path: root/Practical_C_Programming/Chapter_2_3.ipynb

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   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 2:. Convective Mass Transfer"
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.1,Page number94"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "\n",
      "T = 310  \t\t\t\t\t# [K]\n",
      "\t# Since the solubility of oxygen in water at 310 K is extremely low, we are dealing with \tdilute solutions\n",
      "k_L = 3.3*10**-5  \t\t\t\t# [coefficient based on the oxygen concentration \t\t\t\t\t\tdifference in the water, m/s]\n",
      "row = 993  \t\t\t\t\t# [kg/cubic m]\n",
      "M_b = 18  \t\t\t\t\t# [gram/mole]\n",
      "\n",
      "\n",
      "#Calculation\n",
      " \n",
      "\t# Since we are dealing with very dilute solutions\n",
      "\t# Therefore, c = (row/M_avg) = row/M_b\n",
      "c = row/M_b  \t\t\t\t\t# [kmole/cubic m]\n",
      "\t# Using equation 2.10\n",
      "k_x = k_L*c  \t\t\t\t\t# [kmole/square m.s]\n",
      "\n",
      "#Result\n",
      "\n",
      "print\"The mass-transfer coefficient based on the mole fraction of oxygen in the liquid is\",round(k_x,5),\" kmole/square m.s\" \n",
      "\n",
      " "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The mass-transfer coefficient based on the mole fraction of oxygen in the liquid is 0.00182  kmole/square m.s\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.2,Page number:95"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "#Variable declaration\n",
      "\n",
      "\t#  a-ammonia   b-air\n",
      "T = 300  \t\t\t\t\t# [K]\n",
      "P = 1  \t\t\t\t\t\t# [atm]\n",
      "y_a1 = 0.8  \t\t\t\t\t# [ammonia mole fraction in the bulk of the gas \t\t\t\t\t\t\tphase]\n",
      "y_a2 = 0.732  \t\t\t\t\t# [ammonia gas-phase mole fraction on interface]\n",
      "N_a = 4.3*10**-4  \t\t\t\t# [ammonia flux,  kmole/square m.s]\n",
      "\n",
      "#Calculations\n",
      "\n",
      "import math\n",
      "\t# Using equation 2.2\n",
      "F_g = N_a/math.log10((1-y_a2)/(1-y_a1))  \t\t# [kmole/square m.s]\n",
      "\n",
      "#Result\n",
      "\n",
      "print\"The mass-transfer coefficient in the gas phase at that point where flux is 4.3*10**-4 is\",round(F_g,5),\" kmole/square m.s\"\n",
      "print\"\\n\\nNOTE:Calculation mistake in book:\\nAnswer written as 1.47*10^-4,Actually,,...it is 1.47*10^-3.Please REDO last calculation manually to check\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The mass-transfer coefficient in the gas phase at that point where flux is 4.3*10**-4 is 0.00338  kmole/square m.s\n",
        "\n",
        "\n",
        "NOTE:Calculation mistake in book:\n",
        "Answer written as 1.47*10^-4,Actually,,...it is 1.47*10^-3.Please REDO last calculation manually to check\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.3,Page number:96"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "#Variable declaration\n",
      "\n",
      "\t\t# a-methanol   b-water\n",
      "P = 101.3  \t\t\t\t\t\t# [kPa]\n",
      "y_a1 = 0.707  \t\t\t\t\t\t# [mole fraction at interface]\n",
      "y_a2 = 0.656  \t\t\t\t\t\t# [mole fraction at bulk of the gas]\n",
      "k_g = 1.62*10**-5  \t\t\t\t\t# [mass-transfer coefficient at a point \t\t\t\t\t\t\t in the column, kmole/square m.s.kPa]\n",
      "#Calculations\n",
      "\n",
      "\t# Using equation 2.14\n",
      "k_y = k_g*P  \t\t\t\t\t\t# [kmole/square m.s]\n",
      "\t# Using equation 2.12\n",
      "N_a = k_y*(y_a1-y_a2)  \t\t\t\t\t# [kmole/square m.s]\n",
      "\n",
      "#Result\n",
      "\n",
      "print\"The methanol flux at the point of given mass transfer coefficient is\",round(N_a,7),\"kmole/square m.s\" \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The methanol flux at the point of given mass transfer coefficient is 8.37e-05 kmole/square m.s\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.4,Page number:99"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "n = 6  # [number of variables]\n",
      "\n",
      "#Calculations\n",
      "\n",
      "from scipy.optimize import fsolve\n",
      "from numpy import *\n",
      "import math\n",
      "# To determine the number of dimensionless parameters to be formed, we must know the rank, r, of the dimensional matrix.\n",
      "# The dimensional matrix is \n",
