B Lorenz COP of a heat pump and the temperatures of the hot/cold reservoirs

[Devin-M]

@Dale : “The COP of a heat pump and the efficiency of a heat engine both depend strongly on the temperatures of the hot and cold reservoirs. For this calculation you need to go back and check the sources for the temperatures corresponding to each of these numbers. You will find that the Stirling engine numbers are for a much bigger temperature difference than the heat pump numbers.”

I did some new calculations and trying to find the error in my calculation.

I did what was suggested in the other thread and looked at how temperature differences affect the operating efficiency of stirling engines and heat pumps.

I was using this as reference for the heat pump efficiency:

https://backend.orbit.dtu.dk/ws/portalfiles/portal/149827036/Contribution_1380_final.pdf

I was looking at page 4, Table 1 - “Table 1: Operating temperatures for typical applications and theoretical maximum COPCar and COPLor based on source and sink heat exchangers…”

I’m not sure if I understood the table properly, but I was looking at item “Heat recovery from process waste water to heat up tap water” and I believe it indicates a heat source temperature of 28C and an output (hot side) temperature of 90C. It also indicates a lorenz COP of 10.36.

I then went to a stirling engine efficiency calculator located at:

https://www.mide.com/ideal-stirling-cycle-calculator

When I input those temperatures (leaving the other values unchanged) I obtain a predicted efficiency of 17.07%.

When I multiply the lorenz coefficient of performance (10.36) times the stirling engine efficiency (0.1707) I get a value of 176.84%.

Surely this can’t be correct? That would imply that if the compressor in the heat pump was powered by the stirling engine, the stirling engine would be outputting 76.84% more mechanical power than what is used to power the compressor.

 

[Devin-M]

Also, from the same table, the item labeled “milk cooling…”

The hot side (output) after the compression stage (as I understand it) is labeled 65C and the input (heat source) is labeled 20C, with a Lorenz COP of 10.49. When I run those temperatures through the same stirling engine efficiency calculator I get 13.31% efficiency.

10.49 x .1331 = 1.39

Does it mean the stirling engine can output 39% more power under these conditions than is required to operate the heat pump compressor?

 

[Devin-M]

I suspect I found my mistake...

The "logarithmic mean" temperature of the hot side is lower than the "arithmetic mean" temperature of the hot side (same is true for cold side). If I use the logarithmic mean temperatures instead of the arithmetic mean temperatures to calculate the stirling efficiency and then multiply it times the lorenz COP, the answer is always exactly 1, never above 1.

 

[Devin-M]

My only reservation with using the logarithmic mean to calculate the average temperatures of the heat exchangers is this wikipedia article which states one of the limitations with using the logarithmic mean to estimate heat exchanger average temperatures is that it assumes:

“…No phase change during heat transfer…”

…which is not typically the case with a heat pump…

Logarithmic mean temperature difference - Wikipedia​

en.m.wikipedia.org