Star Delta start up problems ?? 1

[Pocko]

Hello all,

I'm after some opinions on the following logged readings i get when "tonging" the current drawn by an induction motor on start up. The motor is directly coupled to a centrifugal fan mounted inside the duct work of a dust extraction system.

The motor's plate rating is 72.2 A FLC. 3 phase 415 vac, 50 Hz.

The starting method utilised is Star/Delta and currently the time delay relay is set at 0.5 seconds.

As the capture method can't log in milliseconds i can't show the inrush current.

The following is data i've captured from a start up (from the red phase only) -

Log # Start Time High Units
1 10:03:36 0 A AC
2 10:03:37 0 A AC
3 10:03:38 0 A AC
4 10:03:39 115 A AC
5 10:03:40 127 A AC
6 10:03:40 128 A AC
7 10:03:41 126 A AC
8 10:03:42 124 A AC
9 10:03:43 353 A AC
10 10:03:43 384 A AC
11 10:03:44 356 A AC
12 10:03:45 130 A AC
13 10:03:45 121 A AC
14 10:03:45 66 A AC
15 10:03:46 50 A AC
16 10:03:47 49 A AC
17 10:03:48 50 A AC
18 10:03:49 50 A AC
19 10:03:50 50 A AC
20 10:03:51 50 A AC

As the voltage during the Star connected period should be 33.33% of the Delta voltage, I was expecting to attain a figure of 33.33% of the run current during the Star period of the start up, or in this case 16.66A and not in the 100's as the log shows! The figure in the 300's 5 seconds in to the start up is another concern!

As a side note in the current state if we were to set the overload using the 58% FLA rule for star/delta starting we would be in trouble, as this figure (41.76 A) would come in under our "normal" operating current and hence trip on overload.

Also as I have no access to the motor curve or the fan's inertia curve. Without this info could/should i determine the correct time for the change over from Star to Delta?

Is it possible this motors Star/Delta wiring has been hooked up to the wrong contactors?

Thanks for your time and if further info is required let me know and i'll do my best to provide it.

Cheers - Chris.

 

[jbartos]

Comments marked ///\\\:

The motor's plate rating is 72.2 A FLC. 3 phase 415 vac, 50 Hz.
///HP=?, kW=?\\The starting method utilised is Star/Delta and currently the time delay relay is set at 0.5 seconds.
///0.5s may need to be adjusted according to the motor acceleration curve for Star and Delta\\As the capture method can't log in milliseconds i can't show the inrush current.
///It may suffice for a slow accelerating motor.\\The following is data i've captured from a start up (from the red phase only) -

Log # Start Time High Units
1 10:03:36 0 A AC
2 10:03:37 0 A AC
3 10:03:38 0 A AC
4 10:03:39 115 A AC
5 10:03:40 127 A AC
6 10:03:40 128 A AC
7 10:03:41 126 A AC
8 10:03:42 124 A AC
9 10:03:43 353 A AC
///This is probably where the delta winding start is in effect.\\10 10:03:43 384 A AC
11 10:03:44 356 A AC
12 10:03:45 130 A AC
13 10:03:45 121 A AC
14 10:03:45 66 A AC
15 10:03:46 50 A AC
16 10:03:47 49 A AC
17 10:03:48 50 A AC
18 10:03:49 50 A AC
19 10:03:50 50 A AC
20 10:03:51 50 A AC
///50A indicates that the motor is not fully loaded. It would show 72.2A if loaded at reated HP or kW\\
As the voltage during the Star connected period should be 33.33% of the Delta voltage,
///Please, notice that the voltage in wye is Van=Vbn=Vcn=Vab/sqrt3=Vbc/sqrt3=Vca/sqrt3
i.e. in your case 415V/1.732=240V=415V x 0.577
57.5% not 33.3% of Delta voltage.\\ I was expecting to attain a figure of 33.33% of the run current during the Star period of the start up, or in this case 16.66A and not in the 100's as the log shows! The figure in the 300's 5 seconds in to the start up is another concern!
///Please, notice that (115A/353A) x 100%=32.6%
which is approximately 1/3 of Delta inrush current in wye connection. This is about right.\\
As a side note in the current state if we were to set the overload using the 58% FLA rule for star/delta starting we would be in trouble, as this figure (41.76 A) would come in under our "normal" operating current and hence trip on overload.
///The overload relays under the 58% rule have to be in delta winding legs where the rated current is 72.2A/sqrt3=42A.\\Also as I have no access to the motor curve or the fan's inertia curve. Without this info could/should i determine the correct time for the change over from Star to Delta?
///Agreed\\Is it possible this motors Star/Delta wiring has been hooked up to the wrong contactors?
///Probably not\\Thanks for your time and if further info is required let me know and i'll do my best to provide it.
///Please, keep on posting info as requested and as it becomes available.\\

