Temperature in Enclosure

[idesign73]

I have what appears to be an easy problem, but I am not getting an answer that makes sense. I have a 60 Watt light bulb in a metallic enclosure (either ASTM B26 Aluminum or 316 stainless steel, I am to compare both). I need to determine two things: 1: The surface temperature on the inside of the enclosure 2: The surface temperature on the outside of the enclosure.
I have assumed that only 5% of the wattage will go toward lighting, 95% going toward heat (57 watts). The ambient temperature outside the enclosure is 40°C. The thermal conductivity of the Al is 100.4 W/mK and the steel is 16.3 W/mK. I assumed the heat transfer coefficient to be 15 W/m^2K for lack of a better number.
For my internal surface temp I get 451K by the following equation:
q/A=h(ts-tf)
where q=57 watts, A=internal surface area=.028m^2, ts=?, tf=40°C, or 313.5K, h=15 W/m^2K
Per a previous post from "25362", I set the conductive heat flow through the wall equal to the convective heat flow from the exterial surface to the ambient air. I come up with an external surface temp of 450.54K for the steel and 451.21K for the aluminum (given that the wall thickness of the enclosure is 0.0064m). A) Shouldn't the surface temp of the Al be lower than that of the steel? B)After taking into account the heat disspating fins, I am still coming up about 70°C too high.
I know this is a long post, and I appologize, but I wanted to give all information possible. Any help would be much appreciated!!

 

[IRstuff]

Not sure where you went wrong, but maybe this was a sanity test.

Since the heat flux is identical in both cases, and going into the same thermal resistance of the air, the correct answer is that the two surfaces have exactly the same temperature, e.g., it's independent of the material.

The temperature inside is different.

TTFN

 

[25362]

For equal heat fluxes and geometries, and because of radiation effects I'd expect the aluminum surface to be quite a few degrees warmer than SS 316. The emissivity of polished aluminum is about 0.04 while that of polished SS 316 would be around 0.6. Kindly recheck.

 

[MintJulep]

IRStuff is correct if you assume that radiation from the eternal surface is the same in both cases.

My CRC Handbook lists the following factors for emissivity:

Highly polished aluminum .02 to .04

Dull smooth clean aluminum or stainless steel .08 to .20

Smooth slightly oxidized aluminum .2 to .4

Given this potential large difference in emissivity, I don't think you can safely assume that radiation is the same for both materials, so you need to factor it into your calculations.

 

[idesign73]

Refering to the post by IRstuff, wouldn't the outside surface have to be cooler because the air temperature is at 40°C as opposed to the temperature inside the enclosure at approximately 177°C.
Could someone please explain to me where the radiation is coming from. I know it seems like a silly question, but it has been about 6 years since I've done heat transfer and I'm a little rusty to say the least. I was under the impression that I had convection from the light bulb to the inner surface, conduction from the inner to the outer surface, then convection from the outer surface to the ambient air (not counting the conduction into the fins).
All help has been and will be appreciated.
thanks

 

[25362]

Thermal radiation is that electromagnetic radiation emitted by a body as a result of its temperature, and propagates at the speed of light. There are various types of electromagnetic radiation, and thermal radiation is only one.

It lies in the range from about 0.1 micrometer to 100 micrometer. The visible light portion of the radiation spectrum is narrow, extending from about 0.35 to 0.75 micrometers.

The propagation of thermal radiation takes place in the form of discrete quanta, each quantum having the energy E=h*f, where h is the Planck constant and has the value h=6.625*10^-34 J.s, and f is the frequency = speed of light/wavelength.

When a plate exposed to a solar flux of say, 700 W/m² is surrounded by a temperature of 25^oC, and the surface is highly polished aluminum, its temperature can be estimated as follows:

(700) a_s=a_ambient(T⁴-298⁴)

a_s=absorptivity=0.15 for solar radiation
a_ambient=long-wave radiation absorptivity exchange with the surroundings=0.04.

(700)*(0.15)=(0.04)*(5.669*10^-8)*(T⁴-298⁴)

T = 482 K = 209^oC.

5.669*10^-8 W/(m²*K⁴) is the Stefan-Boltzman constant.

 

[25362]

The above estimation neglected any convection effects.

 

[davefitz]

a) all 60 Watts are converted to heat within the enclosure so I would use 60 watts as the Q

b)The heat radiated plus convected off the outer wall must equal 60 watts. If you substitute anodized Al for standard oxidized Al you can practically double the overall heat transfer coeeficient as the Al emmissivity will increase from 0.1 (oxidized ) to 0.85 ( anodized), or you can use an emittive paint to raise the outside emmissivity .

Outer natural convection plus mild ambient currents yield an outer convective heat transfer coefficient of about 1.5 btu/hr/ft2/F, then add to that the radiative contribution of 0.133 btu/hr/ft2/F for oxidized or 1.13 ditto for anodized, and you can quickly get tot he resulting outer temp.

 

[MintJulep]

The heat from the light bulb is transfered to the inner surface of the enclosure by radiation and convection in parallel. Possibly also some conduction, depending on how the bulb is mounted inside the enclosure. In any case, ALL of the heat leaving the bulb must reach the inner surface.

Heat transfers through the material of the enclosure via conduction.

Heat exits the outer surface of the enclosure by convection, radiation and conduction in parallel. If the enclosure is levitating conduction can be ignored. If the enclosure is sitting on a surface then there will be condution between the two surfaces.

To further complicate things, the temperature of each wall of the enclosure may not be the same. Depending on the geometry, there may be significant stratification.