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diff --git a/backup/Mass_-_Transfer_Operations_version_backup/Chapter3_1.ipynb b/backup/Mass_-_Transfer_Operations_version_backup/Chapter3_1.ipynb new file mode 100755 index 00000000..36f53f4d --- /dev/null +++ b/backup/Mass_-_Transfer_Operations_version_backup/Chapter3_1.ipynb @@ -0,0 +1,628 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:64d9a62e17838716bd10cd86f93be8c39dc69462337a3d3adc3d6ea158cbc575"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3: Mass-Transfer Coefficients"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.1:Page 53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "\n",
+ "# Illustration 3.1\n",
+ "# Page: 53\n",
+ "\n",
+ "print'Illustration 3.1 - Page: 53\\n\\n'\n",
+ "\n",
+ "# solution\n",
+ "\n",
+ "#****Data*****#\n",
+ "# a = CO2 b = H2O\n",
+ "Ca0 = 0;#[kmol/cubic m]\n",
+ "Cai = 0.0336;#[kmol/cubic m]\n",
+ "Dab = 1.96*10**(-9);# [square m/s]\n",
+ "#*******#\n",
+ "\n",
+ "density = 998.0;# [kg/cubic m]\n",
+ "viscosity = 8.94*10**(-4);#[kg/m.s]\n",
+ "rate = 0.05;#[kg/m.s] mass flow rate of liquid\n",
+ "L = 1;#[m]\n",
+ "g = 9.81;#[m/square s]\n",
+ "# From Eqn. 3.10\n",
+ "Del = ((3*viscosity*rate)/((density**2)*g))**(1.0/3);# [m]\n",
+ "Re = 4*rate/viscosity;\n",
+ "# Flow comes out to be laminar\n",
+ "# From Eqn. 3.19\n",
+ "Kl_avg = ((6*Dab*rate)/(3.141*density*Del*L))**(1.0/2);#[kmol/square m.s.(kmol/cubic m)]\n",
+ "bulk_avg_velocity = rate/(density*Del);#[m/s]\n",
+ "# At the top: Cai-Ca = Cai_Ca0 = Cai\n",
+ "#At the bottom: Cai-Cal\n",
+ "# From Eqn. 3.21 & 3.22\n",
+ "Cal = Cai*(1-(1.0/(exp(Kl_avg/(bulk_avg_velocity*Del)))));# [kmol/cubic m]\n",
+ "rate_absorption = bulk_avg_velocity*Del*(Cal-Ca0);# [kmol/s].(m of width)\n",
+ "print'The rate of absorption is ',round(rate_absorption,8),' kmol/sec.(m of width)'\n",
+ "# The actual value may be substantially larger."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Illustration 3.1 - Page: 53\n",
+ "\n",
+ "\n",
+ "The rate of absorption is 7.2e-07 kmol/sec.(m of width)\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.2: Page 56"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "\n",
+ "# Illustration 3.2\n",
+ "# Page: 56\n",
+ "\n",
+ "print'Illustration 3.2 - Page: 56\\n\\n'\n",
+ "\n",
+ "# solution\n",
+ "\n",
+ "#***Data****#\n",
+ "d = 0.025;# [m]\n",
+ "avg_velocity = 3;# [m/s]\n",
+ "viscosity = 8.937*10**(-4);# [kg/m.s]\n",
+ "density = 997;# [kg/m**3]\n",
+ "#*********#\n",
+ "\n",
+ "kinematic_viscosity = viscosity/density;# [square m/s]\n",
+ "Re = d*avg_velocity*density/viscosity;\n",
+ "# Reynold's number comes out to be 83670\n",
+ "# At this Reynold's number fanning factor = 0.0047\n",
