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Electrical Efficiency Calculations HVAC

  1. Feb 1, 2016 #1
    Hi,

    I am trying to do a comparison of electrical efficiencies between 2 HVAC compressors but struggling to find formulas or calculations to figure it out. Can anyone help me?
     
  2. jcsd
  3. Feb 1, 2016 #2

    russ_watters

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    Welcome to PF!
    Sure, that kind of thing is what I do for a living! Please provide some details about the problem and what you have tried and i'Lloyd be glad to help.
     
  4. Feb 1, 2016 #3
    I am trying to do a direct comparison between a 2015 and a 2010 air conditioning system. I have started by trying to compare the electrical efficiency of the compressors and motors in each system but looking at the manufacturer's specifications I'm not sure which values to use?

    I thought to calculate electrical efficiency was to divide the output power by the input power and multiply it by 100, but I'm not sure which input value to use? The 2015 compressor has an output of 950W and the 2010 compressor has an output of 550W. I will find the specification sheets I'm using and try to upload pictures of them to see if it helps....
     
  5. Feb 1, 2016 #4

    russ_watters

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    Usually they also have efficiency ratings, so the spec sheet may tell you what you need.
     
  6. Feb 1, 2016 #5
    Its not letting me upload photos so I've put in the links to the specifications web pages I'm using. If you are able to pick out the correct information and explain it to me that would be great! The 1st link is the 2015 model and the 2nd is the 2010 model.

    http://www.mitsubishitech.co.uk/Data/M-Series/MSZ-FD-VA/2014/MSZ-FH-VE/MUZ-FH-VE/MUZ-FH-VE_SM_OBH624.pdf [Broken]

    http://www.climaco.se/pdf/Mitsubishi/MUZ-GE253542VA.pdf
     
    Last edited by a moderator: May 7, 2017
  7. Feb 1, 2016 #6

    russ_watters

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    Page 4: coefficient of performance is what you are looking for.
     
  8. Feb 1, 2016 #7
    If COP = Q/W

    where Q is the condenser heat output (kW) and W is the power supplied to the compressor (kW) can I rearrange the formula to find the compressor input?....W=Q/COP

    Then I could calculate the compressor efficiencies as I would have both the inputs and outputs
     
  9. Feb 1, 2016 #8

    russ_watters

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    Yes, you can re-arrange the equation to calculate input power (or use the electrical data), but I guess now I am not sure what you are looking for. Did you really mean "electrical efficiency" (I'm not even sure what that means) or did you mean thermodynamic efficiency? And when comparing equipment of different capacity, input power isn't efficiency: efficiency is efficiency.
     
  10. Feb 1, 2016 #9
    When I looked at the data sheets I couldn't see any value for energy input to the compressor only output.

    So I thought I could rearrange the COP formula to find the energy input value, then use its value in another equation to find out the efficiency of each compressor.

    To calculate each compressor's efficiency I was going to use the formula
    efficiency(n)=output power(W) /input power (W)x 100 = .....% then subtract the 2015 value from the 2010 value to show the difference.

    Does this make sense?
     
  11. Feb 1, 2016 #10

    russ_watters

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    Oh, are you saying you want to exclude the condenser fan and just calculate it for the compressor?

    The spec sheet provides the amperages of the compressor and fan, from which you can calculate power. The one trick is that with any electric power calculation, there is a "power factor", which is why volts times amps doesn't exactly equal the stated power. You can calculate the power factor by taking volts times amps divided by watts, then apply that to the compressor (volts times amps times power factor).

    Or you can just ratio the fan and compressor amperages to get the fraction that is used for the compressor.
    It depends a on what the purpose of this is, but for some problems yes, that is a useful thing...with one caveat:

    You probably don't want to multiply by 100. Air conditioners aren't heat engines and they operate at a COP of greater than 1, which would be an "efficiency" of greater than 100%. That can be confusing, so people tend to use COP directly.
     
  12. Feb 4, 2016 #11
    I understand working out the amperage ratio (2.44/0.24= 10.1667 so ratio is roughly 10:1) but I'm having difficulty working out the power factor for the compressor and fan separately because the specification sheet only gives an overall power supply of 230V.

    To work out the equation power factor = volts x amps / watts I can get values for the amperage and wattage from the table, but what about power because the 230V will be split between everything?

    Are you saying I should apply the amperage ratio to the voltage to calculate power separately for the fan and compressor (so that would be 209.09V and 20.9V)?

    Or, there is a value for overall power factor on the specification sheet (82%) can I apply the amperage ratio to that to get separate power factors (so that would be 74.5% and 7.45%).

    I did try to work out power factor using the 230V but I got a power factor for the compressor >1 (1.02) which I know is incorrect. I do understand the overall concept but can give me a little more guidance please?
     
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