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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 06 : Corona"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.1, Page No 142"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"t=21.0 # air temperature\n",
"b=73.6 # air pressure\n",
"do=3.92*73.6/(273+t)\n",
"m=0.85\n",
"r=0.52\n",
"d=250.0\n",
"\n",
"#Calculations\n",
"Vd=21.1*m *do*r*math.log(250/.52)\n",
"vd=math.sqrt(3)*Vd\n",
"m=0.7\n",
"vv=21.1*m*do*r*(1+(0.3/math.sqrt(r*do)))*math.log(250/0.52)\n",
"Vv=vv*math.sqrt(3)\n",
"Vvg=Vv*0.8/0.7\n",
"\n",
"#Results\n",
"print(\"critical disruptive line to line voltage=%.2f kV \" %vd)\n",
"print(\"visual critical voltage for local corona=%.2f kV \" %vv)\n",
"print(\"visual critical voltage for general corona=%.2f kV \" %Vvg)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"critical disruptive line to line voltage=97.89 kV \n",
"visual critical voltage for local corona=66.09 kV \n",
"visual critical voltage for general corona=130.83 kV \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.2, Page No 142"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables \n",
"d=2.5 \n",
"di=3.0 # internal diameter\n",
"do=9.0 # external diameter\n",
"ri=di/2.0 # internal radius\n",
"ro=do/2.0 # external diameter\n",
"\n",
"#Calculations\n",
"g1max=20/(1.25*math.log(ri/(d/2))+0.208*1.5*math.log(ro/ri)) \n",
"\n",
"#Results\n",
"print(\"g1max=%.0f kV/cm\" %g1max) \n",
"print(\"Since the gradient exceeds 21.1/kV/cm , corona will be present.\")"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"g1max=35 kV/cm\n",
"Since the gradient exceeds 21.1/kV/cm , corona will be present.\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.3, Page No 143"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"m=1.07\n",
"r=0.625\n",
"\n",
"#Calculations\n",
"V=21*m *r*math.log(305.0/0.625)\n",
"Vl=V*math.sqrt(3.0)\n",
"\n",
"#Results\n",
"print(\"critical disruptive voltage=%.0f kV \" %V)\n",
"print(\"since operating voltage is 110 kV , corona loss= 0 \")"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"critical disruptive voltage=87 kV \n",
"since operating voltage is 110 kV , corona loss= 0 \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.4 Page No 143"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"r=0.5\n",
"\n",
"#Calculations\n",
"V=21*r*math.log(100.0/0.5)\n",
"\n",
"#Results\n",
"print(\"critical disruptive voltage=%.1f kV\" %V)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"critical disruptive voltage=55.6 kV\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.5, Page No 146"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"D=1.036 \t # conductor diameter(cm)\n",
"d=2.44\t #delta spacing(m)\n",
"r=D/2 \t #radius(cm)\n",
"\n",
"#Calculations\n",
"Ratio=d*100.0/r \n",
"j=r/(d*100.0) \n",
"Rat2=math.sqrt(j) \n",
"t=26.67 \t #temperature\n",
"b=73.15 # barometric pressure\n",
"mv=0.72 \n",
"V=63.5 \n",
"f=50.0\t #frequency\n",
"do=3.92*b/(273+t) #do=dell\n",
"vd=21.1*.85*do*r*math.log(Ratio) \n",
"print(\"critical disruptive voltage=%.2f kV\" %vd) \n",
"Vv=21.1*mv*do*r*(1+(0.3/math.sqrt(r*do)))*math.log(Ratio) \n",
"Pl=241*(10**-5)*(f+25)*Rat2*((V-vd)**2)/do #power loss\n",
"Vd=0.8*vd \n",
"Pl2=241*(10**-5)*(f+25)*Rat2*((V-Vd)**2)*160/do #loss per phase /km\n",
"Total=3.0*Pl2\n",
"\n",
"#Results \n",
"print(\"visual critical voltage=%.0f kV\" %Vv) \n",
"print(\"Power loss=%.3f kW/phase/km\" %Pl) \n",
"print(\"under foul weather condition ,\") \n",
"print(\"critical disruptive voltage=%.2f kV\" %Vd) \n",
"print(\"Total loss=%.0f kW\\n\"%Total) \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"critical disruptive voltage=54.72 kV\n",
"visual critical voltage=66 kV\n",
"Power loss=0.672 kW/phase/km\n",
"under foul weather condition ,\n",
"critical disruptive voltage=43.77 kV\n",
"Total loss=1626 kW\n",
"\n"
]
}
],
"prompt_number": 5
}
],
"metadata": {}
}
]
}
|
