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| author | nice | 2014-10-09 18:07:00 +0530 |
|---|---|---|
| committer | nice | 2014-10-09 18:07:00 +0530 |
| commit | 36a03d6d76bac315dba73b2ba9555c7e3fe0234f (patch) | |
| tree | 7e46e8873a7c92be2eef962a36e664c775aa6bf2 /Principles_of_Power_System | |
| parent | b8bb8bbfa81499ad7fc3f3508be257da65f543af (diff) | |
| download | Python-Textbook-Companions-36a03d6d76bac315dba73b2ba9555c7e3fe0234f.tar.gz Python-Textbook-Companions-36a03d6d76bac315dba73b2ba9555c7e3fe0234f.tar.bz2 Python-Textbook-Companions-36a03d6d76bac315dba73b2ba9555c7e3fe0234f.zip | |
updated books
Diffstat (limited to 'Principles_of_Power_System')
25 files changed, 0 insertions, 15928 deletions
diff --git a/Principles_of_Power_System/README.txt b/Principles_of_Power_System/README.txt index 2acded4b..2acded4b 100644..100755 --- a/Principles_of_Power_System/README.txt +++ b/Principles_of_Power_System/README.txt diff --git a/Principles_of_Power_System/chapter10_1.ipynb b/Principles_of_Power_System/chapter10_1.ipynb deleted file mode 100644 index 50a55400..00000000 --- a/Principles_of_Power_System/chapter10_1.ipynb +++ /dev/null @@ -1,1146 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:a3dcc0814e961bf1cd3db74adc66300f1f3d1872ea1c97bd01480f29b5cfe9db"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 10: Performance of Transmission\n",
- "Lines"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 10.1, Page Number: 233"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "pf = 0.8 #power factor\n",
- "Z = 10+15j #load impedance(ohm)\n",
- "magVr = 33000 #receiving end voltage(V)\n",
- "P = 1100 #poweer delivered(kW)\n",
- "\n",
- "#Calculation:\n",
- "magI = P*1000/(magVr*pf) #line current(A)\n",
- "phy = math.acos(pf)\n",
- "Vr = magVr+0j #V\n",
- "I = magI*(math.cos(phy)-math.sin(phy)*1j) #A\n",
- "Vs = Vr + I*Z\n",
- "\n",
- "#Angle between Vs and Vr is\n",
- "alpha = math.atan(Vs.imag/Vs.real)\n",
- "phys = phy+alpha\n",
- "pfs = math.cos(phys)\n",
- "Pl = magI**2*Z.real/1000\n",
- "Pi = P+Pl\n",
- "n = P/Pi*100\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) Sending end voltage is\",round(abs(Vs)),\"V\"\n",
- "print \"(ii) sending end power factor is\",round(pfs,4),\"lagging\"\n",
- "print \"(iii)Transmission efficiency is\",round(n,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Sending end voltage is 33709.0 V\n",
- "(ii) sending end power factor is 0.7955 lagging\n",
- "(iii)Transmission efficiency is 98.45 %\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 10.2, Page Number: 235"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable Declaration:\n",
- "a = 0.775 #cross-section of conductor(cm**2)\n",
- "n = 0.9 #transmission efficiency\n",
- "Pr = 200000 #receiving end power(W)\n",
- "pf = 1 #power factor\n",
- "V = 3300 #line voltage(V)\n",
- "ro = 1.725 #specific resistance(micro_ohm-cm)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "Ps = Pr/n #sending end power(W)\n",
- "Pl = Ps-Pr #line loss(W)\n",
- "I = Pr/(V*pf) #line current(A)\n",
- "R = Pl/(2*I**2) #resistance of 1 conductor(ohm)\n",
- "l = R*a/(ro*10**-6) #length of conductor(cm)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The conductor length is\",round(l*10**-5,1),\"km\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The conductor length is 13.6 km\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 10.3, Page Number: 235"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable Declaration:\n",
- "pf = 0.8 #power factor\n",
- "Pr = 5000 #receiving end power(kW)\n",
- "Vr = 22 #receiving end voltage(kV)\n",
- "Z = 4+6j #impedance of each conductor(ohm)\n",
- "\n",
- "#Calculation:\n",
- "phy = math.acos(pf)\n",
- "magVrp = Vr*1000/3**0.5 #sending end voltage/phase(kV)\n",
- "magI = Pr*1000/(3*magVrp*0.8) #Line current(A)\n",
- "Vr = magVrp+1j \n",
- "I = magI*(math.cos(phy)-math.sin(phy)*1j)\n",
- "\n",
- "#(i)Sending end voltage per phase:\n",
- "Vs = magVrp+I*Z\n",
- "Vsl = abs(Vs)*3**0.5 #V\n",
- "\n",
- "#(ii)\n",
- "reg = (abs(Vs)-magVrp)/magVrp*100 #voltage regulation(%)\n",
- "#(iii)\n",
- "Pl = 3*abs(I)**2*Z.real/1000 #line loss(kW)\n",
- "n = Pr/(Pr+Pl)*100 #transmission efficiency(%)\n",
- "\n",
- "#Result:\n",
- "print \"(i) Sending end voltage is\",round(abs(Vsl)/1000,3),\"kV\"\n",
- "print \"(ii)Percentage regulation is\",round(reg,3),\"%\"\n",
- "print \"(iii)Transmission efficiency\",round(n,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Sending end voltage is 23.942 kV\n",
- "(ii)Percentage regulation is 8.825 %\n",
- "(iii)Transmission efficiency 93.93 %\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 10.4, Page Number: 236"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "Pr = 15000 #power delivered(kW)\n",
- "Vl = 132 #line voltage(kV)\n",
- "Ro = 1 #line resistance(ohm/km)\n",
- "pf = 0.8 #power factor\n",
- "\n",
- "#Calculation:\n",
- "I = Pr/(3**0.5*Vl*pf) #line current(A)\n",
- "#the loss in the transmission is to be 5%.\n",
- "Pl = 5*Pr/100 #kW\n",
- "R = Pl*1000/(3*I**2) #line resistance(ohm)\n",
- "d = R/Ro #line length(km)\n",
- "\n",
- "#Result:\n",
- "print \"Length of line is\",round(d,2),\"km\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Length of line is 37.17 km\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 10.5, Page Number: 236"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "import math\n",
- "\n",
- "#Variable Declaration:\n",
- "pf = 0.8 #power factor\n",
- "Pr = 3600 #sending end power(kW)\n",
- "magVs = 33 #receiving end voltage(kV)\n",
- "Z = 5.31+5.54j #impedance of each conductor(ohm)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "R = Z.real #ohm\n",
- "X = Z.imag #ohm\n",
- "phy = math.acos(0.8)\n",
- "magVsp = magVs*1000/(3**0.5) #V/phase\n",
- "magVr = symbols('magVr') #Receiving end voltage(V/phase)\n",
- "magI = Pr*1000/(3*magVr*pf) #line current(A)\n",
- "\n",
- "#(i)Using approximate expression for magVsp,\n",
- "magVr1 = solve((magVr+magI*R*pf+magI*X*math.sin(phy))-magVsp,magVr)[1]\n",
- "\n",
- "#(ii)line current:\n",
- "magI1 = Pr*1000/(3*magVr1*pf) \n",
- "\n",
- "#(iii)Efficiency\n",
- "Pl = 3*magI1**2*R/1000 #kW\n",
- "n = Pr/(Pr+Pl)*100\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) The receiving end voltage \",round(magVr1*3**0.5/1000,2),\"V\"\n",
- "print \"(ii) Line current is\",round(magI1,2),\"A\"\n",
- "print \"(iii)Transmission efficiency is\",round(n,2),\"%\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The receiving end voltage 31.93 V\n",
- "(ii) Line current is 81.36 A\n",
- "(iii)Transmission efficiency is 97.15 %\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 10.6, Page Number: 237"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "Z = 6+8j #line impedance(ohm)\n",
- "magVs = 120 #sending end voltages(kV)\n",
- "magVr = 110 #receiving end voltaes(kV)\n",
- "pf = 0.9 #power factor\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "R = Z.real #ohm\n",
- "X = Z.imag #ohm\n",
- "magVsp = round(120*1000/3**0.5) #V/phase\n",
- "magVrp = round(110*1000/3**0.5) #V/phase\n",
- "phy = math.acos(pf)\n",
- "magI = symbols('magI') #line current(A)\n",
- "magI1 = solve(magVrp+magI*R*math.cos(phy)+magI*X*math.sin(phy)-magVsp,magI)[0]\n",
- "\n",
- "#(i):\n",
- "Po = 3*magVrp*round(magI1)*math.cos(phy)/1000 #kW\n",
- "\n",
- "#(ii):\n",
- "pfs = (magVrp*math.cos(phy)+magI1*R)/magVsp\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) Power output is\",round(Po),\"kW\"\n",
- "print \"(ii)Sending end power factor\",round(pfs,2),\"lagging\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Power output is 111458.0 kW\n",
- "(ii)Sending end power factor 0.88 lagging\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 10.7, Page Number: 237"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable Declaration:\n",
- "Z = 1.5+4j #impedance of the line(ohm)\n",
- "magVr = 11000 #receivig end voltage(V)\n",
- "pf = 0.8 #power factor\n",
- "Pr = 5000 #power delivered()\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "R = Z.real #ohm\n",
- "X = Z.imag #ohm\n",
- "magVrp = magVr/3**0.5 #V/phase\n",
- "phy = math.acos(pf)\n",
- "magI = Pr*1000/(3*magVrp) #line current(A)\n",
- "magVsp = magVrp+magI*R*pf+magI*X*math.sin(phy) #Volt\n",
- "reg = (magVsp - magVrp)/magVrp*100 #voltage regulation(%)\n",
- "Pl = 3*magI**2*R/1000 #line losses(kW)\n",
- "Po = Pr*pf #output power(W)\n",
- "Pi = Po + Pl #Input Power(kW)\n",
- "n = Po/Pi*100 #efficiency(%)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The % regulation is\",round(reg,2),\"%\"\n",
- "print \"The efficiency is\",round(n,1),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The % regulation is 14.88 %\n",
- "The efficiency is 92.8 %\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 10.8, Page Number: 238"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "Pr = 1000 #power delivered(kW)\n",
- "pf = 0.8 #power factor\n",
- "r = 0.03 #line resistance per phase(ohm/km)\n",
- "L = 0.7 #line inductance per phase(mH)\n",
- "l = 16 #line length(km)\n",
- "magVr = 11000 #receiving line voltage(V)\n",
- "f = 50 #power frequency(Hz)\n",
- "\n",
- "#Calculation:\n",
- "R = r*l #line resistance(ohm)\n",
- "X = 2*3.14*f*L/1000*l #line reactance(ohm)\n",
- "magVrp = round(magVr/3**0.5) #receiving end (v/phase)\n",
- "phy = math.acos(pf)\n",
- "magI = round(Pr*1000/(3*magVrp*math.cos(phy)),1) #line current(A)\n",
- "magVsp = magVrp+magI*R*math.cos(phy)+magI*X*math.sin(phy)\n",
- "reg = (magVsp-magVrp)/magVrp*100 #Volt\n",
- "Pl = round(3*magI**2*R/1000,1) #line losses(kW)\n",
- "Pi = Pr + Pl #Input Power(kW)\n",
- "n = Pr/Pi*100 #efficiency(%)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The % regulation is\",round(reg,2),\"%\"\n",
- "print \"The efficiency is\",round(n,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The % regulation is 2.58 %\n",
- "The efficiency is 99.38 %\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 10.9, Page Number: 238"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "\n",
- "#Variable Declaration:\n",
- "Pr = 2000 #load power(kVA)\n",
- "pf = 0.8 #power factor\n",
- "l = 20 #line length(km)\n",
- "r1 = 7.5; x1 = 13.2 #resistance & reactance of transformer primary(ohm)\n",
- "r2 = 0.35; x2 = 0.65 #resistance & reactance of transformer secondary(ohm)\n",
- "r = 0.4; x = 0.5 ##resistance & reactance of line(ohm/km)\n",
- "Vp = 33*1000 #voltage at primary side(kV)\n",
- "Vs = 6.6*1000 #voltage at secondary side(kV)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "R = l*r #resistance of each conuctor(ohm)\n",
- "X = l*x #reactance of each conductor(ohm)\n",
- "phy = math.acos(pf)\n",
- "#Let us transfer the impedance of transformer secondary\n",
- "#to high tension side i.e., 33 kV side.\n",
- "#Equivalent resistance of transformer referred to 33 kV side:\n",
- "R1 = r1 + r2*(Vp/Vs)**2 #ohm\n",
- "\n",
- "#Equivalent resistance of transformer referred to 33 kV side:\n",
- "X1 = x1+ x2*(Vp/Vs)**2 #ohm\n",
- "\n",
- "\n",
- "Rt = R+R1 #Total resistance of line and transformer(ohm)\n",
- "Xt = X+X1 #Total reactance of line and transformer(omh)\n",
- "Vr = Vp/3**0.5 #receiving end voltage(V/phase)\n",
- "I = round(Pr*1000/(3**0.5*Vp)) #line current(A)\n",
- "Vs = Vr+I*Rt*math.cos(phy)+I*Xt*math.sin(phy) #sending end voltage(V)\n",
- "Vsl = 3**0.5*Vs #sending end line voltage(V)\n",
- "pfs = (Vr*pf+I*Rt)/Vs #sending end power factor\n",
- "Pl = 3*I**2*Rt/1000 #line loss(kW)\n",
- "Po = Pr*pf #output power(kW)\n",
- "n = Po/(Po+Pl)*100 #transmission efficiency(%)\n",
- "\n",
- "#Result:\n",
- "print \"Sending end line voltage is\",round(Vsl/1000,1),\"V\"\n",
- "print \"Sending end power factor is\",round(pfs,4)\n",
- "print \"Transmission efficiency is\",round(n,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Sending end line voltage is 35.6 V\n",
- "Sending end power factor is 0.7826\n",
- "Transmission efficiency is 94.72 %\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 10.10, Page Number: 241"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "r = 0.25 #resistance of line(ohm/km)\n",
- "l = 100 #line length(km)\n",
- "x = 0.8 #Reactance(ohm/km)\n",
- "y = 14*10**-6 #susceptance(siemen/km)\n",
- "magVr = 66000 #Receiving end line voltage(V)\n",
- "Pr = 15000 #power delivered(kW)\n",
- "pf = 0.8 #power factor(lagging)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "R = r*l #ohm\n",
- "X = x*l #ohm\n",
- "Y = y*l #siemen\n",
- "magI = Pr*1000/(pf*magVr) #line current(A)\n",
- "phy = math.acos(pf) #phasor angle\n",
- "Vr = magVr+0j #Volt\n",
- "Ir = round(magI*pf)-round(magI*math.sin(phy))*1j #load current(A)\n",
- "Ic = 1j*round(Y*magVr)\n",
- "\n",
- "#(i):\n",
- "Is = Ir+Ic #Sending end current(A)\n",
- "\n",
- "#(ii):\n",
- "delV = Is*(R+X*1j) #voltage rop(V)\n",
- "Vs = Vr+delV #sending end voltage(V)\n",
- "reg = (abs(Vs)-magVr)/magVr*100 #voltage regulation(%)\n",
- "\n",
- "#phase angle between Vr & Ir:\n",
- "theta1 = math.atan(Is.imag/Is.real)\n",
- "\n",
- "#phase angle between Vr & Is:\n",
- "theta2 = math.atan(Vs.imag/Vs.real)\n",
- "\n",
- "phys = abs(theta1)+theta2\n",
- "pfs = math.cos(phys) #supply power factor\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) The sending end current is\",round(abs(Is)),\"A\"\n",
- "print \"(ii) The sending end voltage is\",round(abs(Vs)),\"V\"\n",
- "print \"(iii)Regulation is\",round(reg,2),\"%\"\n",
- "print \"(iv) Supply power factor is\",round(pfs,2),\"lagging\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The sending end current is 240.0 A\n",
- "(ii) The sending end voltage is 79583.0 V\n",
- "(iii)Regulation is 20.58 %\n",
- "(iv) Supply power factor is 0.86 lagging\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 10.11, Page Number: 244"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "l = 100 #line length(km)\n",
- "r = 0.1 #resistance/km/phase(ohm)\n",
- "xl = 0.2 #reactance/km/phase(ohm)\n",
- "b = 0.04*10**-4 #Capacitive susceptance/km/phase(siemen)\n",
- "Pr = 10000 #power delivered(kW)\n",
- "Vrl = 66000 #sending end line volt(V)\n",
- "pf = 0.8 #power factor(lagging)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "R = r*l #Total resistance/phase(ohm)\n",
- "Xl = xl*l #Total reactance/phase(ohm)\n",
- "Y = b*l #Capacitive susceptance(siemen)\n",
- "magVr = round(Vrl/3**0.5) #Receiving end voltage/phase(V)\n",
- "magIr = round(Pr*1000/(3**0.5*Vrl*pf)) #Load current(A)\n",
- "phy = math.acos(pf)\n",
- "Z = R+Xl*1j #Impedance per phase(ohm)\n",
- "#(i) Taking receiving end voltage as the reference phasor,\n",
- "Vr = magVr+0j\n",
- "Ir = magIr*(pf-math.sin(phy)*1j) #A\n",
- "V1 = Vr+Ir*Z/2 #Voltage across C(V)\n",
- "Ic = 1j*Y*V1 #Charging current(A)\n",
- "Is = Ir+Ic #sending end current(A)\n",
- "\n",
- "\n",
- "#(ii) Sending end voltage,\n",
- "Vs = V1+Is*Z/2 #V\n",
- "magVsl = 3**0.5*abs(Vs) #Line value of sending end voltage(V)\n",
- "\n",
- "#(iii) Referring to phasor diagram (iii),\n",
- "theta1 = math.atan(Vs.imag/Vs.real) #angle between Vr & Vs\n",
- "theta2 = math.atan(abs(Is.imag/Is.real)) #angle between Vr & Is\n",
- "phys = theta1+theta2 #angle b/w Vs & Is\n",
- "\n",
- "pfs = math.cos(phys) #Sending end power factor\n",
- "\n",
- "#(iii):\n",
- "Ps = 3*abs(Vs)*abs(Is)*pfs/1000 #Sending end power(kW)\n",
- "n = Pr/Ps*100 #Efficiency(%)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) The sending end current is\",round(abs(Is)),\"A\"\n",
- "print \"(ii) Sending end voltage is\",round(magVsl/1000,3),\"kV\"\n",
- "print \"(iii)Sending end power factor is\",round(pfs,3),\"lagging\"\n",
- "print \"(iv) Transmission efficiency is\",round(n,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The sending end current is 100.0 A\n",
- "(ii) Sending end voltage is 69.532 kV\n",
- "(iii)Sending end power factor is 0.853 lagging\n",
- "(iv) Transmission efficiency is 97.12 %\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 10.12, Page Number: 245"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "l = 100 #line length(km)\n",
- "r = 0.2 #resistance/km/phase(ohm)\n",
- "xl = 0.4 #reactance/km/phase(ohm)\n",
- "b = 2.5*10**-6 #Capacitive susceptance/km/phase(siemen)\n",
- "Pr = 20000 #power delivered(kW)\n",
- "Vrl = 110000 #sending end line volt(V)\n",
- "pf = 0.9 #power factor\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "R = r*l #Total resistance/phase(ohm)\n",
- "Xl = xl*l #Total reactance/phase(ohm)\n",
- "Y = b*l #Capacitive susceptance(siemen)\n",
- "magVr = round(Vrl/3**0.5) #Receiving end voltage/phase(V)\n",
- "magIr = round(Pr*1000/(3**0.5*Vrl*pf)) #Load current(A)\n",
- "phy = math.acos(pf)\n",
- "Z = R+Xl*1j #Impedance per phase(ohm)\n",
- "\n",
- "#(i) Taking receiving end voltage as the reference phasor,\n",
- "Vr = magVr+0j\n",
- "Ir = magIr*(pf-(math.sin(phy))*1j) #A\n",
- "V1 = Vr+Ir*Z/2 #Voltage across C(V)\n",
- "Ic = 1j*Y*V1 #Charging current(A)\n",
- "Is = Ir+Ic #sending end current(A)\n",
- "Vs = V1+Is*Z/2 #V\n",
- "magVsl = 3**0.5*abs(Vs) #Line value of sending end voltage(V)\n",
- "\n",
- "#(ii):\n",
- "Pl = 3*abs(Is)**2*R/2+3*magIr**2*R/2 #line loss(W)\n",
- "n = Pr/(Pr+Pl/1000)*100 #efficiency\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) The current and voltage at the sending end is\",round(magVsl/1000,2),\"kV\"\n",
- "print \"(ii)Efficiency of transmission is\",round(n,2),\"%\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The current and voltage at the sending end is 116.75 kV\n",
- "(ii)Efficiency of transmission is 96.26 %\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 10.13, Page Number: 247"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable Declaration:\n",
- "l = 150 #line length(km)\n",
- "r = 0.1 #resistance/km/phase(ohm)\n",
- "xl = 0.5 #reactance/km/phase(ohm)\n",
- "b = 3*10**-6 #Capacitive susceptance/km/phase(siemen)\n",
- "Pr = 50000 #power delivered(kW)\n",
- "Vrl = 110000 #sending end line volt(V)\n",
- "pf = 0.8 #power factor\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "R = r*l #Total resistance/phase(ohm)\n",
- "Xl = xl*l #Total reactance/phase(ohm)\n",
- "Y = b*l #Capacitive susceptance(siemen)\n",
- "magVr = round(Vrl/3**0.5) #Receiving end voltage/phase(V)\n",
- "magIr = round(Pr*1000/(3**0.5*Vrl*pf)) #Load current(A)\n",
- "phy = math.acos(pf)\n",
- "Z = R+Xl*1j #Impedance per phase(ohm)\n",
- "\n",
- "Vr = magVr+0j\n",
- "Ir = magIr*(pf-(math.sin(phy))*1j) #A\n",
- "Ic1 = Vr*1j*Y/2 #Charging current at the load end(A)\n",
- "Il = Ir+Ic1 #line current(A)\n",
- "Vs = Vr+Il*Z #Sending end voltage(V)\n",
- "magVsl = abs(Vs)*3**0.5 #Line to line sending end voltage(V)\n",
- "Ic2 = 1j*Vs*Y/2 #Charging current at the sending end(A)\n",
- "Is = Il+Ic2 #Sending end current(A)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The sending end voltage is\",round(magVsl/1000,2),\"V\"\n",
- "print \"The sending end current is\",round(abs(Is),1),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The sending end voltage is 143.56 V\n",
- "The sending end current is 306.3 A\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 10.14, Page Number: 248"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "\n",
- "#Variable Declaration:\n",
- "l = 100 #line length(km)\n",
- "r = 0.1 #resistance/km/phase(ohm)\n",
- "xl = 0.5 #reactance/km/phase(ohm)\n",
- "b = 10*10**-6 #Capacitive susceptance/km/phase(siemen)\n",
- "Pr = 20000 #power delivered(kW)\n",
- "Vrl = 66000 #sending end line volt(V)\n",
- "pf = 0.9 #power factor\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "R = r*l #Total resistance/phase(ohm)\n",
- "Xl = xl*l #Total reactance/phase(ohm)\n",
- "Y = b*l #Capacitive susceptance(siemen)\n",
- "magVr = round(Vrl/3**0.5) #Receiving end voltage/phase(V)\n",
- "magIr = round(Pr*1000/(3**0.5*Vrl*pf)) #Load current(A)\n",
- "phy = math.acos(pf)\n",
- "Z = R+Xl*1j #Impedance per phase(ohm)\n",
- "\n",
- "Vr = magVr+0j\n",
- "Ir = magIr*(pf-(math.sin(phy))*1j) #A\n",
- "Ic1 = round(magVr*Y/2)*1j #Charging current at the load end(A)\n",
- "Il = Ir+Ic1 #line current(A)\n",
- "Vs = Vr+Il*Z #Sending end voltage(V)\n",
- "magVsl = abs(Vs)*3**0.5 #Line to line sending end voltage(V)\n",
- "Ic2 = 1j*Vs*Y/2 #Charging current at the sending end(A)\n",
- "Is = Il+Ic2 #Sending end current(A)\n",
- "\n",
- "#(i):\n",
- "theta1 = math.atan(Vs.imag/Vs.real) #angle between Vr & Vs\n",
- "theta2 = math.atan(abs(Is.imag/Is.real)) #angle between Vr & Is\n",
- "phys = theta1+theta2 #angle b/w Vs & Is\n",
- "pfs = math.cos(phys) #Sending end power factor\n",
- "\n",
- "#(ii):\n",
- "reg = (abs(Vs)-magVr)/magVr*100 #voltage regulation(%)\n",
- "\n",
- "#(iii):\n",
- "Ps = 3*abs(Vs)*abs(Is)*pfs/1000 #sending end power(W)\n",
- "n = Pr/Ps*100 #transmission efficiency(%)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) Sending end power factor is\",round(pfs,3),\"lagging\"\n",
- "print \"(ii) Regulation is\",round(reg,2),\"%\"\n",
- "print \"(iii)Transmission efficiency is\",round(n),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Sending end power factor is 0.906 lagging\n",
- "(ii) Regulation is 15.15 %\n",
- "(iii)Transmission efficiency is 95.0 %\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- " Example 10.15, Page Number: 254"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import cmath\n",
- "\n",
- "#Variable Declaration:\n",
- "l = 200 #line length(km)\n",
- "r = 0.16 #resistance/km/phase(ohm)\n",
- "xl = 0.25 #reactance/km/phase(ohm)\n",
- "b = 1.5*10**-6*1j #Capacitive susceptance/km/phase(siemen)\n",
- "Pr = 20000 #power delivered(kW)\n",
- "Vrl = 110000 #sending end line volt(V)\n",
- "pf = 0.8 #power factor\n",
- "\n",
- "#Calculation:\n",
- "R = r*l #Total resistance/phase(ohm)\n",
- "Xl = xl*l #Total reactance/phase(ohm)\n",
- "Y = b*l #Capacitive susceptance(siemen)\n",
- "Z = R+Xl*1j #Series Impedance/phase(ohm)\n",
- "magVr = Vrl/3**0.5 #Receiving end voltage per phase(V)\n",
- "magIr = round(Pr*1000/(3**0.5*Vrl*pf)) #Receiving end current(A)\n",
- "Vs = magVr*cmath.cosh((Y*Z)**0.5)+magIr*(Z/Y)**0.5*cmath.sinh((Z*Y)**0.5) #sending end voltage(V/phase)\n",
- "Is = magVr*(Y/Z)**0.5*cmath.sinh((Y*Z)**0.5)+magIr*cmath.cosh((Y*Z)**0.5)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Sending end line-to-line voltage is\",round(3**0.5*abs(Vs)/1000,1),\"V\"\n",
- "print \"Sending end current is\",round(abs(Is),1),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Sending end line-to-line voltage is 117.0 V\n",
- "Sending end current is 131.5 A\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 10.16, Page Number: 258"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import cmath\n",
- "import math\n",
- "\n",
- "#Variable Declaration:\n",
- "Z = 20+52j #Series line impedance/phase(ohm)\n",
- "Y = 315*10**-6*1j #Shunt admittance/phase(siemen)\n",
- "pf = 0.85 #power factor\n",
- "Pr = 30000 #receiving end power(kW)\n",
- "magVrl = 132000 #receiving end voltage(V)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#(i) Generalised constants of line,\n",
- "A = 1+Z*Y/2\n",
- "D = A\n",
- "B = Z*(1+Z*Y/4)\n",
- "C = Y\n",
- "\n",
- "#(ii) Sending end voltage,\n",
- "magVr = magVrl/3**0.5 #V/phase\n",
- "magIr = Pr*1000/(3**0.5*magVrl*pf) #line current(A)\n",
- "phy = math.acos(pf)\n",
- "Vr = magVr+0j\n",
- "Ir = magIr*(math.cos(phy)-1j*math.sin(phy))\n",
- "Vs = A*Vr+B*Ir\n",
- "magVs = abs(Vs) #sending end voltage(V/phase)\n",
- "magVsl = 3**0.5*magVs #Sending end line-to-line voltage(V)\n",
- "\n",
- "\n",
- "#(iii) Regulation:\n",
- "#At no load, Ir = 0,\n",
- "magVro = abs(Vs/A)\n",
- "reg = (magVro-magVr)/magVr*100 #regulation(%)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i)The A, B, C and D constants of the line are\"\n",
- "print \" A =\",complex(round(A.real,3),round(A.imag,5))\n",
- "print \" B =\",complex(round(B.real,2),round(B.imag,2))\n",
- "print \" C =\",complex(round(C.real,6),round(C.imag,6))\n",
- "print \" D =\",complex(round(D.real,3),round(D.imag,5))\n",
- "\n",
- "print \"(ii) Sending end voltage is\",round(magVs*3**0.5/1000),\"kV\"\n",
- "print \"(iii)Regulation of the line is\",round(reg,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i)The A, B, C and D constants of the line are\n",
- " A = (0.992+0.00315j)\n",
- " B = (19.84+51.82j)\n",
- " C = 0.000315j\n",
- " D = (0.992+0.00315j)\n",
- "(ii) Sending end voltage is 143.0 kV\n",
- "(iii)Regulation of the line is 9.25 %\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 10.17, Page Number: 259"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import cmath\n",
- "import math\n",
- "\n",
- "#Variable Declaration:\n",
- "A = cmath.rect(0.95,math.radians(1.4))\n",
- "B = cmath.rect(96,math.radians(78))\n",
- "C = cmath.rect(0.0015,math.radians(90))\n",
- "D = cmath.rect(0.95,math.radians(1.4))\n",
- "Pr = 50000 #receiving end power(kW)\n",
- "pf = 0.8 #power factor\n",
- "magVrl = 132000 #receiving end voltage(V)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "magVr = magVrl/3**0.5 #Receiving end voltage/phase(V)\n",
- "magIr = Pr*1000/(3**0.5*magVrl*pf) #line current(A)\n",
- "phy = math.acos(pf)\n",
- "Vr = magVr+0j\n",
- "Ir = magIr*(math.cos(phy)-1j*math.sin(phy))\n",
- "Vs = A*Vr+B*Ir #Sending end voltage per phase\n",
- "Is = C*Vr+D*Ir #Sending end current\n",
- "Ic = Is-Ir #Charging current\n",
- "#At no load, Ir = 0,\n",
- "magVro = abs(Vs/A)\n",
- "reg = (magVro-magVr)/magVr*100 #regulation(%)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Charging current is (\",round(abs(Ic)),round(math.degrees(angle(Ic)),1),\") A\"\n",
- "print \"Regulation is\",round(reg),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Charging current is ( 128.0 93.2 ) A\n",
- "Regulation is 30.0 %\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 10.18, Page Number: 260"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import cmath\n",
- "import math\n",
- "\n",
- "#Variable Declaration:\n",
- "A = cmath.rect(0.98,math.radians(3))\n",
- "B = cmath.rect(110,math.radians(75))\n",
- "C = cmath.rect(0.0005,math.radians(80))\n",
- "D = cmath.rect(0.98,math.radians(3))\n",
- "MVA = 50 #receiving end power\n",
- "pf = 0.8 #power factor\n",
- "magVrl = 110 #receiving end voltage(kV)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "Pr = MVA*pf*10**6\n",
- "magVr = round(magVrl/3**0.5,1) #Receiving end voltage/phase(V)\n",
- "magIr = round(MVA*10**6/(3**0.5*magVrl*1000),1) #line current(A)\n",
- "phy = math.acos(pf)\n",
- "Vr = magVr*1000+0j\n",
- "Ir = magIr*(math.cos(phy)-1j*math.sin(phy))\n",
- "#(round(Ir.real)+1j*round(Ir.imag))\n",
- "V1 = round((A*Vr).real)+math.ceil((A*Vr).imag)*1j\n",
- "V2 = round((B*Ir).real)+math.ceil((B*Ir).imag)*1j\n",
- "Vs = V1+V2 #Sending end voltage per phase\n",
- "theta1 = math.atan(Vs.imag/Vs.real)\n",
- "Is = C*Vr+D*Ir #Sending end current\n",
- "theta2 = math.atan(Is.imag/Is.real)\n",
- "phys = theta2-theta1\n",
- "Ps = 3*abs(Vs)*abs(Is)*math.cos(phys) #Sending-end power(W)\n",
- "n = Pr/Ps*100 #efficiency(%)\n",
- "\n",
- "#Result:\n",
- "print \"(i) Sending end voltage is\",round(abs(Vs)),\"V\"\n",
- "print \"(ii) Sending end current is\",round(abs(Is)),\"A\"\n",
- "print \"(iii)Sending-end power is\",round(Ps/10**6,1),\"MW\"\n",
- "print \"(iv) Transmission efficiency is\",round(n,1),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Sending end voltage is 87429.0 V\n",
- "(ii) Sending end current is 246.0 A\n",
- "(iii)Sending-end power is 48.7 MW\n",
- "(iv) Transmission efficiency is 82.2 %\n"
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Principles_of_Power_System/chapter11_1.ipynb b/Principles_of_Power_System/chapter11_1.ipynb deleted file mode 100644 index 2283037e..00000000 --- a/Principles_of_Power_System/chapter11_1.ipynb +++ /dev/null @@ -1,1185 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:1fe403fe815d4353da157ba9f671fac5bd27f5ac7842de08db777bc86330d8e5"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 11: Underground Cables"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.1, Page Number: 273"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "r1 = 0.5 #conductor radius(cm)\n",
- "l = 2000 #conductor length(m)\n",
- "rho = 5*10**12 #Resistivity of insulation(ohm-m)\n",
- "t = 0.4 #insulation thickness(cm)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "r2 = r1+t #Internal sheath radius(cm)\n",
- "R = rho*math.log(r2/r1)/(2*math.pi*l) #Insulation resistance of cable(ohm)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Insulation resistance of cable is\",round(R/10**6),\"Mohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Insulation resistance of cable is 234.0 Mohm\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.2, Page Number: 274"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "r1 = 1.25 #conductor radius(cm)\n",
- "l = 1000 #conductor length(m)\n",
- "rho = 4.5*10**12 #Resistivity of insulation(ohm-m)\n",
- "R = 495*10**6 #Cable insulation resistance(ohm)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Let r2 cm be the internal sheath radius,\n",
- "\n",
- "r2 = r1*math.exp(R*2*math.pi*l/rho)\n",
- "\n",
- "#Result:\n",
- "print \"Insulation thickness is\",round(r2-r1,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Insulation thickness is 1.25 cm\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.3, Page Number: 274"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "r1 = .10 #conductor radius(cm)\n",
- "l = 5000 #conductor length(m)\n",
- "r2 = 0.25 #Internal sheath radius(cm)\n",
- "R = 0.4*10**6 #Cable insulation resistance(ohm)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "rho = R*2*3.14*l/(math.log(r2/r1)*10**9) #resistivity(ohm-m)\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Resistivity of the insulating material is\",round(rho,2),\"* 10**9 ohm-m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Resistivity of the insulating material is 13.71 * 10**9 ohm-m\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.4, Page Number: 275"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "er = 4 #relative permittivity\n",
- "D = 1.8 #internal sheath diameter(cm)\n",
- "l = 1000 #cable length(m)\n",
- "d = 1 #conductor diameter(cm)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "C = er*l/(41.4*math.log10(D/d))*10**-9 #Capacitance(F)\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The capacitance of the cable is\",round(C*10**6,3),\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The capacitance of the cable is 0.378 uF\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.5, Page Number: 276"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "er = 4 #relative permittivity\n",
- "d = 10 #core diameter(cm)\n",
- "l = 1000 #cable length(m)\n",
- "t = 7 #insulation thickness(cm)\n",
- "Vl = 66000 #line voltage(V)\n",
- "f = 50 #frequency(Hz)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "D = d+2*t #conductor diameter(cm)\n",
- "C = 4*1000/(41.4*math.log10(D/d))*10**-3 #Capacitance(uF)\n",
- "\n",
- "Vp = Vl/3**0.5\n",
- "I = 2*3.14*f*C*Vp*10**-6 #Carging current(A)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The capacitance is\",round(C,3),\"uF\"\n",
- "print \"Charging current of a single core cable is\",round(I,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The capacitance is 0.254 uF\n",
- "Charging current of a single core cable is 3.04 A\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.6, Page Number: 276"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Calculation:\n",
- "er = 3 #relative permittivity\n",
- "d = 2.5 #core diameter(cm)\n",
- "l = 4000 #cable length(m)\n",
- "t = 0.5 #insulation thickness(cm)\n",
- "Vl = 33000 #line voltage(V)\n",
- "f = 50 #frequency(Hz)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "D = d+2*t #conductor diameter(cm)\n",
- "C = er*l/(41.4*math.log10(D/d))*10**-3 #Capacitance(uF)\n",
- "Vp = Vl/3**0.5\n",
- "I = 2*3.14*f*C*Vp*10**-6 #Carging current(A)\n",
- "kVAR = 3*Vp*I #Total charging kVAR\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) The capacitance is\",round(C*10**3),\"* 10**-9 F\"\n",
- "print \"(ii) Charging current of a single core cable is\",round(I,2),\"A\"\n",
- "print \"(iii)Total charging kVAR is\",round(kVAR/1000,1),\"* 10**3 kVAR\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The capacitance is 1984.0 * 10**-9 F\n",
- "(ii) Charging current of a single core cable is 11.87 A\n",
- "(iii)Total charging kVAR is 678.3 * 10**3 kVAR\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.7, Page Number: 278"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "\n",
- "#Variable Declaration:\n",
- "V = 33 #voltage of cable(V)\n",
- "d = 1 #conductor diameter(cm)\n",
- "D = 4 #sheath diameter(cm)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "gmax = 2*V/(d*math.log(D/d)) #maximum stress,rms(kV/cm)\n",
- "gmin = gmax*d/D #minimum stress,rms(kV/cm)\n",
- "\n",
- "#Result:\n",
- "print \"The maximum and minimum stress in the insulation are\"\n",
- "print \"gmax =\",round(gmax,2),\"kV/cm rms & gmin =\",round(gmin,2),\"kV/cm rms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum and minimum stress in the insulation are\n",
- "gmax = 47.61 kV/cm rms & gmin = 11.9 kV/cm rms\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.8, Page Number: 278"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "gmax = 40 #kV/cm\n",
- "gmin = 10 #kV/cm\n",
- "d = 2 #conductor diameter(cm)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "D = gmax/gmin*d #cm\n",
- "t = (D-d)/2 #thickness of insulation(cm)\n",
- "V = gmax*d*math.log(D/d)/2\n",
- "\n",
- "#Result:\n",
- "print \"(i) Thickness of insulation is\",t,\"cm\"\n",
- "print \"(ii)Operating voltage is\",round(V,2),\"kV rms\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Thickness of insulation is 3.0 cm\n",
- "(ii)Operating voltage is 55.45 kV rms\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.9, Page Number: 279"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "V = 11 #voltage of cable(V)\n",
- "a = 0.645 #conductor area(cm**2)\n",
- "D = 2.18 #internal diameter of sheath(cm)\n",
- "er = 3.5 #relative permitivity\n",
- "l = 1000 #conductor length(m)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "d = (4*a/3.14)**0.5 #Diameter of the conductor(cm)\n",
- "gmax = 2*V/(d*math.log(D/d)) #Maximum electrostatic stress(kV/cm rms)\n",
- "gmin = 2*V/(D*math.log(D/d)) #Minimum electrostatic stress(kV/cm rms)\n",
- "C = er*l/(41.4*math.log10(D/d))*10**-9 #Capacitance of cable(F)\n",
- "I = 2*3.14*f*C*V*1000 #Carging current(A)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) Maximum electrostatic stress in the cable is\",round(gmax,2),\"kV/cm rms\"\n",
- "print \"(ii) Minimum electrostatic stress in the cable is\",round(gmin,2),\"kV/cm rms\"\n",
- "print \"(iii)Capacitance of the cable per km length is\",round(C*10**6,2),\"* 10**-6 F\"\n",
- "print \"(iv) Charging current is\",round(I,3),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Maximum electrostatic stress in the cable is 27.66 kV/cm rms\n",
- "(ii) Minimum electrostatic stress in the cable is 11.5 kV/cm rms\n",
- "(iii)Capacitance of the cable per km length is 0.22 * 10**-6 F\n",
- "(iv) Charging current is 0.766 A\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.10, Page Number: 280"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "V = 50 #Cable voltage(kV)\n",
- "gmax = 40 #Maximum permissible stress(kV/cm)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "Vp = V*2**0.5 #Peak value of cable voltage(kV)\n",
- "d = 2*Vp/gmax #Most economical conductor diameter(cm)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The most economical value of diameter is\",round(d,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The most economical value of diameter is 3.54 cm\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.11, Page Number: 280"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "Vl = 132 #Cable voltage(kV)\n",
- "gmax = 60 #Maximum permissible stress(kV/cm)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "Vph = Vl/3**0.5 #phase voltage(kV)\n",
- "Vp = Vph*2**0.5 #Peak value of cable voltage(kV)\n",
- "d = 2*Vp/gmax #Most economical conductor diameter(cm)\n",
- "D = 2.718*d #Internal diameter of sheath(cm)\n",
- "\n",
- "#Result:\n",
- "print \"Most economical conductor diameter is\",round(d,1),\"cm\"\n",
- "print \"Internal diameter of sheath, D is\",round(D,2),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Most economical conductor diameter is 3.6 cm\n",
