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+clear;
+clc;
+//Page No.349
+// Example 12.1
+printf('Example 12.1\n\n');
+// Solution
+
+//(a) fig.E12.1a
+F = 10000 ;//[lb/hr]
+//Given
+NaOH_F = 40/100 ;//[wt. fraction]
+NaOH_P1 = 95/100 ;//[wt. fraction of NaOH filter cake]
+NaOH_P2 = (0.05 * 45)/100 ;//[wt. fraction of NaOH in NaOH soln.]
+H2O_P2 = (0.05 * 55)/100 ;//[wt. fraction of H2O in NaOH soln.]
+NaOH_R = 45/100;//[wt. fraction]
+NaOH_G = 50/100;//[wt. fraction]
+//Get P from overall NaOH balance
+P = (NaOH_F * F)/[NaOH_P1 + NaOH_P2] ;//[lb/hr]
+// Get W from overall total balance
+W = F-P ;// [lb/hr]
+
+// Solve following eqn. simultaneously to get G & R
+// NaOH_G * G = F * NaOH_F + NaOH_R * R (NaOH balance on crystallizer)
+//G = R + P (overall balance)
+a = [NaOH_G -NaOH_R;1 -1] ;// matrix formed of coefficients of unknown
+b = [F * NaOH_F;P];// matrix formed by constant
+x = a\b ;// matrix of solutions . x(1) = G, x(2) = R
+G = x(1) ;// [lb/hr]
+R = x(2) ;// [lb/hr]
+printf('(a) Flow rate of water removed by evaporator is %.1f lb/hr\n',W);
+printf(' The recycle rate of the process is %.1f lb/hr\n',R);
+
+// (b) fig.E12.1b
+//given
+NaOH_H = 45/100 ;//[wt fraction]
+H2O_H = 55/100 ;//[wt fraction]
+// Get H & G by solving following eqn. simultaneously
+//NaOH_G * G = [NaOH_P1 + NaOH_P2] * P + NaOH_H * H (NaOH balance on crystallizer)
+//H2O_G * G = H2O_P2 * P + H2O_H * H (H2O balance on crystallizer)
+a1 = [NaOH_G -NaOH_H;NaOH_G -H2O_H] ;// matrix formed of coefficients of unknown
+b1 = [((NaOH_P1 + NaOH_P2) * P);(H2O_P2) * P];// matrix formed by constant
+x1 = ((a1)^-1) * b1 ;// matrix of solutions nw_G = x1(1);H = x1(2)
+nw_G1 = x1(1) ;// [lb/hr]
+H = x1(2);// [lb/hr]
+// By overall NaOH balance
+nw_F = (NaOH_H * H + (NaOH_P1 + NaOH_P2) * P)/NaOH_F ;//[lb/hr]
+printf(' (b) Total feed rate when filterate is not recycled is %.1f lb/hr\n',nw_F); \ No newline at end of file