+ "f = 0.0047;\n",
+ "L = 1;# [m]\n",
+ "press_drop = 2*density*f*L*(avg_velocity**2)/(d);# [N/square m]\n",
+ "P = 3.141*(d**2)*avg_velocity*press_drop/4;# [N.m/s] for 1m pipe\n",
+ "m = 3.141*(d**2)*L*density/4;\n",
+ "# From Eqn. 3.24\n",
+ "Ld = ((kinematic_viscosity**3)*m/P)**(1.0/4);# [m]\n",
+ "# From Eqn. 3.25\n",
+ "Ud = (kinematic_viscosity*P/m)**(1.0/4);# [m/s]\n",
+ "print'Velocity of small eddies is',round(Ud,4),'m/s'\n",
+ "print'Length scale of small eddies is',round(Ld,7),'m'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Illustration 3.2 - Page: 56\n",
+ "\n",
+ "\n",
+ "Velocity of small eddies is 0.0549 m/s\n",
+ "Length scale of small eddies is 1.63e-05 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.3: Page 69"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "\n",
+ "# Illustration 3.3\n",
+ "# Page: 69\n",
+ "\n",
+ "print'Illustration 3.3 - Page: 69\\n\\n'\n",
+ "\n",
+ "# solution\n",
+ "\n",
+ "# Heat transfer analog to Eqn. 3.12\n",
+ "# The Eqn. remains the same with the dimensionless conc. ratio replaced by ((tl-to)/(ti-to))\n",
+ "\n",
+ "# The dimensionless group:\n",
+ "# eta = 2*Dab*L/(3*del**2*velocity);\n",
+ "# eta = (2/3)*(Dab/(del*velocity))*(L/del);\n",
+ "# Ped = Peclet no. for mass transfer\n",
+ "# eta = (2/3)*(1/Ped)*(L/del);\n",
+ "\n",
+ "# For heat transfer is replaced by\n",
+ "# Peh = Peclet no. for heat transfer\n",
+ "# eta = (2/3)*(1/Peh)*(L/del);\n",
+ "# eta = (2/3)*(alpha/(del*velocity))*(L/del);\n",
+ "# eta = (2*alpha*L)/(3*del**2*velocity);\n",
+ "print'Heat transfer analog to Eqn. 3.21 is eta = (2*alpha*L)/(3*del**2*velocity)'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Illustration 3.3 - Page: 69\n",
+ "\n",
+ "\n",
+ "Heat transfer analog to Eqn. 3.21 is eta = (2*alpha*L)/(3*del**2*velocity)\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.4: Page-69"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "\n",
+ "# Illustration 3.4\n",
+ "# Page: 69\n",
+ "\n",
+ "import math\n",
+ "print'Illustration 3.4 - Page: 69\\n\\n'\n",
+ "\n",
+ "# solution\n",
+ "\n",
+ "#***Data****#\n",
+ "# a = UF6 b = air\n",
+ "# The average heat transfer coefficient: Nu_avg = 0.43+0.532(Re^0.5)(Pr^0.31)\n",
+ "# The analogus expression for mass transfer coefficient: Sh_avg = 0.43+0.532(Re^0.5)(Sc^0.31)\n",
+ "d = 0.006;# [m]\n",
+ "velocity = 3.0;# [m/s]\n",
+ "surf_temp = 43.0;# [C]\n",
+ "bulk_temp = 60.0;# [C]\n",
+ "avg_temp = (surf_temp+bulk_temp)/2; #[C]\n",
+ "density = 4.10;# [kg/cubic m]\n",
+ "viscosity = 2.7*10**(-5);# [kg/m.s]\n",
+ "Dab = 9.04*10**(-6);# [square m/s]\n",
+ "press = 53.32;# [kN/square m]\n",
+ "tot_press = 101.33;# [kN/square m]\n",
+ "#******#\n",
+ "\n",
+ "avg_press = press/2.0; # [kN/square m]\n",
