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Diffstat (limited to 'sample_notebooks')
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diff --git a/sample_notebooks/PrashantSahu/Chapter-2-Molecular_Diffusion_-_Principles_of_Mass_Transfer_and_Separation_Process_by_Binay_K_Dutta.ipynb b/sample_notebooks/PrashantSahu/Chapter-2-Molecular_Diffusion_-_Principles_of_Mass_Transfer_and_Separation_Process_by_Binay_K_Dutta.ipynb new file mode 100644 index 00000000..3f6a6b0c --- /dev/null +++ b/sample_notebooks/PrashantSahu/Chapter-2-Molecular_Diffusion_-_Principles_of_Mass_Transfer_and_Separation_Process_by_Binay_K_Dutta.ipynb @@ -0,0 +1,753 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "#Chapter 2 : Molecular Diffusion"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 2.1 Page no. 10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Molar average velocity of gas mixture is: 0.0303 m/s\n",
+ "Mass average velocity of gas mixture is: 0.029 m/s\n"
+ ]
+ }
+ ],
+ "source": [
+ "N2 = 0.05 #mole fraction of Nitrogen denoted as 1\n",
+ "H2 = 0.15 #mole fraction of Hydrogen denoted as 2\n",
+ "NH3 = 0.76 #mole fraction of Ammonia denoted as 3\n",
+ "Ar = 0.04 #mole fraction of Argon denoted as 4\n",
+ "u1 = 0.03\n",
+ "u2 = 0.035\n",
+ "u3 = 0.03\n",
+ "u4 = 0.02\n",
+ "#Calculating molar average velocity\n",
+ "U = N2*u1 + H2*u2 + NH3*u3 + Ar*u4\n",
+ "print 'Molar average velocity of gas mixture is: %.4f m/s'%U\n",
+ "#Calculating of mass average velocity\n",
+ "M1 = 28\n",
+ "M2 = 2\n",
+ "M3 = 17\n",
+ "M4 = 40\n",
+ "M = N2*M1 + H2*M2 + NH3*M3 + Ar*M4\n",
+ "u = (1/M)*(N2*M1*u1 + H2*M2*u2 + NH3*M3*u3 + Ar*M4*u4)\n",
+ "print 'Mass average velocity of gas mixture is: %.3f m/s'%u"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ " ##Example 2.2 Page no. 16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false,
+ "scrolled": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(a) Time for complete evaporation is: 15.93 hours\n",
+ "(b) Time for disappearance of water is: 8.87 hours\n"
+ ]
+ }
+ ],
+ "source": [
+ "from math import log\n",
+ "from math import exp\n",
+ "import numpy as np\n",
+ "\n",
+ "#Calcualtion for (a) part\n",
+ "#calculating vapor pressure of water at 301K\n",
+ "pv = exp(13.8573 - (5160.2/301)) #in bar\n",
+ "#wet-bulb temperature is 22.5 degree centigrade\n",
+ "#calculating mean air-film temperature\n",
+ "Tm = ((28+22.5)/2)+273 #in kelvin\n",
+ "#calculating diffusion coefficient\n",
+ "Dab = ((0.853*(30.48**2))*((298.2/273)**1.75))/(3600*10000) #in m^2/s\n",
+ "l = 2.5e-3 #in m\n",
+ "P = 1.013 #in bar\n",
+ "R = 0.08317 #Gas constant\n",
+ "pAo = exp(13.8573 - (5160.2/295.2)) #vapor pressure of water at the wet-bulb temperature, 22.2C\n",
+ "pAl = 0.6*round(pv,4)\n",
+ "Na = (((round(Dab,7)*P)/(R*298.2*l))*log((P-pAl)/(P-round(pAo,3))))*18 #in kg/m^2s\n",
+ "#amount of water per m^2 of floor area is\n",
+ "thickness = 2e-3\n",
+ "Amount = thickness*1 #in m^3 \n",
+ "#density of water is 1000kg/m^3\n",
+ "#therefore in kg it is\n",
+ "amount = Amount*1000\n",
+ "Time_for_completion = amount/Na #in seconds\n",
+ "Time_for_completion_hours = Time_for_completion/3600\n",
+ "print '(a) Time for complete evaporation is: %.2f'%Time_for_completion_hours,'hours'\n",
+ "\n",
+ "#Calculation for (b) part\n",
+ "water_loss = 0.1 #in kg/m^2.h\n",
+ "water_loss_by_evaporation = Na*3600\n",
+ "total_water_loss = water_loss + water_loss_by_evaporation\n",
+ "time_for_disappearance = amount/total_water_loss\n",
+ "print '(b) Time for disappearance of water is: %0.2f'%time_for_disappearance,'hours'\n",
+ "#Answers may vary due to round off error"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 2.3 Page no. 17"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(a) The molar flux of Ammonia is:1.922E-05 gmol/cm^2.s\n",
