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A Calc method to determine Motor Starting Vdips on Genset systems? 2

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majesus

Electrical
Aug 16, 2007
262
I'm working on a static motor startup problem of a 400HP 575V induction motor. It's power source is four 4.16kV 1100 kW (prime power) parallel gensets (NO utility.) The motor isn't connected directly to the Genset bus, but say 600m via an XFMR and power cables.

I was able to model it perfectly in Etap, and it spitted out an answer. However, I'm curious... can I calculate it by hand?

I did a google search, looked in a bunch of different technical books, read this forum's FAQ on sizing generators and I on went to the IEEE redbook and looked at their graphs on "Minimum generator Voltage due to full voltage starting of a motor." That's fine, but I would like to be able to solve it by myself. I can't seem to find an resource in doing so. Anyone know where I can read up on it, to learn methods for solving it?

Thanks,
Maj
 
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If you just want to do a snapshot calculation at locked rotor conditions, it is really just like any other voltage drop calculation. Convert the motor locked rotor current to an equivalent impedance at a power factor of about 0.2.

Then just draw up the impedance diagram including the source, transformer, cables, etc, and reduce all the series impedances except the motor impedance to a single impedance.

Like any other voltage drop problem, to get an exact answer, you will have to iterate, since the voltage drop depends on the current which depends on the voltage which depends on the voltage drop, etc.

This will give the voltage drop based on symmetrical current, ignoring the very brief asymmetrical inrush current.
 
Hey DPC.
Thanks,

I was trying to do that, I was using R=4.167% pu and X=25% which are typical values for induction motors for Short CCT calcs... But in this case, it was giving me a HUGE Vdrop. I think I've got the wrong base or the wrong Motor impedance values, I'm using the XFMR as my base KVA and the current throught the motor as 6X FLA @ 20% PF.
 
I believe the key work as you stated is:

"Convert the motor locked rotor current to an equivalent impedance at a power factor of about 0.2." So don't use the motor's Xsc and X/R ratio (Xsc/Ra)?
 
Please see article "Equations help determine largest motors that can be connected on utility lines" by D.A. Lentz in Plant Engineering, Sept 1965, p.47.

%V dip =LRM x 100/F,

Where:
LRM= locked rotor MVA at point on the system where motor will be located.
F=maximum three-phase fault in MVA at any point on the system where voltage dip is to be determined.

I have a copy of this article but I do not know how to insert a PDF file into the eng-tips message.
 
Awesome Rob46. Seems to be what I'm looking for :)
Thank you! I owe you a beer. ;)

I was having troubles getting the correct Xpu and Rpu for the system. My Vdrops were way off. I was not getting anything close.

I don't think I was choosing the correct ones for the the motor or the generator. I'll look at it tomorrow and study the method and report back.

Maj
 
If you follow the suggestion of dpc, you might avoid the iterations by using the (well known?) equivalent circuit of the motor instead of "the equivalent impedance at a power factor of about 0.2". Set the slip to one, or almost one in the equivalent circuit. The values of the impedances of the equivalent circuit can be calculated from the "name plate" parameters using the program motdatpr that can be downloaded at
 
Thank you guys, as DPC said it: "Convert the motor locked rotor current to an equivalent impedance at a power factor of about 0.2." However, my method to determine the motor's impedance was waaaaaaaaaay off.


Once I followed rob46's paper it set me straight on how to do that. Makes sense, and easy to do once I learned it.
I then built myself a nice little network with a synch Generator, step up XFMR, transmission line, step down XFMR, one running motor and one start up motor. I determined the impedance values for each component, combined into a simplified equivalent circuit, did a voltage calc drop and got within 2% of Etap's calculations on the motor's terminal.

Sweet. I understand.
Thank you :)

Thanks for pointing me in the right direction.
I appreciate the help as always.
 
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