Selection of Motor Overcurrent relays:

[NCTHAI]

The plant motors are supplied from MCC that has two incomers, two bus sections and bus coupler. MCC incomers has changeover scheme - in the event of any one of the incomer opening on under voltage, respective incomer will open, allow faulty bus voltage to drop below 35% and then close bus coupler. Motor ratings are from 3 KW to 75KW. MCC is installed with conventional Bi-metallic relays – ABB TE 42 Class 10.

Operating Conditions – Plant is operating in normal mode and two incomers are supplying respective bus of the MCC. Incomer 1 is opened on under voltage and bus couple is closed after break in power supply. During the changeover, motors’ contactor remains closed (Control supply is from UPS).

Small capacity motors has less has less (lower) AC voltage time constant – 1.2 S(This is the time taken by motor terminal voltage to drop from rated voltage to @35% of value after loss of power supply).

Assumptions are made that at rated voltage, motors will draw 900% (9 times the rated current) during the start up.

When above facts are combined in the practice, is it possible that motor will experience full cycle start up current (900%)? If yes, how the thermal overload relay will behave? Since it already has some memory (from running at load previously) and when it experiences full starting current, will it trip the motor?

Word from experts will be of great help.

 

[ScottyUK]

Difficult to say without information on the driven load, but unlikely to cause a problem in my opinion. Motors operating on cyclic on/off duty exist in many applications and there generally isn't a problem with the motors tripping in those applications. 900% is pretty high too.

 

[RRaghunath]

High efficiency / premium efficiency motors do have high starting currents such as 900% and even more.
With such high stating currents coupled with allowable run-up time, care needs to be exercised while selecting Bi-metal OL relay, power contactor as well as back-up fuse. Maloperation during starting will be an issue otherwise.
Coming to auto changeover of power source, what is the time taken for changeover?? It depends on that whether the motors would draw starting currents and for how long. There is also possibility of incomer trip when all the connected motors start drawing starting current simultaneously.
Generally speaking, there is no issue in case of fast changeover (no intentional delay between opening of one incomer and closing of bus coupler). In case of delayed changeover, it is the practice to trip the motors before changeover is allowed. In your case, since the control supply is UPS, an UV relay will be required to trip the motor contactors on UV.

 

[NCTHAI]

Dear Scotty UK,
This is process plant and most of the motors are driving centrifugal pumps. Very few are driving the fans. I agree with you that 900% of start up current is pretty high assumption.

Dear RRaghunath,

Thanks for sharing your though and information. I tried to look for the motor catalog and could not find OEM specifications for 900% of starting currents. In my experience, starting current for LV ranges from 450 - 700 % in practice. The scheme is auto change over based on bus voltage of outgoing bus- it waits for bus voltage to drop below 35%. There is good co-ordination between fuse, CB and overload relays.

Based on specifications that AC voltage time constant is 1.2 Sec, changeover will definitely take more than this time. Adding the relay sensing time and CB operating time, I say that changeover delay is 2 Sec.

The question is during the changeover, motor are still running. Is it possible that motors will come to halt with in 2 seconds? If they come to halt within 2 or 3 seconds then only they will experience full starting current. If they are still rotating, the starting current is expected to be lot lower.

Any thoughts?


Does anyone has any published guidelines for selection of bimetallic overload relays? Any standard or OEM published? Please share if available.

 

[edison123]

The starting inrush current starts to drop off only at around 90% of the rated speed. Below that speed, the current would still be 5 to 6 times if you change over even if it is still spinning.

Muthu

http://www.edison.co.in

 

[che12345]

Dear Mr NCTHAI

Q1. "MCC is installed with {conventional Bi-metallic} relays – ABB TE 42 Class 10."
A1. I was unable to find [TE 42] in the ABB catalogue. Could it be [TA 42 DU with
tripping Class 10A to IEC 60947-4-1 ]? The [TA 42 DU ] having three setting
ranges 18...25A, 22..32A, 29...42A respectively. Attention: the lowest setting of 18A
could be (too high) for your motors from 7kW , (depending on the voltage).

A2. According to ABB published tripping curve for [TA 42 DU 3-phase WARM] shows 1S at
9 times the setting current.

A3. Assuming:
the thermal overload setting I1 is set at the motor rated fill-load current
Im and the motor was running at Im, full-load. This would be considered as WARM
condition. When change-over, the motor is still rotating/coasting. The motor would
pick-up speed or re-start but [unlikely] to take (assumed 9x Im for a duration of 1s);
to resume the full-load speed.

