Power cable capacitance

To first state the obvious; unless these cables are multiconductor cables, the capacitance per unit length cannot be determined. But when a manufacturer does not specify cable length capacitance numbers (or any other parameter) this implies that they cannot guarantee any values. This can be due to many factors. So don't expect any controlled capacitance values. Also, the NYY cable specifications I saw on the web have two hold off voltage grades, 600V and 1kV. If you are distributing 1kV then you have the wrong cable to be safe.

cable is only has impedance datas, you can use to calculate the voltage drop & line power loss.

here the impedance datas for ....

positif sequence: 0.3170 (R), 0.0792(X), 0(T)

Zero sequence: 0.49712 (R), 0.20196(X), 0(T)

base on IEC 60245 (rubber insulated cable)

Any of the cable manufacturer does not give the capacitance for any cable they produce. The reason is very simple.

Capacity for cables depends on many factors. Such as capacitance between other phase lines, capacitance between ground, capacitance with other cables in proximity. Of course capacitance is strictly dependent with the cable length, considering it is the electrode of the capacitor.

Unless the transmission is a High voltage line and the lines are not quite long, the capacitance of the cable has no effect.

In the event of 50 0r 60 Hz. power transmission (such as 1000 meter or shorter cable length) and of course below 600 Volt, the effect of the capacitance can be omitted. It is so small that no need to be taken into account.

However you can test and calculate to find it out the real capacitance if you desire for a certain situation.

Loose all the connections and the bridges, of which can estabilish ohmic connections. (Verify there is no connection by using an Ohmmeter or better with a megger.)

Then either directly meter it by an instrument if you have. (By using a good quality digital multimeter of which also capable to measure capacity),

Or calculate using the old method. Apply enough alternative voltage (Possibly as high as mains voltage) across the points to measure the capacitance. (This is U)

Try to meter a readible current value. (Possibly in few milliamps or less). (This is I)

Divide the applied voltage to the measured current. The result is the value of the capacitive reactance of the cable. (Xc) (Xc=U/I)

Use the formula Xc=1/(2xPixFxC ) . Now only C is unknown and this is the value you want to find out. You can calculate it.

Since F is Frequency for the voltage you applied in the test, and C is the Capacity in your question (Pls take cae the value in the formule is Farad).

2 and the Pi are the constant values.

Hope this is the reply for your question.

Kindest Regards.

Hello.

I'd like to slightly disagree. The mA established in the line after the nominal AC voltage is applied between two leads of the cable will also include some leakage in the insulation as well as it would contain inductive component and may simply have induced currents in the leads (having rather long lengths).

Dear Yuri

O.K. There might be some leakage in the cable insulation, due the reasons of humidity or low quality insulation material. This is the point I first recommended first to megger the cable. If there's a certain amount of leakage, then it is not a capacitor, but it is a conductor. The matter of cable capacitance can be negotiated.

The guy, of which the owner of the question, can make sure this by applying Direct Current and an ammeter to the cable. (However the megger directly makes this measurement.)

İn the stuation of leakage, the (DC) ammeter will show continuous current.

In the event of no leakage, first the ammeter will move to show the charge current to the cable, and then drops to zero after the cable is totally charged. (He must take care to discharge the cable later if the voltage is high to prevent an electric shock. A cable, charged with High Voltage is like a snake )

Your second suggestion. He could also read the inductance of the cable is wrong.

The cable itself is the two electrodes of a capacitor. Only charged, but not conducting any circulating current, to create a magnetic field, therefore to create an inductance.

Since there is no Ohmic conductivity, to create a current loop, the inductance effect can not be spoken.

Kindest Regards.

I agree that megger test will show the "resistive" load in the form of insulation leakage, but how can we be sure that the AC current measured is purely capacitive - magnetic field surrounds straight conductors as well.

But, probably, that inductive current would be very small in comparison to capacitive so in general the measured current may be said to be "capacitive".

Best regards.

Dear Yuri B

This is an open circuit measurement.

Both ends of the cable is opened before starting the test. Nothing is fed from the cable.

Therefore no current is passing from the cores of the cables. The only current can be measured is the passing electrons (Coulombs) in view of charging the cable cores.

To create Inductance, a closed loop is needed, of which current is passing inside, to compose a coil form to create magnetic field.

The current must enter from one end of the line and leave from the other end. In such case then we can speak about the current is passing through, and therefore creating a magnetic field. Then an inductance value can be available.

In this measurement, other end of the cable is open. All the bridges and connections are opened in view of achieving this test. Pls take care this is an open circuit.

Hope I could be able to clarify what I mean with my limited English.

Kindest Regards

I meant the "loops" or "bridges" formed by the leakage(s) in insulation, but I missed that the magnetic fields in both leads will be mutually neutralized, so, indeed, only capacitve current remains (besides that of the leakage - "resistive" one).

Best regards.

You're right.

I were always considering zero leakage during calculation.

Thanks for reminding.

Kindest Regards