How to calculate if one transformer of a three phase Star/Delta bank is overloaded?

topgone

Senior Member
Here's another proof of the math being used. Please try ang get yourself a copy of the book by John J. Winders, Jr. of PPL Electric Utilities entitled "Power Transformers and Their Applications". LINK

On page 96:
"Define the following ratios of impedances, w and x, as follows. (Note: In general, these ratios are complex numbers because the impedances are complex numbers.)
w = Zb/Za → Zb  = wZa (3.10.1)
x = Zc/Za → Zc  = xZa (3.10.2)"​

On page 97:
"The voltage drop around the closed Δ path must equal zero:
iaZa + ibZb +  icZc  = iaZa +  ibwZa +  icxZa = 0 (3.10.6)
ia +  w ib + x ic  = 0 (3.10.7)"​
 

Ingenieur

Senior Member
Location
Earth
Here's another proof of the math being used. Please try ang get yourself a copy of the book by John J. Winders, Jr. of PPL Electric Utilities entitled "Power Transformers and Their Applications". LINK

On page 96:
"Define the following ratios of impedances, w and x, as follows. (Note: In general, these ratios are complex numbers because the impedances are complex numbers.)
w = Zb/Za → Zb = wZa (3.10.1)
x = Zc/Za → Zc = xZa (3.10.2)"​

On page 97:
"The voltage drop around the closed Δ path must equal zero:
iaZa + ibZb + icZc = iaZa + ibwZa + icxZa = 0 (3.10.6)
ia + w ib + x ic = 0 (3.10.7)"​

No one is argueing kvl, obviously lol
but Ia x Z = 240 is the basis of your loop (similar for b and c)
crunch the numbers
what is Za in ohms?

you are ignoring the load and only using the xfmr Z
it's the load that determines the line current (and phase) not the xfmr Z
except for a short ckt bolted fault

to do what you propose you have 2 delta in parallel
Xfmr Z
Load Z
combine into one delta
then do your method
But we don't know the load
 
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topgone

Senior Member
No one is argueing kvl, obviously lol
but Ia x Z = 240 is the basis of your loop (similar for b and c)
crunch the numbers
what is Za in ohms?

you are ignoring the load and only using the xfmr Z
it's the load that determines the line current (and phase) not the xfmr Z
except for a short ckt bolted fault

to do what you propose you have 2 delta in parallel
Xfmr Z
Load Z
combine into one delta
then do your method
But we don't know the load
IDK, but that's how iterations are done. You make a guess and then recompute for the differences between your guess and computed new value! Punch back the computed values in, until you arrive at a convergence!:D
I do have a question for you. Can you invert the matrix I posted above?
 

Ingenieur

Senior Member
Location
Earth
IDK, but that's how iterations are done. You make a guess and then recompute for the differences between your guess and computed new value! Punch back the computed values in, until you arrive at a convergence!:D
I do have a question for you. Can you invert the matrix I posted above?
yes, I know how they work
had a course on them last term
implementing on a computer for power grids

you mean the one in you post?
it appears yes
but it is not representative of the system

do this please
your kvl is the sum of Vabc
where Va = Za x Ia, similar for b and c

what is Za in ohms?
what is Va/Za = Ia in amperes? Assume zero ph and 240 vac, valid since we can make any phase the base

what is Ia?
you will see it is the short ckt current
current
50kva / (240 1.73 0.02) x 2 / 0.866 = 13.8 kA

or 240 / 0.0174 = 13.8 kA
Where 0.0174 ohm = 50000/(240^2) x 0.02 or act Z = base Z x puZ
adjusted for other phases contribution and delta-wye phase to line
 

topgone

Senior Member
yes, I know how they work
had a course on them last term
implementing on a computer for power grids

you mean the one in you post?
it appears yes
but it is not representative of the system

do this please
your kvl is the sum of Vabc
where Va = Za x Ia, similar for b and c

what is Za in ohms?
what is Va/Za = Ia in amperes? Assume zero ph and 240 vac, valid since we can make any phase the base

what is Ia?
you will see it is the short ckt current
current
50kva / (240 1.73 0.02) x 2 / 0.866 = 13.8 kA

