ECN Forum Forums Technical Discussion Electrical Theory and Applications 2 phase neutral current?

Draw the the first circuit without a neutral connection at the load, just like a 2 wire 208V baseboard heater. Ignore the source polarity marks.

Without a neutral connection at the load, all of the current "out" of the top source must flow "into" the bottom source. Do you agree?

"Without a neutral connection at the load, all of the current "out" of the top source must flow "into" the bottom source. Do you agree?"
JBD
No. That is a very simple way to describe it. Thats not how it works. Use this example:
You have a 120/208 volt service from the utility. There are 3 phases A,B and C.
The angle betweem the phases is 120 degrees.
120 between A&B, 120 between A&C and 120 between B&C. Take 2 conductors and connect one to A and one to B. Connect these conductors to a 208 volt load. Has the angle between A and B changed. No. Its still 120 degrees and the voltage is 208 volts. Go to the site below and it will show you how the Sine waves look for the 3 phase service. The net voltage between conductors is the net value of adding the
voltage levels on A and B. This is the real way it works.
http://home.comcast.net/~ronaldrc/site1138/55.htm

[This message has been edited by Bob (edited 03-09-2006).]

Bob,

That is not how three phase works. There is no phase angle difference between two ends of the same wire, even if there is impedance in the wire. Do not confuse specific points on a wiring diagram as "phases".

There is a phase angle difference between the three pairs of conductors AB & BC & CA. There is also a phase angle difference between the pairs of AN & BN & CN.

The reference you gave (allaboutcircuits) has very simplified, if not misleading, information on the wiring diagrams.

In their words "Kirchhoff's Current Law tells us that the currents entering and exiting the node between the two loads must be zero" But, they never show actual current flow in any of their multi-wire circuits. They show 100% of the current always leaving the load and entering the source, the real world does not work like that.

"There is no phase angle difference between two ends of the same wire, even if there is impedance in the wire. "
I have no clue what you mean by this. I said
take phase A and B and connect a load between. This is not the same wire. There is a 120 degree phase angle between them.
Read the last post again. Take a look at this site http://home.comcast.net/~ronaldrc/site1138/55.htm . It shows the phases as they are.

Bob, if that last post showed the currents with the phases, I think it would be clearer.

JDB is correct with the 13.4 amps though...assuming this is a multiwire with nuetral pulled...I don't see how you would get 1700w a on one phase and 1500w on another with a straight 208V load....

Dnk...

[This message has been edited by Dnkldorf (edited 03-10-2006).]

I've got to jump in here and 'side' with JBD.

_Voltage_ is always measured between two points. When you have an AC supply, you can graph the voltage versus time. In the ideal case for power distribution, your voltage versus time graph will be a nice sine wave.

If you have more than two nodes in your system, then you can measure the voltage between _any_ pair of nodes. In a system with three nodes, there are _three_ voltages that you can measure; in a system with four nodes then there are _six_ possible pairs of nodes and thus _six_ voltages that you can measure. For each of these possible voltages, you can plot voltage versus time, and get a _different_ sine wave.

When you talk about phase angles in a polyphase system, you are simply describing the timing difference of these sine waves. The phase angles only have meaning if you define your reference zero (which of the possible sine waves is considered as being at zero degrees) and if you define which _pairs_ of points you are measuring between.

In a common three phase wye system, the phase angle of the voltages measured between line and neutral are all 120 degrees apart. In other words, if you were to graph the voltage as measured from neutral to leg A, and graph the voltage as measured from neutral to leg B, you would find a 120 degree phase difference. If, on the other hand you were to measure the voltage from leg A to leg B, and then measure leg A to leg C, you would only see a _60_ degree phase difference between your two measurements. Finally, if you compared the A to neutral measurement with the A to B measurement, you would see both a difference in amplitude, and only a 30 degree phase difference.

If you only ever measured between the A leg and the B leg, say because you had a single phase load connected A to B, then you would only have a single sine wave to look at, and thus _no_ phase angle difference to consider.

If you have two legs of a three phase wye system, then _referenced to neutral_ the two legs will be 120 degrees apart. But if you only connect a single phase load between A and B, at any instant in time the current flowing from phase A will have to exactly match the current flowing to phase B. Referenced to neutral the _voltages_ may be 120 degrees apart, but referenced to each other the current flow is exactly the same (meaning a 0 or 180 degree phase difference depending upon your sign conventions).

