Solar Radiation and surface tempertures 2

[jasno999]

I will use common objects to make this a bit more easy to ask. Lets say I have a medium sized enclosure (say a box that is around 3 feet by 2 feet by 2 feet).

I have a internal heat load (given) and I also have a internal temperature that I would like to maintain (Lets say it is 135F). So I can use Q=m.Cp Delta T to determine the flow rate required to maintain the internal temperature where I want it (based off outdoor air temperatures).

If I know what the internal temperature is because I have a fan sized to maintain that temperature then I assume I can determine the heat transfer through the walls using the equation Q=UA delta T. I know what the build up of the wall is so I can determine the U and I already have Area.

That is great - and once I have my Q I can then use it to determine what the surface temperature of the outside wall will be.

The part I am not sure how to do is- how do I factor solar loading into this so as to determine what the skin temperature will be when it is solar loaded? I was told the number to use for solar loading is 355 BTU/sqft hr but I have no idea how to do this....

I am looking for some detailed instruction on how to determine the true skin temp with solar loading involved..

 

[IRstuff]

You have an added heat load from the solar. That goes into your thermal resistance and changes the temperature.

TTFN

FAQ731-376

 

[25362]

Heat Transfer by JP Holman -McGraw-Hill- chapter 8, 4th Ed., brings a worked out exercise on solar radiation. You'd be able to note that the absortivities/emissivities of exposed surfaces affect the temperatures. BTW, what's the reason for so high a solar heat flux?

 

[IRstuff]

That value is the worst case specified in MIL-HDBK-310 Table IX, but that's only for about 2 hrs during the day.

TTFN

FAQ731-376

 

[MintJulep]

http://www.hvac.okstate.edu/research/Documents/Spitler_Fisher_Pedersen_97.pdf

 

[jasno999]

Can somebody provide me with the equations and show me how to do this? I don't have that book and I am looking for figure this out quickly. I just don't know the equaitons or how to add them into the thermal resistance.

MY surfave emmissivity is assumed to be 0.9 right now.

 

[IRstuff]

??? At some point, you have the heat flow from inside the box to the outside. Add the injected heat there, and your dissipation to ambient should already include both radiative and convective HT.

TTFN

FAQ731-376

 

[jasno999]

Ok as I said above I understand the heat transfer via conduction/convection. I can do that. I jsut don't know how to do the solar part.

For example I have a inside temperature of 140F. My outside temperature is 120F. Typically you would see the heat transfer go thru the wall to the outdside. However if I add solar load into this my skin temperature may go up to somewhere around 150F. I don't knwo how to add the solar component to get to that 150F and that is where I am lookign for help. The other part of this is if the skin is hotter than the inside how would the heat transfer go thru the skin to the outside???

 

[MintJulep]

http://en.wikipedia.org/wiki/Sol-air_temperature

 

[jasno999]

MintJulep - I looked at those equations but I am not sure how to use them. Can you give me an example? FOr instance what are the following:

a = solar radiation absortivity of a surface
I = global solar irradiance
?Qir = infrared radiation transfer of sky temperature to surface

 

[MintJulep]

Personally, I find it easier to simply look up the Sol-air temperature in the tables in the ASHRAE handbook.

 

[vpl]

jasno

I've refrained from answering this question, based on an interaction we had on an earlier question. However, your insistence on somebody just giving you the equations and solving the problem for you (when you aren't giving us the specifics) bothers me. No one is here to do your work for you.

You've asked a number of questions over the last couple years dealing with radiation and have mentioned several times that you're not familiar with dealing with radiative heat transfer. At the risk of seeming impolite, may I suggest that a college level course in heat transfer might be in order. If you've already done that, then you might want to take one dealing specifically with radiative heat transfer.


Patricia Lougheed

Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.

 

[786392]

Dear jasno999 Hello,
Just to add a little bit in my way.
This will be useful if you study with a cool mind the concept of 'Greenhouse(s)' as employed for gardening the delicate/critically senesitive plants in colder/less sunny, mostly chilly climates.Just Google and you should get a lot of info to choose/understand/analyze and surely the answers to your queries.

Best Regards
Qalander(Chem)

 

[jasno999]

VPL - Wow you are very unhelpful. I am asking questions because I thought some people on this board with background in the subject matter could help me out. Really if you don't have an answer or don't like me for some strange reason then don't read my post and don't respond.

What information do you want that I am not providing. I will provide whatever information is needed so that somebody can help me learn how to solve this problem. Look I am not askign anybody to do my work for me as you put it. I am tryign to learn how to do this so I can do it in the future. However I need somebody to show me the equations and how to use them in an example so I can fully understand this.

I just don't understand how to factor solar radiation verses reflective radiation verses radiation out of the surface that is not related to solar.

 

[IRstuff]

We're probably just baffled by how you think you have solved the basic heat transfer problem, and yet are unable to add additional heat input at one spot in the 1-D analysis.

You have Q1 going through the walls, resulting in a deltaT1. You have Q2, the absorbed part of sunlight added to Q1 going out through convection and radiation, resulting in deltaT2. For extra challenge, you'd break up Q1 into the shaded and unshaded walls and process the unshaded walls with no solar loading. It's basically two equations with two unknowns.

TTFN

FAQ731-376

 

[jasno999]

Sounds good. I assume I use the 355 BTU/ sqft hr to get the Q from solar radiation???

I guess I also need to take reflectivity into account with Q = emmesitivity X Stefan-boltmanz eq X (Tsurface - Tsky)^4

I think those are the equatins I am just not sure how to put it all together.

SO I have a Q1 due to the heat transfer from the energy inside the box. I am then gettign Q2 from solar heat goign into the outside surface of the box. However I know I am losing Q1 via convection to the outside via the surface film coefficient and I am losing some Q to the outside via radiation.

Here is the part that confuses me- DO I have radiaiton loss to the outside and also a reflectifity loss or are both of those components contained in one equation??? What is the equation?

 

[IRstuff]

Of course, you have both. Your 355 BTUh/ft is multiplied by the absorbtivity of the surface to get what's actually absorbed by the box. You then use Stefan-Boltzman to determine how much is radiated away, along with the standard convection loss.

TTFN

FAQ731-376

 

[Tunalover]

jasno99-
Note that the "ivity" suffices you use in your terminology is reserved for "optically-pure" surfaces such as mirrors and lenses. The "ance" suffix is used properly in your context of optically impure surfaces (according to NIST).

It is very unwise to assume the solar absorptance (alpha) is the same as the infrared emittance (epsilon). Inexpensive ASTM test methods exist to determine these figures. I don't recall the specifics but any competent thermal test lab should have the equipment to perform the tests (given sufficient size and quantity of samples). Don't let colors fool you either because there are some high-emittance white paints avaiable just as there high-emittance black paints. Common engineering surfaces are far from blackbodies or even graybodies for that matter.


Tunalover

 

[IRstuff]

Tunalover is quite right about being careful about the absorptance and emissivity. The reason is quite simple. Sunlight has a peak wavelength of about 0.55 um, while a room temperature object has a peak wavelength of about 10 um.

TTFN

FAQ731-376