Another high-leg delta problem

[tortuga]

OK, I've thought about the geometry a while, and I came up with the diagram below, where the voltage point A is taken as the fixed reference. Here for simplicity I'm assuming the coil impedances are predominantly resistive, and the the load currents (for three phase and single phase loads separately) are balanced and in phase with the applied voltages. Which means I omitted the single phase neural point N and any unbalanced single phase load; the single phase current L can just be taken to be the average current through the A-C coil.

So the point of the diagram is to show the phase relationships, i.e. the direction of the voltage drops. The direction I've drawn each current is its phase (up to an unspecified choice of +/- 1). The stinger pot resistance brings the high leg voltage point B downward, which is towards the center of the ABC triangle (what would be the neutral point of a wye secondary). Likewise, the 3 phase current brings the voltage point C towards that same center point. And if the single phase current is a net load, it further moves the voltage point C to the left. If it's a net backfeed, the voltage rise would move it to the right instead.

Implicit in this picture is that this is for small effects where we can consider each effect separately and add them up. E.g. if we think of the three phase current causing a large move of the voltage point C, and then add a large resistive single phase A-C load, the single phase voltage drop would move the voltage point C directly towards A, not to the left.

Now what I'm missing is an appropriate measure of "imbalance" for the 3 phase system. The 3 phase loading on an open delta transformer inherently unbalances the system; the voltage angle BAC becomes less than 60 degrees, and all the voltages get smaller but in an unbalanced manner. Adding a single phase load makes the voltage angle BAC even smaller, but it reduces the voltage A-C which reduces the spread of the L-L voltages when X < Y. Whereas with a net single phase backfeed, we get the opposite.

So without such a measure of imbalance, it's not clear to me if single phase backfeed will cause the the imbalance to get worse. But now I agree that it's a possibility.

Cheers, Wayne

View attachment 2569234

Now just connect a 240V water heater across B-C, and model the open wye primary side.

 

[wwhitney]

Now just connect a 240V water heater across B-C, and model the open wye primary side.

That might be interesting, but why do you suggest that particular question? Doesn't seem very relevant. With an open delta, hopefully people aren't powering 2-wire loads from the "phantom" winding.

Cheers, Wayne

 

[Birken Vogt]

That might be interesting, but why do you suggest that particular question? Doesn't seem very relevant. With an open delta, hopefully people aren't powering 2-wire loads from the "phantom" winding.

Cheers, Wayne

It's commonly done here in California delta land, when they run out of regular spaces, for any straight 240 loads that might need power.

Where I work, the neighbor's shop has an AC connected on one of the high leg sides, and a small pump on a different one. I have no idea which is real and which is phantom.

It does not really matter anyway, because the main pot is 50kw, and the wing pot is 25 kw, so those small loads are not going to overload anything.

 

[jaggedben]

OK, I've thought about the geometry a while, and I came up with the diagram below, where the voltage point A is taken as the fixed reference. Here for simplicity I'm assuming the coil impedances are predominantly resistive, and the the load currents (for three phase and single phase loads separately) are balanced and in phase with the applied voltages. Which means I omitted the single phase neural point N and any unbalanced single phase load; the single phase current L can just be taken to be the average current through the A-C coil.

So the point of the diagram is to show the phase relationships, i.e. the direction of the voltage drops. The direction I've drawn each current is its phase (up to an unspecified choice of +/- 1). The stinger pot resistance brings the high leg voltage point B downward, which is towards the center of the ABC triangle (what would be the neutral point of a wye secondary). Likewise, the 3 phase current brings the voltage point C towards that same center point. And if the single phase current is a net load, it further moves the voltage point C to the left. If it's a net backfeed, the voltage rise would move it to the right instead.

Implicit in this picture is that this is for small effects where we can consider each effect separately and add them up. E.g. if we think of the three phase current causing a large move of the voltage point C, and then add a large resistive single phase A-C load, the single phase voltage drop would move the voltage point C directly towards A, not to the left.

Now what I'm missing is an appropriate measure of "imbalance" for the 3 phase system. The 3 phase loading on an open delta transformer inherently unbalances the system; the voltage angle BAC becomes less than 60 degrees, and all the voltages get smaller but in an unbalanced manner. Adding a single phase load makes the voltage angle BAC even smaller, but it reduces the voltage A-C which reduces the spread of the L-L voltages when X < Y. Whereas with a net single phase backfeed, we get the opposite.

