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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /647/CH4/EX4.8 | |
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Diffstat (limited to '647/CH4/EX4.8')
| -rwxr-xr-x | 647/CH4/EX4.8/Example4_8.sce | 34 |
1 files changed, 34 insertions, 0 deletions
diff --git a/647/CH4/EX4.8/Example4_8.sce b/647/CH4/EX4.8/Example4_8.sce new file mode 100755 index 000000000..1c8a60d43 --- /dev/null +++ b/647/CH4/EX4.8/Example4_8.sce @@ -0,0 +1,34 @@ +clear;
+clc;
+
+// Example: 4.8
+// Page: 126
+
+printf("Example: 4.8 - Page: 126\n\n");
+
+// Solution
+
+//*****Data*****//
+Cp_CO2 = 54.56;// [kJ/mol K]
+Cp_O2 = 35.20;// [kJ/mol K]
+Cp_steam = 43.38;// [kJ/mol K]
+Cp_N2 = 33.32;// [kJ/mol K]
+// 2C2H6(g) + 7O2(g) = 4CO2(g) + 6H2O(g)
+deltaH_273 = -1560000;// [kJ/kmol]
+//************//
+
+// Since the air is 25% in excess of the amount required,the combustion may be written as:
+// C2H6(g) + (7/2)O2(g) = 2CO2(g) + 3H2O(g)
+// 25% excess air is supplied.
+// Since the air contains N2 = 79% and O2 = 21%
+// C2H6(g) + 3.5O2(g) + 0.25*3.5O2(g) + (4.375*(79/21))N2 = 2CO2 + 3H2O + 0.875O2 + 16.46N2 .................. (A)
+// Considering the reaction (A),
+// Amount of O2:
+O2 = 3.5 + 3.5*0.25;// [mol]
+// Amount of N2 required:
+N2 = 4.375*(79/21);// [mol]
+// Let the initial temperature of ethane and air be 0 OC and the temperature of products of combustion be T OC
+// Since heat librated by combustion = heat accumulated by combustion products
+Q = -deltaH_273;// [kJ/mol K]
+T = Q/(2*Cp_CO2 + 3*Cp_steam + 0.875*Cp_O2 + N2*Cp_N2);// [OC]
+printf("The theoretical temperature of combustion is %d degree Celsius",T);
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