      "DM =matrix([[0,0,1,1,0,0],[1,1,-3,-1,2,1],[-1,-1,0,0,-1,-1]]) \n",
      "rk= linalg.matrix_rank(DM)\n",
      "print\"Rank of matrix is \",rk \n",
      "\n",
      "#The numbers in the table represent the exponent of M, L, and t in the dimensional expression of each of the six variables involved. For example, the dimensional expression of p is M/Lt  hence the exponents are 1, -1, and -1\n",
      "\n",
      "# From equation 2.16\n",
      "i = n-rk  # [number of dimensional groups]\n",
      "# Let the dimensional groups are pi1, pi2 and pi3\n",
      "# Therefore pi1 = (D_AB)**a*(row)**b*(D)**c*kc\n",
      "#           pi2 = (D_AB)**d*(row)**e*(D)**f*v\n",
      "#           pi3 = (D_AB)**g*(row)**h*(D)**i*u\n",
      "\n",
      "# Solving for pi1\n",
      "# M**0*L**0*t**0 = 1 = (L**2/t)**a*(M/L**3)**b*(L)**c*(L/t)\n",
      "\n",
      "# Solution of simultaneous equation\n",
      "def F(e):\n",
      "    f1 = 2*e[0]-3*e[1]+e[2]+1 \n",
      "    f2 = -e[0]-1 \n",
      "    f3 = -e[1] \n",
      "    return(f1,f2,f3)\n",
      "\n",
      "\n",
      "# Initial guess:\n",
      "e = [0.1,0.8,0.5] \n",
      "y = fsolve(F,e) \n",
      "a = y[0] \n",
      "b = y[1] \n",
      "c = y[2] \n",
      "print\"The coefficients of pi1 are\",a,round(b),c \n",
      "# Similarly the coefficients of pi2 and pi3 are calculated\n",
      "# Therefore we get pi1 = kc*D/D_AB = Sh i.e. Sherwood Number\n",
      "#                  pi2 = v*D/D_AB = P_ed i.e. Peclet Number\n",
      "#                  pi3 = u/(row*D_AB) = Sc i.e. Schmidt Number\n",
      "\n",
      "# Dividing pi2 by pi3 gives\n",
      "#       pi2/pi3 = D*v*row/u = Re i.e. Renoylds number\n",
      "\n",
      "print\"The result of the dimensional analysis of forced-convection mass transfer in a circular conduit indicates that a correlating relation could be of the form\\n Sh = function(Re,Sc)\\n which is analogous to the heat transfer correlation \\n Nu = function(Re,Pr)\" "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Rank of matrix is  3\n",
        "The coefficients of pi1 are -1.0 0.0 1.0\n",
        "The result of the dimensional analysis of forced-convection mass transfer in a circular conduit indicates that a correlating relation could be of the form\n",
        " Sh = function(Re,Sc)\n",
        " which is analogous to the heat transfer correlation \n",
        " Nu = function(Re,Pr)\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.6,Page number:111"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "M_a = 352  \t\t\t\t\t# [molecular weight of UF6, gram/mole]\n",
      "M_b = 29  \t\t\t\t\t# [gram/mole]\n",
      "d = 0.01  \t\t\t\t\t# [diameter, m]\n",
      "x = 0.1  \t\t\t\t\t# [length exposed to air stream, m]\n",
      "v = 1  \t\t\t\t\t\t# [m/s]\n",
      "Ts = 303  \t\t\t\t\t# [surface temperature of solid, K]\n",
      "P_a = 27  \t\t\t\t\t# [vapor pressure of UF6, kPa]\n",
      "Tb = 325  \t\t\t\t\t# [bulk temperature of solid ,K]\n",
      "P = 101.3  \t\t\t\t\t# [kPa]\n",
      "R = 8.314  \t\t\t\t\t# [cubic m.Pa/mole.K]\n",
      "import math\n",
      "\n",
      "y_a1 = P_a/P  \t\t\t\t\t# [mole fraction at point 1]\n",
      "y_a2 = 0  \t\t\t\t\t# [mole fraction at point 2]\n",
      "\n",
      "\t# Along the mass-transfer path-cylinder surface (point 1) to bulk air (point 2)\n",
      "Tavg = (Ts+Tb)/2  \t\t\t\t# [K]\n",
      "\n",
      "\t# At point 1, the gas is saturated with UF6 vapor, while at point 2 the gas is virtually \tfree of UF6\n",
      "\t# Therefore\n",
      "Pavg = (P_a+0)/2  \t\t\t\t# [average partial pressure, kPa]\n",
      "y_a = Pavg/P  \t\t\t\t\t# [mole fraction of UF6]\n",
      "\n",
      "Mavg = M_a*y_a+M_b*(1-y_a)  \t\t\t# [gram/mole]\n",
      "row_avg = P*Mavg/(R*Tavg)  \t\t\t# [kg/cubic m]\n",
      "\n",
      "\t# Parameter for c-O2, d-N2 and a-UF6\n",
      "yi_c = 0.182 \n",
      "yi_d = 0.685   \n",
      "yi_a = 0.133  \n",
      "Tc_c = 154.6    \t\t\t\t# [K]\n",
      "Tc_d = 126.2   \t\t\t\t\t# [K]\n",
      "Tc_a = 505.8  \t\t\t\t\t# [K]\n",
      "Pc_c = 50.4    \t\t\t\t\t# [bar]\n",
      "Pc_d = 33.9   \t\t\t\t\t# [bar] \n",
      "Pc_a = 46.6   \t\t\t\t\t# [bar]\n",
      "M_c = 32        \t\t\t\t# [gram/mole]\n",
      "M_d  = 28    \t\t\t\t\t# [gram/mole]   \n",
