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Can 100 liters of water keep a computer cool?

  1. Oct 13, 2015 #1
    1. The problem statement, all variables and given/known data

    If you had a water-cooled computer that puts out 500W of heat into the loop with a large copper 100 liter reservoir but no radiators, how long can you keep the temperature of the components below 70 degrees Celsius if the initial water and air temperature is 20 degrees Celsius?

    2. Relevant equations

    area x heat transmission coefficient x (water temperature - air temperature)

    Cube length for 100 liters of water = 0.465m
    Heat transmission coefficient for water-copper-air = 13.1W/m^2K
    Room temperature = 20C & maximum water temperature allowed = 70C

    3. The attempt at a solution

    (0.465^2x6) x 13.1 x (70-20) = 849.8W
    (0.465^2x6) x 13.1 x (70-41) = 493W


    So 100 liters of water as well as the copper dissipating the heat should keep the water cooled pc components less than or equal to 70C until the room temperature reaches 41C.


    Is this ok?
     
  2. jcsd
  3. Oct 13, 2015 #2
    Why not just simplify the problem to the time it takes to heat 100 l of water from 20 to 70 deg C with 500 W of power?

    Am I missing something?
     
  4. Oct 13, 2015 #3
    That equation would be:
    ((volume x heat capacity x (final temperature - starting temperature) ) /3600) / 500
    ((100000x 4.185 x (70 - 20) ) /3600) / 500 = 11.625 hours.

    I did look at this equation but it doesn't take into account the heat that would be dissipated by the copper reservoir.
     
  5. Oct 13, 2015 #4

    russ_watters

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    Is the mass of copper or anything about how it would dissipate heat into the air provided?
     
  6. Oct 13, 2015 #5

    CWatters

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    The problem statement says..

    "large copper 100 liter reservoir but no radiators"

    Unless there is more info in the problem statement I would take "no radiators" to mean the 100 litre reservoir is effectively insulated. That's not realistic but the problem statement as written provides no data with which to calculate the losses from the reservoir.
     
  7. Oct 13, 2015 #6
    The mass of the copper tank is not provided and I was thinking that the walls of the tank itself will dissipate the heat into the surrounding air, was this thinking wrong? Would the numbers change a lot depending on the thickness of the copper?

    The original problem wasn't written in English so I had to translate. The reservoir is not insulated, it's basically a copper tank that holds the water and the outside is exposed to the air.
     
  8. Oct 13, 2015 #7

    russ_watters

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    Is the shape provided? It doesn't sound to me like there is enough information to include the reservoir or air in the calculation.
     
  9. Oct 13, 2015 #8
    The shape is a cube that has a length of 46.5cm, what other information is needed?
     
  10. Oct 13, 2015 #9

    russ_watters

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    I see that you calculated it as if it were a cube, but the problem statement doesn't say that. It also doesn't say whether it is covered, insulated, how thick it is or what it is sitting on, all of which will impact heat capacity and dissipation.
     
  11. Oct 13, 2015 #10
    I'll find out and get back to you tomorrow.
    Apart from shape, insulation, wall thickness and resting position is there anything else that will be needed?
     
  12. Oct 14, 2015 #11
    I asked the teacher and he said that the tank can be any shape, there is no insulation (both the air and water have direct contact with copper), the thickness is 5mm and the resting position needs to be realistic (so a part of the area will need to be deducted).

    Is the heat transmission equation that I used in my first post completely wrong?
     
  13. Oct 17, 2015 #12
    Bumping this for some advice.
     
  14. Oct 19, 2015 #13

    CWatters

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    Ok so it looks like they want you to assume there are losses from the copper tank. They say that "both the air and water have direct contact with copper" so I would probably assume they are only interested in conducted losses not radiated or convection losses. Perhaps just assume you have a water-copper-air sandwich where the temperature of the water and air are the same as the surface of the copper.
     
  15. Oct 19, 2015 #14

    CWatters

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    No that looks ok.

    One potential complication is that the temperature of the water isn't constant so the heat loss (power loss) isn't constant.
     
  16. Oct 19, 2015 #15
    Thanks for your reply.

    Isn't that what my initial equation is about?


    Does this need to be worked out to answer the question? I just need to find how much heat the copper will be dissipating into the air when the water is at 70C and if that is higher than 500W then the answer would be 'indefinitely'. I could also then work out the equilibrium temperature where the copper is dissipating 500W.

    People have said that I need to give more information which I did but I'm still not sure about the equation that needs to be used.
     
  17. Oct 19, 2015 #16

    CWatters

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    Yes that would be a reasonable approach to take. Your equation is the correct one to use. Give it a go.
     
  18. Oct 19, 2015 #17
    To find heat dissipation at a water temperature of 70C with one side of the cube covered.
    (0.465^2 x5) x 13.1 x (70-20) = 708W

    When rearranged to find the equilibrium you get a water temperature of 55.3C.

    I'm curious about (not part of my schoolwork) is how the air in a room will react to this, because of air convection 'new' air will constantly be coming in contact with the copper right? I was also wondering about how the temperature of the room would change.

    dry air specific heat = 1.00J/g K
    dry air density = 1275g/m3
    room area = 30m2
    room height = 3m
    temperature change = 5C

    Energy = density x volume x specific heat x temperature difference = 574kJ
    500Wh is 1800kJ

    So does this mean that in about 20 minutes a room with a volume of 90m^3 will go up in temperature by 5C if this computer is putting out 500W?
     
  19. Oct 20, 2015 #18

    CWatters

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    Yes but only if there are no losses through the walls of the room or to ventilation.

    A common problem is working out how big (in Watts) does the boiler/furnace of a house need to be to maintain say 21C indoors when it's -10 outside. The same basic equation you used to calculate the loss through the copper also works for combinations of brick, insulation and glass etc.

    Ventilation has the effect of providing new cold air that has to be heated. So if you know the rate of ventilation (usually specified in air changes per hour) you can also account for the power needed for that.
     
  20. Oct 21, 2015 #19
    Thank you for all your help CWatters, it's very much appreciated.
     
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