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    Why DC offset in fault currents?

    veritas (Electrical) 18 Jul 07 23:20

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

    Skogsgurra (Electrical) 19 Jul 07 0:02

    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 arent cleared after 180 degrees, you still have 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.

    davidbeach (Electrical) 19 Jul 07 1:06I hate to argue with Gunnar, but I don't buy that explanation.One of the basic laws of electricity is that current through an inductordoes not change instantaneously (without application of an infinitevoltage). The inductor in this case being the inductive reactance of

    the system up to the point of the fault. The fault happensinstantaneously and Ohm's law implies that the current change as aresult of the fault should also happen instantaneously. The DC offsetis the only way out of the problem of two laws giving oppositeresults. The Ohm's law effect is an instantaneous change in the (AC)current though the inductance and to keep the current through theinductor from changing instantaneously a DC current of opposite signis created to oppose the AC current. The system X/R determines how

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    rapidly the DC decays. Faults that occur at a current zero crossingdon't produce a DC offset while those that occur at a current max ormin produce the largest DC offset that circuit will produce. The actualDC offset will range between those values and will be different in eachphase.

    RalphChristie (Electrical) 19 Jul 07 3:03

    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 when the voltage is zero, the current cannot changeinstantaniously, 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.

    Skogsgurra (Electrical) 19 Jul 07 4:51OK davidbeach. Maybe I was too simplistic. I should have presentedthe differential equations showing what happens in detail. But, sinceVeritas 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 agood intuitive way out of that dilemma.I hope that you, at least, agree that the DC component stems from thefault happening somewhere else than at 90 or 270 degrees?

    electricpete (Electrical) 19 Jul 07 8:57Assume for simplicity a single-phase pure inductive circuit:

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

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

    If you start at the negative peak of v(t), then you havei(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 andgoes up for 1/2 cycle, then down for 1/2 cycle (back to 0), then up for1/2 cycles, then down for 1/2 cycle etc. This time it is cycling between0 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 fullyoffset waveform that doubles the true peak. That highest possiblepeak value would correspond to 2*sqrt(2)*LRC

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    Now add some resistance back in. The fully offset waveform will looksimilar 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 reach2*sqrt(2)*LRC. The higher the R/L, the more decay and the lower the

    worst-case peak associated with closing at v=0

    electricpete (Electrical) 19 Jul 07 9:07In case the graphical version of the integral is tricky, here is analyticalsolution of the single-phase pure inductive casev = L * di/dti = (1/L)* integral(v(t))Let v(t) = Vmax sin(w*t+theta)i(t) =-(1/L*w)* [cos(w*tau+tteta)]tau = 0...ti(t) =-(1/L*w)* [cos(w*t+theta) - cos(w*0+theta)]

    For theta = -Pi/2i(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 = 0i(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

    Skogsgurra (Electrical) 19 Jul 07 11:51Very good. But remember what the OP asked for: "Can someoneexplain in words why is there DC offset (or a dc transient) in faultcurrents?"

    Homerjs78 (Electrical) 19 Jul 07 12:29Please correct me (anyone) if this is too simplistic or down rightincorrect:The system contains inductance which stores a charge. In the case of

    a fault, the inductance contributes energy in the form of DCcurrent. As with any stored charge / energy, this decays according toa time constant.

    slavag (Electrical) 19 Jul 07 12:46Hi.Homerjs78, I bay your explanation.Gunnar and David also have good explanation, but your is very short

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    and in target.Electricpete, are you remambare all these formules??????Homerjs78, are you approve use your explanation for students?Best Regards.Slava

    supermacc (Electrical) 19 Jul 07 21:16Interesting discussion, I think all posters have provided goodexplanations, Homer has possibly nailed it but I would neverthelesslike to give my explanation. I have to use some equations but theexplanation is IMO simple and it is a slightly different approachcompared to previous posters.

    Assume that the short-circuit appears on the terminals of asynchronous machine (SM). Before the fault, the SM voltage equalsspeed 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 statorcurrent equals zero and then there is no stored magnetic energy inthe stator (but of course in the rotor) before the fault.

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

    There are many simplifications in my explanation, the full picture ismuch more complicated see, for instance, Kovcs: "Transientphenomena in electric machines".

    Skogsgurra (Electrical) 20 Jul 07 4:02I feel somewhat guilty for oversimplifying the problem. And when Ithen see others simplifying it even more - and some going the otherway 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 270degrees, then there is no DC introduced in the fault circuit. Any otherphase angle makes the net "polarity" either positive or negative, hencea DC component. (I should have included in my first explanation that itis the driving EMF that has a DC component and that it is this DCcomponent that creates a DC component in the current. Omitting thatstep was perhaps the great mistake).

