diff options
Diffstat (limited to '1808/CH2/EX2.5/Chapter2_Example5.sce')
-rw-r--r-- | 1808/CH2/EX2.5/Chapter2_Example5.sce | 21 |
1 files changed, 21 insertions, 0 deletions
diff --git a/1808/CH2/EX2.5/Chapter2_Example5.sce b/1808/CH2/EX2.5/Chapter2_Example5.sce new file mode 100644 index 000000000..9cdb45e4b --- /dev/null +++ b/1808/CH2/EX2.5/Chapter2_Example5.sce @@ -0,0 +1,21 @@ +clc
+clear
+//INPUT DATA
+//a.CH4 +b O2 + c N2=10 CO2 + 0.53 CO+2.37 O2 +d H2O +87.1 N2 // Stoichiometric equation for combustion of Methane
+c=87.1;//Nitrogen balance
+b=23.16;//(c/b)=3.76
+d=21.60;//d=2a Hydrogen balance
+a=10.54;//Carbon balance
+//10.54.CH4 +23.16 O2 + 87.1 N2=10 CO2 + 0.53 CO+2.37 O2 +21.06 H2O +87.1 N2 // Stoichiometric equation for combustion of Methane
+//CH4 +2.2 O2 + 8.27 N2=0.95 CO2 + 0.05 CO+0.225 O2 +2 H2O +8.27 N2 // Stoichiometric equation for combustion of Methane with 100 percent of air
+
+//CALCULATIONS
+N=(2.2+8.27)/1;//air fuel ratio on mole basis
+M=N*29/(12+4);//air fuel ratio on mass basis
+Nt=(2+7.52)*(29/16);//theoritical air fuel ratio
+nt=(M/Nt)*100;//Percentage theoritical air
+
+//OUTPUT
+printf('(a) CH4 +2.2 O2 + 8.27 N2=0.95 CO2 + 0.05 CO+0.225 O2 +2 H2O +8.27 N2 \n')
+printf('(b)The air fuel ratio is %3.2f kg of air/kg of fuel \n (c)Percentage theoritical air is %f ',M,nt)
+
|