MCCB takes much longer to trip after installing current limiting reactor, is there a remedy? 4

[bdn2004]

A new Control Panel is fed by a 400A molded case circuit in a panelboard.
It was determined that the available fault current was about 9 kA at a new Control Panel. The SCCR of the panel is 5kA. So the Control Panel is under dutied for the available fault current.

A current limiting reactor is proposed. After installed the available fault current will be limited to 1kA. However, such low fault current on the 400A breaker eliminates the breaker instantaneous region on the breaker curve (the lowest setting is 2000A). Which in turn makes the arc-flash energy much higher.

Is there a remedy for this situation or is this just the trade off you must live with?

 

[ScottyUK]

Smaller current-limiting reactor? For example, a 4kA fault should have you well into the mag trip region of the breaker response while remaining within the panel capability.

 

[bdn2004]

Unless there are other reactors out there, I’m not aware of this is the smallest reactor you can buy off the shelf with ample current rating. We sized it slightly over the FLA of our load ... like 350A, as it’s gotta carry that, and this is the result.

 

[dpc]

Reducing the fault current from 9 kA to 1 kA would take a lot of reactance. If the load is anywhere near 400 A, I'd take a hard look at the voltage drop during normal operation as well as resistance losses in the reactor.

The continuous current rating is one issue, but you're overlooking the amount of reactance. It should be possible to reduce the reactance of the reactor as Scotty proposed. Contact a reactor manufacturer.

 

[bdn2004]

I appreciate your guy's help. I made a mistake the first time in the math and was off a decimal point in the kVA.
After I corrected it the fault current went to about 5500A, and your right, the arcing fault went into the instantaneous range.
Just as an aside ... I also tried this with an isolation transformer - and this correct number is much closer to that value. It's a decent check.

Just for the record here is the calculation I used (the correct one):

Nameplate information on the reactor:
Induction: 0.03 MH, 50/60 HZ, 343 Amps, 0.6kV Class

Convert 0.03 MH to ohms: 2 x pi x f x L = (2)(3.1472)(60)(0.03 MH)(1 H / 1000 MH) = 0.01133992
S = I^2 x Z = 343A x 343A x 0.01133992 Ohms = 1334 VA / ph = 4002 VA, 4.002 kVA