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+{
+ "metadata": {
+  "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 2 : Molecular transport mechanisms"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.1 - Page No :28\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculate the steady-state heat flux across a copper block\n",
+      "\n",
+      "# Variables\n",
+      "deltax = 0.1;  \t\t\t #[m] - thickness of copper block\n",
+      "T2 = 100.;  \t\t\t #[degC] - temp on one side of copper block\n",
+      "T1 = 0.;  \t\t\t #[degC] - temp on other side of the copper block\n",
+      "k = 380.;  \t\t\t #[W/mK] - thermal conductivity\n",
+      "\n",
+      "# Calculations\n",
+      "# using the formula (q/A)*deltax = -k*(T2-T1)\n",
+      "g = -k*(T2-T1)/deltax;\n",
+      "g1 = (g/(4.184*10000));\n",
+      "\n",
+      "# Results\n",
+      "print \" The steady state heat flux across the copper block is q/A = %.1e W/m**2 \\\n",
+      "\\n or in alternate units is  q/A = %.1f cal/cm*sec\"%(g,g1);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The steady state heat flux across the copper block is q/A = -3.8e+05 W/m**2 \n",
+        " or in alternate units is  q/A = -9.1 cal/cm*sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.2 - Page No :29\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculate the momentum flux and force per unit area necessary to maintain a plate in motion\n",
+      "\n",
+      "# Variables\n",
+      "dely = 0.1;  \t\t\t #[m] - distance between two parralel plates\n",
+      "delUx = 0.3;  \t\t\t #[m/sec] - velocity of a plate\n",
+      "mu = 0.001;  \t\t\t #[kg/m*sec] - viscosity\n",
+      "\n",
+      "# Calculations\n",
+      "# using the formula tauyx = F/A = -mu*(delUx/dely)\n",
+      "tauyx = -mu*(delUx/dely);\n",
+      "\n",
+      "# Results\n",
+      "print \"The momentum flux and the the force per unit area, \\nwhich are the same thing \\\n",
+      " is tauyx = F/A = %.3f N/m**2\"%(tauyx);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The momentum flux and the the force per unit area, \n",
+        "which are the same thing  is tauyx = F/A = -0.003 N/m**2\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.3 - Page No :30\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# find the new velocity of the top plate\n",
+      "\n",
+      "# Variables\n",
+      "tauyx = -0.003;  \t\t #[N/m**2] - momentum flux\n",
+      "dely = 0.1;  \t\t\t #[m] - distance between two parallel plates\n",
+      "mu = 0.01;  \t\t\t #[kg/m*sec] - viscosity\n",
+      "\n",
+      "# Calculations\n",
+      "# using the formula tauyx = F/A = -mu*(delUx/dely)\n",
+      "delUx = -((tauyx*dely)/mu)*100;\n",
+      "\n",
+      "# Results\n",
+      "print \" Velocity of the top plate is  deltaUx = %d cm/sec\"%(delUx);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " Velocity of the top plate is  deltaUx = 3 cm/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 4
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.5 - Page No :31\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "'''\n",
+      "Find:\n",
+      "(a) the amount of heat transferred in joules per second;\n",
+      "(b) the number of grams of ice that melt in 30 minutes.\n",
+      "'''\n",
+      "\n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "d = 0.0013;  \t\t\t #[m] - diameter of the tube\n",
+      "delx = 1.;  \t\t\t #[m] - length of the glass tube\n",
+      "T2 = 110.6;  \t\t\t #[degC] - temperature on one end of the rod\n",
+      "T1 = 0.;      \t\t\t #[degC] - temperature on other side of the rod\n",