- "Internal diameter of sheath, D is 9.76 cm\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.12, Page Number: 282"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "d = 2 #conductor diameter(cm)\n",
- "e3 = 3 #relative permittivity\n",
- "e2 = 4\n",
- "e1 = 5\n",
- "D = 8 #overall diameter(cm)\n",
- "gmax = 40 #kV/cm\n",
- "\n",
- "#Calculation:\n",
- "#Graded cable: As the maximum stress in the three dielectrics is the same,\n",
- "d1 = e1*d/e2 #diameter of 1st layer(cm)\n",
- "d2 = e1*d/e3 #diameter of 2nd layer(cm)\n",
- "#Permissible peak voltage for the cable:\n",
- "Vp1 = gmax/2*(d*math.log(d1/d)+d1*math.log(d2/d1)+d2*math.log(D/d2)) #kV\n",
- "Vs1 = Vp1/2**0.5 #Safe working voltage (r.m.s.) for cable(kV)\n",
- "\n",
- "#Ungraded cable:\n",
- "Vp2 = gmax/2*d*log(D/d) #kV\n",
- "Vs2 = Vp2/(2**0.5) #kV\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"For Graded cable, safe working voltage is\",round(Vs1,2),\"kV\"\n",
- "print \"For Ungraded cable, safe working voltage is\",round(Vs2,1),\"kV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "For Graded cable, safe working voltage is 57.75 kV\n",
- "For Ungraded cable, safe working voltage is 39.2 kV\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.13, Page Number: 283"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "d = 3 #conductor diameter(cm)\n",
- "e2 = 4 #relative permittivity\n",
- "e1 = 5\n",
- "D = 9 #overall diameter(cm)\n",
- "g1max = 30 #kV/cm\n",
- "g2max = 20 #kV/cm\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "d1 = g1max/g2max*e1*d/e2 #cm\n",
- "t1 = (d1-d)/2 #Radial thickness of inner dielectric(cm)\n",
- "t2 = (D-d1)/2 #Radial thickness of outer dielectric(cm)\n",
- "Vp = g1max/2*d*math.log(d1/d)+g2max/2*d1*math.log(D/d1)\n",
- "Vsf = Vp/2**0.5 #Safe working voltage(r.m.s.)for the cable(kV)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Radial thickness of inner dielectric is\",round(t1,3),\"cm\"\n",
- "print \"Radial thickness of outer dielectric is\",round(t2,2),\"cm\"\n",
- "print \"Safe working voltage (r.m.s.) for the cable is\",round(Vsf,2),\"kV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Radial thickness of inner dielectric is 1.313 cm\n",
- "Radial thickness of outer dielectric is 1.69 cm\n",
- "Safe working voltage (r.m.s.) for the cable is 38.7 kV\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.14, Page Number: 284"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable Declaration:\n",
- "e1 = 5 #relative permittivity\n",
- "e2 = 3\n",
- "t = 1 #thickness(cm)\n",
- "d = 2 #core diameter(cm)\n",
- "V = 66 #cable voltage(kV)\n",
- "\n",
- "#Calculation:\n",
- "d1 = d+2*t\n",
- "D = d+4*t #total diameter of cable(cm)\n",
- "Vpk = V/3**0.5*2**0.5 #Peak voltage per phase(kV)\n",
- "g1max = 2*Vpk/(d*(math.log(d1/d)+e1/e2*math.log(D/d1))) #kV/cm\n",
- "g2max = 2*Vpk/(d1*(e2/e1*math.log(d1/d)+math.log(D/d1))) #kV/cm\n",
- "\n",
- "#Result:\n",
- "print \"Maximum stresses in two dielectrics are:\"\n",
- "print \"g1max =\",round(g1max,2),\"kV/cm g2max =\",round(g2max,2),\"kV/cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum stresses in two dielectrics are:\n",
- "g1max = 39.37 kV/cm g2max = 32.8 kV/cm\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.15, Page Number: 285"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "d = 2 #cm\n",
- "d1 = 3.1 #cm\n",
- "d2 = 4.2 #cm\n",
- "D = 5.3 #cm\n",
- "V = 66 #cable voltage(kV)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "Vpk = V/3**0.5*2**0.5 #peak voltage/phase(kV)\n",
- "#let V1,V2 & V3 are the voltages at different grades.\n",
- "V1,V2,V3 = symbols('V1 V2 V3')\n",
- "g1max = V1/(d/2*math.log(d1,d))\n",
- "g2max = V2/(d1/2*math.log(d2,d1))\n",
- "g3max = V3/(d2/2*math.log(D,d2))\n",
- "\n",
- "#As the maximum stress in the layers is the same,\n",
- "#\u2234 g1max = g2max = g3max\n",
- "#or 2\u00b728 V1 = 2\u00b712 V2 = 2\u00b704 V3\n",
- "#\u2234 V2 = (2\u00b728/2\u00b712) V1 = 1\u00b7075 V1\n",
- "#and V3 = (2\u00b728/2\u00b704) V1 = 1\u00b7117 V1\n",
- "#Now V1 + V2 + V3 = Vpk\n",
- "#or V1 + 1\u00b7075 V1 + 1\u00b7117 V1 = 53\u00b79\n",
- "V1 = 53.9/3.192\n",
- "V2 = 1.075*V1\n",
- "V1s = Vpk-V1 #Voltage on first intersheath(near to core)(kV)\n",
- "V2s = Vpk-V1-V2 #Voltage on second intersheath(kV)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Voltage on first intersheath is\",round(V1s,2),\"kV\"\n",
- "print \"Voltage on second intersheath is\",round(V2s,2),\"kV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Voltage on first intersheath is 37.0 kV\n",
- "Voltage on second intersheath is 18.85 kV\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.16, Page Number: 286"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "d = 2 #core diameter(cm)\n",
- "D = 5.3 #cm\n",
- "V = 66 #cable voltage(kV)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "Vpk = V/3**0.5*2**0.5 #peak voltage/phase(kV)\n",
- "#(i) Positions of intersheaths.\n",
- "\n",
- "#Let the diameters of intersheaths are d1 and d2 cm respectively.\n",
- "#Let V1 = voltage b/w conductor & intersheath 1\n",
- "# V2 = voltage b/w intersheaths 1 and 2\n",
- "# V3 = voltage b/w intersheath 2 & outer lead sheath\n",
- "\n",
- "\n",
- "#Given the maximum stress in the three layers is the same,\n",
- "#we get the relation as given below:\n",
- "# d1**2 = d * d2 = 2*d2 [\u2235 d = 2 cm]\n",
- "#or d2 = d1**2/2\n",
- "#and d1*d2 = D * d = 5\u00b73 \u00d7 2 = 10.6 cm\n",
- "#or d1 * d1**2/2 = 10\u00b76\n",
- "d1 = 21.2**(1/3) #cm\n",
- "d2 = d1**2/2 #cm\n",
- "\n",
- "\n",
- "#(ii) Voltage on intersheaths,\n",
- "# V = V1 + V2 + V3\n",
- "#or 53\u00b79 = V1+d1/d*V1+d2/d*V1\n",
- "# = 4.28*V1\n",
- "V1 = 53.9/4.28 #kV\n",
- "V2 = d1/d*V1 #kV\n",
- "\n",
- "#(iii) Stresses in dielectrics,\n",
- "gmax = V1/(d/2*math.log(d1/d)) #max stress(kV/cm)\n",
- "gmin = V1/(d1/2*math.log(d1/d)) #min stress(kV/cm\n",
- "\n",
- "#Result:\n",
- "print \"(i) Positions of intersheaths are:\"\n",
- "print \"\\td1 =\",round(d1,2),\"cm d2 =\",round(d2,1),\"cm\"\n",
- "print \"(ii) Voltage on the intersheaths are:\"\n",
- "print \"\\tVoltage on first intersheath is\",round(Vpk-V1,2),\"kV\"\n",
- "print \"\\tVoltage on second intersheath is\",round(Vpk-V1-V2,1),\"kV\"\n",
- "print \"(iii) Maximum and minimum stress are:\"\n",
- "print \"\\tgmax =\",round(gmax),\"kV/cm\\tgmin =\",round(gmin,2),\"kV/cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Positions of intersheaths are:\n",
- "\td1 = 2.77 cm d2 = 3.8 cm\n",
- "(ii) Voltage on the intersheaths are:\n",
- "\tVoltage on first intersheath is 41.3 kV\n",
- "\tVoltage on second intersheath is 23.9 kV\n",
- "(iii) Maximum and minimum stress are:\n",
- "\tgmax = 39.0 kV/cm\tgmin = 28.01 kV/cm\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.17, Page Number: 289"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable Declaration:\n",
- "c = 0.3 #capacitance per kilometre(uF/km)\n",
- "V = 11 #line voltage(kV)\n",
- "l = 5 #length of the cable(km)\n",
- "f = 50 #Hz\n",
- "\n",
- "#Calculation:\n",
- "C3 = c*l #capacitance between a pair of cores with third core\n",
- " #earthed for a length of 5 km (uF)\n",
- "Vph = V*1000/3**0.5 #phase voltage(V)\n",
- "#core to neutral capacitance Cn of this cable is given by :\n",
- "Cn = 2*C3 #uF\n",
- "Ic = 2*math.pi*f*Vph*Cn*10**-6\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \" The charging current is\",round(Ic,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " The charging current is 5.99 A\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.18, Page Number: 290"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable Declaration:\n",
- "V = 66 #line voltage(kV)\n",
- "C1 = 12.6 #uF\n",
- "C2 = 7.4 #uF\n",
- "f = 50 #Hz\n",
- "\n",
- "#Calculation:\n",
- "Vph = V*1000/3**0.5 #phase voltage(V)\n",
- "Ce = C1/3 #core-earth capacitances(uF)\n",
- "Cc = (C2-Ce)/2 #core-core capacitances(uF)\n",
- "Cn = Ce+3*Cc #Core to neutral capacitance(uF)\n",
- "Ic = 2*math.pi*f*Vph*Cn*10**-6\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The charging current is\",round(Ic,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The charging current is 107.74 A\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.19, Page Number: 290"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable Declaration:\n",
- "c = 0.18 #capacitance per kilometre(uF/km)\n",
- "V = 3300 #line voltage(V)\n",
- "l = 20 #length of the cable(km)\n",
- "f = 50 #Hz\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "C3 = c*l #capacitance between a pair of cores with third core\n",
- " #earthed for a length of 20 km (uF)\n",
- "Vph = V/3**0.5 #phase voltage(V)\n",
- "Cn = 2*C3 #Core to neutral capacitance(uF)\n",
- "Ic = 2*math.pi*f*Cn*Vph*10**-6 #charging current(A)\n",
- "kVA = 3*Vph*Ic/1000 #kVA taken by the cable\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The kVA taken by 20 km long cable is\",round(kVA,2),\"kVA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The kVA taken by 20 km long cable is 24.63 kVA\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.20, Page Number: 292"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "k = 5 #thermal resistivity of the dielectric(ohm-m)\n",
- "S2 = 0.45 #thermal resistance b/w the sheath and the ground surface\n",
- "R = 110 #electrical resistance of the cable(u-ohm)\n",
- "r = 15 #core radius(mm)\n",
- "t = 40 #dielectric thickness(mm)\n",
- "T = 55 #temperature(deg. C)\n",
- "n = 1 #no. of conductors\n",
- "\n",
- "#Calculation:\n",
- "r1 = r+t #mm\n",
- "S1 = k/(2*math.pi)*math.log(r1/r) #ohm/m\n",
- "S = S1+S2 #ohm/m\n",
- "I = (T/(n*R*10**-6*S))**0.5 #current loading(A)\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Maximum permissible current loading is\",round(I,1),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum permissible current loading is 580.5 A\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.21, Page Number: 296"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "Q = 15 #ohm\n",
- "P = 45 #ohm\n",
- "l = 300 #length of faulty cable(m)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "L = 2*l #loop length(m)\n",
- "d = Q/(P+Q)*L #Distance of the fault point from test end(m)\n",
- "\n",
- "#Result:\n",
- "print \" The distance of the fault point from the test end is\",d,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " The distance of the fault point from the test end is 150.0 m\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- " Example 11.22, Page Number: 297"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "l = 500 #length of faulty cable(m)\n",
- "#P:Q = 3\n",
- "\n",
- "#Calculation:\n",
- "#Let:\n",
- "P = 3; Q = 1 #ohm\n",
- "#then,\n",
- "d = Q/(P+Q)*2*l #Distance of the fault point from test end(m)\n",
- "\n",
- "#Result:\n",
- "print \"The distance of the fault from the testing end of cables\",d,\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The distance of the fault from the testing end of cables 250.0 m\n"
- ]
- }
- ],
- "prompt_number": 24
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.23, Page Number: 297"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "l = 500 #length of faulty cable(m)\n",
- "rp = 0.001 #resistance of cable(ohm/m)\n",
- "rq = 0.00225 #resistance of sound cable(ohm/m)\n",
- "#P:Q = 2.75:1\n",
- "\n",
- "#Calculation:\n",
- "#Let:\n",
- "P = 2.75; Q = 1 #ohm\n",
- "#then,\n",
- "R = rp*l+rq*l #Resistance of loop(ohm)\n",
- "X = Q/(P+Q)*R #Resistance of faulty cable from test end upto fault point(ohm)\n",
- "d = X/rp #Distance of the fault point from test end(m)\n",
- "\n",
- "#Result:\n",
- "print \"The distance of the fault from the testing end of cables\",round(d),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The distance of the fault from the testing end of cables 433.0 m\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 11.24, Page Number: 297"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "S = 200 #ohm\n",
- "r = 20 #resistance per km(ohm)\n",
- "l = 20 #length of cable(km)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "# R+X = 20*(20+20) #ohm\n",
- "# P = Q\n",
- "X = (800-200)/2 #ohm\n",
- "d = X/r #m\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The distance of the fault from the test end is\",d,\"km\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The distance of the fault from the test end is 15.0 km\n"
- ]
- }
- ],
- "prompt_number": 33
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Principles_of_Power_System/chapter13_1.ipynb b/Principles_of_Power_System/chapter13_1.ipynb deleted file mode 100644 index 9f2fa9af..00000000 --- a/Principles_of_Power_System/chapter13_1.ipynb +++ /dev/null @@ -1,1972 +0,0 @@ -{
- "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 diff --git a/Principles_of_Power_System/chapter14_1.ipynb b/Principles_of_Power_System/chapter14_1.ipynb deleted file mode 100644 index 9abb2d75..00000000 --- a/Principles_of_Power_System/chapter14_1.ipynb +++ /dev/null @@ -1,746 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:0910cfded5ae40d0ea2b9576604d8007ec8328215058593fcec218762063deff"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 14: A.C. Distribution"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 14.1, Page Number: 359"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "l = 300 #line length(m)\n",
- "magI1 = 100 #current at load 1(A)\n",
- "pf1 = 0.707 #power factor at load 1\n",
- "l1 = 200 #line length till load 1(m)\n",
- "magI2 = 200 #current drawn at load 2(A)\n",
- "pf2 = 0.8 #power factor at laod 2\n",
- "l2 = 300 #line length till load 2(m)\n",
- "R = 0.2 #total resistance of line(ohm/km)\n",
- "X = 0.1 #total reactance of the line(ohm/km)\n",
- "\n",
- "#Calculation:\n",
- "phy1 = math.acos(pf1)\n",
- "phy2 = math.acos(pf2)\n",
- "Zac = (R+X*1j)*200/1000 #Impedance of section AC(ohm)\n",
- "Zcb = (R+X*1j)*100/1000 #Impedance of section CB(ohm)\n",
- "#Taking voltage at the far end B as the reference vector, we have\n",
- "I2 = magI2*(pf2-1j*math.sin(phy2)) #Load current at point B(A)\n",
- "I1 = magI1*(pf1-1j*math.sin(phy1)) #Load current at point C(A)\n",
- "Icb = I2 #A\n",
- "Iac = I1+I2 #A\n",
- "Vcb = Icb*Zcb #V\n",
- "Vac = Iac*Zac #V\n",
- "V = Vac+Vcb #Voltage drop in the distributor(V)\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Voltage drop in the distributor is\",round(abs(V),2),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Voltage drop in the distributor is 17.85 V\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 14.2, Page Number: 359"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "l = 2000 #line length(m)\n",
- "magI1 = 80 #current at load 1(A)\n",
- "pf1 = 0.9 #power factor at load 1\n",
- "l1 = 1000 #line length till load 1(m)\n",
- "magI2 = 120 #current drawn at load 2(A)\n",
- "pf2 = 0.8 #power factor at laod 2\n",
- "l2 = 2000 #line length till load 2(m)\n",
- "R = 0.05 #total resistance of line(ohm/km)\n",
- "X = 0.1 #total reactance of the line(ohm/km)\n",
- "magVb = 230 #voltage maintained at point B(V)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "phy1 = math.acos(pf1)\n",
- "phy2 = math.acos(pf2)\n",
- "Zac = (R+X*1j)*1000/1000 #Impedance of section AC(ohm)\n",
- "Zcb = (R+X*1j)*1000/1000 #Impedance of section CB(ohm)\n",
- "#Taking voltage at the far end B as the reference vector, we have\n",
- "I2 = magI2*(pf2-1j*math.sin(phy2)) #Load current at point B(A)\n",
- "I1 = magI1*(pf1-1j*math.sin(phy1)) #Load current at point C(A)\n",
- "Icb = I2 #A\n",
- "Iac = I1+I2 #A\n",
- "Vcb = Icb*Zcb #V\n",
- "Vac = Iac*Zac #V\n",
- "V = Vac+Vcb+magVb*(1+0j) #Voltage drop in the distributor(V)\n",
- "theta = math.atan(V.imag/V.real)\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) Voltage drop in the distributor is\",round(abs(V),2),\"V\"\n",
- "print \"(ii)The phase difference between Va and Vb is \",round(math.degrees(theta),2),\"degrees\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Voltage drop in the distributor is 261.67 V\n",
- "(ii)The phase difference between Va and Vb is 3.83 degrees\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 14.3, Page Number: 360"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "magI1 = 100 #current at load 1(A)\n",
- "pf1 = 0.6 #power factor at load 1\n",
- "magI2 = 100 #current drawn at load 2(A)\n",
- "pf2 = 0.8 #power factor at laod 2\n",
- "R = 0.1 #total resistance of line(ohm/km)\n",
- "X = 0.15 #total reactance of the line(ohm/km)\n",
- "magVb = 200 #voltage maintained at point B(V)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "phy1 = math.acos(pf1)\n",
- "phy2 = math.acos(pf2)\n",
- "Zam = (R+X*1j) #Impedance of section AM(ohm)\n",
- "Zmb = (R+X*1j) #Impedance of section MB(ohm)\n",
- "#Taking voltage at the far end B as the reference vector, we have\n",
- "I2 = magI2*(pf2-1j*math.sin(phy2)) #Load current at point B(A)\n",
- "Imb = I2 \n",
- "Vb = magVb*(1+0j) #V\n",
- "Vmb = Imb*Zmb #V\n",
- "Vm = Vb+Vmb #V\n",
- "alpha = math.atan(Vm.imag/Vm.real) #V\n",
- "#The load current I1 has a lagging p.f. of 0\u00b76 w.r.t. VM. It lags \n",
- "#behind Vm by an angle phy1.\n",
- "#Phase angle between I1 and Vb\n",
- "phy11 = phy1-alpha\n",
- "I1 = magI1*(math.cos(phy11)-math.sin(phy11)*1j) #A\n",
- "Iam = I1+I2 #A\n",
- "Vam = Iam*Zam #V\n",
- "Va = Vm+Vam #V\n",
- "theta = math.atan(Va.imag/Va.real)\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i)Voltage at mid-point is\",round(abs(Vm),1),\"V\"\n",
- "print \"(ii) Sending end voltage Va is\",round(abs(Va),2),\"V\"\n",
- "print \"(iii)The phase difference between Va and Vb is \",round(math.degrees(theta),2),\"degrees\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i)Voltage at mid-point is 217.1 V\n",
- "(ii) Sending end voltage Va is 252.33 V\n",
- "(iii)The phase difference between Va and Vb is 3.07 degrees\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 14.4, Page Number: 362"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "Zab = 1+1j #ohm\n",
- "Zbc = 1+2j #ohm\n",
- "Zac = 1+3j #ohm\n",
- "Ib = 20 #load current at B(A)\n",
- "pfb = 0.8 #power factor at A\n",
- "Ic = 15 #load current at C(A)\n",
- "pfc = 0.6 #power factor at B\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "Iab = Ib*(pfb-1j*math.sin(math.acos(pfb))) #Current in section AB(A)\n",
- "Iac = Ic*(pfc-1j*math.sin(math.acos(pfc))) #Current in section AB(A)\n",
- "Vab = Iab*Zab #Voltage drop in section AB(V)\n",
- "Vac = Iac*Zac #Voltage drop in section AC(V)\n",
- "#point B is at higher potential than point C. The p.d. between B and C\n",
- "#is Thevenin\u2019s equivalent circuit e.m.f. Eo i.e.\n",
- "Eo = Vac-Vab #volt\n",
- "Zo = Zab+Zac #Thevenin\u2019s equivalent impedance(ohm)\n",
- "Ibc = Eo/(Zo+Zbc) #A\n",
- "Iab1 = Iab+Ibc #A\n",
- "Iac1 = Iac-Ibc #A\n",
- "Ia = Iab+Iac #Current fed at A(A)\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The total current fed at A is\",Ia,\"A\"\n",
- "print \"\\nCurrent in AB is\",Iab1.real+round(Iab1.imag,2)*1j,\"A\"\n",
- "print \"\\nCurrent in BC is\",Ibc.real+round(Ibc.imag,2)*1j,\"A\"\n",
- "print \"\\nCurrent in AC is\",Iac1.real+round(Iac1.imag,2)*1j,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The total current fed at A is (25-24j) A\n",
- "\n",
- "Current in AB is (18.6-13.53j) A\n",
- "\n",
- "Current in BC is (2.6-1.53j) A\n",
- "\n",
- "Current in AC is (6.4-10.47j) A\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 14.5, Page Number: 363"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "l = 1000 #line length(m)\n",
- "magI1 = 5 #current at load 1(A)\n",
- "pf1 = 0.8 #power factor at load 1\n",
- "l1 = 600 #line length till load 1(m)\n",
- "pf2 = 0.85 #power factor at motor load B\n",
- "Po = 10 #power output at B(H.P)\n",
- "n = 0.9 #efficiency\n",
- "l2 = 400 #line length till load 2(m)\n",
- "R = 1 #total resistance of line(ohm/km)\n",
- "X = 0.5 #total reactance of the line(ohm/km)\n",
- "magVb = 400 #voltage maintained at point B(V)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "Zac = (R+X*1j)*l1/l #ohm\n",
- "Zcb = (R+X*1j)*l2/l #ohm\n",
- "magVbp = magVb/3**0.5 #volt per phase\n",
- "Vbp = magVbp*(1+0j) #V\n",
- "magIb = Po*746/(3**0.5*magVb*pf2*n) #Line current at B(A)\n",
- "magI2p = magIb\n",
- "I2p = magI2p*(pf2-1j*math.sin(math.atan(pf2))) #A\n",
- "I1p = magI1*(pf1-1j*math.sin(math.atan(pf1))) #A\n",
- "Iac = I1p+I2p #Current in section AC(A)\n",
- "Icb = I2p #Current in section CB(A)\n",
- "Vcb = Icb*Zcb #V\n",
- "Vac = Iac*Zac #V\n",
- "Va = Vbp+Vcb+Vac #V\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Line voltage at A is\",round(abs(Va)*3**0.5),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Line voltage at A is 434.0 V\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 14.6, Page Number: 364"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "magVa = 11000 #volt\n",
- "magIb = 50 #load current at B(A)\n",
- "pf2 = 0.8 #power factor(lagging)\n",
- "magIc = 120 #load current at C(A)\n",
- "pf3 = 1.0 #power factor(lagging)\n",
- "magId = 70 #load current at D(A)\n",
- "pf4 = 0.866 #power factor(lagging)\n",
- "Zab = 1+0.6j #ohm\n",
- "Zbc = 1.2+0.9j #ohm\n",
- "Zcd = 0.8+0.5j #ohm\n",
- "Zda = 3+2j #ohm\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Let current in section AB be (x + j*y).\n",
- "x,y = symbols('x,y')\n",
- "Iab = x+1j*y\n",
- "x,y = symbols('x,y')\n",
- "Ibc = Iab-magIb*(pf2-math.sin(math.atan(pf2))) #A\n",
- "Icd = (x-40+1j*(y+30))-(120+0j) #A\n",
- "Ida = ((x-160)+1j*(y+30))-(70*(0.866-0.5j)) #A\n",
- "Vab = Iab*Zab #Drop in section AB(V)\n",
- "Vbc = Ibc*Zbc #Drop in section BC(V)\n",
- "Vcd = Icd*Zcd #Drop in section CD(V)\n",
- "Vda = Ida*Zda #Drop in section DA(V)\n",
- "# Vab+Vbc+Vcd+Vda = 0\n",
- "# As the real and imaginary parts have to be separately zero,\n",
- "# 6*x-4*y-1009.8 = 0\n",
- "# 4*x+6*y-302.2 = 0\n",
- "x1 = solve(6*x-4*y-1009.8,x)[0]\n",
- "y1 = round(solve(4*x1+6*y-302.2,y)[0],1)\n",
- "x11 = round(solve(6*x-4*y1-1009.8,x)[0],1)\n",
- "#now putting the values of x11 and y1 in above equationa,\n",
- "Iab1 = x11+1j*y1 #A\n",
- "Ibc1 = (x11-40)+1j*(y1+30) #A\n",
- "Icd1 = (x11-40+1j*(y1+30))-(120+0j) #A\n",
- "Ida1 = ((x11-160)+1j*(y1+30))-(70*(0.866-0.5j)) #A\n",
- "magVap = round(magVa/3**0.5) #Voltage at supply end A(V)\n",
- "Vb = magVap*(1+0j)-Iab1*Zab #Voltage at station B(V/phase)\n",
- "Vc = Vb-Ibc1*Zbc #Voltage at station C(V/phase)\n",
- "Vd = Vc-Icd1*Zcd #Voltage at station D(V/phase)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Current in section AB is\",Iab1,\"A\"\n",
- "print \"Current in section BC is\",Ibc1,\"A\"\n",
- "print \"Current in section CD is\",Icd1,\"A\"\n",
- "print \"Current in section DA is\",Ida1,\"A\"\n",
- "print \"Voltage at A is\",magVap*(1+0j),\"V/phase\"\n",
- "print \"Voltage at B is\",Vb,\"V/phase\"\n",
- "print \"Voltage at C is\",Vc,\"V/phase\"\n",
- "print \"Voltage at D is\",Vd,\"V/phase\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current in section AB is (139.8-42.8j) A\n",
- "Current in section BC is (99.8-12.8j) A\n",
- "Current in section CD is (-20.2-12.8j) A\n",
- "Current in section DA is (-80.82+22.2j) A\n",
- "Voltage at A is (6351+0j) V/phase\n",
- "Voltage at B is (6185.52-41.08j) V/phase\n",
- "Voltage at C is (6054.24-115.54j) V/phase\n",
- "Voltage at D is (6064-95.2j) V/phase\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 14.7, Page Number: 368"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "Pr = 10 #load connected to line R(kW)\n",
- "Py = 8 #load connected to line Y(kW)\n",
- "Pb = 5 #load connected to line B(kW)\n",
- "Vl = 400 #line voltage(V)\n",
- "#the loads are non-reactive.\n",
- "\n",
- "#Calculation:\n",
- "Vp = round(Vl/3**0.5) #phase voltage(V)\n",
- "Ir = Pr*1000/Vp #A\n",
- "Iy = Py*1000/Vp #A\n",
- "Ib = Pb*1000/Vp #A\n",
- "\n",
- "#Resolving the three currents along x-axis and y-axis, we have,\n",
- "Ih = Iy*math.cos(math.pi/6)-Ib*math.cos(math.pi/6) #Resultant horizontal component(A)\n",
- "Iv = Ir-Iy*math.cos(math.pi/3)-Ib*math.cos(math.pi/3) #Resultant vertical component(A)\n",
- "In = (Ih**2+Iv**2)**0.5 #current in neutral wire(A)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) Ir =\",round(Ir,1),\"A\"\n",
- "print \" Iy =\",round(Iy,1),\"A\"\n",
- "print \" Ib =\",round(Ib,2),\"A\"\n",
- "print \"(ii) Current in neutral wire is\",round(In,1),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Ir = 43.3 A\n",
- " Iy = 34.6 A\n",
- " Ib = 21.65 A\n",
- "(ii) Current in neutral wire is 18.9 A\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 14.8, Page Number: 369"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "Vl = 400 #line voltage(V)\n",
- "Vp = 230 #voltage across lamp(V)\n",
- "I1 = 70 #current in load RN(A)\n",
- "I2 = 84 #current in load YN(A)\n",
- "I3 = 33 #current in load BN(A)\n",
- "Im = 200 #current taken by the motor(A)\n",
- "pf = 0.2 #power factor(lagging)\n",
- "\n",
- "#Calculation:\n",
- "#Lamp load alone:\n",
- "Ih = I2*math.cos(math.pi/6)-I3*math.cos(math.pi/6) #Resultant H-component(A)\n",
- "Iv = I1-I3*math.cos(math.pi/3)-84*math.cos(math.pi/3) #Resultant V-component(A)\n",
- "In = math.sqrt(Ih**2+Iv**2) #Neutral current(A)\n",
- "\n",
- "\n",
- "#Both lamp load and motor load:\n",
- "Ir = Im*pf #Active component of motor current(A)\n",
- "Ix = Im*math.sin(math.acos(pf)) #Reactive component of motor current(A)\n",
- "IR = ((Ir+I1)**2+Ix**2)**0.5 #A\n",
- "IY = ((Ir+I2)**2+Ix**2)**0.5 #A\n",
- "IB = ((Ir+I3)**2+Ix**2)**0.5 #A\n",
- "P = Vp*(I1+I2+I3)*1 #Watt #( cos phy_L = 1)\n",
- "Pm = 3**0.5*Vl*Im*pf #Power supplied to motor(W)\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Lamp load alone: neutral curent is\",round(In,2),\"A\"\n",
- "print \"\\nWhen Both lamp load and motor load is present:\"\n",
- "print \"The current components are:\"\n",
- "print \"Neutral current is\",round(In,2),\"A\"\n",
- "print \"IR =\",round(IR,1),\"A;\\tIY =\",round(IY,0),\"A;\\tIB =\",round(IB,2),\"A\"\n",
- "print \"Power supplied to the lamp is\",P,\"W\"\n",
- "print \"Power supplied to the motor is\",round(Pm),\"W\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Lamp load alone: neutral curent is 45.64 A\n",
- "\n",
- "When Both lamp load and motor load is present:\n",
- "The current components are:\n",
- "Neutral current is 45.64 A\n",
- "IR = 224.7 A;\tIY = 232.0 A;\tIB = 209.11 A\n",
- "Power supplied to the lamp is 43010 W\n",
- "Power supplied to the motor is 27713.0 W\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 14.9, Page Number: 370"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "Prn = 20 #kW\n",
- "pf1 = 1 #power factor of loaf RN\n",
- "kVAyn = 28.75 #kVA of load YN\n",
- "kVAbn = 28.75 #kVA of load BN\n",
- "pf2 = 0.866 #power factor of laod YN & BN each.(lagging)\n",
- "Vl = 400 #line voltage(V)\n",
- "Vp = 230 #phase voltage(V)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "phy1 = math.acos(pf1)\n",
- "phy2 = math.acos(pf2)\n",
- "phy3 = phy2\n",
- "Ir = Prn*1000/Vp #A\n",
- "Iy = kVAyn*1000/Vp #A\n",
- "Ib = kVAbn*1000/Vp #A\n",
- "Ih = Ir-Iy*math.cos(phy2)-Ib*math.cos(phy2) #A\n",
- "Iv = 0+Iy*math.sin(phy2)-Iy*math.sin(phy3) #A\n",
- "In = math.sqrt(Ih**2+Iv**2) #A\n",
- "#When load from B to N removed.:\n",
- "#When the load from B to N is removed, the various line currents are:\n",
- "#Ir in phase with Vrn; Iy lagging by 30 deg.; Ib = 0.\n",
- "Ir1 = Ir\n",
- "Iy1 = Iy; Ib1 = 0 #A\n",
- "Ih1 = Ir1-Iy1*math.cos(math.pi/6) #A\n",
- "Iv1 = 0-Iy1*math.sin(math.pi/6) #A\n",
- "In1 = math.sqrt(Ih1**2+Iv1**2) #A\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"When no changes were made, the various currents are:\"\n",
- "print \"Ir =\",round(Ir,2),\"A;\\tIy =\",Iy,\"A;\\tIb =\",Ib,\"A;\\tIn =\",round(In,2),\"A\"\n",
- "print \"\\nWhen load from B to N removed, the various currents are:\"\n",
- "print \"Ir =\",round(Ir1,2),\"A;\\tIy =\",Iy1,\"A;\\tIb =\",Ib1,\"A;\\tIn =\",round(In1,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "When no changes were made, the various currents are:\n",
- "Ir = 86.96 A;\tIy = 125.0 A;\tIb = 125.0 A;\tIn = 129.54 A\n",
- "\n",
- "When load from B to N removed, the various currents are:\n",
- "Ir = 86.96 A;\tIy = 125.0 A;\tIb = 0 A;\tIn = 66.03 A\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 14.10, Page Number: 371"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import cmath\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "Vl = 400 #line voltage(V)\n",
- "Vp = 230 #phase voltage(V)\n",
- "Ir = 30 #load current at R-phase(A)\n",
- "pf1 = 0.866 #power factor for R-phase(lagging)\n",
- "Iy = 30 ##load current at Y-phase(A)\n",
- "pf2 = 0.866 #power factor for R-phase(lagging)\n",
- "Ib = 30 ##load current at R-phase(A)\n",
- "pf3 = 1.0 #power factor for R-phase(lagging)\n",
- "R = 0.2 #resistance of each line conductor(ohm)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "phy1 = math.acos(pf1)\n",
- "phy2 = math.acos(pf2)\n",
- "phy3 = math.acos(pf3)\n",
- "\n",
- "VR = Vp*(1+0j) #V\n",
- "VY = Vp*(math.cos(-2*math.pi/3)+math.sin(-2*math.pi/3)) #V\n",
- "VB = Vp*(math.cos(2*math.pi/3)+math.sin(-2*math.pi/3)) #V\n",
- "\n",
- "#the line currents can be expressed as :\n",
- "IR = cmath.rect(30,-math.pi/6) #A\n",
- "IY = cmath.rect(30,-math.pi/2) #A\n",
- "IB = cmath.rect(30,2*math.pi/3) #A\n",
- "IN = IR+IY+IB #A\n",
- "\n",
- "#Since, the area of X-section of neutral is half of any line conductor.\n",
- "Rn = 2*R #resistance of neutral(ohm)\n",
- "#ER = VR + Drop in R phase + Drop in neutral\n",
- "ER = VR+R*IR+IN*2*R #V\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The supply end voltage for R phase is\",round(ER.real,3)+1j*round(ER.imag,3),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The supply end voltage for R phase is (239.588-10.608j) V\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 14.11, Page Number: 371"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "Vl = 400 #line voltage(V)\n",
- "Vp = 230 #phase voltage(A)\n",
- "Pln = 100 #load connected b/n LN(W)\n",
- "Pyn = 150 #load connected b/n YN(W)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#before disconnecting the neutral wire,\n",
- "R1 = Vp**2/Pln #Resistance of lamp L1(ohm)\n",
- "R2 = Vp**2/Pyn #Resistance of lamp L2(ohm)\n",
- "\n",
- "#When the neutral wire is disconnected,\n",
- "EL = 400 #V\n",
- "I = EL/(R1+R2) #A\n",
- "V1 = I*R1 #Voltage across lamp L1(V)\n",
- "V2 = I*R2 #Voltage across lamp L2(V)\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The voltage across the lamps are:\"\n",
- "print \"Lamp 1, Voltage =\",V1,\"V ;\\tLamp 2, voltage =\",V2,\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The voltage across the lamps are:\n",
- "Lamp 1, Voltage = 240.0 V ;\tLamp 2, voltage = 160.0 V\n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Principles_of_Power_System/chapter15_1.ipynb b/Principles_of_Power_System/chapter15_1.ipynb deleted file mode 100644 index 35b4fcc7..00000000 --- a/Principles_of_Power_System/chapter15_1.ipynb +++ /dev/null @@ -1,126 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:99f2463b3650514f9c88ec63c2c7f7ef832ae42348e5f3fe2b1b870ab9434349"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 15: Voltage Control"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 15.1, Page Number: 384"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "Pl = 10000 #load(kW)\n",
- "pf = 0.8 #power factor(lag)\n",
- "Vl = 33000 #receiving end line voltage(V)\n",
- "R = 5 #line resistance(ohm)\n",
- "X = 10 #line reactance(ohm)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "I2 = math.floor(Pl*1000/(3**0.5*Vl*0.8)) #load current(A)\n",
- "Ip = I2*pf #A\n",
- "Iq = I2*math.sin(math.acos(pf)) #A\n",
- "V1 = round(Vl/3**0.5) #sending end voltage(V)\n",
- "#Let Im be the current taken by the synchronous condenser.\n",
- "Im = symbols('Im') #A\n",
- "Im1 = solve((V1+Ip*R-X*(Im-Iq))**2+(Ip*X+(Im-Iq)*R)**2-V1**2,Im)[0]\n",
- "C = 3*V1*round(Im1)/1000 #kVAR\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Capacity of synchronous condenser is\",math.floor(C),\"kVAR\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Capacity of synchronous condenser is 13203.0 kVAR\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 15.2 Page Number: 385"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "Pl = 25000 #load(kW)\n",
- "pf = 0.8 #power factor(lag)\n",
- "Vl = 33000 #receiving end line voltage(V)\n",
- "R = 5 #line resistance(ohm)\n",
- "X = 20 #line reactance(ohm)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "I2 = round(Pl*1000/(3**0.5*Vl*pf),1) #load current(A)\n",
- "Ip = I2*pf #A\n",
- "Iq = I2*math.sin(math.acos(pf)) #A\n",
- "V1 = Vl/3**0.5 #sending end voltage(V)\n",
- "#Let Im be the current taken by the synchronous condenser.\n",
- "Im = symbols('Im') #A\n",
- "Im1 = solve((V1+Ip*R-X*(Im-Iq))**2+(Ip*X+(Im-Iq)*R)**2-V1**2,Im)[0]\n",
- "C = 3*V1*Im1/1000 #kVAR\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Capacity of synchronous condenser is\",round(C/1000,2),\"MVAR\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Capacity of synchronous condenser is 33.11 MVAR\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Principles_of_Power_System/chapter17_1.ipynb b/Principles_of_Power_System/chapter17_1.ipynb deleted file mode 100644 index f0b77b2e..00000000 --- a/Principles_of_Power_System/chapter17_1.ipynb +++ /dev/null @@ -1,842 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:85e6b6fc08cf389e51cb8c88e6ef45df0ab76627422a51dd5811f81a969b3ed6"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 17: Symmetrical Fault Calculations"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 17.1, Page Number: 402"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "kVAa = 15000\n",
- "kVAb = 20000\n",
- "V = 12000\n",
- "kVA_base = 35000\n",
- "Xa = 30 #%reactance of alternator A(%)\n",
- "Xb = 50 #%reactance of alternator B(%)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "Xa1 = kVA_base/kVAa*Xa #% Reactance of alternator A at the base kVA\n",
- "Xb1 = kVA_base/kVAb*Xb #% Reactance of alternator B at the base kVA\n",
- "I = kVA_base*1000/(3**0.5*V) #Line current corresponding to 35000 kVA at 12 kV\n",
- "\n",
- "X = Xa1*Xb1/(Xa1+Xb1) #Total % reactance from generator neutral up to fault point\n",
- "Isc = I*100/X #Short-circuit current(A)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The short-circuit current is\",round(Isc),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The short-circuit current is 4330.0 A\n"
- ]
- }
- ],
- "prompt_number": 48
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 17.2, Page Number: 404"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "kVA = 20000 #kVA rating of alternator\n",
- "V = 10000 #voltage rating of alternator\n",
- "Xa = 5 # % reactance of alternator(ohm)\n",
- "\n",
- "#Calculation:\n",
- "I = kVA*1000/(3**0.5*V) #full load current(A)\n",
- "Vp = V/3**0.5 #phase voltage(A)\n",
- "#As the short-circuit current is to be 8 times the full-load current,\n",
- "Xr = 1/8*100\n",
- "Xe = Xr-Xa #External % reactance required\n",
- "X = Xe*Vp/(I*100) #per phase external reactance required(ohm)\n",
- "\n",
- "#Result:\n",
- "print \"Per phase external reactance required is\",X,\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Per phase external reactance required is 0.375 ohm\n"
- ]
- }
- ],
- "prompt_number": 49
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 17.3, Page Number: 404"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "Vl = 10 #transmission line voltage(kV)\n",
- "Rl = 1 #line resistance(ohm)\n",
- "Xl = 4 #line reactance(ohm)\n",
- "MVAtf = 5 #transformer rating(MVA)\n",
- "Xtf = 5 #reactance of transformer(%)\n",
- "MVAal = 10 #rating of alternator(MVA)\n",
- "Xal = 10 #reactance of alternator(%)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Let 10,000 kVA be the base kVA\n",
- "kVAb = 10000 #base kVA\n",
- "Xalb = kVAb/(MVAal*1000)*Xal #% reactance of alternator on base kVA\n",
- "Xtfb = kVAb/(MVAtf*1000)*Xtf #% reactance of transformer on base kVA\n",
- "Xl1 = kVAb*Xl/(10*Vl**2) #% reactance of transmission line\n",
- "Rl1 = kVAb*Rl/(10*Vl**2) #% resistance of transmission line\n",
- "\n",
- "\n",
- "#(i)For a fault at the end of a transmission line (point F2),\n",
- "Xt = Xalb+Xtfb+Xl1 #Total % reactance\n",
- "Z = (Xt**2+Rl1**2)**0.5 #% impedance from generator neutral upto fault point F2\n",
- "SCkVA1 = kVAb*100/Z #Short-circuit kVA\n",
- "\n",
- "\n",
- "#(ii)For a fault at the high voltage terminals of the transformer (point F1),\n",
- "#Total % reactance from generator neutral upto fault point F1:\n",