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    Second: The other part of the problem is why the DC component staysfor a few, I still say 50 - 100 periods (time constants of a typicaltransformer secondary are in the seconds). And that is classical DCelectricity. The well-known L/R decay - same as RC decay - butopposite.

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

    slavag (Electrical) 20 Jul 07 5:07Hello 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 EHVand 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))*cosAA=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 (Electrical) 20 Jul 07 6:48Yes, we do agree. Time constant is one thing. The time to where thecurrent is (practically) zero is another thing. One convention is thatcurrent is close enough to zero after three time constants. Another isfive 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 justwanted to know how and why the DC component gets into the system.

    waross (Electrical) 20 Jul 07 7:42First I will comment on power.Real power we are familiar with.Reactive power, apparent power, KVA. This is often called power thatdoes 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

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    energized. When the circuit is de-energized The field decays and theenergy of the field is dissipated in the resistance of the circuit. If thecircuit is interrupted by an open circuit much of the energy may bedissipated in the resistance of the arc across the opening contacts. Ifthe 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 areincreasing and decreasing and then increasing and decreasing in theopposite direction in a manner that may almost be described by a sinewave, because an iron core inductor is almost linear.When there is a short circuit on an inductor, such as a transformer, inan AC circuit the discharge of the magnetic energy appears as a DCoffset. Although the time to discharge may be described in cycles, it isa function of the RL constant of the circuit and is not related to the ACcycles 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 thesame no matter the magnitude of the original current. The time forthe current to decay to a negligible value is much less with lessercurrents.hope this helps.respectfully

    slavag (Electrical) 20 Jul 07 9:05Gunnar, sorry, you are right!!!. You write down about time constantand I'm about clearing time, are different things.

    Waross , great!!!!.Regards.Slava

    Skogsgurra (Electrical) 20 Jul 07 9:09Great discussion - and needed.

    My point is that you do get an initial DC component even if you do NOThave any inductance in the circuit. The concept of inductance anddischarge may obscure that fundamental fact.

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

    waross (Electrical) 20 Jul 07 11:01I 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.

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    Sorry. This would be better if I had said;"Inductive power is a transfer of real energy. In the case of aninductive circuit, real energy is continually being added into and thenremoved from the magnetic field of the inductor."Although real energy is being transferred, the net power is zero.

    respectfully

    oldfieldguy (Electrical) 20 Jul 07 11:56Okay, thinking of this in a purely graphical representation, let's talkabout 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 +800amps. Fourth is -700. Still an offset of +100. The +100 difference isthe 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 adverselyaffected by this current profile to the extent that its secondary currentdid not match properly with the other window-type CT in a differentialcircuit. The result was chaotic until we figured out the problem.

    The bar CT remained partially magnetized by the DC offset on startingfor several cycles after the motor was running, but during thistransition currents in its secondary circuit were markedly different thanthose 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 graphicrepresentation.

    veritas (Electrical) 22 Jul 07 19:34Hi all

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

    Great discussion - and needed.

    My point is that you do get an initial DC component even if you doNOT have any inductance in the circuit. The concept of inductanceand discharge may obscure that fundamental fact.

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

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    therefore often used.Gunnar Englund

    If hypothetically the circuit has no reactance but only resistance, anda 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 isthat the current WILL start off at maximum value precisely becausethere is no reactance in the system. There is thus no need for a DCoffset.

    Am I correct?

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

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

    Truth - what is truth?

    electricpete (Electrical) 22 Jul 07 20:44I haven't read all the posts. I'm sure lots of good info. I did notice thatGunnar had some objection to my response which I don't understand. Ithink he is objecting to the fact that I analysed the situation of applyingvoltage to an inductive circuit, while the question was about a fault. It

    is the same situation. If you have a voltage source supplying a loadthrough an inductive impedance, the voltage drop occurs primarily onthe load. If you apply a short circuit accross the load, the voltage dropnow appears accross the inductance. Same thing as closing the switchto increase voltage accross the inductance.

    electricpete (Electrical) 22 Jul 07 20:51I agree with veritas on the point that inductance is requiredsomewhere in the circuit for the dc current offset to occur... eitherupon energization or during a fault.

    electricpete (Electrical) 22 Jul 07 20:57If the objection to my original post has to do with equations rather thanwords, then I would suggest to do the integration graphically. It tellsthe story on a pretty intuitive level imo.

    hold6448 (Electrical) 22 Jul 07 22:28

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    If you interested in a good reference, take a look at "Transients inPower Systems" by Greenwood.

    Skogsgurra (Electrical) 23 Jul 07 3:32Hmm..

    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 dctransient) 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 ifstarted at any other phase angle than 90 or 270 degrees. That was all Isaid. And I thought that was what the OP asked for.

