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+clear;
+clc;
+
+// Illustration 3.7
+// Page: 80
+
+printf('Illustration 3.7 - Page: 80\n\n');
+
+// solution
+
+//****Data*****//
+// a = water b = air
+out_dia = 0.0254;// [m]
+wall_thick = 0.00165;// [m]
+avg_velocity = 4.6;// [m/s]
+T1 = 66;// [C]
+P = 1;// [atm]
+Pa1 = 0.24;// [atm]
+k1 = 11400;// [W/(square m.K)]
+T2 = 24;// [C]
+k2 = 570;// [W/square m.K]
+k_Cu = 381;// [w/square m.K]
+//******//
+
+// For the metal tube
+int_dia = out_dia-(2*wall_thick);// [m]
+avg_dia = (out_dia+int_dia)/2;// [mm]
+Nb = 0;
+Flux_a = 1;
+Ya1 = 0.24;
+Yb1 = 1-Ya1;
+Mav = (Ya1*18.02)+(Yb1*29);// [kg/kmol]
+density = (Mav/22.41)*(273/(273+T1));// [kg/cubic m]
+viscosity = 1.75*10^(-5);// [kg/m.s]
+Cpa = 1880;// [J/kg.K]
+Cpmix = 1145;// [J/kg.K]
+Sc = 0.6;
+Pr = 0.75;
+G_prime = avg_velocity*density;// [kg/square m.s]
+G = G_prime/Mav;// [kmol/square m.s]
+Re = avg_dia*G_prime/viscosity;
+// From Table 3.3:
+// Jd = Std*Sc^(2/3) = (F/G)*Sc^(2/3) = 0.023*Re^(-0.17);
+Jd = 0.023*Re^(-0.17);
+F = (0.023*G)*(Re^(-0.17)/Sc^(2/3));
+
+// The heat transfer coeffecient in the absence of mass transfer will be estimated through Jd = Jh
+// Jh = Sth*Pr^(2/3) = (h/Cp*G_prime)*(Pr^(2/3)) = Jd
+h = Jd*Cpmix*G_prime/(Pr^(2/3));
+
+U = 1/((1/k1)+((wall_thick/k_Cu)*(int_dia/avg_dia))+((1/k2)*(int_dia/out_dia)));// W/square m.K
+
+// Using Eqn. 3.70 & 3.71 with Nb = 0
+// Qt = (Na*18.02*Cpa/1-exp(-(Na*18.02*Cpa/h)))*(T1-Ti)+(Lambda_a*Na);
+// Qt = 618*(Ti-T2);
+// Using Eqn. 3.67, with Nb = 0, Cai/C = pai, Ca1/C = Ya1 = 0.24;
+// Na = F*log(((Flux_a)-(pai))/((Flux_a)-(Ya1));
+
+// Solving above three Eqn. simultaneously:
+Ti = 42.2;// [C]
+pai = 0.0806;// [atm]
+Lambda_a = 43.4*10^6;// [J/kmol]
+Na = F*log(((Flux_a)-(pai))/((Flux_a)-(Ya1)));// [kmol/square m.s]
+Qt1 = 618*(Ti-T2);// [W/square m]
+Qt2 = ((Na*18.02*Cpa/(1-exp(-(Na*18.02*Cpa/h))))*(T1-Ti))+(Lambda_a*Na);// [W/square m]
+
+// since the value of Qt1 & Qt2 are relatively close
+printf('The local rate of condensation of water is %e kmol/square m.s',Na);
+\ No newline at end of file