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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 12 : Thyristors"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 12.1, Page No 278"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"L=2*(10**-7)*math.log(100/.75)\t\t#inductance per unit length\n",
"C=2*math.pi*(10**-9)/(36*math.pi*math.log(100.0/0.75))\t#Capacitance per phase per unit length (F/m)\n",
"Z1=math.sqrt(L/C)\n",
"E=11000.0\n",
"\n",
"#Calculations\n",
"print(\"(i)the natural impedence of line=%.0f ohms\" %Z1)\n",
"Il=E/(math.sqrt(3)*Z1) #line current(amps)\n",
"print(\"(ii)line current =%.1f amps\" %Il)\n",
"R=1000.0\n",
"Z2=R\n",
"E1=2*Z2*E/((Z1+Z2)*math.sqrt(3))\n",
"Pr=3*E1*E1/(R*1000) #Rate of power consumption\n",
"Vr=(Z2-Z1)*E/(math.sqrt(3)*(Z2+Z1)*1000)\t\t#Reflected voltage\n",
"Er=3*Vr*Vr*1000/Z1\t\t\t\t\t\t\t\t#rate of reflected voltage\n",
"print(\"(iii)rate of energy absorption =%.1f kW\" %Pr)\n",
"print(\"rate of reflected energy =%.1f kW\" %Er)\n",
"print(\"(iv)Terminating resistance should be equal to surge impedence of line =%.0f ohms\" %Z1)\n",
"L=.5*(10**-8)\n",
"C=10**-12\n",
"Z=math.sqrt(L/C)\t\t# surge impedence\n",
"VR=2*Z*11/((Z1+Z)*math.sqrt(3))\n",
"Vrl=(Z-Z1)*11/((Z1+Z)*math.sqrt(3))\n",
"PR1=3*VR*VR*1000/(Z)\n",
"d=Vrl\n",
"Prl=3*d*d*1000.0/Z1\n",
"\n",
"#Results\n",
"print(\"(v)Refracted power =%.1f kW\" %PR1)\n",
"print(\"Reflected power =%.1f kW\" %Prl)\n",
"##Answer don't match exactly due to difference in rounding off of digits i between calculations\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(i)the natural impedence of line=294 ohms\n",
"(ii)line current =21.6 amps\n",
"(iii)rate of energy absorption =289.2 kW\n",
"rate of reflected energy =122.9 kW\n",
"(iv)Terminating resistance should be equal to surge impedence of line =294 ohms\n",
"(v)Refracted power =257.9 kW\n",
"Reflected power =154.3 kW\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 12.2, Page No 280"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"Xlc=0.3*(10**-3)\t\t# inductance of cable(H)\n",
"Xcc=0.4*(10**-6)\t\t# capacitance of cable (F)\n",
"Xlo=1.5*(10**-3)\t\t#inductance of overhead line(H)\n",
"Xco=.012*(10**-6)\t\t# capacitance of overhead line (F)\n",
"\n",
"#Calculations\n",
"Znc=math.sqrt((Xlc/Xcc))\n",
"Znl=math.sqrt((Xlo/Xco))\n",
"\n",
"#Results\n",
"print(\"Natural impedence of cable=%.2f ohms \" %Znc)\n",
"print(\"Natural impedence of overhead line=%.1f ohms \" %Znl)\n",
"E=2*Znl*15/(353+27)\n",
"print(\"voltage rise at the junction due to surge =%.2f kV \" %E)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Natural impedence of cable=27.39 ohms \n",
"Natural impedence of overhead line=353.6 ohms \n",
"voltage rise at the junction due to surge =27.91 kV \n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 12.3, Page No 280"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"Z1=600.0\n",
"Z2=800.0\n",
"Z3=200.0\n",
"E=100.0\n",
"\n",
"#Calculations\n",
"E1=2*E/(Z1*((1/Z1)+(1/Z2)+(1/Z3)))\n",
"Iz2=E1*1000.0/Z2\n",
"Iz3=E1*1000.0/Z3\n",
"\n",
"#Results\n",
"print(\"Transmitted voltage =%.2f kV \" %E1)\n",
"print(\"The transmitted current in line Z2=%.2f amps \" %Iz2)\n",
"print(\"The transmitted current in line Z3=%.1f amps \" %Iz3)\n",
"\n",
"#Answer don't match exactly due to difference in rounding off of digits i between calculations"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Transmitted voltage =42.11 kV \n",
"The transmitted current in line Z2=52.63 amps \n",
"The transmitted current in line Z3=210.5 amps \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 12.4 Page No 283"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"Z=350.0 \t#surge impedencr (ohms)\n",
"\n",
"#Calculations\n",
"C=3000.0*(10**-12)\t# earth capacitance(F) \n",
"t=2.0*(10**-6)\n",
"E=500.0\n",
"E1=2*E*(1-math.exp((-1*t/(Z*C))))\n",
"\n",
"#Results\n",
"print(\"the maximum value of transmitted voltage=%.2f kV \" %E1)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"the maximum value of transmitted voltage=851.14 kV \n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 12.5, Page No 283"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"Z=350.0 \t\t#surge impedencr (ohms)\n",
"L=800*(10**-6) \n",
"t=2*(10**-6)\n",
"E=500.0\n",
"\n",
"#Calculations\n",
"E1=E*(1-math.exp((-1*t*2*Z/L)))\n",
"\n",
"#Results\n",
"print(\"The maximum value of transmitted voltage=%.1f kV \" %E1)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum value of transmitted voltage=413.1 kV \n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 12.6, Page No 285"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"eo=50.0\n",
"x=50.0\n",
"R=6.0\n",
"Z=400.0\n",
"G=0\n",
"v=3*(10**5)\n",
"e=2.68\n",
"\n",
"#Calculations\n",
"e1=(eo*(e**((-1/2)*R*x/Z)))\n",
"# answess does not match due to the difference in rounding off of digits. \n",
"print(\"(i)the value of the Voltage wave when it has travelled through a distance 50 Km=%.1f kV \" %e1)\n",
"Pl=e1*e1*1000.0/400\n",
"io=eo*1000.0/Z\n",
"t=x/v\n",
"H=-(50*125*400*((e**-0.75)-1))/(6.0*3*10**5)\n",
"\n",
"#Results\n",
"print(\"(ii)Power loss=%.3fkW \\n heat loss=%.3f kJ\" %(Pl,H))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(i)the value of the Voltage wave when it has travelled through a distance 50 Km=23.9 kV \n",
"(ii)Power loss=1424.550kW \n",
" heat loss=0.726 kJ\n"
]
}
],
"prompt_number": 6
}
],
"metadata": {}
}
]
}
|
