1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Volume of water required to cool thermal/nuclear plants?

  1. Mar 5, 2017 #1
    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. jcsd
  3. Mar 5, 2017 #2
    Let's see more details of your calculation.
     
  4. Mar 5, 2017 #3
    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.
     
  5. Mar 6, 2017 #4
    Check your multiplication.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Volume of water required to cool thermal/nuclear plants?
Loading...