{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 3: Mass Transfer Coeffecients"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.1: Mass_Transfer_Coeffecient_in_Laminar_Flow.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Illustration 3.1\n",
"// Page: 53\n",
"\n",
"printf('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;// [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))^(1/3);// [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))^(1/2);//[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/(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",
"printf('The rate of absorption is %e',rate_absorption);\n",
"// The actual value may be substantially larger."
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.2: Eddy_Diffusio.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Illustration 3.2\n",
"// Page: 56\n",
"\n",
"printf('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)^(1/4);// [m]\n",
"// From Eqn. 3.25\n",
"Ud = (kinematic_viscosity*P/m)^(1/4);// [m/s]\n",
"printf('Velocity of small eddies is %f m/s\n',Ud);\n",
"printf('Length scale of small eddies is %e m',Ld);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.3: Mass_Heat_And_Momentum_Transfer_Analogies.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Illustration 3.3\n",
"// Page: 69\n",
"\n",
"printf('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",
"printf('Heat transfer analog to Eqn. 3.21 is eta = (2*alpha*L)/(3*del^2*velocity)');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.4: Mass_Heat_And_Momentum_Transfer_Analogies.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Illustration 3.4\n",
"// Page: 69\n",
"\n",
"printf('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;// [m/s]\n",
"surf_temp = 43;// [C]\n",
"bulk_temp = 60;// [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; // [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;\n",
"Ca1_by_C = press/tot_press;\n",
"Ca2_by_C = 0;\n",
"Flux_a = 1;\n",
"// Using Eqn. 3.1\n",
"Na = Flux_a*F_avg*log((Flux_a-Ca2_by_C)/(Flux_a-Ca1_by_C));//[kmol UF6/square m.s]\n",
"printf('Rate of sublimation is %e kmol UF6/square m.s',Na);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.5: Flux_Variation_with_Concentration.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Illustration 3.5\n",
"// Page: 73\n",
"\n",
"printf('Illustration 3.5 - Page: 73\n\n');\n",
"\n",
"// solution\n",
"\n",
"//****Data****//\n",
"velocity = 15;// [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^(2/3)/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^(2/3));// [kmol/square m.s]\n",
"printf('The required mass transfer: %f kmol/square m.s',F);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.6: Calculation_of_Bed_depth.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Illustration 3.6\n",
"// Page: 77\n",
"\n",
"printf('Illustration 3.6 - Page: 77\n\n');\n",
"\n",
"// solution\n",
"\n",
"//***Data***//\n",
"Dp = 0.0125;// [m]\n",
"viscosity = 2.4*10^(-5);// [kg/m.s]\n",
"Sc = 2;\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 = 4/3;\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",
"Z = integrate('-((4*Go)/((4-(3*Ya))^2*a))*(3/4)*(E*(Sc^(2/3))*(4-(3*Ya))/(2.06*Go)*(1/log(4/(4-(3*Ya)))))*(((Dp/viscosity)*(Go*(47.6-(19.6/(4-(3*Ya))))))^0.575)','Ya',1,0.005);// [m]\n",
"printf('The bed depth required to reduce the CO content to 0.005 is %f m', Z);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.7: Local_rate_of_condensation.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Illustration 3.7\n",
"// Page: 80\n",
"\n",
"printf('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;// [C]\n",
"P = 1;// [atm]\n",
"Pa1 = 0.24;// [atm]\n",
"k1 = 11400;// [W/(square m.K)]\n",
"T2 = 24;// [C]\n",
"k2 = 570;// [W/square m.K]\n",
"k_Cu = 381;// [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;// [J/kg.K]\n",
"Cpmix = 1145;// [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^(2/3));\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^(2/3));\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",
"printf('The local rate of condensation of water is %e kmol/square m.s',Na);"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.8: Simultaneous_Heat_and_Mass_Transfer.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"\n",
"// Illustration 3.8\n",
"// Page: 81\n",
"printf('Illustration 3.8 - Page: 81\n\n');\n",
"printf('Illustration 3.8 (a)\n\n');\n",
"\n",
"// Solution (a)\n",
"\n",
"//***Data****//\n",
"// a = water b = air\n",
"Nb = 0;\n",
"h = 1100;// [W/square m]\n",
"//*****//\n",
"\n",
"Ma = 18.02;// [kg/kmol]\n",
"Cpa = 2090;// [J/kg.K]\n",
"T1 = 600;// [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",
"printf('The rate of steam flow reqd. is %f kmol/square m.s\n\n',Na);\n",
"// negative sign indicates that the mass flux is into the gas\n",
"\n",
"printf('Illustration 3.8 (b)\n\n');\n",
" \n",
"// Solution (b)\n",
"\n",
"//***Data****//\n",
"// a  =  water b  =  air\n",
"h  =  572;// [W/square m]\n",
"T1  =  25;// [C]\n",
"//******//\n",
"\n",
"Ti  =  260;// [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))*log(1-((Cpa*(T1-Ti))/(Has-Hai))));// [kmol/square m.s]\n",
"printf('The rate of steam flow reqd. is %f kmol/square m.s',Na);\n",
"// negative sign indicates that the mass flux is into "
   ]
   }
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