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

 

[Skogsgurra]

Hello.

It is very simple, actually. But you have to look at the details of a sine wave. Like this:

1. A sine wave has two half-waves. One positive and one negative.

2. A short can occur any time (from 0 to 360 degrees) during the sine-wave cycle.

3. It can also be cleared any time during the cycle.

4. Now, let's assume that the fault happens at 0 degrees. Current will start rising in positive direction.

5. Let's also assume that the fault is cleared after 180 degrees (half a period).

6. If you draw the resulting wave-form on a paper, you will see a positive half-wave. Nothing more. No negative current.

7. The mean value of this positive half-wave is a positive DC current. That is how the DC component enters the scene.

Now, if the faults isn't cleared after 180 degrees, you still has that DC component in the system and it takes some time to get it out of the system. Typically up to 50 or 100 cycles in a large transformer and bolted short.

Look! No Math!

Gunnar Englund

http://www.gke.org

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[davidbeach]

I hate to argue with Gunnar, but I don't buy that explanation.

One of the basic laws of electricity is that current through an inductor does not change instantaneously (without application of an infinite voltage). The inductor in this case being the inductive reactance of the system up to the point of the fault. The fault happens instantaneously and Ohm's law implies that the current change as a result of the fault should also happen instantaneously. The DC offset is the only way out of the problem of two laws giving opposite results. The Ohm's law effect is an instantaneous change in the (AC) current though the inductance and to keep the current through the inductor from changing instantaneously a DC current of opposite sign is created to oppose the AC current. The system X/R determines how rapidly the DC decays. Faults that occur at a current zero crossing don't produce a DC offset while those that occur at a current max or min produce the largest DC offset that circuit will produce. The actual DC offset will range between those values and will be different in each phase.

 

[RalphChristie]

Consider a sine wave. In a purely inductive system, the current is lagging the voltage by 90°. If a fault occur when the voltage is at a maximum, the current starts from zero, and it is symetrical. If the fault occurs if the voltage is zero, the current cannot change instantaniously, it must start from zero. Since it also must lag the voltage by 90°, it becomes asymetrical or offset. This offset is called the dc component. In practice, all circuits have resistance and therefore the d.c. component decays to zero in a few cycles.

Regards
Ralph

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[Skogsgurra]

OK davidbeach. Maybe I was too simplistic. I should have presented the differential equations showing what happens in detail. But, since Veritas already seems to have been through that part and still doesn't "get it", I thought that the positive half sine-wave picture would be a good intuitive way out of that dilemma.

I hope that you, at least, agree that the DC component stems from the fault happening somewhere else than at 90 or 270 degrees?

Gunnar Englund

http://www.gke.org

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[electricpete]

Assume for simplicity a single-phase pure inductive circuit:

v = L * di/dt
i = (1/L)* integral(v(t))
,

Do the integral graphically assuming v(t) is sinusoidal. In all cases i(0)=0

If you start at the negative peak of v(t), then you have
i(t) goes up for 1/4 cycle to it's peak, down for 1/2 cycle to negative peak, up for 1/2 cycle to positive peak etc. No dc offset.

Repeat but this time start integration at v(t)=0. i(t) starts from 0 and goes up for 1/2 cycle, then down for 1/2 cycle (back to 0), then up for 1/2 cycles, then down for 1/2 cycle etc. This time it is cycling between 0 and a peak value which is twice the peak value of the previous non-offset case. This is the highest possible offset which results in a fully offset waveform that doubles the true peak. That highest possible peak value would correspond to 2*sqrt(2)*LRC

Now add some resistance back in. The fully offset waveform will look similar to before except the offset component will decay as exp(-t*L/R). By the time you get to the worst-case first peak at 1/2 cycle, the dc has already decayed some so you will never quite reach 2*sqrt(2)*LRC. The higher the R/L, the more decay and the lower the worst-case peak associated with closing at v=0

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[electricpete]

In case the graphical version of the integral is tricky, here is analytical solution of the single-phase pure inductive case
v = L * di/dt
i = (1/L)* integral(v(t))
Let v(t) = Vmax sin(w*t+theta)
i(t) =-(1/L*w)* [cos(w*tau+tteta)]tau = 0...t
i(t) =-(1/L*w)* [cos(w*t+theta) - cos(w*0+theta)]

For theta = -Pi/2
i(t) =-(1/L*w)* [cos(w*t-Pi/2) - cos(w*0-Pi/2)]
i(t) =(1/L*w)* [sin(w*t)]
no dc offset

For theta = 0
i(t) =-(1/L*w)* [cos(w*t-0) - cos(w*0-Pi/2)]
i(t) =-(1/L*w)* [cos(w*t) + 1]
this is the fully offset case.
Don't forget to try the integration graphically

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[Skogsgurra]

Very good. But remember what the OP asked for: "Can someone explain in words why is there DC offset (or a dc transient) in fault currents?"