+      "k = 0.86;  \t    \t\t #[W/m*K] - thermal conductivity\n",
+      "Hf = 333.5;  \t\t\t #[J/g] - heat of fusion of ice\n",
+      "\n",
+      "# Calculations\n",
+      "# (a)using the equation (q/A) = -k*(delt/delx)\n",
+      "A = (math.pi*d**2)/4;\n",
+      "q = A*(-k*(T2-T1)/delx);\n",
+      "\n",
+      "# Results\n",
+      "print \"a) the heat flow is  q = %.2e J/sec\"%(q);\n",
+      "\n",
+      "# (b) dividing the total heat transfer in 30minutes by the amount of heat required to melt 1g of ice\n",
+      "a = abs((q*30*60)/333.5);\n",
+      "print \"b) the amount or grams of ice melted in 30 minutes is %.1e g\"%(a);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) the heat flow is  q = -1.26e-04 J/sec\n",
+        "b) the amount or grams of ice melted in 30 minutes is 6.8e-04 g\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.6 - Page No :36\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "'''\n",
+      "(a) Find the molar flux of CO,.\n",
+      "(b) Find the number of pounds of CO, that pass through the hole in the iron in\n",
+      "one hour.\n",
+      "'''\n",
+      "\n",
+      "import math \n",
+      "from scipy.integrate import quad \n",
+      "\n",
+      "# Variables\n",
+      "d = 1.2*10**-2;  \t\t #[m] - diameter of the hole\n",
+      "Ca1 = 0.083;  \t\t\t #[kmol/m**3]\n",
+      "Ca2 = 0.;  \t\t\t     #[kmol/m**3]\n",
+      "L = 0.04;  \t\t\t     #[m] - thickness of the iron piece \n",
+      "Dab = 1.56*10**-3;  \t #[m**2/sec] - diffusion coefficient of CO2\n",
+      "A = (math.pi*d**2)/4;  \t #area\n",
+      "\n",
+      "# Calculations\n",
+      "# (a)using the formula (Na/)A = (Ja/A) = -Dab(delCa/delx)\n",
+      "def f0(Ca): \n",
+      "\t return 1\n",
+      "\n",
+      "intdCa =  quad(f0,Ca2,Ca1)[0]\n",
+      "\n",
+      "def f1(x): \n",
+      "\t return 1\n",
+      "\n",
+      "intdx =  quad(f1,0,0.04)[0]\n",
+      "\n",
+      "g = (intdCa/intdx)*Dab;\n",
+      "\n",
+      "# Results\n",
+      "print \"a) The molar flux with respect to stationary coordinates is Na/A) = %.3e kmol/m**2*sec\"%(g);\n",
+      "\n",
+      "# using the formula na/A = (Na/A)*Ma\n",
+      "Ma = 44.01;  \t\t\t #[kg/mol] - molcular weight of co2\n",
+      "na = (intdCa/intdx)*Dab*Ma*A*(3600/0.4539);\n",
+      "print \"b) The mass flow rate is %.3f lb/hr\"%(na);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The molar flux with respect to stationary coordinates is Na/A) = 3.237e-03 kmol/m**2*sec\n",
+        "b) The mass flow rate is 0.128 lb/hr\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.7 - Page No :38\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "'''\n",
+      "(a) Calculate the concentration of CO1 entering in kg me3.\n",
+      "(b) Calculate the concentration of NZ entering in the air stream in lb mol fim3.\n",
+      "(c) Calculate the concentration of CO, exiting in mol cmm3.\n",
+      "(d) Calculate the concentration of Oz exiting in kmol me3.\n",
+      "'''\n",
+      "\n",
+      "# Variables\n",
+      "T = 30+273.15;  \t\t\t #[K] temperature\n",
+      "pA = 3.;          \t\t\t #[atm] partial pressure of the component A\n",
+      "R = 0.082057;  \t    \t\t #[atm*m**3*/kmol*K] gas constant\n",
+      "\n",
+      "# Calculation and Results\n",
+      "# (a) using the equation Ca = n/V = pA/(R*T)\n",
+      "Cco2 = pA/(R*T);\n",
+      "Cco2 = Cco2*(44.01);\n",
+      "print \" a) The concentarion of Co2 entering is %.2f kg/m**3\"%(Cco2);\n",
+      "\n",
+      "# (b) using the same equation as above\n",