or 240 / 0.0174 = 13.8 kA
Where 0.0174 ohm = 50000/(240^2) x 0.02 or act Z = base Z x puZ
adjusted for other phases contribution and delta-wye phase to line
Now, that's revealing something! Skirting the issue doesn't help here.
BTW, let's go back to school.
Here's what you failed to consider/know:
Voltage out of each phase equals the internal transformed values minus the voltage drops on the coils. The correct assumption is that those voltage are displaced 120 degrees apart. Lemme show why Mr. Winders' equation is correct:
Vab = V1 - ia.Za
Vbc = V2 - ib.Zb, and
Vca = V3 - ic.Zc​
[V1] = [V2] = [V3], all displaced by 120 degrees from each other

Taking the KVL around the loop:
(240/0 - ia.Za) + (240/-120 - ib.Zb) + (240/120 - ic.Zc) = 0

Since the voltages are of the same magnitude, the only differences are their respective angles, when summed up, they zero out!
(240/0 + 240/-120 + 240/120) - ia.Za - ib.Zb - ic.Zc = 0 -----> 0 -ia.Za - ib.Zb - ic.Zc = 0
So, you are left with:
-ia.Za - ib.Zb - ic.Zc = 0​
substituting the ratios defined earlier as w and x:
-ia.Za - w.ib.Za - x.ic.Za = 0​
Dividing by -Za:
ia + w.ib + x.ic = 0. -----> IMHO, this equation defines the relationship of the phase currents inside the transformer bank!​
 

wwhitney

Senior Member
Location
Berkeley, CA
You missed a lot. If you only backread, the formulation of the second equation is not Ia + Ib + Ic =0.
OK, I think I've caught up. Here's my understanding:

1) KCL gives us the equations discussed in my first post today.
2) Those equations alone have an infinite number of solutions, with no bound on the magnitudes. We could identify one unique solution (Ia0, Ib0, Ic0) by temporarily assuming Ia + Ib + Ic = 0, Ia is real, and assuming Im Ib > 0. [I'll compute that below]. Then up to choice of phase angle and CW/CCW rotation, all solutions will be of the form Ia = Ia0 + Id ; Ib = Ib0 + Id ; Ic = Ic0 + Id for some complex number Id.
3) With one additional equation, such as some information on the load or coil impedances in the system, we can then find Id and a unique solution.

I don't understand the earlier comments that one of the measured currents must be wrong. If you add an equation of the form Ia + w Ib + x Ic = 0 for known complex numbers w and x, then you should be able to solve for a unique solution. In my terminology:

Ia0 + Id + w Ib0 + w Id + x Ic0 + x Id = 0 gives
Id = -(Ia0 + w Ib0 + x Ic0) / (1 + w + x).

If people are interested, I could compute (Ia0, Ib0, Ic0) by, e.g. the law of cosines.

Cheers, Wayne
 

topgone

Senior Member
OK, I think I've caught up. Here's my understanding:

1) KCL gives us the equations discussed in my first post today.
2) Those equations alone have an infinite number of solutions, with no bound on the magnitudes. We could identify one unique solution (Ia0, Ib0, Ic0) by temporarily assuming Ia + Ib + Ic = 0, Ia is real, and assuming Im Ib > 0. [I'll compute that below]. Then up to choice of phase angle and CW/CCW rotation, all solutions will be of the form Ia = Ia0 + Id ; Ib = Ib0 + Id ; Ic = Ic0 + Id for some complex number Id.
3) With one additional equation, such as some information on the load or coil impedances in the system, we can then find Id and a unique solution.

I don't understand the earlier comments that one of the measured currents must be wrong. If you add an equation of the form Ia + w Ib + x Ic = 0 for known complex numbers w and x, then you should be able to solve for a unique solution. In my terminology:

Ia0 + Id + w Ib0 + w Id + x Ic0 + x Id = 0 gives
Id = -(Ia0 + w Ib0 + x Ic0) / (1 + w + x).

If people are interested, I could compute (Ia0, Ib0, Ic0) by, e.g. the law of cosines.

Cheers, Wayne
Please do. I want to compare your figures with those results I posted a few posts back.
I have done it using the complex matrix inversion and then matrix multiplication!
 

