# COP of a heat pump

1. Jul 8, 2013

### TSN79

The COP of a heat pump can be expressed as $COP_{heating}=\frac{T_{hot}}{T_{hot}-{T_{cold}}}$
As higher is better, the value of this can be improved by reducing the temperature gap at which the system works. I'm just a little unsure about which temperatures are being talked about. The heat pump in question is air-to-water. Does $T_{cold}$ refer to the outside air temp? And is $T_{hot}$ the water leaving the heat pump on the other end? These are my guesses anyway. Thx :)

2. Jul 8, 2013

All correct.

3. Jul 8, 2013

### TSN79

Well, there is obviously something I'm not getting. Take the attached heat pump for instance. The smallest one has a listed COP of 3,14. $T_{hot}$ is said to be 45°C and $T_{cold}$ is 7°C. Using the formula from above this becomes 1,18 - which is obviously not right. What am I not getting?

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4. Jul 8, 2013

Kelvin...

5. Jul 8, 2013

### TSN79

Thx, I should have seen that one :)
Still, 8,4 is the theoretical maximum COP for these temperatures, isn't that right? So how do they arrive at 3,14?

6. Jul 8, 2013

### Staff: Mentor

8.4 COP would require a perfectly efficient compressor, expansion valve, no fan for the condenser, etc., so the lower efficiency comes from taking into account those extra losses.

Though if you are asking how they calculate it, they don't; they measure it.

7. Aug 1, 2013

### philrainey

the equation you gave is for the carnot COP that for a perfect heat pump where there is no temperature differences across heat exchanges to the heat sinks. The COP of a real heat pump is the heat output divided by the energy input and the carnot efficiently is the ratio of the real COP divided by the theoretical perfect carnot COP. And yes for the perfect carnot COP equation to work you have to enter the temperatures in the absolute kelvin scale. That is degrees C plus 273 as absolute o is -273 degrees C.