{
"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 × 105; fuel cost (annual) = Rs. 5 × 106 ; \n",
"maintenance and repairs (annual) Rs. 2 × 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^–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 Kelvin’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 ; η at full-load=90% ; at half-load = 86%.\n",
"Motor B. Cost = Rs. 4000 ; η 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 × 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. are—\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. are—\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 − 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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