maximum internal temp of computer enclosure

[sparkie93]

Help! I haven't touched my heat transfer book since 1985!

I'd like to know if I'm approaching this correctly. We have a 15" panel mount PC to be mounted to the lid of a polycarbonate enclosure (Hoffman type enclosure) and want to be sure it doesn't get too hot.

Dimensions: 22" x 15" x 7" (1178 in3 surface area) enclosure 3/16" thick. Heat transfer rate is 30 watts. Maximum internal temp allowed is 60deg C. Ambient temp (meat processing plant) could range from 0 - 30 degC.

Working backwards through the formula q = k*A*(T1-T2)/L and assuming 60degC for a worse case internal temp gives me 59.87degC. So as long as ambient is under this temp I'm fine.


Is this the correct approach? Does the formula really apply to a volume rather than just a basic area (I used the external area of the enclosure and ignored the fact that the face of the PC will be exposed to the ambient environment), or is there a more correct formula(s) to being using?

Thanks,

Robert

 

[IRstuff]

You're taking too much area credit.

Only the single face that's actually exposed to the ambient air would be counted. Assuming it's the largest face, the delta temperature is more like 70°C. If it's one of the smaller faces, then it's even worse.

A more detailed analysis would allow you to assume some additional heat loss from the case itself, but being polycarbonate, it'll be much worse than from the one exposed face.

TTFN

FAQ731-376

 

[Zesti]

I calculate the heat transfer coefficient to be close to 3 W/m2K (plastic sheet 5mm thickness).

Front plate 22" x 15" = 0,2 m2

Heat transfer is 30 Watts

This gives a temperature difference of 50K. Which is more than is allowed since 30+50=80 degrees C internal.

I think you should have a small fan drawing air through the casing...

PS: Where's the "/L" in your equation coming from?

 

[sparkie93]

For clarity I've attached a file showing the touchscreen mounted into the enclosure. We're cutting an opening in the enclosure lid to mount it. The lid will always remain closed. For simplicity and worse case scenario, I was treating the problem as if the entire touchsreen was fully mounted inside the enclosure.

For "L" I used the thickness of the enclosure: 3/16" (4.7625mm).

For the area I used the entire surface area (all sides) of the enclosure rather than the largest face. I'm not sure whether this is the correct approach or not.

For thermal conductivity of polycarbonate I found ranges of .13 - .24 W/m-K online. I used .152 in my calculations.

We manufacture products for beef/poultry/pork plants. This unit is for data collection and resides in the production area, which is generally cool to very cold. We haven't had problems with them, but want to reduce size/weight of the enclosure as much as possible and wonder at what point to anticipate heat problems. The PC manufacturer has specified a maximum temp of 60degC max and a 30W heat dissipation.

Thanks for the assistance.

 

[IRstuff]

You are using the conductive heat transfer equations, which is only partly correct for this application. The BOX is cooled by CONVECTION. Conduction just gets the heat to the surface.

With 3W/m^2-K, you are already at 77°C at the case surface. With 0.156 W/m-K and 3/16", there's a 4°C rise across the case. That puts in the internal ambient at 81°C, 20°C above the spec. That potentially reduces reliability by a factor of 100 or so.

TTFN

FAQ731-376