So without such a measure of imbalance, it's not clear to me if single phase backfeed will cause the the imbalance to get worse. But now I agree that it's a possibility.

Cheers, Wayne

View attachment 2569234

So if I'm following correctly, my intuition was correct. A three-phase load in combination with backfeed on the split phase will result in a lower ratio of the A-B to A-C voltages than either by themselves. (With the B-C voltage somewhere in between.)

 

[wwhitney]

So if I'm following correctly, my intuition was correct. A three-phase load in combination with backfeed on the split phase will result in a lower ratio of the A-B to A-C voltages than either by themselves. (With the B-C voltage somewhere in between.)

Let's try putting some numbers to post #40. I'll just make all the labeled voltage offsets 5V (which implies the two transformers have identical resistances, X=Y).

Unloaded
A = (-120,0) ; B = (0, 207.8) ; C = (120,0)
AB = BC = AC = 240V

Loaded YK = XK = 5V and loaded XL = 5V
A = (-120,0) ; B = (0, 202.8) ; C = (110.7, 2.5)
AB = 235.6V ; BC = 228.9V ; AC = 230.7V

Loaded just YK = XK = 5V
A = (-120,0) ; B = (0, 202.8) ; C = (115.7, 2.5)
AB = 235.6V ; BC = 231.3V ; AC = 235.7V

Loaded YK = XK = 5V, and backfed XL = 5V
A = (-120,0) ; B = (0, 202.8) ; C = (120.7, 2.5)
AB = 235.6V ; BC = 233.9V ; AC = 240.7

So which imbalance is worst?

Cheers, Wayne

 

[wwhitney]

It does not really matter anyway, because the main pot is 50kw, and the wing pot is 25 kw, so those small loads are not going to overload anything.

For a typical configuration like the above, how does the impedance of the main pot compare to the impedance of the wing pot? In ohms, rather than as %Z.

Thanks,
Wayne

 

[Birken Vogt]

For a typical configuration like the above, how does the impedance of the main pot compare to the impedance of the wing pot? In ohms, rather than as %Z.

That is far beyond what I have the resources to find out.

 

[jaggedben]

For a typical configuration like the above, how does the impedance of the main pot compare to the impedance of the wing pot? In ohms, rather than as %Z.

Thanks,
Wayne

That is far beyond what I have the resources to find out.

But I think we know that the wing pot's resistance is typically higher.

 

[wwhitney]

But I think we know that the wing pot's resistance is typically higher.

OK, if you like below are the comparable numbers to post #45 where Y (the wing pot resistance) is twice X (the lighting pot resistance). But I also will assume the single phase current L is twice the three phase current K. Then we can change the example to YK = 5V, XK = 2.5V, XL = 5V.

Now the spread (maximum difference of voltages) for the single phase backfed case is 7.2V, while for the single phase load case it's 2.5V, and the for the no single phase load/backfeed case it's 2.2V. Easy to argue that the single phase backfed case is most unbalanced.

For the post #45 case, the voltage spreads are, respectively, 6.8V, 6.7V, and 4.4V. Here the spread is about the same between the single phase load case and the single phase backfed case.

This suggests the upshot is that, depending on transformer sizes and impedances, and relative loading, single phase backfeed has the potential to imbalance the 3 phase voltages, but won't always. But checking for such possible imbalance would be an additional constraint on open delta services compared to single phase services, as jaggedben suggested.

Cheers, Wayne


Unloaded
A = (-120,0) ; B = (0, 207.8) ; C = (120,0)
AB = BC = AC = 240V

YK = 5V ; XK = 2.5V; XL = 5V
A = (-120,0) ; B = (0, 202.8) ; C = (112.8, 1.25)
AB = 235.6V ; BC = 233.1 ; AC = 232.8V

YK = 5V ; XK = 2.5V
A = (-120,0) ; B = (0, 202.8) ; C = (117.8, 1.25)
AB = 235.6V ; BC = 235.6V ; AC = 237.8V

YK = 5V ; XK = 2.5V ; XL = -5V (backfed)
A = (-120,0) ; B = (0, 202.8) ; C = (122.8, 1.25)
AB = 235.6V ; BC = 238.1V ; AC = 242.8V

 

[kwired]

It's commonly done here in California delta land, when they run out of regular spaces, for any straight 240 loads that might need power.

Where I work, the neighbor's shop has an AC connected on one of the high leg sides, and a small pump on a different one. I have no idea which is real and which is phantom.

