From 4a1f703f1c1808d390ebf80e80659fe161f69fab Mon Sep 17 00:00:00 2001 From: Thomas Stephen Lee Date: Fri, 28 Aug 2015 16:53:23 +0530 Subject: add books --- sample_notebooks/sureshp/CHAPTER1.ipynb | 206 ++++++++++++++++++++++++++++++++ 1 file changed, 206 insertions(+) create mode 100755 sample_notebooks/sureshp/CHAPTER1.ipynb (limited to 'sample_notebooks/sureshp/CHAPTER1.ipynb') 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": {} + } + ] +} \ No newline at end of file -- cgit