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+clear;
+clc;
+
+// Illustration 6.4
+// Page: 183
+
+printf('Illustration 6.4 - Page: 183\n\n');
+
+// solution
+
+//****Data****//
+//From Illustrtion 6.3:
+G = 0.100;// [kmol/s]
+Density_G = 0.679;// [kg/cubic m]
+q = 5*10^(-3);// [cubic m/s]
+Va = 3.827;// [m/s]
+z = 1.063;// [m]
+L = 0.25;// [kmol/s]
+hL = 0.0106;// [m]
+hW = 0.05;// [m]
+Z = 0.824;// [m]
+E = 0.05;
+ya = 0.18;// [mole fraction methanol]
+
+// a:CH3OH b:H2O
+Ma = 32;// [kg/kmol]
+Mb = 18;// [kg/kmol]
+// From Chapter 2:
+ScG = 0.865;
+Dl = 5.94*10^(-9);// [square m/s]
+// From Eqn. 6.61:
+NtG = (0.776+(4.57*hW)-(0.238*Va*Density_G^0.5)+(104.6*q/Z))/ScG^0.5;
+DE = ((3.93*10^(-3))+(0.0171*Va)+(3.67*q/Z)+(0.1800*hW))^2;// [square m/s]
+thethaL = hL*z*Z/q;// [s]
+NtL = 40000*Dl^0.5*((0.213*Va*Density_G^0.5)+0.15)*thethaL;
+// For 15 mass% methanol:
+xa = (15/Ma)/((15/Ma)+(85/Mb));
+// From Fig 6.23 (Pg 184)
+mAC = -(NtL*L)/(NtG*G);// [Slope of AC line]
+meqb = 2.50;// [slope of equilibrium line]
+// From Eqn. 6.52:
+NtoG = 1/((1/NtG)+(meqb*G/L)*(1/NtL));
+// From Eqn. 6.51:
+EOG = 1-exp(-NtoG);
+// From Eqn. 6.59:
+Pe = Z^2/(DE*thethaL);
+// From Eqn. 6.58:
+eta = (Pe/2)*((1+(4*meqb*G*EOG/(L*Pe)))^0.5-1);
+// From Eqn. 6.57:
+EMG = EOG*(((1-exp(-(eta+Pe)))/((eta+Pe)*(1+(eta+Pe)/eta)))+(exp(eta)-1)/(eta*(1+eta/(eta+Pe))));
+// From Eqn. 6.60:
+EMGE = EMG/(1+(EMG*E/(1-E)));
+printf("Effeciency of Sieve trays: %f",EMGE);
+\ No newline at end of file