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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 3: Principles of Momentum Transfer and Applications"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.1-1, Page number 117"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Force on submerged sphere \n",
      "from math import pi \n",
      "\n",
      "# Variable declaration\n",
      "Rho_air = 1.137             #Density of air at 37.8 degC (kg/m3)\n",
      "u = 0.000019                #Viscosity of air (Pa.s)\n",
      "Dp = 0.042                  #Diameter of the sphere (m)\n",
      "V = 23.0                    #velocity of the sphere (m/s)\n",
      "  \n",
      "# Data SI Units \n",
      "#Calculation\n",
      "Nre = Dp*Rho_air*V/(u)\n",
      "Cd = 0.47                   #Drag coefficient from fig 3.1-2\n",
      "Ap = pi*Dp**2/4\n",
      "Fd = Cd*V**2*Rho_air*Ap/2\n",
      "#Result\n",
      "print 'Reynolds number %4.3e'%Nre\n",
      "print \"The drag coefficient of the sphere is\",Cd\n",
      "print \"The force on the cylinder is\", round(Fd,4),\"N\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Reynolds number 5.781e+04\n",
        "The drag coefficient of the sphere is 0.47\n",
        "The force on the cylinder is 0.1958 N\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.1-2, Page number 117"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Force on a cylinder in a tunnel\n",
      "# Variable declaration\n",
      "V = 1.                            #Velocity of water in the tunnel (m/s)\n",
      "d = 0.09                          #Diameter of cylinder (m)\n",
      "Rho_H2O = 997.2                   #Density of water (kg/m3)\n",
      "L = 1.                            #Length of the tunnel (m)\n",
      "mu = 0.0009142                     #Viscosity of water (Pa.s)\n",
      "T = 24.                           #Temerature of water (\u00b0C)\n",
      "\n",
      "# Data SI Units \n",
      "#Calculation\n",
      "Nre = d*Rho_H2O*V/(mu)\n",
      "Cd = 1.4                          #Drag Coefficient for cylinder from fig 3.1-2\n",
      "Ap = L*d\n",
      "Fd = Cd*V**2*Rho_H2O*Ap/2\n",
      "#Result\n",
      "print 'Reynolds number %4.3e'%Nre\n",
      "print \"The drag coefficient of the cylinder is\",Cd\n",
      "print \"The force on the sphere is \",round(Fd,2),\"N\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Reynolds number 9.817e+04\n",
        "The drag coefficient of the cylinder is 1.4\n",
        "The force on the sphere is  62.82 N\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.1-3, Page number 119"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Surface area in Packed Bed of Cylinders\n",
      "# Variable declaration\n",
      "D = 0.02                   #Diamter of the cylinder (m)\n",
      "h = 0.02                   #Length of the cylinder (m)\n",
      "d_pb = 962.                #Density of packed bed (kg/m3)\n",
      "d_sc = 1600.               #Density of solid cylinders (kg/m3)\n",
      "m_pb = 962.                #Mass of the packed bed (kg)\n",
      "v_pb = 1.                  #Volume of packed bed (m3)\n",
      "v_t = 1.                   #Total volume of the bed (m3)\n",
      "    # Calculation\n",
      "v_sc = m_pb/d_sc\n",
      "    #(a) Calculate void fraction\n",
      "    #e = Volume of voids in bed/total volume of bed\n",
      "e = (v_pb - v_sc)/v_t\n",
      "    #(b)Calculate the effective Diameter of the particle \n",
      "    #Dp = 6/(6/D)\n",
      "Dp = 6/(6/D)\n",
      "    #(c) Calculate the value of \"a\"\n",
      "    # a = 6*(1-e)/Dp\n",
