# Feasibility of sub-zero Peltier cooler

I want to build a micro-cooler (cooling volume of 1 l) to cool a block of tofu from 25°C to -5°C within 1-2 hours.

I'd like to do this with Peltier coolers and I wanted to get some insight from anyone that has attempted this before.

Using this heat load calculator I anticipate about 5 W (assuming losses) of heat load to maintain my temperature differential.

Putting that load into this Peltier calculator I'm given a few different models that should meet my requirements. Does this make sense or did I miss something?

• Don't forget that you'll also have to remove the heat from the tofu, and if it is going to freeze you will need to find the latent heat of fusion of tofu, which will be the killer for the amount of heat needed to be removed in just 2 hours. Commented Oct 14, 2020 at 8:18
• Or the percentage of water in tofu, and the latent heat of fusion for water, which will be easier to find.
– user16324
Commented Oct 14, 2020 at 10:17
• You did not tell us the mass, which is absolutely crucial. You are ~~~= cooling water. This takes 4.2 Joule/gram/degree C. Adjust to suit. The transition to frozen (assuming that Tofu freezes to ice at -5C - which it may not is 334 J/gram for water. For the cooling, with water, you need 4.2J x 30C = 126 J/gram. For say 1000 g that's 126,000 J. In 2 hours that's 126,000/(2 x 3600) = 35 Watts. 5 Watts cools about 140 gram. BUT that's at 100% efficient with perfect insulation. If freezing add 334,000J/kg or about another 46 Watts per kg. Commented Oct 14, 2020 at 10:22
• That's not really a micro cooler, more in bar style mini-fridge territory or beyond I would say. Most small Peltier coolers struggle to moderately chill a few cans of soda in an hour or two, let alone freezing (phase change) something like a liter of water. Commented Oct 14, 2020 at 13:55

You did not tell us the mass, which is absolutely crucial.

Assuming

• Tofu and water have same latent heat and same latent heat of fusion.

• Perfect insulation.

Water takes 4.2 Joule/gram/degree C. Adjust to suit for Tofu.
The transition to frozen (assuming that Tofu freezes to ice at about 0 degrees C - which it may not, is 334 J/gram for water.

For the cooling, with water, you need 4.2J x 30C = 126 J/gram.
For say 1000 g that's 126,000 J.
In 2 hours that's 126,000/(2 x 3600) = 35 Watts.
5 Watts cools about 140 gram.

BUT that's at 100% efficient with perfect insulation.
If freezing to Tofu-ice add 334,000J/kg or about another 46 Watts per kg.

Assume 50% water and 1 kg total mass.

For 1kg of Tofu assuming 50% water and minimal other thermal material (a bad assumption probably) = 500 ml water equivalent.
Cooling 20 to -5 = 63 kJ. Freezing = 117 kJ. = 180 kJ total
180,000 / 3600 / 2 hours = 25 Watts mean over 2 hours.

BUT you cannot freeze it until you have reached 0 C OR whatever the freezing point of Tofu is. (eg saturated salt (NaCl) solution freezes at -20.5 C so freezing point depression MAY happen.

Fortunately, Tofu appears to freeze in the 0 to -5 C region (you probably already know that).
See this paper for this graph. This shows Tofu being cooled to final temperatures of -10, -30

So at 25 Watts

• you cool it to 0C in 120 x 63/180 = about 40 minutes

• and then freeze it over the next 80 minutes.

This assumes good heat transfer throughout the mass.

Note that this is the refrigeration power - NOT the electrical input power.
At say 5% efficiency that would need about 500 Watts of electrical input (!). Or 250 Watts at 10% efficiency,... .
Less if you have higher efficiency. Peltier effect devices tend to have lower efficiencies for higher cooling differentials. You need to carefully check all parameters. Also, you specify 20C as the hot temperature - presumably this is ambient temperature.

For say 250 Watts thermal output and 5 degrees C temperature rise you need a 5/250 = 0.02 C/watt heatsink. To get the hot sink anywhere near ambient at these sorts of power levels you will need a very good heat sink indeed - usually only accomplished with forced air cooling (fan). This need not take a vast amount of extra power, but needs to be allowed for. Even with fan cooling the heatsink will still be impressive".

• Ah realized I missed a lot haha. Assuming firm tofu is 50% water and I'm only looking for the water to freeze that should be 500g of water so 17.5W. Commented Oct 14, 2020 at 17:24
• @GurshanDeol For 1kg of Tofu assuming 50% water and minimal other thermal material (a bad assumption probably): So 500 ml water - Cooling 20 to -5 = 63 kJ. Freezing = 117 kJ. = 180 kJ total = 25 Watts mean over 2 hours . BUT you cannot freeze it until you have reached 0 C OR whatever the freezing point of Tofu is. Saturated salt (NaCl) solution freezes at -20.5 C . || Tofu appears to freeze in the 0 to -5 C region (you probably already know that) See this paper ... Commented Oct 14, 2020 at 23:32
• ...for this graph . || So at 25 Watts you cool it to 0C in 120 x 63/180 = about 40 minutes and then freeze it over the next 80 minutes. This assumes good heat transfer throughout the mass. Commented Oct 14, 2020 at 23:33

So the other answer which delves into the calculation side is very good, and use that for a theoretical basis for your design. I'm not very good at theory, so I experiment instead.

I'm currently designing a printbed for a commercial 3D-printer that is supposed to be able to cool the print down to 0 degrees C. In practice this means that the surface of the printbed will need to reach sub-zero temperatures.

I can tell you straight away that for a block of tofu that large, if you only cool from a single direction (i.e. from the bottom as you do with a printbed for example), your main problem will be that you will be able to cool to bottom part of the tofu to sub-zero without any larger issues, but the top part will not become very cold. I can get my printbed surface down to < -10 degrees C without much issues, and then I'm feeding ~120W into it, with a good heatsnik and a 10W fan on the bottom side of the Peltier element. But since the object I'm trying to cool is exposed to the ambient surrounding, the top part isn't affected much by the cooling.

If you really want to coold the entire block of it you will need to cool from several sides. I've done Peltier cooling with two blocks that squeeze around the object to be cooled, but for your relatively large volume you might even need to cool from even more directions, or build a completely enclosed volume to cool it in (like a refridgerator).

When it comes to component selection, don't underestimate the value of a really good heatsink. Heatsinks, in my experience matter more than fans, increased airflow across the heatsink has very rapidly diminishing returns. If you want to increase the performance even more, go for liquid cooling. We did an experimental build with PC liquid cooling on our Peltier elements, and the performance gain was quite massive over air cooling, but you will always have some leakage somewhere with liquid cooling.

For the Peltier elements themselves, due to construction & physics, Peltier elements designed to run at ~12V seem to be slightly more efficient than elements designed for 24V. I'm not entirely certain why, an expert explained it to me but I didn't fully grasp it, more than that "this is just the way it is, due to how they are designed & built".

For electrical components, if you want to get started really quickly with experimenting, get a somewhat beefy H-bridge evaluation board that you can feed PWM control inputs to. Hook it up to an Arduino and a power supply, and then start to experiment. Also, never run a Peltier element without cooling it, if none of the sides are cooled it will VERY quickly self heat and destroy itself. Peltiers are very fragile.

EDIT: Also, remember to use a thermal paste that is suitable for sub-zero temperatures. We use Thermal Grizzly Kryonaut for example.