    The L/R time constant in the circuit, of course, then dictates whathappens to the DC component. No discussiuon there. I may haveexpressed myself unclear (the "contradiction" part) when I, as veritassays, talk about a circuit with no inductance and then, in the nextsentence, mention that the L/R determines what happens to the DCcomponent. But, if you read that sentence, you will see that there is nocontradiction 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 DCcomponent also without any inductance. I did not not say that there isno inductance. Only that you can get a DC component also with noinductance. What then happens to that component is dependent onthe 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 thinkthat I try to tell lies. That was not my intention. All I wanted to do wasto 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.

    electricpete (Electrical) 23 Jul 07 8:32Gunnar - It was never my intent to rank one explanation againstanother. You are the one who objected to my explanation.

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    I have the highest respect for your opinion in general, but I think youhave missed the mark on this particular item. The dc offset is acharacteristic of change in configuration of an inductive circuitimo. Your explanation 7/19/07 as I understand it is dependent upontime 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 faultinitiation. 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 smoothlyinto the steady state solution i(t) = -/+ cos(w*t) * V/(L*w). If you startintegrating at t=0 or Pi, you have the worst case fully offsetsinusoid. 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/Lcircuit fed by a DC voltage source switched on at t=0. The response toclosing the switch at t=0 is i(t) = V/(R) - V/(R) * exp(-t*L/R). Thatresponse has two term. The first of these terms is the steady stateresponse (which happens to be dc in this dc case, but in general is notnecessarily dc).. The second of these terms is the transientresponse. This transient response is analogous to what we're callingthe dc offset during a fault IMO.

    electricpete (Electrical) 23 Jul 07 9:50One correction:

    "If you start integrating at t = +/- Pi/2 you jump smoothly into thesteady state solution i(t) = -/+ cos(w*t) * V/(L*w).'

    should be:

    "If you start integrating at t = +/- Pi/2 you jump smoothly into thesteady state solution i(t) = -cos(w*t) * V/(L*w).'

    jghrist (Electrical) 23 Jul 07 10:01To say that there is a dc offset in a resistive circuit just because thecurrent starts off positive is misleading IMO. In a resistive circuit, thecurrent will be sinusoidal. Let's say the current is Icos(wt). This ispositive at time t=0, but to say that there is a dc offset is not correct.

    It is easier to visualize the problem with a purely inductive system. Inthis case, the voltage across the inductance equals the driving voltagewhich is a sinusoid. The voltage is Ldi/dt. The current will also be a

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    sinusoid, with or without a dc component.

    Because the current across the inductance cannot changeinstantaneously, 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 ispositive and at a maximum. So current is increasing with the highestslope. When the voltage reaches zero, di/dt must be zero, and thecurrent reaches its peak. If you follow this through a whole cycle andplot 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 iszero. As the voltage increases, di/dt and i increase. When the voltagereaches a maximum, di/dt is a maximum, with the current stillincreasing. When the voltage is gets to zero, di/dt is zero and thecurrent is at a maximum. When the voltage goes negative, di/dt is

    negative, but the current is still positive. If you follow this through awhole cycle, you see that you get a sinusoid with an offset equal tothe peak current so that the minimum current is zero.

    electricpete (Electrical) 23 Jul 07 14:19

    I agree with jgrist and many others.

    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)). Bydefinition it always has zero-crossing at t=0.

    The voltage source is connected to the inductor (undamped) at somevariable angle theta (or delay time) after t=0

    You can vary the angle in the attached spreadsheet and watch theoffset appear for theta=0, 180, disappear for theta = 90, 270.

    (How this relates to fault: fault causes a sudden decrease in loadimpedance and sudden increase in voltage seen across the inductor.)

    electricpete (Electrical) 24 Jul 07 23:34fwiw, I revised the spreadsheet above so you can specify L and R inaddition to the time delay (in degrees) to see a damped response.

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

    For example, you can see the initial input data with X/R ~ 13, gives atrue 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 seenon a fault, but also during motor starting and during transformerinrush... except for the transformer inrush the high exciting currentdue to dc offset creates saturation which leads to significant harmonicsin the current waveform (not modeled here).

    If anyone is interested, the method I used to simulate the LR circuit isderived as follows:v = L di/dt + IR

    L di/dt = v - IRdi/dt = v/L - IR/ Ldi = v/L dt - IR/L dtI = Int v/L dt - Int IR/L dtLet 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 dtIk (1+0.5*R/L dt) = Ip + 0.5*(Vk+Vp)/L dt - 0.5*Ip*R/L dtIk = {Ip + 0.5(Vk+Vp)/L dt - 0.5 IpR/L dt} / (1+0.5*R/L dt) [THIS ISTHE 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]

    slavag (Electrical) 25 Jul 07 2:22Dear Electricpete.Thank you!! Your material is very intresting and helpfull. As Gunnarsaied "Great discussion - and needed".Thank to all of you.Best Regards.Slava

    RalphChristie (Electrical) 25 Jul 07 3:06

    Pete:

    PLS for you

    I do not think anyone was totally wrong in the discussion, it was just explained fromdifferent angles and viewpoints and with different examples. With the lack of pictures it

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    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 isgoing on.

    rbulsara (Electrical) 25 Jul 07 13:40Homersjs78 deserves a star for simple and accurate explanation.

    electricpete (Electrical) 25 Jul 07 20:50Quote: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, theinductance contributes energy in the form of DC current. As with any stored charge /energy, this decays according to a time constant.

