Starting Last Motor - Transformer Voltage drop

[NickParker]

I would like the experts to do a sanity check on the attached excel sheet. The transformer voltage drop % is calculated when the third motor is started with already 2 Motors running. Is the calculated %voltage drop accurate one?

 

[waross]

You are using a 10000 KVA transformer.
My local code will allow a minimum transformer size of the sum of the motor currents plus 25% of the current of the largest motor.
That would be a minimum transformer size of 6123 KVA.
My local code allows an exception to the voltage drop rules for motor starting.
If the client finds the voltage drop during the starting of the third motor unacceptable, you may consider reduced current starting of one or more motors.
The voltage drop in the feeders to the motors will experience a voltage drop when subject to 6 p.u. starting current.This will often serve to lower the starting current.
Some random notes:
If your motor feeders are lengthy, the feeder voltage drop will render your calculations conservative.
Motors are often not loaded to 100% of nameplate rating. If the running motors are at less than 100% of rated load, this will render your calculations conservative.
If this transformer is dedicated to motor loads then there should be no problem with the voltage drop when starting the third motor.
If the transformer is feeding voltage sensitive loads in addition to the motor loads, you may consider reducing the motor starting voltage drop, or, you may consider protecting the sensitive loads with a constant voltage transformer or with a UPS.
In some applications it is possible to unload the running motors prior to starting the third motor.


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

 

[nhcf]

waross-

what local codes require the transformer sizing you reference, or the voltage drop requirements? (I don't think NEC 'requires' either, even though they are prudent design consideration).

 

[waross]

I work under the Canadian Electrical Code.
I don't have a copy of the code on this computer.
Off the top of my head:
Maximum voltage drop from source to load = 5%, exception for motor starting.
Sizing for components supplying more than one motor are typically the sum of the motor currents plus 25% of the current of the largest motor, There may be an exception if the largest motor is not the last to start. I'll check when I get to my other computer.


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

 

[NickParker]

Hi Waross,

I searched this forum on this topic and found your response which is restated below, thread link is

https://www.eng-tips.com/viewthread.cfm?qid=306432

Could you provide me with an example of "rigorous" method and the arithmetic method? Electricpete had done some calculations on that old thread, but I couldn't understand it.

Below is your response from the old thread;

"The current of the running motors may be about 80% PF.
The current of the starting motor may be about 20% PF.
The total current will be much less than the sum of the running and starting currents.
Voltage drop calculations based on the arithmetic sum of the currents will be quite conservative.
The available short circuit current as determined by the impedance of the transformer is valid only for bolted fault conditions. For a rigorous calculation of voltage drop under motor starting conditions, consideration must be given to the phase angle of the current and the X:R ratio of the transformer. Again calculations based on impedance alone may be quite conservative.
Pete, how about crunching some numbers for an example:
Take 6 equal motors running loaded at 80% PF.
Take a similar motor starting at 600% current at 20% PF.
Assume a transformer equal to the KVA load of 10 motors.
Give the transformer an impedance of 6% and an X:R ratio of 6:1
What will be the voltage drop using the simple sum of the currents without regards to phase angles and the PU impedance.
What will be the actual voltage drop considering phase angles and the X:R ratio.
If you take up the challenge I will bribe you with an lps. grin"

 

[7anoter4]

In my opinion, including some 2.3% voltage drop on supply system the voltage drop at transformer terminals will be 10.4% at third motor start.

 

[7anoter4]

Actually, since the starting current will decrease proportionally with the voltage the starting current will be 901 A only and the voltage drop will be only 8.8%.Now,as the supply voltage has to be a bit above the rated [considered at secondary level] in order to produce a rated voltage at secondary terminals at full load, the difference from rated has to be different. In my opinion an accurate solution it is difficult to achieve. An error of +/-2% it will be o.k.
I use to calculate at first the motor starting current as follows:
Imst=(Vhvsys-sqrt(3)*(Ixfrat-nomot*Imrat))*(Zs+Zxfr+Zcbt))/sqrt(3)*(nomot*(Zs+Zxfr+Zcbt)+Zcbm+Zmot)
Where:
Imst=motor start current
Vhvsys=supply system[ far away] voltage=Vrated^2/Ssys[Ssys=short-circuit system power]
Ixfrat=transformer rated current [actually in your case 3*165 A]
nomot=number of starting motors [considered all motors the same]
Imrat=motor rated current [165 A for instance]
Zs=system impedance translated in secondary transformer voltage level Zs=Zsys*(Vlv/Vhv)^2
Zxfr=transformer impedance
Zcbt=transformer to switchgear cable impedance
Zcbm=switchgear to motor cable impedance
Zmot=locked rotor motor impedance=Vrated/Iratedstartmotor/sqrt(3)
I'd prefer to use complex numbers- in excel also.

 

[7anoter4]

Regarding my post of december 8, forget the system voltage drop.Since no voltage was assigned at high voltage side of the transformer the computer took 6600 V as system voltage.Then the short-circuit power was low and system impedance high so the system voltage drop.