A4. In general, it is [unlikely] to cause any problem especially the assumed 9s Im during starting is on the (high side) and certainly [unlikely] for it to maintaining at this (high current) for [a duration of 1s].

Che Kuan Yau (Singapore)

 

[RRaghunath]

Did you have any case of motor trip during bus transfer?
If you could describe the incident with its background, in detail, more specific advice can be given.

 

[dpc]

Bear in mind that there will be dc offset current in addition to the symmetrical locked rotor current. We generally use a multiplier of 1.6 times the locked rotor current to account for this. Assuming of course that the relay is sensitive to asymmetrical current NEC allows short circuit protection to be set as high as 13X the motor FLA, IIRC.

 

[NCTHAI]

Muthu,

Thanks for the advice. My experience in practice is different from this. Motor starting currents are less if motor is restarted when it is already rotating.

Che12345,

A1. The correct relay type is TF42. Regret for typo error and confusion created.

A2. TF42 is Class 10 relay. This means it takes 10 Sec to trip for 600% of current. From catalog, it takes about 3.2 sec for 9 times current.

A3. I agree with you but Muthu has different opinion. See his post above.

A4. From my practical experience, I have the same opinion but I am unable to convince engineering personal and need to have sound engineering calculations / documentation to prove this.

RRaghunath,

I do not have any history of tripping. This is proactive actions. We are selecting the protection relay for new MCC and need to ensure that all the convinced with the selection of relays.

There is opinion that TF42 (Or any other thermal Overload Class 10 relay) when used for High Efficiency motors will not survive the changeover and may trip due to start-up currents when change over occurs.

DPC,

Thanks for the advice. Your comments are more related to coordination between Short circuit and / or locked rotor protection coordination. We have taken care of that during selection of short circuit protection – Fuse / MCCB.

Thanks all for your inputs.

 

[RRaghunath]

"There is opinion that TF42 (Or any other thermal Overload Class 10 relay) when used for High Efficiency motors will not survive the changeover and may trip due to start-up currents when change over occurs." - I guess the reason for that opinion could be when all the motors are drawing starting current simultaneously, the bus voltage is going to dip, increasing the reacceleration time. But, this argument overlooks two aspects - one, that the motor is not at standstill so the reacceleration time will be lower than rated motor starting time from zero speed; two, that the voltage dip means lower starting current and thus longer operating time for the OL relay.
With regard to use of UPS supply for Power contactors and need to trip the motors on UV before changeover, please see my earlier reply.
Shell standard doesn't recommend simultaneous reacceleration of all motors when the power loss is of over 200ms in duration.

 

[aussiejohn2]

Hi NCTHAI,
1.2 seconds is a relatively long time for the voltage to decay. Is it measured or calculated? You can estimate the open circuit time constant from the rotor circuit parameters - search of this site for motor bus transfer (or fast transfer) should turn up old posts where electricpete covers it. Nearby capacitors (including cable capacitance) can prolong the decay and maybe that applies in your case.
Your belief is correct - if the change-over time is seconds and your load is mostly centrifugal pumps (low inertia) there can be significant speed decay hence starting currents. Are you able to do a controlled trip next time you have a plant shutdown and measure (or if the 1.2 seconds was measured, do you have voltage waveforms rather than just magnitude so you can measure frequency)? Otherwise add the rotor and pump inertias (water normally neglected) and from the pump torque-speed curve you can estimate the coast-down time.
Worth modelling the dynamics of both buses to evaluate the bus voltage during the re-acceleration.
AJ

 

[che12345]

Dear Mr NCTHAI

1. I refer to the ABB publication ISAZ700506F0015...(0022) respectively for TF 42- xx thermal
overloads with Class 10 tripping characteristic. It is stated in English a)[from cold state],
b)for 3-phase exceeding 3x Iset, tripping time is +- 20%.
Attention: As for your case, it would be [from Warm state] which is expected to trip at a
[shorter time]. Unfortunately, 3-phase Warm curve was not published.

2. From above publication, at 9x Iset the tripping time is about 3.8s [from cold state]. Tripping
time is expected [to be shorter], [from warm state].

3. In general, it is [unlikely] to cause any problem especially the assumed 9x Im during starting
is on the (high side) and certainly [unlikely] for it to maintaining at this (high current) for
[a duration of say 3...s], upon voltage resumption, irrespective of whether the motor has stopped
or is costing down.

