I am currently doing a small experiment using a Peltier to cool 250ml of water. My aim is to achieve the cooling of 250ml from 23degC to 8degC in under 20minuets which I have done by strapping the Peltier to the underside of an aluminium cup and a large air cooled heat sink and fan to the hot side of the Peltier.

The Peltier was consuming 4A at 6V The heatsink reached 31degC (average from top to bottom)

I am now looking at the theory behind the efficiency of it. I have read that (currently) peltiers can only achieve a maximum efficiency of 15% to me that means for every watt of power in you get 0.15watts cooling power, the rest is wasted in heat and work. When I researched how they got to that figure it came up with co-efficiency of performance (COP). I took the figures from my experiment and applied the COP formula and got somewhere around 0.47COP as COP is a ratio that means I should be getting somewhere close to 50% efficiency. Can someone explain this to me please??

I would also like to rerun the experiment after tweaking some values to find its optimum performance, but I need to address the efficiency thing first.


1 Answer 1


I am not an expert in thermodynamics but I think the following is reasonable:

When heat flows from A to B (temperature $T_a$ and $T_b$) then you could theoretically do work - efficiency given by the ratio of temperatures.

A Peltier is an inefficient heat engine running in reverse (a heat pump), and I thought the efficiency is the ratio between the work you need to do and the work you did do to move the heat.

So taking water from 23 to 8 C with a heat sink at 31C means working against an average temperature difference of 15.5 C. If principle that should mean that you can move about 300 Joules while expending only 15 (20x). In reality the Peltier is at most 15% "efficient", and can move only 45 Joules for 15 Joules expended.

Now your numbers. 250 ml is cooled by 15 degrees, requiring 15*250*4.2 ~ 15 kJ in 20 minutes (1200 seconds) or about 13 W of cooling power. A truly efficient heat pump could move that energy with 0.65 W. A Peltier that is 15% efficient would require about 3W for the job. You are applying 24 W (4*6) to the Peltier - this is 8x less than the most efficient.

Which seems reasonable.

  • $\begingroup$ Are you sure that's 1.3w, not 13? I can't get that figure... Takes 4180j to cool 1ltr 1degc, as I have a quarter of that .. (4180/4) / 20 = 13.125watts right? $\endgroup$
    – Edward
    Commented Jul 8, 2014 at 19:32
  • 1
    $\begingroup$ Sorry you are right - I was off by a factor 10. But thermodynamically you should need only 1/20th of the heat transported - so when you use 2x your real efficiency is about 2.5% (1/40). I updated my answer - still with back of the envelope calculation. $\endgroup$
    – Floris
    Commented Jul 8, 2014 at 19:51

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