I understand that when cooling water with LN2 directly (through pipes for example), that it can be too effective, freezing the water (and blocking the pipe).

From what I read, regulation of the heat transfer is considered complicated to control. Often employing at least a fixed insulation to regulate the rate of heat transfer.

If I immersed a fixed copper heat sink into LN2, the copper interface would be below $0C$ and still risk freezing the water.

My question is directed at those familiar particularly with Peltier cooling devices. Considering an aparatus, where a Peltier is:

  • A) interfaced directly to a flat interface on a heatsink; or
  • B) interfaced with a fixed insulation

The following sub-questions apply:

  1. Does a cooler heatsink on the heated side, improve the max. cooling rate?
  2. If turned off, would the peltier device exhibit any insulating effect?
  3. Would there be a minimum operating temperature, in the case where the Peltier was turned off and cooled toward LN2 temperatures?

Given favourable answers to the sub-questions, and any additional comments, the title question would be answered. Some helpful additional comments may relate to industry standard heat-exchanger/regulator apparatus and their performance/cost.


To reiterate the specifics of my theoretical apparatus. The "heatsink" is in the LN2 canister, the Peltier device outside with the hot interface on the "heatsink". When the peltier voltage drive is 0, the peltier would have an insulation effect R1, resulting in a Rate of heat transfer (ROHT) on the coldside of C1, changing the voltage to full on the Peltier changes the insulation effect to a negative value R2, and increases the ROHT to C2. For intermediatary voltages, the insulation effect is between R1 and R2, and the RoHT is between C1 and C2.

  1. What is R1 (the insulation effect) of a non-specific Peltier device with no voltage applied?
  2. What would be a guestimate for C1 given an LN2 heatsink?

Peltier elements available in trade are commonly rated for operating down to to vicinity of -50 °C This is still much higher than LN2. While the electric properties shouldn't change much (though the efficiency drops), such thermal stress may simply damage them mechanically.

Their primary profit in cooling applications is to raise the temperature threshold versus the ambient level: it is much easier to get a heatsink down from 120°C to 100°C in room temperature, than from 60°C to 40°C so while the heatsink runs really hot, the CPU enjoys nice, low temperature despite the additional heat produced by the element.

Now, applying LN2 to the heatsink completely negates these benefits. Your peltier element is just a fancy heater vulnerable to low temperatures.

I suggest you get a normal resistive heater instead for regulation. They are vastly more cold-resistant and you have such a surplus of "cold" that you really don't need the option to go "even cooler".


The thermal coefficient of unpowered (and shorted, as it will power itself up with Seebeck voltage even unconnected) Peltier module is around 2W/(m*K) (2)

With a typical 36x36x4mm module that would give heat transfer of about 130 Watt at temperature difference 200K (assuming -196°C of LN2, +4°C of CPU)

This is also about the limit of heat the module of this size is able to transfer when powered to Vmax, which would mean this temperature difference (200K) would be about what you'd be able to maintain between the heatsink and the CPU while powering up the module for blocking heat transfer, and so the whole contraption would work just right under the dubious assumption that the module will perform correctly.

Why dubious?

The modules are rated for operating completely under temperatures of above -50°C and with ΔТ not exceeding 100°C

They will be operating with one side in temperature -196°C and ΔТ in vicinity of 200°C You exceed both the minimum temperature rating and the temperature difference between the two sides by 100%. While the semiconductors shouldn't be affected, the plastic that binds them is not nearly so frost-resistant.

The most likely outcome is that the module will come apart at its seams, the binding plastic will shatter and crumble, massive condensation in dehermetized inside will short the circuit and the whole device will fail completely.

  • $\begingroup$ I believe you have misread my question (or I was careless in writing it). I have added another paragraph to provide another summary. $\endgroup$ – Todd Dec 19 '12 at 5:43
  • $\begingroup$ @Todd: Added some extra info. $\endgroup$ – SF. Dec 19 '12 at 9:53
  • $\begingroup$ Thanks, some helpful information there. CPU is an incorrect assumption as the device to be cooled - best to describe it generically as the "Heat Load". I think you answered my question, one could insert an additional static insulator to enable a warmer Peltier, however such discussion is not related to the posted question. Thanks. $\endgroup$ – Todd Dec 20 '12 at 1:16
  • $\begingroup$ @Todd: ...or replace it with a resistive heater. It will be easier to regulate, less expensive, more failsafe, possibly somewhat less energy-efficient and allowing even more heat transfer than the Peltier element in case of maximum throughput; Or go with dry ice, the −78.5°C should be much more survivable for the element and still a plenty to absorb the heat. $\endgroup$ – SF. Dec 20 '12 at 14:28
  • $\begingroup$ A resistive heater isn't so crazy, however this would at least be coupled with a static insulator, so such a device would be used minimally. The problem I see with dry ice is it is solid, so the interface with heat sink is a problem without an intermediary liquid. LN2 is considered the best heat sink with respect to density. $\endgroup$ – Todd Dec 21 '12 at 1:11

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