{
"metadata": {
"name": "",
"signature": "sha256:9774a5727e94c3e9d7f6de1d46ee1710f0f45bee5b4624a7cce654515da8e820"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 8: Mechanical Design of Overhead Lines"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.1, Page Number: 171"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#Variable declaration:\n",
"k = 0.11 #ratio of shunt-capatance to self capacitance\n",
"V = 33/math.sqrt(3) #Voltage across string(kV)\n",
"n = 3 #no. of insulators\n",
"\n",
"#Calculation:\n",
"\n",
"# At Junction A\n",
"# I2 = I1 + i1\n",
"# or V2*w*C = V1*w*C + V1*K*\u03c9*C\n",
"# or V2 = V1*(1 + K) = V1*(1 + 0\u00b711)\n",
"# or V2 = 1\u00b711*V1 ...(i)\n",
"\n",
"# At Junction B\n",
"# I3 = I2 + i2\n",
"# or V3*w*C = V2*w*C + (V1 + V2) K*w*C\n",
"# or V3 = V2 + (V1 + V2)*K\n",
"# = 1\u00b711*V1 + (V1 + 1\u00b711*V1)*0\u00b711\n",
"# V3 = 1\u00b7342 V1\n",
"# Voltage across whole unit,\n",
"# V = V1+V2+V3\n",
"# or V = V1+1.11*V1+1.342*V1\n",
"\n",
"V1 = V/3.452 #Voltage across top unit(V)\n",
"V2 = 1.11*V1 #Voltage across top unit(V)\n",
"V3 = 1.342*V1 #Voltage across top unit(V)\n",
"\n",
"e = V/(n*V3)*100 #string efficiency\n",
"\n",
"#Result:\n",
"print \"(i)\\tVoltage across top unit,V1 is\",round(V1,2),\"kV\"\n",
"print \" \\tVoltage across middle unit, V2 is\",round(V2,2),\"kV\"\n",
"print \" \\tVoltage across bottom unit, V3 is\",round(V3,1),\"kV\"\n",
"print \"(ii)\\tThe string efficiency is \",round(e,1),\"%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(i)\tVoltage across top unit,V1 is 5.52 kV\n",
" \tVoltage across middle unit, V2 is 6.13 kV\n",
" \tVoltage across bottom unit, V3 is 7.4 kV\n",
"(ii)\tThe string efficiency is 85.7 %\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.2, Page Number: 172"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variable declaration:\n",
"V1 = 8 #voltage across top unit(kV)\n",
"V2 = 11 #voltage across middle unit(kV)\n",
"n = 3 #no. of insulators\n",
"#Calculation:\n",
"\n",
"#(i)\n",
"# Let K be the ratio of capacitance between pin and the earth\n",
"# to self capacitance. \n",
"# Let C farad be the self capacitance of each unit, \n",
"# then, capacitance between pin and earth = K*C.\n",
"# Applying Kirchoff\u2019s current law to Junction A,\n",
"# I2 = I1 + i1\n",
"# or V2*w*C = V1*w*C + V1*K*w*C\n",
"# or V2 = V1 (1 + K)\n",
"\n",
"k = (V2-V1)/V1\n",
"\n",
"#(ii)\n",
"# Applying Kirchoff\u2019s current law to Junction B,\n",
"# I3 = I2 + i2\n",
"# or V3*w*C = V2*w*C + (V1 + V2)*K*w*C\n",
"# or V3 = V2 + (V1 + V2)*K = 11 + (8 + 11) * 0\u00b7375\n",
"V3 = 18.12 #kV\n",
"V = V1 + V2+ V3 #kV\n",
"Vl = math.sqrt(3)*V #line voltage(kV)\n",
"\n",
"#(iii)\n",
"e = V/(n*V3)*100 #stirng efficiency(%)\n",
"\n",
"\n",
"#Result:\n",
"print \"(i) Ratio of capacitance b/w pin & earth is\",k\n",
"print \"(ii) The line voltage is\",round(Vl,2),\"kV\"\n",
"print \"(iii)String efficiency is\",round(e,2),\"%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(i) Ratio of capacitance b/w pin & earth is 0.375\n",
"(ii) The line voltage is 64.29 kV\n",
"(iii)String efficiency is 68.29 %\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.3, Page Number: 172"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"\n",
"#Variable declaration:\n",
"n = 3 #no. of insulators\n",
"V3 = 17.5 #voltage across the line unit(V)\n",
"\n",
"# At Junction A\n",
"# I2 = I1 + i1\n",
"# V2 \u03c9 C = V1 \u03c9 C + V1 K \u03c9 C\n",
"# or V2 = V1 (1 + K) = V1 (1 + 0.125)\n",
"# \u2234 V2 = 1\u00b7125 V1\n",
"\n",
"# At Junction B\n",
"# I3 = I2 + i2\n",
"# or V3 \u03c9 C = V2 \u03c9 C + (V1 + V2) K \u03c9 C\n",
"# or V3 = V2 + (V1 + V2) K\n",
