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MV Capacitor Bank Sizing

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Overview

Sizes medium-voltage shunt capacitor banks for power factor correction per IEEE 18-2012, IEC 60871, and IEEE 519. The calculator determines the required kVAr rating, analyses parallel resonance with system harmonics, selects a detuning reactor when needed, specifies voltage and current ratings, designs discharge and protection circuits, and performs an economic payback analysis.

Key capabilities:

  • Power factor correction from existing to target PF
  • Parallel resonance of the installed bank at the point of common coupling
  • Detuning reactor chosen by the lowest significant harmonic (14%, 7% or 5.67%), or forced by the user
  • Capacitor voltage, capacitor kVAr and nameplate rating of a detuned (capacitor + series reactor) bank
  • Achieved power factor from the installed output, with over-correction to leading reported as a failure
  • Circuit breaker and overload protection sizing
  • Discharge resistor and inrush current estimation
  • Stored energy and system loss/efficiency analysis
  • Economic payback with demand and energy savings

Input Parameters

System Parameters

Parameter Type Default Range Description
v_system Number 11.0 kV 0.4 – 33.0 kV System voltage, line-to-line
frequency Enum: 50 / 60 50 Hz — Electrical system frequency
s_sc Number 250 MVA 10 – 10,000 MVA Short-circuit capacity at point of common coupling

Load Parameters

Parameter Type Default Range Description
p_max Number 1000 kW 1 – 100,000 kW Maximum active power demand
pf_existing Number 0.80 0.50 – 1.00 Existing (uncorrected) power factor
pf_target Number 0.95 0.80 – 1.00 Desired power factor after correction

Harmonic Parameters

Parameter Type Default Range Description
h3 Number 0% 0 – 100% 3rd harmonic voltage distortion (single-phase non-linear loads)
h5 Number 0% 0 – 100% 5th harmonic voltage distortion
h7 Number 0% 0 – 100% 7th harmonic voltage distortion
h11 Number 0% 0 – 100% 11th harmonic voltage distortion
h13 Number 0% 0 – 100% 13th harmonic voltage distortion
reactor_selection Enum: auto / none / 5.67 / 7 / 14 auto — Detuning reactor. auto applies the selection rule below; any other value forces the choice, which the resonance check then verifies

Economic Parameters

Parameter Type Default Range Description
energy_rate Number $0.10/kWh 0 – 1.00 Electricity energy charge
demand_rate Number $10/kVA/month 0 – 100 Electricity demand charge
hours_per_year Number 8760 1 – 8,760 Annual operating hours
total_investment Number $50,000 0 – 10,000,000 Total equipment and installation cost

Output Results

Output Unit Description
q_c_required kVAr Required reactive power for PF correction
q_c_final kVAr Installed bank output at system voltage (rounded up to 50 kVAr steps)
q_c_capacitor kVAr Reactive power of the capacitors at their operating voltage, \(Q/(1-p)\)
q_c_nameplate kVAr Capacitor nameplate rating at the rated voltage \(1.1\,V_C\)
achieved.pf / achieved.leading — Power factor at maximum demand with the installed bank, and whether it is leading
s_existing / s_new kVA Apparent power before and after correction
s_reduction kVA Demand reduction achieved
f_r Hz Parallel resonance frequency of the installed bank without reactor
n_r — Harmonic order at resonance
requires_detuning Boolean Whether a detuning reactor is required
f_r_new Hz Series tuning frequency \(f_n/\sqrt{p}\) when detuned
v_cap kV Voltage across the capacitors, \(V/(1-p)\)
v_standard kV Selected standard capacitor voltage rating
i_rated A Rated current per phase
i_max A Maximum current including harmonic factor and margin
x_c Ω Capacitive reactance at fundamental frequency
c_per_phase μF Capacitance per phase (Wye connection)
x_l Ω Reactor reactance
l_per_phase mH Reactor inductance per phase
i_cb_standard A Selected standard circuit breaker rating
i_overload A Overload protection setting (1.25 × Irated)
i_inrush_typical A Typical switching inrush current (15 × Irated)
r_discharge Ω Discharge resistor value
e_total_kj kJ Total stored energy (three-phase)
efficiency % System efficiency including all losses
payback_years years Simple payback period
is_compliant Boolean Resonance check passed and target power factor met (lagging)

Formulas

Required Reactive Power

\[ Q_c = P \times \left(\tan\varphi_1 - \tan\varphi_2\right) \]

Where \(P\) is maximum active power demand (kW), \(\varphi_1 = \cos^{-1}(\text{pf}_{existing})\), and \(\varphi_2 = \cos^{-1}(\text{pf}_{target})\).

Apparent Power and Demand Reduction

\[ S_{existing} = \frac{P}{\text{pf}_{existing}} \qquad S_{new} = \frac{P}{\text{pf}_{target}} \qquad \Delta S = S_{existing} - S_{new} \]

Installed Bank

\[ Q_{c,inst} = \left\lceil \frac{Q_c}{50} \right\rceil \times 50 \]

The bank is specified by its output at system voltage, in 50 kVAr steps — also for a detuned bank, whose steps are sold by effective output. Every result below is computed from \(Q_{c,inst}\).

Parallel Resonance Frequency

\[ f_r = f_n \times \sqrt{\frac{S_{sc}}{Q_{c,inst}}} \]

Where \(S_{sc}\) is the short-circuit capacity at the PCC (kVA) and \(Q_{c,inst}\) is the installed bank output (kVAr). This is the resonance of the bank without a reactor; the harmonic order is \(n_r = f_r / f_n\).

Detuning Reactor Selection

A harmonic above 3% is significant. With reactor_selection = auto, detuning is required when \(f_r\) lies within ±20% of a significant harmonic (3rd, 5th, 7th, 11th or 13th). The reactor is then chosen by the lowest significant harmonic, so that the bank is tuned below every significant harmonic with margin for capacitance tolerance and ageing (both raise the tuning frequency):

Lowest significant harmonic Reactor \(p = X_L/X_C\) Tuning order \(1/\sqrt{p}\) At 50 Hz At 60 Hz
3rd14%2.67134 Hz160 Hz
5th7%3.78189 Hz227 Hz
7th or above5.67%4.20210 Hz252 Hz

Practice per Schneider Electric Electrical Installation Guide ch. L (detuned reactors) and ABB Power factor correction and harmonic filtering in electrical plants. Any other reactor_selection forces the choice. The tuning frequency of the capacitor + reactor branch is:

\[ f_t = \frac{f_n}{\sqrt{p}} \]

Resonance Check

Undetuned bank: \(f_r\) must not lie within ±20% of any significant harmonic. Detuned bank: \(f_t\) must lie below the lowest significant harmonic, so the bank is inductive at every significant harmonic (IEEE 1036). With auto the selection rule satisfies the check; a forced choice (e.g. no reactor near a significant harmonic, or 5.67% with a significant 3rd) fails it.

Detuned Bank: Capacitor Voltage and Rating

Per phase, the capacitor (\(X_C\)) and reactor (\(X_L = pX_C\)) are in series:

\[ Q_{eff} = \frac{V^2}{X_C - X_L} = \frac{V^2}{X_C(1-p)} \qquad V_C = \frac{V}{1-p} \qquad Q_C = \frac{V_C^2}{X_C} = \frac{Q_{eff}}{1-p} \]

The bank output at system voltage is \(Q_{eff} = Q_{c,inst}\). The capacitors run at \(V_C\) and generate \(Q_C\); the reactor absorbs \(pQ_C\). With a rated voltage \(V_{rated} = 1.1\,V_C\) the capacitor nameplate rating is

\[ Q_{N} = Q_C\left(\frac{V_{rated}}{V_C}\right)^2 = \frac{1.21\,Q_{c,inst}}{1-p} \]

(IEEE 1036; equivalently \(Q_N = Q_{eff}(1-p)(V_N/V)^2\) as used in manufacturers' detuned-filter tables). Without a reactor \(p = 0\): \(V_C = V\), \(Q_C = Q_{c,inst}\), \(Q_N = 1.21\,Q_{c,inst}\). The standard voltage is the next value at or above \(V_{rated}\) in 3.3, 6.6, 7.2, 11, 12, 13.2, 13.8, 22, 24, 33 kV, continuing above 33 kV with the IEC 60038 \(U_m\) values 36, 40.5, 52, 72.5 kV.

Achieved Power Factor

\[ Q_{after} = P\tan\varphi_1 - Q_{c,inst} \qquad \text{pf} = \frac{P}{\sqrt{P^2 + Q_{after}^2}} \]

The target is met when the power factor is lagging (\(Q_{after} \ge 0\)) and at or above the target. A negative \(Q_{after}\) means the bank over-corrects to a leading power factor: this fails the target whatever its magnitude, because it raises the bus voltage, gets worse at lighter load and risks motor self-excitation. Use smaller steps or an automatic stepped bank.

Current Ratings

\[ I_{rated} = \frac{Q_{c,actual}}{\sqrt{3} \times V_{LL}} \]

The harmonic current multiplier for a detuned bank is \(k_h = 1/(1 - p)\). Maximum expected current with a 30% margin:

\[ I_{max} = I_{rated} \times k_h \times 1.3 \]

Capacitance and Reactance

\[ X_C = \frac{V_C^2}{Q_C} \qquad C_{phase} = \frac{1}{2\pi f_n X_C} \]

Where \(X_C\) is the capacitive reactance from the capacitor voltage and the capacitors' own reactive power (for an undetuned bank \(V_C = V_{LL}\), \(Q_C = Q_{c,inst}\)), and \(C_{phase}\) is the capacitance per phase for a Wye-connected bank.

Reactor Parameters

\[ X_L = p \times X_c \qquad L_{phase} = \frac{X_L}{2\pi f_n} \]

Reactor resistance is estimated assuming \(X/R = 50\) (typical for MV detuning reactors):

\[ R_L = \frac{X_L}{50} \]

Protection Settings

\[ I_{CB} = 1.35 \times I_{rated} \]

The calculated value is rounded up to the next standard breaker size (16 … 2000 A, then the IEC 60059 R10 series 2500, 3150, 4000, 5000, 6300 A …). Overload setting: \(1.25 \times I_{rated}\). Unbalance alarm at 5% and trip at 10% of \(I_{rated}\).

Discharge Resistor

\[ R_{discharge} = \frac{-t} {C_{phase} \times \ln\!\left(\dfrac{V_{target}}{V_{peak}}\right)} \]

Where \(V_{peak} = V_{cap} \times \sqrt{2}\) (converted to volts), \(t = 300\) s (5 minutes per IEEE 18), and \(V_{target} = 50\) V.

Inrush Current

\[ I_{inrush} \approx 15 \times I_{rated} \]

Typical range is 10–20 × rated current. A pre-insertion resistor to limit inrush to 10× rated:

\[ R_{pre} = \frac{V_{cap} \times 1000}{10 \times I_{rated}} \]

Stored Energy

\[ E_{phase} = \tfrac{1}{2}\,C \,V^2 \qquad E_{total} = 3 \times E_{phase} \]

Where \(C\) is in farads and \(V\) is in volts.

Economic Analysis

\[ \text{Demand savings} = 12 \times \Delta S \times r_{demand} \] \[ \text{Energy savings} = \Delta S \times 0.02 \times h_{year} \times r_{energy} \] \[ \text{Payback} = \frac{\text{Investment}} {\text{Demand savings} + \text{Energy savings}} \]

Where \(\Delta S\) is the kVA demand reduction, \(r_{demand}\) is the demand rate ($/kVA/month), \(r_{energy}\) is the energy rate ($/kWh), and \(h_{year}\) is annual operating hours. The 2% factor approximates reduced system losses.

Worked Examples

The 29 examples below are the validation cases for this calculation. Each one was produced by the calculation engine from the inputs shown, every figure was checked against the independent reference quoted at the end of the example, and the same cases are run against every engine build (see validation scorecard, tool fe_calc_pfc; worked examples for every calculation).

#CaseV (kV)P (kW)pfQc (kVAr)Installed (kVAr)hrReactorpf achievedResult
111 kV, 1 MW, 0.80 → 0.95, no harmonics111,0000.8 → 0.95421.345023.57none0.958PASS
23.3 kV, 500 kW, 0.70 → 0.983.35000.7 → 0.98408.64509.43none0.993PASS
3415 V, 200 kW, 0.85 → 0.950.4152000.85 → 0.9558.210014.14none0.993PASS
4415 V, 40 kW — one 50 kVAr step over-corrects — FAIL0.415400.7 → 0.921.45017.32none0.975 leadFAIL
5400 V, power factor 0.50 → 0.900.41000.5 → 0.9124.815012.91none0.974PASS
66.6 kV, 2 MW, target unity — FAIL6.62,0000.85 → 11,239.51,25010.95none1.000 leadFAIL
711 kV, 1.5 MW, 0.70 → 0.80111,5000.7 → 0.8405.345021.08none0.811PASS
822 kV, 10 MW, 0.85 → 0.972210,0000.85 → 0.973,691.23,70014.70none0.970PASS
9480 V, 60 Hz, 750 kW0.487500.8 → 0.95316.03509.26none0.962PASS
1013.8 kV, 60 Hz, 4 MW13.84,0000.8 → 0.951,685.31,70013.28none0.951PASS
1111 kV, 6 % 5th present but resonance at the 29th112,0000.85 → 0.95582.160028.87none0.952PASS
12415 V, resonance at the 5th with only 3 % 5th0.4151,0000.8 → 0.95421.34504.71none0.958PASS
13415 V, 500 kW, resonance near the 7th (4 %)0.4155000.75 → 0.95276.63005.775.67 %0.962PASS
1411 kV, 3 MW, 10 % 5th and 5 % 7th113,0000.78 → 0.951,420.81,4508.307 %0.953PASS
15400 V, 800 kW, resonance near the 5th0.48000.75 → 0.95442.64504.717 %0.953PASS
16400 V, 300 kW, resonance near the 11th0.43000.8 → 0.95126.415011.555.67 %0.970PASS
17400 V, 100 kW, resonance near the 13th0.41000.5 → 0.9124.815012.915.67 %0.974PASS
186.6 kV, 2.5 MW, resonance near the 7th6.62,5000.82 → 0.95923.39507.955.67 %0.953PASS
19480 V, 60 Hz, resonance near the 5th0.481,5000.8 → 0.95632.06504.807 %0.953PASS
20600 V, 60 Hz, resonance near the 7th0.61,0000.85 → 0.97369.14007.075.67 %0.977PASS
21690 V, 1.2 MW, 7 % detuned — capacitor voltage and nameplate0.691,2000.78 → 0.95568.36005.007 %0.957PASS
22400 V, significant 3rd harmonic — 14 % reactor0.47000.75 → 0.95387.34005.0014 %0.955PASS
23400 V, resonance near the 11th but the 5th is significant — 7 %0.43000.8 → 0.95126.415011.557 %0.970PASS
24415 V, reactor forced to none near a 4 % 7th — FAIL0.4155000.75 → 0.95276.63005.77none0.962FAIL
25400 V, 5.67 % forced with a significant 3rd — FAIL0.47000.75 → 0.95387.34005.005.67 %0.955FAIL
26415 V, rounding up moves the resonance onto the 5th0.4155200.8 → 0.97259.73005.777 %0.985PASS
27400 V, 60 kW, 0.80 → 0.95 — leading — FAIL0.4600.8 → 0.9525.35014.14none0.997 leadFAIL
2833 kV, 1.1 × V above 33 kV3320,0000.85 → 0.955,821.25,85013.07none0.950PASS
29400 V, 1.5 MVAr bank — breaker above 2000 A0.43,0000.8 → 0.971,498.11,5005.77none0.970PASS

Example 1 — 11 kV, 1 MW, 0.80 → 0.95, no harmonics

The basic sizing: Q = P(tan φ1 − tan φ2), rounded up to 50 kVAr, with resonance far above any harmonic.

Given: 11 kV, 50 Hz bus with a 250 MVA fault level; maximum demand 1,000 kW at power factor 0.8 lagging, target 0.95; no harmonic distortion.

Reactive powerφ1 = arccos 0.8 = 36.87°, tan φ1 = 0.75000; φ2 = arccos 0.95 = 18.19°, tan φ2 = 0.32868; Qc = P (tan φ1 − tan φ2) = 1,000 × (0.75000 − 0.32868) = 421.32 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈421.32 / 50⌉ × 50 = 450 kVAr
Apparent powerSbefore = 1,000 / 0.8 = 1,250.00 kVA; Safter = 1,000 / 0.95 = 1,052.63 kVA; reduction 197.37 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(250,000 kVA / 450 kVAr) = 1,178.51 Hz; hr = 1,178.51 / 50 = 23.570
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: no harmonic above 3 % → no reactor
Capacitor voltageVC = V = 11 kV; Vrated = 1.1 × 11 = 12.1000 kV → next standard rating 13.2 kV
Capacitor ratingNo reactor: QC = Qinst = 450 kVAr; nameplate at 1.1 V: 1.21 × 450 = 544.50 kVAr
CurrentIrated = Qinst / (√3 V) = 450 / (√3 × 11) = 23.62 A; Imax = Irated × kh × 1.3 = 23.62 × 1.0000 × 1.3 = 30.70 A
CapacitanceXC = VC² × 1000 / QC = 11.00² × 1000 / 450.000 = 268.9 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 268.9) = 11.838 µF per phase (star)
BreakerICB = 1.35 × 23.62 = 31.9 A → next standard size 32 A; overload setting 1.25 × Irated = 29.5 A
Discharge resistorVpk = √2 × 11,000.0 V = 15,556.3 V; R = t / (C ln(Vpk / 50 V)) = 300 / (11.838 × 10−6 × ln(15,556.3 / 50)) = 4,414.86 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 23.62 = 354.3 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 450.00 = 0.225 kW + reactor 3 I² RL = 0.000 kW + discharge 0.010 kW = 0.235 kW
Target power factorQafter = P tan φ1 − Qinst = 750.00 − 450 = 300.00 kVAr (lagging); pf = P / √(P² + Qafter²) = 1,000 / √(1,000² + 300.00²) = 0.9578 → 0.9578 vs limit 0.9500 PASS
ResonanceNo harmonic above 3 %; undetuned resonance 1,178.51 Hz is not excited PASS
Result2/2 checks pass → PASS; 450 kVAr bank, no reactor, 23.6 A, 32 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.8 = 0.75000, tan φ2 = tan arccos 0.95 = 0.32868; Q = 1000 × (0.75000 − 0.32868) = 421.32 kVAr → 450 kVAr output in 50 kVAr steps; f_r = 50 × √(250000 kVA / 450 kVAr) = 1178.51 Hz (h = 23.570); significant (> 3 %) harmonics: none; no significant harmonic within ±20 % → no reactor; V_C = 11 kV, nameplate 1.21 × 450 = 544.50 kVAr; X_C = 11² × 1000 / 450 = 268.8889 Ω; C = 10⁶ / (2π × 50 × X_C) = 11.838 µF/phase; I = 450 / (√3 × 11) = 23.62 A; CB 1.35 × I = 31.9 A → 32 A; inrush 15 × I = 354.3 A; R_d = 300 / (C ln(√2 V_C / 50)) = 4414.86 kΩ; losses 0.0005 × 450.00 + 0.01 = 0.235 kW; 1.1 × 11.0000 = 12.1000 kV → 13.2 kV standard; Q_after = 750.00 − 450 = 300.00 kVAr, pf = 1000 / √(1000² + 300.00²) = 0.9578 lagging ≥ 0.95; resonance check: 1178.51 Hz clear of every significant harmonic → pass (validation case c0016-11kv-1mw-undetuned)

Example 2 — 3.3 kV, 500 kW, 0.70 → 0.98

A poor starting power factor corrected to 0.98 on a 3.3 kV board; the resonance near the 10th order needs no reactor because no harmonics are present.