+ "Xa = avg_press/tot_press;\n",
+ "Xb = 1-Xa;\n",
+ "Re = d*velocity*density/viscosity;\n",
+ "Sc = viscosity/(density*Dab);\n",
+ "Sh_avg = 0.43+(0.532*(2733**0.5)*(0.728**0.5));\n",
+ "c = 273.2/(22.41*(273.2+avg_temp));# [kmol/cubic m]\n",
+ "F_avg = Sh_avg*c*Dab/d;#[kmol/cubic m]\n",
+ "Nb = 0.0;\n",
+ "Ca1_by_C = press/tot_press;\n",
+ "Ca2_by_C = 0.0;\n",
+ "Flux_a = 1.0;\n",
+ "# Using Eqn. 3.1\n",
+ "Na = Flux_a*F_avg*math.log((Flux_a-Ca2_by_C)/(Flux_a-Ca1_by_C));#[kmol UF6/square m.s]\n",
+ "print'Rate of sublimation is',round(Na,8),' kmol UF6/square m.s'\n",
+ "# the answer is slightly different in textbook due to approximation"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Illustration 3.4 - Page: 69\n",
+ "\n",
+ "\n",
+ "Rate of sublimation is 0.00102088 kmol UF6/square m.s\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.5: Page 73"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "\n",
+ "# Illustration 3.5\n",
+ "# Page: 73\n",
+ "\n",
+ "print'Illustration 3.5 - Page: 73\\n\\n'\n",
+ "\n",
+ "# solution\n",
+ "\n",
+ "#****Data****#\n",
+ "velocity = 15.0;# [m/s]\n",
+ "G = 21.3;# [kg/square m.s]\n",
+ "#******#\n",
+ "\n",
+ "# Since the experimental data do not include the effects of changing Prandtl number.\n",
+ "\n",
+ "# Jh = (h/(Cp*density*viscosity)) = (h/Cp*G)*(Pr^(2/3)) = Shi(Re);\n",
+ "\n",
+ "# Shi(Re) must be compatible with 21.3*(G**0.6);\n",
+ "# Let Shi(Re) = b*(Re**n);\n",
+ "# Re = (l*G)/viscosity;\n",
+ "\n",
+ "# h = (Cp*G/(Pr**(2/3)))*b*(Re**n);\n",
+ "# h = (Cp*G/(Pr**(2/3)))*b*((l*b/viscosity)**n) = 21.3*(G**0.6);\n",
+ "\n",
+ "n = 0.6-1;\n",
+ "# b = 21.3*((Pr**(2/3))/Cp)*((l/viscosity)**(-n));\n",
+ "\n",
+ "# Using data for air at 38 C & 1 std atm.\n",
+ "Cp1 = 1002;# [kJ/kg.K]\n",
+ "viscosity1 = 1.85*10**(-5);#[kg/m.s]\n",
+ "k1 = 0.0273;#[W/m.K]\n",
+ "Pr1 = (Cp1*viscosity1)/k1;\n",
+ "b_prime = 21.3*(Pr1**(2.0/3)/Cp1)*((1/viscosity1)**0.4);\n",
+ "# b = b_prime*l**(0.4);\n",
+ "# Jh = (h/(Cp*G))*Pr**(2/3) = b_prime*((l/Re)**(0.4)) = Shi(Re);\n",
+ "\n",
+ "# The heat mass transfer analogy will be used to estimate the mass transfer coefficient. (Jd = Jh)\n",
+ "\n",
+ "# Jd = (KG*Pbm*Mav*Sc**(2/3))/(density*viscosity) = Shi(Re) = b_prime*((l/Re)**0.4);\n",
+ "\n",
+ "# KG*Pbm = F = (b_prime*density*viscosity)/(Re^0.4*Mav*Sc**(2/3)) = (b_prime*(density*velocity)**0.6*(viscosity^0.4))/(Mav*Sc**(2/3));\n",
+ "\n",
+ "# For H2-H20, 38 C, 1std atm\n",
+ "viscosity2 = 9*10**(-6);# [kg/m.s]\n",
+ "density2 = 0.0794;# [kg/cubic m]\n",
+ "Dab = 7.75*10**(-5);# [square m/s]\n",
+ "Sc = viscosity2/(density2*Dab);\n",
+ "\n",
+ "# Assuming desity, Molecular weight and viscosity of the gas are essentially those of H2\n",
+ "\n",
