Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

COP of Air-Conditioning

  1. Feb 26, 2010 #1
    Hi All,

    I'm doing a study on effects of outside temperature on the energy efficiency of air-conditioning systems in computer server farms/data centers.

    It will be a simulated study so I cannot use the standard definition of COP;

    COP = heat removed / work

    At the same time I also want to examine the effect of the outside temperature. There for I've been using the following derivable definition for COP;

    COP = 1 / (Th/Tl - 1)

    where Th = the temperature of the hot, outside environment
    and tl is the temperature of the inside. cool environment.

    The problem I have is that in certain places, the outside temperature will be lower than the inside temperature, giving me a negative value for COP?? Any comment on this, is a negative COP realistic given that I'm doing work to move heat from a warm environment to a cooler one? or is this incorrect?

    any help is appreciated.
  2. jcsd
  3. Feb 26, 2010 #2


    User Avatar

    Staff: Mentor

    Welcome to PF.
    I don't understand that sentence: what does it being a simulated study have to do with using the definition of CoP?
    If you are doing a study and you want to make the results real-world relevant, I'd take data from a real air conditioner and build a performance model based on its real performance curve.
  4. Feb 26, 2010 #3
    This means that I cannot use the more atypical definition of COP (heat removed/work) as I have no actual data regarding how much work it takes to remove a given amount of heat for a given AC, under certain conditions.

    It would be excellent to get this data, but I have had no luck locating the data for this, especially one that factors the condenser temperature. Thats why I was trying to base my work on the more general model, with possible rough adjustments to account in some way for
    real-world ac's.
  5. Feb 26, 2010 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    No work is required for heat to move from a warm to a cooler environment. Since the air conditioner would simply be turned off in these circumstances, there is no need or sense in trying to come up with a COP in that situation.

    (Unless I am missing something here -- Russ knows this stuff better than I do.)
  6. Feb 26, 2010 #5


    User Avatar

    Staff: Mentor

    Ok. One assumption that might be useful is that two air conditioners using the same refrigerant will have similar performance curves. The difficulty is in establishing the peak efficiency, but if you know roughly what type of AC unit you are talking about, you can find similar ones online.

    I've gone into more detail on the subject recently, in this thread (including links to some manufacturer data and my own analysis): https://www.physicsforums.com/showthread.php?t=379749
  7. Feb 26, 2010 #6


    User Avatar

    Staff: Mentor

    Well, there is a difference is between the theory and the reality. The reality is that if the outside temperature is only a little below the inside temperature, heat won't flow fast enough, so you still may leave the AC on. To be more specific, a typical air conditioner delivers 55F air, so the outside temperature needs to be below 55F to turn off the A/C (assuming the AC unit can use 100% outside air), even though you are only shooting for an inside temperature of 70F.

    Also, there may be a situation where there is no path for heat to flow from inside to outside except through the air conditioner (ie, for a network closet the middle of a middle floor of an office building).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook