# Power for rig

Discussion in 'Question of the Day' started by derekleffew, Jan 21, 2009.

1. ### derekleffewResident CurmudgeonSenior TeamPremium Member

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It's been too long between questions.

I have a rig I'm taking into a ballroom at East Spider'sBreath Resort. The plot contains:
13- S4-26° 575s, 3 with Iris
9- S4-36° 575s, 6 with gobos
2- S4-50° 575s, for backdrop
5- 1kPAR64-WFLs, short nose
7- 1kPAR64-MFLs, on floor bases
21- 6"Fresnels 750s, 11 with Barn Doors, 10 with Top Hats

I'm using a 48x2.4 Touring Rack. Assuming I balance the legs correctly, how many amps per leg do I need, (at 120/208VAC, 3Ø, Wye-Connected, 4wire+ground) ?

As always, those who KNOW the answer are encouraged to wait until others have had a chance to respond. You must show your work to receive full credit.

2. ### photoatdvActive Member

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So are the lights 120v or 208v?

3. ### derekleffewResident CurmudgeonSenior TeamPremium Member

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All lamps can be assumed to be 120V.

4. ### joeboo46Member

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Is this a test???

5. ### lieperjpWell-Known Member

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Yes, it is a test. Derek likes to do this to us sometimes. It's a good exercise. I just try to follow along since I don't know much about electricity.

6. ### photoatdvActive Member

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Taking a stab here- find total watts, divide by 120, then divide by 3 legs (however this is why I don't touch 3-phase without somebody who knows what they are doing standing over my shoulder)

(575*24+12*1000+21*750)/120/3=
13800+21000+15750=50550/120=421.25/3=140.4

So I'd say 150-200A per leg equally distributed

7. ### dhornMember

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photoatdv's reply has the correct formula but has made a small mistake in the 12 PARs at 1000 watts each; should be +12000+ NOT +21000+

The rack will need at least 116 amps per leg.

8. ### David AshtonActive Member

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Could we ask the lighting designer what the maximum number of lights up at any one time is planned to be?

9. ### icewolf08CBModCB Mods

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This is a particularly bad idea, LDs are notorious for changing their minds, not to mention the fact that the director might want more light. You can't plan on what the LD thinks he will need at one time.

Here is how I would do it:
24 @ 575W
- 8 per phase
- 4600W @ 120V per phase = 38.333A

12 @ 1kW
- 4 per phase
- 4000W @ 120V per phase = 33.333A

21 @ 750W
- 7 per phase
- 5250W @ 120V per phase = 43.75A

With this division you would have 115.416A on each phase if you run everything at full.

Here is a total calculation:
Total: 41550W @ 120V = 346.25A
-This still works out to 115.416666A per phase

Consider that the rack can support up to 115200W @ 120V which is 960A. So we are only utilizing 36.068% of the rack's capacity. On 400A service you would have 400A per phase which would give you a total of 1200A. So even if you loaded the rack to full capacity you are only at 80% capacity of the feed.

Since in the US it is pretty common to find 400A disconnects to tie into, this load shouldn't be a problem.

10. ### JeroenMember

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I'm not used to 120/208v more used to European 240/400v (old 220/380),
but shouldn't it be:

I(per phase) = P / [ U * V3 ]
I = 41550 \ [208 * 1,73 ] = A
I per phase = 115.46 A

Since it's 'star' and 208 between phases, although almost the same as calculating per phase for 120v

11. ### gordonmcleodActive Member

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Also remember that per the National Electrical Code your load must not exceed 80% of the fuse rating that protects it
Also some inspectors may insist that the feeders to a dimmer rack be rated not for the load connected to the dimmers but what the load would be if the dimmer systme was run at its rating
This being more of a problem on a permanent system than a temporary one but somem inspectors have interpretations of the code book that would drive one to drink

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Err. We're talking about one of Derek's exam questions here... so whatever makes for the most complicated maths.

