lets say one applies a 50 Hz sinusoidal voltage to a conductor, the resultant a.c resistance is slightly higher than that of it's d.c resistance due to the skin effect. now let's suppose that we superimpose (in effect connecting in series) a dc supply. we will now have 2 components of current, a dc one and an ac one. will the ratio of (ac volts/ac amps) be equal to (dc volts/dc amps) or will the dc resistance still be it's original (slightly lower) value. i ask this due to the fact that accepted explanation for the skin effect is that the current carriers increasingly move away from the core centre as the frequency increases....thus reducing effectively the csa of the cable core. if this is so then this condition would still exist for the dc component current as the electron current carriers will not be at the core centre any more. it's interesting since we have what is theoretically a linear problem i.e 1 that lends itself to superposition....when in fact it may not be the case.
On 7 Aug 2003 19:32:33 -0700, snipped-for-privacy@energy.com.au (Daniel Indyk) Gave us:
Not at 50Hz. The skin is so "thick" at that cycle rate that the resistance is virtually the same. Negligibly different.
Its current carrying capacity MAY be slightly lower, but I kinda doubt it at that frequency.
Wire resistance values are given as DC values, and AC values are called impedance, AND only work if the wire is formed into an inductor via coiling, and the addition of a core other than air, in this case. In other words a load element should drop just as much with the AC RMS value as it does with the DC stimulus.
I see it as the current carriers do not move away from the core, they never have time to propagate that deep to begin with. At 50Hz, however, they do propagate deeply. Technically, not as deep as your DC resistance though, so in the strictest terms, you are correct. There would be slightly less copper available to the source to drop voltage onto. I don't know how little that is though. I think that it's very small.
It is a field effect. I find no need to use Litz wire in my primaries at the switching frequencies I utilize, which are from 30 to a hundred or so kHz. The DC I squared R factor for bigger wire is far more useful than the gains provided by the litz wire. Of course, I am talking about miniaturized power circuits, not big, industrial grid sources or such.
You can test you theory. A bundle of litz wire at the same length as a single conductor should allow for some good characterizations. If 50Hz does have an effect, you should be able to lose less at the end of a length of Litz than you would an identical circular mil equivalent single conductor. The litz configuration must be coated, separate wires, only connected to each other at the ends.
Wire size doubles every three full gauge numbers for AWG sized wire.
So take eight or twelve as a multiplier (strand count), and make a litz wire that consists of 8 or 12 wires of a size that ends up matching the #12 area.
Then, may the tests begin. The difference you would see in the litz wire over the single strand would be an amplified effect of the one you describe. If the effect isn't much with a litz wire, it will be even less on the single wire tests.
For what it's worth, at 50 Hz the observed resistance (yes, ohmic resistrance) of a circular conductor will be about 1.0013 times the dc resistance for a non magnetic circular conductor. At 500 Hz, the observed resistance is about 1.5% greater, and at 1000, 8% more.
in article L4JYa.20232$ snipped-for-privacy@news-server.bigpond.net.au, Daniel Indyk at snipped-for-privacy@mail.bigpond.com wrote on 8/8/03 1:22 AM:
This repost is as ridiculous as the original post. Nevertheless, it may make some sense.
The system is still linear. That is, if the response to V1 is I1 and the response to V2 is I2, the response V1 + V2 is I1 + I2. V1 can be at dc and V2 can be ac. The impedance Z1 = V1/I1 will not equal the impedance Z2 = V2/I2. V2 can be frequency dependent upon frequency. In addition to an increase series resistance by skin effect on Z2, there also is an increase in inductive reactance from skin effect.
in article snipped-for-privacy@mb-m07.wmconnect.com, tony at snipped-for-privacy@wmconnect.comremoove wrote on 8/8/03 4:31 AM:
This is certainly not true in general. These changes depend upon the dimensions of the conductor. The skin depth is proportional to the square root of the frequency. When the depth is large compared to the size of the conductor, a more detailed calculation is necessary.
For 60Hz, the depth is 8.52 mm. Thus at 50 Hz, it is 9.3 mm. For 500 Hz it is 0.93 mm. For 1000 Hz, it is 0.66 mm. How does that compare to your rule of thumb?
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