- "Xt1 = Xalb+Xtfb\n",
- "SCkVA2 = kVAb*100/Xt1 #Short-circuit kVA\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) SC kVA for 1st case = \",round(SCkVA1),\"kVA\"\n",
- "print \"(ii)SC kVA for 2nd case = \",round(SCkVA2),\"kVA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) SC kVA for 1st case = 16440.0 kVA\n",
- "(ii)SC kVA for 2nd case = 50000.0 kVA\n"
- ]
- }
- ],
- "prompt_number": 50
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 17.4, Page Number: 405"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "kVAb = 10000 #kVA base\n",
- "Xt = 5 #reactance of each transformer(%)\n",
- "kVAt = 5000 #kVA rating of each transformer\n",
- "kVA1 = 10000 #kVA of generator A & B each\n",
- "kVA3 = 5000 #kVA of generator C\n",
- "Xa = 12 #reactance of generator A & B each(%)\n",
- "Xc = 18 #reactance of generator c(%)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#The % reactance of generators A, B and C and that of\n",
- "#each transformer on the selected base kVA will be:\n",
- "XA = Xa*kVAb/kVA1\n",
- "XB = Xa*kVAb/kVA1\n",
- "XC = Xc*kVAb/kVA3\n",
- "XT = Xt*kVAb/kVAt\n",
- "\n",
- "#(i) When the fault occurs on the low voltage side of the\n",
- "#transformer,\n",
- "#Total % reactance from generator neutral upto fault point F1\n",
- "XT1 = XA/2*XC/(XA/2+XC) #%\n",
- "MVAf1 = kVAb*100/XT1/1000 #Fault MVA\n",
- "\n",
- "#(i) When the fault occurs on the high voltage side of the\n",
- "#transformer\n",
- "#Total % reactance from generator neutral upto fault point F2\n",
- "XT2 = XT1+XT #%\n",
- "MVAf2 = kVAb*100/XT2/1000 #Fault MVA\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Maximum fault MVA which the circuit breakers on:\"\n",
- "print \"(i) low voltage side is\",round(MVAf1,1),\"MVA\"\n",
- "print \"(ii)high voltage side is\",round(MVAf2),\"MVA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum fault MVA which the circuit breakers on:\n",
- "(i) low voltage side is 194.4 MVA\n",
- "(ii)high voltage side is 66.0 MVA\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 17.5, Page Number: 407"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "#Variable declaration:\n",
- "X1 = 10 #reactance of generator 1 & 2 each(%)\n",
- "X3 = 12 #reactance of generator 3 & 4 each(%)\n",
- "kVA1 = 10000 #kVA rating of generator 1 & 2 each\n",
- "kVA3 = 8000 #kVA rating of generator 3 & 4 each\n",
- "Xr = 10 #reactance of reactor(%)\n",
- "kVAr = 5000 #kVA of reactor\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Fig. above shows the single line diagram of the network.\n",
- "kVAb = 10000 #base kVA\n",
- "X1b = X1*kVAb/kVA1 #% Reactance of generator 1 or 2 on the base kVA\n",
- "X3b = X3*kVAb/kVA3 #% Reactance of generator 3 or 4 on the base kVA\n",
- "Xrb = Xr*kVAb/kVAr #% Reactance of bus-bar reactor on the base kVA\n",
- "\n",
- "#After the fault occurs,\n",
- "Xt = ((X1b/2+Xrb)*X3b/2)/(X1b/2+Xrb+X3b/2)\n",
- "kVAf = kVAb*100/Xt\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Required fault MVA is\",round(kVAf/1000,2),\"MVA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Required fault MVA is 173.33 MVA\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 17.6, Page Number: 408"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "kVAa = 3000 #kVA rating of generator A\n",
- "kVAb = 4500 #kVA rating of generator B\n",
- "Xa = 7 #% reactance of gen A\n",
- "Xb = 8 #% reactance of gen A\n",
- "kVAt = 7500 #kVA rating of transformer\n",
- "Xt = 7.5 #% reactance of transformer\n",
- "Vb = 3.3 #bus voltage(kV)\n",
- "\n",
- "#Calculation:\n",
- "kVAbs = 7500 #base kVA(say)\n",
- "XA = Xa*kVAbs/kVAa #% Reactance of generator A on the base kVA\n",
- "XB = Xb*kVAbs/kVAb #% Reactance of generator B on the base kVA\n",
- "XT = Xt*kVAbs/kVAt #% Reactance of transformer on the base kVA\n",
- "X = symbols('X') #percentage reactance of the bus-bar reactor\n",
- "Xt1 = (XA*XB/(XA+XB))*(X+XT)/((XA*XB/(XA+XB))+(X+XT))\n",
- "SCkVA = kVAt*100*Xt1 #Short-circuit kVA\n",
- "#But the short-circuit kVA should not exceed 150 * 10**3 kVA,\n",
- "#the rupturing capacity of the breaker.\n",
- "\n",
- "X1 = abs(solve(SCkVA-150*1000,X)[0])\n",
- "x = X1*10*Vb**2/kVAt #reactance of the reactor(ohm)\n",
- "\n",
- "#Result:\n",
- "print \"Reactance of the reactor per phase is\",round(x,3),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Reactance of the reactor per phase is 0.106 ohm\n"
- ]
- }
- ],
- "prompt_number": 53
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 17.7, Page Number: 409"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaratioon:\n",
- "MVAbase = 100 #base MVA\n",
- "MVAa = 1500 #estimated short-circuit MVA at busbar A\n",
- "MVAb = 1200 #estimated short-circuit MVA at busbar B\n",
- "kV = 33 #generated voltage at each station(kV)\n",
- "x = 1 #transmission line reactance(ohm)\n",
- "\n",
- "#Calculation:\n",
- "Xa = MVAbase/MVAa*100 #% Reactance of station A on the base MVA\n",
- "Xb = MVAbase/MVAb*100 #% Reactance of station B on the base MVA\n",
- "Xt = MVAbase*1000*x/(10*kV**2)\n",
- "\n",
- "#Fault on station A.\n",
- "Xt1 = (Xb+Xt)*Xa/(Xa+Xb+Xt) #Total % reactance upto fault point F1\n",
- "SCMVA1 = MVAbase*100/Xt1 #Short-circuit MVA\n",
- "\n",
- "#Fault on station B.\n",
- "Xt2 = (Xa+Xt)*Xb/(Xb+Xa+Xt) #Total % reactance upto fault point F2\n",
- "SCMVA2 = MVAbase*100/Xt2 #Short-circuit MVA\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Fault on station A, the short circuit MVA is\",round(SCMVA1),\"MVA\"\n",
- "print \"Fault on station B, the short circuit MVA is\",round(SCMVA2),\"MVA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Fault on station A, the short circuit MVA is 2071.0 MVA\n",
- "Fault on station B, the short circuit MVA is 1831.0 MVA\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 17.8, Page Number: 410"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "kVAbase = 5000 #base kVA\n",
- "Xr = 6 #% reactance of reactor\n",
- "Xg = 12 #% reactance of each generator\n",
- "kVAg = 5000 #given generator rating(kVA)\n",
- "kVAr = 5000 #given reactor rating(kVA)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#(i) With reactors:\n",
- "#Suppose a 3-phase short-circuit fault occurs on section 3 of the bus-bar.\n",
- "\n",
- "Xt1 = round(((Xr+Xg)/2+Xr)*Xg/(((Xr+Xg)/2+Xr)+Xg),2) #% reactance from gen. neutral upto fault point F \n",
- "SCkVA1 = kVAbase*100/Xt1 #Short-circuit input\n",
- "\n",
- "#(ii) Without reactors:\n",
- "Xt2 = Xg/3 #Total % reactance upto fault point F\n",
- "SCkVA2 = kVAbase*100/Xt2\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) With reactors, short circuit MVA\",round(SCkVA1/1000,3),\"MVA\"\n",
- "print \"(ii) With reactors, short circuit MVA\",SCkVA2/1000,\"MVA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) With reactors, short circuit MVA 74.963 MVA\n",
- "(ii) With reactors, short circuit MVA 125.0 MVA\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 17.9, Page Number: 411"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "MVAbase = 5 #base MVA\n",
- "MVAg = 10 #MVA of generator\n",
- "MVAr = 10 #MVA of reactor\n",
- "MVAtr = 5 #MVA of transformer\n",
- "xr = 10 #% reactance of each reactor\n",
- "xg = 30 #% reactance of each generator\n",
- "xtr = 5 #% reactance of each transformer\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "Xg = xg*MVAbase/MVAg #%age reactance of each generator on the base MVA\n",
- "Xr = xr*MVAbase/MVAr #%age reactance of each reactor on the base MVA\n",
- "Xtr = xtr*MVAbase/MVAtr #%age reactance of each transformer on the base MVA\n",
- "\n",
- "#Total %age reactance from generator neutral upto fault point F\n",
- "Xt = ((Xg+Xr)/2+Xtr)*Xg/(((Xg+Xr)/2+Xtr)+Xg)+Xr\n",
- "SCMVA = MVAbase*100/Xt #short circuit MVA\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Short circuit MVA is\",SCMVA,\"MVA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Short circuit MVA is 40.0 MVA\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 17.10, Page Number: 412"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "#Let N = no. of section in bus bar\n",
- "#Q = kVA rating of generator\n",
- "#x = reactance of reactance\n",
- "#b = bus reactances.\n",
- "N,Q,x,b = symbols('N Q x b')\n",
- "\n",
- "#Calculation:\n",
- "X1 = (b+(x+b)/(N-1))*x/(b+(x+b)/(N-1)+x) # %\n",
- "SCkVA = Q*100/X1 #short circuit kVA\n",
- "\n",
- "\n",
- "#Now putting values:\n",
- "Q = 50000 #kVA\n",
- "x = 20 # %\n",
- "b = 10 # %\n",
- "\n",
- "#(i) With 3 sections\n",
- "N1 = 3\n",
- "X1 = (b+(x+b)/(N1-1))*x/(b+(x+b)/(N1-1)+x)\n",
- "SCkVA1 = Q*100/X1\n",
- "\n",
- "#(ii) With 9 sections\n",
- "N2 = 9\n",
- "X2 = (b+(x+b)/(N2-1))*x/(b+(x+b)/(N2-1)+x)\n",
- "SCkVA2 = Q*100/X2\n",
- "\n",
- "#(ii) When N is very large\n",
- "N3 = 9999999999999 #say\n",
- "X3 = (b+(x+b)/(N3-1))*x/(b+(x+b)/(N3-1)+x)\n",
- "SCkVA3 = Q*100/X3\n",
- "\n",
- "#Result:\n",
- "print \"Short circuit kVA\"\n",
- "print \"(i) For 3 sections is\",SCkVA1,\"kVA\"\n",
- "print \"(ii) For 9 sections is\",round(SCkVA2),\"kVA\"\n",
- "print \"(iii)For large N is\",round(SCkVA3),\"kVA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Short circuit kVA\n",
- "(i) For 3 sections is 450000.0 kVA\n",
- "(ii) For 9 sections is 613636.0 kVA\n",
- "(iii)For large N is 750000.0 kVA\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 17.11, Page Number: 414"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "MVAbs = 50 #base MVA\n",
- "MVAg = 10 #MVA of each generators\n",
- "xg = 20 #% reactance\n",
- "xt = 10 #reactance of transformer(%)\n",
- "MVAt = 50 #MVA of transformer\n",
- "kV = 33 #bus voltage\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "Xg = MVAbs/MVAg*xg #% reactance of each of the generator on base MVA\n",
- "Xt = MVAbs/MVAt*xt #% reactance of the transformer on base MVA\n",
- "\n",
- "#Suppose the required reactance of the reactor is X % on 50 MVA base.\n",
- "X = symbols('X')\n",
- "#The reactances of the four generators are in parallel\n",
- "#& their equivalent reactance = 100/4 = 25%.\n",
- "\n",
- "Xtt = (Xg/4*(X+10))/(Xg/4+(X+10))\n",
- "\n",
- "#Now fault MVA at F is not to exceed 500 MVA.\n",
- "Xreq = MVAbs*100/500 #required reactance(%)\n",
- "X1 = solve(Xtt-Xreq,X)[0]\n",
- "Xrt = 10*kV**2*X1/(MVAbs*10**3) #Reactance of the reactor(ohm)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Reactance of the reactor is\",round(Xrt,3),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Reactance of the reactor is 1.452 ohm\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 17.12, Page Number: 415"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "kVAg = 5000 #kVA rating of generator\n",
- "V = 6600 #voltage rating(V)\n",
- "x = 6 #reactance of generator(%)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "X = symbols('X') #% reactance of the reactor\n",
- "kVAbs = 5000 #base kVA\n",
- "#The short-circuit kVA is not to exceed 5 \u00d7 5000 kVA.\n",
- "X1 = solve(kVAbs*100/(X+x)-5*kVAg)[0]\n",
- "\n",
- "X11 = X1*10*(V/1000)**2/kVAg #reactance in ohm\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The required reactance is\",round(X11,2),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The required reactance is 1.22 ohm\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 17.13, Page Number: 416"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "MVAg1 = 15 #MVA rating of A & B generators\n",
- "x1 = 12 #reactance of A & B(%)\n",
- "MVAg3 = 8 #MVA rating of generator C\n",
- "x3 = 10 #reactance of C(%)\n",
- "MVAt = 5 #MVA rating of each transformer\n",
- "xt = 4 #reactance of each transformer(%)\n",
- "MVAr = 10 #MVA of reactor\n",
- "xr = 15 #reactance of reactor(%)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Let 10 MVA be the base MVA.\n",
- "MVAbs = 10\n",
- "#The percentage reactance of various elements on the selected\n",
- "#base MVA will be :\n",
- "Xa = MVAbs/MVAg1*x1\n",
- "Xb = MVAbs/MVAg1*x1\n",
- "Xc = MVAbs/MVAg3*x3\n",
- "Xt = MVAbs/MVAt*xt\n",
- "\n",
- "\n",
- "\n",
- "#After the fault occurs,\n",
- "#The reactances of generators A and B are in parallel & their\n",
- "#equivalent reactance is 8%/2 = 4%.\n",
- "\n",
- "#Total reactance upto fault point F:\n",
- "XT = ((Xa*Xb)/(Xa+Xb)+Xt)*(Xc+xr)/(((Xa*Xb)/(Xa+Xb)+Xt)+(Xc+xr))\n",
- "MVA = MVAbs*100/XT #Fault MVA\n",
- "\n",
- "#Result:\n",
- "print \"Total reactance upto fault point F is\",round(MVA,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total reactance upto fault point F is 119.7 %\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 17.14, Page Number: 417"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "MVAa = 10 #MVA ratng of alterator\n",
- "xa = 20 #reactance of alterator(%)\n",
- "MVAt = 5 #MVA of tranformer\n",
- "xt = 10 #reactance of transformer(%)\n",
- "V1 = 6.6 #voltage on alterator side(kV)\n",
- "V2 = 33 #voltage on transmission line side(kV)\n",
- "xl = 50 #line reactance(ohm)\n",
- "rl = 10 #line resistance(ohm)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "MVAbs = 10 #base MVA\n",
- "Xa = MVAbs/MVAa*xa #% reactance of the alternator on base MVA\n",
- "Xt = MVAbs/MVAt*xt #% reactance of the transformer on base MVA\n",
- "Xl = MVAbs*1000*xl/(10*V2**2) #% reactance of the transmission line\n",
- "Rl = MVAbs*1000*rl/(10*V2**2) #% resistance of the transmission line\n",
- "#When the symmetrical fault occurs at point F on the transmission line (50 km away), then\n",
- "XT = Xa+Xt+Xl #Total % reactance upto the point of fault F\n",
- "Z = math.sqrt(XT**2+Rl**2) #% impedance from generator neutral upto fault point F\n",
- "SCMVA = MVAbs*100/Z #Short-circuit MVA\n",
- "Isc = SCMVA*10**6/(3**0.5*V1*1000) #Short-circuit current fed to the fault by the alternator\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Short-circuit current fed to the fault by the alternator is\",round(Isc),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Short-circuit current fed to the fault by the alternator is 1012.0 A\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 17.15, Page Number: 418"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "#the ratings of the machines and equipments are shown in fig above.\n",
- "\n",
- "MVAbs = 10\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "X1 = MVAbs/10*12 #% reactance of each generator (A, B, C and D) on the base MVA\n",
- "X2 = MVAbs/10*24 #% reactance of the reactor on the base MVA\n",
- "X3 = MVAbs/6*3 #% reactance of the transformer on the base MVA\n",
- "\n",
- "#When fault occurs at point F,\n",
- "XT = (30*6/(30+6))+5 #% reactance from generator neutral upto fault point F\n",
- "MVAf = MVAbs*100/XT #Fault MVA\n",
- "Isc = 100*10**6/(3**0.5*66000) #Short-circuit current\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The fault current is\",round(Isc),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The fault current is 875.0 A\n"
- ]
- }
- ],
- "prompt_number": 62
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Principles_of_Power_System/chapter18_1.ipynb b/Principles_of_Power_System/chapter18_1.ipynb deleted file mode 100644 index d291f15a..00000000 --- a/Principles_of_Power_System/chapter18_1.ipynb +++ /dev/null @@ -1,1223 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:c30652f9881ee18245a1635c17c8cb6587983fc4f38ea35cc91611c084543096"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 18: Unsymmetrical Fault Calculations"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 18.2, Page Number: 429"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from pylab import *\n",
- "import cmath\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "Ir = cmath.rect(100,30*math.pi/180) #current in phase R(A)\n",
- "Iy = cmath.rect(50,300*math.pi/180) #current in phase Y(A)\n",
- "Ib = cmath.rect(30,180*math.pi/180) #current in phase B(A)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "a = cmath.rect(1,120*math.pi/180)\n",
- "I0 = 1/3*(Ir+Iy+Ib) #A\n",
- "I1 = 1/3*(Ir+a*Iy+a**2*Ib) #A\n",
- "I2 = 1/3*(Ir+a**2*Iy+a*Ib) #A\n",
- "In = Ir+Iy+Ib #A\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The positive, negative and zero sequence currents in the R-line are\"\n",
- "print \"I0 = (\",round(abs(I0),2),round(rad2deg(angle(I0)),2),\") A\"\n",
- "print \"I1 = (\",round(abs(I1),2),round(rad2deg(angle(I1)),1),\") A\"\n",
- "print \"I2 = (\",round(abs(I2),2),round(rad2deg(angle(I2)),2),\") A\"\n",
- "print \"Return current in the neutral wire is\"\n",
- "print \"In =(\",round(abs(In),2),round(rad2deg(angle(In)),1),\") A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The positive, negative and zero sequence currents in the R-line are\n",
- "I0 = ( 27.29 4.69 ) A\n",
- "I1 = ( 57.98 43.3 ) A\n",
- "I2 = ( 18.97 24.96 ) A\n",
- "Return current in the neutral wire is\n",
- "In =( 81.88 4.7 ) A\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 18.3, Page Number: 430"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from pylab import *\n",
- "import cmath\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "Ir = 12+6j #current in phase A(A)\n",
- "Iy = 12-12j #current in phase Y(A)\n",
- "Ib = -15+10j #current in phase B(A)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "a = cmath.rect(1,120*math.pi/180)\n",
- "#Red phase\n",
- "Ir0 = 1/3*(Ir+Iy+Ib) #A\n",
- "Ir1 = 1/3*(Ir+a*Iy+a**2*Ib) #A\n",
- "Ir2 = 1/3*(Ir+a**2*Iy+a*Ib) #A\n",
- "\n",
- "#Yellow phase:\n",
- "Iy0 = Ir0 #A\n",
- "Iy1 = a**2*Ir1 #A\n",
- "Iy2 = a*Ir2 #A\n",
- "\n",
- "\n",
- "#Blue phase:\n",
- "Ib0 = Ir0 #A\n",
- "Ib1 = a*Ir1 #A\n",
- "Ib2 = a**2*Ir2 #A\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The sequence currents of red phase are:\"\n",
- "print \"Ir0 =\",Ir0.real+round(Ir0.imag,2)*1j,\"A\"\n",
- "print \"Ir1 =\",round(Ir1.real,2)+round(Ir1.imag,2)*1j,\"A\"\n",
- "print \"Ir2 =\",round(Ir2.real,2)+round(Ir2.imag,2)*1j,\"A\"\n",
- "print \"\\nThe sequence currents of red phase are:\"\n",
- "print \"Iy0 =\",Iy0.real+round(Iy0.imag,2)*1j,\"A\"\n",
- "print \"Iy1 =\",round(Iy1.real,2)+round(Iy1.imag,1)*1j,\"A\"\n",
- "print \"Iy2 =\",round(Iy2.real,1)+round(Iy2.imag,2)*1j,\"A\"\n",
- "print \"\\nThe sequence currents of red phase are:\"\n",
- "print \"Ib0 =\",Ib0.real+round(Ib0.imag,2)*1j,\"A\"\n",
- "print \"Ib1 =\",round(Ib1.real,2)+round(Ib1.imag,2)*1j,\"A\"\n",
- "print \"Ib2 =\",round(Ib2.real,2)+round(Ib2.imag,2)*1j,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The sequence currents of red phase are:\n",
- "Ir0 = (3+1.33j) A\n",
- "Ir1 = (10.85+10.13j) A\n",
- "Ir2 = (-1.85-5.46j) A\n",
- "\n",
- "The sequence currents of red phase are:\n",
- "Iy0 = (3+1.33j) A\n",
- "Iy1 = (3.35-14.5j) A\n",
- "Iy2 = (5.7+1.13j) A\n",
- "\n",
- "The sequence currents of red phase are:\n",
- "Ib0 = (3+1.33j) A\n",
- "Ib1 = (-14.2+4.33j) A\n",
- "Ib2 = (-3.8+4.33j) A\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 18.4, Page Number: 430"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from pylab import *\n",
- "import cmath\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "Er0 = 100 #zero sequence voltage of phase R(V)\n",
- "Er1 = 200-100j #positive sequence voltage of phase R(V)\n",
- "Er2 = -100 #negative sequence voltage of phase R(V)\n",
- "\n",
- "#Calculation:\n",
- "a = cmath.rect(1,120*math.pi/180)\n",
- "Er = Er0+Er1+Er2 #V\n",
- "Ey = Er0+a**2*Er1+a*Er2 #V\n",
- "Eb = Er0+a*Er1+a**2*Er2 #V\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The phase voltages are:\"\n",
- "print \"Er =\",(round(abs(Er),2),round(rad2deg(angle(Er)),2)),\"V\"\n",
- "print \"Ey =\",(round(abs(Ey)),round(rad2deg(angle(Ey)),2)),\"V\"\n",
- "print \"Eb =\",(round(abs(Eb),2),round(rad2deg(angle(Eb)),1)),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The phase voltages are:\n",
- "Er = (223.61, -26.57) V\n",
- "Ey = (213.0, -99.9) V\n",
- "Eb = (338.59, 66.2) V\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 18.5, Page Number: 431"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from pylab import *\n",
- "import cmath\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "Er0 = 0.5-0.866j #zero sequence components of red phase(V)\n",
- "Er1 = 2+0j #positive sequence components of red phase(V)\n",
- "Er = 3+0j #phase R voltage(V)\n",
- "\n",
- "#Calculation:\n",
- "a = cmath.rect(1,120*math.pi/180)\n",
- "Er2 = Er-(Er0+Er1) #negative sequence voltage of red phase(V)\n",
- "Ey = Er0+a**2*Er1+a*Er2 #phase Y voltage(V)\n",
- "Eb = Er0+a*Er1+a**2*Er2 #phase B voltage(V)\n",
- "\n",
- "#Result:\n",
- "print \"The negative sequence component of red phase is\"\n",
- "print \"Ir2 = (\",round(abs(Er2)),round(rad2deg(angle(Er2))),\") V\"\n",
- "print \"\\nThe phase voltages are:\"\n",
- "print \"Iy = (\",round(abs(Ey)),round(rad2deg(angle(Ey))),\") V\"\n",
- "print \"Ib = (\",round(abs(Eb)),round(rad2deg(angle(Eb))),\") V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The negative sequence component of red phase is\n",
- "Ir2 = ( 1.0 60.0 ) V\n",
- "\n",
- "The phase voltages are:\n",
- "Iy = ( 3.0 -120.0 ) V\n",
- "Ib = ( 0.0 120.0 ) V\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 18.6, Page Number: 431"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "In = 12 #current from neutral to ground(A)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "I0 = In/3 #A\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The zero phase sequence components in phases are\",I0,\"A each\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The zero phase sequence components in phases are 4.0 A each\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 18.7, Page Number: 432"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import cmath\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "#(i) Before removal of fuses.\n",
- "Iri = cmath.rect(90,0) #phase R current(A)\n",
- "Iyi = cmath.rect(90,math.pi/180*240) #phase Y current(A)\n",
- "Ibi = cmath.rect(90,math.pi/180*120) #phase B current(A)\n",
- "\n",
- "#(ii) After removal of fuses.\n",
- "Irii = 90+0j #A\n",
- "Iyii = 0 #A\n",
- "Ibii = 0 #A\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "a = cmath.rect(1,120*math.pi/180)\n",
- "#(i) Before removal of fuses.\n",
- "#Since the system is balanced, it will have only positive sequence currents.\n",
- "Ir1i = Iri #A\n",
- "Iy1i = Iyi #A\n",
- "Ib1i = Ibi #A\n",
- "\n",
- "\n",
- "\n",
- "#(ii) After removal of fuses.\n",
- "#The sequence currents in the three lines will be:\n",
- "Ir0ii = 1/3*(Irii+Iyii+Ibii) #A\n",
- "Iy0ii = Ir0ii #A\n",
- "Ib0ii = Ir0ii #A\n",
- "\n",
- "Ir1ii = 1/3*(Irii+Iyii+Ibii) #A\n",
- "Iy1ii = a**2*Ir1ii #A\n",
- "Ib1ii = a*Ir1ii #A\n",
- "\n",
- "Ir2ii = 1/3*(Irii+a**2*Iyii+a*Ibii) #A\n",
- "Iy2ii = a*Ir2ii #A\n",
- "Ib2ii = a**2*Ir2ii #A\n",
- "\n",
- "#Result:\n",
- "print \"(i) Before removal of fuses, the symmetrical components are:\"\n",
- "print \"\\tIr1 = (\",round(abs(Ir1i)),round(rad2deg(angle(Ir1i))),\") A\"\n",
- "print \"\\tIy1 = (\",round(abs(Iy1i)),360+round(rad2deg(angle(Iy1i))),\") A\"\n",
- "print \"\\tIb1 = (\",round(abs(Ib1i)),round(rad2deg(angle(Ib1i))),\") A\"\n",
- "print \"\\n(ii) After the removal of fuses, the symmetrical components are:\"\n",
- "print \"\\tIr0 = (\",round(abs(Ir0ii)),round(rad2deg(angle(Ir0ii))),\") A\"\n",
- "print \"\\tIr1 = (\",round(abs(Ir1ii)),round(rad2deg(angle(Ir1ii))),\") A\"\n",
- "print \"\\tIr2 = (\",round(abs(Ir2ii)),round(rad2deg(angle(Ir2ii))),\") A\"\n",
- "print \"\\n\\tIy0 = (\",round(abs(Iy0ii)),round(rad2deg(angle(Iy0ii))),\") A\"\n",
- "print \"\\tIy1 = (\",round(abs(Iy1ii)),360+round(rad2deg(angle(Iy1ii))),\") A\"\n",
- "print \"\\tIy2 = (\",round(abs(Iy2ii)),round(rad2deg(angle(Iy2ii))),\") A\"\n",
- "print \"\\n\\tIb0 = (\",round(abs(Ib0ii)),round(rad2deg(angle(Ib0ii))),\") A\"\n",
- "print \"\\tIb1 = (\",round(abs(Ib1ii)),round(rad2deg(angle(Ib1ii))),\") A\"\n",
- "print \"\\tIb2 = (\",round(abs(Ib2ii)),360+round(rad2deg(angle(Ib2ii))),\") A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Before removal of fuses, the symmetrical components are:\n",
- "\tIr1 = ( 90.0 0.0 ) A\n",
- "\tIy1 = ( 90.0 240.0 ) A\n",
- "\tIb1 = ( 90.0 120.0 ) A\n",
- "\n",
- "(ii) After the removal of fuses, the symmetrical components are:\n",
- "\tIr0 = ( 30.0 0.0 ) A\n",
- "\tIr1 = ( 30.0 0.0 ) A\n",
- "\tIr2 = ( 30.0 0.0 ) A\n",
- "\n",
- "\tIy0 = ( 30.0 0.0 ) A\n",
- "\tIy1 = ( 30.0 240.0 ) A\n",
- "\tIy2 = ( 30.0 120.0 ) A\n",
- "\n",
- "\tIb0 = ( 30.0 0.0 ) A\n",
- "\tIb1 = ( 30.0 120.0 ) A\n",
- "\tIb2 = ( 30.0 240.0 ) A\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 18.8, Page Number: 433"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import cmath\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "Ir1 = cmath.rect(200,0) #positive phase sequence component of current(A)\n",
- "Ir2 = cmath.rect(100,60*math.pi/180) #negative phase sequence component of current(A)\n",
- "In = cmath.rect(300,300*math.pi/180) #current in the neutral conductor(A)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "a = cmath.rect(1,120*math.pi/180)\n",
- "Ir0 = 1/3*In #Zero phase sequence current in R-line(A)\n",
- "Ir = Ir0+Ir1+Ir2 #Current in the R-line(A)\n",
- "Iy = Ir0+a**2*Ir1+a*Ir2 #Current in the Y-line(A)\n",
- "Ib = Ir0+a*Ir1+a**2*Ir2 #Current in the B-line(A)\n",
- "\n",
- "#Result:\n",
- "print \"Current in the R-line is\"\n",
- "print \"\\tIr = (\",round(abs(Ir)),round(rad2deg(angle(Ir))),\") A\"\n",
- "print \"Current in the Y-line is\"\n",
- "print \"\\tIy = (\",round(abs(Iy)),round(rad2deg(angle(Iy))),\") A\"\n",
- "print \"Current in the B-line is\"\n",
- "print \"\\tIb = (\",round(abs(Ib)),round(rad2deg(angle(Ib))),\") A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Current in the R-line is\n",
- "\tIr = ( 300.0 0.0 ) A\n",
- "Current in the Y-line is\n",
- "\tIy = ( 300.0 -120.0 ) A\n",
- "Current in the B-line is\n",
- "\tIb = ( 0.0 180.0 ) A\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 18.9, Page Number: 434"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import cmath\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "Ir = cmath.rect(10,0) #current in line R(A)\n",
- "Iy = cmath.rect(10,180*math.pi/180) #current in line Y(A)\n",
- "Ib = cmath.rect(0,0) #current in line B(A)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "a = cmath.rect(1,120*math.pi/180)\n",
- "#For R-line:\n",
- "Ir0 = 1/3*(Ir+Iy+Ib) #zero sequence component of Ir(A)\n",
- "Ir1 = 1/3*(Ir+a*Iy+a**2*Ib) #positive sequence component of Ir(A)\n",
- "Ir2 = 1/3*(Ir+a**2*Iy+a*Ib) #negative sequence component of Iy(A)\n",
- "\n",
- "#For Y-line:\n",
- "Iy0 = Ir0 #zero sequence component of Ir(A)\n",
- "Iy1 = a**2*Ir1 #positive sequence component of Ir(A)\n",
- "Iy2 = a*Ir2 #negative sequence component of Iy(A)\n",
- "\n",
- "#For B-line:\n",
- "Ib0 = Ir0 #zero sequence component of Ir(A)\n",
- "Ib1 = a*Ir1 #positive sequence component of Ir(A)\n",
- "Ib2 = a**2*Ir2 #negative sequence component of Iy(A)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The sequence components of R-line are:\"\n",
- "print \"\\tIr0 = (\",round(abs(Ir0),2),round(rad2deg(angle(Ir0))),\") A\"\n",
- "print \"\\tIr0 = (\",round(abs(Ir1),2),round(rad2deg(angle(Ir1))),\") A\"\n",
- "print \"\\tIr0 = (\",round(abs(Ir2),2),round(rad2deg(angle(Ir2))),\") A\"\n",
- "print \"The sequence components of Y-line are:\"\n",
- "print \"\\tIy0 = (\",round(abs(Iy0),2),round(rad2deg(angle(Iy0))),\") A\"\n",
- "print \"\\tIy1 = (\",round(abs(Iy1),2),round(rad2deg(angle(Iy1))),\") A\"\n",
- "print \"\\tIy2 = (\",round(abs(Iy2),2),round(rad2deg(angle(Iy2))),\") A\"\n",
- "print \"The sequence components of B-line are:\"\n",
- "print \"\\tIb0 = (\",round(abs(Ib0),2),round(rad2deg(angle(Ib0))),\") A\"\n",
- "print \"\\tIb1 = (\",round(abs(Ib1),2),round(rad2deg(angle(Ib1))),\") A\"\n",
- "print \"\\tIb2 = (\",round(abs(Ib2),2),round(rad2deg(angle(Ib2))),\") A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The sequence components of R-line are:\n",
- "\tIr0 = ( 0.0 90.0 ) A\n",
- "\tIr0 = ( 5.77 -30.0 ) A\n",
- "\tIr0 = ( 5.77 30.0 ) A\n",
- "The sequence components of Y-line are:\n",
- "\tIy0 = ( 0.0 90.0 ) A\n",
- "\tIy1 = ( 5.77 -150.0 ) A\n",
- "\tIy2 = ( 5.77 150.0 ) A\n",
- "The sequence components of B-line are:\n",
- "\tIb0 = ( 0.0 90.0 ) A\n",
- "\tIb1 = ( 5.77 90.0 ) A\n",
- "\tIb2 = ( 5.77 -90.0 ) A\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 18.10, Page Number: 435"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import cmath\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "Rrb = 10 #resistance of branch RB(ohm)\n",
- "Rry = 20 #resistance of branch RY(ohm)\n",
- "Rby = 5 #resistance of branch BY(ohm)\n",
- "Er = cmath.rect(-100,0) #voltage across BY(V)\n",
- "Ey = cmath.rect(100,60*math.pi/180) #voltage across BR(V)\n",
- "Eb = cmath.rect(100,-60*math.pi/180) #voltage across RY(V)\n",
- "\n",
- "#Calculation:\n",
- "a = cmath.rect(1,120*math.pi/180)\n",
- "IR = Er/Rby #Current in in branch BY(A)\n",
- "IY = Ey/Rrb #Current in in branch RB(A)\n",
- "IB = Eb/Rry #Current in in branch RY(A)\n",
- "\n",
- "#Sequence currents in resistors:\n",
- "IR0 = 1/3*(IR+IY+IB) #Zero sequence component of IR(A)\n",
- "IR1 = 1/3*(IR+a*IY+a**2*IB) #Positive sequence component of IR(A)\n",
- "IR2 = 1/3*(IR+a**2*IY+a*IB) #Negative sequence component of IR(A)\n",
- "\n",
- "IY0 = IR0 #Zero sequence component of IY(A)\n",
- "IY1 = a**2*IR1 #Positive sequence component of IY(A)\n",
- "IY2 = a*IR2 #Negative sequence component of IY(A)\n",
- "\n",
- "IB0 = IR0 #Zero sequence component of IB(A)\n",
- "IB1 = a*IR1 #Positive sequence component of IB(A)\n",
- "IB2 = a**2*IR2 #Negative sequence component of IB(A)\n",
- "\n",
- "#Sequence currents in supply lines:\n",
- "Ir = IB-IY #Line current in R-line(A)\n",
- "Iy = IR-IB #Line current in Y-line(A)\n",
- "Ib = IY-IR #Line current in R-line(A)\n",
- "\n",
- "Ir0 = 1/3*(Ir+Iy+Ib) #Zero sequence component of Ir(A)\n",
- "Ir1 = 1/3*(Ir+a*Iy+a**2*Ib) #Positive sequence component of Ir(A)\n",
- "Ir2 = 1/3*(Ir+a**2*Iy+a*Ib) #Negative sequence component of Ir(A)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The sequence components of R-line are:\"\n",
- "print \"\\tIR0 = (\",round(abs(IR0),2),round(rad2deg(angle(IR0)),1),\") A\"\n",
- "print \"\\tIR1 = (\",round(abs(IR1),2),round(rad2deg(angle(IR1))),\") A\"\n",
- "print \"\\tIR2 = (\",round(abs(IR2),2),round(rad2deg(angle(IR2)),1),\") A\"\n",
- "print \"\\nThe sequence components of Y-line are:\"\n",
- "print \"\\tIY0 = (\",round(abs(IY0),2),round(rad2deg(angle(IY0)),1),\") A\"\n",
- "print \"\\tIY1 = (\",round(abs(IY1),2),round(rad2deg(angle(IY1))),\") A\"\n",
- "print \"\\tIY2 = (\",round(abs(IY2),1),round(rad2deg(angle(IY2)),1),\") A\"\n",
- "print \"\\nThe sequence components of B-line are:\"\n",
- "print \"\\tIB0 = (\",round(abs(IB0),2),round(rad2deg(angle(IB0)),1),\") A\"\n",
- "print \"\\tIB1 = (\",round(abs(IB1),2),360+round(rad2deg(angle(IB1))),\") A\"\n",
- "print \"\\tIB2 = (\",round(abs(IB2),1),round(rad2deg(angle(IB2)),1),\") A\"\n",
- "print \"\\nThe Sequence currents in supply lines are:\"\n",
- "print \"\\tIr0 = (\",round(abs(Ir0),2),round(rad2deg(angle(Ir0))),\") A\"\n",
- "print \"\\tIr1 = (\",round(abs(Ir1),1),round(rad2deg(angle(Ir1))),\") A\"\n",
- "print \"\\tIr2 = (\",round(abs(Ir2),2),round(rad2deg(angle(Ir2)),1),\") A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The sequence components of R-line are:\n",
- "\tIR0 = ( 4.41 160.9 ) A\n",
- "\tIR1 = ( 11.67 180.0 ) A\n",
- "\tIR2 = ( 4.41 -160.9 ) A\n",
- "\n",
- "The sequence components of Y-line are:\n",
- "\tIY0 = ( 4.41 160.9 ) A\n",
- "\tIY1 = ( 11.67 60.0 ) A\n",
- "\tIY2 = ( 4.4 -40.9 ) A\n",
- "\n",
- "The sequence components of B-line are:\n",
- "\tIB0 = ( 4.41 160.9 ) A\n",
- "\tIB1 = ( 11.67 300.0 ) A\n",
- "\tIB2 = ( 4.4 79.1 ) A\n",
- "\n",
- "The Sequence currents in supply lines are:\n",
- "\tIr0 = ( 0.0 0.0 ) A\n",
- "\tIr1 = ( 20.2 -90.0 ) A\n",
- "\tIr2 = ( 7.64 109.1 ) A\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 18.11, Page Number: 437"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import cmath\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "I = 20 #current in each line(A)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "a = cmath.rect(1,120*math.pi/180)\n",
- "#Let R, Y and B be the supply lines. When fuse in the \n",
- "#line B is removed, the various line currents are :\n",
- "Ir = cmath.rect(I,0) #A\n",
- "Iy = cmath.rect(I,math.pi) #A\n",
- "Ib = cmath.rect(0,0) #A\n",
- "\n",
- "#For R-line:\n",
- "Ir0 = 1/3*(Ir+Iy+Ib) #zero sequence component of Ir(A)\n",
- "Ir1 = 1/3*(Ir+a*Iy+a**2*Ib) #positive sequence component of Ir(A)\n",
- "Ir2 = 1/3*(Ir+a**2*Iy+a*Ib) #negative sequence component of Iy(A)\n",
- "\n",
- "#For Y-line:\n",
- "Iy0 = Ir0 #zero sequence component of Ir(A)\n",
- "Iy1 = a**2*Ir1 #positive sequence component of Ir(A)\n",
- "Iy2 = a*Ir2 #negative sequence component of Iy(A)\n",
- "\n",
- "#For B-line:\n",
- "Ib0 = Ir0 #zero sequence component of Ir(A)\n",
- "Ib1 = a*Ir1 #positive sequence component of Ir(A)\n",
- "Ib2 = a**2*Ir2 #negative sequence component of Iy(A)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The sequence components of R-line are:\"\n",
- "print \"\\tIr0 = (\",round(abs(Ir0),2),round(rad2deg(angle(Ir0))),\") A\"\n",
- "print \"\\tIr0 = (\",round(abs(Ir1),2),round(rad2deg(angle(Ir1))),\") A\"\n",
- "print \"\\tIr0 = (\",round(abs(Ir2),2),round(rad2deg(angle(Ir2))),\") A\"\n",
- "print \"\\nThe sequence components of Y-line are:\"\n",
- "print \"\\tIy0 = (\",round(abs(Iy0),2),round(rad2deg(angle(Iy0))),\") A\"\n",
- "print \"\\tIy1 = (\",round(abs(Iy1),2),360+round(rad2deg(angle(Iy1))),\") A\"\n",
- "print \"\\tIy2 = (\",round(abs(Iy2),2),round(rad2deg(angle(Iy2))),\") A\"\n",
- "print \"\\nThe sequence components of B-line are:\"\n",
- "print \"\\tIb0 = (\",round(abs(Ib0),2),round(rad2deg(angle(Ib0))),\") A\"\n",
- "print \"\\tIb1 = (\",round(abs(Ib1),2),round(rad2deg(angle(Ib1))),\") A\"\n",
- "print \"\\tIb2 = (\",round(abs(Ib2),2),360+round(rad2deg(angle(Ib2))),\") A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The sequence components of R-line are:\n",
- "\tIr0 = ( 0.0 90.0 ) A\n",
- "\tIr0 = ( 11.55 -30.0 ) A\n",
- "\tIr0 = ( 11.55 30.0 ) A\n",
- "\n",
- "The sequence components of Y-line are:\n",
- "\tIy0 = ( 0.0 90.0 ) A\n",
- "\tIy1 = ( 11.55 210.0 ) A\n",
- "\tIy2 = ( 11.55 150.0 ) A\n",
- "\n",
- "The sequence components of B-line are:\n",
- "\tIb0 = ( 0.0 90.0 ) A\n",
- "\tIb1 = ( 11.55 90.0 ) A\n",
- "\tIb2 = ( 11.55 270.0 ) A\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 18.12, Page Number: 438"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from pylab import *\n",
- "import cmath\n",
- "from sympy import *\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "Zr = 5-10j #impedance connected to line R(ohm)\n",
- "Zy = 6+5j #impedance connected to line Y(ohm)\n",
- "Zb = 3+15j #impedance connected to line B(ohm)\n",
- "VRY = cmath.rect(3300,0) #line RY voltage(V)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "a = cmath.rect(1,120*math.pi/180)\n",
- "VYB = a**2*VRY #line YB voltage(V)\n",
- "\n",
- "#Let VR, VY, and VB be the voltages across impedances in R, Y and B\n",
- "#phases respectively and IR, IY, and IB the resulting line currents.\n",
- "IR, IY,IB,IR0,IR1,IR2, = symbols('IR IY IB IR0 IR1 IR2')\n",
- "\n",
- "#IR+IY+IB = 0\n",
- "IR0 = 0; IY0 = 0; IB0 = 0;\n",
- "VR = Zr*(IR0+IR1+IR2) #V\n",
- "VY = Zy*(a**2*IR1+a*IR2) #V\n",
- "VB = Zb*(a*IR1+a**2*IR2) #V\n",
- "#solving the equations:\n",
- "IR11 = solve(VR-VY-VRY,IR1)[0]\n",
- "VY = Zy*(a**2*IR11+a*IR2)\n",
- "VB = Zb*(a*IR11+a**2*IR2)\n",
- "IR22 = solve(VY-VB-VYB,IR2)[0]\n",
- "\n",
- "VR = Zr*(IR0+IR1+IR22)\n",
- "VY = Zy*(a**2*IR1+a*IR22)\n",
- "IR11 = solve(VR-VY-VRY,IR1)[0]\n",
- "IR = IR11+IR22 #A\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The line current IR is (\",round(abs(IR)),round(-rad2deg(cmath.phase(IR)),1),\") A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The line current IR is ( 241.0 -18.5 ) A\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 18.13, Page Number: 440"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import cmath\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "R = 1 #resistance of each load(ohm)\n",
- "magVRY = 200 #line RY voltage(V)\n",
- "magVBR = 400 #line BR voltage(V)\n",
- "magVYB = 346 #line YB voltage(V)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "a = cmath.rect(1,120*math.pi/180)\n",
- "#This is a case of a balanced star-connected load supplied\n",
