Question

The following results were obtained in a trial on a boiler fitted with economiser :

	CO2	CO	O2	N2
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 : c_p for air = 1.005 kJ/kg K and c_p for flue gas = 1.05 kJ/kg K

Answer 1

(i) Air supplied = \(\cfrac{N_2\times C}{33(CO_2+CO)}\)

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

= \(\cfrac{80.3\times80}{33(8.3+0)}\) = 23.45 kg

Air applied on the basis of conditions at exit

= \(\cfrac{80.3\times80}{33(7.9+0)}\) = 24.73 kg

∴ Air leakage = 24.73 – 23.45 = 1.28 kg of air per kg of fuel.

For each kg of fuel burnt, the ash collected is 15% i.e., 0.15 kg. 

∴ Weight of fuel passing up the chimney = 1 – 0.15 = 0.85 kg 

∴ Total weight of products

= Weight of air supplied per kg of fuel + Weight of fuel passing through chimney per kg of fuel 

= 23.45 + 0.85 = 24.3 kg

Heat in flue gases per kg of coal 

= Weight of flue gases × Specific heat × Temperature rise above 0°C 

= 24.3 × 1.05 × (410 – 0) = 10461 kJ 

Heat in leakage air = Weight of leakage air × Specific heat × Temperature rise of air above 0°C

= 1.28 × 1.005 × (20 – 0) = 25.73 kJ. 

We can still consider, in the mixture, the gas and the air as separate and having their own specific heats, but sharing a common temperature t. 

For heat balance : 

(1.005 × 1.28 + 24.3 × 1.05) t = 10461 + 25.73

26.8 t = 10486.73 

∴ t = 391.3°C 

∴ Fall in temperature as a result of the air leakage into the economiser 

= 410 – 391.3 = 18.7°C.