+ "(b) and (c)\n",
+ "Velocity of A is 0.522 cm/s\n",
+ "Velocity of B is 0.000 cm/s\n",
+ "Mass average velocity of A is 0.439 cm/s\n",
+ "Molar average velocity of A is 0.47 cm/s\n",
+ "(d) Molar flux of NH3 is 3.062E-06 gmol/cm^2.s\n"
+ ]
+ },
+ {
+ "data": {
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+ "text/plain": [
+ "<matplotlib.figure.Figure at 0x3fa6940>"
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "%matplotlib inline\n",
+ "from math import log\n",
+ "from math import exp\n",
+ "import numpy as np\n",
+ "from matplotlib import pyplot as plt\n",
+ "#calculation for (a) part\n",
+ "l = 1 #thickness of air in cm\n",
+ "pAo = 0.9 #in atm\n",
+ "pAl = 0.1 #in atm\n",
+ "Dab = 0.214 #in cm^2/s\n",
+ "T = 298 #in K\n",
+ "P = 1 #in atm\n",
+ "R = 82.1 #in (cm^3)(atm)/(K)(gmol)\n",
+ "#calculating molar flux of ammonia\n",
+ "Na = ((Dab*P)/(R*T*l))*log((P-pAl)/(P-pAo))\n",
+ "print '(a) The molar flux of Ammonia is:%0.3E'%Na,'gmol/cm^2.s'\n",
+ "\n",
+ "#calculation for (b) and (c) part\n",
+ "Nb = 0 #air is non-diffusing\n",
+ "U = (Na/(P/(R*T))) #molar average velocity\n",
+ "yA = pAo/P\n",
+ "yB = pAl/P\n",
+ "uA = U/yA #\n",
+ "uB = 0 #since Nb=0\n",
+ "Ma = 17\n",
+ "Mb = 29\n",
+ "M = Ma*yA + Mb*yB\n",
+ "u = uA*yA*Ma/M #since u =(uA*phoA + uB*phoB)/pho\n",
+ "print '(b) and (c)'\n",
+ "print 'Velocity of A is %0.3f'%uA,'cm/s'\n",
+ "print 'Velocity of B is %0.3f'%uB,'cm/s'\n",
+ "print 'Mass average velocity of A is %0.3f'%u,'cm/s'\n",
+ "print 'Molar average velocity of A is %0.2f'%U,'cm/s'\n",
+ "\n",
+ "#calculation for (d) part\n",
+ "Ca = pAo/(R*T)\n",
+ "Ia = Ca*(uA - u) #molar flux of NH3 relative to an observer moving\n",
+ " #with the mass average velocity \n",
+ "print '(d) Molar flux of NH3 is %0.3E'%Ia,'gmol/cm^2.s'\n",
+ "\n",
+ "z = []\n",
+ "pa =[]\n",
+ "for i in np.arange(0,1,0.01):\n",
+ " z.append(i)\n",
+ " \n",
+ "for i in range(0,len(z)):\n",
+ " pa.append(1-(0.1*exp(2.197*z[i])))\n",
+ " \n",
+ "from matplotlib.pyplot import*\n",
+ "plot(z,pa);\n",
+ "plt.xlabel('z(cm)');\n",
+ "plt.ylabel('pA(atm)');\n",
+ "#Answers may vary due to round off error"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 2.4 Page no. 19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(a) Rate of diffusion of oxygen 7.151E-10 kmol/s\n",
+ "(b) The partial pressure gradient of oxygen at midway in diffusion path is: -4.25 bar/m\n",
+ "(c)\n",
+ "Molar average velocity and diffusion velocities at \"midway\"\n",
+ "Molar average velocity in z-direction is 9.9E-05 m/s\n",
+ "The diffusion velocity of oxygen 7.9E-04 m/s\n",
+ "The diffusion velocity of Nitrogen -9.9E-05 m/s\n",
+ "Molar average velocity and diffusion velocities at \"top of tube\"\n",
+ "Molar average velocity in z-direction is 9.9E-05 m/s\n",
+ "The diffusion velocity of oxygen 3.72E-04 m/s\n",
+ "The diffusion velocity of Nitrogen -9.9E-05 m/s\n",
+ "Molar average velocity and diffusion velocities at \"bottom of tube\"\n",
+ "Molar average velocity in z-direction is 9.9E-05 m/s\n",
+ "The diffusion velocity of oxygen is not infinity\n",
+ "The diffusion velocity of Nitrogen -9.9E-05 m/s\n",
+ "(d)\n",
+ "New molar flux of (A) 4.95E-06 kmol/m^2.s\n",
+ "New molar flux of (B) 7.19E-06 kmol/m^2.s\n"
+ ]
+ },
+ {
+ "data": {
+ "image/png": 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ogmzw96am8BARkYyS3PQkIiIJoEQhIiIZKVGIiEhGShQiIpKREoWIiGSkRCEi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\nAABJRU5ErkJggg==\n",
+ "text/plain": [
+ "<matplotlib.figure.Figure at 0x3f74d68>"
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "#calculation of (a) part\n",
