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{
"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": {}
}
]
}
|