- "#from an unbalanced 3-phase supply.\n",
- "\n",
- "#Now,\n",
- "\n",
- "# (2)**22 = (1 + 1\u00b775 cos(theta))**2 + (1\u00b775 sin(theta))**2\n",
- "# theta = math.acos(4-1+3*1)/3.5\n",
- "theta = math.pi/2\n",
- "# alpha = math.acos(1+1.75*math.cos(theta))\n",
- "alpha = 60*math.pi/180\n",
- "\n",
- "#As the phase sequence is RYB, therefore, various line voltages are :\n",
- "VRY = cmath.rect(200,math.pi) #V\n",
- "VYB = cmath.rect(346,theta) #V\n",
- "VBR = cmath.rect(400,-alpha) #V\n",
- "\n",
- "#The current in any phase (or line) is equal to phase voltage \n",
- "#divided by resistance in that phase.\n",
- "IR = VRY/(1*3**0.5) #A\n",
- "IY = VYB/(1*3**0.5) #A\n",
- "IB = VBR/(1*3**0.5) #A\n",
- "\n",
- "#Sequence components in red phase are :\n",
- "IR0 = 1/3*(IR+IY+IB) #A\n",
- "IR1 = 1/3*(IR+a*IY+a**2*IB) #A\n",
- "IR2 = 1/3*(IR+a**2*IY+a*IB) #A\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The magnitude of current in any phase is:\"\n",
- "print \"\\tIR0 = (\",round(abs(IR0)),round(rad2deg(angle(IR0))),\") A\"\n",
- "print \"\\tIR1 = (\",round(abs(IR1),1),round(rad2deg(angle(IR1))),\") A\"\n",
- "print \"\\tIR2 = (\",round(abs(IR2),2),round(rad2deg(angle(IR2))),\") A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The magnitude of current in any phase is:\n",
- "\tIR0 = ( 0.0 -90.0 ) A\n",
- "\tIR1 = ( 176.3 -169.0 ) A\n",
- "\tIR2 = ( 66.63 30.0 ) A\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 18.14, Page Number: 449"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "MVAg = 10 #MVA rating of generator\n",
- "Vg = 11 #voltage rating of generator(kV)\n",
- "Zg1 = 1.2j #Positive sequence impedance of generator(ohm)\n",
- "Zg2 = 0.9j #Negative sequence impedance of generator(ohm)\n",
- "Zg0 = 0.4j #Zero sequence impedance of generator(ohm)\n",
- "Zf1 = 1.0j #Positive sequence impedance of feeder(ohm)\n",
- "Zf2 = 1.0j #Negative sequence impedance of feeder(ohm)\n",
- "Zf0 = 3.0j #Zero sequence impedance of feeder(ohm)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Suppose the fault is occured on the red phase. \n",
- "#Taking red phase as the reference,\n",
- "ER = Vg*1000/3**0.5 #Phase e.m.f. of R-phase(V)\n",
- "\n",
- "#(i)The total impedance to any sequence current is given by\n",
- "#the sum of generator and feeder impedances to that sequence current.\n",
- "Z1 = Zg1+Zf1 #ohm\n",
- "Z2 = Zg2+Zf2 #ohm\n",
- "Z0 = Zg0+Zf0 #ohm\n",
- "\n",
- "#For a line-to-ground fault,\n",
- "#I1 = I2 = I0\n",
- "I0 = ER/(Z1+Z2+Z0) #A\n",
- "IR = 3*I0 #fault current(A)\n",
- "\n",
- "#(ii)Line-to-neutral voltage of R-phase,\n",
- "VR = ER-I0*(Zg1+Zg2+Zg0) #V\n",
- "\n",
- "#Result:\n",
- "print \"(i) The magnitude of fault current is\",round(IR.imag)*1j,\"A\"\n",
- "print \"(ii) Line to neutral voltage at the generator terminal is\",round(abs(VR)),\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The magnitude of fault current is (-0-2540j) A\n",
- "(ii) Line to neutral voltage at the generator terminal is 4234.0 V\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 18.15, Page Number: 450"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "Vg = 11000 #voltage rating of the alternator(V)\n",
- "MVAg = 10 #MVA rating of generator\n",
- "X0 = 0.05 #zero sequence reactance(p.u)\n",
- "X1 = 0.15 #positive sequence reactance(p.u)\n",
- "X2 = 0.15 #negative sequence reactance(p.u)\n",
- "\n",
- "#Calculation:\n",
- "#Taking red phase as the reference, let its phase e.m.f. be\n",
- "ER = 1 #p.u\n",
- "\n",
- "#Line-to-ground fault: \n",
- "#Suppose the fault occurs on the red phase.\n",
- "#I1 = I2 = I0\n",
- "I0 = ER/(1j*(X0+X1+X2))\n",
- "IR = 3*I0 #fault current(A)\n",
- "\n",
- "#Three phase fault: \n",
- "#the fault current (say Ish) is limited by the positive sequence reactance only.\n",
- "Ish = ER/(1j*X1) #fault current(A)\n",
- "r = IR/Ish #ratio of the two fault currents\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The ratio of two fault currents is\",round(abs(r),3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The ratio of two fault currents is 1.286\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 18.16, Page Number: 450"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "Vg = 11 #voltage rating of the alternator(kV)\n",
- "MVAg = 25 #MVA rating of generator\n",
- "X0 = 0.05 #zero sequence reactance(p.u)\n",
- "X1 = 0.2 #positive sequence reactance(p.u)\n",
- "X2 = 0.2 #negative sequence reactance(p.u)\n",
- "Xn = 0.3 #neutral to ground reactance(ohm)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Assume that the fault occurs on the red phase.\n",
- "#Taking red phase as the reference, let its phase e.m.f. be \n",
- "ER = 1 #p.u.\n",
- "#First of all, convert the reactance Xn into p.u.\n",
- "Xnp = Xn*MVAg*1000/(Vg**2*1000) #p.u\n",
- "#For a line-to-ground fault,\n",
- "#I1 = I2 = I0\n",
- "I0 = ER/(X1+X2+X0+3*Xnp) #p.u\n",
- "IR = 3*I0 #fault current(p.u)\n",
- "IRa = MVAg*10**6/(3**0.5*Vg*1000) * IR #fault current in amperes\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \" The fault current for a single line to ground fault is\",round(IRa),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " The fault current for a single line to ground fault is 6190.0 A\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 18.17, Page Number: 451"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "V = 10.4 #voltage between the lines(kV)\n",
- "X0 = 0.2j #zero sequence reactance(p.u)\n",
- "X1 = 0.6j #positive sequence reactance(p.u)\n",
- "X2 = 0.5j #negative sequence reactance(p.u)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Taking red phase as the reference, its phase e.m.f. is :\n",
- "ER = round(V*1000/3**0.5) #Phase e.m.f. of R-phase(V)\n",
- "IF = 3**0.5*ER/(X1+X2) #fault current(A)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The fault current is\",round(abs(IF),1),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The fault current is 9453.8 A\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 18.18, Page Number: 451"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "X0 = 0.05j #zero sequence reactance(p.u)\n",
- "X1 = 0.08j #positive sequence reactance(p.u)\n",
- "X2 = 0.07j #negative sequence reactance(p.u)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Taking red phase as the reference, let its phase e.m.f. be \n",
- "ER = 1 #p.u\n",
- "#For a double line-to-ground fault,\n",
- "#IF = IY+IB\n",
- "IF = -3*X2*ER/(X1*X2+X2*X0+X2*X0) #fault current(A)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The fault current is\",round(IF.imag,1)*1j,\"p.u\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The fault current is 16.7j p.u\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 18.19, Page Number: 452"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "MVAg = 20 #MVA rating of generator\n",
- "V = 11 #voltage rating of generator(V)\n",
- "X1 = 20 #positive sequence reactance of generator(%)\n",
- "X2 = 10 #negative sequence reactance of generator(%)\n",
- "X0 = 15 #zero sequence reactance of generator(%)\n",
- "Xn = 5 #generator 1 neutral reactance(%)\n",
- "\n",
- "#Calculation:\n",
- "ER = V*1000/3**0.5 #V\n",
- "#The % reactances in Fig. 18.19b can be converted into ohmic values as under:\n",
- "x1 = X1*V**2*10/(MVAg*1000) #ohm\n",
- "x2 = X2*V**2*10/(MVAg*1000) #ohm\n",
- "x0 = X0*V**2*10/(MVAg*1000) #ohm\n",
- "IR = 3*ER/(x1+x2+x0)*(-1j) #ohm\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Fault current is\",round(IR.imag)*1j,\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Fault current is (-0-6998j) A\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 18.20, Page Number: 453"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "IF1 = 2000 #3 phase fault current(A)\n",
- "IF2 = 2600 #line-to-line fault(A)\n",
- "IF0 = 4200 #line-toground fault(A)\n",
- "MVAg = 50 #MVA rating of the generator\n",
- "V = 11 #voltage rating of the generator(V)\n",
- "\n",
- "#Calculation:\n",
- "#Let X1 = positive sequence reactance,\n",
- "# X2 = negative sequence reactance,and\n",
- "# X0 = zero sequence reactance of the alternator.\n",
- "X1,X2,X0 = symbols('X1 X2 X0')\n",
- "Eph = V*1000/3**0.5 #phase voltage(V)\n",
- "X11 = solve(Eph/X1-IF1)[0] #positive sequence reactance(ohm)\n",
- "X22 = solve(3**0.5*Eph/(X11+X2)-IF2,X2)[0] #negative sequence reactance(ohm)\n",
- "X00 = solve(3*Eph/(X11+X22+X0)-IF0)[0] #zero sequence reactance(ohm)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The positive sequence reactance is\",round(X11,3),\"ohm\"\n",
- "print \"The negative sequence reactance is\",round(X22,3),\"ohm\"\n",
- "print \"The zero sequence reactance is\",round(X00,3),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The positive sequence reactance is 3.175 ohm\n",
- "The negative sequence reactance is 1.055 ohm\n",
- "The zero sequence reactance is 0.306 ohm\n"
- ]
- }
- ],
- "prompt_number": 20
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Principles_of_Power_System/chapter19_1.ipynb b/Principles_of_Power_System/chapter19_1.ipynb deleted file mode 100644 index e6813a0f..00000000 --- a/Principles_of_Power_System/chapter19_1.ipynb +++ /dev/null @@ -1,212 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:34f21e34d56021d11e412854716405b885521e02b7e263afa0b62d127f14da0d"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 19: Circuit Breakers"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 19.1, Page Number: 483"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "I = 1500 #rated current of circuit breaker(A)\n",
- "MVA = 1000 #MVA rating of CB\n",
- "V = 33 #voltage rating of circuit breaker(kV)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "Is = MVA*10**6/(3**0.5*33*1000) #Rated symmetrical breaking current(A,rms)\n",
- "Im = 2.55*Is #Rated making current(A,peak)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) Rated normal current is\",I,\"A\"\n",
- "print \"(ii) Breaking capacity is\",MVA,\"MVA\"\n",
- "print \"(iii) Rated symmetrical breaking current is\",round(Is),\"A (peak)\"\n",
- "print \"(iv) Rated making current is\",round(Im),\"A (peak)\"\n",
- "print \"(v) Short-time rating is\",round(Is),\"for 3 seconds\"\n",
- "print \"(vi) Rated service voltage is\",V,\"kV (r.m.s)\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Rated normal current is 1500 A\n",
- "(ii) Breaking capacity is 1000 MVA\n",
- "(iii) Rated symmetrical breaking current is 17495.0 A (peak)\n",
- "(iv) Rated making current is 44613.0 A (peak)\n",
- "(v) Short-time rating is 17495.0 for 3 seconds\n",
- "(vi) Rated service voltage is 33 kV (r.m.s)\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 19.2, Page Number: 484"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "f = 50 #supply frequency(Hz)\n",
- "V = 11 #voltage rating of generator(V)\n",
- "C = 0.01 #distributed capacitance upto ckt breaker b/w phase and neutral(uF)\n",
- "XL = 5 #reactance of neatral to earth(ohm)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "L = round(XL/(2*math.pi*f),4) #Inductance per phase(H)\n",
- "Emax = round(2**0.5*V/3**0.5,2) #Maximum value of recovery voltage (phase to neutral)(V\n",
- "E = 2*Emax #Peak re-striking voltage(kV)\n",
- "fn = 1/(2*3.14*(L*C*10**-6)**0.5) #frequency of oscillations(Hz)\n",
- "\n",
- "#Since peak re-striking voltage occurs at a time t given by\n",
- "t = 1/(2*fn) #s\n",
- "r = E/t #kV/sec\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) Peak re-striking voltage is\",round(E,2),\"kV\"\n",
- "print \"(ii) Frequency of oscillations is\",round(fn),\"Hz\"\n",
- "print \"(iii)The average rate of rise of re-striking voltage\"\n",
- "print \" upto the first peak is\",math.floor(r/1000),\"*10**3 kV/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Peak re-striking voltage is 17.96 kV\n",
- "(ii) Frequency of oscillations is 12628.0 Hz\n",
- "(iii)The average rate of rise of re-striking voltage\n",
- " upto the first peak is 453.0 *10**3 kV/sec\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 19.3, Page Number: 484"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable delaration:\n",
- "t = 50*10**-6 #time to reach the peak re-striking voltage(s)\n",
- "Vp = 100 #the peak re-striking voltage(kV)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "R = Vp/t #Average RRRV(kV/sec)\n",
- "fn = 1/(2*t) #Natural frequency of oscillations(Hz)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Average RRRV is\",R/10**6,\"* 10**6 kV/sec\"\n",
- "print \"Natural frequency of oscillations is\",fn,\"Hz\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Average RRRV is 2.0 * 10**6 kV/sec\n",
- "Natural frequency of oscillations is 10000.0 Hz\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 19.4, Page Number: 485"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "Im = 11 #magnetising current of transformer(A)\n",
- "Ic = 7 #chopped instantaneous value of current(A)\n",
- "L = 35.2 #inductance (H)\n",
- "C = 0.0023 #capacitance(uF)\n",
- "\n",
- "\n",
- "#Result:\n",
- "e = Ic*(L/(C*10**-6))**0.5 #Voltage across breaker contacts at chopping(V)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Voltage across breaker contacts at chopping is\",round(e/1000),\"kV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Voltage across breaker contacts at chopping is 866.0 kV\n"
- ]
- }
- ],
- "prompt_number": 10
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Principles_of_Power_System/chapter1_1.ipynb b/Principles_of_Power_System/chapter1_1.ipynb deleted file mode 100644 index 95ebdafe..00000000 --- a/Principles_of_Power_System/chapter1_1.ipynb +++ /dev/null @@ -1,68 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 1: Introduction"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 1.1, Page Number: 6"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "Pi = 4200 #input power(W)\n",
- "E = 120 #voltage supply of generator(V)\n",
- "I = 32.2 #supply current(A)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "Po = E*I #output power(W)\n",
- "n = Po/Pi*100 #efficiency(%)\n",
- "Pl = Pi-Po #power lost(W)\n",
- "El = Pl*60 #energy lost per minute(J)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The percentage efficiency of the generator is\",n,\"%\"\n",
- "print \"Energy lost per minute of operation is\",El,\"J\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The percentage efficiency of the generator is 92.0 %\n",
- "Energy lost per minute of operation is 20160.0 J\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Principles_of_Power_System/chapter20_1.ipynb b/Principles_of_Power_System/chapter20_1.ipynb deleted file mode 100644 index 768f8fa6..00000000 --- a/Principles_of_Power_System/chapter20_1.ipynb +++ /dev/null @@ -1,64 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:37bb87f048a2d4a670b09cf05bff6f5c622ef838c4da8caa7225a316e1b26b5e"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 20: Fuses"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 20.1, Page Number: 495"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "r1 = 0.8 #radius of fuse wire(mm)\n",
- "I1 = 8 #blow current at r1 radius(A)\n",
- "I2 = 1 #blow current(A)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "r2 = r1*(I2/I1)**(2/3) #radius for bearing 1 A of blow current(mm)\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The radius of the wire that will blow off at a current of 1A is\",r2,\"mm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The radius of the wire that will blow off at a current of 1A is 0.2 mm\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Principles_of_Power_System/chapter21_1.ipynb b/Principles_of_Power_System/chapter21_1.ipynb deleted file mode 100644 index dbd05a4b..00000000 --- a/Principles_of_Power_System/chapter21_1.ipynb +++ /dev/null @@ -1,69 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:74ef32d0e67f3d77b60188556b39e6627674afc8f6e4feb1b0bbc3ae8dbaee2a"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 21: Protective Relays"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 21.1, Page Number: 507"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "Is = 5 #rated secondary current of C.T\n",
- "CS = 1.25 #current setting\n",
- "TSM = 0.6 #time setting multiplier\n",
- "IF = 4000 #fault current(A)\n",
- "CR = 400/5 #current ratio of CT\n",
- "\n",
- "#Calculation:\n",
- "Ipc = Is*CS #pick-up current(A)\n",
- "If = IF*CR #Fault current in relay coil(A)\n",
- "PSM = If/Ipc #plug setting multiplier\n",
- "#Corresponding to the plug-setting multiplier of 8 given in fig.21.16,\n",
- "#the time of operation is 3.5 seconds.\n",
- "top = 3.5*TSM #Actual relay operating time(s)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The relay operating time is\",top,\"seconds\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The relay operating time is 2.1 seconds\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Principles_of_Power_System/chapter22_1.ipynb b/Principles_of_Power_System/chapter22_1.ipynb deleted file mode 100644 index 38d23b90..00000000 --- a/Principles_of_Power_System/chapter22_1.ipynb +++ /dev/null @@ -1,338 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:a5275b4f1bdd430cf4ed3e203b7e975d86c74b17fe608ee8e07ebac2f3a39a9f"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 22: Protection of Alternators and Transformers"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 22.1, Page Number: 529"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "MVA = 10 #MVA rating of alternator\n",
- "V = 6.6 #voltage rating of alternator(V)\n",
- "X = 10 #per phase reactance of lternator(%)\n",
- "Iop = 175 #operating current(A)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "Vph = V*1000/3**0.5 #phase voltage(kV)\n",
- "I = round(MVA*10**6/(3**0.5*V*1000)) #full load current(A)\n",
- "\n",
- "#Let the reactance per phase be x ohms.\n",
- "r,x = symbols('r x') #r = earthing resistance required to leave 10% of\n",
- " #the winding unprotected\n",
- "x1 = solve(3**0.5*x*I/(6.6*1000)*100-10,x)[0]\n",
- "X1 = x1*0.1 #Reactance of 10% winding\n",
- "E = Vph*0.1 #E.M.F. induced in 10% winding\n",
- "Zf = (X1**2+r**2)**0.5\n",
- "Ief = E/Zf #Earth-fault current due to 10% winding\n",
- "\n",
- "#When this fault current becomes 175 A, the relay will trip\n",
- "r1 = solve(Ief-175,r)[1] #A\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Required value of earth resistance is\",round(r1,3),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Required value of earth resistance is 2.177 ohm\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 22.2, Page Number: 530"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "MVA = 10 #MVA rating of alternator\n",
- "V = 6.6 #voltage rating of alternator(V)\n",
- "CR = 1000/5 #current ratio of CT\n",
- "Rn = 7.5 #resistance of star-point to earth(ohm)\n",
- "Iop = 0.5 #operating current of the relay(A)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Let x % of the winding be unprotected.\n",
- "x = symbols('x')\n",
- "Vph = V*1000/3**0.5 #phase voltage(kV)\n",
- "If = 1000/5*Iop #minimum fault current which will operate the relay(A)\n",
- "E = Vph*x/100 #E.M.F. induced in x% winding(V)\n",
- "Ief = E/Rn #Earth fault current which x% winding will cause(A)\n",
- "#This current must be equal to 100 A.\n",
- "x1 = solve(Ief-If,x)[0]\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Percentage of unprotected winding is\",round(x1,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Percentage of unprotected winding is 19.68 %\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 22.3, Page Number: 530"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "MVA = 10 #MVA rating of alternator\n",
- "V = 6.6 #voltage rating of alternator(V)\n",
- "CR = 1000/5 #current ratio of CT\n",
- "Rn = 6 #resistance of star-point to earth(ohm)\n",
- "Iop = 0.75 #operating current of the relay(A)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Let x % of the winding be unprotected.\n",
- "x = symbols('x')\n",
- "Vph = V*1000/3**0.5 #phase voltage(kV)\n",
- "If = 1000/5*Iop #minimum fault current which will operate the relay(A)\n",
- "E = Vph*x/100 #E.M.F. induced in x% winding(V)\n",
- "Ief = E/Rn #Earth fault current which x% winding will cause(A)\n",
- "#This current must be equal to 100 A.\n",
- "x1 = solve(Ief-If,x)[0]\n",
- "\n",
- "\n",
- "#(ii) Let r2 = the minimum earthing resistance required to \n",
- "#provide protection for 90% of stator winding. \n",
- "#Then, 10% winding would be unprotected\n",
- "x2 = 10 #%\n",
- "r2 = Vph*x2/If*0.01 #ohm\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) The percentage of each of the stator windings is\",round(x1,1),\"%\"\n",
- "print \"(ii)The minimum resistance to provide protection for 90% of\"\n",
- "print \" the stator winding is\",round(r2,2),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The percentage of each of the stator windings is 23.6 %\n",
- "(ii)The minimum resistance to provide protection for 90% of\n",
- " the stator winding is 2.54 ohm\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 22.4, Page Number: 531"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "MVA = 10 #MVA rating of alternator\n",
- "V = 6.6 #voltage rating of alternator(V)\n",
- "CR = 1000/5 #current ratio of CT\n",
- "s = 20 #earth-fault setting(%)\n",
- "Iop = 0.75 #operating current of the relay(A)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Since 85% winding is to be protected, 15% would be unprotected\n",
- "r = symbols('r') #earthing resistance reqd. to leave 15% of winding unprotected(ohm)\n",
- "x = 15 #%\n",
- "Ifl = MVA*10**6/(3**0.5*V*1000) #Full load current(A)\n",
- "IF = s*Ifl/100 #Minimum fault current which will operate the relay\n",
- "Vu = x/100*V*1000/3**0.5 #Voltage induced in 15% of winding(kV)\n",
- "Ief = Vu/r #Earth fault current which 15% winding will cause(A)\n",
- "#This current must be equal to IF.\n",
- "r1 = solve(Ief-IF,r)[0] #ohm\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The value of earthing resistor is\",round(r1,2),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The value of earthing resistor is 3.27 ohm\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 22.5, Page Number: 538"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "r1 = 220/11000 #voltage ratio of transformer\n",
- "r2 = 600/5 #current ratio of protective transformer on 220V side\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Suppose that line current on 220 V side is 600 A\n",
- "\n",
- "Ipd = 5 #Phase current of delta connected CTs on 220V side(A)\n",
- "Ild = 3**0.5*Ipd #Line current of delta connected CTs on 220 V side(A)\n",
- "\n",
- "#This Ild will flow through the pilot wires.\n",
- "Ips = 5*3**0.5 #Phase current of star connected CTs on 11,000 V side(A)\n",
- "\n",
- "#Now, using this relation: Primary apparent power = Secondary apparent power\n",
- "I = 3**0.5*220*600/(3**0.5*11000) #A\n",
- "r3 = I/Ips #Turn-ratio of CTs on 11000 V side\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Turn-ratio of CTs on 11000 V side is (\",round(r3,3),\": 1 )\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Turn-ratio of CTs on 11000 V side is ( 1.386 : 1 )\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 22.6, Page Number: 538"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "r1 = 0.4/11 #line voltage(in kV) ratio of transformer\n",
- "r2 = 500/5 #current ratio of protective transformer\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Suppose the line current on 400 V side is 500 A.\n",
- "Ipd = 5 #Phase current of delta connected CTs on 400 V side(A)\n",
- "Ild = Ipd*3**0.5 #Line current of delta connected CTs on 400 V side(A)\n",
- "#This Ild will flow through the pilot wires.\n",
- "Ips = 5*3**0.5 #Phase current of star-connected CTs on 11000 V side(A)\n",
- "\n",
- "#Primary apparent power = Secondary apparent power\n",
- "I = 3**0.5*400*500/(3**0.5*11000) #A\n",
- "r3 = I/Ips\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \" The ratio of the protective transformers on 11kV side is\",round(r3,3),\"i.e, 10.5:5\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " The ratio of the protective transformers on 11kV side is 2.099 i.e, 10.5:5\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Principles_of_Power_System/chapter26_1.ipynb b/Principles_of_Power_System/chapter26_1.ipynb deleted file mode 100644 index d4334600..00000000 --- a/Principles_of_Power_System/chapter26_1.ipynb +++ /dev/null @@ -1,224 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:260bb8beffe2e4d2155fdaec0c7462f8cf5127d60c79f0bddb469ea9aad62673"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 26: Neutral Grounding"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 26.1, Page Number: 599"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "f = 50 #Supply frequency(Hz)\n",
- "C = 4.5*10**-6 #Line to earth capacitance(F)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "XL = 1/(3*2*math.pi*f*C) #ohm\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The reactance of Peterson coil is\",round(XL,1),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reactance of Peterson coil is 235.8 ohm\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 26.2, Page Number: 599"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "l = 200 #length of transmission line(km)\n",
- "f = 50 #Supply frequency(Hz)\n",
- "c = 0.02*10**-6 #Line to earth capacitance(F/km)\n",
- "V = 230 #voltage rating of line(kV)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "C = c*l #capacitance of coil(F)\n",
- "L = round(1/(3*(2*3.14*f)**2*C),2) #Required inductance of Peterson coil(H)\n",
- "Vph = round(V*1000/1.732) #Voltage across Peterson coil(kV)\n",
- "IF = math.ceil(Vph/(2*3.13*f*L)) #Current through Peterson coil(A)\n",
- "kVA = Vph*IF/1000 #Rating of Peterson coil\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "print \"Inductance of Peterson coil is\",L,\"H\"\n",
- "print \"Rating of Peterson coil is\",round(kVA),\"kVA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Inductance of Peterson coil is 0.85 H\n",
- "Rating of Peterson coil is 66397.0 kVA\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 26.3, Page Number: 600"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "f = 50 #fequency of supply(Hz)\n",
- "C = 1.2*10**-6 #line-to-earth capacitance(F)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#(i) To neutralize capacitance of 100% of the length of the line,\n",
- "#Inductive reactance of the coilis given by\n",
- "XL1 = 1/(3*2*math.pi*f*C) #ohm\n",
- "\n",
- "#(ii) To neutralize capacitance of 90% of the length of the line,\n",
- "#Inductive reactance of the coilis given by\n",
- "XL2 = 1/(3*2*math.pi*f*0.9*C) #ohm\n",
- "\n",
- "\n",
- "#(iii) To neutralize capacitance of 80% of the length of the line,\n",
- "#Inductive reactance of the coilis given by\n",
- "XL3 = 1/(3*2*math.pi*f*0.8*C) #ohm\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) Inductive reactance of the coil to neutralize capacitance\"\n",
- "print \" of 100% of the length of the line is\",round(XL1,2),\"ohm\"\n",
- "print \"\\n(ii) Inductive reactance of the coil to neutralize capacitance\"\n",
- "print \" of 100% of the length of 90% of the line is\",round(XL2,2),\"ohm\"\n",
- "print \"\\n(iii)Inductive reactance of the coil to neutralize capacitance\"\n",
- "print \" of 100% of the length of 80% of the line is\",round(XL3,2),\"ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Inductive reactance of the coil to neutralize capacitance\n",
- " of 100% of the length of the line is 884.19 ohm\n",
- "\n",
- "(ii) Inductive reactance of the coil to neutralize capacitance\n",
- " of 100% of the length of 90% of the line is 982.44 ohm\n",
- "\n",
- "(iii)Inductive reactance of the coil to neutralize capacitance\n",
- " of 100% of the length of 80% of the line is 1105.24 ohm\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 26.4, Page Number: 600"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "r = 0.01 #radius of conductor(m)\n",
- "d = 4 #conductor spacing(m)\n",
- "V = 132 #voltage of the line(kV)\n",
- "f = 50 #supply frequency(Hz)\n",
- "l = 200 #line length(km)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "c = round(2*math.pi*8.885/math.log(d/r),1)*10**-12 #capacitance per unit length(F/m)\n",
- "C = c*l*1000 #Capacitance between phase and earth for 200 km line(F)\n",
- "L = round(1/(3*(2*math.pi*f)**2*C),2) #required inductance L of the arc suppression coil(H)\n",
- "IF = V*1000/(10*math.ceil(3**0.5*2*math.pi*f*L/10)) #Current through the coil(A)\n",
- "kVA = V/3**0.5*IF #kVA\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The inductance of the coil is\",L,\"H\"\n",
- "print \"Rating of the coil is\",round(kVA),\"kVA\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The inductance of the coil is 1.82 H\n",
- "Rating of the coil is 10060.0 kVA\n"
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Principles_of_Power_System/chapter2_1.ipynb b/Principles_of_Power_System/chapter2_1.ipynb deleted file mode 100644 index daf8974b..00000000 --- a/Principles_of_Power_System/chapter2_1.ipynb +++ /dev/null @@ -1,857 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:a62a0cfa22010723ae84accbfbc8f3f3c553c1279f7bb65e42ceaaad44de0807"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 2:Generating Stations"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.1, Page Number: 16"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "n=20 #overall efficiency of plant\n",
- "h=860 #kcal\n",
- "m=0.6 #Mass of fuel burnt(kg) per KW of electrical energy generated\n",
- "\n",
- "#Calculations:\n",
- "x=h*100/(m*n) #Calorific value of fuel(kcal/kg)\n",
- "\n",
- "#Results:\n",
- "print \"Calorific value of fuel =\",round(x,2),\"kcal/kg\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Calorific value of fuel = 7166.67 kcal/kg\n"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.2, Page Number: 17"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "M = 20000 #maximum demand(kW)\n",
- "n_b= 85 #boiler efficiency(%)\n",
- "m=0.9 #coal consumption(kg/kWh)\n",
- "LF=40 #load factor(%)\n",
- "n_t=90 #turbine efficiency(%)\n",
- "c=300 #cost of 1 tonne of coal(Rs)\n",
- "\n",
- "#Calculations:\n",
- "n_th = n_b * n_t/100 #in %\n",
- "cb = LF*M*m*c*24*365/(1000*100)\n",
- "\n",
- "#Results:\n",
- "print \"Thermal efficiency = \",n_th,\"%\"\n",
- "print \"Coal bill per annum = Rs\",cb"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Thermal efficiency = 76.5 %\n",
- "Coal bill per annum = Rs 18921600.0\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.3, Page Number: 17"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "ct=3000000 #annul cost of coal(Rs)\n",
- "cv=5000 #Calorific value of coal(kcal/kg)\n",
- "c=300 #cost of coal per tonne(Rs)\n",
- "n_th=33 #thermal efficiency(%)\n",
- "n_elec=90 #electrical efficiency(%)\n",
- "\n",
- "#Calculations:\n",
- "n_t=n_th*n_elec/100 #overall efficiency(%)\n",
- "h=ct*cv*1000/c #heat of combustion(kcal)\n",
- "ho=n_t*h/(100*860) #heat output(kWh)\n",
- "L=ho/8760 #kW\n",
- "\n",
- "\n",
- "#Results:\n",
- "print \"Avg load on the station=\",round(L),\"kW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Avg load on the station= 1971.0 kW\n"
- ]
- }
- ],
- "prompt_number": 33
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.4, Page Number: 17"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "kWh= symbols('kWh')\n",
- "W = 13500 + 7.5 * kWh #Water evaporated in kg\n",
- "C = 5000 + 2.9 * kWh #coal cumsumption in kg\n",
- "\n",
- "\n",
- "#Calculations:\n",
- "#part (i):\n",
- "#As the station output (i.e., kWh) increases towards infinity,\n",
- "#the limiting value of W/C approaches\n",
- "L1= 7.5/2.9 #in kg\n",
- "\n",
- "#part (ii):\n",
- "#at no load\n",
- "kWh=0\n",
- "c=(5000+2.9*kWh)/8 #coal per hour in kg\n",
- "#Results:\n",
- "print \"Limiting value of water/kg of coal=\",round(L1,1),\"kg\"\n",
- "print \"Required Coal per hour\",c,\"kg\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Limiting value of water/kg of coal= 2.6 kg\n",
- "Required Coal per hour 625.0 kg\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.5, Page Number: 18"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "C=100 #Capacity of station in MW\n",
- "cv=6400 #kcal/kg\n",
- "n_th=0.3 #thermal efficiency\n",
- "n_elec=0.92 #electrical efficiency\n",
- "\n",
- "\n",
- "#Calculations:\n",
- "n_t=n_th*n_elec #overall efficiency\n",
- "U=C*1*10**3 #units generated/hr in kWh\n",
- "H=U*860/n_t #total heat of combustion(kcal)\n",
- "w=H/cv #Coal consumption in kg\n",
- "\n",
- "#Results:\n",
- "print \"The coal consumption per hour =\",round(w),\"kg\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The coal consumption per hour = 48687.0 kg\n"
- ]
- }
- ],
- "prompt_number": 35
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.6, Page Number: 23"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "C=5*10**6 #reservoir capacity in m^3\n",
- "H=200 #water head in m\n",
- "n_t=0.75 #overall efficiency\n",
- "d=1000 #density of water in kg/m^3\n",
- "\n",
- "#Calculations:\n",
- "W=C*d*9.81 #weight of water in Newton\n",
- "E=W*H*n_t/(3600*1000) #electrical energy available(kWh)\n",
- "\n",
- "\n",
- "#Results:\n",
- "print \"The total energy available=\",round(E/10**6,3),\"* 10^6 kWh\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The total energy available= 2.044 * 10^6 kWh\n"
- ]
- }
- ],
- "prompt_number": 36
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.7, Page Number: 23"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "V=94 #volume of water in m^3/sec\n",
- "d=1000 #density of water in kg/m^3\n",
- "H=39 #head of water in m\n",
- "nt=0.80 #overall efficiency\n",
- "\n",
- "\n",
- "#Calculations:\n",
- "W=V*d #weight of water in kg/sec\n",
- "w=W*H*9.81/1000 #work done per sec in kW\n",
- "FC=nt*w #firm capacity in kW\n",
- "YGO=FC*8760 #yearly gross capacity in kWh\n",
- "\n",
- "\n",
- "#Results:\n",
- "print \"Firm capacity=\",FC,\"kW\"\n",
- "print \"Yearly gross output\",round(YGO/10**6),\"* 10^6 kWh\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Firm capacity= 28770.768 kW\n",
- "Yearly gross output 252.0 * 10^6 kWh\n"
- ]
- }
- ],
- "prompt_number": 37
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.8, Page Number: 23"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable Declaration:\n",
- "H=100 #Water head in m\n",
- "Q=1 #discharge, m^3/sec\n",
- "nh=0.86 #hydraulc efficiency\n",
- "nelec=0.92 #electrical efficiency\n",
- "d=1000 #density of water, kg/m^3\n",
- "\n",
- "#Calculations:\n",
- "W=Q*d*9.81 #weight of water in N\n",
- "Po=W*H*nh*nelec/1000 #power produced, kW\n",
- "E=Po*1 #in kWh\n",
- "\n",
- "#Results:\n",
- "print \"Electrical energy generated per hr=\",round(E),\"kWh\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Electrical energy generated per hr= 776.0 kWh\n"
- ]
- }
- ],
- "prompt_number": 38
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.9, Page Number: 23"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "CA=5*10**9 #Catchment area in m^2\n",
- "H=30 #head in m\n",
- "F=1.25 #Annual rainfall in m\n",
- "K=0.80 #yeild factor\n",
- "n=0.70 #overall efficiency\n",
- "LF=0.40 #Load factor\n",
- "d=1000 #density of water(kg/m^3)\n",
- "\n",
- "#Calculations:\n",
- "V=CA*F*K #volume of water utilised per annum(m^3)\n",
- "W=V*d*9.81 #Weight of water available per annum (N)\n",
- "E=round(W*H*n/(10**11*3600),2)*10**8 #Electrical energy available per annum(kWh)\n",
- "Pav=E/8760 #average power(kW)\n",
- "Dmax=Pav/LF #Maximum demand\n",
- "\n",
- "#Results:\n",
- "print \"Average power generated is \",round(Pav),\"kW\"\n",
- "print \"Rating of generators is\",round(Dmax),\"kW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Average power generated is 32648.0 kW\n",
- "Rating of generators is 81621.0 kW\n"
- ]
- }
- ],
- "prompt_number": 39
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.10, Page Number: 24"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable Declaration:\n",
- "A=2.4 #Area of reservoir(km^2)\n",
- "V=5*10**6 #Capacity of reservoir(m^3)\n",
- "H=100 #in m\n",
- "np=0.95 #penstock efficiency\n",
- "nt=0.90 #turbine efficiency\n",
- "ng=0.85 #generation efficiency\n",
- "L=15000 #load supplied in kW\n",
- "\n",
- "#Calculations:\n",
- "W=V*1000*9.81 #in Newton\n",
- "n=int(np*nt*ng*1000)/1000 #overall efficiency\n",
- "E=W*H*n/(1000*3600) #Elecctrical energy generated(kWh)\n",
