Asymmetrical fault current

[lyledunn]

Take a 400v three phase and neutral supply fed from a 11Kv transformer, ( common in UK for small commercial installations).Can the asymmetrical fault current exceed the symmetrical current? If so how?

 

[mpparent]

The assymmetric fault currents can exceed the symmetrical, it really depends on your x/r ratios.

 

[dpc]

As mpparent notes, the total asymmetrical current is a function of the system X/R ratio at the fault point. The higher the X/R ratio, the greater the possible asymmetry. The asymmetrical current ALWAYS reaches a maximum value in the first half cycle of the fault. This is a transient that damps out after few cycles. In a highly inductive circuit such as a power distribution system (neglecting the loads), the current lags the voltage by nearly 90 degrees. When the fault occurs at a zero voltage crossing, this forces the current waveform to be displaced from the normal balanced waveform. This is essentially a dc transient.

Breakers have a symmetrical rating for fault currents but this is based on a maximum X/R ratio (or power factor) used during breaker testing. If the X/R ratio of your system exceeds the test value, the breaker must be de-rated.

Also, the magnitude of the asymmetrical current also depends on the phase angle of the voltage at the time the fault is initiated. The worst case is when the fault occurs at a voltage zero. If you're lucky, there is no asymmetrical current, but of course, you can't count on this.

dpc

 

[230842]

For a given transformer and when 11 kV supply network short circuit power is much higher than nominal transformer nominal power (generally it is so), then no appreciable difference does exist between balanced s.c. (three phases to neutral) and an umbalanced s.c. (one single phase to neutral) while both s.c. currents are higher than that of an unbalance s.c. (phase to phase)
When s.c is considered from any line to ground, then grounding impedance plays a very important role on umbalanced or assymetric s.c. in the case of one single phase to ground, as grounding impedance increases the s.c. impedance path reducing thus s.c. current values Julian

 

[electricpete]

I'm no expert. I think responses from mparent and dpc are correct. I'm sure you are already aware of the definition, but there is some info in Powell Tech Brief #22 at

http://www.powellelectric.com/

I'll cut and paste below:

The figure below shows a typical short circuit current wage form and defines the various component parts of this wave. At the moment of initiation of a short circuit the ac current wave, which is normally symmetrical about the zero axis BX is offset by some value, creating a waveform which is symmetrical about another axis, CC'. The degree of asymmetry is a function of several variables, including the parameters of the power system up to the point of the short circuit and the point on the ac wave at which the short circuit was initiated. In a 3-phase circuit, there is usually one phase which is offset significantly more than the other two phases.

It is convenient to analyze this asymmetrical waveform as consisting of a symmetrical ac wave superimposed on a dc current. CC' represents the dc current, and the value of that current at any instant is represented by the ordinate of CC'. The dc component of the current normally decays rapidly, and reaches an insignificant value within 0.1 s in most power systems. The rate of decay is a function of the system parameters. When the initial value of the dc current is equal to the initial peak value of the ac current, the resulting waveform is said to be fully offset, or to have a 100% dc component. It is possible, in some power systems, to have an offset in excess of 100%, which may result in a waveform that has no current zeros for one or more cycles of the ac power frequency.

The ac component of the short circuit current will also decay, at a rate dependant on the system parameters. In general, the closer the fault is to generators or other large rotating machinery, the faster the decay will be.

In the figure, IMC is the crest, or peak, value of the short circuit current. It is the maximum instantaneous current in the major loop of the first cycle of short circuit current.

The rms symmetrical value of the short circuit current at any instant, such as EE', is the rms value of the ac portion of the current wave. Its value is equal to , and it is shown graphically by the distance from CC' to DD'. The rms asymmetrical value of the short circuit current is the rms value of the combined ac and dc waves, and it is calculated by the formula: I = sqrt((Iac(2)/2 + Idc).

From the last line we see the assymetrical current is higher than symmetrical whenever dc is present. DC is present whevever load is increased (due to fault) in a step fashion. If we look at the fact that voltage accross system impedance will be increasedn suddently, we see I=1/L*Integral(Vmaxsin(2pift))dt which will have dc offset depending upon the phase at which the fault occurs.

So we have not addressed the path in which the dc current flows, which I suspect was your real question. Does anyone have more comments on that?

 

[230842]

I think we are referring to different concepts, hence the so different answers to lyledunn question.
In case we are talking about asymmetrical transient currents,Electricpete reference text is right,and the máximum current peak could reach twice the value of steady s.c. current peak (even more when transformer is close to generator).
On the contrary, if we refer to asymmetrical faults, i.e short circuits of one or two lines to neutral, ground or each other, then my former post sould be taken into consideration Julian

 

[peterb]

Just a couple of points here -
- The term "asymmetrical" in this context is usually taken to have the meaning that electricpete has assumed
- The term "unbalanced" could be used to describe the meaning assumed by 230842
- Depending on the system grounding arrangements, it is entirely possible for the ground fault current level to exceed the 3-phase fault current

 

[Electrifier]

hi friends,

I am not much competent like you all but i have read some where that Asymmetrical current is vector sum of AC and DC components of current at the time of inrush current.Symmetrical current is vector sum is positive sequence, negative sequenceand zero sequence currents.