    I would respectfully disagree. To me it sounds like the idea beingportrayed is that there is initially energy stored in the inductor at thetime of the fault, and the fault allows this initially stored energy todecay 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 feedingthrough an inductance (perhaps a transformer) to a line which has noloads prior to the fault. The initial energy stored in theinductance prior to the fault is zero. Now close in a low-resistance representing a fault on the transformer secondary at themoment the voltage is zero. We get the maximum dc offset. It didn'tcome from the energy initially stored in the inductor, because thatenergy was zero in this case.

    At the risk of being really obnoxious, let me try another explanationexpanding 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 sinusoidalresponse plus a transient response.

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    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 isif Isteadystate(0+) = 0

    8 - Isteadystate(0+) = 0 occurs only if we close the fault at a phaseangle where the steady state solution for current is 0. This occurs onlywhen 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 thanthe maximum or the minimum of the supply voltage, you will have atransient component of the current (the exponentially decaying dccomponent). That transient current occurs as a direct result of thecharacteristic of an inductor v=L di/dt and we can calculate it directlyfrom the corresponding integral equation i = Integral v(t) dt + i(0)

    It is not a dissipation of energy stored before the fault. It is atransient response. In general we expect a transient response

    when the system configuration changes. The transientresponse in a simple L/R system will always be of the formI(0)*exp(-t*L/R)

    I am a hearty supporter of the idea that there is not only one rightanswer. There were good answers given by many others. Mine isprobably not the simple answer some people are looking for and mineis not without equations as was requested in the original post. Myapologies. Feel free to comment. Otherwise I will shut up.

    electricpete (Electrical) 25 Jul 07 20:59

    some obvious errors:"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)"

    sslobodan (Electrical) 26 Jul 07 7:32Sorry i got this late into conversation BUT i think that the guy withpost above hit the spot. When you have integral I(x) you have

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    constant that is result of beginning state (the thing you neglect) andthat component is constant, as we all know from basic ofelectrotehnics constant current is usually called DC current which isanswer to original question. That i(x)=C (as a constant) is aconsequence of beginning state when malfunction occurred and has

    some to do with the type of malfunction (symetric or asymetric ) andis one of tree components (Id - direct Ii - inverse Io - DC component ) Ithas been a long time since I had this subject on college so to make amore profound explanation I would have to read some 200 pages ofmy book from Analysis of EE systems, so I believe who wants themore detailed explanation of DC component would have to do thesame. Unfortunately I don't know any book of such kind in NorthAmerica because I never lived there so if anyone knows such heshould post it here ....

    rbulsara (Electrical) 26 Jul 07 11:12

    Electricpete:I have high regard for your math skills. But this is not about math butgrasping a concept. Poor electricity does not undestand or even knowmath. Math is a means to analyse many phenomena. So it does notmatter how it is explained by math. A waveform can be resolved intoany number of compoenets to arrive at a final waveform. AssumingDC component fits the mathematical explanation for the Asymmetryin fault current wavefors. (This is very akin to resolvoing vectors in xand y rectangular component for anaysis). The fact is the systembehaves as if it has a DC component which can only come from somesort 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.

    jghrist (Electrical) 26 Jul 07 12:08Quote:The fact is the system behaves as if it has a DC component which can only come fromsome sort of stored enegy being discharged.

    But if there is no current before the fault, as Electricpete noted before,where is the stored energy? There will still be a DC component if thefault occurs at the right time (0 for a purely inductive line). If there is

    current before the fault, you could say there is stored energy in theline inductance, but if the fault occurs at 90, there is no dccomponent. What happened to the stored energy?

    I agree that these discussions do not give a good physical descriptionas the OP wanted, but it's hard to describe physics without math. Ithink Davidbeach's original post gives the best physical explanation.

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    davidbeach (Electrical) 26 Jul 07 12:36Thanks jghist, I was feeling rather ignored.

