ECN Forum
Posted By: winnie '80%' breaker rating - 09/25/03 01:08 PM
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
Posted By: Pinemarten Re: '80%' breaker rating - 09/25/03 01:47 PM
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.
Posted By: JBD Re: '80%' breaker rating - 09/25/03 04:33 PM
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.
Posted By: winnie Re: '80%' breaker rating - 09/25/03 06:35 PM
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
Posted By: JBD Re: '80%' breaker rating - 09/26/03 02:04 PM
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.
Posted By: Scott35 Re: '80%' breaker rating - 09/27/03 09:47 AM
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!
Posted By: winnie Re: '80%' breaker rating - 09/27/03 06:42 PM
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).]
Posted By: PCBelarge Re: '80%' breaker rating - 09/27/03 10:09 PM
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
Posted By: C-H Re: '80%' breaker rating - 10/02/03 07:17 PM
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]
Posted By: nesparky Re: '80%' breaker rating - 10/14/03 04:41 AM
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).]
Posted By: SirJaxx Re: '80%' breaker rating - 10/17/03 03:31 AM
Circuit breakers do not trip on Amperage. Circuit Breakers Trip on Heat! If you haven't applied a temperature correction (T310.16) factor into your equation, and attempt to use any type of Circuit Breaker and conductor not rated for that application, you will have begun the what is commonly known as " Insulation Degradation " (310.10). The reason for the 80% allowance is to prevent insulation damage to the conductor, not the breaker.

[This message has been edited by SirJaxx (edited 10-16-2003).]
Posted By: winnie Re: '80%' breaker rating - 10/17/03 05:54 PM
Quote

SirJaxx wrote:

The reason for the 80% allowance is to prevent insulation damage to the conductor, not the breaker.

This is explicitly not correct, and the reason why I asked the question in the first place.

The calculations and deratings in Article 310 provide for the ampacity of conductors which will prevent undue degradation to the insulation. These calculations consider the temperature rating of the insulation, the ambient air flow, the thermal conductivity of the surroundings, the presence of other current carrying conductors, the ambient temperature, etc. The ampacity described for a given conductor in article 310 is the ampacity that can be carried 100% of the time without excessive insulation degradation.

When a normal circuit breaker is used, it must be rated 125% of the continuous load. In order for the conductors to be properly protected by this circuit breaker, they too must be 125% of the continuous load. However if a "100% rated" circuit breaker is used in an appropriate assembly, then it may be sized for 100% of the continuous load, and the conductors as well may be sized (using Article 310) to be at 100% for the continuous load.

Now, it is clear that operating conductors at only 80% of their continuous load is 'better' from the point of view of heat generation and insulation degradation, but operating them at 100% of their continuous load (as determined by 310) is clearly allowed if the proper circuit breaker is used.

See, for example the Exceptions in 215.2(a)(1) and 215.3

-Jon
Posted By: SirJaxx Re: '80%' breaker rating - 10/19/03 06:36 AM
Quote:
See, for example the Exceptions in 215.2(a)(1) and 215.3
Reply:
This section applies to Feeders, not branch circuits as you had specified in your earlier question.
Quote:
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%.
Reply:
Try to turn on 24, 100 watts Halogens lamps, and see for yourself what happens. It Will trip the 20 amp Breaker.( or 18, 100 w on a 15 amp breaker) Continous load or not.

Look when you say a Continous Load, you are implying foreknowledge of a particular load that needs to be served. Residential Branch Circuit receptacle loads are different and have different requirements depending on what room you are in. 210.19 & 240.3
See 230.13 for Non-Dwelling receptacle loads.

But Circuit Breakers are NEVER WEAK. They are designed to trip out depending on the amount of HEAT they encounter. They will trip out in your Oven.

Then you refocused the Question to read as Follows: So the question is: in what way does an 80% rated circuit breaker overheat when loaded at 100%?

And Again as we have all stated here in one or another is this, A BREAKER TRIPS ON HEAT! Not Amperage. Amperage makes Heat, and so does resistance but Regardless of how the Heat reaches the Breaker, either thru the Conductors, the BusBars, or thru a Fire in Loadcenter. Breakers trip on Heat

[This message has been edited by SirJaxx (edited 10-19-2003).]
Posted By: winnie Re: '80%' breaker rating - 10/19/03 10:37 AM
SirJaxx wrote:
Quote

This section applies to Feeders, not branch circuits as you had specified in your earlier question.
Whoops; I was lax in flipping through the NEC to find the section I was referencing. I should have referenced 210.19(A)(1) Exception, which uses the essentially the same text, and presumably is in the code for exactly the same reason.

