Motor resistance

[tryinghard]

I looked at an air compressor motor the other day that has a nameplate:
240V, 1ph, 1.5 hp, & 9.7A.

It was tripping a previously replace solid-state overload, Furnas brand, even though the LRA was 13.5 & FLA 12 to 11. As it turned out the overload is for 3ph and it has phase loss protection so it would not hold past 3 seconds. So we will be replacing it with a single-phase overload.

I did not have a megger with me but I did take an ohm reading between the motor leads and it is 1.5 ohms. Shouldn't the ohm reading between the motor leads be near 24.7?

 

[winnie]

No, the DC resistance should be much lower than 24 ohms. The DC resistance is not what keeps the current down at 10A. In a 100% efficient motor the DC resistance would be 0.

When running on AC, the magnetic windings of the motor induce self induce a voltage nearly equal to the supply voltage. This is what prevents your motor from looking like a 1.5 ohm load.

-Jon

 

[gar]

080501-0845 EST

tryinghard:

To expand a little further.

If you removed the rotor you still have an inductor and assume it is only a single coil (ingnore any starting technique), then one equivalent circuit is a resistance in series with an ideal inductor. The resistance is approximately the DC resistance you measured. Depending upon the magnitude of the inductor (in this case it is smaller than when the rotor is installed) there is an inductive reactance that opposes current flow in addition to the resistance. You then have an impedance that is the combination of the resistance and inductive reactance.

When the rotor is installed that impedance increases because the leakage flux is reduced. This is somewhat what the input impedance would be in a locked rotor condition. Moderately higher than the DC resistance.

Next get the rotor up to speed and a counter EMF is induced back to the stator and this subtracts from the input voltage. So a rough approximation for an equivalent circuit of the motor in a very simplified fashion is an ideal voltage source, an impedance, and an opposing induced voltage. So an approximation of the current of the running motor is the source voltage minus the induced voltage divided by the impedance. And obviously inter-related are different phase angles.

In many ways it is best to learn about DC motors first because then it is easier to see the voltage difference and understand it and its relation to the aramature resistance. I could go into this DC motor theory more if it was of interest. I would not want to get into AC motor theory. For that see Alternating-Current Machinery by Bailey and Gault, McGraw-Hill, 1951.

.

 

[GeorgeB]

It was tripping a previously replace solid-state overload, Furnas brand, even though the LRA was 13.5 & FLA 12 to 11. As it turned out the overload is for 3ph and it has phase loss protection so it would not hold past 3 seconds. So we will be replacing it with a single-phase overload.

Check Furnas specs, but in single phase motor loads, we just ran the current through all 3 overloads, putting them in series. Standardization of components ...

 

[tryinghard]

Thanks guy's