Thermal conductivity λ, perforated plate.

[Anonymous]

I would like to designate the thermal conductivity λ of a perforated polycarbonate plate. I added a new material with natural properties. I uploaded a model in which the holes were filled with cylinders to which I assigned air. I have given the boundary conditions, 10 and 30 degrees Celsius, and I am trying to determine the thermal conductivity. Unfortunately, the results I get do not match those I got in the lab. Does anyone know the solution? 

For this tile, the thermal conductivity coefficient is 0.16, in the calculations from the program I got the result of 0.34

If anyone knows a better way to find the thermal conductivity coefficient for the entire plate, please share it. I calculated it from the formula λ = q d / ΔT. The thermal conductivity coefficient should decrease relative to the full plate, for which λ = 0.2

[Anonymous]

I would like to designate the thermal conductivity λ of a perforated polycarbonate plate. I added a new material with natural properties. I uploaded a model in which the holes were filled with cylinders to which I assigned air. I have given the boundary conditions, 10 and 30 degrees Celsius, and I am trying to determine the thermal conductivity. Unfortunately, the results I get do not match those I got in the lab. Does anyone know the solution? 

For this tile, the thermal conductivity coefficient is 0.16, in the calculations from the program I got the result of 0.34

If anyone knows a better way to find the thermal conductivity coefficient for the entire plate, please share it. I calculated it from the formula λ = q d / ΔT. The thermal conductivity coefficient should decrease relative to the full plate, for which λ = 0.2

[marwan_azzam]

Hello,

I'm not sure what you mean by "thermal conductivity coefficient" .

Is it Thermal Conductivity (W/m.C) or something else?

The file you attached must be corrupt and it didn't unzip so I'm not sure what your setup looks like.

I created the attached simple model.  It's just a Heat Transfer analysis, no need for air and fluid flow.

Based on the results of the Wall Calculator providing us the Heat Flux, Q, we get, we can back calculate Thermal Conductivity based on your equation above.

When the plate is perforated, Q is scaled back based on the surface area.

As such, there would be no need to do any analysis and you can calculate the thermal conductivity based on the area.

Please let us know if that's what you are after and if your question is answered.

cheers,

Marwan

[marwan_azzam]

Hello,

I'm not sure what you mean by "thermal conductivity coefficient" .

Is it Thermal Conductivity (W/m.C) or something else?

The file you attached must be corrupt and it didn't unzip so I'm not sure what your setup looks like.

I created the attached simple model.  It's just a Heat Transfer analysis, no need for air and fluid flow.

Based on the results of the Wall Calculator providing us the Heat Flux, Q, we get, we can back calculate Thermal Conductivity based on your equation above.

When the plate is perforated, Q is scaled back based on the surface area.

As such, there would be no need to do any analysis and you can calculate the thermal conductivity based on the area.

Please let us know if that's what you are after and if your question is answered.

cheers,

Marwan

[Anonymous]

I'm not sure what you mean by "thermal conductivity coefficient" .

Is it Thermal Conductivity (W/m.C) or something else?

Yes it is Thermal Conducitivity which for polycarbone, should be 0,19-0,22 W/(mK)

The file you attached must be corrupt and it didn't unzip so I'm not sure what your setup looks like.

I'm attaching it once again

I created the attached simple model.  It's just a Heat Transfer analysis, no need for air and fluid flow.

Based on the results of the Wall Calculator providing us the Heat Flux, Q, we get, we can back calculate Thermal Conductivity based on your equation above.

When the plate is perforated, Q is scaled back based on the surface area.

As such, there would be no need to do any analysis and you can calculate the thermal conductivity based on the area.

Air has a greater thermal resistance than polycarbonate. By drilling holes, we increase the volume of air in the sample volume, we should increase the thermal resistance of the entire plate.
It seems to me that the sample with the holes only, not treated as areas filled with air, does not reflect the actual conditions, and the program will always measure the thermal resistance of the polycarbonate itself.

Summarizing, I would like to determine from the equation, Thermal Conductivity, a plate consisting of, for example, 90% polycarbonate and 10% air that is in the holes.

[Anonymous]

I'm not sure what you mean by "thermal conductivity coefficient" .

Is it Thermal Conductivity (W/m.C) or something else?

Yes it is Thermal Conducitivity which for polycarbone, should be 0,19-0,22 W/(mK)

The file you attached must be corrupt and it didn't unzip so I'm not sure what your setup looks like.

I'm attaching it once again

I created the attached simple model.  It's just a Heat Transfer analysis, no need for air and fluid flow.

Based on the results of the Wall Calculator providing us the Heat Flux, Q, we get, we can back calculate Thermal Conductivity based on your equation above.

When the plate is perforated, Q is scaled back based on the surface area.

As such, there would be no need to do any analysis and you can calculate the thermal conductivity based on the area.

Air has a greater thermal resistance than polycarbonate. By drilling holes, we increase the volume of air in the sample volume, we should increase the thermal resistance of the entire plate.
It seems to me that the sample with the holes only, not treated as areas filled with air, does not reflect the actual conditions, and the program will always measure the thermal resistance of the polycarbonate itself.

Summarizing, I would like to determine from the equation, Thermal Conductivity, a plate consisting of, for example, 90% polycarbonate and 10% air that is in the holes.

[marwan_azzam · Accepted solution]

Hello,

Thank you for posting the cfz file.

Please use Solid Air as the material for the holes as in attached model.

The numbers should work out.

Marwan

[marwan_azzam · Accepted solution]

Hello,

Thank you for posting the cfz file.

Please use Solid Air as the material for the holes as in attached model.

The numbers should work out.

Marwan

[Anonymous]

Yes, thank you, it finally works! However, when I try to replicate your conditions for the remaining tiles, I do not get the heat flux value. What am I doing wrong? I am attaching a file with 5mm holes.

[Anonymous]

Yes, thank you, it finally works! However, when I try to replicate your conditions for the remaining tiles, I do not get the heat flux value. What am I doing wrong? I am attaching a file with 5mm holes.

[marwan_azzam]

Hello,

It is acting strange depending on whether the material is from Local or Default material library.  Not sure why that is.

Marwan

[marwan_azzam]

Hello,

It is acting strange depending on whether the material is from Local or Default material library.  Not sure why that is.

Marwan

[Anonymous]

Ok, thank you very much anyway.

[Anonymous]

Ok, thank you very much anyway.

[marwan_azzam]

Hi,

This zip file you sent is missing all the required folders.

Please send a zip file of the support share file (*.cfz).

Marwan

[marwan_azzam]

Hi,

This zip file you sent is missing all the required folders.

Please send a zip file of the support share file (*.cfz).

Marwan