Shunt capacitor voltage rise calculation.

[anteg]

I was told that the kvars on a capacitor bank decay by a square of the line voltage. ex. if the voltage feeding the capacitors drops by 1/2, then the capacitors are only capable of delivering 1/4 of their rated capacitance, is this true? is there a formula for this?

I am trying to determine the voltage rise for different line voltage values. I know the basic formula %V= MVAR/MVAsc, but if my line voltage drops so will my MVAR and MVAsc values.

So far I know, Voltage rise = 6.3 MVAR / 75 MVA = 0.084. This means that my end voltage (customer ) 120Vac would rise to (120)(1.084)=130.08Vac with the 6.3 MVARs. Line voltage is 12.4 kV. But how is this affected when the line voltae 12.4 kv drops to a lower value?. I am trying to solve it in the per unit system.

 

[dpc]

A very basic formula. Capacitor banks are, well, capacitors.

Draw a circuit with an ac source connected to a capacitor.

The capacitor is represented by a capacitive reactance that is a function of the capacitance and the frequency.

As voltage goes up, current goes up. Voltage goes down, current goes down.

kVAR rating is simply the current times the voltage through the capacitor. Voltage cut in half, current cut in half, product of current and voltage cut by 4.

Just like power through a shunt resistor.

"An 'expert' is someone who has made every possible mistake in a very narrow field of study." -- Edward Teller

 

[anteg]

When you put it this way it makes sense. So my MVAsc goes down by a factor of 4 if my voltage goes down by 1/2. Same applies to the capacitors, they go down by 1/4 when the voltage goes down by 1/2. Therefore my ratio for my voltage rise = MVAR/MVA does not change with voltage, correct.

Thank you dpc

 

[dpc]

Your voltage rise calculation is a rough approximation only and will let you down.

You need to look at the total circuit, including inductive elements to determine the voltage rise (or drop) that might be caused by addition of shunt capacitors. Voltage doesn't always go up.

"An 'expert' is someone who has made every possible mistake in a very narrow field of study." -- Edward Teller

 

[anteg]

Can you refer to any link where the process or calculations to obtain a closer estimate is discussed. I have tried looking everywhere but have not found anything useful.

 

[jghrist]

The reactive power (var, kvar, or Mvar) varies with the square of the voltage, not the capacitance. Capacitance is constant.

Considering 1Ø for simplicity:

(1) var = volts x amps
(2) Ohm's law: volts = amps x impedance
(3) impedance = 1/(2·pi·f·C)
(4) substitute (3) in (2) volts = amps/(2·pi·f·C)
(5) invert (4) amps = volts·2·pi·f·C
(6) substitute (5) in (1) var = volts²·2·pi·f·C

It can be shown that % voltage rise = kvar·X_L/(10·kV²)
where X_L is the reactance of the line or transformer that the current is flowing through, kvar is the 3Ø capacitor size, and kV is the line-line voltage

 

[anteg]

jghrist,

Capacitance constant? but what if this capacitance is shunted to a line where the voltage drops by 1/2? I don't believe the available KVar would be constant if its line voltage is dropped by 1/2?...

Also, I had seen this formula you mention. Is this one more accurate than MVAR/MVA?. Is the reactance of the line from the substation to the cap bank enough based on the spec for the cable of ohms / mile?

Thank you

 

[dpc]

Capacitance is not the same as kVAR.

Capacitance is a physical constant, independent of voltage or frequency (well, ideally). Capacitance is measured in farads, not kVAR.

"An 'expert' is someone who has made every possible mistake in a very narrow field of study." -- Edward Teller