Okay, thanks Dave and Zekeman. So you're saying not to really worry about the center since the heat transfer is negligible there?
With that being the case would you recommend using an inner radius that is oh say... 5% of the total radius?
Now, sorry to repeat earlier portions of the thread...
So you're not using:
ln(r3/r2) ln(r2/r1)
Req = -------------------- + ------------------
2*pi*K1*L1 2*pi*K2*L2
?
Instead you're saying to use r3-r2 and r2-r1 in the numerators?
This would solve the center radius of 0 problem but I'm not...
Dave,
What are your comments on what I said about the inner radius being zero? I know in all actuality that it's not really zero and that there really is an inner radius but it just seems strange to me how much the result changes the closer you get to zero.
It also seems strange to me that when using the 3D (x,y,z) formula the hot-spot is higher. The heat has 3 directions to go, whereas with the 1D (r) formula only has one.
Granted, the two formulas take generation into account in very different ways so this may not be a very good comparison...
Actually if anything's off I'd think it would be Q'''. The formula I think we settled on was:
Q'''=Power/Length/(PI()*Radius^2)
However since we're considering the problem as an infinitely long cylinder, I'm not convinced we should be using Length.
I do agree that it does not make sense to...
zekeman,
I have been using the same formula as you to get the Keq. I used that value in the 1D equation you mention but that's the one I said is undershooting what the actual hotspot temperature is. The 3D equation I mentioned above is getting much closer to the test results so I'm thinking...
Dave,
Thanks a lot for your extensive reply.
First to answer your question, yes it comes pretty close to what we usually observe for outer surface heat rise for individual windings in an open environment. In cases where the winding is embedded in a potting material, we generally get this...
The full equation would be:
dx*dy*dz
T = Ta + Tr + ----------------------------------------
k(dx*dy*D*L+pi*r^2*(dx+dy)*dz)
Where
T = Hotspot temp
Ta = Ambient temp
Tr = Calculated surface temperature rise
Tr is another approximated...
The cap is not the "can" form. It is just the winding element. Also the the thickness of the current carrying element and the dielectric is extremely thin in comparison to the total diameter, so I think the homogeneous equation should give a pretty good approximation.
Also I've got some...
Also, wouldn't it simplify down even more when substituting .5L for z?
To + q'''L²/8k + 4L²q'''/[k*pi³] *
? sin[(n+.5)*pi]
? [--------------------------------]
n=0 (2n+1)³*Io[(2n+1)*pi*a/L]
Now that I look at it again is the pi supposed to be inside or outside the sine function?
sin[(n+.5)*pi]
--------------------------------
(2n+1)^3*Io[(2n+1)*pi*a/L]
Thanks. And yeah that's pretty complicated.
Just a couple of questions:
1) Is the attached equation the form of the bessel function you're using? It's been a while since I've worked with bessel functions (college calc 2 I think).
2) I'm a little lost in this portion of it...
Thanks again guys.
Dave,
When you say q''' would be "watts of heat generated per unit length divided by the cross sectional area", that would effectively be P/(L*pi*(D/2)^2) correct?
Zekeman and Ione,
You mention convective heat transfer which I wasn't real sure whether to include or not...
Thanks Dave,
I just figured this 1D equation for an infinite cylinder would be similar to a 2D equation for a finite cylinder. I'm also afraid I can't assume uniform heat since the goal is to try and calculate the hotspot temperature based on surface temp and the power dissipated in the cap...
