Risk of resonance due to power-system harmonics

Considering the simplified circuit represented on Figure L29 (no PFC capacitors connected):

The voltage distortion V_h at the busbar level results from two different factors:

• connection of non-linear loads generating harmonic currents I_h,
• voltage distortion U_h present on the supply network due to non-linear loads outside of the considered circuit (incoming harmonic voltage).

A significant indicator of harmonic importance is the percentage of non-linear loads N_LL, calculated by the formula:

[math]\displaystyle{ N_{LL(\%)}= \frac {\text{Power of non-linear loads}}{\text{Power of supply transformer}} }[/math]

Fig. L29 – Simplified circuit diagram

The connection of PFC capacitors (without reactors) results in the amplification of harmonic currents at the busbar level, and an increase of the voltage distortion.

Capacitors are linear reactive devices, and consequently do not generate harmonics. The installation of capacitors in a power system (in which the impedances are predominantly inductive) can, however, result in total or partial resonance occurring at one of the harmonic frequencies.

Because of harmonics, the current I_C circulating through the PFC capacitors is higher compared to the situation where only the fundamental current I₁ is present.

If the natural frequency of the capacitor bank/ power-system reactance combination is close to a particular harmonic, then partial resonance will occur, with amplified values of voltage and current at the harmonic frequency concerned. In this particular case, the elevated current will cause overheating of the capacitor, with degradation of the dielectric, which may result in its eventual failure.

The order h₀ of the natural resonant frequency between the system inductance and the capacitor bank is given by:

[math]\displaystyle{ h_0= \sqrt{\frac {S_{sc} }{Q} } }[/math]

Where:

S_sc = the level of system short-circuit power (kVA) at the point of connection of the capacitor
Q = capacitor bank rating in kvar
h₀ = the order of the natural frequency f₀, i.e. f₀/50 for a 50 Hz system, or f₀/60 for a 60 Hz system.

For example:

Transformer power rating :	S = 630kVA
Short-circuit voltage :	Usc = 6%
Short-circuit power at the busbar level :	Ssc ~ 10 MVA
Reactive power of the capacitor bank :	Q = 350 kvar

Then:

[math]\displaystyle{ h_0= \sqrt{\frac {S_{sc} }{Q} } = \sqrt{\frac {10.10^3}{350} } = 5.5 }[/math]

The natural frequency of the capacitor/system-inductance combination is close to the 5th harmonic frequency of the system.

For a 50Hz system, the natural frequency f₀ is then equal to f₀ = 50 x h₀ = 50 x 5.5 = 275 Hz

Fig. L30 – Simplified circuit diagram