Motor starting

[bob]

Hi,

 

We are in the process of installing one 3 MW , 11 k V slip ring motor. However, our 11 k V network will be ready only next year which means that we will be starting th motor now on our existing 6,6 k V network. Obviously, we have ordered the motor for start connection at 11 k V and delta connected at 6.6 k V. We also know that at 6,6 k V , only one third that is 1 MW will be available at the output shaft to drive the load.

As far as the network is concerned , we know that it is capable of withstanding the starting current of a 1.5 MW motor.

Now my worries are :

1. Will the starting current be a function of a 1.0 MW motor or a 3.0 MW motor bearing in mind that the impedance of the motor is designed for a 3.0 MW application.

2. The inertia of the load is unchanged , I mean that the load is designed for a 3.0 MW application but in the mean time we will only be running at below one third of the design load.

I am concerned whether the motor will be capable of starting the load. According to calculation , it seems to be but some people are arguing that according to their experience it will be very harsh to start the load.

Any comment welcomed.

 

Bob

[marke]

Hi Bob

 

If your motor is designed to be star connected at 11KV, then the voltage across each winding is 6.35KV.

 

If you now delta connect the motor and connect it to 6.35KV, the flux in the iron would be the same as the voltage across the windings will be the same. This means that you do not derate the motor at all, it is still a 3MW motor operating on a lower voltage.

 

If you leave the motor connected in star and reduce the voltage to 6.6KV, then the effective motor size is reduced to one third, i.e. a 1MW motor.

 

On 6.6KV, if the motor is connected in star, it will start like a 1MW motor, but if connected in delta, it will start like a 3MW motor.

 

As the motor is a slip ring motor, you can customise the start characterisitics to suit your network

 

If you need more than 1MW and you network on 6.6KV can withstand higher than 1MW, you should be able to connect the motor in delta and select/configure the secondary resistors to limit the current to the network ratings.

 

The rotor characteristics, (open circuit voltage and short circuit current) are a function of the stator flux. Therefore if you design your secondary resistances for the 11KV star connected installation, you will get the same performance with the 6.6KV delta connected installation.

In the 6.6KV star connected case, the open circuit voltage will reduce by rt 3 as will the short circuit current. In theory, this will not affect the resistor selection, but for a given resistance the toque will be one third and the power dissipation will also be one third. The lower flux density will increase the leakage reactance and I would expect that this will cause a shift in the slip for peak torque for any given rotor resistance.

 

What is the driven load? This will determine the best starting characteristics that you can achieve.

 

Best regards,

[Guest]

Hi Marke ,

 

A STAR for you. The moment of inertia of the load is 4000 kg M2.

 

Regards.

 

Bob

[marke]

Hi Bob

 

The moment of inertia will determine the starting time. In order to determine the start current, we need to have an understanding of the start torque. What sort o machine is the motor driving?

 

Best regards,

[Subhashish]

Hi bob

 

The MI and the accelerating torque shall decide the starting time .

Can try plotting motor speed torque characteristic at particular

voltage level and along with the load torque speed characteristic .Adeqate accelerating gap is necessary to prevent prolonged starting time .

 

regards

[Guest]

how to calculate starting time of motor

[marke]

You need to know the start torque curves of the motor and the driven load, plus the inertia and speed of the driven load.

 

You can then take the difference between the motor torque and driven load torque at all speeds from zero to full speed to develop an acceleration torque curve.

Applying this acceleration torque to the load inertia will give you the rate of acceleration at that speed. This can be integrated to get the toatl acceleration time.

 

Download the Electrical Calculations software from http://www.LMPhotonics.com/busbar32.zip and you can use this software to do the calculations for you.

 

Best regards,

[bob]

Hi Marke ,

 

My concern is that at reduced voltage 6.3 k V , in star connection , would the torque be sufficient to start the load ? The speed of rotation is 750 rpm and the load is a sort of rock crusher.

W are have requested speed and torque curves from the load manufacturer.

 

Regards.

 

Bob

[marke]

Hello Bob

 

Rock Crushers are generally an inertial start rather than a high torque start except for machines such as hammer mills which can be a bit nasty.

If the crusher is required to start full, there could also be problems, so it does depend on the type of crusher and starting load.

 

You are using a slip ring motor, so you can get around 200% torque from zero speed to full speed if you size the start resistors correctly and set the timers to operate in reasonable times.

If ou leave the motor in star at 6.5KV, then you are gettin 200% of 1MW torque which is the equivilent of around 67% torque of the fully rated machine.

My biggest concern would be the start time rather than the start current except where the machine is stopped with load still on it. Fortunately, with a slip ring motor, the slip power during start is dissipated in the rotor resistors rather than in the motor. Provided that the resistors will withstand the inertia at the 11KV setup, they will also withstand the inertia and the 6.3 KV setup. The only issue may be that the values may not be optimum in both setups.

 

If the crusher starts as an inertial load only, I believe that there should be no problems starting in star on 6.3KV.

 

I am interested in any comments that Jraef may have.

 

Best regards,

[Guest]

Mark

I have supplied many liquid starters for various types of crushers, ranging min power from 40 HP to several Megawatts and have found that 1.5 - 2 times FLT is more than adequate.

I have some torque/sped curves if anybody is interested.

Best Regards

Fintan Lawlor

[marke]

Hi Fintan

 

Yes, I agree, provided that you keep the motor operating in the low slip region of the curve (RHS of the torque peak), then you can develop 150% torque at not much more than 150% current. If you operate on the wrong side of the curve, the current will be much higher for the same torque. This requires that your first step sets the maximum torque below zero speed.

 

Best regards,

[Guest]

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