Temp Inside a Solar Thermal Panel

You're not going to get "a" formula in this case... you're right that it's complicated. You will have all three types of heat transfer going on - conduction, convection and radiation. All have simplified formulas that "govern" their behaviour, but they require you to know certain constants. Then a further complication would be that the copper collector is being cooled by water flowing through at non-constant temperature and mass-flow-rate, but you would assume some steady state behaviour.

Someone may have developed a rule-of-thumb method for solar collectors, but without that you will either have to use rough estimates of the type offered by other forum members, or get yourself ready for a crash course in thermodynamics.

For the formulas you will need values for the conductivity and emissivity of the copper, the conductivity of the air, the size of the air gap (which will govern the impact of convection currents within the air, plus the conductivity and emissivity of the glass panel. (you'll find approximate values online for these).

The most simplified form of the formulas are these are

Convection: q(convection) = hA(Ts-Tinf)

Conduction: q(cond)=kA dT/dx

Radiation: q(rad) = εσA(Ts⁴ - Tinf⁴)

where

• q is the heat transfer rate (in Watts)
• A = surface area
• h = heat transfer coefficient
• dT/dx is the rate of change of temp wrt distance through the medium
• Ts is the surface temperature
• Tinf is the temperature distant from your surface (this will be atmospheric temperature outside the glass, and will be the air temperature inside the glass - obviously won't be constant).
• ε = surface emissivity (one value for glass one for copper)
• σ = Stefan Boltzmann constant (google it - it's just a constant)
• k = conductivity constant

Whenever you try to work with these things get very messy, and you only end up with an answer as good as your approximations for the constants, but the way you would probably start this would be to:

1. estimate the power coming into your system from the sun (there are estimates for this for various global locations - but let's say it's 100W for the area of your collector) and

2. calculate the power leaving via the heated water (in watts that's Power =mCΔT where m = mass_flow_rate, ΔT= change in temp in Celsius or Kelvin, C = 4180 J/kg.K) Say that comes out to 45W.

3. then the remainder must be leaving the system, and we could assume that's mainly by convection and radiation out from the glass surface.

So q(convection) + q(rad) ≈ P(solar) - P(water) = 55W

Assuming you have values for your constants, the only unknown in these two equations on the LHS is the outer surface temperature of the glass. Which is one of the things you wanted to know. Knowing that you can use the formulas to work back to the inside temperature, through the air and toward the copper pipe and get some rough idea of temperatures along the way.

But warning Will Robinson - this all risks a highly approximate result. (and of course a slightly dirty pane of glass will change its properties).

In industry this analysis would almost certainly be done by field trials - measuring the temperatures of various types of exchanger in order to estimate some constants, then the data plugged into a software package.

Another way to approximate your problem - maybe simpler - (it's derived from the above stuff anyway so no more or less accurate) would be to use thermal insulation values.

Air has a k value k(air) of about 0.026W/mK. The conductive thermal resistance, Rth, of air is

Rth = L/kA (where L is the distance through the air, A is the area)

Then q(cond) = (T2-T1)/Rth (A cancels out giving per unit area result) - gross simplification, but it would give you an approximate air temperature at any distance from the pipe of a known temperature - but you still have to find or estimate the q value or otherwise estimate the T2 value, but that's what you're trying to find anyway...

Maybe someone in the industry has some simpler tricks - good luck!

(ps - if you're in the US you may have to change all the units over into Rankine-cubits-per-foot-slug-furlong-fortnights or something).