Coal fired plant Thermodynamics help

[Dx]

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

 

[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...)

 

[quantumdude]

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?)

 

[Dx]

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%?

 

[Dx]

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.