      "M_a  = 352   \t\t\t\t\t# [gram/mole]\n",
      "V_c = 73.4      \t\t\t\t# [cubic cm/mole]\n",
      "V_d  = 89.8   \t\t\t\t\t# [cubic cm/mole]\n",
      "V_a  = 250  \t\t\t\t\t# [cubic cm/mole]\n",
      "Z_c = 0.288     \n",
      "Z_d  = 0.290   \n",
      "Z_a  = 0.277 \n",
      "\n",
      "#Calculations\n",
      "\n",
      "\n",
      "\t# From equation 2.52 and 2.53\n",
      "Tcm = yi_c*Tc_c+yi_d*Tc_d+yi_a*Tc_a  \t\t# [K]\n",
      "Pcm = 10**6*R*Tcm*(yi_c*Z_c+yi_d*Z_d+yi_a*Z_a)/((yi_c*V_c+yi_d*V_d+yi_a*V_a)*100000)  \t# [bar]\n",
      "M_avg = yi_c*M_c+yi_d*M_d+yi_a*M_a  \t\t# [gram/mole]\n",
      "\n",
      "\t# From equation 2.50\n",
      "Em =  0.176*(Tcm/(M_avg**3*Pcm**4))**(1.0/6.0)  \t# [uP]**-1\n",
      "\n",
      "\t# From equation 2.51\n",
      "Trm = Tavg/Tcm \n",
      "f_Trm = (0.807*Trm**0.618)-(0.357*math.exp(-0.449*Trm))+(0.340*math.exp(-4.058*Trm))+0.018  \n",
      "\t# From equation 2.49 \n",
      "u = f_Trm/Em  \t\t\t\t\t\t# [uP]\n",
      "u = u*10**-7  \t\t\t\t\t\t# [viscosity, kg/m.s]\n",
      "\n",
      "Re = d*v*row_avg/u  \t\t\t\t\t# [Renoylds number]\n",
      "\n",
      "\t\t# Diffusivity of UF6 vapors in air at 314 K and 1 atm from equation 1.49\n",
      "D_ab = 0.0904  \t\t\t\t\t\t# [square cm/s]\n",
      "\n",
      "Sc = u/(row_avg*D_ab*10**-4)  \t\t\t\t# [Schmidt number]\n",
      "\n",
      "Sh_avg = 0.43 + 0.532*Re**0.5*Sc**0.31  \t\t# [Sherwood number]\n",
      "\t\t# From equation 1.7\n",
      "c = P/(R*Tavg)  \t\t\t\t\t# [kmole/cubic m]\n",
      "\t\t# From Table 2.1 \n",
      "F_av = Sh_avg*D_ab*c*10**-4/d  \t\t\t\t# [kmole/square m.s]\n",
      "\n",
      "\t\t# From equation 2.2\n",
      "N_avg = F_av*math.log((1-y_a2)/(1-y_a1))  \t\t# [kmole/square m.s]\n",
      "S = 2*math.pi*d**2/4 +math.pi*d*x  \t\t\t# [total surface area of the cylinder, \t\t\t\t\t\t\tsquare m]\n",
      "\n",
      "w_a = N_avg*S*M_a  \t\t\t\t\t# [rate of sublimation of the solid, \t\t\t\t\t\t\tkg/s] \n",
      "\n",
      "#Result\n",
      "print\"Rate of sublimation of a cylinder of UF6 is\",round(w_a,5),\"kg/s\\n\\n\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Rate of sublimation of a cylinder of UF6 is 0.00023 kg/s\n",
        "\n",
        "\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.7,Page number:116"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "import math\n",
      "#Variable declaration\n",
      "\n",
      "\t# a-benzene      b-nitrogen\n",
      "T = 300  \t\t\t\t\t# [K]\n",
      "P = 101.3  \t\t\t\t\t# [kPa]\n",
      "v =10  \t\t\t\t\t\t# [m/s]\n",
      "R = 8.314  \t\t\t\t\t# [cubic m.Pa/mole.K]\n",
      "\n",
      "\n",
      "n = -0.5 \n",
      "\t# Data on the properties of C02 at 300 K and 1 bar\n",
      "u = 1.5*10**-5  \t\t\t\t# [viscosity, P]\n",
      "Pr = 0.77  \t\t\t\t\t# [Prandtl number]\n",
      "Cp = 853  \t\t\t\t\t# [J/kg.K]\n",
      "\t# Therefore\n",
      "\t# b = 5.086*l**0.5\n",
      "\t#           j_D = j_H = f(Re) = 5.086*(l**0.5)*Re**-0.5\n",
      "\t# From Table 2.1\n",
      "\t#           F = j_D*c*v/Sc**(2/3) = 5.086*(l**0.5)*c*v/(Re**0.5*Sc**(2/3)) = \t\t5.086*(row*v*u)**0.5/(Mavg*Sc**(2.0/3.0))\n",
      "\n",
      "#Calculations\n",
      "\n",
      "\t# Vapor pressure of benzene\n",
      "P_a = math.exp(15.9008-(2788.51/(T-52.36)))  \t# [mm of Hg]\n",
      "P_a = P_a*101.3/760  \t\t\t\t\t# [kPa]\n",
      "\n",
      "\t# Parameter for a-benzene, b-nitrogen \n",
      "yi_a = 0.07      \n",
      "yi_b = 0.93    \n",
      "Tc_a = 562.2     \n",
      "Tc_b = 126.2   \t\t\t\t\t\t# [K]\n",
      "Pc_a = 48.9      \n",
      "Pc_b = 33.9  \t\t\t\t\t\t# [bar]\n",
      "M_a = 78.1       \n",
      "M_b  = 28    \t\t\t\t\t\t# [gram/mole]\n",
      "V_a = 259        \n",
      "V_b  = 89.8  \t\t\t\t\t\t# [cubic cm/mole]\n",
      "Z_a = 0.271      \n",
      "Z_b  = 0.290 \n",
      "sigma_a = 5.349  \n",
      "sigma_b = 3.798  \t\t\t\t\t# [Angstrom]\n",
      "ek_a = 412.3     \n",
      "ek_b = 71.4  \t\t\t\t\t\t# [E/k, K]\n",
      "\n",
      "\n",
      "\t# From equation 2.52 and 2.53\n",
      "Tcm = yi_b*Tc_b+yi_a*Tc_a  \t\t\t\t# [K]\n",
      "Pcm = 10**6*R*Tcm*(yi_b*Z_b+yi_a*Z_a)/((yi_b*V_b+yi_a*V_a)*100000) \t # [bar]\n",
      "M_avg = yi_b*M_b+yi_a*M_a  \t\t\t\t\t\t# [kg/kmole]\n",
      "\n",
      "#RESULT\n",
