Three-Phase Motor: Phase Failure

I disagree with the statement that the voltage rises, it is the current that rises and not the voltage because the supply voltage is tied to the supply transformer ratio of primary to secondary voltage which may droop but not rise under the fault conditions.

The results are greatly dependent on the point at which the failure occurs and is also dependent on the connections be they delta or star.

For Delta the voltage on one winding will be the phase to phase line voltage with the other two windings being in series across the remaining good phases if the supply is lost. Should the winding connection point be open circuit then the motor will only have one winding energised. The motor slip will increase, the motor will slow down until it stalls and enters locked rotor condition.

For star windings the two good phases will return current via the neutral point while the coil with the loss of supply will have no voltage across it.

Phase loss protection, undervoltage protection and motor stall protection should be employed to stop this occurrence. Stall current can be 6x full load current and normally stall protection is set at 2.5 x full load current. Undervoltage protection is usually set at 85% open circuit voltage.

Phase loss can detected by a core balance transformer which generates an out of balance signal proportional to the unbalanced current flowing in the phase conductors as well it monitors the actual phase voltage. A detection device monitors these unbalance conditions and trips the supply when a set threshold is reached.

thank you for your replay

please are there a mathematical law to clarify the issue further? I would like to know

where this value came from (173%).

thanks

Please re-read #2 and #3⇑ for the explanation is therein.

Phase loss usually results in the motor stalling. If it has been installed correctly then (a) protective device(s) will operate under those circumstances, thereby disconnecting it from the supply.

Rixter has provided you with the theoretical mathematics for the increase, but be aware that the actual increase can be considerably higher than 173% due to power factor changes and increased slippage.

This can be very much the case where a motor is heavily loaded when a phase is lost. The remaining phase currents can approach locked rotor values.

...at which juncture (a) protective device(s) will operate, thereby disconnecting it from the supply.

Is it exactly 173%? I think it could be ~150%. Mathematical explanation is,

It is a Delta connected motor and when one phase is lost, the impedance presented across the remaining two phases goes down to 2/3's of one winding of the motor.

That is because one phase winding is in parallel with two remaining phase windings that are now in series. Thus the net impedance is Z in parallel with 2Z (Z is impedance of one winding).

With net impedance across two phases coming down to 2/3, the current in those two phases increases by 3/2 times or 150%.

The above assumes that the motor is continuing to run at rated speed which may not be the case. Any reduction in speed will make the motor impedance to go down, resulting in increase in the current in those two surviving phases.

...at which juncture (a) protective device(s) will operate, thereby disconnecting it from the supply.

Once a motor loses a phase it will not continue to run as it can not supply the load needed to maintain the torque or the speed, so the current will be any value between full load current 100%, stall trip current nominal 250% and locked rotor current 600% a value which will occur in a few cycles as the motor stops.

There is no running at rated speed after loss of phase so the mathematical explanation depends on stopping time, current drawn and may be a linear graph between 100% and 600% over the number of cycles to trip or stall. Slip frequency goes from approx 4% for an induction motor at rated speed to 100% at stall.

I wish I had someone to help with my homework when I was in school.

CR4 is a relatively recent phenomenon...