costB):\n",
" print \"Motor B would be recommended.\"\n",
"else:\n",
" print \"Motor A would be recommended.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.71 , PAGE NO :- 1998"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Annual Savings = Rs. 7396.89\n"
]
}
],
"source": [
"'''An industrial load takes 106 kWh a year, the power factor being 0.707 lagging.The maximum demand is 500 kVA.\n",
"The tariff is Rs. 75 per annum per kVA maximum demand plus 3 paise per unit. Calculate the yearly cost of supply\n",
"and find the annual saving in cost by installing phase advancing plant costing Rs. 45 per kVA which raises the plant\n",
"power factor from 0.707 to 0.9 lagging. Allow 10% per annum on the cost of the phase advancing plant to cover all\n",
"additional costs.'''\n",
"\n",
"import math as m\n",
"\n",
"#Given\n",
"units = 1.0e+6 #kWh (No. of units consumed in a year)\n",
"max_dem = 500.0 #kVA (Max. demand charge per annum)\n",
"#Max. demand charge per annum\n",
"charge1 = max_dem*75 #Rs\n",
"#Annual energy charges\n",
"charge2 = units*(3.0/100) #Rs\n",
"#Total cost of supply\n",
"tot_charge1 = charge1 + charge2 #Rs\n",
"\n",
"#Now when pf is changed from 0.707 to 0.9 then Max. kVA demand is\n",
"max_dem = max_dem*(0.707/0.9) #kVA\n",
"\n",
"#Max. demand charge per annum\n",
"charge1 = max_dem*75 #Rs\n",
"#Annual energy charges\n",
"charge2 = units*(3.0/100) #Rs\n",
"#Total cost of supply\n",
"tot_charge2 = charge1 + charge2 #Rs\n",
"\n",
"#Now, as we know pf = cosQ\n",
"phi1 = m.acos(0.707)\n",
"phi2 = m.acos(0.9)\n",
"tan1 = m.tan(phi1)\n",
"tan2 = m.tan(phi2)\n",
"\n",
"#kVAR to be supplied is (kW demand)*(tan1 - tan2)\n",
"tot_kvar = max_dem*0.707*(tan1 - tan2) #kVAR\n",
"\n",
"#Annual cost of interest and depreciation\n",
"charge3 =tot_kvar*45*(10.0/100) #Rs\n",
"tot_charge2 = tot_charge2 + charge3 #Rs\n",
"\n",
"#Annual Savings\n",
"savings = tot_charge1 - tot_charge2 #Rs\n",
"\n",
"print \"Annual Savings = Rs. \",round(savings,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.72 , PAGE NO :- 1999"
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Annual charge of Transformer 1 is = Rs. 7884.0\n",
"Annual charge of Transformer 2 is = Rs. 8409.6\n",
"Transformer 1 is chosen.\n",
"To reverse the situation , Capital cost of transformer is > Rs. 5256.0\n"
]
}
],
"source": [
"'''It is necessary to choose a transformer to supply a load which varies over 24\n",
"hour period in the manner given below :\n",
"500 kVA for 4 hours, 1000 kVA for 6 hours, 1500 kVA for 12 hours and 2000 kVA for the rest of the period.\n",
"Two transformers each rated at 1500 kVA have been quoted. Transformer I has iron loss of\n",
"2.7 kW and full-load copper loss of 8.1 kW while transformer II has an iron loss and full-load copper\n",
"loss of 5.4 kW each.\n",
"(i) Calculate the annual cost of supplying losses for each transformer if electrical energy costs\n",
"10 paise per kWh.\n",
"(ii) Determine which transformer should be chosen if the capital cost of the transformer I is\n",
"Rs. 1000 more than that of the transformer II and annual charges of interest and depreciation\n",
"are 10%.\n",
"(iii) What difference in capital cost will reverse the decision made in (ii) above ? '''\n",
