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Diffstat (limited to '1808/CH2/EX2.10/Chapter2_Example10.sce')
| -rw-r--r-- | 1808/CH2/EX2.10/Chapter2_Example10.sce | 17 |
1 files changed, 17 insertions, 0 deletions
diff --git a/1808/CH2/EX2.10/Chapter2_Example10.sce b/1808/CH2/EX2.10/Chapter2_Example10.sce new file mode 100644 index 000000000..77b1fe9c0 --- /dev/null +++ b/1808/CH2/EX2.10/Chapter2_Example10.sce @@ -0,0 +1,17 @@ +clc
+clear
+//INPUT DATA
+//0.8062 CH4+0.0541 C2H6 +0.0187C3H8+0.0160C4H10+0.1050N2+a (O2+3.76 N2)=b (0.078 CO2+0.002 CO +0.07 O2 +0.85N2)+ c H2O ;//Combustion equation for 1 kmol of fuel mixture
+//b*(0.078+0.002)=0.8062+2*(0.0541)+3*(0.0160);//by carbon balance
+c=1.93;//Carbon balance
+a=2.892;//Oxygen balance
+
+//CALCULATIONS
+//(0.8062 CH4 + 0.0541 C2H6 + 0.0187 C3H8 + 0.0160 C4H10 + 0.1050 N2 )+ 2.892 (O2+3.76 N2) = 12.93 (0.078 CO2 )+0.002(0+0.07 O2 +0.85 N2)+1.93 H2O ;//Balanced chemical equation
+xm=a*4.76/1;//Air fuel ratio on molar basis
+//(0.8062 CH4 + 0.0541 C2H6 + 0.0187 C3H8 + 0.0160 C4H10 + 0.1050 N2 )+ 2.892 (O2+3.76 N2) = 1.0345 CO2+1.93 H2O+7.625 N2 ;//Balanced chemical equation
+xth=2*4.76;//Theoretical air fuel ratio
+nth=(xm/xth)*100;//Percentage of theoretical air
+
+//OUTPUT
+printf('(a)The air fuel ratio on molar basis %3.2f kmol of air/kmol of fuel \n (b)Percentage of theoretical air %3.1f percentage ',xm,nth)
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