Question

For a constant-volume stoichiometric propane-air mixture initially at 298K, determine the adiabatic flame temperature and final pressure assuming constant specific heats evaluated at 298K.

Answer 1

Given : constant-volume stoichiometric propane-air mixture

Temprature = 298 K

The comubstion equation will be

C3H8 + 5O2   + 18.8N2   ---> 3CO2 + 4H2O   + 18.8N2

Specific heats are

C3H8 = 50.35 kJ/kmolK

CO2 = 55.396 KJ / Kmole

H2O = 42.44 kJ/kmolK

N2 = 33.0 kJ/kmolK

O2 = -

Enthalpy of formation

C3H8 = −104.7 kJ/mol

CO2 = -394.0 KJ / mole

H2O = -244.5 KJ / mole

N2 = O2 = 0

Q = 0 (for constant volume)

Delta H reactants - Delta H products - V(Pinit - Pfinal) = 0

Pinit X V = Sum of (moles of reactants X Temperature X gas constant)

Pfinal X V = sum fo (moles of products X temeperature x gas constant)

Also we know that

molecular weight of reactants = Mass of mixture / Moles of reactant

molecular weight of products = Mass of mixture / Moles of products

So Delta H reactants - Delta H products - R ( Tinitial / Mw of reactants - Tadiabatic / Mol wt of products) = 0

Delta H reactants = -104700 +2(0) + 8.8(0) = -104700 J / mol or KJ / KMole

Delta H products = [-394000 + 55.396(Tadiab - 298) ] X 3 + [-244500 + 42.44(Tadiab - 298) ] X 4 + [0 + 33(Tadiab - 298) ]X 18.8

Delta H products = -1182000 - 978000 + 368.94 (Tadiab - 298) =

At constant pressure

Delta H reactants = -104700 J / mol or KJ / KMole = Delta H products = -1182000 - 978000 + 368.94 (Tadiab - 298)

On solving

Tadiabatic =5868.82 K

At constant volume

The delta H products = -1182000 - 978000 + 368.94 (Tadiab - 298) + R (moles of reactants Ti - moles of products X Tadiabtic )

R = 8.314

number of moles of products = 7 +18.8 = 25.8

Moles of reactsnts = 6 + 18.8 = 24.8

Delat H products = -1182000 - 978000 + 368.94 (Tadiab - 298) + 8.314 (24.8 X 298 - 25.8 X Tadiabatic)

Delta H products = -2160000 + 368.94 X Tadiab - 109944 + 61443.7 - 214.5 Tadiabatic

= Delta H reactants = -104700 J / mol

This shows

(368.94 - 214.5 ) Tadiabatic = 2004800

Tadiabatic = 12981 K

Also

P1T1 / n1 = P2T2 / n2

Pfinal = 1 X 298 X 25.8 / 12981 X 24.8 = 0.0926 atm