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#128677 09/25/03 09:08 AM
Joined: Sep 2003
Posts: 650
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winnie Offline OP
Member
I have been trying to understand the rules about the allowable loading on a circuit breaker, specifically the requirement that the circuit breaker rating be 125% of the total continuous load placed on a branch circuit. (The same language is used in several places, and seems to apply both to branch circuits and to sub-panel feeders, and presumably in other areas.)

I can understand this as providing a useful bit of safety margin, and (probably more importantly given the various safety margins already built into component ampacity) a useful bit of avoidance of nuisance tripping. However I have also seen it suggested that the circuit breakers themselves are only listed for application up to 80% continuous loading or 100% non-continuous loading, and that the circuit breakers themselves could not tolerate a 100% continuous load unless specifically rated.

What I want to figure out is what to expect when a high continuous load is placed on a circuit breaker. Say, for example that a receptacle branch circuit is loaded to 100% capacity with halogen lamps, which are simply left on. Because the branch circuit has no permanently connected loads, no attempt is made to limit the load to 80%.

I would expect the various components in the circuit (wires, receptacles, circuit breaker, etc) to heat up, but that this heating would be expected and designed for in the selection of components.

Will the (presumably 80% rated) circuit breaker to heat up, do the temperature compensation thing, and eventually simply (and safely) trip? Or will the circuit breaker not trip, but instead overheat in some component part? Or will the circuit breaker eventually trip, but suffer some accelerated wear in doing so?

For the small molded case breakers used in residential applications (15-30A), are these devices usually 80% rated or are they usually 100% rated? The same size cases are often used for much larger breakers, and the terminals on these breakers can often take much larger wire than is needed for the circuits being protected.

Thanks
Jon

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#128678 09/25/03 09:47 AM
Joined: Aug 2003
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The way I always think of breakers/wires/loads is:
The breaker only protects the wire from overheating.
Most loads will take care of themselves.

#128679 09/25/03 12:33 PM
Joined: Jul 2001
Posts: 599
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JBD Offline
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All standard overcurrent protective devices (OCPD) mounted in enclosures(i.e. breakers in panels or fuses in safety switches) are tested to protect conductors which have been sized at 125% of their continuous loading. Enclosed OCPDs intended for 100% loading require "special" ventilated enclosures. Open air testing/rating of OCPD are usually not applicable in the real world.

The application of OCDPs mimic the rating and application of conductors in regards to raceways versus open air as well as different ambient temperatures.

#128680 09/25/03 02:35 PM
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winnie Offline OP
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That part I get; given a 40A continuous load you select conductors and overcurrent protection rated at 50A (in general, ignoring the special cases, etc. etc.) Thanks for confirming that standard circuit breakers are tested to this 80% load requirement.

My question is, given the above, what happens when the breaker is utilized at 100% of its continuous load. This is utilization which is in violation of the listing, and I understand this. What I don't understand is what sort of failure should be expected.

One possibility is that the 'rating' of the breaker is really just a point on its time/temperature curve which will cause the breaker to trip after some number of hours (given standard ambient conditions and air flow, etc.) In this case, if I attempt to _use_ a 20A breaker at 20A, I would expect it to trip after some amount of time, with no damage to the wire or to the breaker itself. In other words, the 'failure' is simply the circuit safely opening.

I guess another way to ask the question: Given the testing on standard molded case circuit breakers, such as those used in residential applications. If a 100% load is applied for an extended period of time, will the molded case circuit breaker protect _itself_ as well as the conductors which it is supposed to be protecting?

-Jon

#128681 09/26/03 10:04 AM
Joined: Jul 2001
Posts: 599
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JBD Offline
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Part of what the OCPD listing does is confirm a specific Time vs Current performance. OCPDs are tested at various current levels to insure they will "trip" in the correct time frame, based on a specific ambient temperature.

