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

#### wwhitney

##### Senior Member
You have six unknowns (three currents with three phase angles or three vector currents with two components each).

If you know phase of the load currents as well as magnitude you have six equations.
If you can approximate load currents as PF = 1, then there is a unique solution. If not, then the solution is undetermined but you can still place upper limits on the delta currents by assuming worst case or worst credible case phasing.
I'm a little weak on the physics, but if you could express the problem as 3 equations and 6 unknowns (or whatever it works out to be) and identify the variables of interest, I'd be happy to try to figure out the maxima and minima over the solution space. My understanding is that since the magnitudes of the load currents are known, the delta currents will be bounded in magnitude as well.

Cheers, Wayne

#### GoldDigger

##### Moderator
Staff member
Yes, that is a basic premise of what is called linear algebra in the optimization and analysis domain, and the optimization process is sometimes as simple as checking corner points.

There are some constraints on how the quantity to be maximized or minimized can depend on the free variables though if you are looking for an easy iterative solution.

I agree that given a limiting value on PF you can find max and min current for each coil in the delta and so sometimes know for sure there is or is not an overload. But there can also be cases where the load limit falls between min and max values.
To be technically picky there can be, in specific cases, an answer to the OP's question even though there is not a unique solution to the algebraic problem.
Thanks for pointing that out.

#### wwhitney

##### Senior Member
Yes, that is a basic premise of what is called linear algebra in the optimization and analysis domain
Yes, my background is in mathematics, so if you or someone else could do the physics and write down the equations in a post, I'd take a stab at the math to see what the possible values of the magnitude of the delta current can be.

Cheers, Wayne

#### Sahib

##### Senior Member
Here is another solution.

The line currents are 256A, 294A, 341A. Per IEEE C57.105-1978, the lighting leg should have all lighting load and share one third of three phase load. The lowest line current is 256A and lowest multiple of 3 is 255. So the three phase load shared by three legs is 255A each. So the per phase current is 255/1.4=182A. So the minimum KVA for each transformer is 182*240=43.68KVA. Assuming same p.f, single phase load on one power leg =294-255=39A. So the lighting KVA (Upper limit)=39*240=9.36KVA. So the total KVA on power legs= 53.04. So one 50KVA transformer is likely to be overloaded. But it may not be serious.

#### Sahib

##### Senior Member
Here is another solution.The line currents are 256A, 294A, 341A. Per IEEE C57.105-1978, the lighting leg should have all lighting load and share one third of three phase load. The lowest line current is 256A and lowest multiple of 3 is 255. So the three phase load shared by three legs is 255A each. So the per phase current is 255/1.4=182A. So the minimum KVA for each transformer is 182*240=43.68KVA. Assuming same p.f, single phase load on one power leg =294-255=39A. So the lighting KVA (Upper limit)=39*240=9.36KVA. So the total KVA on power legs= 53.04. So one 50KVA transformer is likely to be overloaded. But it may not be serious.
Actually, voltage for single phase loads on power legs are 208v not 240v. So no overload at all.

#### Phil Corso

##### Senior Member
Gentlepeople...

Quoting Michael Corleone, “Just when I thought I was out... they pull me back in!”

Wayne… (at least you don't hide behind anonymity) a caveat... "mathematics" is a word 'engineers' on this forum abhor!

I don’t know if it will help you or not, but the mathematical procedure I used is outlined in Post # 62. Unfortunately, no one saw fit to address it!

Regards, Phil Corso

#### topgone

##### Senior Member
Gentlepeople...

Quoting Michael Corleone, “Just when I thought I was out... they pull me back in!”

Wayne… (at least you don't hide behind anonymity) a caveat... "mathematics" is a word 'engineers' on this forum abhor!

I don’t know if it will help you or not, but the mathematical procedure I used is outlined in Post # 62. Unfortunately, no one saw fit to address it!

Regards, Phil Corso
I beg to disagree with that one, Phil. Mathematics is the universal language of engineers, I was told. We love to communicate thru numbers!

Your choice of solving the problem by having the details of the load first is just one of the available avenues. One can also arrive at a solution set by iterative methods. Given the magnitudes, people can play around with different angles and arrive at a convergence. The fact that electrical systems almost always have non-zero diagonals (strictly diagonal dominant) when expressed in matrix form tells one that there will be a singular point where everything fits-->convergence for any intitial guess value. Could be tedious but it feels good to get figures in the end!

#### Phil Corso

##### Senior Member
TopgGone...

What you described is exactly what I did! It required 3-steps!

#### Ingenieur

##### Senior Member
I beg to disagree with that one, Phil. Mathematics is the universal language of engineers, I was told. We love to communicate thru numbers!