"\n",
"#(i.a)Transformer No. 1\n",
"\n",
"iron_loss = 2.7*24 #kWh (Iron loss/day)\n",
"\n",
"cu_loss = 8.1*(((500.0/1500)**2.0)*4 + ((1000.0/1500)**2.0)*6 + ((1500.0/1500)**2.0)*12 + ((2000.0/1500)**2.0)*2) \n",
"#kWh(Copper loss/day)\n",
"\n",
"#Annual energy loss\n",
"annual_loss = 365.0*(iron_loss + cu_loss) #kWh \n",
" \n",
"#Annual cost of both cases \n",
"charge1 = annual_loss*(10.0/100) #Rs\n",
"#-----------------------------------------------------------------------#\n",
"#(i.b)Transformer No. 2\n",
"\n",
"iron_loss = 5.4*24 #kWh (Iron loss/day)\n",
"cu_loss = 5.4*(((500.0/1500)**2.0)*4 + ((1000.0/1500)**2.0)*6 + ((1500.0/1500)**2.0)*12 + ((2000.0/1500)**2.0)*2)\n",
"#kWh(Copper loss/day)\n",
"\n",
"#Annual energy loss\n",
"annual_loss = 365.0*(iron_loss + cu_loss) #kWh \n",
" \n",
"#Annual cost of both cases \n",
"charge2 = annual_loss*(10.0/100) #Rs\n",
"\n",
"print \"Annual charge of Transformer 1 is = Rs.\",round(charge1,2)\n",
"print \"Annual charge of Transformer 2 is = Rs.\",round(charge2,2)\n",
"\n",
"#------------------------------------------------------------------------------------------------------------------------------#\n",
"#(ii)Sice cost of transformer1 is Rs. 1000 more with 10% interest\n",
"#Total annual cost for Transformer 1 is\n",
"charge1a = charge1 + (10.0/100)*1000 #Rs\n",
"\n",
"if(charge1a>charge2):\n",
" print \"Transformer 2 is chosen.\"\n",
"else:\n",
" print \"Transformer 1 is chosen.\"\n",
"#------------------------------------------------------------------------------------------------------------------------------#\n",
"#(iii)Total savings in Transformer 1 is\n",
"savings = charge2 - charge1 #Rs\n",
"#Let x be the capital cost of transformer 1 . Then x*0.1 > savings\n",
"x = savings/0.1 #Rs\n",
"print \"To reverse the situation , Capital cost of transformer is > Rs. \",round(x,2) \n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.73 , PAGE NO :- 1999"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Net savings per annum = Rs. 8590.0\n"
]
}
],
"source": [
"'''Three-phase 50-Hz power is supplied to a mill, the voltage being stepped down to 460-V before use.The monthly power rate\n",
"is 7.50 per kVA. It is found that the average power factor is 0.745 while the monthly demand is 611 kVA.To improve power\n",
"factor, 210 kVA capacitors are installed in which there is negligible power loss. The installed cost of the equipment is\n",
"Rs. 11,600 and fixed charges are estimated at 15% per year. What is the yearly saving introduced by the capacitors.'''\n",
"\n",
"import math as m\n",
"\n",
"#Monthly demand\n",
"max_kva = 611.0 #kVA (maximum demand)\n",
"pf = 0.745 # (power factor)\n",
"#Now, as we know\n",
"phi = m.acos(pf) #angle\n",
"sin1 = m.sin(phi)\n",
"\n",
"#Max demand in kW\n",
"max_kw = max_kva*pf #kW\n",
"#Max demand in kVAR\n",
"max_kvar = max_kva*sin1 #kVAR (lagging)\n",
"#Leading kVAR\n",
"kvar2 = 210.0 #kVAR\n",
"\n",
"#kVAR after pf improvement\n",
"tot_kvar = max_kvar - kvar2 #kVAR\n",
"\n",
"#kVA after power factor improvement is\n",
"new_kva = m.sqrt( max_kw**2 + tot_kvar**2 ) #kVA\n",
"\n",
"#Reduction in kVA\n",
"red_kva = max_kva - new_kva #kVA\n",
"#Monthly Savings in kVA charge\n",
