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Joined: Jan 2005
Posts: 2
D
DJT Offline OP
Junior Member
I AM PROVIDING TRAINING FOR OUR PLANT ELECTRICIANS, AND I WANT TO CONFIRM A COUPLE BITS OF POWER THEORY PRIOR TO GOING LIVE.

EXAMPLE: 13,800V DELTA, 480-277V WYE, 2000KVA
TRANSFORMER.
X1 HAS 1000 AMPS
X2 HAS 1200 AMPS
X3 HAS 1400 AMPS

I HAVE ASKED THEM TO CALCULATE:
NEUTRAL CURRENT
ALL 3 DELTA LINE CURRENTS
ALL 3 DELTA PHASE CURRENTS

TO SOLVE FOR NEUTRAL CURRENT;
THE SQUARE ROOT OF...
1000 SQUARED + 1200 SQUARED + 1400 SQUARED
= 2098 AMPS
DELTA PHASE A CURRENT = 1000A/49.819 = 20AMPS
............B CURRENT = 1200A/49.819 = 24AMPS
............C CURRENT = 1400A/49.819 = 28AMPS
49.819 IS 13800/277.
DELTA A LINE CURRENT = 20 X 1.732 = 35AMP
......B LINE CURRENT = 24 X 1.732 = 42AMP
......C LINE CURRENT = 28 X 1.732 = 48AMP

I REALIZE THAT WITH THESE IMBALANCES, THERE WILL BE PHASE SHIFTS, BUT 120 DEGREE ANGLES WERE USED TO SIMPLIFY.
I HAVE A 50/50 CONFIDENCE LEVEL WITH MY CALCULATIONS.

Joined: Oct 2003
Posts: 147
C
Member
I got 346.4A for the neutral current. The other figures seem to be correct.

The neutral current is calculated by adding the line currents in a vectoral manor.

[This message has been edited by crash (edited 01-25-2005).]

Joined: Sep 2003
Posts: 650
W
Member
The Delta line currents are the vector sum of the phase currents at each node, so you can't simply multiply the phase currents by root three. Even if you are making the simplifying assumption of 120 degrees between each phase, I get 38.16A, 41.62A, 45.08A...I've not double checked this.

-Jon

Joined: May 2004
Posts: 162
C
Member
Try this it' always worked for me.

I N =Sq root of A2 + B2 + C2 – A.B-B.C-C.A

Essentially this provides the vector sum of the three currents. It works if one phase is not utilized also.

Joined: Feb 2004
Posts: 75
S
Member
Here is a great explanation of the nuetral currents on wye systems. It even has the formula to calculate the netrual current.
http://www.electrician.com/electa1/electa3htm.htm


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