diff options
Diffstat (limited to '647/CH12/EX12.12/Example12_12.sce')
-rwxr-xr-x | 647/CH12/EX12.12/Example12_12.sce | 39 |
1 files changed, 39 insertions, 0 deletions
diff --git a/647/CH12/EX12.12/Example12_12.sce b/647/CH12/EX12.12/Example12_12.sce new file mode 100755 index 000000000..d04ad310d --- /dev/null +++ b/647/CH12/EX12.12/Example12_12.sce @@ -0,0 +1,39 @@ +clear;
+clc;
+
+// Example: 12.12
+// Page: 490
+
+printf("Example: 12.12 - Page: 490\n\n");
+
+// Solution
+
+//*****Data******//
+// Reaction: SO2(g) + (1/2)O2 ------> SO3(g)
+P = 1;// [bar]
+T = 750;// [K]
+K = 74;
+//************//
+
+// Moles in Feed:
+nSO2 = 1;
+nO2 = 0.5;
+// Let e be the degree of completion at equilibrium.
+// Moles at Equilibrium:
+// nSO2_eqb = 10 - e;
+// nO2_eqb = 8 - 0.5*e;
+// nSO3_eqb = e;
+// Total no. of moles = 10 - e + 8 - 0.5*e + e = 18 -0.5*e;
+// Mole fraction at Equilibrium:
+// ySO2_eqb = (10 - e)/(18 - 0.5*e);
+// yO2_eqb = (8 - 0.5*e)/(18 - 0.5*e);
+// ySO3_eqb = e/(18 - 0.8*e);
+// Sum of stoichometric coeffecient:
+v = 1 - 1 -0.5;
+Ky = K*P^(-v);
+deff('[y] = f(e)','y = Ky - (e*(18 - (0.5*e)))/((10 - e)*(8 - 0.5*e))');
+e = fsolve(7,f);
+printf("Molar Composition of the gases\n");
+printf("nSO2 = %.2f\n",(10 - e));
+printf("nO2 = %.2f\n",(8 - 0.5*e));
+printf("nSO3 = %.2f\n",e);
\ No newline at end of file |