+ "Mav = 2.02;# [kg/kmol]\n",
+ "F = (b_prime*(density2*velocity)**0.6*(viscosity2**0.4))/(Mav*Sc**(2.0/3));# [kmol/square m.s]\n",
+ "print'The required mass transfer: ',round(F,5),' kmol/square m.s'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Illustration 3.5 - Page: 73\n",
+ "\n",
+ "\n",
+ "The required mass transfer: 0.00525 kmol/square m.s\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.6:Page 77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "\n",
+ "# Illustration 3.6\n",
+ "# Page: 77\n",
+ "\n",
+ "print'Illustration 3.6 - Page: 77\\n\\n'\n",
+ "\n",
+ "# solution\n",
+ "from scipy import integrate\n",
+ "import math \n",
+ "#***Data***#\n",
+ "Dp = 0.0125;# [m]\n",
+ "viscosity = 2.4*10**(-5);# [kg/m.s]\n",
+ "Sc = 2.0;\n",
+ "E = 0.3;\n",
+ "Go = (2*10**(-3))/0.1;# molar superficial mass velocity [kmol/square m.s]\n",
+ "#********#\n",
+ "\n",
+ "# a = CO b = Ni(CO)4\n",
+ "# Nb = -(Na/4);\n",
+ "Flux_a = 4.0/3;\n",
+ "Ca2_by_C = 0;# At the metal interface\n",
+ "# Ca1_by_C = Ya #mole fraction of CO in the bulk\n",
+ "\n",
+ "# Eqn. 3.1 becomes: Na = (4/3)*F*log((4/3)/((4/3)-Ya));\n",
+ "\n",
+ "# Let G = kmol gas/(square m bed cross section).s\n",
+ "# a = specific metal surface\n",
+ "# z = depth \n",
+ "# Therefore, Na = -(diff(Ya*G))/(a*diff(z));# [kmol/((square m metal surface).s)];\n",
+ "# For each kmol of CO consumed, (1/4)kmol Ni(CO)4 forms, representing a loss of (3/4) kmol per kmol of CO consumed.\n",
+ "# The CO consumed through bed depth dz is therefore (Go-G)(4/3) kmol;\n",
+ "# Ya = (Go-(Go-G)*(4/3))/G;\n",
+ "# G = Go/(4-(3*Ya));\n",
+ "# diff(YaG) = ((4*Go)/(4-3*Ya)**2)*diff(Ya);\n",
+ "\n",
+ "# Substituting in Eqn. 3.64\n",
+ "# -(4*Go/((4-3*Ya)**2*a))*(diff(Ya)/diff(z)) = (4/3)*F*log(4/(4-3*Ya));\n",
+ "\n",
+ "# At depth z:\n",
+ "# Mass velocity of CO = (Go-(Go-G)/(4/3))*28;\n",
+ "# Mass velocity of Ni(CO)4 = ((Go-G)*(1/3))*170.7;\n",
+ "# G_prime = 47.6*Go-19.6G; # total mass velocity [kg/square m.s]\n",
+ "# Substituting G leads to:\n",
+ "# G_prime = Go*(47.6-19.6*(4-3*Ya));# [kg/m.s]\n",
+ "# Re = (Dp*G')/viscosity\n",
+ "\n",
+ "# With Go = 0.002 kmol/square m.s & Ya in the range 1-0.005, the range of Re is 292-444;\n",
+ "# From table 3.3:\n",
+ "# Jd = (F/G)*(Sc**(2/3)) = (2.06/E)*Re**(-0.575);\n",
+ "# F = (2.06/E*(Sc)**(2/3))*(Go/(4-3*Ya))*Re**(-0.575);\n",
+ "\n",
+ "a = 6*(1-E)/Dp;\n",
+ "\n",
+ "# Result after arrangement:\n",
+ "\n",
+ "X2=lambda Ya:-((4*Go)/((4-(3*Ya))**2.0*a))*(3.0/4)*(E*(Sc**(2.0/3))*(4-(3*Ya))/(2.06*Go)*(1/math.log(4.0/(4-(3*Ya)))))*(((Dp/viscosity)*(Go*(47.6-(19.6/(4.0-(3*Ya))))))**0.575);# [m]\n",
+ "Z = integrate.quad(X2,1,0.005);\n",
+ "print'The bed depth required to reduce the CO content to 0.005 is',round(Z[0],3),'m'\n",