+ "#given data\n",
+ "from math import log\n",
+ "from math import pi\n",
+ "from math import exp\n",
+ "import numpy as np\n",
+ "T = 298 #in kelvin\n",
+ "P = 1.013 #in bar\n",
+ "pAl = 0 #partial pressure of oxygen(A) at liquid surface\n",
+ "pAo = 0.21*1.013 #partial pressure of oxygen at open mouth\n",
+ "l = 0.05 #length of diffusion path in m\n",
+ "Dab = 2.1e-5 #diffusivity in m^2/s\n",
+ "R = 0.08317 #in m^3.bar.kmol.K\n",
+ "Na = Dab*P*log((P-pAl)/(P-pAo))/(R*T*l) #in kmol/m^2.s\n",
+ "area = (pi/4)*(0.015)**2\n",
+ "rate = area*Na\n",
+ "print '(a) Rate of diffusion of oxygen %0.3E'%rate,'kmol/s'\n",
+ "z = []\n",
+ "pa =[]\n",
+ "for i in np.arange(0,1,0.01):\n",
+ " z.append(i)\n",
+ " \n",
+ "for i in range(0,len(z)):\n",
+ " pa.append(P-(P-pAo)*exp((R*T*Na*z[i])/(Dab*P)))\n",
+ " \n",
+ "from matplotlib.pyplot import*\n",
+ "plot(z,pa);\n",
+ "plt.xlabel('z(cm)');\n",
+ "plt.ylabel('pA(atm)');\n",
+ "\n",
+ "#calculation of (b) part\n",
+ "z = 0.025 #diffusion path\n",
+ "pA = 0.113 #in bar\n",
+ "#we have to find partial pressure gradient of oxygen at mid way of diffusion path\n",
+ "#let dpA/dz = ppd\n",
+ "ppd = -(R*T*round(Na,8)*(P-pA))/(Dab*P)\n",
+ "print '(b) The partial pressure gradient of oxygen at midway in diffusion path is: %0.2f'%ppd,'bar/m'\n",
+ "\n",
+ "#calculation of (c) part\n",
+ "uA = Na*(R*T/pA) #velocity of oxygen\n",
+ "uB = 0 #since nitrogen is non-diffusing hence Nb = 0\n",
+ "U = pA*uA/P #since U=1/C*(uA*Ca + uB*Cb)\n",
+ "vAd = uA - U #diffusion velocity of oxygen\n",
+ "vBd = uB - U #diffusion velocity of nitrogen\n",
+ "print '(c)'\n",
+ "print 'Molar average velocity and diffusion velocities at \"midway\"'\n",
+ "print 'Molar average velocity in z-direction is %0.1E'%U,'m/s'\n",
+ "print 'The diffusion velocity of oxygen %0.1E'%vAd,'m/s'\n",
+ "print 'The diffusion velocity of Nitrogen %0.1E'%vBd,'m/s'\n",
+ "#at z=0(at top of tube)\n",
+ "uA = Na*(R*T/pAo)\n",
+ "uB = 0\n",
+ "U = pAo*uA/P\n",
+ "vAd = uA - U\n",
+ "vBd = uB - U\n",
+ "print 'Molar average velocity and diffusion velocities at \"top of tube\"'\n",
+ "print 'Molar average velocity in z-direction is %0.1E'%U,'m/s'\n",
+ "print 'The diffusion velocity of oxygen %0.2E'%vAd,'m/s'\n",
+ "print 'The diffusion velocity of Nitrogen %0.1E'%vBd,'m/s'\n",
+ "#at z=0.05(at bottom of tube)\n",
+ "#uA = inf\n",
+ "uB = 0\n",
+ "U = pAo*uA/P\n",
+ "vAd = uA - U\n",
+ "vBd = uB - U\n",
+ "print 'Molar average velocity and diffusion velocities at \"bottom of tube\"'\n",
+ "print 'Molar average velocity in z-direction is %0.1E'%U,'m/s'\n",
+ "print 'The diffusion velocity of oxygen is not infinity'\n",
+ "print 'The diffusion velocity of Nitrogen %0.1E'%vBd,'m/s'\n",
+ "\n",
+ "#calculation of (d) part\n",
+ "V = -2*U\n",
+ "pA = 0.113\n",
+ "Nad = round(Na,8) - V*(pA/(R*T))\n",
+ "Nbd = 0 - (P - pA)*V/(R*T)\n",
+ "print '(d)'\n",
+ "print 'New molar flux of (A) %0.2E'%Nad,'kmol/m^2.s'\n",
+ "print 'New molar flux of (B) %0.2E'%Nbd,'kmol/m^2.s'\n",
+ "#Answers may vary due to round off errors"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 2.5 Page no. 21"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 25,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(a) The air-film thickness is :0.00193 m\n"
+ ]
+ }
+ ],
+ "source": [
+ "from math import log\n",
+ "#given data\n",
+ "area = 3*4 #in m^2\n",
+ "mperarea = 3.0/12 #in kg/m^2\n",
+ "#part (a)\n",
+ "P = 1.013 #in bar\n",
+ "Dab = 9.95e-6 #in m^2/s\n",
+ "R = 0.08317 #in m^3.bar./K.kmol\n",
+ "T = 273+27 #in K\n",
+ "#let d=1\n",
+ "d = 1 #in m\n",
+ "pAo = 0.065 #partial pressure of alcohol on liquid surface\n",
+ "pAd = 0 #partial pressure over d length of stagnant film of air\n",
