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diff --git a/Principles_of_Power_System/chapter22_1.ipynb b/Principles_of_Power_System/chapter22_1.ipynb new file mode 100644 index 00000000..38d23b90 --- /dev/null +++ b/Principles_of_Power_System/chapter22_1.ipynb @@ -0,0 +1,338 @@ +{
+ "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": {}
+ }
+ ]
+}
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