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diff --git a/Principles_of_Power_System/chapter17.ipynb b/Principles_of_Power_System/chapter17.ipynb new file mode 100755 index 00000000..f0b77b2e --- /dev/null +++ b/Principles_of_Power_System/chapter17.ipynb @@ -0,0 +1,842 @@ +{
+ "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": {}
+ }
+ ]
+}
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