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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /905/CH9/EX9.2/9_2.sce | |
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diff --git a/905/CH9/EX9.2/9_2.sce b/905/CH9/EX9.2/9_2.sce new file mode 100755 index 000000000..3d5c327c9 --- /dev/null +++ b/905/CH9/EX9.2/9_2.sce @@ -0,0 +1,95 @@ +clear;
+clc;
+
+// Illustration 9.2
+// Page: 511
+
+printf('Illustration 9.2 - Page: 511\n\n');
+
+// solution
+//*****Data*****//
+// A-oxygen B-nitrogen
+t = 0.2*10^-6; // [m]
+qA = 3.97*10^-13; // [mole/m.s.kPa]
+qB = 0.76*10^-13; // [mole/m.s.kPa]
+v = 1; // [Air flow rate at STP, cubic m/s]
+Pp = 0.1*10^6; // [Pa]
+R = 8.314 // [cubic m.Pa/mole.K]
+T = 298; // [K]
+Pf = 1*10^6; // [Pa]
+//*****//
+// Using equation 9.14
+alphaA = qA/qB;
+QA = qA/t; // [mole/square m.s.kPa]
+// molar flow rate
+nf = v*1000/(22.4); // [mole/s]
+r = Pp/Pf; // [pressure ratio]
+QB = qB/t; // [mole/square m.s.kPa]
+alphaid = QA/QB;
+xFa = 0.21;
+xFb = 0.79;
+
+// For Q = 0.1
+Q1 = 0.1
+ // Solution of simultaneous equation
+function[f]=F(e)
+ f(1) = e(1) - (e(3)*(1-e(2)))/((e(2)*(1-e(3))));
+ f(2) = e(2) - (xFa - (e(3)*Q1))/(1-Q1);
+ f(3) = e(1) - (alphaid*(e(2)*(e(1)-1)+1- (r*e(1))))/(e(2)*(e(1)-1)+1 - r);
+ funcprot(0);
+endfunction
+// Initial guess
+e = [4 0.13 0.4];
+y = fsolve(e,F);
+alpha1 = y(1);
+Xa1 = y(2);
+Ya1 = y(3);
+Am1 = Ya1*Q1*nf/(QA*(Xa1*Pf-Ya1*Pp))*1000; // [square m]
+
+// For Q = 0.2
+Q2 = 0.2
+ // Solution of simultaneous equation
+function[f]=F(e)
+ f(1) = e(1) - (e(3)*(1-e(2)))/((e(2)*(1-e(3))));
+ f(2) = e(2) - (xFa - (e(3)*Q2))/(1-Q2);
+ f(3) = e(1) - (alphaid*(e(2)*(e(1)-1)+1- (r*e(1))))/(e(2)*(e(1)-1)+1 - r);
+ funcprot(0);
+endfunction
+// Initial guess
+e = [4 0.13 0.4];
+y = fsolve(e,F);
+alpha2 = y(1);
+Xa2 = y(2);
+Ya2 = y(3);
+Am2 = Ya2*Q2*nf/(QA*(Xa2*Pf-Ya2*Pp))*1000; // [square m]
+
+// For Q = 0.9
+Q9 = 0.9
+ // Solution of simultaneous equation
+function[f]=F(e)
+ f(1) = e(1) - (e(3)*(1-e(2)))/((e(2)*(1-e(3))));
+ f(2) = e(2) - (xFa - (e(3)*Q9))/(1-Q9);
+ f(3) = e(1) - (alphaid*(e(2)*(e(1)-1)+1- (r*e(1))))/(e(2)*(e(1)-1)+1 - r);
+ funcprot(0);
+endfunction
+// Initial guess
+e = [4 0.13 0.4];
+y = fsolve(e,F);
+alpha9 = y(1);
+Xa9 = y(2);
+Ya9 = y(3);
+Am9 = Ya2*Q9*nf/(QA*(Xa9*Pf-Ya9*Pp))*1000; // [square m]
+
+// Similarly for Q =0.3......0.9, Xa, Ya, alpha and Am are calculated
+// Therefore we obtained
+// Solution = [Q,alpha,Xa,Ya]
+Solution = zeros(9,4);
+Solution = [0.1 4.112 0.181 0.475;0.2 4.062 0.156 0.428;0.3 4.018 0.135 0.385;0.4 3.98 0.118 0.348;0.5 3.949 0.105 0.315;0.6 3.922 0.093 0.288;0.7 3.9 0.084 0.264;0.8 3.881 0.077 0.243;0.9 3.864 0.07 0.226];
+Am = [8037;17074;26963;37531;48618;60099;71876;83879;96056;];
+disp(Solution);
+disp(Am);
+
+printf("The maximum oxygen content of the permeate (%f percent) occurs with the smallest cut (Q = 0.1).\n\n",Ya1*100);
+printf("The maximum nitrogen content of the retentate (%f percent) occurs at the largest cut (Q = 0.9).\n\n",(1-Xa9)*100);
+
+printf('The membrane area requirements are very large (e.g, Am = 60,100 square m for Q = 0.6) even though the volumetric flow rate of air is relatively small)');
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