"# = 1\u00b7125 V1 + (V1 + 1\u00b7125 V1) \u00d7 0.125\n",
"# \u2234 V3 = 1\u00b739 V1\n",
"\n",
"V1 = V3/1.39 #Voltage across top unit(kV)\n",
"V2 = 1.125*V1 #Voltage across middle unit(kV)\n",
"\n",
"V = V1+V2+V3 #voltage across line unit(kV)\n",
"e = V/(n*V3)*100 #string efficiency(%)\n",
"\n",
"\n",
"#Result:\n",
"print \"Line to neutral voltage is\",round(V,2),\"kV\"\n",
"print \"String efficiency is\",round(e,2),\"%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Line to neutral voltage is 44.25 kV\n",
"String efficiency is 84.29 %\n"
]
}
],
"prompt_number": 36
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.4, Page Number: 173"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from pylab import *\n",
"from sympy import *\n",
"\n",
"#Variable declaration:\n",
"V3 = 13.1 #voltage across the lowest insulator(kV)\n",
"V2 = 11 #voltage across the middle insulator(kV)\n",
"K = symbols('K')\n",
"# Applying Kirchhoff\u2019s current law to Junctions A and B, we can easily\n",
"# derive the following equations:\n",
"\n",
"# V2 = V1 (1 + K)\n",
"# or V1 = V2/(1 + K) ...(i)\n",
"# and V3 = V2 + (V1 + V2)*K ...(ii)\n",
"# Putting the value of V1 = V2/(1 + K) in eq. (ii), we get,\n",
"\n",
"#V3 = V2 + (V2/(1+k)+ V2)*K\n",
"K1 = solve(V3-(V2 + (V2/(1+K)+ V2)*K),K)[1]\n",
"\n",
"V1 = V2/(1+K1)\n",
"V = V1+V2+V3 #kV\n",
"\n",
"#Result:\n",
"print \"The voltage b/w the bus bars is\",round(float(math.sqrt(3)*V)),\"kV\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The voltage b/w the bus bars is 59.0 kV\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.5, Page Number: 173"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"\n",
"\n",
"#Variable declaration:\n",
"n = 3 #no.of insulators\n",
"k = 1/8 #ratio of self-capacitance to shunt capacitance\n",
"V3 = 15 #safe working voltage of each insulator(kV)\n",
"\n",
"# Applying Kirchhoff\u2019s current law to junction A, we get,\n",
"# V2 = V1*(1 + K)\n",
"# or V1 = V2/(1 + K) = V2/(1 + 0\u00b7125) = 0\u00b789*V2\n",
"\n",
"# Applying Kirchhoff\u2019s current law to Junction B, we get,\n",
"# V3 = V2 + (V1 + V2)*K = V2 + (0.89*V2 + V2) * 0.125\n",
"# V3 = 1.236*V2 \n",
"V2 = V3/1.236 #kV\n",
"V1 = 0.89*V2 #kV\n",
"V = V1+V2+V3 #voltage across the string(kV)\n",
"e = V/(n*V3) *100 #string efficiency(%)\n",
"\n",
"#Result:\n",
"print \"Voltage across the string is\",round(V,3),\"kV\"\n",
"print \"The string efficiency is\",round(e,2),\"%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Voltage across the string is 37.937 kV\n",
"The string efficiency is 84.3 %\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.6, Page Number: 174"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"\n",
"#Variable declaration:\n",
"n = 4 #no. of unit\n",
"\n",
"\n",
"#Calculation:\n",
"# Suppose Xc = 1 \u03a9. Given, the ratio of self-capacitance to shunt\n",
"# capacitance is 10, so, Xc1 = 10 \u03a9 .\n",
"# Suppose that potential V across the string is such that 1 A current \n",
"# flows in the top insulator.\n",
"#Thus,\n",
"V1 = 1*1 #Voltage across top unit(V)\n",
"V2 = 1*1.1 #Voltage across 2nd unit(V)\n",
"V3 = 1*1.31 #Voltage across 3rd unit(V)\n",
"V4 = 1*1.65 #Voltage across 4th unit(V)\n",
"V = V1+V2+V3+V4 #Voltage obtained across the string(V)\n",
"\n",
"#The voltage across each unit expressed as a % of V becomes:\n",
"f1 = V1/V*100\n",
"f2 = V2/V*100\n",
"f3 = V3/V*100\n",
"f4 = V4/V*100\n",
"\n",
"e = V/(n*V4)*100 #string efficiency(%)\n",
"\n",
"#Result:\n",
"print \"(i) The voltage distributation from top unit is \"\n",
"print \"\\t\",round(f1,2),\"%\\t\",round(f2,2),\"%\\t\",round(f3,1),\"%\\t\",round(f4,1),\"%\"\n",
