Transformer Overloading of Center-Tapped Delta Bank Supplying Both 3ph and 1ph Power

[kenth619]

Good day,


I have a complex problem involving a three phase bank of three single phase transformers. The phase voltage on the star primary side is 6.9kV (12kV line-line) and the phase voltage on the secondary center tapped delta is 115 / 230 V using a 4 wire system. The center tap is between the red and white phases.


The bank consists of 2 power transformers and 1 lighting transformer. The lighting transformer kVA rating is less than or equal to 2 times the rating of the power transformer. The possible combination of transformers are 25 kVa, 37.5 kVA, 50 kVA and 75 kVA. For example a bank can have one 50kVA transformer (lighting TF) as the max size with two 25kVA power transformers.


I can work out what each transformer coil is rated at (ie 50kVA / 230, 25kVA / 230). How do I calculate the % current overloading in terms of of each transformer's coil (phase) current given the line current output of the delta secondary and knowing that the phase voltage is 230V (115 / 115 V at the center tap)? This would of course involve calculating the phase currents in the delta secondary and comparing each phase current with that transformer's rated coil current. But what complicates the problem is the fact that the bank is used to supply single phase loads as well.


(1)The three phase load is shared equally among the 3 TFs i.e. 1/3:1/3:1/3, (2)the single phase load 230 V is shared in the ratio 1/3:2/3:1/3 and (3)the single phase load 115V is shared in the ratio 1/6:5/6:1/6.
Points 2 & 3 do not add up to total of 1. In the case of Pt. 2 it is meant to signify that the center TF lighting load is twice that of the other two. In the case of Pt. 3 the center TF single phase (lighting load) is 5 times the other two.
I have included a diagram to help explain. The center TF is the lighting TF.

Normally one would assume that all of the single phase 115V load is being supplied only by the center transformer. However, I mentioned earlier that the secondary windings of the three single phase transformers are connected in a center-tapped DELTA configuration. Wouldn't this mean that the currents flowing through the coils of the power transformers be drawn by the center transformer in supplying the single phase 115V load because this is delta? Correct me if I am wrong but that is how I envisioned it.

The current readings are taken from the secondary side of the transformer as illustrated below.

The aim is to find delta phase currents I_RB, I_BW and I_RW using measured line current values I_R, I_W and I_B for all combinations of transformer sizes to determine if these phase currents exceed the rated coil current of the individual single phase transformers (based on rated kVA and secondary phase voltage 230V) to determine if each transformer is being overloaded.

 

[mivey]

Normally one would assume that all of the single phase 115V load is being supplied only by the center transformer.

For an open wye - open delta, the single phase load kVA is only in the lighting pot but not so with a closed delta.

The kVA in the non-lighting pots is:
1/3*sqrt[3phkVA^2 + 1phkVA^2 - 2*3phkVA*1phkVA*COS(120 +/- (3phpowerfactorangle - 1phpowerfactorangle))]

The kVA in the lighting pot is:
1/3*sqrt[3phkVA^2 + 4*1phkVA^2 + 4*3phkVA*1phkVA*COS(3phpowerfactorangle - 1phpowerfactorangle)]

 

[kenth619]

For an open wye - open delta, the single phase load kVA is only in the lighting pot but not so with a closed delta.

The kVA in the non-lighting pots is:
1/3*sqrt[3phkVA^2 + 1phkVA^2 - 2*3phkVA*1phkVA*COS(120 +/- (3phpowerfactorangle - 1phpowerfactorangle))]

The kVA in the lighting pot is:
1/3*sqrt[3phkVA^2 + 4*1phkVA^2 + 4*3phkVA*1phkVA*COS(3phpowerfactorangle - 1phpowerfactorangle)]

Thank you for the reply. I assume the single phase power and phase angle are for the non-lighting or lighting transformer depending on which equation you are using. But what is the 3 ph kVA and Phase angle? How do we calculate these values?

 

[bob]

For an open wye - open delta, the single phase load kVA is only in the lighting pot but not so with a closed delta.

The kVA in the non-lighting pots is:
1/3*sqrt[3phkVA^2 + 1phkVA^2 - 2*3phkVA*1phkVA*COS(120 +/- (3phpowerfactorangle - 1phpowerfactorangle))]

The kVA in the lighting pot is:
1/3*sqrt[3phkVA^2 + 4*1phkVA^2 + 4*3phkVA*1phkVA*COS(3phpowerfactorangle - 1phpowerfactorangle)]

Is this close to your post? (1)The three phase load is shared equally among the 3 TFs i.e. 1/3:1/3:1/3, (2)the single phase load 230 V is shared in the ratio 1/3:2/3:1/3

 

[mivey]

Thank you for the reply. I assume the single phase power and phase angle are for the non-lighting or lighting transformer depending on which equation you are using. But what is the 3 ph kVA and Phase angle? How do we calculate these values?

