{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 4 : molecular transport and the general property balance" ] }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.1 - Page No :99\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math\n", "\n", "# Variables\n", "# given\n", "id_ = 2.067; \t\t\t #[in] - inside diameter\n", "t = 0.154; \t\t\t #[in] - wall thickness\n", "od = id_+2*t; \t\t\t #[in] - outer diameter\n", "a = 1.075; \t\t\t #[in**2] - wall sectional area of metal\n", "A = a*(1./144); \t\t #[ft**2] - wall sectional area of metal in ft**2\n", "deltaz = 5./12; \t\t #[ft] - length of transfer in z direction\n", "T2 = 10+273.15; \t\t #[K] - temperature at the top\n", "T1 = 0+273.15; \t\t #[K] - temperature at the bottom\n", "q = -3.2; \t\t\t #[Btu/hr] - heat transferred\n", "\n", "# Calculations\n", "deltaT = (T2-T1)+8; \t\t\t #[degF]\n", "k = round(-(q/A)/(deltaT/deltaz),2);\n", "\n", "# Results\n", "print \"Thermal conductivity = %.2f Btu h**-1 ft**-1 degF**-1\"%(k);\n", "Alm = round((2*math.pi*deltaz*((od-id_)/(2*12)))/math.log(od/id_),3); \t\t\t #[ft**2] log-mean area\n", "kincorrect = round(k*(A/Alm),3);\n", "print \"kincorrect = %.3f Btu h**-1 ft**-1 degF**-1 \"%(kincorrect);\n", "print \"The error is a factor of %.1f\"%(32.4)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Thermal conductivity = 9.92 Btu h**-1 ft**-1 degF**-1\n", "kincorrect = 0.306 Btu h**-1 ft**-1 degF**-1 \n", "The error is a factor of 32.4\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.2 - Page No :100\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math \n", "\n", "\n", "# Variables\n", "# given\n", "T1 = 0.; \t\t\t #[degC]\n", "T2 = 10.; \t \t\t #[degC]\n", "km = 17.17; \t\t\t #[W/m*K]\n", "l = 1.; \t\t \t #[m]\n", "r2 = 1.1875;\n", "r1 = 1.0335;\n", "deltaT = T1-T2;\n", "\n", "# Calculations\n", "# umath.sing the formula Qr = -km*((2*pi*l)/ln(r2/r1))*deltaT;\n", "Qr = -km*((2*math.pi*l)/math.log(r2/r1))*deltaT;\n", "\n", "# Results\n", "print \"Heat loss = %.0f W \\nThe plus sign indicates that the heat flow is radially out from the center\"%(Qr);\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat loss = 7767 W \n", "The plus sign indicates that the heat flow is radially out from the center\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.3 - Page No :100\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "# given\n", "km = 9.92; \t \t\t #[Btu/h*ft*degF]\n", "Alm = round(0.242*(12./5),3); \t\t\t #[ft**2]\n", "T1 = 0.; \t\t\t #[degC]\n", "T2 = 10.; \t\t\t #[degC]\n", "deltaT = (T1-T2)*1.8; \t\t\t #[degF]\n", "r2 = 1.1875;\n", "r1 = 1.0335;\n", "deltar = round((r2-r1)/12,3); \t\t\t #[ft]\n", "\n", "# Calculations\n", "# using the formula Qr/Alm = -km*(deltaT/deltar)\n", "Qr = (-km*Alm*(deltaT/deltar));\n", "\n", "# Results\n", "print \" qr by log-mean area method = %.0f Btu/h\"%(Qr);\n", "\n", "\n", "# in SI units \n", "Alm = 0.177; \t\t\t #[m**2]\n", "T1 = 0; \t\t\t #[degC]\n", "T2 = 10; \t\t\t #[degC]\n", "km = 17.17; \t\t\t #[W/m*K]\n", "r2 = 1.1875;\n", "r1 = 1.0335;\n", "deltaT = T1-T2;\n", "deltar = (r2-r1)*0.0254; \t\t\t #[m]\n", "\n", "# umath.sing the same formula\n", "Qr = (-km*(deltaT/deltar))*Alm;\n", "print \" qr in SI units = %.0f W\"%(Qr);\n", "\n", "# Note : Answers are wrong in book. Please calculate manually." