take this scenerio(before you guys think I'm just asking for answers, I actually did the work, but the answer just doesn't sound right):
At a steam power plant, steam engines work in pairs, the heat output of the first one being the approx heat input of the second. The operating temps of the first are 680 C, and 430 C. The second temps being 415 C, and 280 C. If the heat of combustion of coal is 2.8 x 10^7 j/kg, at what rate must coal be burned if the plant is to put out 900 MW of power? assume the efficiency of the engines is 65% of the ideal(Carnot) efficiency.
so from this, this is what I got.
900 mw is the output, therefore, it is Q(L).
e is = 65%, or .65.
What I'm solving for, is the rate, which is work(W).
In order to find work, I need to find Q(h), the amount of heat coming in.
so we have a formula, Q(L)/Q(h) = 1- e, which is equal to 900/Q(h) = .35.
solve for Q(h), and we get Q(h) = .00038.
so we have Q(h), we can use teh formula Q(h) = Work/efficiency. This leads to .00038 = work/.65.
Solve for W, and the work is equal to .00247.
So I get the answer as .00247 J/sec. (I guess).
Now, I did not use he temperatures at all, and I did not use the heat of combustion for Coal at all, for this problem. This is what bugs me. I dont' think the answer is right....
can you guys help?
At a steam power plant, steam engines work in pairs, the heat output of the first one being the approx heat input of the second. The operating temps of the first are 680 C, and 430 C. The second temps being 415 C, and 280 C. If the heat of combustion of coal is 2.8 x 10^7 j/kg, at what rate must coal be burned if the plant is to put out 900 MW of power? assume the efficiency of the engines is 65% of the ideal(Carnot) efficiency.
so from this, this is what I got.
900 mw is the output, therefore, it is Q(L).
e is = 65%, or .65.
What I'm solving for, is the rate, which is work(W).
In order to find work, I need to find Q(h), the amount of heat coming in.
so we have a formula, Q(L)/Q(h) = 1- e, which is equal to 900/Q(h) = .35.
solve for Q(h), and we get Q(h) = .00038.
so we have Q(h), we can use teh formula Q(h) = Work/efficiency. This leads to .00038 = work/.65.
Solve for W, and the work is equal to .00247.
So I get the answer as .00247 J/sec. (I guess).
Now, I did not use he temperatures at all, and I did not use the heat of combustion for Coal at all, for this problem. This is what bugs me. I dont' think the answer is right....
can you guys help?
