clear; clc; // Illustration 8.8 // Page: 317 printf('Illustration 8.8 - Page: 317\n\n'); // Solution //***Data*** // a:NH3 b:air c:H2O ya = 0.416;// [mole fraction] yb = 0.584;// [mole fraction] G1 = 0.0339;// [kmol/square m.s] L1 = 0.271;// [kmol/square m.s] TempG1 = 20;// [OC] //********// // At 20 OC Ca = 36390;// [J/kmol] Cb = 29100;// [J/kmol] Cc = 33960;// [J/kmol] lambda_c = 44.24*10^6;// [J/kmol] // Enthalpy base = NH3 gas, H2O liquid, air at 1 std atm. Tempo = 20;// [OC] lambda_Ao = 0;// [J/kmol] lambda_Co = 44.24*10^6;// [J/kmol] // Gas in: Gb = G1*yb;// [kmol air/square m.s] Ya1 = ya/(1-ya);// [kmol NH3/kmol air] yc1 = 0;// [mole fraction] Yc1 = yc1/(1-yc1);// [kmol air/kmol NH] // By Eqn 8.58: Hg1 = (Cb*(TempG1-Tempo))+(Ya1*(Ca*(TempG1-Tempo))+lambda_Ao)+(Yc1*(Cc*(TempG1-Tempo)+lambda_Co));// [J/kmol air] // Liquid in: xa1 = 0;// [mole fraction] xc1 = 1;// [mole fraction] Hl1 = 0;// [J/kmol air] //Gas out: Ya2 = Ya1*(1-0.99);// [kmol NH3/kmol air] // Assume: TempG2 = 23.9;// [OC] yc2 = 0.0293; deff('[y] = f(Yc2)','y = yc2-(Yc2/(Yc2+Ya2+1))'); Yc2 = fsolve(0.002,f);// [kmol H2O/kmol air] Hg2 = (Cb*(TempG2-Tempo))+(Ya2*(Ca*(TempG2-Tempo))+lambda_Ao)+(Yc2*(Cc*(TempG2-Tempo)+lambda_Co));// [J/kmol air] // Liquid out: Lc = L1-(Yc1*Gb);// [kmol/square m.s] La = Gb*(Ya1-Ya2);// [kmol/square m.s] L2 = La+Lc;// [kmol/square m.s] xa = La/L2; xc = Lc/L2; // At xa & tempo = 20 OC delta_Hs = -1709.6*1000;// [J/kmol soln] // Condition at the bottom of the tower: // Assume: TempL = 41.3;// {OC} // At(TempL+TempG1)/2: Cl = 75481;// [J/kmol] deff('[y] = f40(Cl)','y = Hl1+Hg1-((Gb*Hg2)+(L2*(Cl*(TempL-Tempo)+delta_Hs)))'); Cl = fsolve(7,f40);// [J/kmol.K] // For the Gas: MavG = 24.02;// [kg/kmol] Density_G = 0.999;// [kg/cubic m] viscosity_G = 1.517*10^(-5);// [kg/m.s] kG = 0.0261;// [W/m.K] CpG = 1336;// [J/kg.K] Dab = 2.297*10^(-5);// [square m/s] Dac = 3.084*10^(-5);// [square m/s] Dcb = 2.488*10^(-5);// [square m/s] PrG = CpG*viscosity_G/kG; // For the liquid: MavL = 17.97;// [kg/kmol] Density_L = 953.1;// [kg/cubic m] viscosity_L = 6.408*10^(-4);// [kg/m.s] Dal = 3.317*10^(-9);// [square m/s] kl = 0.4777;// [W/m.K] ScL = viscosity_L/(Density_L*Dal); PrL = 5.72; sigma = 3*10^(-4); G_prime = G1*MavG;// [kg/square m.s] L_prime = L2*MavL;// [kg/square m.s] // From data of Chapter 6: Ds = 0.0472;// [m] a = 57.57;// [square m/cubic m] shiLt = 0.054; e = 0.75; // By Eqn. 6.71: eLo = e-shiLt; // By Eqn. 6.72: kL = (25.1*Dal/Ds)*(Ds*L_prime/viscosity_L)^0.45*ScL^0.5;// [m/s] c = Density_L/MavL;// [kmol/cubic m] Fl = kL*c;// [kmol/cubic m] // The heat mass transfer analogy of Eqn. 6.72: hL = (25.1*kl/Ds)*(Ds*L_prime/viscosity_L)^0.45*PrL^0.5;// [m/s] // The heat transfer analogy of Eqn. 6.69: hG = (1.195*G_prime*CpG/PrG^(2/3))*(Ds*G_prime/(viscosity_G*(1-eLo)))^(-0.36);// [W/square m.K] // To obtain the mass transfer coeffecients: Ra = 1.4; Rc = 1-Ra; // From Eqn. 8.83: Dam = (Ra-ya)/(Ra*((yb/Dab)+((ya+yc1)/Dac))-(ya/Dac));// [square m/s] Dcm = (Rc-yc1)/(Rc*((yb/Dcb)+((ya+yc1)/Dac))-(yc1/Dac));// [square m/s] ScGa = viscosity_G/(Density_G*Dam); ScGc = viscosity_G/(Density_G*Dcm); // By Eqn. 6.69: FGa = (1.195*G1/ScGa^(2/3))*(Ds*G_prime/(viscosity_G*(1-eLo)))^(-0.36);// [kmol/square m.K] FGc = (1.195*G1/ScGc^(2/3))*(Ds*G_prime/(viscosity_G*(1-eLo)))^(-0.36);// [kmol/square m.K] Ra = Ra-0.1; // From Eqn. 8.80: scf(14); for i = 1:3 deff('[yai] = f41(xai)','yai = Ra-(Ra-ya)*((Ra-xa)/(Ra-xai))^(Fl/FGa)'); xai = xa:0.01:0.10; plot(xai,f41) Ra = Ra+0.1; end xgrid(); xlabel("Mole fraction NH3 in the liquid, xa"); ylabel("Mole fraction NH3 in the gas ya"); title("Operating Line curves"); Rc = Rc-0.1; // From Eqn. 8.81: scf(15); for i = 1:3 deff('[yci] = f42(xci)','yci = Rc-(Rc-yc1)*((Rc-xc)/(Rc-xci))^(Fl/FGc)'); xci = xc:-0.01:0.85; plot(xci,f42) Rc = Rc+0.1; end xgrid(); xlabel("Mole fraction H2O in the liquid, xc"); ylabel("Mole fraction H2O in the gas, yc"); title("Operating line Curves"); // Assume: Tempi = 42.7;// [OC] // The data of Fig. 8.2 (Pg 279) & Fig 8.4 (Pg 319) are used to draw the eqb curve of Fig 8.25 (Pg 320). // By interpolation of operating line curves with eqb line and the condition: xai+xci = 1; Ra = 1.38; Rc = 1-Ra; xai = 0.0786; yai = f41(xai); xci = 1-xai; yci = f42(xci); // From Eqn. 8.77: dYa_By_dZ = -(Ra*FGa*a/Gb)*log((Ra-yai)/(Ra-ya));// [kmol H2O/kmol air] // From Eqn. 8.78: dYc_By_dZ = -(Rc*FGc*a/Gb)*log((Rc-yci)/(Rc-yc1));// [kmol H2O/kmol air] // From Eqn. 8.82: hGa_prime = -(Gb*((Ca*dYa_By_dZ)+(Cc*dYc_By_dZ)))/(1-exp(Gb*((Ca*dYa_By_dZ)+(Cc*dYc_By_dZ))/(hG*a)));// [W/cubic m.K] // From Eqn. 8.79: dtG_By_dZ = -(hGa_prime*(TempG1-Tempi))/(Gb*(Cb+(Ya1*Ca)+(Yc1*Cc)));// [K/m] // When the curves of Fig. 8.2 (pg 279) & 8.24 (Pg 319) are interpolated for concentration xai and xci, the slopes are: mar = 0.771; mcr = 1.02; lambda_c = 43.33*10^6;// [J/kmol] // From Eqn. 8.3: Hai = Ca*(Tempi-Tempo)+lambda_Ao-(mar*lambda_c);// [J/kmol] Hci = Cc*(Tempi-Tempo)+lambda_Co-(mcr*lambda_c);// [J/kmol] // From Eqn. 8.76 Tempi2 = TempL+(Gb/(hL*a))*(((Hai-Ca*(TempG1-Tempo)-lambda_Ao)*dYa_By_dZ)+((Hci-Cc*(TempG1-Tempo)-lambda_Co)*dYc_By_dZ)-((Cb+(Ya1*Ca)+(Yc1*Cc))*dtG_By_dZ));// [OC] // The value of Tempi obtained is sufficiently close to the value assumed earlier. deltaYa=-0.05; // An interval of deltaYa up the tower deltaZ = deltaYa/(dYa_By_dZ);// [m] deltaYc = (dYc_By_dZ*deltaZ); // At this level: Ya_next = Ya1+deltaYa;// [kmol/kmol air] Yc_next = Yc1+deltaYc;// [kmol H2O/kmol air] tG_next = TempG1+(dtG_By_dZ*deltaZ);// [OC] L_next = L1+Gb*(deltaYa+deltaYc);// [kmol/square m.s] xa_next = ((Gb*deltaYa)+(L1*xa))/L_next;// [mole fraction NH3] Hg_next = (Cb*(tG_next-Tempo))+(Ya_next*(Ca*(tG_next-Tempo))+lambda_Ao)+(Yc_next*(Cc*(tG_next-Tempo)+lambda_Co));// [J/kmol air] Hl_next = (L1*Hl1)+(Gb*(Hg_next-Hg2)/L_next);// [J/kmol] // The calculation are continued where the specified gas outlet composition are reached. // The packed depth is sum of all deltaZ Z = 1.58;// [m] printf("The packed depth is: %f m\n",Z);