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+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 1 : Introduction"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.1 page no : 6\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "'''Finding the density'''\n",
+      "\n",
+      "#let the total mass of mud be 100lbm\n",
+      "#variables\n",
+      "m_total=100.0;                           #lbm\n",
+      "#70% by wt of mud is sand(SiO2)and remaining is water\n",
+      "m_sand=0.7*m_total;                      #lbm\n",
+      "m_water=0.3*m_total;                     #lbm\n",
+      "rho_sand=165.0;                          #lbm/ft^3\n",
+      "rho_water=62.3;                          #lbm/ft^3\n",
+      "\n",
+      "#calculation\n",
+      "#rho=mass/volume\n",
+      "rho_mud=m_total/((m_sand/rho_sand)+(m_water/rho_water));\n",
+      "\n",
+      "#result\n",
+      "print \"The density of mud=\" ,rho_mud, \"lbm/ft^3\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The density of mud= 110.401675438 lbm/ft^3\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.2 page no : 8\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "'''\n",
+      "Calculate the shear stress at the surface of the inner cylinder\n",
+      "'''\n",
+      "import math\n",
+      "\n",
+      "# variables\n",
+      "D1=25.15                                  #mm\n",
+      "D2=27.62                                  #mm\n",
+      "dr=0.5*(D2-D1)                            #mm\n",
+      "f=10.                                     #rpm\n",
+      "\n",
+      "# calculations\n",
+      "Vo=math.pi*D1*f/60.                       #mm/s\n",
+      "#Let D denote d/dr\n",
+      "DV=Vo/dr                                  #s^-1\n",
+      "tow=0.005                                 #Nm\n",
+      "L=92.37                                   #mm\n",
+      "s=2*tow/D1**2/(math.pi)/L*(10**6) #N/m^2\n",
+      "\n",
+      "# result\n",
+      "print \"The stress at the surface of the inner cylinder is %f N/m^2\"%s "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The stress at the surface of the inner cylinder is 0.054481 N/m^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.3 page no : 15\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#problem on surface tension\n",
+      "# variablees\n",
+      "l=0.10;                                #m (length of sliding part)\n",
+      "f=0.00589;                             #N (pull due to 0.6 gm of mass)\n",
+      "\n",
+      "#calculation\n",
+      "f_onefilm=f/2;                         #N\n",
+      "#surface tension=(force for one film)/(length)\n",
+      "sigma=f_onefilm/l;\n",
+      "\n",
+      "# result\n",
+      "print \"The surface tension of fluid is\",sigma,\"N/m\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The surface tension of fluid is 0.02945 N/m\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.4 page no : 20\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert 327 miles/hr into ft/s\n",
+      "\n",
+      "# variables\n",
+      "V=327.               #miles/hr\n",
+      "#1 mile = 5280 ft\n",
+      "#1 hour = 3600 sec\n",
+      "\n",
+      "# calculation\n",
+      "V1=V*5280/3600.0#ft/s\n",
+      "\n",
+      "# result\n",
+      "print \"327 miles/hr = %f ft/s\"%V1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "327 miles/hr = 479.600000 ft/s\n"
+       ]
+      }
+     ],
+     "prompt_number": 6
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.5 page no : 21\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Convert 2.6 hours into seconds\n",
+      "\n",
+      "# variables\n",
+      "t=2.6            #hr\n",
+      "#1 hr = 3600 s\n",
+      "\n",
+      "# calculations\n",
+      "t1=2.6*3600      #s\n",
+      "\n",
+      "# result\n",
+      "print \"2.6 hours = %f seconds\"%t1"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "2.6 hours = 9360.000000 seconds\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.6 page no : 24\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the acceleration in ft/min^2\n",
+      "\n",
+      "# variables\n",
+      "m=10.                #lbm\n",
+      "F=3.5                #lbf\n",
+      "#1 lbf.s^2 = 32.2 lbm.ft\n",
+      "#1 min = 60 sec\n",
+      "\n",
+      "# calculations\n",
+      "a=F*32.2*60**2/m     #ft/min^2\n",
+      "\n",
+      "# result\n",
+      "print \"The acceleration provided is %f ft/min^2\" % a"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The acceleration provided is 40572.000000 ft/min^2\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 1.7 page no : 24\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "#Calculate the wt of metallic aluminium deposited in an electrolytic cell\n",
+      "\n",
+      "# variables\n",
+      "I=50000.                            #Ampere or Coulumbs/sec\n",
+      "#1 hr = 3600 sec\n",
+      "I1=50000*3600.                      #C/hr\n",
+      "\n",
+      "#calculation\n",
+      "#96500 C = 1 gm.eq\n",
+      "#1 mole of aluminium = 3 gm.eq\n",
+      "#1 mole of aluminium = 27 gm\n",
+      "m=I1*(1.0/96500)*(27/3.0)/1000.0    #Kg/hr\n",
+      "\n",
+      "#result\n",
+      "print \"the wt of metallic aluminium deposited in an electrolytic cell is %f Kg/hr\"%m"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "the wt of metallic aluminium deposited in an electrolytic cell is 16.787565 Kg/hr\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
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