A 700 MW coal-fired power plant has a thermal efficiency of 35%. The coal has a
ID: 892287 • Letter: A
Question
A 700 MW coal-fired power plant has a thermal efficiency of 35%. The coal has a heating value of 22,000 kJ/kg, an ash content of 5.5%, and a sulfur content of 3.8%.
a.) What is the daily input rate of coal (kg/day)?
b.) If the efficiency of SO2 capture in an APC device is 95%, how much SO2 is emitted into the air, kg/day?
c.) If 30% of the ash that comes in with the coal drops out in the boiler as bottom ash, and if the APC device that treats the exhaust gases is 98.5% efficient at capturing the fly ash, what are the total fly ash emissions to the atmoshere, kg/day?
Final answers: (a) 7.85 (10)^6 kg/day
(b) 2.98 (10)^4 kg/day
(c) 4,536 kg/day
Explanation / Answer
700 MW= 700*1000 *1000 W= 7*105 Kw=7*105*3600 Kj/hr=2520000000 Kj/hr
heating value of coal = 22000 Kj/Kg
Coal required= 2520000000/22000= kg/hr=114545.455kg/hr =2749090.91kg/day
but the power plant is 35% efficient hnece coal required= 2749090.91/0.35=7854545 kg/day
sulfur content 7854545*3.8/100= 298472.7 kg/day
S+O2-->SO2
32 kg sulfur gives 64 kg of SO2
298472.7 gives= 298472.7*2 = 596945.5kg/day SO2
So2 removed =95% sulfur dioxide not removed is 5% = 596945*5/100=29847 kg/day
c)Ash in coal ( 5.5%)= 7854545*5.5/100=432000 kg/day
ash settled at the bottom =30% =432000*30/100=129600 kg/day
98.5 % ash is removed, fly ash mot removed =1.5% = 129600*1.5/100=1944 kg/day
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