Why DC offset in fault currents?
Why DC offset in fault currents?
(OP)
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
Thanks in advance.
Veritas






RE: Why DC offset in fault currents?
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
www.gke.org
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RE: Why DC offset in fault currents?
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.
RE: Why DC offset in fault currents?
Regards
Ralph
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RE: Why DC offset in fault currents?
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
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RE: Why DC offset in fault currents?
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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RE: Why DC offset in fault currents?
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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RE: Why DC offset in fault currents?
Gunnar Englund
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RE: Why DC offset in fault currents?
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.
RE: Why DC offset in fault currents?
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
RE: Why DC offset in fault currents?
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".
RE: Why DC offset in fault currents?
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
www.gke.org
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RE: Why DC offset in fault currents?
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
RE: Why DC offset in fault currents?
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
www.gke.org
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RE: Why DC offset in fault currents?
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
RE: Why DC offset in fault currents?
Waross , great!!!!.
Regards.
Slava
RE: Why DC offset in fault currents?
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
www.gke.org
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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: Why DC offset in fault currents?
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
RE: Why DC offset in fault currents?
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
RE: Why DC offset in fault currents?
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?
RE: Why DC offset in fault currents?
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RE: Why DC offset in fault currents?
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RE: Why DC offset in fault currents?
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RE: Why DC offset in fault currents?
RE: Why DC offset in fault currents?
I have to think about what you are actually saying.
I also think that some should think about what I am actually saying.
The OP was about how a DC components can exist in a fault situation (Quote: " Can someone explain in words why is there DC offset (or a dc transient) in fault currents?").
My very simplistic (that seemed to be what the OP asked for) explanation focused on the reason why there is a DC component. And I showed that the average value of a sine does get a DC component if started at any other phase angle than 90 or 270 degrees. That was all I said. And I thought that was what the OP asked for.
The L/R time constant in the circuit, of course, then dictates what happens to the DC component. No discussiuon there. I may have expressed myself unclear (the "contradiction" part) when I, as veritas says, talk about a circuit with no inductance and then, in the next sentence, mention that the L/R determines what happens to the DC component. But, if you read that sentence, you will see that there is no contradiction in it. (Sorry to be JB-ish here, but when I feel misunderstood, I need to explain myself). I have said that you get a DC component also without any inductance. I did not not say that there is no inductance. Only that you can get a DC component also with no inductance. What then happens to that component is dependent on the L/R ratio. If it is zero, then the DC dies away instantly. If it is >0, then it decays according to 1-exp(-L/R).
Sorry if you either think that I am completely ignorant or if you think that I try to tell lies. That was not my intention. All I wanted to do was to explain in a simple manner from where the DC component stems. And I still think that I did that. And I do not think that I am completely ignorant either.
This is, I hope, my last say in this thread.
Gunnar Englund
www.gke.org
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RE: Why DC offset in fault currents?
I have the highest respect for your opinion in general, but I think you have missed the mark on this particular item. The dc offset is a characteristic of change in configuration of an inductive circuit imo. Your explanation 7/19/07 as I understand it is dependent upon time of fault clearing... but dc offset occurs during change of configuration of an inductive circuit even when the fault doesn't clear.
You are correct that the dc offset depends on the time of fault initiation. That can be seen by integrating a sine wave I = (1/L) * Integral (vl dt) where vl is voltage across the inductor. Assume vl(t) = V sin(w*t). If you start integrating at t = +/- Pi/2 you jump smoothly into the steady state solution i(t) = -/+ cos(w*t) * V/(L*w). If you start integrating at t=0 or Pi, you have the worst case fully offset sinusoid. If there is a series resistance, the offset decays away.
