Motor Starting

[Guardiano]

Hi,
I have a 5 MW 6,6 k V wound asynchronous motor and the supply from the utility is at 22 k V. What's the minimum rating of the transformer that is required to start this motor ? The starting time of the load is around 15 seconds.
Thanks
Guardiano

 

[LionelHutz]

For a dedicated transformer, the minimum size must power the continuous MVA and have a low enough voltage drop that the motor will start.

This is a tips site, not a free engineering site. Do you expect anyone to tell you all the data for purchasing a transformer in the 5MVA to 7.5MVA range, especially with the limited data you provided?

 

[Guardiano]

OK LionelHutz, I did not give enough informations. This 5 MW motor forms part of a plant and was only to be started after completion of a power plant project. Now, there is a need to test this motor and the utility transformer is too weak to get it started, only 2 MVA.
A basic calculation gives the following
5 MW= 5/0.85/0.92=6.4 MVA, 0.85 power factor and 0.92 efficiency
Basic rule of thumb, 2-3 times starting MVA for a secondary resistor starting, that is 13-19 MVA
Assuming 5 % percentage impedance and 10 % volt drop gives 5/10*19 that is 9.5 MVA in the worst case.
Guardiano

 

[waross]

Wound asynchronous motor. Is this a wound rotor motor or a standard induction motor?
Is the motor going to be loaded or does it just have to be run up unloaded?
If the motor is the only load on the transformer a much greater voltage drop is acceptable. The transformer itself may form part of a series resistance/impedance starting system.
When the motor is started, the overload on the transformer will cause a voltage drop. The reduced voltage will result in reduced starting current.
I suggest selecting a transformer that will meet the motors running demands. Possibly around 6 MVA. Then check the transformer's short time damage curve to see if it may safely withstand the motor starting duty.

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

 

[edison123]

You want to test the motor before project completion for what purpose? To check mechanical parameters of the motor alone?

If it is a wound induction motor, what is the rotor voltage? With wound rotors of such MW, we use the rotor winding (which is normally of much lower voltage) with normal 415 V in our shop floor via a variac with a water resistance connected to the stator to get it to full speed in motor solo condition. It may take some time but it will eventually get to full speed with a very low MVA demand from the utility. I also use parallel 415 V capacitors to reduce the steady state current to a still lower value.

Muthu

http://www.edison.co.in

 

[LionelHutz]

Knowing the type of motor makes a difference. If this is a wound rotor motor then the right secondary resistance should allow it to start on that 2MVA transformer unless you are starting it loaded. You're also assuming you need to limit the voltage drop to 10% with a dedicated transformer. I've seen dedicated transformers have a 25% drop and the motor started fine.

 

[waross]

I mostly agree with Lionel.
A wound rotor motor will probably start and run unloaded on a 2 MVA transformer. I expect that the voltage drop may be a little high.
Three issues to evaluate:
1> What is the base load on the transformer? How much of the 2 MVA capacity is available to start the motor?
2> Will other loads on the transformer be able to accommodate a large voltage drop during a planned test start of the motor?
3> the utility transformer is too weak to get it started, only 2 MVA.Will the utility's protection and control allow the motor to start without tripping?

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

 

[ScottyUK]

Bill, at full voltage a 2MVA transformer doesn't even cover the magnetising Vars for the motor. The magnetising current is sourced from the transformer even if it doesn't consume much power.

 

[waross]

I stand corrected. Thanks for the heads up Scotty.
I just did some calcs and it looks as if with an assumed PF of about 85%, the magnetizing MVARs will be about 50% of the full load MW.
Browsing some wound rotor motor specs suggests that 85% to 87% may be the best PF we can expect. The actual PF may be lower, eg: 75%

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

 

[ScottyUK]

Guardiano, a few thoughts to consider:

The absolute minimum size is that which can carry the steady-state running MVA load. The starting period of a motor isn't likely to trouble the relatively long thermal time constant of a transformer, but if the sizes of transformer and motor are close you should consider getting additional bracing on the windings.

After that the characteristics of the load come into play - something like a fan or centrifugal pump might start OK from a marginally-sized transformer with very deep voltage depression at the transformer terminals, but a recip compressor almost certainly won't. You need to establish the minimum motor terminal voltage which will allow your load to start, with due consideration to any permitted variation on the utility supply.

The transformer impedance obviously interacts with the transformer rating here - for a captive transformer dedicated to the motor you'd ideally keep the impedance fairly low to minimise volt-drop but you also need to stay within the 6.6kV switchgear capability if that is what you are proposing. An alternative would be to switch it from the high side if the upstream breaker can handle the relatively high duty cycle of motor duty, although my guess is a 22kV breaker won't be ideal from this perspective.

 

[7anoter4]

First of all I agree with Scotty. However, here are my thoughts.
In my opinion [better speaking :"my feeling is"], a 5 MW induction motor and 6.6 kV ratings has to be 3000 rpm[2 poles 50 Hz] .
In this case the moment of inertia has to be 100-150 kg.m^2.Let’s take the total moment of inertia [motor and the load] as 1500 kg.m^2.
In my opinion, if no-load torque is involved the motor velocity reaches the top in 1-2 seconds.
Let's say the starting torque could be arranged-if we choose the right rotor resistances-from 0.8 to
1.2 and an average of 1 of rated torque, if the stator is supplied with rated voltage[6.6 kV].
In my opinion also, a transformer of 2 MVA short-circuit voltage could be 6-8% and the utility short-circuit apparent power could be 250-500 MVA.
If the x/r =5 for both [transformer and system 22 kV] then total X=1.88 to 1.37 ohm and R=0.38 to 0.27 ohm[at 6.6 side].
DV=sqrt(3)*Imot*(R*cos(fi)+X*sin(fi))
The problem is the power factor. The exterior resistance will increase the power factor initially[0.9?]
as it is the large resistance inserted-even the rotor reactance is elevated too-and will be close to 0.2[?] at the start of last stage.
In this case the voltage drop will be 17-22% at the beginning and will be 30-35% at the final stage.
I think it is a chance to succeed the start.
If the no-load motor current will be not more than 30% the transformer could manage this for a while.

 

[GroovyGuy]

I would ask a transformer manufacturer wrt a recommended size of transformer, but I would guess that a 7.5MVA is the smallest standard size unit available (in NA) that would do the job. Next standard size up is 10MVA.
I would also highly recommend that you undertake a motor starting/acceleration study. With a starting time of 15s, which I will assume is at NP voltage, you must have a high inertia-load, n'est pas? With the voltage-drop thru the transformer, during a motor start, the motor's terminal voltage will be reduced. This will significantly reduce the torque capabilities of the motor, to the point where the motor cannot accelerate. You may even exceed the thermal capabilities of the motor.

BTW, the motor is a standard induction motor, correct?