"saving = red_kva*7.5 #Rs\n",
"#Yearly Savings in kVA charge\n",
"saving = 12*saving #Rs\n",
"\n",
"#Fixed charge per annum due to Capital cost on capacitors\n",
"fxd_charge = 0.15*11600 #Rs\n",
"\n",
"#Net savings\n",
"net_saving = saving - fxd_charge #Rs\n",
"\n",
"print \"Net savings per annum = Rs. \",round(net_saving)"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 50.74 , PAGE NO :- 2000"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Minimum number of power units consumed per month = 880.0 units.\n"
]
}
],
"source": [
"'''A supply undertaking is offering the following two tariffs to prospective customers :\n",
"Tariff A : Lighting : 20 paise per unit; domestic power: 5 paise per unit, meter rent: 30 paise per meter per month.\n",
"Tariff B : 12 per cent on the rateable value of the customer's premises plus 3 paise per unit for all purposes.\n",
"If the annual rateable value of the customer's premises is Rs. 2,500 and his normal consumption for lighting per month\n",
"is 40 units, determine what amount of domestic power consumption will make both the tariffs equally advantageous.'''\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"\n",
"#Let x = minimum number of power units consumed per month\n",
"x = Symbol('x')\n",
"#(i) Tariff A\n",
"#Total cost per month = meter rent + lighting charges + power charges\n",
"charge1 = 2*30.0 + 40.0*20.0 + 5*x #Paise\n",
"#(ii) Tariff B\n",
"#Total cost per month = 12% of 2500 + energy charges for al purposes\n",
"charge2 = 0.12*2500.0*(100.0/12) + 3*(40.0 + x) #Paise\n",
"\n",
"#Equating tariff A and B\n",
"eq = Eq(charge1,charge2)\n",
"x = solve(eq)\n",
"x1 = x[0]\n",
"print \"Minimum number of power units consumed per month = \",round(x1,2),\"units.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.75 , PAGE NO :- 2000"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The price of h.t motors per output kW = Rs. 36.2\n"
]
}
],
"source": [
"'''Transformers and low-tension motors of a certain size can be purchased at Rs. 12 per kVA of full output and Rs. 24 per kW \n",
"output respectively.If their respective efficiencies are 98% and 90%, what price per kW output could be paid for high-tension\n",
"motors of the same size but of average efficiency only 89%? Assume an annual load factor of 30%, the cost of energy per unit as\n",
"7 paise and interest and depreciation at the rate of 8% for low-tension motors.'''\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"\n",
"#Let S => The price of h.t motors per output kW in rupees\n",
"S = Symbol('S')\n",
"\n",
"#-------Low-tension Motors with Transformers-----#\n",
"#Interest and depreciation charges of motors\n",
"charge1 = 24.0*0.08/0.9 #Rs\n",
"#Interest and depreciation charges of transformers\n",
"charge2 = 12.0*0.9/0.99*0.08 #Rs\n",
"#Running charges\n",
"charge3 = (8760*0.3)*7/(0.98*0.9*100) #Rs\n",
"\n",
"#Total cost of low tension motors with transformers\n",
"tot_chargeA = charge1 + charge2 + charge3 #Rs -----------(i)\n",
"\n",
"#------High-tension Motors with Transformers----#\n",
"\n",
"#Standing Charges/output kW\n",
"charge1 = S*(0.12)/0.89 #Rs\n",
"#Running charges\n",
"charge2 = (8760*0.3)*7/(0.89*100) #Rs\n",
"\n",
"#Total cost of high tension motors with transformers\n",
