Water cooling with Peltier – Part II

I recently explored the world of Peltier modules with water cooling in this article: https://deavid.wordpress.com/2019/11/18/water-cooling-with-peltier-worth-it/

In it I explained a bit about Peltiers and did some basic calculations for a basic setup. But I promised more, so here it is. If you haven’t read the previous one, go for it now.

As I outlined in the previous article, I tried lots of arrangements for cooling with Peltier modules to see if I can get something better, or at least, different from the usual setup.

The typical setup, which I called the closed cycle, looks like this:

After trying and trying different arrangements, I found one that is orthogonal, I call it the open cycle:

Don’t be fooled, it’s not good at all. But hey, it’s different! And because of this I think it’s fun to see the math (it’s a bit more complicated here) to see what’s the real problem.

Despite it’s simpler looks, it’s quite hard getting it right. Hope you followed the previous article on calculating the closed cycle, I’ll be doing the same for this one here.

As before, I’ll be targeting 100W of cooling power, but with a slight difference: Instead of targeting 20ºC dT, I’ll be targeting 5ºC dT or less and cool it in several stages to reach the same 20ºC dT on the CPU. The reason is the COP can reach up to 6.00 in those scenarios, and we don’t have to account for losses of stacking Peltier modules in this design.

With the closed cycle, if you want more dT, you need to stack peltier modules, and the overall performance of all of the modules gets divided by the number used. Let’s see what happens in this “open” design.

100W with four peltier modules

To be fair, let’s use exactly the same amount of peltier modules as before, so just four. The temperature from the radiator is assumed also to be 35ºC. The target is getting the CPU down to 15ºC.

Now we know that 4 modules need to remove 20ºC, so that’s 5ºC each. Here appears our first problem: the amount of dT here is not related to the dT used on the Peltier, because we’re not comparing hot sides with cold sides. In fact, it’s the amount of heat removed from the water.

If we apply 100W of cooling power to water, how many degrees it will drop? It depends on three things: how much water is there to cool down, how long it stays inside the peltier and the specific heat of water.

Water is one of the liquids with highest specific heat, meaning that one kilogram takes a lot of energy to heat up compared to other materials, like cooper. Cooper is best for heat transfer, but heats up quickly compared to water.

If we go to Wikipedia, it has a nice page with a Table of specific heat capacities:

MaterialSpecific Heat (Joules/gram/Kelvin)
Air1.012
Aluminium0.897
Copper0.385
Water4.181

We have Watts, but we need Joules. Joules is just Watts*second, so we need to get how much time this water spends on the Peltier; assuming a CPU block on top of it, how much time takes to exit it.

Also, we need grams, so we need how much water is inside the water block.

I’m going to do some wild guesses but we’ll see in a moment how it doesn’t matter because everything cancels out:

Assume a cooling block of 40x40x16mm, internal space: 36x36x12mm. Fraction of this volume that is used by Water: 70%

This gives a volume of roughly 10cm³.

For the formula we need grams, not volume. Luckily, the density of water is more or less 1gram/cm³, so it’s just 10 grams of water.

Because the same water goes through all four peltiers, and all four would have the same size, it is safe to calculate just one cooling block with 100W of cooling capacity. But for this, we need to know how much time it takes for the water to go through it.

What moves the water? the pump. So knowing how fast our pump is we’ll know how much time it takes to exit the cooling block. I researched for a bit and I settled for a conservative 200L/h for the pump. That’s 55.56cm³/s. So plug in the numbers and we get:

10cm³ / 55.56cm³/s = 180ms

This is the time spent in the cooling block. So now we can get the amount of cooling in Joules:

100W * 180ms = 18J

Now we add the Specific heat, and we get the change in temperature:

18J / 4.181J/g/K / 10g = 420.52mK

This is a change of temperature of 0.42ºC; So it goes from 35.0ºC to 34.58ºC, not a great deal, right?

Why is it so bad? Aside of the poor design choice, can we improve it tweaking something?

