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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 /647/CH12/EX12.18/Example12_18.sce | |
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Diffstat (limited to '647/CH12/EX12.18/Example12_18.sce')
| -rwxr-xr-x | 647/CH12/EX12.18/Example12_18.sce | 42 |
1 files changed, 42 insertions, 0 deletions
diff --git a/647/CH12/EX12.18/Example12_18.sce b/647/CH12/EX12.18/Example12_18.sce new file mode 100755 index 000000000..4c1dea6fb --- /dev/null +++ b/647/CH12/EX12.18/Example12_18.sce @@ -0,0 +1,42 @@ +clear;
+clc;
+
+// Example: 12.18
+// Page: 502
+
+printf("Example: 12.18 - Page: 502\n\n");
+
+// Solution
+
+//*****Data******//
+T = 875;// [K]
+K = 0.514;
+P = 1;// [bar]
+//*************//
+
+// Reaction: Fe(s) + H2O(g) --------->FeO(s) + H2(g)
+// K = a_FeO*a_H2/(a_Fe*a_H2O)
+// Since the activities of the solid components Fe & FeO at equilibrium may be taken as unity.
+// a_Fe = a_FeO = 1
+// Ka = a_H2/a_H2O;
+// Feed:
+nH2O_feed = 1;
+nH2feed = 0;
+// Let e be the degree of completion at equilibrium.
+// Moles at Equilibrium:
+// nH2O_eqb = 1 - e;
+// nH2_eqb = e;
+// Total moles at equilibrium = 1 - e + e = 1
+// Mole Fractions at Equilibrium:
+// yH20_eqb = 1 - e;
+// yH2_eqb = e;
+// Sum of stoichometric coeffecient:
+v = 1 - 1;
+Ky = K*P^(-v);
+// Ky = e/(1 - e)
+e = Ky/(Ky + 1);
+yH2_eqb = e;
+yH2O_eqb = 1 - e;
+printf("Equilibrium Composition\n");
+printf("yH2 = %.2f\n",yH2_eqb);
+printf("y_H2O = %.2f\n",yH2O_eqb);
\ No newline at end of file |