It does not really matter anyway, because the main pot is 50kw, and the wing pot is 25 kw, so those small loads are not going to overload anything.

If you are connected to the "high leg" you are connected to the "phantom" leg. An occasional light load on it probably isn't going to hurt anything, that was kind of the intent when they designed it with a small pot though they mostly had limited three phase loads in mind when doing so. Also keep in mind you need a straight 240 volt rated breaker for any load that connects to the high leg, 3 pole breakers already are straight 240 rated so no problem with those, it's the two poles that you need to watch out with, and it has been said many times not to use single pole 208 to neutral anyway, but you not finding a single pole breaker rated for this unless it is a 277 volt breaker which likely doesn't fit the 120/240 panel design.

 

[jaggedben]

If you are connected to the "high leg" you are connected to the "phantom" leg. ...

He's using 'phantom' to mean the phase (not the leg) that doesn't have a winding. Whereas the 'real' phases are the ones with windings.

 

[kwired]

He's using 'phantom' to mean the phase (not the leg) that doesn't have a winding. Whereas the 'real' phases are the ones with windings.

 

[wwhitney]

Now just connect a 240V water heater across B-C, and model the open wye primary side.

So I think it is interesting to determine the coil currents (in magnitude and phase) from the various ways of linearly loading a high leg open delta. I'm going to express the phasors (sinewaves of a fixed period) in polar form X < Y, where X is the magnitude and Y is the phase angle in degrees (i.e. the sinewave f(x) = X * sin(2*pi*(t/T + Y/360)), where T is the fixed period, t is time).

For the high leg open delta as pictured in post #40 (with the undepicted neutral point N as the midpoint of A and C), and V_XY meaning the voltage from X to Y, i.e the voltage at Y minus the voltage at X, we have:

V_AB = 240V < 60
V_BC = 240V < -60 = 240V < 300
V_CN = V_NA = 120V < 180
V_CA = 240V < 180
V_NB = 208V < 90

If we apply an impedance Z across a voltage V_XY, we get a current through the impedance of I_yx = V_XY / Z. Here if Z is just a resistance, its magnitude divides the magnitude of V, and the angle of I matches the angle of V; if Z has a non-zero phase angle associated with it (it has a reactive component), then the phase angle of I is the phase angle of V minus the phase angle of Z. (This is just complex multiplication/division in polar form.)

Now if the voltage V_XY is supplied by a coil directly, the coil current I_XY = the load current I_yx. Otherwise the current I_yx travels through multiple coils to get from X to Y, and we add I_yx to each of those coil currents.

Then with those conventions, and for the case that the impedances are all resistances, we get (if I tracked all the angles correctly; the most likely error would be a sign error, equivalently having one of the angles 180 degrees off):

I_AB = |I_ba| < 60 + |I_cb| < 120 + |I_bn| < 90
I_CN = |I_nc| < 180 + |I_cb| < 120
I_NA = |I_an| < 180 + |I_cb| < 120 + |I_bn| < 90

Here |X| = magnitude of X, and so |X| < Y is just a way to explicitly specify the phase angle of X. The above just says that each coil sees the current from loads connected directly across it, plus the current from B-C connected loads, plus the current from N-B connected loads (except the C-N coil doesn't see N-B connected loads). Also, if the impedances aren't purely resistive, their phase angles would subtract from the associated fixed phase angle specified above, but that would make the formulas a lot messier.

In applying the above to a balanced 3-phase load, we need to consider it as 3 identical impedances delta connected A-B, B-C, and C-A. Likewise, the load current from a C-A connected load shows up in both I_nc and I_an.

Cheers, Wayne

 

[wwhitney]

So let's apply the above to a few simple examples.

1) Say we have a 3 phase (resistive) load with a line current of 50A, and we have single phase (resistive) loadings of 20A N-C and 40A A-C. What size transformers do we need?

The 3 phase line current is of magnitude sqrt(3) times the individual load currents of delta connected 2-wire loads, so the 3-phase load contributes 50/sqrt(3) A to each of I_ba, I_cb, I_nc and I_an. Thus we have:

|I_ba| = |I_cb| = 50/sqrt(3) = 28.9A
|I_nc| = 50/sqrt(3) + 40 + 20 = 88.9A
|I_an| = 50/sqrt(3) + 40 = 68.9A

Then the coil currents are given by:

I_AB = 28.9 < 60 + 28.9 < 120 = 28.9 * sqrt(3) < 90 = 50A < 90

I_CN = 88.9 < 180 + 28.9 < 120 = 60A < 180 + 50A < 150
|I_CN|= sqrt(60² + 50² + 2*60*50*cos 30) = 106.3A

I_NA = 68.9 < 180 + 28.9 < 120 = 40A < 180 + 50A < 150
|I_NA|= sqrt(40² + 50² + 2*40*50*cos 30) = 87.0A

The computations for |I_CN| and |I_NA| are done using the law of cosines.

