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

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JFletcher

Senior Member
Location
Williamsburg, VA
I'm not getting any of the answers listed. Breakers 1,2,7,8,13 and 14 are all A phase, yes? So

Circuit #1 = 2100VA/
Circuit #2 = 41,3000VA
Circuit #7 = Not used
Circuit #8 = Not used
Circuit #13 = 31,300VA
Circuit #14 = 22,000VA
-------------------------------------

Circuit #1 is L-N, so 2100/120 = 17.5A
Circuit #2 3ph, so 41,300/208/rt3 = 114.64A
cIRCUIT #13 is 3ph, so 31,300/208/rt3 = 86.88A
Circuit #14 is L-L, so 22,000/208 = 105.77A

total 324.79A, which is close to F, but not exact... is rt3 1.73, 1.732, or actual value, only rounded at the end? Is F even the right answer?
 
I

Ingenieur

Senior Member
Location
Earth
Smart $ said:
For the feeder...

The current on A is not 30A.
The current on B is not 20A.

The current on C is 10A.
The current on N is 10A
Click to expand...

you can mess with phasors
go ahead and crunch them
offset l-l by the ph ang
offset l-n by the ph ang
get the magnitudes sum and get the mag for each at pf = 1
see what the difference is lol
that is why people have disdain for the smugness of engineers rotflmao

ot get a clamp on
 
I

Ingenieur

Senior Member
Location
Earth
JFletcher said:
I'm not getting any of the answers listed. Breakers 1,2,7,8,13 and 14 are all A phase, yes? So

Circuit #1 = 2100VA/
Circuit #2 = 41,3000VA
Circuit #7 = Not used
Circuit #8 = Not used
Circuit #13 = 31,300VA
Circuit #14 = 22,000VA
-------------------------------------

Circuit #1 is L-N, so 2100/120 = 17.5A
Circuit #2 3ph, so 41,300/208/rt3 = 114.64A
cIRCUIT #13 is 3ph, so 31,300/208/rt3 = 86.88A
Circuit #14 is L-L, so 22,000/208 = 105.77A

total 324.79A, which is close to F, but not exact... is rt3 1.73, 1.732, or actual value, only rounded at the end? Is F even the right answer?
Click to expand...


F is the closest answer
 
G

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
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...
Actually an L-L load causes current on each line which is identical in magnitude to the load current but is 60, not 30 degrees out of phase with the line to neutral voltage. The current for L-N loads will be in phase with the L-N voltage.
When you add two currents which are 60 degrees out of phase the difference is not neglegible.
Not as big as when adding two currents that are 120 degrees out of phase, but important.
 
T

topgone

Senior Member
JFletcher said:
I'm not getting any of the answers listed. Breakers 1,2,7,8,13 and 14 are all A phase, yes? So

Circuit #1 = 2100VA/
Circuit #2 = 41,3000VA
Circuit #7 = Not used
Circuit #8 = Not used
Circuit #13 = 31,300VA
Circuit #14 = 22,000VA
-------------------------------------

Circuit #1 is L-N, so 2100/120 = 17.5A
Circuit #2 3ph, so 41,300/208/rt3 = 114.64A
cIRCUIT #13 is 3ph, so 31,300/208/rt3 = 86.88A
Circuit #14 is L-L, so 22,000/208 = 105.77A

total 324.79A, which is close to F, but not exact... is rt3 1.73, 1.732, or actual value, only rounded at the end? Is F even the right answer?
Click to expand...
I did my own schedule of loads and have the phase loads as : A = 37.3 kVA; B= 31.4 kVA; and C = 39.2 kVA. The corresponding line to neutral amps are : Ia = 310.6; Ib = 261.1; and Ic = 326.0 amperes! Nothing matched.

The average three-phase amps using 107,800VA/(1.732 x 208) = 299.2, which fits answer D) 299.2 amps!
 