We could make the cooling block half its size, so we have half the water to cool, so it should be double the cooling power, right? Well, no.

If we have 5cm³ in volume and 5g of mass, then it will exit the block twice as fast, in 90ms. These two cancel out. What else can we tweak?

Obviously we could add several kilowatts of cooling power, but that’s not the point here. What else?

If instead of water we used other liquid that had less specific heat, it could be cooled down much quicker. Just I didn’t found any from Wikipedia that’s not highly toxic or corrosive. This just leaves one thing remaining: The pump.

Exactly, if we pumped much, much slower, the water would cool down to almost any value.

For example, if we bring our pump to a 1% load, the water will take to exit the block 18 seconds, and when it passes all four blocks it will drop by 42ºC, bringing it to -7ºC, converting it into an ice block. To avoid frosting we’ll need some kind of special coolant, like the ones used on cars maybe. Or…

Do not try these things at home!

So well, then we’re set and this is awesome, right? Frosty CPU, this design rocks!… er, no. Still sucks.

If 100W of cooling power moved the liquid down by 42ºC, by how much do you think it will heat up when it passes through the CPU, adding 100W? As the CPU block size does not matter, the answer is 42!

So leaves the CPU at the same 35ºC that came from the radiator; this is still an improvement, as usually would have been hotter. But then it returns to the hot side of the peltier modules where it receives the 100W of heat we transferred from the cold side, and we get another 42ºC up! But wait, we haven’t finished, we need to add the Peltier losses, which are?… well, because the hot side is now 42ºC hotter than the cold side our COP is around 0.4, so another 250W of energy into the water, which are 105ºC on top of the 42ºC, so 147ºC.

Resultado de imagen de nuclear explosion
Thought this design was safer…

What happened? The pump ran too slow and allowed too much temperature difference between the sides of the peltier. To catch up, the Peltier had to run at COP=0.4, which created a huge amount of heat that in turn, because the pump was going too slow, it accumulated, boiled the liquid and the system probably exploded.

“Fixing” the disaster

This needs some tweaking. We want to target a COP of 5.0, so the CPU is only allowed to heat up the liquid by 5.0ºC; If the peltier modules need to cool down the same liquid by 20ºC, it means they need at least four times the cooling performance of the CPU thermal output.

The problem is that we will need to run the modules at 4V, giving around 30W or maybe less. Our target for cooling is now 400W, so at least 14 modules are needed. Let’s make it 15 to get 450W of cooling power.

As we will be running at a COP of 5.0, the amount of energy used by the modules will be 90W. It’s not that much compared to the 71W of the build in the previous article. And this one is rated to drop 22.5ºC in temperature (because of the extra peltier added)

Now we will need to work out which is the optimal pump speed so 100W equates to 5ºC dT.

If 100W @ 200L/h => 0.43ºC, then if we set the pump at 17.2L/h will do 5ºC dT for 100W.

The waste heat from the peltier modules will be 90W (4.5ºC) divided across all 15 modules, so 0.30ºC of extra heat for each one.

Let me draw a diagram of what’s happening:

As you can see, the liquid is cooled in one direction and then reused as a capture system for the transferred heat. The CPU gets the cold part and the radiator the hot one. This way the radiator is able to cool down the CPU bellow room temperature.

Isn’t this bulky and expensive? Yes, of course. Unfeasible for simply 100W of cooling power. Again, if interested, check the previous article where I explore the other design which is the typical one, cheaper and at least it made some sense.

I have chosen 15 peltier modules instead of 14 because is divisible by 3, and the power the Peltier needs here is 4V, which is 1/3 of the 12V rail, so it’s easier to group them in batches of three at 12V per batch, so no controller is needed. At this low voltage, if the modules reach their limit dT (20ºC) they barely work, releasing almost no waste heat.

The peltier modules in this design can be turned off and the pump back to 100% if we don’t need the extra cooling power, for example when idling.