"print \"\\n(ii)String efficiency is\",round(e,1),\"%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(i) The voltage distributation from top unit is \n",
"\t19.76 %\t21.74 %\t25.9 %\t32.6 %\n",
"\n",
"(ii)String efficiency is 76.7 %\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.7, Page Number: 175"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variable declaration:\n",
"n = 5 #no.os insulators\n",
"Vl = 100 #line voltage(kV)\n",
"\n",
"#Suppose Xc = 1 \u03a9.\n",
"#Ratio of self capacitance to shunt capacitance is 10, so, Xc1 = 10 \u03a9. \n",
"#Let V be the voltage s.t 1A current flows in top insulator.\n",
"\n",
"#Calculation:\n",
"#calculation for fraction of voltage drop:\n",
"v1 = 1*1 #Voltage across top unit(V)\n",
"v2 = 1*1.1 #Voltage across 2nd unit(V)\n",
"v3 = 1*1.31 #Voltage across 3rd unit(V)\n",
"v4 = 1*1.65 #Voltage across 4th unit(V)\n",
"v5 = 1*2.16 #voltage across 5th unit(V)\n",
"v = v1+v2+v3+v4+v5 #total voltage(V)\n",
"V = 100/math.sqrt(3) #string voltage(V)\n",
"V1 = round(v1/v*V,3) #kV\n",
"V2 = v2/v*V #kV\n",
"V3 = v3/v*V #kV\n",
"V4 = v4/v*V #kV\n",
"V5 = v5/v*V #kV\n",
"e = V/(n*V5)*100 #string efficiency(%)\n",
"\n",
"\n",
"#Result:\n",
"print \"(i)The distribution of voltage on the insulator discs are\"\n",
"print \"V1 = \",round(V1,3),\"kV \",\"\\tV2 =\",round(V2,2),\"kV \\n\",\n",
"print \"V3 =\",round(V3,1),\"kV\",\"\\t\\tV4 =\",round(V4,2),\"kV\",\n",
"print \"\\nV5 =\",round(V5,1),\"kV\"\n",
"print \"\\n(ii)The string efficiency\",round(e,1),\"%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(i)The distribution of voltage on the insulator discs are\n",
"V1 = 7.997 kV \tV2 = 8.8 kV \n",
"V3 = 10.5 kV \t\tV4 = 13.19 kV \n",
"V5 = 17.3 kV\n",
"\n",
"(ii)The string efficiency 66.9 %\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.8, Page Number: 176"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"from sympy import *\n",
"\n",
"#Variable declaration:\n",
"n = 4 #no.os insulators\n",
"V2 = 13.2 #Voltage across 2th unit\n",
"V3 = 18 #Voltage across 3th unit\n",
"#Suppose K is the ratio of shunt-capacitance to self-capacitance\n",
"k = symbols('k')\n",
"\n",
"# Referring to Fig.(ii), we have,\n",
"# V2/V1 = ( 1 + K)/1\n",
"# V2 = V1 (1 + K)\n",
"# V3/V1 = (1 + 3K + K2)/1\n",
"# V3 = V1 (1 + 3K + K2)\n",
"# putting values of V2 & V3\n",
"\n",
"k1 = solve(13.2*k**2+21.6*k-4.8,k)[1]\n",
"V1 = V2/(1+k1)\n",
"V4 = V1*(1 + k1**33 + 5*k1**2 + 6*k1)\n",
"V = V1+V2+V3+V4 #Voltage between line and earth(kV)\n",
"Vl = math.sqrt(3)*V #Voltage between conductors(kV)\n",
"\n",
"#Result:\n",
"print \"Voltage between conductors is\",round(float(Vl)),\"kV\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Voltage between conductors is 119.0 kV\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.9, Page Number: 177"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"\n",
"\n",
"#Variable declaration:\n",
"k = 5 #ratio of self-capacitance(C) to pin-earth capacitance(C1)\n",
"V = 1+1.2+1.64+2.408 #voltage obtained across the string(V)\n",
"\n",
"\n",
"#Calculation:\n",
"#(i) The voltage across each unit expressed as a % of V\n",
"v1 = 1/V*100\n",
"v2 = 1.2/V*100\n",
"v3 = 1.6/V*100\n",
"v4 = 2.408/V*100\n",
"\n",
"#(ii)String efficiency\n",
"e = V/(4*2.408)*100\n",
"\n",
"#Result:\n",
"print \"(i) The voltage across each unit expressed as a % of V are:\"\n",
"print \"\\tTop Unit: \",round(v1),\"%\"\n",
"print \"\\tSecond from top: \",round(v2,1),\"%\"\n",
"print \"\\tThird from top: \",round(v3),\"%\"\n",
"print \"\\tFourth from top: \",round(v4,1),\"%\"\n",
"print \"(ii)String efficiency is\",round(e,2),\"%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(i) The voltage across each unit expressed as a % of V are:\n",
