380v/660v delta/star meaning 2

[abdulqadir82]

a sew motor has the following name plat e data:
V 380/660 delta/star
A 9.2/4.3
kW 4

I want to clerify that above motor should be essentially connected in delta at 380 and to star at 660?
or it can be connected to star 380v? and the current mentioned as 9.2A refers to delta connection at 380v
and 4.3 A to star or otherwise.[/sup][/sup]

 

[waross]

Delta at 380 volts.
Star at 660 Volts
9.2 Amps at 380 Volts delta connected.
4.3 Amps at 660 Volts star connected. But the ratio is incorrect. Are you sure that the nameplate is not A 9.2/5.3
That would be a correct star/delta ratio.

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

 

[abdulqadir82]

yes 9.2/5.3 A is right.

another question is that if it has been connected in star at 380v for 12 hours what impact would be occured upon the motor?

 

[LionelHutz]

If the motor been operating without a failure then it's likely still OK.

 

[crshears]

Ditto what Bill & Lionel said, although "connected in star at 380v for 12 hours" doesn't provide a complete enough picture, in my view; was the motor running idle for the entire duration, or was it loaded? If it was carrying load, how heavily was it loaded? Was the loading pulsed or continuous?

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[jraef]

When connected in Star and applied 380V, the effective voltage applied to the windings is 58% of the design voltage. The motor output torque would be severely compromised. Full Load Torque capability of the motor would be reduced accordingly to 58%. But possibly worse, PEAK torque, that used for accelerating at startup and re-accelerating after a change in load, will be curtailed by the SQUARE of the voltage change, meaning it will be cut to 33% of normal peak torque. So under the full rated load of that motor, there is a risk of it not having enough torque to accelerate and thereby stall, or if it manages to get to full speed because the load allows it (for example a centrifugal pump or unloaded compressor), once the load is fully coupled there would be a severe increase in slip, which would make it draw more current and overheat.

So as mentioned, with no load on the motor, it usually will accelerate and run, appearing normal. But as soon as you apply a load, it I'll begin to fail.


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

 

[abdulqadir82]

was the motor running idle for the entire duration, or was it loaded? If it was carrying load, how heavily was it loaded? Was the loading pulsed or continuous



the motor was run on load most of time and its protection operated at 7 A and made the motor trip two times. Before tripping when it seems to be normal it was drawing current 3A. It was loaded fully like as 4kw motor can bear load. load was continuous and the motor was run through Frequency Inverter.

 

[Marke]

Operation under load while star connected at 380 volts will result in the motor operating at increased slip and therefore increased rotor losses. The rotor could be compromised by excess heating if it happened for too long.
As Jeff mentioned earlier, the torque will be reduced and the motor may tend to stall.

Mark Empson
Advanced Motor Control Ltd

 

[abdulqadir82]

the said motor is running ok at delta 380v now. before this it was stalling at 380 star.