ECN Forum Forums Technical Discussion Electrical Theory and Applications Calculations

The way I figured this is as follows:

<OL TYPE=1>

[*] 3 Phase Load = 192.70 Amps (80 KVA),


[*] 1 Phase "240 Volt" Load = 141.67 Amps (34 KVA),


[*] 1 Phase "120 Volt" Load = 58.34 Amps (7 KVA).
</OL>

Nominal Totals Per Line ("Phase"):
<OL TYPE=A>

[*] Line "A": 334.37 Amps (80 KVA @ 240v 3Ø + 34 KVA @ 240v),


[*] Line "B": 192.70 Amps (80 KVA @ 240v 3Ø),


[*] Line "C": 392.71 Amps (80 KVA @ 240v 3Ø + 34 KVA @ 240v + 7 KVA @ 120v).
</OL>

Panelboard + Feeder Size

Nominal Load Values WITH LCL:

* Panelboard: 600 Amps - 240/120V 3Ø 4 Wire
* Feeder: 1,000 MCM THHN/THWN Cu. (or equivalent Parallel Feeders, protected by 500 Amp 3 Pole OCPD).


Nominal Load Values WITHOUT LCL:

* Panelboard: 400 Amps - 240/120v 3Ø 4 Wire
* Feeder: 600 MCM THHN/THWN Cu. (or equivalent Parallel Feeders, protected by 400 Amp 3 Pole OCPD).

BTW: Load on the Grounded (Neutral) Conductor would be 7 KVA (58.34 Amps)

Scott35

I've never been involved with delta systems, so perhaps my figuring is way off base, but I'm with Radar.

The way I see it, although there is a basic 120-degree difference between phase currents, the A-C winding with its center-tap neutral is introducing another factor to the overall picture.

Whenever you have a tapped transformer winding, if you use that tap as your reference point, then the voltages at each end will be 180 degrees out of phase with each other.
Let's build up a system:

Start with just a single center-tapped winding, like a 3-wire 1120/240V service, and let's still call the hot legs A and C. We know that the neutral current will be the difference between A and C. Now add one extra winding from A to provide a B-phase in the form of an open delta, but don't connect any load to that phase. Obviously there can be no current in that winding, so the neutral current will remain unchanged.

Now add a third winding B-to-C to close the delta, but still leave no load on the B-phase. Will the neutral current change then? I don't think so, because the basic voltage/current relationships between the phases are still the same. There will be very little (in a theoretically perfect arrangement zero) current flowing in the A-B and B-C windings, will there not? So the A-C winding is still acting just like a 3-wire 120/240V service.

Now add a 3-ph delta-connected load to A-B-C. Obviously there will now be current flowing in all three windings, but will that affect the neutral?

I do not believe so, since the 3-ph load has no neutral connection. The current in the A-C winding will increase, but as the voltages on the A and C phases have not changed with respect to neutral/ground, the single-phase loads connected A-N and C-N are still drawing the same amount of current, and thus the neutral will still carry just the difference of the single-phase loads on A and C.

I will TRY (it is rather lengthy, and somewhat confusing, but then again, I'm dumb) to recount the explanation in the above-mentione book when I get home from work today, assuming I do not stop for beer & wings (wink wink Dink).
That is a big assumption, based on my recent workload and resulting inability to feed my beer & wing habit for several weeks!

Whoops, I missed that this was a _delta_ system with a center tap.

Dnk, The equation that you were using, and which I said was 'spot on', applies when you have a 120 degree phase angle to phases A,B, and C around the neutral. If the power factors deviate at all from being the same, so that the current flows are no longer 120 degrees out of phase, then the equation no longer applies.

Remember that voltage is always measured relative to some reference zero. If you pick a different reference point, then you will measure _different_ voltages. No matter which reference point you select, if you do all of the math and calculate the total work involved to move an electron from on point to another, the overall result will come out the same, but some of the numbers in the middle will look different.

Generally we use the grounded transformer terminal as our reference zero. But remember the discussions of corner grounded delta systems; _any_ transformer terminal could be grounded, and any transformer terminal could be our reference zero for calculations.

From the 'point of view' of the center tap in the delta system, phases A and C _are_ 180 degrees out of phase. If we created a virtual neutral (say by using a zig-zag transformer) and used that virtual neutral as our reference zero, then phases A and C would be 120 degrees out of phase. If we picked phase B as our reference zero, then phases A and C are only 60 degrees out of phase.

It probably makes the most sense to pick the center tap as the voltage reference zero. *grin*

I am going to assume that all of the loads are resistive, just to go through the calculation. This is almost certainly an incorrect assumption; there are probably motors in that three phase load. But if you know the power factors for the various loads, then you can adjust the phase angles for the individual current flows appropriately.

