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[SOLVED] Thermodynamics |
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| May16-03, 02:13 PM | #1 |
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[SOLVED] Thermodynamics
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 [;)] |
| May16-03, 03:27 PM | #2 |
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[8)]
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...) |
| May16-03, 05:17 PM | #3 |
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I don't have a calculator on me, but your method is right.
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| May17-03, 08:57 PM | #4 |
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[SOLVED] Thermodynamics
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?) |
| May18-03, 09:44 AM | #5 |
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Good Guess, Gnome![:D] Thanks Tom! its right. |
| May18-03, 10:05 AM | #6 |
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So please enlighten me. Exactly how did you get 37%?
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| May18-03, 10:18 AM | #7 |
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