To make an attempt to simplify my previous explanation, we simplify the circuit and get rid of the ac. Consider a simple series switched R/L circuit fed by a DC voltage source switched on at t=0. The response to closing the switch at t=0 is i(t) = V/(R) - V/(R) * exp(-t*L/R). That response has two term. The first of these terms is the steady state response (which happens to be dc in this dc case, but in general is not necessarily dc).. The second of these terms is the transient response. This transient response is analogous to what we're calling the dc offset during a fault IMO.
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RE: Why DC offset in fault currents?
"If you start integrating at t = +/- Pi/2 you jump smoothly into the steady state solution i(t) = -/+ cos(w*t) * V/(L*w).'
should be:
"If you start integrating at t = +/- Pi/2 you jump smoothly into the steady state solution i(t) = -cos(w*t) * V/(L*w).'
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RE: Why DC offset in fault currents?
It is easier to visualize the problem with a purely inductive system. In this case, the voltage across the inductance equals the driving voltage which is a sinusoid. The voltage is L·di/dt. The current will also be a sinusoid, with or without a dc component.
Because the current across the inductance cannot change instantaneously, it starts out at zero no matter when the fault occurs.
If at the point of the fault, the voltage is at a maximum, then di/dt is positive and at a maximum. So current is increasing with the highest slope. When the voltage reaches zero, di/dt must be zero, and the current reaches its peak. If you follow this through a whole cycle and plot out the current, you see that you get a sinusoid with no dc offset.
If at the point of the fault, the voltage is zero and rising, then di/dt is zero. As the voltage increases, di/dt and i increase. When the voltage reaches a maximum, di/dt is a maximum, with the current still increasing. When the voltage is gets to zero, di/dt is zero and the current is at a maximum. When the voltage goes negative, di/dt is negative, but the current is still positive. If you follow this through a whole cycle, you see that you get a sinusoid with an offset equal to the peak current so that the minimum current is zero.
RE: Why DC offset in fault currents?
To revisit the integration approach one last time:
I have made the graphical integration easy.
The scenario is: sinusoidal voltage source (Vmax*sin(w*t)). By definition it always has zero-crossing at t=0.
The voltage source is connected to the inductor (undamped) at some variable angle theta (or delay time) after t=0
You can vary the angle in the attached spreadsheet and watch the offset appear for theta=0, 180, disappear for theta = 90, 270.
(How this relates to fault: fault causes a sudden decrease in load impedance and sudden increase in voltage seen across the inductor.)
h
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RE: Why DC offset in fault currents?
http://hom
For example, you can see the initial input data with X/R ~ 13, gives a true peak at first half-cycle of 178% as high as the "steady state" peak that would exist after the transient is gone.
In my mind, this represents not only the type of transient that is seen on a fault, but also during motor starting and during transformer inrush... except for the transformer inrush the high exciting current due to dc offset creates saturation which leads to significant harmonics in the current waveform (not modeled here).
If anyone is interested, the method I used to simulate the LR circuit is derived as follows:
v = L di/dt + IR
L di/dt = v - IR
di/dt = v/L - IR/ L
di = v/L dt - IR/L dt
I = Int v/L dt - Int IR/L dt
Let k = current time, p = previous time (p = k-1)
Apply trapezoidal rule: Xk = Xp + 0.5(X'k + X'p)dt
Ik = Ip + 0.5*(Vk+Vp)/L dt - 0.5*(Ik+Ip)R/L dt
Ik (1+0.5*R/L dt) = Ip + 0.5*(Vk+Vp)/L dt - 0.5*Ip*R/L dt
Ik = {Ip + 0.5(Vk+Vp)/L dt - 0.5 IpR/L dt} / (1+0.5*R/L dt) [THIS IS THE EQUATION USED IN SPREADSHEET]
Ik = {Ip + 0.5(Vk+Vp-Ip*R)/L dt } / (1+0.5*R/L dt) [SIMPLER FORMAT OF SAME EQUATION]
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RE: Why DC offset in fault currents?
Thank you!! Your material is very intresting and helpfull. As Gunnar saied "Great discussion - and needed".
Thank to all of you.
Best Regards.