Given: 3.3 kV, 50 Hz bus with a 40 MVA fault level; maximum demand 500 kW at power factor 0.7 lagging, target 0.98; no harmonic distortion.

Reactive powerφ1 = arccos 0.7 = 45.57°, tan φ1 = 1.02020; φ2 = arccos 0.98 = 11.48°, tan φ2 = 0.20306; Qc = P (tan φ1 − tan φ2) = 500 × (1.02020 − 0.20306) = 408.57 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈408.57 / 50⌉ × 50 = 450 kVAr
Apparent powerSbefore = 500 / 0.7 = 714.29 kVA; Safter = 500 / 0.98 = 510.20 kVA; reduction 204.09 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(40,000 kVA / 450 kVAr) = 471.40 Hz; hr = 471.40 / 50 = 9.428
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: no harmonic above 3 % → no reactor
Capacitor voltageVC = V = 3.3 kV; Vrated = 1.1 × 3.3 = 3.6300 kV → next standard rating 6.6 kV
Capacitor ratingNo reactor: QC = Qinst = 450 kVAr; nameplate at 1.1 V: 1.21 × 450 = 544.50 kVAr
CurrentIrated = Qinst / (√3 V) = 450 / (√3 × 3.3) = 78.73 A; Imax = Irated × kh × 1.3 = 78.73 × 1.0000 × 1.3 = 102.35 A
CapacitanceXC = VC² × 1000 / QC = 3.300² × 1000 / 450.000 = 24.20 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 24.20) = 131.533 µF per phase (star)
BreakerICB = 1.35 × 78.73 = 106.3 A → next standard size 125 A; overload setting 1.25 × Irated = 98.4 A
Discharge resistorVpk = √2 × 3,300.0 V = 4,666.9 V; R = t / (C ln(Vpk / 50 V)) = 300 / (131.533 × 10−6 × ln(4,666.9 / 50)) = 502.80 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 78.73 = 1,180.9 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 450.00 = 0.225 kW + reactor 3 I² RL = 0.000 kW + discharge 0.010 kW = 0.235 kW
Target power factorQafter = P tan φ1 − Qinst = 510.10 − 450 = 60.10 kVAr (lagging); pf = P / √(P² + Qafter²) = 500 / √(500² + 60.10²) = 0.9929 → 0.9929 vs limit 0.9800 PASS
ResonanceNo harmonic above 3 %; undetuned resonance 471.40 Hz is not excited PASS
Result2/2 checks pass → PASS; 450 kVAr bank, no reactor, 78.7 A, 125 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.7 = 1.02020, tan φ2 = tan arccos 0.98 = 0.20306; Q = 500 × (1.02020 − 0.20306) = 408.57 kVAr → 450 kVAr output in 50 kVAr steps; f_r = 50 × √(40000 kVA / 450 kVAr) = 471.40 Hz (h = 9.428); significant (> 3 %) harmonics: none; no significant harmonic within ±20 % → no reactor; V_C = 3.3 kV, nameplate 1.21 × 450 = 544.50 kVAr; X_C = 3.3² × 1000 / 450 = 24.2000 Ω; C = 10⁶ / (2π × 50 × X_C) = 131.533 µF/phase; I = 450 / (√3 × 3.3) = 78.73 A; CB 1.35 × I = 106.3 A → 125 A; inrush 15 × I = 1180.9 A; R_d = 300 / (C ln(√2 V_C / 50)) = 502.80 kΩ; losses 0.0005 × 450.00 + 0.01 = 0.235 kW; 1.1 × 3.3000 = 3.6300 kV → 6.6 kV standard; Q_after = 510.10 − 450 = 60.10 kVAr, pf = 500 / √(500² + 60.10²) = 0.9929 lagging ≥ 0.98; resonance check: 471.40 Hz clear of every significant harmonic → pass (validation case c0016-3300v-500kw-pf098)

Example 3 — 415 V, 200 kW, 0.85 → 0.95

A small LV correction where the 50 kVAr step nearly doubles the required 58 kVAr.

Given: 0.415 kV, 50 Hz bus with a 20 MVA fault level; maximum demand 200 kW at power factor 0.85 lagging, target 0.95; no harmonic distortion.

Reactive powerφ1 = arccos 0.85 = 31.79°, tan φ1 = 0.61974; φ2 = arccos 0.95 = 18.19°, tan φ2 = 0.32868; Qc = P (tan φ1 − tan φ2) = 200 × (0.61974 − 0.32868) = 58.21 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈58.21 / 50⌉ × 50 = 100 kVAr
Apparent powerSbefore = 200 / 0.85 = 235.29 kVA; Safter = 200 / 0.95 = 210.53 kVA; reduction 24.76 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(20,000 kVA / 100 kVAr) = 707.11 Hz; hr = 707.11 / 50 = 14.142
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: no harmonic above 3 % → no reactor
Capacitor voltageVC = V = 0.415 kV; Vrated = 1.1 × 0.415 = 0.4565 kV (LV: no standard HV rating applies)
Capacitor ratingNo reactor: QC = Qinst = 100 kVAr; nameplate at 1.1 V: 1.21 × 100 = 121.00 kVAr
CurrentIrated = Qinst / (√3 V) = 100 / (√3 × 0.415) = 139.12 A; Imax = Irated × kh × 1.3 = 139.12 × 1.0000 × 1.3 = 180.86 A
CapacitanceXC = VC² × 1000 / QC = 0.4150² × 1000 / 100.000 = 1.722 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 1.722) = 1,848.221 µF per phase (star)
BreakerICB = 1.35 × 139.12 = 187.8 A → next standard size 200 A; overload setting 1.25 × Irated = 173.9 A
Discharge resistorVpk = √2 × 415.0 V = 586.9 V; R = t / (C ln(Vpk / 50 V)) = 300 / (1,848.221 × 10−6 × ln(586.9 / 50)) = 65.91 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 139.12 = 2,086.8 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 100.00 = 0.050 kW + reactor 3 I² RL = 0.000 kW + discharge 0.010 kW = 0.060 kW
Target power factorQafter = P tan φ1 − Qinst = 123.95 − 100 = 23.95 kVAr (lagging); pf = P / √(P² + Qafter²) = 200 / √(200² + 23.95²) = 0.9929 → 0.9929 vs limit 0.9500 PASS
ResonanceNo harmonic above 3 %; undetuned resonance 707.11 Hz is not excited PASS
Result2/2 checks pass → PASS; 100 kVAr bank, no reactor, 139.1 A, 200 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.85 = 0.61974, tan φ2 = tan arccos 0.95 = 0.32868; Q = 200 × (0.61974 − 0.32868) = 58.21 kVAr → 100 kVAr output in 50 kVAr steps; f_r = 50 × √(20000 kVA / 100 kVAr) = 707.11 Hz (h = 14.142); significant (> 3 %) harmonics: none; no significant harmonic within ±20 % → no reactor; V_C = 0.415 kV, nameplate 1.21 × 100 = 121.00 kVAr; X_C = 0.415² × 1000 / 100 = 1.7222 Ω; C = 10⁶ / (2π × 50 × X_C) = 1848.221 µF/phase; I = 100 / (√3 × 0.415) = 139.12 A; CB 1.35 × I = 187.8 A → 200 A; inrush 15 × I = 2086.8 A; R_d = 300 / (C ln(√2 V_C / 50)) = 65.91 kΩ; losses 0.0005 × 100.00 + 0.01 = 0.060 kW; 1.1 × 0.4150 = 0.4565 kV; Q_after = 123.95 − 100 = 23.95 kVAr, pf = 200 / √(200² + 23.95²) = 0.9929 lagging ≥ 0.95; resonance check: 707.11 Hz clear of every significant harmonic → pass (validation case c0016-415v-200kw-pf085)

Example 4 — 415 V, 40 kW — one 50 kVAr step over-corrects — FAIL

The 50 kVAr minimum step exceeds the load's own 40.8 kVAr, so the bank drives the power factor to 0.975 LEADING: above 0.90 by magnitude, but an over-corrected (leading) power factor fails the target.

Given: 0.415 kV, 50 Hz bus with a 15 MVA fault level; maximum demand 40 kW at power factor 0.7 lagging, target 0.9; no harmonic distortion.

Reactive powerφ1 = arccos 0.7 = 45.57°, tan φ1 = 1.02020; φ2 = arccos 0.9 = 25.84°, tan φ2 = 0.48432; Qc = P (tan φ1 − tan φ2) = 40 × (1.02020 − 0.48432) = 21.44 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈21.44 / 50⌉ × 50 = 50 kVAr
Apparent powerSbefore = 40 / 0.7 = 57.14 kVA; Safter = 40 / 0.9 = 44.44 kVA; reduction 12.70 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(15,000 kVA / 50 kVAr) = 866.03 Hz; hr = 866.03 / 50 = 17.321
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: no harmonic above 3 % → no reactor
Capacitor voltageVC = V = 0.415 kV; Vrated = 1.1 × 0.415 = 0.4565 kV (LV: no standard HV rating applies)
Capacitor ratingNo reactor: QC = Qinst = 50 kVAr; nameplate at 1.1 V: 1.21 × 50 = 60.50 kVAr
CurrentIrated = Qinst / (√3 V) = 50 / (√3 × 0.415) = 69.56 A; Imax = Irated × kh × 1.3 = 69.56 × 1.0000 × 1.3 = 90.43 A
CapacitanceXC = VC² × 1000 / QC = 0.4150² × 1000 / 50.0000 = 3.444 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 3.444) = 924.111 µF per phase (star)
BreakerICB = 1.35 × 69.56 = 93.9 A → next standard size 100 A; overload setting 1.25 × Irated = 87.0 A
Discharge resistorVpk = √2 × 415.0 V = 586.9 V; R = t / (C ln(Vpk / 50 V)) = 300 / (924.111 × 10−6 × ln(586.9 / 50)) = 131.81 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 69.56 = 1,043.4 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 50.00 = 0.025 kW + reactor 3 I² RL = 0.000 kW + discharge 0.010 kW = 0.035 kW
Target power factorQafter = P tan φ1 − Qinst = 40.81 − 50 = −9.19 kVAr — negative: the bank over-corrects and the power factor is leading, which fails the target whatever its magnitude; pf = P / √(P² + Qafter²) = 40 / √(40² + (−9.19)²) = 0.9746 → 0.9746 vs limit 0.9000 FAIL
ResonanceNo harmonic above 3 %; undetuned resonance 866.03 Hz is not excited PASS
Result1/2 checks pass → FAIL; 50 kVAr bank, no reactor, 69.6 A, 100 A breaker; governing: target power factor. Installed 50 kVAr exceeds the load's 40.8 kVAr: power factor 0.9746 LEADING at maximum demand, and more leading at lighter load. Use smaller steps, an automatic stepped bank or a lower target.
ReferenceHand calculation: tan φ1 = tan arccos 0.7 = 1.02020, tan φ2 = tan arccos 0.9 = 0.48432; Q = 40 × (1.02020 − 0.48432) = 21.44 kVAr → 50 kVAr output in 50 kVAr steps; f_r = 50 × √(15000 kVA / 50 kVAr) = 866.03 Hz (h = 17.321); significant (> 3 %) harmonics: none; no significant harmonic within ±20 % → no reactor; V_C = 0.415 kV, nameplate 1.21 × 50 = 60.50 kVAr; X_C = 0.415² × 1000 / 50 = 3.4445 Ω; C = 10⁶ / (2π × 50 × X_C) = 924.111 µF/phase; I = 50 / (√3 × 0.415) = 69.56 A; CB 1.35 × I = 93.9 A → 100 A; inrush 15 × I = 1043.4 A; R_d = 300 / (C ln(√2 V_C / 50)) = 131.81 kΩ; losses 0.0005 × 50.00 + 0.01 = 0.035 kW; 1.1 × 0.4150 = 0.4565 kV; Q_after = 40.81 − 50 = -9.19 kVAr, pf = 40 / √(40² + -9.19²) = 0.9746 LEADING → fails the target; resonance check: 866.03 Hz clear of every significant harmonic → pass (validation case c0016-415v-40kw-leading-step)

Example 5 — 400 V, power factor 0.50 → 0.90

The lowest accepted starting power factor (0.50), where tan φ1 = 1.732 dominates the kVAr.

Given: 0.4 kV, 50 Hz bus with a 25 MVA fault level; maximum demand 100 kW at power factor 0.5 lagging, target 0.9; no harmonic distortion.

Reactive powerφ1 = arccos 0.5 = 60.00°, tan φ1 = 1.73205; φ2 = arccos 0.9 = 25.84°, tan φ2 = 0.48432; Qc = P (tan φ1 − tan φ2) = 100 × (1.73205 − 0.48432) = 124.77 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈124.77 / 50⌉ × 50 = 150 kVAr
Apparent powerSbefore = 100 / 0.5 = 200.00 kVA; Safter = 100 / 0.9 = 111.11 kVA; reduction 88.89 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(25,000 kVA / 150 kVAr) = 645.50 Hz; hr = 645.50 / 50 = 12.910
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: no harmonic above 3 % → no reactor
Capacitor voltageVC = V = 0.4 kV; Vrated = 1.1 × 0.4 = 0.4400 kV (LV: no standard HV rating applies)
Capacitor ratingNo reactor: QC = Qinst = 150 kVAr; nameplate at 1.1 V: 1.21 × 150 = 181.50 kVAr
CurrentIrated = Qinst / (√3 V) = 150 / (√3 × 0.4) = 216.51 A; Imax = Irated × kh × 1.3 = 216.51 × 1.0000 × 1.3 = 281.46 A
CapacitanceXC = VC² × 1000 / QC = 0.4000² × 1000 / 150.000 = 1.067 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 1.067) = 2,984.155 µF per phase (star)
BreakerICB = 1.35 × 216.51 = 292.3 A → next standard size 315 A; overload setting 1.25 × Irated = 270.6 A
Discharge resistorVpk = √2 × 400.0 V = 565.7 V; R = t / (C ln(Vpk / 50 V)) = 300 / (2,984.155 × 10−6 × ln(565.7 / 50)) = 41.44 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 216.51 = 3,247.6 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 150.00 = 0.075 kW + reactor 3 I² RL = 0.000 kW + discharge 0.010 kW = 0.085 kW
Target power factorQafter = P tan φ1 − Qinst = 173.21 − 150 = 23.21 kVAr (lagging); pf = P / √(P² + Qafter²) = 100 / √(100² + 23.21²) = 0.9741 → 0.9741 vs limit 0.9000 PASS
ResonanceNo harmonic above 3 %; undetuned resonance 645.50 Hz is not excited PASS
Result2/2 checks pass → PASS; 150 kVAr bank, no reactor, 216.5 A, 315 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.5 = 1.73205, tan φ2 = tan arccos 0.9 = 0.48432; Q = 100 × (1.73205 − 0.48432) = 124.77 kVAr → 150 kVAr output in 50 kVAr steps; f_r = 50 × √(25000 kVA / 150 kVAr) = 645.50 Hz (h = 12.910); significant (> 3 %) harmonics: none; no significant harmonic within ±20 % → no reactor; V_C = 0.4 kV, nameplate 1.21 × 150 = 181.50 kVAr; X_C = 0.4² × 1000 / 150 = 1.0667 Ω; C = 10⁶ / (2π × 50 × X_C) = 2984.155 µF/phase; I = 150 / (√3 × 0.4) = 216.51 A; CB 1.35 × I = 292.3 A → 315 A; inrush 15 × I = 3247.6 A; R_d = 300 / (C ln(√2 V_C / 50)) = 41.44 kΩ; losses 0.0005 × 150.00 + 0.01 = 0.085 kW; 1.1 × 0.4000 = 0.4400 kV; Q_after = 173.21 − 150 = 23.21 kVAr, pf = 100 / √(100² + 23.21²) = 0.9741 lagging ≥ 0.9; resonance check: 645.50 Hz clear of every significant harmonic → pass (validation case c0016-400v-pf05-to-09)

Example 6 — 6.6 kV, 2 MW, target unity — FAIL

A target of 1.00 (tan φ2 = 0) cannot be met with 50 kVAr steps: 1,250 kVAr overshoots the 1,239.5 kVAr load, leaving the power factor leading, which fails the target check.

Given: 6.6 kV, 50 Hz bus with a 150 MVA fault level; maximum demand 2,000 kW at power factor 0.85 lagging, target 1; no harmonic distortion.

Reactive powerφ1 = arccos 0.85 = 31.79°, tan φ1 = 0.61974; φ2 = arccos 1 = 0.00°, tan φ2 = 0.00000; Qc = P (tan φ1 − tan φ2) = 2,000 × (0.61974 − 0.00000) = 1,239.49 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈1,239.49 / 50⌉ × 50 = 1,250 kVAr
Apparent powerSbefore = 2,000 / 0.85 = 2,352.94 kVA; Safter = 2,000 / 1 = 2,000.00 kVA; reduction 352.94 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(150,000 kVA / 1,250 kVAr) = 547.72 Hz; hr = 547.72 / 50 = 10.954
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: no harmonic above 3 % → no reactor
Capacitor voltageVC = V = 6.6 kV; Vrated = 1.1 × 6.6 = 7.2600 kV → next standard rating 11 kV
Capacitor ratingNo reactor: QC = Qinst = 1,250 kVAr; nameplate at 1.1 V: 1.21 × 1,250 = 1,512.50 kVAr
CurrentIrated = Qinst / (√3 V) = 1,250 / (√3 × 6.6) = 109.35 A; Imax = Irated × kh × 1.3 = 109.35 × 1.0000 × 1.3 = 142.15 A
CapacitanceXC = VC² × 1000 / QC = 6.600² × 1000 / 1250.00 = 34.85 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 34.85) = 91.342 µF per phase (star)
BreakerICB = 1.35 × 109.35 = 147.6 A → next standard size 160 A; overload setting 1.25 × Irated = 136.7 A
Discharge resistorVpk = √2 × 6,600.0 V = 9,333.8 V; R = t / (C ln(Vpk / 50 V)) = 300 / (91.342 × 10−6 × ln(9,333.8 / 50)) = 628.06 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 109.35 = 1,640.2 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 1,250.00 = 0.625 kW + reactor 3 I² RL = 0.000 kW + discharge 0.010 kW = 0.635 kW
Target power factorQafter = P tan φ1 − Qinst = 1,239.49 − 1,250 = −10.51 kVAr — negative: the bank over-corrects and the power factor is leading, which fails the target whatever its magnitude; pf = P / √(P² + Qafter²) = 2,000 / √(2,000² + (−10.51)²) = 1.0000 → 1.0000 vs limit 1.0000 FAIL
ResonanceNo harmonic above 3 %; undetuned resonance 547.72 Hz is not excited PASS
Result1/2 checks pass → FAIL; 1,250 kVAr bank, no reactor, 109.3 A, 160 A breaker; governing: target power factor. Installed 1250 kVAr exceeds the load's 1239.5 kVAr: power factor 1 LEADING at maximum demand, and more leading at lighter load. Use smaller steps, an automatic stepped bank or a lower target.
ReferenceHand calculation: tan φ1 = tan arccos 0.85 = 0.61974, tan φ2 = tan arccos 1.0 = 0.00000; Q = 2000 × (0.61974 − 0.00000) = 1239.49 kVAr → 1250 kVAr output in 50 kVAr steps; f_r = 50 × √(150000 kVA / 1250 kVAr) = 547.72 Hz (h = 10.954); significant (> 3 %) harmonics: none; no significant harmonic within ±20 % → no reactor; V_C = 6.6 kV, nameplate 1.21 × 1250 = 1512.50 kVAr; X_C = 6.6² × 1000 / 1250 = 34.8480 Ω; C = 10⁶ / (2π × 50 × X_C) = 91.342 µF/phase; I = 1250 / (√3 × 6.6) = 109.35 A; CB 1.35 × I = 147.6 A → 160 A; inrush 15 × I = 1640.2 A; R_d = 300 / (C ln(√2 V_C / 50)) = 628.06 kΩ; losses 0.0005 × 1250.00 + 0.01 = 0.635 kW; 1.1 × 6.6000 = 7.2600 kV → 11 kV standard; Q_after = 1239.49 − 1250 = -10.51 kVAr, pf = 2000 / √(2000² + -10.51²) = 1.0000 LEADING → fails the target; resonance check: 547.72 Hz clear of every significant harmonic → pass (validation case c0016-6600v-2mw-unity)

Example 7 — 11 kV, 1.5 MW, 0.70 → 0.80

The lowest accepted target (0.80): a modest step that still needs 405 kVAr.

