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Diffstat (limited to 'Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch2.ipynb')
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diff --git a/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch2.ipynb b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch2.ipynb new file mode 100755 index 00000000..f2865431 --- /dev/null +++ b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch2.ipynb @@ -0,0 +1,861 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:45c2bf36a13c4358e12d368504a2e6b1e7cd12f738f3e5c55995bd4950ad15ca" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 2 :\n", + "Load Characteristics" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.1 Page No : 39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "from numpy import linspace,array\n", + "from matplotlib.pyplot import bar,suptitle,xlabel,ylabel\n", + "\n", + "# Variables\n", + "t = linspace(0,24,25)\n", + "SL = array([100,100,100,100,100,100,100,100,0,0,0,0,0,0,0,0,0,0,100,100,100,100,100,100,100]);\n", + "R = array([200,200,200,200,200,200,200,300,400,500,500,500,500,500,500,500,500,600,700,800,1000,1000,800,600,300]);\n", + "C = array([200,200,200,200,200,200,200,200,300,500,1000,1000,1000,1000,1200,1200,1200,1200,800,400,400,400,200,200,200]);\n", + "\n", + "# Calculations\n", + "Tl = SL+R+C;\n", + "\n", + "# Results\n", + "#To print lay the Load bar curve diagram\n", + "bar(t,Tl,color='red')#,0.5,'red')\n", + "suptitle('Example 2.1')\n", + "xlabel(\"Time in hrs\")\n", + "ylabel(\"Load in kW\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "metadata": {}, + "output_type": "pyout", + "prompt_number": 7, + "text": [ + "<matplotlib.text.Text at 0x1098963d0>" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x107575d10>" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.2 Page No : 46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "Fls = 0.15;\n", + "Ppl = 80*(10**3); #Power Loss at peak load.\n", + "\n", + "# Calculations\n", + "Avgpl = Fls*Ppl; #Average Power Loss\n", + "TAELCu = Avgpl*8760; #Total annual energy loss\n", + "\n", + "# Results\n", + "print 'a) The average annual power loss = %g kW'%(Avgpl/1000)\n", + "print ' b) The total annual energy loss due to the copper losses of the feeder circuits = %g kWh'%(TAELCu/1000)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a) The average annual power loss = 12 kW\n", + " b) The total annual energy loss due to the copper losses of the feeder circuits = 105120 kWh\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.3 Page No : 47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "TCDi = [9,9,9,9,9,9]; #Load for each house all in kilowatt\n", + "DFi = 0.65; #Demand factor\n", + "Fd = 1.1; #Diversity factor\n", + "\n", + "# Calculations\n", + "Dg = sum(TCDi)*DFi/Fd;\n", + "\n", + "# Results\n", + "print 'The diversified demand of the group on the distribution transformer is %g kW'%(Dg)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The diversified demand of the group on the distribution transformer is 31.9091 kW\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.4 Page No : 48" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "SP = 3000*(10**3); #System peak in kVA per phase\n", + "Cl = 0.5/100; #Percentage of copper loss\n", + "\n", + "# Calculations\n", + "I2R = Cl*SP; #Copper loss of the feeder per phase\n", + "I2R3 = 3*I2R; #Copper losses of the feeder per 3 phase\n", + "\n", + "# Results\n", + "print 'a) The copper loss of the feeder per phase = %g kW'%(I2R/1000)\n", + "print ' b) The total coper losses of the feeder per three phase = %g kW'%(I2R3/1000)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a) The copper loss of the feeder per phase = 15 kW\n", + " b) The total coper losses of the feeder per three phase = 45 kW\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.5 Page No : 48" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "Pi = 2000.