Sizing a 3 Phase Breaker when multiple single phase loads are connected Line to Line

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Harold H

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Here is the problem.... at least it's my problem:
4 single phase heater loads are connected line to line on a 3 Phase 240VAC feed. They are arranged as follows:
4 amp load connected across phase A and B
5 amp load connected across phase B and C
11 amp load connected across phase C and A
13 amp load connected across phase B and A

What is maximum current in any leg so I can size the required 3 phase circuit breaker to protect the wiring for the entire group of loads?

I've searched the forums all over the web and cann't get a clear explanation. I get the 3 phase motor load calcs and the 1 phase calcs, but this line to line single phase current on a 3 phase system is still a head scratcher.

To get the currents in each leg, I originally thought I needed to add all the load currents for each phase (4+11+13 for phase A), then I got word to divide this by 1.73 to get the actual current in each leg. The more I think about this, it is not making sense to me, ergo the call for help. I'm not just looking for the "right answer" but the reason why?
 

Smart $

Esteemed Member
Location
Ohio
The easiest way to do this is to just add "future, but will never be installed loads" to balance out the circuit...

You have a 17A on A-B, so add 12A to B-C, and 6A to C-A. Now you have three line-to-line loads at 17A. Because currents add up at the junctions (where A-B load connects to the same conductor as the C-A load) you multiply 17A times 1.732 (the sqrt of 3) to get 29.4A on each line. This method, however, exagerrates the line current.

If you want to get a more accurate value, you have to calculate using vector math, or draw out the vectors. Using vector math reveals the current on line A will be 24.4A.



Welcome to the forum!
 

Smart $

Esteemed Member
Location
Ohio
BTW, why not arrange the loads differently...

4, 4, and 5 on AB
11 on BC
13 on CA

A lot less imbalance and you can wire for 13*sqrt(3) amps
 

Smart $

Esteemed Member
Location
Ohio
... I'm not just looking for the "right answer" but the reason why?
Guess I forgot the pertinent details on that part :roll:

The currents add up at the junctions, but they only add up arithmetically at each instant in time. Because the currents are out of phase with each other, the instantaneous sum varies at a phase different than the contributing currents' phases. For resistive loads connected to voltages at 0? (reference voltage phase angle) and 120? out of phase, the current that adds to the zero reference phase is actuall the 120? current inverted, which is of the same magnitude but the opposing direction, namely -60? in this case. If both contributing currents were of the same magnitude, they'd sum to the square root of three times either's magnitude, while the phase angle would be halfway between 0? and -60? at -30?. If you change either the magnitude or angle of either contributing current, not only does the resulting magnitude change, so does the phase angle.

Imagine a graph of two sinusoidal waveforms only 60? out of phase rather than 120? like you see in 3? educational pub's. If you align a straightedge with the Y axis (i.e. vertical axis) and slide it in the positive X axis direction, the sum of instantaneous values where the straightedge crosses the waveforms would create a third sinusoidal waveform, with its peak value where the two contributing waveforms cross above the X axis (The negative minimum would be the symmetrical opposite).
 

Dennis Alwon

Moderator
Staff member
Location
Chapel Hill, NC
Occupation
Retired Electrical Contractor
Guess I forgot the pertinent details on that part :roll:

The currents add up at the junctions, but they only add up arithmetically at each instant in time. Because the currents are out of phase with each other, the instantaneous sum varies at a phase different than the contributing currents' phases. For resistive loads connected to voltages at 0? (reference voltage phase angle) and 120? out of phase, the current that adds to the zero reference phase is actuall the 120? current inverted, which is of the same magnitude but the opposing direction, namely -60? in this case. If both contributing currents were of the same magnitude, they'd sum to the square root of three times either's magnitude, while the phase angle would be halfway between 0? and -60? at -30?. If you change either the magnitude or angle of either contributing current, not only does the resulting magnitude change, so does the phase angle.

Imagine a graph of two sinusoidal waveforms only 60? out of phase rather than 120? like you see in 3? educational pub's. If you align a straightedge with the Y axis (i.e. vertical axis) and slide it in the positive X axis direction, the sum of instantaneous values where the straightedge crosses the waveforms would create a third sinusoidal waveform, with its peak value where the two contributing waveforms cross above the X axis (The negative minimum would be the symmetrical opposite).

Sometimes I am quite certain I speak a different language then some of you. :grin:
 

Smart $

Esteemed Member
Location
Ohio
Sometimes I am quite certain I speak a different language then some of you. :grin:
Graphics transcend all language barriers???

Vectoradditionvssubtraction.gif


Vectoradditionvssubtraction2.gif
 

Smart $

Esteemed Member
Location
Ohio
that cleared it up for me ;)
I take your statement as made tongue-in-cheek :confused:

Perhaps some additional info on the graphics might help...

The vector illustration uses rms current magnitudes. We append direction (phase angle) to determine how one current interacts with another at junctions. Otherwise there is no need to know the phase angle. For instance, if 1A-rms passes through a load, we have no need to know the phase angle of the current. Yet if we have 1A-rms flowing through each of two loads with a common connection junction, we need to know the phase angle to determine how those currents interact with each other and a third conductor at the junction (Kirchoff's Current Law).

Representing the rms current with a phase angle permits us to calculate the interaction with vector math (or graphics, as depicted above in the first illustration). It is essentially a shortcut method to determining the current values at each instant and arithmetically adding (or subtracting; as depicted above in the second illustration), then converting the result back to rms value with a phase angle (not shown).

In the second illustration, the magenta lines simply represent two instances in time. Load currents at each instant sum to equal line current at each instant. Plot the sum for every instance and you get a third sinusoidal waveform representing the line current.

Is it clearer than mud now? :roll:
 
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Smart $

Esteemed Member
Location
Ohio
so...

If CA=3.45
AB=1

line current A=4.04

Well I'll be. Put that on my simple cad program, drew a few lines and danged if I didn't get the answer.

I am an exspurt now.

If I know the line current of A is 29.5, and the load on AB is 22.65, then the load of AC would be 10.71. yes?

Just using my cad program, no spreadsheet yet.

Above @ 120 deg angle
Yes, you are now an exspurt :cool:
 
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