You may load a breaker to any amperage level you want (as long as it is below the breaker AIR level) for a time period which will not cause it to trip, and in my opinion it is not a violation. In fact there is nothing that says you cannot trip a breaker. But, if the breaker is tripped regularly you will have to also consider it's mechanical and electrical life.

However, you may not overload the wiring.

Basically, the NEC says: size the load; then pick the wiring; then pick an OCPD which is not larger than the wiring.

#128682 09/27/03 05:47 AM
Joined: Oct 2000
Posts: 2,723
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I'll try to cover this topic as much as possible. If you already know of a certain item, cool! [Linked Image] ; if not, I hope this text
makes sense and you are able to learn from it.
Also, others curious to the topic may benefit from a somewhat complete description.

*** Terms, Acronyms and Abbreviations used:
* VAC = Volts, Alternating Current,
* LCL = Long, Continuous Load(s),
* Current = Amperes, Amps,
* MCCB = Molded Case Circuit Breaker,
* OCPD = Over Current Protective / Protection Device,
* SCA = Short Circuit Amperes,
* G.P. = General Purpose,
* DED = Dedicated,
* HID = High Intensity Discharge Lighting,
* Spec. Ckt. = Specific Load Circuit,
* Trip Rating = Rating, in Amperes, for a given OCPD (i.e. 20 Amp circuit breaker is 20 Amp trip rated),
* THD = Total Harmonic Distortion,
* SMPS = Switch-Mode Power Supply.

To start things off, the definition of Non-Continuous and Continuous Loads should be covered.

* Continuous Load: A Load that is drawing a fixed level of Current for 3 Hours or More (180 Minutes or more).

* Non-Continuous Load: A Load drawing a fixed level of Current for less than 3 Hours (<180 Minutes).

Using the above example, take a given load, run it for 179 minutes then turn it off for 30 minutes, then turn it back on
for 179 minutes, etc... and this will be a Non-Continuous Load. Although this is a very crude example, it makes the
necessary points of a Non-Continuous load.

Same load run for 180 minutes, turned off, run once again for 180 minutes (or even just run once for 180+ minutes), is
now a Continuous load.

Next, for Circuitry (and load calcs + Panel Schedule calcs), any load that is classified as a Continuous Load will
receive an "LCL Adder". This means the Load Current on a Continuous Load receives an additional 25% rating,
making it 125% of the actual load Current.
Example: a Continuous Load draws 10.0 Amperes. With the LCL adder, the load is now figured as 12.5 Amperes -
even though there will only be 10.0 Amperes drawn by the Load.

LCL is only figured once - meaning that if the Branch circuit having the Continuous Load is calculated at 125% (I ×
1.25), this is the only place to figure LCL. The LCL will be "Automatically" carried along with the Subfeeders, to the
next Panel, and so on.
If the Branch Circuit is not figured with LCL added, then it may be added to the Subfeeder. Once again, it only needs
to be added once, but it gets carried over through the system.
FYI: It's better to figure the Branch circuit with LCL, so proper calcs may be done to the given circuit.

These calcs are for System Performance and overall Equipment rating, so the equipment may handle the Continuous
loads properly.
Main item of concern here (and for any LCL Load) is HEAT!!! I²R losses increase with heat! Component failure
increases with heat! Insulation breakdown situations increase with heat!
Heat is the key "to avoid" item in all Electrical Designs and Installations.

Now, for the OCPD and the "80%" thing:

Generally speaking, most of the MCCBs dealt with in normal Construction Electrical Installations are "Not 100%
Rated Frames".
A "100% Rated Frame" will be able to carry a Continuous Load Current for >3 Hours, which is at the rating of the
device's Maximum, without "Derating" the circuit to "80%" of the Maximum.
On "Common", non-100% frames, if an LCL circuit is run through it, the LCL load can only be 80% of the device's
trip rating (current rating). Remember when I said to only add LCL once, this also applies here.

Examples of OCPD loads with and without LCL - for standard types and 100% rating types:
Loads will be 20.0 Amps steady.