Slava
RE: Why DC offset in fault currents?
PLS for you
I do not think anyone was totally wrong in the discussion, it was just explained from different angles and viewpoints and with different examples. With the lack of pictures it is sometimes difficult to visualize the examples and you have done well by creating a spreadsheet for better understanding. Words are good but words and pictures together are great! Well done!
By using any example that fits into what you are used to, the concept can be clearly understood, and in the end it is the important thing - a clear understanding of what is going on.
Regards
Ralph
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RE: Why DC offset in fault currents?
RE: Why DC offset in fault currents?
I would respectfully disagree. To me it sounds like the idea being portrayed is that there is initially energy stored in the inductor at the time of the fault, and the fault allows this initially stored energy to decay with the L/R time constant.
I would like to show a counterexample which proves this wrong.
Start with an assumption that we have an ideal power supply feeding through an inductance (perhaps a transformer) to a line which has no loads prior to the fault. The initial energy stored in the inductance prior to the fault is zero. Now close in a low-resistance representing a fault on the transformer secondary at the moment the voltage is zero. We get the maximum dc offset. It didn't come from the energy initially stored in the inductor, because that energy was zero in this case.
At the risk of being really obnoxious, let me try another explanation expanding on what others have already said:
1 – ASSUME for simplicity the post-fault circuit has no resistance.
2 – ASSUME for simplicity the prefault current is negligible, i.e. i(0-) = 0.
3 - The total post-fault response consists of a steady state sinusoidal response plus a transient response.
4 - The current cannot jump when the fault is applied. i(0+)=i(0-) (since it is flowing through an inductor).
5 – Putting together 2 and 4, we see that the total response at t=0+ is 0. Itotal(0+)=0.
6 – Putting together 5 and 3, Itotal(0+) = Itransient(0+) + Isteadystate(0+) = 0
7 – Examining 6, the only way we can have NO transient is if Isteadystate(0+) = 0
8 - Isteadystate(0+) = 0 occurs only if we close the fault at a phase angle where the steady state solution for current is 0. This occurs only when the voltage is at its peak (since current lags voltage by 90 degrees in steady state inductive circuit).
9 – Examining 7 and 8, if you apply the fault at any time other than the maximum or the minimum of the supply voltage, you will have a transient component of the current (the exponentially decaying dc component). That transient current occurs as a direct result of the characteristic of an inductor v=L di/dt and we can calculate it directly from the corresponding integral equation i = Integral v(t) dt + i(0)
It is not a dissipation of energy stored before the fault. It is a transient response. In general we expect a transient response when the system configuration changes. The transient response in a simple L/R system will always be of the form I(0)*exp(-t*L/R)
I am a hearty supporter of the idea that there is not only one right answer. There were good answers given by many others. Mine is probably not the simple answer some people are looking for and mine is not without equations as was requested in the original post. My apologies. Feel free to comment. Otherwise I will shut up.
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RE: Why DC offset in fault currents?
"i = Integral v(t) dt + i(0)" should be "i = Integral v(t) dt/L + i(0)"
"I(0)*exp(-t*L/R)" should be "I(0)*exp(-t*R/L)"
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RE: Why DC offset in fault currents?
RE: Why DC offset in fault currents?
I have high regard for your math skills. But this is not about math but grasping a concept. Poor electricity does not undestand or even know math. Math is a means to analyse many phenomena. So it does not matter how it is explained by math. A waveform can be resolved into any number of compoenets to arrive at a final waveform. Assuming DC component fits the mathematical explanation for the Asymmetry in fault current wavefors. (This is very akin to resolvoing vectors in x and y rectangular component for anaysis). The fact is the system behaves as if it has a DC component which can only come from some sort of stored enegy being discharged. Also math is not what OP was looking for.
I still believe that homer's explanation is valid enough for a concept.
RE: Why DC offset in fault currents?
I agree that these discussions do not give a good physical description as the OP wanted, but it's hard to describe physics without math. I think Davidbeach's original post gives the best physical explanation.
