Relationship between CB break duties and break times in IEC60909

[ahmedradi]

Hi all,

I've got this software package that calculates the fault levels for our transmission network according to IEC60909 Method-C, which is the equivalent frequency method. Such as the make duty is calculated using instantaneous peak with DC component adjusted X/R is calculated at an equivalent frequency of 20Hz.
For a 50Hz system, such as our, the break duties (symmetrical and asymmetrical) are calculated based on the specified break time (this supposedly includes the breaker opening time plus arcing time). The break times (below in seconds) are translated to an equivalent frequency to adjust the X/R to accurately calculate the DC component as follows:
0.00s <= Tbk < 0.01s Freq = 20.00 Hz
0.01s <= Tbk < 0.02s Freq = 13.50 Hz
0.02s <= Tbk < 0.05s Freq = 7.50 Hz
0.05s <= Tbk < 0.10s Freq = 4.60 Hz
0.10s <= Tbk <= 0.25s Freq = 2.75 Hz

Q1. Are my equivalent frequencies above correct?

Q2. Are there any CB that you are aware of that is faster than 0.01sec or too slow to be beyond the 0.25sec?

Q3. This is the most important of all. According to the software package, I've ended up with a very puzzling break duties, which I couldn't explain from an engineering point of view or mathematically, and that is if I increase the break time, e.g. from 0.01 to 0.25, my break duties are increasing, as if slower breaker would need to be of higher interrupt capabilities!
To make matters worse, the break duties are even higher than make duties. I suspect that there is a bug in the software, but let me first hear from you based on your experience.

The data looks alright, yes there are some anomalies as in every utility, especially at generation sites, e.g. the transient and sub-transient reactances are of the same value, but further out at purely load substations, it is same problem!

Many thanks in advance for your inputs.

 

[ahmedradi]

After reading my post, I realized that I have not provided any relevant information that may be useful for you to help me.

One of the surprising results that I got, though not really a shocking one, the 1ph faults have higher currents that 3ph faults.

Anyway, at one of my generation sites, I have 24 CBs at the 132kV busbar with the following rates:
APM (Asymmetrical Peak Make Duty) = 100 kA
SB (Symmetrical Break Duty) = 40 kA
ASB (Asymmetrical Break Duty) = 51 kA
Break Time = 55 ms

Without applying any frequency equivalent analysis on the CBs to calculate the make/break currents, I get the fault currents at the 132kV busbar as follows:
1ph: sub-transient = 18.898 kA transient = 18.565 kA
3ph: sub-transient = 15.041 kA transient = 14.423 kA

If I apply frequency equivalent analysis to calculate the make/break currents using the above mentioned frequencies against 55ms, I get the following puzzling numbers for the fault currents at the 132kV busbar:
1ph: sub-transient = 20.807 kA transient = 33.744 kA
3ph: sub-transient = 15.041 kA transient = 89.147 kA

And the make/break currents for my CBs @ 55ms are:
1ph: APM=50.83kA (X/R=9.20) SB=30.47kA ASB= 52.14kA (X/R=14.65)
3ph: APM=36.62kA (x/R=9.98) SB=70.41kA ASB=105.51kA (x/R=13.54)

If I adjust the break time to 45ms, then:
1ph: APM=50.83kA (X/R=9.20) SB=29.54kA ASB= 51.12kA (X/R=12.32)
3ph: APM=36.62kA (x/R=9.98) SB=65.08kA ASB= 98.34kA (x/R=11.63)

Obviously, there is something seriously wrong, either in the software calculations or how the results are presented.

 

[ijl]

The equivalent frequencies seem to be correct. But there is something strange in the values of the currents you report. The subtransient or transient currents should not depend on the breaking time or equivalent frequency. Maybe you should check these first.

And yes, the phase to ground fault current may be larger than the three phase fault current. See the equations used to calculate the currents.

 

[ahmedradi]

Yep, they got the transient current completely wrong... And that causes a lot of errors else where. In the software manual, they calculate the symmetrical (SB) and asymmetrical (ASB) breaking currents as follows:

I(sb) = I" + (I" - I') x e(-25 x Tb)

And the asymmetrical breaking current (ASB) as follows:

I(asb) = sqrt(2) x I(bk) + I(dc)

I(dc) = sqrt(2) x I" x e(-2 x Pi x Tb x R/X)

And if the I' is greater than I", then the SB is wrong and subsequently the ASB is wrong as well. The APM (make duty) looks alright.

I'm not quite sure about how they calculate the SB and ASB currents. Are those equations look fine? Or do we have alternative ones to use?

 

[ijl]

The equation for I(sb) contains familiar elements, but on the whole it does not seem to be correct, printing error? By the way, I' should always be smaller than I".

The equation for I(asb) seems to be correct, although not identical with the standards.

I(dc) is correct. R/X should be calculated using the equivalent frequency.