600 C-9 Bulbs

On 29 Nov 2006 11:30:35 -0800 snipped-for-privacy@electrician2.com wrote:
| The facts are the NEC load calculations over kill by 200 to 250 per
| cent. No. 12 wire is tested per the UL standard using 40 amperes. | There is a substantial safety margin built into the code rules that | allow for a lot of mistakes, because over the life of an electrical | system that lasts from 30 to 50 years many mistakes are anticipated. | Who knows who is going to try to modify these systems? So the 80 per | cent rule is a good engineering practice. Likewise, the NEC does not | have built in reserves for expanding an electrical system. I | personally have wire a restaurant el cheapo, only to see the expansion | take place before opening. The service had to be redone. I think a 50 | per cent load should be maximum for circuits to allow for expansion.
I'm not disputing the value of having a reserve. I do question the need for the 80 percent rule for single permanent load dedicated circuits. I would suggest that multi-outlet circuits certainly have a planned max single load limit of 50 percent (e.g. if there are anticipated loads over that amount, then either provide a higher capacity multi-outlet circuit or use a dedicated circuit). What I'm objecting to is the way such a reserve is provided for, particularly in its labeling. If no more than X amps is to put on a circuit, it should be labeled as an X amps circuit, not X*1.25 amps.
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| Phil Howard KA9WGN (ka9wgn.ham.org) / Do not send to the address below |
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On 29 Nov 2006 23:09:22 GMT, snipped-for-privacy@ipal.net wrote:

This BS is just code panels run amok. Nobody can control what a user will plug into a receptacle. It is just dillusions of power that lets a CMP write this drivel. We just limit the O/C device to a safe level and hope it operates when it should.
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Quebec a separate country? Date:Thurs, Nov 30 2006 10:52am

From: snipped-for-privacy@electrician2.com

snipped-for-privacy@electrician2.com

George Fuller wrote:

I attended the very first meeting of the Alaskans for Independence in Fairbanks way back in the early 80's or late 70's as I recall. JopeVogler was there and I got to know him in later years. He was personally offended when the National Park Service cut off access to his mine. However, all he had to do was to apply for a permit and he refused to do so. The Park Service send in a swat team to stop his cat
train and Joe never got over it leaving his cat train in the middle of the wilderness forever. I think it is still there. Joe refused to be buried in the USA and was buried down by Whitehorse from what I hear. The fact is he was just getting revenge and attention with his independence battle and I think he realized it would never happen. Today Alaskans receive much more in federal funds than we give. We would be a darn fool to secede from the USA. Besides with all the military bases here it will never happen. It is my understanding that Alaska voted in statehood because of the military vote in the first place. I was here in those days when Alaska had only 130,000 people and many did not want statehood.

snipped-for-privacy@electrician2.com wrote:

snipped-for-privacy@electrician2.com wrote:

----|
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snipped-for-privacy@electrician.com wrote:

The previous post was a mistake.
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On 29 Nov 2006 15:26:56 GMT, snipped-for-privacy@ipal.net wrote:

Phil the 80% rule is in 240.4(D) A 20a circuit (12ga copper) is good for 25a @ 60c (310.16) and 240.4(D) limits the breaker to 20a. There is your 80% Why make this complicated?
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On Wed, 29 Nov 2006 20:04:05 -0500 snipped-for-privacy@aol.com wrote: | On 29 Nov 2006 15:26:56 GMT, snipped-for-privacy@ipal.net wrote: | |>The answer to my question should focus on why the 80% rule exists | | Phil the 80% rule is in 240.4(D) | A 20a circuit (12ga copper) is good for 25a @ 60c (310.16) and | 240.4(D) limits the breaker to 20a. There is your 80% | Why make this complicated?
So what about 10ga?
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On 30 Nov 2006 03:22:39 GMT, snipped-for-privacy@ipal.net wrote:

30a circuits don't usually have multiple outlets
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On Thu, 30 Nov 2006 02:26:27 -0500 snipped-for-privacy@aol.com wrote: | On 30 Nov 2006 03:22:39 GMT, snipped-for-privacy@ipal.net wrote: |
|>| 240.4(D) limits the breaker to 20a. There is your 80% |>| Why make this complicated? |> |>So what about 10ga? | | | 30a circuits don't usually have multiple outlets
Suppose I have a welder that uses just shy of 30 amps. On a dedicated single outlet I can use it fine. But to roll it around the shop I'll need to use a long unsafe extension cord because the NEC wants to limit me to 24 amps if I make multiple outlets for it (and don't want to waste the money putting in separate branch circuits for each outlet). All because of the risk that I might sneak in a 2nd welder and try to run both on the same circuit and pop a breaker. :-)
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On 30 Nov 2006 18:33:07 GMT, snipped-for-privacy@ipal.net wrote:

