Current flow in delta configuration

[Grouch1980]

Say you have a 3 phase delta configuration on the primary of a transformer. How does the current flow through the phase coils as the AC current on the feeder flips back and forth in direction?

I understand that at any instantaneous time on the feeder, current flows towards the delta wiring on two of the phase wires, while returning on the 3rd phase wire, and this constantly changes... you can see it on a 3 phase sine wave diagram.

How does this affect the current flow through the delta connection? Does the current flow in one direction, counterclockwise, and then the other direction, clockwise? I can't find anything on Google.

 

[wwhitney]

A couple graphics that are useful are below. The first is just to establish a consistent sense of positive for the currents for discussion. The second is a graph of 3 sinewaves of equal magnitude but pairwise offset 120 degrees. For a typical loading, both the line currents as a group (Ia, Ib, Ic) and the coil currents as a group (Iac, Iba, Icb) will look like the 3 sinewave graph.

As the graph shows, at any given point in time 1 or 2 of the coil currents will be positive, and then the other coil currents will be negative. In order to have instantaneous loop current flow around the 3 coils, you'd need all 3 to be positive or negative at one time. So that doesn't happen.

Hope this helps.

Cheers, Wayne

 

[Carultch]

Say you have a 3 phase delta configuration on the primary of a transformer. How does the current flow through the phase coils as the AC current on the feeder flips back and forth in direction?

I understand that at any instantaneous time on the feeder, current flows towards the delta wiring on two of the phase wires, while returning on the 3rd phase wire, and this constantly changes... you can see it on a 3 phase sine wave diagram.

How does this affect the current flow through the delta connection? Does the current flow in one direction, counterclockwise, and then the other direction, clockwise? I can't find anything on Google.

Construct KCL equations at each terminal of the transformer diagram.
Iba + Iac = Ia
Icb + Iba = Ib
Iac + Icb = Ic

Solve the system for Iba, Iac, and Icb:
Iba = Ia/2 + Ib/2 - Ic/2
Iac = Ia/2 - Ib/2 + Ic/2
Icb = -Ia/2 + Ib/2 + Ic/2

Set up sine waves for each of Ia, Ib, and Ic, with 120 degree or 2*π/3 radian phase shifts. Keep it simple, with unit amplitude and unit frequency.
Ia = sin(t)
Ib = sin(t - 2*π/3)
Ic = sin(t + 2*π/3)

After substituting and simplifying, we get the following for the currents in the transformer winding:
Iba = sin(t + 5*π/3)
Iac = sin(t - 5*π/3)
Icb = -sin(t)

As you can see, the currents within the windings are just like the currents entering/exiting the terminals. It is three waveforms that are 120 degrees apart, and that add up to zero at any given snapshot in time. Circulating currents in delta transformers will happen with triplen 180 hz harmonic currents, but not with the fundamental 60 Hz waveforms.

Here are the plots of the winding currents within the Delta network:



And for comparison, here are the line currents as the transformer terminals:

 

[wwhitney]

Construct KCL equations at each terminal of the transformer diagram.
Iba + Iac = Ia
Icb + Iba = Ib
Iac + Icb = Ic

Solve the system for Iba, Iac, and Icb:
Iba = Ia/2 + Ib/2 - Ic/2
Iac = Ia/2 - Ib/2 + Ic/2
Icb = -Ia/2 + Ib/2 + Ic/2

You have a sign error in each of your KCL equations (should be Iba - Iac = Ia, etc), which makes the series of equations degenerate and not invertible, so you can't solve for the coil currents given just the line currents. And without an impedance in the delta loop of coils, you can add an arbitrary loop current to any given solution (Iac, Iba, Icb) and get another solution.

Cheers, Wayne

 

[Carultch]

You have a sign error in each of your KCL equations (should be Iba - Iac = Ia, etc), which makes the series of equations degenerate and not invertible, so you can't solve for the coil currents given just the line currents. And without an impedance in the delta loop of coils, you can add an arbitrary loop current to any given solution (Iac, Iba, Icb) and get another solution.

Cheers, Wayne

Wow, I should've paid more attention to the diagram. I got led astray by a false sense of closure, when the original system I constructed was solvable, and looked like the solution I was expecting. I'm surprised it is a degenerate system.

 

[wwhitney]

For a typical loading, both the line currents as a group (Ia, Ib, Ic) and the coil currents as a group (Iac, Iba, Icb) will look like the 3 sinewave graph.

The nice graphs that Calrutch and I posted of 3 sine waves pair-wise 120 degrees apart are appropriate for a balanced 3 phase load, were the phase shift of the currents will be the same in each coil.

