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diff --git a/905/CH2/EX2.11/2_11.sce b/905/CH2/EX2.11/2_11.sce
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+clear;
+clc;
+
+// Illustration 2.11
+// Page: 129
+
+printf('Illustration 2.11 -  Page: 129\n\n');
+
+// solution
+//*****Data*****//
+// a-water   b-air
+D = 25.4*10^-3; // [diameter of wetted wall tower, m]
+Gy = 10; // [mass velocity, kg/square m.s]
+T1 = 308; // [K]
+P = 101.3; // [kPa]
+p_a1 = 1.95; // [partial pressure of water vapor, kPa]
+R = 8.314; // [cubic m.Pa/mole.K]
+M_a = 18; // [gram/mole]
+Cpa = 1.88; // [kJ/kg.K]
+//*****//
+
+// Properties of dry air at 308 K and 1 atm pressure are
+u = 1.92*10^-5; // [kg/m.s]
+row = 1.14; // [kg/cubic m]
+D_ab = 0.242*10^-4; // [square m/s]
+Sc = 0.696; // [Schmidt number]
+Cp = 1.007; // [kJ/kg.K]
+k = 0.027; // [W/m.K]
+Pr = 0.7; // [Prandtl number]
+
+Re = D*Gy/u; // [Renoylds number]
+// From equation 2,74
+Sh = 0.023*Re^0.83*Sc^0.44; // [Sherwood number]
+// From Table 2.1
+kg = Sh*D_ab/(R*T1*D)*1000; // [mole/square m.s.kPa]
+printf("kg is %e\n",kg);
+// To estimate the heat-transfer coefficient, we use the Dittus-Boelter equation for cooling, equation 2.80
+Nu = 0.023*Re^0.8*Pr^0.3; // [Nusselt number]
+// From Table 2.1
+h = Nu*k/D; // [W/square m.K]
+printf("h is %f\n",h);
+T =373.15; // [K]
+lambda_a = 40.63; // [kJ/mole]
+Tc = 647.1; // [K]
+
+// Solution of simultaneous equation 2.78 and 2.79
+function[f]=F(e) 
+    f(1) = kg*(p_a1 - exp(16.3872-(3885.7/(e(1)-42.98))))-e(2);
+    f(2) = e(2)*M_a*Cpa*(T1-e(1))/(1-exp(-e(2)*M_a*Cpa/h)) + 1000*e(2)*lambda_a*((1-(e(1)/Tc))/(1-(T/Tc)))^0.38;
+    funcprot(0);
+endfunction
+
+// Initial guess
+e = [292 -0.003];
+y = fsolve(e,F);
+Ti = y(1);
+Na = y(2);
+
+printf("The temperature of the liquid water and the rate of water evaporation is %f K and %e mole/square m.s respectively",Ti,Na);
+\ No newline at end of file