Calculating Power Consumption

Hello,
I was wondering if I was right in my calculation.

1.

I will use a simple example to make it more simple.

Let's say I have 9 moving lights at 1000watts each on a 120/208v system. If all the units were well balanced on the distro and work on 208v. I am not taking the Power Factor in condideration here.

1000w x 9 = 9000w
9000w / 208v = 43.26a
43.26a x 1.73 = 74.84a
74.84a / 3 = 25a

i would need 25a on each legs of my power system.

If well balanced, my neutral would technically show 0 amp ( technically ) .. If not well balanced, hot am I calculating the power coming back on the neutral ?


2.

On the same kind of power system, I have 4 Lekos 750w working under 120v

As I couldn't balance the system perfectly, I would have 1 leg presenting 12.5amp and the 2 others at 6.25amp.

How can I calculate the power that would go through my neutral cable?





Thanks

McCready00 said:

How can I calculate the power that would go through my neutral cable?

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For what purpose? I believe the neutral always needs to be sized to the full capacity of the other current carrying conductors (sometimes the neutral needs to be larger due to triplen hamonics). If the 208v instruments are not connected to the neutral, they cannot put any current on it. For the equipment you listed, I believe the most balanced would be 3 of your moving lights on each 208v phase pair and one elipsoidal on each of two legs and two elipsoidals on the third leg. The current actually going through your neutral should be a function of the current going through your elipsoidals.

Keep in mind the inrush from the movers depending on light source could trip the breaker if you go that tight.

Funny but I calculated it at 9000 watts or 3000 watts a phase and divided by 120v = 25 amps - but I'd be more likely to recommend three each movers on three separate 20 amp circuits. But then I just play electrician.

Why does this feel like a homework assignment...

Current flowing on the neutral can be approximated using this formula. It's only an approximation since it assumes steady-state sinusoidal loads with identical power factor on each phase. Theatre lighting is way messier.



A more accurate approximation can be derived by knowing the current and phase angle of each leg, with the neutral current derived by summing the currents on the other 3 legs.

To know the power on the neutral you would need to know the nominal impedance of the wire. However, most power is dissipated in the load and not in the wire connecting the load to the source.

Why are you not calculating using the power factor? This is all very much pointless without that consideration. Second, on your first example (or homework assignment...) if all your movers are @ 208v, you won't see anything on your neutral because nothing is connected to the neutral.

Footer said:

Why are you not calculating using the power factor? This is all very much pointless without that consideration.

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Simply because it was more easy to give the example.



Thanks

sk8rsdad said:

View attachment 11751

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Ok, I totally understand the part why it cannot be exactly right.

but basicly, would I be wrong to say that a well balanced circuit containing 3 lekos 750w ( 1 leko on each legs ) would send "nothing" back thru the neutral cable?

McCready00 said:

Would I be wrong to say that a well balanced circuit containing 3 lekos 750w ( 1 leko on each legs ) would send "nothing" back thru the neutral cable?

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If all three were running at full, then the current on the neutral should be very close to nothing. This is why a loose neutral (a dangerous situation) might not be noticed right away - when the load is balenced the neutral is not active.

robartsd said:

If all three were running at full, then the current on the neutral should be very close to nothing. This is why a loose neutral (a dangerous situation) might not be noticed right away - when the load is balenced the neutral is not active.

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great thanks... it anwers all my questions..

I didn't see the derating in your calculations. For example, on the movers drawing 25 amps: That load is a continuous load so your supply line OCPD would have to produce 25 amps
continuous. The derate is 20%, so even a 30 amp breaker falls a little short (24 amps) but would probably work. Technically, it would have to be a 35 amp circuit, which would be good to 28 amps.

JD said:

I didn't see the derating in your calculations. For example, on the movers drawing 25 amps: That load is a continuous load so your supply line OCPD would have to produce 25 amps
continuous. The derate is 20%, so even a 30 amp breaker falls a little short (24 amps) but would probably work. Technically, it would have to be a 35 amp circuit, which would be good to 28 amps.

Click to expand...

Maybe they never do shows longer than 2 or 2 1/2 hours, so not continuous per NEC. I rarely consider a cord and plug connected stage and studio fixture a continuous load. That said, I'd still look at load being near half the circuit ocpd.