Charging Current in High Voltage Cable

You can't without knowing the source impedance and the impedance between the source and the line. Even then it's tricky, as (obviously) the impedance between the source and and any point along the line will vary with the distance.

The same way one calculates the current it takes to charge a lumped element capacitor with the same capacitance.

It requires an understanding in differential equations. Circuit theory would help, too. If your source is AC instead of DC then you could even get into transmission line theory that Oliver Heaviside developed. The reflected wave from an open load will blow your mind, along with some possible fuses.

Go forth and educate thyself.

Refer to "Conductor Properties" (NEC Chapter 9 Table 8) for identifying the impedance of the cable. Be sure to read all of the footnotes and references.

Once you determine DC resistance/impedance per foot then multiply the total length of the cable by that factor to get total impedance.

Apply Ohms Law of I=E/R to determine charging current.

This method will only be as accurate as the parameters entered and only if DC voltage is used to charge the cable.

A call to the cable manufacturer will yield more accurate numbers on the impedance factors which will drastically improve results of the calculations.

Since you only seek the steady state solution Ohm's Law works fine, just remember to use the correct voltage (which is the entire point of this homework/quiz/interview question). How did you answer it?

Ahem, Ohm's law is for resistance not capacitance. There is no "C" in Ohm's law.

In AC there is, just substitute impedance for resistance as in I=V/Z, and as a reminder, they're all vector quantities.

Yes, all three variables are complex vector quantities. As I said, Oliver Heaviside's transmission line equations will be needed. Remember it was Oliver that coined the term "impedance".

Victor knows his vectors. Ask him.

Roger Murdock: Flight 2-0-9'er, you are cleared for take-off.
Captain Oveur: Roger!
Roger Murdock: Huh?
Tower voice: L.A. departure frequency, 123 point 9'er.
Captain Oveur: Roger!
Roger Murdock: Huh?
Victor Basta: Request vector, over.
Captain Oveur: What?
Tower voice: Flight 2-0-9'er cleared for vector 324.
Roger Murdock: We have clearance, Clarence.
Captain Oveur: Roger, Roger. What's our vector, Victor?
Tower voice: Tower's radio clearance, over!
Captain Oveur: That's Clarence Oveur. Over.

"How do we calculate the no load charging current in Amps"

What other units would it be?

nano, micro, milli, kilo, mega, giga, tera, peta

• If it is less than the current needed to satisfy the load requirements, then don't worry about it.
• If it is more than the current that the correctly-sized protective devices can withstand, then the devices will operate. So don't worry about it.
• If the cable fails before the protective devices operate, then the installation is faulty and will need to be reassessed and rebuilt. Proper Electrical Engineering beforehand will obviate that possibility. If a qualified Electrical Engineer has designed it, then don't worry about it.

In my opinion, in a 300 kV cable[173 kV phase-to-ground, See for instance NEXANS Voltage 160/275 (300)kV.]:

http://www.nexans.com/Corporate/2013/60-500_kV_High_Voltage_full_BD2.pdf

the capacitance does not depend on cable layout configuration but the resistance does.

Also the inductive reactance. If no-load is connected at the end of cable the resistance, reactance and

capacity will be in series. For a short length cable the simple "Ohm Law" could be applicable.

So, as already said, for a precise answer you have to know more about this cable.

However, for an approximate result you may neglect the resistance and even the inductive reactance, in my opinion. About 150 to 300 A per km[0.15 to 3 A per meter] it could be possible.

300 Amp per km??? So a 10km cable draws 3,000A before it carries any load? Especially curious when the largest 300kV cable in the Nexan catalog you cite (and makes no mention of charging current) has an ampacity of only 2,400A.

I would leave opinions to the Presidential debates, the engineering facts are as follows:

For single- conductor and three-conductor shielded cables

I= (0.323*f*k*kv)/(1,000*G) A/phase/mile, where f=frequency in Hz, k=dielectric constant of the insulation (XLPE is around 2.3), kv=line to line system voltage in kV, G= 2.303*log10 (2ri/d) where ri = inside radius of sheath and d = outside diameter of the conductorgeometric factor (which varies between 0.5 to 1.5) based on conductor shape, insulation thickness, and mean periphery of insulation.

Using 50Hz, 2.3, 300kV, and 1.0 in the above yields about 11.1A/ph/mile or approximately 7.0 A/ph/km, and can vary between 5 and 14 A/ph/km depending on the actual cable type. NB- inductive and capacitive reactances are very dependent upon the physical layout of the cables, while resistance isn't (except for the small proximity effect)

Thank you RAMConsult, my mistake: I divided by 18 the capacitance but also the reactance!

However, 9 to 14 A per km it is close to actual. If the capacitance-as declared- it is from 0.14 to 0.22 microF/km the the capacitive reactance has to be Xc=1/(2*pi()*f*Cap) f=60 Hz Xc=18947-12057ohm I=300000/SQRT(3)/Xc I=9.14-14.36 A/km

Calculated the capacitance is less a bit [0.14 it is 0.12 and 0.22 it is 0.19] but a think the manufacturer knows it better.