simple skin effect resistance ratio calculation using S.H.E.E. 3

[electricpete]

Let's say I have a 1 mm solid copper conductor (assume conductivity 58 MS / m) carrying 10khz.

By two different means (finite element and copper conductor, I calculate the same skin effect resistance ratio: Rac / Rdc = 1.0068

Here is the finite element solution:

http://home.comcast.net/~electricpete/eng-tips/SingleConductorPost.ppt

Here is an analytical solution based on a textbook.

http://home.comcast.net/~electricpete/eng-tips/SkinEffectPost.pdf

I cannot come close to recreating that result using the Standard Handbook For Electrical Engineers (S.H.E.E.) formula in chapter 4 Table 4-2. Can anyone recreate this result using the SHEE? x should be around 1.l but I can't get that.

One thing I stumble over is the units. I notice they request resistivity in ab-ohm centimeters. That is a crazy unit even though it is defined their (ab-ohm = 1E-9 ohm). The annoying thing is they should either let us pick our our units or else tell us the units for all the variables but they don't. I guess it should be cgs ? (cm gives me that clue although ab-ohms doesn't). But I remember mu0 (and epsilon0 for that matter) in cgs are a little strange (I think that starting with SI mu0=4*piE-7 and applying unit conversions doesn't necessarily get you to cgs mu0). I tried it as mu0=1 but stilll didn't work.

What is the physical significance of this x parameter they use? If you work it out it has units like ohm-m (what is the physical significance).

Last gripe - Why the heck didn't they use a dimensionless formulation? I have a dimensionless solution in my pdf file – see the plot Rac/Rdc vs Rn where Rn = radius over skin depth. Should be very easy to enter the table using a simple argument and let the user pick his preferred units. And the argument has a very simple phsycial signficiance (radius as multiple of skin depth) instaed of some undefined x. Am I missing something?


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[electricpete]

Must have been late when I was typing last night
Correction in bold:

By two different means (finite element and copper conductor, I calculate the same skin effect resistance ratio

should be

By two different means (finite element and analytical calculation, I calculate the same skin effect resistance ratio

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[jghrist]

Which Standard Handbook For Electrical Engineers? Table 4-2 in Fink & Beaty 14th Edition is not related. What result is in your S.H.E.E.?

 

[electricpete]

Here is the relevant excerpt

http://home.comcast.net/~electricpete/eng-tips/SkinEffect.ppt

For the example that I gave above (1 mm solid copper wire at 10khz), if my calculations are correct we should expect K = R / Rdc = 1.0067. To get this value from table 4-2, we would need x between 1.0 and 1.1 (around 1.07). I have tried every combination of units I can think of, but I can't come up with x anywhere near 1.07



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[7anoter4]

Hi electricpete,
It is a very interesting issue and I have to declare that since I was student I never used Bessel or related Bessel functions. Then we used to calculate skin effect in squirrel cage deep bar and also once for ground fault current return impedance.
For skin effect I always used IEC287 formulae.
Now in thread238-235556, bacon4life gave a link:

http://www.copperinfo.co.uk/busbars/pub22-copper-for-busbars/sec4.htm

The skin effect formula is close to yours:
X=pi*d*sqrt(2*f*miu*10^(-5)/ro)
Where
d = diameter of rod, mm
f = frequency, Hz
r = resistivity, micro ohm* cm
miu = permeability[relative] of copper (=1)
For HC copper at 20°C, r = 1.724 micro ohm cm, hence:
X=1.069*d*sqrt(f)
So for d=1mm and f=10000 Hz x=1.069 indeed.
K is S in this article and the result match the table.
The IEC 287 formulas are:
Xs^2= 8*PI()*FRQ/R(TC@)/10^7*KSKIN
R(TC@)=Conductor resistance ohm/m for TC temperature.
Kskin depends upon conductor type. Usually kskin=1.
If scu =pi*d^2/4=0.785
R(TC@)= 1/58*1[m]/0.785=0.022
Xs^2=8*pi*10000/0.022/10^7=1.145
Ys=( xs^2)^2 /(192+.8*(xs^2)^2)= 0.006789673
Skin factor =1+ys=1.0068
Best Regards

 

[jghrist]

I was able to reproduce both Table 4-2 (Table 5 in Westinghouse T&D) and K=1.00679 for a radius of 1/2 mm. See attached Mathcad calcs.