How to use CL CB's let-through curves

[veritas]

I have a question regarding let-trhough curves of fuses and circuit breakers. I have researched this forum but not found any thread that exactly addresses the question I have. Consider the attached pdf.

The top let-through curve is that of a fuse. We are dealing with a 80A fuse and the prospective fault current is 30kA. According to these curves, the fuse limits the rms fault current to around 6kA which is 8.4kA peak. Thus the current limiting (CL) action of the fuse allows the downstream network to be designed to 6kA and not 30kA.

The bottom curves are those for Schneider NSX range of CL breakers. Schneider, like most LV breaker manufacturers, provide cascading curves where the CL action and let-through energy allows for lesser rated cb’s downstream than the prospective fault level at the downstream location.

So let’s say I have a cascading combination in place, breaker A upstream and lower rated breaker B downstream. What should the fault rating of the cable between them be? Let’s say I have a NSX630A breaker upstream, Iprospective = 100kA. If the peak let-through is 46kA does that mean the rms let-through is 46/1.414 = 32.5kA? So my cable only need be rated for up to 32.5kA and not 100kA? Is this correct?
I am aware that there are let-through energy curves for CL cb’s as well but I have no idea if they’re any use when it comes to determing the reduced symmetrical fault current due to the action of the CL breaker.
Am keen to hear some thoughts regarding how to use these cb let-through curves to determine the reduced symmetrical fault current, if indeed this can be done!

 

[FreddyNurk]

Bear in mind that manufacturer's data is critical in determining the downstream breaker, as unlike fuses, each possible breaker - breaker combination must be tested to ensure the current limiting is as designed. This effectively means that using breakers from different manufacturers is not possible if you're intending to take advantage of the current limiting.

I'll admit I've never considered downsizing the cable in these cases, in all my applications it wasn't a governing factor in selection of a cable.

EDMS Australia

 

[waross]

There was a recent thread that touched on the thermal and mechanical withstand capability of cables during a fault. I understood the consensus to be that fault currents are generally short enough that thermal issues during a fault are generally not an issue.
In regards to your formula:
46/1.414 = 32.5kA?
I was under the impression that the formula for the RMS equivalent of a fully offset asymmetric current was;
46/1.414x2 = 16.25kA?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[veritas]

Yes, I am aware of the cascading requirements and to only use manufacturer recommended tables. I'm sitting with an application though where space is a big constraint and fault levels are rather high (around 60kA) at the MCC. MCC loads are of various lengths but most are not enough to result in enough impedance to lower the fault level sufficiently for a reasonably sized cable. So for a 10kW load I need a 185mm2 cable at 380V due to the fault level requirements.

Using reduced fault levels due to the CL action of the mccb seems very attractive, but the golden question is how to determine the reduced fault level?

 

[veritas]

waross,

Interesting! I was under the impression that the vertical scale of the cb curve peak kA is that for the symmetrical current. The formula you are referring to which yields 16.25kA is the steady state symmetrical rms current which will be correct if 46kA is the peak of the asymmetrical current.

Actually, not sure now if that vertical axis is asym or symmetrical peak.

 

[sibeen]

Veritas, look at the 100 kA point and go up to the orange line and then across, it looks like you get approximately 220 kA projected onto the left hand scale. As this is way higher than the calculated symetrical fault current it can only lead to one conclusion

 

[veritas]

sibeen,

Sorry, you lost me! What are you trying to say?

 

[dpc]

I don't have an answer since I've never found the SC capability of a cable to be the limiting factor for sizing the cable in NEC land. I don't think there is any way to predict the maximum "let-through" current of a current-limiting fuse and current-limiting CB in series other than by testing it.

Note that in the graph, the vertical axis is PEAK current and the horizontal is rms symmetrical.

 

[veritas]

I don't have an answer since I've never found the SC capability of a cable to be the limiting factor for sizing the cable in NEC land.

I recently completed a project in the US where the minimum cable size had to be AWG 2/0 to ensure the cable is not damaged with worst case throughfaults whereas load requirements would have been happy with AWG 2. I have actually more often than not found that fault current being the determining factor for cable sizing.

Anyway, the mystery of the let-through peak curves still remains. I've contacted 3 suppliers and am still awaiting their "experts" to get back to me.