+ "Na = (Dab*P*log((P-pAd)/(P-pAo)))/(R*T*d) #in kmol/m^2.s\n",
+ "Na = Na*60 #in kg/m^2.s\n",
+ "flux = mperarea/(5*60) #since the liquid evaporates completely in 5 minutes\n",
+ "#now we have to find the value of d\n",
+ "d = Na/flux\n",
+ "print '(a) The air-film thickness is :%0.5f'%d,'m'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 2.6 Page no. 24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "(a)\n",
+ "The steady-state flux is: 3.35E-06 kmol/m^2.s\n",
+ "The rate of transport of N2 from vessel 1 to 2: 6.6E-09 kmol/s\n",
+ "(b)\n",
+ "The flux and the rate of transport of oxygen is: -3.35E-06 kmol/m^2.s\n",
+ "(c)\n",
+ "Partial pressure at a point 0.05m from vessel 1 is: 1.2 atm\n",
+ "(d)\n",
+ "Net or total mass flux: -1.340E-05 kmol/m^2.s\n"
+ ]
+ }
+ ],
+ "source": [
+ "from math import pi\n",
+ "#given data\n",
+ "#part (a)\n",
+ "Dab = 0.23e-4*0.5*(293.0/316)**1.75 #in m^2/s\n",
+ "pA1 = 2*0.8 #in atm\n",
+ "pA2 = 2*0.2 #in atm\n",
+ "l = 0.15 #in m\n",
+ "R = 0.0821 #in m^3.atm./K.kmol\n",
+ "T = 293 #in K\n",
+ "Ma = 28\n",
+ "Mb = 32\n",
+ "Na = Dab*(pA1-pA2)/(R*T*l) #in kmol/m^2.s\n",
+ "area = pi/4*(0.05)**2 #in m^2\n",
+ "rate = area*Na\n",
+ "print '(a)'\n",
+ "print 'The steady-state flux is: %0.2E'%Na,'kmol/m^2.s'\n",
+ "print 'The rate of transport of N2 from vessel 1 to 2: %0.1E'%rate,'kmol/s'\n",
+ "\n",
+ "#part (b)\n",
+ "Nb = -Na\n",
+ "print '(b)'\n",
+ "print 'The flux and the rate of transport of oxygen is: %0.2E'%Nb,'kmol/m^2.s'\n",
+ "\n",
+ "#part (c)\n",
+ "#let dpA/dz = ppg\n",
+ "dz = 0.05 #in m\n",
+ "ppg = (pA2 - pA1)/l #in atm/m\n",
+ "pA = pA1 + (ppg)*dz #in atm\n",
+ "print '(c)'\n",
+ "print 'Partial pressure at a point 0.05m from vessel 1 is: %0.1f'%pA,'atm'\n",
+ "\n",
+ "#part (d)\n",
+ "nt = Ma*Na + Mb*Nb\n",
+ "print '(d)'\n",
+ "print 'Net or total mass flux: %0.3E'%nt,'kmol/m^2.s'\n",
+ "#Answers may vary due to round off errors"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 2.7 Page no. 25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Methanol flux: 4.64e-05 kmol/m^2.s\n",
+ "Water flux: -4.06e-05 kmol/m^2.s\n"
+ ]
+ }
+ ],
+ "source": [
+ "from math import log\n",
+ "#given data\n",
+ "Ha = 274.6*32 #molar latent heat of methanol(a)\n",
+ "Hb = 557.7*18 #molar latent heat of water(b)\n",
+ "yAl = 0.76 #mole fraction of methanol in the vapour\n",
+ "yAo = 0.825 #mole fraction of methanol in the vapour at the liquid-vapour interface\n",
+ "P = 1 #in atm\n",
+ "l = 1e-3 #in m\n",
+ "T =344.2 #in K\n",
+ "R = 0.0821 #m^3.atm./K.kmol\n",
+ "Dab = 1.816e-5 #in m^2/s\n",
+ "Na = Dab*P*log((1-0.1247*yAl)/(1-0.1247*yAo))/(0.1247*R*T*l)\n",
+ "print 'Methanol flux: %0.2e'%Na,'kmol/m^2.s'\n",
+ "Nb = -(Ha/Hb)*Na\n",
+ "print 'Water flux: %0.2e'%Nb,'kmol/m^2.s'\n",
+ "#Answers may vary due to round off errors"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 2.8 Page no. 27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Value of pA1 is 0.937 atm\n"
+ ]
+ }
+ ],
+ "source": [
+ "#given values\n",
+ "from math import exp\n",
+ "V1 = 3000 #in cm^3\n",
+ "V2 = 4000 #in cm^3\n",
+ "Dab = 0.23 #in cm^2/s\n",
+ "Dba = 0.23 #in cm^2/s\n",
+ "l1 = 4 #in cm\n",
+ "d1 = 0.5 #in cm\n",
+ "l2 = 2 #in cm\n",
+ "d2 = 0.3 #in cm\n",
+ "pA3 = 1 #in atm\n",
+ "#unknowns\n",
+ "# pA1 and pA2\n",
+ "# dpA1bydt = (Dab/V1*l1)*((pA1)-(pA2))*((math.pi*(d1**2))/4)\n",
+ "#on integrating using Laplace trandformation\n",
+ "# initial conditions\n",
+ "t=18000 #in seconds\n",
+ "pA1 = 1-0.57*(exp((-1.005)*(10**(-6))*t)-exp((-7.615)*(10**(-6))*t))\n",
+ "print 'Value of pA1 is %0.3f'%pA1,'atm'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 2.10 Page no.34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Value of flux of water vapour: 2.96E-05 kmol/m^2.s\n"