    Skogsgurra (Electrical) 26 Jul 07 12:43What about me then? Not a single PLS...

    slavag (Electrical) 26 Jul 07 13:21Folks.Many PLS to all of you.Special to Gunnar:):):)

    rbulsara (Electrical) 26 Jul 07 13:45Jghrist:

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

    circuits(inductors and capacitors), which would discharge upon seeingpath of adequately low impedance such as fault. Just like an ACcapacitor when shorted to ground will show DC current decaying,untill then its fat, dumb and happy as the load impedance is not lowenough for it to discharge appreciably.

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

    rbulsara (Electrical) 26 Jul 07 13:52Or a better explanation could be the inductor is constantly chargingand discharging (changing polarities) in normal state. So there arepoint in time where there is no net charge stored in the inductor and ifthe fault occurs then there is no DC offset. Any other instances therewill be some charge stored in the indcutors (or capacitors) which willdischarge as DC current.

    Skogsgurra (Electrical) 26 Jul 07 17:10It might interest you to have a look at thread248-144465:

    Transformer Inrush Currents and especially the recordings of inrushcurrent in a transformer. As many posters have said, it is an analoguesituation.

    electricpete (Electrical) 26 Jul 07 20:22rbulsara - I would respectfully suggest that you are mistaken.Quote:There is always "energy" stored in reactive devices in AC circuits

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

    Quote: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 primarybreaker opened during a heavy inrush/transient. That proves the inductor discharge.It proves that an inductor discharges with a transient responseI(0)*exp(-t*R/L) and in this case it is very accurate to say that itrepresents decay of initial inductor current and/or dissipation of energyinitially stored in the inductor.The decaying DC offset which appears during a fault has exactly thesame functional form. (If you recall, I mentioned 25 Jul 07 20:50 thatthis same functional form applies to the transient component for alltransients of the simple LR system). But in this case it is no longeraccurate to state that it represents decay of initial current. As has

    been stated twice already, you can have a transient decaying dc offsetwhen the initial current at t=0 is 0! (that is three times now).

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

    First, my understanding is that you mean to use an analogy betweencapacitor/charge/voltage and inductor/flux/current. If there is nocurrent or flux in th inductor when the fault occurs, we can still have adc 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 stillopen).

    When you open it up, change theta_degrees to 360. I hope you'll getthe picture that the current (and stored energy) in the inductor is 0 fora full cycle of the graph before the fault occurs. By your logic there isno offset. But in fact there is an offset.

    bjenks (Electrical) 29 Jul 07 0:34Lots of good stuff here, but I would have to refer back to IEEE Redbook Chapter 4 as the best explaination on DC offset. It even talksabout switching effects of relays and the mutual inductance betweencables. It give an example of what happens when you turn your waterhose on how there is a surge at first. I always looked at the fact thatthere is stored energy in the form of capacitance between wires andinductance of the cables. When a short happens you have a collaping

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    magnetic field from the energy in the system causing a Faraday DCvoltage.

    jghrist (Electrical) 29 Jul 07 12:24Quote:

    I always looked at the fact that there is stored energy in the form of capacitance betweenwires and inductance of the cables.Then why isn't capacitance between the wires in the equations usedto calculate dc offset?

    Skogsgurra (Electrical) 29 Jul 07 14:15This 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 bedistorted the way it has been doesn't really make me happier. Bringingin capacitance between wires and postulating that the stored energy inan inductance causes a high fault current to flow is not physics. Nor isit 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 toexplain why there is sometimes a DC component in the EMF. And it isso simply because any other phase angle than 90 or 270 degrees doesintroduce a DC component. I left out the part about DC current buildingup and flowing in the circuit, decaying according to the L/R timeconstant. Davidbeach's complementary information is correct and was,perhaps, necessary for those who cannot draw her/his ownconclusions.

    But, as is often the case, it opened a Pandora's box. And it attractedsome posters that found it interesting to post their ideas on the

    subject. And that didn't make it any clearer. Physics is not politics. Youcannot have an "opinion" on things like this. Facts are facts and thereare 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.

    rbulsara (Electrical) 29 Jul 07 22:24

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    I still would still say nature (here electricity) does know math, math isfitted to rationlize what we know to be happening. Also I did not implythat 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, evenafter the source is removed and therefore the practice of dischargingthem before any maintenance.

    davidbeach (Electrical) 30 Jul 07 0:47While I am well aware of trapped charge (voltage) on a capacitor, I'venever heard of trapped current in an inductor. Interrupting deviceshave a tendency to open at a current zero in AC circuits leaving nocurrent to be trapped. In DC circuits or AC circuits where the currentis chopped there can be nasty voltage transients as the current in the

    inductor tries to continue to flow, but once the circuit is interruptedthe inductor has nothing trapped.