Given: 11 kV, 50 Hz bus with a 200 MVA fault level; maximum demand 1,500 kW at power factor 0.7 lagging, target 0.8; no harmonic distortion.

Reactive powerφ1 = arccos 0.7 = 45.57°, tan φ1 = 1.02020; φ2 = arccos 0.8 = 36.87°, tan φ2 = 0.75000; Qc = P (tan φ1 − tan φ2) = 1,500 × (1.02020 − 0.75000) = 405.31 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈405.31 / 50⌉ × 50 = 450 kVAr
Apparent powerSbefore = 1,500 / 0.7 = 2,142.86 kVA; Safter = 1,500 / 0.8 = 1,875.00 kVA; reduction 267.86 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(200,000 kVA / 450 kVAr) = 1,054.09 Hz; hr = 1,054.09 / 50 = 21.082
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: no harmonic above 3 % → no reactor
Capacitor voltageVC = V = 11 kV; Vrated = 1.1 × 11 = 12.1000 kV → next standard rating 13.2 kV
Capacitor ratingNo reactor: QC = Qinst = 450 kVAr; nameplate at 1.1 V: 1.21 × 450 = 544.50 kVAr
CurrentIrated = Qinst / (√3 V) = 450 / (√3 × 11) = 23.62 A; Imax = Irated × kh × 1.3 = 23.62 × 1.0000 × 1.3 = 30.70 A
CapacitanceXC = VC² × 1000 / QC = 11.00² × 1000 / 450.000 = 268.9 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 268.9) = 11.838 µF per phase (star)
BreakerICB = 1.35 × 23.62 = 31.9 A → next standard size 32 A; overload setting 1.25 × Irated = 29.5 A
Discharge resistorVpk = √2 × 11,000.0 V = 15,556.3 V; R = t / (C ln(Vpk / 50 V)) = 300 / (11.838 × 10−6 × ln(15,556.3 / 50)) = 4,414.86 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 23.62 = 354.3 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 450.00 = 0.225 kW + reactor 3 I² RL = 0.000 kW + discharge 0.010 kW = 0.235 kW
Target power factorQafter = P tan φ1 − Qinst = 1,530.31 − 450 = 1,080.31 kVAr (lagging); pf = P / √(P² + Qafter²) = 1,500 / √(1,500² + 1,080.31²) = 0.8115 → 0.8115 vs limit 0.8000 PASS
ResonanceNo harmonic above 3 %; undetuned resonance 1,054.09 Hz is not excited PASS
Result2/2 checks pass → PASS; 450 kVAr bank, no reactor, 23.6 A, 32 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.7 = 1.02020, tan φ2 = tan arccos 0.8 = 0.75000; Q = 1500 × (1.02020 − 0.75000) = 405.31 kVAr → 450 kVAr output in 50 kVAr steps; f_r = 50 × √(200000 kVA / 450 kVAr) = 1054.09 Hz (h = 21.082); significant (> 3 %) harmonics: none; no significant harmonic within ±20 % → no reactor; V_C = 11 kV, nameplate 1.21 × 450 = 544.50 kVAr; X_C = 11² × 1000 / 450 = 268.8889 Ω; C = 10⁶ / (2π × 50 × X_C) = 11.838 µF/phase; I = 450 / (√3 × 11) = 23.62 A; CB 1.35 × I = 31.9 A → 32 A; inrush 15 × I = 354.3 A; R_d = 300 / (C ln(√2 V_C / 50)) = 4414.86 kΩ; losses 0.0005 × 450.00 + 0.01 = 0.235 kW; 1.1 × 11.0000 = 12.1000 kV → 13.2 kV standard; Q_after = 1530.31 − 450 = 1080.31 kVAr, pf = 1500 / √(1500² + 1080.31²) = 0.8115 lagging ≥ 0.8; resonance check: 1054.09 Hz clear of every significant harmonic → pass (validation case c0016-11kv-1500kw-target-080)

Example 8 — 22 kV, 10 MW, 0.85 → 0.97

A large MV bank: 3.7 MVAr, 1.1 × 22 kV = 24.2 kV selects the 33 kV standard rating.

Given: 22 kV, 50 Hz bus with a 800 MVA fault level; maximum demand 10,000 kW at power factor 0.85 lagging, target 0.97; no harmonic distortion.

Reactive powerφ1 = arccos 0.85 = 31.79°, tan φ1 = 0.61974; φ2 = arccos 0.97 = 14.07°, tan φ2 = 0.25062; Qc = P (tan φ1 − tan φ2) = 10,000 × (0.61974 − 0.25062) = 3,691.21 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈3,691.21 / 50⌉ × 50 = 3,700 kVAr
Apparent powerSbefore = 10,000 / 0.85 = 11,764.71 kVA; Safter = 10,000 / 0.97 = 10,309.28 kVA; reduction 1,455.43 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(800,000 kVA / 3,700 kVAr) = 735.21 Hz; hr = 735.21 / 50 = 14.704
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: no harmonic above 3 % → no reactor
Capacitor voltageVC = V = 22 kV; Vrated = 1.1 × 22 = 24.2000 kV → next standard rating 33 kV
Capacitor ratingNo reactor: QC = Qinst = 3,700 kVAr; nameplate at 1.1 V: 1.21 × 3,700 = 4,477.00 kVAr
CurrentIrated = Qinst / (√3 V) = 3,700 / (√3 × 22) = 97.10 A; Imax = Irated × kh × 1.3 = 97.10 × 1.0000 × 1.3 = 126.23 A
CapacitanceXC = VC² × 1000 / QC = 22.00² × 1000 / 3700.00 = 130.8 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 130.8) = 24.334 µF per phase (star)
BreakerICB = 1.35 × 97.10 = 131.1 A → next standard size 160 A; overload setting 1.25 × Irated = 121.4 A
Discharge resistorVpk = √2 × 22,000.0 V = 31,112.7 V; R = t / (C ln(Vpk / 50 V)) = 300 / (24.334 × 10−6 × ln(31,112.7 / 50)) = 1,916.36 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 97.10 = 1,456.5 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 3,700.00 = 1.850 kW + reactor 3 I² RL = 0.000 kW + discharge 0.010 kW = 1.860 kW
Target power factorQafter = P tan φ1 − Qinst = 6,197.44 − 3,700 = 2,497.44 kVAr (lagging); pf = P / √(P² + Qafter²) = 10,000 / √(10,000² + 2,497.44²) = 0.9702 → 0.9702 vs limit 0.9700 PASS
ResonanceNo harmonic above 3 %; undetuned resonance 735.21 Hz is not excited PASS
Result2/2 checks pass → PASS; 3,700 kVAr bank, no reactor, 97.1 A, 160 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.85 = 0.61974, tan φ2 = tan arccos 0.97 = 0.25062; Q = 10000 × (0.61974 − 0.25062) = 3691.21 kVAr → 3700 kVAr output in 50 kVAr steps; f_r = 50 × √(800000 kVA / 3700 kVAr) = 735.21 Hz (h = 14.704); significant (> 3 %) harmonics: none; no significant harmonic within ±20 % → no reactor; V_C = 22 kV, nameplate 1.21 × 3700 = 4477.00 kVAr; X_C = 22² × 1000 / 3700 = 130.8108 Ω; C = 10⁶ / (2π × 50 × X_C) = 24.334 µF/phase; I = 3700 / (√3 × 22) = 97.10 A; CB 1.35 × I = 131.1 A → 160 A; inrush 15 × I = 1456.5 A; R_d = 300 / (C ln(√2 V_C / 50)) = 1916.36 kΩ; losses 0.0005 × 3700.00 + 0.01 = 1.860 kW; 1.1 × 22.0000 = 24.2000 kV → 33 kV standard; Q_after = 6197.44 − 3700 = 2497.44 kVAr, pf = 10000 / √(10000² + 2497.44²) = 0.9702 lagging ≥ 0.97; resonance check: 735.21 Hz clear of every significant harmonic → pass (validation case c0016-22kv-10mw-pf097)

Example 9 — 480 V, 60 Hz, 750 kW

A 60 Hz LV system: capacitance and resonance use f = 60 Hz.

Given: 0.48 kV, 60 Hz bus with a 30 MVA fault level; maximum demand 750 kW at power factor 0.8 lagging, target 0.95; no harmonic distortion.

Reactive powerφ1 = arccos 0.8 = 36.87°, tan φ1 = 0.75000; φ2 = arccos 0.95 = 18.19°, tan φ2 = 0.32868; Qc = P (tan φ1 − tan φ2) = 750 × (0.75000 − 0.32868) = 315.99 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈315.99 / 50⌉ × 50 = 350 kVAr
Apparent powerSbefore = 750 / 0.8 = 937.50 kVA; Safter = 750 / 0.95 = 789.47 kVA; reduction 148.03 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 60 × √(30,000 kVA / 350 kVAr) = 555.49 Hz; hr = 555.49 / 60 = 9.258
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: no harmonic above 3 % → no reactor
Capacitor voltageVC = V = 0.48 kV; Vrated = 1.1 × 0.48 = 0.5280 kV (LV: no standard HV rating applies)
Capacitor ratingNo reactor: QC = Qinst = 350 kVAr; nameplate at 1.1 V: 1.21 × 350 = 423.50 kVAr
CurrentIrated = Qinst / (√3 V) = 350 / (√3 × 0.48) = 420.98 A; Imax = Irated × kh × 1.3 = 420.98 × 1.0000 × 1.3 = 547.28 A
CapacitanceXC = VC² × 1000 / QC = 0.4800² × 1000 / 350.000 = 0.6583 Ω; C = 106 / (2π f XC) = 106 / (2π × 60 × 0.6583) = 4,029.531 µF per phase (star)
BreakerICB = 1.35 × 420.98 = 568.3 A → next standard size 630 A; overload setting 1.25 × Irated = 526.2 A
Discharge resistorVpk = √2 × 480.0 V = 678.8 V; R = t / (C ln(Vpk / 50 V)) = 300 / (4,029.531 × 10−6 × ln(678.8 / 50)) = 28.54 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 420.98 = 6,314.8 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 350.00 = 0.175 kW + reactor 3 I² RL = 0.000 kW + discharge 0.010 kW = 0.185 kW
Target power factorQafter = P tan φ1 − Qinst = 562.50 − 350 = 212.50 kVAr (lagging); pf = P / √(P² + Qafter²) = 750 / √(750² + 212.50²) = 0.9621 → 0.9621 vs limit 0.9500 PASS
ResonanceNo harmonic above 3 %; undetuned resonance 555.49 Hz is not excited PASS
Result2/2 checks pass → PASS; 350 kVAr bank, no reactor, 421.0 A, 630 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.8 = 0.75000, tan φ2 = tan arccos 0.95 = 0.32868; Q = 750 × (0.75000 − 0.32868) = 315.99 kVAr → 350 kVAr output in 50 kVAr steps; f_r = 60 × √(30000 kVA / 350 kVAr) = 555.49 Hz (h = 9.258); significant (> 3 %) harmonics: none; no significant harmonic within ±20 % → no reactor; V_C = 0.48 kV, nameplate 1.21 × 350 = 423.50 kVAr; X_C = 0.48² × 1000 / 350 = 0.6583 Ω; C = 10⁶ / (2π × 60 × X_C) = 4029.531 µF/phase; I = 350 / (√3 × 0.48) = 420.98 A; CB 1.35 × I = 568.3 A → 630 A; inrush 15 × I = 6314.8 A; R_d = 300 / (C ln(√2 V_C / 50)) = 28.54 kΩ; losses 0.0005 × 350.00 + 0.01 = 0.185 kW; 1.1 × 0.4800 = 0.5280 kV; Q_after = 562.50 − 350 = 212.50 kVAr, pf = 750 / √(750² + 212.50²) = 0.9621 lagging ≥ 0.95; resonance check: 555.49 Hz clear of every significant harmonic → pass (validation case c0016-60hz-480v-750kw)

Example 10 — 13.8 kV, 60 Hz, 4 MW

A 60 Hz MV bank; 1.1 × 13.8 kV = 15.18 kV selects the 22 kV standard rating.

Given: 13.8 kV, 60 Hz bus with a 300 MVA fault level; maximum demand 4,000 kW at power factor 0.8 lagging, target 0.95; no harmonic distortion.

Reactive powerφ1 = arccos 0.8 = 36.87°, tan φ1 = 0.75000; φ2 = arccos 0.95 = 18.19°, tan φ2 = 0.32868; Qc = P (tan φ1 − tan φ2) = 4,000 × (0.75000 − 0.32868) = 1,685.26 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈1,685.26 / 50⌉ × 50 = 1,700 kVAr
Apparent powerSbefore = 4,000 / 0.8 = 5,000.00 kVA; Safter = 4,000 / 0.95 = 4,210.53 kVA; reduction 789.47 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 60 × √(300,000 kVA / 1,700 kVAr) = 797.05 Hz; hr = 797.05 / 60 = 13.284
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: no harmonic above 3 % → no reactor
Capacitor voltageVC = V = 13.8 kV; Vrated = 1.1 × 13.8 = 15.1800 kV → next standard rating 22 kV
Capacitor ratingNo reactor: QC = Qinst = 1,700 kVAr; nameplate at 1.1 V: 1.21 × 1,700 = 2,057.00 kVAr
CurrentIrated = Qinst / (√3 V) = 1,700 / (√3 × 13.8) = 71.12 A; Imax = Irated × kh × 1.3 = 71.12 × 1.0000 × 1.3 = 92.46 A
CapacitanceXC = VC² × 1000 / QC = 13.80² × 1000 / 1700.00 = 112.0 Ω; C = 106 / (2π f XC) = 106 / (2π × 60 × 112.0) = 23.679 µF per phase (star)
BreakerICB = 1.35 × 71.12 = 96.0 A → next standard size 100 A; overload setting 1.25 × Irated = 88.9 A
Discharge resistorVpk = √2 × 13,800.0 V = 19,516.1 V; R = t / (C ln(Vpk / 50 V)) = 300 / (23.679 × 10−6 × ln(19,516.1 / 50)) = 2,123.28 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 71.12 = 1,066.8 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 1,700.00 = 0.850 kW + reactor 3 I² RL = 0.000 kW + discharge 0.010 kW = 0.860 kW
Target power factorQafter = P tan φ1 − Qinst = 3,000.00 − 1,700 = 1,300.00 kVAr (lagging); pf = P / √(P² + Qafter²) = 4,000 / √(4,000² + 1,300.00²) = 0.9510 → 0.9510 vs limit 0.9500 PASS
ResonanceNo harmonic above 3 %; undetuned resonance 797.05 Hz is not excited PASS
Result2/2 checks pass → PASS; 1,700 kVAr bank, no reactor, 71.1 A, 100 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.8 = 0.75000, tan φ2 = tan arccos 0.95 = 0.32868; Q = 4000 × (0.75000 − 0.32868) = 1685.26 kVAr → 1700 kVAr output in 50 kVAr steps; f_r = 60 × √(300000 kVA / 1700 kVAr) = 797.05 Hz (h = 13.284); significant (> 3 %) harmonics: none; no significant harmonic within ±20 % → no reactor; V_C = 13.8 kV, nameplate 1.21 × 1700 = 2057.00 kVAr; X_C = 13.8² × 1000 / 1700 = 112.0235 Ω; C = 10⁶ / (2π × 60 × X_C) = 23.679 µF/phase; I = 1700 / (√3 × 13.8) = 71.12 A; CB 1.35 × I = 96.0 A → 100 A; inrush 15 × I = 1066.8 A; R_d = 300 / (C ln(√2 V_C / 50)) = 2123.28 kΩ; losses 0.0005 × 1700.00 + 0.01 = 0.860 kW; 1.1 × 13.8000 = 15.1800 kV → 22 kV standard; Q_after = 3000.00 − 1700 = 1300.00 kVAr, pf = 4000 / √(4000² + 1300.00²) = 0.9510 lagging ≥ 0.95; resonance check: 797.05 Hz clear of every significant harmonic → pass (validation case c0016-60hz-13800v-4mw)

Example 11 — 11 kV, 6 % 5th present but resonance at the 29th

Significant 5th and 7th harmonics do not need a reactor when the bank's resonance is far from them.

Given: 11 kV, 50 Hz bus with a 500 MVA fault level; maximum demand 2,000 kW at power factor 0.85 lagging, target 0.95; harmonic distortion 5th 6 %, 7th 4 %.

Reactive powerφ1 = arccos 0.85 = 31.79°, tan φ1 = 0.61974; φ2 = arccos 0.95 = 18.19°, tan φ2 = 0.32868; Qc = P (tan φ1 − tan φ2) = 2,000 × (0.61974 − 0.32868) = 582.12 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈582.12 / 50⌉ × 50 = 600 kVAr
Apparent powerSbefore = 2,000 / 0.85 = 2,352.94 kVA; Safter = 2,000 / 0.95 = 2,105.26 kVA; reduction 247.68 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(500,000 kVA / 600 kVAr) = 1,443.38 Hz; hr = 1,443.38 / 50 = 28.868
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: 5th (250 Hz, 6 % > 3 %): |1,443.4 − 250| / 250 = 477.4 % (outside ±20 %); 7th (350 Hz, 4 % > 3 %): |1,443.4 − 350| / 350 = 312.4 % (outside ±20 %); no significant harmonic within ±20 % → no reactor
Capacitor voltageVC = V = 11 kV; Vrated = 1.1 × 11 = 12.1000 kV → next standard rating 13.2 kV
Capacitor ratingNo reactor: QC = Qinst = 600 kVAr; nameplate at 1.1 V: 1.21 × 600 = 726.00 kVAr
CurrentIrated = Qinst / (√3 V) = 600 / (√3 × 11) = 31.49 A; Imax = Irated × kh × 1.3 = 31.49 × 1.0000 × 1.3 = 40.94 A
CapacitanceXC = VC² × 1000 / QC = 11.00² × 1000 / 600.000 = 201.7 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 201.7) = 15.784 µF per phase (star)
BreakerICB = 1.35 × 31.49 = 42.5 A → next standard size 50 A; overload setting 1.25 × Irated = 39.4 A
Discharge resistorVpk = √2 × 11,000.0 V = 15,556.3 V; R = t / (C ln(Vpk / 50 V)) = 300 / (15.784 × 10−6 × ln(15,556.3 / 50)) = 3,311.14 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 31.49 = 472.4 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 600.00 = 0.300 kW + reactor 3 I² RL = 0.000 kW + discharge 0.010 kW = 0.310 kW
Target power factorQafter = P tan φ1 − Qinst = 1,239.49 − 600 = 639.49 kVAr (lagging); pf = P / √(P² + Qafter²) = 2,000 / √(2,000² + 639.49²) = 0.9525 → 0.9525 vs limit 0.9500 PASS
ResonanceUndetuned resonance 1,443.38 Hz must be outside ±20 % of every significant harmonic (5th, 7th) → 1,443.38 vs limit 350.00 Hz PASS
Result2/2 checks pass → PASS; 600 kVAr bank, no reactor, 31.5 A, 50 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.85 = 0.61974, tan φ2 = tan arccos 0.95 = 0.32868; Q = 2000 × (0.61974 − 0.32868) = 582.12 kVAr → 600 kVAr output in 50 kVAr steps; f_r = 50 × √(500000 kVA / 600 kVAr) = 1443.38 Hz (h = 28.868); significant (> 3 %) harmonics: 5th 6 %, 7th 4 %; no significant harmonic within ±20 % → no reactor; V_C = 11 kV, nameplate 1.21 × 600 = 726.00 kVAr; X_C = 11² × 1000 / 600 = 201.6667 Ω; C = 10⁶ / (2π × 50 × X_C) = 15.784 µF/phase; I = 600 / (√3 × 11) = 31.49 A; CB 1.35 × I = 42.5 A → 50 A; inrush 15 × I = 472.4 A; R_d = 300 / (C ln(√2 V_C / 50)) = 3311.14 kΩ; losses 0.0005 × 600.00 + 0.01 = 0.310 kW; 1.1 × 11.0000 = 12.1000 kV → 13.2 kV standard; Q_after = 1239.49 − 600 = 639.49 kVAr, pf = 2000 / √(2000² + 639.49²) = 0.9525 lagging ≥ 0.95; resonance check: 1443.38 Hz clear of every significant harmonic → pass (validation case c0016-11kv-harmonics-far-from-resonance)

Example 12 — 415 V, resonance at the 5th with only 3 % 5th

The installed bank resonates at the 5th harmonic but 3 % is not 'significant' (> 3 %), so no reactor is fitted and a warning asks for a harmonic survey.

