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calculation of solar heat gain

calculation of solar heat gain

calculation of solar heat gain

(OP)
I have been trying to calculate the solar heat gain of a piece of equipment in the sun.  ie the temperature rise.  

If the object absorbs all the solar flux and reflects nothing, is there a simple way to calculate the delta T ?  

I thought I knew how to do it, but with a solar flux of 1200w/m2 the temperature rise I calculate is several hundred degrees which cannot be correct.

RE: calculation of solar heat gain

Why not?  A 120W filament glows nearly white hot, which puts the temperature of the filament at over 1000°C.

The point of fact is that thermodynamically, it's impossible to absorb perfectly and emit or convect nothing back, short of being a black hole.

You must apply at least convective cooling and possibly radiative cooling to get a half-way decent answer.

TTFN

RE: calculation of solar heat gain

Absorptivities for solar radiation of various surfaces differ. For example, for highly polished aluminum, 0.15; cast iron, 0.94; asphalt, 0.9; various white paint pigments, 0.12-0.16; red brick, 0.75; polished '301' type st. steel, 0.37; tarnished copper, 0.65 (for low temp. radiation, 0.75).

If, for example the surroundings are at 25oC and your 'equipment' is a plate of tarnished copper, neglecting any convection effects:

(1200 W/m2)(0.65) = (0.75)(5.669 x10-8)(T4 -2984)

Thus T = 314 K = 41o

If it were asphalt with both absorptivities, for solar radiation and for low temp. radiation = 0.9.

(1200 W/m2)(0.9) = (0.9)(5.669 x 10-8)(T4 -2984)

T = 316.2 K = 43.2 oC.

The value of 5.669 x 10-8 is the Stefan-Boltzman constant expressed in W/[(m2)(oK4)]

I hope the above exercises may be of help.

RE: calculation of solar heat gain

The temperature of several hundred degrees will not be seen as convective and radiative losses will quickly balance the heat gains.

I have developed an accurate calculation for solar radiation and ambient heat transfer into pipes of various materials and surface finishes. This was in order to determine water temperatures going to safety showers.

25362 is on the right track but the result can vary quite with even small changes in materials or surfaces. For example dust loading on reflective surfaces can quickly destroy the properties.

If you send me some details to denniskb@ozemail.com.au I can run a couple of cases for you.

Dennis Kirk Engineering
www.ozemail.com.au/~denniskb

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