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author | priyanka | 2015-06-24 15:03:17 +0530 |
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committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /572/CH7/EX7.1 | |
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initial commit / add all books
Diffstat (limited to '572/CH7/EX7.1')
-rwxr-xr-x | 572/CH7/EX7.1/c7_1.sce | 21 |
1 files changed, 21 insertions, 0 deletions
diff --git a/572/CH7/EX7.1/c7_1.sce b/572/CH7/EX7.1/c7_1.sce new file mode 100755 index 000000000..23015cfec --- /dev/null +++ b/572/CH7/EX7.1/c7_1.sce @@ -0,0 +1,21 @@ +// (7.1) A cylinder of an internal combustion engine contains 2450 cm3 of gaseous combustion products at a pressure of 7 bar and a temperature of 867C just before the exhaust valve opens. Determine the specific exergy of the gas, in kJ/kg. Ignore the effects of motion and gravity, and model the combustion products as air as an ideal gas. Take T0 = 300 K (27C) and p0= 1.013 bar.
+
+
+//solution
+
+
+//variable initialization
+v = 2450 //volume of gaseous products in cm^3
+P = 7 //pressure of gaseous product in bar
+T = 867 //temperature of gaseous product in degree celcius
+T0 = 300 //in kelvin
+P0 = 1.013 //in bar
+
+//from table A-22
+u = 880.35 //in kj/kg
+u0 = 214.07 //in kj/kg
+s0(T) = 3.11883 //in kj/kg.k
+s0(T0) = 1.70203 //in kj/kg.k
+
+e = (u-u0) + (P0*(8.314/28.97)*[((T+273)/P)-(T0/P0)]) - T0*[s0(T)-s0(T0)-(8.314/28.97)*log(P/P0)] //in kj/kg
+printf('the specific exergy of the gas, in kJ/kg is \n\t e = %f',e)
\ No newline at end of file |