Question

Determine the adiabatic flame temperature of propane burned in 100% excess air. Use a heat of formation of propane of -45000 Btu/lb-mole.

Answer 1

Solution:

Propane : C3H8

For stochiometric condition, number of moles of air per mole of propane = a = (3 + 8/4) = 5

Since 100% air is present, no. of mole of air per mole of propane = 10

Chemical reaction:

C3H8 + 10 (O2 + 3.76 N2) --- > 3 CO2 + 4 H2O + 37.6 N2 + 2.5 O2 ------ > (1)

Reactant are at standard state, i.e at T = 298 K

Product are a adiabatic temperature = Tad (unknown)

Heat transfer = 0

So, HR = HP

hC3H8 + 10* hO2 + 37.6 hN2 = 3*[ hCO2 + cp,CO2* (Tad - 298) ] + 4*[hH2O + cp,H2O*(Tad - 298)] + 37.6*[hN2 + cp,N2*(Tad - 2098)] + 2.5*[hO2 + cp,O2*(Tad - 298)] -------------- > (2)

Here, h is enthalpy of formation.

h is zero for N2, O2

Cp is specific at average temperature = (Tad + 298)/2 ~ 1200 K (assuming Tad ~ 2100 K)

Value are taken from standard tables,

hCO2 = -393,546 kJ/kmol ; cp,CO2 = 56.205 kJ/kmol-k

hH2O = -241,845 kJ/kmol ; cp,H2O = 43.874 kJ/kmol-K

cp,N2 = 13.707 kJ/kmol-k ; cp,O2 = 35.593 kJ/kmol-K

given hC3H8 = - 45000 Btu/lb-mol = -45000 * 2.326 = -104670 kJ/kmol

Subtituting these values in equation (1), we get

-104670 = 3*[-393546 + 56.205*(Tad - 298)] + 4*[-241845 + 43.874*(Tad - 298)] +10*3.76*13.707*(Tad - 298) + 2.5*35.593*(Tad - 298)

2043348 = 948.47*(Tad - 298)

Tad = 2452.3 K (Ans)