MV SOFT STARTER HARMONICS LIMITS AND CAPTIVE TRANSFORMER

[Parkpower]

Hi everyone,

We will install solid state soft starter to drive 2MW 6KV motors. The soft starter will fed by 13.8kV switchgear via 13.8/6kV captive transformers.

The solid state reduced voltage starter switches on-off during starting and causes harmonic component in the network. I know VFD generally needs input/output filter to limit THD. However, I do not find any filter equipped in the catalogue of soft starter manufacturer.

I wonder how we can control input/output THD of soft starter? If we do not limit THD, shall the captive transformer be designed specially to withstands harmonics ?



Thanks,

Parkpower

 

[burnt2x]

You just answered your question with the last statement. You need to choose a bigger supply transformer or specify K-rated transformers to handle your high THD.

 

[LionelHutz]

A soft-starter will produce harmonics for the 10-30 seconds that the motor is accelerating. The filter would be very large and expensive to filter these harmonics.

The transformer will not require any special considerations because of the harmonics. The duration of the start is short enough that the harmonics won't have much of a heating effect. You need to be more concerned about the overload MVA draw during the start and excessive voltage drop.

 

[dpc]

Agree with LionelHutz. The harmonics only occur during starting (and perhaps stopping). I would not be concerned about the captive transformer.

Are you getting a bypass contactor?

"An 'expert' is someone who has made every possible mistake in a very narrow field of study." -- Edward Teller

 

[Parkpower]

Thanks everyone for prompt reply.

Soft starter is c/w bypass contactor. Max starting time is 40 seconds.

K-rated transformer is expensive, and order of harmonics components is not easy to decided to choose K-factor. So I tend to specify a captive transformer.

The captive transformer is rated as 5MVA, 2 times the rating of motor. The normal starting current limit is 30% FLA of motor.

Does any one has experience under similar conditions? Is the rating of transformer good or not?

Parkpower

 

[burnt2x]

There maybe a typo here. Is it really 30% of motor FLA or 30% of normal starting current?
30% of normal starting current is basically a reduced starting voltage of around 55%.
There are more data needed to compute for the transformer size. What is the locked-rotor code of your 2MW motor?

Use this formula:
trafo kVA = motor starting kVA; in your case:
Motor Starting kVA = 55% of motor kVA x locked-rotor factor
For example: NEMA locked-rotor Code E = 4.8 times motor FLA
trafo kVA = 0.55 X 2000 MW/0.8pf X 4.8 = 6,600 kVA = 6.6 MVA.
Therefore, you 5 MVA is too small. 5 MVA is good if your motor is NEMA Code B (3.55 time FLA)

 

[burnt2x]

Parkpower,
Opps!
Talking about errors, pls disregard earlier post:
The calculation should have been:
Code E = 4.8 kVA per HP so;
Motor starting kVA = 0.55 X 2,000,000/746 X 4.8 = 7.078 MVA; say 7.5 MVA for a motor with NEMA Code letter E. Still, 5 MVA is ok if your motor is NEMA Code B.
Hope this helps

 

[jraef]

The captive transformer is rated as 5MVA, 2 times the rating of motor. The normal starting current limit is 30% FLA of motor.

My 2 cents worth on the two issues raised in this statement:


1) 2 times the motor power is typically within the minimum acceptable transformer size range for starting with a soft starter, but this is by no means a comprehensive study, just a "rule of thumb" (educated guess backed by empirical experience).

2) 30% FLA as a current limit setting will never start that motor. I have never personally seen a motor (coupled to a load) start with a soft start with anything less than 200% FLA, and that was an extremely light load. If you assume the motor LRC is 600%, the 200% would represent 33.3% of the LRC, so even if you made that mistake, it will not likely be enough. Has anyone done a thorough evaluation of the application to see if the soft starter is even a viable solution?

By the way, this is all assuming that you are referring to a standard phase-angle fired thyristor soft starter. If you are using an LCI inverter (which many people refer to as a "soft starter" for very large motors such as yours), that's a different story altogether.


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[Parkpower]

Hi,

I made mistake in reply. The starting current limit is 300% FLA, not 30%. Motor locked rotor factor is 6 times FLA. Soft starter is the type of phase-angle fired thyristor.

Then my transformer size is :
Starting voltage is about 70% rated voltage.
Starting kva = 70% . 2MW / 0.9 (PF)/0.95(EFFICIENCY)* 6 = 10MVA

Is it right?

But generally size of captive transformer for direct on line starting motors is only 2 times the size of motors and starting current is 6 times FLA. According to the above calculation, we need a 15MVA captive transformer not a 5MVA one for the 2MW motors DOL starting.

Can somenbody advise related document to refer to?

Parkpower

 

[burnt2x]

Park,
If you want, please refer to Standard Handbook for Electrical Engineers by Fink and Beaty, p. 20-48 (12th Edition):

          1.732 X locked-rotor current X line volts
kVA/HP = -------------------------------------------
                    HP X 1000

Using your data, and solving for kVA/HP, I arrived at a value of 5.233 kVA/HP (FLA =225 amps; LRA = 1350 amps; PF = 0.9, eff = 0.95).
Solving for starting kVA at 70% voltage:
Starting kVA = 0.7 X (2,000,000/746)HP X 5.23 kVA/HP
= 9820 kVA = say 10MVA