In any case, while you are preaching to the choir that 'breakers trip on heat', are you in a position to explain the difference between an 80% rated breaker and a 100% rated breaker? Everyone here (including me, in the post that opened this thread) agrees that breakers trip on heat. But what is the way in which this heat affects an 80% breaker _different_ from the way it affects a 100% breaker.

Again, I have not been trying to argue this point (do breakers trip on heat or not), nor am I intentionally trying to be thick headed. I said in my initial question:
Quote

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?
I really do get that the breaker trips because of heating.

What I am trying to understand is why an 80% breaker is different from a 100% breaker. Both presumably trip on heat, and both will trip at a lower current if they are in a warmer ambient environment. Both will heat up when current flows through them, and both will heat up more when enclosed in a panel, rather than sitting in free air.

Remember, when you say that a 'breaker trips on heat', what you really mean is that the breaker trips when its thermal trip element reaches a certain temperature. When I run current through a breaker, all current carrying components will generate heat, including the thermal trip element. When an 80% breaker is operated at 100% of its trip rating, does the heat generated in the thermal trip element combined with the heat conducted to the thermal trip element, cause the circuit breaker to open _before_ or _after_ other parts of the circuit breaker are damaged or subjected to premature degradation?

The obvious, most sensible, intuitively correct answer is that 'of course a circuit breaker is designed so that it will open before any part of it gets too hot' (if the heating is caused by current flowing through the circuit breaker in a uniform ambient). I no longer trust this answer because I have seen enough references that suggest that an 80% rated breaker _will_ be damaged if used in this fashion.

I believe that at this point the only thing to do is follow PCBelarge's suggestion, and ask the manufacturers.

Regards,
Jonathan Edelson
Posted By: SirJaxx Re: '80%' breaker rating - 10/19/03 03:06 PM
Quote:
In any case, while you are preaching to the choir that 'breakers trip on heat', are you in a position to explain the difference between an 80% rated breaker and a 100% rated breaker?
Reply:
Yes I well try too, but you will need to open your Hymnal as we don't seem to be singing the same song. First off in my 28 years in the field, I haven't encountered an 80% Regular Inverse Time Circuit breaker, and since the code requires all Circuit breakers to be listed with their Ampere Trip rating on them, 240.6. I don't see the type you are suggesting listed there. I can't find a 12 or a 16 or a 24 amp rating for breakers.
But for the sake of argument, let's try this! Instead of Heat, let's use Water. Let's say we have 1oo gallons of water, and what you are wanting to do is put that water in a Bucket that will only hold 80 gallons. Well what happens to the 81st thru 100th gallons of water once the capacity of that bucket is reached? Is the Bucket ruined? No. One a Specified rating or capacity is reached, that's it, end of story. Trip.
But let me ask you this, please tell me the correct size Conductor for feeding a 25 amp, 240v 1p AC compressor unit that is located outside of a House in Texas. Use NMC and route 50' of the cable thru the Attic over to and down the exterior wall, and out to the Disconnect switch. Please list the Circuit Breaker, wire and disconnect sizes.

[This message has been edited by SirJaxx (edited 10-19-2003).]
Posted By: winnie Re: '80%' breaker rating - 10/19/03 09:48 PM
SirJaxx wrote:
Quote

But let me ask you this, please tell me the correct size Conductor for feeding a 25 amp, 240v 1p AC compressor unit that is located outside of a House in Texas. Use NMC and route 50' of the cable thru the Attic over to and down the exterior wall, and out to the Disconnect switch. Please list the Circuit Breaker, wire and disconnect sizes.
This quiz is not particularly applicable to the 80% versus 100% rating question, since the exception that is provided in for example 210.19(A)(1) is _not_ provided in 440.32 for Air-conditioning and Refrigerating Equipment, nor is it provided in 430.22 for Motors, Motor Circuits, and Controllers. For these loads you must _always_ rate your conductors at 125% of the rated-load current (or the 'selection current' which may be greater than the full load current, and indicates an overload current which would be permitted by the motor protection for an extended period of time).

