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Diffstat (limited to '572/CH13/EX13.14/c13_14.sce')
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diff --git a/572/CH13/EX13.14/c13_14.sce b/572/CH13/EX13.14/c13_14.sce new file mode 100755 index 000000000..b440a84b0 --- /dev/null +++ b/572/CH13/EX13.14/c13_14.sce @@ -0,0 +1,43 @@ +//(13.14) Methane gas enters a reactor and burns completely with 140% theoretical air. Combustion products exit as a mixture at temperature T and a pressure of 1 atm. For T = 480 and 1560 K, evaluate the flow exergy of the combustion products, in kJ per kmol of fuel. Perform calculations relative to an environment consisting of an ideal gas mixture at 25C, 1 atm with the molar analysis, yeN2 = 0.7567, yeO2 = 0.2035, yeH2O = 0.0303, yeCO2 = 0.0003.
+
+
+
+//solution
+
+//For 140% theoretical air, the reaction equation for complete combustion of methane is
+//CH4 + 2.8(O2 + 3.76N2) -------> CO2 + 2H2O + 10.53N2 + .8O2
+
+//for product side
+yCO2p = 1/(1+2+10.53+.8)
+yH2Op = 2/(1+2+10.53+.8)
+yN2p = 10.53/(1+2+10.53+.8)
+yO2p = .8/(1+2+10.53+.8)
+
+Rbar = 8.314 //universal gas constant in SI units
+Tnot = 298.15 //in kelvin
+
+yeN2 = .7567
+yeO2 = .2035
+yeH2O = .0303
+yeCO2 = .0003
+
+ebarch = Rbar*Tnot*(log(yCO2p/yeCO2) + 2*log(yH2Op/yeH2O) + 10.53*log(yN2p/yeN2) + .8*log(yO2p/yeO2))
+
+//with data from tables A-23 at 480 and 1560 kelvin,the thermomechanical contribution to the flow exergy, per mole of fuel, is
+contri480 = 17712 //kJ per kmol of fuel
+contri1560 = 390853 //kJ per kmol of fuel
+
+efbar480 = contri480 + ebarch //kJ per kmol of fuel
+efbar1560 = contri1560 + ebarch //kJ per kmol of fuel
+
+printf('at T= 480k, the flow exergy of the combustion products, in kJ per kmol of fuel is: %f',efbar480)
+printf('\nat T = 1560K, the flow exergy of the combustion products, in kJ per kmol of fuel is: %f',efbar1560)
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