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| author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
| commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
| tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3843/CH12/EX12.3 | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
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initial commit / add all books
Diffstat (limited to '3843/CH12/EX12.3')
| -rw-r--r-- | 3843/CH12/EX12.3/Ex12_3.sce | 20 |
1 files changed, 20 insertions, 0 deletions
diff --git a/3843/CH12/EX12.3/Ex12_3.sce b/3843/CH12/EX12.3/Ex12_3.sce new file mode 100644 index 000000000..46ce821ac --- /dev/null +++ b/3843/CH12/EX12.3/Ex12_3.sce @@ -0,0 +1,20 @@ +// Example 12_3
+clc;funcprot(0);
+// Given data
+// The volumetric analysis of the products on dry basis
+CO_2=11.0// %
+CO=1.0;// %
+O_2=3.5;// %
+N_2=84.5;// %
+
+// Calculation
+// The chemical equation is aC_4H_10+b(O_2+3.76N_2)-->11CO_2+1CO+3.5O_2+84.5N_2+cH_2O
+// Balancing each element,
+a=(11+1)/4;// (C)
+c=(10*a)/2;// (H)
+b=(22+1+7+c)/2;// (O)
+printf("\nDividing through the chemical equation by the value of a so that we hve 1 mol fuel is %1.0fC_4H_10+%1.1f(O_2+3.76N_2)-->%1.2fCO_2+%0.2fCO+%1.2fO_2+%2.2fN_2+%1.0fH_2O",a/a,b/a,11/a,1/a,3.5/a,84.5/a,c/a);
+// From example 12.1
+b_1=6.5;// The stoichiometric coefficient
+P_ta=((b/a)/(b_1))*100;// The percent theoretical air in %
+printf("\nThe percent theoretical air=%3.1f percentage",P_ta);
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