Coal fired plant Thermodynamics help

Hello!

A coal fired plant generates 600MW of electric power. The plant uses 4.8x10^6kg of coal a day. The heat of combustion is 3.3x10^7 J/kg. The steam that drives the turbines is at temp of 300 degrees C and the exhaust water is at 37 degrees C. Whar is the overall effiecy of the plant for generating electric power?

I have a formula of and have substituted like so e = W/Q_H = (Q_H - Q_L)/ Q_H = 1- Q_L / Q_H.

I came up with 37% is that correct?

Thanks!
Dx

Last edited by a moderator:

gnome
That's amazing! That plant generates enough power to light 6 bulbs, and uses ONLY 4.8 million kilograms of coal per day.

Maybe they should consider switching to windmills.

(I'm guessing it's supposed to be 600 megaWatts. Try again; show YOUR work, not just the formula you copied out of the book...)

Last edited:
Staff Emeritus
Gold Member
I don't have a calculator on me, but your method is right.

gnome
It's a good formula, but how are you using it?

Why not just:
(600*10^3 kW * 24 h * 3.6 * 10^6 J/kWh)/(4.8*10^6 kg * 3.3*10^7 J/kg) = 32.7%

(In this problem, aren't the steam and exhaust temperatures red herrings?)

Originally posted by gnome

That's amazing! That plant generates enough power to light 6 bulbs, and uses ONLY 4.8 million kilograms of coal per day.

Maybe they should consider switching to windmills.

LMFAO!

Good Guess, Gnome!

Thanks Tom!
its right.

gnome
So please enlighten me. Exactly how did you get 37%?

Originally posted by gnome
It's a good formula, but how are you using it?

Why not just:
(600*10^3 kW * 24 h * 3.6 * 10^6 J/kWh)/(4.8*10^6 kg * 3.3*10^7 J/kg) = 32.7%

(In this problem, aren't the steam and exhaust temperatures red herrings?)

I did it just the way you showed me here, gnome.