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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 |