Gunnar Englund

http://www.gke.org

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

 

[Homerjs78]

Please correct me (anyone) if this is too simplistic or down right incorrect:
The system contains inductance which stores a charge. In the case of a fault, the inductance contributes energy in the form of DC current. As with any stored charge / energy, this decays according to a time constant.

 

[slavag]

Hi.
Homerjs78, I bay your explanation.
Gunnar and David also have good explanation, but your is very short and in target.
Electricpete, are you remambare all these formules??????
Homerjs78, are you approve use your explanation for students?
Best Regards.
Slava

 

[supermacc]

Interesting discussion, I think all posters have provided good explanations, Homer has possibly nailed it but I would nevertheless like to give my explanation. I have to use some equations but the explanation is IMO simple and it is a slightly different approach compared to previous posters.

Assume that the short-circuit appears on the terminals of a synchronous machine (SM). Before the fault, the SM voltage equals speed x (stator flux) and the stator flux equals (stator inductance) x (stator current) + (magnetizing inductance) x (field current). Thus, voltage = w x (Ls x Is + Lm x If). If the SM is unloaded then the stator current equals zero and then there is no stored magnetic energy in the stator (but of course in the rotor) before the fault.

Moments after the fault, the voltage equals zero. The field current and the speed have large time constants and cannot change quickly, so the only way for the right-hand side of the above equation to equal zero is a (negative) dc-step in the stator current, in order to oppose the flux coming from the rotor.

There are many simplifications in my explanation, the full picture is much more complicated see, for instance, Kovács: "Transient phenomena in electric machines".

 

[Skogsgurra]

I feel somewhat guilty for oversimplifying the problem. And when I then see others simplifying it even more - and some going the other way as well - I think that it would be good to divide and conquer. Divide the problem, that is.

First: The OP asks from where the DC comes. My take is that the DC is a result of when the fault occurs. If it occurs at exactly 90 or 270 degrees, then there is no DC introduced in the fault circuit. Any other phase angle makes the net "polarity" either positive or negative, hence a DC component. (I should have included in my first explanation that it is the driving EMF that has a DC component and that it is this DC component that creates a DC component in the current. Omitting that step was perhaps the great mistake).

Second: The other part of the problem is why the DC component stays for a few, I still say 50 - 100 periods (time constants of a typical transformer secondary are in the seconds). And that is classical DC electricity. The well-known L/R decay - same as RC decay - but opposite.

There are, as we have seen, interesting philosophical twists to the problem. Like having a contradiction between two laws of nature and introducing a DC component being the only way out of that dilemma. I am not sure if such a philosophical point of view helps the understanding. To me, it is more a "class-room classic" that everyone remembers but not many understand the deeper meaning of.

Gunnar Englund

http://www.gke.org

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[slavag]

Hello Gunnar.
I'm not so agree with second part of your post.
DC component stays for a few 50-100 periods.
For MV Tn is about few tens miliseconds, for HV and EHV
and near to large generators it's around 400 miliseconds.
Bay the way, Tn is always reduced by case of some ground faults.
For small calculation I'm use formula:
DC copm= e^(-t/Tn( it's L/R))*cosA
A=B-arctan(wL/R)
B it's phase angle of the source voltages in the start fault.
But prefer w/o formulas.
Regards.
Slava

 

[Skogsgurra]

Yes, we do agree. Time constant is one thing. The time to where the current is (practically) zero is another thing. One convention is that current is close enough to zero after three time constants. Another is five time constants.

Of course, the size and type of generator/transformer also plays a role. Do we need to discuss every type of equipment? The OP still just wanted to know how and why the DC component gets into the system.

