Why DC offset in fault currents? 9

[veritas]

This may seem like a very basic question but it is one I struggle to conceptually understand. Can someone explain in words why is there DC offset (or a dc transient) in fault currents? I know all the formulas to calculate maximum offset, etc. but must admit I do not find the physics behind it easy. I know it has to do with the point on the voltage wave at which the fault occurrs and also the X/R of the system but to put it all together...

Thanks in advance.

Veritas

 

[slavag]

Folks.
Many PLS to all of you.
Special to Gunnar

 

[rbulsara]

Jghrist:

There is always "energy" stored in reactive devices in AC circuits(inductors and capacitors), which would discharge upon seeing path of adequately low impedance such as fault. Just like an AC capacitor when shorted to ground will show DC current decaying, untill then its fat, dumb and happy as the load impedance is not low enough for it to discharge appreciably.

I have oscilloscope print out of tranfomer transients (very akin to short circuit currents) which shows the DC decay only, AC waveform having disappeared once the primary breaker opened during a heavy inrush/transient. That proves the inductor discharge.

 

[rbulsara]

Or a better explanation could be the inductor is constantly charging and discharging (changing polarities) in normal state. So there are point in time where there is no net charge stored in the inductor and if the fault occurs then there is no DC offset. Any other instances there will be some charge stored in the indcutors (or capacitors) which will discharge as DC current.

 

[Skogsgurra]

It might interest you to have a look at thread248-144465 and especially the recordings of inrush current in a transformer. As many posters have said, it is an analogue situation.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[electricpete]

rbulsara - I would respectfully suggest that you are mistaken.

There is always "energy" stored in reactive devices in AC circuits

No. Not when the current is 0 or passes through 0 in an inductor (Energy(t) = 0.5*L*i(t)^2)

I have oscilloscope print out of tranfomer transients (very akin to short circuit currents) which shows the DC decay only, AC waveform having disappeared once the primary breaker opened during a heavy inrush/transient. That proves the inductor discharge.

It proves that an inductor discharges with a transient response I(0)*exp(-t*R/L) and in this case it is very accurate to say that it represents decay of initial inductor current and/or dissipation of energy initially stored in the inductor.

The decaying DC offset which appears during a fault has exactly the same functional form. (If you recall, I mentioned 25 Jul 07 20:50 that this same functional form applies to the transient component for all transients of the simple LR system). But in this case it is no longer accurate to state that it represents decay of initial current. As has been stated twice already, you can have a transient decaying dc offset when the initial current at t=0 is 0! (that is three times now).

So there are point in time where there is no net charge stored in the inductor and if the fault occurs then there is no DC offset.

First, my understanding is that you mean to use an analogy between capacitor/charge/voltage and inductor/flux/current. If there is no current or flux in th inductor when the fault occurs, we can still have a dc offset.

Again see this spreadsheet:

http://home.houston.rr.com/electricpete/eng-tips/ApplySinusoidVoltageToDAMPED_Inductor.xls

(press cancel if it asks you for a password... the spreadsheet will still open).

When you open it up, change theta_degrees to 360. I hope you'll get the picture that the current (and stored energy) in the inductor is 0 for a full cycle of the graph before the fault occurs. By your logic there is no offset. But in fact there is an offset.

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Eng-tips forums: The best place on the web for engineering discussions.

 

[bjenks]

Lots of good stuff here, but I would have to refer back to IEEE Red book Chapter 4 as the best explaination on DC offset. It even talks about switching effects of relays and the mutual inductance between cables. It give an example of what happens when you turn your water hose on how there is a surge at first. I always looked at the fact that there is stored energy in the form of capacitance between wires and inductance of the cables. When a short happens you have a collaping magnetic field from the energy in the system causing a Faraday DC voltage.

 

[jghrist]

I always looked at the fact that there is stored energy in the form of capacitance between wires and inductance of the cables.

Then why isn't capacitance between the wires in the equations used to calculate dc offset?

 

[Skogsgurra]

This thread has gone all wrong.

Now it is around 50 % facts and 50 % fiction.

