Subject: Re: Thermal loss through the back of panel?

From: [email-masked] (Nick Pine)
Date: 4 Jul 1995 10:51:02 -0400

Horacio Gasquet  <[email-masked]> wrote:

>     What I want to know is whether anyone has factual information 
>quantifying how much energy is re-radiated out the front of a solar 
>collector vs the losses through the back side due to conduction and 
>radiation of the heat that builds up in the box. 

You can find some facts and more elaborate calculations in the second
edition of Duffie and Beckman's _Solar Engineering of Thermal Processes_,
but you might look at it this way, for starters:

Assume the absorber plate heats up to a certain temperature T
       the ambient temp is Ta, say Ta = 0F, for simplicity
       the box has a surface area of A on the front and A on the back,
           ignoring the sides, say A = 1 ft^2
       the box has an R-value on the back side of, say R10
       there is no heat Qu, being withdrawn from the box by liquid flow
           through the tubing, ie Qu = 0 
       full sun puts about 300 Btu/ft^2/hr into the front of the box
	   (if Qu is not zero, you might want to subtract it from this
	   solar input, using 300 - Qu in the calc below, instead of 300)
       the single glazing has an R-value of 1, to the still outside air

Then in full sun, 300 Btu/hr goes into the box thru the glass, and 300 Btu/hr
goes out, thru the front and back. What must T be to make this happen?

(T-0)x(1 ft^2/R1 + 1 ft^2/R10) = T/1.1 = 300 ==> T = 272F.

In this case, about 90% of the heat flows out of the box thru the front. 

The model above could be further simplified by ignoring the heatflow
out of the back. Or it could be further complicated and refined by
considering that a hot surface radiates heat according to the equation
U = Aes(T+460)^4, where U is in Btu/hr, A is the area of the surface in ft^2,
e is the emissivity of the surface, eg 1, s = 0.174 x 10^-8 is the
Stefan-Boltzman constant and T + 460 is the absolute temperature of
the surface in Rankine degrees. In this next step, I'd assume that all
of the heat that is radiated by the absorber is absorbed by the glass,
which then heats up and loses heat by convection to the outside world...
This would probably reduce the calculated absorber plate temperature.

Let's see, ignoring the heat loss thru the back this time...

    0.174 E-8 ((T+460)^4 - (0 + 460)^4)) = 300 
                            ^--the outside world radiates heat to the
                               collector too, which warms it up...

so, 0.174 E-8 (T+460)^4 = 300 + 0.174E-8 x (460^4) = 378, 

or  (T + 460)^4 = 378/0.174E-8 = 2.17 E11,

or   T + 460 = (2.17 E11)^(1/4) = 683,

so   T = 683 - 460 = 223F.

>It seems that insulation on all sides except the glass would greatly
>increase the temperature of the interior of the collector.  

True, but the front glass is the big heat leaker. Adding more insulation
to the back and sides may not help much. Coating the absorber plate
with a selective surface might help a lot.

Nick