From 37d315828bbfc0f5cabee669d2b9dd8cd17b5154 Mon Sep 17 00:00:00 2001
From: hardythe1
Date: Wed, 17 Jun 2015 11:14:34 +0530
Subject: add books

---
 sample_notebooks/kartiksankhla/Chapter2.ipynb | 159 ++++++++++++++++++++++++++
 1 file changed, 159 insertions(+)
 create mode 100755 sample_notebooks/kartiksankhla/Chapter2.ipynb

(limited to 'sample_notebooks/kartiksankhla/Chapter2.ipynb')

diff --git a/sample_notebooks/kartiksankhla/Chapter2.ipynb b/sample_notebooks/kartiksankhla/Chapter2.ipynb
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@@ -0,0 +1,159 @@
+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:e984fee9b841dd6e9b7eedf1533b0a0d297cd9f484c047f051ce48a09b156826"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter2-Nuclear Engineering"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex1-pg54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "## Example 2.1\n",
+      "import math\n",
+      "#determine atoms in deuterium\n",
+      "## Given data\n",
+      "atom_h = 6.6*10**24;                      ## Number of atoms in Hydrogen\n",
+      "## Using the data given in Table II.2, Appendix II for isotropic abundance of deuterium\n",
+      "isoab_H2 = 0.015;                        ## Isotropic abundance of deuterium\n",
+      "## Calculation\n",
+      "totatom_d=(isoab_H2*atom_h)/100.;\n",
+      "## Result\n",
+      "print\"%s %.2e %s \"%('\\n Number of deuterium atoms = ',totatom_d,'');\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Number of deuterium atoms =  9.90e+20  \n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex2-pg54"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "## Example 2.2\n",
+      "import math\n",
+      "#determine atomic weight of oxygen\n",
+      "## Given data \n",
+      "## Using the data given in the example 2.2\n",
+      "atwt_O16 = 15.99492;                      ## Atomic weight of O-16 isotope\n",
+      "isoab_O16 = 99.759;                       ## Abundance of O-16 isotope\n",
+      "atwt_O17 = 16.99913;                      ## Atomic weight of O-17 isotope\n",
+      "isoab_O17 = 0.037;                        ## Abundance of O-17 isotope\n",
+      "atwt_O18 = 17.99916;                      ## Atomic weight of O-18 isotope\n",
+      "isoab_O18 = 0.204;                        ## Abundance of O-18 isotope\n",
+      "## Calculation\n",
+      "atwt_O=(isoab_O16*atwt_O16 + isoab_O17*atwt_O17 + isoab_O18*atwt_O18)/100.;\n",
+      "## Result\n",
+      "print\"%s %.2f %s \"%('\\n Atomic Weight of Oxygen = ',atwt_O,'');\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Atomic Weight of Oxygen =  16.00  \n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Ex3-pg55"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "## Example 2.3\n",
+      "import math\n",
+      "#determine rest mass energy of electron\n",
+      "## Given data\n",
+      "me = 9.1095*10**(-28);                     ## Mass of electron in grams\n",
+      "c = 2.9979*10**10;                         ## Speed of light in vacuum in cm/sec\n",
+      "## Calculation\n",
+      "rest_mass = me*c**2;\n",
+      "## Result\n",
+      "print\"%s %.2e %s \"%('\\n Rest mass energy of electron = ',rest_mass,' ergs\\n');\n",
+      "print('Expressing the result in joules')\n",
+      "## 1 Joule = 10^(-7)ergs\n",
+      "rest_mass_j = rest_mass*10**(-7);\n",
+      "print\"%s %.2e %s \"%('\\n Rest mass energy of electron = ',rest_mass_j,' joules\\n');\n",
+      "print('Expressing the result in MeV')\n",
+      "## 1 MeV = 1.6022*10^(-13)joules\n",
+      "rest_mass_mev = rest_mass_j/(1.6022*10**(-13));\n",
+      "print\"%s %.2f %s \"%('\\n Rest mass energy of electron = ',rest_mass_mev,' MeV\\n');\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "\n",
+        " Rest mass energy of electron =  8.19e-07  ergs\n",
+        " \n",
+        "Expressing the result in joules\n",
+        "\n",
+        " Rest mass energy of electron =  8.19e-14  joules\n",
+        " \n",
+        "Expressing the result in MeV\n",
+        "\n",
+        " Rest mass energy of electron =  0.51  MeV\n",
+        " \n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file
-- 
cgit