WHAT DETERMINES STARTING CURRENT

[JamesDUK]

I have a 4.5MW 11kV 3ph 50HZ squirrel cage electric motor driving a gas compressor on an offshore platform. If the onboard GT is down the compressor will be run from a sub sea cable, I am told we cannot start the motor with the SS cable because the starting current is too high. Our specification called for a start current of 450%FLC.
Is the starting current on the motor influenced by the load it sees, i.e. if the compressors is depressurised for a re-start will the unit be more likely to start. Requests to vendors and in house electrical engineers have produced the following: The start current is given as a percentage of FLC and is related to the design power of the motor, when the power is determined the start current cannot be changed. One criteria to determine the motor power is the start-up torque on the shaft. This implies to me that the motor is sized to start with the compressor pressurised or not and once that decision has been made it the start current won't change. Am I right?

 

[Alex68]

The starting current of a motor is also called "locked rotor current" and is absolutely indipendent from the mechanical load. It only depends on the motor design and usually is about 5 times the FLC.

This big current causes a voltage drop on the cable. So the voltage at the motor terminals is reduced.

The motor torque is proportional to the squared terminal voltage. So a vltg decrease of 10% produces a torque decrease of 20%! If the voltage drop is too big, the motor could stall.

Concerning the cable, the starting current is more or less constant during the starting period, which usually lasts some secs for a compressor. Probably the starting current will not significantly heat the cable, but it must be verified.

 

[RRaghunath]

jamesduk,

I can add the following to what Alex68 has said above.

The starting current is not affected by the load connected to the motor shaft but the starting duration is certainly affected.

To illustrate the point, if the compressor is depressurised, the torque demand on the motor is less and thus the accelerating torque available is correspondingly greater resulting in motor accelerating to full speed in a shorter time.

It can also be said that the torque being proportional to Square of V, the chances of motor accelerating to full speed (so to say successful start), even with higher voltage drop, are better if the motor is started with compressor depressurised than otherwise. This is because at lower terminal voltage, though the available torque from the motor is less it may still be adequate for a successful start.

 

[Marke]

The Full voltage start current and start torque are a function primarily of the rotor characteristics. A shallow bar, low resistance rotor (generally refered to as a design A rotor) exhibits a high Locked Rotor current and a low locked rotor torque. A deeper bar with higher resistance (Desing B rotor), exhibits a reduced Locked Rotor Current and an increased Locked Rotor Torque. i.e. is much more efficient at start. The down side to this is that a Design A rotor has a higher efficiency under running conditions. There are many different variations of rotor design and it is possible to achieve an improvement i both starting conditions and running conditions with the correct rotor design.
Under full voltage starting, the current is independant of the load. The start current is initially equal to Locked Rotor Current and remains high until the motor is at 85 - 90% full speed when the current begins to fall quickly.
Reducing the voltage during start with reduce the current (proportional to voltage reduction) and the torque (proportional to voltage reduction squared). The driven load determines the minimum starting torque required, and the rotor design determines the minimum current required to achieve that torque. see

http://www.lmphotonics.com/m_control.htm

and

http://www.lmphotonics.com/m_start.htm

Best regards, Mark Empson

http://www.lmphotonics.com

 

[JamesDUK]

Thank you all, between you covered my queries. It has been very useful background to the problem and put me in a much stronger position.

 

[jbartos]

Suggestion: Normally, there is a clear distinction on the induction motors complying with Nema standards in terms of Nema Code letter, and Nema Design Letter. Visit

http://www.joliet-equipment.com/glossary.htm

for definitions.

 

[swgrmfg]

Adding to Marke's answer, a major component in the starting current is the air gap between the stator and the rotor. One can actually bring down the starting current to as low as 2.5PU with a larger air gap. Of course this requires careful consideration when calculating the accelerating torque of the load (motor torque minus load torque) and the acceleration time (for relay settings).

 

[electricpete]

That's interesting. I guess ny a simple linear model, the larger air gap would increase leakage reactance (good for reducing starting current), at the expense of lower magnetizing reactance (bad for running power factor and efficiency).

I remember reading that motors are designed to push leakage reactances into saturation during starting. I never understood what that meant. Wouldn't a saturated leakage reactance allow a higher current than an unsaturated leakage reactance?

 

[edison123]

Increasing the air gap will definitely increase the no load current (stable current). Will it not also increase flux density which in turn will aid in a higher starting torque ? Regarding the reduction in starting current, I am not so sure but starting duration will be reduced due to higher starting torque with higher flux density.

 

[electricpete]

Thanks Edison. I agree increasing airgap results in increased exciting current. I don't think that increasing airgap results in increased flux density.

 

[jbartos]

Suggestion: Dual squirrel cages as reducers of the squirrel cage induction motor starting current and locked rotor current should not be overlooked.