Given: 0.415 kV, 50 Hz bus with a 10 MVA fault level; maximum demand 1,000 kW at power factor 0.8 lagging, target 0.95; harmonic distortion 5th 3 %.

Reactive powerφ1 = arccos 0.8 = 36.87°, tan φ1 = 0.75000; φ2 = arccos 0.95 = 18.19°, tan φ2 = 0.32868; Qc = P (tan φ1 − tan φ2) = 1,000 × (0.75000 − 0.32868) = 421.32 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈421.32 / 50⌉ × 50 = 450 kVAr
Apparent powerSbefore = 1,000 / 0.8 = 1,250.00 kVA; Safter = 1,000 / 0.95 = 1,052.63 kVA; reduction 197.37 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(10,000 kVA / 450 kVAr) = 235.70 Hz; hr = 235.70 / 50 = 4.714
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: 5th (250 Hz, 3 % ≤ 3 %, not significant): |235.7 − 250| / 250 = 5.7 % (within ±20 %); no harmonic above 3 % → no reactor
Capacitor voltageVC = V = 0.415 kV; Vrated = 1.1 × 0.415 = 0.4565 kV (LV: no standard HV rating applies)
Capacitor ratingNo reactor: QC = Qinst = 450 kVAr; nameplate at 1.1 V: 1.21 × 450 = 544.50 kVAr
CurrentIrated = Qinst / (√3 V) = 450 / (√3 × 0.415) = 626.04 A; Imax = Irated × kh × 1.3 = 626.04 × 1.0000 × 1.3 = 813.86 A
CapacitanceXC = VC² × 1000 / QC = 0.4150² × 1000 / 450.000 = 0.3827 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 0.3827) = 8,316.995 µF per phase (star)
BreakerICB = 1.35 × 626.04 = 845.2 A → next standard size 1,000 A; overload setting 1.25 × Irated = 782.6 A
Discharge resistorVpk = √2 × 415.0 V = 586.9 V; R = t / (C ln(Vpk / 50 V)) = 300 / (8,316.995 × 10−6 × ln(586.9 / 50)) = 14.65 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 626.04 = 9,390.6 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 450.00 = 0.225 kW + reactor 3 I² RL = 0.000 kW + discharge 0.010 kW = 0.235 kW
Target power factorQafter = P tan φ1 − Qinst = 750.00 − 450 = 300.00 kVAr (lagging); pf = P / √(P² + Qafter²) = 1,000 / √(1,000² + 300.00²) = 0.9578 → 0.9578 vs limit 0.9500 PASS
ResonanceNo harmonic above 3 %; undetuned resonance 235.70 Hz is not excited PASS
Result2/2 checks pass → PASS; 450 kVAr bank, no reactor, 626.0 A, 1,000 A breaker; governing: target power factor. Resonance 236 Hz is within ±20 % of harmonic(s) [5] but none exceeds 3 %; confirm the harmonic survey before installing an undetuned bank.
ReferenceHand calculation: tan φ1 = tan arccos 0.8 = 0.75000, tan φ2 = tan arccos 0.95 = 0.32868; Q = 1000 × (0.75000 − 0.32868) = 421.32 kVAr → 450 kVAr output in 50 kVAr steps; f_r = 50 × √(10000 kVA / 450 kVAr) = 235.70 Hz (h = 4.714); significant (> 3 %) harmonics: none; no significant harmonic within ±20 % → no reactor; V_C = 0.415 kV, nameplate 1.21 × 450 = 544.50 kVAr; X_C = 0.415² × 1000 / 450 = 0.3827 Ω; C = 10⁶ / (2π × 50 × X_C) = 8316.995 µF/phase; I = 450 / (√3 × 0.415) = 626.04 A; CB 1.35 × I = 845.2 A → 1000 A; inrush 15 × I = 9390.6 A; R_d = 300 / (C ln(√2 V_C / 50)) = 14.65 kΩ; losses 0.0005 × 450.00 + 0.01 = 0.235 kW; 1.1 × 0.4150 = 0.4565 kV; Q_after = 750.00 − 450 = 300.00 kVAr, pf = 1000 / √(1000² + 300.00²) = 0.9578 lagging ≥ 0.95; resonance check: 235.70 Hz clear of every significant harmonic → pass (validation case c0016-415v-near-5th-3pct-warning)

Example 13 — 415 V, 500 kW, resonance near the 7th (4 %)

The installed 300 kVAr resonates at 289 Hz, within 20 % of a 4 % 7th harmonic; the 7th is the lowest significant harmonic, so a 5.67 % reactor tunes the bank to 210 Hz.

Given: 0.415 kV, 50 Hz bus with a 10 MVA fault level; maximum demand 500 kW at power factor 0.75 lagging, target 0.95; harmonic distortion 7th 4 %.

Reactive powerφ1 = arccos 0.75 = 41.41°, tan φ1 = 0.88192; φ2 = arccos 0.95 = 18.19°, tan φ2 = 0.32868; Qc = P (tan φ1 − tan φ2) = 500 × (0.88192 − 0.32868) = 276.62 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈276.62 / 50⌉ × 50 = 300 kVAr
Apparent powerSbefore = 500 / 0.75 = 666.67 kVA; Safter = 500 / 0.95 = 526.32 kVA; reduction 140.35 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(10,000 kVA / 300 kVAr) = 288.68 Hz; hr = 288.68 / 50 = 5.774
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: 7th (350 Hz, 4 % > 3 %): |288.7 − 350| / 350 = 17.5 % (within ±20 %); Resonance near 7th harmonic → reactor by the lowest significant harmonic (7th; 3rd → 14 %, 5th → 7 %, otherwise 5.67 %) = 5.67 % detuning reactor
Tuned frequencyp = XL/XC = 0.0567; ftuned = f / √p = 50 / √0.0567 = 209.98 Hz (h = 4.20)
Capacitor voltageSeries branch: VC = V / (1 − p) = 0.415 / 0.9433 = 0.4399 kV (6.00 % rise across the capacitors); Vrated = 1.1 × 0.4399 = 0.4839 kV (LV: no standard HV rating applies)
Capacitor ratingQeff = V² / (XC(1 − p)), so the capacitors generate QC = Qinst / (1 − p) = 300 / 0.9433 = 318.03 kVAr at VC; nameplate at 1.1 VC: 1.21 × 318.03 = 384.82 kVAr
CurrentIrated = Qinst / (√3 V) = 300 / (√3 × 0.415) = 417.36 A; Imax = Irated × kh × 1.3 = 417.36 × 1.0601 × 1.3 = 575.18 A
CapacitanceXC = VC² × 1000 / QC = 0.4399² × 1000 / 318.030 = 0.6086 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 0.6086) = 5,230.281 µF per phase (star)
ReactorXL = p XC = 0.0567 × 0.6086 = 0.03451 Ω; L = 1000 XL / (2π f) = 0.1098 mH; RL = XL / 50 = 0.0006901 Ω
BreakerICB = 1.35 × 417.36 = 563.4 A → next standard size 630 A; overload setting 1.25 × Irated = 521.7 A
Discharge resistorVpk = √2 × 439.9 V = 622.2 V; R = t / (C ln(Vpk / 50 V)) = 300 / (5,230.281 × 10−6 × ln(622.2 / 50)) = 22.75 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 417.36 = 6,260.4 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 318.03 = 0.159 kW + reactor 3 I² RL = 0.361 kW + discharge 0.010 kW = 0.530 kW
Target power factorQafter = P tan φ1 − Qinst = 440.96 − 300 = 140.96 kVAr (lagging); pf = P / √(P² + Qafter²) = 500 / √(500² + 140.96²) = 0.9625 → 0.9625 vs limit 0.9500 PASS
ResonanceTuned 209.98 Hz must lie below the lowest significant harmonic (7th = 350 Hz) → 209.98 vs limit 350.00 Hz PASS
Result2/2 checks pass → PASS; 300 kVAr bank, 5.67 % reactor tuned to 210.0 Hz, 417.4 A, 630 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.75 = 0.88192, tan φ2 = tan arccos 0.95 = 0.32868; Q = 500 × (0.88192 − 0.32868) = 276.62 kVAr → 300 kVAr output in 50 kVAr steps; f_r = 50 × √(10000 kVA / 300 kVAr) = 288.68 Hz (h = 5.774); significant (> 3 %) harmonics: 7th 4 %; within 20 % of the 7th (350 Hz) → reactor by the lowest significant harmonic (7th) = 5.67 %; f_t = 50 / √0.0567 = 209.98 Hz; V_C = 0.415 / (1 − 0.0567) = 0.4399 kV; Q_C = 300 / 0.9433 = 318.03 kVAr; nameplate 1.21 × Q_C = 384.82 kVAr; X_C = 0.4399² × 1000 / 318.03 = 0.6086 Ω; X_L = 0.0567 X_C = 0.0345 Ω, L = 0.1098 mH; C = 10⁶ / (2π × 50 × X_C) = 5230.281 µF/phase; I = 300 / (√3 × 0.415) = 417.36 A; CB 1.35 × I = 563.4 A → 630 A; inrush 15 × I = 6260.4 A; R_d = 300 / (C ln(√2 V_C / 50)) = 22.75 kΩ; losses 0.0005 × 318.03 + 3 I² X_L/50 + 0.01 = 0.530 kW; 1.1 × 0.4399 = 0.4839 kV; Q_after = 440.96 − 300 = 140.96 kVAr, pf = 500 / √(500² + 140.96²) = 0.9625 lagging ≥ 0.95; resonance check: f_t 209.98 Hz < 350 Hz → pass (validation case c0016-415v-detuned-7th)

Example 14 — 11 kV, 3 MW, 10 % 5th and 5 % 7th

The installed bank resonates near the 7th, but the 5th is also significant, so the 7 % reactor (tuned 189 Hz) is chosen; capacitor voltage 11 / 0.93 = 11.83 kV → 13.2 kV rating.

Given: 11 kV, 50 Hz bus with a 100 MVA fault level; maximum demand 3,000 kW at power factor 0.78 lagging, target 0.95; harmonic distortion 5th 10 %, 7th 5 %.

Reactive powerφ1 = arccos 0.78 = 38.74°, tan φ1 = 0.80228; φ2 = arccos 0.95 = 18.19°, tan φ2 = 0.32868; Qc = P (tan φ1 − tan φ2) = 3,000 × (0.80228 − 0.32868) = 1,420.79 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈1,420.79 / 50⌉ × 50 = 1,450 kVAr
Apparent powerSbefore = 3,000 / 0.78 = 3,846.15 kVA; Safter = 3,000 / 0.95 = 3,157.89 kVA; reduction 688.26 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(100,000 kVA / 1,450 kVAr) = 415.23 Hz; hr = 415.23 / 50 = 8.305
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: 5th (250 Hz, 10 % > 3 %): |415.2 − 250| / 250 = 66.1 % (outside ±20 %); 7th (350 Hz, 5 % > 3 %): |415.2 − 350| / 350 = 18.6 % (within ±20 %); Resonance near 7th harmonic → reactor by the lowest significant harmonic (5th; 3rd → 14 %, 5th → 7 %, otherwise 5.67 %) = 7 % detuning reactor
Tuned frequencyp = XL/XC = 0.0700; ftuned = f / √p = 50 / √0.0700 = 188.98 Hz (h = 3.78)
Capacitor voltageSeries branch: VC = V / (1 − p) = 11 / 0.9300 = 11.8280 kV (7.53 % rise across the capacitors); Vrated = 1.1 × 11.8280 = 13.0108 kV → next standard rating 13.2 kV
Capacitor ratingQeff = V² / (XC(1 − p)), so the capacitors generate QC = Qinst / (1 − p) = 1,450 / 0.9300 = 1,559.14 kVAr at VC; nameplate at 1.1 VC: 1.21 × 1,559.14 = 1,886.56 kVAr
CurrentIrated = Qinst / (√3 V) = 1,450 / (√3 × 11) = 76.11 A; Imax = Irated × kh × 1.3 = 76.11 × 1.0753 × 1.3 = 106.38 A
CapacitanceXC = VC² × 1000 / QC = 11.83² × 1000 / 1559.14 = 89.73 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 89.73) = 35.474 µF per phase (star)
ReactorXL = p XC = 0.0700 × 89.73 = 6.281 Ω; L = 1000 XL / (2π f) = 19.9932 mH; RL = XL / 50 = 0.1256 Ω
BreakerICB = 1.35 × 76.11 = 102.7 A → next standard size 125 A; overload setting 1.25 × Irated = 95.1 A
Discharge resistorVpk = √2 × 11,828.0 V = 16,727.3 V; R = t / (C ln(Vpk / 50 V)) = 300 / (35.474 × 10−6 × ln(16,727.3 / 50)) = 1,454.86 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 76.11 = 1,141.6 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 1,559.14 = 0.780 kW + reactor 3 I² RL = 2.183 kW + discharge 0.010 kW = 2.972 kW
Target power factorQafter = P tan φ1 − Qinst = 2,406.84 − 1,450 = 956.84 kVAr (lagging); pf = P / √(P² + Qafter²) = 3,000 / √(3,000² + 956.84²) = 0.9527 → 0.9527 vs limit 0.9500 PASS
ResonanceTuned 188.98 Hz must lie below the lowest significant harmonic (5th = 250 Hz) → 188.98 vs limit 250.00 Hz PASS
Result2/2 checks pass → PASS; 1,450 kVAr bank, 7 % reactor tuned to 189.0 Hz, 76.1 A, 125 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.78 = 0.80228, tan φ2 = tan arccos 0.95 = 0.32868; Q = 3000 × (0.80228 − 0.32868) = 1420.79 kVAr → 1450 kVAr output in 50 kVAr steps; f_r = 50 × √(100000 kVA / 1450 kVAr) = 415.23 Hz (h = 8.305); significant (> 3 %) harmonics: 5th 10 %, 7th 5 %; within 20 % of the 7th (350 Hz) → reactor by the lowest significant harmonic (5th) = 7 %; f_t = 50 / √0.07 = 188.98 Hz; V_C = 11 / (1 − 0.07) = 11.8280 kV; Q_C = 1450 / 0.9300 = 1559.14 kVAr; nameplate 1.21 × Q_C = 1886.56 kVAr; X_C = 11.8280² × 1000 / 1559.14 = 89.7293 Ω; X_L = 0.07 X_C = 6.2811 Ω, L = 19.9932 mH; C = 10⁶ / (2π × 50 × X_C) = 35.474 µF/phase; I = 1450 / (√3 × 11) = 76.11 A; CB 1.35 × I = 102.7 A → 125 A; inrush 15 × I = 1141.6 A; R_d = 300 / (C ln(√2 V_C / 50)) = 1454.86 kΩ; losses 0.0005 × 1559.14 + 3 I² X_L/50 + 0.01 = 2.972 kW; 1.1 × 11.8280 = 13.0108 kV → 13.2 kV standard; Q_after = 2406.84 − 1450 = 956.84 kVAr, pf = 3000 / √(3000² + 956.84²) = 0.9527 lagging ≥ 0.95; resonance check: f_t 188.98 Hz < 250 Hz → pass (validation case c0016-11kv-3mw-detuned-7th)

Example 15 — 400 V, 800 kW, resonance near the 5th

The installed 450 kVAr resonates at 236 Hz next to an 8 % 5th harmonic: a 7 % reactor tunes it to 189 Hz, below the 5th.

Given: 0.4 kV, 50 Hz bus with a 10 MVA fault level; maximum demand 800 kW at power factor 0.75 lagging, target 0.95; harmonic distortion 5th 8 %, 7th 5 %.

Reactive powerφ1 = arccos 0.75 = 41.41°, tan φ1 = 0.88192; φ2 = arccos 0.95 = 18.19°, tan φ2 = 0.32868; Qc = P (tan φ1 − tan φ2) = 800 × (0.88192 − 0.32868) = 442.59 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈442.59 / 50⌉ × 50 = 450 kVAr
Apparent powerSbefore = 800 / 0.75 = 1,066.67 kVA; Safter = 800 / 0.95 = 842.11 kVA; reduction 224.56 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(10,000 kVA / 450 kVAr) = 235.70 Hz; hr = 235.70 / 50 = 4.714
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: 5th (250 Hz, 8 % > 3 %): |235.7 − 250| / 250 = 5.7 % (within ±20 %); 7th (350 Hz, 5 % > 3 %): |235.7 − 350| / 350 = 32.7 % (outside ±20 %); Resonance near 5th harmonic → reactor by the lowest significant harmonic (5th; 3rd → 14 %, 5th → 7 %, otherwise 5.67 %) = 7 % detuning reactor
Tuned frequencyp = XL/XC = 0.0700; ftuned = f / √p = 50 / √0.0700 = 188.98 Hz (h = 3.78)
Capacitor voltageSeries branch: VC = V / (1 − p) = 0.4 / 0.9300 = 0.4301 kV (7.52 % rise across the capacitors); Vrated = 1.1 × 0.4301 = 0.4731 kV (LV: no standard HV rating applies)
Capacitor ratingQeff = V² / (XC(1 − p)), so the capacitors generate QC = Qinst / (1 − p) = 450 / 0.9300 = 483.87 kVAr at VC; nameplate at 1.1 VC: 1.21 × 483.87 = 585.48 kVAr
CurrentIrated = Qinst / (√3 V) = 450 / (√3 × 0.4) = 649.52 A; Imax = Irated × kh × 1.3 = 649.52 × 1.0753 × 1.3 = 907.93 A
CapacitanceXC = VC² × 1000 / QC = 0.4301² × 1000 / 483.870 = 0.3823 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 0.3823) = 8,325.793 µF per phase (star)
ReactorXL = p XC = 0.0700 × 0.3823 = 0.02676 Ω; L = 1000 XL / (2π f) = 0.0852 mH; RL = XL / 50 = 0.0005352 Ω
BreakerICB = 1.35 × 649.52 = 876.9 A → next standard size 1,000 A; overload setting 1.25 × Irated = 811.9 A
Discharge resistorVpk = √2 × 430.1 V = 608.3 V; R = t / (C ln(Vpk / 50 V)) = 300 / (8,325.793 × 10−6 × ln(608.3 / 50)) = 14.42 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 649.52 = 9,742.8 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 483.87 = 0.242 kW + reactor 3 I² RL = 0.677 kW + discharge 0.010 kW = 0.929 kW
Target power factorQafter = P tan φ1 − Qinst = 705.53 − 450 = 255.53 kVAr (lagging); pf = P / √(P² + Qafter²) = 800 / √(800² + 255.53²) = 0.9526 → 0.9526 vs limit 0.9500 PASS
ResonanceTuned 188.98 Hz must lie below the lowest significant harmonic (5th = 250 Hz) → 188.98 vs limit 250.00 Hz PASS
Result2/2 checks pass → PASS; 450 kVAr bank, 7 % reactor tuned to 189.0 Hz, 649.5 A, 1,000 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.75 = 0.88192, tan φ2 = tan arccos 0.95 = 0.32868; Q = 800 × (0.88192 − 0.32868) = 442.59 kVAr → 450 kVAr output in 50 kVAr steps; f_r = 50 × √(10000 kVA / 450 kVAr) = 235.70 Hz (h = 4.714); significant (> 3 %) harmonics: 5th 8 %, 7th 5 %; within 20 % of the 5th (250 Hz) → reactor by the lowest significant harmonic (5th) = 7 %; f_t = 50 / √0.07 = 188.98 Hz; V_C = 0.4 / (1 − 0.07) = 0.4301 kV; Q_C = 450 / 0.9300 = 483.87 kVAr; nameplate 1.21 × Q_C = 585.48 kVAr; X_C = 0.4301² × 1000 / 483.87 = 0.3823 Ω; X_L = 0.07 X_C = 0.0268 Ω, L = 0.0852 mH; C = 10⁶ / (2π × 50 × X_C) = 8325.793 µF/phase; I = 450 / (√3 × 0.4) = 649.52 A; CB 1.35 × I = 876.9 A → 1000 A; inrush 15 × I = 9742.8 A; R_d = 300 / (C ln(√2 V_C / 50)) = 14.42 kΩ; losses 0.0005 × 483.87 + 3 I² X_L/50 + 0.01 = 0.929 kW; 1.1 × 0.4301 = 0.4731 kV; Q_after = 705.53 − 450 = 255.53 kVAr, pf = 800 / √(800² + 255.53²) = 0.9526 lagging ≥ 0.95; resonance check: f_t 188.98 Hz < 250 Hz → pass (validation case c0016-400v-detuned-5th)

Example 16 — 400 V, 300 kW, resonance near the 11th

The installed bank resonates within 20 % of a 4 % 11th harmonic; detuning is needed even though the 5th and 7th are clean, and the 5.67 % reactor suffices.

