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

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Discussion Overview

The discussion revolves around the thermal load that solar heat will impose on an air conditioning system due to an electronics enclosure with a specific surface area and insulation properties. Participants explore the calculations involved in determining how much solar energy will be absorbed and how it will affect the temperature and cooling requirements of the enclosure.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • B Kelly presents initial calculations indicating that a white-painted enclosure top absorbs approximately 316.414 watts of solar radiation, converting to about 1079.9 BTU per hour.
  • Q suggests that to fully understand the impact on the air conditioner, the equilibrium temperature of the panel must be determined, as it influences how much heat is lost to the atmosphere and how much is conducted into the enclosure.
  • Bryan expresses uncertainty about calculating the equilibrium temperature but notes that the heat generated by the electronics may exceed the solar gain, estimating an average heat output of about 2000 BTU per hour, with spikes up to 5700 BTU.
  • Another participant provides equations for calculating heat transfer through conduction and convection, indicating that the total heat absorbed must equal the sum of heat rejected to the atmosphere and heat entering the enclosure.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the method for calculating the equilibrium temperature or the exact impact of solar heat on the air conditioning load. There are multiple viewpoints on how to approach the problem, and uncertainty remains regarding the calculations and assumptions involved.

Contextual Notes

Limitations include the need for specific temperature values and the dependence on accurate definitions of heat transfer coefficients and insulation properties. The discussion does not resolve the mathematical steps necessary to find the equilibrium temperature.

bkelly
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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
 
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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'.
 
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
 
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.
 

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