Using a "Standard" rated type frame (non-100%):

20 Amp load on 20 Amp circuit (breaker + #12 THW / THHN cu) for <180 Minutes: No Derating needed.

20 Amp Load for 180 minutes or more: Derate circuit to 80% Capacity.
This results in either reducing the total load to 16 Amps (20 × 0.8 = 16) and using a 20 Amp circuit (20 Amp breaker
+ #12 THW / THHN cu),
or;
Installing a 30 Amp circuit for the 20 Amp continuous Load (30 Amp breaker + #10 THW / THHN cu).
The 30 Amp circuit has a Maximum of 24 Amps for a Continuous Load.

In this design situation, it would help to figure the branch circuit with LCL added in first, then size circuit per the total
load figure.

Using a 100% Rated Frame:

Circuit may be driven with a Continuous Load of 20 Amps for >180 minutes, and still use 20 Amp circuit.

In this design situation, best to NOT figure branch circuit with LCL first (need to know the "real" continuous load
current), add the LCL to the Feeders instead.

As to the trip rating, here's some information (Standard Overload only, not the AIC ratings or the Time-Current curve
figures...these are way beyond the scope of this discussion):

A circuit breaker / fuse may have 100% of it's rating drawn across it for eons, yet it will not trip (unless the breaker is
weak and sucks really bad!). If excessive heat is concentrated within the Panelboard, this may effect the maximum trip
points of breakers - making them trip at levels below their maximum rating - like 85-90% of the trip rating.

Trip ratings are for 100% of the rating, at a given system VAC.
Example: a Square D QOB 320 (3 pole 20 amp 240 VAC max) can carry 20 Amps for an infinite time, and will not
trip under normal circumstances.
The same breaker may allow 21 amps to flow across it for a few days before it trips (if it trips at all!).
Same breaker may allow 25 Amps for 3-5 hours, 30 amps for 1-3 hours, 40 amps for upto an hour, 60 amps for 15
minutes, 100 amps for 1 - 5 minutes, and so on. (this is the Time-Current trip Curve).
Under fault situations, the same breaker may trip in 10 seconds with a SCA of 300 amps, 5 seconds with an SCA of
500 amps, and never trip (even explode) with an SCA of 25,000 amps. (this is the AIC rating and figures of
Time-Current trip too).

Now to applications!

Mainly, an LCL load is kind of obvious. Lighting is one obvious one. Certain Ded. and Spec. Ckt. loads may fall into
the LCL realm.
For HID Lighting, there is also the need to figure starting current in the circuit, for certain Ballast types. On Fluorescent
Lighting using high Frequency Electronic Ballasts with THD >10%, the circuitry needs to be adjusted accordingly for
the excessive Harmonic Load Current.
Same goes for Computer equipment and other equipment using SMPSs.

Along with the Ungrounded Conductors having an LCL and Harmonic load calc, the Common Grounded Conductor -
when used - should also get equal calcs.

On G.P. Receptacle circuits, it's very difficult (actually impossible) to determine what LCL devices might be used.
At times, it may be "Assumed" - like portable floor heaters, but this is such a variable.
Proper System designing (along with Installation) would reflect the Client's needs and load requirements.

I'll finish this off here, since it's running very long!
Let me know if this was helpful, if something is questionable, if something needs to be edited, or if I should add more
on the subject later.

Scott35

p.s. feel free to bring up another topic for discussion!


Scott " 35 " Thompson
Just Say NO To Green Eggs And Ham!
#128683 09/27/03 02:42 PM
Joined: Sep 2003
Posts: 650
W
winnie Offline OP
Member
Scott,

Thanks for the analysis. I do want to keep pressing the issue, however, as I belive that you've not answered my question, but helped me to refine it. I am not trying to understand _how_ the 80% rule (or the 125% rule) is applied; this is spelled out in the code and explained in the code guide books. The above example helps as well. I am trying to understand the _why_ of the 80% rule and the implications of not following it.

