X/R ratio for L-L-G Fault 2

[stretched2thin]

I'm trying to calculate the assymetrical offset for a L-L-G fault and am not totally sure I'm understanding things correctly. If you look at the sequence diagram for a L-G fault and the Thevenin equivalent circuit for a 3-phase fault, it's easy to see how the X/R ratio is defined for each. However, when you look at the sequence diagram for a L-L-G fault, it looks like I1, I2, and I0 would have different X/R ratios according to their respective Thevenen impedances.

Am I thinking about this correctly?

 

[stevenal]

I think so.

Assuming Z2=Z1,one faulted phase should see an impedance of (Z0-Z1*a^2)/(Z1^2+2*Z1*Z0) and the other (Z0-Z1*a)/(Z1^2+2*Z1*Z0). Should be able to get the effective X/R ratios from those equations. I adapted the formulas from the Cooper Electrical Distribution-System Protection book.

 

[rsherry]

I would suggest not trying to apply X/R ratios to sequence networks at all for a couple of reasons :

1) The X/R ratio at the point of fault is only a crude way of trying to assess the DC decrement of a transient current. This transient is a point on wave phenomenon and is not truly comparable to positive, negative or zero sequence currents. In considering transient components of fault current (and hence X/R ratios) we are assuming worst case conditions for one phase for any particular fault.

2) The errors in using a single X/R ratio will always be an order of magnitude above any different X/R ratio you may get from working out the sequence networks X/R values.

I would suggest calculating the X/R ratio at the point of fault in the usual way (Thevenin equivalent of positive sequence) and using this value for all fault currents at this place in the network.

 

[jbartos]

Comment: Normally, X/R ratios are not considered in the symmetrical component analysis of power distribution networks since the fault magnitudes are determined at time equal to zero.
Please, would you mention in which context or possibly Reference, the original posting is dealing with?

 

[jghrist]

Stevenal's approach is not to find the X/R of the sequence currents, but to find the X/R of the resultant phase currents. In the case of line-to-line or double line-to-ground faults, however, this approach gives strange answers that I don't think can be interpreted correctly to get a decrement.

For an example 100 kV fault, I calculate the following for a double line-to-ground fault:

I_b = 4297A X/R = -2.27
I_c = 4206A X/R = 0.61

For the same impedances, 3Ø fault X/R = 3.25 and 1Ø fault X/R = 3.46

 

[stretched2thin]

If I'm interested in sizing equipment based on symmetrical and assymetrical fault data, then it seems to me that I need to know the assymetrical offset for each type of fault. I know that 3-phase faults are usually the highest magnitude, but not always. So, I think I probably need to know the offset for each type.

Also, it doesn't seem that it would be correct to use the same X/R ratio to calculate the offset for different faults since the thevenin impedance for a 3-phase fault is different than the thevenin impedance for a L-G fault.

Finally, why is using the X/R ratio a crude way to calculate offset? How else would you do it? I believe that the X/R method is arrived at by solving the differential equation describing fault conditions, which I would think would be pretty accurate. I'm not sure how else to do it. Any advice would be appreciated.

 

[stevenal]

In defense of my method, IEEE 367 uses the "effective" X/R in calculating GPR for single line to ground faults. "Effective X/R" not well defined , but I've seen it interpreted as the X/R of 2*Z1+Z0. I believe the fault analysis program I use, Aspen, uses a similar approach. I admit that jghrist's negative ratio throws me. Maybe an absolute value is needed. Maybe if it's the R term that's negative, it means maximum offset. I agree that it's a crude method. It's a way of keeping calculations in the frequency domain where everything is nice and neat. An EMTP approach would be much more accurate. But using X/R seems to be an ANSI/IEEE acceptable approach. I believe ANSI also has something about how X/R is calculated for your task of sizing equipment.

In reviewing Cooper, I see I an error I made yesterday. The expressions should be inverted. There's also a -j in the current calculation that might need to be included in the impedance terms for calculating X/R. I'll be studying this some more.

 

[stretched2thin]

Stevenval:

I think you're right. If you look at the sequence network diagram for a L-G fault, you can see that the thevenin fault impedance is Z1+Z2+Z0 which is the same as 2*Z1+Z0 if Z1=Z2.

My questions were directed more to rsherry. I'd like to know what method she uses to calc offset.

I'm admittedly not an expert on this, so I'm open to any info anyone has.

 

[jghrist]

Stevenal,

Actually, I couldn't reproduce your equation for impedance, so I divided V_b by I_b to get an impedance of -9.39 + 21.29j ohms.

With X/R = -2.27, the decrement factor is 3.4 for 1/2 cycle or 1407 for 3 cycles! I don't think this is in the realm of reasonable accuracy.