 

[jbartos]

Suggestion: The asymmetrical current must contain the dc component called dc offset. This is causing the asymmetrical current to be greater than symmetrical. If there is no dc offset then the current is only symmetrical; however, it may contain a symmetrical transient. DC offsets are often present in faults that are very close to generators or transformers. Further away from generators or transformers, the dc offsets can be negligible. Normally, the asymmetrical current is also expressed in asymmetrical rms values.
The electricpete reference link has the following:

http://www.powellelectric.com/

click on technical briefs and then on PTB#22

 

[ijl]

When the initial short circuit current Ik" (without the DC component) is considered, then some experimentation with a short circuit program confirmed julian's findings. An additional observation is that in the case of a line to line to neutral fault, the current can be larger than in the case of a 3-phase fault, depending on the impedances in the network. You can check these points with a short circuit program, for example with the one at

http://pp.kpnet.fi/ijl

 

[ijl]

Well, once more. I started to wonder, how did I get that result. With the help of some more calculations and a textbook I realized the obvious: If the zero sequence impedance is smaller than the other impedances, then the line to neutral and line to line to neutral fault currents can be larger than the 3-phase fault current (always the Ik", without the DC component, that is). But such a situation is rather unlikely in a distribution network. Or is it ? (That was a good refreshing homework assignment. Oops.., maybe it really was one

 

[cuky2000]

Here is some possible explanation of the asymmetrical and symmetrical SC values on a three phase SC:

The peak asymmetrical value of the current after initiation of a SC could be calculated as a function of the symmetrical rms value of the initial short circuit current at the instant the SC occurs as follow:
Ipeak_asymm = k.Irms_symm Eq (A)

Approximately the following relation could model the asymmetrical factor for a three-phase SC as a function of X/R:
k=1.0220+0.96899.e^{-[3.0301/(X/R)]} Eq (B)

Considering two extreme cases:
a- X/R=infinite...........k(inf)=2 (perfect asymmetrical)
b- X/R=0....................k(0)~1 (perfect symmetrical)

SC in real life should be between the two cases above. Therefore, the asymmetrical three-phase SC should be greater than the symmetrical SC of the same circuit.

Special case:
c- X/R < 0................k(<0)>0 <sub>Result from Eq (A).
NOTE: I am not sure if this result represent a realistic situation. (Any further comment is welcome).

Hypothetical R & X values:

For R < 0. This value may not be realistic even for circuit with superconductor with Rmax=0.
For X< 0 . We know that capacitor do not contribute to SC. Is the DC component in this case negative ?. Even if this is the case, the other two phase may not be negative.</sub>

 

[skeandhu]

The short answer is no the Earth Fault or Asymmetric Fault current will never be higher than the Short Circuit or Symmetric fault current speaking from a practical point of view. Earth fault protection of equipment using Core Balance Transformers on the individual machines, and current transformers monitoring the fault current in the neutral point of the transformer can detect fault currents of a few milliamps. We can even limit the amount of Earth fault current returning to the transformer through the neutral point by connecting a resistance or impedance between the transformer neutral and earth, usually rated for up to 10 seconds to allow the fault protection time to operate. Its only fairly recently that symmetrical fault current protection passed the HRC fuse level.

 

[jwerthman]

It isn't clear what the original post is asking.

By definition, the asymmetrical current is always higher than the symmetrical current because of the DC offset. As has been said already, the amount of the offset is a function of the system X/R ratio and the time of the fault on the sinusoidal waveform.

If the question is related to balanced three phase faults versus unbalanced single-line-to-ground faults, then the answer depends on the location of the fault and the configuration of the transformer windings.

For a delta-wye transformer with the secondary neutral solidly grounded, the maximum current for a bolted SLG fault at the transformer terminals will always exceed the maximum current for a bolted three phase fault at the terminals. The difference becomes more pronounced as the primary system impedance increases with respect to the transformer impedance. The reason is that for a fault at the transformer terminals, the zero sequence network only includes the transformer impedance. The positive sequence network includes the primary system impedance plus the transformer impedance. As the fault moves away from the transformer secondary terminals, then the zero sequence network also includes the secondary system zero sequence impedance to the fault. At some location, usually not far from the transformer, the maximum three phase fault current will exceed the maximum SLG current.

If the transformer is grounded wye-grounded wye connected, then the maximum three phase fault current will nearly always be higher than the maximum SLG fault current because the primary system zero sequence is included in the SLG fault network.

 

[cuky2000]

If the sequences impedances are known at the fault location, a quick empirical way to determine the worst-case of short circuit fault is suggested as follow:

For Three Phase: Z2/Z1<2.15-4.3(Z2/Zo)+2.3(Z2/Zo)^2

For Single Phase: Z2/Z1 > 1.41 (Z2/Zo)-1.5(Z2/Zo)^2 + 1.1(Z2/Zo)^3

If Z2/Z1 result between the two cases above the worst SC is phase-to-phase.