Transformer sizing and voltage regulation

Introduction

Choosing a distribution transformer is a few lines of arithmetic that are easy to get wrong in the details: the load must be converted to apparent power at its power factor, allowed to grow, and kept below the loading the owner accepts; the unit then has to hold its secondary voltage under load and not waste too much energy in iron and copper. This calculator sizes the transformer to the IEC 60076 preferred ratings (or checks a given one), splits the impedance voltage into its resistive and reactive parts from the nameplate losses, computes the voltage regulation and the secondary voltage at the actual load and power factor, the losses and efficiency with the maximum-efficiency point, the annual energy loss and the secondary short-circuit current.

It is for electrical engineers specifying shore-connection, shipyard and industrial transformers, for operators checking an existing unit against a new load, and for students.

Formulas

S = P / cos φ,   S_req = S (1 + g) / k_load   → next preferred rating S_r (25 … 5 000 kVA, R10 series)
I_HV = S_r/(√3 U_HV),   I_LV = S_r/(√3 U_LV),   x = S/S_r
u_r = P_k / S_r,   u_x = √(u_k² − u_r²)
ε = x (u_r cos φ + u_x sin φ) + x² (u_x cos φ − u_r sin φ)² / 2    (per unit; sin φ > 0 lagging)
U_LV,load = U_LV (1 − ε)
P_loss = P_0 + x² P_k,   η = P/(P + P_loss),   x_η,max = √(P_0/P_k)
I_sc ≈ I_LV / u_k

Assumptions and limitations

• Three-phase two-winding transformer at nominal tap, ONAN; diversity already in the load; no cyclic overload credit (IEC 60076-7).
• Regulation by the standard second-order approximation; the no-load secondary voltage is the rated value.
• Nameplate losses as entered; Ecodesign Tier 2 units have much lower losses than older ones.
• Harmonics (K-factor), altitude and ambient derating, parallel operation and tap ranges are not covered; the short-circuit current is the transformer-only estimate (see the IEC 60909 calculator).

Worked example

A 400 V switchboard with 800 kW of load at power factor 0.8 lagging, 20 % growth expected, transformers to be loaded at most 80 %; candidate unit 11/0.4 kV, u_k 6 %, P_0 1.1 kW, P_k 10.5 kW.

1. S = 800/0.8 = 1 000 kVA; S_req = 1 000 × 1.2/0.8 = 1 500 kVA → preferred 1 600 kVA (62.5 % loaded now, 75 % after growth).
2. Checking an existing 1 000 kVA unit at the present load (x = 1): u_r = 10.5/1 000 = 1.05 %, u_x = √(6² − 1.05²) = 5.91 %.
3. Regulation ε = 1.05 × 0.8 + 5.91 × 0.6 + (5.91 × 0.8 − 1.05 × 0.6)²/200 = 4.47 % → secondary 382 V at full load.
4. Losses 1.1 + 10.5 = 11.6 kW, efficiency 800/811.6 = 98.6 %; the unit is most efficient at √(1.1/10.5) = 32 % load; 69.6 MWh lost per year at 6 000 h.
5. Rated currents 52.5 A HV and 1 443 A LV; secondary short-circuit current about 1 443/0.06 = 24 kA; inrush 12–17 kA for the first cycles.

With the 20 % growth the 1 000 kVA unit would run at 120 %: the calculator flags it and names 1 250 kVA as the next rating.

FAQ

Related calculators

• Short-circuit current (IEC 60909) — fault levels with network, transformer and cables.
• Three-phase power and motor current — the loads that make up P and cos φ.
• Generator load, kVA and fuel — the generating side of the same plant.
• Cable sizing and voltage drop — the feeders downstream of the transformer.