I guess that this is not a code question at all, but instead a UL listing question, which means that what I really need to do is dig through the UL listing requirements for overcurrent protection devices and molded case circuit breakers to find the answer...but those books are _expensive_ [Linked Image] I'll guess I'll have to find a suitable library. [Linked Image]

From my online searching, I have two UL links:

UL guide info for Molded Case Circuit Breakers

Scope of UL Standard 489

A further reference is in the McGraw-Hill National Electrical Code Handbook, 24th edition page 142 bottom paragraph ( referencing NEC 210.19(A)) which states "...load limitation to 80 percent of the rating of the protective device are based on the inability of the protective device itself to handle continuous load without overheating."

The first specifically repeats the 80% loading requirements for continuous loads, and makes it clear that most circuit breakers used for branch circuit and feeder protection are explicitly _not_ intended to be used at 100% load for extended periods of time. This is quite at odds with what JBD describes above, since it strongly suggests that certain loading conditions damage the circuit breaker. It strikes me as totally against common sense that this is the case, and in my gut I _want_ to agree with JBD, but the wording in the code, in the UL guide info, and in the code handbook all seem to suggest otherwise.

As Scott says, the problem is _heat_, and clearly significantly more heat will be generated in a circuit breaker at 100% loading than at 80% loading...since this is I^2R heating, I would expect nearly 60% more heat production and a similarly increased temperature rise.

The concern is 'where is this heating occurring', and does this 'overheating' equate to damage. In most cases, I would expect a component that overheats to be damaged, but because thermal circuit breakers are temperature sensitive devices, overheating could present itself in a different fashion.

If the component that 'overheats' is the bimetal thermal trip element, then 'overheating' is simply that component operating at excessive temperature for the current being carried. The result of this overheating would be that the breaker simply trips at lower than rated trip current. In this case, there would no more damage than the ordinary wear and tear of interrupting a circuit under load.

However if the component that 'overheats' is another part of the circuit breaker, then it could be damaged. One then ends up with the rather disturbing possibility that a breaker is fine under normal loading (it remains closed) and fine under overload (it opens the circuit and protects itself and the wires), but subject to damage under excessive but not particularly high loading.

So the question is: in what way does an 80% rated circuit breaker overheat when loaded at 100%?

I hope that this better focuses the question, and that I'm not just being obtuse and beating a dead horse [Linked Image] I also expect that the answer will actually be different for different models and sizes of breaker. As I said above, my particular focus is on the 15 and 20 amp breakers which would be used in residential service in residential panelboards, where it is not uncommon to keep loading a receptacle branch circuit until the breaker trips.

Thanks
Jon

[This message has been edited by winnie (edited 09-27-2003).]

[This message has been edited by winnie (edited 09-27-2003).]

#128684 09/27/03 06:09 PM
Joined: Jun 2003
Posts: 681
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Winnie

I myself would like to know the answer in language electricians can understand. I think at this point that a breaker manufacturer rep is the one we need to contact for further info.

Pierre


Pierre Belarge
#128685 10/02/03 03:17 PM
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C-H Offline
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This is for European breakers, but I think the physics is similar on the other side of the Atlantic.

Taken from http://www.hager.co.uk/techServ/pdf/circuitprot.pdf

[Linked Image from i.kth.se]

#128686 10/14/03 12:41 AM
Joined: Jun 2001
Posts: 642
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Member
As scott said in his post, the reason is heat build up.
Unless listed for 100% loading a circuit breaker will eventually trip if operated at >80% of it's rating. When it trips depends on several factors ie. heat in the panel, age of breaker, load on breaker, manufacturer etc. The trip/time charts the manufacturers publish are a guide only.
If too much load is on the circuit, problems will start to appear soon enough.
A good design leaves room for future added load. How much room is another question.
As A general rule of thumb this also applies to fuses in much the same way.

[This message has been edited by nesparky (edited 10-14-2003).]


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