wwhitney
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- Berkeley, CA
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But ampacity wasn't the question, it was voltage drop. Why is the reactance of the 500 kcmil just a little less than that of the 250 kcmil?
Cheers ,Wayne
Single Conductor or Two in Parallel: 3/0 or 500 kcmil for 400A
- Thread starter paullmullen
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Cheers ,Wayne
kwired
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If the voltage drop is less then Ohm's law says the resistance (actually the overall impedance and not just resistance) must be less doesn't it?
electrofelon
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- Cherry Valley NY, Seattle, WA
Are proximity effect and skin effect considered part of the impedance?
Tulsa Electrician
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I would like more information on the run before I decide which I would do.
synchro
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I'm not a physicist, but I did take two years of physics and a following year in electromagnetic fields, etc. But that was long ago.
To understand why the reactance drops slowly with conductor size, consider the case of two adjacent conductors of the same gauge. The voltages developed across a given length of conductors 1 and 2, where each has a resistance R and with applied currents I1 and I2 respectively in the same direction, are:
V1 = I1 (R + jωL1) + I2 ( jωM )
V2 = I2 (R + jωL2) + I1 ( jωM )
L1 and L2 are the self inductances of wires 1 and 2, and M is the mutual inductance between them.
If the two conductors are connected in parallel then their individual voltage drops would be the same and so let V = V1 = V2. We can assume L1 = L2 = L and an applied current I splits evenly between the two conductors such that I1 = I2 = I/2 because of symmetry.
Substituting in the equations above, we get:
V = I/2 (R + jωL) + I /2 ( jωM)
= I [R/2 + jω( L+M )/2 ]
The mutual inductance M between the two conductors will be less than their self inductance L, but it can be significant when the conductors are physically close and therefore surrounded by approximately the same magnetic field.The currents are in the same direction and so they are each contributing to the magnetic field instead of cancelling it.
You can see that the mutual inductance M above prevents the inductive reactance ω( L+M )/2 from dropping by a factor of two from the reactance ωL with one conductor, which it would if the mutual inductance were not present and M was equal to zero. And so going to a larger diameter conductor will also have a similar effect. Currents flowing along the conductor at different locations on its cross sectional area will have mutual couplings with each other, and therefore develop additional emf's (induced voltages) that add to those from the self inductance. As a result, the inductance does not fall in direct proportion to the cross sectional area of the conductor.
By the way, in Static and Dynamic Electricity by Smythe the inductance per unit length of two parallel conductors of radius a, and center-to-center spacing b is:
L = μ/4π (1 + 4 ln(b/a) ) where μ is the permeability of the medium, usually free space.
The current density in the conductors is assumed to be uniform. This is a reasonable assumption if the skin and proximity effects are small.
Note that if the ratio of conductor center-to-center spacing to the diameter is maintained, the inductance from this formula does not get lower as the conductor diameter is increased. In practical installations, this spacing may not increase proportionally to the conductor diameter. For one thing, the insulation thickness does not change much when going to larger conductor sizes. And so that might be why the reactance is reduced on larger conductors in the tables of the NEC and the manufacturers.
paullmullen
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Short question. Lot's of nuance. Thanks everyone.
wwhitney
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Based on the physics behind Note 2 to Chapter 9 Table 9, it turns out that any difference is not significant for the special case when the power factor is 1. That is:
The voltage change in the wire is I * Z, where Z is the impedance, Z = (R, X) in vector form. To first order, only the component of I * Z that is in phase with (parallel to as vectors) the voltage will affect the magnitude of the voltage; the component of I * Z that is 90 degrees out of phase with (perpendicular to as vectors) the voltage will instead induce a phase shift in the voltage, without changing the magnitude. [This is an approximation for the case I * Z is small compared to V.] Note 2 calls the parallel component of I * Z the effective impedance Zeff.
As the power factor of an AC circuit varies from 1 down to 0, the phase angle between the current and the voltage varies from 0 up to 90 degrees. As this happens, Zeff changes from being purely the resistance at PF=1 to purely the inductance at PF=0. At intermediate power factors, if A is the power factor angle given by PF = cos A, then Zeff = R cos A + X sin A, which blends the resistance and reactance.
Looking at Chapter 9, Table 9, the inductance of a wire is not too far from constant as the wire size changes (for PVC conduit, in ohms/km, 0.190 for #14 AWG vs 0.121 at 1000 MCM, a change of less than a factor of 1.6), while the resistance of the wire changes inversely with area (for copper in PVC, again in ohms/km, 10.2 for #14 AWG vs 0.049 for 1000 MCM, a 200 fold change).
