**COP** or * Coefficient Of Performance* is the most

**basic energy-efficiency**metric of any heat engine. It’s instrumental when comparing heat pumps, fridges, and air conditioners.

Definition: COP is the ratio of how much useful heat (or cold) a heat pump will produce if we give it certain energy input.

Basically, it tells us how much heat we can generate with every watt of energy.

*Example:*We have a 1000W heat pump with a

**COP of 3.5**. That means that we power it with 1000W, and the heat pump gives us back 3500W worth of heat. That’s a highly energy-efficient heat pump. It will boil almost

**10 gallons of water per hour**.

*For comparison:*A 1000W heat pump with a

**COP of 2**will boil less than

**6 gallons of water per hour**.

Do keep in mind that the amount of electricity used is the same in both cases. Running a 1000W heater for an hour costs about 13 cents.

On average, a COP 3.5 heater will boil a gallon of water for less than 1 cent. COP 2 heater will boil a gallon of water for more than 1 cent.

It’s much better to have a COP 3.5 water heater than a COP 2 water heater. For example, the best tankless water heaters are known to have a high coefficient of efficiency.

**Example:** Even a small electric tankless water heater is powered by 9,000W. Bigger ones with 15+ GPM can consume as much as 36 kWh every hour. The high COP of such a heat pump is essential for electricity cost-optimization. On the other hand, devices that don’t spend a lot of energy – something like battery-powered portable air conditioners – have a low COP.

Let’s look at how COP is calculated, what is the highest possible COP of a heat pump, and how the electricity bill is affected by HVAC units with different COP values.

Table of Contents

## How To Calculate COP? COP Formula

Here is the COP formula *(an equation that calculates the coefficient of performance for any heat pump)*:

**COP = Q/W**

where Q is the heat the heater generates if we give it a certain amount of work (W).

For cooling, Q represents the heat that we take away from a cold reservoir. Air conditioner, for example, takes heat away from a room (cold reservoir).

Note: COP for heating and COP for cooling may be different. The best mini-split heat pumps, for example, are capable of cooling down space as well as heating it.

A good mini-split system will usually have a cooling COP ratio of 2 or more, and a heating COP ratio of 3 or more.

If we apply the **1st law of thermodynamic** and do a bit of derivation, we can see the COP values for a theoretically 100% ideal heat pump and ideal air conditioner (we also call that a Carnot machine). Let’s do the heat pump first:

### COP Of A Heat Pump

Here’s how you can calculate the theoretically maximum COP of a heat pump:

COP_{heat pump }= T_{hot}/(T_{hot}-T_{cold})

T_{hot} is that cozy hot temperature we want to have during cold winters (let’s say 95F; that’s 308 in Kelvins).

T_{cold} is the cold temperature at which the heat pump starts to operate (57F or 287K).

Theoretical maximum COP is calculated like this:

COP_{heat pump }= 308K/(308K-287K) = 14,67

So, in theory, a heat pump can have a COP even above 10. Nonetheless, the real COP of a heat pump in practice is a lot lower.

Golden standard: The standard test to measure the COP of a heat pump is conducted with T_{hot} = 95F (308K) and T_{cold} = 32F (273K). That means that in 100% ideal case the maximum COP is 8.8. But in practice, it’s lower.

In fact, the highest COP a heat pump can achieve is about 4.5. Any heat pump with a COP of above 3 has a very high energy efficiency.

Here is a graph of how much more efficient high COP heat pumps are. We put the COP 2 as zero and calculated by how many percent more efficient the higher COP heat pumps are.

You can see, for example, that a 3.2 COP pump is 60% more energy-efficient than 2 COP pumps.

### Coefficient Of Efficiency Of Air Conditioners

OK, let’s calculate the maximal theoretical efficiency of a cooling device. Namely, an air conditioner or a refrigerator. By applying the 1st law of thermodynamics, we can deduce the ‘Carnot COP’ for a cooling device is calculated as such:

COP_{cooling}= T_{cold}/(T_{hot}-T_{cold})

T_{cold} is the chill temperature you want to have in your room during the summer. T_{hot} is the high heat wave temperature.

