{
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"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": {}
}
]
}