Excessive current at motor startup 2

[sry110]

Hopefully this will be a simple one for you electrical guys:
I have a 4-pole 3.0kW motor, 380V/3-ph/50Hz, whose data sheet shows D (delta) connection and starting current of 40 Amps. The Customer reports that when they started the motor, it pulled 70 Amps which consequently caused the motor trip out on over-current.

I noticed that 70 divided by 40 is 1.75, which looks close to 1.732 (square root of 3). Then I looked at the catalog for the motor and found that the rated current for this motor in 400V Delta connection is a factor of approx. 1.73 higher than the same motor in 690V Wye connection. Also, 690 divided by 400 is 1.73.

I'm not drawing any conclusions, just providing some observations. Can anyone explain what could be causing our motor to draw approx. 1.75x its rated starting current at start-up?

 

[xnuke]

Perhaps you're trying to start a load that is too large for the motor.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[sry110]

@ xnuke - our Customer suggested that the motor is not strong enough to start the load; however, my understanding is that the motor will not (or cannot?) draw higher amps than the Starting Current (Istart) listed on the data sheet. In other words, my understanding is that the Starting Current (Istart) is the IEC nomenclature equivalent to 'Locked Rotor Current' in NEC/NEMA world, and my understanding of 'Locked Rotor Current' is that the motor will sit there and pull that amperage value if it comes up against a load it cannot move, causing the rotor to be 'locked'. Is that not correct?

 

[waross]

Most overloads have an inverse time effect. They allow higher than rated current for a short time. When the load is to great, the starting time may exceed the time that the O/L relay will allow for that current level.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[sry110]

@ waross: Is there any condition where a motor can physically pull higher amps than its rated Locked Rotor Current (or Starting Current as they say in IEC world). What the Customer is reporting is that the motor is drawing 70 Amps at startup, while the data sheet for the motor shows that the Starting Current (Locked Rotor Current) is 40 Amps. I'm trying to figure out how this is possible.

 

[itsmoked]

The only condition I can think of would be if the motor is spinning backwards when starting is attempted, as then it can be generating. This happens with fans and water pumps.

Otherwise sry110, I would agree that you should not see greater than LRA. Meaning the supply is too high a voltage, the frequency is too low, or there is a wiring error. Possibly the LRA is specified for a wye connection and somehow the motor is wired delta.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[sry110]

@ itsmoked: it also crossed my mind that the motor might be intended for wye connection but was wired delta by the customer. However, the data sheet for the motor states that it is intended for a delta connection. But running with the idea that the motor data sheet is incorrect: if I apply 400V connected in Delta to a motor that is rated for the same voltage with connection in Wye, is it true that the resulting current would be a factor of squareroot(3) higher than the rated values?

 

[itsmoked]

I doubt it. My reasoning is that if you do anything to significantly mis-wire the motor, as your example would provide, the motor wouldn't behave in a linear manner. It's not a well behaved "resistor" but would experience magnetic saturation, V/Hz issues, or under-voltage causing 'obscene' overcurrent. In all cases not a tidy √3 result.

Have you confirmed the 'customers' reading of this transient event is correct?

Keith Cress
kcress -

http://www.flaminsystems.com

 

[sry110]

Thanks for the additional feedback. Getting information from this customer is very difficult, but we are trying to get more details.
The story they gave us is this: the Customer used the 40A locked rotor current from the motor data sheet to set the current limit on their motor starter. When they started the motor they saw 70A which tripped the motor due to high current. The Customer concluded that we need to supply them a larger motor (4kW instead of the existing 3kW) so that the motor can start the load turning without encountering the high current condition.

It doesn't make sense to me. Under normal conditions, the highest current a motor can draw is its locked rotor current. Putting a higher power motor in place of the existing one will result in higher locked rotor current than the existing motor. Am I correct?

 

[Parchie]

Sry110,
Did you review your load torque versus motor torque capacity? (motor torque capacity greater than the load torque). You may just be needing a change in drive sheeve diameter, imo.