Sahib

Senior Member
Location
India
topgone:
Why further argument to prove you are right? You have got phase currents and line currents per your calculations. They should satisfy KCL at each node of the delta. Just show it is so to quieten the doubters (me included).:)

Please use Kirchoff current law to verify the result.
 

wwhitney

Senior Member
Location
Berkeley, CA
Please do. I want to compare your figures with those results I posted a few posts back.
I have done it using the complex matrix inversion and then matrix multiplication!
I'm going to bed now but I'll do it in the morning. The arbitrary choice of (Ia0, Ib0, Ic0) I made won't match your solution. But if you remind me what complex values of w and x you used, then I can solve for Ia, Ib, and Ic and compare results.

Cheers, Wayne
 

topgone

Senior Member
I'm going to bed now but I'll do it in the morning. The arbitrary choice of (Ia0, Ib0, Ic0) I made won't match your solution. But if you remind me what complex values of w and x you used, then I can solve for Ia, Ib, and Ic and compare results.

Cheers, Wayne
As you wish:
TRANSFORMER - TRAFO IMPEDANCE
TR. "A" - 0.012672+j0.019584 ohms
TR. "B" - 0.012672+j0.019584 ohms
TR. "C" - 0.006912+j0.012288 ohms

w = 1 + J0
X = 0.603252032520325+j0.0373983739837398 ohms
 

wwhitney

Senior Member
Location
Berkeley, CA
To begin at the beginning, the delta coil currents are waveforms which may be represented as complex numbers Ia, Ib, Ic, assuming that they all are pure sinewaves of the same frequency. I'll refer to a triple of complex numbers as a solution or just I, I0, I1, etc., with the understanding that I = (Ia, Ib, Ic).

The measured magnitudes of the line currents impose three real constraints on I by KCL:

|Ia - Ib| = 256
|Ib - Ic| = 294
|Ic - Ia| = 341.

So every solution I forms a triangle in the complex plane with side lengths 256, 294, 341. There is exactly one such solution where furthermore Ia + Ib + Ic = 0, Ia is real, and the imaginary part of Ib is positive. Call that solution I0.

To construct I0, I find it easiest to construction a couple intermediate solutions. So let I1 be the solution where Ic1 = 0, Ia1 is real and positive (and thus Ia1=341), and Ib1 has positive imaginary part. That is, our triangle has the side of length 341 on the real axis and points up. We know further that |Ib1| = 294, so we just need to constrain the polar angle of Ib1, call that Theta1. The law of cosines tells us that

256^2 = 294^2 + 341^2 - 2*294*341*cos(Theta1)

Solving for cos(Theta1) gives cos(Theta1) = 0.684, which means sin(Theta1) = 0.729. Thus Ib1 = 294 * (0.684 + 0.729i) = 201.1 + 214.4i

That means our first intermediate solution is I1 = (341, 201.1 + 214.4i, 0). This obviously doesn't have the property that Ia + Ib + Ic = 0, but that is easily fixed by subtracting (Ia+Ib+Ic)/3 = 180.7 + 71.5i from each value. So our second intermediate solution is I2 = (160.3 - 71.5i, 20.4 + 142.9i, -180.7 - 71.5i).

This is almost the desired solution, except Ia is no longer real. This can be remedied by rotating the solution I2. In polar form, Ia2 = 175.5 * e^(-0.41955 i). So we can multiple the solution I2 by e^(0.41955 i) to get I0, yielding I0 = (175.5, -39.6 + 138.8i, -135.9 - 138.9i). Here you can see that I've accumulated an error of 0.1i by rounding.

Cheers, Wayne
 

wwhitney

Senior Member
Location
Berkeley, CA
w = 1 + J0
X = 0.603252032520325+j0.0373983739837398 ohms
Great, thanks. I thought that w would be 1 since two of the transformers are the same.

A quick question--suppose all the transformers were the same. Then the KVL constraint you have would reduce to Ia + Ib + Ic = 0, so the solution should match the I0 solution I just computed. Is that right? That means that in a delta bank with identical coils, if there are no harmonics, you can't have any circulating current, regardless of the characteristics of the load?