Gunnar Englund

http://www.gke.org

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

 

[waross]

First I will comment on power.
Real power we are familiar with.
Reactive power, apparent power, KVA. This is often called power that does no useful work.
However reactive power does work. In the case of an inductive circuit, real energy is stored in the magnetic field of the inductor.
In a DC circuit the magnetic field is created when the circuit is energized. When the circuit is de-energized The field decays and the energy of the field is dissipated in the resistance of the circuit. If the circuit is interrupted by an open circuit much of the energy may be dissipated in the resistance of the arc across the opening contacts. If the current is interrupted by a short across the inductor, the energy of the magnetic field is dissipated in the resistance of the inductor.
In an AC circuit the magnetic field and the stored energy are increasing and decreasing and then increasing and decreasing in the opposite direction in a manner that may almost be described by a sine wave, because an iron core inductor is almost linear.
When there is a short circuit on an inductor, such as a transformer, in an AC circuit the discharge of the magnetic energy appears as a DC offset. Although the time to discharge may be described in cycles, it is a function of the RL constant of the circuit and is not related to the AC cycles or frequency.
As others have mentioned, the current depends on the instantaneous value of the magnetic field.
The time for the current to decay to 1% (or any other PU value) is the same no matter the magnitude of the original current. The time for the current to decay to a negligible value is much less with lesser currents.
hope this helps.
respectfully

 

[slavag]

Gunnar, sorry, you are right!!!. You write down about time constant and I'm about clearing time, are different things.
Waross , great!!!!.
Regards.
Slava

 

[Skogsgurra]

Great discussion - and needed.

My point is that you do get an initial DC component even if you do NOT have any inductance in the circuit. The concept of inductance and discharge may obscure that fundamental fact.

The decay is then entirely a function of L/R in the circuit and has, as said, nothing to do with frequency. But it is a convenient measure and therefore often used.

Gunnar Englund

http://www.gke.org

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

 

[waross]

I have created some confusion between power and energy.
However reactive power does work. In the case of an inductive circuit, real energy is stored in the magnetic field of the inductor.
Sorry. This would be better if I had said;
"Inductive power is a transfer of real energy. In the case of an inductive circuit, real energy is continually being added into and then removed from the magnetic field of the inductor."
Although real energy is being transferred, the net power is zero.
respectfully

 

[oldfieldguy]

Okay, thinking of this in a purely graphical representation, let's talk about a fairly rapidly decaying current applied to a CT. The first half-cycle is +1000 amps. The second half-cycle is -900 amps. +1000 and -900 gives us an offset of 100 amps. Third half-cycle is +800 amps. Fourth is -700. Still an offset of +100. The +100 difference is the DC offset.

I'm recalling a real-world application here where we had a bar-type CT subjected to the inrush current of motor starting, and it was adversely affected by this current profile to the extent that its secondary current did not match properly with the other window-type CT in a differential circuit. The result was chaotic until we figured out the problem.

The bar CT remained partially magnetized by the DC offset on starting for several cycles after the motor was running, but during this transition currents in its secondary circuit were markedly different than those from the window CT of the identical ratio.

We tracked all these festivities by oscillography of the outputs of the suspect devices. It's really impressive to see it in graphic representation.

old field guy

 

[veritas]

Hi all

I want to thank you all for your posts so far. Indeed it makes for very interesting reading and I find it very educational - even after 14 yrs of being in the business! - I guess I'd rather learn it now than never! A point Skogsgurra touched on:

Great discussion - and needed.

My point is that you do get an initial DC component even if you do NOT have any inductance in the circuit. The concept of inductance and discharge may obscure that fundamental fact.

The decay is then entirely a function of L/R in the circuit and has, as said, nothing to do with frequency. But it is a convenient measure and therefore often used.
Gunnar Englund

If hypothetically the circuit has no reactance but only resistance, and a fault occurs at voltage maximum, then the current would also have to be at max since it must be in phase with the voltage. My view is that the current WILL start off at maximum value precisely because there is no reactance in the system. There is thus no need for a DC offset.

Am I correct?

I'm afraid Skogskurra, this contradicts what you stated. The second part of your statement also appears to contradict the first part - i.e. you say offset still needed even if no inductance but then the offset decays as per the L/R of the circuit - but circuit has no L.

But thanks for your valuable contributions and those of all the others so far.

Truth - what is truth?