severe voltage drop 1

[pumpo]

We are experiencing a severe voltage drop at motor start. The motor 500Kw 3ph 2pole. 2300V
rated. Generator, cummins 2Mw set to 460 approx, 4-1 transformer, resistive load bank between gen/trans. we get 36%Vd at motor start. estimated start current is fine it is at or just above estimated value. this seems to be the magic number around here. I am a computer guy with limited electrical so any help would be appreciated.

 

[aolalde]

Pumpo:

36% voltage drop seems to be high, but if it does not affects other loads connected to the same generator and the motor starts and reaches full speed you could live with it.

When you have limited capacity to start motors try to start first the larger motor so you will have full capacity of your MG set, after the large motor has started schedule the next larger inrush current load and so on.

 

[pumpo]

The gen is soley for the purpose of starting and running the pump nothing else is online with the pump. that is why it seems so high.

 

[Marke]

Hello pumpo

You do not mention your starting method.
Induction motors draw a very high start current when started on full voltage. This start current is primarily a function of the rotor design. The start current is independant of the load connected to the motor shaft. The start current can be reduced by reducing the start voltage. Reducing the start voltage will also reduce the start torque by the voltage reduction squared, therefore the minimum start voltage/current is a function of the load torque curve and the full voltage start curves of the motor.
It is probable that you could reduce the voltage drop by using a reduced voltage starter such as a solid state soft starter. I wold not recommend a primary resistance starter in this case.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[pumpo]

We are starting DOL. in this case. The start current fro this motor is estimated and not tested. we calculate this value from the Irms Vrms. so the start current is actually unknown at initial testing of the pump. I have been given this equation for motor start conditions
Astart/Ameasured = Vstart/Vmeasured is that correct at start only?

 

[aolalde]

That formula is correct only during the starting condition.

Are you having problems starting the motor?

Other way as this is a dedicated generator, the voltage drop will help to reduce the inrush current, lowering the kVA and kW demand at inrush condition and reducing the stress for the system, certainly the motor will take longer to accelerate but I think that is not important for a pump application.

 

[jbartos]

Suggestion: Considering the size of the motor, 500kW, it may be a good idea to simulate the motor starting before selecting hardware to reduce voltage drop during the motor starting. There is more than one solution feasible from the hardware standpoint.

 

[pumpo]

What software would you guys recomend??

 

[pumpo]

One more question i was told that if our start time for motor to reach Rpm is increased by 36% that that will be proportionaly correct to the Vd. That sounds a little too simple for me.

 

[Marke]

Hello pumpo
The start current for an induction motor started DOL is equal to the locked rotor current. This current drops gradually as the motor accelerates, and usually only falls significantly as the motor approaches full speed.
I am not sure how you estimated the start current, but typically, the locked rotor current is in the range 600% - 900% of the motor rating. The LRC is a function of the motor design.

Reducing the start voltage will reduce the start current in proportion to the voltage reduction, and will reduce the torque by the square of the voltage reduction.

The acceleration time is a function of the accelerati0on torque and the inertia of the system (motor and pump in this case)
The acceleration torque is the difference between the torque produced by the motor and the torque required by the load at that speed.
If the acceleration torque is much greater than the shaft torque, the start time will vary with the voltage variation squared. As the start torque approaches the shaft torque, the variation in time is increases.

Best regards,

Mark Empson

http://www.lmphotonics.com