Given: 0.4 kV, 50 Hz bus with a 20 MVA fault level; maximum demand 300 kW at power factor 0.8 lagging, target 0.95; harmonic distortion 11th 4 %.

Reactive powerφ1 = arccos 0.8 = 36.87°, tan φ1 = 0.75000; φ2 = arccos 0.95 = 18.19°, tan φ2 = 0.32868; Qc = P (tan φ1 − tan φ2) = 300 × (0.75000 − 0.32868) = 126.39 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈126.39 / 50⌉ × 50 = 150 kVAr
Apparent powerSbefore = 300 / 0.8 = 375.00 kVA; Safter = 300 / 0.95 = 315.79 kVA; reduction 59.21 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(20,000 kVA / 150 kVAr) = 577.35 Hz; hr = 577.35 / 50 = 11.547
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: 11th (550 Hz, 4 % > 3 %): |577.4 − 550| / 550 = 5.0 % (within ±20 %); Resonance near 11th harmonic → reactor by the lowest significant harmonic (11th; 3rd → 14 %, 5th → 7 %, otherwise 5.67 %) = 5.67 % detuning reactor
Tuned frequencyp = XL/XC = 0.0567; ftuned = f / √p = 50 / √0.0567 = 209.98 Hz (h = 4.20)
Capacitor voltageSeries branch: VC = V / (1 − p) = 0.4 / 0.9433 = 0.4240 kV (6.00 % rise across the capacitors); Vrated = 1.1 × 0.4240 = 0.4664 kV (LV: no standard HV rating applies)
Capacitor ratingQeff = V² / (XC(1 − p)), so the capacitors generate QC = Qinst / (1 − p) = 150 / 0.9433 = 159.02 kVAr at VC; nameplate at 1.1 VC: 1.21 × 159.02 = 192.41 kVAr
CurrentIrated = Qinst / (√3 V) = 150 / (√3 × 0.4) = 216.51 A; Imax = Irated × kh × 1.3 = 216.51 × 1.0601 × 1.3 = 298.38 A
CapacitanceXC = VC² × 1000 / QC = 0.4240² × 1000 / 159.020 = 1.131 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 1.131) = 2,814.954 µF per phase (star)
ReactorXL = p XC = 0.0567 × 1.131 = 0.06412 Ω; L = 1000 XL / (2π f) = 0.2041 mH; RL = XL / 50 = 0.001282 Ω
BreakerICB = 1.35 × 216.51 = 292.3 A → next standard size 315 A; overload setting 1.25 × Irated = 270.6 A
Discharge resistorVpk = √2 × 424.0 V = 599.7 V; R = t / (C ln(Vpk / 50 V)) = 300 / (2,814.954 × 10−6 × ln(599.7 / 50)) = 42.90 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 216.51 = 3,247.6 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 159.02 = 0.080 kW + reactor 3 I² RL = 0.180 kW + discharge 0.010 kW = 0.270 kW
Target power factorQafter = P tan φ1 − Qinst = 225.00 − 150 = 75.00 kVAr (lagging); pf = P / √(P² + Qafter²) = 300 / √(300² + 75.00²) = 0.9701 → 0.9701 vs limit 0.9500 PASS
ResonanceTuned 209.98 Hz must lie below the lowest significant harmonic (11th = 550 Hz) → 209.98 vs limit 550.00 Hz PASS
Result2/2 checks pass → PASS; 150 kVAr bank, 5.67 % reactor tuned to 210.0 Hz, 216.5 A, 315 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.8 = 0.75000, tan φ2 = tan arccos 0.95 = 0.32868; Q = 300 × (0.75000 − 0.32868) = 126.39 kVAr → 150 kVAr output in 50 kVAr steps; f_r = 50 × √(20000 kVA / 150 kVAr) = 577.35 Hz (h = 11.547); significant (> 3 %) harmonics: 11th 4 %; within 20 % of the 11th (550 Hz) → reactor by the lowest significant harmonic (11th) = 5.67 %; f_t = 50 / √0.0567 = 209.98 Hz; V_C = 0.4 / (1 − 0.0567) = 0.4240 kV; Q_C = 150 / 0.9433 = 159.02 kVAr; nameplate 1.21 × Q_C = 192.41 kVAr; X_C = 0.4240² × 1000 / 159.02 = 1.1308 Ω; X_L = 0.0567 X_C = 0.0641 Ω, L = 0.2041 mH; C = 10⁶ / (2π × 50 × X_C) = 2814.954 µF/phase; I = 150 / (√3 × 0.4) = 216.51 A; CB 1.35 × I = 292.3 A → 315 A; inrush 15 × I = 3247.6 A; R_d = 300 / (C ln(√2 V_C / 50)) = 42.90 kΩ; losses 0.0005 × 159.02 + 3 I² X_L/50 + 0.01 = 0.270 kW; 1.1 × 0.4240 = 0.4664 kV; Q_after = 225.00 − 150 = 75.00 kVAr, pf = 300 / √(300² + 75.00²) = 0.9701 lagging ≥ 0.95; resonance check: f_t 209.98 Hz < 550 Hz → pass (validation case c0016-400v-detuned-11th)

Example 17 — 400 V, 100 kW, resonance near the 13th

The installed 150 kVAr resonates at 645 Hz, within 1 % of a 5 % 13th harmonic; the 5.67 % reactor tunes to 210 Hz, well below it.

Given: 0.4 kV, 50 Hz bus with a 25 MVA fault level; maximum demand 100 kW at power factor 0.5 lagging, target 0.9; harmonic distortion 13th 5 %.

Reactive powerφ1 = arccos 0.5 = 60.00°, tan φ1 = 1.73205; φ2 = arccos 0.9 = 25.84°, tan φ2 = 0.48432; Qc = P (tan φ1 − tan φ2) = 100 × (1.73205 − 0.48432) = 124.77 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈124.77 / 50⌉ × 50 = 150 kVAr
Apparent powerSbefore = 100 / 0.5 = 200.00 kVA; Safter = 100 / 0.9 = 111.11 kVA; reduction 88.89 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(25,000 kVA / 150 kVAr) = 645.50 Hz; hr = 645.50 / 50 = 12.910
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: 13th (650 Hz, 5 % > 3 %): |645.5 − 650| / 650 = 0.7 % (within ±20 %); Resonance near 13th harmonic → reactor by the lowest significant harmonic (13th; 3rd → 14 %, 5th → 7 %, otherwise 5.67 %) = 5.67 % detuning reactor
Tuned frequencyp = XL/XC = 0.0567; ftuned = f / √p = 50 / √0.0567 = 209.98 Hz (h = 4.20)
Capacitor voltageSeries branch: VC = V / (1 − p) = 0.4 / 0.9433 = 0.4240 kV (6.00 % rise across the capacitors); Vrated = 1.1 × 0.4240 = 0.4664 kV (LV: no standard HV rating applies)
Capacitor ratingQeff = V² / (XC(1 − p)), so the capacitors generate QC = Qinst / (1 − p) = 150 / 0.9433 = 159.02 kVAr at VC; nameplate at 1.1 VC: 1.21 × 159.02 = 192.41 kVAr
CurrentIrated = Qinst / (√3 V) = 150 / (√3 × 0.4) = 216.51 A; Imax = Irated × kh × 1.3 = 216.51 × 1.0601 × 1.3 = 298.38 A
CapacitanceXC = VC² × 1000 / QC = 0.4240² × 1000 / 159.020 = 1.131 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 1.131) = 2,814.954 µF per phase (star)
ReactorXL = p XC = 0.0567 × 1.131 = 0.06412 Ω; L = 1000 XL / (2π f) = 0.2041 mH; RL = XL / 50 = 0.001282 Ω
BreakerICB = 1.35 × 216.51 = 292.3 A → next standard size 315 A; overload setting 1.25 × Irated = 270.6 A
Discharge resistorVpk = √2 × 424.0 V = 599.7 V; R = t / (C ln(Vpk / 50 V)) = 300 / (2,814.954 × 10−6 × ln(599.7 / 50)) = 42.90 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 216.51 = 3,247.6 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 159.02 = 0.080 kW + reactor 3 I² RL = 0.180 kW + discharge 0.010 kW = 0.270 kW
Target power factorQafter = P tan φ1 − Qinst = 173.21 − 150 = 23.21 kVAr (lagging); pf = P / √(P² + Qafter²) = 100 / √(100² + 23.21²) = 0.9741 → 0.9741 vs limit 0.9000 PASS
ResonanceTuned 209.98 Hz must lie below the lowest significant harmonic (13th = 650 Hz) → 209.98 vs limit 650.00 Hz PASS
Result2/2 checks pass → PASS; 150 kVAr bank, 5.67 % reactor tuned to 210.0 Hz, 216.5 A, 315 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.5 = 1.73205, tan φ2 = tan arccos 0.9 = 0.48432; Q = 100 × (1.73205 − 0.48432) = 124.77 kVAr → 150 kVAr output in 50 kVAr steps; f_r = 50 × √(25000 kVA / 150 kVAr) = 645.50 Hz (h = 12.910); significant (> 3 %) harmonics: 13th 5 %; within 20 % of the 13th (650 Hz) → reactor by the lowest significant harmonic (13th) = 5.67 %; f_t = 50 / √0.0567 = 209.98 Hz; V_C = 0.4 / (1 − 0.0567) = 0.4240 kV; Q_C = 150 / 0.9433 = 159.02 kVAr; nameplate 1.21 × Q_C = 192.41 kVAr; X_C = 0.4240² × 1000 / 159.02 = 1.1308 Ω; X_L = 0.0567 X_C = 0.0641 Ω, L = 0.2041 mH; C = 10⁶ / (2π × 50 × X_C) = 2814.954 µF/phase; I = 150 / (√3 × 0.4) = 216.51 A; CB 1.35 × I = 292.3 A → 315 A; inrush 15 × I = 3247.6 A; R_d = 300 / (C ln(√2 V_C / 50)) = 42.90 kΩ; losses 0.0005 × 159.02 + 3 I² X_L/50 + 0.01 = 0.270 kW; 1.1 × 0.4240 = 0.4664 kV; Q_after = 173.21 − 150 = 23.21 kVAr, pf = 100 / √(100² + 23.21²) = 0.9741 lagging ≥ 0.9; resonance check: f_t 209.98 Hz < 650 Hz → pass (validation case c0016-400v-detuned-13th)

Example 18 — 6.6 kV, 2.5 MW, resonance near the 7th

An MV detuned bank: 1.1 × V/(1 − p) = 1.1 × 6.6 / 0.9433 = 7.70 kV is above 7.2 kV, so the 11 kV standard rating is chosen.

Given: 6.6 kV, 50 Hz bus with a 60 MVA fault level; maximum demand 2,500 kW at power factor 0.82 lagging, target 0.95; harmonic distortion 7th 5 %.

Reactive powerφ1 = arccos 0.82 = 34.92°, tan φ1 = 0.69800; φ2 = arccos 0.95 = 18.19°, tan φ2 = 0.32868; Qc = P (tan φ1 − tan φ2) = 2,500 × (0.69800 − 0.32868) = 923.30 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈923.30 / 50⌉ × 50 = 950 kVAr
Apparent powerSbefore = 2,500 / 0.82 = 3,048.78 kVA; Safter = 2,500 / 0.95 = 2,631.58 kVA; reduction 417.20 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(60,000 kVA / 950 kVAr) = 397.36 Hz; hr = 397.36 / 50 = 7.947
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: 7th (350 Hz, 5 % > 3 %): |397.4 − 350| / 350 = 13.5 % (within ±20 %); Resonance near 7th harmonic → reactor by the lowest significant harmonic (7th; 3rd → 14 %, 5th → 7 %, otherwise 5.67 %) = 5.67 % detuning reactor
Tuned frequencyp = XL/XC = 0.0567; ftuned = f / √p = 50 / √0.0567 = 209.98 Hz (h = 4.20)
Capacitor voltageSeries branch: VC = V / (1 − p) = 6.6 / 0.9433 = 6.9967 kV (6.01 % rise across the capacitors); Vrated = 1.1 × 6.9967 = 7.6964 kV → next standard rating 11 kV
Capacitor ratingQeff = V² / (XC(1 − p)), so the capacitors generate QC = Qinst / (1 − p) = 950 / 0.9433 = 1,007.10 kVAr at VC; nameplate at 1.1 VC: 1.21 × 1,007.10 = 1,218.59 kVAr
CurrentIrated = Qinst / (√3 V) = 950 / (√3 × 6.6) = 83.10 A; Imax = Irated × kh × 1.3 = 83.10 × 1.0601 × 1.3 = 114.53 A
CapacitanceXC = VC² × 1000 / QC = 6.997² × 1000 / 1007.10 = 48.61 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 48.61) = 65.484 µF per phase (star)
ReactorXL = p XC = 0.0567 × 48.61 = 2.756 Ω; L = 1000 XL / (2π f) = 8.7730 mH; RL = XL / 50 = 0.05512 Ω
BreakerICB = 1.35 × 83.10 = 112.2 A → next standard size 125 A; overload setting 1.25 × Irated = 103.9 A
Discharge resistorVpk = √2 × 6,996.7 V = 9,894.8 V; R = t / (C ln(Vpk / 50 V)) = 300 / (65.484 × 10−6 × ln(9,894.8 / 50)) = 866.39 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 83.10 = 1,246.6 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 1,007.10 = 0.504 kW + reactor 3 I² RL = 1.142 kW + discharge 0.010 kW = 1.656 kW
Target power factorQafter = P tan φ1 − Qinst = 1,745.01 − 950 = 795.01 kVAr (lagging); pf = P / √(P² + Qafter²) = 2,500 / √(2,500² + 795.01²) = 0.9530 → 0.9530 vs limit 0.9500 PASS
ResonanceTuned 209.98 Hz must lie below the lowest significant harmonic (7th = 350 Hz) → 209.98 vs limit 350.00 Hz PASS
Result2/2 checks pass → PASS; 950 kVAr bank, 5.67 % reactor tuned to 210.0 Hz, 83.1 A, 125 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.82 = 0.69800, tan φ2 = tan arccos 0.95 = 0.32868; Q = 2500 × (0.69800 − 0.32868) = 923.30 kVAr → 950 kVAr output in 50 kVAr steps; f_r = 50 × √(60000 kVA / 950 kVAr) = 397.36 Hz (h = 7.947); significant (> 3 %) harmonics: 7th 5 %; within 20 % of the 7th (350 Hz) → reactor by the lowest significant harmonic (7th) = 5.67 %; f_t = 50 / √0.0567 = 209.98 Hz; V_C = 6.6 / (1 − 0.0567) = 6.9967 kV; Q_C = 950 / 0.9433 = 1007.10 kVAr; nameplate 1.21 × Q_C = 1218.59 kVAr; X_C = 6.9967² × 1000 / 1007.10 = 48.6087 Ω; X_L = 0.0567 X_C = 2.7561 Ω, L = 8.7730 mH; C = 10⁶ / (2π × 50 × X_C) = 65.484 µF/phase; I = 950 / (√3 × 6.6) = 83.10 A; CB 1.35 × I = 112.2 A → 125 A; inrush 15 × I = 1246.6 A; R_d = 300 / (C ln(√2 V_C / 50)) = 866.39 kΩ; losses 0.0005 × 1007.10 + 3 I² X_L/50 + 0.01 = 1.656 kW; 1.1 × 6.9967 = 7.6964 kV → 11 kV standard; Q_after = 1745.01 − 950 = 795.01 kVAr, pf = 2500 / √(2500² + 795.01²) = 0.9530 lagging ≥ 0.95; resonance check: f_t 209.98 Hz < 350 Hz → pass (validation case c0016-6600v-detuned-7th)

Example 19 — 480 V, 60 Hz, resonance near the 5th

At 60 Hz the 7 % reactor chosen for a significant 5th tunes to 60/√0.07 = 227 Hz, below the 300 Hz 5th harmonic.

Given: 0.48 kV, 60 Hz bus with a 15 MVA fault level; maximum demand 1,500 kW at power factor 0.8 lagging, target 0.95; harmonic distortion 5th 6 %.