One thing I missed on this analysis is the radiator efficiency. Because the pump goes 10x times slower, the liquid now has 10x more time to cool down, so it’s probable that the output temperature of the radiator is much closer to ambient than would have used to. I don’t know where to get data from to calculate it, so I think its best to just assume 35ºC as before.

If we look carefully it’s a strange coincidence that we get numbers so close to the previous time. It looks like it’s more or less the same but performed in the opposite, hard way.

The conclusion for this thought experiment is: math is nice, it let’s you play with imaginary systems and predict the outcome with some degree of precision, without actually investing money and time to build those and ultimately fail. Hope you enjoyed as much as I did!

Water cooling with Peltier, worth it?

I have been intrigued again why Peltier modules are never used in builds for refrigerating computers bellow ambient temperatures and I checked again the web and YouTube with not much results. No one explains why these aren’t useful, all I see is: They’re not efficient. Of course they’re not, so what?

Linus shows what would happen to the average consumer if tries to get into Peltier
der8auer shows how to build it if you’re into electronics

As I am very lazy and I don’t want to start and spend lots of time and money trying to make them work. So I thought “let’s work out the math and see what the theory says”. And… oh crap they’re right; it’s not worth the hassle. And if the theory is bad, in practice is going to get worse.

But I’m getting ahead of myself. Let’s get back to the basics.

What is a Peltier module?

Peltier modules, oversimplifying, are refrigerators without moving parts. Just apply some current on the cables, and the heat will move across, effectively cooling one side and heating the other.

A common misconception is that refrigerators or peltier modules create cold. Instead we should say that they move or transfer heat. Peltiers have two sides, one will heat up and the other will cool down. Because they’re far from 100% efficiency, the hot side will heat up way faster than the cold one gets cold.

Speaking of efficiency, one term that appears a lot in datasheets for these is the COP, or coefficient of performance. The easy way to understand it is, for every Watt consumed by the peltier, how many Watts of heat move from cold to the hot side. Don’t be fooled if you see a COP of 3 or 5; We’re used to see efficiency from 0% to 100% but COP goes from 0 to Infinity those being 0% and 100% efficiency respectively.

I’ll put down a table for reference:

COPEfficiency
0.2520%
0.5033%
1.0050%
2.0066%
3.0075%
4.0080%
5.2584%
9.0090%
99.099%

And also a graph, I love graphs!

Actual formula: 1/(1-x/100)-1

I guess this is clear. COP goes to infinity when efficiency is 100%.

Next thing I want to talk is about “dT” or temperature differential. It’s the difference in Kelvin between sides of the peltier. Because Kelvin and Celsius are the same scale with different starting points, I just use Celsius when talking about dT.

Peltier modules efficiency heavily depends on the temperature differential. If you try to make both sides too different in temperature, the COP will reach zero, meaning your peltier is wasting electricity and outputting heat without actually doing work (i.e. cooling). For example, most Peltier modules can handle up to 50ºC dT more or less. This means that if the hot side is 40ºC, the cold barely will reach -10ºC, and from there you’ll be using a lot of power without being able to cool anything.

But it gets more complicated, as the dT increases, the COP will decrease, and at some point before the peltier limit it will not make any sense to cool it further down. Does it make sense to use 1000W to cool down 100W? (COP=~0.1)

I want to insist that regardless of adding a cooling power of, say, 100W, you still have to cool those down, plus the extra heat produced by the Peltier itself using conventional methods. Heat is transferred, not removed. You’ll need radiators to transfer this heat into the air regardless.

The other factor that affects the efficiency is the load of the module, or the amount of heat per second we want to transfer. As we go higher, the efficiency decreases a lot.

Actual performance on Peltiers

These modules are known to be less efficient than conventional gas compression, like it is used in A/C, refrigerators, freezers, etc. No one is thinking on creating Peltier version of these, unless they’re meant to be really portable and work under batteries, it’s pointless.

