Choosing the between Inverse-time and Instantaneous-trip Breaker.

[Jraef]

And yet another thread goes off into the weeds discussing oddball circumstances and details not germane to the original posting...

 

[templdl]

And yet another thread goes off into the weeds discussing oddball circumstances and details not germane to the original posting...
View attachment 15755

Ditto as this is not unusual.

 

[ActionDave]

And yet another thread goes off into the weeds discussing oddball circumstances and details not germane to the original posting...

Ditto as this is not unusual.

Threadjacking is always something to be wary of. In this case the OPs question was answered satisfactorily and the conversation continued on to other things. I don't see a problem.

 

[Ingenieur]

The problem is if the instant element also in inverse breaker is fixed type (permitted by code), it may compromise motor protection.

So an instant and it are both unsuitable?
both are fine if code compliant

Even if fixed at 1700% or 578 A it will trip with a conservative i failt of 760 A

give a scenario with numbers that illustrates your concerns

 

[Sahib]

So an instant and it are both unsuitable?
both are fine if code compliant

Even if fixed at 1700% or 578 A it will trip with a conservative i failt of 760 A

give a scenario with numbers that illustrates your concerns

Considera 460Vac, 40 horsepower energy efficient motor with 52 FLA. A thermal magneticbreaker sized to 250% FLA is selected. Since a 130A (2.5 x52 = 130A) rating is not a standard ampere rating, a 150Afixed instantaneous breaker is selected. The actual instantaneouspick up is 2250 symmetrical rms amperes or 15 timesthe breaker ampere rating. This breaker will not respond instantaneously until a current of 2250 symmetrical rmsamperes or 3181 peak amperes flows. The 40 horsepowermotor’s first half cycle inrush is 629 peak amperes. So there will be no problem starting the motor, but a short circuitcurrent of, for example, 1900 amperes will not becleared until the thermal portion of the trip unit responds in, typically, several seconds. This may cause FURTHER damage due to delay in protective device operation.

 

[ActionDave]

Considera 460Vac, 40 horsepower energy efficient motor with 52 FLA. A thermal magneticbreaker sized to 250% FLA is selected. Since a 130A (2.5 x52 = 130A) rating is not a standard ampere rating, a 150Afixed instantaneous breaker is selected. The actual instantaneouspick up is 2250 symmetrical rms amperes or 15 timesthe breaker ampere rating. This breaker will not respond instantaneously until a current of 2250 symmetrical rmsamperes or 3181 peak amperes flows. The 40 horsepowermotor’s first half cycle inrush is 629 peak amperes. So there will be no problem starting the motor, but a short circuitcurrent of, for example, 1900 amperes will not becleared until the thermal portion of the trip unit responds in, typically, several seconds. This may cause FURTHER damage due to delay in protective device operation.

Yep. You have a real world problem to solve right there. What do you think is the best solution?

 

[Ingenieur]

Considera 460Vac, 40 horsepower energy efficient motor with 52 FLA. A thermal magnetic breaker sized to 250% FLA is selected. Since a 130A (2.5 x52 = 130A) rating is not a standard ampere rating, a 150A fixed instantaneous breaker is selected. The actual instantaneous pick up is 2250 symmetrical rms amperes or 15 timesthe breaker ampere rating. This breaker will not respond instantaneously until a current of 2250 symmetrical rmsamperes or 3181 peak amperes flows. The 40 horsepower motor’s first half cycle inrush is 629 peak amperes. So there will be no problem starting the motor, but a short circuit current of, for example, 1900 amperes will not be cleared until the thermal portion of the trip unit responds in, typically, several seconds. This may cause FURTHER damage due to delay in protective device operation.

first I would choose a 125, not a 150, but let's use your 150
how was the 1900 fault i value derived? what location in the circuit?
the 2250 A / 1500% is too high
an Eaton Series C F Frame 150 A
at 1900 A (1900/150 = 1250%) min .008 max 0.025 sec, at 1500% it is in the mag/instant. region
the motor will not see 1900 if the fault is before it? it is shorted out of the fault path
even in you example not an issue
so what if the CB takes a few sec to trip?: the OL can handle it (see below), the conductor can handle it, and obviously the CB can handle it, and the motor is not involved

if an mag/instant only is used set at 800% ~425 A it will trip no issue
your original point was a mag only causes issues, it seems to eliminate them?

