Posted by Maurice Taylor on 25/06/2021 18:54:26:
Hi,A 1/2 hp Brook motor on an ML7 takes nearly 50A for 100mS on start ,this will sometimes trip a domestic mcb but probably wouldn’t trip an inductive load mcb ,as previously mentioned .An earth leakage fault would trip the elcb not the mcb.
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I’m just a bit mystified of that graph/screenshot of AC current.
It would appear, from the y axis, that the current after half a second is approx 5A (likely close to 6?).
As it is peak current RMS current (equivalent to DC) would be 5A/root 2 = 5A/1.414 = 3.5(4)A. At 230V (RMS value of the grid) that equates to 813W If the motor is rated at 370W, it shows that the motor is only 46% efficient.
That would be 370W at full load, not at idle – like these figures may actually be (motor achieves full speed in only slightly over 1/4 of a second).
I am wondering what the current (and therefore power) would be at full motor load?
Now, if the peak current was actually 6A and the typical UK grid voltage is 240V (often it exceeds that) the input Wattage would be in excess of 1000 which would indicate a motor efficiency of around only 1/3rd.
I reckon that 1/2HP at, say 0.8 Power Factor and 70% efficiency, would draw only 660W (including the Wattless power) at full load.
Perhaps the power factor is making that great discrepancy? Anyone out there explain, please?
Re stopping a much larger generator – I expect the generator would be a stator wound machine with a small electronic AVR. An armature wound (slip ring) genny would easily cope with such transient spikes, wheras the new fangled things are pretty hopeless on that score.