# Volume of water required to cool thermal/nuclear plants?

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1. Mar 5, 2017

### kokodile

1. The problem statement, all variables and given/known data
In the year 2004 the USA produced 1787 TWh of electrical energy in conventional thermal plants and 476 TWh in nuclear plants. Assuming 30% efficiency for nuclear plants and 40% for conventional thermal plants, determine the (annual) volume of cooling water required to cool these plants in once-through cooling if the cooling water undergoes a temperature increase of 10°C. (Neglect the heat lost up the chimney in conventional plants, and assume two significant digits in given data.)
ηthermal=.4
Ethermal=1787 TWh
ηnuclear=.3
Enuclear=476 TWh

2. Relevant equations

η=Useful (electrical) energy/total energy in
Q=mcΔT
Volume=mass/ρ
1 TWh=10^9 KWh
1 kWh = 3.6*10^6 J

3. The attempt at a solution
I did each plant one separately (of course) and converted the 1787 TWh of the thermal plant to Joules so that I could use it in the specific heat equation. When I converted it to Joules I got 9.5*10^24 Joules. I took that number and plugged it into my specific heat equation to get 2.3*10^20 kg. I plugged that mass into the Volume equation and got a volume of 2.3*10^17 m^3

I did the same steps for the nuclear plant and got 4*10^16 m^3.

Because the book only has one answer, I assumed they added the two up and got 2.7*10^17 m^3. The answer in the book is 2.7*10^11 m^3. So I'm close, but my exponent is off. I'm pretty sure I'm converting it correctly from TWh to Joules.

I also found that when I converted the TWh to MJ instead of Joules, I get the right answer, but I can't have it in MJ because the specific heat equation uses 4186 Jkg-1K-1

Last edited: Mar 5, 2017
2. Mar 5, 2017

### Staff: Mentor

Let's see more details of your calculation.

3. Mar 5, 2017

### kokodile

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 before I find the waste heat energy?)

Okay, here is my conversion from TWh to Joules.

2680.5 TWh x 1012W/1 TW x 1 J/s/1 W x 3600s/1 h = 9.65*1024 J (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*1024J/4186J*10K = 2.3*1020 kg

2.3*1020 kg/1000 = 2.3*1017m3

For the second plant, I took the exact same steps.

Using the efficiency equation, I did 476 TWh/.3 and got 1586.7 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 1012W/1 TW x 1 J/s/1 W x 3600s/1 h = 4*1024 J

4*1024J/4186J*10K = 9.6*1019

9.6*1019 kg/1000 = 9.56*1016

If I add those two up, I get (2.3*1017) + (9.56*1016) = 3.25*1017m3

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.

4. Mar 6, 2017
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