heat gain problem

I am about to design a storage shed to be placed on a playground and would prefer to make the shed completely out of steel, including the
siding. Due to its location, kids will likely touch its surface and I am concerned with it being painfully hot to the touch in the summer sun. I have several questions. First, why does steel surfaces appear to get hotter to the touch than comparable wood, masonry surfaces? What techniques, materials, coatings would anyone recommend to reduce the surface temperature due to the sun? thanks for any and all input. rick
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semidemiurge wrote:

Heat conductivity is much higher.
Michael Dahms
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Nope!
This is not a materials property question. This is an energy balance question.
The steady state energy balance on a free standing wall exposed to the sun is
J_in = J_out
where J_in is the incoming heat flux and Jout is the outgoing heat flux.
The incoming and outgoing heat fluxes are
J_in = qc_air + qr_sun J_out = qc_wall + qr_wall
where qc_air is the conduction/convection term from the air to the wall, qr_sun is the radiation term from the sun to the wall, qc_wall is the conduction/convection term from the wall to the air, and qr_wall is the radiation term from the wall to the air.
The conduction/convection terms have the form
qc_air = h_air A (T_air - T_wall) (only when T_air > T_wall) qc_wall = h_wall A (T_wall - T_air) (only when T_wall > T_air)
The radiation terms have the form
qr_sun = e_wall S A (T_sun^4 - T_wall^4) qr_wall = e_air S A (T_wall^4 - T_air^4) (only when T_wall > T_air) qr_wall = e_wall S A (T_air^4 - T_wall^4) (only when T_air > T_wall)
Solution for T_wall > T_air (to see why this happens!)
qc_air = 0 (air transfers no heat to the wall by conduction/convection)
e_wall S A (T_sun^4 - T_wall^4) h_wall A (T_wall - T_air) + e_air S A (T_wall^4 - T_air^4)
In general, the radiative term qr_sun is large (note the T_sun to the fourth power!!!) and the term qr_wall is small (T_wall is only somewhat larger than T_air and the emissivity of air is small). Therefore, to a first approximation, the steady-state temperature of the wall is found from ...
e_wall (T_sun^4 - T_wall^4) ~ h_wall (T_wall - T_air)
A first order solution to this is ...
T_wall ~ (T_sun^4 - (h_wall/e_wall)dT)^(1/4)
where dT = T_wall - T_air.
To lower the temperature of the wall, you can ...
1) Raise the conduction/convection of heat away from the wall. This could be accomplished by putting big fans around the building that blow a constant flux of air along the outside of the walls.
2) Lower the emissivity of the wall. This could be done by painting the walls in a highly reflective color (white).
Alternatively or in addition, you can extend the roof line of your building out sufficiently on the east, south, and west sides (or east, north, and west sides if you live in the southern hemisphere) to assure that those walls are always in the shade. In this way, the value T_sun in the above equations essentially becomes T_air.
HTH
-- JJW
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Jeffrey, Thank you for the thorough treatment of the thermodynamics of the problem. I still think there is a difference between materials. I have observed different surface temperatures of objects painted the same color (Trex decking - made from cellulose and polymers vs. mild steel painted a very similar shade of gray with a similar sheen). I think emissivity varies due to chemical composition, so despite the surface colors appearing very similar (visual spectrum), maybe they are different at other wavelengths (infrared, ultraviolet)? Just a thought. thanks again, that took some time:) rick
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Yes, I see now where your question was directed, and differences in materials thermal properties are important to consider in how hot they "feel" when at the same temperature.
As I pointed out in a different post, I wonder what role thermal diffusivity plays relative to thermal conductivity to answer this question (although, for the most part, thermal conductivity and thermal diffusivity are nearly directly related anyway).

I think emissivities also vary by surface roughness????
Certainly, the energy that a surface absorbs is absorbed across the entire spectrum of electromagnetic radiation. What we see as "visible" light is not the entire amount of energy. Therefore, a difference in chemical composition of the absorber could change its emissivity in a non-visible wavelength range, causing lower or higher heat absorption and therefore lower or higher net temperature.
-- JJW
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Jeffrey J Weimer wrote:

Take peaces of wood, aluminum and steel. Heat them to 100 C, then touch them! You'll feel the difference.
Michael Dahms
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Hmmm ....
My apologies. I think I see now.
Assuming all materials have the same temperature, which "feels" hotter and why? I was focusing on the fact that some materials get hotter than others under the same conditions.
Is a useful analogy, that cooking oil at 100 oC will cause a greater burn than water at 100 oC?
In this case, since touching the surface initiates a non-steady state heat transfer, it seems that thermal conductivity and thermal capacity both come into play.
In other words, how hot something "feels" (when at a given temperature) is not just dependent on the rate of heat transfer out of the material (through its thermal conductivity k), it is also dependent on the amount of heat available within a given mass of material (through its density rho and specific heat capacity Cp).
Isn't this the thermal diffusivity (k / rho Cp)?
-- JJW
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Jeffrey J Weimer wrote:

Right! An additional point is the heat transfer coeffcient. Altogether, aluminum-burns are the worst.
Michael Dahms
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Yes, because it is defining the transfer of heat through the "film" between the material surface and the skin surface. This film would be different for different textures and/or coatings on a material.

Interestingly, aluminum has a lower thermal diffusivity (and conductivity) than steel. However, aluminum has a thicker, adhering oxide layer as compared to steel.
-- JJW
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Jeffrey J Weimer wrote:

You are wrong. http://de.wikipedia.org/wiki/Temperaturleitf%C3%A4higkeit
(I didn't find such a nice table in english.)
Michael Dahms
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Ooops! You're right! I was reading my Ashby (Materials Selection in Mechanical Design, 3rd Ed) charts backwards! That would account for why Al feels hotter than steel. The charts also say, if you must use steel, use stainless.

(dass war aber kein Problem)
-- JJW
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Jeffrey J Weimer wrote:

My experience says the same.
Michael Dahms
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semidemiurge wrote, On 6/6/2007 11:09 PM:

Rick, Plant some pyracantha, berberis, or rose bushes around your shed. The plant's thorns will discourage the kids from touching the hot metal. ;-)
But seriously Rick, do you really think that kids touching the hot metal is going to to be a major problem?
We had a metal shed in the back yard of the house that I grew up in. And yes, that sheet metal got very hot in the summertime. But somehow I managed to survive this hazard (along with my brothers, my sister, and all the rest of the neighborhood kids).
--

Paul D Oosterhout
I work for SAIC (but I don't speak for SAIC)
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Paul, Yes it is a problem. Metal slides are now not allowed on school playgrounds due to burns caused by south facing slides getting extremely hot. That is why you are seeing less and less metal playstructures (unless coated with a thick layer of vinyl or powdercoat). Slides in particular are usually made of plastic. rick

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