      "a = 6*(1-e)/Dp\n",
      "#Result\n",
      "\n",
      "print \"(a) The void fraction of bed is\",round(e,3)\n",
      "print \"(b) The effective diameter of the particle is\",round(Dp,4),\"m\"\n",
      "print \"(c) The calculated value of 'a' is \",round(a,2), \"1/m\"\n",
      "print 'The answers are correct because of book uses rounded numbers'"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a) The void fraction of bed is 0.399\n",
        "(b) The effective diameter of the particle is 0.02 m\n",
        "(c) The calculated value of 'a' is  180.38 1/m\n",
        "The answers are correct because of book uses rounded numbers\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.1-4, Page number 121"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Pressure drop and flow of gases through packed beds\n",
      "from math import pi\n",
      "from scipy.optimize import root\n",
      "# Variable declaration\n",
      "d = 0.0127            #Diameter of the pipe (m)\n",
      "e = 0.38              #Void fraction\n",
      "d_bed = 0.61          #Bed diameter (m)\n",
      "h = 2.44              #Height of bed (m)\n",
      "p1 = 1.1              #Pressure P1 (atm)\n",
      "T = 311.              #Temperature of water (K)\n",
      "R = 8314.34           #Gas constant \n",
      "M = 28.97             #Molecular weight of air \n",
      "L = 2.44              #Length of the bed (m)\n",
      "Mdot = 0.358          #Mass rate of air, kg/s\n",
      "\n",
      "# Data\n",
      "mu = 0.000019          #Viscosity of water (Pa.s)\n",
      "# Calculation\n",
      "A = pi*d_bed**2/4\n",
      "G = Mdot/A\n",
      "Nre = d*G/((1-e)*mu)\n",
      "    #Rho_av = M*Pav/(R*T)\n",
      "p1 = p1*101325.       #convert to pascal\n",
      "er = 0.1\n",
      "\n",
      "f = lambda dp:dp*(M*(p1+0.5*dp)/(R*T))*d*e**3/(G**2*L*(1-e))-(150./Nre + 1.75)\n",
      "sol = root(f,0.049e5)\n",
      "dp = sol.x[0]\n",
      "\n",
      "#Result\n",
      "print \"The calclated Reynolds number is\",round(Nre,0)\n",
      "print 'The calculated pressure drop is %6.5e Pa'%(dp)\n",
      "print 'The answer is difference than book because of method of solution, and is more correct.\\nBook assumes a pressure drop and calculates density at average pressure'"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The calclated Reynolds number is 1321.0\n",
        "The calculated pressure drop is 4.75988e+03 Pa\n",
        "The answer is differencethan book because of method of solution and is more correct.\n",
        "Book assumes a pressure drop and calculates density at average pressure\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.1-5, Page number 122"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Mean Diameter for a Particle\n",
      "# Variable declaration\n",
      "# Data  \n",
      "x1 = 0.25\n",
      "x2 = 0.40\n",
      "x3 = 0.35\n",
      "Dp1 = 25.\n",
      "Dp2 = 50.\n",
      "Dp3 = 75.\n",
      "fi = 0.68\n",
      "\n",
      "# Calculation\n",
      "    #Dpm = 1/(x1/(fi*Dp1) + x2/(fi*Dp2) + x3/(x3/(fi*Dp3))\n",
      "Dpm = 1/((x1/(fi*Dp1)) + (x2/(fi*Dp2)) + (x3/(fi*Dp3)))\n",
      "#Result\n",
      "print \"The calculated effective mean diameter is \",round(Dpm,4),\"mm\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The calculated effective mean diameter is  30.0 mm\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.1-6, Page number 125 "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Minimum Velocity for Fluidization\n",
      "from scipy.optimize import root\n",
      "\n",
      "# Variable declaration\n",
