I am not asking about the energy that is actually stored inside the BESS, or the percentage that can be recovered from it. Neither am I asking about the efficiency of inverters, or transmission losses.

I'm interested in the amount of energy that is additionally needed to operate the system, for example for cooling. Any technology welcome!

I've found the following information online:

A 1.5 MWh lithium battery for research built to minimise cooling requirements, with a claimed average cooling need of 120 W: https://www.batterietechnikum.kit.edu/downloads/2023_Datenblatt_EN_Li-Gro%c3%9fspeicher.pdf

As for the production of waste heat, I found this article and advertisement by a company that builds battery enclosures, claiming that lithium BESS produce roughly 1% to 2% of charged / discharged power as waste heat. As an example, a 1 MWh battery being discharged at 250 kW is claimed to produce about 5 kW of heat: https://www.solarpowerworldonline.com/2019/04/the-importance-of-thermal-management-of-stationary-lithium-ion-energy-storage-enclosures/

This question was prompted by a colleague who mentioned that a specific lithium BESS with a capacity of 20 MWh located in Central Europe would in February discharge at a rate of about 250 kW to supply power for its own cooling even when not doing anything else.

Given the above numbers, these 250 kW seem excessive.

  • 2
    $\begingroup$ It would be educational for some of us if you edit the question to explain "BESS". I'm guessing "battery energy storage system". $\endgroup$
    – Transistor
    Feb 16 at 0:00
  • $\begingroup$ Your question seems to conflate the power that batteries generate as heat with the power needed to actively remove that heat. These can be very different numbers depending on the way that the system designers choose to deal with heat from the cells. Can you please edit your question to separate the concepts of waste heat flow (which happens to be in watts) from power needed to deal with waste heat flow (which also happens to be in watts, but is still markedly different). $\endgroup$
    – TimWescott
    Feb 16 at 14:57

1 Answer 1


So how much extra energy does it take to collect, transport, and dump 5000 J/s of heat from the battery bank? Closed loop liquid cooling systems tend to have a COP of about 3.5. COP is kW/ton. So about 420W of power to provide 5kW of cooling - or an EER of 12ish. Well-sited systems of decent size can beat this depending on the cost and complexity of the system.

For the sake of completeness, EER is here defined as heat transfer (kW) / power consumption (kW). Both COP and EER are defined differently in different places and different disciplines. EER is often expressed as Btu/kw, and COP is sometimes expressed as kW/kW (this last one was news to me).

  • 1
    $\begingroup$ Is something mixed up here, Phil? Isn't heat pump CoP defined as "the ratio of useful heat energy pumped to electrical energy consumption". That would be kW/kW giving a unitless result. 5 kW of cooling for 420 W of power would be a CoP of 11.9, would it not? $\endgroup$
    – Transistor
    Feb 19 at 23:33
  • $\begingroup$ @Transistor That's the EER definition. Nice timing, by the way. $\endgroup$
    – Phil Sweet
    Feb 19 at 23:34
  • $\begingroup$ @Transistor Okay, something is amiss here, you are right about that. I was pulling data from different sites, and it looks like they may have used different conventions. I'll go back and try to figure out what I did. $\endgroup$
    – Phil Sweet
    Feb 19 at 23:49
  • $\begingroup$ That's fine and thanks for the reply. I keep a few "sanity check" numbers in my head and these include "a CoP of 3.5 to 5 is generally pretty good" and "a ton of cooling is about 3.5 kW of cooling continuously". $\endgroup$
    – Transistor
    Feb 20 at 0:00

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