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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 1:Fluid properties"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.1"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "## Finding Specific weight,Density,Specific Gravity\n",
      "##Given\n",
      "V = 0.001                   ##volume in m^3\n",
      "w = 9.6                  ##weight in Newton\n",
      "g = 9.81                 ##gravitational force in m/s^2\n",
      "\n",
      "##calculation\n",
      "spwt = (w/V)               ##Specific weight in N/m^3\n",
      "rho = (spwt/g)                ##density in kg/m^3\n",
      "spgr = (rho/1000)          ##Specific gravity no units\n",
      "\n",
      "#Results\n",
      "print \"specific weight = \",round(spwt),\"N/m^3\"\n",
      "print \"density = \",round(rho),\"kg/m^3\"\n",
      "print \"specific gravity = \",round(spgr,2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "specific weight =  9600.0 N/m^3\n",
        "density =  979.0 kg/m^3\n",
        "specific gravity =  0.98\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.3"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##Finding of Viscosity\n",
      "##Given\n",
      "dy=0.025E-3            ##distance in meter\n",
      "du=0.5                    ##velocity  in m/s \n",
      "tau=1.471                 ##shear stress in N/m^2\n",
      "##To Find\n",
      "mu=tau*dy/du              ##viscosity in Ns/m^2 \n",
      "mu1=mu*10                 ## Viscosity in poise\n",
      "print \"Viscosity =\",mu,\" in Ns/m^2\"\n",
      "print \"Viscosity =\",mu1,\" in poise\"  \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Viscosity = 7.355e-05  in Ns/m^2\n",
        "Viscosity = 0.0007355  in poise\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.4"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##Finding of Diameter of water droplet\n",
      "##Given\n",
      "st=0.716                  ##Surface Tension in N/m\n",
      "p=0.147E4               ##Pressure in N/m^2\n",
      "##To Find \n",
      "d=4*st/p                   ##Diameter in meter \n",
      "d1=d*1E2                  ##Diameter in centimeter \n",
      "d2=d*1E3                  ##Diameter in millimeter\n",
      "print \"Diameter =\",round(d,5),\"m\"\n",
      "print \"Diameter =\",round(d1,5),\"cm\"\n",
      "print \"Diameter =\",round(d2,5),\"mm\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Diameter = 0.00195 m\n",
        "Diameter = 0.19483 cm\n",
        "Diameter = 1.9483 mm\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.5"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##Finding of Shear Stress\n",
      "##Given\n",
      "##du/dy = vg\n",
      "vg=0.25            ##Velocity gradient in m/sec/meter\n",
      "nu=6.30E-4         ##Kinematic viscosity in m^2/sec\n",
      "rho=1268.4             ##Mass density in Kg/m^3\n",
      "mu=rho*nu              ##Dynamic Viscosity\n",
      "##To Find\n",
      "tau=mu*vg          ##Shear stress in N/m^2\n",
      "print \"Shear Stress =\",round(tau,3),\"N/m^2\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Shear Stress = 0.2 N/m^2\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.6"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##Finding of increase of Pressure\n",
      "##Given\n",
      "k=2.07*1E6             ## Bulk Modulus in kN/m^2\n",
      "dv=0.01              ##Change in Volume\n",
      "##To Find\n",
      "p=k*(dv)            ##Change in pressure\n",
      "print \"Increase in Pressure =\",p,\"kN/m^2\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Increase in Pressure = 20700.0 kN/m^2\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.7"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import numpy as np\n",
      "##Finding of Cappilary rise\n",
      "##Given\n",
      "d=0.03*1E-2                ##Diameter in meter\n",
      "st=0.0735                   ##Surface Tension in N/m\n",
      "x=0                     ##contact angle in degree\n",
      "w=1000*9.81\n",
      "##To Find\n",
      "h=(4*st)*np.cos(x)/(w*d)\n",
      "h1=h*1E2\n",
      "print \"Capillary rise =\",round(h,4),\"m\"\n",
      "print \"Capillary rise =\",round(h1,4),\"cm\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Capillary rise = 0.0999 m\n",
        "Capillary rise = 9.9898 cm\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1.8"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "##Finding of Kinematic Viscosity\n",
      "##Given\n",
      "tau=0.2158                 ##Shear stress in N/m^2\n",
      "vg=0.218                ##Velocity Gradient in sec^-1\n",
      "rho=959.5                  ##Density in Kg/m^3;\n",
      "##To Find \n",
      "mu=tau*1/vg\n",
      "print \"Dynamic Viscosity =\",round(mu,5),\"Ns/m^2\"\n",
      "nu=mu/rho\n",
      "print \"Kinematic Viscosity =\",round(nu,5),\"m^2/sec\"\n",
      "nu1=nu*1E4\n",
      "print \"Kinematic Viscosity =\",round(nu1,5),\"cm^2/sec\"\n",
      "nu2=nu1*1E-4\n",
      "print \"Kinematic Viscosity =\",round(nu2,5),\"strokes\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Dynamic Viscosity = 0.98991 Ns/m^2\n",
        "Kinematic Viscosity = 0.00103 m^2/sec\n",
        "Kinematic Viscosity = 10.31692 cm^2/sec\n",
        "Kinematic Viscosity = 0.00103 strokes\n"
       ]
      }
     ],
     "prompt_number": 15
    }
   ],
   "metadata": {}
  }
 ]
}