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diff --git a/Transport_Phenomena/ch13.ipynb b/Transport_Phenomena/ch13.ipynb new file mode 100644 index 00000000..760170dd --- /dev/null +++ b/Transport_Phenomena/ch13.ipynb @@ -0,0 +1,314 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 13 : Unsteady-state transport" + ] + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 13.1 - Page No :651\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "find the Biot number for: (a) a copper sphere\n", + "of radius 5 cm; (b) a copper cylinder of radius 5 cm and length 30 cm; and (c) a\n", + "square copper rod of length 40 cm and cross sectional area the same as the\n", + "cylinder of radius 5 cm.\n", + "'''\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "# given\n", + "h = 12.; \t\t\t #[W/m**2*K] - heat transfer coefficeint\n", + "k = 400.; \t\t\t #[W/m*K] - thermal conductivity\n", + "\n", + "# Calculation and Results\n", + "# (a) for sphere\n", + "r = 5.*10**-2; \t\t\t #[m] - radius of copper sphere\n", + "Lc = ((4*math.pi*((r)**3))/3)/(4*math.pi*((r)**2));\n", + "Nbi = h*Lc*(1./k);\n", + "print \" a) The biot no. is Nbi = %.0e\"%(Nbi);\n", + "\n", + "# (b) for cyclinder\n", + "r = 0.05; \t\t\t #[m] - radius of cyclinder\n", + "L = 0.3; \t\t\t #[m] - height of cyclinder\n", + "Lc = (math.pi*((r)**2)*L)/(2*math.pi*r*L);\n", + "Nbi = h*Lc*(1./k);\n", + "print \" b) The biot no. is Nbi = %.1e\"%(Nbi);\n", + "\n", + "# (c) for a long square rod\n", + "L = .4; \t\t\t #[m] - length of copper rod\n", + "r = 0.05; \t\t\t #[m] - radius of a cyclinder havimg same cross sectional area as that of square\n", + "x = ((math.pi*r**2)**(1./2));\n", + "Lc = ((x**2)*L)/(4*x*L);\n", + "Nbi = h*Lc*(1./k);\n", + "print \" c) The biot no. is Nbi = %.3e\"%(Nbi);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " a) The biot no. is Nbi = 5e-04\n", + " b) The biot no. is Nbi = 7.5e-04\n", + " c) The biot no. is Nbi = 6.647e-04\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 13.6 - Page No :684\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Calculate the temperature at the center after\n", + "1.2s if the heat transfer coefficient is 20 W m-*K-l\n", + "'''\n", + "\n", + "# Variables\n", + "# given\n", + "d = 1*0.0254; \t\t #[m] banana diameter\n", + "Lr = d/2; \t\t\t #[m]; \n", + "Lz = (1.2/2)*(0.0254); \n", + "x = Lz;\n", + "r = Lr;\n", + "k = 0.481; # thermal conductivity\n", + "h = 20.; # heat coefficient\n", + "mr = k/(h*Lr);\n", + "mz = k/(h*Lz);\n", + "nr = r/Lr;\n", + "nz = x/Lz;\n", + "t = 1.2; \t\t\t #[sec]\n", + "\n", + "# Calculations\n", + "alpha = 1.454*10**-4;\n", + "Xr = (alpha*t)/(Lr**2);\n", + "Xz = (alpha*t)/(Lz**2);\n", + "\n", + "# using the above value of m,n,X the value for Ycz and Ycr from fig 13.14 is\n", + "Ycr = 0.42;\n", + "Ycz = 0.75;\n", + "Yc = Ycr*Ycz;\n", + "T_infinity = 400.; \t\t\t #[K]\n", + "To = 295.;\n", + "Tc = T_infinity-(Yc*(T_infinity-To));\n", + "\n", + "# Results\n", + "print \" The temperature t the centre is Tc = %.0f K\"%(Tc);\n", + "\n", + "\n", + "# Answer is vary because of rounding error." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The temperature t the centre is Tc = 367 K\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 13.7 - Page No :688\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Consider the agitation vessel in Example 13.5. Divide the\n", + "thickness into 10 increments and perform two iterations of the explicit method.