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Diffstat (limited to '572/CH12/EX12.1/c12_1.sce')
| -rwxr-xr-x | 572/CH12/EX12.1/c12_1.sce | 29 |
1 files changed, 29 insertions, 0 deletions
diff --git a/572/CH12/EX12.1/c12_1.sce b/572/CH12/EX12.1/c12_1.sce new file mode 100755 index 000000000..c07e95203 --- /dev/null +++ b/572/CH12/EX12.1/c12_1.sce @@ -0,0 +1,29 @@ +//(12.1) The molar analysis of the gaseous products of combustion of a certain hydrocarbon fuel is CO2, 0.08; H2O, 0.11; O2, 0.07; N2, 0.74. (a) Determine the apparent molecular weight of the mixture. (b) Determine the composition in terms of mass fractions (gravimetric analysis).
+
+//solution
+
+//variable initialization
+n1 = .08 // mole fraction of CO2
+n2 = .11 // mole fraction of H2O
+n3 = .07 //mole fraction of O2
+n4 = .74 //mole fraction of N2
+
+//part(a)
+M1 = 44 //molar mass of CO2 in kg/kmol
+M2 = 18 //molar mass of H2O in kg/kmol
+M3 = 32 //molar mass of O2 in kg/kmol
+M4 = 28 //molar mass of N2 in kg/kmol
+
+M = M1*n1 + M2*n2 + M3*n3 + M4*n4 //in kg/kmol
+printf('the apparent molecular weight of the mixture in kg/kmol is: %f',M)
+
+//part(b)
+mf1 = (M1*n1/M)*100 //mass fraction of CO2 in percentage
+mf2 = (M2*n2/M)*100 //mass fraction of H2O in percentage
+mf3 = (M3*n3/M)*100 //mass fraction of O2 in percentage
+mf4 = (M4*n4/M)*100 //mass fraction of N2 in percentage
+
+printf('\n\nthe mass fraction of CO2 in percentage is: %f',mf1)
+printf('\nthe mass fraction of H2O in percentage is: %f',mf2)
+printf('\nthe mass fraction of O2 in percentage is: %f',mf3)
+printf('\nthe mass fraction of N2 in percentage is: %f',mf4)
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