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diff --git a/Principles_of_Power_System/chapter13.ipynb b/Principles_of_Power_System/chapter13.ipynb new file mode 100755 index 00000000..9f2fa9af --- /dev/null +++ b/Principles_of_Power_System/chapter13.ipynb @@ -0,0 +1,1972 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:9a20e0c71265585c31a8df4c18dfac853044b054341d1794943bfacb3c219678"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 13: D.C. Distribution"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.1, Page number: 313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable Declaration:\n",
+ "#given a 2-wire d.c. distributor cable AB\n",
+ "\n",
+ "r = 0.01*2 #Resistance per 1000m of distributor\n",
+ "Va = 300 #p.d at point A(V)\n",
+ "AC = 500 #m\n",
+ "CD = 500 #m\n",
+ "DE = 600 #m\n",
+ "EB = 400 #m\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "Rac = r*AC/1000 #Resistance of section AC\n",
+ "Rcd = r*CD/1000 #Resistance of section CD\n",
+ "Rde = r*DE/1000 #Resistance of section DE\n",
+ "Reb = r*EB/1000 #Resistance of section EB\n",
+ "Ieb = 50 #current in branch EB(A)\n",
+ "Ide = Ieb+200 #current in branch DE(A)\n",
+ "Icd = Ide+150 #current in branch CD(A)\n",
+ "Iac = Icd+100 #current in branch AC(A)\n",
+ "Vc = Va - Iac*Rac #P.D. at load point C(V)\n",
+ "Vd = Vc - Icd*Rcd #P.D. at load point D(V)\n",
+ "Ve = Vd - Ide*Rde #P.D. at load point E(V)\n",
+ "Vb = Ve - Ieb*Reb #P.D. at load point B(V)\n",
+ "\n",
+ "#Result:\n",
+ "print \"P.D. at load point C is\",Vc,\"V\"\n",
+ "print \"P.D. at load point D is\",Vd,\"V\"\n",
+ "print \"P.D. at load point E is\",Ve,\"V\"\n",
+ "print \"P.D. at load point B is\",Vb,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "P.D. at load point C is 295.0 V\n",
+ "P.D. at load point D is 291.0 V\n",
+ "P.D. at load point E is 288.0 V\n",
+ "P.D. at load point B is 287.6 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.2, Page number: 314"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from sympy import *\n",
+ "\n",
+ "\n",
+ "#Variable declaration:\n",
+ "AB = 300 #m\n",
+ "AC = 40 #m\n",
+ "CD = 60 #m\n",
+ "DE = 50 #m\n",
+ "EF = 100 #m\n",
+ "FB = 50 #m\n",
+ "Vm = 10 #max. permissible volatge(V)\n",
+ "rho = 1.78*10**-8 #resistivity of cable(ohm)\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "Iac = 220 #A\n",
+ "Icd = 190 #A\n",
+ "Ide = 150 #A\n",
+ "Ief = 50 #A\n",
+ "#Suppose that resistance of 100m length of the distributor is r ohms.\n",
+ "l = 100 #m\n",
+ "r = symbols('r')\n",
+ "Rac = AC*r/100 #ohm\n",
+ "Rcd = CD*r/100 #ohm\n",
+ "Rde = DE*r/100 #ohm\n",
+ "Ref = EF*r/100 #ohm\n",
+ "\n",
+ "Vt = Iac*Rac+Icd*Rcd+Ide*Rde+Ief*Ref\n",
+ "r1 = solve(Vt-Vm,r)[0]\n",
+ "a = float(rho*l/(r1/2)) #cm**2\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"X-sectional area of conductor is\",round(a*10**4,3),\"cm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "X-sectional area of conductor is 1.164 cm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.3, Page number: 315"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable Declaration:\n",
+ "r1 = 0.25 #resistance of trolley wire(ohm/km)\n",
+ "r2 = 0.03 #resistance of of track(ohm/km)\n",
+ "V = 600 #sub-station voltage(V)\n",
+ "SA = 2 #km\n",
+ "AB = 4 #km\n",
+ "\n",
+ "#Calculation:\n",
+ "R = r1+r2 #ohm/km\n",
+ "Isa = 40+20 #A\n",
+ "Iab = 20 #A\n",
+ "Vsa = Isa*R*SA #V\n",
+ "Vab = Iab*R*AB #V\n",
+ "Va = V-Vsa #Voltage across tram A\n",
+ "Vb = Va-Vab #Voltage across tram A\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"Voltage across tram A is\",Va,\"V\"\n",
+ "print \"Voltage across tram B is\",Vb,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage across tram A is 566.4 V\n",
+ "Voltage across tram B is 544.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.4, Page number: 315"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "a = 0.27 #cross-sectional area of each conductor(cm**2)\n",
+ "V = 250 #supply voltage at point A(V)\n",
+ "rho = 1.78*10**-6 #Resistivity of the wire(ohm-cm)\n",
+ "AB = 75 #m\n",
+ "BC = 100 #m\n",
+ "CD = 50 #m\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "Icd = 20 #curent in CD(A)\n",
+ "Ibc = 20+15 #current in BC(A)\n",
+ "Iab = 20+15+12 #current in AB(A)\n",
+ "R = round(rho*BC*100/a,3) #Single-core resistance of the section of 100 m length(ohm)\n",
+ "Rab = R*AB/100*2 #ohm\n",
+ "Rbc = R*BC/100*2 #ohm\n",
+ "Rcd = R*CD/100*2 #ohm\n",
+ "Vb = V-Iab*Rab #Voltage at tapping point B(V)\n",
+ "Vc = Vb-Ibc*Rbc #Voltage at tapping point C(V)\n",
+ "Vd = Vc-Icd*Rcd #Voltage at tapping point D(V)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i)The current in various sections of the conductor are:\"\n",
+ "print \"\\tIcd =\",Icd,\"A \\tIbc =\",Ibc,\"A\\tIab =\",Iab,\"A\"\n",
+ "print \"\\n(ii)The resistances of the various sections are :\"\n",
+ "print \"\\tRab =\",Rab,\"ohm\\tRbc =\",Rbc,\"ohm\\t\",Rcd,\"ohm\"\n",
+ "print \"\\n(iii)Voltage at tapping point B is\",round(Vb,2),\"V\"\n",
+ "print \" Voltage at tapping point C is\",round(Vc,2),\"V\"\n",
+ "print \" Voltage at tapping point D is\",round(Vd,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i)The current in various sections of the conductor are:\n",
+ "\tIcd = 20 A \tIbc = 35 A\tIab = 47 A\n",
+ "\n",
+ "(ii)The resistances of the various sections are :\n",
+ "\tRab = 0.099 ohm\tRbc = 0.132 ohm\t0.066 ohm\n",
+ "\n",
+ "(iii)Voltage at tapping point B is 245.35 V\n",
+ " Voltage at tapping point C is 240.73 V\n",
