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+clear;
+clc;
+
+// Illustration 4.5
+// Page: 245
+
+printf('Illustration 4.5 -  Page: 245\n\n');
+
+// solution
+//*****Data*****//
+//  a-chloroform    b-water    c-air
+T = 298; // [K]
+Dv = 1; // [vessel diameter, m]
+Vb = 10; // [kg/s]
+ca = 240*10^-6; // [gram/l]
+xr = 0.9; // [chloroform which is to be removed]
+m = 220;
+Ds = 0.5; // [diameter of sparger, m]
+no = 90; // [number of orifices]
+Do = 3*10^-3; // [diameter of orifice, m]
+nm = 0.6; // [mechanical efficiency]
+rowb = 1000; // [kg/cubic m]
+R = 8.314;
+Mc = 29; // [gram/mole]
+Mb = 18; // [gram/mole]
+g = 9.8; // [square m/s]
+//*****//
+
+Vair = 0.1; // [kg/s as calculated in chapter 3]
+mg = Vair/no; // [mass flow rate through each orifice, kg/s]
+ug = 1.8*10^-5; // [kg/m.s]
+Reo = 25940; // [Renoylds number]
+// From equ. 4.20
+dp = 0.0071*Reo^-0.05; // [m]
+
+// Since the water column height is not known, therefore an iterative procedure must be implemented.
+// Assuming column height, Z = 0.5 m
+Z = 0.5; // [m]
+// For Z = 0.5 m
+rowl = 1000; // [kg/cubic m]
+Ps = 101.3; // [kPa]
+Po = Ps +  (1000*9.8*0.5/1000); // [kPa]
+Pavg = (Po+Ps)/2; // [kPa]
+rowg = Pavg*Mc/(R*T); // [kg/cubic m]
+
+area = %pi*Dv^2/4; // [square m]
+vg = Vair/(rowg*area); // [m/s]
+// In this case rowl = rowg and sigma = sigmaAW
+// From equation 4.22
+// Vg = vg
+// vg/vs = 0.182
+vs = vg/0.182; // [m/s]
+vl = -Vb/(rowl*area); // [negative because water flows downward, m/s]
+// From equ 4.21
+
+deff('[y] = f12(phig)','y = vs - (vg/phig)-(-vl/(1-phig))');
+phig = fsolve(0.1,f12);
+// Now in this case
+S = vl/(1-phig);
+// Value of 'S' comes out to be less than 0.15 m/s
+// Therefore 
+dp = (dp^3*Po/Pavg)^(1/3); // [m]
+// From equ 4.23
+a = 6*phig/dp; // [m^-1]
+// Now we calculate diffusivity of chloroform
+Vba = 88.6; // [cubic cm/mole]
+u = 0.9*10^-3; // [Pa-s]
+e = (9.58/(Vba)-1.12);
+// From equation 1.53
+Dl = 1.25*10^-8*((Vba)^-0.19 - 0.292)*T^1.52*(u*10^3)^e; // [square cm/s]
+
+// And Schmidt number is 
+Scl = 833; // [Schmidt Number]
+
+// Now we calculate  dp*g^(1/3)/Dl^(2/3) = J
+J = dp*g^(1/3)/(Dl*10^-4)^(2/3)
+Reg = dp*vs*rowl/u; // [Gas bubble Renoylds number]
+// From equ 4.25
+Shl = 2 + 0.0187*Reg^0.779*Scl^0.546*J^0.116;
+
+// For dilute solution xbm = 1 or c = 55.5 kmole/cubic m
+// Then for Nb = 0
+c = 55.5; // [kmole/cubic m]
+kx = Shl*c*Dl*10^-4/dp; // [kmole/square m.s]
+mtc = kx*a; // [kmole/cubic m.s]
+
+L = Vb/Mb; // [kmole/s]
+Gmx = L/area; // [kmole/square m.s]
+V = Vair/Mc; // [kmole/s]
+A = L/(m*V); // [absorption factor]
+
+// From equ 4.28
+ // For, xin/xout = x = 10
+ x = 10;
+Z = (Gmx/(kx*a*(1-A)))*log(x*(1-A)+A);
+
+// With this new estimated Z ,we again calculate average pressure in the   // column of water
+Po1 = 110.1; // [kPa]
+Pavg1 = 105.7; // [kPa]
+rowg1 = Pavg1*Mc/(R*T);
+// Now value of rowg1 obtained is very close to value used in the first   // iteration. Therefore on three iteractions we achieve a value of 'Z'
+Z1 = 0.904; // [m]
+
+rowgo = Po1*Mc/(R*T); // [kg/cubic m]
+vo1 = 4*mg/(%pi*Do^2*rowgo); // [m/s]
+// Therefore,    vo1^2/(2*gc) = F
+gc = 1;
+F = vo1^2/(2*gc); // [J/kg]
+// And           R*T*log(Po/Ps)/Mc = G
+G = R*T*1000*log(Po1/Ps)/Mc; // [J/kg]
+Zs = 0
+// And   (Z1-Zs)*g/gc = H
+H = (Z1-Zs)*g/gc; // [J/kg]
+// From equ 4.27
+W = F+G+H; // [J/kg]
+// Now the air compressor power is
+W1 = W*Vair*10^-3/nm; // [kW]
+
+printf("The depth of the water column required to achieve the specified 90percent removal efficiency is %f m\n\n",Z1);
+printf("The power required to operate the air compressor is %f kW\n\n",W1);
+\ No newline at end of file