+ "#the answers are slightly different in textbook due to approximation while here answers are precise"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Illustration 3.6 - Page: 77\n",
+ "\n",
+ "\n",
+ "The bed depth required to reduce the CO content to 0.005 is 0.132 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.7: Page 80"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "\n",
+ "# Illustration 3.7\n",
+ "# Page: 80\n",
+ "\n",
+ "print'Illustration 3.7 - Page: 80\\n\\n'\n",
+ "\n",
+ "# solution\n",
+ "\n",
+ "#****Data*****#\n",
+ "# a = water b = air\n",
+ "out_dia = 0.0254;# [m]\n",
+ "wall_thick = 0.00165;# [m]\n",
+ "avg_velocity = 4.6;# [m/s]\n",
+ "T1 = 66.0;# [C]\n",
+ "P = 1.0;# [atm]\n",
+ "Pa1 = 0.24;# [atm]\n",
+ "k1 = 11400.0;# [W/(square m.K)]\n",
+ "T2 = 24.0;# [C]\n",
+ "k2 = 570.0;# [W/square m.K]\n",
+ "k_Cu = 381.0;# [w/square m.K]\n",
+ "#******#\n",
+ "\n",
+ "# For the metal tube\n",
+ "int_dia = out_dia-(2*wall_thick);# [m]\n",
+ "avg_dia = (out_dia+int_dia)/2;# [mm]\n",
+ "Nb = 0;\n",
+ "Flux_a = 1;\n",
+ "Ya1 = 0.24;\n",
+ "Yb1 = 1-Ya1;\n",
+ "Mav = (Ya1*18.02)+(Yb1*29);# [kg/kmol]\n",
+ "density = (Mav/22.41)*(273/(273+T1));# [kg/cubic m]\n",
+ "viscosity = 1.75*10**(-5);# [kg/m.s]\n",
+ "Cpa = 1880.0;# [J/kg.K]\n",
+ "Cpmix = 1145.0;# [J/kg.K]\n",
+ "Sc = 0.6;\n",
+ "Pr = 0.75;\n",
+ "G_prime = avg_velocity*density;# [kg/square m.s]\n",
+ "G = G_prime/Mav;# [kmol/square m.s]\n",
+ "Re = avg_dia*G_prime/viscosity;\n",
+ "# From Table 3.3:\n",
+ "# Jd = Std*Sc**(2/3) = (F/G)*Sc**(2/3) = 0.023*Re**(-0.17);\n",
+ "Jd = 0.023*Re**(-0.17);\n",
+ "F = (0.023*G)*(Re**(-0.17)/Sc**(2.0/3));\n",
+ "\n",
+ "# The heat transfer coeffecient in the absence of mass transfer will be estimated through Jd = Jh\n",
+ "# Jh = Sth*Pr^(2/3) = (h/Cp*G_prime)*(Pr^(2/3)) = Jd\n",
+ "h = Jd*Cpmix*G_prime/(Pr**(2.0/3));\n",
+ "\n",
+ "U = 1/((1/k1)+((wall_thick/k_Cu)*(int_dia/avg_dia))+((1/k2)*(int_dia/out_dia)));# W/square m.K\n",
+ "\n",
+ "# Using Eqn. 3.70 & 3.71 with Nb = 0\n",
+ "# Qt = (Na*18.02*Cpa/1-exp(-(Na*18.02*Cpa/h)))*(T1-Ti)+(Lambda_a*Na);\n",
+ "# Qt = 618*(Ti-T2);\n",
+ "# Using Eqn. 3.67, with Nb = 0, Cai/C = pai, Ca1/C = Ya1 = 0.24;\n",
+ "# Na = F*log(((Flux_a)-(pai))/((Flux_a)-(Ya1));\n",
+ "\n",
+ "# Solving above three Eqn. simultaneously:\n",
+ "Ti = 42.2;# [C]\n",
+ "pai = 0.0806;# [atm]\n",
+ "Lambda_a = 43.4*10**6;# [J/kmol]\n",
+ "Na = F*log(((Flux_a)-(pai))/((Flux_a)-(Ya1)));# [kmol/square m.s]\n",
+ "Qt1 = 618*(Ti-T2);# [W/square m]\n",
+ "Qt2 = ((Na*18.02*Cpa/(1-exp(-(Na*18.02*Cpa/h))))*(T1-Ti))+(Lambda_a*Na);# [W/square m]\n",
+ "\n",
+ "# since the value of Qt1 & Qt2 are relatively close\n",