One conclusion from this is that for smaller wire sizes, X sin A will be much smaller than R cos A, and in calculating Zeff the inductance can almost be ignored. Another conclusion is that for larger sizes, when the power factor is not very close to 1 and the reactance matters for calculating Zeff, the larger sizes do comparably worse as far as voltage drop, because while their resistance drops with size, their inductance hardly does.
Cheers, Wayne
wwhitney
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Thanks very much for your response. I'm not sure I followed all the details of the first part, but the above is very clear if I'm willing to accept that formula. Lengths in it appear only as the ratio b/a, so as you say as a first approximation you would expect the inductance L to be independent of diameter.
Cheers, Wayne
ggunn
PE (Electrical), NABCEP certified
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Did any of it show enough of a difference to drive the decision?
Tulsa Electrician
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- Tulsa
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- Electrician
Bummer.
I was going at it from a D or Distance and I, load in amps
Anyway, the 2- 3/0 can not be ran as far as a single 500 based on CM.
I was hoping you would have shared voltage and distance.
The two 3/0 will give you around 335 Mcm and this is where one may choose to use 500.
If short runs two 3/0 may be favored over a single 500
Cost being one of them.
For a hundred foot run the two 3/0 may be favored over that the 500 may be favored or a increase in size to a 4/0.
I would do two things before deciding. Do a basic CM required.
Then run the cost of one run or two based on where your running the raceway.
I also would use approximate K 12.9 for CU. Which would be close enough for the CM or D.
I would also consider the pulling of the wire. Sometimes it's easier to the 3/0 with out pulling equipment.
Tulsa Electrician
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For a hundred foot run the two 3/0 may be favored over that the 500 may be favored or a increase in size to a 4/0
I should have wrote , for a 200' run an increase in size may be favored.
Depending on voltage of circuit.
Sorry for that.
I should have wrote , for a 200' run an increase in size may be favored.
Depending on voltage of circuit.
Sorry for that.
electrofelon
Senior Member
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- Cherry Valley NY, Seattle, WA
I never stated it, but my "the 2x250's have a lower voltage drop" statement is based on the results from the southwire VD calculator. 100A, 500 feet, Al conductors:
for .8 PF:
2x250=.0847 resistance, .041 reactance, VD 4.28V
1x500= .0424 resistance, .039 reactance, VD 5.09V
for 1.o PF:
2x250= .0847 resistance, .041 reactance, VD 5.13V
1x500= .0424 resistance, .039 reactance, VD 5.94V
For the parallel case, they seem to give the resistance results per conductor, but the reactance number for the set? Odd.
wwhitney
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I figured. Also, their results for PF 1.0 do not match what Chapter 9, Table 9, Note 2 would give.
No, that is still the reactance per conductor. That's part of the point, double the area, the resistance basically halves but the reactance hardly changes.
Cheers, Wayne
- Location
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Well, 250 kcmil is certainly getting into the diameter where skin effect becomes noticeable, even at 60 Hz. Look at the tables which show both the AC and the DC resistive component of impedance. Figuring in the effect of inductance on voltage drop only makes it worse.
kwired
Electron manager
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Proximity effect sounds to me like capacitance - so yes.
Skin effect, I haven't studied all that much but is more of an issue at higher frequency. I suppose you can say it adds impedance whether it is resistive, inductive, capacitive IDK. I don't think it should add anything to ampacity calculations for 60 hertz applications using NEC T310.16 as that table likely has already factored this in if it even matters at 60 Hz.
wwhitney
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If we look at Chapter 9, Tables 8 and 9 for the DC resistance vs the resistive component of the 60 Hz AC impedance (in PVC conduit) for 250 kcmil Al vs 500 kcmil Al, we find (ohms/km):
250 kcmil Al 0.2778 DC 0.279 AC
500 kcmil Al 0.1391 DC 0.141 AC
So if "skin effect" is the ratio of the AC to DC resistances, that's 1.004 at 250 kcmil Al and 1.014 at 500 kcmil Al. Or an increase of 0.4% at 250 kcmil and 1.4% at 500 kcmil.
But the reactance effect on AC impedance as the load power factor changes is much larger than that at these sizes.
Cheers, Wayne
ggunn
PE (Electrical), NABCEP certified
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- Consulting Electrical Engineer - Photovoltaic Systems
Aren't skin effect and reactance (capacitance and inductive) two different things?
wwhitney
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Yes, as I said at these sizes the reactance effect as the load power factor varies is a much bigger effect that any skin effect on resistance.
Cheers, Wayne
zbang
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IIRC, the "skin effect" at 60Hz is something like a quarter inch, so unless you're using rather large conductors it's won't matter much.
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Which is why I have said skin effect is basically a negligible consideration when using the NEC table values.
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