Worst-Case Scenario for Asymmetrical Current

This might help...

http://peguru.com/2011/06/what-do-symmetrical-asymmetrical-momentary-interrupting-close-latch-ratings-mean/

You need to talk with your 1st level professor. There is something you are missing. Really.

Looking at the graph, the DC current is maximum at the origin and the symmetrical current at this time is ~-0.85. I cannot understand your statement about assymetric current curve on the graph. Please clarify.

Looking at the graph you provided, the symmetrical current is at maximum peak value of 1 p.u. at approximately the 0.385 cycles time period and the asymmetrical current waveforms are at maximum peak value of 1.25 p.u. at approximately the 0.375 cycles time period.

The time period you have indicated with the green arrow at .5 cycles indicates the symmetrical current maximum value is at 0.87 p.u. and the asymmetrical current value is at approximately 1.1 p.u..

The graph clearly illustrates both symmetrical and asymmetrical current values peaked well before the 0.5 cycles time period and have begun to decrease rapidly.

These facts render your argument incorrect.

The next time period wherein the symmetrical and asymmetrical current reach maximum values is in the negative voltage region time period of 0.75-0.85 cycles however the maximum value for both are less than 1 p.u.

You very much need to review interpretation of graphs and master the procedure if you want to be successful in this field of study.

Thank you all for your help. We know, Asymmetrical Current = Symmetrical Current + DC offset. Now the higher the value of Symmetrical Current, the higher the value of Asymmetrical current. When we do arc flash study, we try to assume the worst case scenario for asymmetrical current, which happens to be at voltage zero and symmetrical current at negative peak (according to multiple souces). My questioin is, why don't we consider the very next peak when the voltage is zero but the symmetrical current is at it's positive peak? Shouldn't we get a higher asymmetrical current for the latter case?

I think I am overlooking something. If someone can please point out what, I'd be grateful.

Dear Sadia27,

I guess you are the person who made the "Original Post" question as "Anonymous", so welcome to CR4.

I have a feeling that when you look at the graph, your mind tells you the "DC current" is asymmetric.

Well, it is, since it is positive polarity only - the other curves are cyclic and swing positive and negative!

Confusingly, the actual current (from the mathematical transient solution, it is the sum of the steady-state sinusoid and the exponential current which is positive only - on your example graph) is labelled "Asymmetric current".

There are some, often used, habits of "use of words" here which can be confusing.

One is that DC is often used to describe a current or voltage of one polarity, even if it has complicated changes of magnitude - to contrast it with alternating values which look "sinusoid" in power engineering (speech or music waveforms rarely do).

Another is that waveforms which look like a distorted "sine wave" with one half cycle different to the other are often called "asymmetric" (a wave can have identical positive and negative shapes - symmetric, even though it has the appearance of a distorted sine wave).

I guess these uses of words have become a habit in power engineering because the voltage and current are supposed to be sinusoidal at 50 or 60 Hz and trouble results when they are not.

One example, needing the kind of analysis you have posted, is the effect on breakers of the DC current which can occur, depending on the point in the voltage wave at which the contacts close [or a short circuit fault occurs].

Another is that, in the early days of AC power, difficulty was found in parallel operation or even connection of generators of different size and design, which led to the specification of very close to sinusoidal waveforms for generators rather than the distorted outputs causing the trouble. If you wonder how they managed to determine waveform without ultra-violet chart recorders or oscilloscopes, I can tell you (because I have an ancient textbook that told me!).

Thinking of your words in this post "worst case scenario....at zero voltage and symmetrical current at negative peak", I wonder if you are thinking the DC component and the symmetrical component can be independently chosen.

This is not true, they both come from the mathematical solution.

Thinking fundamentally, the initial current in a newly closed circuit must always be zero - and in a mostly inductive circuit it can only increase at a limited rate. So it is essential that at time zero the theoretical DC current and symmetric sinusoid have equal and opposite magnitudes.

Note that the symmetrical current is at its peak at voltage zero only for an inductor with no resistance, your graphical example is for low L/R values applying to normal loads of high power factor - the DC component is negligible by the second half cycle.

You are correct to wonder if a later half cycle might be worse than the first, but since the peaks of the sinusoid are always the same magnitude and the total current only increased by the exponential which always falls with time, I think you will agree the first half cycle current peak is always greatest.

I suggest you investigate the currents when energising transformer primaries or short circuiting generators at their terminals. In both these cases the L/R time constant for the DC current is much longer than for usual loads and several cycles can be very offset by the DC component.

67model

If you apply AC voltage to an inductor, after steady state is achieved, the current will lag the voltage by 90 degrees (symmetrical current). On startup, the current is initially in phase as shown (asymmetrical current), and the difference is a "dc current" as shown. It's not "at the same time" as the symmetrical current. One is shown on startup and the other is steady state in relationship to the voltage waveform.

Another homework question because the OP slept in class. No thanks!!

This does indeed sound like a student's question, but sleeping in class is not the only reason for seeking outside help. I still remember, from well over 50 years ago, a professor saying: "And therefore, obviously, ,,,". Whatever it was that followed, was NOT obvious to me, and I was too meek and/or stupid to speak up and say so. If the internet had existed then, I would have definitely have looked there for assistance.

If this were really homework, some value would have to be calculated. As I see it, the student is simply seeking help understanding a concept, and I see no problem with that! At 75, I'm definitely a student when it comes to large reactive loads on AC circuits, so I was/am looking at the thread, hoping to add a bit more knowledge to my level of understanding.

I still don't know what the vertical axis "p.u." stands for.

Thanks! After reading that three times, it is starting to make sense!

Dick