Adiabatic flame temperature

Adiabatic flame temperature

1. Estimate the constant-pressure adiabatic flame temperature for the combustion of a stoichiometric C3H8–air mixture. The pressure is 1 bar and the initial reactant temperature is 298 K. Use the following assumptions:

• “Complete combustion”, i.e., the product mixture consists only of CO2, H2O, and N2.

• The enthalpy of the product mixture is estimated using constant specific heats evaluated at 1200 K  (Ti + Tad)/2, where Tad is guessed to be about 2100 K.

2. Under the same assumptions, but obviously considering excess O2 in the products, estimate the adiabatic flame temperature for a C3H8–air mixture with equivalence ratio = 0.8.

--- Answers

1.

The mixture composition follows from balancing the chemical equation for a stoichiometric mixture of propane and air:

( )

3 8 2 2 2 2 2

C H +5 O +3.76N →3CO +4H O+18.58N where in the product mixture we find

2 2 2

CO 3, H O=4, N =18.58

N = N N

Properties:

Species Enthalpy of formation at 298 K

0 ,

h (kJ/kmol) f i

Specific Heat at 1200 K

,

cp i (kJ/kmol-K)

C3H8 –103847 (Table B.1, Turns’ textbook) –

CO2 –393546 (Table A.2) 56.21 (Table A.2)

H2O –241845 (Table A.6) 43.87 (Table A.6)

N2 0 33.71 (Table A.7)

O2 0 –

By applying the first law of thermodynamics:

react prod

react prod

= =

i i i i

H =





( )( ) ( ) ( )

3 8

react

C H

1 103847 5 0 18.58 0 103847 kJ/mol

H = − + +

= −

( )

( ) ( )

( ) ( )

( ) ( )

0

prod , ,

prod

C H

298.15 3 393546 56.21 298.15 4 241845 43.87 298.15

18.58 0 33.71 298.15 kJ/mol .

i f i p i ad

ad

ad

ad

H N h c T

T T T

 

=  + − 

 

= − + − 

 

+ − + − 

 

+  + − 



Equating (first law) H_reactto H_prodand solving for T yields _ad

T = 2404 K for ad  =1 (stoichiometric conditions)

2.

In this case, the mixture composition follows from balancing the chemical equation for a mixture of propane and air, with excess air. Suppose that we provide 4.76a moles of air, with a>5 (5 is the value for a stoichiometric mixture). We can write the species balance as:

Another way is:

0

react ,

react react

(in our case)

i i i f i

H =





( )

0

prod , ,

prod prod

i f i ad 298 i p i

H =





If we define, on a one-mole-of-fuel basis, the enthalpy of reaction H_R⁰ and the average heat capacity C_p of the product mixture, respectively as:

0 0 0

, ,

prod react

R i f i i f i

H N h N h

 =





prod

p i p i

C =



0

298.15 ^R

ad

p

T H

C

= +− .

Thus,

( ) ( ) ( ) _{3 8}

0

3 393546 4 241845 103847 2044171 kJ/molC H

HR

 =  − +  − − − = −

3 8

3 56.21 4 43.87 18.58 33.71 970.4 kJ/molC H K

Cp =  +  +  =

then

( )

2044171 K 298.15

9 . 2404

ad 70 4

T − −

= + = .

( )

3 8 2 2 2 2 2 2

C H +a O +3.76N →3CO +4H O+3.76 Na +bO

where a and b must be computed from the knowledge of the equivalence ratio. From the definition of the equivalence ratio  we can compute the value of a:

( )

( ) ( )

( )

stoic

stoic

A F F A

A F F A

 = =

where

( )

1

air stoic air

stoic

fuel stoic fuel

m a MW

A F m MW

   

=  =  ;

( )

1

air air

fuel fuel

m a MW

A F m MW

   

= =  

hence

0.8 5 6.25

0.8 0.8

stoic stoic

a a

a a

=  = = =

From the balance of oxygen atoms (or by simply observing that b= −a a_stoic) we find

2 3 2 4 1 2

a =  +  +  b that is

5 1.25 b= − =a . We can now rewrite the species balance for  =0.8:

( )

3 8 2 2 2 2 2 2

C H +6.25 O +3.76N →3CO +4H O 1.25O+ +23.5N where in the products we find

2 2 2 2

CO 3, H O=4, N =23.5, O =1.25

N = N N N

In addition to the property values used in b.1, we need the Specific Heat at 1200 K for O2 : Species Enthalpy of formation at 298.15 K

0 ,

h (kJ/kmol) f i

Specific Heat at 1200 K

,

cp i (kJ/kmol-K)

O2 0 35.593 (Table A.11)

By applying the first law: _{react prod}

react prod

= =

i i i i

H =





( )(

)

( )

( )

103847 kJ

H = − + +

= −

( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

0

prod , ,

prod

298.15

3 393546 56.21 298.15 4 241845 43.87 298.15 23.5 0 33.71 298.15 1.25 0 35.59 298.15

3 393546 4 241845

3 56.21 4 43.87 23.5 33.71 1.25 3

i f i p i ad

ad

ad

ad

ad

H N h c T

T T T T

 

=  + − 

 

= − + − 

 

+ − + − 

 

+  + − 

 

+  + − 

= −  − 

−  +  +  + 



( )

( )

5.59 298.15 3 56.21 4 43.87 23.5 33.71 1.25 35.59 T_ad



+  +  +  + 

that is:

Equating H_reactto H_prodand solving for T yields _ad

( )

( )

103847 kJ 3 393546 4 241845

3 56.21 4 43.87 23.5 33.71 1.25 35.59 298.15 3 56.21 4 43.87 23.5 33.71 1.25 35.59 T_ad

− = −  − 

−  +  +  +  

+  +  +  + 

103847 3 393546 4 241845 298.15

3 56.21 4 43.87 23.5 33.71 1.25 35.59

2044171

298.15 2029

1180.78 Tad

K

− +  + 

= +

 +  +  + 

= + =

T = 2029 K for ad  =0.8