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+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:918b48292f00614bbba5ee6b8d2f7b2b929a3939332547fe2fb6585d1e0c7729"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2:THERMODYNAMIC CONCEPTS AND THE ZEROTH LAW"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.1, Page No:33"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Variable declaration\n",
+      "patm = 14.5 ;                                   # atmospheric pressure in psia\n",
+      "pgauge = 2.5;                                   # gauge pressure in psia\n",
+      "A = 10;                                         # Area of the piston in in^2\n",
+      "g=9.80665;                                      # Acceleration due to gravity in m/s^2\n",
+      "#Calculation for (a)\n",
+      "p = patm + pgauge;                              #total pressure of gas\n",
+      "m=(p-patm)*A ;                                  # mass of the piston\n",
+      "#Result for (a)\n",
+      "print \"In English units\"\n",
+      "print \"Mass of the piston =\",m,\"lbm\"\n",
+      "#Calculation for (b)\n",
+      "p=(p*0.454*g)/(0.0254**2);                      # conversion of English unit to SI units\n",
+      "patm=(patm*0.454*g)/(0.0254**2);                # conversion of English unit to SI units\n",
+      "m = ((p-patm)*(A*2.54**2*10**-4))/g;            # Mass of the piston\n",
+      "#Result for (b)\n",
+      "print \"\\nIn SI units \\n\",\"Mass of the piston =\",m,\"kg\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "In English units\n",
+        "Mass of the piston = 25.0 lbm\n",
+        "\n",
+        "In SI units \n",
+        "Mass of the piston = 11.35 kg\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 2.2, Page No:34"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math\n",
+      "#Variable declaration\n",
+      "d_r = 13600;                                   # Density of manometric fluid (mercury) in kg/m^*3\n",
+      "g = 8.92;                                      # Gravitational acceleration in m/s^2\n",
+      "#Variable declaration\n",
+      "z1=0.589*math.sin(math.radians(60));           # vertical height of fluid at section 1\n",
+      "z2=2*math.sin(math.radians(30));               # vertical height of fluid at section 2\n",
+      "z=z2-z1;                                       # Difference in vertical heights of fluid\n",
+      "patm = 14.7;                                   # Atmospheric pressure in lbf/in^2\n",
+      "patm=(patm*4.44822*144/0.3048**2);             # conversion of lbf/in^2 unit to N/m^2 unit\n",
+      "p=patm + (d_r*g*(z2-z1));                      # Balance of force at A\n",
+      "#Result\n",
+      "print \"Difference in vertical heights of fluid = \",round(z,2),\"m\"\n",
+      "print \"\\nThe pressure of fluid in the vessel = \",round(p/1000,0),\"kPa\" \n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Difference in vertical heights of fluid =  0.49 m\n",
+        "\n",
+        "The pressure of fluid in the vessel =  161.0 kPa\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file