As commented previously, any solution of the KCL constraints (up to a rotation by a factor of e^i Theta) with CCW phase rotation will be of the form I = I0 + (Id, Id, Id) for some complex number Id. If we impose a KVL constraint of the form Ia + wIb + xIc = 0, then we can find a unique (up to rotation) solution by solving for Id:

Id = -(Ia0 + w Ib0 + x Ic0) / (1 + w + x).

For I0 = (175.5, -39.6 + 138.8i, -135.9 - 138.9i), w = 1 and x = 0.60325 + 0.037398i, this gives Id = -23.0 - 18.8i. So the solution up to rotation would be I = (152.5 - 18.8i, -62.6 + 120.0i, -158.9 - 157.7i).

As previously, we can make Ia real by the appropriate rotation (multiplying by e^(0.12266i)) to get I = (153.7, -76.8 + 111.4i, -138.4 - 176.0i). And I double checked that this satisfies the original KCL constraints as well as the new KVL constraint for the assumed values of w and x, at least up to rounding error.

Cheers, Wayne
 

Ingenieur

Senior Member
Location
Earth
Now, that's revealing something! Skirting the issue doesn't help here.
BTW, let's go back to school.
Here's what you failed to consider/know:
Voltage out of each phase equals the internal transformed values minus the voltage drops on the coils. The correct assumption is that those voltage are displaced 120 degrees apart. Lemme show why Mr. Winders' equation is correct:
Vab = V1 - ia.Za
Vbc = V2 - ib.Zb, and
Vca = V3 - ic.Zc​
[V1] = [V2] = [V3], all displaced by 120 degrees from each other

Taking the KVL around the loop:
(240/0 - ia.Za) + (240/-120 - ib.Zb) + (240/120 - ic.Zc) = 0

Since the voltages are of the same magnitude, the only differences are their respective angles, when summed up, they zero out!
(240/0 + 240/-120 + 240/120) - ia.Za - ib.Zb - ic.Zc = 0 -----> 0 -ia.Za - ib.Zb - ic.Zc = 0
So, you are left with:
-ia.Za - ib.Zb - ic.Zc = 0​
substituting the ratios defined earlier as w and x:
-ia.Za - w.ib.Za - x.ic.Za = 0​
Dividing by -Za:
ia + w.ib + x.ic = 0. -----> IMHO, this equation defines the relationship of the phase currents inside the transformer bank!​

Talk about skirting the issue lol
what is Va vac
what is Za Ohms
what is Ia = Va/Za

the formula is flawed
that is fault current
not load current
 

Ingenieur

Senior Member
Location
Earth
Please do. I want to compare your figures with those results I posted a few posts back.
I have done it using the complex matrix inversion and then matrix multiplication!
It is incorrect
can't be solved without
some phase information ... We have none
or phase assumptions...results in a forced solution
 

Ingenieur

Senior Member
Location
Earth
Now, that's revealing something! Skirting the issue doesn't help here.
BTW, let's go back to school.
Here's what you failed to consider/know:
Voltage out of each phase equals the internal transformed values minus the voltage drops on the coils. The correct assumption is that those voltage are displaced 120 degrees apart. Lemme show why Mr. Winders' equation is correct:
Vab = V1 - ia.Za
Vbc = V2 - ib.Zb, and
Vca = V3 - ic.Zc​
[V1] = [V2] = [V3], all displaced by 120 degrees from each other (that is an assumption)

Taking the KVL around the loop:
(240/0 - ia.Za) + (240/-120 - ib.Zb) + (240/120 - ic.Zc) = 0

Since the voltages are of the same magnitude, the only differences are their respective angles, when summed up, they zero out!
(240/0 + 240/-120 + 240/120) - ia.Za - ib.Zb - ic.Zc = 0 -----> 0 -ia.Za - ib.Zb - ic.Zc = 0
So, you are left with:
-ia.Za - ib.Zb - ic.Zc = 0 This assumes that the currents are balanced
substituting the ratios defined earlier as w and x:
-ia.Za - w.ib.Za - x.ic.Za = 0​
Dividing by -Za:
ia + w.ib + x.ic = 0. -----> IMHO, this equation defines the relationship of the phase currents inside the transformer bank!​

-ia.Za - ib.Zb - ic.Zc = 0
not valid
assume Za=Zb=Zc and PU 1, not unreasonable
then ia + ib + ic = 0
not valid, you are assuming the phase currents are balanced
they are not, the net unbalance will circulate in the delta, and will not be zero if the lines currents are not balanced
 
