Refrigeration using Peltier device

Emanuel · October 24, 2012

Hello everyone, I'm carrying out a project which provides the cooling of a fluid inside a tube. In particular I should cool water or a glycolytic solution from a temperature of 25 °C to about 0 °C. I thought to do this through the use of the thermoelectric effect, then the Peltier cells. I have two ideas. The first involves the use of a fluid reservoir and the cold face of the Peltier cells is in contact with the bottom of the container (figure). I thought I could get at the bottom not just one but multiple cells . The second one is to use a system which is a kind of enclosure of the tube in which the fluid flows (figure). I can give you some more information: Case 1) deltaT = 25 °C -> 25 °C to about 0 °C fluid volume / container -> 1 liter container size -> might be 10 cm x 10 cm x10 cm thick container -> 1mm steel Calculating the heat subtracted from the volume of water: Q = 4186 * 1 * 25 = 104kJ. So if I want that the refrigeration is about 10 minutes I have to subtract cooling power of 104kJ/600s = 174W. Because of the dispersion may well need more than enough, say 200W (but it is a random number, I have not done any calculation). With cells taken, for example, I run the following calculations: I took that model because one side is 47mm and the bottom of the container there are 4. Their power is 72W and the maximum temperature difference of 74K, which means that for each kelvin of difference between the temperatures of the two faces is lost approximately 1W. The thermal resistance of the steel wall is negligible. 4-cell 72W make 288W, 200W serve to us, means that we have 88W for the internal conduction of the cells, which means 22W per cell; then the maximum temperature difference between the two faces must be of approximately 22K. Since the bottom of the container is at about 0C, the other face of the cells at the maximum can go to 22C. If the air is at 20C, I have a temperature difference of 2K to dissipate from the area to 1dm^2 a power equal to 4x (13.1Ax8.8V +50 W) = 660W. Then I need a heatsink with thermal resistance of 2K/165W = 12mK / W for each cell. Or a single heatsink of 2K/660W = 3mK / W. CALCULATION ARE RIGHT? Case 2) flow -> 1l/min deltaT = 25 ° C -> 25 ° C to about 0 ° C wall thickness -> 0.1mm In this case I would understand if it is possible and if the cooling depends on the length of the tube and such other factors. I know that in the market there are some chillers that use the traditional refrigeration cycle but also the thermoelectric effect. Also, if someone knew other methods to achieve my goal can tell me please. Thank you to all, Emanuel

Excuse me for the previous message.. This is the correct message..

Hello everyone,

I'm carrying out a project which provides the cooling of a fluid inside a tube. In particular I should cool water or a glycolytic solution from a temperature of 25 °C to about 0 °C. I thought to do this through the use of the thermoelectric effect, then the Peltier cells.

I have two ideas. The first involves the use of a fluid reservoir and the cold face of the Peltier cells is in contact with the bottom of the container (figure). I thought I could get at the bottom not just one but multiple cells .

The second one is to use a system which is a kind of enclosure of the tube in which the fluid flows (figure).

I can give you some more information:

Case 1) deltaT = 25 °C -> 25 °C to about 0 °C

fluid volume / container -> 1 liter

container size -> might be 10 cm x 10 cm x10 cm

thick container -> 1mm steel

Calculating the heat subtracted from the volume of water: Q = 4186 * 1 * 25 = 104kJ. So if I want that the refrigeration is about 10 minutes I have to subtract cooling power of 104kJ/600s = 174W. Because of the dispersion may well need more than enough, say 200W (but it is a random number, I have not done any calculation).

With cells taken, for example, I run the following calculations: I took that model because one side is 47mm and the bottom of the container there are 4.

Their power is 72W and the maximum temperature difference of 74K, which means that for each kelvin of difference between the temperatures of the two faces is lost approximately 1W.

The thermal resistance of the steel wall is negligible.

4-cell 72W make 288W, 200W serve to us, means that we have 88W for the internal conduction of the cells, which means 22W per cell; then the maximum temperature difference between the two faces must be of approximately 22K.

Since the bottom of the container is at about 0C, the other face of the cells at the maximum can go to 22C. If the air is at 20C, I have a temperature difference of 2K to dissipate from the area to 1dm^2 a power equal to 4x (13.1Ax8.8V +50 W) = 660W.

Then I need a heatsink with thermal resistance of 2K/165W = 12mK / W for each cell. Or a single heatsink of 2K/660W = 3mK / W.

CALCULATION ARE RIGHT?

Case 2) flow -> 1l/min

deltaT = 25 ° C -> 25 ° C to about 0 ° C

wall thickness -> 0.1mm

In this case I would understand if it is possible and if the cooling depends on the length of the tube and such other factors.

I know that in the market there are some chillers that use the traditional refrigeration cycle but also the thermoelectric effect.

Also, if someone knew other methods to achieve my goal can tell me please.

Thank you to all, Emanuel

Enthalpy · October 26, 2012

Hi Emanuel, welcome here!

I agree approximately with your computation of Case 1...

- 660W sink for useful 174W explain why Peltier elements fit this purpose badly. They serve for sensitive cameras, where vibrations aren't acceptable. Can you get the cold machine from an old fridge?

- This design is marginal! 2K could be -2K and anyway, they are too little for the sink. Alone the contacts between the metal and the Peltier will drop more. Accept more than 10min, or put more Peltier, to get margins...

- Where does the 13.1A*8.8V+50W come from? The doc of your Peltier element probably gives a voltage that depends on the temperature difference, and this should integrate the heat soaked from the cold face. Check if it gives a resistance as well.

- Heat exchange within the liquid is bad unless you mix it. Especially if you cool the bottom, you stop convection. Difficult to compute, but much more important than the metal wall.

In case 2, is the 0.1mm wall thickness corrrect? Anyway, heat exchange within the liquid, not wall thickness, will limit the cooling. This is a seriously difficult topic, and alas, a necessary one. Either you experiment, or you take a hard time learning it. The only book I know is in French: "Introduction aux transferts thermiques" by Sacadura, and it's nothing simple. In fact, it's fluid mechanics, but worse. Expect weeks of headaches if you're a physics-oriented engineer, before you give up learning properly and just suppose to understand how to exploit the diagrams.

The experimental option means: make internal corrugation in your tube, for instance by drilling many narrow holes in a piece of copper or aluminium as a replacement for the tube; observe and improve if needed. This shape allows you to put more Peltier elements if needed.

Remember the heat conductivity of an alloy (especialy stainless steel) is much lower than the pure base element, and remember that corrosion depends on the alloy and most aluminium alloys are very bad against corrosion.

As for the size of your cold sink: the heat power you want to remove is similar to a house fridge, and these accept a temperature drop much bigger than 2K, so have a look a their size! I regret to tell the pictures are far off. A fan can improve that.

Edited October 26, 2012 by Enthalpy

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