+ " Voltage at tapping point D is 239.41 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.5, Page number: 317"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "\n",
+ "\n",
+ "#Variable declaration:\n",
+ "i = 2 #A/m\n",
+ "l = 200 #length of distributor(m)\n",
+ "r1 = 0.3 #Resistance of single wire is 0\u00b73 ohm/km.\n",
+ "x = 150 #m\n",
+ "\n",
+ "#Calculation:\n",
+ "r = 2*r1/1000 #Resistance of distributor per metre run(ohm)\n",
+ "V = i*r*(l*x-x**2/2) #volt\n",
+ "I = i*l #Total current entering the distributor(A)\n",
+ "R = r*l #Total resistance of the distributor(ohm)\n",
+ "V1 = 1/2*I*R #Total drop over the distributor(V)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i)The voltage drop upto a distance of 150m from \"\n",
+ "print \"\\tthe feeding point is\",V,\"V\"\n",
+ "print \"(ii)The maximum voltage drop is\",V1,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i)The voltage drop upto a distance of 150m from \n",
+ "\tthe feeding point is 22.5 V\n",
+ "(ii)The maximum voltage drop is 24.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.6, Page number: 317"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration:\n",
+ "i = 0.4 #A/m\n",
+ "l = 500 #length of distributor(m)\n",
+ "Vm = 10 #maximum permissible voltage drop(V)\n",
+ "rho = 1.7*10**-6 #ohm-cm\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "I = i*l #Current entering the distributor(A)\n",
+ "#Let r ohm be the resistance per metre length of the distributor (both wires).\n",
+ "r = 2*Vm/(I*l) #resistance per metre length(ohm)\n",
+ "a = rho*100/(r/2) #Area of cross-section of the distributor conductor(ohm)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The cross-sectional area of the distributor conductor is\",a,\"cm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The cross-sectional area of the distributor conductor is 1.7 cm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.7, Page number: 318"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration:\n",
+ "i = 1.6 #A/m\n",
+ "l = 250 #length of distributor(m)\n",
+ "r1 = 0.0002 #Resistance of single wire is 0\u00b73 ohm/m.\n",
+ "x = 150 #m\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "I = i*l #Current entering the distributor(A)\n",
+ "r = 2*r1 #Resistance of the distributor per metre run\n",
+ "R = r*l #Total resistance of distributor(ohm)\n",
+ "V1 = 1/2*I*R #Voltage drop over the entire distributor(ohm)\n",
+ "#the voltage necessary at feed point to maintain 250 V at the far end\n",
+ "V2 = V1+250 #Voltage at feeding point(V)\n",
+ "#Voltage drop upto a distance of 150 metres from feeding point\n",
+ "V3 = i*r*(l*125-125**2/2) #Volt\n",
+ "##Voltage necessary to maintain 250V at the mid-point of the distributor.\n",
+ "V4 = 250+V3\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i) Voltage drop over the entire distributor is\",V2,\"V\"\n",
+ "print \"(ii)The necessary voltage is\",V4,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Voltage drop over the entire distributor is 270.0 V\n",
+ "(ii)The necessary voltage is 265.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.9, Page number: 319"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration:\n",
+ "i = 0.75 #current loading(A/m)\n",
+ "l = 300 #distributor length(m)\n",
+ "x = 200 #m\n",
+ "r = 0.00018 #resistance of distributor(go and return)(ohm/m)\n",
+ "V = 250 #voltage fed at 1 end(V)\n",
+ "\n",
+ "#Calculation:\n",
+ "V1 = i*r*(l*x-x**2/2) #Voltage drop(V)\n",
+ "#Voltage at a distance of 200 m from supply end\n",
+ "V2 = V-V1 #V\n",
+ "P = i**2*r*l**3/3 #Power loss in the distributor(W)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"Volatage at a distance of 200m is\",V2,\"V\"\n",
+ "print \"Power loss in the distributor is\",P,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Volatage at a distance of 200m is 244.6 V\n",
+ "Power loss in the distributor is 911.25 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.10, Page number: 321"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from sympy import *\n",
+ "\n",
+ "\n",
+ "#Variable declaration:\n",
+ "AB = 600 #m\n",
+ "Va = 220 #end voltage(V)\n",
+ "a = 1 #cross-section of conductor(cm**2)\n",
+ "rho = 1.7*10**-6 #resistvity of conductor(ohm-cm)\n",
+ "\n",
+ "#Calculation:\n",
+ "\n",
+ "#Let Ia amperes be the current supplied from the feeding\n",
+ "#end A.\n",
+ "r = 2*rho*100/a #ohm\n",
+ "Rac = r*100 #ohm\n",
+ "Rcd = r*150 #ohm\n",
+ "Rde = r*150 #ohm\n",
+ "Ref = r*100 #ohm\n",
+ "Rfb = r*100 #ohm\n",
+ "#Voltage at B = Voltage at A \u2212 Drop over length AB\n",
+ "Ia = symbols('Ia')\n",
+ "Ia1 = solve(Va-(Ia*Rac+(Ia-20)*Rcd+(Ia-60)*Rde+(Ia-110)*Ref+(Ia-140)*Rfb)-Va,Ia)[0]\n",
+ "#We can see that currents are coming to load point E from \n",
+ "#both sides i.e. from point D and point F. \n",
+ "#So, E will be the point of minimum potential.\n",
+ "\n",
+ "Ve = Va-(61.7*Rac+41.7*Rcd+1.7*Rde) #Minimum consumer voltage(V)\n",
+ "\n",
+ "#Result:\n",
+ "print \"Minimum consumer voltage is\",round(Ve,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum consumer voltage is 215.69 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.11, Page number: 322"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "from sympy import *\n",
+ "\n",
+ "#Variable declaration:\n",
+ "AB = 200 #m\n",
+ "Va = 230 #1 end voltage(V)\n",
+ "Vb = 235 #2 end voltage(V)\n",
+ "r1 = 0.3 #The resistance per km of one conductor(ohm)\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "#Let Ia amperes be the current supplied from the feeding\n",
+ "#end A.\n",
+ "r = 2*r1 #Resistance of 1000m length of distributor(both wires)\n",
