{
"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": {}
}
]
}