 

[Marke]

Hello Pocko

An induction motor has a very high locked rotor current, typically 6 - 8 times the rated full load current of that motor.
When you apply full voltage starting to that motor, it will draw a very high overload current until it reaches almost full speed. The initial start current will equal the Locked Rotor Current and will slowly fall as the motor accelerates. At half speed, the start current will still be very close to locked rotor current and the current only begins to fall significantly once the motor reaches around 85% speed. The actual shape of the start current curve varies from motor to motor. The full voltage start current is independant of the shaft load. The start time is dependant on shaft load.
When the motor is connected in star, the same rules apply except that the start current is one third of the full voltage start current.
The results that you have shown agree with this.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[aolalde]

Your problem is the time required for the fan to accelerate from zero to full load speed (it will be better to provide complete nameplate data).

This motor has an inrush current of 384 amperes (very normal 5.33*72.2 Amps) when delta connected).
The inrush current will reduce to 1/3 when wye connected; 384/3= 128 Amperes.

So the expected inrush current is 128 Amperes. The initial reduction (115 Amps) should be due to the momentary line voltage drop.

After 3 seconds or 10:03:42 (not 0.5 sec) the starter change to delta connection, since the motor speed is still low the 384 Amperes inrush are demanded. Again a momentary line voltage drop reduces the inrush to 353 Amps momentarily.

The motor at full torque takes 4 more seconds (10:03:46) to accelerate to full speed and the current drops to 50 amperes.

This performance shows that the inertia of the fan and the load torque curve of the fan demand a larger motor torque to accelerate to full load speed.

The motor torque is reduced to 1/3 of the full voltage torque too, so at least it will take 12 seconds more to accelerate ( 15 seconds total) on wye connection ( if the load torque is not to close to the reduced motor torque curve).

The accelerating torque is the difference from Motor Torque minus the Load torque.

If this is a new application the inertia and load torque should be matched to the motor ratings and type of starting method.

If this application worked before a new set up of time is required. Other way start at full voltage since you are not having the benefits of a wye-start delta runs system.

 

[shape]

Just a question with regards to your problem.Do you have a damper on your ducting system this should also aid in start up current reduction

 

[jbartos]

Comment on the previous posting: The "start-up current" will be divided into two time intervals:
1. Damper closed. The motor will start with the regular inrush current; however, it will have its steady state current leveled off at lower current than needed for the damper open motor operation.
2. Damper opens. The motor will add more HP on its output to the load. It will draw more current. There may be an abrupt increase of the motor line current depending how fast the damper is opened (there may be another current transient). Then, the current will level off for the design motor-load conditions.

 

[jraef]

Pocko
Just to offer some quicker clarification to what is being said above, I think your initial mistake was in this part of your post:

"I was expecting to attain a figure of 33.33% of the run current during the Star period of the start up, or in this case 16.66A ..."

You were thinking that the Star current would be 33% of the RUNNING current. It is 33% of the STARTING current, which is as much as 200% of Full Load current, which in this case is even higher than you Running current.

"Venditori de oleum-vipera non vigere excordis populi"

 

[jraef]

Oops, I meant 600%, not 200%. I was thinking torque when I was typing... my bad.

"Venditori de oleum-vipera non vigere excordis populi"