+ ]
+ }
+ ],
+ "source": [
+ "#given values\n",
+ "from math import log\n",
+ "y1l = 0 #mol fraction of dry air\n",
+ "y10 = (17.53/760) #mol fraction of water\n",
+ "l = 1.5 #in mm\n",
+ "C = 0.0409 #in kmol/m^3 : calculated by P/RT\n",
+ "D12 = 0.923 #Diffusivity of hydrogen over water\n",
+ "D13 = 0.267 #Diffusivity of oxygen over water\n",
+ "y2 = 0.6 #mole fraction of hydrogen\n",
+ "y3 = 0.4 #mole fraction of oxygen\n",
+ "D1m = 1/((y2/D12)+(y3/D13)) #calculating mean diffusivity\n",
+ "Ni = (D1m*C*1000/(l*10000))*log((1-y1l)/(1-y10))\n",
+ "print 'Value of flux of water vapour: %0.2E'%Ni,'kmol/m^2.s'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 2.11 Page no. 35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 32,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Flux of ethane 4.804E-05 gmol/cm^2.s\n"
+ ]
+ }
+ ],
+ "source": [
+ "from math import log\n",
+ "#given data\n",
+ "y1 = 0.4 #mole fraction of ethane(1)\n",
+ "y2 = 0.3 #mole fraction of ethylene(2)\n",
+ "y3 = 0.3 #mole fraction of hydrogen(3)\n",
+ "#calculating D13\n",
+ "#The Lennard-Jones parameters are\n",
+ "sigma1 = 4.443 #in angstrom\n",
+ "sigma2 = 4.163 #in angstrom\n",
+ "sigma3 = 2.827 #in angstrom\n",
+ "e1byk = 215.7\n",
+ "e2byk = 224.7\n",
+ "e3byk = 59.7\n",
+ "sigma13 = (sigma1 + sigma3)/2 #in angstrom\n",
+ "e13byk = (e1byk*e3byk)**0.5\n",
+ "kTbye13 = 993/113.5\n",
+ "ohmD13 = 0.76 #from collision integral table\n",
+ "D13 = ((0.001858)*(993**1.5)*((1.0/30)+(1.0/2))**0.5)/((2)*(sigma13**2)*(ohmD13))\n",
+ "#calculating D23\n",
+ "sigma23 = (sigma2+sigma3)/2\n",
+ "kTbye23 = ((993/224.7)*(993/59.7))*0.5\n",
+ "ohmD23 = 0.762\n",
+ "D23 = (0.001858*(993**1.5)*((1.0/28)+(1.0/2))**0.5)/(2*(sigma23**2)*ohmD23)\n",
+ "D = (D13+D23)/2 #in cm^2/s\n",
+ "l = 0.15 #in cm\n",
+ "#at z=0 (bulk gas)\n",
+ "y10 = 0.6\n",
+ "y20 = 0.2\n",
+ "y30 = 0.2\n",
+ "#at z=l (catalyst surface)\n",
+ "y1l = 0.4\n",
+ "y2l = 0.3\n",
+ "y3l = 0.3\n",
+ "C = 2.0/(82.1*993) #calculated by P/RT\n",
+ "N1 = (D*C/l)*log((y10+y20)/(y1l+y2l))\n",
+ "print 'Flux of ethane %0.3E'%N1,'gmol/cm^2.s'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 2.12 Page no. 43"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The liquid-film thickness is: 0.0004 m\n"
+ ]
+ }
+ ],
+ "source": [
+ "#given data\n",
+ "from math import pi\n",
+ "rc = 5e-4 #in m\n",
+ "D = 7e-10 #in m^2/s\n",
+ "Cab = 1 #in kmol/m^3\n",
+ "Na = 3.15e-6 #in kmol/m^2.s\n",
+ "W = 4*pi*(rc**2)*Na #the rate of reaction\n",
+ "#let (rc+delta)/delta = 1\n",
+ "w1 = 4*pi*D*Cab*rc*1 #flux of the reactant to the surface of the catalyst\n",
+ "rcplusdelta = W/w1\n",
+ "delta = rc/(rcplusdelta-1) #stagnant liquid-film thickness \n",
+ "print 'The liquid-film thickness is: ',delta,'m'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 2.13 Page no. 46"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Tortuosity factor is: 2.5\n"
+ ]
+ }
+ ],
+ "source": [
+ "#given data\n",
+ "from math import log\n",
+ "V1 = 60.2 #in cm^3; volume of compartment 1\n",
+ "V2 = 59.3 #volume of compartment 2 in cm^3\n",
+ "Ca1i = 0.3 #initial concentration of KCl in compartment 1\n",
+ "Ca2i = 0 #initial concentration of KCl in compartment 2\n",
+ "Ca1f = 0.215 #final concentration of KCl in compartment 1\n",
+ "Ca2f = 0.0863 #final concentration of KCl in compartment 2\n",
+ "D = 1.51e-5 #diffusivity of KCl in cm^2/s\n",
+ "tf = 55.2*3600 #time of the experiment in s\n",
+ "#calcutaling cell constant\n",
+ "beta = (1/(D*tf))*log((Ca1i - Ca2i)/(Ca1f - Ca2f))\n",
+ "#diffusion of propionic acid\n",
+ "Cpa1i = 0.4 #initial concentration of propionic acid in compartment 1\n",
+ "Cpa2i = 0 #initial concentration of propionic acid in compartment 2\n",
