120/208 3-ph, 4W panelboard amps on A phase?

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JasonCo

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Houston, Texas
On a 120/208 3-ph, 4W panelboard, the total current on the A phase feeder that supplies the panel is ___ A?

The panelboard looks like this
Circuit 1 = 2.1kVA
Circuit 3-5 = 7.9kVA
Circuit 2-4-6 = 41.3 kW
Circuit 10 = 3.2 KVA
Circuit 12-14 = 22 KW
Circuit 9-11-13 = 31.3 kVA

Knowing this, I did the calculations
A phase circuits will have 17.5 amps + 114.6 amps + 183.3 amps + 86.8 amps = 402.2 amps on A phase.

Answer choices are...
A. 234.2 amps
B. 266.2 amps
C. 283.2 amps
D. 299.2 amps
E. 345.3 amps
F. 323.8 amps

What am I doing wrong? I used Watts / (208 x 1.732) for the 3-pole breakers, and I used equation Watts / 120 for my 2-pole and 1-pole breakers.. No idea how I am messing this one up. Any help is greatly appreciated, been stuck on this for a while now. Thank you very much for your time, really.
 
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Ingenieur

Senior Member
Location
Earth
Try this
draw a panel schedule
assign the kva to each ckt no.
Add up the a phase alone divide by 120
add up the a phase 2 pole shared divide by 208
the 3p will be as you calculated
sum those
 
J

JasonCo

Senior Member
Location
Houston, Texas
ohhhhhhhh okay I messed up on the 2-pole breakers then... My equation would be watts / 208 for the 2-pole. I was doing watts / 120.. Duh! Thanks man :)
 
T

topgone

Senior Member
JasonCo said:
D? I got F as my answer. The total current in A phase would be 324.8 amps.
A phase loads would come out to 17.5 amps, 114.6 amps, 105.7 amps, 86.9 amps = 324.8 amps
Click to expand...
My bad. I was thinking about the average line amps for a total connected load of 107.8 kVA.
 
J

JasonCo

Senior Member
Location
Houston, Texas
topgone said:
My bad. I was thinking about the average line amps for a total connected load of 107.8 kVA.
Click to expand...

Its no problem! Yeah I had to take into consideration the 3-pole 2-pole and 1-pole calculations, which alters the average line amps for 107.8 kVA. I appreciate your help though with this, I'm not done with all the math I have to get done before next Tuesday, will probably see another thread by me soon! These Delta Wye calculations can get a bit confusing sometimes
 
Smart $

Smart $

Esteemed Member
Location
Ohio
JasonCo said:
D? I got F as my answer. The total current in A phase would be 324.8 amps.
A phase loads would come out to 17.5 amps, 114.6 amps, 105.7 amps, 86.9 amps = 324.8 amps
Click to expand...
This is another no correct choice Q&A.

You can't arithmetically add Line A amperes when you have a mix of 3Ø and 1Ø L-N with 1Ø L-L.
 
J

JasonCo

Senior Member
Location
Houston, Texas
Smart $ said:
This is another no correct choice Q&A.

You can't arithmetically add Line A amperes when you have a mix of 3Ø and 1Ø L-N with 1Ø L-L.
Click to expand...

In a way if you are trying to find Line to Neutral, then forget the 3-pole and 2-pole amps because they all cancel each other out because they are all the same values so amps in the neutral would = 0. Which leaves you with a 1-pole breaker where the only path the amps can take is back through the neutral, there is no other amp current for it to cancel out in. So Line to Neutral would be 5.3KVA / 120 = 44.2amps on the neutral in this panel. Think I'm correct with this way of thinking hopefully..

But for adding amps on A phase, I thought you just add them together. I guess for test purposes that is what you would do, but you are saying this is incorrect in the real world, wouldn't work?
 
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GoldDigger

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Just adding the magnitudes is conservative in that the magnitude of the vector sum will always be less than or equal to the sum of the magnitudes. But a conservative upper limit is not necessarily the "right" answer.
 
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Ingenieur

Senior Member
Location
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Aren' all loads on ph a 'in phase' so the vector magnitude = scalar magnitude?
only real components once phase shift is offset/ignored

the issue I see is mixing P and S without a pf to put everything in S
 
Smart $

Smart $

Esteemed Member
Location
Ohio
JasonCo said:
...but you are saying this is incorrect in the real world, wouldn't work?
Click to expand...
Depends on what you call real. For test purposes is just as real as, well, a real circuit. They are just different in nature. But I know your meaning. As GoldDigger points out, it will yield a cap to approximating an actual circuit value.

Conventional kVA addition is one-third 3Ø loads, one-half L-L loads, and full L-N loads. You will see that is less than arithmetic addition of ampere values.

If you use vector addition, you will see the result is less than both when there is unity power factor. The result could anywhere from 0 to the cap previously mentioned if power factors are not at unity.
 
Smart $

Smart $

Esteemed Member
Location
Ohio
Ingenieur said:
Aren' all loads on ph a 'in phase' so the vector magnitude = scalar magnitude?
only real components once phase shift is offset/ignored

the issue I see is mixing P and S without a pf to put everything in S
Click to expand...
Assuming unity power factor for all loads, L-L loads are ±30° out of phase with 3Ø and L-N load currents
 
I

Ingenieur

Senior Member
Location
Earth
Smart $ said:
Assuming unity power factor for all loads, L-L loads are ±30° out of phase with 3Ø and L-N load currents
Click to expand...

doesn't matter
each is calculated independently for magnitude only
empirically integrated so all loads, 1, l-l, 3 phase are in phase for the same/given phase
 
Smart $

Smart $

Esteemed Member
Location
Ohio
Ingenieur said:
doesn't matter
each is calculated independently for magnitude only
empirically integrated so all loads, 1, l-l, 3 phase are in phase for the same/given phase
Click to expand...

A-B circuit 10A
A-N circuit 10A
A-B-C circuit 10A

What is the vector magnitude of Line A current?
 
I

Ingenieur

Senior Member
Location
Earth
Smart $ said:
A-B circuit 10A
A-N circuit 10A
A-B-C circuit 10A

What is the vector magnitude of Line A current?
Click to expand...

Ph ang does not matter
30 on a
20 on b
10 on c
10 plus the unbalance on n

engineers have a tendency to over-think and complicate things with math complexities not required
 
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