Efficiency of parallel and sequential heat pumps

Question

Consider two identical heat pumps, for example, split-system air conditioners. There're two ways to make them work - in parallel or sequential. Parallel means that "hot" radiators of the machines are put outside the room, while the "cold" radiators are put inside the room. Sequential means that the cold radiator of the first machine is inside the room, while the hot radiator is put inside, say, a large isolated box, where it is cooled by the cold radiator of the second machine, and the hot radiator of the second machine is outside the room.

In which case - parallel or sequential - the efficiency of the cooling would be better?

Answer 1

The COP of a heat pump is a function of temperature lift (change from source to destination). The higher the lift the lower the COP. 30C lift at 300% COP a practical rule of thumb. In paralel you can get twice the heat at a constant lift. In series twice the lift for the same total heat generated at the same efficiency. Higher (perhaps double) efficiency but the same total heat as a single heat pump can be achieved serially, by halving each's lift.

High lift is generally a limitation of heat pumps and so a serial arrangement is a method of overcomming those limitations.

Answer 2

Efficiency of one heat pump in cooling can be defined by expression

$$\eta = \frac{Q_C}{W},$$

that is heat that is taken from cooler reservoir divided by the work put into the pump.

If you have two pumps in parallel efficiency shall be the same, as you will have heat twice as large and work twice as large

$$\eta = \frac{2Q_C}{2W}.$$

If you however have two pumps in sequence equation reads as

$$\eta = \frac{Q_C}{2W}.$$

So in order to obtain at least same efficiency you should extract heat twice as large by each pump. So the efficiency of each pump should be twice as large for about half of the temperature change. So far so easy.

In the next step we must take into consideration exact cyclical thermodynamical processes. There are plenty of them that you can use and no can be exactly theoretically calculated. In such cases it is useful to observe the most efficient thermodynamical process, that is Carnot cycle and extract conclusions from it.

Efficiency of the heat engine based on Carnot cycle can be shown to be

$$\eta = \frac{T_C}{T_H-T_C}.$$

In case of two pumps in the first iteration intermediate temperature $T' = \frac{T_C+T_H}{2}$ in exactly in the middle. Efficiency of two pumps shall be

$$\eta_1 = \frac{T_C}{T'-T_C}, \eta_2 = \frac{T'}{T_H-T'}$$

Obviously $\eta_1 = 2 \eta$ and $\eta_2 > 2 \eta$, therefore two sequential pumps in cooling should be more efficient.

It is interesting that two sequential pumps in warming should be less efficient using the same arguments! I cannot find the error in the reasoning, so before accepting the answer, please wait some time that others check and give their comments.

Answer 3

Let's say that we have a target temperature of $T_c$ and an exterior temperature of $T_c+\Delta T$ and we want to use two heat pumps to reach the target in roughly equal steps. We assume that the amount of fluid used is the same in the two to keep the change in Q and T to be proportional. The temperature at the input to the second stage is $T_c + \Delta T/2$. The energy used for the compressor is W. The heat removed in each step is $Q/2$.

The efficiency at each step is: $$ \eta_1 = \frac{Q/2}{W_1}, \eta_2 = \frac{Q/2}{W2}$$ The overall efficiency is: $$ \eta_{12} = \frac{Q}{W_1+W_2} = \frac{Q}{Q/(2\eta_1)+Q/(2\eta_2)} =\frac{2\eta_1\eta_2}{\eta_1+\eta_2}$$

One can already see that this will turn out well because, if the two efficiencies are roughly the same, then the overall efficiency is about that efficiency, and we know that taking smaller temperature steps is more efficient.

If we assume that the process follows closely to a perfect Carnot cycle, then: $$ \eta_1 = \frac{T_c + \Delta T/2}{\Delta T/2}=\frac{2T_c}{\Delta T}+1 = 2\eta + 1$$ $$\eta_2 = \frac{T_c}{\Delta T/2}=\frac{2T_c}{\Delta T} = 2\eta$$ where $\eta$ is the efficiency (COP) of a single heat pump cooling to the target temperature alone.

Substituting back into the overall efficiency: $$\eta_{12} = \left(2 + \frac{2}{4\eta+1}\right) \eta $$

An improvement of better than a factor of 2!

Answer 4

the heating COPtotal of the sequential heat pumps (hp) will be:

COPtotal = COP1*COP2/(COP1+COP2-1)

imagine a hp1(cop1 = 10) in series with hp2(cop2=10) with the resulting Q=1000W.

the work of hp2 = Q/cop2 = 1000/10=100W

This means the Qin of hp2 = 1000W-100W=900W

Qin for hp2 is Qout for hp1, the work of hp1 = Qin/cop1 = 900/10 = 90W

Now we see that for 1000W we put in 190W so 1000/190 = 5.26

So putting them sequential is not a matter of increasing the COP. You only do this because the temperature lift is to high for a single heat pump.

Answer 5

My answer is different: The equation COPtotal = COP1*COP2/(COP1+COP2-1) is not relevant to heat pumps in series, but relevant only to heat pumps in CASCADE. In heat pumps in CASCADE, the first heat pump extract heat from the Winter air... the secondary heat pump (water to water) only concentrates the heat... it does not ADD heat from the winter air... instead it extracts heat from the water already pre-heated and reinjects that heat to obtain higher temperature but the same about of energy (plus the waste heat consumed by the compressor of the water to water heat pump). In other words, heat pumps in Cascade CONCENTRATE heat, (steals heat to returns it).

With heat pumps in series, 6 heat pumps in series all take air from the winter sky, add it to a stream of water that gets hotter by 5 degrees C at every stage... if there are 6 stages, that 30 degrees C of heat is added in 5 degree steps. This heat is partly from the winter's air, and partly the heat given up by the compressor (useful waste heat). The first heat pump doesn't work as hard.... good COP because it lifts from 7°C air to 35°C water.... but the last heat pump works really hard (from 7°C to 60°C water).

You can assume the COP is only 45% of the "Carnot" efficiency. You then perform an iterative calculation in 6 steps (6 lines in Excel) calculation.... adding up to a heat pump consumption....and a resulting COP