Transformer Sizing Calculator
Select a distribution transformer kVA rating from connected load, diversity and future spare capacity, with primary and secondary full-load currents.
Inputs
Fraction of load running together
Formula
- kVA = kW × diversity / cos φ × (1 + spare)
- I_FL = kVA × 1000 / (√3 × V)
- I_sc ≈ I_FL / (Z% / 100)
A transformer is rated in apparent power because its losses depend on current, not on the real power delivered. The connected load is first reduced by a diversity factor (not everything runs at once), converted to kVA using the site power factor, then increased by a spare margin so the unit is not replaced when the load grows. Loading a distribution transformer beyond 80% continuously shortens insulation life.
Step-by-step calculation
Diversified load
kW × diversity
200 kW
Apparent power
kW / cos φ
235.29 kVA
With spare
× (1 + spare)
282.35 kVA
Standard size
next size up the IEC series
315 kVA
Fault current
I_FL / (Z%/100)
8,765 A
Worked example
- 250 kW at 0.85 pf, diversity 0.8, 20% spare.
- kVA = 250 × 0.8 / 0.85 × 1.2 = 282 kVA → select a standard 315 kVA unit.
- LV full-load current = 315 000 / (1.732 × 415) = 438 A; with 5% impedance, fault level ≈ 8.8 kA.
Assumptions
- Three-phase, 50 Hz oil-filled distribution transformer; balanced loading.
Tips
- Improving the power factor from 0.85 to 0.95 cuts the required kVA by more than 10%.
- Two smaller transformers in parallel give redundancy and better part-load efficiency than one large one.
Warnings
- Continuous loading above 80% of rating accelerates insulation ageing.
- The downstream switchgear must be rated for the calculated short-circuit current.
Standards & references
- IEC 60076
- IS 1180 / IS 2026
- IEC 60909 (fault levels)
Frequently asked questions
How do I convert kW to kVA?
Divide the real power by the power factor: kVA = kW / cos φ. At 0.85 pf, 250 kW needs 294 kVA.
What loading should a transformer run at?
Design for 70–80% of nameplate rating at peak. That leaves headroom for growth and keeps insulation temperature within design life.
What does 5% impedance mean?
It is the primary voltage, as a percentage of rated voltage, needed to circulate full-load current with the secondary shorted. It sets the short-circuit current — roughly 20 × full load at 5%.
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