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Calculate % dry flue gas losses
A process industry consuming 10 TPH of saturated steam at 10 kg/sq.cm(g) pressure has been using coal as fuel in boiler.
Typical ultimate analysis of the coal:
Carbon : 41.11%
Hydrogen : 2.76 %
Nitrogen : 1.22 %
Oxygen : 9.89 %
Sulphur : 0.41%
Ash : 38.63
Water : 5.89
Flue gas temperature = 200°C
Ambient temperature = 30°C
Enthalpy of steam = 668 kcal/kg
Feed water temperature = 80°C
Specific heat of flue gases = 0.23 kcal/kgoC
Boiler efficiency with Indian coal =72 %
GCV of coal = 4,000 kCal/kg
Oxygen content in dry flue gases = 10%
Annual Hours of operation = 8000 hrs.
Determine:
(i) Quantity of annual coal requirement in tonnes/year
(ii) Calculate % dry flue gas losses
Solution:
a)
Coal requirement Q = Steam (q) x (hg – hf)/( Efficiency x GCV)
= 10 x (668-80) / (0.72 x 4000)
= 2.042 T/Hr
= 2.042 x 8000 hrs
= 16336 Tonnes/year
Theoretical air requirement for coal
= [(11.6 x C%) + {34.8 x(H2% - O2%/8)} + (4.35 x S%)] kg / kg of coal
100
= [(11.6 x 41.11) + {34.8 x(2.76 – 9.89/8)} + (4.35 x 0.41)]
100
= 5.31 kg / kg of coal
(Or)
C+ O2 = CO2
12+32= 44
(C%*32)/12
2H2+ O2 = 2H2O
4+32=36
(H%*32)/4
S+ O2 = SO2
32+32=64
(S%*32)/32
Total oxygen required = (41.11* 32/12) + (2.76*32/4)+ (0.41*32/32)
= (109.63) + (22.08) + (0.41)
= 132.1 kg/ 100 kg fuel
Oxygen already present in 100 kg fuel = 9.89 kg/ 100 kg fuel
Additional oxygen required = 132.1 – 9.89 kg/ 100 kg fuel
= 122.21 kg/ 100 kg fuel
Quantity of dry air required
(Air contains 23% O2 by weight ) = 122.1/ 0.23
= 531.35 kg/ 100 kg fuel
Theoretical air required = 531.35/100
= 5.31 kg air/ kg fuel
Excess air = O2x100/(21-O2)
Excess air = 10 x 100/ (21 – 10) = 90.9%
Actual air= 5.31 * (100+90.9)/100
=10.137 kg air/kg coal
Heat loss in dry flue gas = m x CP (Tf – Ta) x 100 /GCV
= (10.137+ 1) x 0.23 x (200 – 30) x 100
4000
= 10.89 %
Very good
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