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diff --git a/839/CH26/EX26.1/Example_26_1.sce b/839/CH26/EX26.1/Example_26_1.sce
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+++ b/839/CH26/EX26.1/Example_26_1.sce
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+//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') 

diff --git a/839/CH26/EX26.4/Example_26_4.sce b/839/CH26/EX26.4/Example_26_4.sce
new file mode 100755
index 000000000..5da677be8
--- /dev/null
+++ b/839/CH26/EX26.4/Example_26_4.sce
@@ -0,0 +1,28 @@
+//clear//

+clear;

+clc;

+

+//Example 26.4

+//Given

+F = 10; //[gal/day-ft^3]

+Do = 300*10^-6; //[m]

+Di = 200*10^-6; //[m]

+vi = 0.5; //[cm/s]

+rho = 1; //[g/cm^3]

+mu = 0.01; //[g/cm-s], assumed

+f = 0.97;

+

+//Solution

+//For 10 gal/day-ft^2

+Jw = F*231*16.3871/(24*3600*929); //[cm/s]

+Nre = Do*100*vi*rho/mu;

+Ds = 1.6*10^-5; //[cm^2/s]

+Nsc = mu/(rho*Ds);

+

+//Using Eq.(12.69), Analogously to mass transfer

+Nsh = (0.35+0.56*Nre^0.52)/Nsc^-0.3;

+kc = Nsh*Ds/(Do*100); //[cm/s]

+//From Eq.(26.49)

+gama = Jw*f/kc;

+disp('A concentration differnce of 12 percent will not be significant till good flow distribution is maintained');

+

diff --git a/839/CH26/EX26.5/Example_26_5.sce b/839/CH26/EX26.5/Example_26_5.sce
new file mode 100755
index 000000000..4656b61de
--- /dev/null
+++ b/839/CH26/EX26.5/Example_26_5.sce
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+//clear//

+clear;

+clc;

+

+//Example 26.5

+//Given (from Example 26.4)

+F = 10; //[gal/day-ft^2], based on external area

+Do = 300*10^-6; //[m]

+Di = 200*10^-6; //[m]

+vi = 0.5; //[cm/s]

+rho = 1; //[g/cm^3]

+mu = 10^-3; //[Pa-s], assumed

+f = 0.97;

+L = 3; //[m]

+

+//Solution

+//(a)

+//Jw based on area

+Jw = 4.72*10^-4*Do/Di*10^-2; //[m/s]

+dt = 200*10^-6; //[m]

+D = dt; //[m]

+//From Eq.(26.53)

+Vbar = 4*(Jw)*L/Di; //[m/s]

+//From Eq.(26.56)

+delta_ps = (Vbar*32*mu*L)/(D)^2*(1/2)/10^5; //[atm]

+disp('atm',delta_ps,'pressure drop = ','m/s',Vbar,'exit velocity = ');

+

+//(b)

+//If the fibres are open at both ends, the effective length is 1.5m and

+//the exit velocity is half as great. The pressure drop is one-fourth as 

+//large as it was:

+deltaP = delta_ps/4; //[atm]

+disp('atm',deltaP,'pressure drop (if both ends are open) = ')