+ "print'The local rate of condensation of water is ',round(Na,6),' kmol/square m.s'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Illustration 3.7 - Page: 80\n",
+ "\n",
+ "\n",
+ "The local rate of condensation of water is 0.000232 kmol/square m.s\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.8: Page 81"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "\n",
+ "# Illustration 3.8\n",
+ "# Page: 81\n",
+ "\n",
+ "import math\n",
+ "print'Illustration 3.8 - Page: 81\\n\\n'\n",
+ "print'Illustration 3.8 (a)\\n\\n'\n",
+ "\n",
+ "# Solution (a)\n",
+ "\n",
+ "#***Data****#\n",
+ "# a = water b = air\n",
+ "Nb = 0;\n",
+ "h = 1100.0;# [W/square m]\n",
+ "#*****#\n",
+ "\n",
+ "Ma = 18.02;# [kg/kmol]\n",
+ "Cpa = 2090;# [J/kg.K]\n",
+ "T1 = 600.0;# [C]\n",
+ "Ti = 260;# [C]\n",
+ "# The positive dirn. is taken to be from the bulk gas to the surface.\n",
+ "Has = 2.684*(10**6);# enthapy of saturated steam at 1.2 std atm, rel. to the liquid at 0 C in [J/kg]\n",
+ "Hai = 2.994*(10**6);# enthalpy of steam at 1 std atm, 260 C in [J/kg]\n",
+ "\n",
+ "# Radiation contributions to the heat transfer from the gas to the surface are negligible. Eqn. 3.70 reduces to\n",
+ "Na = -((h/(Ma*Cpa))*log(1-((Cpa*(T1-Ti))/(Has-Hai))));# [kmol/square m.s]\n",
+ "print'The rate of steam flow reqd. is',round(Na,4),' kmol/square m.s\\n\\n'\n",
+ "# negative sign indicates that the mass flux is into the gas\n",
+ "\n",
+ "print'Illustration 3.8 (b)\\n\\n'\n",
+ " \n",
+ "# Solution (b)\n",
+ "\n",
+ "#***Data****#\n",
+ "# a = water b = air\n",
+ "h = 572.0;# [W/square m]\n",
+ "T1 = 25.0;# [C]\n",
+ "#******#\n",
+ "\n",
+ "Ti = 260.0;# [C]\n",
+ "# The positive dirn. is taken to be from the bulk gas to the surface.\n",
+ "Has = 1.047*10**(5);# enthapy of saturated steam at 1.2 std atm, rel. to the liquid at 0 C in [J/kg]\n",
+ "Hai = 2.994*(10**6);# enthalpy of steam at 1 std atm, 260 C in [J/kg]\n",
+ "\n",
+ "# Radiation contributions to the heat transfer from the gas to the surface are negligible. Eqn. 3.70 reduces to\n",
+ "Na = -((h/(Ma*Cpa))*math.log(1-((Cpa*(T1-Ti))/(Has-Hai))));# [kmol/square m.s]\n",
+ "print'The rate of steam flow reqd. is',round(Na,4),' kmol/square m.s'\n",
+ "# negative sign indicates that the mass flux is into \n",
+ "# the answer of part B in textbook is incorrect"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Illustration 3.8 - Page: 81\n",
+ "\n",
+ "\n",
+ "Illustration 3.8 (a)\n",
+ "\n",
+ "\n",
+ "The rate of steam flow reqd. is -0.0348 kmol/square m.s\n",
+ "\n",
+ "\n",
+ "Illustration 3.8 (b)\n",
+ "\n",
+ "\n",
+ "The rate of steam flow reqd. is 0.0028 kmol/square m.s\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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