      "d = 0.12                      #Size of the particle (mm)\n",
      "fi = 0.88                     #Shape of the particle \n",
      "P = 2.                        #Pressure of the fluidized bed (atm)\n",
      "v = 0.42                      #Voidage \n",
      "A = 0.30                      #Cross section of the bed (m2)\n",
      "M = 300.                      #Weight of the solid (kg)\n",
      "ep1 = 0.\n",
      "ep_mf = 0.42                  #voidage at minimum fluidisation\n",
      "    #Data\n",
      "mua = 1.85e-5                 #Viscosity of air (Pa.s)\n",
      "rhoa = 2.374                  #Density of air at 2 atm, kg/m3\n",
      "p = 2.0165e5                  #Pressure, Pa\n",
      "rhop = 1000                   #Density of particle, kg/m3\n",
      "Dp = 0.00012                  #Particle diameter, m\n",
      "g = 9.80665                   #m/s2\n",
      "# Calculation\n",
      "\n",
      "#Part A \n",
      "V = M/rhop\n",
      "L1 = V/A\n",
      "Lmf = L1*(1. - ep1)/(1 - ep_mf)    \n",
      "\n",
      "#Part B\n",
      "delp = Lmf*(1.-ep_mf)*(rhop-rhoa)*g\n",
      "\n",
      "#Part C\n",
      "#Nremf = d*umf*rhoa/mua\n",
      "Nrea = Dp*rhoa/mua\n",
      "\n",
      "f = lambda u: 1.75*(Nrea*u)**2/(fi*ep_mf**3)+150*(1.-ep_mf)*(Nrea*u)/(fi**2*ep_mf**3)-Dp**3*rhoa*(rhop-rhoa)*g/mua**2\n",
      "sol = root(f,0.001)\n",
      "umf = sol.x[0]\n",
      "Nremf = Dp*rhoa*umf/mua\n",
      "\n",
      "f1 = lambda u: Dp*rhoa*u/mua - sqrt(33.7**2 + 0.0408*Dp**3*rhoa*(rhop-rhoa)*g/mua**2) + 33.7\n",
      "sol = root(f1,umf)\n",
      "umfn = sol.x[0]\n",
      "#Result\n",
      "print \"(a) The minimum height of the fluidized bed is\",round(Lmf,3),\"m\"\n",
      "print '(b) Pressure drop under conditions of minimum fluidization is %5.4e'%(delp),\"Pa\"\n",
      "print \"(c) Minimum fluidozation velocity\",round(umf,6), \"m/s\"\n",
      "print \"    Reynold number at Minimum fluidization is\",round(Nremf,5)\n",
      "print '(d) Minimum fluidozation velocity using simplified relation %7.6f'%umfn\n",
      "print 'The answer is correct because of book uses rounded numbers'"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a) The minimum height of the fluidized bed is 1.724 m\n",
        "(b) Pressure drop under conditions of minimum fluidization is 9.7834e+03 Pa\n",
        "(c) Minimum fluidozation velocity 0.005015 m/s\n",
        "    Reynold number at Minimum fluidization is 0.07723\n",
        "(d) Minimum fluidozation velocity using simplified relation 0.004605\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.1-7, Page number 126"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Expansion of Fluidized Bed \n",
      "from scipy.optimize import root\n",
      "\n",
      "# Variable declaration\n",
      "Nre_mf = 0.07723            #From previous problem\n",
      "vmf = 0.005015\n",
      "epmf = 0.42\n",
      "\n",
      "#Calculation\n",
      "f = lambda K:vmf-K*epmf**3/(1-epmf)\n",
      "sol = root(f,0.01)\n",
      "K = sol.x[0]\n",
      "vop = 3*vmf\n",
      "\n",
      "f1 = lambda ep:vop-K*ep**3/(1-ep)\n",
      "sol = root(f1,0.1)\n",
      "ep1 = sol.x[0]\n",
      "#Result\n",
      "print 'Operating velocity %6.5f m/s'%vop\n",
      "print 'Voidage of bed at operating velocity %4.3f m/s'%(ep1)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Operating velocity 0.01505 m/s\n",
        "Voidage of bed at operating velocity 0.555 m/s\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.2-1, Page number 128"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Flow Measurement Using a Pitot Tube\n",
