For those who work heat transfer issues, perhaps for IT equipment: I have an outside box with heat-dissipating equipment inside cooled by a fan pulling in ambient air across the box. Can you help me think through what the heat balance on this box will look like?
a) First, the outer wall’s net solar load is found by balancing (radiative losses + convective losses) = solar radiation gain
b) Inside the box, there is the heat dissipated by the equipment.
c1) There is forced convection on the wall’s inner surface, and (c2) I believe there is also free convection from the wall’s outer surface (being warmed by Step c1 over and above the results from Step a)
d) There is conduction heat loss across the wall.
e) Last, there is the heat removed by the fans’ flowrate (cfm).
Do these steps sound correct? Did I forget any steps? Am I over-accounting for anything? Are there any general assumptions I can apply to simplify the analysis? My goal is to find the steady-state temp inside the box.
[FYI: I am aware it’s an iterative appraoch to balance all these steps and find a steady-state temp inside the box (at Tmax data for my location, max solar angle, and assume no breeze)]. Thanks in advance for your thoughts!
I think you don’t have to consider the conduction through the wall if it’s a metal box. There would be heat transfer coefficients on the inner and outer surfaces, and the heat transfer coefficient for conduction through the wall would be the conductivity of the metal over its thickness. I figure the inner and outer heat transfer coefficients would be a few dozen at best, whereas it’d be 10,000 or more for the wall conduction at least (all in SI units). Since the net heat transfer coefficient inside to outside would be the inverse of the sum of the inverses, the conduction one won’t matter.
I don’t think you have to do anything iterative. You are guaranteed you are going to export all the heat generated by the equipment and absorbing the sunlight. You just have to divide that energy by the box external area and by the net heat transfer coefficient.
If you want to deal with stuff like how the temperature in the box won’t be the same everywhere and so the box exterior won’t have a uniform flux, you might need a finite element approach. How fancy do you want it?
[oops, rereading and editing - you have the air heating by the through fan to account for, and it’s a separate loss mechanism in parallel with the box surface. I guess that’s what you meant about iterating. Still, maybe a method where you assume a result and then back calculate, which gives you the correct result instead - what’s that called, tabla falsa, something like that?]
Napier, since it’s been awhile since I’ve done heat transfer, help me think this through as I am getting myself confused. Let’s focus on the wall of the box. If I had the steady-state temp in the box (assumed greater than the exterior temp), wouldn’t I analyze thermal conductivity of the metal to find how much heat is lost through the wall? Then, that heat would add to the outer wall temp which I’d then have to calculate how much is lost to free convection. Is that correct?
Also, I am trying to better picture the forced convention. The heat generated in the box will be carried by the airstream, but we believe the airstream is not sufficient, so the temp will keep rising in the box. When I think of forced convention, I think of an airstream that is cooling a hotplate. But, in this case, the situation is reversed! The airstream is hotter than the wall! Do I really have forced convection?
For well designed industrial air coolers, the temperature approach is in the 10-15 deg F range. This means :
If the air temperature is 60 F, the hot fluid will get cooled to 70F or 75F for a well designed exchanger. I am not very sure what you are trying to do so a little picture will help a lot.
For a air cooled exchanger where the hot fluid is a gas, the overall heat transfer coefficient is 10 (+/- 20%) of Btu/(ft2 deg.F h). When the hot fluid is liquid (like water) the overall coefficient is 100 (+/- 20%) Btu/(ft2 deg.F h).
I appreciate the thought, but to picture my case…imagine a metal box housing electrical equipment and located outside exposed to all kinds of weather. There is no cooling by liquid, such as a binary loop design.