RE: Why DC offset in fault currents?
RE: Why DC offset in fault currents?
Gunnar Englund
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RE: Why DC offset in fault currents?
Many PLS to all of you.
Special to Gunnar:):):)
RE: Why DC offset in fault currents?
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.
RE: Why DC offset in fault currents?
RE: Why DC offset in fault currents?
Gunnar Englund
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RE: Why DC offset in fault currents?
No. Not when the current is 0 or passes through 0 in an inductor (Energy(t) = 0.5*L*i(t)^2)
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).
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://hom
(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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RE: Why DC offset in fault currents?
RE: Why DC offset in fault currents?
RE: Why DC offset in fault currents?
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
www.gke.org
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RE: Why DC offset in fault currents?
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.
RE: Why DC offset in fault currents?
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.
RE: Why DC offset in fault currents?
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.
RE: Why DC offset in fault currents?
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?
RE: Why DC offset in fault currents?
RE: Why DC offset in fault currents?
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
RE: Why DC offset in fault currents?
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.
RE: Why DC offset in fault currents?
"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.
RE: Why DC offset in fault currents?
RE: Why DC offset in fault currents?
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?
RE: Why DC offset in fault currents?
Iteration. (s) See Iteration.
Gunnar Englund
www.gke.org
--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...
RE: Why DC offset in fault currents?
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.
RE: Why DC offset in fault currents?
to my opinion you are just commenting the result of the differential equations governing this phenomenon. You are also commenting the Lenz laws.
But to my opinion you are not explaining, with easy words, where the dc current is coming from.
We can maybe use the Maxwell's equations anc commenting them, in order to explain this.
But I think we are not mastering this issue untill we are able to explain with easy words where this current is coming from.
And nobody has done it untill now, neither me.
For instance: the fact that the dc decays, should suggest a dissipation. Of what?
Explaining with easy words is for instance the explanations that a boy at elementary school can understand. No differential equations.
A boy at elementary school can understand oscillation between L and C, if we make a comparison with water, tanks and pumps.
A boy at elemtary school can understand the charge of a capacitor, if we tell him that the capacitor is a tank, current is water and it needs time to fill a tank. Etc.
I am sorry, but if we are not able to explain with easy words this phenomenon, my opinion is that we are not mastering this issue.
RE: Why DC offset in fault currents?
But if you say that energy came from energy stored within the circuit (as y ou implied previously), that is the part I disagree with. Quote follows:
[quote]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.[quote]
Again I disagree with that last quote and that incorrect simplification is the whole reason I objected to the way energy had been discussed.
If you wanted to include energy in the description of where the dc offset ORIGINATES from (not how it decays), you might start by examining the sinusoidal steady state energy storage of an inductor. We know current lags voltage by 90 degrees. There are four quadrants of the cycle
Let's say the voltage source is sin(theta) where theta = w*t
1 - theta=0-90: VL > 0, IL < 0 (energy coming out of inductor)
2 - theta=90-180:VL > 0, IL > 0 (energy going into inductor)
3 - theta=180-270: VL < 0, IL > 0 (energy coming out of inductor)
4 - theta=270-360: VL < 0, IL < 0 (energy going into inductor
At the end of period 1 we have 0 energy. At the end of period 2 we have max energy. At the end of period 3 we have 0 energy. At the end of period 4, we have max energy.
Now let's return to our simplest situation, fault occurs with 0 initial current and angle theta=0. The system does not begin in steady state. For the first half cycle the voltage is positive. But guess what, current remains positive through and beyond the first half cycle as well. We add energy for a full half cycle from the voltage source. This is longer then we ever add in steady state. Since we have added energy for longer than steady state, the final energy level at the end of the first half cycle is higher than it will become in the post-fault steady state condition.
So I would say that the energy associated with the worst-case dc offset is added to the LR system by the power source during the first half cycle after the fault (it is not present before the fault).
