Data Center COP

Hi All,

I'm a complete newbie to the whole area really, I studying a PhD in computer science and as part of the work I'm currently doing I'm trying to understand how to calculate the energy efficiency of air-conditioning systems relative to the outside temperature.

The equation I have been using to calculate the COP is this:

COP = 1 / (Th/Tl -1)

Where Th is the outside temperature and Tl is the (cooled) indoor temperature (reversed carnot)

The question I have about this is whether this model assumes that the heat being removed by the aircon is being generated by heat transfer from the outside?

What I am looking at though is computer server-farms (typically well insulated) where the main heat source is generated by the equipment and not the outdoor temperature. How does this effect the above calculation of COP? how do I factor in the 3rd value i.e. the heat generated by the computer servers?

To put this another way, if I were to use the above equation to measure the COP of 2 refrigerators with the same Th and Tl, but placed a heat source in one, surely the COP would not be the same??



(Sorry if this is not the correct forum for this but any help would be greatly appreciated)

Hi Ray and welcome to the forum.

I don't have an answer to your query but I'm sure someone will have an input.

Our normal CoP is based on the linked formulae.

http://en.wikipedia.org/wiki/Coefficient_of_performance

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Originally Posted by rayc

The equation I have been using to calculate the COP is this:

COP = 1 / (Th/Tl -1)

Where Th is the outside temperature and Tl is the (cooled) indoor temperature (reversed carnot)

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You cannot use the COP equation for Carnot cycle refrigerators or heat pumps as a measure of real life systems.
The Carnot cycle is a theoretical cycle without losses, against which real life systems can be compared.

For an ideal closed vapour compression cycle, a better COP equation would be

COPR = (h1 - h4)/(h2 - h1)

where h1 and h4 are the evaporator points on the Mollier chart and h2 and h1 are the compressor points on the mollier chart, basically, refrigeration effect divided by power input.

This is based on steady flow conditions

Are you familiar with the Mollier chart?

Hi Ray
To calculate OAT verses power consumption of an A/C system energy efficiency is not easy. To measure this from an existing system is not so hard. You require two data parameters - 1, actual power consumption 2 degree day data. Then plot them both on a chart. This is how an energy audit would show it for a site requiring heating. But I worked at a huge data centre which used cooling towers. Where cooling towers are used humidity has a biger effect than the temp. The above is not the answer to your 1st Question but should point you in the right direction.
For the second part a measure of system effiency is output over input (simple as that). So power consumption required to remove the heat over the heat input from conditioned space including computers, fans, lighting, and heat transfer from envelope. The data centre I worked in there was no real differance in the load on chillers between night and day temps as the computer load was way bigger. Hope this helps a little.

The efficiency of the cooling system in a data center mainly depends on how the cooling is done. There are many ways to achieve proper data centers cooling. The most common is the CRAC (computer room air conditioner), talking of which the most common is by means of chilled water, while the second most common is direct expansion system.
Chilled water systems cool down pumped water with a water chiller. In this case water is a secondary fluid.
In direct expansion systems, there is no secondary fluid; in fact the refrigerant is expanded directly in an air cooled coil.

The measure of performance is a direct function of the evaporating temperature of refrigerant, and the condensing temperature of it. In other words the higher the difference between the warm and the cold side of your considered cycle, the higher the heat transfer, but the lower the performance of the compressor (less mass flow for the same pressure ratio).

You can also have other means of cooling, like adiabatic or free cooling, exploiting times of the year where the ambient temperature is lower than computer room temperature.

As you see, it takes much more than a simple fraction to evaluate data center power consumption.

Thanks for the replies guys.

First off, after running myself in circles yesterday I think my question/analogy is wrong... if you put a heat source in a refrigerator it does not (in the ideal system) effect the COP as more work is done to compensate for the additional heat, thereby maintaining the COP.

In general my problem is that I cannot use a real system to calculate the COP as I'm running a simulation to 'estimate' how much energy could be saved. So where COP = Heat Removed/Work, I don't actually know the work value.

