TITLE.PM5

FUELS AND COMBUSTION 519

dharm

\M-therm\Th11-2.pm5

For 200% theoretical air the combustion equation would be

C 8 H 18 + (2) (12.5) O 2 + (2) (12.5)

79

21

F

HG

I

KJ N^2

→ 8CO 2 + 9H 2 O + (1) (12.5) O 2 + (2) (12.5)

79

21

F

HG

I

KJ N^2

Mass of fuel = (1) (8 × 12 + 1 × 18) = 114 kg/mole

Mass of air = (2) (12.5) 1 79

21

F +

HG

I

KJ

28.97 = 3448.8 kg/mole of fuel

(i)Air-fuel ratio :

Air-fuel ratio, A/F = Mass of air

Mass of fuel

= 3448 8

114

. = 30.25

i.e., A/F = 30.25. (Ans.)

(ii)Dew point of the products, tdp :

Total number of moles of products

= 8 + 9 + 12.5 + (2) (12.5)^79

21

F

HG

I

KJ

= 123.5 moles/mole fuel

Mole fraction of H 2 O =
9
123 5.
= 0.0728
Partial pressure of H 2 O = 100 × 0.0728 = 7.28 kPa
The saturation temperature corresponding to this pressure is 39.7°C which is also the dew-
point temperature.
Hence tdp = 39.7°C. (Ans.)
Note. The water condensed from the products of combustion usually contains some dissolved gases and
therefore may be quite corrosive. For this reason the products of combustion are often kept above the dew point
until discharged to the atmosphere.
Example 11.17. One kg of ethane (C 2 H 6 ) is burned with 90% of theoretical air. Assuming
complete combustion of hydrogen in the fuel determine the volumetric analysis of the dry prod-
ucts of combustion.
Solution. The complete combustion equation for C 2 H 6 is written as :
C 2 H 6 + 3.5O 2 → 2CO 2 + 3H 2 O
The combustion equation for C 2 H 6 for 90% theoretical air is written as :

C 2 H 6 + (0.9) (3.5) O 2 + (0.9) (3.5)

79

21

F

HG

I

KJ N^2 →^ a CO^2 + b CO + 3H^2 O + (0.9) (3.5)

79

21

F

HG

I

KJ N^2

By balancing carbon atoms on both the sides, we get

2 = a + b ...(i)

By balancing oxygen atoms on both the sides, we get

(0.9) (3.5) (2) = 2a + b + 3 ...(ii)

Substituting the value of b ( = 2 – a) from eqn. (i) in eqn. (ii), we get

(0.9) (3.5) (2) = 2a + 2 – a + 3

6.3 = a + 5

∴ a = 1.3