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ASdrawn

Motor power & current calculator

Three-phase current I = P / (√3 × U × cos φ × η). A 4 kW four-pole motor on 400 V at cos φ 0.81 and 89.1 % efficiency draws 8.00 A, which is the full-load current WEG prints for it, and turns 26.3 N·m at 1450 rpm.

Enter the shaft power from the nameplate (or the torque and speed), the supply, the power factor and the efficiency. The power triangle is drawn to the power factor, beside the motor.

Know
Supply
RATIO TRIANGLE, S FIXED
Current I
Torque T
Input power P1
Apparent power S
How it’s calculated

Three-phase I = P / (√3 × U × cos φ × η), single-phase I = P / (U × cos φ × η), with P the shaft power and U the line voltage. Input P1 = P / η; apparent S = P1 / cos φ. Torque T [N·m] = 9550 × P [kW] / n [rpm].

Four-pole IE3 motors at 400 V 50 Hz, full load

Output, kWFrameSpeed, rpmη, %cos φI, A (WEG)I, A (ABB)I, A (Siemens)
0.7580142082.50.821.61.681.75
1.590L145086.00.803.153.23.15
2.2100L143587.00.804.564.34.4
4112M145089.10.8188.67.9
7.5132M146591.50.8513.914.714.3
11160M147092.20.8320.721.120.5
15160L147093.00.8228.428.528.5
22180L147093.60.8340.941.541
37225S/M148094.60.8665.669.866
55250S/M148095.30.8696.997.896
75280S/M148595.60.87130133133

Why does a motor draw more than its kW?

The nameplate kW is what the shaft gives out. The supply also pays for the losses (the efficiency takes those) and carries a magnetising current that flows back and forth without doing work (the power factor takes that). So the volt-amps are the kW divided by both, and the current is the volt-amps over √3 × the line voltage on a three-phase supply.

How do you get torque from kW and rpm?

T = 9550 × P / n, with T in N·m, P in kW and n in rpm: 9550 is 60 000 / 2π, rounded. A four-pole motor on 50 Hz runs a little under 1500 rpm at full load, so its torque in N·m is a little over 6.4 times its kW.