+ "Rac = r*50/1000 #ohm\n",
+ "Rcd = r*25/1000 #ohm\n",
+ "Rde = r*25/1000 #ohm\n",
+ "Ref = r*50/1000 #ohm\n",
+ "Rfb = r*50/1000 #ohm\n",
+ "Ia = symbols('Ia')\n",
+ "Ia1 = round(solve(Va-(Ia*Rac+(Ia-25)*Rcd+(Ia-75)*Rde+(Ia-105)*Ref+(Ia-145)*Rfb)-Vb,Ia)[0],2)\n",
+ "Iac = Ia1 #Current in section AC(A)\n",
+ "Icd = Ia1-25 #Current in section CD(A)\n",
+ "Ide = Ia1-75 #Current in section DE(A)\n",
+ "Ief = Ia1-105 #Current in section EF(A)\n",
+ "Ifb = Ia1 -145 #Current in section FB(A)\n",
+ "#The currents are coming to load point D from both sides of \n",
+ "#the distributor. Therefore, load point D is the point of minimum potential.\n",
+ "Vd = Va-(Iac*Rac+Icd*Rcd) #volt\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i) Currents in various sections of the distributor are:\"\n",
+ "print \"\\tCurrent in section AC is\",Iac,\"A\"\n",
+ "print \"\\tCurrent in section AC is\",Icd,\"A\"\n",
+ "print \"\\tCurrent in section AC is\",Ide,\"A\"\n",
+ "print \"\\tCurrent in section AC is\",Ief,\"A\"\n",
+ "print \"\\tCurrent in section AC is\",Ifb,\"A\"\n",
+ "print \"\\n(ii)Voltage at D is\",round(Vd,3),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Currents in various sections of the distributor are:\n",
+ "\tCurrent in section AC is 33.33 A\n",
+ "\tCurrent in section AC is 8.33 A\n",
+ "\tCurrent in section AC is -41.67 A\n",
+ "\tCurrent in section AC is -71.67 A\n",
+ "\tCurrent in section AC is -111.67 A\n",
+ "\n",
+ "(ii)Voltage at D is 228.875 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.12, Page number: 323"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "from sympy import *\n",
+ "\n",
+ "\n",
+ "#Variable declaration:\n",
+ "AB = 600 #m\n",
+ "Va = 440 #1 end voltage(V)\n",
+ "Vb = 430 #2 end voltage(V)\n",
+ "r1 = 0.01 #The resistance per 100m of one conductor(ohm)\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "#Let Ia amperes be the current supplied from the feeding\n",
+ "#end A.\n",
+ "r = 2*r1 #Resistance of 1000m length of distributor(both wires)\n",
+ "Rac = r*150/100 #ohm\n",
+ "Rcd = r*150/100 #ohm\n",
+ "Rde = r*50/100 #ohm\n",
+ "Ref = r*100/100 #ohm\n",
+ "Rfb = r*150/100 #ohm\n",
+ "Ia = symbols('Ia')\n",
+ "Ia1 = round(solve(Va-(Ia*Rac+(Ia-100)*Rcd+(Ia-300)*Rde+(Ia-550)*Ref+(Ia-850)*Rfb)-Vb,Ia)[0],2)\n",
+ "Iac = Ia1 #Current in section AC(A)\n",
+ "Icd = Ia1-100 #Current in section CD(A)\n",
+ "Ide = Ia1-300 #Current in section DE(A)\n",
+ "Ief = Ia1-550 #Current in section EF(A)\n",
+ "Ifb = Ia1-850 #Current in section FB(A)\n",
+ "Pl = Iac**2*Rac+Icd**2*Rcd+Ide**2*Rde+Ief**2*Ref+Ifb**2*Rfb #power loss(W)\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i) The currents supplied from A to B are:\\n\\tIa =\",Ia1,\"A\\tIb =\",abs(Ifb),\"A\" \n",
+ "print \"(ii) The power dissipated in the distributor is\",round(Pl/1000,3),\"kW\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The currents supplied from A to B are:\n",
+ "\tIa = 437.5 A\tIb = 412.5 A\n",
+ "(ii) The power dissipated in the distributor is 14.706 kW\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.13, Page number: 325"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "from sympy import *\n",
+ "\n",
+ "#Variable declaration:\n",
+ "Va = 600 #V\n",
+ "Vb = 590 #V\n",
+ "r = 0.04 #track resistance of go and return path(ohm/km)\n",
+ "AB = 6 #distance b/w the sub-stations(km)\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "#Let min. potential occurs at point M at a distance x km from the substation A.\n",
+ "\n",
+ "\n",
+ "Ia,x = symbols('Ia x') #Ia is current supplied by the sub-station A\n",
+ "Ram = r*x #Track resistance for section AM(ohm)\n",
+ "Rmb = r*(AB-x) #Track resistance for section MB(ohm)\n",
+ "Vm = Va-Ia*Ram #Potential at M(V) ...(i)\n",
+ "Vm1 = Vb-(300-Ia)*Rmb #also,Potential at M(V) ...(ii)\n",
+ "\n",
+ "#from (i) & (ii):\n",
+ "Ia = 341.7-50*x\n",
+ "Vm = Va-(341.7 - 50*x)*0.04*x\n",
+ "Vm2 = diff(Vm,x)\n",
+ "x1 = round(solve(Vm2,x)[0],2)\n",
+ "Ia1 = 341.7-50*x1 #A\n",
+ "Ib = 300-Ia1 #A\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i) The point along the track where minimum potential occurs is\",x1,\"km\"\n",
+ "print \"(ii)Current supplied by sub-ation A is\",Ia1,\"A\"\n",
+ "print \" Current supplied by sub-station B is\",Ib,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The point along the track where minimum potential occurs is 3.42 km\n",
+ "(ii)Current supplied by sub-ation A is 170.7 A\n",
+ " Current supplied by sub-station B is 129.3 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.14, Page number: 327"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "\n",
+ "\n",
+ "#Variable declaration:\n",
+ "i = 0.5 #current loading(A/m)\n",
+ "l = 1000 #cable length(m)\n",
+ "V = 220 #end voltages(V)\n",
+ "r1 = 0.05 #Resistance of each of 2 conductors(ohm/km)\n",
+ "\n",
+ "#Calculation:\n",
+ "r = 2*r1/1000 #ohm/m\n",
+ "I = i*l #A\n",
+ "R = r*l #ohm\n",
+ "Vm = I*R/8 #Max. voltage drop(V)\n",
+ "#Minimum voltage will occur at the mid-point of the distributor & its value is\n",
+ "Vmin = V-Vm #V\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The maximum voltage drop is\",Vm,\"V\"\n",
+ "print \"The minimum voltage is\",Vmin,\"V at the midpoint of the distributor.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum voltage drop is 6.25 V\n",
+ "The minimum voltage is 213.75 V at the midpoint of the distributor.\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.15, Page number: 327"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration:\n",
