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IEEE 519 · IEEE C57.110

Harmonic Distortion Loss Calculator

Enter the measured current spectrum rather than a single THD figure, and get what it costs — extra copper loss in money, the neutral current triplens produce, and the K-factor your transformer is being asked to live with.

The load

Current spectrum

Each harmonic as a percentage of the fundamental, from a power quality analyser. The defaults are typical of a plant that is mostly six-pulse VFD load.

Valuing it

Current THD
23.3%
Above the 8 % IEEE 519 limit for most industrial consumers. Expect the utility to raise it eventually, and expect the symptoms sooner.
Extra losses, per year
₹13,080
0.27 kW of additional copper loss — 5.5 % on top of the fundamental loss — for 1,635 kWh a year that does no work.
RMS current349.1A
Above the fundamental by2.7%
Neutral current81.6A
Neutral vs line current24%
K-factor2.75
Transformer capability96.1%
Fundamental copper loss5.00kW
ProjectCircuit ref
Prepared byDate
Checked byDate

Harmonic Distortion Loss Calculator · IEEE 519 · IEEE C57.110 · EnergyCalcHQ · energycalchq.com
Preliminary calculation. The figures behind it are representative values for the stated conditions, not a substitute for the current edition of the standard or the manufacturer's published data. Verify before issuing for construction. Not a substitute for a qualified engineer or a protection study.

For page numbers, keep Headers and footers ticked under More settings in the print dialog.

Why the spectrum, not just THD

A single THD number tells you how much distortion there is, not what it will do. Two plants at 20 % THD behave completely differently if one is 20 % fifth harmonic from drives and the other is 20 % third harmonic from single-phase electronics — the first heats the transformer, the second overloads the neutral.

Everything below follows from the spectrum by definition, so it is worth ten minutes with a power quality analyser rather than an assumption.

THD    = √(Σ Ih²) / I₁
I_rms  = I₁ × √(1 + Σ(Ih/I₁)²)

Losses go with the square, so they add up quickly

Copper loss is I²R, and R does not care whether the current is doing useful work. Since the harmonic currents add in quadrature, the extra loss over the clean-sine case is exactly the sum of the squared harmonic ratios:

extra loss = base loss × Σ(Ih/I₁)²

At 25 % THD that is a 6 % increase in copper loss — modest as a percentage, and a real number in kilowatt-hours on a feeder running six thousand hours a year. It is also 3 % more RMS current in a cable you sized for the fundamental.

Skin effect makes it slightly worse than this, because higher frequencies see more resistance than the 50 Hz figure. The calculation here is the floor.

The neutral is where triplens go

Third harmonic — and its multiples, the triplens — is in phase across all three lines. In the neutral the three contributions do not cancel like the fundamental does. They add:

I_neutral ≈ 3 × I₃

So 35 % third harmonic gives a neutral current slightly above the line current, on a balanced load. A neutral carrying more than the phases is not a fault condition — it is arithmetic — and it is why a 3.5-core cable with a reduced neutral is the wrong cable for an office block full of LED drivers and switch-mode supplies.

Where third harmonic exceeds about 33 %, IEC 60364-5-52 has you size the cable on the neutral current rather than the line current. That is the case this calculator flags, and the derating detail is in cable derating factors.

K-factor and transformer derating

Harmonic currents drive eddy-current losses in a transformer's windings that rise with the square of frequency. The K-factor captures that:

K = Σ (Ih / I_rms)² × h²

A pure sine wave gives K = 1. Typical office and drive loads land between 4 and 13. A K-rated transformer is built to carry that without derating; a standard transformer has to be derated instead, and IEEE C57.110 puts the capability at:

derating = √[(1 + P_EC) / (1 + K × P_EC)]

with P_EC the eddy loss at rated current, taken here as 0.05 for a typical oil-filled distribution transformer. Dry-type units have higher eddy losses and derate harder — if the transformer is dry-type, treat the figure shown as optimistic.

Feed the result into the transformer calculator: a unit at 92 % capability that you had planned to load to 85 % is actually running at 92 % of what it can now do.

The symptoms, in the order they appear

  • Capacitors failing repeatedly. The first and most expensive symptom. Capacitors are a low impedance to high frequencies, and a plain bank resonating with the supply inductance amplifies whatever harmonic sits near the resonant point. The fix is detuned reactors — APFC panels.
  • Neutral conductors and terminals running hot, on a circuit whose phases are comfortable.
  • Transformer humming and running warm at loads it used to handle.
  • Nuisance tripping of thermal-magnetic devices responding to true RMS current the design never allowed for.
  • Metering disagreement. Different meter designs measure distorted current differently, which is one reason a submeter total drifts from the utility bill — metering accuracy.

What to do about it

  1. Measure first. At the point of common coupling and at the largest non-linear loads. IEEE 519 sets 8 % current THD for most industrial consumers at the PCC, and the limits are stricter on stiffer supplies.
  2. Line reactors on drives — 3 % or 5 % impedance. Cheap, passive, and typically halves a drive's current distortion.
  3. Detuned reactors on capacitor banks, 7 % as standard, before harmonics destroy them.
  4. Full-size or oversized neutrals where triplens are significant, and separate neutrals rather than shared ones.
  5. Twelve-pulse or active front-end drives on large loads, which cancel the 5th and 7th at source.
  6. Active harmonic filters where distortion is severe and the load varies. Expensive, effective, and the last resort rather than the first.