Reactive powerφ1 = arccos 0.8 = 36.87°, tan φ1 = 0.75000; φ2 = arccos 0.95 = 18.19°, tan φ2 = 0.32868; Qc = P (tan φ1 − tan φ2) = 1,500 × (0.75000 − 0.32868) = 631.97 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈631.97 / 50⌉ × 50 = 650 kVAr
Apparent powerSbefore = 1,500 / 0.8 = 1,875.00 kVA; Safter = 1,500 / 0.95 = 1,578.95 kVA; reduction 296.05 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 60 × √(15,000 kVA / 650 kVAr) = 288.23 Hz; hr = 288.23 / 60 = 4.804
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: 5th (300 Hz, 6 % > 3 %): |288.2 − 300| / 300 = 3.9 % (within ±20 %); Resonance near 5th harmonic → reactor by the lowest significant harmonic (5th; 3rd → 14 %, 5th → 7 %, otherwise 5.67 %) = 7 % detuning reactor
Tuned frequencyp = XL/XC = 0.0700; ftuned = f / √p = 60 / √0.0700 = 226.78 Hz (h = 3.78)
Capacitor voltageSeries branch: VC = V / (1 − p) = 0.48 / 0.9300 = 0.5161 kV (7.52 % rise across the capacitors); Vrated = 1.1 × 0.5161 = 0.5677 kV (LV: no standard HV rating applies)
Capacitor ratingQeff = V² / (XC(1 − p)), so the capacitors generate QC = Qinst / (1 − p) = 650 / 0.9300 = 698.92 kVAr at VC; nameplate at 1.1 VC: 1.21 × 698.92 = 845.70 kVAr
CurrentIrated = Qinst / (√3 V) = 650 / (√3 × 0.48) = 781.83 A; Imax = Irated × kh × 1.3 = 781.83 × 1.0753 × 1.3 = 1,092.88 A
CapacitanceXC = VC² × 1000 / QC = 0.5161² × 1000 / 698.920 = 0.3811 Ω; C = 106 / (2π f XC) = 106 / (2π × 60 × 0.3811) = 6,959.575 µF per phase (star)
ReactorXL = p XC = 0.0700 × 0.3811 = 0.02668 Ω; L = 1000 XL / (2π f) = 0.0708 mH; RL = XL / 50 = 0.0005336 Ω
BreakerICB = 1.35 × 781.83 = 1,055.5 A → next standard size 1,250 A; overload setting 1.25 × Irated = 977.3 A
Discharge resistorVpk = √2 × 516.1 V = 729.9 V; R = t / (C ln(Vpk / 50 V)) = 300 / (6,959.575 × 10−6 × ln(729.9 / 50)) = 16.08 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 781.83 = 11,727.4 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 698.92 = 0.349 kW + reactor 3 I² RL = 0.978 kW + discharge 0.010 kW = 1.338 kW
Target power factorQafter = P tan φ1 − Qinst = 1,125.00 − 650 = 475.00 kVAr (lagging); pf = P / √(P² + Qafter²) = 1,500 / √(1,500² + 475.00²) = 0.9533 → 0.9533 vs limit 0.9500 PASS
ResonanceTuned 226.78 Hz must lie below the lowest significant harmonic (5th = 300 Hz) → 226.78 vs limit 300.00 Hz PASS
Result2/2 checks pass → PASS; 650 kVAr bank, 7 % reactor tuned to 226.8 Hz, 781.8 A, 1,250 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.8 = 0.75000, tan φ2 = tan arccos 0.95 = 0.32868; Q = 1500 × (0.75000 − 0.32868) = 631.97 kVAr → 650 kVAr output in 50 kVAr steps; f_r = 60 × √(15000 kVA / 650 kVAr) = 288.23 Hz (h = 4.804); significant (> 3 %) harmonics: 5th 6 %; within 20 % of the 5th (300 Hz) → reactor by the lowest significant harmonic (5th) = 7 %; f_t = 60 / √0.07 = 226.78 Hz; V_C = 0.48 / (1 − 0.07) = 0.5161 kV; Q_C = 650 / 0.9300 = 698.92 kVAr; nameplate 1.21 × Q_C = 845.70 kVAr; X_C = 0.5161² × 1000 / 698.92 = 0.3811 Ω; X_L = 0.07 X_C = 0.0267 Ω, L = 0.0708 mH; C = 10⁶ / (2π × 60 × X_C) = 6959.575 µF/phase; I = 650 / (√3 × 0.48) = 781.83 A; CB 1.35 × I = 1055.5 A → 1250 A; inrush 15 × I = 11727.4 A; R_d = 300 / (C ln(√2 V_C / 50)) = 16.08 kΩ; losses 0.0005 × 698.92 + 3 I² X_L/50 + 0.01 = 1.338 kW; 1.1 × 0.5161 = 0.5677 kV; Q_after = 1125.00 − 650 = 475.00 kVAr, pf = 1500 / √(1500² + 475.00²) = 0.9533 lagging ≥ 0.95; resonance check: f_t 226.78 Hz < 300 Hz → pass (validation case c0016-60hz-480v-detuned-5th)

Example 20 — 600 V, 60 Hz, resonance near the 7th

A 600 V, 60 Hz bank resonating near the 420 Hz 7th harmonic; detuned with 5.67 % to 252 Hz.

Given: 0.6 kV, 60 Hz bus with a 20 MVA fault level; maximum demand 1,000 kW at power factor 0.85 lagging, target 0.97; harmonic distortion 7th 4 %.

Reactive powerφ1 = arccos 0.85 = 31.79°, tan φ1 = 0.61974; φ2 = arccos 0.97 = 14.07°, tan φ2 = 0.25062; Qc = P (tan φ1 − tan φ2) = 1,000 × (0.61974 − 0.25062) = 369.12 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈369.12 / 50⌉ × 50 = 400 kVAr
Apparent powerSbefore = 1,000 / 0.85 = 1,176.47 kVA; Safter = 1,000 / 0.97 = 1,030.93 kVA; reduction 145.54 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 60 × √(20,000 kVA / 400 kVAr) = 424.26 Hz; hr = 424.26 / 60 = 7.071
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: 7th (420 Hz, 4 % > 3 %): |424.3 − 420| / 420 = 1.0 % (within ±20 %); Resonance near 7th harmonic → reactor by the lowest significant harmonic (7th; 3rd → 14 %, 5th → 7 %, otherwise 5.67 %) = 5.67 % detuning reactor
Tuned frequencyp = XL/XC = 0.0567; ftuned = f / √p = 60 / √0.0567 = 251.98 Hz (h = 4.20)
Capacitor voltageSeries branch: VC = V / (1 − p) = 0.6 / 0.9433 = 0.6361 kV (6.02 % rise across the capacitors); Vrated = 1.1 × 0.6361 = 0.6997 kV (LV: no standard HV rating applies)
Capacitor ratingQeff = V² / (XC(1 − p)), so the capacitors generate QC = Qinst / (1 − p) = 400 / 0.9433 = 424.04 kVAr at VC; nameplate at 1.1 VC: 1.21 × 424.04 = 513.09 kVAr
CurrentIrated = Qinst / (√3 V) = 400 / (√3 × 0.6) = 384.90 A; Imax = Irated × kh × 1.3 = 384.90 × 1.0601 × 1.3 = 530.45 A
CapacitanceXC = VC² × 1000 / QC = 0.6361² × 1000 / 424.040 = 0.9541 Ω; C = 106 / (2π f XC) = 106 / (2π × 60 × 0.9541) = 2,780.201 µF per phase (star)
ReactorXL = p XC = 0.0567 × 0.9541 = 0.05410 Ω; L = 1000 XL / (2π f) = 0.1435 mH; RL = XL / 50 = 0.001082 Ω
BreakerICB = 1.35 × 384.90 = 519.6 A → next standard size 630 A; overload setting 1.25 × Irated = 481.1 A
Discharge resistorVpk = √2 × 636.1 V = 899.5 V; R = t / (C ln(Vpk / 50 V)) = 300 / (2,780.201 × 10−6 × ln(899.5 / 50)) = 37.34 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 384.90 = 5,773.5 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 424.04 = 0.212 kW + reactor 3 I² RL = 0.481 kW + discharge 0.010 kW = 0.703 kW
Target power factorQafter = P tan φ1 − Qinst = 619.74 − 400 = 219.74 kVAr (lagging); pf = P / √(P² + Qafter²) = 1,000 / √(1,000² + 219.74²) = 0.9767 → 0.9767 vs limit 0.9700 PASS
ResonanceTuned 251.98 Hz must lie below the lowest significant harmonic (7th = 420 Hz) → 251.98 vs limit 420.00 Hz PASS
Result2/2 checks pass → PASS; 400 kVAr bank, 5.67 % reactor tuned to 252.0 Hz, 384.9 A, 630 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.85 = 0.61974, tan φ2 = tan arccos 0.97 = 0.25062; Q = 1000 × (0.61974 − 0.25062) = 369.12 kVAr → 400 kVAr output in 50 kVAr steps; f_r = 60 × √(20000 kVA / 400 kVAr) = 424.26 Hz (h = 7.071); significant (> 3 %) harmonics: 7th 4 %; within 20 % of the 7th (420 Hz) → reactor by the lowest significant harmonic (7th) = 5.67 %; f_t = 60 / √0.0567 = 251.98 Hz; V_C = 0.6 / (1 − 0.0567) = 0.6361 kV; Q_C = 400 / 0.9433 = 424.04 kVAr; nameplate 1.21 × Q_C = 513.09 kVAr; X_C = 0.6361² × 1000 / 424.04 = 0.9541 Ω; X_L = 0.0567 X_C = 0.0541 Ω, L = 0.1435 mH; C = 10⁶ / (2π × 60 × X_C) = 2780.201 µF/phase; I = 400 / (√3 × 0.6) = 384.90 A; CB 1.35 × I = 519.6 A → 630 A; inrush 15 × I = 5773.5 A; R_d = 300 / (C ln(√2 V_C / 50)) = 37.34 kΩ; losses 0.0005 × 424.04 + 3 I² X_L/50 + 0.01 = 0.703 kW; 1.1 × 0.6361 = 0.6997 kV; Q_after = 619.74 − 400 = 219.74 kVAr, pf = 1000 / √(1000² + 219.74²) = 0.9767 lagging ≥ 0.97; resonance check: f_t 251.98 Hz < 420 Hz → pass (validation case c0016-60hz-600v-detuned-7th)

Example 21 — 690 V, 1.2 MW, 7 % detuned — capacitor voltage and nameplate

A detuned bank is specified by its output at system voltage. The series reactor raises the capacitor voltage to V/(1 − p), so the capacitors generate Q/(1 − p) and, rated at 1.1 × V_C, carry a nameplate of 1.21 Q/(1 − p); the power factor is achieved with the installed output, not an uprated one.

Given: 0.69 kV, 50 Hz bus with a 15 MVA fault level; maximum demand 1,200 kW at power factor 0.78 lagging, target 0.95; harmonic distortion 5th 6 %.

Reactive powerφ1 = arccos 0.78 = 38.74°, tan φ1 = 0.80228; φ2 = arccos 0.95 = 18.19°, tan φ2 = 0.32868; Qc = P (tan φ1 − tan φ2) = 1,200 × (0.80228 − 0.32868) = 568.32 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈568.32 / 50⌉ × 50 = 600 kVAr
Apparent powerSbefore = 1,200 / 0.78 = 1,538.46 kVA; Safter = 1,200 / 0.95 = 1,263.16 kVA; reduction 275.30 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(15,000 kVA / 600 kVAr) = 250.00 Hz; hr = 250.00 / 50 = 5.000
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: 5th (250 Hz, 6 % > 3 %): |250.0 − 250| / 250 = 0.0 % (within ±20 %); Resonance near 5th harmonic → reactor by the lowest significant harmonic (5th; 3rd → 14 %, 5th → 7 %, otherwise 5.67 %) = 7 % detuning reactor
Tuned frequencyp = XL/XC = 0.0700; ftuned = f / √p = 50 / √0.0700 = 188.98 Hz (h = 3.78)
Capacitor voltageSeries branch: VC = V / (1 − p) = 0.69 / 0.9300 = 0.7419 kV (7.52 % rise across the capacitors); Vrated = 1.1 × 0.7419 = 0.8161 kV (LV: no standard HV rating applies)
Capacitor ratingQeff = V² / (XC(1 − p)), so the capacitors generate QC = Qinst / (1 − p) = 600 / 0.9300 = 645.16 kVAr at VC; nameplate at 1.1 VC: 1.21 × 645.16 = 780.65 kVAr
CurrentIrated = Qinst / (√3 V) = 600 / (√3 × 0.69) = 502.04 A; Imax = Irated × kh × 1.3 = 502.04 × 1.0753 × 1.3 = 701.78 A
CapacitanceXC = VC² × 1000 / QC = 0.7419² × 1000 / 645.160 = 0.8532 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 0.8532) = 3,730.664 µF per phase (star)
ReactorXL = p XC = 0.0700 × 0.8532 = 0.05973 Ω; L = 1000 XL / (2π f) = 0.1901 mH; RL = XL / 50 = 0.001195 Ω
BreakerICB = 1.35 × 502.04 = 677.8 A → next standard size 800 A; overload setting 1.25 × Irated = 627.6 A
Discharge resistorVpk = √2 × 741.9 V = 1,049.3 V; R = t / (C ln(Vpk / 50 V)) = 300 / (3,730.664 × 10−6 × ln(1,049.3 / 50)) = 26.42 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 502.04 = 7,530.7 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 645.16 = 0.323 kW + reactor 3 I² RL = 0.903 kW + discharge 0.010 kW = 1.236 kW
Target power factorQafter = P tan φ1 − Qinst = 962.74 − 600 = 362.74 kVAr (lagging); pf = P / √(P² + Qafter²) = 1,200 / √(1,200² + 362.74²) = 0.9572 → 0.9572 vs limit 0.9500 PASS
ResonanceTuned 188.98 Hz must lie below the lowest significant harmonic (5th = 250 Hz) → 188.98 vs limit 250.00 Hz PASS
Result2/2 checks pass → PASS; 600 kVAr bank, 7 % reactor tuned to 189.0 Hz, 502.0 A, 800 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.78 = 0.80228, tan φ2 = tan arccos 0.95 = 0.32868; Q = 1200 × (0.80228 − 0.32868) = 568.32 kVAr → 600 kVAr output in 50 kVAr steps; f_r = 50 × √(15000 kVA / 600 kVAr) = 250.00 Hz (h = 5.000); significant (> 3 %) harmonics: 5th 6 %; within 20 % of the 5th (250 Hz) → reactor by the lowest significant harmonic (5th) = 7 %; f_t = 50 / √0.07 = 188.98 Hz; V_C = 0.69 / (1 − 0.07) = 0.7419 kV; Q_C = 600 / 0.9300 = 645.16 kVAr; nameplate 1.21 × Q_C = 780.65 kVAr; X_C = 0.7419² × 1000 / 645.16 = 0.8532 Ω; X_L = 0.07 X_C = 0.0597 Ω, L = 0.1901 mH; C = 10⁶ / (2π × 50 × X_C) = 3730.664 µF/phase; I = 600 / (√3 × 0.69) = 502.04 A; CB 1.35 × I = 677.8 A → 800 A; inrush 15 × I = 7530.7 A; R_d = 300 / (C ln(√2 V_C / 50)) = 26.42 kΩ; losses 0.0005 × 645.16 + 3 I² X_L/50 + 0.01 = 1.236 kW; 1.1 × 0.7419 = 0.8161 kV; Q_after = 962.74 − 600 = 362.74 kVAr, pf = 1200 / √(1200² + 362.74²) = 0.9572 lagging ≥ 0.95; resonance check: f_t 188.98 Hz < 250 Hz → pass (validation case c0016-690v-detuned-capacitor-rating)

Example 22 — 400 V, significant 3rd harmonic — 14 % reactor

With a 5 % 3rd harmonic the bank must be tuned below 150 Hz: the 14 % reactor (tuned 50 / √0.14 = 134 Hz) is chosen.

Given: 0.4 kV, 50 Hz bus with a 10 MVA fault level; maximum demand 700 kW at power factor 0.75 lagging, target 0.95; harmonic distortion 3rd 5 %, 5th 6 %.

Reactive powerφ1 = arccos 0.75 = 41.41°, tan φ1 = 0.88192; φ2 = arccos 0.95 = 18.19°, tan φ2 = 0.32868; Qc = P (tan φ1 − tan φ2) = 700 × (0.88192 − 0.32868) = 387.26 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈387.26 / 50⌉ × 50 = 400 kVAr
Apparent powerSbefore = 700 / 0.75 = 933.33 kVA; Safter = 700 / 0.95 = 736.84 kVA; reduction 196.49 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(10,000 kVA / 400 kVAr) = 250.00 Hz; hr = 250.00 / 50 = 5.000
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: 3rd (150 Hz, 5 % > 3 %): |250.0 − 150| / 150 = 66.7 % (outside ±20 %); 5th (250 Hz, 6 % > 3 %): |250.0 − 250| / 250 = 0.0 % (within ±20 %); Resonance near 5th harmonic → reactor by the lowest significant harmonic (3rd; 3rd → 14 %, 5th → 7 %, otherwise 5.67 %) = 14 % detuning reactor
Tuned frequencyp = XL/XC = 0.1400; ftuned = f / √p = 50 / √0.1400 = 133.63 Hz (h = 2.67)
Capacitor voltageSeries branch: VC = V / (1 − p) = 0.4 / 0.8600 = 0.4651 kV (16.27 % rise across the capacitors); Vrated = 1.1 × 0.4651 = 0.5116 kV (LV: no standard HV rating applies)
Capacitor ratingQeff = V² / (XC(1 − p)), so the capacitors generate QC = Qinst / (1 − p) = 400 / 0.8600 = 465.12 kVAr at VC; nameplate at 1.1 VC: 1.21 × 465.12 = 562.79 kVAr
CurrentIrated = Qinst / (√3 V) = 400 / (√3 × 0.4) = 577.35 A; Imax = Irated × kh × 1.3 = 577.35 × 1.1628 × 1.3 = 872.74 A
CapacitanceXC = VC² × 1000 / QC = 0.4651² × 1000 / 465.120 = 0.4651 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 0.4651) = 6,843.663 µF per phase (star)
ReactorXL = p XC = 0.1400 × 0.4651 = 0.06512 Ω; L = 1000 XL / (2π f) = 0.2073 mH; RL = XL / 50 = 0.001302 Ω
BreakerICB = 1.35 × 577.35 = 779.4 A → next standard size 800 A; overload setting 1.25 × Irated = 721.7 A
Discharge resistorVpk = √2 × 465.1 V = 657.8 V; R = t / (C ln(Vpk / 50 V)) = 300 / (6,843.663 × 10−6 × ln(657.8 / 50)) = 17.01 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 577.35 = 8,660.3 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 465.12 = 0.233 kW + reactor 3 I² RL = 1.302 kW + discharge 0.010 kW = 1.545 kW
Target power factorQafter = P tan φ1 − Qinst = 617.34 − 400 = 217.34 kVAr (lagging); pf = P / √(P² + Qafter²) = 700 / √(700² + 217.34²) = 0.9550 → 0.9550 vs limit 0.9500 PASS
ResonanceTuned 133.63 Hz must lie below the lowest significant harmonic (3rd = 150 Hz) → 133.63 vs limit 150.00 Hz PASS
Result2/2 checks pass → PASS; 400 kVAr bank, 14 % reactor tuned to 133.6 Hz, 577.4 A, 800 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.75 = 0.88192, tan φ2 = tan arccos 0.95 = 0.32868; Q = 700 × (0.88192 − 0.32868) = 387.26 kVAr → 400 kVAr output in 50 kVAr steps; f_r = 50 × √(10000 kVA / 400 kVAr) = 250.00 Hz (h = 5.000); significant (> 3 %) harmonics: 3rd 5 %, 5th 6 %; within 20 % of the 5th (250 Hz) → reactor by the lowest significant harmonic (3rd) = 14 %; f_t = 50 / √0.14 = 133.63 Hz; V_C = 0.4 / (1 − 0.14) = 0.4651 kV; Q_C = 400 / 0.8600 = 465.12 kVAr; nameplate 1.21 × Q_C = 562.79 kVAr; X_C = 0.4651² × 1000 / 465.12 = 0.4651 Ω; X_L = 0.14 X_C = 0.0651 Ω, L = 0.2073 mH; C = 10⁶ / (2π × 50 × X_C) = 6843.663 µF/phase; I = 400 / (√3 × 0.4) = 577.35 A; CB 1.35 × I = 779.4 A → 800 A; inrush 15 × I = 8660.3 A; R_d = 300 / (C ln(√2 V_C / 50)) = 17.01 kΩ; losses 0.0005 × 465.12 + 3 I² X_L/50 + 0.01 = 1.545 kW; 1.1 × 0.4651 = 0.5116 kV; Q_after = 617.34 − 400 = 217.34 kVAr, pf = 700 / √(700² + 217.34²) = 0.9550 lagging ≥ 0.95; resonance check: f_t 133.63 Hz < 150 Hz → pass (validation case c0016-400v-3rd-harmonic-14pct)

Example 23 — 400 V, resonance near the 11th but the 5th is significant — 7 %

Detuning is triggered by the 11th, but the reactor is chosen by the LOWEST significant harmonic (5th): 7 %, so the bank stays inductive at the 5th too.