The problem is that you have to relate the current angle to a voltage angle, and it isn't clear for a double line-to-ground fault what voltage angle to relate it to. The -2.27 comes from relating it to phase b voltage.

 

[rsherry]

Well i've checked references back to Lackey and Edith Clarke (1950s) and not found a clear explanation... so to just expand on my thoughts a little for now :
On the X/R ratio to use :
For a 3 phase fault the DC component is not the same in all 3 phases - it's not a positive sequence current, it's not part of a rotating set of phasors and it's not solvable using symmetric component analysis. In symmetrical component analysis a 3 phase fault is the same for all 3 phases and we add in the DC offset element separately.
Given the one-phase nature of the phenomenon the effective impedance of a single phase (which is the positive sequence impedance) seems the best source for the X/R ratio needed to estimate its rate of decay.
For an unbalanced fault symmetrical component analysis will provide the now unbalanced phase current values but it still cannot solve the offset. For a 2 phase to ground fault one faulted phase will see a higher current than the other due to the DC offset but the symmetrical component analysis will not show this.
Symmetrical component analysis is only a tool to solve the unbalanced AC currents - the DC component does not flow in these 'sequence networks'. They cannot be used to calculate its value so why would you use them to calculate its decay. I would suggest the one phase impedance is still the appropriate one to use.

The second point is easier - as this is covered in most references.
On why X/R is not an accurate calculation :
I was a bit sweeping in this statement - a single X/R value can give a very good estimate of the real decay rate for the DC component but in many practical situations the infeeds to a fault have different X/R ratios and decay at different rates - the combined decay is not the same as the decay from the thevenin equivalent impedance. The error is particularly large for faults close to generators or grid transformers (paths with high X/R ratios). ANSI and IEC standards specify a margin to apply to fault currents calculated with a single X/R value to cover for this.

As stevenal suggested only an EMTP type study would give an accurate result, but this is usually considered unnecessary as symmetrical component fault calculations (with margins applied for their known limitations) are good enough.

 

[stretched2thin]

Thanks, rsherry. I hadn't considered the fact that symmetrical components requires balances phase components for each positive, negative, and zero sequence quantity.

 

[stevenal]

I checked the Cooper formulas against Aspen. I can reproduce Aspen's X/R from the Cooper impedances for all faults but the DLG. The current formulas work, but X/R values were all much lower than Aspen. Even tried swinging the angles around a bit per jghrist's comment above. Sherman Chan is smarter than me. His DLG X/R was a little higher than the three phase value and quite a bit smaller than the SLG value.

 

[jghrist]

Symmetrical component analysis of double line-to-ground faults will give different currents in the faulted phases. This is independent of dc offset.

For a Ø-Ø fault not involving ground, you can develop an equivalent fault impedance by dividing I_b by the V_cb. V_cbwill have an angle of -90° with respect to phase a. Doing this with my previous example gives a Z = 5.26 + 17.06j ohms and an X/R = 3.25, exactly the same as for the 3Ø fault case. This is logical considering there is no zero sequence involved.

It gets complicated for a double line-to-ground fault. It's hard to say what voltage to relate to the current to get an equivalent Z.

 

[stretched2thin]

Yes. I agree. It does indeed get complicated.

Thank you all for some great input.

 

[jbartos]

Suggestion to the original posting: Please, would you post voltage level since there are different ANSI standards, voltage dependent, covering faults as related to switchgear and circuit breakers.

 

[stretched2thin]

jbartos:

I'm not interested in any particular level at the moment. I'm interested in knowing how to do this for all voltage levels from 120 V panels to 500 kV circuit breakers.

 

[stevenal]

From C37.010:

For determinating the system X/R ratio, it should be noted that no completely accurate way exists of
combining two parallel circuits with different values of X/R into a single circuit with one value of X/R. The
current from the several circuits will be the sum of several exponentially decaying terms, usually with
different exponents, while that from a single circuit contains just one such term. Investigation has shown that
by reducing reactance to a single value with complete disregard for the resistances and reducing the resistance
to a single value with complete disregard for the reactances gives, in general, more accurate results than any
other reasonably simple procedure (including the phasor representation used at system frequency). In
addition, the error for practical cases is on the conservative side. For these reasons, this procedure is
recommended.

 

[jbartos]

Comment on stretched2thin (Electrical) May 20, 2004 marked ///\\jbartos:
I'm not interested in any particular level at the moment. I'm interested in knowing how to do this for all voltage levels from 120 V panels to 500 kV circuit breakers.
///ANSI C37 series dealing with switchgear will do. It will reveal differences among voltage levels.\\\