- "x=L*3*3600/(A*10**6*9.81*H*n)\n",
- "\n",
- "#Results:\n",
- "print \"Total electrical energy generated is \",round(E),\"kWh\"\n",
- "print \"Fall in reservoir level is\",round(x*100,3),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total electrical energy generated is 989175.0 kWh\n",
- "Fall in reservoir level is 9.478 cm\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.11, Page Number: 25"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "H=25 #head of reservoir in m\n",
- "Pr=400 #power required by factory(kW)\n",
- "n=0.80 #overall efficiency of plant\n",
- "\n",
- "#Calculations:\n",
- "#part (i):\n",
- "#(a):\n",
- "d1 = 10 #discharge in m^3/sec\n",
- "w1 = d1*1000*9.81 #weight of water in N\n",
- "P1 = w1*H*n/1000 #power developed(kW)\n",
- "\n",
- "#(b)\n",
- "d2 = 6 #in m^3/sec\n",
- "P2 = P1*d2/d1 #kW\n",
- "\n",
- "#(c)\n",
- "d3 = 1.5 #in m^3/sec\n",
- "P3 = P1*d3/d1 #kW\n",
- "\n",
- "Ps = Pr-P3 #standby power(kW)\n",
- "\n",
- "#part(ii):\n",
- "Dav = (d1*4+d2*2+d3*6)/12 #avg discharge(m^3/sec)\n",
- "P = P1*Dav/d1 #power developed(kW)\n",
- "Pex = P-Pr #Excess power available(kW)\n",
- "\n",
- "\n",
- "#Results:\n",
- "print \"(i) Standby power is \",round(Ps),\"kW\"\n",
- "print \"(ii) Excess power available is \",round(Pex,1),\"kW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Standby power is 106.0 kW\n",
- "(ii) Excess power available is 597.4 kW\n"
- ]
- }
- ],
- "prompt_number": 41
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {
- "slideshow": {
- "slide_type": "-"
- }
- },
- "source": [
- "Example 2.12, Page Number: 25"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "#for part (i):\n",
- "C = 10 #Installed capacity(MW)\n",
- "H = 20 #head of reservoir(m)\n",
- "n = 0.80 #overall efficiency\n",
- "LF = 0.40 #load factor\n",
- "#for part (ii):\n",
- "Q2 = 20 #discharge\n",
- "\n",
- "\n",
- "#Calculations:\n",
- "#for part(i):\n",
- "U = C*LF*24*7*10**3 #units generated per week(kWh)\n",
- "Q = U/(H*n*9.81*24*7) #Discharge in m^3/sec\n",
- "\n",
- "#for part(ii):\n",
- "U2 = Q2*9.81*1000*n*H*24/1000 #units generated per day(kWh)\n",
- "LF2 = U2/(C*10**3*24)\n",
- "\n",
- "#Results:\n",
- "print \"(i) The river discharge is \",round(Q,2),\"m^3/sec\"\n",
- "print \"(ii) The load factor is \", round(LF2*100,1),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The river discharge is 25.48 m^3/sec\n",
- "(ii) The load factor is 31.4 %\n"
- ]
- }
- ],
- "prompt_number": 42
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.13, Page Number: 26"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "H = 15 #head of reservoir(m)\n",
- "n = 0.85 #efficiency\n",
- "L = 0.40 #load factor\n",
- "\n",
- "\n",
- "#Calculations:\n",
- "Qavg = (500+520+850+800+875+900+546)/7 #in m^3/sec\n",
- "\n",
- "#It is clear from graph that on three dyas \n",
- "#(viz., Sun, Mon. and Sat.), the discharge is less than\n",
- "#the average discharge.\n",
- "\n",
- "V1 = (500+520+546)*24*3600 #Actual volume available in these 3 days(m^3/s)\n",
- "V2 = 3*Qavg*24*3600 #Vol. of water required in these 3 days(m^3/s)\n",
- "Pr = V2-V1 #Pondage required(m^3/sec)\n",
- "Po = Qavg*9.81*1000*H*n #Avg output produced(W)\n",
- "C = Po/L #Capacity of plant(W)\n",
- "\n",
- "#Results:\n",
- "print \"(i) The average daily discharge is \",Qavg,\"m^3/sec\"\n",
- "print \"(ii) Pondage required is (\",round(Pr/10**5),\"* 10^5) m^3\"\n",
- "print \"(iii)Installed capacity of plant is \",round(C/10**6),\"MW\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The average daily discharge is 713.0 m^3/sec\n",
- "(ii) Pondage required is ( 495.0 * 10^5) m^3\n",
- "(iii)Installed capacity of plant is 223.0 MW\n"
- ]
- }
- ],
- "prompt_number": 43
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.14, Page Number: 29"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "w = 0.28 #fuel consumption in kg/kWh\n",
- "C = 10000 #calorific value of fuel(kcal/kWh)\n",
- "na = 0.95 #efficiency of alternator\n",
- "\n",
- "\n",
- "#Calculations:\n",
- "H = w*C/860 #heat produced by 0.28 kg/kWh of fuel\n",
- "\n",
- "no = 1/H #Overall efficiency\n",
- "ne = no/na #Efficiency of engine\n",
- "\n",
- "#Results:\n",
- "print \"(i) The overall efficiency is \",round(no*100,1),\"%\"\n",
- "print \"(ii)The efficiency of the engine\",round(ne*100,1),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The overall efficiency is 30.7 %\n",
- "(ii)The efficiency of the engine 32.3 %\n"
- ]
- }
- ],
- "prompt_number": 44
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.15, Page Number: 30"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "w = 1000 #fuel consumption in kg/day\n",
- "E = 4000 #Units generated in kWh/day\n",
- "C = 10000 #calorific value in kcal/kg\n",
- "na = 0.96 #Alternator efficiency\n",
- "nem = 0.95 #engine mechanical efficiency\n",
- "\n",
- "\n",
- "#Calculations:\n",
- "s = w/E #specific fuel consumption(kg/kWh)\n",
- "E2 = w*C #energy input per day(kcal/day)\n",
- "no = E*860/E2 #overall efficiency\n",
- "ne = no/na #engine efficiency\n",
- "net = ne/nem #engine thermal efficiency\n",
- "\n",
- "#Results:\n",
- "print \"Specific fuel consumption is \",s,\"kg/kWh\"\n",
- "print \"Overall efficiency is \",round(no*100,1),\"%\"\n",
- "print \"Thermal efficiency of the engine is \",round(net*100,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Specific fuel consumption is 0.25 kg/kWh\n",
- "Overall efficiency is 34.4 %\n",
- "Thermal efficiency of the engine is 37.72 %\n"
- ]
- }
- ],
- "prompt_number": 45
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.16, Page Number: 30"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "C1 = 700 #capacity of plant 1(kW)\n",
- "C2 = 2*500 #capacity of plant 2(kW)\n",
- "pcf = 0.40 #plant capacity factor\n",
- "w = 0.28 #fuel cunsumption in kg/kWh\n",
- "H = 10200 #specific heat of fuel in kcal/kg\n",
- "\n",
- "#Calculatios:\n",
- "M = (C1+C2)*30*24 #max energy can be produced in 30 days(kWh)\n",
- "E = pcf*M #Actual energy produced in 30 days(kWh)\n",
- "W = E*w #actual fuel consumption in kg\n",
- "\n",
- "Po = E*860 #output energy in kWh\n",
- "Pin = W*H #Input energy in kWh\n",
- "n = Po/Pin #Overall efficiency\n",
- "\n",
- "#Results:\n",
- "print \"The fuel oil required is \",W,\"kg\"\n",
- "print \"Ovreall efficiency is\",round(n*100),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The fuel oil required is 137088.0 kg\n",
- "Ovreall efficiency is 30.0 %\n"
- ]
- }
- ],
- "prompt_number": 46
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.17, Page Number: 34"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "M = 300 #Energy received from reactor(MW)\n",
- "E1 = 200 #Energy released fron each atom(MeV)\n",
- "\n",
- "\n",
- "#Calculations:\n",
- "E2 = M*10**6*3600 #Energy released per hour(J)\n",
- "E3 = E1*1.6*10**-19*10**6 #Energy released per fission(J)\n",
- "N = E2/E3 #No of atoms fissioned\n",
- "m = N*235/(6.022*10**23) #mass of uranium fissioned per hr(g)\n",
- "\n",
- "#Results:\n",
- "print \"Mass of Uranium fissioned per hour is\",round(m,2),\"g\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Mass of Uranium fissioned per hour is 13.17 g\n"
- ]
- }
- ],
- "prompt_number": 47
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 2.18, Page Number: 35"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "t = 30 #days\n",
- "w = 2 #weight of uranium(kg)\n",
- "Eo = 200 #energy released per fission(MeV)\n",
- "\n",
- "\n",
- "#Calculations:\n",
- "N = 2*1000*6.022*10**23/235 #No of atoms fissioned in 2kg of fuel\n",
- "Po = N*Eo*(1.6*10**-19)*10**6/(24*60*60*30) #Watt\n",
- "\n",
- "\n",
- "#Results:\n",
- "print \"Power output is\",round(Po*10**-6,1),\"MW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power output is 63.3 MW\n"
- ]
- }
- ],
- "prompt_number": 48
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Principles_of_Power_System/chapter3_1.ipynb b/Principles_of_Power_System/chapter3_1.ipynb deleted file mode 100644 index 42073d28..00000000 --- a/Principles_of_Power_System/chapter3_1.ipynb +++ /dev/null @@ -1,1355 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:c36425eab1371d7c77f7bd1775095f254e3df888559e80a11117b968dfc1b5b4"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 3: Principles of Power System"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.1, Page Number: 50"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "M = 100 #Maximum demand on power station(MW)\n",
- "L = 0.40 #annual load factor\n",
- "\n",
- "#Calculation:\n",
- "E = M*L*8760*10**6 #Energy generated in a year(kWh)\n",
- "\n",
- "#Results:\n",
- "print \"Energy generated in a year is (\",E/10**8,\"* 10^5) kWh\" "
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Energy generated in a year is ( 3504.0 * 10^5) kWh\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.2, Page Number: 50"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "L = 43 #connected load(MW)\n",
- "M = 20 #Maximum demand(MW)\n",
- "E = 61.5*10**6 #Units genearted per annum(kWh)\n",
- "\n",
- "#calculation:\n",
- "DF = M/L #demand factor\n",
- "LF = E/(M*8760*1000) #load factor\n",
- "\n",
- "#Results:\n",
- "print \"The demand factor is\",round(DF,3)\n",
- "print \"Load factor is \",round(LF*100,1),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The demand factor is 0.465\n",
- "Load factor is 35.1 %\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.3, Page Number: 50"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "\n",
- "#variable declaration:\n",
- "\n",
- "#power station delivers 100 MW for 2 hours, \n",
- "#50 MW for 6 hours and is shut down for the rest\n",
- "#of each day.\n",
- "M = 100 #Maximum capacity of station(MW)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "E1 = 100*2+50*6 #Energy supplied per day(MWh)\n",
- "n = 365-45 #No.of working days\n",
- "LF = E1*320/(M*n*24) #load factor\n",
- "\n",
- "print \"Annual load factor is\",round(LF*100,1),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Annual load factor is 20.8 %\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.4, Page Number: 50"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "M = 25 #Maximum demand(MW)\n",
- "LF = 0.60 #Load factor\n",
- "PCF = 0.50 #plant capacity factor\n",
- "PUF = 0.72 #plant use factor\n",
- "\n",
- "\n",
- "#Calculations:\n",
- "L = LF*M #Average load(MW)\n",
- "PC = L/PCF #Plant gapacity(MW)\n",
- "R = PC-M #Reserve capacity(MW)\n",
- "E = L*24 #daily energy produced(MWh)\n",
- "ME = E/PUF #Maximum energy produced(MWh/day)\n",
- "\n",
- "#Results:\n",
- "print \"Reserve capacity is \",R,\"MW\"\n",
- "print \"Daily energy produced is\",E,\"MWh\"\n",
- "print \"Maximum energy that could be produced dauly is \",ME,\"MWh/day\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Reserve capacity is 5.0 MW\n",
- "Daily energy produced is 360.0 MWh\n",
- "Maximum energy that could be produced dauly is 500.0 MWh/day\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.5, Page Number: 51"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "M = 2500 #Maximum demand(kW)\n",
- "E = 45*10**5 #kWh/year\n",
- "\n",
- "#Calculation:\n",
- "D = (1500+750+100+450)/M #diversity factor\n",
- "LF = E/(M*8760) #Annual load factor\n",
- "\n",
- "#Results:\n",
- "print \"Diversity factor is\",D\n",
- "print \"Annual load factor is\",round(LF*100,1),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Diversity factor is 1.12\n",
- "Annual load factor is 20.5 %\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.6, Page Number: 51"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "M = 15000 #Max demand(kW)\n",
- "LF = 0.50 #Annual load factor\n",
- "PCF = 0.4 #Plant capacity factor\n",
- "\n",
- "\n",
- "#Calculations:\n",
- "L = LF*M #load factor(kW)\n",
- "PC = L/PCF #plant capacity(kW)\n",
- "R = PC-M #Reserve capacity(kW)\n",
- "\n",
- "#Results:\n",
- "print \"The reserve capacity is \",R,\"kW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The reserve capacity is 3750.0 kW\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.7, Page Number: 51"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "DFo = 1.35 #Overall diversity factor of system\n",
- "\n",
- "\n",
- "\n",
- "#Calculations:\n",
- "M = (1500+2000+10000)/DFo #max demand(kW)\n",
- "#for domestic load:\n",
- "CL1 = (1500*1.2)/0.8 #kW\n",
- "\n",
- "#for Commercial load:\n",
- "CL2 = 2000*1.1/0.9 #kW\n",
- "\n",
- "#for Industrial load:\n",
- "CL3 = 10000*1.25/1 #kW\n",
- "\n",
- "\n",
- "#Results:\n",
- "print \"Maximum demand is \",M,\"kW\"\n",
- "print \"Connected loads for each type are:\"\n",
- "print CL1,\"kW \",round(CL2),\"kW \",CL3,\"kW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum demand is 10000.0 kW\n",
- "Connected loads for each type are:\n",
- "2250.0 kW 2444.0 kW 12500.0 kW\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.8, Page Number: 52"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "Dt = 1.3 #Diversity among transformers\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#for transformer 1:\n",
- "M1 = (10*0.65)/1.5 #Max demand (kW)\n",
- "\n",
- "#for transformer 2:\n",
- "M2 = (12*0.6)/3.5 #max demand(kW)\n",
- "\n",
- "#for transformer:\n",
- "M3 = (15*0.7)/1.5 #max demand(kW)\n",
- "\n",
- "Mf = (M1+M2+M3)/Dt #Maximum demand on feeder(kW)\n",
- "\n",
- "#Results:\n",
- "print \"Maximum load on the feeder is \",round(Mf,1),\"kW\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Maximum load on the feeder is 10.3 kW\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.9, Page Number: 52"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "n1 = 1000 #No of houses\n",
- "CL1 = 1.5 #Avg connected load in each house(kW)\n",
- "DF1 = 0.4 #Demand factor\n",
- "DvF = 2.5 #Diversity factor\n",
- "\n",
- "n2 = 10 #no. of factories\n",
- "M2 = 90 #factory overall max demand(kW)\n",
- "\n",
- "n3 = 7 #no of tubewells\n",
- "C3 = 7 #capacity of tubewells(kW)\n",
- "\n",
- "DF = 1.2 #Diversity factor among above types of consumers\n",
- "\n",
- "#Calculations:\n",
- "Ms = ((DF1*n1*CL1)/DvF)+M2+n3*C3 #Sum of maximum demand on the station(kW)\n",
- "M = Ms/DF #maximum demand on the station(kW)\n",
- "\n",
- "#Results:\n",
- "print \"Minimum capacity on the power station is\",round(M),\"kW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Minimum capacity on the power station is 316.0 kW\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.10, Page Number: 53"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "%pylab inline\n",
- "\n",
- "\n",
- "#Calculations:\n",
- "M = 70.0 #Max demand, from load curve(MW)\n",
- "E = 40*6+50*4+60*2+50*4+70*4+40*4 #Units generated per day(MWh)\n",
- "L = E/24 #Avg load(MW)\n",
- "LF = L/M #Load factor\n",
- "\n",
- "n1 = linspace(0,6,10);\n",
- "M1 = linspace(40,40,10);\n",
- "plot(n1,M1);\n",
- "hold(True);\n",
- "\n",
- "n2 = linspace(6,10,10);\n",
- "M2 = linspace(50,50,10);\n",
- "plot(n2,M2,'b');\n",
- "\n",
- "n3 = linspace(10,12,10);\n",
- "M3 = linspace(60,60,10);\n",
- "plot(n3,M3,'b');\n",
- "\n",
- "n4 = linspace(12,16,10);\n",
- "M4 = linspace(50,50,10);\n",
- "plot(n4,M4,'b');\n",
- "\n",
- "n5 = linspace(16,20,10);\n",
- "M5 = linspace(70,70,10);\n",
- "plot(n5,M5,'b');\n",
- "\n",
- "n6 = linspace(20,24,10);\n",
- "M6 = linspace(40,40,10);\n",
- "plot(n6,M6,'b');\n",
- "\n",
- "ylim(0,100);\n",
- "xlim(0,24);\n",
- "grid(1,linewidth=0.5);\n",
- "ylabel(\"Load in MW ------>\");\n",
- "xlabel(\"Time in hours ----->\");\n",
- "title(\"Daily Load Curve\")\n",
- "\n",
- "#Results:\n",
- "print \"Maximum demand is \",M,\"MW\"\n",
- "print \"Units generated per day is (\",E/100,\"* 10^5) kWh\"\n",
- "print \"Average load is\",L*1000,\"kW\"\n",
- "print \"Load factor is \",round(LF*100,1),\"%\"\n",
- "print \"The Daily load curve is shown below:\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Populating the interactive namespace from numpy and matplotlib\n",
- "Maximum demand is "
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " 70.0 MW\n",
- "Units generated per day is ( 12.0 * 10^5) kWh\n",
- "Average load is 50000.0 kW\n",
- "Load factor is 71.4 %\n",
- "The Daily load curve is shown below:\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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cXGztEmwSc1HGXPSJnInNNIHc3Fw0b95ct+3r64vc3FwrVkREZP9spglIkmTt\nEu5bWq3W2iXYJOaijLnoEzmT+tYu4A4fHx/k5OTotnNycuDr61vpOq1bt2azMCA+Pt7aJdgk5qKM\nueiz50z8/f0N/sxmjhMoKytDu3bt8P3336NZs2YIDAzE2rVr0aFDB2uXRkRkt2zmlUD9+vXx2Wef\n4cknn0R5eTkmTpzIBkBEZGY280qAiIgsz2YGw1XhgWTK1Go1unTpgq5duyIwMNDa5VjNhAkT4Onp\nic6dO+suKywsRGhoKNq2bYsBAwYI9zZApUxmz54NX19fdO3aFV27dkVqaqoVK7SOnJwcBAcHo2PH\njujUqRM+/fRTAOI+Xu6LJsADyQyTJAlpaWnIyMjA/v37rV2O1URFRek9oWk0GoSGhiIrKwshISHQ\naDRWqs46lDKRJAnTpk1DRkYGMjIy8NRTT1mpOutxdHTEwoULceTIEezduxdLly7F0aNHhX283BdN\ngAeSGccVPaBPnz5wc3OrdFlSUhIiIiIAABEREdi4caM1SrMapUwAPl68vLwQEBAAAHByckKHDh2Q\nm5sr7OPlvmgCPJDMMEmS0L9/f3Tv3h3Lly+3djk2paCgAJ6engAAT09PFBQUWLki27BkyRL4+/tj\n4sSJwix5GKLVapGRkYHHHntM2MfLfdEEeGyAYbt370ZGRga2bNmCpUuXYufOndYuySZJksTHEYAp\nU6YgOzsbBw8ehLe3N15//XVrl2Q1V69exYgRI7B48WI4OztX+plIj5f7oglU50AyUXl7ewMAmjZt\niuHDhws9F7iXp6cn8vPzAQB5eXnw8PCwckXW5+HhoXuCmzRpkrCPl1u3bmHEiBEIDw/HsGHDAIj7\neLkvmkD37t1x4sQJaLVa3Lx5E+vXr8fQoUOtXZbVlZaW4sqVKwCAa9eu4dtvv630ThDRDR06VHcU\naHx8vO5/dpHl5eXpvv/f//4n5ONFlmVMnDgRfn5+iImJ0V0u7ONFvk9s3rxZbtu2rdy6dWv5ww8/\ntHY5NuH06dOyv7+/7O/vL3fs2FHoXMaMGSN7e3vLjo6Osq+vr7xy5Ur50qVLckhIiNymTRs5NDRU\nLioqsnaZFnVvJrGxsXJ4eLjcuXNnuUuXLvIzzzwj5+fnW7tMi9u5c6csSZLs7+8vBwQEyAEBAfKW\nLVuEfbzwYDEiIoHdF8tBRERkHmwCREQCYxMgIhIYmwARkcDYBIiIBMYmQEQkMDYBsppLly7pTmns\n7e2tO8VgaeccAAAEiElEQVSxs7Mzpk6dWuf7W7ZsGVavXl3t66elpWHIkCF1XgeRLbGZTxYj8bi7\nuyMjIwMAMGfOHDg7O2PatGlm298LL7xgtvs2RVlZGerXt/7/esXFxVCpVNYug6yMrwTIZtw5bvHu\nv8Bnz56NiIgI9O3bF2q1Gl9//TXeeOMNdOnSBQMHDkRZWRkA4Oeff0ZQUBC6d++Op556SncOmLvN\nnj0b8+fPBwAEBQVhxowZeOyxx9CuXTvs2rVL7/qSJOHq1asYNWoUOnTogHHjxul+9v3336Nbt27o\n0qULJk6ciJs3bwKo+JCfwsJCAMCBAwcQHBys23d4eDh69+6NiIgIHDlyBIGBgejatSv8/f1x8uTJ\nuoqx2tatW4fOnTtjwYIF+P333y2+f7INbAJk87Kzs7F9+3YkJSVh3LhxCA0NxaFDh9CwYUOkpKTg\n1q1bePnll/HVV1/hwIEDiIqKwv/93//p3c/dZ4aUJAnl5eXYt28fFi1ahDlz5uhdX5ZlZGRkYPHi\nxcjMzMTp06eRnp6O69evIyoqComJiTh06BDKysrwxRdf6O7XkGPHjuH777/HmjVrsGzZMsTExCAj\nIwM///yzVU6IOHnyZGzZsgWlpaXo27cvRo0aha1btwr/eQOiYRMgmyZJEgYOHIh69eqhU6dOuH37\nNp588kkAQOfOnaHVapGVlYUjR46gf//+6Nq1K/7xj39U6/Mmnn32WQBAt27doNVqFa8TGBiIZs2a\nQZIkBAQEIDs7G8ePH0erVq3wyCOPAKj4AJIff/yxyt9j6NCheOCBBwAAjz/+OD788EN8/PHH0Gq1\nePDBB6sbSZ3y9fXFrFmzkJmZiaioKERFRWH48OFWqYWsw/oLk0RVaNCgAQDAwcEBjo6OussdHBxQ\nVlYGWZbRsWNHpKenm3S/d56Q69Wrp1tWMnSdu69371/7sizrLqtfvz5u374NALh+/Xql6zVq1Ej3\n/dixY9GzZ08kJydj0KBBWLZsmW7pyJicnBzdGXQnT56M8vJyLF++HJIkISUlBZGRkbhw4QJ69OiB\nSZMm6eYgc+fOxb59+5CSkgJJkvDLL7/o7nP//v1YtWoVtm3bhjFjxiA6OrrKOsh+sAmQTavO0kS7\ndu1w8eJF7N27Fz179sStW7dw4sQJ+Pn51ej+jJEkCe3atYNWq8WpU6fQunVrrF69Gv369QNQMRM4\ncOAAnnrqKXz11VcG95udnY1WrVrh5ZdfxtmzZ3H48OFqNYHmzZvrhul3vPjii7rvt27dWulnd193\nyJAh+OCDD3Tb3377Ld588014e3tj0qRJWLJkiU0MrMmy+F+cbMbd6/VK3999nbu3HR0dsWHDBrzy\nyiu4fPkyysrK8Nprryk2AUNr9kqXG/p0qQceeACrVq3CqFGjUFZWhsDAQEyePBkA8N5772HixIlw\ncXFBUFCQwd8jMTERq1evhqOjI7y9vRVnGOb20EMPITk5udJHt5J4eCppIiKBcTBMRCQwNgEiIoGx\nCRARCYxNgIhIYGwCREQCYxMgIhIYmwARkcDYBIiIBPb/pozSJIWf61MAAAAASUVORK5CYII=\n",
- "text": [
- "<matplotlib.figure.Figure at 0x4a7dfb0>"
- ]
- }
- ],
- "prompt_number": 31
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.11, Page Number: 54"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "%pylab inline\n",
- "\n",
- "#Calculation:\n",
- "M = 350.0 #Maximum demand(kW)\n",
- "DF = (200+100+50+100)/M #Diversity factor\n",
- "E = 100*6+(100+50)*2+(200+100+50)*2+(200+100)*8+100*6 #kWh/day\n",
- "LF = E/(24*M) #load factor\n",
- "\n",
- "n1 = linspace(0,6,10);\n",
- "M1 = linspace(100,100,10);\n",
- "plot(n1,M1);\n",
- "\n",
- "\n",
- "hold(True);\n",
- "\n",
- "n2 = linspace(6,8,10);\n",
- "M2 = linspace(150,150,10);\n",
- "plot(n2,M2,'b');\n",
- "\n",
- "n3 = linspace(8,10,10);\n",
- "M3 = linspace(350,350,10);\n",
- "plot(n3,M3,'b');\n",
- "\n",
- "n4 = linspace(10,18,10);\n",
- "M4 = linspace(300,300,10);\n",
- "plot(n4,M4,'b');\n",
- "\n",
- "n5 = linspace(18,24,10);\n",
- "M5 = linspace(100,100,10);\n",
- "plot(n5,M5,'b');\n",
- "\n",
- "ylim(0,400);\n",
- "xlim(0,24);\n",
- "grid(linewidth=0.5);\n",
- "ylabel(\"Load in kW ------>\");\n",
- "xlabel(\"Time in hours ----->\");\n",
- "title(\"Daily Load Curve\")\n",
- "\n",
- "#Results:\n",
- "print \"Diversity factor is \",round(DF,3)\n",
- "print \"Unis generated per day is \",E,\"kWh\"\n",
- "print \"Load factor is \",round(LF*100,1),\"%\"\n",
- "print \"The daily load curve is shown below:\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Populating the interactive namespace from numpy and matplotlib\n",
- "Diversity factor is "
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " 1.286\n",
- "Unis generated per day is 4600 kWh\n",
- "Load factor is 54.8 %\n",
- "The daily load curve is shown below:\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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8kszMTNq3b09CQoLb72Px4sXMmzeP4OBgIiMjXc40fK1jx44sW7as\nyW1gxXx0GWoREZPTsFhExORUCERETE6FQETE5FQIRERMToVARMTkVAhERExOhUBExORUCERETO7/\nAV3udrfsY+WJAAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x49b18b0>"
- ]
- }
- ],
- "prompt_number": 30
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.12, Page Number: 54"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "%pylab inline\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "M1 = 800 #max demand of consumer 1(Watts)\n",
- "M2 = 1000 #max demand of consumer 2(Watts)\n",
- "M3 = 1200 #max demand of consumer 3(Watts)\n",
- "\n",
- "n1 = linspace(0,8,10);\n",
- "m1 = linspace(200,200,10);\n",
- "plot(n1,m1);\n",
- "\n",
- "hold(True);\n",
- "n2 = linspace(8,14,10);\n",
- "m2 = linspace(800,800,10);\n",
- "plot(n2,m2,'b');\n",
- "\n",
- "n3 = linspace(14,16,10);\n",
- "m3 = linspace(2400,2400,10);\n",
- "plot(n3,m3,'b');\n",
- "\n",
- "n4 = linspace(16,22,10);\n",
- "m4 = linspace(800,800,10);\n",
- "plot(n4,m4,'b');\n",
- "\n",
- "n5 = linspace(22,24,10);\n",
- "m5 = linspace(400,400,10);\n",
- "plot(n5,m5,'b');\n",
- "\n",
- "ylim(0,2500);\n",
- "xlim(0,24);\n",
- "grid(linewidth=0.5);\n",
- "ylabel(\"Load in Watts ------>\");\n",
- "xlabel(\"Time in hours ----->\");\n",
- "title(\"Daily Load Curve\");\n",
- "annotate(\"(Midnight)\",xy=(0,0));\n",
- "annotate(\"(Midnight)\",xy=(24,0));\n",
- "annotate(\"(Noon)\",xy=(12,0));\n",
- "\n",
- "#load factors of each consumers:\n",
- "LF1 = (600*6+200*2+800*6)/(24*M1)\n",
- "LF2 = (200*8+1000*2+200*2)/(24*M2)\n",
- "LF3 = (200*6+1200*2+200*2)/(24*M3)\n",
- "#The simultaneous maximum demand on the station is 200 + 1000 + 1200 = 2400 W\n",
- "#and occurs from 2 P.M. to 4 P.M.\n",
- "DF = (M1+M2+M3)/2400.0 #Diversity factor\n",
- "LF = (200*8+800*6+2400*2+800*6+400*2)/(24*2400.0) #load factor\n",
- "\n",
- "#Results:\n",
- "print \"(i) The maximum demand of individual consumers are:\"\n",
- "print \"\\tConsumer 1 =\",M1,\"W, Consumer 2 =\",M2,\"W, Consumer 3 =\",M3,\"W\"\n",
- "print \"\\n(ii)Load Factors of individual consumers are:\"\n",
- "print \"\\tConsumer 1 =\",round(LF1*100,1),\"%, Consumer 2 =\",round(LF2*100,1),\"\"\"%,\n",
- "\\tConsumer 3 =\"\"\",round(LF3*100,1),\"%\"\n",
- "print \"\\n(iii)Diversity factor is \",DF\n",
- "print \"\\n(iv) Load factor of the station is\",round(LF*100,1),\"%\"\n",
- "print \"\\n The load curve is shown below:\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Populating the interactive namespace from numpy and matplotlib\n",
- "(i) The maximum demand of individual consumers are:"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\tConsumer 1 = 800 W, Consumer 2 = 1000 W, Consumer 3 = 1200 W\n",
- "\n",
- "(ii)Load Factors of individual consumers are:\n",
- "\tConsumer 1 = 45.8 %, Consumer 2 = 16.7 %,\n",
- "\tConsumer 3 = 13.9 %\n",
- "\n",
- "(iii)Diversity factor is 1.25\n",
- "\n",
- "(iv) Load factor of the station is 29.2 %\n",
- "\n",
- " The load curve is shown below:\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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mJg2KjIfXAMljXgwxJ/KYF8tX44dRMjIyEBYWJvXlHB0dsW7dugY3q1NDQ5WI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\n/wAb/9x823/nZQAAAABJRU5ErkJggg==\n",
- "text": [
- "<matplotlib.figure.Figure at 0x49b1ed0>"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.13, Page Number: 55"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "A = 12 #Area of load curve(cm**2)\n",
- "Lo = 1000 #load under 1cm length(kW)\n",
- "To = 2 #time under 1cm length(hr)\n",
- "M = 3000 #maximum demand(kW)\n",
- "\n",
- "#Calculation:\n",
- "L = Lo*To*A/24 #Average load(kW)\n",
- "LF = L/M #load factor\n",
- "\n",
- "\n",
- "#Results:\n",
- "print \"Load factor is\",round(LF*100,1),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Load factor is 33.3 %\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.14, Page Number: 56"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "C = 75 #Capacity of each generator(MW)\n",
- "n = 4 #No. of generators\n",
- "CV = 10000 #Calorific value of oil used(kcal/kg)\n",
- "H = 2860 #Avg heat rate(kcal/kWh)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "E = (260*6+200*8+160*4+100*6) #Units generated per day(MWh)\n",
- "L = E/24 #Avg load(MW)\n",
- "LF =L/260 #Load factor\n",
- "PCF = L/(n*C) #Plant capacity\n",
- "TH = H*E*10**3 #heat generated per day(kcal)\n",
- "w = TH/CV #daily fuel requirement(kg)\n",
- "\n",
- "print \"Daily load factor is\",round(LF*100,1),\"%\"\n",
- "print \"Plant capacity factor is\",round(PCF*100,1),\"%\"\n",
- "print \"Daily fuel requirement is (\",w/1000,\"* 10^3) kg\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Daily load factor is 70.5 %\n",
- "Plant capacity factor is 61.1 %\n",
- "Daily fuel requirement is ( 1258.4 * 10^3) kg\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.15, Page Number: 56"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "%pylab inline\n",
- "\n",
- "#Calculations:\n",
- "E = 20*2+40*4+60*4+20*4+50*4+20*6 #Units generated per day(MWh)\n",
- "\n",
- "#1st plot\n",
- "subplot(1,3,1)\n",
- "n1 = linspace(0,8,10);\n",
- "m1 = linspace(20,20,10);\n",
- "plot(n1,m1);\n",
- "\n",
- "hold(True);\n",
- "n2 = linspace(8,12,10);\n",
- "m2 = linspace(40,40,10);\n",
- "plot(n2,m2,'b');\n",
- "\n",
- "n3 = linspace(12,16,10);\n",
- "m3 = linspace(60,60,10);\n",
- "plot(n3,m3,'b');\n",
- "\n",
- "n4 = linspace(16,20,10);\n",
- "m4 = linspace(20,20,10);\n",
- "plot(n4,m4,'b');\n",
- "\n",
- "n5 = linspace(20,24,10);\n",
- "m5 = linspace(40,40,10);\n",
- "plot(n5,m5,'b');\n",
- "\n",
- "ylim(0,100);\n",
- "xlim(0,24);\n",
- "grid(linewidth=0.5);\n",
- "ylabel(\"Load in MW ------>\");\n",
- "xlabel(\"Time in hours ----->\");\n",
- "title(\"$(i)$ Daily Load Curve\");\n",
- "\n",
- "#next plot\n",
- "subplot(1,3,3)\n",
- "n1 = linspace(0,4,10);\n",
- "m1 = linspace(60,60,10);\n",
- "plot(n1,m1);\n",
- "\n",
- "hold(True);\n",
- "n2 = linspace(4,8,10);\n",
- "m2 = linspace(50,50,10);\n",
- "plot(n2,m2,'b');\n",
- "\n",
- "n3 = linspace(8,12,10);\n",
- "m3 = linspace(40,40,10);\n",
- "plot(n3,m3,'b');\n",
- "\n",
- "n4 = linspace(12,24,10);\n",
- "m4 = linspace(20,20,10);\n",
- "plot(n4,m4,'b');\n",
- "\n",
- "ylim(0,100);\n",
- "xlim(0,24);\n",
- "grid(linewidth=0.5);\n",
- "ylabel(\"Load in MW ------>\");\n",
- "xlabel(\"Hours duration ----->\");\n",
- "title(\"$(ii)$ Load Duration Curve\");\n",
- "\n",
- "\n",
- "#Results:\n",
- "print \"Energy generated per day is (\",E,\"* 10^3) kWh\"\n",
- "print \"\\nThe daily load curve and load duration curves are shown below:\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Populating the interactive namespace from numpy and matplotlib\n",
- "Energy generated per day is ("
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " 840 * 10^3) kWh\n",
- "\n",
- "The daily load curve and load duration curves are shown below:\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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qiFFOtc7JJCQk4MqVK9i6dSv69OmDgoICju8SKRhzlpSk1uGyiIgI5OTkICws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\nyBARkWT+H5YlVY/jFClJAAAAAElFTkSuQmCC\n",
- "text": [
- "<matplotlib.figure.Figure at 0x49c3450>"
- ]
- }
- ],
- "prompt_number": 28
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.16, Page Number: 57"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "c1 = 10 #Capacity of each of 2 generators(MW)\n",
- "c2 = 5 #Capacity of 3rd generator(MW)\n",
- "\n",
- "#Calculation:\n",
- "C = 2*c1+c2 #Installed capacity(MW)\n",
- "A = (1/2)*(20+4) #Avg load, seen from the curve(MW)\n",
- "PF = A/C #plant factor\n",
- "E = 8760*A #Units generated per annum(MWh)\n",
- "LF = A/20 #Load factor\n",
- "PUF = 20/C #Plant capacity factor\n",
- "\n",
- "#Results:\n",
- "print \"(i) Installed capacity is \",C,\"MW\"\n",
- "print \"(ii) Plant factor is \",round(PF*100,1),\"%\"\n",
- "print \"(iii) Units generated per annum is \",(E/1000),\"* 10^3 kWh\"\n",
- "print \"(iv) Load factor is \",round(LF*100,1),\"%\"\n",
- "print \"(V) Utilisation factor is \",round(PUF*100,1),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Installed capacity is 25 MW\n",
- "(ii) Plant factor is 48.0 %\n",
- "(iii) Units generated per annum is 105.12 * 10^3 kWh\n",
- "(iv) Load factor is 60.0 %\n",
- "(V) Utilisation factor is 80.0 %\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.17, Page Number: 57 "
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "n = 0.72 #efficiency of the load\n",
- "\n",
- "#Calculation:\n",
- "Ma = (10*0.746/0.72)* 0.65 + 5 * 0.60 #kW\n",
- "Mb = (7.5*0.746/0.72)*0.75+4*0.60 #kW\n",
- "Mc = (15*0.746/0.72)*0.65 #kW\n",
- "Md = (5*0.746/0.72)*0.75+2*0.60 #kW\n",
- "\n",
- "SM = Ma+Mb+Mc+Md #Sum of maximum demands(kW)\n",
- "\n",
- "#The diversity factor between consumers of this type of service is 1\u00b75 from above table.\n",
- "M1 = SM/1.5 #Maximum demand on transformer 1(kW)\n",
- "\n",
- "#In a similar manner, the other transformer loads are determined to be\n",
- "# Total Simultaneous\n",
- "#Transformer 2 26 kW 7\u00b743 kW\n",
- "#Transformer 3 29\u00b713 kW 19\u00b740 kW\n",
- "\n",
- "#The diversity factor between transformers is 1\u00b73\n",
- "M2 = 7.43; M3 = 19.40\n",
- "MF = (M1+M2+M3)/1.3 #Maximum load on feeder(kW)\n",
- "\n",
- "\n",
- "#Results:\n",
- "print \"The maximum load on the feeder is \",round(MF,2),\"kW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum load on the feeder is 37.64 kW\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.18, Page Number: 61"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "%pylab inline\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "E = 10**3*(20*2+40*3+50*5+35*3+70*3+40*2+20*6) #Units generated per day(kWh)\n",
- "A = E/24 #Average load(kW)\n",
- "LF = A/(70*10**3) #Load factor\n",
- "\n",
- "n1 = linspace(0,8,10);\n",
- "m1 = linspace(20,20,10);\n",
- "plot(n1,m1);\n",
- "\n",
- "hold(True);\n",
- "n2 = linspace(8,11,10);\n",
- "m2 = linspace(40,40,10);\n",
- "plot(n2,m2,'b');\n",
- "\n",
- "n3 = linspace(11,16,10);\n",
- "m3 = linspace(50,50,10);\n",
- "plot(n3,m3,'b');\n",
- "\n",
- "n4 = linspace(16,19,10);\n",
- "m4 = linspace(35,35,10);\n",
- "plot(n4,m4,'b');\n",
- "\n",
- "n5 = linspace(19,22,10);\n",
- "m5 = linspace(70,70,10);\n",
- "plot(n5,m5,'b');\n",
- "\n",
- "n6 = linspace(22,24,10);\n",
- "m6 = linspace(40,40,10);\n",
- "plot(n6,m6,'b');\n",
- "\n",
- "\n",
- "ylim(0,100);\n",
- "xlim(0,24);\n",
- "grid(linewidth=0.5);\n",
- "ylabel(\"Load in MW ------>\");\n",
- "xlabel(\"Time in hours ----->\");\n",
- "title(\"Daily Load Curve\");\n",
- "\n",
- "#Results:\n",
- "print \"Load factor is \",round(LF*100,2),\"%\"\n",
- "print \"\"\"\\nOperational schedule: Referring to the load curve shown\n",
- "below, the operational schedule will be as under :\n",
- " (i) Set No. 1 will run for 24 hours.\n",
- " (ii) Set No. 2 will run from 8.00 hours to midnight.\n",
- " (iii)Set No. 3 will run from 11.00 hours to 16 hours and \n",
- " again from 19 hours to 22 hours.\"\"\"\n",
- "print \"\\nThe Load curve is shown below:\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Populating the interactive namespace from numpy and matplotlib\n",
- "Load factor is "
- ]
- },
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " 55.06 %\n",
- "\n",
- "Operational schedule: Referring to the load curve shown\n",
- "below, the operational schedule will be as under :\n",
- " (i) Set No. 1 will run for 24 hours.\n",
- " (ii) Set No. 2 will run from 8.00 hours to midnight.\n",
- " (iii)Set No. 3 will run from 11.00 hours to 16 hours and \n",
- " again from 19 hours to 22 hours.\n",
- "\n",
- "The Load curve is shown below:\n"
- ]
- },
- {
- "output_type": "stream",
- "stream": "stderr",
- "text": [
- "WARNING: pylab import has clobbered these variables: ['tri', 'axes', 'legend', 'stackplot', 'figure', 'rc_context', 'quiver', 'streamplot', 'rc', 'text', 'colors', 'contour', 'axis', 'colorbar', 'test', 'rcdefaults', 'table']\n",
- "`%pylab --no-import-all` prevents importing * from pylab and numpy\n"
- ]
- },
- {
- "metadata": {},
- "output_type": "display_data",
- "png": 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jI8aNG4ctW7ZYu6x6obi42Nol2CTmooy56BM5E5tpArm5uWjZsqVu29fXF7m5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\nAABJRU5ErkJggg==\n",
- "text": [
- "<matplotlib.figure.Figure at 0x4c39f30>"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- " Example 3.19, Page Number: 61"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "#peak loads(MW) are:\n",
- "P1 = 10\n",
- "P2 = 5\n",
- "P3 = 8\n",
- "P4 = 7\n",
- "\n",
- "DF = 1.5 #Diversity factor\n",
- "LF = 0.60 #Avg annual load factor\n",
- "\n",
- "#Calculation:\n",
- "M = (P1+P2+P3+P4)/DF #Max demand on the station(MW)\n",
- "E = (LF*M)*8760 #Annual energy supplied(MWh)\n",
- "#the installed capacity should be 15% to 20% more than the\n",
- "#maximum demand to meet the future needs.\n",
- "#Taking Installed capacity to be 20% more than the maximum demand,\n",
- "IC = 1.2*M #MW\n",
- "\n",
- "\n",
- "#Results:\n",
- "print \"(i) The maximum demand on the station is\",M,\"MW\"\n",
- "print \"(ii) Annual energy supplied by the station is (\",E/1000,\"* 10^6) kWh\"\n",
- "print \"(iii)Installed capacity is\",IC,\"MW\"\n",
- "print \" Suitable unit sizes are 4, each of 6 MW capacity.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The maximum demand on the station is 20.0 MW\n",
- "(ii) Annual energy supplied by the station is ( 105.12 * 10^6) kWh\n",
- "(iii)Installed capacity is 24.0 MW\n",
- " Suitable unit sizes are 4, each of 6 MW capacity.\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.20, Page Number: 64"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "c1 = 20 #capacity of standby station(MW)\n",
- "c2 = 18 #Capacity of base load station(MW)\n",
- "E1 = 7.35*10**6 #Annual standby output(kWh)\n",
- "E2 = 101.35*10**6 #Annual base load station output(kWh)\n",
- "P3 = 12 #peak load on standby station(MW)\n",
- "t = 2190 #hours of use by standby station/year\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "LF1 = E1/(t*P3*10**3) #Annual load factor of standby station\n",
- "PCF1 = E1/(8760*c1*10**3) #Plant capacity factor of standby station\n",
- "LF2 = E2/(8760*c2*10**3) #Annual load factor of base load station\n",
- "#for base load station,\n",
- "PCF2 = LF2 #Plant capacity factor of base load station\n",
- "\n",
- "#Results:\n",
- "print \"Annual load factor of standby station is\",round(LF1*100,1),\"%\"\n",
- "print \"Plant capacity factor of standby station is\",round(PCF1*100,1),\"%\"\n",
- "print \"Annual load factor of base load station is\",round(LF2*100,1),\"%\"\n",
- "print \"Plant capacity factor of base load station is\",round(PCF2*100,1),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Annual load factor of standby station is 28.0 %\n",
- "Plant capacity factor of standby station is 4.2 %\n",
- "Annual load factor of base load station is 64.3 %\n",
- "Plant capacity factor of base load station is 64.3 %\n"
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.21, Page Number: 64"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "M = 60000 #maximum load of hydro plant(kW)\n",
- "m = 20000 #minimum load of hydro pland(kW)\n",
- "P = 50000 #peak load of plant per day(kW)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Let OE = Capacity of steam plant\n",
- "#EC = Capacity of hydro plant\n",