*(10**3); #Peak for industrial load\n", + "Pr = 2000.*(10**3); #Peak for residential load\n", + "Dg = 3000.*(10**3); #System peak load as specified in the diagram\n", + "P = [Pi,Pr]; #System peaks for various loads \n", + "\n", + "# Calculations\n", + "Fd = sum(P)/Dg; #Diversity factor\n", + "LD = sum(P)-Dg; #Load diversity factor\n", + "Fc = 1/Fd; # Coincidence factor\n", + "\n", + "# Results\n", + "print 'a) The diversity factor of the load is %g'%(Fd)\n", + "print ' b) The load diversity of the load is %g kW'%(LD/1000)\n", + "print ' c) The coincidence factor of the load is %g'%(Fc)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a) The diversity factor of the load is 1.33333\n", + " b) The load diversity of the load is 1000 kW\n", + " c) The coincidence factor of the load is 0.75\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.6 Page No : 50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import array,transpose\n", + "\n", + "# Variables\n", + "#Refer diagram of the first example of this chapter\n", + "Ps = 100.; #Peak load for street lighting in kW\n", + "Pr = 1000.; #Peak load for Residential in kW\n", + "Pc = 1200.; #Peak Commercial load in kW\n", + "P = array([Ps,Pr,Pc]) #Peaks of various loads\n", + "\n", + "Ls5 = 0.; #Street lighting load at 5.00 PM in kW\n", + "Lr5 = 600.; #Residential load at 5.00 PM in kW\n", + "Lc5 = 1200.; #Commercial Load at 5.00 PM in kW\n", + "\n", + "# Calculations\n", + "Cstreet = Ls5/Ps;\n", + "Cresidential = Lr5/Pr;\n", + "Ccommercial = Lc5/Pc;\n", + "C = array([Cstreet,Cresidential,Ccommercial]); #Class distribution for various factors\n", + "\n", + "Fd = (sum(P))/(sum(P* transpose(C)));\n", + "Dg = (sum(P* transpose(C)));\n", + "Fc = 1/Fd;\n", + "\n", + "print 'a The class distribution factors factor of:'\n", + "print ' i) Street lighting = %g \\\n", + "\\nii) Residential = %g \\\n", + "\\niii) Commercial = %g'%(C[0],C[1],C[2])\n", + "print ' b) The diversity factor for the primary feeder = %g'%(Fd)\n", + "print ' c) The diversified maximum demand of the load group = %g kW'%(Dg) \n", + "print ' d) The coincidence factor of the load group = %g'%(Fc)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a The class distribution factors factor of:\n", + " i) Street lighting = 0 \n", + "ii) Residential = 0.6 \n", + "iii) Commercial = 1\n", + " b) The diversity factor for the primary feeder = 1.27778\n", + " c) The diversified maximum demand of the load group = 1800 kW\n", + " d) The coincidence factor of the load group = 0.782609\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.7 Page No : 55" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "print 'Assuming a monthly load curve as shown in the figure attached\\\n", + " to this code'\n", + "\n", + "# Variables\n", + "TAE = 10.**7; # Total annual energy in kW\n", + "APL = 3500.; #Annual peak load in kW\n", + "\n", + "# Calculations\n", + "Pav = TAE/8760; #Annual average power demand\n", + "Fld = Pav/APL; #Annual load factor\n", + "\n", + "# Results\n", + "print 'a) The annual power demand is %g kW'%(Pav)\n", + "print 'b) The annual load factor is %g'%(Fld)\n", + "print 'The unsold energy, as shown in the figure is a measure of capacity and investment math.cost. Ideally it should be kept at a minimum'\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Assuming a monthly load curve as shown in the figure attached to this code\n", + "a) The annual power demand is 1141.55 kW\n", + "b) The annual load factor is 0.326158\n", + "The unsold energy, as shown in the figure is a measure of capacity and investment math.cost. Ideally it should be kept at a minimum\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.8 Page No : 57" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "print 'Assuming a monthly load curve as shown in the figure attached to this code'\n", + "\n", + "# Variables\n", + "Nl = 100.; #100% percent load to be supplied\n", + "TAE = 10.