Thermal conductivity of thermal insulation pads

[Faringdon]

Hi
Please advise on the meaning of Thermal conductivity.
The units are Watts per (Metre.Kelvin)

So if Thermal conductivity is 1,  then it means that if a one metre long copper bar of uniform area is in contact with a heat source at one of its ends, then that means that for every 1 watt dissipated, there is a  1 degree reduction in temperature at the other end of the 1 metre bar?

And if its 1000W/(m.K), then that means that for a uniform area copper bar connected at one of a heat source, the source would have to dissipate 1000W to get a 1 kelvin drop across the length of that 1 metre bar?

[jpanhalt]

Understanding the units is important, and I am one who believes it is particularly important to carry them along during calculations as a check.  Did you bother to look it up for examples?
https://en.wikipedia.org/wiki/Thermal_conductivity#/media/File:Thermal_conductivity.svg

Judging from the plethora of recent posts, I am assuming you have a lot of time on your hands again or are you trying to solve real problems assigned to you by a new employer?

[Kean]

Someone really needs to stop posting on forums and read some engineering books.  That would be me, but even more so the OP.

[ConKbot]

W/(m k) is in a simplified/reduced form that doesn't make intuitive sense. (W m)/(m^2 K) is the full form and is more intuitive, with distance on top, area on the bottom.

[Siwastaja]

If you get the units right without cheating, you got the calculation right, too. I often forget formulas or they are sometimes hard to find from a simple Wikipedia lookup, but units are easier to remember. Work out the formula from the units.

Unit of length is m, and unit of area is m^2, so if you forget on which sides of division they are, resulting unit ends up wrong. Repeat until units are correct, and your temperature difference does not contain 1/m, m or m^2 in it; unit of temperature is just K. Not 1/K, for example!

[Mechatrommer]

recap younger dave...

[Faringdon]

W/(m k) is in a simplified/reduced form that doesn't make intuitive sense. (W m)/(m^2 K) is the full form and is more intuitive, with distance on top, area on the bottom.

Thanks for this.
Yes, you can see that for every 1 metre you go along a bar of unit area........you loose a Kelvin for every watt at the source end.