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diff --git a/839/CH26/EX26.1/Example_26_1.sce b/839/CH26/EX26.1/Example_26_1.sce new file mode 100755 index 000000000..0137dd78b --- /dev/null +++ b/839/CH26/EX26.1/Example_26_1.sce @@ -0,0 +1,82 @@ +//clear//
+clear;
+clc;
+
+//Example 26.1
+//Given
+alpha = 5;
+per = 0.2; //[scf/ft^2-h-atm]
+Pf = 150; //[lbf/in.^2]
+Pp = 15; //[lbf/in.^2]
+
+//Solution
+//(a)
+R = Pp/Pf;
+//At the feed inlet
+xin = 0.209;
+//Using Eq.(26.17)
+A = alpha-1;
+B = 1-alpha-1/R-xin*(alpha-1)/R;
+C = alpha*xin/R;
+yi_in = (-B-sqrt(B^2-4*A*C))/(2*A);
+//At the discharge end
+xd = 0.05;
+//Using Eq.(26.17)
+A = alpha-1;
+B = 1-alpha-1/R-xd*(alpha-1)/R;
+C = alpha*xd/R;
+yi_d = (-B-sqrt(B^2-4*A*C))/(2*A);
+
+//For an approximate solution, these terminal compositions are
+//averaged to give
+ybar = (yi_in+yi_d)/2;
+//From an overall material balance
+//Basis
+Lin = 100; //[scfh]
+V = (Lin*xin-Lin*xd)/(ybar-xd);
+//disp(ybar,'and permeate composition is','percent',V/Lin*100,'The permeate in the feed is');
+
+
+//For more accurate calculation
+j = 2;
+yi_in(1) = 0.5148;
+x(1) = 0.209;
+y(1)= 0.5148;
+L = Lin;
+deltaV = [];
+deltaVybar = [];
+ybar = [];
+for i = 0.2:-0.01:xd
+x(j) = i;
+A = alpha-1;
+B = 1-alpha-1/R-x(j)*(alpha-1)/R;
+C = alpha*x(j)/R;
+yi_in(j) = (-B-sqrt(B^2-4*A*C))/(2*A);
+ybar(j-1) = (yi_in(j-1)+yi_in(j))/2;
+deltaV(j) = L*(x(j-1)-x(j))/(ybar(j-1)-x(j));
+V = sum(deltaV);
+L = Lin - V;
+deltaVybar(j) = deltaV(j-1)*ybar(j-1);
+deltaVybarsum = sum(deltaVybar);
+y(j-1) = deltaVybarsum/V;
+j = j+1;
+end
+disp(y($),'and permeate composition is','percent',V/Lin*100,'The permeate recovered');;
+
+
+//(b)
+//The membrane area obtained from the flux of A using
+//Eq.(26.29) and (26.13)
+//for the first increment x = 0.209 to x = 0.2
+deltaybar1 = 1.4856; //[scfh], for Lin = 100 scfh
+//At x = 0.209
+A1 = 0.209-0.1*0.5148;
+//At x = 0.2
+A2 = 0.2-0.1*(0.50);
+Aavg = (A1+A2)/2
+QAP1 = 0.2*10; //scfh/ft^3
+//for specified flow of 300 scfh
+deltaA = 1/2*1.486/Aavg*180; //[ft^2]
+//The calculation continued with increments of 0.01
+A = 211/2.0*180; //[ft^2]
+disp('ft^2',A,'The membrane area needed is')
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