Control/Dimming Calculating power requirements on 3 phase systems

[hobbsies]

Let me preface by saying I'm studying for the ETCP Lx exam and I made up this question.

I want to calculate how much power I will use/need for this system.

Say I have this inventory in my rig:
24 Mac 2ks at 1200w each requiring 208v
48 575w S4 LEKOs on 120v
200A 3phase service
24 way 208 distro
48 2.4k dimmer rack

Assuming power factor = 1

The 24way distro will use 1200w/208v/1.732= 3.33A*8 = 26.64A/phase.
The dimmer rack will use 575w/120v*16=76.67A/phase

Do I need to be factoring in the phase offset for the dimmer rack too? Or is it safe to to say that I'll be using 103.31A/leg of the service?

EDIT:
Found the answer to this question. This seems to be the correct answer. HNG fixtures are just calculated using Ohm's law, like I did.

 

[DanH]

Double-check your math on the 208V line there...should be more like 80A/leg (1200*24)/120/3....

 

[JD]

The only wildcard is that on the pure dimming load, your neutral can be conducting more than your hots due to Triplen Harmonics at some dimmer settings. Since the neutral is only being used for the dimmers, and your (dimmer) loading on the hot legs is 76 amps, and your service and feeders are rated for 200 amps, this would not be a factor.
The only time you can consider a dimmer to be operating at a power factor of 1 is when all of your dimmers are at 100%. (or 0%)

 

[derekleffew]

The 24way distro will use 1200w/208v/1.732= 3.33A*8 = 26.64A/phase.

Arc sources don't work that way. https://www.controlbooth.com/threads/208v-power-load-balancing.35498/#post-308063

 

[STEVETERRY]

Let me preface by saying I'm studying for the ETCP Lx exam and I made up this question.

I want to calculate how much power I will use/need for this system.

Say I have this inventory in my rig:
24 Mac 2ks at 1200w each requiring 208v
48 575w S4 LEKOs on 120v
200A 3phase service
24 way 208 distro
48 2.4k dimmer rack

Assuming power factor = 1

The 24way distro will use 1200w/208v/1.732= 3.33A*8 = 26.64A/phase.
The dimmer rack will use 575w/120v*16=76.67A/phase

Do I need to be factoring in the phase offset for the dimmer rack too? Or is it safe to to say that I'll be using 103.31A/leg of the service?

EDIT:
Found the answer to this question. This seems to be the correct answer. HNG fixtures are just calculated using Ohm's law, like I did.

Your math is incorrect. It should be:

(1200/208) * 1.732=9.976A per Mac 2K, or 240A/3 or 80A each for L1, L2, and L3

plus

575/120=4.8A * 48/3 = 77A per phase

80A +77A = 157A per phase

157/.8= 196A per phase with 0.8 power factor, which is more realistic.

Don't forget that thermal/magnetic main breakers are rated for only 80% continuous loading, so a 200A service may not be large enough.

ST

 

[hobbsies]

Your math is incorrect. It should be:

(1200/208) * 1.732=9.976A per Mac 2K, or 240A/3 or 80A each for L1, L2, and L3

plus

575/120=4.8A * 48/3 = 77A per phase

80A +77A = 157A per phase

157/.8= 196A per phase with 0.8 power factor, which is more realistic.

Don't forget that thermal/magnetic main breakers are rated for only 80% continuous loading, so a 200A service may not be large enough.

ST

According to Cadena's Electricity for the Entertainment Electrician & Technician, the formula for calculating current on a 3 phase system is:

I3-Θ = W/V/PF/1.732

Since no parenthesis are included in that equation, what is the correct way to process that equation? I'm assuming PF = 1 for both simplicity and because PF is not offered in the practice exam question which I based this question off of.

 

[STEVETERRY]

According to Cadena's Electricity for the Entertainment Electrician & Technician, the formula for calculating current on a 3 phase system is:

I3-Θ = W/V/PF/1.732

Since no parenthesis are included in that equation, what is the correct way to process that equation? I'm assuming PF = 1 for both simplicity and because PF is not offered in the practice exam question which I based this question off of.