+ "Va = 255 #Voltage at feeding point A(V)\n",
+ "Vb = 250 #Voltage at feeding point B(V)\n",
+ "l = 500 #Length of distributor(m)\n",
+ "i = 1 #Current loading(A/m)\n",
+ "r1 = 0.1 #resistance of each conductor(ohm/km)\n",
+ "\n",
+ "#Calculation:\n",
+ "r = 2*r1/1000\n",
+ "#(i) Let the minimum potential occur at a point C distant x \n",
+ "# metres from the feeding point A.\n",
+ "\n",
+ "x = (Va-Vb)/(i*r*l)+l/2 #m\n",
+ "Vc = Va-i*r*x**2/2 #minimum voltage(V)\n",
+ "\n",
+ "\n",
+ "#(ii)\n",
+ "ia = i*x #Current supplied from A(A)\n",
+ "ib = i*(l-x) #Current supplied from B(A)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i) The minimum voltage is\",Vc,\"V at\",x,\"m from point A\"\n",
+ "print \"(ii)Current supplied from A is\",ia,\"A\"\n",
+ "print \" Current supplied from B is\",ib,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The minimum voltage is 246.0 V at 300.0 m from point A\n",
+ "(ii)Current supplied from A is 300.0 A\n",
+ " Current supplied from B is 200.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.16, Page number: 328"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration:\n",
+ "l = 800 #Length of distributor(m)\n",
+ "i = 1.25 #Current loading(A/m)\n",
+ "r1 = 0.05 #resistance of each conductor(ohm/km)\n",
+ "Vc = 220 #minimum voltage(V)\n",
+ "x = 450 #Distance of point C from A(m)\n",
+ "\n",
+ "#Calculation:\n",
+ "r = 2*r1/1000\n",
+ "Vac = i*r*x**2/2 #Voltage drop in section AC(V)\n",
+ "Va = Vac+Vc #Voltage at feeding point A(V)\n",
+ "Vbc = i*r*(l-x)**2/2 #Voltage drop in section BC(V)\n",
+ "Vb = Vc+Vbc #Voltage at feeding point B(V)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"Voltage at feeding point A is\",round(Va,2),\"V\"\n",
+ "print \"Voltage at feeding point B is\",round(Vb,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage at feeding point A is 232.66 V\n",
+ "Voltage at feeding point B is 227.66 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.17, Page number: 329"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration:\n",
+ "i = 1.25 #current loading (A/m)\n",
+ "l = 1000 #length of distributor(m)\n",
+ "r1 = 0.05 #resistance of each conductor(ohm/km)\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "I = i*l #Total current fed to the distributor(A)\n",
+ "R = r*l #Total resistance of the distributor(ohm)\n",
+ "V = I*R/8 #Max. voltage drop(V)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"Maximum voltage drop in the distributor is\",round(V,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum voltage drop in the distributor is 15.63 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.19, Page number: 330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration:\n",
+ "l = 900 #distributor length(m)\n",
+ "Va = 400 #voltage fed at a A(V)\n",
+ "i = 0.5 #current loading(A/m)\n",
+ "r = 0.0001 #resistance of distributor per m (go and return)(ohm)\n",
+ "x = 500 #length AD(m)\n",
+ "\n",
+ "#Calculation:\n",
+ "#Drops due to concentrated loads.\n",
+ "Iac = 300 #A\n",
+ "Icd = 250 #A\n",
+ "Ide = 150 #A\n",
+ "Vac = Iac*200*r #Drop in section AC(V)\n",
+ "Vcd = Icd*300*r #Drop in section CD(V)\n",
+ "Vde = Ide*300*r #Drop in section DE(V)\n",
+ "Vab1 = Vac+Vcd+Vde #Total drop over AB(V)\n",
+ "\n",
+ "\n",
+ "#Drops due to uniform loading:\n",
+ "Vab2 = i*r*l**2/2 #V\n",
+ "Vad = i*r*(l*x-x**2/2) #V\n",
+ "Vb = Va-(Vab1+Vab2) #V\n",
+ "Vd = Va-(Vac+Vcd+Vad) #V\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i) Voltage at point B is\",Vb,\"V\"\n",
+ "print \"(ii)Voltage at point D is\",Vd,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Voltage at point B is 361.75 V\n",
+ "(ii)Voltage at point D is 370.25 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.20, Page number: 331"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "from sympy import *\n",
+ "\n",
+ "#Variable declaration:\n",
+ "r = 0.1*10**-3 #Distributor resistance per metre length\n",
+ "i = 0.5 #uniform current loading(A/m)\n",
+ "Va = 240 #end voltage at A(V)\n",
+ "Vb = 240 #end voltage at B(V)\n",
+ "l = 1000 #length of distributor(m)\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "#(i) Point of minimum potential:\n",
+ "#We know that, the point of minimum potential is not affected\n",
+ "#by the uniform loading of the distributor. \n",
+ "#Considering the concentrated loads first as shown below:\n",
+ "\n",
+ "I = symbols('I')\n",
+ "Iac = I\n",
+ "Icd = I-120\n",
+ "Ide = I-180\n",
+ "Ief = I-280\n",
+ "Ifb = I-320\n",
+ "I1 = solve((Iac*200+Icd*200+Ide*300+Ief*200+Ifb*100)*10**-4-Va+Vb,I)[0]\n",
+ "\n",
+ "#Showing the current distributions, we can see that D is \n",
+ "#the point of minimum potential.\n",
+ "\n",
+ "#(ii) The feeding point A will supply I1 A due to concentrated\n",
+ "# loading plus 0\u00b75 \u00d7 400 = 200 A due to uniform loading.\n",
+ "\n",
+ "Ia = I1+200 #Current supplied by A(A)\n",
+ "\n",
+ "#The feeding point B will supply a current of 154 A due to \n",
+ "#concentrated loading plus 0\u00b75 \u00d7 600 = 300 A due to uniform loading.\n",
+ "\n",
+ "Ib = 320-I1+300 #Current supplied byB(A)\n",
+ "\n",
+ "\n",
+ "\n",
+ "#(iii) We got that D is the point of minimum potential.\n",
+ "Iac = I1; Icd = I1-120 #ampere\n",
+ "Vad1 = Iac*200*10**-4+Icd*200*10**-4 #Drop in AD due to conc. loading(V)\n",
+ "AD = 400 #for uniform loading(V)\n",
+ "Vad2 = i*r*AD**2/2 #Drop in AD due to uniform. loading(V)\n",
+ "Vd = 240-Vad1-Vad2 #Minimum potential(V)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i) The point of minimum potential is D\"\n",