Let’s look at what would a COP of an air conditioner be in a standardized T_{hot} = 95F (308K) and T_{cold} = 32F (273K) temperature interval. Plugging the temperature in the COP cooling equation above, we get 7.8.

If you remember, the maximum coefficient of performance of a heat pump was 8.8. If you are in the market for an air conditioner, make sure to get one with a COP value of above 2. That’s a very high COP for a practical HVAC cooling device.

Usually, the problem is that you won’t find a COP value anywhere, not even on a specification sheet. Energy-efficiency is usually represented with such metrics as EER and SEER; all these as based on the COP of an HVAC divide. For example, we have compared the best portable air conditioners by comparing their EER ratings.

### Seasonal COP Or SCOP

In 2013, SCOP or *Seasonal Coefficient Of Performance* was introduced. We know that COP is a measure of energy-efficiency for a heating or cooling device. What measuring SCOP is trying to achieve is to objectively measure energy-efficiency over the winter season (for heating) and summer season (for cooling).

Basically, the relationship between SCOP and COP is the same as with SEER and EER.

SCOP would give a much more realistic idea about how energy efficient an HVAC device is in practice, i.e., in a real summer season.

Nonetheless, the SCOP has still considered a very new methodology of measuring seasonal cooling and heating efficiency. As such, you will rarely find the SCOP ratio on older devices. In fact, even the new devices rarely include SCOP in their specification sheets, primarily because they have yet to measure it.

I liked that you pointed out that you will want to consider what the heat pumps you have max output. It does seem like a good thing to know if you feel like your heat pump isn’t keeping your house warm enough during the winter. I know that I hate being cold so I would want to get a good pump in my house.

Insulated homes really don’t need much. Infact, modern building standards we’re having problems getting small enough ones.

COP is defined as the relationship between the power (kW) that is drawn out of the heat pump as cooling or heat, and the power (kW) that is supplied to the compressor. For example: A given heat pump used for air cooling has a COP = 2.

Yes, Clover, this is correct.

Hello LearnMetrics, the COP for heat pumps article is straight simple, and clear. However, I find it most useful to indicate that the Carnot efficiency is a known overestimation of the actual measured COP. Furthermost Carnot formulas have been revised to include the non-reversible reality of processes, with formulas called Endoreversible Efficiency. So in conclusion one can calculate 3 different COPs:

Carnot > Endoreversible > Measured.

To have an even clearer understanding, heat pumps are the reverse Rankine thermodynamic cycle, and different than the Rankine that is usually in steady operating conditions, the reverse Rankine heat pumps operate normally inside of an envelope of different range of conditions. A revised article considering the above would be also appreciated I am happy to facilitate other parameters and units that are not so easy to understand, in part for having been developed about 2 centuries ago.

Hello Hugo, thank you for your informative input. You’re exactly correct; thank you for the additional info you shared with all our visitors.

If I am creating a heat pump system, which is basically an airconditioner, where I have a chilled tank of water, at say 4C as my source of cold, and a room which I want to maintain at 22C as my hot side, can I calculate the expected COP, if I know it takes 3kW to maintain the losses in the system – ie, I need to remove heat at 3kJ/sec to keep the room at 22C ? I can’t find anyone who will explain the math for this!

Thanks

Steve

Hello Steve, the theoretical COP depends only on the temperature difference, ie. COP = 277K / (295K – 277K) = 15.4. Now, the practical expected COP needs to take into account the losses in the system, as you have correctly proposed. How to actually calculate that is a bit beyond our knowledge; some textbooks might come in handy here.

I’m looking at pool heat pumps and one of the models claims a COP of 14. Are you saying this is a lie?

Hello Nick, this article is primarily focused on air source heat pumps. But you are correct; the pool heat pumps can have a higher COP. Department of Energy specifically states that ‘COPs usually range from 3.0 to 7.0’ with an average of 5.0 COP. We’re not experts on pool heat pumps here but COP of 14 seems a bit unrealistic nonetheless. Hope this helps.