- "x,y = symbols('x,y') #two variables as indicated in given curve\n",
- "# As steam electric conversion is 60%,\n",
- "#\u2234 Area HEFI = 0\u00b76 \u00d7 Area FGB ... (i)\n",
- "#But Area HEFI = Area CFE \u2212 Area CHI\n",
- "# = 1/2*(x*y-50000)\n",
- "# Now Area FGB = 1/2*(FG*GB) = 1/2*(24-x)*(40000-y)\n",
- "# or 0.2*x*y + 12000*x + 7.2*y \u2212 338000 = 0\n",
- "#Also from similar triangles CEF and CDB, we get,\n",
- "# y/40000 = x/24\n",
- "y = x/24*40000\n",
- "x1 = round(solve(0.2*x*y + 12000*x + 7.2*y-338000,x)[1])\n",
- "y1 = x1/24*40000\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Capacity of the hydro plant is\",y1,\"kW\"\n",
- "print \"Capacity of the steam plant is\",60000-y1,\"kW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Capacity of the hydro plant is 20000.0 kW\n",
- "Capacity of the steam plant is 40000.0 kW\n"
- ]
- }
- ],
- "prompt_number": 32
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 3.22, Page Number: 65"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "#Suppose the maximum demand of reservoir plant is y MW and it \n",
- "#operates for x hours\n",
- "x,y = symbols('x y')\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "E = 10**3*(1/2*(320+160)*8760) #Units generated/annum\n",
- "#As the steam plant, run-of-river plant and hydro plant generate \n",
- "#units in the ratio of 7 : 4 : 1, so, units generated by \n",
- "#each plant are:\n",
- "\n",
- "E1 = E*7/12 #by steam plant\n",
- "E2 = E*4/12 #by run-of-river plant\n",
- "E3 = E/1/12 #by reservoir plant\n",
- "\n",
- "#(i) Maximum demand on run-of-river plant = Area OEBA/OA\n",
- "M2 = 700.8*10**6/8760 #kW\n",
- "\n",
- "#Now,\n",
- "x = 8760*y/160\n",
- "E33 = 10**3*1/2*(x*y)\n",
- "#But the units generated by reservoir plant are E3 kWh\n",
- "y1 = solve(E33-E3,y)[1]\n",
- "M3 = y1 #maximum demand on reservoir station(MW)\n",
- "M1 = 320-80-M2/1000 #Maximum demand on steam station(MW)\n",
- "LF2 = 100 #load factor of river plant(%), because it acts \n",
- " #as a base station.\n",
- "\n",
- "LF1 = E1*100/(M1*1000*8760) #load factor of of run-of-river plant(%)\n",
- "LF3 = E3*100/(M3*1000*8760) #load factor of reservoir plant(%)\n",
- "\n",
- " \n",
- "\n",
- "#Result:\n",
- "print \"(i)Maximum demand on run-of-river plant is\",round(M1),\"MW\"\n",
- "print \" Maximum demand on steam station is\",round(M2),\"MW\"\n",
- "print \" Maximum demand on reservoir station is\",round(y1),\"MW\"\n",
- "print \"\\n(ii)Load factor of steam plant plant is\",LF1,\"%\"\n",
- "print \" Load factor of of run-of-river plant is\",LF2,\"%\"\n",
- "print \" Load factor of of reservoir plant is\",round(LF3),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i)Maximum demand on run-of-river plant is 160.0 MW\n",
- " Maximum demand on steam station is 80000.0 MW\n",
- " Maximum demand on reservoir station is 80.0 MW\n",
- "\n",
- "(ii)Load factor of steam plant plant is 87.5 %\n",
- " Load factor of of run-of-river plant is 100 %\n",
- " Load factor of of reservoir plant is 25.0 %\n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Principles_of_Power_System/chapter4_1.ipynb b/Principles_of_Power_System/chapter4_1.ipynb deleted file mode 100644 index c67cf3f8..00000000 --- a/Principles_of_Power_System/chapter4_1.ipynb +++ /dev/null @@ -1,873 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:674aadd4593cd639b4637a0cb17a9ace907a7cf82297146653a75d9ee7f7f19b"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 4: Economics of Power Generation"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 4.1, Page Number: 74"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable Declaration:\n",
- "P = 90000 #initial cost of transformer(Rs)\n",
- "n = 20 #20 years\n",
- "S = 10000 #Salvage value(Rs)\n",
- "\n",
- "#Calculation:\n",
- "D = (P-S)/n #Annual depreciation charge(Rs)\n",
- "\n",
- "print \"The annual depreciation charge is Rs\",D"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The annual depreciation charge is Rs 4000.0\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 4.2, Page Number: 74"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "P = 200000 #Initial cost of transformer(Rs)\n",
- "S = 10000 #Salvage value of transformer(Rs)\n",
- "n = 20 #Useful life(years)\n",
- "r = 0.08 #annual interest rate(%)\n",
- "\n",
- "#Calculation:\n",
- "q = (P-S)*r/(((1+r)**n)-1) #Annual payment(Rs)\n",
- "\n",
- "#Results:\n",
- "print \"The annual amount to be saved is Rs\",round(q)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The annual amount to be saved is Rs 4152.0\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 4.3, Page Number: 75"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "P = 1560000 #Initial cost of equipment(Rs)\n",
- "S = 60000 #salvage value of equipment(Rs)\n",
- "n = 25 #useful life(years)\n",
- "t = 20 #years\n",
- "#Calculation:\n",
- "\n",
- "#(i) Straight line method:\n",
- "D1 = (P-S)/n #Annual depreciation(Rs)\n",
- "V1 = P-D1*t #Value of equipment after 't' years\n",
- "\n",
- "#(ii)Diminishing value method:\n",
- "D2 = 1-(S/P)**(1/n) #Annual unit depreciation(Rs)\n",
- "V2 = P*(1-round(D2,3))**t #Value of equipment after 't' years\n",
- "\n",
- "#(iii)Sinking fund method:\n",
- "r = 0.05 #rate of interest\n",
- "q = (P-S)*(r/((1+r)**n-1)) #Annual deposit in the sinking fund(Rs)\n",
- "q1 = 31433*(((1+r)**20-1)/r) #Sinking fund at the end of 20 years(Rs)\n",
- "V = P-q1 #Value of plant after 20 years(Rs)\n",
- "\n",
- "#Results:\n",
- "print \"The depreciated values using:\"\n",
- "print \"(i) Straight line method: Rs\",V1\n",
- "print \"(ii)Diminishing value method Rs:\",round(V2)\n",
- "print \"(iii)Sinking fund method: Rs\",round(V)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The depreciated values using:\n",
- "(i) Straight line method: Rs 360000.0\n",
- "(ii)Diminishing value method Rs: 115615.0\n",
- "(iii)Sinking fund method: Rs 520638.0\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 4.4, Page Number: 76"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "M = 50000 #Max demand(kW)\n",
- "CC = 95*10**6 #Capital cost(Rs)\n",
- "LF = 0.4 #annual load factor\n",
- "C1 = 9*10**6 #Annual cost of fuel and oil(Rs)\n",
- "C2 = 7.5*10**6 #taxes wages salaries etc(Rs)\n",
- "i = 12 #interest & depreciation(%)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "E = M*LF*8760 #units generated(kWh/year)\n",
- "AFC = i*CC/100 #Annual fixed charges(Rs)\n",
- "T = C1+C2 #Total annual running charges(Rs)\n",
- "TAC = AFC+T #Total annual charges(Rs)\n",
- "c = TAC/E #cost per unit(Rs)\n",
- "\n",
- "#Results:\n",
- "print \"Cost per unit generated = \",round(c,2)*100,\"paise\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Cost per unit generated = 16.0 paise\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 4.5, Page Number: 76"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "C = 50000 #Installed capacity(kW)\n",
- "E = 220*10**6 #units generated(kWh/year)\n",
- "AFC1 = 160 #annual fixed charges per kW of C(Rs/kW)\n",
- "RC = 0.04 #running charges(Rs)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "AFC = AFC1*C #annual fixed charges(Rs)\n",
- "ARC = RC*E #annual running charges(Rs)\n",
- "c = (AFC+ARC)/E #cost per unit(Rs)\n",
- "\n",
- "#Results:\n",
- "print \"Cost per unit generated is \",round(c,4)*100,\"paise\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Cost per unit generated is 7.64 paise\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 4.6, Page Number: 76"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "C = 160000 #cost of plant(Rs)\n",
- "r = 12 #annual fixed charges(%)\n",
- "r1 = 5 #interest(%)\n",
- "r2 = 5 #depreciation(%)\n",
- "r3 = 2 #taxes(%)\n",
- "M = 100 #max demand(kW)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "AFC = C*r/100 #annual fixed charges(Rs)\n",
- "\n",
- "#(i)when load factor is 100%\n",
- "E1 = M*1*8760 #kWh/year\n",
- "c1 = AFC/E1 #Rs\n",
- "\n",
- "#(i)when load factor is 50%\n",
- "E2 = M*0.5*8760 #kWh/year\n",
- "c2 = AFC/E2 #Rs\n",
- "\n",
- "\n",
- "#Results:\n",
- "print \"Fixed charges when:\"\n",
- "print \"(i) load factor is 100% :\",round(c1*100,2),\"paise\"\n",
- "print \"(ii)load factor is 50% :\",round(c2*100,2),\"paise\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Fixed charges when:\n",
- "(i) load factor is 100% : 2.19 paise\n",
- "(ii)load factor is 50% : 4.38 paise\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 4.7, Page Number: 77"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "PC = 50 #plant capacity(MW)\n",
- "LF = 0.4 #annual load factor\n",
- "C1 = 1.2*10**7 #capital cost(Rs)\n",
- "C2 = 4*10**5 #annual cost of wages, taxation etc.(Rs)\n",
- "C3 = 1.0 #cost of fuel,lubrication, maintenance etc.(paise/kWh)\n",
- "r1 = 5 #interest rate(%)\n",
- "r2 = 6 #depreciation(%)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "T1 = (C1*(r1+r2)/100)+C2 #Total annual fixed charges(Rs)\n",
- "E = PC*10**3*LF*8760 #units generated(kWh/year)\n",
- "C4 = E*C3/100 #Cost of fuel, lubrication etc.(Rs)\n",
- "T = T1+C4 #total annual charges(Rs)\n",
- "c = T/E #generating cost(Rs/kWh)\n",
- "\n",
- "\n",
- "#Results:\n",
- "print \"Cost per kWh is \",round(c*100),\"paise\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Cost per kWh is 2.0 paise\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 4.8, Page Number: 77"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "PC = 300 #plant capacity(MW)\n",
- "CF = 0.5 #capacity factor\n",
- "LF = 0.6 #annual load factor\n",
- "C1 = 9*10**7 #Annual cost of fuel, oil etc(Rs)\n",
- "C2 = 10**9 #capital cost(Rs)\n",
- "r1 = 10 #annual interest and depreciation(%)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "L = CF*PC #avg load(MW)\n",
- "M = L/LF #max demand(MW)\n",
- "RC = PC-M #Reserve capacity(MW)\n",
- "TC = C1+C2*r1/100 #total cost(Rs)\n",
- "E = L*10**3*8760 #units generated(kWh/year)\n",
- "c = TC/E #cost per kWh\n",
- "\n",
- "#Results:\n",
- "print \"The minimum reserve capacity of the station is\",RC,\"MW\"\n",
- "print \"Cost per kWh generated is \",round(c*100),\"paise\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The minimum reserve capacity of the station is 50.0 MW\n",
- "Cost per kWh generated is 14.0 paise\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 4.9, Page Number: 78"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "#Variable declaration:\n",
- "PC = 50 #plant capacity(MW)\n",
- "CC = 1000 #capital cost(Rs/kW)\n",
- "r1 = 10 #annual depreciation charges(%)\n",
- "r2 = 20 #part of salaries, maitenance to fixed charges(%)\n",
- "M = 40 #max demand(MW)\n",
- "LF = 0.60 #load factor\n",
- "C1 = 700000 #Annual cost of salaries, maintenance charges etc.(Rs)\n",
- "R1 = 1 #royalty(Re/(kW*year))\n",
- "R2 = 0.01 #royalty paid for using the river water for generation(Re/kWh)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#for annual fixed cost\n",
- "E = M*10**3*LF*8760 #kWh/year\n",
- "C = CC*PC*10**3 #capital cost(Rs)\n",
- "T1 = r1*C/100+r2*C1/100 #total annual fixed charges(Rs)\n",
- "\n",
- "C2 = T1/(M*10**3)+R1 #Cost per kW(Rs)\n",
- "\n",
- "#for running cost:\n",
- "C3 = ((1-r2/100)*C1)/E+R2 #Cost per kW(Rs)\n",
- "\n",
- "\n",
- "#Results:\n",
- "print \"Total generation cost in two part form is given by:\"\n",
- "print \"Rs (\",C2,\"* kW +\",round(C3,4),\"* kWh)\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total generation cost in two part form is given by:\n",
- "Rs ( 129.5 * kW + 0.0127 * kWh)\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 4.10, Page Number: 78"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "M = 60 #plant capacity(MW)\n",
- "LF = 0.5 #load factor\n",
- "C1 = 5*10**6 #capital cost of building and equipment(Rs)\n",
- "C2 = 900000 #annual cost of fuel,oil,taxation and wages(Rs)\n",
- "r1 = 10 #interest and depreciation(%)\n",
- "C3 = 5000000 #annual cost of organisation and \n",
- " #interest on cost of site etc.\n",
- "\n",
- "#Calculation:\n",
- "E = M*10**3*LF*8760 #kWh/year\n",
- "a = C3 #Rs\n",
- "T2 = r1*C3/100 #Annual semi-fixed cost(Rs)\n",
- "b = T2/(M*10**3) #cost per kW\n",
- "c = C2/E #cost per kWh\n",
- "\n",
- "#Results:\n",
- "print \"The required values are:\"\n",
- "print \"a = Rs\",a,\", b = \",round(b,2),\", c = Re\",round(c,4)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The required values are:\n",
- "a = Rs 5000000 , b = 8.33 , c = Re 0.0034\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 4.11, Page Number: 79"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "PC = 100 #installed capacity(kW)\n",
- "CC = 3000 #plant cost(Rs/kW of PC)\n",
- "r1 = 5 #interest(%)\n",
- "r2 = 2 #depreciation(2%)\n",
- "r3 = 2 #operation & maintenance(%)\n",
- "r4 = 1.5 #insurance, rent(%)\n",
- "r = 12.5 #losses in transmission and distribution(%)\n",
- "DF = 1.25 #diversity factor\n",
- "LF = 0.4 #load factor\n",
- "M = 0.8*PC #max demand(kW)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "L = M*LF #avg demand(kW)\n",
- "C1 = CC*PC #Capital cost(Rs)\n",
- "T1 = (r1+r2)*C1/100 #annual fixed charges(Rs)\n",
- "T11 = T1/(DF*M) #annual fixed charges(Rs/kW)\n",
- "RC = C1*(r3+r4)/100 #annual running charges(Rs)\n",
- "E = L*8760 #kWh/year\n",
- "E1 = E*(1-r/100) #units reaching the consumer(kWh)\n",
- "RC1 = RC/E1 #annual running charges(Rs/kWh)\n",
- "T = T1+RC #total charges(Rs)\n",
- "C2 = T/E1 #cost per kWh(Rs)\n",
- "\n",
- "#Results:\n",
- "print \"Total generation cost in two part form is given by:\"\n",
- "print \"Rs (\",T11,\"* kW +\",round(RC1,3),\"* kWh)\"\n",
- "print \"Overall cost of generation per kWh is \",round(C2*100,1),\"paise\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total generation cost in two part form is given by:\n",
- "Rs ( 210.0 * kW + 0.043 * kWh)\n",
- "Overall cost of generation per kWh is 12.8 paise\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 4.12, Page Number: 80"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#variable declaration:\n",
- "M = 1000 #max demand(kW)\n",
- "LF = 0.5 #load factor\n",
- "#for (i)a private oil engine generating plant:\n",
- "\n",
- "CC1 = 12*10**5 #capital cost(Rs)\n",
- "c1 = 0.005 #Cost of repair and maintenance(Rs/kWh)\n",
- "c2 = 1600 #Cost of fuel(Rs/1000kg)\n",
- "r1 = 10 #Interest and depreciation (%)\n",
- "w = 0.3 #fuel consumption(kg/kWh)\n",
- "c3 = 50000 #wages(Rs)\n",
- "\n",
- "#for (i)Public supply company:\n",
- "#Rs 150 per kW of maximum demand plus 15 paise per kWh\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "E = M*LF*8760 #kWh/year\n",
- "\n",
- "#for(i) Private oil engine generating plant:\n",
- "W = w*E #annual fuel consumption(kg)\n",
- "C1= W*c2/1000 #cost of fuel(Rs)\n",
- "T1 = C1+c1*E+r1*CC1/100+c3 #total annual cost(Rs)\n",
- "\n",
- "#for(ii)Public supply company:\n",
- "T2 = 150*M+.15*E #total annual cost(Rs)\n",
- "\n",
- "\n",
- "#Results:\n",
- "print \"(i)For Private oil engine generating plant,\" \n",
- "print \" total annual charges is Rs\",T1\n",
- "print \"(ii)Public supply company, total annual charges is Rs\",T2"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i)For Private oil engine generating plant,\n",
- " total annual charges is Rs 2294300.0\n",
- "(ii)Public supply company, total annual charges is Rs 807000.0\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- " Example 4.13, Page Number: 80"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "LF = 0.3 #load factor\n",
- "M = 100 #max demand(MW)\n",
- "\n",
- "#steam: \n",
- "cc1 = 1250 # Capital cost/kW installed(Rs)\n",
- "r11 = 12 # Interest and depreciation(%)\n",
- "c11 = 5 # Operating cost/kWh, paise\n",
- "tc11 = 0 # Transmission cost/kWh\n",
- "\n",
- "#hydro:\n",
- "cc2 = 2500 # Capital cost/kW installed(Rs)\n",
- "r21 = 10 # Interest and depreciation(%)\n",
- "c21 = 1.5 # Operating cost/kWh,paise\n",
- "tc21 = 0.2 # Transmission cost/kWh,paise\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "E = M*LF*8760*10**3 #kWh/year\n",
- "\n",
- "#(i)steam station in conjunction with a hydro station:\n",
- "Eh = 100*10**6 #units supplied by hydro station(kWh)\n",
- "Es = E-Eh #units supplied by steam station(kWh)\n",
- "Pmh = 40 #max o/p of hydro stn.(MW)\n",
- "Pms = 60 #max o/p of steam stn(MW)\n",
- "\n",
- "#(i)(a)for steam station:\n",
- "Cs1 = cc1*Pms*10**3 #capital cost(Rs)\n",
- "Ts1 = r11*Cs1/100+c11*Es/100+0 #total cost(Rs)\n",
- "\n",
- "# (b)for hydro station:\n",
- "Ch1 = cc2*Pmh*10**3 #capital cost(Rs)\n",
- "Th1 = r21*Ch1/100+c21*Eh/100+tc21*Eh/100 #total cost(Rs)\n",
- "\n",
- "Ta = Ts1+Th1 #total annual cost(Rs)\n",
- "OC1 = Ta/E #kWh\n",
- "\n",
- "\n",
- "\n",
- "#(ii)Steam station:\n",
- "Pm2 = 100*10**3 #max o/p of steam plant(kW)\n",
- "Cs2 = cc1*Pm2 #capital cost(Rs)\n",
- "Ts2 = (Cs2*r11/100)+c11*E/100 #Rs\n",
- "OC2 = Ts2/E #overall cost(Rs)\n",
- "\n",
- "#(iii)Hydro station:\n",
- "Ch3 = cc2*Pm2 #capital cost(Rs)\n",
- "Th3 = Ch3*r21/100+c21*E/100 #Rs\n",
- "OC3 = Th3/E+tc21/100 ##overall cost(Rs)\n",
- "\n",
- "#Results:\n",
- "print \"(i)steam station in conjunction with a hydro station:\"\n",
- "print \" for steam stn., overall cost is \",round(OC1*100,2),\"paise\"\n",
- "print \"(ii) Steam station, overall cost is \",round(OC2*100,2),\"paise\"\n",
- "print \"(ii) Hydro station, overall cost is \",round(OC3*100,2),\"paise\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i)steam station in conjunction with a hydro station:\n",
- " for steam stn., overall cost is 10.97 paise\n",
- "(ii) Steam station, overall cost is 10.71 paise\n",
- "(ii) Hydro station, overall cost is 11.21 paise\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 4.14, Page Number: 81"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "M = 150*10**3 #maximum demand(kW)\n",
- "\n",
- "#Steam plant:\n",
- "cc1 = 1600 #Rs/kW\n",
- "oc1 = 0.06 #operating cost(Rs/kWh)\n",
- "r1 = 7 #interest(%)\n",
- "\n",
- "#Hydro plant:\n",
- "cc2 = 3000 #Rs/kW\n",
- "oc2 = 0.03 #operating cost(Rs/kWh)\n",
- "r2 = 7 #interest(%)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "x = symbols('x') #total no. of units generated\n",
- "#for steam plant:\n",
- "cs = cc1*M #capital cost(Rs)\n",
- "Ts = r1*cs/(x*100)+oc1 #total cost(Rs)\n",
- "Ts1 = Ts #Rs/kWh\n",
- "\n",
- "#for hydro plant:\n",
- "ch = cc2*M #capital cost(Rs)\n",
- "Th = r2*ch/(100*x)+oc2 #total cost(Rs)\n",
- "Th1 = Th #Rs/kWh\n",
- "L = solve(Ts1-Th1,x)\n",
- "LF = (L[0])/(M*8760)\n",
- "\n",
- "print \"Load factor is \",round(LF*100,1),\"%\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Load factor is 37.3 %\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 4.15, Page Number: 82"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "Eo = 40*10**6 #units needed to generate\n",
- "\n",
- "#Hydro Steam\n",
- "cc1 = 2100; cc2 = 1200 #Capital cost(Rs/kW) \n",
- "rc1 = 3.2; rc2 = 5 #Running cost(paise/kWh) \n",
- "r1 = 7.5; r2 = 9 #Interest & depreciation(%)\n",
- "RC1 = 33; RC2 = 25 #Reserve capacity(%)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Let x kW be the maximum demand. \n",
- "#Let y be the annual load factor at which cost/unit\n",
- "#of steam and hydro stations is the same.\n",
- "\n",
- "x,y = symbols('x y')\n",
- "E = x*y*8760 #units generated per annum(kWh)\n",
- "C2 = x+RC2*x/100 #installed capacity of steam station(kW)\n",
- "C1 = x+RC1*x/100 #installed capacity of hydro station(kW)\n",
- "\n",
- "#steam station:\n",
- "CC2 = cc2*C2 #capital cost(Rs)\n",
- "OC2 = (r2*CC2/100+rc2*E/100)/E #Overall cost/kWh\n",
- "\n",
- "#hydro station:\n",
- "CC1 = cc1*C1 #capital cost(Rs)\n",
- "OC1 = (r1*CC1/100+rc1*E/100)/E #Overall cost/kWh\n",
- "\n",
- "LF = solve(OC1-OC2,y)[0] #load factor\n",
- "E2 = 8760*x*LF #kWh\n",
- "M = solve(E2-Eo,x)[0] #max. demand(kW)\n",
- "C = (135*M + 438*M*LF) #cost of generation(Rs)\n",
- "\n",
- "#Results\n",
- "print \"Load factor is \",round(LF*100,2),\"%\"\n",
- "print \"Required cost of generation is Rs (\",round(C/10**6,1),\"* 10**6 )\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Load factor is 47.23 %\n",
- "Required cost of generation is Rs ( 3.3 * 10**6 )\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 4.16, Page Number: 83"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "\n",
- "#Let x = Installed capacity of station B in kW\n",
- "#y = Hours of operation of station B\n",
- "\n",
- "x,y = symbols('x y')\n",
- "M = 50000 #max demand(kW)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "PCa = M-x #installed capacity of stationA(kW)\n",
- "Eb = (1/2)*x*(8760*x/M) #Units generated/annum by station B\n",
- "Ea = (1/2)*8760*M-Eb #Units generated/annum by station A\n",
- "Cb = 50000+50*x+0.03*Eb #Rs\n",
- "Ca = 75000+80*(50000-x)+0.02*Ea #Rs\n",
- "C = Ca+Cb #total operating cost(Rs)\n",
- "#After differentiating C w.r.t x, we get C1 as\n",
- "C1 = -30+0.00174*x\n",
- "x1 = solve(C1,x)[0]\n",
- "\n",
- "PCb = x1 #kW\n",
- "PCa = M-PCb #kW\n",
- "t = 8760*PCb/M #hours of operation of station B\n",
- "\n",
- "#Results:\n",
- "print \"Installed capacity of station A is \",round(PCa),\"MW\"\n",
- "print \"Installed capacity of station B is \",round(PCb),\"MW\"\n",
- "print \"No. of hours of operation of plant B is\",round(t/10)*10,\"hrs\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Installed capacity of station A is 32759.0 MW\n",
- "Installed capacity of station B is 17241.0 MW\n",
- "No. of hours of operation of plant B is 3020.0 hrs\n"
- ]
- }
- ],
- "prompt_number": 14
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Principles_of_Power_System/chapter5_1.ipynb b/Principles_of_Power_System/chapter5_1.ipynb deleted file mode 100644 index 195b1538..00000000 --- a/Principles_of_Power_System/chapter5_1.ipynb +++ /dev/null @@ -1,762 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:95fdfaa4e5e5dbdce2d60b65f4e1e462efff574f2746529cb62b6b5d733a0dc4"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 5: Tariff"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 5.1, Page Number: 91"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "M = 200 #max demand(kW)\n",
- "LF = 0.4 #load factor\n",
- "c1 = 100 #tarif(Rs/kW)\n",
- "c2 = 10 #tariff(pais/kWh)\n",
- "\n",
- "#Calculation:\n",
- "E = M*LF*8760 #units consumed/year\n",
- "T = c1*M+E*c2/100 #annual charges(Rs)\n",
- "OC = T/E #overall cost(Rs/kWh)\n",
- "\n",
- "#Results:\n",
- "print \"Overall cost per kWh is \",round(OC*100,2),\"paise\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Overall cost per kWh is 12.85 paise\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 5.2, Page Number: 91"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "V = 220 #voltage(V)\n",
- "I = 20 #current(A)\n",
- "E = 8760 #kWh\n",
- "c1 = 20 #tariff part1(paise/unit for 500hrs)\n",
- "c2 = 10 #tariff part2 for additional unit(paise/unit)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#assuming power factor to be unity.\n",
- "M = V*I/1000 #max demand(kW)\n",
- "\n",
- "#part (i):\n",
- "E1 = M*500 #kWh\n",
- "C1 = c1*E1/100 #Rs\n",
- "E2 = E-E1 #kWh\n",
- "C2 = 10*E2/100 #kWh\n",
- "T = C1+C2 #total annual bill(Rs)\n",
- "T2 = T/E #equivalent flat rate(Rs/kWh)\n",
- "\n",
- "#Results:\n",
- "print \"(i) Annual bill is Rs\",T\n",
- "print \"(ii)Eqv flat rate is \",round(T2*100,1),\"paise\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Annual bill is Rs 1096.0\n",
- "(ii)Eqv flat rate is 12.5 paise\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 5.3, Page Number: 92"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "\n",
- "#for tariff (a):\n",
- "c1 = 100 #tariff part1(Rs)\n",
- "c11 = 15 #tariff part2(paise/kWh)\n",
- "\n",
- "#for tariff (b):\n",
- "c2 = 30 #paise/kWh\n",
- "\n",
- "#Calculation:\n",
- "#Let x be the number of units at which charges \n",
- "#due to both tariffs become equal.\n",
- "\n",
- "x = symbols('x')\n",
- "x1 = solve(c1+c11*x/100 - c2*x/100 , x)[0]\n",
- "\n",
- "#Results:\n",
- "print \"Tariff(a) is economical if consumption is more than\",round(float(x1),2),\"units.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Tariff(a) is economical if consumption is more than 666.67 units.\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 5.4, Page Number: 92"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "#for 1t tariff:\n",
- "c11 = 30 #Rs/annum\n",
- "c12 = 3 #paise/unit\n",
- "\n",
- "#for 2nd tariff:\n",
- "c21 = 6 #paise/unit for 1st 400 units\n",
- "c22 = 5 #paise/unit for extra units\n",
- "\n",
- "#Calculation:\n",
- "#Let x (> 400) be the number of units taken per annum \n",
- "#for which the annual charges due to both tariffs become equal.\n",
- "\n",
- "x=symbols('x')\n",
- "T1 = c11+c12*x/100 #charges due to 1st tariff(Rs)\n",
- "T2 = c21*400/100+c22*(x-400)/100 #charges due to 2nd tariff(Rs)\n",
- "x1 = solve(T1-T2,x)[0]\n",
- "\n",
- "#Results:\n",
- "print \"Required no. of units are \",round(x1),\"kWh\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Required no. of units are 1300.0 kWh\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 5.5, Page Number: 92"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "M = 50 #max load on the station(MW)\n",
- "AD = 75 #aggregate demand by consumers(MW)\n",
- "E = 18*10**7 #units/annum\n",
- "\n",
- "#for annual fixed charges:\n",
- "c11 = 28*10**5 #for generation(Rs)\n",
- "c12 = 32*10**5 #for transmission & distribution(Rs)\n",
- "c13 = 90*10**5 #for fuel(Rs)\n",
- "\n",
- "#for running charges:\n",
- "c21 = 0.9*90*10**5 #fuel cost(Rs)\n",
- "r = 85 #% of power transmitted\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "T1 = c11+c12+c13*0.1 #10% of fuel used for fixed charges(Rs)\n",
- "C1 = T1/(AD*10**3) #Rs/kW\n",
- "\n",
- "\n",
- "E1 = r*E/100 #units delivered to consumers\n",
- "C2 = c21/E1 #cost per kWh\n",
- "\n",
- "#Results:\n",
- "print \"Tariff is\",C1 ,\"Rs/kW of maximum demand plus\",round(C2*100,1),\"paise per kWh.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Tariff is 92.0 Rs/kW of maximum demand plus 5.3 paise per kWh.\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 5.6, Page Number: 93"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "M = 75*10**3 #Max emand(kW)\n",
- "LF = 0.4 #load factor\n",
- "\n",
- "c1 = 60 #1st part of generating cost(Rs/kW)\n",
- "c2 = 4 #2nd part of generating cost(paise/kW)\n",
- "\n",
- "CT = 2000000 #annual capital charges for transmission system(Rs)\n",
- "CD = 1500000 #annual capital charges for distribution system(Rs)\n",
- "\n",
- "dt = 1.2 #diversity factor of tr. system\n",
- "dd = 1.25 #diversity factor of tr. system\n",
- "\n",
- "nt = 0.9 #efficiency of tr system\n",
- "nd = 0.85 ##efficiency of distribution system\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#(i) Cost at substation:\n",
- "#(a)Annual fixed charges:\n",
- "\n",
- "Tafc1 = c1*M+CT #total annual fixed cost(Rs)\n",
- "S1 = M*dt #sum of all the max demands(kW)\n",
- "AC1 = Tafc1/S1 #Annual cost per kW of max. demand(Rs)\n",
- "\n",
- "#(b) Running Charges:\n",
- "Cs1 = c2/nt #Cost/kWh at substation(paise)\n",
- "\n",
- "#(ii) Cost at consumer\u2019s premises:\n",
- "Tafc2 = Tafc1+CD #Total annual fixed charges at consumer\u2019s premises(Rs)\n",
- "S2 = S1*dd #sum of of maximum demands of all consumers(kW)\n",
- "AC2 = Tafc2/S2 #Annual cost per kW of maximum demand(Rs)\n",
- "#As the distribution efficiency is 85%, therefore, for each kWh delivered from\n",
- "#substation, only 0\u00b785 kWh reaches the consumer\u2019s premises\n",
- "Cs2 = Cs1/nd #Cost/kWh at consumer premises(paise)\n",
- "\n",
- "#Result:\n",
- "print \"(i)At sub-station, the cost is Rs\",round(AC1,2),\"per annum per kW maximum demand \"\n",
- "print \" plus\",round(Cs1,2),\"paise per kWh\"\n",
- "print \"\\n(ii)At consumer\u2019s premises, the cost is\",round(AC2,2),\"per annum per kW maximum demand\"\n",
- "print \" plus\",round(Cs2,2),\"paise per kWh.\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i)At sub-station, the cost is Rs 72.22 per annum per kW maximum demand \n",
- " plus 4.44 paise per kWh\n",
- "\n",
- "(ii)At consumer\u2019s premises, the cost is 71.11 per annum per kW maximum demand\n",
- " plus 5.23 paise per kWh.\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 5.7, Page Number: 94"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "# Fixed charges Running charges #Station \n",
- "# (per kW) (paise/kWh)\n",
- "Cf1 = 300; Cr1 = 25 #Diesel \n",
- "Cf2 = 1200; Cr2 = 6.25 #Steam \n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Suppose energy supplied in one year is 100 units i.e., 100 kWh.\n",
- "\n",
- "#Diesel Station:\n",
- "L = symbols('L') #load factor\n",
- "E = 100 #kWh(say)\n",
- "P = E/8760 #avg power, kW\n",
- "M = P/L #max deamnd(kW)\n",
- "C1 = Cf1*M+E*Cr1/100 #Fixed and running charges for 100 kWh\n",
- "\n",
- "#Steam station\n",
- "C2 = Cf2*M+E*Cr2/100 #Fixed and running charges for 100 kWh\n",
- "\n",
- "L1 = solve(C1-C2,L)[0]\n",
- "\n",
- "#Result:\n",
- "print \"The load fctor is \",round(L1*100,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The load fctor is 54.79 %\n"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 5.8, Page Number: 95"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "M = 100 #max demand(kW)\n",
- "LF = 0.6 #load factor\n",
- "pf = 0.8 #power factor\n",
- "c1 = 75 #1st part tariff(Rs/kVA)\n",
- "c2 = 15 #2nd part tariff(paise/kWh)\n",
- "\n",
- "#Calculation:\n",
- "E = M*LF*8760 #units consumed/year\n",
- "M1 = M/pf #max demand in kVA\n",
- "AB = M1*c1+E*c2/100 #annual bill(Rs)\n",
- "\n",
- "#Result:\n",
- "print \"Annual bill is Rs\",AB"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Annual bill is Rs 88215.0\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 5.9, Page Number: 95"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "M = 240 #max load(kW)\n",
- "pf = 0.8 #power factor\n",
- "E = 50000 #annual units consumption(kW)\n",
- "c1 = 50 #1st part tariff(Rs/KVA)\n",
- "c2 = 10 #2nd part tariff(paise/unit)\n",
- "\n",
- "#Calculation:\n",
- "M1 = M/pf #KVA\n",
- "AB = M1*c1+E*c2/100 #annual bill(Rs)\n",
- "FR = AB/E #Rs\n",
- "\n",
- "#now\n",
- "pf1 = 1\n",
- "M2 = M\n",
- "AB1 = M2*c1+E*c2/100 #Rs\n",
- "S = AB-AB1 #annual saving(Rs)\n",
- "\n",
- "#Result:\n",
- "print \"Flat rate of energy consumption is \",FR*100,\"paise\"\n",
- "print \"Annual saving is Rs\",S"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Flat rate of energy consumption is 40.0 paise\n",
- "Annual saving is Rs 3000.0\n"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 5.10, Page Number: 96"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "M = 50 #max demand(kW)\n",
- "E = 36000 #energy consume(kWh)\n",
- "R = 23400 #reactive power(KVAR)\n",
- "c1 = 80 #1st part tariff(Rs/kW)\n",
- "c2 = 8 #2nd part tariff(paise/unit)\n",
- "c3 = 0.5 #3rd part tariff(p/kWh)for each 1% of pf below 86%\n",
- "\n",
- "#Calculation:\n",
- "L = E/(24*30) #avg load(kW)\n",
- "RP = R/(24*30) #avg reactive power(kVAR)\n",
- "\n",
- "theta = math.atan(RP/L) #power factor angle\n",
- "pf = math.cos(theta) \n",
- "PFS = E*c3*(0.86-pf) #power factor surcharge(Rs)\n",
- "MB = c1*L+c2*E/100+PFS #monthly bill(Rs)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The monthly bill is Rs\",round(MB,1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The monthly bill is Rs 7268.0\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 5.11, Page Number: 96"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "c1 = 150 #1st part tariff(Rs/KVA)\n",
- "c2 = 8 #2nd part tariff(paise/unit)\n",
- "LF = 0.3 #load factor\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#suppose max demand is 1kVA\n",
- "\n",
- "#(i)When p.f. is unity:\n",
- "pf = 1\n",
- "OC1 = c1*100/(8760*LF)+c2 #operating cost(Rs)\n",
- "\n",
- "#(ii) When p.f. is 0\u00b77\n",
- "pf1 = 0.7\n",
- "OC2 = c1*100/(8760*LF*pf1)+c2 #operating cost(Rs)\n",
- "\n",
- "#Result:\n",
- "print \"At unity p. f., overall cost is Rs\",round(OC1,2)\n",
- "print \"At 0.7 p. f., overall cost is Rs\",round(OC2,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "At unity p. f., overall cost is Rs 13.71\n",
- "At 0.7 p. f., overall cost is Rs 16.15\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 5.12, Page Number: 97"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#variable declaration\n",
- "L = 200 #avg load(kW)\n",
- "pf = 0.8 #power factor\n",
- "M = 250 #max demand(kW)\n",
- "l = 4 #losses(%)\n",
- "r = 12 #interest & depreciation(%)\n",
- "C = 50 #high voltage equipment cost(Rs)\n",
- "t = 8 #working hours\n",
- "n = 300 #no. of working working\n",
- "\n",
- "#for system(i)high voltage supply:\n",
- "c11 = 5 #1st part tariff(paise/unit)\n",
- "c12 = 4.50 #2nd part tariff(per month per kVA)\n",
- "\n",
- "#for system(ii)low voltage supply:\n",
- "c21 = 5.5 #1st part tariff(paise/unit)\n",
- "c22 = 5 #2nd part tariff(Rs per month per kVA)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#(i) High voltage supply:\n",
- "\n",
- "M1 = M/pf #Max. demand in kVA\n",
- "#As the losses in h.v. equipment are 4%, therefore, \n",
- "#capacity of h.v. equipment:\n",
- "Cap = round(M1/(1-l/100),1) #capacity of h.v. equipment(kVA)\n",
- "C1 = C*Cap #Capital investment on h.v. equipment(Rs)\n",
- "E1 = L*t*n/(1-l/100) #units consumed(kWh)\n",
- "T1 = C1*r/100+Cap*c12*12+c11*E1/100 #Total annual cost(Rs)\n",
- "\n",
- "#(i) low voltage supply:\n",
- "M2 = M/pf #Max. demand in kVA\n",
- "E2 = L*t*n #units consumed(kWh)\n",
- "T2 = M2*c22*12+E2*c21/100 #kWh\n",
- "\n",
- "T = T2 - T1\n",
- "\n",
- "#Results:\n",
- "print \"Difference in the annual costs of two systems is Rs\",T\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Difference in the annual costs of two systems is Rs 620.0\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 5.13, Page Number: 97"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "#(i) Purchasing diesel set:\n",
- "M1 = 1000 #kW\n",
- "C1 = 400 #Rs/kW\n",
- "r1 = 10 #annual interest depreciation(%)\n",
- "c11 = 75 #Rs/kW\n",
- "c12 = 5 #paise/unit\n",
- "\n",
- "#(ii) Purchasing from grid supply:\n",
- "r1 = 10 #annual interest depreciation(%)\n",
- "c21 =120 #Rs/kW\n",
- "c22 = 3 #paise/unit\n",
- "#after 2 years:\n",
- "M2 = 2500 #kW\n",
- "E = 5.5*10**6 #units reached\n",
- "\n",
- "#Calculation:\n",
- "#(i) Purchasing diesel set:\n",
- "CC = M1*C1 #Rs\n",
- "#The present capacity of the station is 2000 kW and the expected\n",
- "#maximum demand after two years is 2500 kW.\n",
- "P = 2500-2000 #extra power to be generated(kW)\n",
- "T1 = CC*r1/100+P*c11+E*c12/100 #total annual cost(Rs)\n",
- "\n",
- "#(ii) Purchasing from grid supply:\n",
- "T2 = P*c21+E*c22/100 #total annual cost(Rs)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Alternative (ii) is cheaper by Rs\",T1-T2,\"per annum\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Alternative (ii) is cheaper by Rs 127500.0 per annum\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 5.14, Page Number: 98"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from pylab import *\n",
- "from sympy import *\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "#H.V supply:\n",
- "c11 = 70 #1st part tariff(Rs/kVA)\n",
- "c12 = 3 #2nd part tariff(paise/kWh)\n",
- "\n",
- "#L.V supply:\n",
- "c21 = 65 #1st part tariff(Rs/kVA)\n",
- "c22 = 4 #2nd part tariff(paise/kWh)\n",
- "\n",
- "c = 50 #cost of transformer & switchgear for HV side(Rs/kVA)\n",
- "r1= 2 #transformer losses(%)\n",
- "r2 = 15 #annual fixed charges(%) of transformer & switchgear\n",
- "n = 6 #no of working hours\n",
- "\n",
- "#Calculation:\n",
- "(x,y) = symbols('x y') #say x = Factory load in kW\n",
- " #y = No. of working days above which H.V.\n",
- " #supply is cheaper\n",
- "#for HV side: \n",
- "r = x*round(1/(1-r1/100),4) #rating of transformer & switchgear(kVA)\n",
- "E1 = x*y*round(n*1/(1-r1/100),2) #units consumed per annnum\n",
- "T11 = x*math.floor(1/(1-r1/100)*c11*100)/100+x*round(1/(1-r1/100)*r2*c/100,2) #total fixed charges(Rs)\n",
- "T12 = E1*c12/100 #total running charges(Rs)\n",
- "T1 = T11+T12 #total annual charges(Rs)\n",
- "\n",
- "#for LV side:\n",
- "E2 = x*y*n #units consumed per annnum\n",
- "T21 = c21*x #total fixed charges(Rs)\n",
- "T22 = c22*E2/100 #total running charges(Rs)\n",
- "T2 = T21+T22 #total annual charges(Rs)\n",
- "y11 = solve(T1-T2,y)[0]\n",
- "\n",
- "#Result:\n",
- "print \"If the factory is run for more than\",math.floor(y11),'days' #the ans. is different from that in book\n",
- "print \"then H. V. supply will be cheaper.\" #due to calculation using improper rounding."