\n", + "'''\n", + "\n", + "from numpy import *\n", + "# Variables\n", + "# given\n", + "T_x0 = 300.; \t\t\t #[K]\n", + "Tw = 400.; \t\t\t #[K]\n", + "L = 0.013; \t\t\t #[m]\n", + "alpha = 2.476*(10**-5); \t\t\t #[m**/sec]\n", + "h = 600.; \t\t\t #[W/m**2*K]\n", + "pcp = 3.393*(10**6); \t\t\t #[J/m**3*K]\n", + "L = 0.013; \t\t\t #[m]\n", + "del_tax = L/10.;\n", + "betaa = 0.5;\n", + "del_tat = 0.03;\n", + "\n", + "# Calculations\n", + "del_tat = betaa*((del_tax)**2)*(1./alpha);\n", + "T_infinity = 400.; \t\t\t #[K]\n", + "\n", + "# to be sure that the solution is stable, it is customary to truncate this number\n", + "del_tat = 0.03; \t\t\t #[sec]\n", + "# betaa = alpha*del_tat*((1./del_tax)**2);\n", + "Told = zeros(11)\n", + "for i in range(11):\n", + " Told[i] = 300.;\n", + "\n", + "a = ((2*h*del_tat)/(pcp*del_tax));\n", + "b = ((2*alpha*del_tat)/(pcp*((del_tax)**2)));\n", + "\n", + "Tnew = zeros(11)\n", + "for j in range(11):\n", + " Tnew[0] = (T_infinity*0.08162)+(Told[0]*(1-0.08162-0.8791))+(Told[1]*0.8791)\n", + " for k in range(9):\n", + " Tnew[k+1] = (betaa*Told[k+2])+((1.-2*betaa)*(Told[k+1]))+(betaa*Told[k]);\n", + " Tnew[10] = ((2*betaa)*(Told[9]))\n", + " Told = Tnew;\n", + "# Results\n", + "print \"Told values : \" ,(Told);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Told values : [ 325.54820838 319.78194857 315.05971328 311.28295197 308.32959437\n", + " 306.07276601 304.39590474 303.20406441 302.43143939 302.04512688\n", + " 302.04512688]\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 13.9 - Page No :700\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Estimate the depth that corresponds to a\n", + "temperature of -5\u00b0C if it is assumed that the period of time is 4 months.\n", + "'''\n", + "\n", + "# Variables\n", + "p = 2050.; \t\t\t #[kg/m**3] - density of soil\n", + "cp = 1840.; \t\t\t #[J/kg*K] - heat cpapacity of soil\n", + "k = 0.52; \t\t\t #[W/m*K] - thermal conductivity of soil\n", + "alpha = 0.138*10**-6; \t\t\t #[m**2/sec]\n", + "t = 4*30*24*3600; \t\t\t #[sec] - no. of seconds in 4 months\n", + "Tx = -5.; \t\t\t #[degC]\n", + "Tinf = -20.; \t\t\t #[degC]\n", + "T0 = 20.; \t\t\t #[degC]\n", + "\n", + "# from the fig 13.24 the dimensionless dismath.tance Z is \n", + "Z = 0.46;\n", + "\n", + "# Calculations\n", + "# then the depth is\n", + "x = 2*((alpha*t)**(1./2))*Z\n", + "\n", + "# Results\n", + "print \" the depth is x = %.1f m = %.1f ft\"%(x,x*3.6/1.10);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the depth is x = 1.1 m = 3.6 ft\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 13.10 - Page No :701\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# find the concentration of KC1 at the center after 60min.\n", + "\n", + "# Variables\n", + "d = 0.01; \t\t\t #[m] - diameter of cyclindrical porous plug\n", + "D = 2.*10**-9; \t\t\t #[m**2/sec] - diffusion coefficient\n", + "t = 60.*60; \t\t\t #[sec]\n", + "r = d/2.;\n", + "m = 0.;\n", + "Ca_inf = 0.;\n", + "Ca_0 = 10.;\n", + "X = (D*t)/((r)**2);\n", + "# from fig 13.14 the ordinate is\n", + "Y = 0.7;\n", + "\n", + "# Calculations\n", + "Ca_c = Ca_inf-Y*(Ca_inf-Ca_0);\n", + "\n", + "# Results\n", + "print \" the concentration of KCL at the centre after 60 min is Ca = %.2f kg/m**3\"%(Ca_c);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " the concentration of KCL at the centre after 60 min is Ca = 7.00 kg/m**3\n" + ] + } + ], + "prompt_number": 8 + } + ], + "metadata": {} + } + ] +}
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