##### Capacitor Bank Protections

The current of the protection fuses should be selected as If≅2In.

51 relay should be set with a delay for 0.1 seconds between 4-6In (short circuit protection)

50 relay should be set with a delay for 4 seconds for 1.3In (overload protection)

50N relay is recommended to be set with a delay for 4 seconds at the setting of 0.05In (overload protection)

The value of resistor R (kΩ) required to be connected across the capacitor to drop the voltage of the capacitor battery with a capacitance of C (μF) to under 75 V after 10 minutes (600 seconds), can be calculated as below:

For Delta connection

For Star connection

U:System voltage (V)

In: Capacitor nominal current (A)

##### Inrush Current (I_C) Calculation When a Single Battery is Connected to the Circuit

- - U: Phase-Neutral voltage (V)
- - Xc: Phase-Neutral capacitive reactance (Ω)
- - XL: Total inductive reactance between batteries (Ω)
- - Q; Q_1; Q_2: Battery powers (kVAr)
- - S_SC: Short circuit power (kVA) at the point where the capacitors are connected
- - I_N : Nominal current (A_rms) of the battery.
- - I_SC : Short circuit current (A_rms) at the point where the capacitor bank is connected.

The value of the inductor to be connected in series with the capacitor to limit the inrush current down to I_C ≤100I_N:

###### Example:

Given than:

Q=200 kVAr U=5000 V f-f

S=1000 kVA Z=5%

Inrush current I_C =I_N √(2 S_SC/Q)

I_N=Q/(√3.U)=200/(√3 x5)=23 A_rms

S_SC=S/Z_SC =1000/(5/100)=20.000 kVA

I_C=23√(2 20.000/200)=325 A smaller than 100 x 23 A

Inrush current reactor not required.

##### Inrush Current (I_C) Calculation When (n+1) Number of Capacitor Batteries are Connected in Parallel:

When (n) number batteries are energized, (n+1)st step will be energized.

Q (kVAr) : Power of a single step battery

U (kV) : Grid voltage (phase to phase)

ω (rad/s) : 2πf

C (μF) : Capacitance of the capacitor

I (μH/m) : Inductance of bars and cables between the batteries

f_r (Hz) : Resonance frequency

L (μH) : Inrush reactor connected in series to the battery

I_C (A): The peak value of the initial charging current

I_N (A_rms) : Nominal current of the battery

Q =U^2.C.ω= √3.U.I_N

I_C= √(2/3)U.n/(n+1).√(C/I)

f_r=1/(2π√(I.C))

The required reactor to satisfy the expression I_C≤100I_N :

L (μH)=(2.10^6)/3 x Q/(2πf) x (n/(n+1))^2 x 1/((I_C )^2)

If inrush reactor (L) is added,

I_C= √((2 x 10^6)/3 x Q/(2πf) x (n/(n+1))^2 x 1/L)

###### Example:

For a capacitor bank with 0,5 𝜇H/m inductance, 5 meters height, U=5000 V (phase to phase) with (n+1) = 3 steps, each of which has Q = 200 kVAr power;

- I_N=Q/(√3U)=200/(1.73 x 5) = 23 A_rms

C= √3(U x I_N)/(U^2 x 2πf) = 1,73(23 x 5000)/(5000^2.314)=25,3x10^-6

C=25,3 μF

Inrush current I_C=√(2/3)U x n/(n+1)√(C/I)

- I_C = 0,81 x 5000 x 2/3 √(25,3/(0,5 x 5)) =>

I_C=8589 A_p = 8,59 kA ≥ 100 x 23 A Reactor required!

- Reactor inductance L(𝜇H)

L ≥(2 x 10^6)/3 x (Q.10^(-3))/ω x (n/(n+1))^2 x 1/((I_C )^2)

= 2x10^6 x 0,2/(2π50)(2/3)^2 x 1/(8590)^2 = 7,67 μH

If 50 𝜇H reactor is connected instead of 7,67 𝜇H, then the inrush current will be:

I_C=√(2/3) x 5000 x 2/3 x √(25,3/50) = 1935 A_p

- Resonance frequency f_r = 1/(2π√(L.C))

= 1/(2π√(50 x 10^(-6)x 25,3 x 10^(-6))) = 4475 Hz

##### Calculations Related to Capacitor Banks:

In capacitors I_max=1,3I_n

- V_max=1.1 V_n - 12 hours / day
- V_max=1.2 V_n - 5 min
- V_max=1.3 V_n - 1 min

When a capacitor bank with a power of Q (kVAr) is connected to a system with a short circuit power of S_sc (kVA), the resonance frequency is:

- S: Power (kVA) of the transformer supplying the capacitor
- S_SC: Short circuit power (kVA) of the transformer supplying the capacitor
- Z_SC: Short circuit impedance of the transformer supplying the capacitor (%)

Determining the Q_N of the capacitor required in order to provide a capacitive power of Q_s to a system with a voltage of (U_s):