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Get ready for a question about an absurd idea I'm trying to determine the feasibility of. I want to both heat and cool a flat 3mm aluminum sheet of dimensions 50x50cm (half meter squared). If cost isn't an issue, and power considerations aren't either, is there a way to wire 100 Peltier plates (with dimensions 50x50mm each) together in a grid to provide temperature control over the aluminum sheet? I was thinking of using 12V, 30W ones.

The aluminum sheet would have to be cooled to 10 degrees C and also have the capability to heat the aluminum sheet to 80 degrees C. There needs to be uniform temperature on the entire surface of the plate within plus or minus 0.2 degrees C. Ideally it should reach the temperature set by the control system in approximately 15 minutes (no longer than an hour)

What kind of heat sink system would I need to have both cooling and heating capability (possible to swap hot and cold sides with reversing current direction)? Could I make a system that has the aluminum sheet as the top layer, then one side of the Peltier connected to the sheet by thermal paste, then on the other side a heat sink system with fans that pull air away from the Peltier?

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  • \$\begingroup\$ So, is the whole aluminium sheet at the same temperature? How quickly does its temperature need to change? \$\endgroup\$
    – gbulmer
    Commented Oct 22, 2015 at 18:57
  • \$\begingroup\$ There needs to be uniform temperature on the entire surface of the plate within plus or minus 0.2 degrees C. Ideally it should reach the temperature set by the control system in approximately 15 minutes (no longer than an hour). \$\endgroup\$
    – Toledoj
    Commented Oct 22, 2015 at 19:02
  • \$\begingroup\$ Edit your question, by adding all that extra information, including time; don't leave it on the comment. It'll help the community to be 'on the same page'. \$\endgroup\$
    – gbulmer
    Commented Oct 22, 2015 at 19:04
  • \$\begingroup\$ Getting rid of 3kW with acceptable temperature rise will require a big fan. What is the controlled side of the aluminum plate thermally 'connected' to? What range of ambient temperatures do you anticipate? Do you want to control the 100 Peltiers individually? Reversing current will do what you expect- reverse the direction of heat flow. \$\endgroup\$ Commented Oct 22, 2015 at 19:16
  • \$\begingroup\$ 30W can be arbitrary. Maybe install 30W rated ones but run them at a lower voltage if I can achieve temperatures required on the plate. The plate will interface thermally with flowing air with temperatures ranging from 50 degrees C (when a 10 degree plate temp) and 20 degrees C (when a 80 degree plate temp). No need to control each Peltier individually within the plate - the only thing which needs to have control is the plate temperature as a whole. \$\endgroup\$
    – Toledoj
    Commented Oct 22, 2015 at 19:25

2 Answers 2

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The simple answer is: maybe. There are a few things you've skipped over. First is the plate thermally insulated? If the plate is covered in thermal insulation, you can certainly cool it to 10 C, and probably get down to about -20 C if the insulation is very good.

100 30-watt TECs will draw 3 kW. A general rule of thumb with Peltiers is that you can get 10% of that in cooling. So. Let's run things backwards. 10 C is about 15 C below room temperature. If you were to run 300 watts of heating into your plate, could you get a 15 C rise in temperature? If so, you have a pretty good shot at cooling it to your specifications. On the other hand, if it's exposed to air, it may be hard to get the effect you need.

It's not reasonable to think that you can pack your TECs perfectly. As KalleMP has pointed out, you need to leave some room to run wires, but let's say you used a 9x9 array of coolers, with about 5 mm between units. Then you would have 81 coolers, with a power dissipation of about 2400 watts, and about 240 watts cooling. Let's assume your controlled plate is perfectly insulated. Let's also assume your plate is 6 mm thick, rather than 3, so as to let the metal spread the temperature evenly. Then the volume of the plate is .015 cu meters, and the weight is 40.5 kg. The specific heat of aluminum is .9 J/deg-kg. To drop the temperature of the plate 15 C will take an energy of $$E = 0.9 \times 15 \times 40.5 = 547 \text{J} $$ and at 240 watts it will only take about 2.3 seconds to drop the temperature you need. The less the insulation, the more power you'll need, and keep in mind that Peltiers do not produce much temperature differential when run at high current.

Your construction approach is appropriate, but as has been mentioned, in principle you might need a pretty good cooling system to handle 2.4 to 3 kW of waste heat.

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  • \$\begingroup\$ I would take those assumtions and start optimizing towards your goal. I thing the 2.3s to 15 min is a huge margin and I think i'ts quite realistic, so start by using a much thicker Plate instead of a sheet, say 10-15mm, this also helps mechanically. With the gained thermal capacity you'll get away with fewer elements, as the heat wil be spreaded towards the surface. \$\endgroup\$
    – wiebel
    Commented Jun 22, 2017 at 8:05
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Unless there is some special cunning reason for purchasing 100 Peltier devices I would try to simplify this.

If you want to have the capability to pump 3000W of heat then Peltier seems to not be the way. You would need to leave room for the wires between the modules if wiring them in parallel or at need for the connections if in series (or series parallel). You would need to have a heatsink on the 'back' capable of dumping about 6000W if under full load.

I would suggest you fit heat pipes to the back of the plate brought out to a common header on one edge of the plate and have this temperature controlled by a refrigeration system. Have a passive resistive heating system over the back of the plate if heating is required and you don't want to reverse the heat pump.

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  • \$\begingroup\$ Oh, I forgot to mention, clamping that many fragile ceramic elements evenly by a pretty flexible sheet of aluminium will require a lot of ingenuity as well. \$\endgroup\$
    – KalleMP
    Commented Oct 24, 2015 at 16:53

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