either way, a non-issue imho
apparently not one in the NEC eyes either
if code compliant not an issue

a better estimate of the fault current
assume a ckt with a 20 v drop (480-460) or 4%
v drop = 52 x R = 20 so R = 0.39 Ohm, basically the minimum, would be lower if motor windings were involved
I fault = 460/0.39 = 1180 A
on the Eaton it curve ~ 0.01 to 4 sec

a Class 20 OL (most curves go out to 1300% of setting or 1.25 x 1300 = 1600% x fla)
at 10x the setting (1250% fla) it trips in 6-8 sec
this means the relay is rated to carry that current without damage (the curve flattens out, time ~6 sec, up to 1600% x fla)

 

[Sahib]

Yep. You have a real world problem to solve right there. What do you think is the best solution?

An inverse breaker with adjustable instant element will do.

 

[Sahib]

first I would choose a 125, not a 150, but let's use your 150

Are you sure it will not trip on O/L?

how was the 1900 fault i value derived? what location in the circuit?
the 2250 A / 1500% is too high
an Eaton Series C F Frame 150 A
at 1900 A (1900/150 = 1250%) min .008 max 0.025 sec, at 1500% it is in the mag/instant. region
the motor will not see 1900 if the fault is before it? it is shorted out of the fault path
even in you example not an issue
so what if the CB takes a few sec to trip?: the OL can handle it (see below), the conductor can handle it, and obviously the CB can handle it, and the motor is not involved

The 1900A short circuit current is a guess. It is possible with high
system fault capacity and short to ground via motor core.

if an mag/instant only is used set at 800% ~425 A it will trip no issue

The inrush current of the 40 hp energy efficient motor in this case is more than 425A and so the breaker would trip.

 

[kwired]

To prevent further damage.
You have no idea.

Further damage to life and other property, the faulted motor already needs replaced or rewound.

 

[kwired]

I've got to install a main breaker on a main panel for a 117A dwelling unit. I've doubts on which breaker should I install: an Inversed-time one or an Instantaneous-trip one.

At first, I decided to install a 125A inversed-time breaker by NEC 240.6, but I got confused if an instantaneous-trip should be use in this case. This is mainly because I yet don't know the characteristics that should be met in order to use any.

Which one do you think I should install?

If your panel is a typical "loadcenter" your only choices of breakers that fit it will all be inverse time breakers AFAIK.

Besides the amp rating about your only other choice will be interrupt rating.

 

[Ingenieur]

Are you sure it will not trip on O/L?

The 1900A short circuit current is a guess. It is possible with high
system fault capacity and short to ground via motor core.

The inrush current of the 40 hp energy efficient motor in this case is more than 425A and so the breaker would trip.

it will not trip on thermal 125/52 = 240% assuming motor is 100% loaded, unlikely

unlikely, the conductors will limit it

You would then increase it, that is why it is allowed to be 1100% for that motor type
but stays at 250% for it, nuisance tripping would be more likely
but unlikely in either case

typical starting kva code prem eff K
max 8.99 kva/hp
450 A

the inrush would be suppressed by conductor Z
typically 20-40%
V drop 20 v
so R = 20/52 ~ 0.4 Ohm
motor R = 460/450 ~ 1.0
new staring i = 460/1.4 ~ 330, 27 % decrease
an instant/mag set at 425 (800%) would not trip
you wll not have 460 vac when starting: larger i = larger v drop
If i is 8 times as large motor v = 460 - (0.4 x 450) = 280
starting i drops more 280/1
starting load condition has more to do with it:?load, unloaded, inertia, etc

not an issue if code compliant

 

[Sahib]

Ingenieur:
I considered worst case scenario in post#25 and the Code does not prohibit the conditions mentioned in it. Thus the 150A inverse breaker with fixed instant element set at 15 times would operate only when the short circuit fault escalates to around 40 times motor FLA.

 

[Ingenieur]

Ingenieur:
I considered worst case scenario in post#25 and the Code does not prohibit the conditions mentioned in it. Thus the 150A inverse breaker with fixed instant element set at 15 times would operate only when the short circuit fault escalates to around 40 times motor FLA.

the response to post 25 made in post 27 proves it is a not an issue
if code compliant, not an issue