1.73...

[McCready00]

Well.. I was just wondering why we don't use it when only 1 fixture is connected to the circuit..

1 Fixture of 750 watts..
750 / 208 = 3.6
phase 1 - 3.6
phase 2 - 3.6
phase 3 - 0

2 fixtures of 750 watts..
(750 x 2) / 208
phase 1 - 6.2
phase 2 - 3.6
phase 3 - 3.6

are my calculs right? why am I only using 1.73 when I have more than 1 fixture on the same phase?

 

[epimetheus]

I'm not exactly following where you are using the 1.73 factor (which is actually the square root of 3). The root 3 factor is the difference between phase-to-phase voltage and phase-to-neutral voltage in a 3 phase power system. For example, in a 3 phase, 4 wire, 208Y/120V system, your phase-to-phase voltage is 208V and your phase-to-neutral is 208/sqrt(3) = 120V. In the example you show above, each fixture draws 3.6A @ 208V. This is assuming your fixtures are connected phase-to-phase. Your amperages are correct if the first fixture is connected phase 1-2, and the second phase 3-1.

 

[McCready00]

thanks for replying..

Well I knew about the square thing.. but my question was more in the direction of adding the current on the same phase..

Let me give you an exemple..

Let's say I follow the second calculation I've given on the original thread.

On the first phase... It gave me 6.2.. which was 3.6 x 1.73..
Why not simply adding another 3.6 and then getting 7.2 amps on the first phase, and 3.6 and the second and third phases..

Is that clear enough?

 

[epimetheus]

My bad, brain fart on my part. You've got phase-to-phase currents and you want to convert those to phase-to-netural currents. This always trips me up when I'm trying to do it on the fly. I just had to do this for a project at work recently, let me see if I can find my notes...

Short answer - the phase-to-netural currents don't directly add because of the phase angle. I don't have time ATM to go into more detail. Maybe somebody else can, if not I'll pick it up tonight.

 

[STEVETERRY]

My bad, brain fart on my part. You've got phase-to-phase currents and you want to convert those to phase-to-netural currents. This always trips me up when I'm trying to do it on the fly. I just had to do this for a project at work recently, let me see if I can find my notes...

Err...I don't think that is what is being asked.

When using 208V loads that are delta (phase to phase) connected, each of the three lines (L1, L2, L3) has current made up of two current components:

L1 A-B, C-A
L2 B-A, B-C
L3 C-B, C-A

Therefore, for an equal distribution of load on the three phases, the line current for L1, L2, L3 is 1.73 x the phase current.

In this case, the neutral current is zero, since no loads are Y-connected.

ST

 

[epimetheus]

Err...I don't think that is what is being asked.

When using 208V loads that are delta (phase to phase) connected, each of the three lines (L1, L2, L3) has current made up of two current components:

L1 A-B, C-A
L2 B-A, B-C
L3 C-B, C-A

Therefore, for an equal distribution of load on the three phases, the line current for L1, L2, L3 is 1.73 x the phase current.

In this case, the neutral current is zero, since no loads are Y-connected.

ST

My read of the OP was that he was asking about unbalanced loads, i.e. 2 phase-to-phase fixtures on a 3 phase system...

 

[McCready00]

thanks for the informations..

well.. this is not exactly what I wanted to know..

Let me try to be more specific about my needs.

Let's say i got a 750 watts fixture on a 208volt circuit, on phases X and Y.. nothing else is connected to the circuit.. Here is what I get :

X = 3.6amps
Y = 3.6amps
Z = 0 amp

Now, let's say i gotta add another fixture of the same type.. Although I put this last on phases Z and X. I get ;

X = 6.2amps
Y = 3.6amps
Z = 3.6amps


Well..... One more exemple that might help you to understand my question;

1 fixture = 750watts / 208v = 3.6 (amps through 2 phases)
3 fixtures = (((2250watts / 208v) x 1.73 ) = 18.71amps (total on 3 phases)

So my question is why am I not using the 1.73 in the "1 fixture" formula?

 

[FMEng]

So my question is why am I not using the 1.73 in the "1 fixture" formula?

Because the multiplier of 1.73 only works when all three phases are carrying the same amount of current. In your question, the currents are not balanced, therefore you need to compute the power by using the voltage to ground of each phase, times its current, then add them together.

For balanced load three-phase:
P = E x I x 1.73, where E is volts phase to phase, and I is equal in all phases.
This is basically a short cut.

For un-balanced loads:
Ptotal = P1 + P2 + P3, where P1 is the load of phase one, etc.
and P1= E1 x I1, where E1 is volts phase to ground, and I1 is the current in phase one.

Of course, most stage lighting loads are connected line to neutral for 120 V, so your example is theoretical.

 

[McCready00]

ok then am I good if I say:

I got 4 750w fixtures working under 208v
+
2 Leko 575

--------

3 x 750 = 2250
(((2250 / 208) x 1.73) / 3) = amps per legs
+
750 / 208 = amps on 2 used legs
+
(575 x 2) / 120 = amps for two leko on two single legs..

thanks

 

[sk8rsdad]

Let's see if I can simplify this a bit. In your first case, a single load on a single phase, the math reduces to the single phase equation. That is, the current on the neutral is equal to the current on the single leg that is feeding the load.

 

[McCready00]

thanks.. although I already know that..

I don't know how to clearly ask the question. I actually trying to understand when you got to add the 1.73 to the formula.

I understand where it comes from in the simewave and everything.. But someone said it only has to be used when the phases are well balanced. Since, when you get a number of fixtures (taking the same power) that you can't devide by 3, your phases can't be balanced.

Here is where my question appears.. If you get 4 fixtures at 750watts each, working on three phases circuits, could I use this next formula;

3 x 750 = 2250w
(((2250w / 208) x 1.73) / 3) = 6.24 amps on each phases

+

750 / 208 = 3.6 amps on phases X and Y

then I would get on each phases ;

X ; 9.85 amps
Y ; 9.85 amps
Z ; 6,24 amps


right?