1 files changed, 48 insertions, 0 deletions
diff --git a/572/CH13/EX13.1/c13_1.sce b/572/CH13/EX13.1/c13_1.sce
new file mode 100755
index 000000000..2c875ba9c
--- /dev/null
+++ b/572/CH13/EX13.1/c13_1.sce
@@ -0,0 +1,48 @@
+//(13.1)  Determine the air–fuel ratio on both a molar and mass basis for the complete combustion of octane, C8H18, with (a) the theoretical amount of air, (b) 150% theoretical air (50% excess air).
+
+//solution
+
+//part(a)
+//the combustion equation can be written in the form of 
+//C8H18 + a(O2 + 3.76N2) --> b CO2 + c H2O + d N2
+//using conservation of mass principle
+b = 8
+c = 18/2
+a = (2*b+c)/2
+d = 3.76*a
+
+//The air–fuel ratio on a molar basis is
+AFbar = a*(1+3.76)/1
+
+Ma = 28.97                                                                      //molar mass of air
+MC8H18 = 114.22                                                                 //molar mass of C8H18
+//The air–fuel ratio expressed on a mass basis is
+AF = AFbar*[Ma/MC8H18]
+
+printf('The air–fuel ratio on a molar basis is:  %f',AFbar)
+printf('\nThe air–fuel ratio expressed on a mass basis is:  %f',AF)
+
+//part(b)
+//For 150% theoretical air, the chemical equation for complete combustion takes the form
+//c8H18 + 1.5*12.5*(O2 + 3.76N2) --->  b CO2 + c H2O + d N2 + e O2
+//using conservation of mass
+b = 8
+c =18/2
+e = (1.5*12.5*2 - c -2*b)/2
+d = 1.5*12.5*3.76
+//The air–fuel ratio on a molar basis is
+AFbar = 1.5*12.5*(1+3.76)/1
+//The air–fuel ratio expressed on a mass basis is
+AF = AFbar*[Ma/MC8H18]
+printf('\n\nThe air–fuel ratio on a molar basis is:  %f',AFbar)
+printf('\nThe air–fuel ratio expressed on a mass basis is:  %f',AF)
+
+
+
+
+
+
+
+
+
+