That is my best attempt to discuss the subject with energy. I don't think it made it any simpler. The simplest approach IMO:
v = L di/dt
i = (1/L) int v dt
Start integrating a V=V0sin(w*t) at t=0 and you will see the dc offset. If you haven't done it yet, get out your pencil and paper and do the integration graphically.
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RE: Why DC offset in fault currents?
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RE: Why DC offset in fault currents?
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RE: Why DC offset in fault currents?
I am surprised to read your comments.
I have read many books that describe a coil surounded by a magnetic field and the energy released when the field collapses.
In the event of a closed circuit, the collapsing magnetic field will induce a current and the resulting voltage will be dependent on the circuit resistance. The circuit resistance in the event of a short circuit will equal the coil resistance. When the voltage is removed from an induction coil, the DC current will continue for an easily calculated time. Is this not a form of stored energy?
When an induction coil (that may also be a transformer winding) is energised with AC there will be an alternating magnetic field surrounding the coil. This field will store and return energy to the circuit as the current alternates. If the coil is short circuited any energy stored in the magnetic field at that instant will be disipated in the resistance of the circuit, that may be soley the resistance of the coil or transformer winding.
A magnetic field will store energy whether the cause of the field is AC or DC. A short circuit or fault has no commutating mechanism so that energy must be disipated as a DC offset.
If your calculus seems to indicate an absence of stored energy, I suggest that you disect your arguments. I think that you will find the stored energy quietly hiding behind one of your derived terms.
KISS
Energy stored in a DC induced magnetic field is acceped.
Energy is alternatively stored and returned in a manner that may be approximated by a sine wave in an AC induced magnetic field.
In the event of a fault the instanteneous stored energy will be disipated in the coil resistance following the same laws as a DC induced field.
Respectfully
RE: Why DC offset in fault currents?
Stored Energy = 0.5 * L * i^2
d/dt (Energy) = v i = 0.5 * L * d/dt(i^2)
Using the chain rule for differentiaion:d/dt(i^2)=2 i di/dt
v i = 0.5 * L * 2 i di/dt
Simplifying:
v = L di/dt
I didn't interpret David to state anything to the contrary. I heard him express an opinion that energy does not add much to the question of where does the energy come from. That is the same sentiment I expressed above.
Agreed 100%.
Scenario 1 - (waross' scenario) If you short out the voltage source of an inductor, the initial current will continue to flow and decay as the energy is disippated in the inductor. You will note in this type of transient, the current is never larger than the initial current.
Scenario 2 - The transient associted with worst case dc offset after a fault. The voltage accross the inductor INCREASES (it is not shorted to 0). The current flowing the inductor can INCREASE beyond the value that it had at the time of the short. It is a whole different transient.
The two scenario's have similarities. The similarity is that the energy associated with the dc comopnent decays away as it is dissipated in the circuit resistance giving a dc current of exp(-t*R/L).
The two scenario's have an important difference in terms of where the energy associated with the dc offset came from. In scenario 1 where I short out the votlage source, the energy came from the prefault circuit. In scenario 2 where I applied a fault, the energy associated with the dc offset was delivered into the circuit by the power suply within the first half cycle after the fault.
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RE: Why DC offset in fault currents?
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RE: Why DC offset in fault currents?
We continue.
Electricpete, now you are realy great!!!!!!
Storage energy or accamulated energy in inductive element (magnetic field)initial current will continiue TO FLOW IN CASE OF AC SOURCE SHORT, this is a point.
Possible iteration with water, valve and pipe.
You close valve in begining of pipe and water continue flow
according to kinetic energy.
Second case of Electricpete it's: CLOSE OF AC CIRCUIT with AC source, L and R.
My English is not so good for write this iteration, but maybe we try together write it.
Same pipe is closed and we have valve, possible open this valve with angels from 0 up to 90 deg:
Only this open angle influence on ARC of water (DC offset).
with 0 we have max arc with 90 we don't have any arc.