 

[dpc]

I'd be interested in how you determined the maximum allowable throughfault current and duration. I've been doing this a long time and it's never been an issue based on the cable damage curves that are published by cable suppliers and used in commercial software.

I really doubt if anyone is going to give you a definitive answer unless they have actual test data. Current-limiting circuit breakers are rarely used in the US in my experience. I've never seen one in the wild.

Good luck.

Dave

 

[veritas]

With any cable installation one of the 1st questions I address is whether the cable needs to withstand the maximum throughfault current of the max fault current at its source. Often the situation arises where the cable is direct buried under walkways, roads, concrete structures, etc. The length of the cable of course is another important factor. In this situation it may be very expensive to replace the whole cable with a cable fault.

Best is to locate the fault and replace only the faulted section of cable. This means the cable needs to be rated for source fault current level.

 

[veritas]

Then I had a small pump (around 45kW) fed from an mcc (60kA fault level) by a 15m 25mm2 XLPE cable. Sufficient to meet the load requirements. However, no protection in the world could protect the cable for a throughfault of around 21.5kA. I had to go to 150mm2 to satisfy the fault-trip time requirement. A bit overkill for such a small load.

Thus if I knew what the reduced fault level would be due to the current limitng action of the breaker then I could save on cable size. Or maybe I'm barking up the wrong tree and I've lost the plot?

 

[FreddyNurk]

For the cable to be exposed to such a fault for the likely duration that you're using, then as its a final subcircuit load then I'd expect that at least the circuit breaker protecting it has failed to trip, as well as the circuit breaker above it.

Or, do you really need at least a 1 second withstand duration if the protection is expected to trip in about 5 cycles?


EDMS Australia

 

[veritas]

Freddy,

You're 100% correct! The 500ms allows for backup protection operating time as well. So, motor protection on that circuit should clear the fault in < 100ms. Then allow 500ms for the backup to trip. Backup has to co-ordinate with other protection as well.

 

[ScottyUK]

It's unusual to size a final circuit cable on an LV system to survive a fault cleared by the backup protection, at least in my experience.

Don't forget that the very high fault levels only exist close to the MCC or switchboard: on a low voltage circuit you don't need much cable length before the fault level drops away. Try calculating the fault level at the motor terminals for your 45kW motor if you use your 25mm² cable and compare it to the fault level at the board. A cable fault close to the board may destroy the cable, but one further out probably won't.

 

[veritas]

That's exactly what I did. However, the cable run of 15 is so short that it does not allow the fault level to drop sufficiently below the cable damage curve to allow the protection to operate. Another option might be to loop the cable so as to extend it's length but there may be practical impediments to this one. Will see.

I'm still interested as to how to use those peak-let through curves. They must have been compiled for some useful purpose! Will let know when I find out.

 

[wroggent]

You probably realize this but coordinated backup protection is not going to be current limiting.

 

[veritas]

Finally managed to crack this! (I think!) Had some very good input from Siemens and would like to acknowledge their contribution. Decided to write it all up for the benefit of all those who contributed to this post and others who might be interested.

It is essentially written for the IEC world where I am based but is easily extended to the NEMA world as indicated at the end.

So please find attached notes I made.

Any comments will be most welcome!

 

[veritas]

Has an example where a 16m,25mm2 XLPE cable was installed. It has a 7.14kA withstand rating for 0.25s. Fault current at cable source was 23.75kA and 22.74kA at load end. No way cable would be able to handle a 23kA fault within reasonable protection operating time.

K²S² for cable is 12758121A²t. I used 80% of this so around 10million. Maximum let-through energy of upstream current limiting breaker is 1.9million. As this is less than the breaker's 10million, problem solved - in theory at least!

 

[wroggent]

Are you plugging in trip times that are longer than what the TCC indicates?


As I understand it, the I^2*t let through curves for current limiting breakers (or fuses) represent the maximum amount of energy that will pass through the breaker for a given symmetrical current value and this value is to be less than the amount of energy that could pass through the breaker during the first half cycle with no current limiting action. From this you can assume that a current limiting breaker is going to (completely) interrupt current usually less than but in the extreme case very little more than a half cycle. Half a cycle at 50HZ is 0.01 seconds. Much less than 0.25s

I think your conclusion is correct (the cable will be protected) but your analysis doesn't make sense to me. I think the correct approach would be to take the actual trip time from the TCC for your fault level and then plug it into the adiabatic equation.