Power for rig2

[derekleffew]

120K PAR rig, plus 18 MAC2000Profiles.

1. How many amps per leg do I need, assuming a 120/208Y 3Ø service?
2. For feeder, do I need two grounded conductors?
3. For feeder, what is the minimum AWG of the grounding conductor?
4. This is for an outdoor festival. If I have to rent a generator, what size kVA do I need?

Standard disclaimers apply: students only, show your work, cite your references, spelling and grammar count, et cetera.

 

[TupeloTechie]

I'll take a stab at this, but I hardly know anything about this stuff.
(1)
120k of Pars = 120,000w
18 Mac2k @ 1200w = 21,600w

120,000w + 21,600w = 141,600w

141,600w / 120v = 1180I

1180I / 3 = 393.34I

So 400amps per leg should cut it right?

(2)
I have no idea, but because you mentioned it I would guess that yes you need 2 neutrals. I looked around here on the booth and saw a thread asking about this, after reading it I am still kind of confused but still think that you would need the 2 neutrals because your dealing with 120k in dimming.

(3)
Again something I have no clue, I thought 4/0 was the min for 400amp service, however I know the grounding conductor (or green) always seems to be of a smaller gauge. Tried looking around by can't find this, could someone give me a hint as to where this would be, electrical code?

(4)
I know nothing at all about generators, but I found this formula on the internet.
400I / (1000 / 208v / 1.73) = kVA

400I / 2.779 = 143.93 kVA

So I guess a 145kVA generator?

 

[n1ist]

Just my quick guesses:

120K PARs @ 120V = 40KA total / 3 phases = 333A per phase
18 MAC @ 208V 1.6KW = 28.8KW = 80A per phase
So the total is 413A

For single feeders, that looks like 350mcm. As for the ground, it's smaller than the feeder; I think #2 is OK here.

I'm guessing you would want to double or upsize the neutral due to all of the dimmer loads.

But than again, I have never worked with that size loads before.
/mike

 

[headcrab]

I agree with n1ist.
For EGC sizing look at table 250.122.
I will quote the NEC:
"Where single-conductor feeder cables, not installed in raceways, are used on multiphase circuits feeding portable switchboards containing solid-state phase-control dimmers, the neutral conductor shall have an ampacity of at least 130 percent of the ungrounded circuit conductors feeding the portable switchboard. (Article 520.63)"
Therefore the neutral must be at least 600kcmil, and the ground 2AWG.

 

[TupeloTechie]

crap.. mac 2k's use 208v well then disregard my entire post

 

[derekleffew]

MAC2000s (with electronic ballast) can run on any voltage between 100-250V, but in this case we'll be using 208V.

From the Owner's Manual, under Maximum Power and Current:
"208 V / 60 Hz: . . . . . . 1470 W, 10.3 A".

Feel free to give it another try.

 

[n1ist]

Stab 2:

120K PARs @ 120V = 40KA total / 3 phases = 333A per phase
18 MAC @ 208V 10.3A = (18 * 10.3 * 1.73)/3 = 105A per phase
So the total is 439A

 

[derekleffew]

The LD just called to tell me he "forgot" some things. The 120K is actually 575W S4 PARs. He's also added six ACL bars (GE 4552 lamps), six 8-lights (DWE lamps), and six VL2500Washes on the floor.

5) Now how much power do I need?
6) What size generator?
7) How many sets of 4/0 90°C Feeder with E1016 Cam-Lok connectors?

 

[Derrick]

Is the LD planning on lighting all this up at the same time? In the LD's lighting plot, what can be determined to be the max load during the show?

 

[derekleffew]

See this post: http://www.controlbooth.com/forums/question-day/10774-power-rig.html#post118761. One must base power needs on the assumption that the LD will run everything at Full.

 

[Footer]

Is the LD planning on lighting all this up at the same time? In the LD's lighting plot, what can be determined to be the max load during the show?

You don't want to be the dimmer tech resposible for trying to find which breaker blew, and what door it is behind that you don't have a key to, especially when there is 3,000 people screaming to hear the end of freebird.

 

[Derrick]

Dang it! Was hoping Derek was posing a trick question, still learning him though.

pours coffee, sharpens pencils...

