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diff --git a/409/CH12/EX12.1/Example12_1.sce b/409/CH12/EX12.1/Example12_1.sce new file mode 100755 index 000000000..c73207b26 --- /dev/null +++ b/409/CH12/EX12.1/Example12_1.sce @@ -0,0 +1,47 @@ +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);
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