Why DC offset in fault currents? 9

[veritas]

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

Thanks in advance.

Veritas

 

[521AB]

Davisbeach,
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.

 

[electricpete]

If you wanted to say that the exponentially decaying dc pattern represents dissipation of energy in the dc component of the current, I think that is fairly accurate.

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:

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.

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.


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

 

[electricpete]

A very important correction in bold:

"But if you say that energy came from energy stored within the circuit before the fault(as you implied previously), that is the part I disagree with."

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

 

[electricpete]

Also, in my discussion above I have assumed for simplicity that the circuit is inductive with negligible resistance (except where I mentioned decay).

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

 

[waross]

Hi David.
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

 

[electricpete]

I'm sure it is known to all the participants, but I will mention that the energy 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^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.

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?

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.

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

 

[electricpete]

correction in bold: "Scenario 1... as the energy is disippated in the resistor. You will note in this type of transient, the current is never larger than the initial current."

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

 

[slavag]

Hi.
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

 

[521AB]

I have written several times that I have not been able to explain the initial question. Why are you continuing to underline it? If this makes you feel better, feel free to continue. By the way, I also wrote, in my first post: "... Wrong?".
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... :-(

 

[521AB]

Or another idea: what happens if instead of a sinus we consider a square wave, symmetrical to zero?

 

[slavag]

Hi 521AB.
Is not so simple Q.
Slava

 

[davidbeach]

waross, energy storage in an inductor is a function of current. If the prefault current is zero there is no stored energy. Lack of prefault current does not impact the direction or magnitude of the DC offset following initiation of the fault compared to non-zero prefault current. What would consideration of energy add to that?

 

[electricpete]

To 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 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.


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

 

[electricpete]

minor correcction:
"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"

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

 

[electricpete]

Another correction. In the description of steady state operation:
"The average temperature is 100."

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

 

[jghrist]

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.

I agree with this. I respectfully submit that anyone who finds this temperature analogy easier to understand is reading the wrong Eng-Tips forum. They should be reading the Heat Transfer & Thermodynamics engineering Forum.

 

[electricpete]

In case the final conclusion of the analogy wasn't obvious, I should have mentioned that in this case the offset would correspont to the 5F increase in average temperature after turn-on (105F average during the first few cycles after turn-on) compared to the average temperature during steady state (100F).

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