How to sizing an starter autotransformer

[ricartengenheiro]

I am trying to use an AutoTransformer to start a 5MW 13.8 kV squirel cage induction motor. The FLA of this motor is 242A, the LRC is 1452A, power factory ranges from 0.1 until 0.88. This motor has 800 kgm? of moment of inertia (rotor + load) and its stalling time is 11s @ full nominal voltage. I've made some calculations and considering three starts per hour (one each 20 minutes) I found 40 seconds to start this motor and 1,4 MVA necessary in autotransformer using 50% of full nominal voltage. However I don't really know if my calculations are correct. Could anyone plese explain how to calculate the nominal power of this autotransformer ?

 

[bob]

http://www.google.com/search?source...93&q=How+to+sizing+an+starter+autotransformer

 

[Jraef]

On something this big, you don't want to guess.

The best way to do this is to hire the services of someone who has access to software that will do what is called a "Transient Motor Starting" analysis. The two most popular software packages worldwide are SKM Power Tools and ETAP. These software packages are VERY expensive, i.e. $15,000US, so it is not worth buying it just for one project. But many professional engineers will have this and will do the calculations for you for a much more reasonable fee. I'm sure if you contact whomever sells these software packages in Brazil, they will know who has it and will do the TMS analysis for you.

http://www.skm.com/distributors.html

http://etap.com/company/etap-reps.htm#

 

[ricartengenheiro]

Here is what I'm thinking about it.

Considering side (1) as line connection and side (2) as load connection, we know that:

S_t = I₁(V₁-V₂) (1)
(N₁/N₂)=(I₂/I₁)=(V₁/V₂) (2)

Where S_t is the power of the autotransformer. Combining (1) and (2):

S_t = (I₂.V₂/V₁).(V1-V2) = S_L.(V₁-V₂)/V₁ (3)

Where S_L is the load power.

Since the relation (LRC/FLA) is 5 and considering 50% tap:

V₂=V_n/2
I₂ = (I_n/2).(5)

where V_n and I_n is the nominal voltage and current of the motor. So

S_L = V₂.I₂ = (V_n/2).(I_n/2).(5) = (5/4).S_n (4)

From equation (3):

S_t = S_L.(1/2) (5)

where S_n is the nominal motor power. Combining (4) and (5):

S_t = (5/8).S_n

So, considering 50% of full nominal voltage, the autotransformer shall have 62,5% of the nominal motor power.

Am I right ?

Regards.

 

[Phil Corso]

Ricartengenheiro...

Your approach contains several errors: 1) the Start to FLA Current ratio is 6 not 5; 2) your derivation of the short-time equivalent of the AT?s capacity fails to consider a basic characteristic of electrical apparatus, i.e., Root-Mean-Square; and 3) the #/sequence of ?starts? proposed! I suggest the following:

A) Determine the Continuous-Duty capacity of the AT.

B) Determine the RMS value of the starting-current profile. I would use a the profile represented by the positive-quadrant of a circle, with the y-axis representing starting-current, and the x-axis representing starting-time(s) plus cool-down!

C) For a motor of the size cited NEMA does not recommend 3-starts per hour, but, instead, two successive starts, including an appropriate cool-down or coast-down increment between the first and second starts!

I haven't questioned your intention(s!) And I stress this approach is approximate but should put you in the ball-park. However as my caveat, I'm reminded of Mark Twain's reply to the question, "Do you know where the Mississippi's sand-bars are?" His reply was, " No, but I know where they aint!"

Regards, Phil Corso

 

[Phil Corso]

"R"

Correction: 1st paragraph, penultimate sentance, after word "proposed" insert "is questionable"!

Phil

 

[Besoeker]

I am trying to use an AutoTransformer to start a 5MW 13.8 kV squirel cage induction motor. The FLA of this motor is 242A, the LRC is 1452A, power factory ranges from 0.1 until 0.88. This motor has 800 kgm? of moment of inertia (rotor + load) and its stalling time is 11s @ full nominal voltage. I've made some calculations and considering three starts per hour (one each 20 minutes) I found 40 seconds to start this motor and 1,4 MVA necessary in autotransformer using 50% of full nominal voltage. However I don't really know if my calculations are correct. Could anyone plese explain how to calculate the nominal power of this autotransformer ?

A 50% tap will likely not give you enough torque. Bear in mind the the torque developed is approximately proportional to the square of the voltage. So, for half the voltage you get a quarter of the torque all other things being equal.

ATX starters are not something I get into in a big way. The last, at 6.6MW, was a bit bigger than yours. It ended up on the 80% tap.
On Phil's point about the ratio of start current to FLC, in that particular case it was 6.5:1 at rated voltage. But I have seen it range from just over five times to seven times.
You need the data for your motor.