Original set

120K par @ 120 volt = 1000 amp /3 phases = 334 amps per phase or 40,000 watts per phase. 120,000 watts total.

18 MAC2000 @ 208 = 186 amp / 3 phases = 62 amps per phase or 8820 watts per phase. 26,460 watts total

The "oops I forgot"

ACL bar, 4 lamps. 250 watt x 4 = 1000 watt @ 120 = 8.3 amps x 6 units = 50 amps / 3 phases = 17 amps per phase or 2000 watts per phase. 6000 watts total

8 lights. 650 watt x 8 = 5200 watt @ 120 = 44 amps x 6 units = 264 amps / 3 phases = 88 amps per phase or 10,400 watts per phase. 31,200 watts total

VL2500W (assuming 208) 1000 watt @ 208 = 5 amps x 6 units = 30 amps / 3phases = 10 amps per phase or 2000 watts per phase. 6000 watts total

I calculate the load to be around 210KW, mostly lighting, not really considering the motors in the movers so PF would be 100%. Couldn't find a PF rating for the ballast in the MAC200 or the VL2500W, so I am assuming 100%

With the load evenly distributed across the phases, each leg would need to be cabable of carrying 511 amps. TWO sets of 4/0 feeder or a larger feeder needed.

A 250Kva generator would be needed.

I'm not as familiar with the code as you folks are. But I do know or seem to remember that at some point you are required to have TWO neutrals if the feeder is such n such. That's the part I can't remember.

 

[n1ist]

But the ACL 4-bars are 28V with 4 in series, giving 250W@120V per bar. 2.1A * 6 or 4.2A per phase.

Also, isn't there a square root of 3 involved in those delta/Y 208V calculations?

/mike

 

[Derrick]

Thanks Mike, you are correct. Here is the revised er..emmm guess.

Original set

120K par @ 120 volt = 1000 amp /3 phases = 334 amps per phase or 40,000 watts per phase. 120,000 watts total.

***18 MAC2000 @ 208 = 186 amp / 3 phases = 62 amps per phase or 12,896 watts per phase. 38,688 watts total

The "oops I forgot"

***ACL strips, 4 lamps. 250 watt @ 120 = 2.1 amps x 6 units = 12.6 amps / 3 phases = 4.2 amps per phase or 500 watts per phase. 1500 watts total

8 lights. 650 watt x 8 = 5200 watt @ 120 = 44 amps x 6 units = 264 amps / 3 phases = 88 amps per phase or 10,400 watts per phase. 31,200 watts total

VL2500W (assuming 208) 1000 watt @ 208 = 5 amps x 6 units = 30 amps / 3phases = 10 amps per phase or 2000 watts per phase. 6000 watts total

208v total load is 44,688 watts or 216 amps
44,688 x 1.73 = 77,310 watts
216 x 1.73 = 374 amps

120v total load is 152,700 watts or 1273 amps


***I now, calculate the load to be around 230KW

With the load evenly distributed across the phases, each leg would need to be cabable of carrying 549 amps. TWO sets of 4/0 feeder or a larger feeder needed.

***A generator capable of 288Kva would be needed.

 

[derekleffew]

... The 120K is actually 575W S4 PARs. ...

So instead of 120x 1000W lamps, it's now 120x 575W lamps.

 

[Derrick]

I figured that the "120k" was a total par lighting system wattage, ie; 120,000 watts. 120 PAR @ 1000 watt or 208 S4 at 575 watt each = 119600 watts.

So, it is 120 S4 @575 watts each?

 

[derekleffew]

...So, it is 120 S4 @575 watts each?

Total of twenty bars of six 575W SourceFour PARs each. MFLs on the front truss and NSPs on the US truss.

 

[Derrick]

Revised for the S4

Original set

***120, 575 watt S4 par @ 120 volt = 575 amp /3 phases = 192 amps per phase or 23,000 watts per phase. 69,000 watts total.