      "from math import pi\n",
      "\n",
      "# Variable declaration\n",
      "d = 600.          #Diameter of circular duct (m)\n",
      "dh1 = 10.7        #Pitot tube reading (mm)\n",
      "dh2 = 205.        #Pitot tube reading (mm)\n",
      "Cp = 0.98         #Pitot tube coefficient \n",
      "mu = 0.0000203    #Viscosity of air (Pa.s)\n",
      "rhoa = 1.043      #Density of air (kg/m3)\n",
      "delh = 0.205      #Head loss in height of water (m)\n",
      "rhow = 1000.      #Density of water (kg/m3)\n",
      "g = 9.80665       #Gravitational accleration (m/s2)\n",
      "\n",
      "#Calculation \n",
      "d = 600./1e3\n",
      "dp1 = (dh2/1e3)*(rhow - rhoa)*g\n",
      "p1abs = 101325 + dp1\n",
      "rhoac = rhoa*p1abs/101325\n",
      "dp = (dh1/1e3)*g*(rhow - rhoa)\n",
      "v = Cp*sqrt(2*dp/rhoa)\n",
      "Nre = d*v*rhoa/mu\n",
      "A = 3.14*d**2/4\n",
      "vav = 0.85*v\n",
      "Q = A*vav\n",
      "\n",
      "#Result \n",
      "print 'Total absolute pressure %7.1f Pa'%p1abs\n",
      "print 'Corrected density of air %5.4f kg/m3'%rhoac\n",
      "print 'Velocity at centre %5.3f m/s'%v\n",
      "print 'Average velocity through pipe %3.2f m/s'%vav\n",
      "print 'Reynolds number %5.3f'%Nre\n",
      "print \"The Volumetric flow rate is\",round(Q,4),\"m/s\"\n",
      "print 'The answers are correct because of book uses rounded numbers'"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Total absolute pressure 103333.3 Pa\n",
        "Corrected density of air 1.0637 kg/m3\n",
        "Velocity at centre 13.894 m/s\n",
        "Average velocity through pipe 11.81 m/s\n",
        "Reynolds number 428315.769\n",
        "The Volumetric flow rate is 3.3375 m/s\n"
       ]
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.2-2, Page number 132"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Metering Oil Flow by an Orifice\n",
      "from math import pi\n",
      "\n",
      "#Variable declaration\n",
      "d0 = 0.0566                     #Diameter of orifice (m)\n",
      "d1 = 0.1541                     #Diameter of pipe (m)\n",
      "Rhooil = 878.                   #Density of oil (kg/m3)\n",
      "mu = 0.0041                     #Viscoity of oil (cp)\n",
      "dP = 93200.                     #Pressure diffrence across orifice (kN/m2)\n",
      "Co = 0.61                       #Coeficient of discharge of orifice \n",
      "\n",
      "#Calculations\n",
      "beta = d0/d1\n",
      "v = Co*sqrt(2*dP/Rhooil)/sqrt(1 - beta**4)\n",
      "Q = v*pi*d0**2/4\n",
      "Nre = d0*v*Rhooil/mu\n",
      "\n",
      "#Result\n",
      "print 'Velocity of oil %5.3f m/s'%v\n",
      "print \"The volumetric flowrate\",round(Q,5), \"m3/s\"\n",
      "print 'Reynolds number %5.4e'%Nre\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Velocity of oil 8.970 m/s\n",
        "The volumetric flowrate 0.02257 m3/s\n",
        "Reynolds number 1.0872e+05\n"
       ]
      }
     ],
     "prompt_number": 26
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.3-1, Page number 135"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Calculation of Brake Horse Power of a Pump\n",
      "\n",
      "#Variable declaration\n",
      "m = 40.                 #Flowrate for the Pump (gal/min)\n",
      "mfps = 5.56             #Flowrate of the Pump in fps units (lbm/s)\n",
      "D = 62.4                #Density of water in fps units (lbm/ft3)\n",
      "n = .60                 #Efficiency of the pump \n",
      "H = 38.5                #Developed Head (ft)\n",
      "\n",
      "#Calculations\n",
      "mfps = m*(1./60)*(1./7.481)*62.4\n",
      "Ws = -H\n",
