when should circuit breakers operate? 2

[wonderchi]

Hi all,

If a fault occurs in the low voltage system, and if the LV circuit breakers operate at around 8th cycle (rough estimate), what happens before the 8th cycle...will the circuit breaker blow up?

Why shouldnt low voltage circuit breakers operate in earlier cycles...2 or 3cycle or even 1/2cycle (if technology permits)..(since the asymmetrical fault will be much greater)?

Thanks

 

[davidbeach]

What are the interrupting ratings of the breakers and what are the settings?

 

[DanDel]

The 'instantaneous' trip function of the C/B, whether it's a thermo/magnetic or electronic trip, will trip at a certain level of current with no intended time delay (say, 4000A for a 400A rated C/B), but since nothing really happens instantaneously, there is a few-cycle delay during which the C/B carries the current.
If it has an interrupting rating of, say, 65kA, then it can safely carry and interrupt fault current at that level. If the fault current is higher than it's rating, it will probably 'blow up'.

 

[wonderchi]

The circuit breaker will trip with no intended time delays, but because of the mechanical features, it takes a few cycle for it to operate.

So...case scenario:

if a breaker is size to handle 100kA fault, when a fault occurs, the current is 110kA at 1/2cycle, but after 1 cycle, the fault is now 90kA...

1. is there any rule of thumb or standards which shows how much the mechanical features inside the breaker can handle faults greater than the rated size for a ? period of time.

 

[rbulsara]

wonder:

Yes, there are no rule of thumb but actual curves are available. Its called TCC (Time current curve) of a protective device such as circuit breaker and fuses. It will show how much current it can withstand for how long (RMS value). You have to draw a cutoff line at its rated short circuit current.

For excample, a TCC for a CB rated for 100A, and 10KAIC shall be read only for current up to 10,000A.

Besides breaker are built and tested to applicable ANSI/IEEE and UL standards which
allow for asymmetrical peak current withstand capabilites at 1/2 cycle as well as actual breaking capacity. Momentary duties without opening are higher (in amps) than acutal making or breaking capacities. Beacuse making and breaking capacities invariably occur after highest peak during a fault.

Heavy duty breakers usually used as mains or upstream in the system has to have capacities to withstand a fault current little longer than downstream devices to give the downstream breaker (closer to the fault) a chance to open before the upstream device opens. This is the very basic idea of device coordination.

 

[davidbeach]

Circuit breakers are rated in symmetrical amps, with an X/R value used for the test. Any current in excess of that will over duty the breaker, possibly with catastrophic results. If the X/R of the fault is greater than the test X/R, the breaker has to be derated to account for the increased offset (asymmetry) of the fault.

If you have a breaker rated less than the available fault current, you have a code violation. There is no way around it. You either need to reduce the fault current (increase the impedance in the system ahead of the breaker), or you need to replace the breaker with one rated for the conditions of use. Don't try to game the system.

 

[wonderchi]

david:

Although the circuit breakers are rated in symmetrical amps with an x/r value test values, is it true to say that the asymmetrical amps can be found by looking in to a asymmetrical/symmetrical vs x/r graph...therefore if the fault x/r is greater than the test x/r therefore you will have a bigger asymmetrical fault current than the circuit breakers asymmetrical values...Thanks

rbulsara:

in relation to TCC graph...can you expend a little more on it...Thanks

 

[davidbeach]

The point is that you don't need to worry about asymmetrical current, the whole rating scheme works around symmetrical current and the X/R value. X/R determines the maximum DC offset, but for any given fault the amount of offset can be anywhere from zero to the maximum, all depending on where on the wave the fault happens.

The IEEE Blue Book would be a good reference to start with, followed by the IEEE Red Book.

 

[rbulsara]

wonderchi:

I am not sure what part you want expanded on, but here is an example of a TCC for a GE molded case breaker. Please review it. As you can see it will open in less than .02s at 10,000A (x- axis is in multiple of current rating of breaker) and somewhere around 20 seconds for 700A fora 100A breaker.

http://www.geindustrial.com/publibrary/checkout/38652.30055.26427.27925/PDF/GES-6124A.pdf

Also follow what davidbeach said, you need to refer to those IEEE books for more information.

 

[DanDel]

Actually, only the manufacturer's TCC for a device ends at its rated interrupting current. In a coordination study, these curves typically end at the calculated fault current from the associated short circuit study for the bus at which it is connected.
The horizontal line at the top of the Instantaneous trip curve (at just under the 0.02sec line for the listed curve from rbulsara) is supposed to be the maximum time that that C/B will take to clear a fault at that current (from about 35x to just over 100x the C/B rating). This is not really true, since it typically will take 0.06sec (3-4 cycles) to clear the fault.
If a C/B is rated for 10kA interrupting, then that means it can handle the slightly higher fault current previous to the 10kA at 3-4 cycles.
For more info on this, check out IEEE 242, chapter 6, 'LV Circuit Breakers'

 

[wonderchi]

Dandel:

When you say:
"If a C/B is rated for 10kA interrupting, then that means it can handle the slightly higher fault current previous to the 10kA at 3-4 cycles. "

But how much slightly are we talking about?

 

[DanDel]

The Momentary and Interrupting fault currents are calculated from the symmetrical fault currents using the appropriate ANSI multiplying factors (MF), which are based on the equivalent system X/R ratio. The equations are available in the appropriate C37 IEEE references for each type of device, like LV C/Bs, HV C/Bs, etc.

I can try to reporduce a couple of them here.

The momentary MF is equal to: the square root of (1 + 2e^(-(2pi/(X/R))).

The interrupting MF for a HV C/B is: the square root of (1 + 2e^(-t(4pi/(X/R))), where t= the C/B contact parting time in cycles.

The fused and unfused LV C/Bs have different calculations, a little too complicated to write in here.

 

[rbulsara]

Wonderchi:

IEEE red book says LV breakers or fuses(less than 1000V) are rated for symmetical currents at 1/2 cycle and incorporate asymmetrical capabiltiy as necessary for cirucuit X/R ratio of 6.6 or less. This will satisfy most installations fed by say 1500kVA transformers. For systems fed by larger or multiple transformers some derating will appply per IEEE C37.13.