TITLE.PM5

524 ENGINEERING THERMODYNAMICS

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\M-therm\Th11-2.pm5

Solution. The example is solved by a tabular method ; a specimen calculation is given below :
For CH 4 :
CH 4 + 2O 2 → CO 2 + 2H 2 O
i.e., 1 mole CH 4 + 2 moles O 2 → 1 mole CO 2 + 2 moles H 2 O
There are 0.2 moles of CH 4 per mole of the coal gas, hence
0.2 moles CH 4 + 0.2 × 2 moles O 2 → 0.2 moles CO 2 + 0.2 × 2 moles H 2 O
∴ O 2 required for the CH 4 in the coal gas = 0.4 moles per mole of coal gas.
The oxygen in the fuel (0.004 moles) is included in column 4 as a negative quantity.

Product Moles/mole Combustion equation O 2 moles/ Products

fuel mole fuel CO 2 H 2 O

12 3 456

H 2 O 0.504 2H 2 + O 2 → 2H 2 O 0.252 — 0.504

CO 0.17 2CO + O 2 → 2CO 2 0.085 0.17 —

CH 4 0.20 CH 4 + 2O 2 → CO 2 + 2H 2 O 0.400 0.20 0.40

C 4 H 8 0.02 C 4 H 8 + 6O 2 → 4CO 2 + 4H 2 O 0.120 0.08 0.08

O 2 0.004 — – 0.004 — —

N 2 0.062 — — — —

CO 2 0.04 — — 0.04 —

Total = 0.853 0.49 0.984

(i)Stoichiometric A/F ratio :

Air required =

0853

021

.
.
= 4.06 moles/mole of fuel
(where air is assumed to contain 21% O 2 by volume)
∴ Stoichiometric A/F ratio = 4.06/1 by volume. (Ans.)
(ii)Wet and dry analyses of the products of combustion if the actual mixture is
30% weak :
Actual A/F ratio with 30% weak mixture
= 4.06 +

30

100 × 4.06 = 1.3 × 4.06 = 5.278/1

Associated N 2 = 0.79 × 5.278 = 4.17 moles/mole fuel

Excess oxygen = 0.21 × 5.278 – 0.853 = 0.255 moles

Total moles of N 2 in products = 4.17 + 0.062 = 4.232 moles/mole fuel.

Analysis by volume of wet and dry products :

Product Moles/mole fuel % by vol. (dry) % by vol. (wet)

CO 2 0.490 9.97 8.31

H 2 O 0.984 — 16.68

O 2 0.255 5.19 4.32

N 2 4.170 84.84 70.69

Total wet = 5.899 100.00 100.00

• H 2 O = 0.984 (Ans.)
  Total dry = 4.915