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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+//clc()
+//CO2(g) + C(s) = 2CO(g) H1298 = 170kJ/mol
+//O2(g) + 2C(s) = 2CO(g) H2298 = -221.2kJ/mol
+T2 = 1298;//K
+T1 = 298;//K
+Hc = 0.02;//kJ/molK
+Ho = 0.03;//kJ/molK
+Hco = 0.03;//kJ/molK
+Hco2 = 0.05;//kJ/molK
+//let the flue gas contain x mol CO2 per mole of oxygen, product contains 2(1+x)mol CO. Nitrogen in reactant and product remain the same
+//enthalpy of cooling xmol CO2, 1 mol O2 and 2 + xmol carbon from 1298 to 298K is given as, H1 = (Hco2 * x + Ho * 1 + Hc * (2 + x)) * (298 - 1298)
+//H1 = (-70x - 70)kJ
+//enthalpy of heating the product, H2 = 2 * ( 1 + x )* Hco * (1298 - 298)
+//H2 = 60 + 60x kJ
+//Hr = 170x - 221.2
+//Htotal = 0 = H1 + H2 + Hr
+x = (221.2 + 70 - 60)/(170 + 60 - 70);
+disp("mol",x,"moles of CO2 present per mol of oxygen in feed stream = ") \ No newline at end of file