On an HHV basis, a 600 MW coal fired power plant has a heat rate of 9700 Btu/Kwh

Question

On an HHV basis, a 600 MW coal fired power plant has a heat rate of 9700 Btu/Kwh. The particular coal being burned has an LHV of 5957 Btu/lbm and an HHV of 6440 Btu/lbm.

a.) What is its HHV?

I calculated this to be 35.18% using the equation: effeciency = 3412 (Btu/kwh)/ Heat rate (Btu input/Kwh output)

b.) What is its LHV?

I calculated this to be 38.03% using the equation: thermal effeciency (HHV) = thermal effeciency (LHV) x (LHV/HHV)

c.) At what rate will coal have to be supplied to the plant (tons/hr)?

Please show your work on this step, this is the part that I don't understand how to do

Explanation / Answer

a)

thermal effeciency (HHV) = (3412/9700)*100 = 35.18 %

b)

thermal effeciency (LHV) = thermal effeciency (HHV)*HHV/LHV

thermal effeciency (LHV) = 35.18*(6440/5957) = 38.03 %

c)

Power = 600 MW = 2047.08*10^6 Btu/hr

Let m be the mass of coal

Power = Mass*HHV

Mass = 2047.08*10^6/6440 = 31.787*10^3 lbm/hr

1 metric ton = 2204.63 lbm

Mass = 14.42 tons/hr

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.