How to calculate outside surface temperature of a metal box with hot air flowing through it

[natebott]

First off, I want to apologize for the elementary nature of this question. It has only been about 13 years since I did any heat transfer/thermo type calculations.

I have a boxed in area with a know inlet and outlet. I have a flow rate (in CFM) and temperature of the heated air entering this box, as well as, the ambient temperature of the surroundings. The box is made of steel. I need to determine the surface temperature of the outside of the box.

Any guidance is greatly appreciated.

NB

 

[IRstuff]

The delta temperature times the joule mass flow gives you the heat transferred. That heat must therefore go through natural convection to escape the box, which results in the surface temperature

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[ione]

You have 3 thermal resistances in series as the heat transfer process foresees forced convention inside the box + conduction through the box wall + natural convection on the outside of the box (you might also add radiation depending on temperature involved). It’s an iterative process as you have to make assumption on wall temperature. As stated by IRstuff consider that when you reach an equilibrium the heat transferred from air flowing inside the box must equate that transferred from the box to the ambient.

 

[AWloo]

Just to add to what ione said, you should consider radiation effects when dealing with natural convection along the outside of the duct, until you know the wall temperature. Consider radiation and natural convection in parallel after the two series resistors.

 

[racookpe1978]

You could also assume worse case and assume the uninsulated steel box will approximately equal the air temperature inside, but will always be slightly less than air temperature. Actual temperature will be slightly less depending on heat transfer losses, those losses calculated as above due to resistance across the boundary layer and wall, and losses outside from radiation, convection, and a little bit of conduction at the wall supports.

You can spend a lot of effort and dollars trying to get all the approximations "just right", and not need that much accuracy in the real world. Or go measure a similar facility with an IR thermometer, about 60.00 at the hardware store. 1/2 hour of design time. 8<)

 

[natebott]

My specific problem is that we are going to be putting air through this boxed in area to keep wood particles from freezing to the top of the box. Ambient air temp as low as -40 C and inner air temp at 510 C. In that, "worst case" is actually "best case" in my case! Not sure if that points anyone in a different direction or not.

Thanks again for all the replies!!

 

[IRstuff]

510C !!??

SAE AIR 1168-4 Ice, Rain, Fog, and Frost Protection has the information on how to keep surfaces ice free. If your air is really 510C, I doubt that anything will stick, except for waxy substanaces and styrofoam-like materials, since they'll actually melt onto the box.

What is the purpose of all this?

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[Tunalover]

510C! How'd you come up with that? I doubt if the fan(s) will survive that temperature.

Tunalover

 

[natebott]

It is the exhaust air temperature from a diesel engine. We are routing the exhaust under a portion of our machinery in order melt ice/frozen wood debris. Diesel manufacturer data shows exhaust temps of 950 deg. F. There are no fans in this system.

 

[MortenA]

Second that - if you box is only steel (no inside insulation or stone or anything, then i think the surface will be close to 510 C

 

[Tunalover]

At that temperature you CANNOT ignore the heat dissipation by radiation from the box to the environment!

Tunalover

 

[racookpe1978]

Draw your problem! You have a lot of them.

Heat loss across the thin box wall is almost neglibile compared to all of the rest.

What you are describing is a simple very hot box with an interior gas temperature of 510 C ENTERING at one side at 510 C and leaving at something quite a bit lower at the other side. The low gas flow rate (because it is only an exhaust) and relatively low speed in the large box increase heat loss from gas to the four box walls and floor and supports and outside air under, around and over the box walls.
The four box walls will lose heat by convection to the air differently on the bottom, sides and top because of natural convection flow rates differ in the three locations. No outside fan, right?
The four walls will have radiation losses different on the bottom, sides, and top because of the different wall temperatures. Further, on each side => with different radiation losses up the sides of the two walls are to the 4th power, the sides will not have a uniform radiation loss vertically. The side walls will probably have a single emissivity, but it will be different than the top and the bottom walls because of different dirt and paint layers over time.
Conduction gains will occur from upstream back from the engine (unless it is isolated by a flex hose connection or long pipe from the manifold), but will have conduction losses at each box support and the outlet assembly.
Your heat "gain" to the wood particles in some container above the box will have to be evaluated in the same manner as the heat heat losses from the box.

You somehow were thinking in the original question that the heat was simply transmitted from gas to wood through a simple box wall, and that heat loss was the entire part of your problem. You don't even know the wall temperatures of the top of the box yet.

What assumptions have you made for heat transfer from bottom-of-wood container up to the irregular wood particles?