"tot_chargeB = charge1 + charge2 #Rs -----------(ii)\n",
"\n",
"#Equating (i) and (ii)\n",
"eq = Eq(tot_chargeA,tot_chargeB)\n",
"S = solve(eq)\n",
"S1 = S[0]\n",
"print \"The price of h.t motors per output kW = Rs.\",round(S1,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.76 , PAGE NO :- 2001"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"number of hours per week above which the H.V. supply is cheaper = 36.0\n"
]
}
],
"source": [
"'''An industrial load can be supplied on the following alternative tariffs (a) highvoltage supply at Rs. 45 per kVA per annum \n",
"plus 1.5 paise per kWh or (b) low-voltage supply at Rs. 50 per annum plus 1.8 paise per kWh. Transformers and switchgear etc.\n",
"for the H.V. supply cost Rs. 35 per kVA, the full-load transformer losses being 2%. The fixed charges on the capital cost of\n",
"the high-voltage plant are 25% and the installation works at full-load. If there are 50 working weeks in a year, find the number\n",
"of working hours per week above which the H.V. supply is cheaper.'''\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"\n",
"#Let x be the number or working hours per week above which H.V. supply is cheaper than the L.V. supply.\n",
"x = Symbol('x')\n",
"\n",
"load = 100.0 #kW (load)\n",
"rload = load/(98.0/100) #kW (rated load with 2% losses)\n",
"\n",
"#Cost of switchgear & trasformer\n",
"cost = rload*35.0 #Rs \n",
"#annual fixed charge\n",
"charge1 = cost*0.25 #Rs\n",
"\n",
"#Annual energy consumption\n",
"units = 100.0*x*50.0 #kWh\n",
"\n",
"#(a) H.V. Supply\n",
"#Total annual cost = 45 * kVA + energy charges + charge on H.V. plant\n",
"charge_hv = 45*(rload) + units/0.98*(1.5/100) + charge1 #Rs\n",
"\n",
"#(b) L.V. Supply\n",
"#Total annual cost = Rs. 50 Ã— kVA + energy charges\n",
"charge_lv = 50*load + units*(1.8/100) #Rs\n",
"\n",
"#If two annual costs are equal then\n",
"eq = Eq(charge_hv,charge_lv)\n",
"x = solve(eq)\n",
"x1 = x[0] \n",
"print \"number of hours per week above which the H.V. supply is cheaper = \",round(x1)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.77 , PAGE NO :- 2002"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Motor X is cheaper by = Rs. 34.8\n"
]
}
],
"source": [
"'''For a particular drive in a factory requiring 10 h.p. (7.46 kW) motors, following tenders have been received. Which one\n",
"will you select ?\n",
" cost efficiency\n",
"Motor X Rs. 1,150 86%\n",
"Motor Y Rs. 1,000 85%\n",
"Electrical tariff is Rs. 50 per kW + 5 paise per kWh. Assume interest and depreciation as 10% .'''\n",
"\n",
"#(i) Motor X\n",
"pin = 7.46/0.86 #kW (Full load power input)\n",
"units = pin*8760.0 #kWh (Units consumed/year)\n",
"#kW charges\n",
"charge1 = 50.0*pin #Rs\n",
"#kWh charges\n",
"charge2 = (5.0/100)*units #Rs\n",
"#Fixed charges\n",
"charge3 = (10.0/100)*1150.0 #Rs\n",
"\n",
"#Total annual charges\n",
"tot_chargeX = charge1 + charge2 + charge3 #Rs\n",
"\n",
"#-------------------------------------------------------------------------------#\n",
"#(ii) Motor Y\n",
"pin = 7.46/0.85 #kW (Full load power input)\n",
"units = pin*8760.0 #kWh (Units consumed/year)\n",
"#kW charges\n",
"charge1 = 50.0*pin #Rs\n",
"#kWh charges\n",
"charge2 = (5.0/100)*units #Rs\n",
