Effect of solar heat on air conditioning

[bkelly13]

I have the task of desiging the cooling for an electronics enclosure. I have done a bunch of calculations for heat load from the electronics and from heat transfered for outside air through the walls. Now I need to figure in solar heat.

This box has a top with 23.3 square feet. It is painted white and I have read that a glossy white paint will absorb only 14 percent of the solar heat. That tells me that of the the 97 watts per square foot from solar radiation this top will will absorb 13.58 watts for a total of 316.414 watts. I understand that 316.414 watts is converted to btu via the constant 3.413 resulting in 1079.9 btu per hour.

The top has an R value of 9. How many watts of this solar radiation will get though the top? In other words, how much load will the solar energy put on my air conditioner?

 

[CESSNA1]

BKELLY13: ASHRAE (

http://www.ashrae.org)

has a book on heating and air conditioning design that includes solar heat gain.

Regards
Dave

 

[KiwiMace]

You are confusing solar with conductive heat gain - The R-value is a red herring. Heat not reflected is essentially heat gained. R-value only works with conductive heat in differences of temperature. What you should be interested in is the Sol-Air temperature, an effective outdoor air temp amended by the solar gains. Look it up in Cessna1's book.

 

[bkelly13]

RE: You are confusing solar with conductive heat gain - The R-value is a red herring.

Recognizing that I am far from an HVAC expert, I don't agree with that. Solar radiation will heat the outside surface. To heat the inside, the heat must move through the wall or ceiling. In doing that, it meets resistance, measured in R-value. If the wall or ceiling has an r-value of 5, more of the solar heat will get through that if the r-value is 10.

The equation I found says in essence, divide the temperature difference by the r-value to get heat flow. But it does not say how to figure the temperature difference due to solar radiation.

Regarding ashrae.org and other sources, I am certain the answer is there, I am just having a hard time recognizing it.

Thanks for your time.

 

[willard3]

Heat from solar is radiant heat and obeys the same physical laws as light....if the wall is opaque, radiant doesn't get inside.

The skin will heat up, but it's primary heat transfer will be to re-radiate to the surround and not through the wall.

Air moving over the surface will be the other significant heat transfer component.

Next time you are near the sunny side of the house, put your hand on it....lots less than ~110F and that's on a hot day in summer. In winter, it will be lots less.

 

[KiwiMace]

bkelly13: Apologies, I was too brief. You are bang on correct. The solar affects the outer surface temp, the outer surface temp drives more heat transfer.

Here's what I meant:

You can't use the R-value directly on direct (absorbed component) of incident solar heat gain. The R-value (W/mK) is only relevant when you are talking about conduction driven by a temperature difference. ie. Q=R.x.dT. The R-value is important, yes, but not where you appeared to be looking to use it.

So having said that:

The important concept you are looking for is Sol-air temp. The outside temp is modified with the sol-air factors to give a more appropriate outside temp that accounts for solar loads, and can be used with the R-value.

Hope this is more helpful.

 

[CESSNA1]

BKELLY13: It appears you are looking for a one number answer. Unfortunately it does not exist. The heat gain and A/C requirements are dependent on the location on the earth, type of construction, orientation, ventilation, and other factors. It also requires a basic knowledge of how to apply the methods. I suggest that you contact someone in your area with knowledge of HVAC or try your local college and/or public library for books on the subject. You may also want to contact the local ASHRAE CHAPTER to get some help.

Regards
Dave

 

[stanlsimon]

Maybe this is a dumb question. Why not put a piece of tin over it to create a shaded environment?

 

[tombmech]

If the piece of tin is shiny, problem solved.

However, I think the answer bkelly is looking for is somewhat more simple: the 1080 Btuh will increase the surface temperature of the metal over ambient. That increase will be dictated by the specific heat of the metal and the mass:

Q= m * Cp * (T1-T2). In this case, T2 would be ambient air, solving for that quantity:

T1 = (Q/(m*Cp)) + T2.

T1 can then be used as the outside temperature for the conduction gradient - especially since any metal has an R value of essentially -0-. Note that the mass may well be the entire enclosure, not just the part exposed to the sun.

I agree with the others that this is kind of going at it backwards, especially when there are so many empirical tables available that give you an effective Cooling Load Temperature Difference, given the combined effects of radiation, ambient conduction, and convective heat transfer on the surface.

If the white paint only absorbs 14% of the solar radiation, that is most likely well within the safety factor of any applied AC load. If in doubt - like stanlsimon says: shade the thing.