+ "print \"(ii) Current supplied by A is\",round(Ia),\"A\"\n",
+ "print \" Current supplied by B is\",round(Ib),\"A\"\n",
+ "print \"(iii)The value of minimum potential is\",round(Vd,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The point of minimum potential is D\n",
+ "(ii) Current supplied by A is 366.0 A\n",
+ " Current supplied by B is 454.0 A\n",
+ "(iii)The value of minimum potential is 231.76 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.21, Page number: 332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "from sympy import *\n",
+ "\n",
+ "\n",
+ "#Variable declaration:\n",
+ "l = 500 #distributor length(m)\n",
+ "Va = 240 #voltage at end A(V)\n",
+ "Vb = 240 #voltage at end B(V)\n",
+ "R = 0.001 #resistance of the distributor(ohm/m)\n",
+ "i = 1 #current loading(A/m)\n",
+ "\n",
+ "#Calculation:\n",
+ "#Let D be the point of minimum potential\n",
+ "x = symbols('x') #x is current flowing in section CD(A)\n",
+ "\n",
+ "#(i) If r is the resistance of the distributor (go and return)\n",
+ "#per metre,\n",
+ "r = symbols('r')\n",
+ "Vad = (100+x)*100*r+x*150*r #Voltage drop in length AD\n",
+ "Vbd = 1*r*200**2/2+(60-x)*250*r #Voltage drop in length BD\n",
+ "x1 = solve(Vad-Vbd,x)[0] #A\n",
+ "\n",
+ "#(ii)\n",
+ "I = 60+1*200 #Total current(A)\n",
+ "Ia = 100+x1 #Current supplied by A(A)\n",
+ "Ib = 360-150 #Current supplied by B(V)\n",
+ "Vd = Va - Ia*100*0.001-50*150*0.001 #Minimum potential(V)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i) The point of minimum voltage is D.\"\n",
+ "print \"(ii)The value of minimum voltage is\",round(Vd,1),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) The point of minimum voltage is D.\n",
+ "(ii)The value of minimum voltage is 217.5 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.22, Page number: 334"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "from sympy import *\n",
+ "\n",
+ "#Variable declaration:\n",
+ "l = 300 #distributor length(m)\n",
+ "r1 = 0.03/100 #resistance of single conductor(ohm/m)\n",
+ "Va = 240 #volt\n",
+ "\n",
+ "#Calculation:\n",
+ "r = 2*r1 #resistance of both wires(ohm/m)\n",
+ "Rab = r*150 #ohm\n",
+ "Rbc = r*50 #ohm\n",
+ "Rca = r*100 #ohm\n",
+ "#suppose a current IA flows in section AB of the distributor.\n",
+ "#currents in sections BC and CA will be (Ia \u2212 120) & (Ia \u2212 200)\n",
+ "Ia = symbols('Ia')\n",
+ "Ia1 = solve(0.09*Ia+0.03*(Ia-120)+0.06*(Ia-200),Ia)[0]\n",
+ "\n",
+ "#The actual distribution of currents ca be drawn easily\n",
+ "#from where it is seen that B is the point of min. potential.\n",
+ "Iab = Ia1 #Current in section AB(A)\n",
+ "Ibc = Ia1-120 #Current in section BC(A)\n",
+ "Ica = Ia1-200 #current in section CA(A)\n",
+ "Vb = Va-Iab*Rab #Voltage at point B(V)\n",
+ "Vc = Vb-Ibc*Rbc #Voltage at point C(V)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i) Current in section AB is\",round(Iab,2),\"A from A to B\"\n",
+ "print \" Current in section BC is\",round(abs(Ibc),2),\"A from C to B\"\n",
+ "print \" Current in section ca is\",round(abs(Ica),2),\"A from A to C\"\n",
+ "print \"(ii)Voltage at point B is\",round(Vb,1),\"V\"\n",
+ "print \" Voltage at point C is\",round(Vc,1),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current in section AB is 86.67 A from A to B\n",
+ " Current in section BC is 33.33 A from C to B\n",
+ " Current in section ca is 113.33 A from A to C\n",
+ "(ii)Voltage at point B is 232.2 V\n",
+ " Voltage at point C is 233.2 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.23, Page number: 335"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "from sympy import *\n",
+ "\n",
+ "#Variable declaration:\n",
+ "Va = 220 #voltage at point A(V)\n",
+ "Rab = 0.1 #ohm\n",
+ "Rbc = 0.05 #ohm\n",
+ "Rcd = 0.01 #ohm\n",
+ "Rde = 0.025 #ohm\n",
+ "Rea = 0.075 #ohm\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "#(i)\n",
+ "I1 = solve(0.1*I+0.05*(I-10)+0.01*(I-30)+0.025*(I-60)+0.075*(I-70),I)[0]\n",
+ "\n",
+ "#(ii)\n",
+ "Iab = I1 #A\n",
+ "Ibc = I1-10 #A\n",
+ "Icd = I1-30 #A\n",
+ "Ide = I1-60 #A\n",
+ "Iea = I1-70 #A\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i) C is the point of minimum potential.\"\n",
+ "print \"(ii)Current in section AB is\",round(Iab,2),\"A from A to B\"\n",
+ "print \" Current in section BC is\",round(Ibc,2),\"A from B to C\"\n",
+ "print \" Current in section CD is\",round(abs(Icd),2),\"A from D to C\"\n",
+ "print \" Current in section DE is\",round(abs(Ide),2),\"A from E to D\"\n",
+ "print \" Current in section EA is\",round(abs(Iea),2),\"A from A to E\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) C is the point of minimum potential.\n",
+ "(ii)Current in section AB is 29.04 A from A to B\n",
+ " Current in section BC is 19.04 A from B to C\n",
+ " Current in section CD is 0.96 A from D to C\n",
+ " Current in section DE is 30.96 A from E to D\n",
+ " Current in section EA is 40.96 A from A to E\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.24, Page number: 336"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "from sympy import *\n",
+ "\n",
+ "#Variable declaration:\n",
+ "Rab = 0.02 #ohm\n",
+ "Rbc = 0.018 #ohm\n",
+ "Rcd = 0.025 #ohm\n",
+ "Rda = 0.02 #ohm\n",
+ "Va = 250 #voltage ate point A(V)\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "I = symbols('I')\n",
+ "I1 = solve(0.02*I+0.018*(I-150)+0.025*(I-450)+0.02*(I-700),I)[0]\n",
+ "Vab = 336.75*0.02 #V\n",
+ "Vbc = 186.75*0.018 #V\n",
+ "Vcd = 113.25*0.025 #V\n",
+ "Vda = 363.25*0.02 #V\n",
+ "Vb = Va-Vab #V\n",
+ "Vc = Vb-Vbc #V\n",
+ "Vd = Vc+Vcd #V\n",
