# Calculating coefficient of performance (COP) for Heat Pump

Here I am considering the following cycle; I know that the coefficient of performance for a heat pump can be calculated by; $$COP=\frac{Q_h}{W}$$ Where $$Q_h$$ is the heat exhausted into the hot reservoir. I am running into a problem with this... From here I know that the work $$W$$ for this cycle is given by; $$W=nkT_h\ln{(V_f/V_i)}-nkT_h\left(1-\frac{V_i}{V_f}\right)$$ And then given that; $$\frac{V_f}{V_i}=\frac{T_h}{T_c}$$ $$W=nkT_h\ln{(\frac{T_h}{T_c})}-nkT_h\left(1-\frac{T_c}{T_h}\right)$$ Where we take $$T_c$$ to be the temperature at point 1. What I am confused about is when we calculate, correct me if I'm wrong, but isn't the process 2-3 the only segment that releases heat into the hot reservoir? and thus, $$Q_h=Q_{2\to3}$$? When I carry out the calculations, I then get; $$Q_h=nkT_h\ln{(\frac{T_h}{T_c})}$$ If I then calculate the coefficient of performace of this heat pump, I obtain the relation; $$COP=\frac{\ln{(\frac{T_h}{T_c})}}{\ln{(\frac{T_h}{T_c})}-\left(1-\frac{T_c}{T_h}\right)}$$ Which in fact is greater than the carnot COP for a heat pump, $$COP_{C}=\frac{T_h}{T_h-T_c}$$ Have I gone somewhere in my calculations? or am I missing something else?

• isn't heat also given off in segment 3,1? It isn't all at Th, but it is given off. An isn't only a tiny amount of heat received by the working fluid at T1? Jun 20, 2020 at 12:50
• Yes, @ChetMiller heat is being removed from the system from 3-1. I couldn't however justify whether this heat was being dumped into the hot reservoir or not... Jun 21, 2020 at 1:57
• This suggestion then makes the COP; $$COP=\frac{T_h\ln{(\frac{T_c}{T_h})}-\frac{f}{2}(T_c-T_h)}{T_h\ln{(\frac{T_h}{T_c})}-T_h(1-\frac{T_c}{T_h})}$$ Jun 21, 2020 at 2:06
• @ChetMiller This, however, wouldn't make sense. I don't think the isochoric process in the cycle actually dumps heat into the hot reservoir but rather the cold and thus is not relevant to the analysis. Jun 21, 2020 at 6:25
• In my judgment, no. Who says you can do this reversibly with just one hot reservoir? And who says you can do this with only one cold reservoir? Jun 21, 2020 at 12:00

Note that the step $$1 \to 2$$ takes heat from the hot reservoir. (It cannot be from the cold reservoir, as this would require heat flowing from cold to hot.) Since the goal of the cycle is to send heat to the hot reservoir, accounting for this lowers the efficiency.

• How are you defining the "hot reservoir" in connection with process 1-2?. Heat is being transferred from a series of thermal reservoirs ranging from a low of $T_1$ to a high of $T_h$. Jun 24, 2020 at 17:07
• @BobD In order to talk about the COP as a single number, we have to have a single hot reservoir, since the efficiency is proportional to the heat sent to it. Sure you could have multiple reservoirs, but then I wouldn't know how to talk about a COP at all. Jun 24, 2020 at 17:11
• I'm just trying to understand what is the temperature of the "hot reservoir" referenced in the first sentence. Because only an infinitesimal amount of heat is absorbed at $T_h$ in process 1-2. Jun 24, 2020 at 17:20
• @BobD Assuming there only is one hot reservoir, it can be any temperature at all, as long as it's above $T_h$. During the $1 \to 2$ process, heat is transferred from it through a finite temperature difference. (As a result, the cycle is necessarily not reversible, but that's fine because OP is trying to show that it's less efficient than a reversible cycle would be.) Jun 24, 2020 at 17:24
• OK, but I was going under the assumption that the cycle was reversible. I will ask for clarification. Thanks. Jun 24, 2020 at 17:30

Have I gone somewhere in my calculations?

Yes, I believe some of your equations are incorrect. First of all, it appears you have used the wrong equation for work. You have used.

$$W=nkT_h\ln{(V_f/V_i)}-nkT_h\left(1-\frac{V_i}{V_f}\right)$$

This is the work done for the reversible heat engine cycle described in your original post. The isothermal expansion work is positive (done by the gas) and the isobaric compression work is negative (work done on the gas). However, for the heat pump the cycle reverses. Work done during the isothermal compression process is negative (done on the gas) and during the isobaric expansion process is positive (done by the gas). I believe equation should then be:

$$W=nkT_h\ln{(V_i/V_f)}+nkT_h\left(1-\frac{V_i}{V_f}\right)$$

It also appears you used the wrong equation for the heat rejected in the isothermal compression. Your equation is:

$$Q_h=nkT_h\ln{(\frac{T_h}{T_c})}$$

Which would yield a positive value for $$Q_h$$, whereas it should be negative as heat is out of the system. I believe it should be

$$Q_h=nkT_h\ln{(\frac{T_c}{T_h})}$$

Since it appears you have used the wrong equations for both work and heat, that would lead you to an in incorrect equation for the COP. I suggest you review the calculations.

or am I missing something else?

I believe so. I think there is a fundamental problem with considering your cycle as a refrigeration/heat pump cycle. To my knowledge (which is admittedly limited re refrigeration) all such cycles normally involve two fixed temperature environments, the environment to be cooled and the environment to be heated. That is the basis of the Carnot COP that you are comparing your cycle with.

In your cycle, if it is to be considered reversible as your previous heat cycle post, it appears there are multiple temperature environments, which makes it unclear what the objective (environment for the desired heat transfer) is.

Heat is transferred out in both the isothermal and isochoric processes. Which is the desired heat transfer to be used in the COP?

During the heat absorption (isobaric expansion), heat is obtained from a higher (than $$T_c$$) temperature environment. Is this the same environment where heat is rejected during the isothermal compression? If so, then it would seem it must be subtracted from $$Q_h$$ reducing the COP as pointed out by @knzhou.

Hope this helps.