Interpretation of Short Circuit Parameters

[cuky2000]

We are in process trying understanding the design parameters provided by a metropolitan utility to calculate the grounding requirement on a 138kV/13.8kV Area Substation

Some of the parameter are:

* Design LGF SC current of 46kA for 2 seconds (120 cycles) follow by 10 kA for the next second (180 cycles total)
* X/R ratio of 38.8

QUESTIONS:
1- Should we need to calculate the assymetric equivalent RMS value or are usually included in the above data?
2- There is any reference or standard that address step function as describe above?.
3- Is the energy equivalent a suitable model to determine the allowable step and touch potential based in RMS or should the peack current values play a roll in this case?

Thanks

 

[dpc]

I cannot answer all your questions, but as for Question #1, I would certainly assume that the magnitudes of current provided are symmetrical rms values. The X/R ratio can be used to estimate the asymmetrical rms current.

The step change in current could be due to multiple sources with different clearing times.

My assumption is that the maximum current would be used for calculation of step and touch potentials.



"The more the universe seems comprehensible, the more it also seems pointless." -- Steven Weinberg

 

[7anoter4]

I agree with dpc that for touch and step potential [as per IEEE-80] you need only rms currents.
But I'll try to guess the following:
According to IEC 60909 standards:
I"k=initial symmetrical short-circuit current
Ik=steady state short-circuit current
ip= peak short-circuit current due to d.c. component
ip=kappa*sqrt(2)*I"k
kappa=1.02+0.98*exp(-3*R/X)
ipeak is the maximum fault current value used for dynamical stability of installation [it is not rms].
I think:
I"k=46 kA [rms]
Ik=10 Ka [rms]
See [from ABB Switchgear Manual]:

But I don't think the total duration of the fault [fault clearing time] may be so long.
According to IEEE 80 the recommended duration for thermal stability of the Grounding Grid is 1 sec.
For touch and step usually is 6-10 cycles [0.1sec] as the total duration of breaker trip [without delay] may be. A decrement [dcr] factor is taken into consideration .For calculation of touch and step Ifault=I"k*dcr
DCR = Sqrt(1 + TA / T * ( 1 - Exp(-T / TA))) where : T = fault clearing time [sec]
TA= 10 / PI() / F F=60 or 50 Hz
If T=>1 dcr=1[approx.]

 

[jghrist]

The time is used both in determining the thermal limits of conductors and in the allowable step- and touch-potential.

For thermal limits, an RMS calculation would be appropriate. Ieq = sqrt(46^2·2 + 10^2·1)/3 = 22kA for 3 seconds. Check that 46kA for 2 seconds does not exceed the conductor thermal limit.

It isn't nearly a clear what to use for allowable potentials. These are based on empirical formulas for fibrillation currents that are of a constant I²t form, so the same RMS calculation should work. 46kA for 2 seconds is an awful long time - you are going to have very low allowable potentials.

 

[cuky2000]

Thanks all for your input.

Our initial approach was to use a larger uniform function of 46 kA for 3 s that certainly covers safely the case of the step function of 46 kA for 2 sec follows by 10 kA for 1 sec. However, the amount of copper required and the cost associated was prohibited requiring additional refine engineering studies.

One of the dilemma using lower current value such the one calculating the equivalent RMS uniform function(ex. 22 kA or 38kA @ 3 s) from the given step function (46 @ 2 +10@1s) is the lock of backup support by standard or credible experimental data for two consecutive electrical shocks.

Using less time than 3 second (180 Cy) in calculating the allowable step and touch potential appears to be very risky proposition from the contractual point of view.

We are challenging our colleagues in the utility to provide an explanation and I hope to get the OK to share their rational of such long clearing time.

PS: Jghrist, please verify if the value 3 should be inside the SQRT.

 

[cuky2000]

Enclosed is the explanation provided by the utility relay and protection department justifying the three seconds clearing time:

• The distribution system is designed for double contingency (n-2). Therefore, the design should be based assuming 2 equipment failures.

• For 2 equipment failure would cause 2 stages of clearing: First 46 kA for 2 seconds, followed by 10 kA for another 1 second.

• The system also has auto reclosing capability that also causes 2 fault stages of clearing.

Another interesting feature is that the concentric neutral of the 13.8 kV cable UG cables is grounded at the source sending-end substation, cross bonded in the street manhole, ungrounded at the receiving-end substation and continuously grounded to the low voltage distribution system.

This arrangement only allow for 70% current division factor.

The dominant effect is still the 3 seconds clearing time.

Any suggestions?