Smart $

Smart $

Esteemed Member
Location
Ohio
topgone said:
I did my own schedule of loads and have the phase loads as : A = 37.3 kVA; B= 31.4 kVA; and C = 39.2 kVA. The corresponding line to neutral amps are : Ia = 310.6; Ib = 261.1; and Ic = 326.0 amperes! Nothing matched.

The average three-phase amps using 107,800VA/(1.732 x 208) = 299.2, which fits answer D) 299.2 amps!
Click to expand...
I got the same kVA values.

Conventional kVA amperes:
A=310.8A, B=261.3A, C=326.3A (differences probably from rounding)

Vector amperes:
A=315.3A, B=261.9A, C=328.0A, N=23.5A
 
Smart $

Smart $

Esteemed Member
Location
Ohio
Smart $ said:
...
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.
Click to expand...
After revisiting the use of vectors, I realize these clauses are incorrect.

The entire statement should have been:
If you use vector addition, the result could anywhere from 0 to the magnitude cap previously mentioned.
 
Smart $

Smart $

Esteemed Member
Location
Ohio
GoldDigger said:
Actually an L-L load causes current on each line which is identical in magnitude to the load current but is 60, not 30 degrees out of phase with the line to neutral voltage. ...
Click to expand...
Please check again.

Wye%20angles.gif
 
Smart $

Smart $

Esteemed Member
Location
Ohio
Smart $ said:
Ingenieur said:
is 29< a <30 ?
Click to expand...
No.
Click to expand...
I'm sorry about earlier reply. The answer should have been yes. I was looking at the value of the kVA method.

Ingenieur said:
Is 29<A<30 or ~30
Is 17<B<20 or ~20
is that close enough for an apprentice electrician just starting his training
Click to expand...
I think an apprentice electrician at this stage of learning should be made aware of the kVA method.

A=28.7A B=18.7A C=10.0A

Vector values:

A=29.1A B=19.3A C=10.0A


Both calculations using the NEC 220.5(B) rounding rule would yield...

A=29A B=19A C=10A


...but as loads get larger and/or greater in quantity, and power factor(s) enter the equation, the discrepancy can grow. However, the NEC does not proscribe using vector sum(s) to determine load or load balancing. Probably the reason the NEC is rather conservative in its method of ampacity determination.

PS: here's a link to the latest version of my Excel "calculator":
https://drive.google.com/file/d/0By0rzU3UuNh7Y1V2NHhwM1BwQzA/view?usp=sharing
 
I

Ingenieur

Senior Member
Location
Earth
Smart $ said:
I'm sorry about earlier reply. The answer should have been yes. I was looking at the value of the kVA method.


I think an apprentice electrician at this stage of learning should be made aware of the kVA method.

A=28.7A B=18.7A C=10.0A

Vector values:

A=29.1A B=19.3A C=10.0A


Both calculations using the NEC 220.5(B) rounding rule would yield...

A=29A B=19A C=10A


...but as loads get larger and/or greater in quantity, and power factor(s) enter the equation, the discrepancy can grow. However, the NEC does not proscribe using vector sum(s) to determine load or load balancing. Probably the reason the NEC is rather conservative in its method of ampacity determination.

PS: here's a link to the latest version of my Excel "calculator":
https://drive.google.com/file/d/0By0rzU3UuNh7Y1V2NHhwM1BwQzA/view?usp=sharing
Click to expand...

Thanks for the clarification
vector B 19.3 (got the 29.1 for A)
is that correct for B?
by hand I got 17.32
won't be the first time I screwed up
won't be the last lol
 
Last edited:
Smart $

Smart $

Esteemed Member
Location
Ohio
Ingenieur said:
Thanks for the clarification
vector B 19.3 (got the 29.1 for A)
is that correct for B?
by hand I got 17.32
won't be the first time I screwed up
won't be the last lol
Click to expand...
Yes...
|i|θij
10-120-5-8.66025
10-150-8.66025-5
19.31852-135-13.6603-13.6603
 
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