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:

Material | Specific Heat (Joules/gram/Kelvin) |

Air | 1.012 |

Aluminium | 0.897 |

Copper | 0.385 |

Water | 4.181 |

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…

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**.

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!