+ "\n",
+ "#With interconnector\n",
+ "Eo = Va-Vc #Voltage between points A and C(V)\n",
+ "Ro = (Rab+Rbc)*(Rcd+Rda)/((Rab+Rbc)+(Rcd+Rda)) #Resistance viewed from points A & C(ohm)\n",
+ "Rac = 0.02 #Resistance of interconnector(ohm)\n",
+ "Iac = Eo/(Ro+Rac) #Current in interconnector AC(A)\n",
+ "\n",
+ "#Let ABCD be the ring distribution system.\n",
+ "#Let the current in section AB is I1. Then,\n",
+ "#current in section BC will be I1 \u2212 150. \n",
+ "#As the voltage drop round the closed mesh ABCA is zero.\n",
+ "I11 = symbols('I11')\n",
+ "I2 = solve(0.02*I11+0.018*(I11-150)-0.02*252.4,I11)[0]\n",
+ "\n",
+ "Vab1 = I2*0.02 #V\n",
+ "Vbc1 = 53.15*0.018 #V\n",
+ "Vad1 = 244.45*0.02 #V\n",
+ "Vb1 = Va-Vab1 #V\n",
+ "Vc1 = Vb1-Vbc1 #V\n",
+ "Vd1 = Va-Vad1 #V\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"Before adding interconnector,\"\n",
+ "print \"\\tVoltage at point B is\",Vb,\"V\"\n",
+ "print \"\\tVoltage at point C is\",round(Vc,3),\"V\"\n",
+ "print \"\\tVoltage at point D is\",round(Vd,3),\"V\"\n",
+ "print \"After adding interconnector,\"\n",
+ "print \"\\tPotential of B is\",round(Vb1,2),\"V\"\n",
+ "print \"\\tPotential of C is\",round(Vc1,2),\"V\"\n",
+ "print \"\\tPotential of D is\",round(Vd1,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Before adding interconnector,\n",
+ "\tVoltage at point B is 243.265 V\n",
+ "\tVoltage at point C is 239.903 V\n",
+ "\tVoltage at point D is 242.735 V\n",
+ "After adding interconnector,\n",
+ "\tPotential of B is 245.92 V\n",
+ "\tPotential of C is 244.97 V\n",
+ "\tPotential of D is 245.11 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.25, Page number: 338"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "from sympy import *\n",
+ "\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "Rbd = 0.05 #resistance of interconnector(ohm)\n",
+ "Rab = 0.075 #resistance of branch AB(ohm)\n",
+ "Rbc = 0.025 #resistance of branch BC(ohm)\n",
+ "Rcd = 0.01 #resistance of branch CD(ohm)\n",
+ "Rde = 0.05 #resistance of branch DE(ohm)\n",
+ "Rea = 0.1 #resistance of branch EA(ohm)\n",
+ "\n",
+ "#Calculation:\n",
+ "#Let ABCDE be the ring distribution system.\n",
+ "#When interconnector BD is removed, let the current in branch \n",
+ "#AB be I.\n",
+ "# Current in BC = I-10;\n",
+ "# Current in CD = I-40;\n",
+ "# Current in DE = I-60;\n",
+ "# Current in EA = I-70;\n",
+ "\n",
+ "\n",
+ "#As the total drop round the ring ABCDEA is zero.\n",
+ "I = symbols('I')\n",
+ "\n",
+ "I1 = solve(Rab*I+Rbc*(I-10)+Rcd*(I-40)+Rde*(I-60)+Rea*(I-70),I)[0]\n",
+ "\n",
+ "#The actual distribution of currents will be as shown in Fig.(ii) above.\n",
+ "Vbcd = 30.96*0.025+0.96*0.01 #V\n",
+ "Eo = 0.7836 #V\n",
+ "Ro = (0.075+0.1+0.05)*(0.025+0.01)/((0.075+0.1+0.05)+(0.025+0.01))\n",
+ "\n",
+ "#(i)\n",
+ "Ibd = Eo/(Ro+Rbd) #A\n",
+ "Vbd = Ibd*0.05 #V\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i) Current in interconnector BD is\",round(Ibd,1),\"A\"\n",
+ "print \"(ii)Voltage drop along interconnector BD is\",round(Vbd,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current in interconnector BD is 9.8 A\n",
+ "(ii)Voltage drop along interconnector BD is 0.49 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.26, Page number: 341"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration:\n",
+ "I1 = 50 #current on +ve side(A)\n",
+ "I2 = 40 #current on -ve side(A)\n",
+ "Rae = 0.1 #resistance of outer wire(ohm)\n",
+ "Rbg = 0.1 \n",
+ "Vab = 500 #V\n",
+ "Van = 250 #V\n",
+ "Vbn = 250 #V\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "I3 = I1-I2 #current in the neutral wire(A)\n",
+ "Rnl = 2*Rae #resistance of neutral wire(ohm)\n",
+ "Vel = Van-I1*Rae-(I1-I2)*Rnl #V\n",
+ "Vlg = Vbn+(I1-I2)*Rnl-I2*Rbg #V\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"Voltage at the load end on the positive side is\",Vel,\"V\"\n",
+ "print \"Voltage at the load end on the negative side is\",Vlg,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage at the load end on the positive side is 243.0 V\n",
+ "Voltage at the load end on the negative side is 248.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.27, Page number: 342"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "\n",
+ "#Vairable declaration:\n",
+ "r = 0.1 #resistance of each conductor(ohm)\n",
+ "Rel = 5 #ohm\n",
+ "Rcl = 6 #ohm\n",
+ "Vel = 240 #line to neutral voltage(V)\n",
+ "Vlc = 240 #line to neutral voltage(V)\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "I1 = Vel/5 #Current on +ve outer(A)\n",
+ "I2 = Vlc/6 #Current on \u2212ve outer(A)\n",
+ "In = I1-I2 #current in neutral(A)\n",
+ "V1 = Vel+I1*r+(I1-I2)*r #V\n",
+ "V2 = Vlc-(I1-I2)*r+I2*r #V\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"Voltage between +ve outer and neutral at feeding end is\",V1,\"V\"\n",
+ "print \"Voltage between -ve outer and neutral at feeding end is\",V2,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage between +ve outer and neutral at feeding end is 245.6 V\n",
+ "Voltage between -ve outer and neutral at feeding end is 243.2 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.28, Page number: 343"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration:\n",
+ "V = 250 #voltage on the two sides of the middle wire(V)\n",
+ "P1 = 35 #power of load 1(kW)\n",
+ "P2 = 20 #power of load 2(kW)\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "R1 = V**2/(P1*1000) #resistance of load 1(ohm)\n",
+ "R2 = V**2/(P2*1000) #resistance of load 2(ohm)\n",
+ "#After breaking of neutral wire,\n",
+ "I = 2*V/(R1+R2) #Circuit current(A)\n",
