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Imitating Solar Load in a Temperature Chamber

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ddelaiarro

Mechanical
May 17, 2004
45
BLUF: Is it possible to mimic a solar loading environment solely in a temperature chamber and, if so, how?

We have an 6061 Aluminum box with a known heat load inside. It needs to be subjected to a solar load IAW MIL-STD-810G, Method 505.4, Proc I. Essentially, it is a 24-hour exposure cycle with varying ranges of temperature and solid load. Our company has a thermal chamber that we'd like to pretest our units in to make sure we pass qualification testing. We can't create a solar load with our equipment, but what I'd like to do is create a temperature profile that would essentially mimic the total heat load our module will experience during solar load.

I know the following:
- Surface area of the module, A
- Reflectance level for the paint used
- Temperature and Thermal Loading (W/m2) profiles for the 24-hour cycle

What I'd like to do is convert the know Power from the solar load (W/m2 * A) at each temperature interval to a Delta T that I can add to the prescribed ambient temperature to essentially mimic the solar loading in my thermal chamber.

Any suggestions?
 
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Go buy some flood light lamps and then create a grid system apparatus to hang the flood lights and then hang that over what you are testing. You can calc out the W/m^2 or get this meter




Tobalcane
"If you avoid failure, you also avoid success."
“Luck is where preparation meets opportunity”
 
MintJulep - thanks for the link. I'm pretty sure I can come up with the convection coefficient (h) and the heat Transfer (Q) would be the Power Dissipated inside the box plus the Power Load due to solar effect (W/m2) times the area (A). The Outer Surface Temperature (Ts) value is where I'm hitting a roadblock. Seems to me that, without the Ambient Temperature (Tsol), finding Ts would be impossible. You would have one equation with two unknowns. Any thoughts?

Twoballcane - thanks for the suggestion and we were actually going to go that route, but the problem is we have no way of controlling the ambient temperature AND having the flood lights in the same space. The module takes up roughly half our thermal chamber and I would not have enough room to put a grid of flood lights in there as well.

Thanks again for the suggestions folks and any more assistance would be appreciated as this is not my area of expertise.
 
ASHRAE handbook provides a through derivation that at some point brings in the conductivity of the enclosure wall and interior temperature and does away with surface convection coefficient and outer surface temperature.

Wiki give the resultant equation, but without the derivation.

 
Haven't your systems engineers already modeled that? You have a peak of 1120 W/m^2 coming in, and energy being radiated and convected away.

For most of our stuff, we add 5ºC to 10ºC to the specified ambient.

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize
 
MintJulep[\b] - Thanks for the additional resource. I think I'll be able to use that and derive the information I need.

IRstuff - Interesting question. We're a pretty small outfit with a small systems engineering group (1-2 people). It seems that our systems group and what you're describing perform completely different tasks, however. Our systems guys deal more with how the different electronic modules we design and build interact with each other (power loading, Ethernet and serial connectivity, etc). They are more EE's by training than anything else.

I appreciate the 5-10oC suggestion but you are correct with your caveat. Our equipment cannot assume a high wind load for cooling. The test procedure we need to follow assume little to no air flow to expedite cooling. We were able to successfully test at 70oC for ~4hrs yesterday and our requirement is a 24hr cycle that maxes out at 49oC ambient temperature and 1120 W/m2 solar load (although, not at the same time). I think we should be OK, but I still would like empirical proof for our team and management.
 
Twocallcane - Absolutely is (DTB is on our list of potential test houses) and we will be performing the test at a similar location. We are trying "pre-wire" the test so that we have a level of confidence with our ability to pass the test on the first shot which would A) give the customer confidence in our capability and B) save us the time and cost of retesting.

MintJulep - that site is very useful. I'm just curious, do you have an idea on how to calculate the value of [Δ]Qir? The equation is:

[Δ]Qir = Fr[·]hr[·][Δ]To-sky

I'm having trouble defining all three of those values (Fr, hr, [Δ]To-sky). Are they just values from my heat exchange course in college that I'm forgetting? A quick look in the Bejan and Kraus Heat Transfer Handbook didn't result in any clarification.
 
Not quite. It looks to be a simplification of the blackbody radiation, os the Fr term should be the view factor, the hr should be the emissivity*Planck's constant*(T0+Tsky), T0-sky should the surface temperature minus the sky temperature, which should be on the order of 250K to 290K, depending on the atmospheric conditions.

TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize
 
First off, thanks a ton for the insight. It's really helping.

The research I've been doing on view factors seem to be for radiation from one body to another and don't seem to be relevant to an object and the sun. Any suggestions on resources (online, preferably) to acquire the appropriate view factor?

Also, in your post above you state

T0-sky should the surface temperature minus the sky temperature, which should be on the order of 250K to 290K, depending on the atmospheric conditions.

Did you mean surface temperature or ambient temperature? Everything I've been reading points to ambient temperature, not surface temperature. I'm also having a tough time understanding the value of Tsky. Any suggestions?
 
The Wikipedia article on Tsol states that:
Tsol = Tamb+(A[⋅]I-?Qir)/ho)
I know the following variables:
[ul][li]Tamb - The chamber temperature. This is a known/given variable that changes over time.[/li]
[li]A - The exposed area of the module. This variable is known and constant.[/li]
[li]I - The solar irradiance. This is a known/given variable that changes over time.[/li]
[li]ho - The convection coefficient of the module. This variable is known (or can be found/looked up) and is constant.[/li][/ul]

The only value I don't know is ?Qir which Wikipedia describes as
extra infrared radiation due to difference between the external air temperature and the apparent sky temperature. = Fr[⋅]hr[⋅]?To ? sky [W/m²]

Since I'm having trouble calculating this variable (values of Fr, hr, & ?To ? sky seem to be fairly ambiguous), what if I essentially make ?Qir = 0.

Would this not give me a worst case scenario?

Having done that, my temperature profile is quite high, higher than most of the components prescribed temperature range. That seems to be a problem.

I guess the main question is: How much of a role does ?Qir really play in this equation?
 
Quite a bit. To recap.

Q.sol + Q.box = Q.conv + Q.rad

i.e., the solar load and power dissipation from the box are the inputs. The outputs are the convected heat transfer and the radiated heat transfer. Zeroing the radiated heat places the entire heat burden on convection. This might be an example calculation in Mathcad (also PDF'd):


TTFN

FAQ731-376
Chinese prisoner wins Nobel Peace Prize
 
Understood - that makes a lot of sense. The Mathcad files helped a little, but I still am not sure of what's going on here.

emissivity*Planck's constant*(T0+Tsky)
I think your equation for hr is incorrect. If you look at ?Qir = Fr?hr??To ? sky and do a unit analysis, you get

?Qir = 1 (Fr is unitless) * {1 (emissivity is unitless) * J * s * K} * K = J*s*K2. No matter how I try, I can't make that W/m2.
 
I can't see where the problem is.

?Qir is in W/m^2 as

Fr is dimensionless
hr is in W/(m^2*K)
?To ? sky is in K
 
After talking it over with some others here, we've decided on a simplified method with some known issues to at least somewhat simulate this test. Here's our solution and thought process:

The convection heat transfer equation is as follows:

P = k?Aconv??T where ?T = Tcase - Tamb, Aconv is the area of the module used for convection and P = power to be dissipated

Rearranged, that equation works out to:

Tcase = Tamb + P/(h?Aconv)

Obviously, as the power dissipation requirement or ambient temperature rise, the case temperature will rise. The power dissipation requirements are defined by P/(h?Aconv). We know that, in a solar loading environment, this section of the equation will rise. However, due to our test limitations, we cannot adjust this part of the equation. Therefore, we will change the Tamb portion of the equation to mimic the change in the P/(h?Aconv) portion of the equation.

The new ambient temperature which will mimic solar exposure, Tsol, will be raised by the product of the solar load, W, and the exposure area, Aexp, which is defined as the area of the module exposed to the solar load. Therefore:

?Tamb = (W?Aexp)/(h?Aconv)

and

Tsol = Tamb + ?Tamb

This solution has obvious shortcomings with the most glaring being localization of heat. In this solution we are taking a localized heat load (W?Aexp) and distributing it over the entire area used for convection (Aconv). This results in a more constant thermal gradient and less localized thermal stresses which may affect performance. Again, this test is being used as a 'pre-test' to a more rigid qualification test and is just a sanity check before going to the lab.

Thoughts on the approach?

 
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