"#Fixed charges\n",
"charge3 = (10.0/100)*1000.0 #Rs\n",
"\n",
"#Total annual charges\n",
"tot_chargeY = charge1 + charge2 + charge3 #Rs\n",
"\n",
"#-------------------------------------------------------------------------------#\n",
"if (tot_chargeX > tot_chargeY):\n",
" savings = tot_chargeX - tot_chargeY #Rs\n",
" print \"Motor Y is cheaper by = Rs.\",round(savings,2)\n",
"else:\n",
" savings = tot_chargeY - tot_chargeX #Rs\n",
" print \"Motor X is cheaper by = Rs.\",round(savings,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.78 , PAGE NO :- 2002"
]
},
{
"cell_type": "code",
"execution_count": 20,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Cost of Motor B is = Rs. 10250.0\n"
]
}
],
"source": [
"'''A 200-h p. (149.2 kW) motor is required to operate at full-load for 1500 hr, at half-load for 3000 hr per year and to\n",
"be shut down for the remainder of the time. Two motors are available.\n",
"\n",
"Motor A : efficiency at full load = 90% ; at half-load = 88%\n",
"Motor B : efficiency at full load = 90% ; at half-load = 89%\n",
"The unit of energy is 5 paise/kWh and interest and depreciation may be taken as 12 per cent per year. If motor A cost\n",
"Rs. 9,000 ; what is the maximum price which could economically be paid for motor B ?'''\n",
"\n",
"from sympy import Symbol,solve,Eq\n",
"#(i)Motor A\n",
"pin_fl = 149.2/0.9 #kW (full-load power input)\n",
"pin_hl = (149.2/2)/0.88 #kW (half-load power input)\n",
"pin_hl = round(pin_hl,1)\n",
"pin_fl = round(pin_fl,1)\n",
"#Total energy consumed in a year\n",
"units = 1500.0*pin_fl + 3000.0*pin_hl #kWh\n",
"#Cost of energy is\n",
"charge1 = (5.0/100)*units #Rs\n",
"#Interest and depreciation on motors\n",
"charge2 = 0.12*9000 #Rs\n",
"tot_chargeA = charge1 + charge2 #Rs ----------(i)\n",
"#------------------------------------------------------------------#\n",
"\n",
"#(ii)Motor B\n",
"pin_fl = 149.2/0.9 #kW (full-load power input)\n",
"pin_hl = (149.2/2)/0.89 #kW (half-load power input)\n",
"pin_hl = round(pin_hl,1)\n",
"pin_fl = round(pin_fl,1)\n",
"#Total energy consumed in a year\n",
"units = 1500.0*pin_fl + 3000.0*pin_hl #kWh\n",
"#Cost of energy is\n",
"charge1 = (5.0/100)*units #Rs\n",
"#Let the cost of motor be x\n",
"x = Symbol('x')\n",
"#Interest and depreciation on motors\n",
"charge2 = 0.12*x #Rs\n",
"tot_chargeB = charge1 + charge2 #Rs-------------(ii)\n",
"\n",
"#Equating (i)&(ii)\n",
"eq = Eq(tot_chargeA,tot_chargeB)\n",
"x = solve(eq)\n",
"x1 = x[0]\n",
"print \"Cost of Motor B is = Rs.\",round(x1,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.79 , PAGE NO :- 2003"
]
},
{
"cell_type": "code",
"execution_count": 20,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total cost of Tender A is = Rs. 938.83\n",
"Total cost of Tender B is = Rs. 916.26\n",
"Tender B is cheaper by = Rs. 22.57\n"
]
}
],
"source": [
"'''Two tenders A and B for a 1000-kVA, 0.8 power factor transformer are : A,full-load efficiency = 98.5% and iron\n",
"loss = 6 kW at rated voltage ; B, 98.8% and iron loss 4 kW but costs Rs. 1,500 more than A. The load cycle is 2000\n",
"hours per annum at full-load, 600 hours at halfload and 400 hours at 25 kVA. Annual charges for interest and\n",
