Answer:see explanation.
Explanation:
(a).The equation of reaction is;
C2H4 + 3(O2 + 3.76N2) -----> 2CO2 + 2H2O + 11.28 N2.
Theoretical air is the amount of air that will allow the complete combustion of the fuel. Air contains 21% of oxygen,79% of Nitrogen. Thus, the molecular mass of air = 29 [Kg/Kmol] . The number of oxygen needed to oxidize the hydrocarbon(C2H4) is 79/21= 3.76 moles of Nitrogen.
(b). N(total) =2+2+11.28 =15.28 kmoles of product
15.28[h(T) - h°] of air
= 15.28(Mass of air) (Cp,1000k)
T(adiabatic) = 802,310 kJ/k. Mole.
When mass of air= 29 Kg/ k.mol
Cp,1000k = 1.142 kJ.k which is the specific heat capacity of air.
T(adiabatic) = 1275k -----> assuming all the products are air.
(C). 2{∆h} co2 + 2{∆h} H20 +11.28 {∆h} N2
After series of calculations and checking of tables:
S(o2) = 320.173 - 8.314 ln 0.12= 337.46 kJ/ K.mol.k
S(H2O)= 273.986- 8.314 ln 0.1406 = 290.30 KJ/ kmol. K
S(N2) = 258.503- 8.314 ln 0.7344= 261.07 kJ/kmol.k
= 2(337.46) + 2(290.30)+11.28(261.07)-360.79-3(218.01)-11.28(193.46)
=1003.3378 kJ/kmol. K