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " qr by log-mean area method = 7980 Btu/h\n", " qr in SI units = 7769 W\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.4 - Page No :101\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "from scipy.integrate import quad \n", "\n", "# Variables\n", "# given\n", "x1 = 0; \t\t\t #[cm]\n", "x2 = 30; \t\t\t #[cm]\n", "p1 = 0.3; \t\t\t #[atm]\n", "p2 = 0.03; \t\t\t #[atm]\n", "D = 0.164; \t\t\t #[am**2/sec]\n", "R = 82.057; \t\t\t #[cm**3*atm/mol*K]\n", "T = 298.15; \t\t\t #[K]\n", "\n", "# Calculations\n", "# using the formula Nax*int(dx/Ax) = -(D/RT)*int(1*dpa)\n", "def f4(x): \n", "\t return 1./((math.pi/4)*(10-(x/6))**2)\n", "\n", "a = quad(f4,x1,x2)[0]\n", "\n", "def f5(p): \n", "\t return 1\n", "\n", "b = quad(f5,p1,p2)[0]\n", "Nax = -((D/(R*T))*b)/a;\n", "\n", "# Results\n", "print \"Mass transfer rate = %.2e mol/sec = %.2e mol/h \\nthe plus sign indicates diffusion to the right\"%(Nax,Nax*3600);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Mass transfer rate = 2.37e-06 mol/sec = 8.53e-03 mol/h \n", "the plus sign indicates diffusion to the right\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.5 - Page No :105\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "from sympy import *\n", "\n", "# Variables\n", "# given\n", "r = Symbol('r')\n", "ro = 0.5; \t\t\t #[inch] - outside radius\n", "ro = 0.0127; \t\t #[m] - outside radius in m\n", "Tg = 2.*10**7; \t #[J/m**3*sec] - heat generated by electric current\n", "Tw = 30.; \t\t\t #[degC] - outside surface temperature\n", "km = 17.3; \t\t #[W/m*K] - mean conductivity\n", "\n", "# Calculations\n", "# using the formula T = Tw+(Tg/4*km)*(ro**2-r**2)\n", "T = Tw+(Tg/(4*km))*(ro**2-r**2);\n", "\n", "# Results\n", "print \"T = \",T,\n", "print \" where r is in meters and T is in degC\"\n", "def t(r):\n", " return Tw+(Tg/(4*km))*(ro**2-r**2);\n", "\n", "print \"At the centre line r = 0, the maximum temperature is %.1f degC. \\\n", "\\nAt the outside the temperature reduces to the boundary condition value of %.2f degC.\\\n", "\\nThe distribution is parabolic between these 2 limits\"%(t(0),t(0.0127));\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "T = -289017.341040462*r**2 + 76.6156069364162 where r is in meters and T is in degC\n", "At the centre line r = 0, the maximum temperature is 76.6 degC. \n", "At the outside the temperature reduces to the boundary condition value of 30.00 degC.\n", "The distribution is parabolic between these 2 limits\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 4.7 - Page No :119\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import math\n", "\n", "# Variables\n", "# given\n", "r = 10.**-3; \t\t\t #[m] - radius\n", "l = 1.; \t\t\t #[m] - length\n", "Q = 10.**-7; \t\t\t #[m**3/s] - flow rate\n", "pressure = 1.01325*10**5\n", "sPage_No = 1.1;\n", "pwater = 1000.; \t\t #[kg/m**3] - density of water at 4degC\n", "\n", "# Calculations\n", "deltap = round((145 * pressure)/14.696,-4)\n", "pfluid = sPage_No *pwater;\n", "mu = abs(r*-(deltap)*(math.pi*r**3))/((4*Q)*(2*l));\n", "mupoise = mu*10;\n", "mucentipoise = mupoise*100;\n", "\n", "# Results\n", "print \" mu = %.3f Ns-m**-2 = %.2f poise = %.0f cP\"%(mu,mupoise,mucentipoise);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " mu = 3.927 Ns-m**-2 = 39.27 poise = 3927 cP\n" ] } ], "prompt_number": 2 } ], "metadata": {} } ] }