"depreciation are 12.5% of capital cost and energy costs 3 paise per kWh. Which tender is better and what would be\n",
"the annual saving.'''\n",
"\n",
"#Tender A\n",
"#Transformer full-load output\n",
"fl_out = 1000 * 0.8 #kW\n",
"eff = (98.5/100) # (efficiency)\n",
"pin = fl_out/eff # (input power)\n",
"#Total losses\n",
"tot_loss = pin - fl_out #kW\n",
"#F.L. Cu losses\n",
"fl_culoss = tot_loss - 6.0 #kW\n",
"#Total losses per year for a running period of 3000 hr. areâ€”\n",
"#Iron loss\n",
"ir_loss = 3000 * 6.0 #kWh\n",
"#F.L. Cu losses for 2000 hours\n",
"fl_units = 2000 * fl_culoss #kWh\n",
"#Cu loss at half-load for 600 hours\n",
"hl_units = 600*(fl_culoss/4) #kWh \n",
"\n",
"#Cu loss at 25 kVA load for 400 hours\n",
"units_kva = ((25.0/1000)**2)*fl_culoss*400.0 #kWh\n",
"\n",
"#Total energy loss per year\n",
"enrgy_loss = ir_loss + fl_units + hl_units + units_kva #kWh\n",
"#Total Cost\n",
"costA = enrgy_loss*(3.0/100) #Rs.\n",
"\n",
"#---------------------------------------------------------------------------#\n",
"\n",
"#Tender B\n",
"#Transformer full-load output\n",
"fl_out = 1000 * 0.8 #kW\n",
"eff = (98.8/100) # (efficiency)\n",
"pin = fl_out/eff # (input power)\n",
"#Total losses\n",
"tot_loss = pin - fl_out #kW\n",
"#F.L. Cu losses\n",
"fl_culoss = tot_loss - 4.0 #kW\n",
"#Total losses per year for a running period of 3000 hr. areâ€”\n",
"#Iron loss\n",
"ir_loss = 3000 * 4 #kWh\n",
"#F.L. Cu losses for 2000 hours\n",
"fl_units = 2000 * fl_culoss #kWh\n",
"#Cu loss at half-load for 600 hours\n",
"hl_units = 600*(fl_culoss/4) #kWh \n",
"\n",
"#Cu loss at 25 kVA load for 400 hours\n",
"units_kva = ((25.0/1000)**2)*fl_culoss*400.0 #kWh\n",
"\n",
"#Total energy loss per year\n",
"enrgy_loss = ir_loss + fl_units + hl_units + units_kva #kWh\n",
"#Total Cost\n",
"costB = enrgy_loss*(3.0/100) + 1500.0*(12.5/100) #Rs.\n",
"\n",
"print \"Total cost of Tender A is = Rs.\",round(costA,2) \n",
"print \"Total cost of Tender B is = Rs.\",round(costB,2)\n",
"\n",
"if (costA > costB):\n",
" diff = costA - costB\n",
" print \"Tender B is cheaper by = Rs.\",round(diff,2)\n",
"else:\n",
" diff = costB - costA\n",
" print \"Tender A is cheaper by = Rs.\",round(diff,2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.80 , PAGE NO :- 2003"
]
},
{
"cell_type": "code",
"execution_count": 21,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Transformer B should cost Rs. 2920.0 less than A.\n"
]
}
],
"source": [
"'''Transformer A has iron loss of 150 kWh and load loss of 140 kWh daily while the corresponding losses of transformer B are 75 kWh\n",
"and 235 kWh. If annual charges are 12.5% of the capital costs and energy costs 5 paise per kWh, what should be the difference in the\n",
"cost of the two transformers so as to make them equally economical ?'''\n",
"\n",
"#Transformer A\n",
"#Annual loss\n",
"annual_lossA = 365.0*(150 + 140) #kWh (Yearly loss)\n",
"#Transformer B\n",
"annual_lossB = 365.0*(75 + 235) #kWh (Yearly loss)\n",
"\n",
"#Difference in yearly loss\n",
"tot_loss = annual_lossB - annual_lossA #kWh\n",
"\n",
"#Value of this loss\n",
"value = tot_loss*(5.0/100) #Rs\n",
"\n",
"\n",
"#As transformer B is costlier .\n",
"#Let the difference in capital cost of transformers be x\n",