"\tTop Unit: 16.0 %\n",
"\tSecond from top: 19.2 %\n",
"\tThird from top: 26.0 %\n",
"\tFourth from top: 38.5 %\n",
"(ii)String efficiency is 64.87 %\n"
]
}
],
"prompt_number": 54
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.10, Page Number: 177"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from sympy import *\n",
"\n",
"#Variable declaration:\n",
"V1,V2,V3,V= symbols('V1 V2 V3 V') #volts (shown in fig(ii))\n",
"\n",
"\n",
"#Calculation:\n",
"\n",
"#We know that in an actual string of insulators, 3 capacitances\n",
"#exist i.e, self-capacitance of each insulator, shunt capacitance \n",
"#and capacitance of each unit to line.\n",
"#But, capacitance of each unit to line is very small and\n",
"#can be neglected.\n",
"\n",
"#At Junction A\n",
"# I2 + i1 = I1 + i1\n",
"#or V2*w*C + (V2 + V3)*0\u00b71*w*C = V1*\u03c9*C + 0\u00b715*C*V1*w\n",
"#or 0\u00b71*V3 = 1\u00b715*V1 \u2212 1\u00b71*V2\n",
"# V3 = 11\u00b75*V1 \u2212 11*V2 ...(i)\n",
"\n",
"#At Junction B\n",
"# I3 + i\u20322 = I2 + i2\n",
"#or V3*w*C + V3*0\u00b71*C*w = V2*w*C + (V1 + V2)*w*0\u00b715 C\n",
"#or 1\u00b71*V3 = 1\u00b715*V2 + 0\u00b715*V1 ...(ii)\n",
"#Putting the value of V3 from exp (i). into exp. (ii), we get,\n",
"# 1\u00b71*(11\u00b75*V1 \u2212 11*V2) = 1\u00b715*V2 + 0\u00b715*V1\n",
"#or 13\u00b725*V2 = 12\u00b75*V1\n",
"# V2 = 12.5/13.25*V1\n",
"# V3 = 11.5*V1-11*V2\n",
"# V = V1+V2+V3\n",
"# V = 40.55/13.25*V1\n",
"V1 = 13.25/40.55*V\n",
"V2 = 12.5/13.25*V1\n",
"V3 = 14.8/13.25*V1\n",
"\n",
"#The voltage across each unit expressed as a % of V becomes:\n",
"p1 = V1/V*100\n",
"p2 = V2/V*100\n",
"p3 = V3/V*100\n",
"\n",
"\n",
"#Results:\n",
"print \"The voltage across each unit expressed as a % of V becomes:\"\n",
"print \"for top unit, v1 =\",round(p1,1),\"%\",\n",
"print \"\\nfor middle unit, v2 =\",round(p2,1),\"%\"\n",
"print \"for 3rd unit, v3 =\",round(p3,1),\"%\"\n",
"\n",
"print \"String efficiency is\",round(V/(3*V3)*100,1),\"%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The voltage across each unit expressed as a % of V becomes:\n",
"for top unit, v1 = 32.7 % \n",
"for middle unit, v2 = 30.8 %\n",
"for 3rd unit, v3 = 36.5 %\n",
"String efficiency is 91.3 %\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.11, Page Number: 179"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from sympy import *\n",
"\n",
"#Variable Declaration:\n",
"V1,V2,V3,V= symbols('V1 V2 V3 V') #volts (shown in fig(i))\n",
"\n",
"#Calculation:\n",
"\n",
"#At Junction A\n",
"# I2 + i\u20321 = I1 + i1\n",
"#or V2*w*C + (V2+V3)*w*0\u00b71*C = V1*w*C+V1*0\u00b72*C*\u03c9 \n",
"# V3 = 12 V1 \u2212 11 V2 # ...(i)\n",
"\n",
"#At Junction B\n",
"# I3 + i\u20322 = I2 + i2\n",
"#or V3 \u03c9 C + V3 \u00d7 0\u00b73 C \u00d7 \u03c9 =V2 \u03c9 C + (V1 + V2) \u03c9 \u00d7 0\u00b72 C\n",
"#or 1\u00b73 V3 = 1\u00b72 V2 + 0\u00b72 V1 ...(ii)\n",
"#Substituting the value of V3 from exp. (i) into exp. (ii), we get,\n",
"# 1\u00b73*(12*V1 \u2212 11*V2) = 1\u00b72*V2 + 0\u00b72*V1\n",
"#or 15\u00b75*V2 = 15\u00b74*V1\n",
"V2 = 15.4*V1/15.5 #...(iii)\n",
"V2 = 0.993*V1\n",
"\n",
"#Substituting the value of V2 from exp. (iii) into exp. (i), we get,\n",
"V3 = 12*V1 - 11*0.993*V1\n",
"V3 = 1.077*V1\n",
"\n",
"#Voltage between conductor and earth (i.e. phase voltage)\n",
"V = V1+V2+V3\n",
"n = V/(3*V3)*100\n",
"\n",
"#Result:\n",
"print \"String efficiency is\",round(n),\"%\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"String efficiency is 95.0 %\n"
]
}
],
"prompt_number": 55
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.13, Page Number: 184"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from pylab import *\n",
