Starting a heavy fan: 3.7 seconds, not 14

3.7 sdirect-on-line start

14 sthe same start in star-delta

188 Astarting current for 3.7 s

The duty

A centrifugal fan absorbs 10 kW at 1,475 rpm. Its impeller and the coupling have a moment of inertia of 4.5 kgm² referred to the motor shaft. Fan torque rises with the square of speed.

The motor

EM3-160M-4-11: 11 kW, M_N 71.3 Nm, M_A 3.0 × M_N, M_K 3.7 × M_N, rotor inertia 0.086 kgm², starting current 9.0 × 20.9 A.

Accelerating torque

Mean motor torque ≈ 0.45 × (M_A + M_K) × M_N = 0.45 × 6.7 × 71.3 = 215 Nm Fan torque at speed = 10,000 ÷ 154.5 rad/s = 64.7 Nm Mean fan torque ≈ 64.7 ÷ 3 = 21.6 Nm (square law) Accelerating torque = 215 − 21.6 = 193 Nm

Starting time

t = J × ω ÷ T_acc = (0.086 + 4.5) × 154.5 ÷ 193 = 3.7 s

For 3.7 s the motor draws about 188 A and the rotor heats rapidly. EMPD checks this time against the permissible starting time of the frame before quotation.

Why star-delta is the wrong answer here

Star torque = 215 ÷ 3 = 72 Nm → accelerating torque 72 − 21.6 = 50 Nm t = 4.586 × 154.5 ÷ 50 = 14 s

Star-delta cuts the current but quadruples the time, so the rotor absorbs far more heat. For high-inertia fans, start direct on line where the supply allows it, or use a soft starter or inverter with a controlled ramp.

Application study: an engineering example worked by EMPD with catalogue data. Your application is confirmed by our engineers before quotation. Send us your duty.

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