Skip to content
Engineerly
Power & current

Motor kW / HP / Amps Converter

Convert a motor rating between kilowatts and horsepower and work out the input power, apparent power and full-load current for a single-phase or three-phase motor from its efficiency and power factor. You can also work backwards from a measured current to the input power and an estimated shaft output. Nameplate values always take precedence over the example defaults.

Last reviewed

Nameplate kW or hp is shaft output. Input power is larger by the efficiency: Pin = Pout / η, and the current is drawn by Pin together with the power factor.

For estimation and education. Not a substitute for a licensed engineer, the applicable codes (IEC 60364, NEC/NFPA 70, BS 7671, local regulations) or manufacturer data. Read the disclaimer.

Inputs

Supply
Start from

Motor

kW

IEC motors are rated in kW of shaft output

V

Nameplate value; 0.85 is an example

Nameplate value as a fraction; 0.90 is an example

How to use this calculator

  1. Choose single-phase or three-phase supply and enter the voltage (line-to-line for three-phase).
  2. Choose where to start: rated output in kW (IEC motors), rated output in hp (NEMA motors) or a measured line current.
  3. Enter the power factor and efficiency from the nameplate or datasheet. PF 0.85 and efficiency 0.90 are only examples.
  4. Read the full-load current, input kW and kVA, or, from a measured current, the input power and estimated shaft output.

How it works: the formula

1 hp = 745.7 W P_in = P_out / η S = P_in / PF Three-phase: I = P_out / (√3 × V × PF × η) Single-phase: I = P_out / (V × PF × η) From current: P_in = √3 × V × I × PF (or V × I × PF), P_out ≈ P_in × η

A motor's rating is the mechanical power at its shaft. NEMA MG 1 motors are rated in horsepower, IEC 60034-1 motors in kilowatts. To deliver that output the motor draws more electrical power, the difference being its losses, so the input power is the output divided by the efficiency. The supply also has to carry the reactive current that magnetises the motor, which the power factor accounts for. Dividing the apparent power by √3 × V (or V for single-phase) gives the line current.

Variables and units

SymbolMeaningUnit
P_outRated (shaft) outputkW or hp
P_inElectrical input powerkW
SInput apparent powerkVA
ηEfficiency at the operating point—
PFPower factor at the operating point—
VSupply voltage (line-to-line for three-phase)V
ILine currentA

Worked example

An 11 kW three-phase motor at 400 V with PF 0.85 and efficiency 0.90 (example values):

P_in = 11 / 0.90 = 12.22 kW, S = 12.22 / 0.85 = 14.38 kVA I = 11 000 / (1.73205 × 400 × 0.85 × 0.90) = 20.75 A Output in hp = 11 000 / 745.7 = 14.75 hp

A 10 hp single-phase motor at 230 V with the same PF and efficiency: P_out = 7.457 kW, P_in = 8.286 kW and I = 8286 / (230 × 0.85) = 42.38 A. Working back from a measured 20 A on a three-phase 400 V motor: S = 13.86 kVA, P_in = 11.78 kW and the estimated shaft output is 10.60 kW (14.22 hp).

Assumptions

  • Sinusoidal supply at the entered voltage; balanced phases for three-phase motors.
  • Power factor and efficiency are those at the operating point, as entered.
  • 1 hp = 745.7 W (mechanical horsepower, as used on NEMA nameplates); metric horsepower (about 735.5 W) is not used.

Limitations

  • Real power factor and efficiency depend on the motor's size, speed and load. Both fall at part load, so the full-load current is not proportional to the shaft load.
  • In NEC jurisdictions, conductors and short-circuit protection for motor circuits are sized from the full-load current tables of NEC Article 430 (Table 430.248 single-phase, Table 430.250 three-phase) rather than from the nameplate; overload protection uses the nameplate current.
  • Starting current is several times the full-load current and is not calculated here; it matters for voltage drop, generator sizing and protection settings.
  • Motors fed from a variable-speed drive draw input current that depends on the drive, not on the motor's power factor.
  • Estimating the shaft output from a measured current is rough at light load, because the magnetising current does not fall with load.

References

Standards and texts that cover this topic. Use the edition adopted in your jurisdiction.

  • IEC 60034-1, Rotating electrical machines — Rating and performance (rated output is the mechanical power at the shaft, in kW).
  • IEC 60034-30-1, Rotating electrical machines — Efficiency classes of line-operated AC motors (IE code): IE1 to IE4.
  • NEMA MG 1, Motors and Generators (ratings in horsepower of shaft output, nameplate data).
  • NFPA 70, National Electrical Code, Article 430 — Motors, motor circuits and controllers (full-load current tables for conductor and protection sizing).

Frequently asked questions

Is the kW on the nameplate the power the motor draws?

No. Under IEC 60034-1 and NEMA MG 1 the rated output is the mechanical power at the shaft. The electrical input is larger by the losses: an 11 kW motor with 90 % efficiency draws about 12.2 kW.

How many kW is 1 hp?

One mechanical horsepower, as used on NEMA nameplates, is 745.7 W, so 1 hp ≈ 0.746 kW and 1 kW ≈ 1.341 hp. Metric horsepower (PS or CV), about 735.5 W, is slightly smaller and is not used here.

What do the efficiency classes IE1 to IE4 mean?

IEC 60034-30-1 groups line-operated AC motors into efficiency classes: IE1 standard, IE2 high, IE3 premium and IE4 super premium efficiency. The minimum efficiency for each class depends on the rated output, the number of poles and the frequency, so take the actual value from the motor's datasheet.

Should I size the cable from the nameplate current or the NEC table?

Where the NEC applies, Article 430 requires the full-load current from its tables for conductor and short-circuit protection sizing, and the nameplate current for overload protection. Under IEC practice designers normally use the maker's rated current. Follow the code that applies to your installation.

Why does a motor at half load draw more than half its full-load current?

Part of the current magnetises the motor and stays almost constant whatever the load. As the load falls, the power factor falls too, so the current falls much less than the shaft power.

Embed this calculator

Paste this code into your page to show the calculator in a frame. The attribution link must stay visible. The embedded page carries no ads and stores nothing.

How these formulas were checked: methodology and testing.