Given: 0.4 kV, 50 Hz bus with a 20 MVA fault level; maximum demand 300 kW at power factor 0.8 lagging, target 0.95; harmonic distortion 5th 4 %, 11th 4 %.

Reactive powerφ1 = arccos 0.8 = 36.87°, tan φ1 = 0.75000; φ2 = arccos 0.95 = 18.19°, tan φ2 = 0.32868; Qc = P (tan φ1 − tan φ2) = 300 × (0.75000 − 0.32868) = 126.39 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈126.39 / 50⌉ × 50 = 150 kVAr
Apparent powerSbefore = 300 / 0.8 = 375.00 kVA; Safter = 300 / 0.95 = 315.79 kVA; reduction 59.21 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(20,000 kVA / 150 kVAr) = 577.35 Hz; hr = 577.35 / 50 = 11.547
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: 5th (250 Hz, 4 % > 3 %): |577.4 − 250| / 250 = 130.9 % (outside ±20 %); 11th (550 Hz, 4 % > 3 %): |577.4 − 550| / 550 = 5.0 % (within ±20 %); Resonance near 11th harmonic → reactor by the lowest significant harmonic (5th; 3rd → 14 %, 5th → 7 %, otherwise 5.67 %) = 7 % detuning reactor
Tuned frequencyp = XL/XC = 0.0700; ftuned = f / √p = 50 / √0.0700 = 188.98 Hz (h = 3.78)
Capacitor voltageSeries branch: VC = V / (1 − p) = 0.4 / 0.9300 = 0.4301 kV (7.52 % rise across the capacitors); Vrated = 1.1 × 0.4301 = 0.4731 kV (LV: no standard HV rating applies)
Capacitor ratingQeff = V² / (XC(1 − p)), so the capacitors generate QC = Qinst / (1 − p) = 150 / 0.9300 = 161.29 kVAr at VC; nameplate at 1.1 VC: 1.21 × 161.29 = 195.16 kVAr
CurrentIrated = Qinst / (√3 V) = 150 / (√3 × 0.4) = 216.51 A; Imax = Irated × kh × 1.3 = 216.51 × 1.0753 × 1.3 = 302.64 A
CapacitanceXC = VC² × 1000 / QC = 0.4301² × 1000 / 161.290 = 1.147 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 1.147) = 2,775.264 µF per phase (star)
ReactorXL = p XC = 0.0700 × 1.147 = 0.08029 Ω; L = 1000 XL / (2π f) = 0.2556 mH; RL = XL / 50 = 0.001606 Ω
BreakerICB = 1.35 × 216.51 = 292.3 A → next standard size 315 A; overload setting 1.25 × Irated = 270.6 A
Discharge resistorVpk = √2 × 430.1 V = 608.3 V; R = t / (C ln(Vpk / 50 V)) = 300 / (2,775.264 × 10−6 × ln(608.3 / 50)) = 43.26 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 216.51 = 3,247.6 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 161.29 = 0.081 kW + reactor 3 I² RL = 0.226 kW + discharge 0.010 kW = 0.316 kW
Target power factorQafter = P tan φ1 − Qinst = 225.00 − 150 = 75.00 kVAr (lagging); pf = P / √(P² + Qafter²) = 300 / √(300² + 75.00²) = 0.9701 → 0.9701 vs limit 0.9500 PASS
ResonanceTuned 188.98 Hz must lie below the lowest significant harmonic (5th = 250 Hz) → 188.98 vs limit 250.00 Hz PASS
Result2/2 checks pass → PASS; 150 kVAr bank, 7 % reactor tuned to 189.0 Hz, 216.5 A, 315 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.8 = 0.75000, tan φ2 = tan arccos 0.95 = 0.32868; Q = 300 × (0.75000 − 0.32868) = 126.39 kVAr → 150 kVAr output in 50 kVAr steps; f_r = 50 × √(20000 kVA / 150 kVAr) = 577.35 Hz (h = 11.547); significant (> 3 %) harmonics: 5th 4 %, 11th 4 %; within 20 % of the 11th (550 Hz) → reactor by the lowest significant harmonic (5th) = 7 %; f_t = 50 / √0.07 = 188.98 Hz; V_C = 0.4 / (1 − 0.07) = 0.4301 kV; Q_C = 150 / 0.9300 = 161.29 kVAr; nameplate 1.21 × Q_C = 195.16 kVAr; X_C = 0.4301² × 1000 / 161.29 = 1.1470 Ω; X_L = 0.07 X_C = 0.0803 Ω, L = 0.2556 mH; C = 10⁶ / (2π × 50 × X_C) = 2775.264 µF/phase; I = 150 / (√3 × 0.4) = 216.51 A; CB 1.35 × I = 292.3 A → 315 A; inrush 15 × I = 3247.6 A; R_d = 300 / (C ln(√2 V_C / 50)) = 43.26 kΩ; losses 0.0005 × 161.29 + 3 I² X_L/50 + 0.01 = 0.316 kW; 1.1 × 0.4301 = 0.4731 kV; Q_after = 225.00 − 150 = 75.00 kVAr, pf = 300 / √(300² + 75.00²) = 0.9701 lagging ≥ 0.95; resonance check: f_t 188.98 Hz < 250 Hz → pass (validation case c0016-400v-11th-trigger-5th-lowest-7pct)

Example 24 — 415 V, reactor forced to none near a 4 % 7th — FAIL

The same bank as the 415 V detuned-7th case with the reactor forced off: the installed bank resonates within 20 % of a significant 7th, so the resonance check fails.

Given: 0.415 kV, 50 Hz bus with a 10 MVA fault level; maximum demand 500 kW at power factor 0.75 lagging, target 0.95; harmonic distortion 7th 4 %; reactor forced to none.

Reactive powerφ1 = arccos 0.75 = 41.41°, tan φ1 = 0.88192; φ2 = arccos 0.95 = 18.19°, tan φ2 = 0.32868; Qc = P (tan φ1 − tan φ2) = 500 × (0.88192 − 0.32868) = 276.62 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈276.62 / 50⌉ × 50 = 300 kVAr
Apparent powerSbefore = 500 / 0.75 = 666.67 kVA; Safter = 500 / 0.95 = 526.32 kVA; reduction 140.35 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(10,000 kVA / 300 kVAr) = 288.68 Hz; hr = 288.68 / 50 = 5.774
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: 7th (350 Hz, 4 % > 3 %): |288.7 − 350| / 350 = 17.5 % (within ±20 %); reactor forced to none (user selection)
Capacitor voltageVC = V = 0.415 kV; Vrated = 1.1 × 0.415 = 0.4565 kV (LV: no standard HV rating applies)
Capacitor ratingNo reactor: QC = Qinst = 300 kVAr; nameplate at 1.1 V: 1.21 × 300 = 363.00 kVAr
CurrentIrated = Qinst / (√3 V) = 300 / (√3 × 0.415) = 417.36 A; Imax = Irated × kh × 1.3 = 417.36 × 1.0000 × 1.3 = 542.57 A
CapacitanceXC = VC² × 1000 / QC = 0.4150² × 1000 / 300.000 = 0.5741 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 0.5741) = 5,544.663 µF per phase (star)
BreakerICB = 1.35 × 417.36 = 563.4 A → next standard size 630 A; overload setting 1.25 × Irated = 521.7 A
Discharge resistorVpk = √2 × 415.0 V = 586.9 V; R = t / (C ln(Vpk / 50 V)) = 300 / (5,544.663 × 10−6 × ln(586.9 / 50)) = 21.97 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 417.36 = 6,260.4 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 300.00 = 0.150 kW + reactor 3 I² RL = 0.000 kW + discharge 0.010 kW = 0.160 kW
Target power factorQafter = P tan φ1 − Qinst = 440.96 − 300 = 140.96 kVAr (lagging); pf = P / √(P² + Qafter²) = 500 / √(500² + 140.96²) = 0.9625 → 0.9625 vs limit 0.9500 PASS
ResonanceUndetuned resonance 288.68 Hz must be outside ±20 % of every significant harmonic (7th): within ±20 % of the 7th → 288.68 vs limit 350.00 Hz FAIL
Result1/2 checks pass → FAIL; 300 kVAr bank, no reactor, 417.4 A, 630 A breaker; governing: resonance
ReferenceHand calculation: tan φ1 = tan arccos 0.75 = 0.88192, tan φ2 = tan arccos 0.95 = 0.32868; Q = 500 × (0.88192 − 0.32868) = 276.62 kVAr → 300 kVAr output in 50 kVAr steps; f_r = 50 × √(10000 kVA / 300 kVAr) = 288.68 Hz (h = 5.774); significant (> 3 %) harmonics: 7th 4 %; reactor forced to none; V_C = 0.415 kV, nameplate 1.21 × 300 = 363.00 kVAr; X_C = 0.415² × 1000 / 300 = 0.5741 Ω; C = 10⁶ / (2π × 50 × X_C) = 5544.663 µF/phase; I = 300 / (√3 × 0.415) = 417.36 A; CB 1.35 × I = 563.4 A → 630 A; inrush 15 × I = 6260.4 A; R_d = 300 / (C ln(√2 V_C / 50)) = 21.97 kΩ; losses 0.0005 × 300.00 + 0.01 = 0.160 kW; 1.1 × 0.4150 = 0.4565 kV; Q_after = 440.96 − 300 = 140.96 kVAr, pf = 500 / √(500² + 140.96²) = 0.9625 lagging ≥ 0.95; resonance check: 288.68 Hz within ±20 % of the 7th → FAIL (validation case c0016-415v-forced-undetuned-fails)

Example 25 — 400 V, 5.67 % forced with a significant 3rd — FAIL

A 5.67 % reactor tunes to 210 Hz, above a 5 % 3rd harmonic (150 Hz): the bank would be capacitive at the 3rd, so the resonance check fails.

Given: 0.4 kV, 50 Hz bus with a 10 MVA fault level; maximum demand 700 kW at power factor 0.75 lagging, target 0.95; harmonic distortion 3rd 5 %, 5th 6 %; reactor forced to 5.67 %.

Reactive powerφ1 = arccos 0.75 = 41.41°, tan φ1 = 0.88192; φ2 = arccos 0.95 = 18.19°, tan φ2 = 0.32868; Qc = P (tan φ1 − tan φ2) = 700 × (0.88192 − 0.32868) = 387.26 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈387.26 / 50⌉ × 50 = 400 kVAr
Apparent powerSbefore = 700 / 0.75 = 933.33 kVA; Safter = 700 / 0.95 = 736.84 kVA; reduction 196.49 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(10,000 kVA / 400 kVAr) = 250.00 Hz; hr = 250.00 / 50 = 5.000
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: 3rd (150 Hz, 5 % > 3 %): |250.0 − 150| / 150 = 66.7 % (outside ±20 %); 5th (250 Hz, 6 % > 3 %): |250.0 − 250| / 250 = 0.0 % (within ±20 %); reactor forced to 5.67 % (user selection)
Tuned frequencyp = XL/XC = 0.0567; ftuned = f / √p = 50 / √0.0567 = 209.98 Hz (h = 4.20)
Capacitor voltageSeries branch: VC = V / (1 − p) = 0.4 / 0.9433 = 0.4240 kV (6.00 % rise across the capacitors); Vrated = 1.1 × 0.4240 = 0.4664 kV (LV: no standard HV rating applies)
Capacitor ratingQeff = V² / (XC(1 − p)), so the capacitors generate QC = Qinst / (1 − p) = 400 / 0.9433 = 424.04 kVAr at VC; nameplate at 1.1 VC: 1.21 × 424.04 = 513.09 kVAr
CurrentIrated = Qinst / (√3 V) = 400 / (√3 × 0.4) = 577.35 A; Imax = Irated × kh × 1.3 = 577.35 × 1.0601 × 1.3 = 795.67 A
CapacitanceXC = VC² × 1000 / QC = 0.4240² × 1000 / 424.040 = 0.4240 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 0.4240) = 7,506.543 µF per phase (star)
ReactorXL = p XC = 0.0567 × 0.4240 = 0.02404 Ω; L = 1000 XL / (2π f) = 0.0765 mH; RL = XL / 50 = 0.0004809 Ω
BreakerICB = 1.35 × 577.35 = 779.4 A → next standard size 800 A; overload setting 1.25 × Irated = 721.7 A
Discharge resistorVpk = √2 × 424.0 V = 599.7 V; R = t / (C ln(Vpk / 50 V)) = 300 / (7,506.543 × 10−6 × ln(599.7 / 50)) = 16.09 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 577.35 = 8,660.3 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 424.04 = 0.212 kW + reactor 3 I² RL = 0.481 kW + discharge 0.010 kW = 0.703 kW
Target power factorQafter = P tan φ1 − Qinst = 617.34 − 400 = 217.34 kVAr (lagging); pf = P / √(P² + Qafter²) = 700 / √(700² + 217.34²) = 0.9550 → 0.9550 vs limit 0.9500 PASS
ResonanceTuned 209.98 Hz must lie below the lowest significant harmonic (3rd = 150 Hz) → 209.98 vs limit 150.00 Hz FAIL
Result1/2 checks pass → FAIL; 400 kVAr bank, 5.67 % reactor tuned to 210.0 Hz, 577.4 A, 800 A breaker; governing: resonance
ReferenceHand calculation: tan φ1 = tan arccos 0.75 = 0.88192, tan φ2 = tan arccos 0.95 = 0.32868; Q = 700 × (0.88192 − 0.32868) = 387.26 kVAr → 400 kVAr output in 50 kVAr steps; f_r = 50 × √(10000 kVA / 400 kVAr) = 250.00 Hz (h = 5.000); significant (> 3 %) harmonics: 3rd 5 %, 5th 6 %; reactor forced to 5.67 %; f_t = 50 / √0.0567 = 209.98 Hz; V_C = 0.4 / (1 − 0.0567) = 0.4240 kV; Q_C = 400 / 0.9433 = 424.04 kVAr; nameplate 1.21 × Q_C = 513.09 kVAr; X_C = 0.4240² × 1000 / 424.04 = 0.4240 Ω; X_L = 0.0567 X_C = 0.0240 Ω, L = 0.0765 mH; C = 10⁶ / (2π × 50 × X_C) = 7506.543 µF/phase; I = 400 / (√3 × 0.4) = 577.35 A; CB 1.35 × I = 779.4 A → 800 A; inrush 15 × I = 8660.3 A; R_d = 300 / (C ln(√2 V_C / 50)) = 16.09 kΩ; losses 0.0005 × 424.04 + 3 I² X_L/50 + 0.01 = 0.703 kW; 1.1 × 0.4240 = 0.4664 kV; Q_after = 617.34 − 400 = 217.34 kVAr, pf = 700 / √(700² + 217.34²) = 0.9550 lagging ≥ 0.95; resonance check: f_t 209.98 Hz ≥ 150 Hz → FAIL (validation case c0016-400v-forced-567-with-3rd-fails)

Example 26 — 415 V, rounding up moves the resonance onto the 5th

The required 259.7 kVAr alone would resonate at 310 Hz, 24 % above the 5th and outside ±20 %; the installed 300 kVAr resonates at 289 Hz, inside it. The resonance is taken from the installed bank, so a 7 % reactor is fitted.

Given: 0.415 kV, 50 Hz bus with a 10 MVA fault level; maximum demand 520 kW at power factor 0.8 lagging, target 0.97; harmonic distortion 5th 5 %.

Reactive powerφ1 = arccos 0.8 = 36.87°, tan φ1 = 0.75000; φ2 = arccos 0.97 = 14.07°, tan φ2 = 0.25062; Qc = P (tan φ1 − tan φ2) = 520 × (0.75000 − 0.25062) = 259.68 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈259.68 / 50⌉ × 50 = 300 kVAr
Apparent powerSbefore = 520 / 0.8 = 650.00 kVA; Safter = 520 / 0.97 = 536.08 kVA; reduction 113.92 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(10,000 kVA / 300 kVAr) = 288.68 Hz; hr = 288.68 / 50 = 5.774
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: 5th (250 Hz, 5 % > 3 %): |288.7 − 250| / 250 = 15.5 % (within ±20 %); Resonance near 5th harmonic → reactor by the lowest significant harmonic (5th; 3rd → 14 %, 5th → 7 %, otherwise 5.67 %) = 7 % detuning reactor
Tuned frequencyp = XL/XC = 0.0700; ftuned = f / √p = 50 / √0.0700 = 188.98 Hz (h = 3.78)
Capacitor voltageSeries branch: VC = V / (1 − p) = 0.415 / 0.9300 = 0.4462 kV (7.52 % rise across the capacitors); Vrated = 1.1 × 0.4462 = 0.4909 kV (LV: no standard HV rating applies)
Capacitor ratingQeff = V² / (XC(1 − p)), so the capacitors generate QC = Qinst / (1 − p) = 300 / 0.9300 = 322.58 kVAr at VC; nameplate at 1.1 VC: 1.21 × 322.58 = 390.32 kVAr
CurrentIrated = Qinst / (√3 V) = 300 / (√3 × 0.415) = 417.36 A; Imax = Irated × kh × 1.3 = 417.36 × 1.0753 × 1.3 = 583.41 A
CapacitanceXC = VC² × 1000 / QC = 0.4462² × 1000 / 322.580 = 0.6173 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 0.6173) = 5,156.537 µF per phase (star)
ReactorXL = p XC = 0.0700 × 0.6173 = 0.04321 Ω; L = 1000 XL / (2π f) = 0.1375 mH; RL = XL / 50 = 0.0008642 Ω
BreakerICB = 1.35 × 417.36 = 563.4 A → next standard size 630 A; overload setting 1.25 × Irated = 521.7 A
Discharge resistorVpk = √2 × 446.2 V = 631.1 V; R = t / (C ln(Vpk / 50 V)) = 300 / (5,156.537 × 10−6 × ln(631.1 / 50)) = 22.95 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 417.36 = 6,260.4 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 322.58 = 0.161 kW + reactor 3 I² RL = 0.452 kW + discharge 0.010 kW = 0.623 kW
Target power factorQafter = P tan φ1 − Qinst = 390.00 − 300 = 90.00 kVAr (lagging); pf = P / √(P² + Qafter²) = 520 / √(520² + 90.00²) = 0.9854 → 0.9854 vs limit 0.9700 PASS
ResonanceTuned 188.98 Hz must lie below the lowest significant harmonic (5th = 250 Hz) → 188.98 vs limit 250.00 Hz PASS
Result2/2 checks pass → PASS; 300 kVAr bank, 7 % reactor tuned to 189.0 Hz, 417.4 A, 630 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.8 = 0.75000, tan φ2 = tan arccos 0.97 = 0.25062; Q = 520 × (0.75000 − 0.25062) = 259.68 kVAr → 300 kVAr output in 50 kVAr steps; f_r = 50 × √(10000 kVA / 300 kVAr) = 288.68 Hz (h = 5.774); significant (> 3 %) harmonics: 5th 5 %; within 20 % of the 5th (250 Hz) → reactor by the lowest significant harmonic (5th) = 7 %; f_t = 50 / √0.07 = 188.98 Hz; V_C = 0.415 / (1 − 0.07) = 0.4462 kV; Q_C = 300 / 0.9300 = 322.58 kVAr; nameplate 1.21 × Q_C = 390.32 kVAr; X_C = 0.4462² × 1000 / 322.58 = 0.6173 Ω; X_L = 0.07 X_C = 0.0432 Ω, L = 0.1375 mH; C = 10⁶ / (2π × 50 × X_C) = 5156.537 µF/phase; I = 300 / (√3 × 0.415) = 417.36 A; CB 1.35 × I = 563.4 A → 630 A; inrush 15 × I = 6260.4 A; R_d = 300 / (C ln(√2 V_C / 50)) = 22.95 kΩ; losses 0.0005 × 322.58 + 3 I² X_L/50 + 0.01 = 0.623 kW; 1.1 × 0.4462 = 0.4909 kV; Q_after = 390.00 − 300 = 90.00 kVAr, pf = 520 / √(520² + 90.00²) = 0.9854 lagging ≥ 0.97; resonance check: f_t 188.98 Hz < 250 Hz → pass (validation case c0016-415v-installed-kvar-resonance)

Example 27 — 400 V, 60 kW, 0.80 → 0.95 — leading — FAIL

The 25 kVAr requirement rounds up to a 50 kVAr step, more than the load's 45 kVAr: power factor 0.9965 leading fails the 0.95 target although its magnitude exceeds it.

