TITLE.PM5

FUELS AND COMBUSTION 517

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\M-therm\Th11-2.pm5

Hence CO 2 produced by fuel =

88

46

× 100 = 1.913 or 191.3%. (Ans.)

H 2 O produced by fuel =
54
46
× 100 = 1.174 or 117.4%. (Ans.)
Example 11.13. Calculate the amount of theoretical air required for the combustion of
1 kg of acetylene (C 2 H 2 ) to CO 2 and H 2 O.
Solution. For combustion of acetylene (C 2 H 2 ) the stoichiometric equation is written as
C 2 H 2 + x O 2 → a CO 2 + b H 2 O...(i)
Balancing the carbon atoms on both sides of the combustion eqn. (i), we get
2C = a C i.e., a = 2
Now balancing hydrogen atoms on both sides, we get
2H = 2bH
b = 1
Thus, eqn. (i) becomes
C 2 H 2 + x O 2 → 2CO 2 + H 2 O
Now, balancing oxygen atoms in the above equation
2 x = 2 × 2 + 1 = 5 i.e., x = 2.5
Hence, the final combustion eqn. (i) is
C 2 H 2 + 2.5O 2 → 2CO 2 + H 2 O ...(ii)
Thus, for combustion of C 2 H 2 in air, we get

C 2 H 2 + 2.5O 2 + 2.5^79

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N 2 → 2CO 2 + H 2 O + 2.5^79

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N 2

On a mass basis, this becomes

(2 × 12 + 2 × 1) C 2 H 2 + 2.5 (2 × 16) O 2 + 2.5^79

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(2 × 14) N 2

→ 2 (12 + 2 × 16) CO 2 + (1 × 2 + 1 × 16) H 2 O + 2.5^79

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(2 × 14) N 2

26 kg C 2 H 2 + 80 kg O 2 + 263.3 N 2 → 88 kg CO 2 + 18 kg H 2 O + 263.3 kg N 2
or 1 kg C 2 H 2 + 3.076 kg O 2 + 10.12 kg N 2 → 3.38 kg CO 2 + 0.69 kg H 2 O + 10.12 kg N 2
i.e., Amount of air = 3.076 + 10.12 = 13.196 kg of air per kg of C 2 H 2
Hence amount of theoretical air required for combustion of 1 kg acetylene
= 13.196 kg. (Ans.)
Example 11.14. Determine the gravimetric analysis of the products of complete combus-
tion of acetylene with 200 per cent stoichiometric air.
Solution. The stoichiometric air equation (Example 11.3) is written as :

C 2 H 2 + 2.5 O 2 + 2.5^79

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N 2 → 2CO 2 + H 2 O + 2.5^79

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KJ

N 2

If 200 per cent stoichiometric air is used, the combustion equation for C 2 H 2 becomes

C 2 H 2 + (2) (2.5) O 2 + (2) (2.5)

79

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KJ N^2 → 2CO^2 + H^2 O + (2) (2.5)

79

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KJ N^2 + (1) (2.5) O^2

(2 × 12 + 2 × 1) kg C 2 H 2 + (2) (2.5) (2 × 16) kg O 2 + (2) (2.5)

79

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KJ (2 × 14) kg N^2