      "Bhp = -Ws*mfps/(n*550)\n",
      "\n",
      "#Result\n",
      "print \"The Calculated Brake Horse Power is \",round(Bhp,2), \"hp\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Calculated Brake Horse Power is  0.65 hp\n"
       ]
      }
     ],
     "prompt_number": 27
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.3-2, Page number 137"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Brake-kQ Power of a Centrifugal Fan\n",
      "#Variable declaration\n",
      "m = 28.32               #Flowrate of the air (m3/min)\n",
      "Ma = 28.97              #Molecular weight of air \n",
      "V = 22.414              #Volume of 1 kg air at 101.3 atm pressure and 273.2 K\n",
      "T1 = 273.2              #Temperature of air at atmospheric pressure (K) \n",
      "T2 = 366.3              #Temperature of air at suction (K)\n",
      "T3 = 294.1\n",
      "P1 = 760.               #Atmospheric pressure of air (mm Hg)\n",
      "P2 = 741.7              #Suction Pressure of air (mm Hg)\n",
      "P3 = 769.6              #Discharge Pressure of air (mm Hg)\n",
      "n = 60.                 #Efficiency of the pump\n",
      "confac = 101325.        #Conversion factor pressure from (mm Hg) to (N/m2) (N/m2.atm)\n",
      "rhow = 0.958            #Density of water (kg/m3)\n",
      "V1 = 0.                 #Velocity of air at suction (m/s2)\n",
      "V2= 45.7                #Velocity of air at discharge (m/s2)\n",
      "z1 = 0.                 #Height at suction (m)\n",
      "z2 = 0.                 #Height at discharge (m)\n",
      "g = 9.8                 #Gravitational accleration (m/s)\n",
      "F = 0.\n",
      "#Calculations\n",
      "n = n/100\n",
      "Rho1 = Ma*T1*P2/(V*T2*P1)\n",
      "Rho2 = Rho1*P3/P2\n",
      "Rhoavg = (Rho1 + Rho2)/2\n",
      "mmks = m*(1./60)*(1./22.414)*(T1/T3)*(Ma)\n",
      "P = (P3 - P2)*confac/(P1*rhow)\n",
      "\n",
      "    #z1*g + V1**2/2 + P1/Rho_avg + Ws = z2*g + V2**2/2 + P2/Rho_avg + F\n",
      "#-Ws = -z1*g - V1**2/2 - P1/Rho_avg +z2*g + V2**2/2 + P2/Rho_avg + F   \n",
      "Ws = -z1*g - V1**2/2 + P +z2*g + V2**2/2 + F \n",
      "BkW = Ws*mmks/(n*1000)\n",
      "\n",
      "#Result\n",
      "print \"The Calculated Brake Horse Power is\",round(BkW,2), \"kW\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Calculated Brake Horse Power is 4.65 kW\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.3-3, Page number 140"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Compression of Methane\n",
      "from math import log\n",
      "#Variable Declaration\n",
      "Q = 0.00756                #Flowrate of methane (kg.mol/s)\n",
      "P1 = 137.9                 #Initial pressure (kPa)\n",
      "P2 = 551.6                 #Final pressure (kPa)\n",
      "T1 = 26.7                  #Temperature of methane (degC)\n",
      "M_CH4 = 16.                #Molecular weight of Methane\n",
      "gama = 1.31    \n",
      "R = 8314.3                 #Gas constant \n",
      "n = .8                     #Efficiency of compressor\n",
      "#Calculation\n",
      "    #(a) Calculation for part (a)\n",
      "T1_K = T1 + 273.2\n",
      "Q_mks = Q*M_CH4\n",
      "Ws = gama*R*T1_K*((P2/P1)**((gama - 1)/gama) - 1)/((gama - 1)*M_CH4)\n",
      "B_kW = Ws*Q_mks/(n*1000)\n",
      "#(b) Calculation for part (b)\n",
      "    # Ws_b = 2.3026*R*T1_K*math.log10(P2/P1)/M_CH4\n",
      "Ws_b = (2.3026/M_CH4)*R*T1_K*log(P2/P1,10)\n",
      "B_kW_b = Ws_b*Q_mks/(n*1000)\n",
      "#Result \n",
      "print \"The power required for adiabatic compression\",round(B_kW,2),\"kW\"\n",
