0
\$\begingroup\$

Diagrams

I am thinking of making a portable refrigerator with the above schematics, but I am doubtful that the aluminium frame would be enough of a Heat Sink on it's own.

All the peltier elements are of same dimension (40x40x3)mm ~ (40x40x3.5)mm with 12VDC operational voltage and 6A max current.

Is there a way to know how much cool it will be and if the heatsink is enough or not?

If the heatsinks are insufficient, how should I improve my design? I would not prefer to add fans/water cooler etc until and unless there is no other way out!

\$\endgroup\$
  • \$\begingroup\$ If i understand your drawing right, you are hoping to use Al that is on the side of the peltier cell as a heatsink, right? \$\endgroup\$ – Vladimir Cravero May 30 '14 at 9:42
  • \$\begingroup\$ Yes, Al on both side as heat conductors for distributing the cooling effect on inside and as heat sink on the outside \$\endgroup\$ – Rijul Gupta May 30 '14 at 9:47
  • \$\begingroup\$ I think I am lost, just bear with me once more. You have this Al box, drill four square holes on the sides, fit four cells so that the hot side faces outwards and the in side faces inwards, is this correct? \$\endgroup\$ – Vladimir Cravero May 30 '14 at 9:50
  • 1
    \$\begingroup\$ Partly correct. There are 2 boxes and no drilling. Layout is outer box, fix peltier on side, fix inner box in contact with peltier. \$\endgroup\$ – Rijul Gupta May 30 '14 at 10:27
  • 1
    \$\begingroup\$ I think there's two, thin, aluminium boxes, one on the inside on the 'cool' side of the element and one on the outside on the 'hot' side of the element acting as the heatsink. I asusme there's insulation between the two? .... ^ what he said ^ :-) \$\endgroup\$ – RJR May 30 '14 at 10:28
1
\$\begingroup\$

No, don't think that works.

In order to see what happens, I plugged some numbers in here. The calculater doesn't support a flat plate, so I used only two 'fins' (the minimum).
Here are the numbers that I used, sort of modeling the long side of you box:

H - 1
L - 200
W - 70
b - 0.5
t - 1
N - 2
T - 25 (ambient air)
natural convection, vertical base
Emissivity - 0.09 (from here)
Ls - 70
Ws - 70
P - 75W

The resulting source temperature is a bit above 650C - not good...

Thermal resistance of the plate is about 6.6C/W, which means that for every watt you dissipate, the plate's temperature will rise 6.6C.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Please explain why this will not work and how to improve? \$\endgroup\$ – Rijul Gupta May 30 '14 at 11:19
  • \$\begingroup\$ What if I line the inner and outer box with fins ? Would it work then? (Would pins be better?) \$\endgroup\$ – Rijul Gupta May 30 '14 at 11:53
  • \$\begingroup\$ Is the source temperature maximum temperature that we can apply or operating temperature? \$\endgroup\$ – Rijul Gupta May 30 '14 at 14:10
  • \$\begingroup\$ The source temperature is how hot you peltier will get. Play around with the calculator to see if fins might work. I'm assuming you don't want the element/outside heat sink to become so hot you'll burn yourself. \$\endgroup\$ – RJR May 31 '14 at 11:13
0
\$\begingroup\$

How will this not work? Let me count the ways.

1 - Your boxes are apparently made of "thin" aluminum, which presumably means ~1 mm thick. These walls will flex as the box is moved, and you cannot guarantee that the two sides of the Peltier devices will remain in contact with the aluminum.

2 - Assembly would be a nightmare, as you would have to mount the TECs on one box, then slide the two together. And since there is no physical contact between the boxes, there is no way to hold the TECs in place.

3 - If thermal contact is made, the thermal conductivity of aluminum is inadequate to spread much heat laterally away from the TECs, given the very small cross-sectional area which the small thickness implies. This means that the immediate areas on each side of the TECs will get very hot and very cold.

4 - These hot and cold areas (roughly 40 mm x 40 mm, or maybe a little larger) have no good way to dissipate their heat and cold - the surface area is just too small.

5 - Finally, there is no insulation between the two boxes, and with a spacing of 3 mm there is no way you will ever provide it. So the heat separated by the TECs will immediately flow back to the cold side.

Those are the problems. What can you do?

1 - You must provide major insulation between the inner and outer boxes. An inch of foam is not too much.

2 - The hot side of each TEC must have a fan-cooled heatsink. The cold side also needs a heatsink, but that can be somewhat smaller than the hot side and may not need to have a fan. I would recommend the outer heat sink be physically connected to the outer box, while the inner heat sink is attached to (and supported by) the outer heatsink, and the inner box floats mechanically. Ideally, the layout of the inside of the unit will prevent food from being placed so as to block the circulation of cold air. The design of the inner heat sink (pins vs fins) probably goes in favor of fins, just for cost reasons. If you are unwilling to provide fans for the outer heatsinks, you must use larger heatsinks, and you must design the unit housing to prevent blockage of air circulation.

And for what it's worth, I'm with you on avoiding water cooling. On a portable unit that would be a nightmare.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.