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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 /409/CH17/EX17.3/Example17_3.sce | |
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diff --git a/409/CH17/EX17.3/Example17_3.sce b/409/CH17/EX17.3/Example17_3.sce new file mode 100755 index 000000000..f522f54e4 --- /dev/null +++ b/409/CH17/EX17.3/Example17_3.sce @@ -0,0 +1,49 @@ +clear ;
+clc;
+// Example 17.3
+printf('Example 17.3\n');
+// Page no. 519
+// Solution Fig E17.3b
+
+// Given
+// coal analysis from handbook
+ex_air = .4 ;// Fraction of excess air required
+w_C = 12 ;// Mol. wt. of C-[g]
+mol_C = 71/w_C ;//[kg mol]
+w_H2 = 2.016 ;// Mol. wt. of H2 - [g]
+mol_H2 = 5.6/w_H2;
+air_O2 = 0.21;// Fraction of O2 in air
+air_N2 = 0.79;// Fraction of N2 in air
+
+// Natural Gas
+// Basis = 1 kg mol C
+// CH4 + 2O2 --> CO2 + 2H2O .... Eqn. (a)
+CO2_1 = 1 ;// By Eqn. (a) CO2 produced -[kg mol]
+H2O_1 = 2 ;// By Eqn. (a) H2O produced -[kg mol]
+Req_O2_1 = 2 ;// By Eqn. (a) -[kg mol]
+ex_O2_1 = Req_O2_1*ex_air ;// Excess O2 required -[kg mol]
+O2_1 = Req_O2_1 + ex_O2_1 ;// Total O2 required - [kg mol]
+N2_1 = O2_1*(air_N2/air_O2) ;//Total N2 required - [kg mol]
+Total_1 = CO2_1 + H2O_1 + N2_1 + ex_O2_1 ;// Total gas produced- [kg mol]
+
+// Coal
+// C + O2 --> CO2 ..eqn (b)
+// H2 + 1/2(O2) --> H2O.... eqn (c)
+CO2_2 = 1 ;// By Eqn. (a) CO2 produced -[kg mol]
+H2O_2 = mol_H2/mol_C ;// By Eqn. (a) H2O produced -[kg mol]
+Req_O2_2 = 1 + (mol_H2/mol_C)*(1/2) ;// By Eqn. (b) and (c) -[kg mol]
+ex_O2_2 = Req_O2_2*ex_air ;// Excess O2 required -[kg mol]
+O2_2 = Req_O2_2 + ex_O2_2; // Total O2 required - [kg mol]
+N2_2 = O2_2*(air_N2/air_O2); //Total N2 required - [kg mol]
+Total_2 = CO2_2 + H2O_2 + N2_2 + ex_O2_2 ;// Total gas produced- [kg mol]
+
+// Let P (total pressure) = 100 kPa
+P = 100 ;// Total pressure -[kPa]
+p1 = P*(H2O_1/Total_1) ;// Partial pressure of water vapour in natural gas - [kPa]
+Eq_T1 = 52.5 ;// Equivalent temperature -[degree C]
+p2 = P*(H2O_2/Total_2) ;// Partial pressure of water vapour in coal - [kPa]
+Eq_T2 = 35 ;// Equivalent temperature -[degree C]
+printf(' Natural gas Coal\n')
+printf(' ---------------------- --------------------\n')
+printf('Partial pressure: %.1f kPa %.1f kPa\n',p1,p2 ) ;
+printf('Equivalent temperature: %.1f C %.1f C\n',Eq_T1,Eq_T2 );
\ No newline at end of file |