      "print \"The power required for isothermal compression\",round(B_kW_b,2),\"kW\"\n",
      "print 'The answers are correct because of book uses rounded numbers'"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The power required for adiabatic compression 38.66 kW\n",
        "The power required for isothermal compression 32.67 kW\n",
        "The answers are correct because of book uses rounded numbers\n"
       ]
      }
     ],
     "prompt_number": 30
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.4-1, Page number 145"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Power Consumption by an Agitator\n",
      "\n",
      "#Variable Declaration\n",
      "Dt = 1.83            #The tank diameter (m)\n",
      "W = 0.122            #Width of the tank (m)\n",
      "Da = 0.61            #The turbine diameter (m)\n",
      "J = 0.15             #Width of baffle (m)\n",
      "Vrpm = 90.           #Speed of the turbine (rpm)\n",
      "Rho = 929.           #Density of liquid (kg/m3)\n",
      "mu = 10.             #Viscosity of liquid (cp)\n",
      "mub = 100000.        #Viscosity of the in part (b)liquid (cp)\n",
      "\n",
      "#Calcualtion\n",
      "mu = mu*1e-3\n",
      "mub = mub*1e-3\n",
      "    #Calculation for (a)\n",
      "Vrps = Vrpm/60.\n",
      "Nre = Da**2*Vrps*Rho/mu\n",
      "NpT = 5. \n",
      "P = NpT*Rho*Vrps**3*Da**5/1000.\n",
      "    #Calculation for (b)\n",
      "Nreb = Da*Vrps*Rho/mub\n",
      "NpL = 14.\n",
      "Pb = NpL*Rho*Vrps**3*Da**5/1000.\n",
      "\n",
      "#Result \n",
      "print \"(a)The power required by the mixer is\",round(P,3),\"kW\"\n",
      "print \"(b)The power requird by the mixer is for higher viscosity\",round(Pb,2),\"kW\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a)The power required by the mixer is 1.324 kW\n",
        "(b)The power requird by the mixer is for higher viscosity 3.71 kW\n"
       ]
      }
     ],
     "prompt_number": 31
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.4-3, Page number 149 "
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Scale-up of Turbine Agitation System\n",
      "from math import  pi\n",
      "#Variable Declaration\n",
      "Dt1 = 1.83                 #Tank diameter (m)\n",
      "Da1 = 0.61                 #Diameter of turbine (m)\n",
      "W1 = 0.122                 #Width of turbine (m)\n",
      "J1 = 0.15                  #Width of baffle (m)\n",
      "N1 = 1.5                   #Speed of the turbine (rev/s)\n",
      "Rho = 929.                 #Density of fluid (kg/m3)\n",
      "mu = 0.01                  #Viscosity of the fluid (Pa.s)\n",
      "H1 = Dt1\n",
      "Np = 5.0         \n",
      "\n",
      "#Calcualtion\n",
      "V1 = pi*Dt1**2*H1/4.\n",
      "V2 = 3.*V1\n",
      "R = (V2/V1)**(1./3)\n",
      "Dt2 = R*Dt1\n",
      "Da2 = R*Da1\n",
      "W2 = R*W1\n",
      "J2 = R*J1\n",
      "P1 = Np*Rho*N1**3*Da1**5\n",
      "P1_kW = P1/1000.\n",
      "    #(a)Calculation for equal rate of mass transfer\n",
      "n = 2./3\n",
      "N2 = N1*(1/R)**n\n",
      "Nre = Da2**2*N2*Rho/mu\n",
      "\n",
      "P2 = Np*Rho*N2**3*Da2**5\n",
      "P2_kW = P2/1000.\n",
      "    #The power per unit volume\n",
      "P_v1 = P1_kW/V1 \n",
      "P_v2 = P2_kW/V2\n",
      "    #(b)Calculation for equal liquid motion\n",
      "n= 1.\n",
      "N2_b = N1*(1./R)**n\n",
      "P2_b = Np*Rho*N2_b**3*Da2**5\n",
      "P2_vb = P2_b/V2\n",
      "\n",
      "#Result \n",
      "print \"(a) For equal mass transfer\"\n",
      "print '    Total power for smaller tank %5.3f W and \\n    power required per unit volume is %5.4f W' %(P1,P_v1)\n",
