Calculating Peltier Sizing

[Mosaic]

Hi:

I have an application with a 6" x 6" x .25" OR a 6 x 12" by 0.125" Aluminum plate at ambient 30 deg C that I need to chill down to -10 deg C. The space limitations indicate a Peltier approach.

I have some questions on the thermal loading calculations.

While I can calc the the energy required to do this using the Spec. Heat Cap of Al. I am not sure how to compensate for losses to atmospheric or thermal coupling due to the plate crossection.

The calcs deal with the Delta T and the AL mass, but what about the heat continually absorbed from the ambient environment?
1) How do I compensate for the area of the plate surface?
2) An equal mass of Al plate at .125" thickness will behave differently due to the amount of radiant surface and thinner cross-section (weaker thermal coupling).
How do I calculate for this app?
3) How do i compensate for the reduced energy requirements when frost is formed on the Aluminum, as this acts as an insulator to ambient etc.

I understand that the Peltier can move its rated power (168W) plus its dissipated power V* I at an optimum Delta T gradient. I am achieving the required 52 Deg C delta T with NO alum Plate attached, with a heatsink temp of 42 deg C.

Thanks much.

 

[ronv]

Is it in open air or an enclosure?

 

[Mosaic]

The cold plate is open air.

 

[Mr RB]

It's just going to get covered in water ie ice. Why do you need a big metal plate out in the open air at -10'C?

You can characterise the plate in 'C per W by heating it with 10W or 20W measured, then measuring it's heat rise. This will still be very dependent on air circulation but gives you a ball park idea, and like you said ice buildup will affect it too.

For proper design of a cooling system you need to analyse where the heat needs to go from-to and where it should not go from-to, then thermal couple one and thermal insulate the other. It's really hard to offer suggestions when you don't mention the application and just give wishy washy info about big metal plates being cooled down...

 

[ronv]

I think you might come close with Degrees C per watt = 50 divided by the square root of the area in centimeters. So your 6x6 plate is about -2.3 C per watt. Depending on what you are trying to do you could insulate the bottom side. This would also help to keep the hot side from talking to the cold side. Big Heat sink, big fan on the hot side. I don't think you have to worry much about the losses if the plate is 1/4 in thick and only 6x6 with a 2 inch peltier in the middle. No idea on the frost. Make sure to seal the edges of the peltier.

 

[Mosaic]

Thanx ronv for that info, and RB , I like the idea to do an empirical to characterise the plate as I realise that Aluminum comes in all sorts of grades and purity. Extruded alum is about half as effective as pure alum. with heat xfer.

 

[ronv]

Let us know what you find out.

 

[chemelec]

I really doubt you will get to -10 C in open air.