HV ckt breaker fault rating

[KnicksJets]

Folks,
Got a question on rating a HV circuit breaker.

Say the fault the breaker sees, as calculated by ASPEN, is:
ISC=54597.4 Amps
3LG_AMPS=60602 Amps
For a given X/R=38.4.

Question: What is the minimum interrupter rating (both symmetrical and asymmetrical) for a breaker to interrupt this fault. Assume that the breaker contact parting time is 2 cycles and interrupting cycle is 3.

Minor discussion:
I am assuming 3LG_AMPS is the symmetrical fault current calculated by multiplying ISC by a factor obtained from IEEE std C37.010 (figure 8) since the X/R is higher than 15. Is my assumption correct?
Then how do you get asymmetrical value from this?

%-) I thought this would be a topic that is already explored but I just couldn’t find a directly related topic when I searched the forum.

 

[davidbeach]

ASPEN has a breaker rating module that will do all of these calculations for you.

 

[KnicksJets]

Thanks but I want to understand what I am seeing.

 

[dpc]

You need to know the methodology being used by ASPEN, since ANSI C37 provides some options for calculating the SC currents and asymmetrical currents. (Complex impedances versus separate R and X reductions, etc).

I'm not familiar with ASPEN breaker duty calcs, so I don't know why 3LG_AMPS is different than Isc.

The ANSI breaker duty also requires/allows adjustments for local generation versus remote source and other factors, breaker contact parting time, etc.

As you say, if the actual system X/R ratio exceeds the test basis for the circuit breaker symmetrical rating, the SC rating of the breaker may need to be reduced.

The best resource for the theoretical basis for the ANSI breaker duty calcs is the book by Conrad St. Pierre "A Practical Guide to Short-Circuit Calculations".



David Castor

http://www.cvoes.com

 

[KnicksJets]

Thanks for the discussion.
ASPEN aside, if I have a Symmetrical current and X/R over 17, then I need to multiply this by that X/R multiplier obtained from figure 8 of IEEE c37.010 right? How do you get Asymmetrical value then?

 

[dpc]

You read C37.010, pick a method and follow that procedure. ANSI C37 has at least four different methods to determine the appropriate X/R ratio at the fault point.

The equation to convert rms fault current to asymmetrical is in ANSI C37.010, I believe. It's an exponential function and not something that easily be written in ASCII.

If you have access to the IEEE Red Book, you should find some additional guidance there. This is not as straightforward as one might think.



David Castor

http://www.cvoes.com