"import math\n",
"\n",
"#Variable declaration:\n",
"r = 1 #conductor radius(cm)\n",
"d = 100 #conductor spacing(cm)\n",
"go = 30/math.sqrt(2) #Dielectric strength of air,(rms)(kV/cm)\n",
"dl = 0.952 #air density factor\n",
"mo = 0.9 #irregularity factor\n",
"\n",
"#Calculation:\n",
"Vc = mo*go*dl*r*math.log(d/r) #kV/phase\n",
"Vl = Vc*3**0.5 #kV/line\n",
"\n",
"#Result:\n",
"print \"The disruptive critical voltage for the line is\",round(float(Vl),2),\"kV\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The disruptive critical voltage for the line is 144.97 kV\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.14, Page Number: 184"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from pylab import *\n",
"\n",
"#Variable declaration:\n",
"r = 0.978 #conductor radius(cm)\n",
"go = round(30/2**0.5,1) #Dielectric strength of air,rms(kV/cm)\n",
"Vc = 210/3**0.5 #Disruptive voltage/phase(kV)\n",
"mo = 1 #irregularity factor(for smooth conductor)\n",
"dl = 1 #air density factor(at std pressure and temperature)\n",
"\n",
"#Calculation:\n",
"d = r*10**(Vc/(2.3*mo*go*dl*r))\n",
"\n",
"#Result:\n",
"print \"The spacing between the conductors is\",round(d),\"cm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The spacing between the conductors is 341.0 cm\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.15, Page Number: 185"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from pylab import *\n",
"import math\n",
"\n",
"#Variable declaration:\n",
"r = 1.5 #conductor radius(cm)\n",
"t = 40 #temperature(deg C)\n",
"b = 76 #atmospheric pressure(cm)\n",
"mo = 0.85 #irregularity factor\n",
"f = 50 #Hz\n",
"d = 200 #conductor spacing(cm)\n",
"go = 30/2**0.5 #Dielectric strength of air,rms(kV/cm)\n",
"Vl = 220 #line voltage(kV)\n",
"\n",
"#Calculation:\n",
"dl = round(3.92*b/(273+t),3)\n",
"\n",
"Vc = round(mo*go*dl*r*round(math.log(d/r),2),1) #Critical disruptive voltage per phase(kV)\n",
"V = round(Vl/3**0.5) #kV/phase\n",
"P = 242.2/dl*(f+25)*(r/d)**0.5*(V-Vc)**2*10**-5 #kW/km/phase\n",
"Pl = 3*P #corona loss(kW)\n",
"\n",
"#Result:\n",
"print \"Total corona loss per km for three phases is\",round(Pl,5),\"kW\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Total corona loss per km for three phases is 0.05998 kW\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.16, Page Number: 185"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from sympy import *\n",
"\n",
"#Variable declaration:\n",
"P1 = 53 #corona loss(kW)\n",
"V1 = 106 #operating voltage(kV)\n",
"P2 = 98 #corona loss(kW)\n",
"V2 = 110.9 #operating voltage(kV)\n",
"V3 = 113 #opeating voltage(kV)\n",
"\n",
"\n",
"#Calculation:\n",
"#As f, \u03b4, r and d are the same for the two cases,\n",
"#therfore, P \u221d (V \u2212 Vc)**2\n",
"k,Vc = symbols('k Vc') #k = proportionality constant\n",
" #Vc = critical disuptive voltage(kV)\n",
"#for 1st case,\n",
"# V1/3**0.5 = k*(64 - Vc)**2\n",
"# V2/3**0.5 = k*(61\u00b72 - Vc)**2\n",
"# Dividing above equations & solving,\n",
"Vc1 = round(solve(((64-Vc)/(61.2-Vc))-math.sqrt(98/53),Vc)[0],2)\n",
"Vc1 = 54\n",
"#Let W kilowatt be the power loss at 113 kV.\n",
"#W = k*(113/3**0.5-Vc1)**2\n",
"W = P1*((65.2-Vc1)/(61.2-Vc1))**2\n",
"\n",
"\n",
"\n",
"#Result:\n",
"print \"The disruptive critical voltage at 113 kV is\",Vc1,\"kV\"\n",
"print \"The corona loss at 113 kV is\",round(W),\"kW\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The disruptive critical voltage at 113 kV is 54 kV\n",
"The corona loss at 113 kV is 128.0 kW\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.17, Page Number: 190"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"\n",