I'd suggest we assume all on for calculating per phase loading and parts partially powered as relevant to create maximum harmonics and minimum cancellation when assessing neutral current...

13. ### STEVETERRYWell-Known Member

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OK, now for extra credit:

For 115.46A per leg (I am not confirming that this is the right answer to the original question since this is still in progress, but use it for my extra credit questions), what would be the minimum breaker or fuse size at the head of the feeder? While we're at it, what would be the minimum single-conductor sizes, temperature ratings, and allowable wire types of each of the conductors on this 4-wire plus ground portable feeder cable if it were 50' in length?

And what would the minimum conductor sizes be if we only had a 400-amp disconnect in the facility and our feeder length was only 15 feet? How about if the length was only 10 feet?

Hint: you will need the NEC, specifically Article 520 and Article 400. Assume a US installation.

Superior ETC swag goes to the first correct and complete answer supported by correct discussion.

ST

Last edited: Jan 26, 2009
14. ### derekleffewResident CurmudgeonSenior TeamPremium Member

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NEC Articles referenced above may be read at no cost; see Free Copy of the 2008 National Electrical Code (NFPA 70) onOne Project Closer for information and links.

Since [user]STEVETERRY[/user] has made the Extra Credit so difficult (though no less important!), another hint:

This document: BSR E1.18-1 - 200x, Standard for the selection, installation, and use of single-conductor portable power feeder cable systems for use at 600 volts nominal or less for the distribution of electrical energy in the entertainment and live-event industries, although in public review until 02/24/09, and thus not an official standard, may contain pertinent information. PDF available for download here.

Furthermore, since "Superior ETC swag" (cash value 1/10th of \$0.01) is now involved, both the Original Question and the Extra Credit portion are now officially open to ALL CB members, except: employees/representatives/dealers of ETC, and members of the CB Senior Team. As stated above, swag offer applies only to the Extra Credit portion.

15. ### rosebudldMember

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Watts Volts Amps Qty Total A
575 120 4.79 13 62.29
575 120 4.79 9 43.13
575 120 4.79 2 9.58
1000 120 8.33 5 41.67
1000 120 8.33 7 58.33
750 120 6.25 21 131.25
346.25
legs 3.00
Amps per leg 115.42
30% Overage 34.63
Safe Phase 150.04
Reality 400A

16. ### SteveBWell-Known Member

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Lets see. I have 3 polo shirts, 2 old and one new. An ETC laptop bag. An ETC Tee-shirt or 2, An ETC hat.....

" what would be the minimum breaker or fuse size at the head of the feeder?

150 amps, yields 120 amps when factoring in 20% de-rating of breaker/fuse.

"While we're at it, what would be the minimum single-conductor sizes, temperature ratings, and allowable wire types of each of the conductors on this 4-wire plus ground portable feeder cable if it were 50' in length?

Umm.. those biggie thick black wires, with single pole connectors that were originally blue on them but somebody's put green electrical tape over it, or Red tape over the white connector ?. I had to scrounge up my turn-arounds

"And what would the minimum conductor sizes be if we only had a 400-amp disconnect in the facility and our feeder length was only 15 feet? How about if the length was only 10 feet?"

Whatever the friggin lighting comany had me drag out of the BOTTOM of the road case full of the 100 and 50ft feeders that I don't need ?.

That's what MY Sat. night was all about !, so theory took a back seat.

Grin

SB

17. ### mnfreelancerActive Member

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Per NFPA 70 520.53 (H)(3) A feeder ten feet or less can be reduced in size if it doesn't penetrate walls or floors. Conductors can be rated for 1/4th of the current limiting device on the SUPPLY side (IE the 400 amp company switch in this example) 1/4th of 400 is 100, so according to table 400.5(B) a 6 AWG SC type cable with a 90 degree C temperature rating would be able to carry 105 amps which is the nearest conductor greater than or equal to 100. However, the question specifies 115.46 amps are needed so 4 AWG cable would be the minimum size, at 140 amps with a relatively narrow safety margain. It would be safer and more appropriate to assume a 20% safety margain, giving us 144.33 amps, therefore necessitating a 3 AWG conductor able to handle 165 Amps.