"x = value/0.125 #Rs\n",
"\n",
"print \"Transformer B should cost Rs.\",round(x,2),\"less than A.\""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## EXAMPLE 50.81 , PAGE NO :- 2004"
]
},
{
"cell_type": "code",
"execution_count": 22,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Total annual charges for Transformer A is = Rs. 24804.0\n",
"Total annual charges for Transformer B is = Rs. 23046.0\n",
"Total annual charges for Transformer C is = Rs. 28601.0\n",
"Transformer B is cheaper.\n"
]
}
],
"source": [
"'''Quotations received from three sources for transformers are :\n",
" Price No-load loss Full-load loss\n",
"A Rs. 41,000 16 kW 50 kW\n",
"B Rs. 45,000 14 kW 45 kW\n",
"C Rs. 38,000 19 kW 60 kW\n",
"If the transformers are kept energized for the whole of day (24 hours), but will be on load for 12 hours per day, the remaining\n",
"period on no-load, the electricity cost being 5 paise per kWh and fixed charges Rs. 125 per kW of loss per annum and if\n",
"depreciation is 10% of the initial cost, which of the transformers would be most economical to purchase ?'''\n",
"\n",
"#Transformer A\n",
"fl_loss = 50.0 #kW (Full load losses) \n",
"nl_loss = 16.0 #kW (No load losses)\n",
"\n",
"fl_culoss = fl_loss - nl_loss #kW (Full load copper loss)\n",
"\n",
"#Cu loss for 12 hours .Therefore units consumed\n",
"cu_units = fl_culoss*12 #kWh\n",
"#Iron loss for 24 hours.Therefore units consumed\n",
"ir_units = nl_loss*24 #kWh\n",
"#Total loss per day\n",
"day_loss = cu_units + ir_units #kWh\n",
"#Annual loss is\n",
"anl_loss = day_loss*365.0 #kWh\n",
"#Cost of this loss is\n",
"loss_charge = anl_loss*(5.0/100) #Rs\n",
"#Annual fixed charges\n",
"fxd_charge = 125.0*fl_loss #Rs\n",
"#Annual depreciation\n",
"dep_charge = 0.1*41000.0 #Rs\n",
"\n",
"#Total annual charges\n",
"tot_chargeA = loss_charge + fxd_charge + dep_charge #Rs\n",
"print \"Total annual charges for Transformer A is = Rs. \",round(tot_chargeA,2)\n",
"#--------------------------------------------------------------------------------------------#\n",
"\n",
"#Transformer B\n",
"fl_loss = 45.0 #kW (Full load losses) \n",
"nl_loss = 14.0 #kW (No load losses)\n",
"\n",
"fl_culoss = fl_loss - nl_loss #kW (Full load copper loss)\n",
"\n",
"#Cu loss for 12 hours .Therefore units consumed\n",
"cu_units = fl_culoss*12 #kWh\n",
"#Iron loss for 24 hours.Therefore units consumed\n",
"ir_units = nl_loss*24 #kWh\n",
"#Total loss per day\n",
"day_loss = cu_units + ir_units #kWh\n",
"#Annual loss is\n",
"anl_loss = day_loss*365.0 #kWh\n",
"#Cost of this loss is\n",
"loss_charge = anl_loss*(5.0/100) #Rs\n",
"#Annual fixed charges\n",
"fxd_charge = 125.0*fl_loss #Rs\n",
"#Annual depreciation\n",
"dep_charge = 0.1*45000.0 #Rs\n",
"\n",
"#Total annual charges\n",
"tot_chargeB = loss_charge + fxd_charge + dep_charge #Rs\n",
"print \"Total annual charges for Transformer B is = Rs. \",round(tot_chargeB,2)\n",
"#--------------------------------------------------------------------------------------------#\n",
"\n",
"#Transformer C\n",
"fl_loss = 60.0 #kW (Full load losses) \n",
"nl_loss = 19.0 #kW (No load losses)\n",
"\n",
"fl_culoss = fl_loss - nl_loss #kW (Full load copper loss)\n",
"\n",
"#Cu loss for 12 hours .Therefore units consumed\n",
