TITLE.PM5

FUELS AND COMBUSTION 527

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\M-therm\Th11-2.pm5

H 2 O =

3

13 678.

× 100 = 21.93%. (Ans.)

N 2 =

8 678

13 678

.

.

× 100 = 63.45%. (Ans.)

The total dry moles = 0.614 + 1.386 + 8.678 = 10.678

Hence dry analysis is :

CO 2 =

0 614

10 678

.

.

× 100 = 5.75%. (Ans.)

CO = 1 386

10 678

.

.

× 100 = 12.98%. (Ans.)

N 2 = 8 678

10 678

.

.

× 100 = 81.27%. (Ans.)
Example 11.23. For the stoichiometric mixture of example 11.22 calculate :
(i)The volume of the mixture per kg of fuel at a temperature of 50°C and a pressure of
1.013 bar.
(ii)The volume of the products of combustion per kg of fuel after cooling to a temperature
of 130°C at a pressure of 1 bar.
Solution. As before,
C 2 H 6 O + 3O 2 + 3 ×^79
21
N 2 → 2CO 2 + 3H 2 O + 3 ×^79
21
N 2

∴ Total moles reactants = 1 + 3 + 3 ×

79

21

= 15.3

From equation, pV = nR 0 T

V = nR T

p

(^0) = 15 3 8 314 10 50 273
1 013 10
3
5
.. ( )
.
×××+
×
= 405.6 m^3 /mole of fuel
In 1 mole of fuel there are (2 × 12 + 6 + 16) = 46 kg
(i)∴ Volume of reactants per kg of fuel = 405 6
46
. = 8.817 m (^3). (Ans.)
When the products are cooled to 130°C the H 2 O exists as steam, since the temperature is
well above the saturation temperature corresponding to the partial pressure of the H 2 O. This must
be so since the saturation temperature corresponding to the total pressure is 99.6°C, and the
saturation temperature decreases with pressure. The total moles of the products is
=^233
79
21
F ++×
HG
I
KJ = 16.3
From equation, pV = nR 0 T
V =
nR T
p
(^0) = 16 3 8 314 10 130 273
110
3
5
..×××+( )
×
= 546.14 m^3 /mole of fuel.
(ii)∴ Volume of products per kg of fuel =
54614
46
.
= 11.87 m^3. (Ans.)
+Example 11.24. The following is the composition of coal gas supplied to a gas engine :
H 2 = 50.6 per cent ; CO = 10 per cent ; CH 4 = 26 per cent ; C 4 H 8 = 4 per cent ; O 2 = 0.4 per
cent ; CO 2 = 3 per cent ; N 2 = 6 per cent.