      "print '    Total power for larger tank %5.3f W and \\n    power required per unit volume is %5.4f W' %(P2,P_v2)\n",
      "print \"(b) The power per unit volume for equal liquid motion is\",round(P2_vb/1000,4),\"kW\"\n",
      "print 'The answers are different than book because book uses rounded numbers'"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(a) For equal mass transfer\n",
        "    Total power for smaller tank 1324.063 W and \n",
        "    power required per unit volume is 0.2751 W\n",
        "    Total power for larger tank 3972.189 W and \n",
        "    power required per unit volume is 0.2751 W\n",
        "(b) The power per unit volume for equal liquid motion is 0.1907 kW\n",
        "The answers are correct because of book uses rounded numbers\n"
       ]
      }
     ],
     "prompt_number": 35
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.5-1, Page number 158"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Pressure Drop of Power-Law Fluid in Laminar Flow\n",
      "#Variable Declaration\n",
      "Rho = 1041.              #Density of fluid (kg/m3)\n",
      "L = 14.9                 #Length of the tube (m)\n",
      "D = 0.0524               #Inside diameter of the tube (m)\n",
      "V = 0.0728               #Average Velocity of the flowing fluid (m/s)\n",
      "K = 15.23                #Rheological properties of fluid \n",
      "n = 0.4\n",
      "\n",
      "#Calcualtion\n",
      "    #Calculation for part (a)\n",
      "delP = (K*4*L/D)*(8*V/D)**n\n",
      "Ff = delP/Rho\n",
      "Nre = ((D**n)*(V**(2-n))*Rho)/(K*8**(n-1))\n",
      "print 'Nre = %5.4f is in Laminar range' %Nre\n",
      "    #Calculation for part (b)\n",
      "f = 16./Nre\n",
      "delPb = 4*f*Rho*L*V**2/(D*2)\n",
      "\n",
      "#Result  \n",
      "print \"(a) The pressure drop calculated is\",round(delP,0),\"N/m2\"\n",
      "print \"    The friction loss calculated is\",round(Ff,2),\"J/kg\"\n",
      "print \"(b) The pressure drop calculated by using the friction factor method is\",round(delP/1000,2),\"kN/m2\"\n",
      "print 'The answers are different than book because book uses rounded numbers'"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Nre = 1.1060 is in Laminar range\n",
        "(a) The pressure drop calculated is 45391.0 N/m2\n",
        "    The friction loss calculated is 43.6 J/kg\n",
        "(b) The pressure drop calculated by using the friction factor method is 45.39 kN/m2\n"
       ]
      }
     ],
     "prompt_number": 36
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.5-2, Page number 160"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Turbulent Flow of Power-Law Fluid\n",
      "#Variable Declaration\n",
      "Rho = 961.              #Density of fluid (kg/m3)\n",
      "Di = 0.0508             #Inner diameter of circular tube (m)\n",
      "V_av = 6.1              #Average velocity of fluid (m/s)\n",
      "K = 2.744               #Consisitency index\n",
      "n = 0.30                #Flow behaiviour index\n",
      "L = 30.5                #Length of tubing, m    \n",
      "f = 0.0032              #Friction factor for turbulent flow\n",
      "#Calcualtion\n",
      "Nre = Di**n*V_av**(2-n)*Rho/K*8**(n-1)\n",
      "delP = 4*f*Rho*L*V_av**2/(2*Di)\n",
      "delP_kN = delP/1000 \n",
      "#Result \n",
      "print \"The frictional pressure drop for turbine is\",round(delP_kN,1), \"kN/m2\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The frictional pressure drop for turbine is 137.4 kN/m2\n"
       ]
      }
     ],
     "prompt_number": 37
    }
   ],
   "metadata": {}
  }
 ]
}