Motor current rating (vs reality)

[Dan Friedrichs]

Maybe a dumb question, but I'm unclear on why motors are rated for currents that would seem to generate well more power than their nameplate says they're good for.

For instance, I have a 2HP dust collector motor rated 220V, 12A. Even assuming 80% efficiency, that's 2.8HP, isn't it?

I have access to some fun tools: a clamp-on current meter and a battery-powered (floating) oscilloscope. For fun, I hooked them up to said motor while it was running (with no restriction on the dust collector input). I measured 238V, 8.6A, and 1454W (calculated by the oscilloscope as a point-by-point multiplication of V*I, then averaged, so it correctly reports real power accounting for power factor).

20200611_171850.jpg

1454W is 1.95HP, which seems right, since it's a 2HP dust collector and I removed the restrictions on the inlet (assuming the manufacturer built a more generous safety margin than 0.05HP in, originally...).

So why is that 8.6A, and not the 12A on the nameplate?


In another vein, I have a motor I thought was 5HP (no nameplate). But using the same measurement equipment, I find it draws 6.2HP under load (power washer), and I've used it like this for years. Would a 5HP motor survive driving a 6.2HP load for long periods?

[Jerry Bruette]

Would the 12 amps be full load rating? Your 8.6 would be with little or no load and 12 would be full load right before you let the smoke out.

[Dan Friedrichs]

Would the 12 amps be full load rating? Your 8.6 would be with little or no load and 12 would be full load right before you let the smoke out.

No, the 8.6A was measured with a 2HP load (full load for that motor). With no load (dust collector inlet blocked), it measures even less.

[Mike Henderson]

In another vein, I have a motor I thought was 5HP (no nameplate). But using the same measurement equipment, I find it draws 6.2HP under load (power washer), and I've used it like this for years. Would a 5HP motor survive driving a 6.2HP load for long periods?

I would suspect the motor is really a 6HP motor. Service factor will allow you to exceed the motor HP rating but that would require a 1.2 service factor to get 6 out of 5. Not impossible but I'd suspect a higher HP rating on the motor first.

Mike

[David L Morse]

Dust Collector

1) It appears that 1.454 kW is the input power. Assuming 80% efficiency that's about 1.55 HP out.

2) If the motor label says 12 A at 220 V then you would expect about 11.1 A when powered by 238 V. Also, if labeled to NEMA standards, the tolerance on the full load current is +/- 10%. (Coincidentally(?), the NEC Table 430.248 current listed for that motor is 12A.)

3) Using typical midrange values of 0.85 power factor and 0.8 efficiency the calculated input for 2 HP out is 2.19 kVA or 9.22 A at 238 V. According to your photo the power factor of the motor is actually 0.756. Using that value and 0.80 efficiency you get 10.4 A input for 2 HP out and 238V in.

4) 1.55/2 = 0.775 or about a 23% margin at your operating conditions. Current increases directly with air density and inversely with input Voltage. That sounds like a good margin given your conditions.

Power Washer

I think you left out efficiency again.

[Dan Friedrichs]

Thanks, David - I was hoping you'd reply. D'oh! Well, I'm embarrassed it slipped my mind that 2HP is the output power. That makes perfect sense. Thanks for the excellent explanation.

[Rod Sheridan]

No, the 8.6A was measured with a 2HP load (full load for that motor). With no load (dust collector inlet blocked), it measures even less.

You’re measuring input power, not output power......Rod