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diff --git a/sample_notebooks/MohdGufran/chapter_10_1.ipynb b/sample_notebooks/MohdGufran/chapter_10_1.ipynb new file mode 100644 index 00000000..fc8eeccf --- /dev/null +++ b/sample_notebooks/MohdGufran/chapter_10_1.ipynb @@ -0,0 +1,358 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter No - 10 : Mass Transfer\n", + " " + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 10.1 - Page No. : 318" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "P1=4 # in bar\n", + "P2=2 # in bar\n", + "T=25 # in degree C\n", + "Dhp=9*10**-8 # in m**2/s\n", + "S=3*10**-3 # in kg mole/m**3 bar\n", + "del_x=0.5*10**-3 # thickness in m\n", + "#(a) The molar concentration of a gas in terms of solubility\n", + "CH1=S*P1 # in kg mole/m**3\n", + "CH2=S*P2 # in kg mole/m**3\n", + "#(b) Molar diffusion flux of hydrogen through plastic memberence is given by Fick's law of diffision\n", + "#N_H= N_h/A = Dhp*(CH1-CH2)/del_x#\n", + "N_H= Dhp*(CH1-CH2)/del_x # in kg mole/s-m**2\n", + "print \"Molar diffusion flux of hydrogen through the membrane = %0.2e kg mole/s-m**2\" %N_H\n", + "#Mass_d_Flux= N_H*Molecular_Weight \n", + "Molecular_Weight=2#\n", + "Mass_d_Flux= N_H*Molecular_Weight \n", + "print \"Molar diffusion flux = %0.3e kg/s-m**2\" %Mass_d_Flux" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Molar diffusion flux of hydrogen through the membrane = 1.08e-06 kg mole/s-m**2\n", + "Molar diffusion flux = 2.160e-06 kg/s-m**2\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 10.2 - Page No. : 322" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "T=25 # in degree C\n", + "T=T+273 # in K\n", + "P=1#\n", + "V1=12 #Molecular volume of H2 in cm**3/gm mole\n", + "V2=30 #Molecular volume of Air in cm**3/gm mole\n", + "M1=2 # Molecular weight of H2\n", + "M2=29 # Molecular weight of Air\n", + "#The diffusion coefficient for gases in terms of molecular volumes may be express as\n", + "D_AB= .0043*T**(3/2)/(P*(V1**(1/3)+V2**(1/3)))*(1/M1+1/M2)**(1/2)#\n", + "print \"The diffusion coefficient for gases in terms of molecular volumes = %0.3f cm**2/sec\" %D_AB" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The diffusion coefficient for gases in terms of molecular volumes = 2.997 cm**2/sec\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 10.3 - Page No. : 322" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "T=300 # temp of gas mixture in K\n", + "D_HN2=18*10**-6 # in m**2/s at 300 K, 1 bar\n", + "T1=300 # in K\n", + "D_HO2=16*10**-6 # in m**2/s at 273 K, 1 bar\n", + "T2=273 # in K\n", + "O_2=0.2#\n", + "N_2=0.7#\n", + "H_2=0.1#\n", + "#The diffusivity at the mixture temperature and pressure are calculated as \n", + "# D1/D2 = (T1/T2)**(3/2)*(P2/P1)\n", + "D_HO2= (T/T2)**(3/2)*1/4*D_HO2#\n", + "D_HN2= (T/T1)**(3/2)*1/4*D_HN2#\n", + "#The composition of oxygen and nitrogen on a H2 free basis is \n", + "x_O= O_2/(1-H_2)#\n", + "x_N= N_2/(1-H_2)#\n", + "\n", + "# The effective diffusivity for the gas mixture at given temperature and pressure is\n", + "D= 1/(x_O/D_HO2+x_N/D_HN2) # in m**2/s\n", + "print \"Effective diffusivity = %0.3e m**2/s\" %D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Effective diffusivity = 4.524e-06 m**2/s\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 10.4 - Page No. : 323" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# Given data\n", + "d=3 # in mm\n", + "d=d*10**-3 # in meter\n", + "T=25 # in \u00b0C\n", + "T=T+273 # in K\n", + "D= 0.4*10**-4 # in m**2/s\n", + "R= 8314#\n", + "P_A1=1 # in atm\n", + "P_A1=P_A1*10**5 # in w/m**2\n", + "P_A2=0#\n", + "C_A2=0#\n", + "x2= 15 # in meter\n", + "x1= 