      "\n",
      "print\"Average molecular weight is\",round(M_avg,1),\"kg/kmole\" \n",
      "\n",
      "row = P*M_avg/(R*T)  \t\t\t\t\t\t\t# [kg/cubic m]\n",
      "\n",
      "#RESULT\n",
      "\n",
      "print\"Density of mixture is\",round(row,2),\"kg/cubic\"\n",
      "\t# From equation 2.50\n",
      "Em =  0.176*(Tcm/(M_avg**3*Pcm**4))**(1.0/6.0)  \t\t\t# [uP]**-1\n",
      "\n",
      "\t# From equation 2.51\n",
      "Trm = T/Tcm \n",
      "f_Trm = (0.807*Trm**0.618)-(0.357*math.exp(-0.449*Trm))+(0.340*math.exp(-4.058*Trm))+0.018  \n",
      "\t# From equation 2.49 \n",
      "u = f_Trm/Em  \t\t\t\t\t\t\t\t# [uP]\n",
      "u = u*10**-7  \t\t\t\t\t\t\t\t# [viscosity, kg/m.s]\n",
      "print\"Average viscosity of mixture is \",round(u,7),\"kg/m.s\\n\\n\" \n",
      "\n",
      "\t# Calculating diffusivity of benzene using equation 1.49\n",
      "D_ab = 0.0986  \t\t\t\t\t\t\t\t# [square cm/s]\n",
      "Sc = u/(row*D_ab*10**-4)  \t\t\t\t\t\t# [Schmidt number]\n",
      "\n",
      "F = 5.086*(row*v*u)**0.5/(M_avg*Sc**(2.0/3.0))  \t\t\t# [kmole/square m.s]\n",
      "\n",
      "\n",
      "#RESULT\n",
      "\n",
      "print\"The required mass transfer coefficient is\",round(F,5),\"kmole/square m.s\" "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Average molecular weight is 31.5 kg/kmole\n",
        "Density of mixture is 1.28 kg/cubic\n",
        "Average viscosity of mixture is  1.64e-05 kg/m.s\n",
        "\n",
        "\n",
        "The required mass transfer coefficient is 0.00196 kmole/square m.s\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.8,Page number:120"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "import math\n",
      "\n",
      "T = 300  \t\t\t\t\t\t# [K]\n",
      "l = 3  \t\t\t\t\t\t\t# [length of vertical plate, m]\n",
      "b = 1.5  \t\t\t\t\t\t# [width of vertical plate, m]\n",
      "P = 101.3  \t\t\t\t\t\t# [kPa]\n",
      "v = 5  \t\t\t\t\t\t\t# [velocity across the width of plate, \t\t\t\t\t\t\tm/s]\n",
      "row_a = 0.88  \t\t\t\t\t\t# [gram/cubic cm]\n",
      "\n",
      "\n",
      "y_a1 = 0.139  \t\t\t\t\t\t# [mole fraction of benzene at inner \t\t\t\t\t\t\tedge]\n",
      "y_a2 = 0  \n",
      "\n",
      "\t# The film conditions, and average properties, are identical to those in Example 2.7, \tonly the geometry is different\n",
      "\t# Therefore\n",
      "M_avg = 31.4  \t\t\t\t\t\t# [kg/kmole]\n",
      "row = 1.2  \t\t\t\t\t\t# [kg/cubic m]\n",
      "u = 161*10**-7  \t\t\t\t\t# [kg/m.s]\n",
      "D_ab = 0.0986 \t\t \t\t\t\t# [square cm/s]\n",
      "Sc = 1.3  \t\t\t\t\t\t# [Schmidt Number]\n",
      "Re = row*v*b/u  \t\t\t\t\t# [Renoylds Number]\n",
      "\n",
      "if Re > 4000:\n",
      "    print\"The flow across the plate is turbulent\\n\" \n",
      "elif Re<2000:\n",
      "    print\"The flow across the plate is laminar\\n\" \n",
      "\t#Using equation 2.57\n",
      "Sh_l = 0.036*Re**0.8*Sc**(1.0/3.0)  \n",
      "\n",
      "\t# Nitrogen (component B) does not react with benzene (component A), neither dissolves in \t\tthe liquid  therefore, NB = 0 and siA = 1. The F-form of the mass-transfer coefficient \t\t\tshould be used    \n",
      "F = Sh_l*1.26*D_ab*10**-4/(M_avg*b)  \t\t\t# [kmole/square m.s]\n",
      "N_a = F*math.log((1-y_a2)/(1-y_a1))  \t\t\t# [kmole/square m.s]\n",
      "\n",
      "\t# The total mass rate of evaporation over the surface of the plate\n",
      "S = 1.5*3  \t\t\t\t\t\t# [square m]\n",
      "M_a = 78.1  \t\t\t\t\t\t# [gram/mole]\n",
      "wa = N_a*S*M_a*60*1000  \t\t\t\t# [gram/min]\n",
      "\n",
      "V = wa/row_a  \t\t\t\t\t\t# [volumetric flow rate, ml/min]\n",
      "\n",
      "print\"Liquid benzene should be supplied at the top of the plate at the rate \",round(V),\"ml/min\\nso that evaporation will just prevent it from reaching the bottom of the plate.\" "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The flow across the plate is turbulent\n",
        "\n",
        "Liquid benzene should be supplied at the top of the plate at the rate  1473.0 ml/min\n",
        "so that evaporation will just prevent it from reaching the bottom of the plate.\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.9,Page number:123"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "dp1 = 10**-3  \t\t\t\t\t# [diameter of spherical drop of water, m]\n",
