1 files changed, 41 insertions, 0 deletions
diff --git a/572/CH13/EX13.8/c13_8.sce b/572/CH13/EX13.8/c13_8.sce
new file mode 100755
index 000000000..065465dd1
--- /dev/null
+++ b/572/CH13/EX13.8/c13_8.sce
@@ -0,0 +1,41 @@
+//(13.8)  Liquid octane at 25C, 1 atm enters a well-insulated reactor and reacts with air entering at the same temperature and pressure. For steady-state operation and negligible effects of kinetic and potential energy, determine the temperature of the combustion products for complete combustion with (a) the theoretical amount of air, (b) 400% theoretical air.
+
+//solution
+
+//part(a)
+//For combustion of liquid octane with the theoretical amount of air, the chemical equation is
+//C8H18(l) + 12.5 O2 + 47N2   -------> 8 CO2 + 9 H2O(g) + 47N2
+//with enthalpy of formation data from Table A-25
+hfbarC8H18 = -249910                                                            //in kj/kmol
+hfbarCO2 = -393520
+hfbarH2O = -241820
+
+RHS = hfbarC8H18 -(8*hfbarCO2 + 9*hfbarH2O)                                     //in kj/kmol
+
+//at temperature 2400k
+LHS1 = 5089337                                                                   //in kj/kmol
+//at temperature 2350 k
+LHS2 = 4955163                                                                   //in kj/kmol
+//Interpolation between these temperatures gives
+Tp = 2400 + [(2400-2350)/(LHS1-LHS2)]*(RHS-LHS1)
+printf('the temperature in kelvin with theoretical amount of air is:  %f',Tp)
+
+//part(b)
+//For complete combustion of liquid octane with 400% theoretical air, the chemical equation is
+//C8H18(l) + 50O2 + 188N2  -------->  8CO2 + 9H2O + 37.5O2 + 188N2
+
+//proceeding iteratively as part(a)
+Tp = 962                                                                        //in kelvin
+printf('\n\nthe temperature in kelvin using 400 percent theoretical air is:  %f',Tp)
+
+
+
+
+
+
+
+
+
+
+
+