Can Outside Temperature Affect the COP of Air-Conditioning Systems?

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In summary, the study is trying to find out if a negative value for COP is realistic given that the outside temperature will be lower than the inside temperature. The difficulty is that the study cannot use the more atypical definition of COP, as it has no actual data regarding how much work it takes to remove a given amount of heat for a given AC, under certain conditions.
  • #1
rayc
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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.
 
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  • #2
Welcome to PF.
rayc said:
It will be a simulated study so I cannot use the standard definition of COP;

COP = heat removed / work
I don't understand that sentence: what does it being a simulated study have to do with using the definition of CoP?
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?
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.
 
  • #3
russ_watters said:
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?

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.

russ_watters said:
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.

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.
 
  • #4
rayc said:
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?
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.)
 
  • #5
rayc said:
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.
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
 
  • #6
Redbelly98 said:
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.)
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).
 

1. What is the definition of COP of Air-Conditioning?

The COP (Coefficient of Performance) of air-conditioning is a measure of the cooling efficiency of an air conditioner. It is calculated by dividing the cooling output (in BTUs or Watts) by the energy input (in Watts).

2. How is COP of Air-Conditioning different from SEER?

COP is a measure of cooling efficiency at a specific temperature, while SEER (Seasonal Energy Efficiency Ratio) is an average measure of efficiency over a range of temperatures. COP is typically used for commercial and industrial air conditioners, while SEER is used for residential air conditioners.

3. What is a good COP for an air conditioner?

A good COP for an air conditioner varies depending on the type and size of the unit. Generally, a higher COP indicates a more efficient air conditioner. For residential air conditioners, a COP of 3 or higher is considered good, while for industrial units, a COP of 5 or higher is considered good.

4. Can the COP of an air conditioner be improved?

Yes, the COP of an air conditioner can be improved by regular maintenance and cleaning, using energy-efficient models, and ensuring proper sizing and installation. Additionally, using a programmable thermostat and setting it to a higher temperature when the AC is not in use can also improve the COP.

5. What factors affect the COP of an air conditioner?

The main factors that affect the COP of an air conditioner include the age and condition of the unit, the type and quality of insulation in the building, the outdoor temperature and humidity, and the size and efficiency of the unit. Other factors such as air flow, location of the unit, and usage patterns can also impact the COP.

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