"cu_units = fl_culoss*12 #kWh\n",
"#Iron loss for 24 hours.Therefore units consumed\n",
"ir_units = nl_loss*24 #kWh\n",
"#Total loss per day\n",
"day_loss = cu_units + ir_units #kWh\n",
"#Annual loss is\n",
"anl_loss = day_loss*365.0 #kWh\n",
"#Cost of this loss is\n",
"loss_charge = anl_loss*(5.0/100) #Rs\n",
"#Annual fixed charges\n",
"fxd_charge = 125.0*fl_loss #Rs\n",
"#Annual depreciation\n",
"dep_charge = 0.1*38000.0 #Rs\n",
"\n",
"#Total annual charges\n",
"tot_chargeC = loss_charge + fxd_charge + dep_charge #Rs\n",
"print \"Total annual charges for Transformer C is = Rs. \",round(tot_chargeC,2)\n",
"#--------------------------------------------------------------------------------------------#\n",
"if(tot_chargeA < tot_chargeB):\n",
" if(tot_chargeA < tot_chargeC):\n",
" print \"Transformer A is cheaper.\"\n",
" else:\n",
" print \"Transformer C is cheaper.\"\n",
"else:\n",
" if(tot_chargeB < tot_chargeC):\n",
" print \"Transformer B is cheaper.\"\n",
" else:\n",
" print \"Transformer C is cheaper.\" \n"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
"## EXAMPLE 50.82 , PAGE NO :- 2005"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Line power factor at full-load = 0.99\n",
"Line power factor at no-load = 0.27\n"
]
}
],
"source": [
"'''A 37.3 kW induction motor has power factor 0.9 and efficiency 0.9 at fullload,power factor 0.6 and efficiency 0.7 at half-load.\n",
"At no-load, the current is 25% of the full-load current and power factor 0.1. Capacitors are supplied to make the line power factor\n",
"0.8 at half-load.With these capacitors in circuit, find the line power factor at (i) full-load and (ii) no-load.'''\n",
"\n",
"import math as m\n",
"from sympy import Symbol\n",
"\n",
"#Full-load motor input P1\n",
"p1 = 37.3/0.9 #kW\n",
"#Lagging kVAR drawn by the motor at full-load,\n",
"kvar1 = p1*(m.tan(m.acos(0.9))) #kVAR\n",
"\n",
"#Half-load motor input P2\n",
"p2 = (37.3/2)/0.7 #kW\n",
"#Lagging kVAR drawn by motor at half-load,\n",
"kvar2 = p2*(m.tan(m.acos(0.6))) #kVAR\n",
"\n",
"#Let the line voltage be Vl\n",
"Vl = Symbol('Vl')\n",
"#Full load current\n",
"I1 = 37.3e+3/(1.73*0.9*0.9*Vl) #A\n",
"\n",
"#Current at no-load\n",
"I0 = 0.25*I1\n",
"\n",
"#Motor Input at no-load P0 = 1.73*Vl*I0*cosQ\n",
"P0 = 1.73*Vl*I0*0.1/1000 #W\n",
"\n",
"#Lagging kVAR drawn by motor at no-load\n",
"kvar0 = P0*m.tan(m.acos(0.1)) #kW\n",
"\n",
"#Lagging kVAR drawn from mains at half-load\n",
"kvar1 = p2*m.tan(m.acos(0.8)) #kW\n",
"\n",
"#kVAR supplied by capacitors, kVARC = kVAR2 âˆ’ kVAR2C\n",
"cap_kvar = kvar2 - kvar1 \n",
"#kVAR drawn from the main at full-load with capacitors\n",
"fl_kvar = kvar1 - cap_kvar #kVAR\n",
"\n",
"#(i) Line power factor at full-load is\n",
"pf_fl = m.cos(m.atan(fl_kvar/p1))\n",
" \n",
"print\"Line power factor at full-load = \",round(pf_fl,2) \n",
"#----------------------------------------------------------------------------#\n",
" \n",
"#(ii) kVAR drawn from mains at no-load with capacitors\n",
"nl_kvar = kvar0 - cap_kvar #kVAR \n",
"#Line power factor at no-load\n",
"pf_nl = m.cos (m.atan(nl_kvar/P0))\n",
"print \"Line power factor at no-load = \",round(pf_nl,2)"
]
}
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