Single Phased Motor with Delta Windings

Status
Not open for further replies.
P

philly

Senior Member
I'm trying to see what would happen with a 480V motor that had its windings connected in Delta when the motor lost a phase. For this I'm going to assume that a phase was lost on the primary side of the motor contactor, and we are taking voltage readings on the secondary side of the contactor.

For this case I would think that when measuring between one of the two good phases and the lost phase we would read 240V. This is because the two windings that are connected between the two good phases and the open phase will now be in series parallel with the winding connected between the two good phases and will have 480V across the series combination. Seeing that each of the impedances of these windings are equal I would think that 240V would drop across each winding and when measuring across each of these windings including the open phase in the measurement you would see 240V.

Would you also see 240V L-G if you read from this open phase to ground?
 
If you energized a three phase motor the way you have, with one of three legs dropped, for an extended period of time, you will damage the windings and MAKE them short to ground. All you need to do is read at the top of the contactor, if you've dropped a phase, DO NOT ENGAGE starter, and do not read through motor windings. You're assumptions are correct, but don't do it that way or you're gonna damage the motor. If no harm has been done, you will not read a short to ground. If you read to ground ON ANY PHASE,,you've let the "smoke out",:roll: and there's no putting it back
 
mcclary's electrical said:
If you energized a three phase motor the way you have, with one of three legs dropped, for an extended period of time, you will damage the windings and MAKE them short to ground. All you need to do is read at the top of the contactor, if you've dropped a phase, DO NOT ENGAGE starter, and do not read through motor windings. You're assumptions are correct, but don't do it that way or you're gonna damage the motor. If no harm has been done, you will not read a short to ground. If you read to ground ON ANY PHASE,,you've let the "smoke out",:roll: and there's no putting it back
Click to expand...

Of course I would never energize a motor with the conditions I described and I know that the effect of this condition on the motor will be detrimental.

I was simply trying to look from a theoretical standpoint at a hypothetical situation that someone in the field may see when taking readings.

Will you read 240V L-G on the lost leg if reading at the location I described?
 
the reading would not be exactly 240, but yes, your assumptions are correct in that you would read thru the motor windings and have a voltage reading on the dropped phase, due to induction, from the other two energized phases. The induction taking place inside the motor.
 
Check attached dwg. Under single phasing conditions You can see the coil AB and AC are now in series and in parallel with coil BC.
You have full voltage across BC and half voltage across AB and AC.
 
Last edited:
mcclary's electrical said:
If you energized a three phase motor the way you have, with one of three legs dropped, for an extended period of time, you will damage the windings and MAKE them short to ground. All you need to do is read at the top of the contactor, if you've dropped a phase, DO NOT ENGAGE starter, and do not read through motor windings. You're assumptions are correct, but don't do it that way or you're gonna damage the motor. If no harm has been done, you will not read a short to ground. If you read to ground ON ANY PHASE,,you've let the "smoke out",:roll: and there's no putting it back
Click to expand...
What doesn't make any sense is that people will deliberately do this when they install a static phase converter on a 3-phase motor. :confused:
 
Start with a model of a delta connected resistor array. In this case, as has been noted, the now 'open' phase will be sitting at the midpoint between the two connected phases. The voltage to ground will depend upon the relation of these phases to ground. To figure out this voltage, you need to use vector math to find the voltage of the midpoint between the two connected phases.

If the supply were a common 480/277 wye with a grounded neutral, then the midpoint between two phases is sitting at about 140V to ground. If the supply were a 'corner grounded delta', and one of the 'hot' legs opened, then the midpoint between the remaining phases is at 240V to ground. If the supply were a corner grounded delta and the 'grounded' leg opened, then the midpoint between the remaining phases is at about 416V to ground. If the supply were a 'high leg' delta (where the midpoint of one of the supply legs is grounded, and the 'high' leg open, then the midpoint between the remaining phases is approximately at ground potential.

Now consider the motor. Instead of being and easily analyzed resistive load, you have a composite of resistance and inductance introducing phase shift and storing energy. I _believe_ that if the shaft is not rotating, then by the symmetry of the situation the results will be similar to the resistive case. However with the shaft rotating you end up with a very interesting situation.

Rick references this with his mention of phase converters. When the motor is spinning, it can absorb energy on the two connected phases, store than energy in the rotating rotor and shaft, and release that energy on the third leg; effectively 'regenerating' that third leg. If you have a lightly loaded motor already spinning, and something happens to open one of the supply legs, then the terminal voltages and terminal to ground voltages will not change much at all. Unloaded three phase motors are intentionally used all the time to create the third phase when only single phase power is available.

-Jon
 
Good Explanation! I understand it all but this one part.

winnie said:
If the supply were a corner grounded delta and the 'grounded' leg opened, then the midpoint between the remaining phases is at about 416V to ground.
Click to expand...

Why would we have 416V to ground in this case?
 
If you draw out the vectors for the corner grounded delta case, you get an equilateral triangle, with ground at one corner and the two phase voltages at the other corners. The 'midpoint' between the two phase voltages is on the side of that triangle, at sqrt(3)/2 *480V away from ground.

This is assuming the resistive (straight line) result.

-Jon
 
Status
Not open for further replies.
Top