      "Tair = 323  \t\t\t\t\t# [K]\n",
      "P = 101.3  \t\t\t\t\t# [kPa]\n",
      "Twater = 293  \t\t\t\t\t# [K]\n",
      "R = 8.314  \t\t\t\t\t# [cubic m.Pa/mole.K]\n",
      "M_a = 18.0  \t\t\t\t\t# [gram/mole]\n",
      "M_b = 29.0  \t\t\t\t\t# [gram/mole]\n",
      "import math\n",
      "\n",
      "\n",
      "#Calculation\n",
      "\n",
      "dp2 = (1.0/2.0)**(1.0/3.0)*dp1  \t\t# [m]\n",
      "dp = (dp1+dp2)/2  \t\t\t\t# [m]\n",
      "\n",
      "row_p = 995  \t\t\t\t\t# [density of water, kg/cubic m]\n",
      "row1b = 1.094  \t\t\t\t\t# [density of air, kg/cubic m]\n",
      "u = 1.95*10**-5  \t\t\t\t# [kg/m.s]\n",
      "row_pr = row_p-row1b  \t\t\t\t# [kg/cubic m]\n",
      "g = 9.8  \t\t\t\t\t# [accleration due to gravity, square m/s]\n",
      "\t# Combining equation 2.68 and 2.69\n",
      "Ga = 4*dp**3*row1b*row_pr*g/(3*u**2)  \t\t# [Galileo Number]\n",
      "\n",
      "\t# Relationship between Re and Cd\n",
      "\t# Re/Cd = Re**3/Ga = 3*row**2*vt**3/(4*g*u*row_pr)\n",
      "\n",
      "\t# The following correlation is used to relate Re/Cd, to Ga\n",
      "\t# ln(Re/Cd)**(1/3) = -3.194 + 2.153*ln(Ga)**(1/3) - 0.238*(ln(Ga)**(1/3))**2 + \t0.01068*(ln(Ga)**(1/3))**3\n",
      "\t# Therefore let A = (Re/Cd)\n",
      "A = math.exp(-3.194 + 2.153*math.log(Ga**(1.0/3.0)) - 0.238*(math.log(Ga**(1.0/3.0)))**2 + 0.01068*(math.log(Ga**(1.0/3.0)))**3) \n",
      "\n",
      "\t# Therefore 'vt' will be\n",
      "vt = A*(4*g*row_pr*u/(3*row1b**2))**(1.0/3.0)  \t# [Terminal velocity of particle, m/s]\n",
      "\n",
      "#Result\n",
      "\n",
      "print\"Terminal velocity of particle is\",round(vt,2),\"m/s\"  \n",
      "\n",
      "\n",
      "#Calculation\n",
      "\n",
      "P_w = 2.34  \t\t\t\t\t# [vapor pressure of water, kPa]\n",
      "y_w = P_w/P  \t\t\t\t\t# [mole fraction of water at the inner edge of \t\t\t\t\t\tthe gas film]\n",
      "M_avg = 18*y_w+29*(1-y_w)  \t\t\t# [gram/mole]\n",
      "\n",
      "row2b = P*M_avg/(R*Twater)  \t\t\t# [kg/cubic.m]\n",
      "delta_row = row2b - row1b  \t\t\t# [kg/cubic.m]\n",
      "\n",
      "Tavg = (Tair+Twater)/2  \t\t\t# [K]\n",
      "\t\t# At Temperature equal to Tavg density and viscosity are\n",
      "row3 = 1.14  \t\t\t\t\t# [kg/cubic.m]\n",
      "u1 = 1.92*10**-5  \t\t\t\t# [kg/m.s]\n",
      "\n",
      "Grd = g*row3*delta_row*(dp**3)/(u1**2) \n",
      "\n",
      "\t\t# Diffusivity of water at Tavg and 1 atm is\n",
      "D_ab = 0.242*10**-4  \t\t\t\t# [square m/s]\n",
      "Sc = u1/(row3*D_ab)  \t\t\t\t# [Schmidt Number]\n",
      "Re = dp*row3*vt/u1  \t\t\t\t# [Renoylds Number]\n",
      "\t\n",
      "\t# From equation 2.65 Re is greater than  0.4*Grd**0.5*Sc**(-1/6)\n",
      "\t# Therfore equation 2.64 can be used to calculate mass transfer coefficient\n",
      "\n",
      "Sh = 2+0.552*(Re**0.5)*Sc**(1.0/3.0)  \t\t# [Sherwood Number]\n",
      "\t# From Table 2.1\n",
      "\t# Sh = kc*P_bm*dp/(P*D_ab), since P_bm is almost equal to P\n",
      "\t# Therefore \n",
      "\t# Sh = kc*dp/D_ab \n",
      "kc = Sh*D_ab/dp  \t\t\t\t# [m/s]\n",
      "\n",
      "ca2 = 0  \t\t\t\t\t# [dry air concentration]\n",
      "ca1 = P_w/(R*Twater)  \t\t\t\t# [interface concentration, kmole/cubic.m]\n",
      "\t# Average rate of evaporation \n",
      "wa = math.pi*dp**2*M_a*kc*(ca1-ca2)*1000  \t# [g/s]\n",
      "\n",
      "\t# Amount of water evaporated\n",
      "m = row_p*math.pi*dp1**3/12*1000  \t\t# [g]\n",
      "\t# Time necessary to reduce the volume by 50%\n",
      "t = m/wa  \t\t\t\t\t# [s]\n",
      "\n",
      "D = t*vt  \t\t\t\t\t# [distance of fall, m]\n",
      "\n",
      "#Result\n",
      "\n",
      "print\"The distance of fall is\",round(D),\"m\"  "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Terminal velocity of particle is 3.59 m/s\n",
        "The distance of fall is 90.0 m\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.10,Page number:127"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "Re = 1223  \t\t\t\t# [Renoylds Number]\n",
      "Sc = 0.905  \t\t\t\t# [Schmidt Number]\n",
      "c = 0.039  \t\t\t\t# [molar density, kg/cubic m]\n",
      "v = 1  \t\t\t\t\t# [gas velocity, m/s]\n",
      "\t# Therefore \n",
      "#Calculations\n",
      "\n",
      "Gm = c*v  \t\t\t\t# [kmole/square m.s]\n",