+ "V1 = I*R1 #V\n",
+ "V2 = I*R2 #V\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"Voltage across +ve outer and middle wire is\",round(V1,1),\"V\"\n",
+ "print \"Voltage across -ve outer and middle wire is\",round(V2,1),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage across +ve outer and middle wire is 181.8 V\n",
+ "Voltage across -ve outer and middle wire is 318.2 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.29, Page number: 343"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "\n",
+ "#Calculation:\n",
+ "Vck = 250-0.75-0.28+0.2 #V\n",
+ "Vdm = Vck-0.3-0.18+0.084 #V\n",
+ "Vjg = 250-0.2-1.2 #V\n",
+ "Vlh = Vjg+0.28-0.084-0.864 #V\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"Voltage across load CK is\",Vck,\"V\"\n",
+ "print \"Voltage across load DM is\",Vdm,\"V\"\n",
+ "print \"Voltage across load JG is\",Vjg,\"V\"\n",
+ "print \"Voltage across load LH is\",Vlh,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage across load CK is 249.17 V\n",
+ "Voltage across load DM is 248.774 V\n",
+ "Voltage across load JG is 248.6 V\n",
+ "Voltage across load LH is 247.932 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.30, Page number: 344"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration:\n",
+ "Vab = 500 #volts\n",
+ "Van = 250 #volts\n",
+ "Vnb = 250 #volts\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "#Voltage across CK = 250 \u2212 Drop in AC \u2212 Drop in KJ \u2212 Drop in JN\n",
+ "Vck = Van-4-0.5-0.1 #volts\n",
+ "\n",
+ "#Voltage across DM = 245\u00b74 \u2212 Drop in CD \u2212 Drop in ML + Drop in KL\n",
+ "Vdm = Vck-1.28-0.5+0.42 #volts\n",
+ "\n",
+ "#Voltage across JG = 250 + Drop in JN \u2212 Drop in GB\n",
+ "Vjg = Van+0.1-1.9 #volts\n",
+ "\n",
+ "#Voltage across LH = 248\u00b72 + Drop in KJ \u2212 Drop in KL \u2212 Drop in HG\n",
+ "Vlh = Vjg+0.5-0.42-2.4 #volts\n",
+ "\n",
+ "#Voltage across PF = 245\u00b788 + Drop in ML \u2212 Drop in MP \u2212 Drop in FH\n",
+ "Vpf = Vlh+0.5-0.72-1.02 #volts\n",
+ "\n",
+ "#Result:\n",
+ "print \"Voltage across CK is\",Vck,\"V\"\n",
+ "print \"Voltage across DM is\",Vdm,\"V\"\n",
+ "print \"Voltage across JG is\",Vjg,\"V\"\n",
+ "print \"Voltage across LH is\",Vlh,\"V\"\n",
+ "print \"Voltage across PF is\",Vpf,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltage across CK is 245.4 V\n",
+ "Voltage across DM is 244.04 V\n",
+ "Voltage across JG is 248.2 V\n",
+ "Voltage across LH is 245.88 V\n",
+ "Voltage across PF is 244.64 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.31, Page number: 346"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration:\n",
+ "Vel = 240 #volt\n",
+ "Vlc = 240 #volt\n",
+ "Rel = 4 #ohm\n",
+ "Rlc = 6 #ohm\n",
+ "r = 0.15 #resistance of each conductor(ohm)\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "I1 = Vel/Rel #Current in the positive outer(A)\n",
+ "I2 = Vlc/Rlc #current in the negative outer(A)\n",
+ "In = I1-I2 #current in the neutral wire(ohm)\n",
+ "V1 = Vel+I1*r+In*r #Voltage between +ve outer and neutral at feeding point(V)\n",
+ "V2 = Vlc-In*r+I2*r #Voltage between -ve outer and neutral at feeding point(V)\n",
+ "\n",
+ "\n",
+ "#(i)When neutral breaks,\n",
+ "#When there is a break in the neutral, the system\n",
+ "#is equivalent to 2- wire d.c. system.Now,\n",
+ "R1 = Rel+Rlc+2*r #ohm\n",
+ "Vt1 = V1+V2 #voltage at the feeding end(V)\n",
+ "I11 = Vt1/R1 #Load current(A)\n",
+ "Vl1 = I11*Rel #Voltage across 4 ohm resistance(V)\n",
+ "Vl2 = I11*Rlc #voltage across 6 ohm resistance(V)\n",
+ "\n",
+ "\n",
+ "#(ii) When +ve outer breaks,\n",
+ "#When there is a break in the +ve outer, there will be\n",
+ "#no current in 4 ohm load.\n",
+ "R2 = Rlc+2*r #Total circuit resistance(ohm)\n",
+ "Vt2 = V2 #volt\n",
+ "I22 = Vt2/R2 #load current(A)\n",
+ "VR2 = I22*Rlc #volt\n",
+ "\n",
+ "\n",
+ "#(iii)When \u2212ve outer breaks,\n",
+ "#When there is a break in the negative outer, there will be no\n",
+ "#current in 6 \u03a9 load.\n",
+ "R3 = 4+2*r #Total circuit resistance(ohm)\n",
+ "Vt3 = V1 #volt\n",
+ "I33 = Vt3/R3 #load current(A)\n",
+ "VR3 = I33*Rel #volt\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"When there is a break in the:\"\n",
+ "print \"(i) neutral wire, current is\",round(I11,2),\"A\" \n",
+ "print \" & voltage is\",round(Vl1,2),\"V and\",round(Vl2,2),\"V\"\n",
+ "print \" across 6 ohm & 4 ohm loads respectively.\"\n",
+ "print \"(ii) positive outer, current is\",round(I22,2),\"A\"\n",
+ "print \" and voltage is\",round(VR2,2),\"V\"\n",
+ "print \"(iii)negative outer, current is\",round(I33,2),\"A\"\n",
+ "print \" and voltage is\",round(VR3,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "When there is a break in the:\n",
+ "(i) neutral wire, current is 48.06 A\n",
+ " & voltage is 192.23 V and 288.35 V\n",
+ " across 6 ohm & 4 ohm loads respectively.\n",
+ "(ii) positive outer, current is 38.57 A\n",
+ " and voltage is 231.43 V\n",
+ "(iii)negative outer, current is 58.6 A\n",
+ " and voltage is 234.42 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.32, Page number: 348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "V = 500 #voltage b/w the outer supplies(V)\n",
+ "P1 = 1500 #load(kW)\n",
+ "P2 = 2000 #load(kW)\n",
+ "Vn = 250 #outer to neutral voltage(V)\n",
+ "\n",
+ "#Calculation:\n",
+ "I1 = P1*1000/Vn #Load current on +ve outer(A)\n",
+ "I2 = P2*1000/Vn #Load current on -ve outer(A)\n",
+ "In = I2-I1 #A\n",
+ "Pl = P1+P2 #kW\n",
+ "Ig = Pl*1000/V #Current supplied by main generator(A)\n",
+ "Ia = Ig-I1 #A\n",