I do not agree with this formula. It should be (total W/V) / PF *1.732 / 3 to arrive at each of the L1, L2, and L3 currents. An easy way to remember this is that three-phase line currents are comprised of adjacent phase currents i.e. those of A-B, B-C, C-A. Therefore, line currents must be greater than phase currents by a 1.732 multiplier, which is why 1.732 is a multiplier rather than a divisor.

ST

 

[danTt]

I do not agree with this formula. It should be (total W/V) / PF *1.732 / 3 to arrive at each of the L1, L2, and L3 currents. An easy way to remember this is that three-phase line currents are comprised of adjacent phase currents i.e. those of A-B, B-C, C-A. Therefore, line currents must be greater than phase currents by a 1.732 multiplier, which is why 1.732 is a multiplier rather than a divisor.

ST

Just to push a bit further, if you had 10 1200w units on XY, and 5 units on YZ (and for the sake of discussion, none on XZ), would the calculation look like (watts per leg/V)/PF*1.732 ? : X: (10*1200/120)/PF*1.732 Y: (15*1200/120)/PF*1.732 Z: (5*1200/120)/PF*1.732? Or does it get messier?

 

[derekleffew]

@danTt , punch your numbers into the formulae at https://www.controlbooth.com/threads/unbalanced-3-phase-equation.40130/#post-347005 and let us know your results.

 

[hobbsies]

@danTt , punch your numbers into the formulae at https://www.controlbooth.com/threads/unbalanced-3-phase-equation.40130/#post-347005 and let us know your results.

The end of that thread is hilarious

Wow. this thread has really gone FAR to the technical side. Let me give you a real world example of how to apply the above in an actually entertainment job setting. Also I have included some shorthand for doing all the complex math.

"Hey, Is a 65kw geni big enough for that street fair?"

"hmm. probably, but I think it will be close"

"Okay, I'll just order a 100kw then."

Using @STEVETERRY 's formula for @danTt 's equation I get
x = 86.6 (1200/2*10)/120*1.732
y = 129.9 (1200/2*15)/120*1.732
z = 43.3 (1200/2*5)/120*1732

^^Should I be doing the above, or 1200*(#of units on leg)/208*1.732?

Using @YesItWillWork 's formula for the above I get:
x = 99.92
y = 132.19
z = 49.96

Note, in order to calculate xy, yz, xz, I had to use Steve's formula (1200*#units)/208*1.732

 

[danTt]

@YesItWillWork s formula assumed that we knew what the line->line current was, however, all of my reading makes it sound like thats much more complicated to figure out when things are not balanced, so I'm not sure if it's as simple as what I asked. I think @MikeJ 's first post in that thread Unbalanced 3-phase equation? is probably the closest to accurate without getting into wild and crazy math, but his last post is probably much more accurate for day to day usage. (Though I don't understand why you'd need to multiple by 1.732 when converting to amps, but not when calculating the wattage on each leg...

 

[STEVETERRY]

The end of that thread is hilarious


Using @STEVETERRY 's formula for @danTt 's equation I get
x = 86.6 (1200/2*10)/120*1.732
y = 129.9 (1200/2*15)/120*1.732
z = 43.3 (1200/2*5)/120*1732

^^Should I be doing the above, or 1200*(#of units on leg)/208*1.732?

Using @YesItWillWork 's formula for the above I get:
x = 99.92
y = 132.19
z = 49.96

Note, in order to calculate xy, yz, xz, I had to use Steve's formula (1200*#units)/208*1.732

First of all, my previously mentioned formula is only applicable to balanced three-phase loads, whether they are made up of three-phase devices or collections of single-phase devices connected line-to-line rather than line-to-neutral, as in the original 208V moving light example.

Second, accurate calculation of unbalanced three-phase load currents requires the use of vector analysis. Other than for the pure intellectual joy of the endeavor, IMHO this is somewhat a waste of time in portable entertainment systems.

Why? As my esteemed colleague Ken Vannice recently pointed out in a classic Duh! moment during a discussion very much like this one, portable power devices such as transformers, generators, circuit breakers, and feeder cables are not available in asymmetrical current ratings for each phase. That means that the system must be sized for the highest phase current.