      "\t# From equation 2.9 \n",
      "\t# Kg*P = ky\n",
      "\t# Therefore substituting in equation 2.73 we obtain\n",
      "ky = 0.281*Gm/(Re**0.4*Sc**0.56)  \t# [kmole/square m.s]\n",
      "\t# Now equation 2.73 were obtained under very dilute concentrations\n",
      "\t# Therefore\n",
      "y_bm = 1  \n",
      "\t# From equation 2.6\n",
      "F = ky*y_bm  \t\t\t\t# [kmole/square m.s]\n",
      "\n",
      "#Result\n",
      "\n",
      "print\"Mass transfer coefficient is \",round(F,6),\"kmol/m.^2-s\" \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Mass transfer coefficient is  0.000675 kmol/m.^2-s\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.11,Page number:129"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "D = 25.4*10**-3  \t\t\t\t# [diameter of wetted wall tower, m]\n",
      "Gy = 10  \t\t\t\t\t# [mass velocity, kg/square m.s]\n",
      "T1 = 308  \t\t\t\t\t# [K]\n",
      "P = 101.3  \t\t\t\t\t# [kPa]\n",
      "p_a1 = 1.95  \t\t\t\t\t# [partial pressure of water vapor, kPa]\n",
      "R = 8.314  \t\t\t\t\t# [cubic m.Pa/mole.K]\n",
      "M_a = 18  \t\t\t\t\t# [gram/mole]\n",
      "Cpa = 1.88  \t\t\t\t\t# [kJ/kg.K]\n",
      "\n",
      "\n",
      "# Properties of dry air at 308 K and 1 atm pressure are\n",
      "u = 1.92*10**-5  \t\t\t\t# [kg/m.s]\n",
      "row = 1.14  \t\t\t\t\t# [kg/cubic m]\n",
      "D_ab = 0.242*10**-4  \t\t\t\t# [square m/s]\n",
      "Sc = 0.696  \t\t\t\t\t# [Schmidt number]\n",
      "Cp = 1.007  \t\t\t\t\t# [kJ/kg.K]\n",
      "k = 0.027  \t\t\t\t\t# [W/m.K]\n",
      "Pr = 0.7  \t\t\t\t\t# [Prandtl number]\n",
      "\n",
      "\n",
      "#Calculations\n",
      "\n",
      "import math\n",
      "from scipy.optimize import fsolve\n",
      "from numpy import *\n",
      "\n",
      "Re = D*Gy/u  \t\t\t\t\t# [Renoylds number]\n",
      "# From equation 2,74\n",
      "Sh = 0.023*Re**0.83*Sc**0.44  \t\t\t#[Sherwood number]\n",
      "# From Table 2.1\n",
      "kg = Sh*D_ab/(R*T1*D)*1000  \t\t\t# [mole/square m.s.kPa]\n",
      "\n",
      "# To estimate the heat-transfer coefficient, we use the Dittus-Boelter equation for cooling, equation 2.80\n",
      "Nu = 0.023*Re**0.8*Pr**0.3  \t\t\t# [Nusselt number]\n",
      "# From Table 2.1\n",
      "h = Nu*k/D  \t\t\t\t\t# [W/square m.K]\n",
      "\n",
      "T =373.15  \t\t\t\t\t# [K]\n",
      "lambda_a = 40.63  \t\t\t\t# [kJ/mole]\n",
      "Tc = 647.1  \t\t\t\t\t# [K]\n",
      "\n",
      "# Solution of simultaneous equation 2.78 and 2.79\n",
      "def F(e): \n",
      "    f1=kg*(p_a1 - math.exp(16.3872-(3885.7/(e[0]-42.98))))-e[1] \n",
      "    f2=e[1]*M_a*Cpa*(T1-e[0])/(1-exp(-e[1]*M_a*Cpa/h)) + 1000*e[1]*lambda_a*((1-(e[0]/Tc))/(1-(T/Tc)))**0.38 \n",
      "    return(f1,f2)    \n",
      "\n",
      "\n",
      "# Initial guess\n",
      "e = [292,-0.003] \n",
      "y = fsolve(F,e) \n",
      "Ti = y[0] \n",
      "Na = y[1] \n",
      "\n",
      "print\"The temperature of the liquid water and the rate of water evaporation is\",round(Ti),\"K and\",round(Na,3),\" mole/square m.s respectively\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The temperature of the liquid water and the rate of water evaporation is 295.0 K and -0.013  mole/square m.s respectively\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.12,Page number:131"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "D = 25.4*10**-3  \t\t\t# [Internal diameter of tower, m]\n",
      "Z = 1.5  \t\t\t\t# [length of the wetted section, m]\n",
      "Gy = 10  \t\t\t\t# [mass velocity of air, kg/square m.s]\n",
      "Tair = 308  \t\t\t\t# [K]\n",
      "Twater = 295  \t\t\t\t# [K]\n",
      "P = 101.3  \t\t\t\t# [kPa]\n",
      "M_a = 18.0  \t\t\t\t# [gram/mole]\n",
      "M_b = 29.0  \t\t\t\t# [gram/mole]\n",
      "R = 8.314  \t\t\t\t# [cubic m.Pa/mole.K]\n",
      "\n",
      "#Calculations\n",
      "\n",
      "import math\n",
      "\n",
      "Pa = 2.64  # [kPa]\n",
      "\n",
      "Gm = Gy/M_b  \t# [Assuming that gas phase is basically dry air, kmole/square m.s]\n",
      "\t\t# The properties of dry air at 308 K and 1 atm are (from example 2.9)\n",
      "row = 1.14  \t\t\t\t# [kg/cubic m]\n",
      "u = 1.92*10**-5  \t\t\t# [kg/m.s]\n",
      "D_ab = 0.242*10**-4  \t\t\t# [square m/s]\n",
      "Sc = 0.692  \t\t\t\t# [Schmidt number]\n",