**7; # Total annual energy in kW\n", + "APL = 3500.; #Annual peak load in kW\n", + "\n", + "# Calculations\n", + "Pav = TAE/8760; #Annual average power demand\n", + "Fld = (Pav+Nl)/(APL+Nl); #Annual load factor\n", + "Cr = 3; #Capacity math.cost\n", + "Er = 0.03; #Energy math.cost\n", + "ACC = Nl*12*Cr; #Additional capacity math.cost per kWh\n", + "AEC = Nl*8760*Er; #Additional energy math.cost per kWh\n", + "TAC = ACC+AEC; #Total annual math.cost\n", + "\n", + "# Results\n", + "print 'a) The new annual load factor on the substation is %g'%(Fld)\n", + "print 'b) The total annual additional costs to NL&NP to serve this load is $%g'%(TAC)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Assuming a monthly load curve as shown in the figure attached to this code\n", + "a) The new annual load factor on the substation is 0.344876\n", + "b) The total annual additional costs to NL&NP to serve this load is $29880\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.9 Page No : 58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "TAE = 5.61*(10**6); #Total annual energy in kW\n", + "APL = 2000.; #Annual peak load in kW\n", + "Lc = 0.03; #Cost of energy per kWh in dollars\n", + "Plp = 100.; #Power at peak load in kW\n", + "\n", + "# Calculations\n", + "Fld = TAE/(APL*8760); \n", + "Fls = (0.3*Fld)+(0.7*(Fld**2));\n", + "AvgEL = Fls*Plp; #Average energy loss\n", + "AEL = AvgEL*8760; #Annual energy loss\n", + "Tlc = AEL*Lc; #Cost of total annual copper loss\n", + "\n", + "# Results\n", + "print 'a) The annual loss factor is %g'%(Fls)\n", + "print ' b) The annual copper loss energy is %g kWh and the cost of total annual copper loss is $%g'%(AEL,Tlc)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a) The annual loss factor is 0.167834\n", + " b) The annual copper loss energy is 147022 kWh and the cost of total annual copper loss is $4410.67\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.10 Page No : 59" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "from numpy import array,arctan,transpose,arccos,radians\n", + "from sympy import Symbol,solve\n", + "\n", + "Fd = 1.15;\n", + "Pi = [1800,2000,2200]; #Demands of various feeders in kW (Real Power)\n", + "PF = [0.95,0.85,0.90]; #Power factor of the respective feeders\n", + "Dg = sum(Pi)/Fd;\n", + "P = Dg;\n", + "theta = arccos(PF);\n", + "\n", + "Q = sum(Pi*(radians(arctan(theta))))/Fd;\n", + "S = math.sqrt((P**2)+(Q**2));\n", + "LD = sum(Pi)-Dg;\n", + "\n", + " #Transformer sizes\n", + "Tp = array([2500,3750,5000,7500]);\n", + "Ts = array([3125,4687,6250,9375]); \n", + "\n", + "Ol = 1.25; #Maximum overload condition\n", + "Eol = Ts*Ol; #Overload voltages of the transformer\n", + "Ed = abs(Eol-S); # Difference between the overload values of the transformers and the P value of the system\n", + "\n", + "A = sorted(Ed); # To sort the differences and choose the best match\n", + "\n", + "T = [Tp[1],Ts[1]]; #Suitable transformer\n", + "\n", + "g = Symbol('g');\n", + "X = (1+g)-pow(2,1./10); #To find out the fans on rating\n", + "R = solve(X)[0];\n", + "g = R*100;\n", + "\n", + "n = Symbol('n');\n", + "Sn = 9375.; # Rating of the to be installed transformer\n", + " # Equation (1+g)**n * S = Sn\n", + " # a = (1+g)\n", + " # b = Sn/S\n", + "\n", + "a = 1+R;\n", + "b = Sn/S;\n", + "n = math.log(b)/math.log(a);\n", + "\n", + "print 'a) The 30 mins annual maximum deman on the substation transformer are %g kW and %g kVA respectively'%(P,S)\n", + "print ' b) The load diversity is %g kW'%(LD)\n", + "print ' c) Suitable transformer size for 25 percent short time over loads is %g/%g kVA'%(T[0],T[1])\n", + "print ' d) Fans on rating is %g percent and it will loaded for %g more year if a 7599/9375 kVA transformer is installed'%(g,math.ceil(n))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a) The 30 mins annual maximum deman on the substation transformer are 5217.39 kW and 5217.53 kVA respectively\n", + " b) The load diversity