Heating from the sun.

[Floyd44]

Hello,

Lets say i have half of a wooden/glass/metal board exposed to the sun and the other half is being covered by something and therefore is in the shade.
How do i calculate how much will the part of the board, that is exposed to the sun, heat up from being exposed to the suns energy? I am interested in the max. temperature difference - from one side of the board, to the other.

Thanks for any advice.

Best regards,

 

[SuperSalad]

First, it will depend on the location on Earth (ie. what level of solar radiation exposure is present at the latitude in question, the air temperature, humidity, etc.).

Next, you can factor in the absorption fraction: Engineering ToolBox: Absorbed Solar Radiation

Finally, the dimensions of the material and the thermal conductivity properties of that material.

With that information you should be able to get a rough approximation of expected temperatures.

Andrew H.

http://www.MotoTribology.com

 

[Floyd44]

Sory if i am asking ''stupid'' questions, but i am really not familiar with thermal physics.

Okay lets set up a hypothethical scenario:
Summer conditions: 30°C air temperature, 70% air humidity, solar radiation 900 W/m2
Lets use the polished Aluminium board (2000 x 200 x 4 mm, weight of 43,2 kg) with ''absorbed'' solar radiation of 0,30.

What is the formula used in these 2 situations:

- For the temperature of the board in the shade,
- For the temperature of the board in the sun.

Best regards,

 

[3DDave]

If only heat transfer was so simple that there was a formula for that. Even cases that look trivial can be difficult.

Here's a text to help out.

https://ahtt.mit.edu

A Heat Transfer Textbook, 5th ed
John H. Lienhard IV, University of Houston
John H. Lienhard V, Massachusetts Institute of Technology

Copyright (c) 2000-2020, John H. Lienhard IV and John H. Lienhard V

This introduction to heat and mass transfer, oriented toward engineering students, may be downloaded without charge. The ebook is fully illustrated, typeset in searchable pdf format, with internal and external links.

 

[Floyd44]

Is there a simple method, for aproximate results?

 

[LittleInch]

The one in the shade will be air temperature surely? so 30C

The one in the sun? - 40-45C for a shiny surface.

Black metal sheets can hit 60-70C in the midday sun so with a shiny surface less.

But temperature is dependant on where the heat goes. For boards/sheets this is mainly convection with some conduction. So wind speed makes a massive difference to the temp which could approach air temp.

The best ROT I've heard of is that the effect on people is about 8 Degrees C in the shade as opposed to being out in the sun, but then we have in built cooling systems...

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[chicopee]

Hey guys don't forget the shaded side will be heated by primarily conduction from the sun metal plate side exposed to the sun. There will be a small amount of radiation heat transfer on the shaded side but for simplicity disregard that effect. Also the OP has said nothing about the ambient temperature which will affect the plate temperature.

 

[Floyd44]

Okay lets make it even easier, lets divide the aluminium plate into two separate parts.

So we have one metal plate fully in the sun, while the other one in the shade.

How do i calculate the difference between the one in the shade and then one in the sun?

 

[SuperSalad]

The one in the shade will be the same as the ambient air temperature (or close enough to it so as not to make a significant difference).

The one in the sun will be the temperature at which your plate is losing heat through convection at a rate of around 270 W/m², assuming the absorbed radiation of 0.3 from the link I provided above.

If you solve for dT in: q = hc A dT , using the 270 W/m² for q/A and the convective coefficient for air ranges from 10-100 depending on flow rate of air over the plate.

the dT value would be added to the ambient air temperature to get your plate temperature.

Andrew H.

http://www.MotoTribology.com

 

[Floyd44]

From what i can tell, you already need the ''dT'' in order to get ''hc'' - how do you get around that?

 

[SuperSalad]

You can approximate the convective heat transfer coefficient for air by the air velocity over the surface. Convective Heat Transfer (the formula it is about halfway down the page in that link, it also contains the unit conversions for your hypothetical)

Andrew H.

http://www.MotoTribology.com

 

[IRstuff]

An object in the "shade" can still get quite a bit of radiative energy from the sky or reflected from the ground; unless the object is in a dark cave, it will likely be warmer than air temperature. The converse is also true, an object exposed to a clear night sky can get colder than the air temperature.

TTFN (ta ta for now)
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