    I agree 100% with Skogsgurra that stored energy has absolutelynothing to do with DC offset in fault currents. And it is exceedinglysimple to prove that. Establish a source with an inductor and leave itopen circuited for a moment, but apply an AV voltage. With it opencircuited there is no current flow. Now apply a fault at a voltagemaxima. Note the DC offset in the fault current. Because three wasno current flow prefault there could not have been any stored energyin the inductor. It doesn't matter where in the AC voltage cycle the

    fault happens; there is never any stored energy in theinductor. Where in the cycle the fault happens does determine theamount of DC offset.

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

    bacon4life (Electrical) 30 Jul 07 23:15The DC component comes from a lackof stored energy rather thanan 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 theinductor is zero. The first quarter of the cycle (second half of positivewave) is spent adding energy to the inductor and the second quarterremoves all the energy and returns the current to zero. The third putsin energy of the oppposite polarity and the fourth removes it.

    During a fault when closing at other than maximum voltage, there willbe no energy stored in the inductor(no current flow). Now the rest of

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    the positive wave charges the inductor instead of just the secondhalf.

    Notice that now the current increases for two quarter cycles in arow(resulting in twice the current). This is the key difference in that

    normally the first part of the positive waveform reduces energy storedin the capacitor. Since there is no energy stored at the beginning, thecurrent 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 thansteady state, the difference is that there is zero energy stored insteadof various amounts thoughout the cycle.

    521AB (Electrical) 31 Jul 07 14:13To 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 theabove energy at the moment of the short. Like a capacitor, theinductance will "discharge" its energy in form of heat (Joule effect) onthe resistor, and this is done by generating a current with anexponential decaying curve.

    Wrong?

    jghrist (Electrical) 31 Jul 07 14:18Quote: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 formof 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 (Electrical) 31 Jul 07 14:56Dear 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

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    DC component : Angle, L and R it's all. Please remamber circuit withAC 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/oany phisical connection to DC offset.

    ( I'm also mistaked in my first post, sorry), but checked myself again.Best Regards.Slava

    davidbeach (Electrical) 31 Jul 07 15:32If energy storage has anything to do with it, and frankly I doubt it,then bacon4life comes far closer than anyone else who mentionedenergy storage. Any talk of energy storage has to deal with the caseof a fault on a previously open circuit where i(t)=0 for all t

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

    521AB (Electrical) 3 Aug 07 5:47Let 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 whenit touches the ground by solving the fundamental motion differentialequation F = m a. The result is a mathematical result that has to beaccepted, 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 intuitivelyyou do not need. It's enough to say that the mass, falling, transformsthe potential energy into kinetik energy. Potential has to do withheigth, kinetik has to do with speed.

    Who is right? Both.

    I think Veritas was looking for an answer similar to the seconddescription, 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 tochange 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 tostore is similar to kinetik energy of a mass? There is one inertia.... (aswell as the capacitor: we can assume it stores potential energy..)That's why I was thinking in terms of energy. But I agree that Icouldn't explain where the dc component is coming from intuitively.

    Where is the dc component coming from?

    Skogsgurra (Electrical) 3 Aug 07 10:34The old lexicon definition of "Iteration" comes to mind:

    Iteration. (s) See Iteration.

    davidbeach (Electrical) 3 Aug 07 11:56Since the phenomena can be entire explained without introducingenergy storage, what could energy storage possibly add except

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    confusion? And, because it can be entirely explained withoutintroducing energy storage, the addition of energy storage is simplywrong. It doesn't matter how much you want it to be about energystorage; this is not about opinions, it is about facts and there are nofacts that support energy storage as a mechanism for explaining DC

    offset.

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

    521AB (Electrical) 4 Aug 07 11:03Davisbeach,to my opinion you are just commenting the result of the differential

    equations governing this phenomenon. You are also commenting theLenz laws.But to my opinion you are not explaining, with easy words, where thedc current is coming from.We can maybe use the Maxwell's equations anc commenting them, inorder to explain this.But I think we are not mastering this issue untill we are able to explainwith 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 boyat elementary school can understand. No differential equations.A boy at elementary school can understand oscillation between L andC, if we make a comparison with water, tanks and pumps.A boy at elemtary school can understand the charge of a capacitor, ifwe tell him that the capacitor is a tank, current is water and it needstime 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.

    electricpete (Electrical) 4 Aug 07 11:44If you wanted to say that the exponentially decaying dc patternrepresents dissipation of energy in the dc component of the current, Ithink that is fairly accurate.

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    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. Quotefollows:

    [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 theabove energy at the moment of the short. Like a capacitor, theinductance will "discharge" its energy in form of heat (Joule effect) onthe resistor, and this is done by generating a current with anexponential decaying curve.[quote]

    Again I disagree with that last quote and that incorrect simplification isthe whole reason I objected to the way energy had been discussed.