+ "Cpa1f = 0.32 #final concentration of propionic acid in compartment 1\n",
+ "Cpa2f = 0.0812 #final concentration of propionic acid in compartment 2 by mass balance\n",
+ "tfp = 56.4*3600 #time for the experiment\n",
+ "Dp = (1/(beta*tfp))*log((Cpa1i-Cpa2i)/(Cpa1f-Cpa2f)) #diffusivity of the propionic acid\n",
+ "#calculating tortusity factor\n",
+ "A= (math.pi/4)*(3.5**2) #area of the diaphragm\n",
+ "epsilon = 0.39 #average porosity of the diaphragm\n",
+ "l = 0.18 #thickness of hte diaphragm\n",
+ "tou = (A*epsilon/(beta*l))*(1/V1 + 1/V2)\n",
+ "print 'Tortuosity factor is: ',round(tou,1)"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.10"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/sample_notebooks/UmangAgarwal/Sample_Notebook.ipynb b/sample_notebooks/UmangAgarwal/Sample_Notebook.ipynb new file mode 100644 index 00000000..34fb4a40 --- /dev/null +++ b/sample_notebooks/UmangAgarwal/Sample_Notebook.ipynb @@ -0,0 +1,128 @@ +Sample Notebook - Heat and Mass Transfer by R.K. Rajput : Chapter 1 - Basic Concepts
+author: Umang Agarwal
+
+
+# Example 1.1 Page 16-17
+
+L=.045; #[m] - Thickness of conducting wall
+delT = 350 - 50; #[C] - Temperature Difference across the Wall
+k=370; #[W/m.C] - Thermal Conductivity of Wall Material
+#calculations
+#Using Fourier's Law eq 1.1
+q = k*delT/(L*10**6); #[MW/m^2] - Heat Flux
+#results
+print '%s %.2f %s' %("\n \n Rate of Heat Transfer per unit area =",q," W");
+#END
+
+# Example 1.2 Page 17
+
+L = .15; #[m] - Thickness of conducting wall
+delT = 150 - 45; #[C] - Temperature Difference across the Wall
+A = 4.5; #[m^2] - Wall Area
+k=9.35; #[W/m.C] - Thermal Conductivity of Wall Material
+#calculations
+#Using Fourier's Law eq 1.1
+Q = k*A*delT/L; #[W] - Heat Transfer
+#Temperature gradient using Fourier's Law
+TG = - Q/(k*A); #[C/m] - Temperature Gradient
+#results
+print '%s %.2f %s' %("\n \n Rate of Heat Transfer per unit area =",Q," W");
+print '%s %.2f %s' %("\n \n The Temperature Gradient in the flow direction =",TG," C/m");
+#END
+
+# Example 1.3 Page 17-18
+
+x = .0825; #[m] - Thickness of side wall of the conducting oven
+delT = 175 - 75; #[C] - Temperature Difference across the Wall
+k=0.044; #[W/m.C] - Thermal Conductivity of Wall Insulation
+Q = 40.5; #[W] - Energy dissipitated by the electric coil withn the oven
+#calculations
+#Using Fourier's Law eq 1.1
+A = (Q*x)/(k*delT); #[m^2] - Area of wall
+#results
+print '%s %.2f %s' %("\n \n Area of the wall =",A," m^2");
+#END
+
+# Example 1.4 Page 18-19
+
+delT = 300-20; #[C] - Temperature Difference across the Wall
+h = 20; #[W/m^2.C] - Convective Heat Transfer Coefficient
+A = 1*1.5; #[m^2] - Wall Area
+#calculations
+#Using Newton's Law of cooling eq 1.6
+Q = h*A*delT; #[W] - Heat Transfer
+#results
+print '%s %.2f %s' %("\n \n Rate of Heat Transfer =",Q," W");
+#END
+
+# Example 1.5 Page 19
+
+L=.15; #[m] - Length of conducting wire
+d = 0.0015; #[m] - Diameter of conducting wire
+A = 22*d*L/7; #[m^2] - Surface Area exposed to Convection
+delT = 120 - 100; #[C] - Temperature Difference across the Wire
+h = 4500; #[W/m^2.C] - Convective Heat Transfer Coefficient
+print 'Electric Power to be supplied = Convective Heat loss';
+#calculations
+#Using Newton's Law of cooling eq 1.6
+Q = h*A*delT; #[W] - Heat Transfer
+Q = round(Q,1);
+#results
+print '%s %.2f %s' %("\n \n Rate of Heat Transfer =",Q," W");
+#END
+
+# Example 1.6 Page 20-21
+
+T1 = 300 + 273; #[K] - Temperature of 1st surface
+T2 = 40 + 273; #[K] - Temperature of 2nd surface
+A = 1.5; #[m^2] - Surface Area
+F = 0.52; #[dimensionless] - The value of Factor due geometric location and emissivity
+sigma = 5.67*(10**-8) #(W/(m^2 * K^4)) - Stephen - Boltzmann Constant