4. Motor [starting current] from [stand-still] for a high-efficiency motor is generally higher than
a standard motor.
However, the [current surge] when voltage is [re-applied] on a [coasting motor] could be very much
higher than the [starting current] of a stationary motor. This applies throughout, irrespective of whether it is a standard or a high-efficiency motor.
Note: this phenomenon is analogous to the [open-transition star-delta] starting, that in some
instances; the [current surge] is very much higher than the [starting current] being observed
during the change-over from star to delta connection.
Attention: [Current surge] which is different from [starting current] is always of very short duration. It is dependent on the [phase difference] between the motor generated back-emf and the re-applied supply voltage.

Che Kuan Yau

 

[waross]

Closing out of phase on a rotating motor with residual voltage of 35% may cause greater than normal locked rotor current.
The safest way may be to allow the motors enough time for the residual voltage to drop to zero and then do a staggered restart.

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

 

[NCTHAI]

RRaghunath,

Thanks for your comments and information shared. I agree with your comments. The motors are not protected by IED and hence do not have auto restart function that can be distributed over the time.

Aussiejohn2,

Per consultants, 1.2 Seconds is the constant from the motor data sheets. From IEEE, this time constant defines the time period for delay in voltage to 37%.

For suggestion of controlled trip, will try this during FAT of switchgear once ready for the test. Thanks for your advice about modeling. Will consider those items.

Che12345,

Relay is conventional relay and I am not aware of difference between warm and cold status of relay. To my understanding, when load current is lower than the set value, the relay had infinite operating time. Definitely, the heater coil will be hot inside the relay operating mechanism but I have not come across any literature that quantifies this. If anyone knows about it, please enlighten me.

Waross,

Your comments are correct. The Change over does not happen at 35% of voltage. We will provide the 35% voltage as triggering event followed by time delay (2 – 3 seconds and then closing the healthy CB. It is correct that if at 35% voltage closing is done, it may result in asynchronous closing (Different voltage and frequency) and may result in severe damage to couplings as worst case scenario. So actual sequence will be open faulty CB at 80% or line under voltage, Allow bus voltage to drop below 35%, wait for definite time delay (2-3 seconds) and then close the bus coupler.

In this entire scheme, motor contactor remains closed and motors are still de-accelerating. It is presumed that due to load inertia, motors will come not to standstill but will keep rotating during the changeover.

Thanks everyone for their inputs.

 

[che12345]

Dear Mr NCTHAI

N1. "... TF42 is Class 10 relay. This means it takes 10 Sec to trip for 600% of current.
From catalog, it takes about 3.2 sec for 9 times current".
C1. TF42 Class 10 relay complied with [IEC 60947-4-1]* per ABB. This [ ]* does NOT stated
that a Class 10 relay takes 10s to trip at 6x current. NO test is conducted at 6x current, irrespective of whether in cold or warm state.
Note: The tripping characteristic value is [not] a " spot-on " defined value at a defined
current; besides dependent on whether in cold or warn state. It is a [band] with [max
and min "check-gates"]. For easy application, the curve is usually drawn as a [single line], instead of a [band]. The [tolerance] is dependent on the multiple of the current setting.
See below C2 for detail on [max and min "check-gates"].

N2. " Relay is ... [I am not aware of difference between warm and cold status...] I have
not come across [any literature that quantifies this]....".
C2. Please refer to [ ]* sub-clause 5.7.3 characteristic values Table 2 Trip class of...
and sub-clause 8.2.1.5.1 Table 3 Limits of operation...
FYI: [ ]* is summarized as following
a) starting from cold, at 1.05x of current setting; tripping [shall not] occur in less
than 2h,
b) when the overload relay terminals have reached thermal equilibrium at the test current; current is subsequently raised to 1.2x current setting, tripping [shall] occur in less
than 2h,
c) starting from thermal equilibrium, energized at 1.5x the current setting; tripping
[shall] occur in less than 4min,
d) starting from cold state, at 7.2x the current setting, tripping [shall] occur within
4 to 10s.
Note:
e.g. 1. starting from cold state, at 1.05x current setting, tripping shall not occur in
[less than] 2h. The 2h is considered as the "min check-gate". Any values higher than 2h
are acceptable but not lower than 2h.
2. similarly, starting from thermal equilibrium, energized at 1.5x the current setting; tripping shall occur in [less than] 4min is considered as the "max check-gate". Any values lower than 4min are acceptable but not higher than 4min.
3. likewise,starting from cold state, at 7.2x the current setting, tripping shall occur [within] 4 to 10s, where 4s is the min and the 10s is the max check-gate. Any values that is higher than 4s but not exceeding 10s are acceptable.

C3. Take note that tripping time in (warm state) is always [shorter than] the (cold state). This " thermal memory " inherent feature is desirable, as it reflects the heating/cooling initial of the motor winding.

Che Kuan Yau(Singapore)