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| author | kinitrupti | 2017-05-12 18:40:35 +0530 |
|---|---|---|
| committer | kinitrupti | 2017-05-12 18:40:35 +0530 |
| commit | d36fc3b8f88cc3108ffff6151e376b619b9abb01 (patch) | |
| tree | 9806b0d68a708d2cfc4efc8ae3751423c56b7721 /Mass_-_Transfer_Operations/Chapter3.ipynb | |
| parent | 1b1bb67e9ea912be5c8591523c8b328766e3680f (diff) | |
| download | Python-Textbook-Companions-d36fc3b8f88cc3108ffff6151e376b619b9abb01.tar.gz Python-Textbook-Companions-d36fc3b8f88cc3108ffff6151e376b619b9abb01.tar.bz2 Python-Textbook-Companions-d36fc3b8f88cc3108ffff6151e376b619b9abb01.zip | |
Revised list of TBCs
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diff --git a/Mass_-_Transfer_Operations/Chapter3.ipynb b/Mass_-_Transfer_Operations/Chapter3.ipynb deleted file mode 100755 index 36f53f4d..00000000 --- a/Mass_-_Transfer_Operations/Chapter3.ipynb +++ /dev/null @@ -1,628 +0,0 @@ -{
- "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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