More pertinent to the present discussion would be to determine the correct size conductor and circuit breaker for a 18A, 120V single phase lighting load expected to operate for the entire business day. Use THHN in appropriately sized EMT; this is the only circuit in the EMT; presume that the length of run is short enough that voltage drop is not a factor. Ambient temperatures for the entire run are below 30C. What conductor ampacity and standard breaker size should be used, and what size conductor provides this ampacity. If there are several possible choices of breaker and conductor which might be used, describe the trade-offs of each. I will attempt to answer your quiz; your answers to my quiz would be instructive.

1) First determine the required ampacity of the conductors.
440.32: Branch circuit conductors supplying a single motor-compressor shall have an ampacity not less than 125 percent of either the motor-compressor rated-load current or the branch-circuit selection current, whichever is greater. Since a branch-circuit selection current is not specified, then the minimum conductor ampacity must be 125% of 25A, or 31.25A.

2) Then determine the size of conductor needed to carry this current.
Even though you specify NMC cable, I presume that you mean NMC-B cable with a 90C temperature rating. The 60C ampacity needs to be used, but the 90C ampacity can be used for derating relative to attic temperatures. From table 310.16 'Allowable Ampacity...Not More that Three...' the minimum size 60C conductor which can carry 31.25A is 8AWG. The allowable 60C ampacity for this 8AWG conductor is 40A.

3) Check for thermal derating of the cable.
The allowable 90C ampacity for the 8AWG conductor is 55A, which permits operation in an ambient of 70C with an ampacity of 31.9A. The temperature in the attic is _probably_ less than 70C, but I don't yet know how to determine this.

4) Determine the circuit breaker needed to protect the conductors from overcurrent and short circuit.
The minimum standard size circuit breaker greater than 31.25A is 35A. This is greater than the permitted value to protect the motor-compressor from overload, but will safely protect the circuit conductors. The motor will require additional overload protection.

5) Double check that the circuit breaker protects the conductors.
Since the circuit breaker will permit 35A to flow, we need to re-check that the conductors in the attic are properly protected. In a 60C ambient, the ampacity of the conductors is 39A, which is greater than 35A. Thus if the attic is cooler than 60C, the conductors will be properly protected by the 35A breaker. If the attic is warmer than 60C, then larger conductors will need to be used. 60C is 140F, and depending upon the insulation and ventilation of the roof in question, I am certain that an attic can get warmer than this, but this would need to be checked.

6) Check for voltage drop.
You do not specify the height of the building. I will assume 30 feet, giving a total cable length of approximately 110 feet. The round trip distance is 220 feet. The resistance of 8AWG wire at 90C (worst case scenario) is approximately 0.9 ohms per 1000 feet. At 25A, the voltage drop to the load is about 5V, which is acceptable on a 240V supply. However the startup load will be considerably greater. According to 440.4(A) the nameplate is supposed to report the locked-rotor current for this compressor, however this current was not provided in your question. A momentary voltage drop as high as 30V may be expected given the cable as described and an estimated locked rotor current of 160A. This is _probably_ acceptable since the compressor is the only load on the circuit, however additional information is needed to evaluate this

7) To determine the disconnect size, use 440.12(A)(1). The disconnect needs to be at least 28.75A, presumably a 30A standard size. A horsepower rating for the compressor could be determined according to 440.12(A)(2) this requires either a nameplate horsepower rating, or a comparison of motor full load current to table 430.148 and a comparison of motor locked rotor current to table 430.151(A) My best guess given the information provided is that a horsepower rated disconnect would be rated for 5 horsepower at 240V.

8) Because of the current rating of the compressor, a standard size circuit breaker does not provide overload protection for the compressor. If the compressor does not already contain overload protection, then additional overload protection would be required.

Comments?

Quote

First off in my 28 years in the field, I haven't encountered an 80% Regular Inverse Time Circuit breaker, and since the code requires all Circuit breakers to be listed with their Ampere Trip rating on them, 240.6. I don't see the type you are suggesting listed there. I can't find a 12 or a 16 or a 24 amp rating for breakers.