0#\n", + "A= pi/4*d**2#\n", + "M_A= D*A/(R*T)*(P_A1-P_A2)/(x2-x1) # in kg mole/sec\n", + "N_B= M_A#\n", + "M_B= M_A*29 # in kg/sec\n", + "print \"Value of N_B = %0.3e kg mole/sec\" %N_B\n", + "print \"Value of M_B = %0.3e kg /sec\" %M_B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Value of N_B = 7.608e-13 kg mole/sec\n", + "Value of M_B = 2.206e-11 kg /sec\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 10.5 - Page No. : 325" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "# Given data\n", + "P=3 # in atm\n", + "P=P*10**5 # in N/m**2\n", + "r1=10 # in mm\n", + "r1=r1*10**-3 # in m\n", + "r2=20 # in mm\n", + "r2=r2*10**-3 # in m\n", + "R=4160 # in J/kg-K\n", + "T=303 # in K\n", + "D=3*10**-8 # in m**2/s\n", + "S=3*0.05# # Solubility of hydrogen at a pressure of 3 atm in m**3/m**3 of rubber tubing\n", + "del_x=r2-r1 # in m\n", + "L=1 # in m\n", + "Am=2*pi*L*del_x/log(r2/r1)#\n", + "#Formula P*V= m*R*T\n", + "V=S#\n", + "m=P*V/(R*T) # in kg/m**3 of rubber tubing at the inner surface of the pipe\n", + "C_A1=m#\n", + "C_A2=0#\n", + "#Diffusion flux through the cylinder is given\n", + "M=D*(C_A1-C_A2)*Am/del_x#\n", + "print \"Diffusion flux through the cylinder = %0.2e kg/sm\" %M" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diffusion flux through the cylinder = 9.71e-09 kg/sm\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 10.6 - Page No. : 329" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "# Given data\n", + "R=4160 # in J/kg-K\n", + "M=2#\n", + "D_AB=1.944*10**-8 # in m**2/s\n", + "R_H2=R/M#\n", + "S=2*0.0532# # Solubility of hydrogen at a pressure of 2 atm in cm**3/cm**3 of pipe\n", + "P=2 # in atm\n", + "P=P*1.03*10**5 # N/m**2\n", + "T=25 # in degree C\n", + "T=T+273 # in K\n", + "r1=2.5 # in mm\n", + "r1=r1*10**-3 # in m\n", + "r2=5 # in mm\n", + "r2=r2*10**-3 # in m\n", + "del_x=r2-r1 # in m\n", + "L=1 # in m\n", + "#Formula P*V= m*R*T\n", + "V=S#\n", + "m=P*V/(R*T) # in kg/m**3 of pipe\n", + "# So, Concentration of H2 at inner surface of the pipe\n", + "C_A1=0.0176 # in kg/m**3\n", + "# The resistance of diffusion of H2 away from the outer surface is negligible i.e.\n", + "C_A2=0#\n", + "Am=2*pi*L*del_x/log(r2/r1)#\n", + "# Loss of H2 by diffusion \n", + "M_A= D_AB*(C_A1-C_A2)*Am/del_x#\n", + "print \"Loss of H2 by diffusion = %0.2ef kg/s\" %M_A" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Loss of H2 by diffusion = 3.10e-09f kg/s\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example No : 10.7 - Page No. : 330" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from numpy import pi\n", + "from math import log\n", + "# Given data\n", + "Px1= 0.14 # in bar\n", + "Px2= 0#\n", + "P=1.013 # in bar\n", + "Py1=P-Px1# # in bar\n", + "Py2=P-Px2# # in bar\n", + "D=8.5*10**-6 # in m**2/s\n", + "d=5 # diameter in meter\n", + "L=1 # in mm\n", + "L=L*10**-3 #in meter\n", + "M=78 # molecular weight\n", + "Am_x= 1/4*pi*d**2*M#\n", + "R=8314#\n", + "del_x=3 # thickness in mm\n", + "del_x=del_x*10**-3 # in m\n", + "T=20 # in degree C\n", + "T=T+273 # in K\n", + "P=P*10**5 # in N/m**2\n", + "m_x= D*Am_x*P*log(Py2/Py1)/(R*T*del_x)#\n", + "# The mass of the benzene to be evaporated\n", + "mass= 1/4*pi*d**2*L#\n", + "density=880 # in kg/m**3\n", + "m_b= mass*density#\n", + "toh=m_b/m_x # in sec\n", + "print \"Time taken for the entire organic compound to evaporate = %0.0f seconds\" %toh\n", + "\n", + "\n", + "# Note: Answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time taken for the entire organic compound to evaporate = 644 seconds\n" + ] + } + ], + "prompt_number": 15 + } + ], + "metadata": {} + } + ] +}
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