If you apply certain unbalanced set of loads, such as a resistive load on one phase, a 45 degree lagging load on another, and a 45 degree leading load on the third, then there will be a period of time in each cycle when all the coil currents are positive or negative, and that will mean that instantaneously the current in the 3 coils is flowing in a loop. But this is the exception rather than the rule.

Cheers, Wayne

 

[wwhitney]

I'm surprised it is a degenerate system.

The line currents have to sum to zero by KCL, so specifying the line currents only provides two independent constraints. While there are 3 independent variables, the 3 coil currents. The L-L load impedances can be arbitrary, allowing you to realize any collection of coil currents.

So that means that there are different sets of load impedances that will give you the same line currents. All you need to do is shift the coil currents by the same vector. I think this is a simple example:

(1) The load impedances are (1, 1, 1), all 1 ohm. The coil currents are all magnitude V_LL and are in phase with the coil voltages.
(2) The load impedances are (1/sqrt(3) < 30 degrees, 1/sqrt(3) < -30 degrees, infinite). At least for one of the two senses of rotation--I have left unspecified the order of listing the load impedances. This subtracts the last coil current in case (1) from all of the coil currents.

Cheers, Wayne

P.S. A lot of my posts on topics like this are me working this out for my own understanding. So corrections are welcome.

 

[kwired]

Say you have a 3 phase delta configuration on the primary of a transformer. How does the current flow through the phase coils as the AC current on the feeder flips back and forth in direction?

I understand that at any instantaneous time on the feeder, current flows towards the delta wiring on two of the phase wires, while returning on the 3rd phase wire, and this constantly changes... you can see it on a 3 phase sine wave diagram.

How does this affect the current flow through the delta connection? Does the current flow in one direction, counterclockwise, and then the other direction, clockwise? I can't find anything on Google.

If you have equal load on all three lines (to keep it simplest for now) taking a measurement on say the A conductor you would have one current that is flowing between A and B and one current that is flowing between A and C but they are 120 deg out of phase with one another Multiply that current (they are both the same with equal load on them per our keeping it simple for now) by square root of three and the result is the effective current you will measure on the A conductor.

Take a look at the plots in post 3, and imagine clamping an ammeter around all three of the current lines there. No matter where you clamp it you get the same net reading but each individual current is at a different level depending on where you clamp it.

Unbalance the load and the net current changes but it still reads the same net wherever you would clamp it.

 

[Grouch1980]

As the graph shows, at any given point in time 1 or 2 of the coil currents will be positive, and then the other coil currents will be negative. In order to have instantaneous loop current flow around the 3 coils, you'd need all 3 to be positive or negative at one time. So that doesn't happen.

View attachment 2568210

So... what this image shows is not accurate, correct? since at any one point, the current on either line A, B, or C should be going the opposite direction. So the counterclockwise loop inside the delta wiring doesn't happen?

 

[wwhitney]

So... what this image shows is not accurate, correct? since at any one point, the current on either line A, B, or C should be going the opposite direction. So the counterclockwise loop inside the delta wiring doesn't happen?

With AC, the arrows don't show the instantaneous direction of current flow, but just show what direction is considered positive.

I.e. the arrow under the notation I_BA shows that the positive sense will be from B to A. Half the cycle, when the current sinewave is positive, the current will flow in the coil from B to A. The other half cycle, when the current sinewave is negative, the current will flow from A to B.

Cheers, Wayne

 

[kwired]

With AC, the arrows don't show the instantaneous direction of current flow, but just show what direction is considered positive.

I.e. the arrow under the notation I_BA shows that the positive sense will be from B to A. Half the cycle, when the current sinewave is positive, the current will flow in the coil from B to A. The other half cycle, when the current sinewave is negative, the current will flow from A to B.

Cheers, Wayne

Now add C into the mix and and you always have current flowing in three directions at all times but at any point in time each of the three currents are at different levels but still results in the same net effect in total current in each conductor, source winding or within the load.

Looking at the sine waves in post 3, anytime one current is at max positive peak the other two are at half negative peak. If you figured out the vector values at any point in the time line the net result (effective output) is the same for every point in the time line and is 1.732 times what the result of a single phase wave with same amplitude would be.

(I think I said that right)

 

[wwhitney]

Looking at the sine waves in post 3, anytime one current is at max positive peak the other two are at half negative peak. If you figured out the vector values at any point in the time line the net result (effective output) is the same for every point in the time line and is 1.732 times what the result of a single phase wave with same amplitude would be.