1phkVA and 3phkVA are the load kVAs for the single-phase and three-phase loads. The angles are the phase angles of the single-phase and three-phase loads

 

[mivey]

Is this close to your post? (1)The three phase load is shared equally among the 3 TFs i.e. 1/3:1/3:1/3, (2)the single phase load 230 V is shared in the ratio 1/3:2/3:1/3

Plug in unity power factors and it will reduce to that.

 

[mivey]

Plug in unity power factors and it will reduce to that.

Actually just plugging in zero for the other load reduces it.

 

[mivey]

How do we calculate these values?

Start by extracting the three-phase current. From the remainder, extract the full-winding single-phase current. The remainder is the half-winding single-phase current. Then apply the 1/3:1/3:1/3 for the three-phase, the 1/3:2/3:1/3 for the full-winding single-phase, and the 1/6:5/6:1/6 for the half-winding single-phase.

FWIW, the single-phase currents split due to parallel impedances (say for a full winding impedance Z). For the full winding load you have Z + Z in parallel with Z so you get 2Z/3Z and 1Z/3Z current division. For the half winding load you have Z + Z + Z/2 in parallel with Z/2 so you get 2.5Z/3Z and 0.5Z/3Z current division (the 5/6 and 1/6 currents).

 

[Smart $]

... For the half winding load you have Z + Z + Z/2 in parallel with Z/2 so you get 2.5Z/3Z and 0.5Z/3Z current division (the 5/6 and 1/6 currents).

When all loads are combined, this last part only applies to the load imbalance on the split phase winding... as the balanced load reduces to 1/3:2/3:1/3 line-to-line load.

@kenth619,

Do not discount the possibility of single phase line-to-line loads connected across to the power winding terminals. They have the same 1Z/2Z relationship with all three windings, but respective to their connections, as the single phase loads connected to the lighting winding terminals.

 

[kenth619]

Thank you mivey, Smart $ and bob for taking the time to reply to my post. I will attempt to work out the solution and post my results here later.

For an open wye - open delta, the single phase load kVA is only in the lighting pot but not so with a closed delta.

The kVA in the non-lighting pots is:
1/3*sqrt[3phkVA^2 + 1phkVA^2 - 2*3phkVA*1phkVA*COS(120 +/- (3phpowerfactorangle - 1phpowerfactorangle))]

The kVA in the lighting pot is:
1/3*sqrt[3phkVA^2 + 4*1phkVA^2 + 4*3phkVA*1phkVA*COS(3phpowerfactorangle - 1phpowerfactorangle)]

mivey, do you have a source for that equation (book or paper)?

 

[kenth619]

Lets see if I understand this so far...

First I use the equations:
kVA for Power Transformers:
1/3*sqrt[3phkVA^2 + 1phkVA^2 - 2*3phkVA*1phkVA*COS(120 +/- (3phpowerfactorangle - 1phpowerfactorangle))]

kVA for Lighting Transformer:
1/3*sqrt[3phkVA^2 + 4*1phkVA^2 + 4*3phkVA*1phkVA*COS(3phpowerfactorangle - 1phpowerfactorangle)]

To work out the 3ph kVA and 1ph kVA in the power transformers and lighting transformers (set 1phKVA to 0 and leave out 1phPF to find 3ph values and vice versa to find 1ph values) having prior knowledge of the 3ph and 1ph kVA and power factors of the load being supplied to the customers.

This will give two kVA values for the transformers to do separate 3ph and 1ph analysis of currents.

Start by extracting the three-phase current. From the remainder, extract the full-winding single-phase current. The remainder is the half-winding single-phase current. Then apply the 1/3:1/3:1/3 for the three-phase, the 1/3:2/3:1/3 for the full-winding single-phase, and the 1/6:5/6:1/6 for the half-winding single-phase.

FWIW, the single-phase currents split due to parallel impedances (say for a full winding impedance Z). For the full winding load you have Z + Z in parallel with Z so you get 2Z/3Z and 1Z/3Z current division. For the half winding load you have Z + Z + Z/2 in parallel with Z/2 so you get 2.5Z/3Z and 0.5Z/3Z current division (the 5/6 and 1/6 currents).

I follow everything you say about the impedance ratios and how the current is split for the single phase analysis. But how do I go about extracting the three-phase currents? And from what remainder is the full-winding and half-winding single-phase current? I didn't quite understand that part.

 

[mivey]

When all loads are combined, this last part only applies to the load imbalance on the split phase winding

Yes, what I called half-winding single-phase.

Do not discount the possibility of single phase line-to-line loads connected across to the power winding terminals. They have the same 1Z/2Z relationship with all three windings, but respective to their connections, as the single phase loads connected to the lighting winding terminals.