On a very similar note NFPA 70 520.53 (H)(4) specifies that a conductor not passing through wall/floor like above and less than 20 feet in length can be reduced in size, specifically to 1/2 of the ampacity of the supply protection. 1/2 of 400 is 200, and a 1 AWG conductor of type SC at 90 degrees C can pass 220 amps. 115.46 amps with a 20% safety margain is once again 144.33 amps which is well within the limits of a 1 AWG conductor.

Still researching the other "extra credit" problem...

EDIT : Per NFP 70 520.53 (O)(2), the neutral feeder should have 130% of the ampacity of the non-grounded conductors. The scenario involving the 10 foot feeders would require (165 amps x 1.30)=214.5 amps, necessitating a 1 AWG conductor for the neutral. The other involving the 20 foot feeders would require 2/0 AWG cable for the neutral.

Last edited: Jan 27, 2009
18. ### STEVETERRYWell-Known Member

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You're well on your way to telling me what size swag you want! Good job so far!

ST

ST

19. ### mnfreelancerActive Member

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Extra credit attempt part II…or maybe it's part one, since I kind of went out of order...

In a purely logical sense the minimum circuit protection rating at the head of the feeder is 115.46 amps, however this is far from a nice, even number typical of fuse/breaker ratings (at least at currents this high) AND unpractical and poorly thought out. The size of the protection at the head of the feeder is determined by the current requirements of the “utilization equipment” and more importantly the ampacity of the feeder used. It is always ok to use circuit protection rated for less current than the conductor ratings however not the other way around.

As I established in my previous post, if you let the 115.46 A represent an 80% load, you would want to plan for 144.33 amps at the hypothetical 100% load. Since the feeder is 50 feet in length the exceptions mentioned in my other post are not applicable, thus 520.53(H)(2) applies, imposing a minimum current-carrying conductor size of 2 AWG. Table 400.5(B) lists 2 AWG single conductor type SC 90 degree C rated cable as having an ampacity of 190 amps. All of that being said, I would guess that the most appropriate short-circuit protection rating for the head of this feeder would be 150 amps per leg. This rating is well inside the conductor’s rating and allows enough head-room and safety margin to be safe and practical.

As I mentioned in my other post, 520.53(O)(2) requires that the neutral have 130% the ampacity of the ungrounded conductors –
(190 amps x 1.30)=247 amps.
90 degree C rated 1/0 AWG cable handles up to 260 A per 400.5(B).
As for the size of the equipment grounding conductor, in the real world it would likely be the same as the supply conductors used in the rig as specified above, however technically it is unusual that it needs to be as large. 250.122(A) specifies use of table 250.122 for determining grounding conductor size. At 200 amps short-circuit protection a 6 AWG COPPER grounding conductor is required. This is also the minimum size allowable for a grounding conductor in this situation as per 520.53(H)(2).

As for the type of cable, NFPA 70 520.53(H)(1) requires use of “extra-hard usage cords or cables”. NFPA 70 table 400.4 lists a number of cables use extra-hard usage including PPE, SC, SCE, SCT, SE, SEW, SEO and W among a few others unlikely to be used in our field. The above selected are reflected in the draft document Derek linked to (BSR E1.18-1). That document also specifies that all feeder used in our field is to have a 90 degree C temp rating. This seems like common sense as it allows more amperage on smaller (read: lighter) conductors.

As a shameless attempt to score more points on this question, for a 50 foot long set of feeders the maximum number of interconnections allowable on these feeders is three (per run). This means 25’ long feeder cables are the shortest you’ll be lugging out of the bottom of the road case. This is specified in 520.53(J).

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