Given: 0.4 kV, 50 Hz bus with a 10 MVA fault level; maximum demand 60 kW at power factor 0.8 lagging, target 0.95; no harmonic distortion.

Reactive powerφ1 = arccos 0.8 = 36.87°, tan φ1 = 0.75000; φ2 = arccos 0.95 = 18.19°, tan φ2 = 0.32868; Qc = P (tan φ1 − tan φ2) = 60 × (0.75000 − 0.32868) = 25.28 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈25.28 / 50⌉ × 50 = 50 kVAr
Apparent powerSbefore = 60 / 0.8 = 75.00 kVA; Safter = 60 / 0.95 = 63.16 kVA; reduction 11.84 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(10,000 kVA / 50 kVAr) = 707.11 Hz; hr = 707.11 / 50 = 14.142
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: no harmonic above 3 % → no reactor
Capacitor voltageVC = V = 0.4 kV; Vrated = 1.1 × 0.4 = 0.4400 kV (LV: no standard HV rating applies)
Capacitor ratingNo reactor: QC = Qinst = 50 kVAr; nameplate at 1.1 V: 1.21 × 50 = 60.50 kVAr
CurrentIrated = Qinst / (√3 V) = 50 / (√3 × 0.4) = 72.17 A; Imax = Irated × kh × 1.3 = 72.17 × 1.0000 × 1.3 = 93.82 A
CapacitanceXC = VC² × 1000 / QC = 0.4000² × 1000 / 50.0000 = 3.200 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 3.200) = 994.718 µF per phase (star)
BreakerICB = 1.35 × 72.17 = 97.4 A → next standard size 100 A; overload setting 1.25 × Irated = 90.2 A
Discharge resistorVpk = √2 × 400.0 V = 565.7 V; R = t / (C ln(Vpk / 50 V)) = 300 / (994.718 × 10−6 × ln(565.7 / 50)) = 124.32 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 72.17 = 1,082.5 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 50.00 = 0.025 kW + reactor 3 I² RL = 0.000 kW + discharge 0.010 kW = 0.035 kW
Target power factorQafter = P tan φ1 − Qinst = 45.00 − 50 = −5.00 kVAr — negative: the bank over-corrects and the power factor is leading, which fails the target whatever its magnitude; pf = P / √(P² + Qafter²) = 60 / √(60² + (−5.00)²) = 0.9965 → 0.9965 vs limit 0.9500 FAIL
ResonanceNo harmonic above 3 %; undetuned resonance 707.11 Hz is not excited PASS
Result1/2 checks pass → FAIL; 50 kVAr bank, no reactor, 72.2 A, 100 A breaker; governing: target power factor. Installed 50 kVAr exceeds the load's 45.0 kVAr: power factor 0.9965 LEADING at maximum demand, and more leading at lighter load. Use smaller steps, an automatic stepped bank or a lower target.
ReferenceHand calculation: tan φ1 = tan arccos 0.8 = 0.75000, tan φ2 = tan arccos 0.95 = 0.32868; Q = 60 × (0.75000 − 0.32868) = 25.28 kVAr → 50 kVAr output in 50 kVAr steps; f_r = 50 × √(10000 kVA / 50 kVAr) = 707.11 Hz (h = 14.142); significant (> 3 %) harmonics: none; no significant harmonic within ±20 % → no reactor; V_C = 0.4 kV, nameplate 1.21 × 50 = 60.50 kVAr; X_C = 0.4² × 1000 / 50 = 3.2000 Ω; C = 10⁶ / (2π × 50 × X_C) = 994.718 µF/phase; I = 50 / (√3 × 0.4) = 72.17 A; CB 1.35 × I = 97.4 A → 100 A; inrush 15 × I = 1082.5 A; R_d = 300 / (C ln(√2 V_C / 50)) = 124.32 kΩ; losses 0.0005 × 50.00 + 0.01 = 0.035 kW; 1.1 × 0.4000 = 0.4400 kV; Q_after = 45.00 − 50 = -5.00 kVAr, pf = 60 / √(60² + -5.00²) = 0.9965 LEADING → fails the target; resonance check: 707.11 Hz clear of every significant harmonic → pass (validation case c0016-400v-60kw-leading)

Example 28 — 33 kV, 1.1 × V above 33 kV

1.1 × 33 kV = 36.3 kV exceeds the 36 kV Um, so the next standard rating (IEC 60038) is 40.5 kV.

Given: 33 kV, 50 Hz bus with a 1,000 MVA fault level; maximum demand 20,000 kW at power factor 0.85 lagging, target 0.95; no harmonic distortion.

Reactive powerφ1 = arccos 0.85 = 31.79°, tan φ1 = 0.61974; φ2 = arccos 0.95 = 18.19°, tan φ2 = 0.32868; Qc = P (tan φ1 − tan φ2) = 20,000 × (0.61974 − 0.32868) = 5,821.20 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈5,821.20 / 50⌉ × 50 = 5,850 kVAr
Apparent powerSbefore = 20,000 / 0.85 = 23,529.41 kVA; Safter = 20,000 / 0.95 = 21,052.63 kVA; reduction 2,476.78 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(1,000,000 kVA / 5,850 kVAr) = 653.72 Hz; hr = 653.72 / 50 = 13.074
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: no harmonic above 3 % → no reactor
Capacitor voltageVC = V = 33 kV; Vrated = 1.1 × 33 = 36.3000 kV → next standard rating 40.5 kV
Capacitor ratingNo reactor: QC = Qinst = 5,850 kVAr; nameplate at 1.1 V: 1.21 × 5,850 = 7,078.50 kVAr
CurrentIrated = Qinst / (√3 V) = 5,850 / (√3 × 33) = 102.35 A; Imax = Irated × kh × 1.3 = 102.35 × 1.0000 × 1.3 = 133.05 A
CapacitanceXC = VC² × 1000 / QC = 33.00² × 1000 / 5850.00 = 186.2 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 186.2) = 17.099 µF per phase (star)
BreakerICB = 1.35 × 102.35 = 138.2 A → next standard size 160 A; overload setting 1.25 × Irated = 127.9 A
Discharge resistorVpk = √2 × 33,000.0 V = 46,669.0 V; R = t / (C ln(Vpk / 50 V)) = 300 / (17.099 × 10−6 × ln(46,669.0 / 50)) = 2,565.44 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 102.35 = 1,535.2 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 5,850.00 = 2.925 kW + reactor 3 I² RL = 0.000 kW + discharge 0.010 kW = 2.935 kW
Target power factorQafter = P tan φ1 − Qinst = 12,394.89 − 5,850 = 6,544.89 kVAr (lagging); pf = P / √(P² + Qafter²) = 20,000 / √(20,000² + 6,544.89²) = 0.9504 → 0.9504 vs limit 0.9500 PASS
ResonanceNo harmonic above 3 %; undetuned resonance 653.72 Hz is not excited PASS
Result2/2 checks pass → PASS; 5,850 kVAr bank, no reactor, 102.3 A, 160 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.85 = 0.61974, tan φ2 = tan arccos 0.95 = 0.32868; Q = 20000 × (0.61974 − 0.32868) = 5821.20 kVAr → 5850 kVAr output in 50 kVAr steps; f_r = 50 × √(1e+06 kVA / 5850 kVAr) = 653.72 Hz (h = 13.074); significant (> 3 %) harmonics: none; no significant harmonic within ±20 % → no reactor; V_C = 33 kV, nameplate 1.21 × 5850 = 7078.50 kVAr; X_C = 33² × 1000 / 5850 = 186.1538 Ω; C = 10⁶ / (2π × 50 × X_C) = 17.099 µF/phase; I = 5850 / (√3 × 33) = 102.35 A; CB 1.35 × I = 138.2 A → 160 A; inrush 15 × I = 1535.2 A; R_d = 300 / (C ln(√2 V_C / 50)) = 2565.44 kΩ; losses 0.0005 × 5850.00 + 0.01 = 2.935 kW; 1.1 × 33.0000 = 36.3000 kV → 40.5 kV standard; Q_after = 12394.89 − 5850 = 6544.89 kVAr, pf = 20000 / √(20000² + 6544.89²) = 0.9504 lagging ≥ 0.95; resonance check: 653.72 Hz clear of every significant harmonic → pass (validation case c0016-33kv-standard-voltage-above-33)

Example 29 — 400 V, 1.5 MVAr bank — breaker above 2000 A

A 1.5 MVAr bank at 400 V: 1.35 × 2,165 A = 2,923 A is above the 2000 A list; the next IEC 60059 R10 rating is 3150 A.

Given: 0.4 kV, 50 Hz bus with a 50 MVA fault level; maximum demand 3,000 kW at power factor 0.8 lagging, target 0.97; no harmonic distortion.

Reactive powerφ1 = arccos 0.8 = 36.87°, tan φ1 = 0.75000; φ2 = arccos 0.97 = 14.07°, tan φ2 = 0.25062; Qc = P (tan φ1 − tan φ2) = 3,000 × (0.75000 − 0.25062) = 1,498.13 kVAr
Installed bankOutput at system voltage in 50 kVAr steps: Qinst = ⌈1,498.13 / 50⌉ × 50 = 1,500 kVAr
Apparent powerSbefore = 3,000 / 0.8 = 3,750.00 kVA; Safter = 3,000 / 0.97 = 3,092.78 kVA; reduction 657.22 kVA
Parallel resonanceInstalled bank without reactor: fr = f √(Ssc / Qinst) = 50 × √(50,000 kVA / 1,500 kVAr) = 288.68 Hz; hr = 288.68 / 50 = 5.774
Detuning testDetune when a harmonic > 3 % lies within ±20 % of fr: no harmonic above 3 % → no reactor
Capacitor voltageVC = V = 0.4 kV; Vrated = 1.1 × 0.4 = 0.4400 kV (LV: no standard HV rating applies)
Capacitor ratingNo reactor: QC = Qinst = 1,500 kVAr; nameplate at 1.1 V: 1.21 × 1,500 = 1,815.00 kVAr
CurrentIrated = Qinst / (√3 V) = 1,500 / (√3 × 0.4) = 2,165.06 A; Imax = Irated × kh × 1.3 = 2,165.06 × 1.0000 × 1.3 = 2,814.58 A
CapacitanceXC = VC² × 1000 / QC = 0.4000² × 1000 / 1500.00 = 0.1067 Ω; C = 106 / (2π f XC) = 106 / (2π × 50 × 0.1067) = 29,841.552 µF per phase (star)
BreakerICB = 1.35 × 2,165.06 = 2,922.8 A → next standard size 3,150 A (IEC 60059 R10 above 2000 A); overload setting 1.25 × Irated = 2,706.3 A
Discharge resistorVpk = √2 × 400.0 V = 565.7 V; R = t / (C ln(Vpk / 50 V)) = 300 / (29,841.552 × 10−6 × ln(565.7 / 50)) = 4.14 kΩ (to 50 V in 5 min)
InrushIinrush ≈ 15 × Irated = 15 × 2,165.06 = 32,476.0 A (range 10–20 ×)
LossesCapacitors 0.0005 × QC = 0.0005 × 1,500.00 = 0.750 kW + reactor 3 I² RL = 0.000 kW + discharge 0.010 kW = 0.760 kW
Target power factorQafter = P tan φ1 − Qinst = 2,250.00 − 1,500 = 750.00 kVAr (lagging); pf = P / √(P² + Qafter²) = 3,000 / √(3,000² + 750.00²) = 0.9701 → 0.9701 vs limit 0.9700 PASS
ResonanceNo harmonic above 3 %; undetuned resonance 288.68 Hz is not excited PASS
Result2/2 checks pass → PASS; 1,500 kVAr bank, no reactor, 2,165.1 A, 3,150 A breaker; governing: target power factor
ReferenceHand calculation: tan φ1 = tan arccos 0.8 = 0.75000, tan φ2 = tan arccos 0.97 = 0.25062; Q = 3000 × (0.75000 − 0.25062) = 1498.13 kVAr → 1500 kVAr output in 50 kVAr steps; f_r = 50 × √(50000 kVA / 1500 kVAr) = 288.68 Hz (h = 5.774); significant (> 3 %) harmonics: none; no significant harmonic within ±20 % → no reactor; V_C = 0.4 kV, nameplate 1.21 × 1500 = 1815.00 kVAr; X_C = 0.4² × 1000 / 1500 = 0.1067 Ω; C = 10⁶ / (2π × 50 × X_C) = 29841.552 µF/phase; I = 1500 / (√3 × 0.4) = 2165.06 A; CB 1.35 × I = 2922.8 A → 3150 A; inrush 15 × I = 32476.0 A; R_d = 300 / (C ln(√2 V_C / 50)) = 4.14 kΩ; losses 0.0005 × 1500.00 + 0.01 = 0.760 kW; 1.1 × 0.4000 = 0.4400 kV; Q_after = 2250.00 − 1500 = 750.00 kVAr, pf = 3000 / √(3000² + 750.00²) = 0.9701 lagging ≥ 0.97; resonance check: 288.68 Hz clear of every significant harmonic → pass (validation case c0016-400v-large-bank-breaker-r10)

Standards & References

  • IEEE 18-2012 — IEEE Standard for Shunt Power Capacitors
  • IEC 60871-1 — Shunt Capacitors for A.C. Power Systems Having a Rated Voltage Above 1000 V
  • IEEE 519-2022 — IEEE Standard for Harmonic Control in Electric Power Systems
  • IEC 60076-6 — Power Transformers — Part 6: Reactors (reactor design reference)
  • IEEE 1036-2010 — IEEE Guide for Application of Shunt Power Capacitors
  • IEEE C37.012 — IEEE Guide for Capacitance Current Switching for AC HV Circuit Breakers

Tips & Limitations

Tips
  • A target power factor of 0.95 is typical for utility tariff compliance. Correcting to unity (1.0) is rarely economic and risks leading power factor at light load.
  • Always measure harmonic distortion at the PCC before installing a capacitor bank. Even small harmonic currents can be amplified dramatically at resonance.
  • With 6-pulse VFDs (significant 5th) the calculator chooses 7%; 5.67% is used when the lowest significant harmonic is the 7th or above, and 14% when the 3rd is significant (single-phase non-linear loads).
  • Demand charge savings typically dominate the payback calculation. Verify that the utility tariff actually penalises low power factor or high kVA demand.
  • Standard capacitor voltage ratings with a detuning reactor are higher than system voltage. Always check that the selected capacitor voltage rating exceeds the calculated \(V_{cap}\).
  • Consider stepped banks (automatic switching) for sites with widely varying load profiles to avoid leading power factor at light load.
Limitations
  • Single-step bank only — the calculator sizes a single fixed bank. Multi-step automatic bank configurations are not modelled.
  • Simplified energy savings — system loss reduction uses a flat 2% impedance assumption. Detailed system modelling would give more accurate results.
  • No back-to-back switching — inrush calculations assume a single bank energising against the system. Back-to-back capacitor bank switching produces significantly higher inrush currents.
  • Reactor choices — only the 5.67%, 7% and 14% detuning reactors are offered; tuned harmonic filters are not designed.
  • No temperature or altitude derating — capacitor and reactor ratings are not adjusted for elevated ambient temperature or high altitude.
  • Balanced three-phase system — unbalanced loads and single-phase capacitor configurations are not modelled.

Developer Guide

Architecture

File layout:

  • components/c0016_capacitor_bank_sizing/component.py — Component registration, parameter schema, calculator class (CapacitorBankSizingCalculator)
  • components/c0016_capacitor_bank_sizing/calculations.py — Pure calculation functions (no Django dependencies)
  • components/c0016_capacitor_bank_sizing/equations.py — Equation builder for MathJax report generation (ARTICLES.md standard)
  • components/c0016_capacitor_bank_sizing/excel_generator.py — Excel workbook report generation

The calculator class is CapacitorBankSizingCalculator(ElectricalCalculator). It delegates all computation to calculate_capacitor_bank_sizing(data) in calculations.py.

Key constants in calculations.py:

  • REACTOR_DETUNING_OPTIONS — dict of 5.67%, 7%, and 14% reactor parameters
  • STANDARD_VOLTAGES — list of standard capacitor voltage ratings (3.3 – 72.5 kV)
  • STANDARD_BREAKER_SIZES / next_standard_breaker() — breaker ratings, IEC 60059 R10 above 2000 A

Code Paths

Key functions in calculations.py:

Function Purpose
calculate_capacitor_bank_sizing(data) Main entry point — orchestrates all 15 sub-calculations
calculate_power_factor_correction(data) PF angles, required kVAr, demand reduction
calculate_resonance_frequency(q_c, s_sc, f_n) Parallel resonance frequency and harmonic order
select_detuning_reactor(f_n, resonance_info, h5, h7, h11, h13, h3, reactor_selection) Detuning decision (harmonic proximity) and reactor by lowest significant harmonic, or forced
check_resonance(f_n, resonance_info, reactor_info, harmonics) Resonance check of the installed bank (parallel or tuning basis)
calculate_adjusted_capacitor_rating(q_c, p) Installed output (50 kVAr steps), capacitor kVAr \(Q/(1-p)\) and nameplate
calculate_achieved_power_factor(p, q_existing, q_inst, pf_target) Power factor with the installed bank; leading flag; target met
calculate_voltage_ratings(v_system, p) Capacitor voltage stress with reactor, standard selection
calculate_current_ratings(q_c, v_ll, p) Rated and maximum current including harmonic factor
calculate_capacitance(q_c, v_ll, f_n) Reactance and capacitance per phase (Wye)
calculate_reactor_parameters(p, x_c, i_rated, f_n) Reactor inductance, resistance, quality factor
calculate_reactor_losses(i_rated, r_l) Copper losses in reactor windings
calculate_protection_settings(i_rated) CB rating, overload, unbalance alarm/trip
calculate_discharge_resistor(v_cap, c, t, v_target) Discharge resistor sizing per IEEE 18
calculate_inrush_current(v_cap, i_rated) Switching inrush and pre-insertion resistor
calculate_stored_energy(c, v_cap) Energy stored per phase and total
calculate_losses_and_efficiency(q_c, reactor_losses) System losses (capacitor + reactor + discharge) and efficiency
calculate_economic_analysis(s_red, e_rate, d_rate, h, inv) Annual savings and simple payback

Testing

Running tests:

pytest calcs/components/c0016_capacitor_bank_sizing/ -v

How to extend:

  1. Add new reactor options by extending REACTOR_DETUNING_OPTIONS in calculations.py.
  2. To support multi-step banks, add step-sizing logic in a new function and call it from calculate_capacitor_bank_sizing().
  3. For additional voltage ratings, update the STANDARD_VOLTAGES list.
  4. Extend the schema in component.py for new parameters (e.g., altitude derating, temperature).
  5. Write tests covering new calculation paths and edge cases (zero harmonics, maximum reactor percentage, boundary kVAr rounding).