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+clear;

+clc;

+

+// Illustration 2.12

+// Page: 131

+

+printf('Illustration 2.12 -  Page: 131\n\n');

+

+// solution

+//*****Data*****//

+// a-water    b-dry air

+D = 25.4*10^-3; // [Internal diameter of tower, m]

+Z = 1.5; // [length of the wetted section, m]

+Gy = 10; // [mass velocity of air, kg/square m.s]

+Tair = 308; // [K]

+Twater = 295; // [K]

+P = 101.3; // [kPa]

+M_a = 18; // [gram/mole]

+M_b = 29; // [gram/mole]

+R = 8.314; // [cubic m.Pa/mole.K]

+//*****//

+

+// The water vapor partial pressure at the interface remains constant at the vapor pressure of liquid water at 295 K, which is pa1 = Pa = 2.64 kPa

+// The water vapor partial pressure at the bulk of the gas phase increases from pA2 = pAin = 0 for the dry inlet air to pa2= pAout for the air leaving the tower 

+Pa = 2.64; // [kPa]

+

+Gm = Gy/M_b; // [Assuming that gas phase is basically dry air, kmole/square m.s]

+// The properties of dry air at 308 K and 1 atm are (from example 2.9)

+row = 1.14; // [kg/cubic m]

+u = 1.92*10^-5; // [kg/m.s]

+D_ab = 0.242*10^-4; // [square m/s]

+Sc = 0.692; // [Schmidt number]

+

+Re = Gy*D/u; // [Renoylds number]

+

+if(Re<35000 & Re>2000)

+// From equation 2.74

+Sh = 0.023*Re^0.83*Sc^0.44; // [Sherwood number]    

+

+printf("Sherwood number is %f\n\n",Sh);

+else()

+    printf('We cannot use equation 2.74')

+end

+

+c = P/(R*Tair); // [kmole/cubic m]

+// Now using equation 2.89

+Pa_out = Pa*(1-exp((-4*Sh*Z*c*D_ab)/(Gm*D^2))); // [kPa]

+printf("The partial pressure of water in the air leaving the tower is %e kPa\n\n",Pa_out);
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