Calculate the Flame temperature

[GeorgeP1]

Homework Statement

A fuel gas consists of 75% butane (C4H10), 10% propane (C3H8) and 15% butene (C4H8) by volume.
It is to be fed to the combustion chamber in 10% excess air at 25°C, where it is completely burnt to carbon dioxide and water. The flue gases produced are to be used to generate 5 bar steam from water at 90°C.

Determine the maximum flame temperature ?.

Homework Equations

[/B]
heat input to process = heat out of process
Heat out of process = ∑ (mass × mean specific heat capacity × TF ) for each product of combustion

The Attempt at a Solution

[/B]
2581 Kj

1987°C

This answer is incorrect.
Any help is much appreciated.

 

[Bystander]

C'mon --- you know the "drill" --- sum the heat capacities, do the division, ...

 

[GeorgeP1]

butane = -2660 kj/mol
propane = -2046 kj/mol
butene = -2540 kj/mol
heat released by 1 mole of fuel
= (0.75x2660)+(0.1x2046)+(0.15x2540)
=2581 kj (heat in)

is this correct so far?

 

[Bystander]

What are the products, and what are their heat capacities?

 

[GeorgeP1]

Flue gas =

H2O = 4.75 moles
CO2 = 3.9 moles
O2 = 0.63 moles
N2 = 26.06 moles
Total = 35.34 moles

assuming a flame temperature of 2000°C

Heat content = enthalpy x no. of molesFor N2, 66.10 x 26.06 = 1722.6 kJ

CO2, 108.32 x 3.9 = 422.5 kJ

O2, 69.65 x 0.63 = 43.9 kJ

H2O, 86.24 x 4.75 = 409.6 kJ

Total = 2598.6 kJ

 

[GeorgeP1]

how does this look?

 

[marognef]

how does this look?

George - looks good to me, did anyone agree?