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| author | Thomas Stephen Lee | 2015-08-28 16:53:23 +0530 |
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| committer | Thomas Stephen Lee | 2015-08-28 16:53:23 +0530 |
| commit | 4a1f703f1c1808d390ebf80e80659fe161f69fab (patch) | |
| tree | 31b43ae8895599f2d13cf19395d84164463615d9 /sample_notebooks/sureshp/CHAPTER1.ipynb | |
| parent | 9d260e6fae7328d816a514130b691fbd0e9ef81d (diff) | |
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diff --git a/sample_notebooks/sureshp/CHAPTER1.ipynb b/sample_notebooks/sureshp/CHAPTER1.ipynb new file mode 100755 index 00000000..6c66d92e --- /dev/null +++ b/sample_notebooks/sureshp/CHAPTER1.ipynb @@ -0,0 +1,206 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:cb7821e52dda76fa27e45154f9c14ca9fa493fbd763be86cc9be92d018411582"
+ },
+ "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": {}
+ }
+ ]
+}
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