TITLE.PM5

FUELS AND COMBUSTION 515

dharm

\M-therm\Th11-2.pm5

Total carbon in the flue gas = a × b × 12 (CO 2 + CO) per kg of carbon in the flue gas

=

ab

ab

×× ×

×× +

28

12

2

2

N

(CO CO ) ...(i)

For every kg of fuel supplied for combustion

C

100

kg of carbon pass out of the flue, whence

the weight of N 2 per kg of fuel is

28

12 100

2

2

N

(CO CO )

C kg

+

× ...(ii)

If all this nitrogen has come from the atmosphere, then air supplied per kg of fuel

=

28 100

12 100 77

2

2

×××

+××

NC

(CO CO )

...(iii)

(since air contains 77% nitrogen by weight)

∴ Air per kg of fuel =

NC

(CO CO )

2

2

100 77 12

28 100

×

××

×

+

= NC

33(CO CO )

2

2

×

+

. (Ans.)

It has been shown above that he total carbon of the flue gas is given by a × b × 12 (CO 2 + CO)

and that in CO by a × b × 12 CO, whence the proportion of carbon burnt to CO is given by

ab

ab

×× ×

×× +

12

(^122)
CO
(CO CO ) =
CO
CO CO+ 2
and if C is the proportion of carbon actually burnt per kg of fuel, then
Carbon burnt to CO per kg of fuel =
(CO)(C)
CO CO+ 2

. Proved.

+ Example 11.11. The following results were obtained in a trial on a boiler fitted with

economiser :

CO 2 CO O 2 N 2

Analysis of gas entering

the economiser 8.3 0 11.4 80.3

Analysis of gas leaving

the economiser 7.9 0 11.5 80.6

(i)Determine the air leakage into the economiser if the carbon content of the fuel is 80 per
cent.
(ii)Determine the reduction in temperature of the gas due to air leakage if atmospheric
temperature is 20°C and flue gas temperature is 410°C. Ash collected from ash pan is 15 per cent
by weight of the fuel fired.
Take : cp for air = 1.005 kJ/kg K and cp for flue gas = 1.05 kJ/kg K.

Solution. (i) Air supplied =

NC

(CO CO)

2 ×

33 2 +

Air supplied on the basis of conditions at entry to the economiser

=

80 3 80

33 8 3 0

.

(. )

×

+

= 23.45 kg