is 782.609 kW\n", + " c) Suitable transformer size for 25 percent short time over loads is 3750/4687 kVA\n", + " d) Fans on rating is 7.17735 percent and it will loaded for 9 more year if a 7599/9375 kVA transformer is installed\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.11 Page No : 62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import array\n", + "\n", + "print 'NOTE'\n", + "print 'The figure 1 attached along with this code is the Maximum diversified 30- min demand characteristics\\\n", + "of various residential loads; A = Clothes dryer; D = range; E = lighting and miscellaneous appliances; \\\n", + "J = refrigerator Only the loads required for this problem have been mentioned '\n", + "\n", + "Ndt = 50.; #Number of distribution transformers\n", + "Nr = 900.; #Number of residences\n", + "\n", + " #When the loads are six.\n", + "PavMax6 = array([1.6,0.8,0.066,0.61]); #Average Maximum diversified demands (in kW) per house for dryer, range, refrigerator, for lighting and misc aapliances respectively according to the figure 1 attached with code. \n", + "\n", + "Mddt = sum(6*PavMax6); #30 min maximum diversified demand on the distribution transformer\n", + "\n", + " #When the loads are 900.\n", + "PavMax900 = array([1.2,0.53,0.044,0.52]); # #Average Maximum diversified demands (in kW) per house for dryer, range, refreigerato, for lighting and misc aapliances respectively according to the figure 1 attached with code.\n", + "\n", + "Mdf = sum(Nr*PavMax900); #30 min maximum diversified demand on the feeder\n", + "\n", + " #From the figure 2 attached to this code\n", + "Hdd4 = array([0.38,0.24,0.9,0.32]); #Hourly variation factor at time 4 PM for dryer, range, refrigerator, lighting and misc appliances\n", + "Hdd5 = array([0.30,0.80,0.9,0.70]); #Hourly variation factor at time 5 PM for dryer, range, refrigerator, lighting and misc appliances\n", + "Hdd6 = array([0.22,1.0,0.9,0.92]); #Hourly variation factor at time 6 PM for dryer, range, refrigerator, lighting and misc appliances\n", + "\n", + "Thdd4 = (6*PavMax6)*(Hdd4.T); #Gives the total hourly diversified demand in kW at time 4 PM\n", + "Thdd5 = (6*PavMax6)*Hdd5; #Gives the total hourly diversified demand in kW at time 5 PM\n", + "Thdd6 = (6*PavMax6)*Hdd6; #Gives the total hourly diversified demand in kW at time 6 PM\n", + "\n", + "print ' a) The 30 min maximum diversified demand on the distribution transformer = %g kW'%(Mddt)\n", + "print ' b) The 30 min maximum diversified demand on the distribution transformer = %g kW'%(Mdf)\n", + "print ' c The total hourly diversified demands at:'\n", + "print ' i) 4.00 PM = %g kW'%(sum(Thdd4))\n", + "print ' ii) 5.00 PM = %g kW'%(sum(Thdd5))\n", + "print ' iii) 6.00 PM = %g kW'%(sum(Thdd6))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "NOTE\n", + "The figure 1 attached along with this code is the Maximum diversified 30- min demand characteristicsof various residential loads; A = Clothes dryer; D = range; E = lighting and miscellaneous appliances; J = refrigerator Only the loads required for this problem have been mentioned \n", + " a) The 30 min maximum diversified demand on the distribution transformer = 18.456 kW\n", + " b) The 30 min maximum diversified demand on the distribution transformer = 2064.6 kW\n", + " c The total hourly diversified demands at:\n", + " i) 4.00 PM = 6.3276 kW\n", + " ii) 5.00 PM = 9.6384 kW\n", + " iii) 6.00 PM = 10.6356 kW\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.12 Page No : 72" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from numpy import array,arctan,radians,degrees,ceil\n", + "\n", + "T = 730; #Average monthly time in hrs\n", + "Pla = 22; #Peak Load for consumer A in kW\n", + "Plb = 39; #Peak load for consumer B in kW\n", + "MEC = array([0.025,0.02,0.015]); #Monthly Energy charges in cents/kWh according to the units consumed\n", + "Uc = array([1000,3000,3000]); #Units consumption according to the Energy