single phase to 3-phase calculations

[oldsparky52]

I'm looking at a project that has 3 single phase transformers (480:120/240) to be installed. Each of these transformers will be fed from a 3-phase 480V [FONT=inherit !important][FONT=inherit !important]MDP[/FONT][/FONT]. The loads stated below are the 480V loads feeding the transformer.

Transformer 1 has a calculated load of 270A connected to A&B phases
Transformer 2 has a calculated load of 393A connected to B&C phases
Transformer 3 has a calculated load of 181A connected to C&A phases

The question is what are the calculated loads for A,B,C phases feeding the MDP.

(Please let's not let this go to other questions, I am looking for the proper math for this calculation).

I came across a formula on the "web" and I came up with
A=393
B=577
C=508

I'm not sure I got the correct formula and I'm not sure I implemented it correctly so I'm looking for either confirmation of my calculations or instructions on what the correct answer is and how it is calculated.

Thanks

 

[mivey]

I'm looking at a project that has 3 single phase transformers (480:120/240) to be installed. Each of these transformers will be fed from a 3-phase 480V [FONT=inherit !important][FONT=inherit !important]MDP[/FONT][/FONT]. The loads stated below are the 480V loads feeding the transformer.

Transformer 1 has a calculated load of 270A connected to A&B phases
Transformer 2 has a calculated load of 393A connected to B&C phases
Transformer 3 has a calculated load of 181A connected to C&A phases

The question is what are the calculated loads for A,B,C phases feeding the MDP.

(Please let's not let this go to other questions, I am looking for the proper math for this calculation).

I came across a formula on the "web" and I came up with
A=393
B=577
C=508

I'm not sure I got the correct formula and I'm not sure I implemented it correctly so I'm looking for either confirmation of my calculations or instructions on what the correct answer is and how it is calculated.

Thanks

You must have the correct formula. I got:
A = 393.1043119
B = 577.4590895
C = 508.2745321

 

[winnie]

Your numbers are approximately correct.

The equation that you found is an approximation which assumes that all current flow is a perfect sine wave (no harmonics) and that the current flow on the different phases has the same power factor. If these assumptions are correct then the formula gives the correct values.

When adding up AC currents, you need to consider the phase angle of the current flow. Two 1A currents can add together to give anything from 0 to 2A, depending on the phase angle. This is how current can 'balance' on the neutral.

If you want to learn more about this, look up 'vector addition'. Current and phase angle can be represented as a vector, which permits them to be added graphically and intuitively. (Of course, even this is an approximation, but it is close enough for most work.)

-Jon

 

[mivey]

In considering Jon's points, the long-hand way would be:

Let A-B = 0d, B-C = 240d, C-A = 120d (or pick the correct angle for the load currents if you know the power factor)

Looking at the terminal currents (1,2,3 for the different loads) gives:
A1 = 270<0d, B1 = 270<180d
and B2 = 393<240d, C2 = 393<60d
and C3 = 181<120d, A3 = 181<300d


Finding the real & imaginary components for summing gives (1,2,3 for the different loads):

Real(A1) = 270*cos(0d) = 270, Im(A1) = 270*sin(0d) = 0
Real(B1) = 270*cos(180d) = -270, Im(B1) = 270*sin(180d) = 0
Real(B2) = 393*cos(240d) = -196.5, Im(B2) = 393*sin(240d) = -340.3480
Real(C2) = 393*cos(60d) = 196.5, Im(C2) = 393*sin(60d) = 340.3480
Real(C3) = 181*cos(120d) = -90.5, Im(C3) = 181*sin(120d) = 156.7506
Real(A3) = 181*cos(300d) = 90.5, Im(A3) = 181*sin(300d) = -156.7506


Summing phase currents gives the following results in complex & polar notation:

A1+A3 = 360.5 -j156.7506 = 393.10431<-23.500d
B1+B2 = -466.5 -j340.3480 = 577.45909<-143.886d
C2+C3 = 106 +j497.0986 = 508.27453<77.963d

watch the quadrants when converting back to polar form (I'm too lazy to detail that but can if you need it).

For harmonics, separate the calcs by frequency.

 

[oldsparky52]

Thank you everyone.