From 69aa63da6cb2bc3f5822533e3f97c46d7efbaafc Mon Sep 17 00:00:00 2001
From: root
Date: Fri, 11 Jul 2014 14:35:18 +0530
Subject: removing books
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Transport_Phenomena/ch13.ipynb | 298 -----------------------------------------
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-{
- "metadata": {
- "name": "",
- "signature": "sha256:855f448243394a83380cc86f4309c883de0251b8f54f502d859f2d887a1b20dc"
- },
- "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",
- "\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",
- "\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",
- "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",
- "# 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": [
- "\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": {}
- }
- ]
-}
\ No newline at end of file
--
cgit