+#calculations
+#Using Stephen-Boltzmann Law eq 1.9
+Q = F*sigma*A*(T1**4 - T2**4) #[W] - Heat Transfer
+#Equivalent Thermal Resistance using eq 1.10
+Rth = (T1-T2)/Q; #[C/W] - Equivalent Thermal Resistance
+#Equivalent convectoin coefficient using h*A*(T1-T2) = Q
+h = Q/(A*(T1-T2)); #[W/(m^2*C)] - Equivalent Convection Coefficient
+#results
+print '%s %.2f %s' %("\n \n Rate of Heat Transfer =",Q," W");
+print '%s %.2f %s' %("\n The equivalent thermal resistance =",Rth," C/W");
+print '%s %.2f %s' %("\n The equivalent convection coefficient =",h," W/(m^2 * C)");
+#END
+
+# Example 1.7 Page 21-22
+
+L = 0.025; #[m] - Thickness of plate
+A = 0.6*0.9; #[m^2] - Area of plate
+Ts = 310; #[C] - Surface Temperature of plate
+Tf = 15; #[C] - Temperature of fluid(air)
+h = 22; #[W/m^2.C] - Convective Heat Transfer Coefficient
+Qr = 250; #[W] - Heat lost from the plate due to radiation
+k = 45; #[W/m.C] - Thermal Conductivity of Plate
+#calculations
+# In this problem, heat conducted by the plate is removed by a combination of convection and radiation
+# Heat conducted through the plate = Convection Heat losses + Radiation Losses
+# If Ti is the internal plate temperature, then heat conducted = k*A*(Ts-Ti)/L
+Qc = h*A*(Ts-Tf); #[W] - Convection Heat Loss
+Ti = Ts + L*(Qc + Qr)/(A*k); #[C] - Inside plate Temperature
+#results
+print '%s %.2f %s' %("\n \n Rate of Heat Transfer =",Ti," C");
+#END
+
+# Example 1.8 Page 22
+
+Ts = 250; #[C] - Surface Temperature
+Tsurr = 110; #[C] - Temperature of surroundings
+h = 75; #[W/m^2.C] - Convective Heat Transfer Coefficient
+F = 1; #[dimensionless] - The value of Factor due geometric location and emissivity
+sigma = 5.67*(10**-8) #(W/(m^2 * K^4)) - Stephen - Boltzmann Constant
+k = 10; #[W/m.C] - Thermal Conductivity of Solid
+#calculations
+# Heat conducted through the plate = Convection Heat losses + Radiation Losses
+qr = F*sigma*((Ts+273)**4-(Tsurr+273)**4) #[W/m^2] - #[W] - Heat lost per unit area from the plate due to radiation
+qc = h*(Ts-Tsurr); #[W/m^2] - Convection Heat Loss per unit area
+TG = -(qc+qr)/k; #[C/m] - Temperature Gradient
+#results
+print '%s %.2f %s' %("\n \n The temperature Gradient =",TG," C/m");
+#END
diff --git a/sample_notebooks/UmangAgarwal/Sample_Notebook_1.ipynb b/sample_notebooks/UmangAgarwal/Sample_Notebook_1.ipynb new file mode 100644 index 00000000..34fb4a40 --- /dev/null +++ b/sample_notebooks/UmangAgarwal/Sample_Notebook_1.ipynb @@ -0,0 +1,128 @@ +Sample Notebook - Heat and Mass Transfer by R.K. Rajput : Chapter 1 - Basic Concepts
+author: Umang Agarwal
+
+
+# Example 1.1 Page 16-17
+
+L=.045; #[m] - Thickness of conducting wall
+delT = 350 - 50; #[C] - Temperature Difference across the Wall
+k=370; #[W/m.C] - Thermal Conductivity of Wall Material
+#calculations
+#Using Fourier's Law eq 1.1
+q = k*delT/(L*10**6); #[MW/m^2] - Heat Flux
+#results
+print '%s %.2f %s' %("\n \n Rate of Heat Transfer per unit area =",q," W");
+#END
+
+# Example 1.2 Page 17
+
+L = .15; #[m] - Thickness of conducting wall
+delT = 150 - 45; #[C] - Temperature Difference across the Wall
+A = 4.5; #[m^2] - Wall Area
+k=9.35; #[W/m.C] - Thermal Conductivity of Wall Material
+#calculations
+#Using Fourier's Law eq 1.1
+Q = k*A*delT/L; #[W] - Heat Transfer
+#Temperature gradient using Fourier's Law
+TG = - Q/(k*A); #[C/m] - Temperature Gradient
+#results
+print '%s %.2f %s' %("\n \n Rate of Heat Transfer per unit area =",Q," W");
+print '%s %.2f %s' %("\n \n The Temperature Gradient in the flow direction =",TG," C/m");
+#END
+
+# Example 1.3 Page 17-18
+
+x = .0825; #[m] - Thickness of side wall of the conducting oven
+delT = 175 - 75; #[C] - Temperature Difference across the Wall
+k=0.044; #[W/m.C] - Thermal Conductivity of Wall Insulation