      "\n",
      "Re = Gy*D/u  \t\t\t\t# [Renoylds number]\n",
      "\n",
      "if Re<35000 and Re>2000:\n",
      "    Sh = 0.023*Re**0.83*Sc**0.44  \t# [Sherwood number]    \n",
      "    print\"Sherwood number is\",round(Sh,1) \n",
      "else:\n",
      "    print\"We cannot use equation 2.74\"\n",
      "\n",
      "c = P/(R*Tair)  \t\t\t# [kmole/cubic m]\n",
      "\t# Now using equation 2.89\n",
      "Pa_out = Pa*(1-math.exp((-4*Sh*Z*c*D_ab)/(Gm*D**2)))  \t\t# [kPa]\n",
      "\n",
      "#Result\n",
      "print\"The partial pressure of water in the air leaving the tower is\",round(Pa_out,2),\"kPa\" "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Sherwood number is 51.6\n",
        "The partial pressure of water in the air leaving the tower is 1.94 kPa\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.13,Page number:134"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "Gy = 10.0  \t\t\t\t# [kg/square m.s]\n",
      "dp = 3.5*10**-3  \t\t\t# [diameter of spherical glass beads, m]\n",
      "D = 25.4*10**-3 \t\t\t# [Internal diameter of tower, m]\n",
      "Tair = 308  \t\t\t\t# [K]\n",
      "Twater = 295  \t\t\t\t# [K]\n",
      "P = 101.3  \t\t\t\t# [kPa]\n",
      "M_a = 18  \t\t\t\t# [gram/mole]\n",
      "M_b = 29  \t\t\t\t# [gram/mole]\n",
      "R = 8.314  \t\t\t\t# [cubic m.Pa/mole.K]\n",
      "\n",
      "#Calculation\n",
      "\n",
      "import math\n",
      "from scipy.optimize import fsolve\n",
      "\t# The properties of dry air at 308 K and 1 atm are (from example 2.12)\n",
      "row = 1.14  \t\t\t\t# [kg/cubic m]\n",
      "u = 1.92*10**-5  \t\t\t# [kg/m.s]\n",
      "D_ab = 0.242*10**-4  \t\t\t# [square m/s]\n",
      "Sc = 0.692  \t\t\t\t# [Schmidt number]\n",
      "c = 0.04 \t \t\t\t# [mole/cubic m]\n",
      "Gm = 0.345  \t\t\t\t# [kmole/square m.s]\n",
      "\n",
      "Re = Gy*dp/u  \t\t\t\t# [Renoylds number]\n",
      "if Re<2500 and Re>10:\n",
      "    \t\t\t\t\t# Subsituting in equation 2.90\n",
      "    jd = 1.17*Re**-0.415 \n",
      "    print\"Renoylds number is \",Re\n",
      "else:\n",
      "    print \" \"\n",
      "Std = 0.052/(Sc**(2.0/3.0)) \n",
      "\t\t# From Table 2.1 \n",
      "Sh = Std*Re*Sc  \t\t\t# [Sherwood number]\n",
      "\t\t# From equation 2.94\n",
      "e = 0.406+0.571*(dp/D)  \t\t# [bed porosity]\n",
      "e=round(e,3)\n",
      "\t#Illustration 2.13(a) \n",
      "\t# Solution(a)\n",
      "\t# Now Paout = 0.99*Pa\n",
      "\t# Using equation 2.93 to calculate 'Z'\n",
      "def f12(Z):\n",
      "    return(0.99 - 1 + math.exp(-6*(1-e)*Sh*c*Z*D_ab/(Gm*dp**2))) \n",
      "Z = fsolve(f12,0.06) \n",
      "\n",
      "#Result\n",
      "Z=round(Z[0],3)\n",
      "print\"The depth of packing required is\",Z,\"m=\",Z*100,\"cm\" \n",
      "\n",
      "\t#Illustration 2.13(b)\n",
      "\t# Solution(b)\n",
      "\t# From equation 2.95\n",
      "deltaP = (150*(1-e)/Re + 1.75)*((1-e)*(Gy**2)*Z)/(dp*row*e**3)  \t# [Pa]\n",
      "\n",
      "#Result\n",
      "print\"The gas pressure drop through the bed is\",round(deltaP),\"Pa  (Approx) \\nDUE TO LACK OF PRECISION IN CALCULATION IN BOOK.\\niF DONE MANUALLY,THIS ANSWER STANDS CORRECT\"  "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Renoylds number is  1822.91666667\n",
        "The depth of packing required is 0.078 m= 7.8 cm\n",
        "The gas pressure drop through the bed is 15817.0 Pa  (Approx) \n",
        "DUE TO LACK OF PRECISION IN CALCULATION IN BOOK.\n",
        "iF DONE MANUALLY,THIS ANSWER STANDS CORRECT\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.14,Page number:138"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "m = 40000.0  \t\t\t\t\t# [kg/hr]\n",
      "Twater = 298  \t\t\t\t\t# [K]\n",
      "v = 0.1  \t\t\t\t\t# [superficial velocity, m/s]\n",
      "P = 101.3 \t\t\t\t\t# [kPa]\n",
      "V = 40*10**-3  \t\t\t\t\t# [Flow rate of nitrogen, cubic m/min]\n",
      "d = 2.90*10**-4  \t\t\t\t# [Outside diameter of fibres, m]\n",
      "pf = 0.4  \t\t\t\t\t# [Packing factor]\n",
      "a = 46.84*100  \t\t\t\t\t# [surface area per unit volume, m**-1]\n",
      "R = 8.314  \t\t\t\t\t# [cubic m.Pa/mole.K]\n",
      "\n",
      "#Calculation\n",
      "\n",
      "import math\n",
      "\n",
      "dw = 1000  \t\t\t\t\t# [density of water, kg/cubic m]\n",