So we need a "stinger" transformer (A-B) of 50A or 12 kVA, and a "lighting" transformer (A-N-C) of 106.3A or 25.5 kVA.

[BTW, the way that I recombined equal currents at 60 degrees apart to get the 3-phase 50A line current in each coil suggests that the previous solution may also have a nice formulation in terms of the 3 phase line currents directly.]

2) Now add a 40A/9.6kW load A-B to example 1.

Only I_AB changes, and now:

I_AB = 68.9 < 60 + 28.9 < 120 = 40A < 60 + 50A < 90
|I_AB| = 87.0A (same computation as |I_NA| above)

So the A-B transformer would need to be upsized to 87A or 20.9 kVA.

3) Instead add the 40A/9.6kW load to B-C.

Now all the coil currents change:

I_AB = 28.9 < 60 + 68.9 < 120 = 50A < 90 + 40A < 120
|I_AB| = 87.0A

I_NA = 68.9 < 180 + 68.9 < 120 = 68.9 * sqrt(3) < 150 = 119.3A < 150

I_CN = 88.9 < 180 + 68.9 < 120 = 20A < 180 + 119.3A < 150
|I_CN|= sqrt(20² + 119.3² + 2*20*119.3*cos 30) = 137.0A

So now the A-B transformer would need to be upsized to 87A / 20.9 kVA, and the A-C transformer would need to be upsized to 137A / 32.9 kVA.

4) Instead add a 9.6 kW load to B-N.

The load currents in Example 1 are unchanged, except we now have |I_nb| = 80/sqrt(3) = 46.2A. This changes I_AB and I_NA:

I_AB = 28.9 < 60 + 28.9 < 120 + 46.2 < 90 = 50A < 90 + 46.2 < 90 = 96.2A < 90

I_NA = 68.9 < 180 + 28.9 < 120 + 46.2 < 90 = (68.9 + 28.9 sin 30) < 180 + (46.2 + 28.9 cos 30) < 90 = 83.4A < 180 + 71.2A < 90
|I_NA| = sqrt(83.4² + 71.2²) = 109.6A

So now the A-B transformer needs to be upsized to 96.2A / 23.1 kVA, while the A-C transformer only needs to be upsized to 109.6A / 26.3 kVA.

Cheers, Wayne

 

[jaggedben]

Just some light reading for the holidays.

 

[jaggedben]

Seriously, I *may* try to understand that when I have some time.

 

[tortuga]

I_NA = 68.9 < 180 + 28.9 < 120 + 46.2 < 90 = (68.9 + 28.9 sin 30) < 180 + (46.2 + 28.9 cos 30) < 90 = 83.4A < 180 + 71.2A < 90
|I_NA| = sqrt(83.4² + 71.2²) = 109.6A

What does the < operator mean?
It would be quite useful if this was in a excel or libreoffice file where one could input combinations of delta and open delta type loads.

 

[synchro]

What does the < operator mean?

There's not a key on a standard keyboard character for it, but I'm sure Wayne meant that < would represent the angle, ( ∠ using the 2220 Unicode symbol) and the number following it would be in degrees.

 

[wwhitney]

What does the < operator mean?

Sentence two of the first post today:

"I'm going to express the phasors (sinewaves of a fixed period) in polar form X < Y, where X is the magnitude and Y is the phase angle in degrees."

So in Cartesian coordinates, X < Y = (X cos Y, X sin Y). Which I used an analogue of in the part of my post that you quoted, except that the basis I used was 1 < 90 and 1 < 180.

Cheers, Wayne

 

[electrofelon]

With an open delta, hopefully people aren't powering 2-wire loads from the "phantom" winding.

Cheers, Wayne

Wayne, I know you are not an electrician so maybe you havnt thought about it, but how do you think we can tell which phases make up the phantom winding (genuine question, not being snarky)? That is not a thing that is normally coordinated or marked on equipment. Frequently we dont even know if its open or closed delta.