A 700 MW coal-fired power plant has a thermal efficiency of 35%. The coal has a
ID: 808294 • 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?
d.) Cooling water flows through the heat exchangers at a rate of 225,000 gal/minute. The water comes into the heat exchanger initially at 75 degree F and geets warmed to 105 degree F. How much heat is being removed by the cooling water (Btu/min)? Note that the Cp for water is 1.0 Btu/(lb-degree F), and the density of water is 8.33 lb/gal.
e.) The heat calculated in part d is transferred to the air by evaporation of some of the cooling water. The latent heat of vaporization for water at this temerature is 1,034 Btu/lb. How much water is evaporated (and must be replaced) each day, lb/day?
Explanation / Answer
700 Mw =700*1000 KW=7*105 KW=7*108 watts
1 watt =3.6 Kj/hr
7*108 watts=7*3.6*108 kj/hr=25.2*108 kj/hr
heating vallue of coal= 22000 Kj/Kg
amout of coal = 25.2X108/22000 = 114545.455Kg/hr=114545.455*24 Kg/day=2749090.91 kg/day
b) Amount of sulfur =3.8% =3.8*2749090.91*3.8/100=104465.45 kg/day=104465.45.32 kg moles/day= 3264.54 kg moles/day
The reaction is S+O2--> SO2
moles of SO2 formed = 3264.54 kg moles/day
the device captures 95 SO2 , hence 5% SO2 is vented , SO2 vented =3264.54*5/100=163.22 kg moles/day or
=163.22*96 kg/day= 15669.8182 kg/day ( Molecular weight of SO2 =96)
c) total ash ( 5.5%) =2749090.91*5.5/100=151200 kg/day
30% drop to the bottom of the boiler this is equal = 151200*30/100=45360 kg/day
remaining ash=151200*0.7=105840 kg/day
98.5 % is captured, ash that is not captured is carried along with exhaust
ash in the exhasut= 105840*1.5/100=1587.6 kg/day
d) Cooling water flow rate= 225000gal/minute
density of water = 8.33lb/gal mass of water/minute=225000*8.33 lb/minnute=1874250 lb/minute
heat reomved =mass of coolng wate/min * temperatue differene* Cp
= 1874250 lb/min* ( 105-75)*1=56227500 btu/min
e) latent heat of vaporization of water= 1034 btu/lb
water evaporated= 56227500/1034 lb/min=54378. 63 lb/min =54378.63*60*24lb/day=78305222 lb/day
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