600 C-9 Bulbs

We are putting light bulbs on a large outside tree in a city park. We have six 100 foot strings of 100 C-9 lights. We have two 20 amp GFI circuits available. I plan to put 300 lights on each circuit. My math

shows 2300 watts available and C-9 bulbs use 7 watts. This comes to

2100 watts. I will use 12 guage extension cords. The light strings and bulbs are commercial grade. This all looks good on paper, but I am

worried because there is not much room for real world error. Does anybody have experience with this many lights? Wasn't there a movie about a guy causing a large scale blackout when he threw the switch on his house decorations? I don't want to be that guy. Any help you can give me would be appreciated. Thanks, Gus

Reply to
gus
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Well Gus here is what you do. Call out a qualified electrician, pay him to do the work and you get to be the hero for not fudging something up.

Reply to
Brian

You appear to have made one or more errors, get advice from the electrician before you do the work. Not just the math, there are many code requirements in regard to the materials and methods.

j
Reply to
operator jay

Depending on voltage drop, you may be okay, or you are going to be tripping the breakers. What country are you in? What did they do last year? How much trouble to get another circuit or reduce the load? You do know that someone makes a LED replacement for the c-9 bulbs which use way less power and last lots longer.

--Dale

Reply to
Dale Farmer

It's not the brownout we should worry about, it's the fire. You realize, of course, that you cannot plug one of your strings into the end of another?

Ed

Reply to
ehsjr

| We are putting light bulbs on a large outside tree in a city park. We | have six 100 foot strings of 100 C-9 lights. We have two 20 amp GFI | circuits available. I plan to put 300 lights on each circuit. My math | | shows 2300 watts available and C-9 bulbs use 7 watts. This comes to | 2100 watts. I will use 12 guage extension cords. The light strings | and bulbs are commercial grade. This all looks good on paper, but I am

Normally the design current for a 20 amp circuit should be 80%, which would be 16 amps here. Your wattage calculation appears to have assumed

115 volts. What matters is what the total current is for the 100 bulbs with the voltage supplied (should be close to 120 volts in the USA).

If the bulbs are truly 7 watts at 120 volts, this is 17.5 amps total for three strings on one circuit. You could get away with this working. If you can find a couple shorter strings and back the number of bulbs down to say 500, you can make it work more within the code requirements. Reducing one string in each of the 2 sets to 74 working bulbs, either by a shorter string or by leaving in some dead bulbs clustered in the back of the tree, would get you under 16 amps.

| worried because there is not much room for real world error. Does | anybody have experience with this many lights? Wasn't there a movie | about a guy causing a large scale blackout when he threw the switch on | his house decorations? I don't want to be that guy. Any help you can | give me would be appreciated.

That's just movie fiction to impress people. There's also the urban legend of the kid hitting a power pole with his bat just as the lights went out in a past big NYC blackout.

You are not going to do much more with a big error in this than burn out some plug or cord. At worst, it could burn down the tree of the hot spot created is close to it.

Surely you have an electrical inspector office around there. While this might not fall under inspection issues by not being a permanent install, maybe they can give you some advice on a short term setup, especially if you are an employee of the same jurisdiction (city?).

Reply to
phil-news-nospam

300 lights x7 watts=2100 watts per three strings of lights. Assuming 120 volts for your voltage, 20x120=2400watts 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 =1920 watts (1920/120=16 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

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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.

Reply to
electrician

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.

Reply to
gfretwell

It is standard practice to limit the continuous current on a circuit breaker to 80 per cent of its rating unless the circuit breaker is listed for continuous loading. Although Table 310.16 lists 25 amperes as the ampacity that a No. 12 AWG copper conductor with 60 degree C. insulation can carry continuously, the asterisk sends us to 240.4(D) that requires that the maximum OCPD be 20 amperes. You know this, I am sure.

The bottom line is long established practices of using the 80 per cent rule has kept buildings safe for many years. Too often these temporary installations become rather permanent and last for years. Just how far do we want to go in shaving the Code rules when we know good and well when we follow them we get a safe installation and escape the liabilities that may come with a burn down, death, or both. And liabilities are a big item. If there is a fire, then there is an investigation, and experts are called in, and if someone has died or been injured the money involved can become very large. I know of one case where the settlement was about $3 million, and the payer was held accountable because they had violated several Code rules. In this case it was the NESC and not the NEC.

Reply to
electrician

I forgot to give the reference to continuous load:

210.20 Overcurrent Protection. Branch-circuit conductors and equipment shall be protected by overcurrent protective devices that have a rating or setting that complies with 210.20(A) through (D).

(A) Continuous and Noncontinuous Loads. Where a branch circuit supplies continuous loads or any combination of continuous and noncontinuous loads, the RATING of the overcurrent device shall not be less than the noncontinuous load plus 125 percent of the continuous load. Exception: Where the assembly, including the overcurrent devices protecting the branch circuit(s), is listed for operation at 100 percent of its rating, the ampere rating of the overcurrent device shall be permitted to be not less than the sum of the continuous load plus the noncontinuous load.

Reply to
electrician

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.

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He'll be ok and code compliant if he follows the advice at the site he referenced.

Ed

Reply to
ehsjr

My contention is the 80% is built into 15 and 20 a circuits by

240.4(D). The writers of the NEC know they have no control over the user and it is expected they may be loaded to the max. BTW where is the hard and fast "rule" that you can't plug in more than 80% of the circuit ampacity on a 20a receptacle circuit? 210.23(A) says (1) Cord-and-Plug-Connected Equipment. The rating of any one cord-and-plug-connected utilization equipment shall not exceed 80 percent of the branch-circuit ampere rating. (2) Utilization Equipment Fastened in Place. The total rating of utilization equipment fastened in place, other than luminaires (lighting fixtures), shall not exceed 50 percent of the branch-circuit ampere rating where lighting units, cord-and-plug-connected utilization equipment not fastened in place, or both, are also supplied.

"One" of his strings is less than 80% (or 50%) of 2400w You might even argue they are 600 "luminaires" which makes (2) moot and none are reallty "fastened in place". They are just draped on a tree.

Reply to
gfretwell

| 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.

Reply to
phil-news-nospam

On Tue, 21 Nov 2006 01:32:38 -0500 snipped-for-privacy@aol.com wrote: | On 20 Nov 2006 16:33:21 -0800, snipped-for-privacy@electrician2.com wrote: | |>The bottom line is long established practices of using the 80 per cent |>rule has kept buildings safe for many years. | | My contention is the 80% is built into 15 and 20 a circuits by | 240.4(D). The writers of the NEC know they have no control over the | user and it is expected they may be loaded to the max. | BTW where is the hard and fast "rule" that you can't plug in more than | 80% of the circuit ampacity on a 20a receptacle circuit? | 210.23(A) says

My bet is that the 15 and 20 amp circuits, the ones most common in homes, and the ones most subject to cheap contractor work, are more clearly specified because otherwise there will be more common errors or bad intentions in choosing what to actually use.

| (1) Cord-and-Plug-Connected Equipment. The rating of any one | cord-and-plug-connected utilization equipment shall not exceed 80 | percent of the branch-circuit ampere rating. | (2) Utilization Equipment Fastened in Place. The total rating of | utilization equipment fastened in place, other than luminaires | (lighting fixtures), shall not exceed 50 percent of the branch-circuit | ampere rating where lighting units, cord-and-plug-connected | utilization equipment not fastened in place, or both, are also | supplied. | | "One" of his strings is less than 80% (or 50%) of 2400w | You might even argue they are 600 "luminaires" which makes (2) moot | and none are reallty "fastened in place". They are just draped on a | tree.

"Utilization Equipment Fastened in Place" isn't easily moved somewhere else to get to a different circuit, and I guess they would frown on a practice of using more extension cords (I'm trying to plan my home wiring so I never need to have extension cords even for temporary utilization).

Reply to
phil-news-nospam

My understanding is that the 80% limits are because a fuse or thermal CB may trip if the load is over 80% and over 3 hrs. I don't think it generally has anything to do with safety of the rest of the circuit.

Matching this from the fount of all knowledge - Wikipedia: fuse

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"Overcurrent devices installed inside of enclosures are "derated" at least per the US NEC. This is a hold-over from the first mounting of electrical devices on the surface of slate slabs. The slate was the insulating material between devices mounted in air. So, rather than change the fuse rating, it became common to allow only 80% of the current value of the overcurrent device when the circuit is in operation for 3 hours or more (continuous loading)."

-------------- A few code cycles back there were several proposals to change the prohibition of plugging in loads over 80% - may have been 240.4-D or maybe 210.21-B-2 which limits plug-in loads to 12A on a 15A receptacle on a 20A circuit. One argument was that this is a limitation on how the system is used after it is wired and inspected. The code generally stays away from such limits (and this one is particularly impractical). As you said, the NEC "has no control over the user". (The CMP response was 'the code is the way we want it' thus explaining the logic behind the rule.) Another argument was that some UL listed devices, like hair dryers, with current draw of 12-15A can have 15A plugs - UL thinks plug-in devices at

100% are OK. (The CMP response was 'we are right, they are wrong'.) (Both arguments for change make sense to me.)

-- bud--

Reply to
Bud--

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.
Reply to
electrician

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.

Reply to
Brian

Right - he posted a link on 11/19 in a subsequent post:

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However, assuming this is so and

You're right on the voltage drop. I assumed the string was like the "drawing" below with 100 feet total wire and 50 feet total length:

AC --b---b---b---b---b---+ | AC ----b---b---b---b---b-+

Your voltage drop is based on 200 feet total wire, 100 feet total length. That's what is shown on the link he posted, so my assumption was wrong.

The original post

I gave the site in my post:

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Table 310.16 lists

Right. Daisy chaining would be a bad idea. The site he referenced indicates you can connect 100 bulb strings together *IF* you use a 3 outlet power adapter. Doing that insures the strings are powered individually, not daisy chained.

Ed

Reply to
ehsjr

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