I'd suggest taking a look at the UL Guide for Molded Case Circuit Breakers (link posted above). You might find this interesting. Look under the section headed 'Maximum Load'. Most breakers are not rated to carry more than 80% of their trip level for periods longer than 3 hours at a time. This means, for example, that a 15A breaker being used to supply a continuous load that is greater than 12A is most likely violating its listing. I expect that _most_ of the breakers that you have installed have been 80% rated breakers.

Regards,
Jonathan Edelson

[This message has been edited by winnie (edited 10-19-2003).]

[This message has been edited by winnie (edited 10-19-2003).]
Posted By: JBD Re: '80%' breaker rating - 10/20/03 02:59 PM
All of the documentation I have found lead me to believe the 80% rating is an application restriction caused by the NEC requirement for sizing conductors at 125% of the circuit current.

No where is an 80% level mentioned in the UL testing of "unenclosed" molded case circuit breakers. The primary difference between an 80% and a 100% application, is the enclosure.

Can you provide any documentation to support your statement "...I have seen enough references that suggest that an 80% rated breaker _will_ be damaged..."?

Selected phrases from NEMA AB3-2001 standard for Molded Case Breakers.

1.5.1
Circuit breakers ... provide overcurrent protection for conductors and equipment... before the current reaches a value that will cause an excessive or dangerous temperature in conductors or conductor insulation...

5.1.3.1
Thermal magnetic molded case circuit breakers are normally calibrated at 100 percent of rated current in open air ...

A1.2
Underwriters Laboratories, Inc... The regular tests ... consist of calibration checks, both 200 percent and 135 percent...
Posted By: winnie Re: '80%' breaker rating - 10/20/03 07:55 PM
JBD writes:
Quote

All of the documentation I have found lead me to believe the 80% rating is an application restriction caused by the NEC requirement for sizing conductors at 125% of the circuit current.
I have reasons for believing that the reverse is true; that the 125% oversizing of the circuit conductors is based upon the rating requirements of the circuit breakers rather than the circuit breaker requirements being set by the conductors.

1) Why have a whole Article dedicated to determining the ampacity of conductors, and then arbitrarily require in other Articles that this ampacity be derated by 20%? Why not simply alter the ampacity tables to incorporate this derating? Or use different insulation degradation characteristics which reduce the allowed operating temperature...if THHN wire were restricted to a maximum operating temperature of 80C rather than 90C, it would have lower ampacity but would last longer.

2) There exist breakers that are listed for continuous operation at 100% of their trip rating. In some articles (for example 210 Branch Circuits 210.19(A)(1) and 210.20(A) ) there are specific exemptions to the 125% rating requirement. If one is using a normal 80% rated circuit breaker, then the circuit breaker and the wire must _both_ be rated to 125% of the continuous load, but if one uses a less common 100% rated circuit breaker, then the circuit breaker _and_ the wire can simply be determined for the continuous load. Clearly if the code wanted to require that wires _always_ be rated 125% of the continuous load placed on them, 210.19(A)(1) Exception would not exist.

3) There clearly are situations in which the code wants the conductors to be oversized; in situations described by these articles, the exception for 100% rated breakers is not found.

4) The McGraw-Hill Handbook for the NEC, in discussing these exceptions, explicitly says that this is for protection of the breaker, not the wire...but I wouldn't trust this information as a sole source.

JBD writes:
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Can you provide any documentation to support your statement "...I have seen enough references that suggest that an 80% rated breaker _will_ be damaged..."?

I have seen a couple of usnet news references; but these are poor enough to be not really be considered supporting documentation. The only serious documentation is the McGraw-Hill Handbook mentioned above, where they explicitly state that normal 80% rated circuit breakers will be damaged if operated at > 80% loading for extended periods. The example that they give is specifically in the context of industrial installations, but the same rules seem to apply to smaller residential circuits and breakers. I can provide a page number reference later in the week.

As I've said above, my engineering gut tells me that a normal 80% breaker, when loaded to >80% (but less than 100%) for an extended period of time will heat up, and might trip at lower than rated trip current because of the higher temperature...but without any damage. I am trying to confirm that this is the case, and that the tale of circuit breaker damage described in the McGraw-Hill handbook is false.

The NEMA quotes that you provide state that the conductors are protected, and that the breakers are calibrated to trip with a particular time/temperature/curve. But I don't think that they say what will happen to the _circuit breaker_ when it heats up inside of an enclosure.

Regards,
Jonathan Edelson
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