      "Ql = m/(3600*1000)  \t\t\t\t# [volumetric water flow rate, cubic m/s]\n",
      "\t# Shell diameter\n",
      "D = (4*Ql/(math.pi*v))**0.5  \t\t\t# [Shell diameter, m]\n",
      "\n",
      "\t# the properties of dilute mixtures of oxygen in water at 298 K\n",
      "u = 0.9  \t\t\t\t\t# [cP]\n",
      "\t# Diffusivity from equation 1.53\n",
      "D_ab = 1.93*10**-9  \t\t\t\t# [square m/s]\n",
      "Sc = 467  \t\t\t\t\t# [Schmidt number]\n",
      "\n",
      "Re = d*v*dw/(u*10**-3)  \t\t\t# [Renoylds number]\n",
      "\n",
      "\t# Substituting in equation (2-97) gives\n",
      "Sh = 0.53*(1-1.1*pf)*((1-pf)/pf)**-0.47*(Re**0.53*Sc**0.33) \n",
      "\n",
      "kl = Sh*D_ab/d  \t\t\t\t# [mass-transfer coefficient on the shell side, \t\t\t\t\t\tm/s]\n",
      "\n",
      "\t# From the specified BFW flow rate\n",
      "L = m/(3600*18)  \t\t\t\t# [kmole/s]\n",
      "\t# From ideal gas law\n",
      "V1 = V*P/(Twater*R*60)  \t\t\t# [kmole/s]\n",
      "\t# From the solubility of oxygen in water at 298 K,\n",
      "M = 4.5*10**4 \n",
      "A = L/(M*V1)  \t\t\t\t\t# [Absorption factor]\n",
      "\n",
      "#Result\n",
      "\n",
      "print\"Absorption factor is\",round(A,3) \n",
      "\n",
      "#Calculation\n",
      "\n",
      "\t# For 99% removal of the dissolved oxygen\n",
      "\t# x_in/x_out = b = 100\n",
      "b = 100 \n",
      "c = 55.5 \t\t\t\t\t# [molar density, kmole/cubic m]\n",
      "\t# Substituting in equation 2.99 yields\n",
      "V_T = (L*math.log(b*(1-A)+A))/(kl*a*c*(1-A)) \t # [cubic m]\n",
      "\n",
      "\t# The module length, Z is\n",
      "Z = V_T/(math.pi*D**2.0/4.0) \n",
      "\n",
      "#Result\n",
      "print\"The shell diameter and module length is\",round(D,3),\"m and\",round(Z,2),\" m respectively\"    "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Absorption factor is 0.503\n",
        "The shell diameter and module length is 0.376 m and 2.15  m respectively\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 2.15,Page number:140"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "#Variable declaration\n",
      "d=2.21/100\t\t\t#[m]\n",
      "mu=8.82*10**-6\t\t#[kg/m-s]\n",
      "rho=2.81\t\t#[kg/m**3]\n",
      "\n",
      "c=34.14\t\t\t#[mol/m**3]\n",
      "D=array([2.228/(10**6),2.065/(10**6),1.832/(10**6)])      #Velocities in [m**2/s]\n",
      "\n",
      "\n",
      "#Calculation\n",
      "#Gy=rho*v\t\t\n",
      "#Re=Gy*d/mu\t\t#Reynolds number\n",
      "Re=21750\n",
      "print \"Reynolds number=\",Re\n",
      "Sc=[]\n",
      "Sh=[]\n",
      "F=[]\n",
      "for i in range(0, 3):\n",
      "    sc=mu/(rho*D[i])                   #Schmidt number\n",
      "    Sc.append(sc)\n",
      "    sh=0.023*Re**(0.83)*sc**(0.44)  #Sherwood number\n",
      "    Sh.append(sh)\n",
      "    f=sh*c*D[i]/d                   #Binary mass transfer coefficient in [mol/m^2-s]\n",
      "    F.append(f)\n",
      "print \"Schmidt number are:\"\n",
      "for i in range(0,3):\n",
      "    print round(Sc[i],3)\n",
      "print \"Sherwood number number are:\"\n",
      "for i in range(0,3):\n",
      "    print round(Sh[i],1)\n",
      "print\"Binary mass transfer coefficients are:\"\n",
      "for i in range(0,3):\n",
      "    print round(F[i],3)\n",
      "#After modifying mathcad program of example 1.17,we have\n",
      "N1=-0.0527                          #[mol/m^2-s]\n",
      "N2=0.0395                          #[mol/m^2-s]\n",
      "N3=0.0132                          #[mol/m^2-s]\n",
      "print\"The program yields the following results:\"\n",
      "print \"N1=\",N1,\"mol/m**2-s\"\n",
      "print \"N2=\",N2,\"mol/m**2-s\"\n",
      "print \"N3=\",N3,\"mol/m**2-s\"\n",
      "    "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Reynolds number= 21750\n",
        "Schmidt number are:\n",
        "1.409\n",
        "1.52\n",
        "1.713\n",
        "Sherwood number number are:\n",
        "106.5\n",
        "110.1\n",
        "116.1\n",
        "Binary mass transfer coefficients are:\n",
        "0.367\n",
        "0.351\n",
        "0.328\n",
        "The program yields the following results:\n",
        "N1= -0.0527 mol/m**2-s\n",
        "N2= 0.0395 mol/m**2-s\n",
        "N3= 0.0132 mol/m**2-s\n"
       ]
      }
     ],
     "prompt_number": 36
    }
   ],
   "metadata": {}
  }
 ]
}