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "If the factory is run for more than 249.0 days\n",
- "then H. V. supply will be cheaper.\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Principles_of_Power_System/chapter6_1.ipynb b/Principles_of_Power_System/chapter6_1.ipynb deleted file mode 100644 index a1cd7a3c..00000000 --- a/Principles_of_Power_System/chapter6_1.ipynb +++ /dev/null @@ -1,1112 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:357707f074867e12cc217076fc35399c9b1e6483f5b92a23e0c3be75457e81b5"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 6: Power Factor Improvement"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.1, Page Number: 108"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Variable declaration:\n",
- "kW1 = 300\n",
- "pf1 = 0.6 #power factor\n",
- "pf2 = 1\n",
- "\n",
- "#Calculation:\n",
- "kVA = kW1/pf1\n",
- "kW2 = kVA*pf2\n",
- "kW = kW2-kW1\n",
- "\n",
- "#Result:\n",
- "print \"Increased power supplied by the alternator is\",kW,\"kW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Increased power supplied by the alternator is 200.0 kW\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.2, Page Number: 109"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "V = 400 #rated voltage(V)\n",
- "Im = 31.7 #motor curtrent(A)\n",
- "pf1 = 0.7 #initial power factor\n",
- "pf2 = 0.9 #raised power factor\n",
- "f = 50 #Hz\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Referring to the phasor diagram,\n",
- "Ima = Im*pf1 #Active component of Im(A)\n",
- "I = Ima/pf2 #total current(A)\n",
- "Imr = Im*math.sqrt(1-pf1**2) #Reactive component of Im(A)\n",
- "Ir = I*math.sqrt(1-pf2**2) #Reactive component of I(A)\n",
- "Ic = Imr-Ir #A\n",
- "\n",
- "C = Ic/(V*2*math.pi*f)\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Required value of capacitor is\",round(C,7),\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Required value of capacitor is 9.46e-05 uF\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.3, Page Number: 110"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "#(i):\n",
- "kW1 = 20\n",
- "pf1 = 1\n",
- "\n",
- "#(ii):\n",
- "kW2 = 100\n",
- "pf2 = 0.707\n",
- "\n",
- "#(iii):\n",
- "kW3 = 50\n",
- "pf3 = 0.9\n",
- "\n",
- "#Calculation:\n",
- "kVA1 = kW1/pf1\n",
- "kVA2 = kW2/pf2\n",
- "kVA3 = kW3/pf3\n",
- "\n",
- "kVAR1 = kVA1*0\n",
- "kVAR2 = -kVA2*math.sin(math.acos(0.707))\n",
- "kVAR3 = kVA3*math.sin(math.acos(0.9))\n",
- "\n",
- "kW = kW1+kW2+kW3\n",
- "kVAR = kVAR1+kVAR2+kVAR3\n",
- "kVA = math.sqrt(kW**2+kVAR**2)\n",
- "pf = kW/kVA\n",
- "\n",
- "#Result:\n",
- "print \"Total kW is\",kW\n",
- "print \"Total kVA is\",round(kVA)\n",
- "print \"Power factor is\",round(pf,3)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total kW is 170\n",
- "Total kVA is 186.0\n",
- "Power factor is 0.913\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.4, Page Number: 111"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "pf1 = 0.75 #Original p.f lag\n",
- "P = 5 #motor input(kW)\n",
- "pf2 = 0.9 #final p.f lag\n",
- "\n",
- "#Calculation:\n",
- "kVAR = P*(math.tan(math.acos(pf1))-math.tan(math.acos(pf2)))\n",
- "R = kVAR/3\n",
- "\n",
- "#Result:\n",
- "print \"Rating of capacitors is\",round(R,3),\"kVAR\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Rating of capacitors is 0.663 kVAR\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.5, Page Number: 111"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "Vph = 400 #rated voltage(V)\n",
- "n = .93 #efficiency\n",
- "P = 74.6 #power output(kW)\n",
- "pf1 = 0.75 #power faactor\n",
- "pf2 = 0.95 #raised power factor\n",
- "n1 = 4 #no. of capacitors in each branch\n",
- "V1 = 100 #rating of capacitors(V)\n",
- "f = 50 #frequency(Hz)\n",
- "\n",
- "#Calculation:\n",
- "Pi = round(P/n) #motor input(kW)\n",
- "phy1 = math.acos(pf1) \n",
- "phy2 = math.acos(pf2)\n",
- "#Leading kVAR taken by the condenser bank\n",
- "kVAR = round(Pi*(math.tan(phy1)-math.tan(phy2)),2) \n",
- "kVAR1 = round(kVAR/3,2) #Leading kVAR taken by each of three sets\n",
- "\n",
- "#Fig. above shows the delta* connected condenser bank.\n",
- "C = symbols('C')\n",
- "Icp = 2*math.pi*f*C*Vph #Phase current of capacitor(A)\n",
- "\n",
- "c = solve(Vph*Icp/1000-kVAR1,C)[0] #capacitance(F)\n",
- "\n",
- "#Result:\n",
- "print \"Capacitance of each capacitor is\",round(c*4*10**6,1),\"uF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Capacitance of each capacitor is 1173.8 uF\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.6, Page Number: 112"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "P = 800 #load(kW)\n",
- "pf1 = 0.8 #initial power factor\n",
- "pf2 = 0.9 #improved power factor\n",
- "t = 3000 #working hours\n",
- "c1 = 100 #1st part tariff(Rs/kVA)\n",
- "c2 = 20 #2nd part tariff(paise/kWh)\n",
- "c3 = 60 #cost of capacitors(Rs/kVAR)\n",
- "n = 10 #interest & depreciation on capacitors(%)\n",
- "\n",
- "#Calculation:\n",
- "phy1 = math.acos(pf1)\n",
- "phy2 = math.acos(pf2)\n",
- "#Leading kVAR taken by the capacitors:\n",
- "kVAR = P*(math.tan(phy1)-math.tan(phy2))\n",
- "\n",
- "#Annual cost before p.f. correction:\n",
- "kVAm1 = P/pf1 #max demand\n",
- "T1 = kVAm1*c1+P*t*c2/100 #total annual cost(Rs)\n",
- "\n",
- "#Annual cost after p.f. correction:\n",
- "kVAm2 = P/pf2 #max demand\n",
- "T2 = c1*kVAm2+P*t*c2/100+n*c3*kVAR/100\n",
- "\n",
- "T = T1-T2 #annual saving(Rs)\n",
- "\n",
- "#Result:\n",
- "print \"Annual saving is Rs\",round(T)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Annual saving is Rs 9836.0\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.7, Page Number: 112"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "P1 = 200 #factory load(kW)\n",
- "pf1 = 0.85 #initial power factor\n",
- "pf2 = 0.9 #improved power factor\n",
- "t = 2500 #working hours\n",
- "c1 = 150 #1st part tariff(Rs/kVA)\n",
- "c2 = 5 #2nd part tariff(paise/kWh)\n",
- "c3 = 420 #cost of capacitors(Rs/kVAR)\n",
- "Pcl = 100 #capacitor loss(W/kVAR)\n",
- "n = 10 #interest & depreciation on capacitors(%)\n",
- "\n",
- "#Calculation:\n",
- "phy1 = math.acos(pf1)\n",
- "phy2 = math.acos(pf2)\n",
- "#suppose the leading kVAR taken by the capacitors is x:\n",
- "x = symbols('x')\n",
- "Pcl1 = Pcl*x/1000 #capacitor loss(kW)\n",
- "P2 = P1+Pcl1 #total power(kW)\n",
- "x1 = solve((P1*math.tan(phy1)-P2*math.tan(phy2))-x,x)[0]\n",
- "\n",
- "\n",
- "#Annual cost before p.f. correction:\n",
- "kVAm1 = P1/pf1 #max demand\n",
- "T1 = kVAm1*c1+P1*t*c2/100 #total annual cost(Rs)\n",
- "\n",
- "\n",
- "#Annual cost after p.f. correction:\n",
- "kVAm2 = P1/pf2 #max demand\n",
- "T2 = c1*kVAm2+P1*t*c2/100+n*c3*x1/100+round(Pcl*x1*t*c2/100000)\n",
- "\n",
- "T = T1-T2 #annual saving(Rs)\n",
- "\n",
- "#Result:\n",
- "print \"Annual saving is Rs\",round(T)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Annual saving is Rs 553.0\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.8, Page Number: 113"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable Declaration:\n",
- "pf = 0.8 #power factor\n",
- "kVA = 750 # monthly demand\n",
- "c1 = 8.50 #monthly power rate(Rs/kVA per month)\n",
- "R = 250 #rating of capacitors(kVA)\n",
- "C2 = 20000 #installed cost of equipment(Rs)\n",
- "r = 10 #fixed charge rate(%)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "phy = math.acos(pf)\n",
- "kW = kVA*math.cos(phy) #kW component of demand\n",
- "kVAR = kVA*math.sin(phy) #kVAR component of demand\n",
- "kVAR1 = kVAR-R #Leading kVAR taken by the capacitors\n",
- "kVA1 = math.sqrt(kVAR1**2+kW**2) #kVA after p.f. improvement\n",
- "kVA11 = round(kVA-kVA1,1) #Reduction in kVA\n",
- "T1 = c1*kVA11*12 #Yearly saving on kVA charges\n",
- "T2 = r*C2/100 #Fixed charges/year(Rs)\n",
- "T = T1-T2 #net saaving(Rs)\n",
- "\n",
- "#Result:\n",
- "print \"The annual saving is Rs\",T"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The annual saving is Rs 9985.0\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.9, Page Number: 113"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "pf1 = 0.8 #initial power factor\n",
- "pf2 = 0.9 #improved power factor\n",
- "L1 = 200 #load(kW)\n",
- "L2 = 80 #motor load(kW)\n",
- "\n",
- "#Calculation:\n",
- "phy1 = math.acos(pf1)\n",
- "phy2 = math.acos(pf2)\n",
- "L = L1+L2 #combined load(kW)\n",
- "#In Fig. above, \u0394 OAB is the power triangle for load,\n",
- "#\u0394 ODC for combined load and \u0394 BEC for the motor.\n",
- "#(i):\n",
- "kVAR = L1*math.tan(phy1)-L*math.tan(phy2) #Leading kVAR taken by the motor\n",
- "#(ii):\n",
- "kVArat = math.sqrt(L2**2+kVAR**2) #kVA rating of motor\n",
- "#(iii):\n",
- "pf = L2/kVArat #p.f. of motor\n",
- "\n",
- "#Result:\n",
- "print \"The leading kVAR taken by the motor is\",round(kVAR,2)\n",
- "print \"kVA rating of the motor is\",round(kVArat,2)\n",
- "print \"Power factor at which the motor operates is\",round(pf,3),\"leading\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The leading kVAR taken by the motor is 14.39\n",
- "kVA rating of the motor is 81.28\n",
- "Power factor at which the motor operates is 0.984 leading\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.10, Page Number: 114"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "#for induction motor:\n",
- "kW1 = 37.3 \n",
- "pf1 = 0.8 #lagging\n",
- "n1 = 0.85\n",
- "\n",
- "#for synchronous motor:\n",
- "kW2 = 18.65\n",
- "pf = 0.9 #leading\n",
- "n2 = 0.9\n",
- "\n",
- "#for lighting load:\n",
- "kW3 = 10\n",
- "pf3 = 1\n",
- "c1 = 60 #Rs/kVA of max demand\n",
- "c2 = 5 #paie per kWh\n",
- "t = 2000 #working hours\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "P1 = kW1/n1 #power input to induction motor(kW)\n",
- "Q1 = P1*math.tan(math.acos(pf1)) #Lagging kVAR taken by induction motor\n",
- "\n",
- "P2 = kW2/n2 #Input power to synchronous motor(kW)\n",
- "Q2 = P2**math.tan(math.acos(pf1)) #Leading kVAR taken by synchronous motor\n",
- "\n",
- "#Since lighting load works at unity p.f., its lagging kVAR = 0.\n",
- "Q3 = 0 \n",
- "\n",
- "kVAR = Q1-Q2 #net kVAR\n",
- "kW = round(P1+P2+kW3,1) #total active power\n",
- "kVA = round(math.sqrt(kW**2+kVAR**2))\n",
- "\n",
- "T = c1*kVA+c2*kW*t/100\n",
- "\n",
- "#Result:\n",
- "print \"The annual electical charges is Rs\",round(T)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The annual electical charges is Rs 12140.0\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.11, Page Number: 114"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "#(i)for a lighting load:\n",
- "P1 = 500 #kW\n",
- "\n",
- "\n",
- "#(ii) for a load:\n",
- "P2 = 400 #kW\n",
- "pf2 = 0.707 #power factor lagging\n",
- "\n",
- "\n",
- "#(iii)for a load:\n",
- "P3 = 800 #kW\n",
- "pf3 = 0.8 #power factor leading\n",
- "\n",
- "#(iv)for a load:\n",
- "P4 = 500 #kW\n",
- "pf4 = 0.6 #power factor lagging\n",
- "\n",
- "\n",
- "#(v)a synchronous motor driving a d.c. generator:\n",
- "P5 = 540 #power rated at generator(kW)\n",
- "n = 0.9 #overall efficiency\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Total lagging kVAR taken by loads (ii) and (iv)\n",
- "kVAR24 = P2*math.tan(math.acos(pf2))+P4*math.tan(math.acos(pf4))\n",
- "\n",
- "#Leading kVAR taken by the load (iii)\n",
- "kVAR3 = P3*math.tan(math.acos(pf3))\n",
- "\n",
- "#Leading kVAR to be taken by synchronous motor\n",
- "kVAR5 = kVAR24-kVAR3\n",
- "Pi = P5/n #motor input(kW)\n",
- "phy = math.atan(kVAR5/Pi) #phase angle of synchronous motor\n",
- "pf = math.cos(phy)\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The power factor of synchronous motor is\",round(pf,2,),\"leading\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The power factor of synchronous motor is 0.79 leading\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.12, Page Number: 115"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "P1 = 100 #power of synchronous motor(hp)\n",
- "P2 = 200 #power aggregated by induction motor(hp)\n",
- "pf2 = 0.707 #lagging power factor for induction motor\n",
- "n2 = 0.82 #efficiency of induction motor\n",
- "P3 = 30 #lighting load(kW)\n",
- "c1 = 100 #1st part tariff(Rs per kVA per annum)\n",
- "c2 = 0.06 #2nd part tariff Rs per kWh\n",
- "pfa = 0.8 #power factor factor initially(lagging)\n",
- "na = 0.93 #efficiency of synchronous motor initially\n",
- "pfb = 0.8 #power factor factor changed later(leading)\n",
- "nb = 0.93 #efficiency of synchronous motor changed later\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#(a) When synchronous motor runs at p.f. 0\u00b78 lagging.\n",
- "kW1 = round(P1*735.5/(na*1000)) #input to synchronous motor(kW)\n",
- "kVAR1 = round(kW1*math.tan(math.acos(pfa)),2) #Lagging kVAR taken by the synchronous motor(kVAR)\n",
- "kW2 = round(P2*735.5/(n2*1000),1) #input to induction motor(kW)\n",
- "kVAR2 = kW2*math.tan(math.acos(pf2)) #Lagging kVAR taken by the induction motor(kVAR) \n",
- "#Since lighting load works at unity p.f., its lagging kVAR is zero.\n",
- "kVARt = kVAR1+kVAR2 #Total lagging kVAR\n",
- "kWt = kW1+kW2+P3 #Total active power(kW)\n",
- "kVAt = round(math.sqrt(kWt**2+kVARt**2),1) #total kVA\n",
- "C1 = kVAt*c1 #annual kVA demand charges\n",
- "E1 = kWt*8760 #Energy consumed/year(kWh)\n",
- "C2 = round(0.06*E1) #annual energy charges(Rs)\n",
- "T1 = C1+C2 #total Annual bill(Rs)\n",
- "\n",
- "\n",
- "#(b) When synchronous motor runs at p.f. 0\u00b78 leading.\n",
- "kVARn = kVAR2-kVAR1 #Net lagging kVAR\n",
- "kWtt = kWt #total active power(kW)\n",
- "kVAtt = math.sqrt(kVARn**2+kWtt**2) #total kVA\n",
- "C11 = c1*kVAtt #annual kVA charges(Rs)\n",
- "C22 = C2 #Annual energy charges(Rs)\n",
- "T2 = C11+C22 #total annual bill(Rs)\n",
- "Ts = T1-T2 #annual saving(Rs)\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The annual saving is Rs\",round(Ts/100)*100"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The annual saving is Rs 6200.0\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.13, Page Number: 118"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "pf1 = 0.75 #Power factor of the factory(lagging)\n",
- "x = 72 #Max. demand charges(Rs per kVA per annum)\n",
- "d = 10 #interest & depreciation of capital investment(%)\n",
- "M = 175 #max. demand(kW)\n",
- "c2 = 120 #cost of phase advancing equipment(Rs/kVAR)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "y = c2*d/100 #Expenditure on phase advancing equipment(Rs/kVAR/annum)\n",
- "pf2 = math.sqrt(1-(y/x)**2) #power factor\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Most economical p.f. at which factory should operate is\",round(pf2,3),\"lagging\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Most economical p.f. at which factory should operate is 0.986 lagging\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.14, Page Number: 119"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "L = 400 #avg. demand(kW)\n",
- "pf1 = 0.8 #original power factor(lagging)\n",
- "LF = 0.5 #annual load factor\n",
- "c1 = 50 #first part tariff(Rs/kVA per annum)\n",
- "c2 = 0.05 #second part tariff(Rs/kWh)\n",
- "pf2 = 0.95 #improved power factor\n",
- "c3 = 100 #phase advancement equipment cost(Rs/kVAR)\n",
- "d = 10 #annual interest and depreciation(%)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "M = L/LF #maximum demand(kW)\n",
- "#Leading kVAR taken by phase advancement equipment:\n",
- "kVAR1 = M*(math.tan(math.acos(pf1))-math.tan(math.acos(pf2)))\n",
- "y = d*c3/100 #Expenditure on phase advancing equipment(Rs/kVAR/annum)\n",
- "kVA1 = M/pf1 #Max. kVA demand at pf1 power fctor\n",
- "kVA2 = round(M/pf2) #Max. kVA demand at pf2 power fctor\n",
- "Cs = c1*(kVA1-kVA2) #Annual saving in maximum demand charges\n",
- "Ce = y*kVAR1 #Annual expenditure on phase advancing equipment\n",
- "NS = Cs-Ce #net annual saving\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) The capacity of the phase advancing equipment is\",round(kVAR1),\"kVAR\"\n",
- "print \"(ii)The annual saving is Rs\",round(NS)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The capacity of the phase advancing equipment is 337.0 kVAR\n",
- "(ii)The annual saving is Rs 4529.0\n"
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.15, Page Number: 119"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "L = 50 #avg. demand(kW)\n",
- "pf1 = 0.75 #original power factor(lagging)\n",
- "LF = 0.5 #annual load factor\n",
- "c1 = 100 #first part tariff(Rs/kVA of max demand per annum)\n",
- "c2 = 0.05 #second part tariff(Rs/kWh)\n",
- "c3 = 600 #cost of loss free capacitors(Rs/kVAR)\n",
- "d = 10 #depreciation & interest(%)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "y = c3*d/100 #Expenditure on capacitors(Rs/kVAR/annum)\n",
- "pfe = math.sqrt(1-(y/c1)**2) #Most economical power factor\n",
- "M = L/LF #Max kW demand\n",
- "kVAm1 = M/pf1 #The maximum kVA demand at pf1 power factor\n",
- "kVAm2 = M/pfe #The maximum kVA demand at pfe power factor\n",
- "Cs = c1*(kVAm1-kVAm2) #annual saving(Rs)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Most economical power factor is\",round(pfe,3),\"lagging\"\n",
- "print \"Annual saving is Rs\",round(Cs)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Most economical power factor is 0.8 lagging\n",
- "Annual saving is Rs 833.0\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.16, Page Number: 120"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "L = 200 #avg. demand(kW)\n",
- "pf1 = 0.8 #original power factor(lagging)\n",
- "LF = 0.5 #annual load factor\n",
- "c1 = 100 #first part tariff(Rs/kVA of max demand per annum)\n",
- "c2 = 0.05 #second part tariff(Rs/kWh)\n",
- "c3 = 500 #cost of phase advancing plant(Rs/kVAR)\n",
- "d = 10 #depreciation & interest(%)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "y = c3*d/100 #Expenditure on capacitors(Rs/kVAR/annum)\n",
- "pfe = math.sqrt(1-(y/c1)**2) #Most economical power factor\n",
- "Pc = L*(math.tan(math.acos(pf1))-math.tan(math.acos(pfe))) #Capacity of phase advancing plant\n",
- "E = L*5000 #units consumed per year(kWh)\n",
- "C2 = c2*E #annual energy charges(Rs)\n",
- "C1 = y*Pc #Annual cost of phase advancing plant(Rs)\n",
- "Cm = c1*L/pfe #Max. demand charge(Rs)\n",
- "C = C2+C1+Cm #annual charge(Rs)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) The value to which the power factor be improved is\",round(pfe,3),\"lagging\"\n",
- "print \"(ii) The capacity of the phase advancing plant is\",round(Pc,2),\"kVAR\"\n",
- "print \"(iii)The new bill for energy is Rs\",round(C,1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The value to which the power factor be improved is 0.866 lagging\n",
- "(ii) The capacity of the phase advancing plant is 34.53 kVAR\n",
- "(iii)The new bill for energy is Rs 74820.5\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.17, Page Number: 120"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "E = 80000 #units consumed per year\n",
- "kVAm1 = 500 #maximum kVA demand\n",
- "pf1 = 0.707 #initial power factor(lagging)\n",
- "c1 = 120 #first part tariff(Rs/kVA of max demand per annum)\n",
- "c2 = 0.025 #second part tariff(Rs/kWh)\n",
- "c3 = 50 #cost of phase advancing plant(Rs/kVAR)\n",
- "pf2 = 0.9 #increased powe factor(Rs)\n",
- "d = 10 #depreciation on advancing plant(%)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "C1 = c1*kVAm1+c2*E #annual cost of supply(Rs)\n",
- "P = kVAm1*pf1 #max. kW demand\n",
- "#Leading kVAR taken by phase advancing equipment:\n",
- "kVARe = P*(math.tan(math.acos(pf1))-math.tan(math.acos(pf2)))\n",
- "C2 = round(kVARe*c3*d/100) #Annual cost of phase advancing equipment(Rs)\n",
- "#When p.f. is raised from 0\u00b7707 lag to 0\u00b79 lag\n",
- "kVAm2 = P/pf2 #new maximum kVA demand\n",
- "kVAr = kVAm1-kVAm2 #Reduction in kVA demand\n",
- "Cs = round(c1*kVAr) #annual saving(Rs)\n",
- "#As the units consumed remain the same, therefore, saving will\n",
- "#be equal to saving in M.D. charges minus annual cost of \n",
- "#phase advancing plant.\n",
- "Cs2 = Cs-C2 #annual charges(Rs)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The annual cost of supply is Rs\",C1\n",
- "print \"Annual saving is Rs\",Cs2"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The annual cost of supply is Rs 62000.0\n",
- "Annual saving is Rs 11955.0\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.18, Page Number: 123"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "pf = 0.7 #power factor when plant working at its max. capacity(lagging)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#For calculating the cost of increasing plant capacity. \n",
- "#We have, referring to Fig. 6.15, the increase in kVA capacity is BD.\n",
- "\n",
- "#Now OE*cos phy2 = OD*cos phy1\n",
- "#or OB*cos phy2 = OD*cos phy1 (Since OE = OB)\n",
- "#or OD = OB * cos phy2/cos phy1 \n",
- "# = OB * 0\u00b785/0\u00b77\n",
- "# = 1\u00b72143 OB\n",
- "\n",
- "#Increase in the kVA capacity of the plant is\n",
- "#BD = OD \u2212 OB = 1\u00b72143 \u00d7 OB \u2212 OB = 0\u00b72143 OB\n",
- "#Total cost of increasing the plant capacity\n",
- "# = Rs 800 \u00d7 0\u00b72143 \u00d7 OB\n",
- "# = Rs 171\u00b744 \u00d7 OB ...(i)\n",
- "\n",
- "cos_phy1 = 0.7; sin_phy1 = 0.714\n",
- "cos_phy1 = 0.85; sin_phy1 = 0.527\n",
- "\n",
- "#Leading kVAR taken by p.f. correction equipment is\n",
- "# ED = CD \u2212 CE = OD*sin phy1 \u2212 OE*sin phy2\n",
- "# = 1\u00b72143 * OB*sin phy1 \u2212 OB*sin phy2\n",
- "# = OB*(1\u00b72143 * 0\u00b7714 \u2212 0\u00b7527) = 0\u00b734 * OB\n",
- "#Suppose the cost per kVAR of the equipment be Rs y.\n",
- "#So, total cost of p.f. correction equipment\n",
- "# = Rs 0\u00b734 \u00d7 OB \u00d7 y ...(ii)\n",
- "#The cost per kVAR of the equipment that would justify its \n",
- "#installation is when exp. (i) = exp. (ii)\n",
- "#i.e.,\n",
- "# 171\u00b744 \u00d7 OB = 0\u00b734 \u00d7 OB \u00d7 y\n",
- "y = 171.44/0.34 #Rs/kVAR\n",
- "\n",
- "#If the losses in p.f. correction equipment are neglected, \n",
- "#then its kVAR = kVA.\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The maximum cost per kVA of p.f. correction equipment\"\n",
- "print \"that can be paid is Rs\",round(y,1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum cost per kVA of p.f. correction equipment\n",
- "that can be paid is Rs 504.2\n"
- ]
- }
- ],
- "prompt_number": 36
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 6.19, Page Number: 124"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "pf = 0.7 #power factor when plant working at its max. capacity(lagging)\n",
- "\n",
- "#Calculation:\n",
- "#Cost of increasing plant capacity\n",
- "# BD = OD \u2212 OB\n",
- "# = OB * 0.866/0.70-OB\n",
- "# = OB*(1\u00b7237 \u2212 1)\n",
- "# = 0\u00b7237 * OB\n",
- "#\u2234 Annual cost of increasing the plant capacity\n",
- "# = Rs 10 * 0\u00b7237 * OB\n",
- "# = Rs. 2.37 * OB ...(i)\n",
- "\n",
- "#Cost of phase advancing equipment. Leading kVAR taken by\n",
- "#phase advancing equipment,\n",
- "# ED = CD \u2212 CE\n",
- "# = OD*sin phy1 \u2212 OE*sin phy2\n",
- "# = 1.237 * OB * sin phy1 \u2212 OB*sin phy2\n",
- "# = OB*(1\u00b7237 * 0\u00b7174 \u2212 0\u00b75) = 0\u00b7383 \u00d7 OB\n",
- "#Let the cost per kVAR of the equipment be Rs y.\n",
- "#Annual cost of phase advancing equipment\n",
- "# = Rs 0.1 * y * 0\u00b7383 * OB ...(ii)\n",
- "#For economical use, the two costs should be equal i.e.,\n",
- "# exp. (i) = exp. (ii).\n",
- "#or 0\u00b71 * y * 0\u00b7383 * OB = 2\u00b737 * OB\n",
- "y = 2.37/(0.1*0.383) #Rs\n",
- "\n",
- "\n",
- "#If the losses in the phase advancing equipment are neglected,\n",
- "#then its kVAR = kVA.\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The maximum cost per kVA of phase advancing equipment\"\n",
- "print \"that can be paid is Rs\",round(y,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The maximum cost per kVA of phase advancing equipment\n",
- "that can be paid is Rs 61.88\n"
- ]
- }
- ],
- "prompt_number": 2
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Principles_of_Power_System/chapter7_1.ipynb b/Principles_of_Power_System/chapter7_1.ipynb deleted file mode 100644 index bcb1dc3f..00000000 --- a/Principles_of_Power_System/chapter7_1.ipynb +++ /dev/null @@ -1,559 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:855ede77f42d7d38570b25f9dfe540c158b87984c52d22c6171f4226086950d9"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 7: Supply Systems"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 7.1, Page Number: 145"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "V1 = 200 #initial volt of system(V)\n",
- "V2 = 400 #final volt of the system(V)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "I1,I2,R1,R2 = symbols('I1 I2 R1 R2')\n",
- "I2 = V1*I1/V2\n",
- "W1 = 2*I1**2*2*R1 #Power loss in 200 V system\n",
- "W2 = 2*I2**2*2*R2 #Power loss in 400 V system\n",
- "# As power loss in the two cases is the same,\n",
- "R22 = solve(W1/W2-1,R2)[0]\n",
- "a = R22/R1 #ratio of cables' cross-section\n",
- "v = 1/a #ratio of cables' volumes\n",
- "s = (1-v)*100\n",
- "\n",
- "#Result:\n",
- "print \"% Saving in feeder copper is \",s,\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "% Saving in feeder copper is 75 %\n"
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 7.2, Page Number: 146"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "#P = power upplied, W = power loss, \n",
- "#I = current, V = voltage, R = resistance\n",
- "\n",
- "R,V = symbols('R V')\n",
- "P1,I1,W1 = symbols('P1,I1,W1') #for 2-wire system\n",
- "P2,I2,W2 = symbols('P2,I2,W2') #for 3-wire system\n",
- "\n",
- "#Calculation:\n",
- "#Two-wire d.c. system:\n",
- "P1 = V*I1\n",
- "W1 = 2*I1**2*R\n",
- "L1 = W1/P1*100 #% power loss\n",
- "\n",
- "#3-phase, 3-wire a.c. system:\n",
- "P2 = math.sqrt(3)*V*I2\n",
- "W2 = 3*I2**2*R\n",
- "L2 = W2/P2*100 #% power loss\n",
- "\n",
- "I2 = solve(L1-L2,I2)[0]\n",
- "#r2 = P2/P1\n",
- "r2 = math.sqrt(3)*I2*V/(I1*V)\n",
- "P1 = 2*P2\n",
- "\n",
- "#Result:\n",
- "print '''Additional power which can be supplied at \n",
- "unity p.f. by 3-phase, 3-wire a.c. system = 100%.'''"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Additional power which can be supplied at \n",
- "unity p.f. by 3-phase, 3-wire a.c. system = 100%.\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 7.3, Page Number: 146"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "#Let R be the resistance per conductor in each case\n",
- "\n",
- "R,V = symbols('R V')\n",
- "P1,I1,W1 = symbols('P1,I1,W1') #for 3-wire dc system\n",
- "P2,I2,W2 = symbols('P2,I2,W2') #for 4-wire ac system\n",
- "\n",
- "#Calculation:\n",
- "\n",
- "#3-wire d.c. system.:\n",
- "#At balanced loads, implying no power loss in the neutral wire\n",
- "P1 = 2*V*I1\n",
- "W1 = 2*I1**2*R\n",
- "L1 = W1/P1*100 #% power loss\n",
- "\n",
- "\n",
- "#3-phase, 4-wire a.c. system.:\n",
- "#At balanced loads, implying no power loss in the neutral wire\n",
- "P2 = 3*V*I2\n",
- "W2 = 3*I2**2*R\n",
- "L2 = W2/P2*100 #% power loss\n",
- "\n",
- "I2 = solve(L1-L2,I2)[0]\n",
- "r = P2/P1\n",
- "#solving r, we get\n",
- "P2 = 1.5*P1\n",
- "\n",
- "#Result:\n",
- "print '''Extra power that can be supplied at unity power \n",
- "factor by 3-phase, 4-wire a.c. system = 50%.'''"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Extra power that can be supplied at unity power \n",
- "factor by 3-phase, 4-wire a.c. system = 50%.\n"
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 7.4, Page Number: 148"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "R,V,phy = symbols('R V phy') #phy is power factor angle\n",
- " #V is voltage between the conductors\n",
- " #R is resistance per conductor \n",
- "P1,I1,W1 = symbols('P1,I1,W1') #for Single phase 2-wire system\n",
- "P2,I2,W2 = symbols('P2,I2,W2') #for 3-phase, 3-wire a.c. system\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#Single phase 2-wire system\n",
- "P1 = 2*V*I1*math.cos(phy)\n",
- "W1 = 2*I1**2*R\n",
- "L1 = W1/P1*100 #% power loss\n",
- "\n",
- "#3-phase, 3-wire a.c. system. \n",
- "P2 = math.sqrt(3)*V*I2*math.cos(phy)\n",
- "W2 = 3*I2**2*R\n",
- "L2 = W2/P2*100 #% power loss\n",
- "\n",
- "I22 = solve(L1-L2,I2)[0]\n",
- "r = P1/(math.sqrt(3)*V*I22*math.cos(phy))\n",
- "#As the power supplied by single phase, 2-wire is 200 kW,\n",
- "P1 = 200\n",
- "\n",
- "#Result:\n",
- "print \"Power supplied by 3-phase, 3-wire a.c. system is\",P1*r,\"kW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Power supplied by 3-phase, 3-wire a.c. system is 400 kW\n"
- ]
- }
- ],
- "prompt_number": 38
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 7.5, Page Number: 149"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "MVA = 5 #power transmitted to load\n",
- "n = 0.9 #efficiency\n",
- "pf = 0.8 #power factor\n",
- "V = 33 #receiving end voltage(V)\n",
- "l = 50*10**3 #length of line(m)\n",
- "s = 2.85*10**-8 #specific resistance of aluminium(ohm-m)\n",
- "\n",
- "#Calculation:\n",
- "P = MVA*pf*10**6 #power transmitted(W)\n",
- "W = 10*P/100 #line loss(W)\n",
- "\n",
- "\n",
- "#(i) Single phase, 2-wire system:\n",
- "I1 = MVA*10**3/V #Apparent power(A)\n",
- "a1 = 2*I1**2*s*l/W #area of cross-section(m**2)\n",
- "v1= 2*a1*l #volume of conductor required(m**3)\n",
- "\n",
- "#(ii) 3-phase, 3-wire system:\n",
- "I2 = MVA*10**3/(math.sqrt(3)*V) #Apparent power(A)\n",
- "a2 = 3*I2**2*s*l/W #area of cross-section(m**2)\n",
- "v2 = 3*a2*l #volume of conductor required(m**3)\n",
- "\n",
- "#Result:\n",
- "print \"(i) Single phase, 2-wire system, Vol. of conductor required is\",round(v1,2),\"m^3\"\n",
- "print \"(ii)3-phase, 3-wire system, Vol. of conductor required\",round(v2,2),\"m^3\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Single phase, 2-wire system, Vol. of conductor required is 16.36 m^3\n",
- "(ii)3-phase, 3-wire system, Vol. of conductor required 12.27 m^3\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 7.6, Page Number: 149"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "V1 = 11*10**3 #Supply voltage(V)\n",
- "pf = 0.8\n",
- "R1 = 0.15 #Total resistance in ac system(ohm)\n",
- "R2 = 0.05 #Total resistance in dc system(ohm)\n",
- "r1 = 0.15 #voltage drop in ac system(%)\n",
- "r2 = 0.25 #voltage drop in dc system(%)\n",
- "\n",
- "#Calculation:\n",
- "#Single phase system:\n",
- "I1 = r1*V1/R1 #line current(A)\n",
- "P1 = V1*I1*pf/1000 #power received by consumer(kW)\n",
- "\n",
- "#D.C. two-wire system:\n",
- "P2 = P1\n",
- "V2 = (P2*10**3*R2/r2)**0.5 #supply voltage(V)\n",
- "\n",
- "#Result:\n",
- "print \"The supply voltage is\",V2,\"V\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The supply voltage is 4400.0 V\n"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 7.7, Page Number: 153"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "I = 200 #current(A)\n",
- "l = 1 #length of cable(km)\n",
- "c1 = 5 #cost per unit(paise/kWh)\n",
- "r1 = 10 #interrest & depreciation(%)\n",
- "ro = 1.73*10**-6 #resistivity(ohm-m)\n",
- "\n",
- "#Calculation:\n",
- "#If 'a' is the area os cross-section of the cnductor,\n",
- "#cost of cable including installation = (20*a + 20) Rs./m.\n",
- "\n",
- "a = symbols('a')\n",
- "R = ro*l*10**5/a\n",
- "E = 2*I**2*R*8760/1000 #energy lost per annum\n",
- "C = c1/100*E #cost of energy lost(Rs)\n",
- "\n",
- "#capital cost for 1 km length of the cable is = Rs. 20,000*a.\n",
- "\n",
- "V = r1/100*20000*a #Variable annual charge(Rs)\n",
- "a1 = solve(C-V,a)[1]\n",
- "\n",
- "#Result:\n",
- "print \"The most economical conductor size is\",round(a1,2),\"cm**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The most economical conductor size is 1.74 cm**2\n"
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 7.8, Page Number: 153"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "P = 5 #line load(MW)]\n",
- "V = 33 #kV\n",
- "pf = 0.8 #power factor\n",
- "c1 = 4 #paise/kWh\n",
- "r1 = 10 #interest and depreciation(%)\n",
- "ro = 10**-6 #ohm-cm\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "a = symbols('a') #size of conductor(cm**2)\n",
- "R = ro*l*10**5/a #ohm\n",
- "I = P*10**3/(math.sqrt(3)*V*pf) #current (A)\n",
- "E = 3*I**2*R*8760/1000 #kWh\n",
- "C = c1/100*E #Annual cost of energy lost(R)\n",
- "#The capital cost (variable) of the cable = 25000*a Rs/km\n",
- "V = r1/100*25000*a #Rs\n",
- "a1 = solve(C-V,a)[1]\n",
- "\n",
- "#Result:\n",
- "print \"The most economical size of conductor is\",round(a1,2),\"cm**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The most economical size of conductor is 0.71 cm**2\n"
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 7.9, Page Number: 154"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "I = 250 #current through the feeder(A)\n",
- "c1 = 5 #Rs/kg\n",
- "r1 = 10 #interest and depreciation(%)\n",
- "c2 = 5 #paise/kWh\n",
- "d = 8.93 #density of wire(gm/cm**3)\n",
- "ro = 1.73*10**-8 #speciafic resistance(ohm-m)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "l = 1 #length of wire(1 m say)\n",
- "a = symbols('a') #conductor size(m**2)\n",
- "R = ro*l/a #ohm\n",
- "E = 2*I**2*R*8760/1000 #kWh\n",
- "C = c2/100*E #annual cost os energy lost(Rs)\n",
- "m = 2*a*1*d*10**3 #Mass of 1 metre feeder\n",
- "CC = c1*m #capital cost(Rs)\n",
- "V = r1*CC/100 #Variable Annual charge(Rs)\n",
- "a1 = solve(C-V)[1]\n",
- "\n",
- "#Result:\n",
- "print \"The most economic size of conductor is\",round(a1*10000,2),\"cm**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The most economic size of conductor is 3.26 cm**2\n"
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 7.10, Page Number: 154"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "l = 1 #line length(km)\n",
- "V = 110 #supply voltage(kV)\n",
- "\n",
- "#Working hours(hrs) Load(MW) Power factor\n",
- "t1 = 6; P1 = 20 ; pf1 = 0.8\n",
- "t2 = 12; P2 = 5 ; pf2 = 0.8\n",
- "t3 = 6; P3 = 6 ; pf3 = 0.8\n",
- "\n",
- "c1 = 6 #\n",
- "n = 365 #no.of days used\n",
- "r1 = 10 #interest and dep-reciation(%)\n",
- "\n",
- "#Calculation:\n",
- "a = symbols(\"a\") #cross-section of conductor(cm**2)\n",
- "R = 0.176/a #Resistance per km of each conductor(ohm)\n",
- "#The load currents at various loads are :\n",
- "#At 20 MW,\n",
- "I1 = P1*1000/(math.sqrt(3)*V*pf1) #A\n",
- "I2 = P2*1000/(math.sqrt(3)*V*pf2) #A\n",
- "I3 = P3*1000/(math.sqrt(3)*V*pf3) #A\n",
- "\n",
- "E1= 3*R*1/1000*(I1**2*t1+I2**2*t2+I3**2*t3) #Energy loss per day in 3-phase line(kWh)\n",
- "E = E1*n #energy lost per annum(kWh)\n",
- "C = c1*E/100 #Annual cost of energy(Rs)\n",
- "v = r1*6000*a/100 #variable annual charge(Rs/kWh)\n",
- "a1 = solve(C-v,a)[1]\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"THe most economical size of the conductor is\",round(a1,2),\"cm**2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "THe most economical size of the conductor is 1.56 cm**2\n"
- ]
- }
- ],
- "prompt_number": 24
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Principles_of_Power_System/chapter8_1.ipynb b/Principles_of_Power_System/chapter8_1.ipynb deleted file mode 100644 index 6b383670..00000000 --- a/Principles_of_Power_System/chapter8_1.ipynb +++ /dev/null @@ -1,1455 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:9774a5727e94c3e9d7f6de1d46ee1710f0f45bee5b4624a7cce654515da8e820"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 8: Mechanical Design of Overhead Lines"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.1, Page Number: 171"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "k = 0.11 #ratio of shunt-capatance to self capacitance\n",
- "V = 33/math.sqrt(3) #Voltage across string(kV)\n",
- "n = 3 #no. of insulators\n",
- "\n",
- "#Calculation:\n",
- "\n",
- "# At Junction A\n",
- "# I2 = I1 + i1\n",
- "# or V2*w*C = V1*w*C + V1*K*\u03c9*C\n",
- "# or V2 = V1*(1 + K) = V1*(1 + 0\u00b711)\n",
- "# or V2 = 1\u00b711*V1 ...(i)\n",
- "\n",
- "# At Junction B\n",
- "# I3 = I2 + i2\n",
- "# or V3*w*C = V2*w*C + (V1 + V2) K*w*C\n",
- "# or V3 = V2 + (V1 + V2)*K\n",
- "# = 1\u00b711*V1 + (V1 + 1\u00b711*V1)*0\u00b711\n",
- "# V3 = 1\u00b7342 V1\n",
- "# Voltage across whole unit,\n",
- "# V = V1+V2+V3\n",
- "# or V = V1+1.11*V1+1.342*V1\n",
- "\n",
- "V1 = V/3.452 #Voltage across top unit(V)\n",
- "V2 = 1.11*V1 #Voltage across top unit(V)\n",
- "V3 = 1.342*V1 #Voltage across top unit(V)\n",
- "\n",
- "e = V/(n*V3)*100 #string efficiency\n",
- "\n",
- "#Result:\n",
- "print \"(i)\\tVoltage across top unit,V1 is\",round(V1,2),\"kV\"\n",
- "print \" \\tVoltage across middle unit, V2 is\",round(V2,2),\"kV\"\n",
- "print \" \\tVoltage across bottom unit, V3 is\",round(V3,1),\"kV\"\n",
- "print \"(ii)\\tThe string efficiency is \",round(e,1),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i)\tVoltage across top unit,V1 is 5.52 kV\n",
- " \tVoltage across middle unit, V2 is 6.13 kV\n",
- " \tVoltage across bottom unit, V3 is 7.4 kV\n",
- "(ii)\tThe string efficiency is 85.7 %\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.2, Page Number: 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "V1 = 8 #voltage across top unit(kV)\n",
- "V2 = 11 #voltage across middle unit(kV)\n",
- "n = 3 #no. of insulators\n",
- "#Calculation:\n",
- "\n",
- "#(i)\n",
- "# Let K be the ratio of capacitance between pin and the earth\n",
- "# to self capacitance. \n",
- "# Let C farad be the self capacitance of each unit, \n",
- "# then, capacitance between pin and earth = K*C.\n",
- "# Applying Kirchoff\u2019s current law to Junction A,\n",
- "# I2 = I1 + i1\n",
- "# or V2*w*C = V1*w*C + V1*K*w*C\n",
- "# or V2 = V1 (1 + K)\n",
- "\n",
- "k = (V2-V1)/V1\n",
- "\n",
- "#(ii)\n",
- "# Applying Kirchoff\u2019s current law to Junction B,\n",
- "# I3 = I2 + i2\n",
- "# or V3*w*C = V2*w*C + (V1 + V2)*K*w*C\n",
- "# or V3 = V2 + (V1 + V2)*K = 11 + (8 + 11) * 0\u00b7375\n",
- "V3 = 18.12 #kV\n",
- "V = V1 + V2+ V3 #kV\n",
- "Vl = math.sqrt(3)*V #line voltage(kV)\n",
- "\n",
- "#(iii)\n",
- "e = V/(n*V3)*100 #stirng efficiency(%)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i) Ratio of capacitance b/w pin & earth is\",k\n",
- "print \"(ii) The line voltage is\",round(Vl,2),\"kV\"\n",
- "print \"(iii)String efficiency is\",round(e,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) Ratio of capacitance b/w pin & earth is 0.375\n",
- "(ii) The line voltage is 64.29 kV\n",
- "(iii)String efficiency is 68.29 %\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.3, Page Number: 172"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "n = 3 #no. of insulators\n",
- "V3 = 17.5 #voltage across the line unit(V)\n",
- "\n",
- "# At Junction A\n",
- "# I2 = I1 + i1\n",
- "# V2 \u03c9 C = V1 \u03c9 C + V1 K \u03c9 C\n",
- "# or V2 = V1 (1 + K) = V1 (1 + 0.125)\n",
- "# \u2234 V2 = 1\u00b7125 V1\n",
- "\n",
- "# At Junction B\n",
- "# I3 = I2 + i2\n",
- "# or V3 \u03c9 C = V2 \u03c9 C + (V1 + V2) K \u03c9 C\n",
- "# or V3 = V2 + (V1 + V2) K\n",