Because of this, if you have an imbalanced three phase load, even if it is perfectly resistive, you will find some 'power factor' (difference in phase angle between current and voltage) when you compare the voltage phase angles and the current phase angles.

Finally, the 30 degree phase angle difference mentioned above shows up when you compare primary and secondary phasing on a delta to wye transformer.

-Jon

DNK
"JDB is correct with the 13.4 amps though...assuming this is a multiwire with nuetral pulled...I don't see how you would get 1700w a on one phase and 1500w on another with a straight 208V load...."
HE is correct with 13.4 amps. His results and mine are the same. The load shown is 2 single phase loads connected phase to neutral with a net neutral current of 13.4 amps.


"If you only ever measured between the A leg and the B leg, say because you had a single phase load connected A to B, then you would only have a single sine wave to look at, and thus _no_ phase angle difference to consider."

Winnie
The single sine wave you refer to is the net result of the addition of phase A and Phase B. There are times when phase A is at
about 170 volts and Phase B is at another value that results in a single sine wave. The phase difference still exists.

"If you have two legs of a three phase wye system, then _referenced to neutral_ the two legs will be 120 degrees apart. But if you only connect a single phase load between A and B, at any instant in time the current flowing from phase A will have to exactly match the current flowing to phase B."
The current wave form follows the voltage wave form if the load is resistive. If its inductive then the current does not follow the voltage. The currents add as vectors just as the voltage and as you say at any point on the sine waves the resultant currnet in both phases are the same.

" Referenced to neutral the _voltages_ may be 120 degrees apart, but referenced to each other the current flow is exactly the same (meaning a 0 or 180 degree phase difference depending upon your sign conventions)."
That is not correct. Again the currents are added as vectors as said above. If you look at my first post I added the current as vectors.




[This message has been edited by Bob (edited 03-10-2006).]

It sounds like we are converging on saying the same thing: you have to do the vector math. But either I am misunderstanding you, or you are not correct about the phasing of the current being supplied to the loads being discussed.

Consider a three phase wye system.

Call the neutral point our voltage zero reference.

Represent the voltage and phase of leg A relative to the neutral point by a vector of unit length and angle 0 degrees.

The voltage and phase of leg B relative to the neutral point will then be a vector of unit length and angle 120 degrees.

The voltage and phase of leg B relative to _leg A_ will then be represented by a vector of length sqrt(3) and angle 150 degrees.

For a resistive load connected between leg B and leg A, the current will follow the applied voltage. Thus the current flowing through this resistive load, from phase B to phase A will be sinusiodal and at phase 150 degrees relative to our reference phase voltage. The current flowing _out_ of phase B is at phase angle 150 degrees, the current flowing out of phase A is at 330 degrees relative to our reference zero.

The current flowing through the resistive load is in phase with the voltage difference between leg B and A, _not_ in phase with the phase to neutral voltage of either phase A or phase B.

If you look at this from the other side: consider a 120/208V wye system, and place a single phase 16 ohm resistive load between phase A and phase B. 13 amps will flow through this resistive load, and 2704W will be delivered to the load. But for 13A to flow through the load, 13A has to flow through both phase A and phase B. This means that 1560VA is being supplied by each phase. If 3120VA is being supplied, but only 2704W delivered to the load, then there has to be a power factor somewhere

-Jon

Winne posted
"If you look at this from the other side: consider a 120/208V wye system, and place a single phase 16 ohm resistive load between phase A and phase B. 13 amps will flow through this resistive load, and 2704W will be delivered to the load. But for 13A to flow through the load, 13A has to flow through both phase A and phase B. This means that 1560VA is being supplied by each phase. If 3120VA is being supplied, but only 2704W delivered to the load, then there has to be a power factor somewhere "

I do not know where you get the 1560 VA or the 3120 VA. None of the values in the problem when properly used can produce this figure.
Power = I²xR = 13²x 16 = 2704 watts.
Power = V x I = 208 x 13 = 2704 watts.

Where did you get these numbers?

The phase A wye connected transformer coil is delivering 120V phase to neutral. 120V * 13A = 1560 VA. Same for phase B.

-Jon