+      "pN2 = (0.79)*3;  \t\t\t #[atm] partial pressure of mitrogen(as nitrogen is 79% in air)\n",
+      "R = 0.7302;  \t\t    \t #[atm*ft**3*lb/mol*R] - gas constant\n",
+      "T = T*(1.8);  \t\t\t     #[R] temperature\n",
+      "CN2 = pN2/(R*T);\n",
+      "print \" b) The concentration of N2 entering is %.2e lb mol/ft**3\"%(CN2);\n",
+      "\n",
+      "# (c) using the same equation as above\n",
+      "nt = 6.;\n",
+      "nCo2 = 4.;\n",
+      "nO2 = 2.*(0.21);\n",
+      "nN2 = 2.*(0.79);\n",
+      "yCo2 = nCo2/nt;\n",
+      "yO2 = nO2/nt;\n",
+      "yN2 = nN2/nt;\n",
+      "R = 82.057;       \t\t\t #[atm*cm**3/mol*K] - gas constant\n",
+      "T = 30+273.15;  \t\t\t #[K] - temperature\n",
+      "pCo2 = 3*yCo2;\n",
+      "Cco2 = pCo2/(R*T);\n",
+      "print \" c) The concentartion of Co2 in the exit is %.2e mol/cm**3\"%(Cco2);\n",
+      "\n",
+      "# (d) using the same equation as above\n",
+      "R = 8.3143;               \t\t\t #[kPa*m**3/kmol*K] - gas constant\n",
+      "pO2 = 3*(yO2)*(101.325);  \t\t\t #[kPa] - partial pressure\n",
+      "CO2 = pO2/(R*T);\n",
+      "print \" d) The concentration of O2 in the exit stream is %.2e kmol/m**3\"%(CO2);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " a) The concentarion of Co2 entering is 5.31 kg/m**3\n",
+        " b) The concentration of N2 entering is 5.95e-03 lb mol/ft**3\n",
+        " c) The concentartion of Co2 in the exit is 8.04e-05 mol/cm**3\n",
+        " d) The concentration of O2 in the exit stream is 8.44e-03 kmol/m**3\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.8 - Page No :39\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculate the initial rate of mass transfer of the CO,. Is the air transferred?\n",
+      "\n",
+      "import math\n",
+      "\n",
+      "# Variables\n",
+      "delx = 0.3-0;     \t    \t\t #[m] - length\n",
+      "d = 0.05-0;  \t    \t    \t #[m] - diameter\n",
+      "A = (math.pi*d**2)/4;  \t\t\t #[m**2] - area;\n",
+      "R = 8.314*10**3;  \t\t\t     #[N*m/kmol*K] - gas constant\n",
+      "xco1 = 0.15;  \t\t\t        # mole prcent of co in one tank\n",
+      "xco2 = 0.;            \t\t\t # mole percent of co in other tank\n",
+      "p2 = 1.;  \t\t            \t #[atm] - pressure in one tank\n",
+      "p1 = p2;              \t\t\t #[atm] - pressure in other tank\n",
+      "D = 0.164*10**-4;  \t    \t\t #[m**2/sec] - diffusion coefficient\n",
+      "T = 298.15;  \t\t\t         #[K] - temperature\n",
+      "\n",
+      "# Calculations\n",
+      "# using the formula (Na/A) = (Ja/A) = -D*(delca/delx) = -(D/R*T)*(delpa/delx);\n",
+      "delpa = (p2*xco2-p1*xco1)*10**5;  \t\t\t #[N/m**2] - pressure difference\n",
+      "Na = -((D*A)/(R*T))*(delpa/delx);\n",
+      "\n",
+      "# Results\n",
+      "print \"The initial rate of mass transfer of co2 is %.1e kmol/sec\"%(Na);\n",
+      "print \"In order for the pressure to remain at 1 atm, a diffusion of air must occur which is in the opposite\\\n",
+      " direction \\nand equal to %.1e kmol/sec\"%(Na);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The initial rate of mass transfer of co2 is 6.5e-10 kmol/sec\n",
+        "In order for the pressure to remain at 1 atm, a diffusion of air must occur which is in the opposite direction \n",
+        "and equal to 6.5e-10 kmol/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.9 - Page No :44\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Calculate the viscosity of toluene in centipoise.\n",
+      "\n",
+      "# Variables\n",
+      "# given\n",