My guess is that nowadays Peltiers are more or less the same regardless the manufacturer, as long they’re not cheap ones. It is an old technology and has seen lots of uses elsewhere. So I’ll use a particular module from a particular provider for examples here, basically because they included nice datasheets where I could extract lots of useful information.

For sake of reference, here is the page where I’m looking for those:
https://tetech.com/peltier-thermoelectric-cooler-modules/high-performance/

And here is the datasheet for the most powerful one:
https://tetech.com/wp-content/uploads/2019/03/HP-199-1.4-0.8.pdf

I have chosen the most powerful one (172W) because the other ones perform more or less the same in terms of efficiency, and because we need to load them to less than 50% to be more or less efficient, this one is the most sensible even for small setups. Big setups will need several of those.

Probably the most important number to get from the datasheet is the maximum efficiency we can get for any given temperature difference:

Note the vertical axis is on logarithmic scale; I love these because it’s easier to read; but the fact that it’s logarithmic means the actual shape is a curve. Also, it’s hard to see on the graph, but it peaks on COP 5-6 (hard to tell exactly on the datasheet) for a dT of 0ºC. From there it falls very quickly. When gets on 45ºC, starts really to move towards zero. The maximum dT for this module is 65ºC.

This graph corresponds to 45W cooling power approximately, which is 26% of the maximum of the module. As most modules have the same behavior, probably you can assume for any Peltier that 26% load is more or less the best for efficiency. The datasheet actually places the optimum load for efficiency at 30W, but there is not much more data to back it up. 45W is already really low, so I’ll go with that.

As you might ask, how worse it gets when it delivers more power? Well, the datasheet offers a good graph for this:

As you can see, the efficiency falls logarithmically as we increase load from a certain point, and this point is the maximum COP I showed earlier. So it’s not really a good idea to increase the power by much, as we will spend a lot of energy.

How much energy? Well, the lines converge at COP=0.25 more or less for a 40ºC dT. This means that for removing 172W of heat we will be spending a whooping 688W of energy, and we will need a power supply and a cooling solution for 860W. Not bad, if our plan is to build an expensive room heater!

For a COP=2.00, the same 172W would have needed only 86W, but because our peltier only cools around 50W at this efficiency for 9.6ºC dT, then only 25W would be needed for powering the peltier. Much better.

With this I think it is clear that we need to stay as close as possible to the maximum COP if we don’t want to build a heater instead of a cooler PC.

Possible Peltier builds

I tried (again, on theory only) lots of different ways of connecting the Peltier. I want to make use of water cooling for optimizing heat extraction and temperature stabilization. Peltiers directly stacked on top of the CPU have the problem of being hard to control because there’s not much thermal mass. Adding a water loop for the cool part of the peltier adds lots of water which it’s hard to warm or heat up, so we should be able to throttle temperatures easily.

From all schemas I tried, I ended with two completely opposite ones. Everything else seemed to be either a variant or less effective. I like to call them “open cycle” and “closed cycle”; probably it’s a pun to rocket engines, but I like the names regardless.

For closed cycle, which is the typical setup, the Peltier is alone removing heat from the CPU in its own loop, and the main loop removes heat of the peltier. Here’s a diagram:

The arrows describe the flow of water; as you can see there are two loops, one for hot water and another for cold water. I didn’t place the reservoir as doesn’t change much and sometimes is included on the pump itself. For the colder loop, a tiny reservoir is recommended, as increasing the surface will leak cold to the hot environment.

You could place the peltier on top of the CPU, but this setup allows you to put several Peltiers one after another, and Peltiers are usually bigger than the CPU socket.

The problem with this setup is, if the Peltier fails, or the CPU outputs more heat constantly than the Peltier can remove, the cold loop will become hotter and hotter until the water boils from the CPU heat. Hopefully the CPU will throttle before that and prevent the whole thing from exploding.