 

[7anoter4]

I don't think the rated current of 3 kW/380 V induction motor could be less than 7 A [80% pf. 80% eff.]
and the start current to rated ratio it has to be 7.The voltage drop for such a small motor could be neglected.
So, the start current at 380 V/50 Hz has to be 7*7=49 A.
The load has no any influence at start. The induction motor behaves at start like a [static] transformer.
In my opinion, the actual voltage could be 420 V [400 V it is now standard in Europe +5% permissible]
and the frequency may be 48 Hz.
The new start current at 420V has to be 49*420/380=54.15 A.
IEC 60034-1 standard states:
"Unless stated otherwise, tolerances on declared values shall be as specified in Table 20”.
Table 20 Pos. 9 "Locked rotor current of cage induction motors with any specified starting apparatus =+
20%"
New starting current could be 54.15*1.2=65 A.
Another 4% for 48 Hz then finally Ist=65*1.04=67 A.
However, it is hard to say, it could be such a chain of coincidences as described above.
Nevertheless, check the actual voltage and frequency and compare with the rating plate.

 

[ScottyUK]

Post a photograph of the motor data plate and one of the motor terminal box wiring.

 

[sry110]

@ Parchie - I should have clarified that this is a directly coupled motor, not a belt drive. Our design calculation shows that the load torque is less than the available torque from the motor. The Customer has concluded that the actual load torque is higher than calculated and exceeding the motor available torque, and the basis of their conclusion is that the motor pulled 70 Amps instead of the 40 Amps locked rotor current shown on the data sheet. What I'm trying to figure out it is this: If I have a motor capable of producing 50 N-m locked rotor torque (and assuming that locked rotor torque is higher than breakdown torque) and it comes up against a load that requires 100 N-m to move, is the motor going to produce any more than 50 N-m? And if not, then is it going to pull any higher current than the locked rotor current advertised on the data sheet (within a reasonable tolerance)?


@ 7anoter4 - Thanks for your analysis. For what it's worth, the relevant motor data (taken from the manufacturer's data sheet) is as follows:
Rated Output P[sub]n[/sub]: 3 kW
Rated voltage U[sub]N[/sub]: 380 VD +/- 5% (IEC 60034-1)
Rated frequency f[sub]N[/sub]: 50 Hz +/- 2% (IEC 60034-1)
Rated current I[sub]N[/sub]: 6.2 A
No Load Current: 2.8 A
Starting Current I[sub]S[/sub]/I[sub]N[/sub]: 6.4

The motor is started using direct-on-line starter.


@ScottyUK - Unfortunately I do not have a photo of the wiring inside the terminal box, but the motor nameplate is attached below.

 

[waross]

How is the customer measuring the starting current?
Ossilograph tests show that at the moment of energization there may be a very high and very short transient current. This transient is so short as to be ignored by most overload protection devices. The magnitude varies and depends on a combination of the residual magnetizm and the point on wave that the voltage is applied.
This transient may be several times the normal starting current.
Some digital clamp meters had a peak lock function that would detect and report this transient rather than the normal starting surge.
How is the current being measured?
What is the load?
Is the motor back spinning when the power is applied?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[sry110]

@ waross - Thank you for the insight. I will ask the Customer to confirm how they are measuring the current.
The 3kW motor is the driver for our turning gear assembly, and the turning gear is driving a steam turbine + compressor train. Assuming no mechanical issues with the steam turbine or compressor, the turning gear will see momentary high load as it breaks the drive train away from rest and accelerates it to steady state turning speed (in this case 20 rpm). There is a 76:1 gear ratio in the turning gear between the motor and the drive train, so the drive train inertia referred to the motor is very small and therefor the motor accelerates the load to full speed very quickly - less than 2 seconds.

The motor is sized such that its locked rotor torque exceeds the required breakaway torque of the train when referred back to the motor through our gearbox. Once the shaft train is rotating the torque to keep it turning at turning gear speed is negligible (i.e. no appreciable aerodynamic load and very low bearing friction) so the load on the turning gear motor is negligible. Under normal conditions, the turning gear motor will pull close to no-load amps when turning the train continuously.

The input shaft of the turning gear, to which the 3kW motor is directly coupled, is fitted with a unidirectional overrunning clutch (a backstop clutch) which prevents reverse rotation of our system, so it is not possible that the motor is backspinning at time of energizing.

 

[waross]

Looks as if I have lost my spell check feature. Sorry.
In NEMA land we generally use a rule of thumb that starting current is about 6 times full load current. However the smaller the HP the larger the ratio and a 3 HP motor may have a starting current of 700%. Still only about 40 Amps.
You say that the O/L is tripping on overcurrent. If that is accurate then you may be tripping on the first cycle transient. A sustained high current will trip on overload rather than over current.
If the applied voltage is somewhat above rated voltage that may be a contributing factor.
Can you supply the make and model number of the O/L device?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[sry110]

Fortunately (or unfortunately, depending how you look at it), our scope of supply for electrical devices is limited to the motor. The Customer provides the motor starter and any other electrical equipment needed to operate and monitor the motor. So in short, I cannot supply the make and model of the overload device.
But if I am understanding you correctly, there may be a transient current at start-up that exceeds the locked rotor current shown on the data sheet? Would this happen at in-rush, or when the motor sees the load? I only ask because in our turning gear system there is a good amount of backlash to be taken up between when the motor starts and when it actually comes up against the load, so if I were to monitor motor amps with a very fine sampling rate I would expect to see a spike at the moment the motor is energized (in-rush), then a quick drop off because there is initially no load, then another spike when the motor comes up against the stationary load.

 

[jraef]

I agree with Bill, my first inclination is to believe this is a measurement issue, and I too have seen the "Peak Hold" feature of some digital meters create a cascading sequence of conclusion jumping that gets everyone away from the real issues. Magnetic inrush current can be, in theory, up to 2000% of the FLA, but only for a fraction of a second, too fast for most protective devices to even see it. But a digital clamp-on ammeter with a sampling rate fast enough to see it, or even partially see it, and that Peak Hold button pressed, will give you a false impression of what the starting current REALLY is. This tends to be the first thing I ask about now when I am given this kind of scenario. Get that out of the way first.

The type of OL relay is relevant as well, because many people are going to Solid State OL relays now, and some of them will react in a similar way if not programmed correctly. They shouldn't, but it happens.

Still, it COULD be that even though the current is NOT really 70A, the OL is is tripping because the motor is NOT accelerating the load fast enough. That might be a number of other issues, all being masked by the initial measurement error that took everyone down a rabbit hole of "70A starting current" which you correctly identified as impossible.

One POSSIBLE issue is related to where you started this thread with, the misconnected motor. But you were thinking the wrong direction. If the motor is SUPPOSED to be connected in Delta, and they connected it in Star, then the motor would draw LESS current than normal, but also produce far less TORQUE. Think Star-Delta starter, but never coming out of Star. What would happen is that initially the current is low, but if the motor cannot fully accelerate with that severely curtailed torque, it sits in limbo at less that Locked Rotor Current, but still more that the OL curve will allow for that period of time, so the OL trips. Then because of the 70A red herring, nobody is seeing the REAL problem.


"You measure the size of the accomplishment by the obstacles you had to overcome to reach your goals" -- Booker T. Washington

 

[Parchie]

I'd give a star to @jraef on this one! Just had that problem where our tech wired a 30HP motor in wye instead of delta. The OL tripped. I was called and inspected. The usual 240 gpm discharge of the driven pump reduced to 120 gpm. I suspected the motor was operating at a reduced speed. I inquired how the motor was wired and the tech told me it was wired wye instead of delta!

 

[sry110]

Thanks all for the advice. Now we have some more ammunition to go back to the customer and hopefully get to the bottom of this. I will follow back if we get any meaningful information from them.