At a windfarm connected at 33kV, we have pole mounted capacitors, rated at 750kVAr, connected permanently in circuit. I would like to know how much current these capacitors take to maintain their charge.

I have tried to calculate the Amperes using the formulae below but think my answer is incorrect:

Q = CV
Q = (750 X 103)(33 X 103)
Q = (2.4 X 10 to the power 10) Coulombs

I = Q/T
I = (2.4 X 10 to the power 10) / (60 X 60)
I = 6.67 MA per hour!

I would appreciate if someone can confirm if the above is correct, and if not, advise the way forward.

Many thanks in advance.

Andy

Edited: 11 October 2011 at 01:22 PM by andymcmanus

I don't think the above is correct.

You're first expression Q=CV describes the charge available from a fully charged capacitor.

But on the second expression,

1. when it's fully charged, it won't admit current.

2. The d.c. charging current is an inverse-exponential function over time. Rather than me typing the theory out in this Forum, you can read about the d.c. charging characteristics here: http://en.wikipedia.org/wiki/Capacitor

Taking this further, what you'd need to calculate are the losses over time from parasitic losses, coupled with any ripple losses, and then integrate (through the charging function) over time to calculate the average charging current.

I assume from your question, through, that this is a d.c. supply?

If the supply they are connected in is a.c., losses are a little more simple to calculate: simply use the parasitic resistance in parallel with the capacitive impedance to calculate your real and reactive power losses through them.

-------------------------
Eur Ing Graham Kenyon CEng MIET

you're using the wrong formula.

kVA = Volts x Amps. so I = kVA / V

So I = (750 x 10e3) x sqrt3/(33 x 10e3)
= 750 x 1.723 / 33
~ 39A per phase

Agreed, but why "to maintain their charge" in the OP?

-------------------------
Eur Ing Graham Kenyon CEng MIET

I agree that 'Maintaining charge' only makes sense in a dc system, and I agree with the rest of your comments.

KVAr is not a measure of capacitance, so it can't be used in the original formula. From the rest of Andy's post, I assume the capacitors are slapped across the line to provide some reactive, so I have taken the question to be how much current are these capacitors taking.

If not, I am sure we will hear

No
For a 3 phase system, kVAr = (root3) x kV x Amps x Sin(theta)
For a capacitor, the current leads the voltage by 90 degrees
Therefore Sin(theta) = 1
Hence Amps = 750 / [ (root 3) x 33] = 13.1 amps/phase

For a single phase system, kVAr = kV x Amps x Sin(theta)
and Amps = 750 / 33 = 22.7 amps
Best wishes
John

Edited: 10 October 2011 at 09:10 PM by williamjohn

Hi,

Thanks very much for all replies, which have made me realise that the initial question wasn't explained very clearly and highlighted some fundamental errors.

Firstly, Mike is correct in his assumption in that this is an AC system. The caps are mounted on poles and connected at 33kV on a windfarm circuit.

The phrase 'maintaining charge' was misleading, sorry ;-).

The question should therefore be:

- How much current does it take to charge the capacitors?

I do not think the formula VAr = root 3 x VI SIN (theta) is correct in this instance because the power factor (COS theta and therefore SIN theta) is continuously changing as the import and export of active power fluctuates. Is it not the case that this formula only returns the phase current and not the current required to charge the capacitor?

Mike pointed out that I was using kVAr in the formula when I should be using Capacitance. The C can be calculated as follows:

C = 1/ 2 pie fXc
C = 1/ 2 x pie x 50 x 750 x 10e3
C = 4.24Pf

Q = CV
Q = (4.24 x 10e-9)(33 X 10e3)
Q = 0.14mC

I = Q/T
I = 0.14 x 10e-3 / (60 X 60)
I = 38.9uA per hour.

Could this be the correct answer??

No.
Sorry, wrong formula again. You're using kVA instead of impedance there. Check out the formula and theory via the link given by gkenyon.

What are you trying to calculate? A capacitor in an ac circuit can only be said to charge up on an instantaneous basis, the current will be an ac waveform, determined by the voltage at the point of connection. Flow of active power does not matter, or only if it affects the voltage.

A capacitor installed as you describe takes a fixed current, at a fixed angle to the voltage, and is there to modify the resultant current in the main conductor. It's rated in kVA to make it easy. Correct answer is given by williamjohn. (I was trying to type without writing it down earlier, and got the 3 and root3 the wrong way round)

Andy
In the formula KVAr = (root3) x kV x Amps x Sin(theta) the theta applies to the circuit involved. To get the current to the capacitors, the circuit is only the circuit to the capacitors where theta is 90 degrees.

This circuit is in parallel with the system load.

If you use the system power factor, you will get the system current not the current to the capacitors.
Best wishes
John