I'm trying to develop a thermoelectric (Peltier tile-based) beverage cooling system. Ideally, I'd also like a device to solve the warm beer problem if the thermodynamics give me any confidence.

First, I'll detail my thought flow in the theory sense and then I'll get to my specific setup that I had in mind.

Obviously, if one immerses a 100 W, 10% efficient Peltier tile in a glass of water at 298 K (let's say, containing 10 mol exactly), if the container system is assumed to be adiabatic in nature, to drop the temperature to 274 K (roughly 35 °F) requires:

$$ Q = m \, c_p \, \Delta T = 180.2 \times 4.1813 \times 24 = 18083.3 \:\mathrm{J} $$

Then, since the efficiency gives the nominal "heat transfer" power to be 10 W:

$$ t = \frac{18083.3\:\mathrm{J}}{10\:\mathrm{J/s}} = 1808.33 \:\mathrm{s} $$

So, the cooling would take about 30 minutes to chill the glass down, much more when you consider that a glass of water doesn't adhere closely to adiabatic behavior.

The actual setup (poorly-drawn hand sketch follows):

The experimental setup

Now, obviously there are four sets of values to consider here:

  • The specific heat and thermal conductivity of the beverage itself
  • The specific heat and thermal conductivity of the glass container holding the beverage, easily approximated by making the assumption that the glass can be characterized by fused silica
  • The specific heat and thermal conductivity of the water transfer fluid, well-established
  • The specific heat and thermal conductivity of the 6061 aluminum working chamber, which is again well-established

My problem is how to model the system and get model calculations for the transfers between each step, so, for example, solving the equation so I know if I use a given cumulative wattage of Peltier tile, it'll reduce the temperature of the system by a given amount in a certain amount of time.

As before, I'm willing to concede, for the purposes of discussion, that this system can be assumed to be adiabatic with respect to the external environment; that is, no heat is absorbed from the environment during the cooling process.

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    $\begingroup$ you are missing how much heat the pelier cooler will radiate on it's hot side, given a tempreature. $\endgroup$
    – mart
    May 3, 2015 at 19:49
  • $\begingroup$ I'm not missing that, I'm just not interested in making that part of the model. Effectively, once heat gets transferred to the Peltier cooler, I don't mind what happens to it as long as it doesn't go backwards. For the sake of discussion, I'm planning on cooling the tiles at potentially a much higher rate than the rate at which they could pull energy from the system. $\endgroup$
    – ecfedele
    May 3, 2015 at 22:58
  • $\begingroup$ FWIW on the cooling side the "figure of merit" will be the K/W (temperature differential per Watt) rating of the heat sink. Even though you do not wish to consider it at present it is a non-trivial part of the real world problem. Heat energy out is about 110 Watt. Delta T desired is 24 K. Every degree K drop across the heatsink comes off the cooling budget for a given external ambient sink. To get 1 degree K delta at the HS you need 0.01 K/W HS which is about undoable. Somewhere in the 0.01 - 0.1 K/W range becomes conceivable but even a 0.1 k/W HS is better than you'd get passively .... $\endgroup$ May 4, 2015 at 9:28
  • $\begingroup$ ... without humungous metal work. So you are into blown air (at east), possibly fluid circulation and probably heat pipes to get the energy to where the HS can get at it. All this is if you want the HS to effectively be able to be ignored. Mere mortals usually accept a more significant delta T and factor it into the requirement. | A starting point for modelling, which is liable to be "not too bad" is to model each portion of the thermal path as an equivalent thermal resistance with Rth proportional to k.t/A (t = thickness of material, A = area , k = constant based on material. I suspect .... $\endgroup$ May 4, 2015 at 9:33
  • $\begingroup$ ....the Al bucket will have negligible effect, the glass not as bad as you might expect as long as as thin as would be sensible, and water non-homogeneity is probably more important than thermal resistance - ie if you really are providing semi infinte cooling then stirring the water probably makes sense. | Finger in air feel suggests a few degrees (say 1 to 10 depending on stage of cooling) differential between water and target fluid. | Something I've not seen done - but I assume it is, is to use heap pipes to get the Peltier coolth to where it belongs. If you are allowed to open the .... $\endgroup$ May 4, 2015 at 9:38

1 Answer 1


There are two wattage ratings for a peltier plates. One is the power consumed, the other is the heat transported across the chip when there is a temperature difference of zero. An important note is that the amount of heat transported across the plate is dependent on the temperature difference across the plate.

Peltier graph

Image courtesy of: http://www.heatsink-guide.com/peltier.htm

Using the graph above as an example, if the hot side of the peltier plate is not attached to a heat sink then it will quickly heat up to 70 degrees hotter than the beverage unit and heat will stop flowing outward. At that point if this plate was 10% efficient it would still be producing out about 400W of heat. That heat would be produced on the hot side of the chip* but that means that without a heat sink of some sort the hot side would quickly become even hotter than 70 degrees above the beverage unit and heat would start to flow back into the beverage unit, heating it rather than cooling it.

Thus the heat sinks on your beverage "cooling" unit will be very, very important to how well your system cools (or heats) your beverage, and without any information on them my guess is that your system will heat your beverage.

*The heat is actually produced throughout the chip but it gets transferred out along with the other heat and the graph is compensated to account for this extra heat transfer, and only show the heat transferred away from the cool side.

  • $\begingroup$ Ah, so the overall system would be only as good as the ability to remove heat from the Peltiers. That makes perfect sense. $\endgroup$
    – ecfedele
    Aug 3, 2015 at 20:04

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