Here's an blatantly obvious illustrative example: If I have a system with four single-phase 208V loads of 5kW each, one phase will need to be twice the load of the other two phases. Can I size the system for the lower-loaded phases? Do I even care about the power contribution of the lower loaded phases? No! I must size the entire portable system for the highest loaded phase multiplied by three, after making a best effort at phase balancing by moving single-phase loads around.

ST

 

[epimetheus]

Check this out. I created this for work to calculate station service loading for the substations I design. I've had it vetted by other engineers in the office, but let me know if you find any discrepancies. There are 3 tabs for 3 phase, high leg delta, and single phase. The system voltage can be changed in the upper right. The sheets are protected, but there is no password. The loads are entered as kVA. If you need to enter them as amps, you'll need to calculate the kVA based on voltage.

 

[hobbsies]

First of all, my previously mentioned formula is only applicable to balanced three-phase loads, whether they are made up of three-phase devices or collections of single-phase devices connected line-to-line rather than line-to-neutral, as in the original 208V moving light example.

Second, accurate calculation of unbalanced three-phase load currents requires the use of vector analysis. Other than for the pure intellectual joy of the endeavor, IMHO this is somewhat a waste of time in portable entertainment systems.

Why? As my esteemed colleague Ken Vannice recently pointed out in a classic Duh! moment during a discussion very much like this one, portable power devices such as transformers, generators, circuit breakers, and feeder cables are not available in asymmetrical current ratings for each phase. That means that the system must be sized for the highest phase current.

Here's an blatantly obvious illustrative example: If I have a system with four single-phase 208V loads of 5kW each, one phase will need to be twice the load of the other two phases. Can I size the system for the lower-loaded phases? Do I even care about the power contribution of the lower loaded phases? No! I must size the entire portable system for the highest loaded phase multiplied by three, after making a best effort at phase balancing by moving single-phase loads around.

ST

Yeah, this makes total sense. I wish I could post the original practice exam question I'm trying to understand how to solve, but I don't want to get in trouble. In the question, it says the venue provides two 200A 3Θ services, which is JUST enough power, and 100% dependent on how you solve for current on each leg for your movers. So I'm just trying to figure out how I would solve that on the exam and get the correct answer. Every person I've shown the question to has gotten the answer wrong and has thought that there wasn't enough power provided for all of the fixtures, and thus un-answerable...

 

[JD]

I think the question itself may be flawed. Look at the @STEVETERRY answer in post #5- 196 amps would be "just enough" IF and only if you did assume a PF of 1.
Coincidentally, the derating of the breaker would be .8 (80%) -OR- the Power Factor would be .8
If they truly ignored the power factor, then the load would be 157 amps, and the breaker would be 200 x .8 = or 160 amps.... just enough.
Conversely, if you calculated the power factor at .8 and ignored the breaker derate, then your draw would be 196 amps on a 200 amp breaker..... just enough.
HOWEVER both cannot be true, and yet in the real world they are. Therefore, it better be a real quick show! (before the lights go out.)
There is one other real world consideration- All of the math assumes that everything would be on at the same time. Although the movers would draw the same no matter what, a blacked out lamp gathers no amps..

 

[hobbsies]

There is one other real world consideration- All of the math assumes that everything would be on at the same time. Although the movers would draw the same no matter what, a blacked out lamp gathers no amps..

Well i guess that begs the question, is ETCP encouraging that type of thinking? I'd hope not, but this question on the practice exam seems to think it's OK.

 

[STEVETERRY]

Well i guess that begs the question, is ETCP encouraging that type of thinking? I'd hope not, but this question on the practice exam seems to think it's OK.

Often, in a standardized exam, a question can arise: "What are the assumptions in this test question?". Sometimes, the exam will purposely state assumptions that might be unlikely in the real world, but serve to clarify the question and do not detract from the key issue that the question is testing. "Assuming unity power factor....." is one of those. I doubt the ETCP program actually encourages the assumption of unity power factor. Thus, it pays to read and re-read test questions for clarity.


ST

 

[RickR]

When I took the test it quickly became apparent that the complexity of the wording is a stand in for 'real world' situations. Overall there was very little assuming. I don't recall anything on show duration as most of the power scenarios felt like film work.

I didn't feel any push other than knowing is a lot better than ignoring.