charges\n", + "MDC = 2; #Monthly demand charge in dollars/kW\n", + "\n", + "Pa = 7000.; #Units served to Consumer A in kWh\n", + "Pb = 7000.; #Units served to Consumer B in kWh\n", + "\n", + "#Power factors\n", + "Pfa = 0.9; # Lag\n", + "Pfb = 0.76; #Lag\n", + "\n", + "#Monthly Load factors\n", + "Flda = Pa/(Pla*T);\n", + "Fldb = Pb/(Plb*T);\n", + "\n", + "#Continous kilovoltamperes for each distribution transformer\n", + "Sa = Pla/Pfa;\n", + "Sb = Plb/Pfb;\n", + "\n", + "#Ratings of the distribution transformers needed\n", + "Ta = round(Sa/5)*5;\n", + "Tb = round(Sb/5)*5;\n", + "\n", + "#Billing Charges\n", + "#For Consumer A\n", + "Mbda = Pla*(0.85/Pfa); # Monthly billing demand\n", + "Mdca = Mbda*MDC; #Monthly demand charge\n", + "#Since the units served are 7000 it can be split according to the rates as 1000, 3000, 3000 excess units.\n", + "Uca = Uc; #Units consumption by A\n", + "Meca = MEC*Uca.T; #Monthly energy charge\n", + "Tmba = Meca+Mdca; #Total monthly bill\n", + "\n", + "#For Consumer B\n", + "Mbdb = Plb*(0.85/Pfb); # Monthly billing demand\n", + "Mdcb = Mbdb*MDC; #Monthly demand charge\n", + "#Since the units served are 7000 it can be split according to the rates as 1000, 3000, 3000 excess units.\n", + "Ucb = Uc; #Units consumption by B\n", + "Mecb = MEC*Ucb.T; #Monthly energy charge\n", + "Tmbb = Mecb+Mdcb; #Total monthly bill\n", + "\n", + "#To find the capacitor size\n", + "Q1 = Pb*math.radians(arctan(math.acos(Pfb))); #For original power factor\n", + "Q2 = Pb*math.radians(arctan(math.acos(0.85))); #For new power factor\n", + "\n", + "dQ = (Q1-Q2)/T; #Capacitor size\n", + "\n", + "#For new power factor\n", + "#For Consumer B\n", + "Mbdbn = Plb*(1); # Monthly billing demand\n", + "Mdcbn = Mbdbn*MDC; #Monthly demand charge\n", + "#Since the units served are 7000 it can be split according to the rates as 1000, 3000, 3000 excess units.\n", + "Ucbn = Uc; #Units consumption by B\n", + "Mecbn = MEC*Ucbn.T; #Monthly energy charge\n", + "Tmbbn = Mecbn+Mdcbn; #Total monthly bill\n", + "\n", + "Saving = abs(Tmbbn-Tmbb); #Saving due to capacitor installation\n", + "Ci = 30; # Cost of capacitor in dollar per kVAr\n", + "Cc = Ci*dQ; #The math.cost of the installed capacitor\n", + "PP = Cc/Saving; #Payback Period\n", + "PPr = 90/Saving; #Realistic Payback period\n", + "\n", + "print 'a Monthly load factor for :'\n", + "print ' i) Consumer A = %g'%(Flda)\n", + "print ' ii) Consumer B = %g'%(Fldb)\n", + "print ' b Rating of the each of the distribution transformer:'\n", + "print ' i) A = %g kVA'%(Ta)\n", + "print ' ii) B = %g kVA'%(Tb)\n", + "print ' c Monthly bil for:'\n", + "print ' i) Consumer A = $%g'%(sum(Tmba)) \n", + "print ' ii) Consumer B = $%g'%(sum(Tmbb))\n", + "print ' d) The capacitor size required is %g kVAr'%(dQ)\n", + "print ' e Payback period:'\n", + "print ' i) Calculated : %g months'%(sum(ceil(PP)))\n", + "print ' ii) Realistic as capacitor size available is 3 kVAr : %g months'%(sum(ceil(PPr)))\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a Monthly load factor for :\n", + " i) Consumer A = 0.435866\n", + " ii) Consumer B = 0.245873\n", + " b Rating of the each of the distribution transformer:\n", + " i) A = 25 kVA\n", + " ii) B = 50 kVA\n", + " c Monthly bil for:\n", + " i) Consumer A = $254.667\n", + " ii) Consumer B = $391.711\n", + " d) The capacitor size required is 0.018276 kVAr\n", + " e Payback period:\n", + " i) Calculated : 3 months\n", + " ii) Realistic as capacitor size available is 3 kVAr : 30 months\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.13 Page No : 84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables \n", + "Kh = 7.2; #Meter consmath.tant\n", + "Kr1 = 32; #Revolutions of the disk in the first reading\n", + "Kr2 = 27; #Revolutions of the disk in the second reading\n", + "T1 = 59; #Time interval for revolutions of disks for the first reading\n", + "T2 = 40; #Time interval for revolutions of disks for the