What I want to do is calculate the COP (based on the carnot ideal system) and use a percentage of this COP value to simulate a 'real' system. Then based on this COP value try to work out the energy used.

so using the equation COP = 1 / TH/TL - 1 I get an estimate of the COP based on the outside and return temperature, and then use

Work = Heat Removed/COP

to work out the estimated work value, where i can guess the heat value based on the workload of the servers in the data centre.

The problem I'm having revolves around when the outside temperature is lower than the inside one. So the servers are creating a lot of heat in a country/area where the outside temparatue is low.

This effects the estimation of the ideal COP, as TH smaller than TL will give a negative COP value. From a practical viewpoint should this not have a high COP value as your moving heat from a high temperature to a low temperature? Do air-conditioners/refrigerators act any differenlty in this situation?

I know this is probably a bit left field but any help/thoughts are greatly appreciated.

P.S. I have not looked at Mollier charts specifically but I am aware of them from other posts on here. I didn't want to go down that route yet as I wasn't looking a specific refrigerant at this point, just the 'general' theory but it is something I might follow.

Hi Tesla, thanks for your input.

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Originally Posted by Tesla

So power consumption required to remove the heat over the heat input from conditioned space including computers, fans, lighting, and heat transfer from envelope.

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This comes back to one of my first questions, do the theoretical models I used above assume that the only heat load is from the environment i.e. heat transfer from the envelope?? Hence only the outside (Th) and cooled (TL) temperature are used.

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Originally Posted by Tesla

The data centre I worked in there was no real differance in the load on chillers between night and day temps as the computer load was way bigger. Hope this helps a little.

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Am I wrong in my understanding then that higher outside temperatures effect the efficiency of how the heat is dissipated at the condenser?

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Originally Posted by rayc

Am I wrong in my understanding then that higher outside temperatures effect the efficiency of how the heat is dissipated at the condenser?

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You are perfectly right, but there is not heat gain from the ambient to the room, provided appropriate insulation is installed.
So, the ambient temperature is not in a direct relationship to heat gain, and limited relationship to power consumption.

The question you asked regarding the equation

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COP = 1 / (Th/Tl -1)

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Is misleading as our definition of COP seems to be wildly different to yours.(see Franks post and search on site for definition)

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This comes back to one of my first questions, do the theoretical models I used above assume that the only heat load is from the environment i.e. heat transfer from the envelope?? Hence only the outside (Th) and cooled (TL) temperature are used.

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Your model assumes that the outside temperature is related to the off "cooling" coil temperature and that the COP is inversely proportional to that relationship (-1)

This doesn't make sense unless you are using 100% outside air rather than localised recirculation.( What is the -1 for????)

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Am I wrong in my understanding then that higher outside temperatures effect the efficiency of how the heat is dissipated at the condenser?

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No you are not wrong but you are not right. Most condensers are designed to operate within a specific operational envelope. This is made possible by various means but a fan and coil is most common. This of course can only operate within its operational envelope.

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Originally Posted by TRASH101

our definition of COP seems to be wildly different to yours.(see Franks post and search on site for definition)

Your model assumes that the outside temperature is related to the off "cooling" coil temperature and that the COP is inversely proportional to that relationship (-1)

This doesn't make sense unless you are using 100% outside air rather than localised recirculation.( What is the -1 for????)

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I understand the standard definition of COP is heat removed/work done. However I cannot determine the work value as it's not a real system. However the equation I'm using is a derivative of this, taken from books on Thermodynamics.

Where Work = Qh -Ql

(Qh = heat dissipated to environment, Ql = heat removed from cool environment)

so then COP becomes Ql / Qh - Ql and then

1 / (Qh/Ql - 1)

where Qh/Ql => Th/Tl

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Originally Posted by rayc

I understand the standard definition of COP is heat removed/work done.

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This is wrong.
For refrigeration, COP is a measure of Work Done (refrigeration effect)/ Power input.

The Mollier chart can be used to plot a system's operating parameters taken from actual readings, or, you can use it to plot from theoretical parameters, or manufacturers data, thereby enabling a comparison to be made between different operating conditions. A very useful tool.