+ "Ib = I2-Ig #A\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i) Current in the neutral is\",In,\"A\"\n",
+ "print \"(ii) total current supplied by main generator is\",Ig,\"A\"\n",
+ "print \"(iii)Current in machine A is\",Ia,\"A\"\n",
+ "print \" Current in machine B\",Ib,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current in the neutral is 2000.0 A\n",
+ "(ii) total current supplied by main generator is 7000.0 A\n",
+ "(iii)Current in machine A is 1000.0 A\n",
+ " Current in machine B 1000.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.33, Page number: 349"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration:\n",
+ "V = 500 #volt\n",
+ "Vn = 250 #volt\n",
+ "P1 = 150 #load(kW)\n",
+ "P2 = 100 #load(kW)\n",
+ "Pl = 3 #loss in each balancer machine(kW)\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "Pt = P1+P2+2*Pl #Total load on the main generator(kW)\n",
+ "Ig = Pt*1000/V #Current supplied by the main generator(A)\n",
+ "I1 = P1*1000/Vn #Load current on +ve side(A)\n",
+ "I2 = P2*1000/Vn #Load current on -ve side(A)\n",
+ "In = I1-I2 #Current in neutral wire(A)\n",
+ "Ia = I1-Ig #Current through machine A(A)\n",
+ "Ib = Ig-I2 #Current through machine B(A)\n",
+ "Pa = Ia*Vn/1000 #Load on machine A(kW)\n",
+ "Pb = Ib*Vn/1000 #Load on machine B(kW)\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i) Total load on the main generator is\",Pt,\"kW\"\n",
+ "print \"(ii)Load on machine A is\",Pa,\"kW\"\n",
+ "print \" Load on machine B is\",Pb,\"kW\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Total load on the main generator is 256 kW\n",
+ "(ii)Load on machine A is 22.0 kW\n",
+ " Load on machine B is 28.0 kW\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.34, Page number: 350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration:\n",
+ "I1 = 1200 #current on +ve side(A)\n",
+ "I2 = 1000 #current on -ve side(A)\n",
+ "Pm = 200 #motor load acoss outer(kW)\n",
+ "Pl = 5 #loss in balancer machine(kW)\n",
+ "V = 500 #voltage across the outers(V)\n",
+ "Vn = 250 #voltage across the outer and the neutral(V)\n",
+ "\n",
+ "#Calculation:\n",
+ "#Since the positive side is more heavily loaded, \n",
+ "#so, machine A acts as a generator and machine B as a motor.\n",
+ "P1 = Vn*I1/1000 #Load on +ve side(kW)\n",
+ "P2 = Vn*I2/1000 #Load on -ve side(kW)\n",
+ "P3 = 200 #load on outers(kW)\n",
+ "Pt = P1+P2+P3+2*Pl #total load(kW)\n",
+ "Ig = Pt*1000/500 #Current of main generator(A)\n",
+ "In = I1-I2 #Current in neutral(A)\n",
+ "Ia = 1600-Ig #Current through machine A(A)\n",
+ "Ib = Ig-1400 #Current through machine B(A)\n",
+ "Pa = Vn*Ia/1000 #Load on machine A(kW)\n",
+ "Pb = Vn*Ib/1000 #Load on machine B(kW)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i) Current of the main generator is\",Ig,\"A\"\n",
+ "print \"(ii)Load on balancer machines: Pa =\",Pa,\"kW & Pb =\",Pb,\"kW\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i) Current of the main generator is 1520.0 A\n",
+ "(ii)Load on balancer machines: Pa = 20.0 kW & Pb = 30.0 kW\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.35, Page number: 350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "from sympy import *\n",
+ "\n",
+ "#Variable declaration:\n",
+ "Ipos = 800 #current on on the positive side(A)\n",
+ "Ineg = 550 #current on the negative side(A)\n",
+ "I = 1500 #curent across the outers(A)\n",
+ "Ra = 0.2 #resistance of each arotary balancer's armature(ohm)\n",
+ "Io = 5 #no load current(A)\n",
+ "Vn = 500 #supply voltage\n",
+ "\n",
+ "\n",
+ "#Calculation:\n",
+ "I1 = Ipos+I #Total current on +ve side(A)\n",
+ "I2 = Ineg+I #Total current on \u2212ve side(A)\n",
+ "In = Ipos-Ineg #Current in neutral wire(A)\n",
+ "\n",
+ "#Let the current through machines A and B be Ia and Ib \n",
+ "Ib = symbols('Ib')\n",
+ "Ia = In-Ib\n",
+ "#Each machine has same value of back e.m.f. E as their field\n",
+ "#currents & speeds are the same.\n",
+ "E = Vn/2-Ra*Io #Back e.m.f(V)\n",
+ "Va = E-Ia*Ra #Terminal p.d. across A(V)\n",
+ "Vb = E+Ib*Ra #Terminal p.d. across B(V)\n",
+ "Ib1 = round(solve(Va*Ia+Vn*Io+Ia**2*Ra+Ib**2*Ra-Vb*Ib,Ib)[0])\n",
+ "Ia1 = In-Ib1\n",
+ "Va1 = E-Ia1*Ra\n",
+ "Vb1 = E+Ib1*Ra\n",
+ "Ig = 2300-Ia1 #Load on main generator(A)\n",
+ "\n",
+ "#Result:\n",
+ "print \"(i)Current loading of each balancer machine: Ia =\",Ia1,\"A & Ib =\",Ib1,\"A\"\n",
+ "print \"(ii)Voltage across machine A is\",Va1,\"V\"\n",
+ "print \" Voltage across machine B is\",Vb1,\"V\"\n",
+ "print \"(iii)Load on main generator is\",Ig,\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(i)Current loading of each balancer machine: Ia = 120.0 A & Ib = 130.0 A\n",
+ "(ii)Voltage across machine A is 225.0 V\n",
+ " Voltage across machine B is 275.0 V\n",
+ "(iii)Load on main generator is 2180.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 13.36, Page number: 352"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration:\n",
+ "V = 500 #voltage at supply end(v)\n",
+ "l = 3 #line length(km)\n",
+ "Ifl = 120 #full loag current(A)\n",
+ "r = 0.5 #resistance of cable(ohm/km)\n",
+ "\n",
+ "#Calculation:\n",
+ "R = r*l #total resistance of the line(ohm)\n",
+ "Vl = R*Ifl #F.L. voltage drop in the line(V)\n",
+ "Vb = Vl #Terminal voltage of booster(V)\n",
+ "Po = Vb*Ifl/1000 #Output of booster(kW)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"Booster Voltage is\",Vl,\"V\"\n",
+ "print \"Output of booster is\",Po,\"kW\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Booster Voltage is 180.0 V\n",
+ "Output of booster is 21.6 kW\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |