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{
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"name": "",
"signature": "sha256:cb7821e52dda76fa27e45154f9c14ca9fa493fbd763be86cc9be92d018411582"
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"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"CHAPTER 1 - Breakdown Mechanism of Gases Liquid and Solid Materials"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"EXAMPLE 1.1 - PG NO.51"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Chapter 1,Example 1.1 Page 51\n",
"import math\n",
"I = 600. # micor amps\n",
"x = 0.5 # distance in cm\n",
"V = 10. # kV\n",
"I2 = 60. # micro amps\n",
"x2 = 0.1 # distance in cm \n",
"#Calculation 600 = I0*exp(0.5*alpha) and 60 = I0*exp(0.1*alpha)\n",
"alpha =math.log(600./60.)/(0.5-0.1)\n",
"print'%s %.3f %s' %(\"Townsends first ionising coefficient = \",alpha,\" ionizing collisions/cm\")\n",
"\n",
"#Answers may vary due to round of error \n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Townsends first ionising coefficient = 5.756 ionizing collisions/cm\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"EXAMPLE 1.2 - PG NO.52"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Chapter 1,Example 1.2 Page 52\n",
"import math\n",
"# Refering the table in example 1.2\n",
"# slope between any two points (math.log(I/I0)/x)\n",
"# taking the gap between 2 and 2.5 mm\n",
"I1= 1.5*10**-12\n",
"I2= 5.6*10**-12\n",
"I0 = 6*10**-14\n",
"gi1 = math.log(I1/I0) # gradual increase when gap is 2\n",
"gi2 = math.log(I2/I0) # gradual increase when gap is 2.5 #claculation in text is wrong\n",
"slope = (gi1-gi2)/0.05\n",
"print'%s %.3f %s' %(\"Slope = \", -slope,'\\n') \n",
"#evaluvating ghama\n",
"e1 = math.exp(-slope*0.5)\n",
"e2 = math.exp(-slope*0.5) # -1 is ignored due to the large magnitude\n",
"ghama = (7*10**7-6*e1)/(e2*7*10**7)\n",
"print'%s %.3f %s' %(\"Ghama for set 1= \", ghama*100000,\"*10^-5 /cm \\n \")\n",
"#Gap between the slope for set 2\n",
"alpha = math.log(12./8.)/0.05\n",
"print'%s %.1f %s' %(\"Alpha = \", alpha,\" collosions/cm \\n\")\n",
"e1 = math.exp(alpha*0.5)\n",
"e2 = math.exp(alpha*0.5) # -1 is ignored due to the large magnitude\n",
"ghama = (2*10**5-e1)/(e2*2*10**5)\n",
"print'%s %.1f %s' %(\"Ghama for set 2=\", ghama*100,\"*10^-2 colissions/cm \\n\")\n",
"\n",
"#Answers may vary due to round of error \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Slope = 26.346 \n",
"\n",
"Ghama for set 1= 0.182 *10^-5 /cm \n",
" \n",
"Alpha = 8.1 collosions/cm \n",
"\n",
"Ghama for set 2= 1.7 *10^-2 colissions/cm \n",
"\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"EXAMPLE 1.3 - PG NO.53"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Chapter 1,Example 1.3 Page 53\n",
"\n",
"#employing equation Vb = K*d**n\n",
"#88 = K*4**n --- eq(1) 165 = K*8**n ---eq(2) \n",
"#dividing eq(2)/q(1)\n",
"Vb1 = 88.\n",
"Vb2 = 165.\n",
"n1 = 0.6286/0.693\n",
"K1 = Vb1/4**n1\n",
"#135 = K*6**n --- eq(1) 212 = K*10**n ---eq(2) \n",
"#dividing eq(2)/q(1) \n",
"Vb1 = 135.\n",
"Vb2 = 212.\n",
"n2 = 0.4513/0.5128\n",
"K2 = Vb1/6.**n2\n",
"n = (n1+n2)/2.\n",
"K = (K1+K2)/2.\n",
"print'%s %.2f %s %.2f' % (\"n =\",n,\"K = \",K,)\n",
"\n",
"#Answer may vary due to round of error \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"n = 0.89 K = 26.46\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"EXAMPLE 1.4 - PG NO.53"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Chapter 1,Example 1.4 Page 53\n",
"# Determine (pd)min Vbmin\n",
"import math\n",
"A = 12.\n",
"B = 365.\n",
"e = 2.718\n",
"ghama = 0.02\n",
"K = 51.\n",
"pd = (e/A)*math.log(1.+(1./ghama))\n",
"Vbmin = (B/A)*e*math.log(K)\n",
"print'%s %.2f %s %d %s' % (\"(pd)min = \",pd,\" Vbmin = \",Vbmin,\"Volts\")\n",
"\n",
"#Answers may vary due to round of error\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(pd)min = 0.89 Vbmin = 325 Volts\n"
]
}
],
"prompt_number": 4
}
],
"metadata": {}
}
]
}
|