"#Variable declaration:\n",
"w = 680 #weigth of conductor(kg/km)\n",
"l = 260 #length ofspan(m)\n",
"u = 3100 #ultimate strength(kg)\n",
"sf = 2 #safety factor\n",
"h = 10 #ground clearance(m)\n",
"\n",
"\n",
"#Calculation:\n",
"T = u/sf #working tension(kg)\n",
"s = w*l**2/(8*T*1000) #span(m)\n",
"H = h+s\n",
"\n",
"\n",
"#Result:\n",
"print \"Conductor should be supported at a height of \",round(H,1),\"m\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Conductor should be supported at a height of 13.7 m\n"
]
}
],
"prompt_number": 56
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.18, Page Number: 190"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variable declaration:\n",
"l = 150 #length ofspan(m)\n",
"a = 2 #conductor cross-section(cm**2)\n",
"go = 9.9 #specific gravity of conductor(gm/cm**3)\n",
"ww = 1.5 #Wind force/m length of conductor(kg)\n",
"T = 2000 #working tension(kg)\n",
"\n",
"#Calculation:\n",
"w = go*a*1/10 #Wt. of conductor/m length(kg)\n",
"wt = round((w**2+ww**2)**0.5,2) #Total wt of 1m length of conductor(kg)\n",
"s = wt*l**2/(8*T) #sag(m)\n",
"#This is the value of slant sag in a direction making \n",
"#an angle theta with the vertical\n",
"\n",
"theta = math.atan(ww/w)\n",
"Vs = s*math.cos(theta)\n",
"\n",
"#Result:\n",
"print \"The sag is\",round(s,2),\"m\"\n",
"print \"The vertical sag is\",round(Vs,2),\"m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The sag is 3.49 m\n",
"The vertical sag is 2.78 m\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.19, Page Number: 191"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variable declaration:\n",
"l = 200 #span length(m)\n",
"w = 1170/1000 #weigth of conductor(kg)\n",
"u = 4218 #Ultimate Strength(kg/cm**2)\n",
"sf = 5 #safety factor\n",
"a = 1.29 #cross-section of conductor(cm**2)\n",
"P = 122 #wind pressure(kg/m**2)\n",
"\n",
"#Calculation:\n",
"T = u/sf*a #working tension(kg)\n",
"d = round((4*a/math.pi)**0.5,2) #diameter of conductor(cm)\n",
"ww = P*d*10**-2 #Wind force/m length,\n",
"wt = (ww**2+w**2)**0.5 #total wt of conductor per metre length\n",
"S = wt*l**2/(8*T)\n",
"theta = math.atan(ww/w)\n",
"VS = S*math.cos(theta)\n",
"\n",
"#Result:\n",
"print \"The sag is\",round(S,2),\"m\"\n",
"print \"The vertical sag is\",round(VS,2),\"m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The sag is 8.97 m\n",
"The vertical sag is 5.38 m\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.20, Page Number: 191"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"\n",
"#Variable declaration:\n",
"l = 275 #span length(m)\n",
"w = 0.865 #weigth of conductor/m(kg)\n",
"d = 1.96 #conductor diameter(cm)\n",
"t = 1.27 #ice coating thickness(cm)\n",
"u = 8060 #Ultimate Strength(kg)\n",
"sf = 2 #safety factor\n",
"a = 1.29 #cross-section of conductor(cm**2)\n",
"wo = 0.91 #Weight of 1 c.c. of ice(gm)\n",
"P = 3.9 #wind pressure(gm/cm**2)\n",
"\n",
"#Calculation:\n",
"T = u/sf #working tension(kg)\n",
"v = 3.14*t*(d+t)*100 #Volume of ice per metre(cm**3)\n",
"wi = v*wo/1000 #kg\n",
"ww = P*(d+2*t)*100/1000 #kg\n",
"wt = ((w+wi)**2+ww**2)**0.5 #total weight(kg)\n",
"S = wt*l**2/(8*T) #sag(m)\n",
"\n",
"\n",
"#Result:\n",
"print \"Sag is\",round(S,1),\"m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Sag is 6.3 m\n"
]
}
],
"prompt_number": 60
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.21, Page Number: 192"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from sympy import *\n",
"\n",
"#Variable declaration:\n",
"Sv = 2.35 #vertical sag(m)\n",
"P = 1.5 #wind pressure(kg/m)\n",
"B = 2540 #breaking stress(kg/cm**2)\n",
"w = 1.125 #wt of conductor(kg/m)\n",
"l = 214 #length of conductor(m)\n",