The good thing on the setup is, if the Peltier outperforms the CPU, the system will get colder and colder until the temperature difference from both sides of the peltier is too big so the CPU outperforms and you get an equilibrium. If you run this at the maximum power for the Peltier, the dT will reach around 40-50ºC eventually (when the CPU idles), and if the hot side is 40ºC, the cold side will be at -10ºC, freezing the water. Even before, you’ll get severe condensation, so be careful.

I wanted something that does not rely entirely on the Peltier so in case of failure the radiator can take care of the heat of the CPU. This is the open cycle:

There is one single, shared loop, where the same water that has been cooled down it’s used to get the heat from the Peltier after the CPU has used the cooling power.

Depending on who you ask, at first glance it will look like an awesome idea or a terrible idea. After doing some math, it seems to be really bad; but I’ll show later because it’s interesting. And no, it’s not wasting energy besides its inability to build up cold on the CPU side. It can be tweaked to work, but feels… strange.

So as just said, the problem here is that it’s unable to build up cold in the CPU when the Peltier outperforms the heat, but also it does a negligible work when the Peltier works bellow the CPU heat output. So not really worth it.

The only good thing is that the Peltier can stop working at any time and the radiator will take over. It is a very safe setup.

From here it looks we have a clear winner, so let’s work out the math for it!

Closed Cycle build

I tried to model a ThreadRipper with 250W TDP, but it gets really crazy. Don’t even bother to ask to add as well two GPU, which will get you near 1000W without peltier, that gets clearly out of range for a peltier build requiring insane amounts of cooling and power. Unless you plan to double it as a room heater and you plan to use a case like this one:

Raijintek Enyo – the biggest monster in cases I’ve ever witnessed

So I settled with a consumer CPU with a 65W to be cooled with our peltier, we want to use the extra room in cooling to give it a bit of overclocking, so our target is 100W of cooling power.

The first thing we need to know is at which temperature the water gets from the radiator to the Peltier. Basically, the lower it is, the better it will work everything; so put the biggest radiator possible. I will assume it will cool down to 35ºC. Then if we target for a 20ºC dT in the Peltier, our CPU should sit on 15ºC. This should be good enough for some extra overclocking while avoiding the dew point on most climates.

At 20ºC dT the Peltier has a maximum COP of 1.4 at 7V, which gives around 25W of cooling power. This is fine, we will just use four Peltiers instead of one for 100W of power. The peltier modules will use 71W additionally, so the radiator has to work with 171W instead. (The radiator has to be capable of several times that so it can bring the temperature down enough in order reach 15ºC on the CPU).

Now we’ll need temperature probes, a controller, and a regulator. When the system reaches the desired temperature it needs to stop to prevent frosting and condensation from happening. Instead of actively regulating the voltage, which is complex, we could just wire each 2 peltier modules in series and if we get power from the 12V rail, each peltier will get 6V. This way the regulator only needs to power on and off the modules based on temperature, which can be done with MOSFET or relays; both are really efficient. Actively regulating the voltage incurs in losses and complex circuitry that we might want to avoid for simplicity; but there are solutions built already:

Temperature Controllers – TE Tech Products

If you go with one of these, it should be able to regulate the power precisely, giving the best efficiency for any scenario.

As you can see, it seems to work, so why everyone says “don’t do it?”. Well, it’s not worth the hassle.

First, if the radiator was able to bring down the 172W to 35ºC, it’s probably capable of getting the raw CPU with 100W to 32ºC, which is really good already.

Second, the cost: 170€ for the Peltiers, 200€ for the regulator and 30€ an extra pump is going to be more or less 400€ that you might have spent on a better CPU that surely would outperform your current overclocked CPU.

So, what if there’s no better CPU? Then we’re targeting to cool down around 400W, so add another 286W for the Peltier. On cost, you’ll need sixteen instead of four, so that’s around 930€ on the Peltier solution. And your electricity supplier is going to be really happy about that as well, that’s definitely a good thing, right?

This type of build will suit only to some enthusiasts, more for curiosity and playing around than raw performance. Hope it helped!

P.S.: I haven’t forgot about the calculations for the open cycle design before, I will add an update just for those soon.