second reading\n", + "\n", + "# Calculations\n", + "# Self contained watthour meter; D = (3.6*Kr*Kh)/T\n", + "\n", + "def Id1(a,b):\n", + " return ((3.6*Kh*a)/b) #Function to calculate the instaneous demand\n", + "\n", + "D1 = Id1(Kr1,T1);\n", + "D2 = Id1(Kr2,T2);\n", + "Dav = (D1+D2)/2;\n", + "\n", + "# Results\n", + "print 'The instantenous demands are %g kW and %g kW for reading 1 and 2 and the average demand is %g kW'%(D1,D2,Dav)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The instantenous demands are 14.0583 kW and 17.496 kW for reading 1 and 2 and the average demand is 15.7772 kW\n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.14 Page No : 84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "#For a transformer type watthour meter; D = (3.6*Kr*Kh*CTR*PTR)/T\n", + "CTR = 200.;\n", + "PTR = 1.;\n", + "Kh = 1.8;\n", + "Kr1 = 32.; #Revolutions of the disk in the first reading\n", + "Kr2 = 27.; #Revolutions of the disk in the second reading\n", + "T1 = 59.; #Time interval for revolutions of disks for the first reading\n", + "T2 = 40.; #Time interval for revolutions of disks for the second reading\n", + "\n", + "# Calculations\n", + "def Id1(a,b):\n", + " return ((3.6*Kh*a*CTR*PTR)/b) #Function to calculate the instaneous demand\n", + "\n", + "D1 = Id1(Kr1,T1);\n", + "D2 = Id1(Kr2,T2);\n", + "Dav = (D1+D2)/2;\n", + "\n", + "# Results\n", + "print 'The instantenous demands are %g kW and %g kW for reading 1 and 2 and the average demand is %g kW'%(D1,D2,Dav)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The instantenous demands are 702.915 kW and 874.8 kW for reading 1 and 2 and the average demand is 788.858 kW\n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.15 Page No : 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "# Variables\n", + "Kh = 1.2;\n", + "CTR = 80.;\n", + "PTR = 20.;\n", + "#Revolutions of the disk in a watthour meter for 1 and 2 readings respectively\n", + "Kr1 = 20.;\n", + "Kr2 = 30.;\n", + "#Revolutions of the disk in a VArhour meter for 1 and 2 readings respectively\n", + "Kr3 = 10.;\n", + "Kr4 = 20.\n", + "#Time interval for revoltion of disks in watthour meter for 1 and 2 readings respectively\n", + "T1 = 50.;\n", + "T2 = 60.;\n", + "#Time interval for revoltion of disks in VArhour meter for 1 and 2 readings respectively\n", + "T3 = 50.;\n", + "T4 = 60.;\n", + "\n", + "def Id1(a,b):\n", + " return ((3.6*Kh*a*CTR*PTR)/b) #Function to calculate the instaneous demand\n", + "\n", + "#instanmath.taneous kilowatt demands for readings 1 and 2\n", + "D1 = Id1(Kr1,T1);\n", + "D2 = Id1(Kr2,T2);\n", + "\n", + "#instanmath.taneous kilovar deamnds for readings 1 and 2\n", + "D3 = Id1(Kr3,T3);\n", + "D4 = Id1(Kr4,T4);\n", + "\n", + "Davp = (D1+D2)/2; #Average kilowatt demand\n", + "Davq = (D3+D4)/2; #Average kilovar demand\n", + "\n", + "Dav = math.sqrt((Davp**2)+(Davq**2)); #Average kilovoltampere demand\n", + "\n", + "# Results\n", + "print 'a) The instanmath.taneous kilowatt hour demands for readings 1 and 2 are %g kW and %g kW respectively'%(D1,D2)\n", + "print ' b) The average kilowatt demand is %g kW'%(Davp)\n", + "print ' c) The instanmath.taneous kilovar hour demands for readings 1 and 2 are %g kVAr and %g kVAr respectively'%(D3,D4)\n", + "print ' d) The average kilovar demand is %g kVAr'%(Davq)\n", + "print ' e) The average kilovoltampere demand is %g kVA'%(Dav)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a) The instanmath.taneous kilowatt hour demands for readings 1 and 2 are 2764.8 kW and 3456 kW respectively\n", + " b) The average kilowatt demand is 3110.4 kW\n", + " c) The instanmath.taneous kilovar hour demands for readings 1 and 2 are 1382.4 kVAr and 2304 kVAr respectively\n", + " d) The average kilovar demand is 1843.2 kVAr\n", + " e) The average kilovoltampere demand is 3615.52 kVA\n" + ] + } + ], + "prompt_number": 43 + } + ], + "metadata": {} + } + ] +}
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