It occurred to me after my initial post that this is a homework question and simplifying assumptions are in play. One would be that single-phase loads are on the lighting pot.

 

[mivey]

mivey, do you have a source for that equation (book or paper)?

It is from a Westinghouse handbook but forget it, your problem is much less complicated.

 

[mivey]

Lets see if I understand this so far...

First I use the equations:

Forget that as it is beyond the scope of what it appears you are doing.

I follow everything you say about the impedance ratios and how the current is split for the single phase analysis. But how do I go about extracting the three-phase currents? And from what remainder is the full-winding and half-winding single-phase current? I didn't quite understand that part.

Because of the way the exercise is set up, we know that Ib>Iw and Ib>Ir. We have either Ir>Iw or Iw>Ir. Let's suppose Ir>Iw. We then have Ib>Ir>Iw.

Ib is the three-phase load current. Subtract Ib from Iw to get Iw'. Iw' is the full-winding single-phase load current. Subtract Iw' from Ir to get Ir'. Ir' is the half-winding single-phase load current.

Do you understand why Ib>Ir>Iw?

I also assume you have no trouble with the allowable combinations of transformers so you should be able to post a set of current ranges that are valid for all combinations.

Work on that and let's see what you get.

 

[mivey]

I also assume you have no trouble with the allowable combinations of transformers so you should be able to post a set of current ranges that are valid for all combinations.

Meaning that each transformer combination will have a set of non-overloading current ranges. It would be useful if you wrote the winding currents in the form of equations with variables something like T, F, H for 3-phase, full-winding and half-winding loads. Don't forget sqrt(3).

Also, rated current is rated current. If half of the winding carries less than rated current that does not mean the other half can carry more than rated current.

 

[kenth619]

Yes, what I called half-winding single phase.
It occurred to me after my initial post that this is a homework question and simplifying assumptions are in play. One would be that single-phase loads are on the lighting pot.

Just to clarify certain points

Please don't view this as a homework question, but rather as a discussion; it is a problem facing the distribution section of the local utility company where I live and I am doing this project with them. I am told this is a problem other utilities face as well and it would be interesting to know how they deal with this problem.

3 phase 230V loads are fed by red, white and blue (all) phases. 230V 1 ph loads can be fed by all line-to-line connections (red-white, white-blue, blue-red). This is because WRT the earthed neutral phase-neutral voltages are 115V, 199.2V and 115V red, white and blue respectively (remember this is a center-tapped delta). If the delta is balanced these voltages will be 0, 90 and 180 degrees apart. 115V 1ph can only be supplied by red-neutral and white-neutral. In a center tap delta V line-line is not equal to V phase-neutral.

Another important point to note is that information about customer loading is only available in the form of total connected loads (ratings) and not instantaneous demand at the point of measuring line currents.So it is difficult to use the equation previous given to calculate kVA of the transformers because instantaneous loading of the customer is not available. The only information available is amp clamp measured output line currents (and neutral current if needed).

The aim really is to determine the line-line currents (notice I did not say phase-neutral currents to avoid confusion) in the delta and then compare these values to the coil current rating of the individual 1ph transformers (kVA rating / 230V voltage rating) to calculate % overloading at the time of taking the line current readings.

I have a hypothesis. I am of the view that regardless of single phase loading, the phase-neutral currents (notice I did not say line-line) in the delta must be 0, 90 and 180 degrees apart WRT neutral for balanced delta operation because of the presence of the center-tapped grounded neutral. And from these phase-neutral currents we can determine the line-line currents.

I smell a trigonometry solution to this problem unless my hypothesis is wrong of course. Something similar to what was done in this post but for a center-tapped delta.

 

[GoldDigger]

Do you expect to see a significant amount of high leg to neutral (208V) loads, or just 240V single and three phase loads?
Also, in a closed delta half-phase 120V loads are supplied both by the half winding and the parallel contribution of the other path around the loop.

 

[mivey]

Just to clarify certain points

Please don't view this as a homework question, but rather as a discussion;

OK. It just seemed like things were getting too simplified and in light of the rest of your post: it is.

it is a problem facing the distribution section of the local utility company where I live and I am doing this project with them. I am told this is a problem other utilities face as well and it would be interesting to know how they deal with this problem.

You have to do a better load analysis or take some real readings. Except for some very consistent loads, you are not going to make an accurate analysis. The best you can do is assume a power factor, assume some worst-case winding loads based on the amps, and read the different service drops to get an idea about how to divvy up the amps.

