Voltage drop calculation

I think what you are really asking is 'is it safe to estimate the AC voltage drop  from the result of a DC test plus one or other small angle approximation that avoids using pythagoras or complex numbers in it completely rigorous form ?' although you have not written it that way of course.

The answer has to be related to how far out  is the answer due to this effect  relative to all the other uncertainties. Clearly for cables sufficiently small cf the 50Hz skin depth, we can ignore the phase angle stuff due to cable inductance, and even when we do not it is a voltage drop without loss - we get the energy back later in the cycle.

For 50Hz the skin depth in copper is about 1cm, so the gently decaying current density with depth into the conductor is such that by 1cm it has fallen to 1/e of the surface value.  A core of 10mm dia is a touch over 75mm2, and by  then the idea of negligible phase angle is starting to fail.

OF course at 400Hz, as used in aircraft, or some higher frequency VFDs one cannot escape the full calculation on small wires as a higher frequencies for a  given inductance the impedance is proportionally higher.

What is a small angle - well let us assume sin x~ x is our approximation here (in rads of course) 1 cycle is 2pi rads, so 1 rad is ~ 57 degrees so we may expect sin of 1 degree to be 1/57 or more easily for mental maths about 1/60 - say 0.0166666 recurring or 0.0167 to 3SF Now my sine tables say sin 1 is 0.01745

Is the sine of ten degrees ten times this, and if it isn't, how far off are we. sin 10 is 0.1736 so down by about 1%

by 20 degrees sin 20= 0.342 we are off by about 4%

for a 10% error, you need to be out to about 40 degrees..

So you can indeed pretend that sin X=X, and for that matter that cos X = 1 - X²  and it works pretty will within +/- 30 to 40 degrees of zero.

So now we know the limits of the approximation we need to decide what given percentage error changes us from success to failure and then we can set some limits.

Without knowing the cable size you have in mind it is hard to be sure, but normally by the time inductance is biting hard at  50Hz, other factors like heat and mechanical impossibility of installation have forced us towards trefoil or quadrfoils of smaller conductors anyway.

I suspect in practice either approximation would do, and when one does not work, neither will the other.

Mike.

Thank you Mike, sterling stuff but a bit too late in the evening to digest. I am playing with a 2-core 240mm2 copper cable as 4D2A / B full current of 476A, 100m. I have been advised that I can ignore the (mv/A/m)r and use ONLY (mv/A/m)x.

Any way, for the normal UK method above I get 11.5v, for the Irish method 11.75v, for the less stringent method, oddly 9.14 and for using sin theta only, 4.3. 

I put my own Excel sheet together but it also checks long hand. Maybe someone with a cable calc programme would run it for a check.

Ah well, at 240mm2 you are well into the territory where the resistance is so low the self inductance cannot be ignored.

Neither can the mutual inductance, or in this case partial cancellation of the magnetic effects by the adjacent core. DC tests will give a silly answer at 50 Hz

Table suggest

millohms per meter

for 240mm there are 3 useful figures
0.180 resistance result on DC

note the slightly different resistance at AC -

0.190 real part AC
(as current is reduced nearer center of conductors so more has to flow in the reduced area at the edges - so skin depth increases resistive loss.)

0.150 imaginary part  AC (this is the energy put into and recovered in generating, destroying and then reversing the magnetic field in the space around the cable and is quarter of a cycle behind the resistive contribution)

0.24 total effect - now that should equal to the vector sum of the above. Lets check that.

     sqrt (0.19² + 0.15²) (full Pythag no approximations at 1AM here..) that is sqrt (0.0361 +0 .0225) = 0.242 milliohms per metre length

So table 3rd col is correct and not really needed.

Now you have 100m and 476A so  24.2 milliohms equivalent over the full length

and then at that current I*Z  = 11,5 V drop.

So either of your approximations is OK to within a few %, as expected.

regards Mike

PS beware of my posts made late at night - there may be beer related errors !!

PPS edited in the morning after to correct punctuation and a late night typo

I would take your capability, even with the beer handicap! Thank you as always for your full and measured response. The forum can count itself lucky to have you!

I forgot to mention in my post that the load current is at a power factor of 0.8 which is almost the same as the power factor of the cable. That is not always the case, so I assume that is why the sin part of the formula is included. 
I therefore assume that multiplying the voltage drop calculated using (mv/A/m)z by the load power factor of 0.8 would be incorrect.

Your point about skin effect is really interesting, I hadn’t noticed that in the tables. It appears to start having an impact around 95mm2.

I would take your capability, even with the beer handicap! Thank you as always for your full and measured response. The forum can count itself lucky to have you!

I forgot to mention in my post that the load current is at a power factor of 0.8 which is almost the same as the power factor of the cable. That is not always the case, so I assume that is why the sin part of the formula is included. 
I therefore assume that multiplying the voltage drop calculated using (mv/A/m)z by the load power factor of 0.8 would be incorrect.

Your point about skin effect is really interesting, I hadn’t noticed that in the tables. It appears to start having an impact around 95mm2.