- "# = 1\u00b7125 V1 + (V1 + 1\u00b7125 V1) \u00d7 0.125\n",
- "# \u2234 V3 = 1\u00b739 V1\n",
- "\n",
- "V1 = V3/1.39 #Voltage across top unit(kV)\n",
- "V2 = 1.125*V1 #Voltage across middle unit(kV)\n",
- "\n",
- "V = V1+V2+V3 #voltage across line unit(kV)\n",
- "e = V/(n*V3)*100 #string efficiency(%)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Line to neutral voltage is\",round(V,2),\"kV\"\n",
- "print \"String efficiency is\",round(e,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Line to neutral voltage is 44.25 kV\n",
- "String efficiency is 84.29 %\n"
- ]
- }
- ],
- "prompt_number": 36
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.4, Page Number: 173"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from pylab import *\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "V3 = 13.1 #voltage across the lowest insulator(kV)\n",
- "V2 = 11 #voltage across the middle insulator(kV)\n",
- "K = symbols('K')\n",
- "# Applying Kirchhoff\u2019s current law to Junctions A and B, we can easily\n",
- "# derive the following equations:\n",
- "\n",
- "# V2 = V1 (1 + K)\n",
- "# or V1 = V2/(1 + K) ...(i)\n",
- "# and V3 = V2 + (V1 + V2)*K ...(ii)\n",
- "# Putting the value of V1 = V2/(1 + K) in eq. (ii), we get,\n",
- "\n",
- "#V3 = V2 + (V2/(1+k)+ V2)*K\n",
- "K1 = solve(V3-(V2 + (V2/(1+K)+ V2)*K),K)[1]\n",
- "\n",
- "V1 = V2/(1+K1)\n",
- "V = V1+V2+V3 #kV\n",
- "\n",
- "#Result:\n",
- "print \"The voltage b/w the bus bars is\",round(float(math.sqrt(3)*V)),\"kV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The voltage b/w the bus bars is 59.0 kV\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.5, Page Number: 173"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "n = 3 #no.of insulators\n",
- "k = 1/8 #ratio of self-capacitance to shunt capacitance\n",
- "V3 = 15 #safe working voltage of each insulator(kV)\n",
- "\n",
- "# Applying Kirchhoff\u2019s current law to junction A, we get,\n",
- "# V2 = V1*(1 + K)\n",
- "# or V1 = V2/(1 + K) = V2/(1 + 0\u00b7125) = 0\u00b789*V2\n",
- "\n",
- "# Applying Kirchhoff\u2019s current law to Junction B, we get,\n",
- "# V3 = V2 + (V1 + V2)*K = V2 + (0.89*V2 + V2) * 0.125\n",
- "# V3 = 1.236*V2 \n",
- "V2 = V3/1.236 #kV\n",
- "V1 = 0.89*V2 #kV\n",
- "V = V1+V2+V3 #voltage across the string(kV)\n",
- "e = V/(n*V3) *100 #string efficiency(%)\n",
- "\n",
- "#Result:\n",
- "print \"Voltage across the string is\",round(V,3),\"kV\"\n",
- "print \"The string efficiency is\",round(e,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Voltage across the string is 37.937 kV\n",
- "The string efficiency is 84.3 %\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.6, Page Number: 174"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "n = 4 #no. of unit\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "# Suppose Xc = 1 \u03a9. Given, the ratio of self-capacitance to shunt\n",
- "# capacitance is 10, so, Xc1 = 10 \u03a9 .\n",
- "# Suppose that potential V across the string is such that 1 A current \n",
- "# flows in the top insulator.\n",
- "#Thus,\n",
- "V1 = 1*1 #Voltage across top unit(V)\n",
- "V2 = 1*1.1 #Voltage across 2nd unit(V)\n",
- "V3 = 1*1.31 #Voltage across 3rd unit(V)\n",
- "V4 = 1*1.65 #Voltage across 4th unit(V)\n",
- "V = V1+V2+V3+V4 #Voltage obtained across the string(V)\n",
- "\n",
- "#The voltage across each unit expressed as a % of V becomes:\n",
- "f1 = V1/V*100\n",
- "f2 = V2/V*100\n",
- "f3 = V3/V*100\n",
- "f4 = V4/V*100\n",
- "\n",
- "e = V/(n*V4)*100 #string efficiency(%)\n",
- "\n",
- "#Result:\n",
- "print \"(i) The voltage distributation from top unit is \"\n",
- "print \"\\t\",round(f1,2),\"%\\t\",round(f2,2),\"%\\t\",round(f3,1),\"%\\t\",round(f4,1),\"%\"\n",
- "print \"\\n(ii)String efficiency is\",round(e,1),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The voltage distributation from top unit is \n",
- "\t19.76 %\t21.74 %\t25.9 %\t32.6 %\n",
- "\n",
- "(ii)String efficiency is 76.7 %\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.7, Page Number: 175"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "n = 5 #no.os insulators\n",
- "Vl = 100 #line voltage(kV)\n",
- "\n",
- "#Suppose Xc = 1 \u03a9.\n",
- "#Ratio of self capacitance to shunt capacitance is 10, so, Xc1 = 10 \u03a9. \n",
- "#Let V be the voltage s.t 1A current flows in top insulator.\n",
- "\n",
- "#Calculation:\n",
- "#calculation for fraction of voltage drop:\n",
- "v1 = 1*1 #Voltage across top unit(V)\n",
- "v2 = 1*1.1 #Voltage across 2nd unit(V)\n",
- "v3 = 1*1.31 #Voltage across 3rd unit(V)\n",
- "v4 = 1*1.65 #Voltage across 4th unit(V)\n",
- "v5 = 1*2.16 #voltage across 5th unit(V)\n",
- "v = v1+v2+v3+v4+v5 #total voltage(V)\n",
- "V = 100/math.sqrt(3) #string voltage(V)\n",
- "V1 = round(v1/v*V,3) #kV\n",
- "V2 = v2/v*V #kV\n",
- "V3 = v3/v*V #kV\n",
- "V4 = v4/v*V #kV\n",
- "V5 = v5/v*V #kV\n",
- "e = V/(n*V5)*100 #string efficiency(%)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"(i)The distribution of voltage on the insulator discs are\"\n",
- "print \"V1 = \",round(V1,3),\"kV \",\"\\tV2 =\",round(V2,2),\"kV \\n\",\n",
- "print \"V3 =\",round(V3,1),\"kV\",\"\\t\\tV4 =\",round(V4,2),\"kV\",\n",
- "print \"\\nV5 =\",round(V5,1),\"kV\"\n",
- "print \"\\n(ii)The string efficiency\",round(e,1),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i)The distribution of voltage on the insulator discs are\n",
- "V1 = 7.997 kV \tV2 = 8.8 kV \n",
- "V3 = 10.5 kV \t\tV4 = 13.19 kV \n",
- "V5 = 17.3 kV\n",
- "\n",
- "(ii)The string efficiency 66.9 %\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.8, Page Number: 176"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "n = 4 #no.os insulators\n",
- "V2 = 13.2 #Voltage across 2th unit\n",
- "V3 = 18 #Voltage across 3th unit\n",
- "#Suppose K is the ratio of shunt-capacitance to self-capacitance\n",
- "k = symbols('k')\n",
- "\n",
- "# Referring to Fig.(ii), we have,\n",
- "# V2/V1 = ( 1 + K)/1\n",
- "# V2 = V1 (1 + K)\n",
- "# V3/V1 = (1 + 3K + K2)/1\n",
- "# V3 = V1 (1 + 3K + K2)\n",
- "# putting values of V2 & V3\n",
- "\n",
- "k1 = solve(13.2*k**2+21.6*k-4.8,k)[1]\n",
- "V1 = V2/(1+k1)\n",
- "V4 = V1*(1 + k1**33 + 5*k1**2 + 6*k1)\n",
- "V = V1+V2+V3+V4 #Voltage between line and earth(kV)\n",
- "Vl = math.sqrt(3)*V #Voltage between conductors(kV)\n",
- "\n",
- "#Result:\n",
- "print \"Voltage between conductors is\",round(float(Vl)),\"kV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Voltage between conductors is 119.0 kV\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.9, Page Number: 177"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "k = 5 #ratio of self-capacitance(C) to pin-earth capacitance(C1)\n",
- "V = 1+1.2+1.64+2.408 #voltage obtained across the string(V)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#(i) The voltage across each unit expressed as a % of V\n",
- "v1 = 1/V*100\n",
- "v2 = 1.2/V*100\n",
- "v3 = 1.6/V*100\n",
- "v4 = 2.408/V*100\n",
- "\n",
- "#(ii)String efficiency\n",
- "e = V/(4*2.408)*100\n",
- "\n",
- "#Result:\n",
- "print \"(i) The voltage across each unit expressed as a % of V are:\"\n",
- "print \"\\tTop Unit: \",round(v1),\"%\"\n",
- "print \"\\tSecond from top: \",round(v2,1),\"%\"\n",
- "print \"\\tThird from top: \",round(v3),\"%\"\n",
- "print \"\\tFourth from top: \",round(v4,1),\"%\"\n",
- "print \"(ii)String efficiency is\",round(e,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(i) The voltage across each unit expressed as a % of V are:\n",
- "\tTop Unit: 16.0 %\n",
- "\tSecond from top: 19.2 %\n",
- "\tThird from top: 26.0 %\n",
- "\tFourth from top: 38.5 %\n",
- "(ii)String efficiency is 64.87 %\n"
- ]
- }
- ],
- "prompt_number": 54
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.10, Page Number: 177"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "V1,V2,V3,V= symbols('V1 V2 V3 V') #volts (shown in fig(ii))\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "\n",
- "#We know that in an actual string of insulators, 3 capacitances\n",
- "#exist i.e, self-capacitance of each insulator, shunt capacitance \n",
- "#and capacitance of each unit to line.\n",
- "#But, capacitance of each unit to line is very small and\n",
- "#can be neglected.\n",
- "\n",
- "#At Junction A\n",
- "# I2 + i1 = I1 + i1\n",
- "#or V2*w*C + (V2 + V3)*0\u00b71*w*C = V1*\u03c9*C + 0\u00b715*C*V1*w\n",
- "#or 0\u00b71*V3 = 1\u00b715*V1 \u2212 1\u00b71*V2\n",
- "# V3 = 11\u00b75*V1 \u2212 11*V2 ...(i)\n",
- "\n",
- "#At Junction B\n",
- "# I3 + i\u20322 = I2 + i2\n",
- "#or V3*w*C + V3*0\u00b71*C*w = V2*w*C + (V1 + V2)*w*0\u00b715 C\n",
- "#or 1\u00b71*V3 = 1\u00b715*V2 + 0\u00b715*V1 ...(ii)\n",
- "#Putting the value of V3 from exp (i). into exp. (ii), we get,\n",
- "# 1\u00b71*(11\u00b75*V1 \u2212 11*V2) = 1\u00b715*V2 + 0\u00b715*V1\n",
- "#or 13\u00b725*V2 = 12\u00b75*V1\n",
- "# V2 = 12.5/13.25*V1\n",
- "# V3 = 11.5*V1-11*V2\n",
- "# V = V1+V2+V3\n",
- "# V = 40.55/13.25*V1\n",
- "V1 = 13.25/40.55*V\n",
- "V2 = 12.5/13.25*V1\n",
- "V3 = 14.8/13.25*V1\n",
- "\n",
- "#The voltage across each unit expressed as a % of V becomes:\n",
- "p1 = V1/V*100\n",
- "p2 = V2/V*100\n",
- "p3 = V3/V*100\n",
- "\n",
- "\n",
- "#Results:\n",
- "print \"The voltage across each unit expressed as a % of V becomes:\"\n",
- "print \"for top unit, v1 =\",round(p1,1),\"%\",\n",
- "print \"\\nfor middle unit, v2 =\",round(p2,1),\"%\"\n",
- "print \"for 3rd unit, v3 =\",round(p3,1),\"%\"\n",
- "\n",
- "print \"String efficiency is\",round(V/(3*V3)*100,1),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The voltage across each unit expressed as a % of V becomes:\n",
- "for top unit, v1 = 32.7 % \n",
- "for middle unit, v2 = 30.8 %\n",
- "for 3rd unit, v3 = 36.5 %\n",
- "String efficiency is 91.3 %\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.11, Page Number: 179"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable Declaration:\n",
- "V1,V2,V3,V= symbols('V1 V2 V3 V') #volts (shown in fig(i))\n",
- "\n",
- "#Calculation:\n",
- "\n",
- "#At Junction A\n",
- "# I2 + i\u20321 = I1 + i1\n",
- "#or V2*w*C + (V2+V3)*w*0\u00b71*C = V1*w*C+V1*0\u00b72*C*\u03c9 \n",
- "# V3 = 12 V1 \u2212 11 V2 # ...(i)\n",
- "\n",
- "#At Junction B\n",
- "# I3 + i\u20322 = I2 + i2\n",
- "#or V3 \u03c9 C + V3 \u00d7 0\u00b73 C \u00d7 \u03c9 =V2 \u03c9 C + (V1 + V2) \u03c9 \u00d7 0\u00b72 C\n",
- "#or 1\u00b73 V3 = 1\u00b72 V2 + 0\u00b72 V1 ...(ii)\n",
- "#Substituting the value of V3 from exp. (i) into exp. (ii), we get,\n",
- "# 1\u00b73*(12*V1 \u2212 11*V2) = 1\u00b72*V2 + 0\u00b72*V1\n",
- "#or 15\u00b75*V2 = 15\u00b74*V1\n",
- "V2 = 15.4*V1/15.5 #...(iii)\n",
- "V2 = 0.993*V1\n",
- "\n",
- "#Substituting the value of V2 from exp. (iii) into exp. (i), we get,\n",
- "V3 = 12*V1 - 11*0.993*V1\n",
- "V3 = 1.077*V1\n",
- "\n",
- "#Voltage between conductor and earth (i.e. phase voltage)\n",
- "V = V1+V2+V3\n",
- "n = V/(3*V3)*100\n",
- "\n",
- "#Result:\n",
- "print \"String efficiency is\",round(n),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "String efficiency is 95.0 %\n"
- ]
- }
- ],
- "prompt_number": 55
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.13, Page Number: 184"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from pylab import *\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "r = 1 #conductor radius(cm)\n",
- "d = 100 #conductor spacing(cm)\n",
- "go = 30/math.sqrt(2) #Dielectric strength of air,(rms)(kV/cm)\n",
- "dl = 0.952 #air density factor\n",
- "mo = 0.9 #irregularity factor\n",
- "\n",
- "#Calculation:\n",
- "Vc = mo*go*dl*r*math.log(d/r) #kV/phase\n",
- "Vl = Vc*3**0.5 #kV/line\n",
- "\n",
- "#Result:\n",
- "print \"The disruptive critical voltage for the line is\",round(float(Vl),2),\"kV\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The disruptive critical voltage for the line is 144.97 kV\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.14, Page Number: 184"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from pylab import *\n",
- "\n",
- "#Variable declaration:\n",
- "r = 0.978 #conductor radius(cm)\n",
- "go = round(30/2**0.5,1) #Dielectric strength of air,rms(kV/cm)\n",
- "Vc = 210/3**0.5 #Disruptive voltage/phase(kV)\n",
- "mo = 1 #irregularity factor(for smooth conductor)\n",
- "dl = 1 #air density factor(at std pressure and temperature)\n",
- "\n",
- "#Calculation:\n",
- "d = r*10**(Vc/(2.3*mo*go*dl*r))\n",
- "\n",
- "#Result:\n",
- "print \"The spacing between the conductors is\",round(d),\"cm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The spacing between the conductors is 341.0 cm\n"
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.15, Page Number: 185"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from pylab import *\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "r = 1.5 #conductor radius(cm)\n",
- "t = 40 #temperature(deg C)\n",
- "b = 76 #atmospheric pressure(cm)\n",
- "mo = 0.85 #irregularity factor\n",
- "f = 50 #Hz\n",
- "d = 200 #conductor spacing(cm)\n",
- "go = 30/2**0.5 #Dielectric strength of air,rms(kV/cm)\n",
- "Vl = 220 #line voltage(kV)\n",
- "\n",
- "#Calculation:\n",
- "dl = round(3.92*b/(273+t),3)\n",
- "\n",
- "Vc = round(mo*go*dl*r*round(math.log(d/r),2),1) #Critical disruptive voltage per phase(kV)\n",
- "V = round(Vl/3**0.5) #kV/phase\n",
- "P = 242.2/dl*(f+25)*(r/d)**0.5*(V-Vc)**2*10**-5 #kW/km/phase\n",
- "Pl = 3*P #corona loss(kW)\n",
- "\n",
- "#Result:\n",
- "print \"Total corona loss per km for three phases is\",round(Pl,5),\"kW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total corona loss per km for three phases is 0.05998 kW\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.16, Page Number: 185"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "P1 = 53 #corona loss(kW)\n",
- "V1 = 106 #operating voltage(kV)\n",
- "P2 = 98 #corona loss(kW)\n",
- "V2 = 110.9 #operating voltage(kV)\n",
- "V3 = 113 #opeating voltage(kV)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#As f, \u03b4, r and d are the same for the two cases,\n",
- "#therfore, P \u221d (V \u2212 Vc)**2\n",
- "k,Vc = symbols('k Vc') #k = proportionality constant\n",
- " #Vc = critical disuptive voltage(kV)\n",
- "#for 1st case,\n",
- "# V1/3**0.5 = k*(64 - Vc)**2\n",
- "# V2/3**0.5 = k*(61\u00b72 - Vc)**2\n",
- "# Dividing above equations & solving,\n",
- "Vc1 = round(solve(((64-Vc)/(61.2-Vc))-math.sqrt(98/53),Vc)[0],2)\n",
- "Vc1 = 54\n",
- "#Let W kilowatt be the power loss at 113 kV.\n",
- "#W = k*(113/3**0.5-Vc1)**2\n",
- "W = P1*((65.2-Vc1)/(61.2-Vc1))**2\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The disruptive critical voltage at 113 kV is\",Vc1,\"kV\"\n",
- "print \"The corona loss at 113 kV is\",round(W),\"kW\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The disruptive critical voltage at 113 kV is 54 kV\n",
- "The corona loss at 113 kV is 128.0 kW\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.17, Page Number: 190"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "w = 680 #weigth of conductor(kg/km)\n",
- "l = 260 #length ofspan(m)\n",
- "u = 3100 #ultimate strength(kg)\n",
- "sf = 2 #safety factor\n",
- "h = 10 #ground clearance(m)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "T = u/sf #working tension(kg)\n",
- "s = w*l**2/(8*T*1000) #span(m)\n",
- "H = h+s\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Conductor should be supported at a height of \",round(H,1),\"m\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Conductor should be supported at a height of 13.7 m\n"
- ]
- }
- ],
- "prompt_number": 56
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.18, Page Number: 190"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "l = 150 #length ofspan(m)\n",
- "a = 2 #conductor cross-section(cm**2)\n",
- "go = 9.9 #specific gravity of conductor(gm/cm**3)\n",
- "ww = 1.5 #Wind force/m length of conductor(kg)\n",
- "T = 2000 #working tension(kg)\n",
- "\n",
- "#Calculation:\n",
- "w = go*a*1/10 #Wt. of conductor/m length(kg)\n",
- "wt = round((w**2+ww**2)**0.5,2) #Total wt of 1m length of conductor(kg)\n",
- "s = wt*l**2/(8*T) #sag(m)\n",
- "#This is the value of slant sag in a direction making \n",
- "#an angle theta with the vertical\n",
- "\n",
- "theta = math.atan(ww/w)\n",
- "Vs = s*math.cos(theta)\n",
- "\n",
- "#Result:\n",
- "print \"The sag is\",round(s,2),\"m\"\n",
- "print \"The vertical sag is\",round(Vs,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The sag is 3.49 m\n",
- "The vertical sag is 2.78 m\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.19, Page Number: 191"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "l = 200 #span length(m)\n",
- "w = 1170/1000 #weigth of conductor(kg)\n",
- "u = 4218 #Ultimate Strength(kg/cm**2)\n",
- "sf = 5 #safety factor\n",
- "a = 1.29 #cross-section of conductor(cm**2)\n",
- "P = 122 #wind pressure(kg/m**2)\n",
- "\n",
- "#Calculation:\n",
- "T = u/sf*a #working tension(kg)\n",
- "d = round((4*a/math.pi)**0.5,2) #diameter of conductor(cm)\n",
- "ww = P*d*10**-2 #Wind force/m length,\n",
- "wt = (ww**2+w**2)**0.5 #total wt of conductor per metre length\n",
- "S = wt*l**2/(8*T)\n",
- "theta = math.atan(ww/w)\n",
- "VS = S*math.cos(theta)\n",
- "\n",
- "#Result:\n",
- "print \"The sag is\",round(S,2),\"m\"\n",
- "print \"The vertical sag is\",round(VS,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The sag is 8.97 m\n",
- "The vertical sag is 5.38 m\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.20, Page Number: 191"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "l = 275 #span length(m)\n",
- "w = 0.865 #weigth of conductor/m(kg)\n",
- "d = 1.96 #conductor diameter(cm)\n",
- "t = 1.27 #ice coating thickness(cm)\n",
- "u = 8060 #Ultimate Strength(kg)\n",
- "sf = 2 #safety factor\n",
- "a = 1.29 #cross-section of conductor(cm**2)\n",
- "wo = 0.91 #Weight of 1 c.c. of ice(gm)\n",
- "P = 3.9 #wind pressure(gm/cm**2)\n",
- "\n",
- "#Calculation:\n",
- "T = u/sf #working tension(kg)\n",
- "v = 3.14*t*(d+t)*100 #Volume of ice per metre(cm**3)\n",
- "wi = v*wo/1000 #kg\n",
- "ww = P*(d+2*t)*100/1000 #kg\n",
- "wt = ((w+wi)**2+ww**2)**0.5 #total weight(kg)\n",
- "S = wt*l**2/(8*T) #sag(m)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Sag is\",round(S,1),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Sag is 6.3 m\n"
- ]
- }
- ],
- "prompt_number": 60
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.21, Page Number: 192"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "Sv = 2.35 #vertical sag(m)\n",
- "P = 1.5 #wind pressure(kg/m)\n",
- "B = 2540 #breaking stress(kg/cm**2)\n",
- "w = 1.125 #wt of conductor(kg/m)\n",
- "l = 214 #length of conductor(m)\n",
- "a = 3.225 #conductor cross-section\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "f = symbols('f') #safety factor\n",
- "wt = (w**2+P**2)**0.5 #Total wt. of one m length of conductor\n",
- "T = B*a/f #working stress(kg)\n",
- "cos_theta = w/wt\n",
- "S = Sv/cos_theta #slant sag\n",
- "T1 = wt*l**2/(8*S) \n",
- "f1 = solve(T-T1,f)[0] #safety factor\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Safety factor is\",round(f1)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Safety factor is 3.0\n"
- ]
- }
- ],
- "prompt_number": 61
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.22, Page Number: 192"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "\n",
- "#Variable declaration:\n",
- "l = 150 #span length(m)\n",
- "u = 5000 #ultimate strength(kg/cm**2)\n",
- "go = 8.9 #specific gravity of material(gm/cc)\n",
- "P = 1.5 #wind pressure(kg/m)\n",
- "sf = 5 #safety factor\n",
- "a = 2 #cross-section of conductor(cm**2)\n",
- "h = 7 #ground clearance(m)\n",
- "\n",
- "#Calculation:\n",
- "w = a*go*100/1000 #weight of conductor per m(kg)\n",
- "T = u*a/sf #working stress per m(kg)\n",
- "wt = (w**2+P**2)**0.5 #Total wt of 1 m length of conductor(kg)\n",
- "S = wt*l**2/(8*T) #slant sag(m)\n",
- "Sv = S*w/wt #conductor vertical sag(m)\n",
- "H = h+Sv #m\n",
- "\n",
- "#Result:\n",
- "print \"Conductor should be supported at a height of\",round(H,1),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Conductor should be supported at a height of 9.5 m\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.23, Page Number: 193"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "l = 500 #distance b/w the two towers(m)\n",
- "T = 1600 #tension in the conductor(kg)\n",
- "w = 1.5 #weight of the conductor(kg/m)\n",
- "h1 = 30 #height of 1st tower(m)\n",
- "h2 = 90 #height of 2nd tower(m)\n",
- "\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "h = h2-h1 #difference in levels(m)\n",
- "x1,x2 = symbols('x1 x2')\n",
- "x2 = l-x1\n",
- "S1 = w*x1**2/(2*T) #m\n",
- "S2 = w*x2**2/(2*T) #m\n",
- "x11 = solve(S2-S1-h,x1)[0]\n",
- "S1 = w*x11**2/(2*T) #m\n",
- "#Let the mid-point P be at a distance x from the lowest point O\n",
- "x = l/2-x11\n",
- "Smid = w*x**2/(2*T) #m\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The min. clearance of the conductor and water is \",round(h1-S1),\"m\"\n",
- "print \"The clearance mid-way b/w the supports is\",round(Smid+h1-S1,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The min. clearance of the conductor and water is 23.0 m\n",
- "The clearance mid-way b/w the supports is 30.7 m\n"
- ]
- }
- ],
- "prompt_number": 62
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.24, Page Number: 194"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "l = 600 #distance b/w the two towers(m)\n",
- "f = 5 #safety factor\n",
- "U = 8000 #kg/cm**2\n",
- "a = 2.2 #area of cross-section(cm**2)\n",
- "w = 1.925 #weight of the conductor(kg/m)\n",
- "h = 15 #difference in towers levels(m)\n",
- "wi = 1 #ice weight(kg)\n",
- "\n",
- "#Calculation:\n",
- "T = U*a/f #tension in the conductor(kg)\n",
- "wt = wi+w #total weight of conductor(kg)\n",
- "x1,x2 = symbols('x1 x2')\n",
- "x2 = l-x1\n",
- "S1 = wt*x1**2/(2*T) #m\n",
- "S2 = wt*x2**2/(2*T) #m\n",
- "x11 = solve(S2-S1-h,x1)[0]\n",
- "x22 = l-x11\n",
- "S22 = wt*x22**2/(2*T) #m\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The sag from the taller of the two supports is\",round(S22,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The sag from the taller of the two supports is 45.27 m\n"
- ]
- }
- ],
- "prompt_number": 63
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.25, Page Number: 195"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "#Variable declaration:\n",
- "l = 400 #distance b/w the two towers(m)\n",
- "T = 2000 #tension in the conductor(kg)\n",
- "w = 1 #weight of the conductor(kg/m)\n",
- "h1 = 40 #height of 1st tower(m)\n",
- "h2 = 90 #height of 2nd tower(m)\n",
- "\n",
- "#Calculation:\n",
- "h = h2-h1 #difference in levels(m)\n",
- "x1,x2 = symbols('x1 x2')\n",
- "x2 = l-x1\n",
- "S1 = w*x1**2/(2*T) #m\n",
- "S2 = w*x2**2/(2*T) #m\n",
- "x11 = solve(S2-S1-h,x1)[0]\n",
- "S1 = w*x11**2/(2*T) #m\n",
- "x_mid = 50+400/2 #Horizontal distance of mid-point P from lowest point O\n",
- "Sp = w*x_mid**2/(2*T) #SAg at P(m)\n",
- "S2 = w*(l-x11)**2/(2*T) #m\n",
- "hc = h2-S2+Sp #Clearance of mid-point P above water level(m)\n",
- "\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Clearance of mid-point P above water level is\",round(hc),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Clearance of mid-point P above water level is 55.0 m\n"
- ]
- }
- ],
- "prompt_number": 127
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.26, Page Number: 195"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from pylab import *\n",
- "from sympy import *\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "T = 1500 #tension in the conductor(kg)\n",
- "w = 1 #weight of the conductor(kg/m)\n",
- "DE = 300 #distance b/w tower's top(m)\n",
- "H = 22 #height of each tower(m)\n",
- "\n",
- "#Calculation:\n",
- "# Suppose, the conductors are supported between towers AD and BE over\n",
- "# a hillside having gradient of 1 : 20.\n",
- "# Let the lowest point on the conductor is O and\n",
- "# sin(theta) = 1/20.\n",
- "# since, the lowest conductor is fixed 2m below the top of each tower.\n",
- "theta = math.asin(1/20)\n",
- "BE = H-2 #Effective height of each tower(m)\n",
- "EC = DE*1/20\n",
- "x1,x2 = symbols('x1 x2')\n",
- "DC = (DE**2-EC**2)**0.5 #Horizontal distance b/w two towers(m)\n",
- "x2 = DC-x1\n",
- "S1 = w*x1**2/(2*T) #m\n",
- "S2 = w*x2**2/(2*T) #m\n",
- "x11 = solve(S2-S1-EC,x1)[0]\n",
- "S2 = w*(DC-x11)**2/(2*T) #m\n",
- "BC = BE+EC\n",
- "OG = BC-S2-x11*math.tan(theta)\n",
- "\n",
- "#Result:\n",
- "print \"Clearance of the lowest point O from the ground is\",round(OG,2),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Clearance of the lowest point O from the ground is 14.4 m\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 8.27, Page Number: 196"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "from sympy import *\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "l = 300 #distance b/w tower(m)\n",
- "S = 10 #span(m)\n",
- "h = 8 #clearance(m)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "#On level ground:\n",
- "r = 8*S/l**2 #ratio w/T\n",
- "\n",
- "\n",
- "#On sloping ground:\n",
- "#The conductors are supported between towers AD and BE \n",
- "#over a sloping ground having a gradient 1 in 15 as shown above.\n",
- "\n",
- "DE = 300\n",
- "sin_theta = 1/15\n",
- "\n",
- "h = DE*sin_theta #Vertical distance b/w the two towers(m)\n",
- "BE = S+h #height of tower(m)\n",
- "x1,x2,x = symbols('x1 x2 x')\n",
- "x2 = l-x1\n",
- "S1 = r*x1**2/2 #m\n",
- "S2 = r*x2**2/2 #m\n",
- "x11 = solve(S2-S1-h,x1)[0]\n",
- "S11 = r*x11**2/(2) #m\n",
- "x22 = l-x11\n",
- "S22 = r*x22**2/(2) #m\n",
- "tan_theta = 1/15\n",
- "GF = x11*tan_theta\n",
- "BC = h-S22-GF\n",
- "#Since O is the origin, the equation of slope of ground is given by :\n",
- "y = x/15-10.5\n",
- "#C = Equation of conductor curve \u2212 y\n",
- "C = r*x**2/2-y\n",
- "C1 = diff(C,x)\n",
- "xm = solve(C1,x)[0]\n",
- "#putting values of xm in above equations,\n",
- "ym = xm/15-10.5\n",
- "Cm = C = r*xm**2/2-ym\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Required minimum clearance is\",round(Cm),\"m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Required minimum clearance is 8.0 m\n"
- ]
- }
- ],
- "prompt_number": 129
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file diff --git a/Principles_of_Power_System/chapter9_1.ipynb b/Principles_of_Power_System/chapter9_1.ipynb deleted file mode 100644 index 925aa356..00000000 --- a/Principles_of_Power_System/chapter9_1.ipynb +++ /dev/null @@ -1,740 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:1c5f89ed0470787a153a235132b85821455bc93d217ca226b9eba94372c90683"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "Chapter 9: Electrical Design of Overhead\n",
- "Lines"
- ]
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 9.1, Page Number: 214"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "d = 200 #Spacing of conductors(cm)\n",
- "r = 1.2/2 #Radius of conductor(cm)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "L = 10**-7*(1+4*math.log(d/r))*10**3 #Loop inductance per m length of the line(H)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \" The loop inductance per km of the line is\",round(L*1000,3),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- " The loop inductance per km of the line is 2.424 mH\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 9.2, Page Number: 214"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "d = 300 #Spacing of conductors(cm)\n",
- "r = 1 #Radius of conductor(cm)\n",
- "u1 = 1 #relative permeability of copper\n",
- "u2 = 100 #relative permeability of steel\n",
- "\n",
- "#Calculation:\n",
- "#(i) With copper conductors,\n",
- "Lc = 10**-7*(u1+4*math.log(d/r))*10**3 #Loop inductance/km\n",
- "\n",
- "#(ii) With steel conductors,\n",
- "Ls = 10**-7*(u2+4*math.log(d/r))*10**3 ##Loop inductance/km\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The loop inductance per km length of the line are:\"\n",
- "print \"(i) for copper, Lc\",round(Lc*1000,2),\"mH and\"\n",
- "print \"(ii) for steel line, Ls =\",round(Ls*1000,2),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The loop inductance per km length of the line are:\n",
- "(i) for copper, Lc 2.38 mH and\n",
- "(ii) for steel line, Ls = 12.28 mH\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 9.3, Page Number: 214"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "d = 200 #cm\n",
- "r = 0.62 #conductor radius(cm)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "L = 10**-7*(0.5+2*math.log(d/r))*10**3 #Inductance/phase/km\n",
- "\n",
- "#Result:\n",
- "print \"The inductance per km of a 3-ph transmission line is\",round(L*10**3,1),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The inductance per km of a 3-ph transmission line is 1.2 mH\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 9.4, Page Number: 214"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "D12 = 2 #m\n",
- "D23 = 2.5 #m\n",
- "D13 = 4.5 #m\n",
- "r = 0.62 #m\n",
- "\n",
- "#Calculation:\n",
- "Deq = (D12*D23*D13)**(1/3)*100 #Equivalent equilateral spacing(cm)\n",
- "L = 10**-7*(0.5+2*math.log(Deq/r))*10**3\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The inductance per km of the line is\",round(L*1000,3),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The inductance per km of the line is 1.274 mH\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 9.5, Page Number: 214"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "r = 1.25 #cm\n",
- "D12 = 2 #m\n",
- "D23 = 2 #m\n",
- "D13 = 4 #m\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "Deq = (D12*D23*D13)**(1/3)*100 #Equivalent equilateral spacing(m)\n",
- "L = 10**-7*(0.5+2*math.log(Deq/r))*10**3\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The inductance per km of the line is\",round(L*1000,2),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The inductance per km of the line is 1.11 mH\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 9.6, Page Number: 214"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "r = 0.5 #radius of conductor(m)\n",
- "Dab = 25 #cm\n",
- "Daa1 = 100 #cm\n",
- "Dbb1 = 100 #cm\n",
- "Dab1 = 103 #cm\n",
- "Da1b = 103 #cm\n",
- "Da1b1 = 25 #cm\n",
- "\n",
- "#Calculation:\n",
- "GMR = 0.7788*r #G.M.R. of conductor(cm)\n",
- "Ds = (GMR*Daa1)**(1/2) #Self G.M.D. of aa1 combination(cm)\n",
- "Dm = (Dab*Dab1*Da1b*Da1b1)**(1/4) #Mutual G.M.D. between a and b(cm)\n",
- "L1 = 2*10**-7*math.log(Dm/Ds) #Loop inductance per conductor per m(H)\n",
- "L = 2*L1*1000\n",
- "\n",
- "#Result:\n",
- "print \"The inductance per km of the line is\",round(L*1000,2),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The inductance per km of the line is 0.84 mH\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 9.7, Page Number: 216"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "\n",
- "#Variable declaration:\n",
- "r = 1.3 #conuctor radius(cm)\n",
- "Dab = 3 #m\n",
- "Dab1 = 6.7 #m\n",
- "Da1b = 6.7 #m\n",
- "Da1b1 = 3 #m\n",
- "Daa1 = 8.48 #m\n",
- "Da1a = 8.48 #m\n",
- "Dbb1 = 6 #m\n",
- "Db1b = 6 #m\n",
- "Dca = 6 #m\n",
- "Dc1a = 6 #m\n",
- "Dca1 = 6 #m\n",
- "Dc1a1 = 6 #m\n",
- "\n",
- "#Calculation:\n",
- "GMR = 1.3*0.7788/100 #m\n",
- "Daa = GMR\n",
- "Da1a1 = GMR\n",
- "Dbb = GMR\n",
- "Db1b1 = GMR\n",
- "Ds1 = (Daa*Daa1*Da1a1*Da1a)**(1/4) #m\n",
- "Ds2 = (Dbb*Dbb1*Db1b1*Db1b)**(1/4) #m\n",
- "Ds = (Ds1*Ds2*Ds1)**(1/3) #m\n",
- "DAB = (Dab*Dab1*Da1b*Da1b1)**(1/4) #m\n",
- "DBC = DAB\n",
- "DCA = (Dca*Dc1a*Dca1*Dc1a1)**(1/4) #m\n",
- "Dm = (DAB*DBC*DCA)**(1/3) #Equivalent mutual G.M.D(m)\n",
- "\n",
- "L1 = 2*10**-7*math.log(Dm/Ds) #Loop inductance per conductor per m(H)\n",
- "L = L1*1000\n",
- "\n",
- "#Result:\n",
- "print \"The inductance per km of the line is\",round(L*1000,2),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The inductance per km of the line is 0.58 mH\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 9.8, Page Number: 217"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable Declaration:\n",
- "r = 0.75 #conductor radius(cm)\n",
- "Db1b = 5.5 #m\n",
- "Dca = 6 #m\n",
- "Dca1 = 4 #m\n",
- "Dc1a = 4 #m\n",
- "Dc1a1 = 6 #m\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "GMR = 0.7788*r #m\n",
- "Dab = (3**2+0.75**2)**0.5 #m\n",
- "Dab1 = (3**2+4.75**2)**0.5 #m\n",
- "Daa1 = (6**2+4**2)**0.5 #m\n",
- "Daa = GMR/100 #m\n",
- "Da1a1 = Daa #m\n",
- "Da1a = Daa1 #m\n",
- "Dbb = GMR/100 #m\n",
- "Dbb1 = 5.5 #m\n",
- "Db1b1 = Dbb #m\n",
- "\n",
- "Ds1 = (Daa*Daa1*Da1a1*Da1a)**(1/4) #m\n",
- "Ds2 = (Dbb*Dbb1*Db1b1*Db1b)**(1/4) #m\n",
- "Ds3 = Ds1\n",
- "Ds = (Ds1*Ds2*Ds3)**(1/3) #Equivalent self G.M.D. of one phase(m)\n",
- "DAB = (Dab*Dab1*Da1b*Da1b1)**(1/4) #m\n",
- "DCA = (Dca*Dca1*Dc1a*Dc1a1)**(1/4) #m\n",
- "DBC = DAB #m\n",
- "Dm = (DAB*DBC*DCA)**(1/3) #Equivalent mutual G.M.D.(m)\n",
- "\n",
- "\n",
- "L1 = 2*10**-7*math.log(Dm/Ds) #Loop inductance per conductor per m(H)\n",
- "L = L1*1000\n",
- "\n",
- "#Result:\n",
- "print \"The inductance per km of the line is\",round(L*1000,3),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The inductance per km of the line is 0.627 mH\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 9.9, Page Number: 218"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable Declaration:\n",
- "r = 5.3/100 #conductor radius(m)\n",
- "\n",
- "#Calculation:\n",
- "GMR = 0.7788*r #m\n",
- "DAA = GMR #m\n",
- "DAA1 = 24 #m\n",
- "DA1A1 = DAA #m\n",
- "DA1A = 24 #m\n",
- "\n",
- "DBB = GMR #m\n",
- "DBB1 = 24 #m\n",
- "DB1B1 = DBB #m\n",
- "DB1B = 24 #m\n",
- "Ds1 = (DAA*DAA1*DA1A1*DA1A)**(1/4) #m\n",
- "Ds2 = (DBB*DBB1*DB1B1*DB1B)**(1/4) #m\n",
- "#Similarly:\n",
- "Ds3 = 0.995 #m\n",
- "Ds = (Ds1*Ds2*Ds3)**(1/3) #Equivalent self-G.M.D. of one phase(m)\n",
- "#DAB = (DAB \u00d7 DAB\u2032 \u00d7 DA\u2032\u0392 \u00d7 DA\u2032B\u2032)**1/4\n",
- "DAB = (8*32*16*8)**(1/4) #m\n",
- "DBC = DAB #m\n",
- "#DCA = (DCA \u00d7 DCA\u2032 \u00d7 DC\u2032A \u00d7DC\u2032A\u2032)1/4 #m\n",
- "DCA = (16*8*40*16)**(1/4) #m\n",
- "\n",
- "Dm = (DAB*DBC*DCA)**(1/3) #m\n",
- "\n",
- "L = 2*10**-7*math.log(Dm/Ds) #Loop inductance per conductor per m(H)\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"The inductance per m of the line is\",round(L*10**7,2),\"* 10**-7 H/m\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The inductance per m of the line is 5.36 * 10**-7 H/m\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 9.10, Page Number: 219"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable Declaration:\n",
- "r = 1 #consuctor radius(cm)\n",
- "\n",
- "#Calculation:\n",
- "#Mutual G.M.D., Dm = (Dab*Dab'*Da'b*Da'b')**(1/4)\n",
- "Dm = (120*140*100*120)**(1/4) #m\n",
- "\n",
- "#Self G.M.D., Ds = (Daa*Daa'*Da'a'*Da'a)**(1/4)\n",
- "#Daa = Da'a' = 0.7788 cm; Daa' = Da'a = 20 cm\n",
- "Ds = (0.7788*20*0.7788*20)**(1/4) #cm\n",
- "L = 4*10**-4*math.log(Dm/Ds) #H/km\n",
- "\n",
- "#Result:\n",
- "print \"The total inductance of the line per km is\",round(L*1000,2),\"mH/km\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The total inductance of the line per km is 1.36 mH/km\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 9.11, Page Number: 224"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "r = 1 #conductor radius(cm)\n",
- "d = 300 #conductor spacing(cm)\n",
- "e = 8.854*10**-12 #permitivity of free space(F/m)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "C = math.pi*e/math.log(d/r) #F/m\n",
- "\n",
- "#Result:\n",
- "print \"Capacitance of the line per km is\",round(C*10**11,4),\"* 10**-2 uF/km\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Capacitance of the line per km is 0.4877 * 10**-2 uF/km\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 9.12, Page Number: 225"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "r = 0.625 #conductor radius(cm)\n",
- "d = 200 #conductor spacing(cm)\n",
- "e = 8.854*10**-12 #permitivity of free space(F/m)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "C = 2*math.pi*e/math.log(d/r) #F/m\n",
- "\n",
- "#Result:\n",
- "print \"Capacitance of the line per km is\",round(C*10**9,4),\" uF/km\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Capacitance of the line per km is 0.0096 uF/km\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 9.13, Page Number: 225"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable Declaration:\n",
- "d1 = 2 #m\n",
- "d2 = 2.5 #m\n",
- "d3 = 4.5 #m\n",
- "r = 0.625 #conductor raius(cm)\n",
- "l = 100 #line length(km)\n",
- "Vl = 66000 #line voltage(V)\n",
- "f = 50 #frequency of current(Hz)\n",
- "e = 8.854*10**-12 #permitivity of free space(F/m)\n",
- "\n",
- "#Calculation:\n",
- "d = (d1*d2*d3)**(1/3)*100 #cm\n",
- "\n",
- "#(i)Line to neutral capacitance,\n",
- "C1 = 2*math.pi*e/math.log(d/r)*10**9 #uF/km\n",
- "C11 = C1*100 #uF\n",
- "\n",
- "#(ii)Charging current per phase,\n",
- "Ic = Vl*2*math.pi*f*C11*10**-6/(3**0.5) #A\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Capacitance per phase is\",round(C11,2),\"uF\"\n",
- "print \"Charging current per phase is\",round(Ic,1),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Capacitance per phase is 0.91 uF\n",
- "Charging current per phase is 10.9 A\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 9.14, Page Number: 225"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable declaration:\n",
- "r = 1 #conductor radius(cm)\n",
- "d = 250 #equilateral conductor spacing(cm)\n",
- "e = 8.854*10**-12 #permitivity of free space(F/m)\n",
- "\n",
- "\n",
- "#Calculation:\n",
- "C = 2*math.pi*e/math.log(d/r)*10**9 #uF/km\n",
- "C1 = C*100 #F\n",
- "\n",
- "#Result:\n",
- "print \"Capacitance of the line per km is\",round(C1,4),\"uF/phase\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Capacitance of the line per km is 1.0075 uF/phase\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 3,
- "metadata": {},
- "source": [
- "Example 9.15, Page Number: 226"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "from __future__ import division\n",
- "import math\n",
- "\n",
- "#Variable Declaration:\n",
- "d1 = 4 #m\n",
- "d2 = 4 #m\n",
- "d3 = 8 #m\n",
- "r = 1 #conductor raius(cm)\n",
- "l = 100 #line length(km)\n",
- "Vl = 132000 #line voltage(V)\n",
- "f = 50 #frequency of current(Hz)\n",
- "e = 8.854*10**-12 #permitivity of free space(F/m)\n",
- "\n",
- "#Calculation:\n",
- "Deq = (d1*d2*d3)**(1/3)*100 #cm\n",
- "C = 2*math.pi*e/math.log(Deq/r)*10**9 #uF/km\n",
- "C1 = C*100 #uF\n",
- "Ic = Vl*2*math.pi*f*C1*10**-6/(3**0.5) #A\n",
- "\n",
- "\n",
- "#Result:\n",
- "print \"Charging current per phase is\",round(Ic,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Charging current per phase is 21.41 A\n"
- ]
- }
- ],
- "prompt_number": 1
- }
- ],
- "metadata": {}
- }
- ]
-}
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