I'm hope my iteration more or less is clear.
521AB please pay attention we dont have energy storage.
Regards.
Slava
Ohhh
RE: Why DC offset in fault currents?
I read the first post (very good one), I got shocked when I realized that I didn't know the reason, I thought some minutes and I answered, according to my intuition.
I am able to solve differential equations, and if I had been at school, and my teacher had asked the same question, I had answered that it is just the transient solution of the differential equation, which depends on the initial conditions etc... Teacher happy, me happy.
I also assume that everybody in the electric engineering field has this knowledge. So, please, no integrations or differentiations anymore, not even graphics or numerics.
Back to the issue: usually energy easily explains thinks that might appear complicated, that's why I went in the energy direction. But I see that it is not helping in this case.
What are we analyzing? Actually an elementary differential equation of the first order, energized by a sinus waveform. This is maybe what is making everything complicated.
Can't we use a machanics model:
Look: one mass energized by the force Force(t) = sin (w * t) in a gas where friction is proportional to the speed by coefficient alpha ( Force (t) - alpha v = m dv/dt ).
This is also a first order equation. Depending on when we "connect" the alternate force "Force (t)" to the mass, we should get a dc component in the speed of the mass. If we remove the friction with the gas (electrically, we assume R = 0), this dc component will never decay.
Still I don't see it...
RE: Why DC offset in fault currents?
RE: Why DC offset in fault currents?
Is not so simple Q.
Slava
RE: Why DC offset in fault currents?
RE: Why DC offset in fault currents?
Here is my attempt at a thermal analogy. I don't think it adds much, but see if you like it.
Start with one pound-mass of water in a well-insulated (but not perfectly insulated container). The ambient temperature outside is 100F and constant.
We arrange a programmable heat pump that adds exactly 1 BTU/minute for 10 minutes, then subtracts one BTU/minute for 10 minutes, then repeats.
We know that during 10 minute half cycle , the 10BTU will equate to a 10F change in the water temperautre.
What does the steady state temperature profile look like? Intuitively we suspect that it cycles from 95F to 105F. The proof comes from considering that since there is zero net heat input from the heat pump, there must zero net heat input from the ambient over a half cycle, which can only happen if the temperautre profile is symmetric with respect to ambient temperature.
So now we know the steady state profile. At 95 heat turns on and heats to 105F. At 105F, coole turns on and cools to 95F. The average temperature is 0.
Now turn off my heat pump for a day and let the system come to equilibrium with the 100F ambient.
Now turn on the heat pump starting with a 10 minute heating cycle. The temperature will go from 100F to 110F. Then the cool pump will come on and bring back to 100F, then back to 110F etc. Over time since the average temperature of this transient system is above 100 ambient temperautre, it will lose energy to the ambient, until it reaches the steady state solution cycling between 95F and 105F.
Comparing the our inductor: heat transfer rate (1btu/minute) corresponds to voltage. Temperature corresponds to current. Ambient temperature (100F) corresponds to 0 current. Small heat leakage to ambient corresponds to series resistance.
If you appreciate that the inductor is just an integrator (current is integral of voltage), this should not be an unnatural analgoy. I suspect some people like better to compare Temperature T to voltage and heat transfer rate (btu/minute) to current. If you do that then the thermal system is the analogy of a capacitor (with large paralle resistance) driven by a current source w/ith current source turned on from zero initial conditions. That is the dual problem of an inductor driven by a voltage source turned on from zero initial condition.
I think it is simpler to leave thermal systems and capacitors out of it and just recognize the sense in which an inductor is an integrator.
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RE: Why DC offset in fault currents?
"The proof comes from considering that since there is zero net heat input from the heat pump, there must zero net heat input from the ambient over a full cycle"
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RE: Why DC offset in fault currents?
"The average temperature is 100."
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RE: Why DC offset in fault currents?
RE: Why DC offset in fault currents?
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