3 phase 230V loads are fed by red, white and blue (all) phases. 230V 1 ph loads can be fed by all line-to-line connections (red-white, white-blue, blue-red). This is because WRT the earthed neutral phase-neutral voltages are 115V, 199.2V and 115V red, white and blue respectively (remember this is a center-tapped delta). If the delta is balanced these voltages will be 0, 90 and 180 degrees apart. 115V 1ph can only be supplied by red-neutral and white-neutral. In a center tap delta V line-line is not equal to V phase-neutral.

That is red, blue, and white respectively.

Another important point to note is that information about customer loading is only available in the form of total connected loads (ratings) and not instantaneous demand at the point of measuring line currents.

Without load data your guess is going to be pretty wild. You need to re-think your measurement method or change your objective. If you just want to be reasonably sure the transformer is not overloaded, then you have a chance. Trying to calculate a real percentage is out with just clamp amps at the transformer.

So it is difficult to use the equation previous given to calculate kVA of the transformers because instantaneous loading of the customer is not available.

Then you have to make some intelligent estimates of the load.

The only information available is amp clamp measured output line currents (and neutral current if needed).

Then you might be able to say it probably is overloaded or probably isn't overloaded. A real percentage is out.

The aim really is to determine the line-line currents (notice I did not say phase-neutral currents to avoid confusion) in the delta and then compare these values to the coil current rating of the individual 1ph transformers (kVA rating / 230V voltage rating) to calculate % overloading at the time of taking the line current readings.

Who are you working with at the utility? The POCO engineer should know better than that.

I have a hypothesis. I am of the view that regardless of single phase loading, the phase-neutral currents (notice I did not say line-line) in the delta must be 0, 90 and 180 degrees apart WRT neutral for balanced delta operation because of the presence of the center-tapped grounded neutral.

Not unless you have all resistive load. Think about it. If it don't click, ask and I'll elaborate.

I smell a trigonometry solution to this problem unless my hypothesis is wrong of course.

It is.

 

[mivey]

You need to re-think your measurement method or change your objective.

My suggestion would just be to change the measurement methodology. It is pretty simple for a lineman to get in the bucket and take a reading of the winding currents directly. On the non-lighting pot, either X1 or X3 would give you the winding current and you could calculate the percent overload as a percent of rated current.

For the lighting pot, you need to read both X1 and X3 then make your overload calculation for each winding half.

Any particular reason you want to use the readings from the secondary cable? If you must use these readings, perhaps you could change the objective to flag a transformer for a direct winding reading if the currents are possibly overloading the transformer.

Here is an example of some transformer and secondary cable readings with a mix of single-phase and three-phase loads where the single-phase loads are not constrained to the lighting pot on the delta:

Secondary cable currents towards the bank (I included the angles even though the clamp-on just gives the magnitudes):
A = 260.4@94.23D = TA_X1 + TC_X3
B = 161.5@-32.13D = TB_X1 + TA_X3
C = 241.4@-132.42D = TC_X1 + TB_X3
N = 45.5@5.54D


Now compare that to what you get if you read the winding currents directly (currents into the bushings):
TA_X1 = 126.2@107.68D

TB_X1 = 104.2@19.27D
TB_X2 = 45.5@5.54D
TB_X3 = 148.8@-164.89D

TC_X1 = 140.7@-97.82D

 

[kenth619]

My suggestion would just be to change the measurement methodology. It is pretty simple for a lineman to get in the bucket and take a reading of the winding currents directly. On the non-lighting pot, either X1 or X3 would give you the winding current and you could calculate the percent overload as a percent of rated current.

For the lighting pot, you need to read both X1 and X3 then make your overload calculation for each winding half.

Any particular reason you want to use the readings from the secondary cable? If you must use these readings, perhaps you could change the objective to flag a transformer for a direct winding reading if the currents are possibly overloading the transformer.

Here is an example of some transformer and secondary cable readings with a mix of single-phase and three-phase loads where the single-phase loads are not constrained to the lighting pot on the delta:

Secondary cable currents towards the bank (I included the angles even though the clamp-on just gives the magnitudes):
A = 260.4@94.23D = TA_X1 + TC_X3
B = 161.5@-32.13D = TB_X1 + TA_X3
C = 241.4@-132.42D = TC_X1 + TB_X3
N = 45.5@5.54D


Now compare that to what you get if you read the winding currents directly (currents into the bushings):
TA_X1 = 126.2@107.68D

TB_X1 = 104.2@19.27D
TB_X2 = 45.5@5.54D
TB_X3 = 148.8@-164.89D

TC_X1 = 140.7@-97.82D

Really good answer. I think this could be the solution as everything you said makes sense looking at the coil connections. I will get back to you if not tomorrow, the following day to let you know how this goes once I speak with the POCO Engineer.

One more question though. What is the correct way to calculate the half winding current rating? Should it be (full winding kVA rating/2)/115V?