    If you wanted to include energy in the description of where the dcoffset ORIGINATES from (not how it decays), you might start byexamining the sinusoidal steady state energy storage of aninductor. We know current lags voltage by 90 degrees. There are fourquadrants 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 wehave max energy. At the end of period 3 we have 0 energy. At theend of period 4, we have max energy.

    Now let's return to our simplest situation, fault occurs with 0 initialcurrent and angle theta=0. The system does not begin in steadystate. For the first half cycle the voltage is positive. But guess what,current remains positive through and beyond the first half cycle aswell. 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 addedenergy for longer than steady state, the final energy level at the end ofthe first half cycle is higher than it will become in the post-fault steadystate condition.

    So I would say that the energy associated with the worst-case dc offsetis added to the LR system by the power source during the first halfcycle after the fault (it is not present before the fault).

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    That is my best attempt to discuss the subject with energy. I don'tthink it made it any simpler. The simplest approach IMO:v = L di/dti = (1/L) int v dt

    Start integrating a V=V0sin(w*t) at t=0 and you will see the dcoffset. If you haven't done it yet, get out your pencil and paper anddo the integration graphically.

    electricpete (Electrical) 4 Aug 07 11:47A very important correction in bold:Quote:"But if you say that energy came from energy stored within the circuitbefore the fault(as

    you implied previously), that is the part I disagree with."

    electricpete (Electrical) 4 Aug 07 11:59Also, in my discussion above I have assumed for simplicity that thecircuit is inductive with negligible resistance (except where Imentioned decay).

    waross (Electrical) 4 Aug 07 12:09Hi David.I am surprised to read your comments.I have read many books that describe a coil surounded by a magneticfield 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 thecircuit resistance. The circuit resistance in the event of a short circuitwill equal the coil resistance. When the voltage is removed from aninduction coil, the DC current will continue for an easily calculatedtime. Is this not a form of stored energy?When an induction coil (that may also be a transformer winding) isenergised with AC there will be an alternating magnetic fieldsurrounding the coil. This field will store and return energy to thecircuit as the current alternates. If the coil is short circuited anyenergy stored in the magnetic field at that instant will be disipated inthe resistance of the circuit, that may be soley the resistance of thecoil or transformer winding.A magnetic field will store energy whether the cause of the field is ACor DC. A short circuit or fault has no commutating mechanism so thatenergy must be disipated as a DC offset.If your calculus seems to indicate an absence of stored energy, Isuggest that you disect your arguments. I think that you will find thestored energy quietly hiding behind one of your derived terms.

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    KISSEnergy stored in a DC induced magnetic field is acceped.Energy is alternatively stored and returned in a manner that may beapproximated 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 DCinduced field.Respectfully

    electricpete (Electrical) 4 Aug 07 12:45I'm sure it is known to all the participants, but I will mention that theenergy storage characteristics of an inductor (Energy = 0.5 * L * i^2)are inextricably tied to the terminal characteristics (v = L di/dt).

    Stored Energy = 0.5 * L * i^2d/dt (Energy) = v i = 0.5 * L * d/dt(i^2)

    Using the chain rule for differentiaion:d/dt(i^2)=2 i di/dtv i = 0.5 * L * 2 i di/dtSimplifying:v = L di/dt

    I didn't interpret David to state anything to the contrary. I heard himexpress an opinion that energy does not add much to the question ofwhere does the energy come from. That is the same sentiment Iexpressed above.

    Quote:

    When the voltage is removed from an induction coil, the DC current will continue for aneasily calculated time. Is this not a form of stored energy?Agreed 100%.

    Scenario 1 - (waross' scenario) If you short out the voltage source of aninductor, the initial current will continue to flow and decay as theenergy is disippated in the inductor. You will note in this type oftransient, the current is never larger than the initial current.

    Scenario 2 - The transient associted with worst case dc offset after afault. The voltage accross the inductor INCREASES (it is not shorted to

    0). The current flowing the inductor can INCREASE beyond the valuethat 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 energyassociated with the dc comopnent decays away as it is dissipated inthe circuit resistance giving a dc current of exp(-t*R/L).

    The two scenario's have an important difference in terms of where the

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    energy associated with the dc offset came from. In scenario 1 where Ishort out the votlage source, the energy came from the prefaultcircuit. In scenario 2 where I applied a fault, the energy associatedwith the dc offset was delivered into the circuit by the power suplywithin the first half cycle after the fault.

    electricpete (Electrical) 4 Aug 07 12:48correction in bold: "Scenario 1... as the energy is disippated in theresistor. You will note in this type of transient, the current is neverlarger than the initial current."

    slavag (Electrical) 4 Aug 07 13:29Hi.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 ACSOURCE SHORT, this is a point.Possible iteration with water, valve and pipe.You close valve in begining of pipe and water continue flowaccording 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 trytogether write it.Same pipe is closed and we have valve, possible open this valve withangels 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.SlavaOhhh

    521AB (Electrical) 4 Aug 07 14:46

    I have written several times that I have not been able to explain theinitial question. Why are you continuing to underline it? If this makesyou feel better, feel free to continue. By the way, I also wrote, in myfirst post: "... Wrong?".I read the first post (very good one), I got shocked when I realized thatI 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,

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    and my teacher had asked the same question, I had answered that itis just the transient solution of the differential equation, whichdepends on the initial conditions etc... Teacher happy, me happy.I also assume that everybody in the electric engineering field has thisknowledge. So, please, no integrations or differentiations anymore,

    not even graphics or numerics.

