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diff --git a/A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter50.ipynb b/A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter50.ipynb new file mode 100644 index 00000000..695fb3bf --- /dev/null +++ b/A_TEXTBOOK_OF_ELECTRICAL_TECHNOLOGY_(VOL-III)_by_B.L.Thareja/chapter50.ipynb @@ -0,0 +1,5204 @@ +{ + "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)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |