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Medium Voltage Motor Overload Trip Issues 2

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Bilegan

Electrical
Oct 7, 2008
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We are doing the commissioning of a direct online starter (DOL) 4.16kV, 778HP, 60Hz. medium voltage motor being used for a reaction furnace air blower. The airflow is controlled by a sensor which is upstream of the blower. The motor is tripping recently due to overload. Motor FLA=93.9A, LRA=564A at rated 4kV voltage and 0.18pf, motor rotor Wk2=40, full load torque (ft-lb)=1548, pull-up torque=60%, breakdown torque=280%, starting time at 100% rated voltage=14sec, and 25sec for 80% RV.

Multilin 369 relay device 49 overload pick-up set at 1.05xFLA, starting current is FLA x 6.7pu and apply the default overload standard curve multiplier value of 8. Acceleration time from start set at 16sec at 100% rated voltage. Device 27 starting U/V set at 0.8xrated, running U/V at 0.85xrated.

Immediately after the motor tripped, I retrieved data from the Multilin 369 motor protection relay and reveals that the motor tripped at the starting current=6.12xFLA, acceleration time=11.9sec, average motor load=394kW, phase current: A=546A, B=547A, C=550A, voltage: Vab=3.98kV, Vbc=3.99kV, Vca=4.08kV, frequency=59.96Hz, power factor=0.22lagging.

What exactly the cause of the overload trip after 11.9sec motor acceleration time. Do I need to account the other motor performance data for the custom overload curve? instead of relying on the default standard overload curve of the 369 MFR. Any ideas on why the motor is tripping on the above values.




Regards,
Bilegan

Murphy's Law - If anything can go wrong..it will.
 
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Check page 153 of the instruction manual at the link. The default setting for this relay will trip after 10 seconds at 6.0xFLA.

Personally I do not think this is a bad thing because that is a long time to be at such a high current. Was the motor even turning?
 
Looks to me as though it tripped on overload because... it was over loaded!

Is the damper closed when you start? On most centrifugal blowers the load is proportional to air flow, so the dampers may need to be closed until AFTER the motor has reached full speed in order to avoid overloading the motor when starting. It's a very common misconception that closing the dampers INCREASES the load because people think of the blower as pushing the air against a solid wall, when in fact the exact opposite is true. If the damper is closed, there is no air flow so there is very little load on the motor.

Happens all the time.

"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
For the best use of Eng-Tips, please click here -> faq731-376
 
All good comments. I'll through in my 2 cents fwiw... one caveat is I have never worked with digital relays.

I think most fans as large as the one mentioned in the original post are squirrel cage fans, and in that case what Jeff said is exactly true (they draw more power at high flow). Just as clarification in general, I wanted to mention that at least in our plant we have many axial flow fans mounted direclty in ducts up to about 75hp, and in that case the axial flow fans draw more power at low flow (losses dominate the power draw at low flow in axial flow fans).

Assuming it is a radial flow fan, if the controller is modulating based on flow, it may well be sensing flow below target prior to start and controlling the damper wide open prior to start, which makes for worst case motor start and it's worth looking at changing the damper controls or lineup for starting.

Then again, some fans have huge inertia and take a long time to start for that reason... we don't know yet.

Assuming you cannot reduce this long starting time (either it is due to inertia or you do not have ability to change the mechanical load), then since it is so much longer than "normal", it makes sense you might not get away with the default settings. If it were an electromechanical relay, the strategy would be to plot time-current curves for calculated motor current during start (at various voltages), for motor relay, and for motor thermal limits (again at several voltages which bound expected voltage range). Preferably choose thermal limits which already assume one start was just completed, and then the motor is protected for 2 starts in a row. With digital relays there is a complication that the protection curve may shift depending on the history of starts, so I imagine the approach is slightly different.




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(2B)+(2B)' ?
 
Of course the objective of plotting those time current curves is to examine whether there is margin between calculated motor starting profile and relay trip profile (margin to prevent trip), as well as margin between motor trip profile and motor thermal limit curve (margin to prevent damage)

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(2B)+(2B)' ?
 
Bilegun,
If you plot the given FLC,LRC & 14 sec starting time @ 100% voltage,369 Thermal multiplier=8 etc, there is no possibility for the starting current curve to HIT the thermal curve.May be the above time is for motor only.As jaref pointed out,closing the outlet damper at start is very important to reduce the acc time.Also there had been instances where an air flow is coming through the inlet dampers thereby turning the motor in the reverse direction before the starting signal is received.Final result is again longer acc time.
 
I am trying to reconcile these 2 quotes:
Kiribanda said:
If you plot the given FLC,LRC & 14 sec starting time @ 100% voltage,369 Thermal multiplier=8 etc, there is no possibility for the starting current curve to HIT the thermal [TOC] curve
Bilegan said:
Immediately after the motor tripped, I retrieved data from the Multilin 369 motor protection relay and reveals that the motor tripped at the starting current=6.12xFLA [exactly 574=LRA], acceleration time=11.9sec,
At first glance they seem to conflict. i.e. the first one suggests the motor should not trip even at 14 seconds, while the second one suggests it did trip at 11.9 seconds. Maybe the motor was started twice in a row so the settings were reduced for the 2nd start? Or some other explanation for this apparent contradiction?



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(2B)+(2B)' ?
 
apply the default overload standard curve multiplier value of 8.

The default multiplier is x4 and will trip in 10 seconds @ 6pu current.

If you set it to x8 then it would trip in approximately 20 seconds @ 6pu current.

So, what is the overload multiplier setting? The default of x4 or was it changed to x8?
 
rhatcher
Nothing in page 153 saying that the relay will trip after 10sec at 6.0pu x FLA. Are you referring to Table 5-1: 369 Standard Overload Curves of page 5-147?

jraef
Yes, the discharge valve for reaction furnace was shut-off during start-up, however, the blow-off valve were required by the operations to be open at 40%.

Kiribanda
The 14sec is the requirement for the motor coupled with the blower to accelerate from start.

LionelHutz
When the motor tripped during start-up, the applied overload multiplier setting is x8.

The operations people are discussing the time set for trip delay of airflow at low-low. They also seeing an option to use the by-pass line for airflow.

I need your expert advice, please help.

Regards,
Bilegan

Murphy's Law - If anything can go wrong..it will.
 
Bilegan,
The operations people are discussing the time set for trip delay of airflow at low-low.
It looks to me that while the fan is accelarating to full speed the low-low air flow switch is activated due to low air flow and trips the fan!!Did you not notice the "LOW-LOW AIR FLOW" alarm also during the each tripping?
 
bilegan said:
rhatcher Nothing in page 153 saying that the relay will trip after 10sec at 6.0pu x FLA. Are you referring to Table 5-1: 369 Standard Overload Curves of page 5-147?
fwiw, I think the numbers that Lionel gave were correct (20 sec at 6pu for multiplier 8) as can be verified from figure 5-7 on page 5-146.

It still leaves the question why did it trip at 11.9sec if the relay is supposed to trip at 20 sec under these conditions (multiplier 8 with just above 6 times FLA). The standard (vs custom) curve was used? No unbalance? No previous starts/runtime just before this? If all no, beats me.... might want to doublecheck all the info about relay settings and conditions retrieved (current and time of the trip).

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(2B)+(2B)' ?
 
kiribanda,
As per the sequence of event recorder at the control room, 4.16kV MCC sent signal to the DCS that the multilin 369MFR tripped the blower motor.

electricpete
Yes, you are correct, LionelHutz info are correct. The figure 5-7 on page 5-146 can be read in conjunction with Table 5-1: 369 Standard Overload Curves of page 5-147.

Yes, it still leaves the question to me why did it trip at 11.9sec. Standard overload curve multiplier = 8 is used. Current unbalance is 0.0%, ground current unbalance is 0.0% after the tripped. We did motor solo run test prior to commissioning and the motor performs well, no abnormal reading in terms of voltage, current, frequency, power, reactive power.

Regards,
Bilegan

Murphy's Law - If anything can go wrong..it will.
 
I suggest you read pages 5-26 to 5-32 in the manual until you understand how the thermal model works.

The 369 will has a running thermal capacity used memory. The relay would not trip in 11.9 seconds if the thermal capacity used was 0% and the starting current was below 6.7pu. Your trip data is not showing any abnormal unbalance that would bias the overload. So, it seems obvious that you started with the thermal capacity used at some value above 0%. Depending on the settings, you could trip after operating for hours or even after starting and then leaving the motor off for an hour.

So, we might be more help if you would actually share all the thermal model settings.
 
Sorry guys, I incorrectly referred to the custom curve settings on page 153 (5-146). At least I got the manual posted for the rest of you to see. The standard curves that should be used are on the next page and do show a tripping time of 20 seconds for curve #8 at 6xFLA.

However, I stick by my comment:

"Personally I do not think this (tripping) is a bad thing because that is a long time to be at such a high current. Was the motor even turning?"

As ElectricPete astutely pointed out, the trip current of 6.12FLA is equal to 98% of the motor LRA of 546 amps...after 11.9 seconds!

This doesn't seem right to me. Am I missing something?
 
Here are the findings that I can share with you guys just in case you will encounter such kind of issue.

During start-up, the starting temperature of the fan (blower) is lower than the gas temperature. When they increased the shaft output, then this is the cause of motor overloading, the 369 MFR tripped the motor at 11.9sec. They should have reduce the degree of opening of the discharge valve to the reaction furnace in that case.
The process engineers proceeded to start the fan (blower) in combine air restriction with bypass operation and the trip delay of airflow at low-low was set at 17seconds. The motor accelerated in 12sec and recorded the running current and voltage as 78A and 4.2kV respectively.

rhatcher,
Sorry, the starting current should be 5.85xFLA instead of 6.12. Cheers!

Thank you rbulsara, rhatcher, jraef, electricpete, Kiribanda, and LionelHutz for your valuable post, I appreciate.

Regards,
Bilegan

Murphy's Law - If anything can go wrong..it will.
 
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