"a = 3.225 #conductor cross-section\n",
"\n",
"\n",
"#Calculation:\n",
"f = symbols('f') #safety factor\n",
"wt = (w**2+P**2)**0.5 #Total wt. of one m length of conductor\n",
"T = B*a/f #working stress(kg)\n",
"cos_theta = w/wt\n",
"S = Sv/cos_theta #slant sag\n",
"T1 = wt*l**2/(8*S) \n",
"f1 = solve(T-T1,f)[0] #safety factor\n",
"\n",
"\n",
"#Result:\n",
"print \"Safety factor is\",round(f1)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Safety factor is 3.0\n"
]
}
],
"prompt_number": 61
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.22, Page Number: 192"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"\n",
"#Variable declaration:\n",
"l = 150 #span length(m)\n",
"u = 5000 #ultimate strength(kg/cm**2)\n",
"go = 8.9 #specific gravity of material(gm/cc)\n",
"P = 1.5 #wind pressure(kg/m)\n",
"sf = 5 #safety factor\n",
"a = 2 #cross-section of conductor(cm**2)\n",
"h = 7 #ground clearance(m)\n",
"\n",
"#Calculation:\n",
"w = a*go*100/1000 #weight of conductor per m(kg)\n",
"T = u*a/sf #working stress per m(kg)\n",
"wt = (w**2+P**2)**0.5 #Total wt of 1 m length of conductor(kg)\n",
"S = wt*l**2/(8*T) #slant sag(m)\n",
"Sv = S*w/wt #conductor vertical sag(m)\n",
"H = h+Sv #m\n",
"\n",
"#Result:\n",
"print \"Conductor should be supported at a height of\",round(H,1),\"m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Conductor should be supported at a height of 9.5 m\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.23, Page Number: 193"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from sympy import *\n",
"\n",
"#Variable declaration:\n",
"l = 500 #distance b/w the two towers(m)\n",
"T = 1600 #tension in the conductor(kg)\n",
"w = 1.5 #weight of the conductor(kg/m)\n",
"h1 = 30 #height of 1st tower(m)\n",
"h2 = 90 #height of 2nd tower(m)\n",
"\n",
"\n",
"\n",
"#Calculation:\n",
"h = h2-h1 #difference in levels(m)\n",
"x1,x2 = symbols('x1 x2')\n",
"x2 = l-x1\n",
"S1 = w*x1**2/(2*T) #m\n",
"S2 = w*x2**2/(2*T) #m\n",
"x11 = solve(S2-S1-h,x1)[0]\n",
"S1 = w*x11**2/(2*T) #m\n",
"#Let the mid-point P be at a distance x from the lowest point O\n",
"x = l/2-x11\n",
"Smid = w*x**2/(2*T) #m\n",
"\n",
"\n",
"#Result:\n",
"print \"The min. clearance of the conductor and water is \",round(h1-S1),\"m\"\n",
"print \"The clearance mid-way b/w the supports is\",round(Smid+h1-S1,2),\"m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The min. clearance of the conductor and water is 23.0 m\n",
"The clearance mid-way b/w the supports is 30.7 m\n"
]
}
],
"prompt_number": 62
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.24, Page Number: 194"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from sympy import *\n",
"\n",
"#Variable declaration:\n",
"l = 600 #distance b/w the two towers(m)\n",
"f = 5 #safety factor\n",
"U = 8000 #kg/cm**2\n",
"a = 2.2 #area of cross-section(cm**2)\n",
"w = 1.925 #weight of the conductor(kg/m)\n",
"h = 15 #difference in towers levels(m)\n",
"wi = 1 #ice weight(kg)\n",
"\n",
"#Calculation:\n",
"T = U*a/f #tension in the conductor(kg)\n",
"wt = wi+w #total weight of conductor(kg)\n",
"x1,x2 = symbols('x1 x2')\n",
"x2 = l-x1\n",
"S1 = wt*x1**2/(2*T) #m\n",
"S2 = wt*x2**2/(2*T) #m\n",
"x11 = solve(S2-S1-h,x1)[0]\n",
"x22 = l-x11\n",
"S22 = wt*x22**2/(2*T) #m\n",
"\n",
"\n",
"#Result:\n",
"print \"The sag from the taller of the two supports is\",round(S22,2),\"m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The sag from the taller of the two supports is 45.27 m\n"
]
}
],
"prompt_number": 63
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.25, Page Number: 195"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from sympy import *\n",
