How much solar heat will my electronics enclosure put on my air conditioner?

[bkelly]

We have an electronics enclosure that 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?

Thank you,
B Kelly

 

[Q_Goest]

I think you're going to have to determine the surface temperature of the panel at equilibrium.

As the surface temperature warms up, part of that 316 watts is lost through convection to the atmosphere. It is going to heat the air as it passes over the panel.

Part of that 316 watts will also be conducted through your insulation and warm the enclosure.

So the total 316 watts can be broken up into heat rejected to atmosphere (A) due to convection and heat rejected into the box (B) which must be removed by your air conditioner. The relative portions of A and B must be determined by calculating the equilibrium temperature of the box surface where the solar energy is 'deposited'.

 

[bkelly]

Hello Q,
Somehow I knew this was more complicated that I was planning on. I don't know how to come up with the equilibrium temperature. The only thing I think I know is that with the small size of the top and its r-value, the heat from the electronics inside overwhelms the solar gain. I will just work it from that angle. I expect to average about 2000 btu per hour during normal operation, but can spike up to about 5700 btu. I figure about 6000 btu unit will do fine.

Thanks for your time,
Bryan

 

[Q_Goest]

Hi Bryan. It's not that hard actually to find that temperature. You have two equations, one for conduction through the top of the box and one for convection where you're rejecting heat to atmosphere.

Q total = Q atm + Q box

Q total = hAdT + k/L dT

Adjust the equations to fit your units, etc...

Your Q total is the 316 watts. So the total heat Q atm and Q box must equal that. You know the temperature inside the box and the temperature of the atmosphere so the only variable is the equilibrium temperature. From that, you get both Q atm and Q box.