# Heat Pumps and COP?

Heat Pump COP (Coefficient of Performance)

$$COP=1-\frac{T_c}{T_H}$$

With Heat Pumps is the efficiency/COP more dependent on the hot or the cold reservoir and why?

• The given equation is incorrect. Look at the link to find out why. Jul 2 '16 at 16:23
• The text mentions both efficiency and COP, but these have different formulas. Usually COP is the expression used for refrigerators/heat pumps (moving heat from a cold place to a warmer), while efficiency is a term usually used in the case where work and not heat transfer is the goal (engines e.g). The formula shown is for efficiency not COP. Also note, the formula gives the extreme / ideal case (the Carnot cycle) and is not the general expression of efficiency. Jul 2 '16 at 16:40

Actually the formula given in your question describes the thermal efficiency $Nth$ of a device , usually a heat engine that transforms thermal energy into mechanical energy and it's always: $Nth<1$.

On the other hand $COP$ or coefficient of performance is the ratio $Q/W$ of the device called heat pump, where $Q$ is the heat removed from the cold reservoir and $W$ the input work required in order to remove this heat (usually electrical energy).

For a heat pump operating at maximum theoretical efficiency (i.e. Carnot efficiency), it can be shown that: $COP=Q/W=Thot/(Thot-Tcold)$.

Because heat pumps do not transform one form of energy to another but in fact they transfer heat from a cold reservoir to a hot reservoir,COP can and usually has a value much greater than $1$.

Notice also that the $COP$ term is commonly used to describe the performance of a heat pump device, when it is operating in heating mode.

Now about your question: Let's assume that a heat pump is operating at $Tcold=273K$ and $Thot=290K$ so $ΔΤ=17K$ and $COP=290/17=17.06$

Now let's assume that the same device operates at the same $ΔΤ=17K$ but the reservoirs temperatures are 10 degrees higher. In this case $COP=300/17=17.67>17.06$.

This means that our device is more efficient if the temperature of cold reservoir is greater (in heating mode), something expected because physically is more difficult to 'pump' or remove heat from a colder reservoir-enviroment.

It's obvious that the COP performance of a certain device depends on the enviroment conditions and it's value usually varies between 2.5 and 5

If you look closely at the working of the heat pump , it follows reverse Carnot cycle. As you know that Carnot cycle is explained from piston and its intraction with source and sink slabs. When piston is placed on sink stand and expanded isothermally the heat is transferred from sink to piston, then the piston is adiabatically compressed so that its temperature increase to source temperature. Then it is placed on source stand and compressed isothermally so that the excess enegy is transferred to source (for heating in room), then it is expanded adiabatically to reach original state.

Now in this process if the temperature of the sink is low, less energy is transferred from sink to piston. If the temperature of source is high more work has to be done during adiabatic compression to increase the piston temperature to source temperature. In both cases coefficient of performance (COP) will go down

According to this hyperphysics article on heat pumps, the COP is given by

$$\rm{COP}=\frac{T_H}{T_H-T_C}$$

which is not quite the equation you gave (actually it is the inverse).

Once could ask the question: if I change either the high or low temperature by 1 degree, which of these will change the value of COP most?

If you lower $T_C$ by 1°C, the denominator gets bigger by 1 degree, and the numerator is unchanged. If you increase $T_H$ by 1°C, the denominator gets bigger, but so does the numerator. The ratio will change less than if just the denominator changed.

You can see this mathematically by taking the partial derivative with respect to each (which I will leave as an exercise for you).

It is obvious, then, that the COP is more sensitive to the temperature of the cold reservoir.