TITLE.PM5

518 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th11-2.pm5

→ 2 (12 + 2 × 16) kg CO 2 + (2 × 1 + 16) kg H 2 O + (2) (2.5)

79

21

F

HG

I

KJ (2 × 14) kg N^2

+ (1) (2.5) (2 × 16) kg O 2

or 26 kg C 2 H 2 + 160 kg O 2 + 526.6 kg N 2 → 88 kg CO 2 + 18 kg H 2 O + 526.6 kg N 2 + 80 kg O 2

or 1 kg C 2 H 2 + 6.15 kg O 2 + 20.25 kg N 2 → 3.38 kg CO 2 + 0.69 kg H 2 O + 20.25 kg N 2 + 3.07 kg O 2

Thus for 1 kg of fuel, the products will consist of

CO 2 = 3.38 kg

H 2 O = 0.69 kg

O 2 = 3.07 kg

N 2 = 20.25 kg

Total mass of products = 27.39 kg

∴ Mass fractions are :

CO 2 =

338

27 39

.

.

= 0.123

H 2 O =^069

27 39

.

.

= 0.025

O 2 =

307

27 39

.

.

= 0.112

N 2 =

20 25

27 39

.
.
= 0.739.
Hence the gravimetric analysis of the complete combustion is :
CO 2 = 12.3%, H 2 O = 2.5%, O 2 = 11.2%, N 2 = 73.9%. (Ans.)
Example 11.15. Calculate the theoretical air-fuel ratio for the combustion of octane, C 8 H 18.
The combustion equation is :

C 8 H 18 + 12.5 O 2 + 12.5

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F

HG

I

KJ^ N^2 →^ 8CO^2 + 9H^2 O + 12.5^

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F

HG

I

KJ^ N^2.

Solution. The air-fuel ratio on a mole basis is

A/F =

12 5 12 5^79

21

1

..+ FHG IKJ
= 59.5 mol air/mol fuel
The theoretical air-fuel ratio on a mass basis is found by introducing the molecular weight
of the air and fuel

A/F =

59 5 28 97

812118

.(. )

()×+×

= 15.08 kg air/kg fuel. (Ans.)

Example 11.16. One kg of octane (C 8 H 18 ) is burned with 200% theoretical air. Assuming

complete combustion determine :

(i)Air-fuel ratio

(ii) Dew point of the products at a total pressure 100 kPa.

Solution. The equation for the combustion of C 8 H 18 with theoretical air is

C 8 H 18 + 12.5 O 2 + 12.5

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21

F

HG

I

KJ N^2 → 8CO^2 + 9H^2 O + 12.5

79

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F

HG

I

KJ N^2