Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

Chapter 15 | 761

The combustion equation for 1 kmol of fuel is obtained by dividing the
above equation by 1.36,

(a) The air–fuel ratio is determined by taking the ratio of the mass of the air
to the mass of the fuel (Eq. 15–3),

(b) To find the percentage of theoretical air used, we need to know the theo-
retical amount of air, which is determined from the theoretical combustion
equation of the fuel,

O 2 :

Then,

That is, 31 percent excess air was used during this combustion process.
Notice that some carbon formed carbon monoxide even though there was
considerably more oxygen than needed for complete combustion.

(c) For each kmol of fuel burned, 7.37 0.65 4.13 61.38  9 
82.53 kmol of products are formed, including 9 kmol of H 2 O. Assuming that
the dew-point temperature of the products is above 25°C, some of the water
vapor will condense as the products are cooled to 25°C. If Nwkmol of H 2 O
condenses, there will be (9
Nw) kmol of water vapor left in the products.
The mole number of the products in the gas phase will also decrease to
82.53
Nwas a result. By treating the product gases (including the remain-
ing water vapor) as ideal gases, Nwis determined by equating the mole frac-
tion of the water vapor to its pressure fraction,

Therefore, the majority of the water vapor in the products (73 percent of it)
condenses as the product gases are cooled to 25°C.

Nw6.59 kmol

9 
Nw

82.53
Nw



3.1698 kPa

100 kPa

Nv

Nprod,gas



Pv

Pprod

131%



1 16.32 21 4.76 2 kmol

1 12.50 21 4.76 2 kmol

Percentage of theoretical air

mair,act

mair,th



Nair,act

Nair,th

ath 8 4.5Sath12.5

C 8 H 18 ath 1 O 2 3.76N 22 S8CO 2 9H 2 O3.76athN 2

19.76 kg air/kg fuel

AF

mair

mfuel



1 16.324.76 kmol 21 29 kg>kmol 2

1 8 kmol 21 12 kg>kmol 2  1 9 kmol 21 2 kg>kmol 2

7.37CO 2 0.65CO4.13O 2 61.38N 2 9H 2 O

C 8 H 18 16.32 1 O 2 3.76N 22 S