(20 marks) Problem (2) A 600-MWe power plant burns Alabama, Jefferson, St. Clair
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(20 marks) Problem (2) A 600-MWe power plant burns Alabama, Jefferson, St. Clair coal with average moisture and ash fractions of 18 and 15 percent, respectively. This plant operates with a heat rate of 9000 Btu/kWh An analysis of the refuse pit gives a higher heating value of 2742 /kg. An orsat analysis of the flue ga gives 12.78% CO2, 5.49% O2, and 1.75% CO. Find: 1. The thermal efficiency of the power plant. 2. The coal rate. 3. The dilution coefficient and percent excess air 4. Compare between the mass rate of the forced draft fan and the mass rate of the induced draft fan if atmospheric conditions are 44 C, 0.93 atm, and a relative humidity of The exhaust gas is at 300 C, 0.88 atm.Explanation / Answer
Problem 2.
1. Efficiency is found from heat rate
2. Coal rate is easily found by considering power output, heat rate and GCV of coal.
Coal rate = 600x103x 8863 x 1.055/2742 = 2046 x103 kg/h
3. Orsat analysis gives CO2, O2, and CO volume percentages. The residue from Orsat analysis will be N2 + SO2 + Ar etc(small amounts).. SO2 +Ar can be neglected in preliminary analysis..For our case N2 % vol is 79.98. C in coal needs to be solved.
To do the combustion calculation we first start with moisture+ash free coal. The combustion reaction is given by
xC+y(O2+ 3.76N2)= 0.1278 CO2 + 0.0549 O2 + 0.0175 CO + 0.7998 N2
Hence x = 0.1278+0.0175 = 0.1453
From O2 balance, y = (2x0.1278+2 x 0.0549 + 0.0175)/2
=0.1915
However from N2 balance, y = 0.7998/3.76 = 0.2127, This difference is likely due to N2 content in coal or H2 content in coal which changes O2 balance. This can be overlooked for prelimimnary analysis.
From combustion calculation 0.1915 moles of air is actually consumed per 0.1453 moles of C or 1.31 moles per mole of C.
If we remove O2 from product, the resultant reaction is stoichiometric.
0.1453C + y(O2+ 3.76N2)= 0.1278 CO2 + 0.175 CO + 3.76yN2
hence y =( 2x0.1278+0.0175)/2 = 0.13655
Hence therritical air required = 013655/0.1453 = 0.93978 per mol of C
Hence excess air = (actual air-theroritical air)/Theoritical air
= (1.31-0.93978)/0.93978 = 39.4 %
Dilution coefficient = Actual air/theoritical air = 1.31/0.93978 = 1.39
4. The FD fan flow is deduced from actual air requirement. Atmospheric moisture to be added via psychrometric chart. Consider C percentage as 67 % and add the coal moisture to the flue gas. So tolal flue gas flow can be calculated. By this way ID fan flow can be determined.
Problem 4.1- Follow similar strategy as with Problem 2
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