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+clear ;
+clc;
+// Example 11.5
+printf('Example 11.5\n\n');
+// Page no.324
+// Solution
+
+// Option 1
+F = 15 ;//[L/hr]
+cs_in = 10 ;//Nutrient conc. input vessel  - [g nutrient/L substrate]
+V1 = 100 ;// [L]
+V2 = 50 ;//[L]
+Yxs = 0.2 ;// [cells/g]
+umax = 0.4 ;//[hr^ - 1]
+Ks = 2 ;//[g/L] - Monod constant
+// Use eqn. 10.1 for balances and Monod eqns. applies to each vessel
+//Cells: 0 - F/V * x_out  + u * x_out - 0 = 0....(a)
+//Nutrient: F/V * cs_in - F/V * cs_out  + 0 - (u * x_out)/(Yxs) = 0.....(b)
+//From eqn.(a) F/V = u(dilution rate)...(c)
+// From eqn. (b)  x_out = Yxs(cs_in - cs_out)....(d)
+u1 = F/V1 ;//[hr^ - 1] //[hr^ - 1]
+cs_out = (Ks * u1/umax)/(1 - (u1/umax))  ;//Nutrient conc. output vessel  - [g nutrient/L substrate]
+// Find x_out by eqn. (d)
+x_out = Yxs * (cs_in - cs_out) ;//[g cells / L substrate]
+
+//Option 2
+//For vessel 1
+u2 = F/V2;
+cs_out1 = (Ks * u2/umax)/(1 - (u2/umax))  ;//Nutrient conc. output vessel  - [g nutrient/L substrate]
+x_out1 = Yxs * (cs_in - cs_out1) ;//[g cells / L substrate]
+// For vessel 2
+// Eqn. (a) is now F/V  * x_out1 - F/V * x_out2 + u3 * x_out2 = 0...(e)
+// Eqn. (b) is now F/V  * cs_out1 - F/V * cs_out2 + (u3 * x_out2)/Yxs = 0...(f)
+// u3 = (umax * cs_out2) / (Ks + cs_out2).. Monod Eqn...(g)
+// (e),(f) and (g) form a non - linear set of equations , solving them we get cs_out2 = 1.35 g nutrient/L substrate and x_out2  = 1.73 g cells/L
+x_out2 =  1.73 ;// From eqn. (e),(f) and (g) - [g cells / L substrate]
+
+printf('g cells/L from option 1 is %.2f.\n',x_out);
+printf(' g cells/L from option 2 is %.2f.\n',x_out2);
+printf(' By comparing option 1 and option 2 the respective answers are essentially the same.\n');
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