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Short Circuit MVA vs. Voltage Drop

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raisinbran

Electrical
Sep 27, 2004
67
Gentlemen,
There should be a relationship between short circuit MVA at any point in a distribution system and the voltage drop that occurs when a motor is started at that same point in the system. Is there a straight forward relationship between the two or does the voltage drop also depend heavily on the system X/R ratio?
Thanks in advance,
Raisinbran
 
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Yes, the impedance is impedance and if it stays in the circuit it has some effect. V=IZ=I(R+jX) apply all the time.

Short circuit MVA helps you determine the source impedance, which is part of the total impedance, along with the supply feeder impedance up to the load terminals while considering the VD. How significant could the effect be? It depends on the magnitude of the current and the power factor and other values involved.










Rafiq Bulsara
 
If you have local generation and the volt-drop (or some fraction of it) occurs at the generator terminals then you may see some response from the AVR as it tries to restore the machine voltage. Depends a little on the operating mode of the AVR. Your model will end up somewhat more complex if you need to factor in local generation.


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If we learn from our mistakes I'm getting a great education!
 
The short circuit current of a circuit depends on the impedance of the circuit. For a transformer with a short on the secondary terminals, the impedance and current depend on both the inductive reactance and the resistance of the transformer. From these values, the X/R ratio may also be determined.
For a short circuit away from the transformer, the circuit impedance is now determined by the vector sum of
A> The sum of the transformer and conductor resistances and B> The sum of the transformer and conductor inductive reactances.
The X/R ratio of the transformer does not apply to a short circuit at a distance from the transformer.
As Scotty says, it gets complicated for a generator. Add the speed stability of the prime mover to the action of the AVR. Then consider whether the machine is self excited, self excited with a current boost module, PMG excited or excited from a source separate to both the electrical output and the prime mover of the gen set.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
If you use the MVA method, the voltage drop = load MVA / (load MVA + source MVA).

In this method, series and parallel MVAs are combined using the same formulas as capacitors. In the case of motor starting, sources use the SC MVA and the motor uses the starting MVA.
 
mivey
If you compare your solutions to rigorous solutions you may find errors. MVAs are not all created equal as your method assumes. You are ignoring the power factors of the MVAs.
Compare the % regulation of a transformer to the % impedance rating to see how MVAs are not equal.
I learned that on this forum several years ago. Was that you that helped me Rafiq?
Or it may have been Lionel or DickDV.
At full load and unity power factor the calculated voltage drop based on the transformer impedance will be more than the actual voltage drop which will be based on the transformer regulation.
Try replacing >- load MVA
With >-- load MW + load MVAR
and replace
(load MVA + source MVA)
With >-- (load MW + load MVAR + source MW + source MVAR)

Did I get that right E-Pete?

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

I use the MVA for a quick check of volt drop & fault, not for a detailed analysis. I use an impedance calc for the details. I have noticed some difference in the results but nothing earth shattering.

I'll add the power factor to my MVA quick calcs and see if the results get appreciably closer to the impedance calcs.
 
Usually at full load and below the load power factor will dominate. At high power factors the transformer reguation may dominate.
At short circuit conditions the transformer X + R comprises the total impedance and is described by the percent impedance.
Check a few transformer specs and compare percent impedance and regulation figures.

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

Me? I do not recall. I generally do not believe in short cuts, I only understand V=IZ.

I did not check validity of mivey's MVA method, but I would assume that ignores R and one has to convert the nameplate MVAs to a given base value.

Rafiq Bulsara
 
The lesson that was shared with me was the difference between impedance voltage and regulation, and the error of using transformer impedance to calculate voltage drop at normal load conditions and power factors.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
The R is not ignored and there are no base conversions.

Here are the formulas to get the MVA values needed for a motor start check:

System: MVA = MVA_base / Z_system_pu
Line: MVA = kV^2 / Z_?
Xfmr: MVA = MVA_rated / Z_pu
Motor: MVA = kVA / X"d
 
The X/R ratio is ignored. Therein is the error.

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