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Introductory Physics Homework Help
Volume of water required to cool thermal/nuclear plants?
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[QUOTE="kokodile, post: 5708841, member: 582829"] I think I also forgot to type out another step I took. Using the efficiency equation, I did 1787 TWh/.4 and got 4467.5 TWh. I used 4467.5 TWh and 1787 TWh to find the waste heat energy. So I did 4467.5-1787 and got 2680.5 TWh. Because this is still in TWh, I converted it to Joules. (Should I convert it to Joules [I]before[/I] I find the waste heat energy?) Okay, here is my conversion from TWh to Joules. 2680.5 TWh x 10[SUP]12[/SUP]W/1 TW x 1 J/s/1 W x 3600s/1 h = [B]9.65*10[SUP]24[/SUP] J[/B] (I originally wrote that I got 9.5*10^24 J, but it's actually 9.65*10^24. I still get the same answer though) 9.65*10[SUP]24[/SUP]J/4186J*10K = [B]2.3*10[SUP]20[/SUP] kg [/B] 2.3*10[SUP]20[/SUP] kg/1000 = [B]2.3*10[SUP]17[/SUP]m[SUP]3[/SUP][/B] For the second plant, I took the exact same steps. Using the efficiency equation, I did 476 TWh/.3 and got [B]1586.7[/B] TWh. I used 1586.7 TWh and 476 TWh to find the waste heat energy. So I did 1586.7-476 and got 1110.7 TWh. This is still in TWh, so I converted it to Joules. 1110.7 TWh x 10[SUP]12[/SUP]W/1 TW x 1 J/s/1 W x 3600s/1 h = [B]4*10[SUP]24[/SUP] J[/B] [B][/B] 4*10[SUP]24[/SUP]J/4186J*10K = [B]9.6*10[SUP]19[/SUP][/B] [B][/B] 9.6*10[SUP]19[/SUP] kg/1000 = [B]9.56*10[SUP]16[/SUP][/B] [B][/B] If I add those two up, I get (2.3*10[SUP]17[/SUP]) + (9.56*10[SUP]16[/SUP]) = [B]3.25*10[SUP]17[/SUP]m[SUP]3[/SUP] [/B] I did realize that I forgot to find the waste heat energy first the second time around, but even with finding the waste heat energy, I still get the wrong answer. [/QUOTE]
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Volume of water required to cool thermal/nuclear plants?
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