cupoftea
Advanced Member level 5
Hi,
This sil pad has thermal conductivity of 6.6 w.m/m^2.k
So if we have a bar of it which is 1 metre long and 1 m^2 in area, and we have a 6.6w source at the source end, then we see a 1 degc drop at the other end.
Regarding the sil pad..
Its size is 15mm by 20mm.
As such its area is 300e-6 m^2
Its thickness is 0.178mm
Since its 6.6 w.m/m^2.k, then this means its also 1 w.m/m^2. (0.15K) [dividing top and bottom by 6.6]
So for a metre squared block, 1 metre long, the temperature would be 0.15degC lower at the end of the block, if there was 1 Watt at the "heat source" end.
So anyway, we only have 0.178mm, not one metre, so in fact, the temperature will be much nearer to the source temp.
In fact, it will be 0.000178 * 0.15degC = 26.7e-6 degC. (ie that s the temp difference)
But we also have less area, so we must scale the temperature figure upward for that.
So we must multiply degC by 1/300e-6....so we get 26.7e-6 * 1/(300e-6) = 0.089degc/WATT.
Bu this sounds too low.
Would you agree?
This sil pad has thermal conductivity of 6.6 w.m/m^2.k
So if we have a bar of it which is 1 metre long and 1 m^2 in area, and we have a 6.6w source at the source end, then we see a 1 degc drop at the other end.
Regarding the sil pad..
Its size is 15mm by 20mm.
As such its area is 300e-6 m^2
Its thickness is 0.178mm
Since its 6.6 w.m/m^2.k, then this means its also 1 w.m/m^2. (0.15K) [dividing top and bottom by 6.6]
So for a metre squared block, 1 metre long, the temperature would be 0.15degC lower at the end of the block, if there was 1 Watt at the "heat source" end.
So anyway, we only have 0.178mm, not one metre, so in fact, the temperature will be much nearer to the source temp.
In fact, it will be 0.000178 * 0.15degC = 26.7e-6 degC. (ie that s the temp difference)
But we also have less area, so we must scale the temperature figure upward for that.
So we must multiply degC by 1/300e-6....so we get 26.7e-6 * 1/(300e-6) = 0.089degc/WATT.
Bu this sounds too low.
Would you agree?