Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations waross on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Conductivity of Lexan

Status
Not open for further replies.

duncanelliot

Civil/Environmental
Feb 8, 2006
10
GB
Hi there,

I have a need to understand the thermal properties of Lexan and was hoping one of you might be able to help.

I hold an Msc in Civil Engineering but only have a basic knowledge of physics (A-level) so please bear that in mind if you reply.

PROBLEM:

1- I have a choice between two thicknesses of Lexan, 6 and 12mm and I have to decide which to use as a roofing material.

2- I want to maximize the amount of heat coming in in terms of solar radiation; which I believe manifests itself in term of infra red light (is visual-spectrum light important when it comes to heating the air inside, bearing in mind that the building is in the UK and we don't get much sunlight).

3- It would be nice to know the relationship is between thickness of the material (M) and the amount of infra red energy which passes through the panel and also to find out what the relationship between thickness (M) and thermal conductivity is.

Thanks,

Duncs
 
Replies continue below

Recommended for you

Both relationships are linear.

There is not any memory with less satisfaction than the memory of some temptation we resisted.
- James Branch Cabell
 
Optical transmission is not linear with thickness. Double the thickness, square the transmission. Thermal conductivity, though, is inversely proportional to thickness.

TTFN

FAQ731-376
 
The Beer-Lambert law says the absorption is proportional to the thickness. Does the transmission behave differently?


There is not any memory with less satisfaction than the memory of some temptation we resisted.
- James Branch Cabell
 
I'm with Demon 3.

Depending on the nature and composition of the surface, light in any wavelength can do one or more of four things when it strikes a sheet.

These are.

Be reflected away with no impact on the surface or anything beyond it.

Be scattered but transmitted in a random pattern. A reasonable portion will pass though the sheet to the other side. Somw will be refracted to the point that it comes out the original side or the edges.

Be transmitted straight through.

Be absorbed.

The heat inside the building comes from light being absorbed and converted to heat according to the laws of conservation of energy. This can happen in the sheet whereas the sheet gets hot and conducts heat to the surface and the surrounding air, or it can happen as light is absorbed by objects inside the room.

To test, place otherwise similar black and white (or at least dark and light coloured) objects in similar positions near a window facing the sun. Leave in direct sunlight for say half an hour. Touch each object.

This is a common mistake made by people who tint glass to reduce heat by applying a dark film to the inside of the glass. The contents of the room do not generate as much heat, but the tint on the inside of the glass does. It is still inside the room.

Regards

eng-tips, by professional engineers for professional engineers
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
 
Absorption coefficient is actually a constant with units of inverse thickness.

But, the Beer-Lambert transmission is given by exp(-alpha*thickness), so double the thickness results in transmission squared.


TTFN

FAQ731-376
 
That's why you add ODs with neutral density filters, but ODs are like dB, OD1 os 10% transmission. OD2 is 1% transmission. OD3 is 0.1% transmission, etc.

TTFN

FAQ731-376
 
most of the solar energy comes from the visible portion of the spectrum, when absorbed it will be converted to thermal energy
{infrared only appears to be somewhat important when you have no VISIBLE energy but can feel the heat}

since the lexan has low absorption in the visible spectrum (assuming no coatings or tints) go for the thicker material which will transmitt almost as much solar energy but provide better insulation (but higher cost)
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor

Back
Top