Why is it that the concept of energy attracts so many people with marginal understanding and a fuzzy, to say the least, notion.

I feel quite maltreated. And seeing that a simple physical fact can be distorted the way it has been doesn't really make me happier. Bringing in capacitance between wires and postulating that the stored energy in an inductance causes a high fault current to flow is not physics. Nor is it engineering.

Can I have your attention again for a short while? Thanks.

The OP wanted to know why there is DC in the fault current. As you all(?) know, an EMF is needed to produce a current. So, I set out to explain why there is sometimes a DC component in the EMF. And it is so simply because any other phase angle than 90 or 270 degrees does introduce a DC component. I left out the part about DC current building up and flowing in the circuit, decaying according to the L/R time constant. Davidbeach's complementary information is correct and was, perhaps, necessary for those who cannot draw her/his own conclusions.

But, as is often the case, it opened a Pandora's box. And it attracted some posters that found it interesting to post their ideas on the subject. And that didn't make it any clearer. Physics is not politics. You cannot have an "opinion" on things like this. Facts are facts and there are no mysteries. Davidbeach and Electricpete have presented them. Can we have an end to this "exchange on opinions" now, please?

It is Eng-Tips, after all.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[rbulsara]

I still would still say nature (here electricity) does know math, math is fitted to rationlize what we know to be happening. Also I did not imply that inductors or capacitors produce "high" fault currents, they "can" produce a DC discharge when shorted which is very akin to DC "component". Its just a way to grasp a concept.

Caps and inductors in AC circuit do "retain" charge and energy, even after the source is removed and therefore the practice of discharging them before any maintenance.

 

[davidbeach]

While I am well aware of trapped charge (voltage) on a capacitor, I've never heard of trapped current in an inductor. Interrupting devices have a tendency to open at a current zero in AC circuits leaving no current to be trapped. In DC circuits or AC circuits where the current is chopped there can be nasty voltage transients as the current in the inductor tries to continue to flow, but once the circuit is interrupted the inductor has nothing trapped.

I agree 100% with Skogsgurra that stored energy has absolutely nothing to do with DC offset in fault currents. And it is exceedingly simple to prove that. Establish a source with an inductor and leave it open circuited for a moment, but apply an AV voltage. With it open circuited there is no current flow. Now apply a fault at a voltage maxima. Note the DC offset in the fault current. Because three was no current flow prefault there could not have been any stored energy in the inductor. It doesn't matter where in the AC voltage cycle the fault happens; there is never any stored energy in the inductor. Where in the cycle the fault happens does determine the amount of DC offset.

This is all freshman circuits class and frankly not that complex.

 

[bacon4life]

The DC component comes from a lack of stored energy rather than an excess of it.

During a normal AC sine wave in an inductive circuit, when the voltage is at a maximum, the energy stored (and current) in the inductor is zero. The first quarter of the cycle (second half of positive wave) is spent adding energy to the inductor and the second quarter removes all the energy and returns the current to zero. The third puts in energy of the oppposite polarity and the fourth removes it.

During a fault when closing at other than maximum voltage, there will be no energy stored in the inductor(no current flow). Now the rest of the positive wave charges the inductor instead of just the second half.

Notice that now the current increases for two quarter cycles in a row(resulting in twice the current). This is the key difference in that normally the first part of the positive waveform reduces energy stored in the capacitor. Since there is no energy stored at the beginning, the current will increase more than the steady state case.

Electricpete was right that it is the transient portion of the response. Thus if you think about how the initial conditions are different than steady state, the difference is that there is zero energy stored instead of various amounts thoughout the cycle.

 

[521AB]

To my opinion:

instantaneous magnetic energy stored in an inductor En(t) = 1/2 * L * ( i(t) )^2.
The short circuit is just the closure of the L/R circuit, containing the above energy at the moment of the short. Like a capacitor, the inductance will "discharge" its energy in form of heat (Joule effect) on the resistor, and this is done by generating a current with an exponential decaying curve.

Wrong?