+      "A = 5.;  \t\t\t     #[m**2] - area of the plates\n",
+      "Ft = 0.083  \t\t\t #[N] - force on the top plate\n",
+      "Fb = -0.027;  \t\t\t #[N] - force on the bottom plate\n",
+      "ut = -0.3;  \t\t\t #[m/sec] - velocity of the top plate\n",
+      "ub = 0.1;  \t\t\t     #[m/sec] - velocity of the bottom plate\n",
+      "dely = 0.01;  \t\t\t #[m]\n",
+      "delux = ut-ub;  \t\t #[m/sec]\n",
+      "\n",
+      "# Calculations\n",
+      "# using the formula tauyx = F/A = -mu(delux/dely)\n",
+      "tauyx = (Ft-Fb)/A;\n",
+      "mu = tauyx/(-delux/dely);  \t\t\t #[Ns/m**2]\n",
+      "mu = mu*10**3;  \t\t\t #[cP]\n",
+      "\n",
+      "# Results\n",
+      "print \" The viscosity of toulene in centipose is %.2f cP\"%(mu);\n",
+      "\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " The viscosity of toulene in centipose is 0.55 cP\n"
+       ]
+      }
+     ],
+     "prompt_number": 15
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.11 - Page No :51\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Predict the diffusion coefficient of water vapor\n",
+      "\n",
+      "# Variables\n",
+      "po = 1.;          \t\t\t #[atm] - pressure\n",
+      "p = 2.;  \t\t        \t #[atm] - pressure\n",
+      "To = 0+273.15;  \t\t\t #[K] - temperature\n",
+      "T = 75+273.15;  \t\t\t #[K] - temperature\n",
+      "Do = 0.219*10**-4;  \t     #[m**2/sec];\n",
+      "n = 1.75;\n",
+      "\n",
+      "# Calculations\n",
+      "# usin the formula D = Do*(po/p)*(T/To)**n\n",
+      "D = Do*(po/p)*(T/To)**n;\n",
+      "\n",
+      "# Results\n",
+      "print \"The diffusion coefficient of water vapour in air at %d atm and %d degC is  \\nD \\\n",
+      " = %.3e m**2/sec\"%(p,T-273.15,D);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The diffusion coefficient of water vapour in air at 2 atm and 75 degC is  \n",
+        "D  = 1.674e-05 m**2/sec\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 3,
+     "metadata": {},
+     "source": [
+      "Example 2.12 - Page No :52\n"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Estimate the viscosity of air and of water\n",
+      "\n",
+      "from scipy.optimize import fsolve \n",
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "# given\n",
+      "T = 53+273.15;  \t\t\t #[K] - temperature\n",
+      "mu1 = 1.91*10**-5;\n",
+      "mu2 = 2.10*10**-5;\n",
+      "T1 = 313.15;  \t\t\t #[K] - temperature \n",
+      "T2 = 347.15;  \t\t\t #[K] - temperature\n",
+      "\n",
+      "# Calculations\n",
+      "# for air\n",
+      "# using linear interpolation of the values in table 2.2\n",
+      "def f(a):\n",
+      "    return  math.log(mu1/a)/math.log(T1);\n",
+      "\n",
+      "def g(a):\n",
+      "    return math.log(mu2)-math.log(a)-f(a)*math.log(T2);\n",
+      "\n",
+      "a1 = 10**-7;\n",
+      "A = fsolve(g,a1)\n",
+      "B = f(A);\n",
+      "\n",
+      "# using the formula ln(mu) = lnA+Bln(t)\n",
+      "mu = math.e**(math.log(A)+B*math.log(T))*10**3;  \t\t\t #[cP]\n",
+      "\n",
+      "# Results\n",
+      "print \" the viscosity of air at %d degC is %.4f cP\"%(T-273.15,mu);\n",
+      "\n",
+      "# similarly for water\n",
+      "BdivR = 1646;\n",
+      "A = 3.336*10**-8;\n",
+      "mu = A*math.e**(BdivR/T)*10**5 \t\t\t #[cP]\n",
+      "print \" the viscosity of water at %d degC is %.3f cP\"%(T-273.15,mu);\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        " the viscosity of air at 53 degC is 0.0198 cP\n",
+        " the viscosity of water at 53 degC is 0.519 cP\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file