"\n",
"#Variable declaration:\n",
"l = 400 #distance b/w the two towers(m)\n",
"T = 2000 #tension in the conductor(kg)\n",
"w = 1 #weight of the conductor(kg/m)\n",
"h1 = 40 #height of 1st tower(m)\n",
"h2 = 90 #height of 2nd tower(m)\n",
"\n",
"#Calculation:\n",
"h = h2-h1 #difference in levels(m)\n",
"x1,x2 = symbols('x1 x2')\n",
"x2 = l-x1\n",
"S1 = w*x1**2/(2*T) #m\n",
"S2 = w*x2**2/(2*T) #m\n",
"x11 = solve(S2-S1-h,x1)[0]\n",
"S1 = w*x11**2/(2*T) #m\n",
"x_mid = 50+400/2 #Horizontal distance of mid-point P from lowest point O\n",
"Sp = w*x_mid**2/(2*T) #SAg at P(m)\n",
"S2 = w*(l-x11)**2/(2*T) #m\n",
"hc = h2-S2+Sp #Clearance of mid-point P above water level(m)\n",
"\n",
"\n",
"\n",
"#Result:\n",
"print \"Clearance of mid-point P above water level is\",round(hc),\"m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Clearance of mid-point P above water level is 55.0 m\n"
]
}
],
"prompt_number": 127
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.26, Page Number: 195"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from pylab import *\n",
"from sympy import *\n",
"import math\n",
"\n",
"#Variable declaration:\n",
"T = 1500 #tension in the conductor(kg)\n",
"w = 1 #weight of the conductor(kg/m)\n",
"DE = 300 #distance b/w tower's top(m)\n",
"H = 22 #height of each tower(m)\n",
"\n",
"#Calculation:\n",
"# Suppose, the conductors are supported between towers AD and BE over\n",
"# a hillside having gradient of 1 : 20.\n",
"# Let the lowest point on the conductor is O and\n",
"# sin(theta) = 1/20.\n",
"# since, the lowest conductor is fixed 2m below the top of each tower.\n",
"theta = math.asin(1/20)\n",
"BE = H-2 #Effective height of each tower(m)\n",
"EC = DE*1/20\n",
"x1,x2 = symbols('x1 x2')\n",
"DC = (DE**2-EC**2)**0.5 #Horizontal distance b/w two towers(m)\n",
"x2 = DC-x1\n",
"S1 = w*x1**2/(2*T) #m\n",
"S2 = w*x2**2/(2*T) #m\n",
"x11 = solve(S2-S1-EC,x1)[0]\n",
"S2 = w*(DC-x11)**2/(2*T) #m\n",
"BC = BE+EC\n",
"OG = BC-S2-x11*math.tan(theta)\n",
"\n",
"#Result:\n",
"print \"Clearance of the lowest point O from the ground is\",round(OG,2),\"m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Clearance of the lowest point O from the ground is 14.4 m\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 8.27, Page Number: 196"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"from sympy import *\n",
"\n",
"\n",
"#Variable declaration:\n",
"l = 300 #distance b/w tower(m)\n",
"S = 10 #span(m)\n",
"h = 8 #clearance(m)\n",
"\n",
"\n",
"#Calculation:\n",
"#On level ground:\n",
"r = 8*S/l**2 #ratio w/T\n",
"\n",
"\n",
"#On sloping ground:\n",
"#The conductors are supported between towers AD and BE \n",
"#over a sloping ground having a gradient 1 in 15 as shown above.\n",
"\n",
"DE = 300\n",
"sin_theta = 1/15\n",
"\n",
"h = DE*sin_theta #Vertical distance b/w the two towers(m)\n",
"BE = S+h #height of tower(m)\n",
"x1,x2,x = symbols('x1 x2 x')\n",
"x2 = l-x1\n",
"S1 = r*x1**2/2 #m\n",
"S2 = r*x2**2/2 #m\n",
"x11 = solve(S2-S1-h,x1)[0]\n",
"S11 = r*x11**2/(2) #m\n",
"x22 = l-x11\n",
"S22 = r*x22**2/(2) #m\n",
"tan_theta = 1/15\n",
"GF = x11*tan_theta\n",
"BC = h-S22-GF\n",
"#Since O is the origin, the equation of slope of ground is given by :\n",
"y = x/15-10.5\n",
"#C = Equation of conductor curve \u2212 y\n",
"C = r*x**2/2-y\n",
"C1 = diff(C,x)\n",
"xm = solve(C1,x)[0]\n",
"#putting values of xm in above equations,\n",
"ym = xm/15-10.5\n",
"Cm = C = r*xm**2/2-ym\n",
"\n",
"\n",
"#Result:\n",
"print \"Required minimum clearance is\",round(Cm),\"m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Required minimum clearance is 8.0 m\n"
]
}
],
"prompt_number": 129
}
],
"metadata": {}
}
]
}