 

[jghrist]

instantaneous magnetic energy stored in an inductor En(t) = 1/2 * L * ( i(t) )^2.
The short circuit is just the closure of the L/R circuit, containing the above energy at the moment of the short. Like a capacitor, the inductance will "discharge" its energy in form of heat (Joule effect) on the resistor, and this is done by generating a current with an exponential decaying curve.

Wrong?

And if i(t)=0 before the short?

 

[slavag]

Dear Colleagues.
Sorry, but we must finished this discussion.
Gunnar, you are right, we opened Pandora's box.
DC offset is fact and low, is not opinion or option.
Please separate two things:
1. Why DC offset ( not only in fault)? Only 3 parameters influence on DC component : Angle, L and R it's all. Please remamber circuit with AC source, L element, R element and switch.
Electricpete, David explained it as well as possible.
2. Accumalated energy in inductive elements, yes we have, but w/o any phisical connection to DC offset.
( I'm also mistaked in my first post, sorry), but checked myself again.
Best Regards.
Slava

 

[davidbeach]

If energy storage has anything to do with it, and frankly I doubt it, then bacon4life comes far closer than anyone else who mentioned energy storage. Any talk of energy storage has to deal with the case of a fault on a previously open circuit where i(t)=0 for all t<0. No energy. And forget capacitors, the whole phenomena can be explained without the need for any capacitance.

This should be laid to rest, there is no room for opinion, the physical laws are what they are; the DC offset is how the AC current is allowed to change instantaneously without the total current through the inductance changing instantaneously. Simple, piece of cake, easy as pie.

 

[521AB]

Quote:
"Quote:
instantaneous magnetic energy stored in an inductor En(t) = 1/2 * L * ( i(t) )^2.
The ............ the inductance will "discharge" its energy in form of heat (Joule effect) on the resistor, and this is done by generating a current with an exponential decaying curve.

Wrong?
And if i(t)=0 before the short?"

If i(t) = 0 before the short, (CB open before the fault), the dc is there because you are "charging" the inductance (storing magnetic energy in the inductance).

I would like to underline that one inductance acts in such a "strange" way exactely because it stores magnetic energy.

 

[davidbeach]

Nothing strange, no energy storage necessary. Forget about energy storage, it has nothing to do with the DC offset.

 

[521AB]

Let me give some comments to this issue.
The original question from Veritas was very interesting.
"Where is this dc component coming from?"

Energy or not energy?

Let's take a mechanicl phenomenon: a falling mass.

You can predict the speed of a mass falling from a certain height when it touches the ground by solving the fundamental motion differential equation F = m a. The result is a mathematical result that has to be accepted, but it is difficult to understand in intuitive terms.

You can also introduce the energy concept (potential and kinetik). Mathematically you can give values to those energies, but intuitively you do not need. It's enough to say that the mass, falling, transforms the potential energy into kinetik energy. Potential has to do with heigth, kinetik has to do with speed.

Who is right? Both.

I think Veritas was looking for an answer similar to the second description, if it is possible to give it. Where is this dc component coming from?
The fact that an inductor does not allow the current through itself to change discontinuosly, has to do with the law: voltage = - L d flux / dt, but can't we say that the magnetic energy that the inductor is able to store is similar to kinetik energy of a mass? There is one inertia.... (as well as the capacitor: we can assume it stores potential energy..)
That's why I was thinking in terms of energy. But I agree that I couldn't explain where the dc component is coming from intuitively.

Where is the dc component coming from?

 

[Skogsgurra]

The old lexicon definition of "Iteration" comes to mind:

Iteration. (s) See Iteration.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[davidbeach]

Since the phenomena can be entire explained without introducing energy storage, what could energy storage possibly add except confusion? And, because it can be entirely explained without introducing energy storage, the addition of energy storage is simply wrong. It doesn't matter how much you want it to be about energy storage; this is not about opinions, it is about facts and there are no facts that support energy storage as a mechanism for explaining DC offset.

The DC comes from exactly the same place as the added AC current at fault onset. However much additional AC current there is through the inductance there is a DC current of equal magnitude and opposite polarity (relative to the instantaneous AC polarity at the fault onset). All is well.