There is a whole section about welders. Those rules are as confusing for the DIY as the motor rules.
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On Thu, 30 Nov 2006 22:27:20 -0500 snipped-for-privacy@aol.com wrote: | On 30 Nov 2006 18:33:07 GMT, snipped-for-privacy@ipal.net wrote: |
|>| |>| 30a circuits don't usually have multiple outlets |> |>Suppose I have a welder that uses just shy of 30 amps. On a dedicated |>single outlet I can use it fine. | | There is a whole section about welders. Those rules are as confusing | for the DIY as the motor rules.
And maybe also for kilns? Of course those don't move around as much and won't as likely need multiple outlets for a dedicated load.
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On Mon, 20 Nov 2006 19:06:18 -0500 snipped-for-privacy@aol.com wrote:
| I read all this scary stuff about terrible things that might happen if | you put 18 amps on a 20 amp circuit but I think it is largely | bullshit. 310.16 tells us 12 ga wire is really good for 25a (at the | 60c rating). That 80% safety factor you want comes in 240.4(D) that | limits the overcurrent device to 20 amps. The branch circuit will be | fine. If he splits his load across 2 cords in each duplex outlet the | down stream load is safe. Why all this hysteria? It is certainly not | backed up by the code.
I do have to agree. I think it can go even more than that. But if the work is in any way involved under code, then it's more about liability issues.
I do know that the tolerance range for circuit breaker trip curves easily puts the infinity time current as high as 125%. So not only could 25 amps be drawn on such a circuit, it might never trip the circuit breaker.
But I won't be the one to recommend it.
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gus wrote:

300 lights x7 watts!00 watts per three strings of lights. Assuming 120 volts for your voltage, 20x120$00watts for the total load at 20 amperes. If your lights stay on for three hours or more you are limited to 80 per cent of this value or .8x2400 20 watts (1920/120 amperes, 2100/120 = 17.5 amperes.) Then 2100 watts is too much load for a continuous load on a 20 ampere circuit breaker unless the circuit breaker is listed for continuous loads. If it is a Square D of commercial grade, it may very well be listed for a continuous load.
For voltage drop the voltage drop is distributed along each 100 foot string handling 7x100 or 700 watts. 700/120=5.8 amperes. The total voltage drop at the last light will be about 1.9 per cent which is acceptable. Your last light should see about 118 volts assuming, again, that there is 120 volts at the supply end. However, this voltage drop assumes an ambient of 86 Degrees F. It is probably cooler outside. Also, the free air ampacity for No. 12 conductors is considerably higher. Taken from Table 310.17 of the NEC, the free air ampacity of No. 12 is from 30 amperes for 60 C insulation to 40 amperes for 90 degree C insulation. Assuming an ambient of 50 degrees F, the voltage drop would be about 1.5 per cent (using my advanced voltage drop calculator at http://www.electrician2.com )
In the final analysis you are a little over the limit by placing 2100 watts of continuous load on a normal 20 ampere circuit breaker for a normal inside environment, but for a cooler outside environment, I would go ahead with installation.
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snipped-for-privacy@electrician2.com wrote:

You missed the the construction of the strings in the site he referenced. The strings themsleves are wired with #18, not #12. The total voltage drop at the end of 1 string will be over 3.7 volts. If he daisy chains, he exceeds the ampacity of #18 (8.1 amps) and increases the voltage drop unacceptably. http://www.thelearningpit.com/elec/tools/tables/Wire_table.htm
He'll be ok and code compliant if he follows the advice at the site he referenced.
Ed

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ehsjr wrote:

I did not see anything about No. 18 AWG conductors being used for the light string in the original post. However, assuming this is so and assuming these are like those light strings sold at Wal-Mart then we have the UL listing. UL DOES NOT USE NEC CODE RULES WHEN DOING A LISTING! This is an anomaly for sure, but UL has been caught time and again for listing products that do not meet the NEC. Anyway, assuming we have 100 lights at 7 watts on a 100 foot string would give an approximat voltage drop of 8 volts or 7.7 per cent. The original post did say there were 2300 watts available for each 20 Ampere circuit which would indicate that the operating voltage is 115 volts, in place of 120 volts. This and the fact that the voltage drop lowers the wattages of each lamp along the string would mean that the actual voltage drop would be slightly less than 8 volts more like 6 or 7 volts, still high, but probably acceptable to UL.
Now where did you get 8.1 amperes for No. 18 AWG? Table 310.16 lists 14 amperes for a 90 degree C. insulation. Also, Table 400.5(A) lists 10 amperes for 2 conductor No. 18 cords. If the strings are daisy chained a new problem is created. It appears the lamps should not be daisy chained due to the voltage drop.
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wrote:

Here is one you guys will love. I was told that a brand "X" breaker, listed to trip at 20a will actually trip at 16a. I told my coworker he was full of $#!% and get back to work. I am told it all has to do with the 80% rule in the code. I assured him that a 20a breaker will trip at 20a.
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Brian wrote:

But that doesn't mean the products are a problem. As you said UL has different sets of rules and products are manufactured under different conditions than construction. Have you seen UL listed products (properly constructed) that were unsafe?

The bulbs are 130V so it is a parallel string with only 1 bulb at 100 ft (200 ft wire). The drop at the end is probably more like 4V. The drop at the start is 0.

A 20A breaker, not rated for continuous duty, may trip at 16A some time over 3 hours. They trip on heat, which may build up over a long time span. Same with fuses. My understanding is that is the basis of the 80% continuous rule.
-- bud--
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Bud-- wrote:

For a voltage of 130 volts, 100x7 watts or 700 watt load, and a No. 18 copper conductor, the per cent voltage drop is approximately 8.6 volts or 7.2 per cent not 4 volts. The 100 feet is one way circuit length. The 2 multiplier in the below equation is for the 200 foot total length.
VD = 2xKxLxI / cma Per cent VD = (130/VD) x 100
k.9 for copper L is one way circuit length of 100 feet I is 700watts/130 volts or 5.4 amperes cma for No. 18 AWG is 1620 circualr mils
This is the way we have been doing this for about 40 years. This voltage drop is distributed along the circuit just like it is in a recreational park lot feeder. At the 50th lamp the drop would be about 3.6 volts or 3.6 per cent.
REF: NEC 551.53(C) FPN: Due to the long circuit lengths typical in most recreational vehicle parks, feeder conductor sizes found in the ampacity tables of Article 310 may be inadequate to maintain the voltage regulation suggested in the fine print note to 210.19. Total circuit voltage drop is a sum of the voltage drops of each serial circuit segment, where the load for each segment is calculated using the load that segment sees and the demand factors of 551.73(A).
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snipped-for-privacy@electrician2.com wrote:

I should explain after pulling out my copy of the Neher-McGrath Paper from 1957 that 12.9 is the circular mil ohms per foot for 100 per cent IACS copper at 75 degrees C. Since resistance is inversely proportional to circular mil ohm area the resistance of one foot of No. 18 AWG copper is 12.9 divided by the circular mil area of 1620 circular mils. This the same way the Neher-McGrath paper does this minus the lay factor. For 100 lights over a 100 foot long circuit we actually have 100 parallel circuits. For each light circuit we have the resistance of the conductor to the light and the resistance of the conductor from the light or 2 times the one-way circuit length.
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snipped-for-privacy@electrician2.com wrote:

Correction: Per cent VD = (VD/130) x 100 not the other way around.
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| Here is one you guys will love. I was told that a brand "X" breaker, listed | to trip at 20a will actually trip at 16a. I told my coworker he was full of | $#!% and get back to work. I am told it all has to do with the 80% rule in | the code. I assured him that a 20a breaker will trip at 20a.
Breakers are not very accurate devices. Since when operating them right around the rated current only involves the thermal element, this needs to take in other considerations such as manufacturing tolerance, ambient temperature, age, etc. The element in one could get a bit hotter than in another for the same current. The point where the element's thermal bend causes the mechanism to trip can also vary.
I think the 80% rule is well justified on the basis that a wide range of tolerance needs to be taken in to account in the design of permanent installations. So, for example, I don't think there is an issue with a 15 amp appliance plugged into into the upper half of a 15 amp duplex receptacle wired to a 20 amp circuit, other than for very little current capacity remaining on the circuit. As for plugging it into the lower half, that makes it easier to plug something else into the same circuit.
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