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Ingenieur

Senior Member
Location
Earth
As you wish:
TRANSFORMER - TRAFO IMPEDANCE
TR. "A" - 0.012672+j0.019584 ohms
TR. "B" - 0.012672+j0.019584 ohms
TR. "C" - 0.006912+j0.012288 ohms

w = 1 + J0
X = 0.603252032520325+j0.0373983739837398 ohms
let's look at xfmr A
coil Z = 0.0127 Ohm (rounded)
that means the current per your equations is Ia = 240/0.0127 = 18.93 kA, similar magnitude when I assumed a PU of 2%

-ia.Za - ib.Zb - ic.Zc = 0
ie, the Vdrop across each coil summed around the delta equals zero

so the drop per coil is ix,Zx wher x = a,b,c
the current thru a coil is NOT 18,94 kA
unless shorted as in bolted fault
and with NO load
 

Ingenieur

Senior Member
Location
Earth
Great, thanks. I thought that w would be 1 since two of the transformers are the same.

A quick question--suppose all the transformers were the same. Then the KVL constraint you have would reduce to Ia + Ib + Ic = 0, so the solution should match the I0 solution I just computed. Is that right? That means that in a delta bank with identical coils, if there are no harmonics, you can't have any circulating current, regardless of the characteristics of the load?

As commented previously, any solution of the KCL constraints (up to a rotation by a factor of e^i Theta) with CCW phase rotation will be of the form I = I0 + (Id, Id, Id) for some complex number Id. If we impose a KVL constraint of the form Ia + wIb + xIc = 0, then we can find a unique (up to rotation) solution by solving for Id:

Id = -(Ia0 + w Ib0 + x Ic0) / (1 + w + x).

For I0 = (175.5, -39.6 + 138.8i, -135.9 - 138.9i), w = 1 and x = 0.60325 + 0.037398i, this gives Id = -23.0 - 18.8i. So the solution up to rotation would be I = (152.5 - 18.8i, -62.6 + 120.0i, -158.9 - 157.7i).

As previously, we can make Ia real by the appropriate rotation (multiplying by e^(0.12266i)) to get I = (153.7, -76.8 + 111.4i, -138.4 - 176.0i). And I double checked that this satisfies the original KCL constraints as well as the new KVL constraint for the assumed values of w and x, at least up to rounding error.

Cheers, Wayne
very observant
the load determines the imbalance, not the xfmr Z (at least by an influence of Z load / Z xfmr typically 100:1)
in this case xfmr Z 0.0127, load 240/(~300/sqrt3) ~ 1.4 Ohm ratio ~ 110:1

look at it this way
you could manipulate the load X/R ratio (reactive/resistive Z) and have the same line current magnitudes, but a completely different phase relationship, yet the currents derived by the method used above would yield the same result?
does that make sense?
at least one phase relationship determined by the LOAD (not transformer) must be known
 
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Phil Corso

Senior Member
Gentlepeople...

Add following (I call it Step 4) to the three presented previously in Post # 67:

a) Assign double lower-case subscripts to Source values; double upper-case subscripts to Load-values; and single lower-case to Line-currents!

b) Convert Delta-arrangement of single-phase Xfmr's to an equivalent Y-arrangement.

c) Transform Xfmr-winding impedances from Delta to Wye!

d) Now current in Phase Xfmr, say A-B, is equal to line-current Ia, etc.

e) Product Volts x Amps clearly reveal the S or kVA carried by each Xfmr.

Phil Corso
 

Phil Corso

Senior Member
LongCircuit...

The answer to your original query is, None are overloaded!

Execute a Delta-to-Wye transformation of the 3 single-phase transformers!

Then, as an example, kVA of Xfmr R-B is [240/SQRT(3)] x |Ib| = |Vbn| x |Ib| or 35.5kVA. Similarly, kVA in Xfmrs W-B and R-W are 40.8 and 47.4, respectively!
Note: ‘n’ represents the centroid of the Wye configured Xfmrs!

Let me know if you want more data, i.e., kW, kVAr, PF, and Xfmr losses using Xfmr parameters suggested by TopGone!

Regards, Phil
 
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