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{
"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",
"\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]\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;\n",
"h = 20.;\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": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 13.7 - Page No :688\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\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",
"\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": {}
}
]
}
|
