Dashboard PE Power Exam Prep Transmission and Distribution Distribution & power factor Power factor correction with capacitor banks

Power factor correction with capacitor banks

Q_cap = P · (tan θ_old − tan θ_new), bank types, where to install, harmonic resonance, and the simple-payback economics.

Sophomore ~11 min

Step 1 — Why correct PF: penalty, capacity, losses, voltage

0.55×
Q_cap — kVAR PF_new save

Reference notes

Use Next → on the narrator above to step through power-factor correction with capacitor banks: why it pays back, how to size kVAR, the four bank types, placement trade-offs, harmonic resonance and detuning, and simple-payback economics.

Why correct PF?

Typical industrial PF-correction project: customer at 0.78 PF lagging corrected to 0.95 lagging, with payback under 18 months from demand-charge avoidance alone.

Sizing the capacitor bank

Real power P does not change with capacitor compensation — only Q does. Before correction Q_old = P · tan θ_old; after Q_new = P · tan θ_new. The capacitor supplies the difference:

Qcap = P · (tan θold − tan θnew)
tan θ = √(1 − PF²) / PF

Worked example: 800 kW at 0.78 PF lagging → 0.95 PF lagging.

Bank types

Placement

Three classical locations, ordered by loss-reduction benefit:

  1. At the load (e.g., capacitor at each motor's terminals): eliminates reactive current all the way upstream. Best loss reduction; most expensive to install.
  2. At the busbar: serves multiple loads from a single bank. Moderate loss reduction; simpler switching.
  3. At the substation: cheapest $/kVAR. Only addresses utility-side demand penalty; reactive current still flows through customer's internal distribution.

Harmonic resonance

A capacitor bank Q_cap on a bus with short-circuit MVA S_SC forms a parallel resonant circuit with the source inductance. The resonant frequency is:

fr = f1 · √(SSC / Qcap)

If f_r happens to land near a harmonic injected by VFDs or rectifiers (5th and 7th are the usual suspects on 6-pulse drives, 11th and 13th on 12-pulse), the harmonic voltage at the bus is amplified — sometimes ten-fold or more, causing damaged capacitors, tripped drives, and overheating transformers.

The fix is a detuning reactor in series with each capacitor, sizing the LC pair to resonate at a frequency just below the lowest troublesome harmonic — typically 4.7×f_1 (just under the 5th) for 6-pulse drive environments. The detuned LC presents low impedance to the harmonic instead of amplifying it.

Three-phase bank topology

Two common 3-φ topologies:

Both deliver the same TOTAL 3-φ kVAR at the same applied V_LL — the difference is in unit cost and the failure-mode behaviour (unbalanced fault detection differs between the two topologies).

Economic analysis

Payback (years) = Cost / Annual savings = (Qcap · $/kVAR) / (demand $/yr + loss $/yr)

Typical LV-bank installed cost: $25–$50 per kVAR. Annual demand-charge avoidance: $30–$60 per kVAR-year (utility-specific). Loss reduction: a few percent of the existing distribution losses. Most industrial PF-correction projects show simple payback under 2 years.

Over-correction caveat: pushing past the utility's deadband to leading PF often triggers a leading-PF penalty too. And during low-load periods a fixed over-sized bank can lift the bus voltage above 1.05 pu — damaging customer equipment. Use automatic switching, or size for slightly under-correction (e.g., target 0.95 instead of 0.98).

Take-away. Size with Q_cap = P · (tan θ_old − tan θ_new). Pick the bank type to match load variability (fixed for steady load, switched/automatic for varying). Install at the load for max loss reduction; at the substation for cheapest $/kVAR. Add detuning reactors if VFDs or rectifiers are on the bus. Calculate payback — most projects come in under 2 years. Don't over-correct.

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