18 MAC2000 @ 208 = 186 amp / 3 phases = 62 amps per phase or 12,896 watts per phase. 38,688 watts total

The "oops I forgot"

ACL strips, 4 lamps. 250 watt @ 120 = 2.1 amps x 6 units = 12.6 amps / 3 phases = 4.2 amps per phase or 500 watts per phase. 1500 watts total

8 lights. 650 watt x 8 = 5200 watt @ 120 = 44 amps x 6 units = 264 amps / 3 phases = 88 amps per phase or 10,400 watts per phase. 31,200 watts total

VL2500W (assuming 208) 1000 watt @ 208 = 5 amps x 6 units = 30 amps / 3phases = 10 amps per phase or 2000 watts per phase. 6000 watts total

208v total load is 44,688 watts or 216 amps
44,688 x 1.73 = 77,310 watts
216 x 1.73 = 374 amps

120v total load is 101,700 watts or 848 amps


***I now, calculate the load to be around 180KW

***With the load evenly distributed across the phases, each leg would need to be cabable of carrying 355 amps. ONE set of 4/0 feeder needed.

***A generator capable of 225Kva would be needed.

 

[icewolf08]

But the ACL 4-bars are 28V with 4 in series, giving 250W@120V per bar. 2.1A * 6 or 4.2A per phase.

Also, isn't there a square root of 3 involved in those delta/Y 208V calculations?

/mike

Thanks for playing, please don't blow yourself up!

Just a quick note, while ACLs use 28V lamps wired in series, that does not mean that your load does not increase. Each lamp operates at 250W (in fact they probably run slightly hotter since in theory they are getting 30V instead of 28V) for a total of 1000W. So you are in fact running 1000W at 120V with an ACL bar. Here is the math:

250W @ 30V = 8.333A
1000W @ 120V =8.333A​

Consider that when you wire ACLs in series so that you can run them on 120V power you are increasing the voltage and the wattage by 4. For that to happen the current draw of the system has to go down by a factor of 4.

IF you were running four ACLs in series on 28V then the wattage would remain constant. This is because with four even loads in series, each would do 1/4 of the total work. Each lamp would operate at 62.5W.

This is the same concept that applies to MR-16 Zip-Strips. A three circuit strip has five windows for each color and two lamps in each window for a total of ten lamps per circuit. Since they use 12V MR-16 lamps, the lamps are wired in series to achieve 120V operation. The standard lamp is a 75W 12V lamp for a total of 750W per circuit.

It is important to remember this when you are working with low voltage lamps because doing the math incorrectly can land you in a world of hurt. If you base your show off under-calculated loads you may not be able to achieve the effects you want. If you assume that an ACL bar only draws 250W and you put for of them on a dimmer it won't work. So once again let this be a disclaimer: electricity is dangerous and can kill. If you are not appropriately trained in proper usage and the appropriate maths involved in working with electricity please seek professional assistance before you start any major electrical projects!

 

[Derrick]

So, back to the original thinking on the ACL....

Original set

120, 575 watt S4 par @ 120 volt = 575 amp /3 phases = 192 amps per phase or 23,000 watts per phase. 69,000 watts total.

18 MAC2000 @ 208 = 186 amp / 3 phases = 62 amps per phase or 12,896 watts per phase. 38,688 watts total

The "oops I forgot"

ACL strips, 4 lamps. 250 watt x 4 = 1000 watt @ 120 = 8.3 amps x 6 units = 50 amps / 3 phases = 17 amps per phase or 2000 watts per phase. 6000 watts total

8 lights. 650 watt x 8 = 5200 watt @ 120 = 44 amps x 6 units = 264 amps / 3 phases = 88 amps per phase or 10,400 watts per phase. 31,200 watts total

VL2500W (assuming 208) 1000 watt @ 208 = 5 amps x 6 units = 30 amps / 3phases = 10 amps per phase or 2000 watts per phase. 6000 watts total

208v total load is 44,688 watts or 216 amps
44,688 x 1.73 = 77,310 watts
216 x 1.73 = 374 amps

120v total load is 106200 watts or 885 amps


***I now, calculate the load to be around 184KW

***With the load evenly distributed across the phases, each leg would need to be cabable of carrying 367 amps. ONE set of 4/0 feeder needed.

***A generator capable of 230Kva would be needed.

I appreciate the disclaimer, I enjoy the math, don't use it nearly as much as I should though.