Voltage imbalance calculations.

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PsychicViking

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Does anyone know how the utility calculates voltage imbalance. I can find lots of different formulas, but want to know which one to use when I send a report to the local utility. My boss is using:


% voltage imbalance = (maximum imbalance)/(average voltage)*100

I think I read there were two formulas mentioned in ANSI c84.1. Two formulas does not help. I need one.
 
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Mayimbe

Senior Member
Location
Horsham, UK
take a meassure of the voltage for a constant period of time, say an hour, plot it, and plot the voltage that should have in the same graph.

that is a very good report by me

that would no leave question about your unbalance problem to the local utility.

i think that your formula should be this way:

% voltage imbalance = (average of the unbalance voltage)/(voltage that you should expect)*100
 
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Mayimbe

Senior Member
Location
Horsham, UK
taking a meassure every hour for a period of time (say 2 or 3 workable days, theres equipment that does this kind of things, dont remember the name but i have used them before). sum all of it then divide it by numbers of meassures. or plot it.

i do would change the name "(average of the unbalance voltage)" for "(average operation voltage)", i was victim of my own "low budget english".

if it is bigger than 5 % is bad or whatever the criteria that you use.
 
mivey

mivey

Senior Member
Mayimbe said:
i think that your formula should be this way:

% voltage imbalance = (average of the unbalance voltage)/(voltage that you should expect)*100
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No.

The correct ANSI formula (C84.1-2006-C.3) is:

% Voltage unbalance = 100 * (max deviation from Mean of {Vab, Vbc, Vca}) / Mean of {Vab, Vbc, Vca}
where
Mean = (|Vab|+|Vbc|+|Vca|)/3
and the deviations are given by:
|Vab - Mean|
|Vbc - Mean|
|Vca - Mean|

If a utility is following ANSI (C84.1-2006-C.2), the maximum unbalance is 3% at the service point with no load. Most utilities I am familiar with try to stay at 2% or less and are usually below 1%.

NEMA does not consider less than 1% unbalance to be an unusual service condition for a motor and you would normally not have to derate the motor (actually about 99% of rated HP at a 1% unbalance). Operation above a 5% unbalance is not recommended.

FWIW, the IEC formula is (not really applicable in the US):
Negative sequence voltage unbalance = V2/V1 = sqrt{[1-sqrt(3-6*B)] / [1+sqrt(3-6*B)]}
where B = (Vab^4+Vbc^4+Vca^4) / (Vab^2+Vbc^2+Vca^2)
 
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Mayimbe

Senior Member
Location
Horsham, UK
agree with you mivey. but that doesnt mean that my method is wrong. in fact i bet my method is more accurate that the method that you show.

because the Vab, Vcb and Vca that you use are puntual or instantaneous data. my data is stadistic. my data have less error than yours. and i can prove it :D

and what if is a single phase Voltage? V-N? or the client dont need or want to know whats happening in the other phases but just the one that he need?
 
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Cold Fusion

Senior Member
Location
way north
Mayimbe said:
agree with you mivey. but that doesnt mean that my method is wrong. in fact i bet my method is more accurate that the method that you show. ---
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May -
If you are trying to describe voltage imbalance, your method doesn't do that.. So, the accuracy may not matter much. :-?

cf
 
mivey

mivey

Senior Member
Mayimbe said:
agree with you mivey. but that doesnt mean that my method is wrong. in fact i bet my method is more accurate that the method that you show.
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I think you would lose that bet
Mayimbe said:
because the Vab, Vcb and Vca that you use are puntual or instantaneous data.
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says who?
Mayimbe said:
my data have less error than yours. and i can prove it
Click to expand...
I would love to see the proof.:)
Mayimbe said:
and what if is a single phase Voltage? V-N? or the client dont need or want to know whats happening in the other phases but just the one that he need?
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You do know that unbalance is referring to the unbalance between phases, right?:D
 
mivey

mivey

Senior Member
Mayimbe said:
is unbalance? or imbalance? :S

is there a diference? what does my method does then?
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Use unbalance.

I'm not sure what you are doing as you have some undefined terms in your equations. Your proof would clear that up.
 
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Mayimbe

Senior Member
Location
Horsham, UK
i cant prove nothing if i dont know how are you getting the values of Vca, Vab and Vbc (or how the op will obtain this data).

what i was trying to say about the Vn voltage, is that if i buy a single phase output service for V= 1 p.u, and i have a V = 0.8 p.u something is wrong with the service, (it can be an unbalance problem or a Vd problem, or neutral problem), i think that was the issue, how to prove that you have a bad voltage service.

excuse me if im wrong.

didnt mean to offend you mivey.
 
mivey

mivey

Senior Member
Mayimbe said:
i cant prove nothing if i dont know how are you getting the values of Vca, Vab and Vbc (or how the op will obtain this data).

what i was trying to say about the Vn voltage, is that if i buy a single phase output service for V= 1 p.u, and i have a V = 0.8 p.u something is wrong with the service, (it can be an unbalance problem or a Vd problem, or neutral problem), i think that was the issue, how to prove that you have a bad voltage service.
Click to expand...
Sounds like you are trying to address the voltage rating and operating limits, not the voltage unbalance. The voltage ratings and acceptable voltage ranges are the main thrust of ANSI C84.1.
 
mivey

mivey

Senior Member
Mayimbe said:
i cant prove nothing if i dont know how are you getting the values of Vca, Vab and Vbc (or how the op will obtain this data).
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BTW, Using the average of the individual readings over an hour may not help you. The average voltage unbalance over an hour might not be too bad, but the motor/chiller/etc can trip long before that.

How I get the values depends on the situation but I use a snapshot of the three phase meter readings if I have my three phase meter with me. This is not quite "instantaneous" as you would need to capture the waveform to get that because the meter reading is damped. I usually take several readings. The preferred reading, of course, is when there is an anomaly.

A single phase meter will work fine if the voltage is fairly stable. If it is moving, several readings may be required to get an average reading, but I would normally take the set of readings in less than a minute.
 
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