I can't find an explanation to determine, or the maths to calculate, the phase and line currents that a motor winding would expect to see during an attempted start if one supply phase is open circuit?
Would it be equivalent to Locked rotor current or less or more?
Can someone shed any light on this?
We're investigating the failure of a 75KW 415V 3ph 4 pole motor which has burnt out due to single phasing during a start attempt. the frame size is abnormally small for 75KW (250s/m) and as such the Locked rotor withstand time is only 18 seconds (cold).
Any opinions would be appreciated.
Assuming it's connected in Delta the worst case coil current will be the locked rotor (coil phase to phase).
It's a very common occurrence for a 3 phase motor to try and start with only 2 phases, the motor starter should have protected it.
The motor will not actually start because there is no phase rotation. So although the rotor is not actually "locked", it is not going to be moving and therefore the current will remain at Locked Rotor Current (600% of FLC) until something clears. If an OL relay is set for Class 10, it is supposed to clear in 10 seconds or less at 600% current (LRC), so it SHOULD have protected the motor against that 18 second thermal damage curve. If the OL relay was bypassed, mis-adjusted or Class 20, it would not.
I find it VERY common for people to mis-adjust OL relays. A common issue is that they read code regulations stating that it is "permissible" for OL relays to begin acting on the motor at 115-120% of FLC, so they adjust them to that value, not understanding that this was ALREADY factored into the design of the OL relay by the mfr. So if you actually read the instructions they TELL YOU to set the dial at 100% of the motor nameplate FLC. If you set it to 120%, then the LRC trip time is affected too.
Thanks for both replies.
Unfortunately (where I am) we're very much living in the past. Overload & out of balance protection is provided by P&B Golds M2 relay's which are characteristically slow to operate.
During bench tests, simulated locked rotor and locked rotor+single phase condition tests using the nameplate locked rotor current found the relay trip time to be 40-50 seconds and 18-20 seconds.
In the case of the locked rotor, fuses would operate within 15 seconds. However, in the case of single-phasing from standstill the the relay or fuse is not fast enough to protect the motor.
Referring to my original question, I wasn't sure what the line current would be for a motor being started with a single-phase fault already present.
When a single-phasing fault occurs on a delta wound motor that is already running, the current in one phase is doubled.
Assuming that when a single-phasing fault is apparent when a motor is started. one phase will pull locked rotor current and in a reversal of the running single phase case, the remaining phases will draw 50% less than locked rotor current. therefore the line current in this scenario would be 1.5x the phase winding with locked rotor current (or 86.5% of the stated locked rotor current).
Does this make sense and is it correct?
You are probably not using the right OL relay. Unbalanced load protection has been built in IEC type overload relays for some 40 years, unless you are using an antique for protection (World war 2 equipment). You should replace it with something newer. If P&B stands for Potter & Brumfield, you got circuit breakers, probably single phase ones, which are not right for this application. Feel free to contact me at email@example.com for further advice. To protect your motor better, there are phase-loss relays too, in case the mains supply misses any phase.
I hope this helps.