There are no vector values at a single point in time; the vector math represents a simplification of the waveform math, i.e. considering a whole cycle rather than just a single point in time.

Also, net result of what? If you add the waveforms in post #3 (equal magnitude, pairwise 120 degrees apart), you'll get zero at any point in time.

If you take the waveforms in post #3 to be both voltage and current (i.e. voltage and current are in phase, magnitudes may differ), and you calculate the waveform I*V for each of the 3 phases, and add them, then you'll get a constant function. I.e. 3 phase can deliver constant power overall. This is not the case for single phase, where the analogous I*V function would be 0 twice each cycle.

The way the sqrt(3) arises in power computations is as follows: for the 3 phase case, if the currents I are take as the coil currents (which equal the load currents), and not the line currents), then the I*V for each coil is the same as the single phase case, and the power is just 3 times the single phase case. But if you take the currents I' to be the line currents (which are the pairwise differences of the coil currents, now in the vector/waveform sense), their magnitude is sqrt(3) times the coil currents, so the total power is now sqrt(3) * I' * V.

Cheers, Wayne

 

[Grouch1980]

With AC, the arrows don't show the instantaneous direction of current flow, but just show what direction is considered positive.

Oh. I thought it was the direction of current, which didn't make sense... since one has to go the opposite direction. That clarifies that part for me.

 

[Grouch1980]

As you can see, the currents within the windings are just like the currents entering/exiting the terminals. It is three waveforms that are 120 degrees apart, and that add up to zero at any given snapshot in time. Circulating currents in delta transformers will happen with triplen 180 hz harmonic currents, but not with the fundamental 60 Hz waveforms.

Here are the plots of the winding currents within the Delta network:
View attachment 2568214

Understood. Thanks for the plot. So the current flow in the windings is the same as the line currents, just shifted in time from the line currents (with all three still 120 degrees apart)

 

[wwhitney]

So the current flow in the windings is the same as the line currents, just shifted in time from the line currents (with all three still 120 degrees apart)

Same shape (60 Hz sinewave), but with different magnitude and shifted in time by a partial cycle. When all the coil currents are equal in magnitude and pairwise 1/3 cycle apart (120 degrees apart), the line currents will have magnitude sqrt(3) times the coil currents' magnitude.

Cheers, Wayne

 

[Grouch1980]

Same shape (60 Hz sinewave), but with different magnitude and shifted in time by a partial cycle. When all the coil currents are equal in magnitude and pairwise 1/3 cycle apart (120 degrees apart), the line currents will have magnitude sqrt(3) times the coil currents' magnitude.

Cheers, Wayne

Got it.

 

[kwired]

There are no vector values at a single point in time; the vector math represents a simplification of the waveform math, i.e. considering a whole cycle rather than just a single point in time.

Also, net result of what? If you add the waveforms in post #3 (equal magnitude, pairwise 120 degrees apart), you'll get zero at any point in time.

If you take the waveforms in post #3 to be both voltage and current (i.e. voltage and current are in phase, magnitudes may differ), and you calculate the waveform I*V for each of the 3 phases, and add them, then you'll get a constant function. I.e. 3 phase can deliver constant power overall. This is not the case for single phase, where the analogous I*V function would be 0 twice each cycle.

The way the sqrt(3) arises in power computations is as follows: for the 3 phase case, if the currents I are take as the coil currents (which equal the load currents), and not the line currents), then the I*V for each coil is the same as the single phase case, and the power is just 3 times the single phase case. But if you take the currents I' to be the line currents (which are the pairwise differences of the coil currents, now in the vector/waveform sense), their magnitude is sqrt(3) times the coil currents, so the total power is now sqrt(3) * I' * V.

Cheers, Wayne

I figured I'd dig myself into a hole on this, I understand it but not well enough to explain everything about it.

I may or may not understood exactly what OP's question is either.

A somewhat simple explanation (that may have holes in it as well) is that with three phase circuits each line sort of has more than one current it is carrying. At the corners of a delta you have current going three ways A-B, B-C and C-A. The A-B and B-C currents individually might be 10 amps but the B phase conductor carries 1.732 times that value so 17.32 amps.

 

[Strathead]

I will also throw in that this is electrical THEORY. Current may very well not flow.

 

[ggunn]

I will also throw in that this is electrical THEORY. Current may very well not flow.

Ah, yes, the Magic Smoke Theory.

 

[Grouch1980]

I may or may not understood exactly what OP's question is either.

I was asking what the direction of current flow is inside the delta configuration. I wasn't sure if clockwise, counterclockwise, or if it flowed the same way as the line currents... 2 in one direction with the 3rd in the opposite direction.