    Back to the issue: usually energy easily explains thinks that mightappear complicated, that's why I went in the energy direction. But Isee that it is not helping in this case.

    What are we analyzing? Actually an elementary differential equationof the first order, energized by a sinus waveform. This is maybe whatis making everything complicated.Can't we use a machanics model:

    Look: one mass energized by the force Force(t) = sin (w * t) in a gaswhere 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 dccomponent in the speed of the mass. If we remove the friction withthe gas (electrically, we assume R = 0), this dc component will neverdecay.

    Still I don't see it...

    521AB (Electrical) 4 Aug 07 14:47Or another idea: what happens if instead of a sinus we consider asquare wave, symmetrical to zero?

    slavag (Electrical) 4 Aug 07 14:56Hi 521AB.Is not so simple Q.Slava

    davidbeach (Electrical) 5 Aug 07 16:59waross, energy storage in an inductor is a function of current. If theprefault current is zero there is no stored energy. Lack of prefaultcurrent does not impact the direction or magnitude of the DC offsetfollowing initiation of the fault compared to non-zero prefaultcurrent. What would consideration of energy add to that?

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    electricpete (Electrical) 5 Aug 07 22:02To 521AB:

    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 notperfectly insulated container). The ambient temperature outside is100F and constant.We arrange a programmable heat pump that adds exactly 1BTU/minute for 10 minutes, then subtracts one BTU/minute for 10minutes, then repeats.We know that during 10 minute half cycle , the 10BTU will equate to a10F change in the water temperautre.What does the steady state temperature profile look like? Intuitivelywe suspect that it cycles from 95F to 105F. The proof comes fromconsidering that since there is zero net heat input from the heat

    pump, there must zero net heat input from the ambient over a halfcycle, which can only happen if the temperautre profile is symmetricwith respect to ambient temperature.So now we know the steady state profile. At 95 heat turns on andheats to 105F. At 105F, coole turns on and cools to 95F. The averagetemperature is 0.Now turn off my heat pump for a day and let the system come toequilibrium with the 100F ambient.Now turn on the heat pump starting with a 10 minute heatingcycle. The temperature will go from 100F to 110F. Then the coolpump will come on and bring back to 100F, then back to 110F

    etc. Over time since the average temperature of this transient systemis above 100 ambient temperautre, it will lose energy to the ambient,until it reaches the steady state solution cycling between 95F and105F.Comparing the our inductor: heat transfer rate (1btu/minute)corresponds to voltage. Temperature corresponds tocurrent. Ambient temperature (100F) corresponds to 0 current. Smallheat leakage to ambient corresponds to series resistance.If you appreciate that the inductor is just an integrator (current isintegral of voltage), this should not be an unnatural analgoy. I suspectsome people like better to compare Temperature T to voltage and heat

    transfer rate (btu/minute) to current. If you do that then the thermalsystem is the analogy of a capacitor (with large paralle resistance)driven by a current source w/ith current source turned on from zeroinitial conditions. That is the dual problem of an inductor driven by avoltage source turned on from zero initial condition.I think it is simpler to leave thermal systems and capacitors out of itand just recognize the sense in which an inductor is an integrator.

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    electricpete (Electrical) 5 Aug 07 22:04minor correcction:"The proof comes from considering that since there is zero net heatinput from the heat pump, there must zero net heat input from theambient over a full cycle"

    electricpete (Electrical) 5 Aug 07 22:05Another correction. In the description of steady state operation:"The average temperature is 100."

    jghrist (Electrical) 5 Aug 07 22:18Quote:I think it is simpler to leave thermal systems and capacitors out of it and just recognizethe sense in which an inductor is an integrator.I agree with this. I respectfully submit that anyone who finds thistemperature analogy easier to understand is reading the wrong Eng-Tips forum. They should be reading the Heat Transfer &Thermodynamics engineering Forum.

    electricpete (Electrical) 5 Aug 07 22:20In case the final conclusion of the analogy wasn't obvious, I shouldhave mentioned that in this case the offset would correspont to the 5Fincrease in average temperature after turn-on (105F average duringthe first few cycles after turn-on) compared to the averagetemperature during steady state (100F).

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