In my calculations below, I calculate the current flowing _into_ each terminal. This gives a current phase angle that is approximately 180 degrees out of phase with the voltage phase angle of that terminal. It is interesting to note that even with the assumption of resistive loads, the current flowing from each terminal is not quite in phase with the voltage at each terminal. An unbalanced three phase resistive load does not have unity power factor.

I am also assuming a total KVA of 315; 240KVA of 3 phase load plus 75KVA of single phase load. You will get different numbers if you read the question differently.

Finally, in my calculations below, the loads are very carefully double counted, because I count the current from phase C to phase A and again as the current from phase A to C. This is compensated by the phase angle calculations; as a check, take the sum of all phase currents times 240/root(3) and you will get roughly the same total VA as the sum of the original specifications.

I will select the voltage measured from phase A to the Grounded center tap as the phase reference zero.
This gives us the following voltage phase angles for the single phase loads:
A to G 0 degrees 120V
G to A 180 degrees 120V
C to G 180 degrees 120V
G to C 0 degrees 120V

For the three phase loads, we assume that 80 KVA is connected between each pair A-B, B-C, C-A, so we need the three phase angles:
A to C 0 degrees 240V
C to B 120 degrees 240V
B to A 240 degrees 240V
A to B 60 degrees 240V
B to C 300 degrees 240V
C to A 180 degrees 240V

First the neutral current:
Our A to G load is 34 KVA at 120V, so 283.33A at 0 degrees
Our C to G load is 41 KVA at 120V, so 341.67A at 180 degrees
283.33A at 0 plus 341.67A at 180 gives
58.33A at 180 degrees

Now the phase A current:
Our G to A current is 283.33A at 180 degrees (just the opposite of the A to G current)
Our B to A load is 80 KVA at 240V, so 333.33A at 240 degrees
Our C to A load is 333.33A at 180 degrees
Now we add up 283.33 at 180 degrees with 333.33 at 240 degrees and 333.33 at 180 gives
834.83A at 200 degrees

Phase B current:
Our G to B current is 0
Our A to B current is 333.33A at 60 degrees
Our C to B current is 333.33A at 120 degrees
gives 577.35A at 90 degrees

Phase C current
G to C is 341.67A at 0 degrees
A to C is 333.33A at 0 degrees
B to C is 333.33A at 300 degrees
gives 889.8A at 341 degrees

I'll leave selection of OCPD and conductor size for someone else

-Jon

If you look at the current flowing through the coil connected between A and C phases, you will see that this is strictly single phase. There can be no phase shift from one end of a transformer winding to the other end(even if there is a center-tap).

The normal 120/240 single phase calculations apply for any load connected A-N, N-C.
I=kVA/E, use L-L kVA if using L-L voltage

Normal three phase calculations can be used for any loads connected A-B-C.
I=kVA/(E*1.73), use L-L kVA if using L-L voltage

Now just add up the line currents.

JBD:

Not quite. Your statement is correct that you calculate the single phase currents as single phase currents and the three phase currents as three phase currents. However the phase angle of the phase A three phase currents will be different from the phase A single phase currents, so you need to use vector addition to 'just add them up'.

-Jon

The phase currents in windings AB and BC are

80KVA/240V = 333A.

These currents combine to yield,

Ib = 333Ax1.732 = 577A for the 80 kva per phase.
Single phase =34 + (34+ 7) kva = 68 + 7 kva.
For winding CA, 68KVA of the single phase load may be treated as a 240V load and added to the 80KVA load to yield 148KVA of 240V load . Phase (not line) current is 148KVA/240 = 616A. To obtain the line current, Ia, we combine vectorially the phase current from winding AB.

Ia = 616A + 333A[cos(60) +jsin(60)] = 783A + j289A yields 835A magnitude.

The remaining 7KVA of single phase load adds 58.3A into node C. Then

Ic = 616A + 58.3A + 333A[cos(60) +jsin(60)] = 841A + j289A yields 890A magnitude.

Ia = 835A
Ib = 577A
Ic = 890A

winnie,

You can not have a phase angle difference on a single conductor. Yes the current flowing in the C-B winding will be at a different phase angle from the current in the C-A winding, but there is still only 1 current in the C phase conductor.

Go ahead and work out the math for the summation of these two winding currents at the C conductor and then tell me if it is substantially different from my method. I don't think it is worth the effort for sizing a conductor and overcurrent protective device. Because if we want to get down and dirty exact, we really need to look at the power factor of each load before we try to perform vector addition.

More questions than answers are generated based on the ambiguously worded question--

However, remember that in a delta system, the phase current (L-N) is less than the line currents(L-L) by a factor of 1.73.
Therefore, in order to determine line currents, and the required OCPDs, wouldn't we need clarification of the terms "single phase load"?

Is there an engineer in the house?

"Is there an engineer in the house?"
I just worked it out for you. So did winnie.
Our answers are the same.

[This message has been edited by Bob (edited 02-11-2006).]