{
"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.\""
]
},
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"cell_type": "code",
"execution_count": null,
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