+Q = 40.5; #[W] - Energy dissipitated by the electric coil withn the oven
+#calculations
+#Using Fourier's Law eq 1.1
+A = (Q*x)/(k*delT); #[m^2] - Area of wall
+#results
+print '%s %.2f %s' %("\n \n Area of the wall =",A," m^2");
+#END
+
+# Example 1.4 Page 18-19
+
+delT = 300-20; #[C] - Temperature Difference across the Wall
+h = 20; #[W/m^2.C] - Convective Heat Transfer Coefficient
+A = 1*1.5; #[m^2] - Wall Area
+#calculations
+#Using Newton's Law of cooling eq 1.6
+Q = h*A*delT; #[W] - Heat Transfer
+#results
+print '%s %.2f %s' %("\n \n Rate of Heat Transfer =",Q," W");
+#END
+
+# Example 1.5 Page 19
+
+L=.15; #[m] - Length of conducting wire
+d = 0.0015; #[m] - Diameter of conducting wire
+A = 22*d*L/7; #[m^2] - Surface Area exposed to Convection
+delT = 120 - 100; #[C] - Temperature Difference across the Wire
+h = 4500; #[W/m^2.C] - Convective Heat Transfer Coefficient
+print 'Electric Power to be supplied = Convective Heat loss';
+#calculations
+#Using Newton's Law of cooling eq 1.6
+Q = h*A*delT; #[W] - Heat Transfer
+Q = round(Q,1);
+#results
+print '%s %.2f %s' %("\n \n Rate of Heat Transfer =",Q," W");
+#END
+
+# Example 1.6 Page 20-21
+
+T1 = 300 + 273; #[K] - Temperature of 1st surface
+T2 = 40 + 273; #[K] - Temperature of 2nd surface
+A = 1.5; #[m^2] - Surface Area
+F = 0.52; #[dimensionless] - The value of Factor due geometric location and emissivity
+sigma = 5.67*(10**-8) #(W/(m^2 * K^4)) - Stephen - Boltzmann Constant
+#calculations
+#Using Stephen-Boltzmann Law eq 1.9
+Q = F*sigma*A*(T1**4 - T2**4) #[W] - Heat Transfer
+#Equivalent Thermal Resistance using eq 1.10
+Rth = (T1-T2)/Q; #[C/W] - Equivalent Thermal Resistance
+#Equivalent convectoin coefficient using h*A*(T1-T2) = Q
+h = Q/(A*(T1-T2)); #[W/(m^2*C)] - Equivalent Convection Coefficient
+#results
+print '%s %.2f %s' %("\n \n Rate of Heat Transfer =",Q," W");
+print '%s %.2f %s' %("\n The equivalent thermal resistance =",Rth," C/W");
+print '%s %.2f %s' %("\n The equivalent convection coefficient =",h," W/(m^2 * C)");
+#END
+
+# Example 1.7 Page 21-22
+
+L = 0.025; #[m] - Thickness of plate
+A = 0.6*0.9; #[m^2] - Area of plate
+Ts = 310; #[C] - Surface Temperature of plate
+Tf = 15; #[C] - Temperature of fluid(air)
+h = 22; #[W/m^2.C] - Convective Heat Transfer Coefficient
+Qr = 250; #[W] - Heat lost from the plate due to radiation
+k = 45; #[W/m.C] - Thermal Conductivity of Plate
+#calculations
+# In this problem, heat conducted by the plate is removed by a combination of convection and radiation
+# Heat conducted through the plate = Convection Heat losses + Radiation Losses
+# If Ti is the internal plate temperature, then heat conducted = k*A*(Ts-Ti)/L
+Qc = h*A*(Ts-Tf); #[W] - Convection Heat Loss
+Ti = Ts + L*(Qc + Qr)/(A*k); #[C] - Inside plate Temperature
+#results
+print '%s %.2f %s' %("\n \n Rate of Heat Transfer =",Ti," C");
+#END
+
+# Example 1.8 Page 22
+
+Ts = 250; #[C] - Surface Temperature
+Tsurr = 110; #[C] - Temperature of surroundings
+h = 75; #[W/m^2.C] - Convective Heat Transfer Coefficient
+F = 1; #[dimensionless] - The value of Factor due geometric location and emissivity
+sigma = 5.67*(10**-8) #(W/(m^2 * K^4)) - Stephen - Boltzmann Constant
+k = 10; #[W/m.C] - Thermal Conductivity of Solid
+#calculations
+# Heat conducted through the plate = Convection Heat losses + Radiation Losses
+qr = F*sigma*((Ts+273)**4-(Tsurr+273)**4) #[W/m^2] - #[W] - Heat lost per unit area from the plate due to radiation
+qc = h*(Ts-Tsurr); #[W/m^2] - Convection Heat Loss per unit area
+TG = -(qc+qr)/k; #[C/m] - Temperature Gradient
+#results
+print '%s %.2f %s' %("\n \n The temperature Gradient =",TG," C/m");
+#END
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