Your choice of solving the problem by having the details of the load first is just one of the available avenues. One can also arrive at a solution set by iterative methods. Given the magnitudes, people can play around with different angles and arrive at a convergence. The fact that electrical systems almost always have non-zero diagonals (strictly diagonal dominant) when expressed in matrix form tells one that there will be a singular point where everything fits-->convergence for any intitial guess value. Could be tedious but it feels good to get figures in the end!
the matrix is indeterminent
I tried 5 different iterative methods, none converged
we need at least one phase relationship, either current or load

we only want S, but S has phase as a component
if all we want is |S| then there have been a dozen methods shown, pick one, lol

#### Phil Corso

##### Senior Member
TopGone,

Would you agree that if all line-currents were 256A, then minimum load carried by the Xfmrs is 106kVA?

Would you agree that if all line-currents were 341A, then maximum load carried by the Xfmrs is 141kVA?

Thus, the load must be between 106 and 141kVA!

With the procedure I provided earlier, sing the OP's values, PF was 0.86, and 0.59 for the min, max cases, respectively!

More later, Phil

Last edited:

#### topgone

##### Senior Member
TopGone,

Would you agree that if all line-currents were 256A, then minimum load carried by the Xfmrs is 106kVA?

Would you agree that if all line-currents were 341A, then maximum load carried by the Xfmrs is 141kVA?

Thus, the load must be between 106 and 141kVA!

With the procedure I provided earlier, sing the OP's values, PF was 0.86, and 0.59 for the min, max cases, respectively!

More later, Phil
@Phil,
My figures are about the same as what you got!
If all line currents were balanced at 256A, the maximum load is 107.09 kVA.
If all line currents were balanced at 341A, the maximum load is 142.64 kVA.

#### Sahib

##### Senior Member
Hi folks!
Without knowing load power factors, it is not possible to find phase currents because there exist equations relating phase KVA to load power factors as below.

KVAl=1/3*SQRT[(K3*K3)+(4*K1*K1)-(4*K1*K3*COS(A))

KVAP=1/3*SQRT[(K3*K3)+(*K1*K1)-(K1*K3*COS(120+/-A))

where
KVAl=single phase load in KVA of lighting leg
KVAP=Three phase load in KVA of each power leg
A=Difference in power factor angle between three phase and single phase loads.

#### topgone

##### Senior Member
Hi folks!
Without knowing load power factors, it is not possible to find phase currents because there exist equations relating phase KVA to load power factors as below.

KVAl=1/3*SQRT[(K3*K3)+(4*K1*K1)-(4*K1*K3*COS(A))

KVAP=1/3*SQRT[(K3*K3)+(*K1*K1)-(K1*K3*COS(120+/-A))

where
KVAl=single phase load in KVA of lighting leg
KVAP=Three phase load in KVA of each power leg
A=Difference in power factor angle between three phase and single phase loads.
Sorry sahib. Just do some backreading and you'll see that this problem is a done thing!
Here are the equations (in matrix form) that will help you solve the problem.
Code:
``````Ia	1	-1	0                                                                   256/ -1.5531º
Ib X	1	1	0.603252032520325+0.0373983739837398i                           =   0/0 º
Ic	-1	0	1                                                                   341/ 121.553 º``````
I am posting the transformer phase currents for you to compare with, here:
Code:
``````PHASE     MAGNITUDE	       ANGLE	   LOADING
Ia =           158.84     	-25.94º	     76%
Ib =           129.23     	-151.05º     62%
Ic =           223.94	         99.15º	     72%``````
The iterated line currents are:
Code:
``````I_1 = 256.00/	-1.5531 º
I_2 = 294.03/	-105.28 º
I_3 = 341.00/	121.55 º``````

#### Sahib

##### Senior Member
Please use Kirchoff current law to verify the result.

#### Ingenieur

##### Senior Member
Sorry sahib. Just do some backreading and you'll see that this problem is a done thing!
Here are the equations (in matrix form) that will help you solve the problem.
Code:
``````Ia    1    -1    0                                                                   256/ -1.5531º
Ib X    1    1    0.603252032520325+0.0373983739837398i                           =   0/0 º
Ic    -1    0    1                                                                   341/ 121.553 º``````
I am posting the transformer phase currents for you to compare with, here:
Code:
``````PHASE     MAGNITUDE           ANGLE       LOADING
Ia =           158.84         -25.94º         76%
Ib =           129.23         -151.05º     62%
Ic =           223.94             99.15º         72%``````
The iterated line currents are:
Code:
``````I_1 = 256.00/    -1.5531 º
I_2 = 294.03/    -105.28 º
I_3 = 341.00/    121.55 º``````
the 0.6 should be 0 and have a corresponding current
there is no solution, fixed or iterative

#### Ingenieur

##### Senior Member
here's the equation that needs solved

A B = C where

A (3x3)
1 -1 0
0 1 -1
-1 0 1

B phase currents, ie, the unknowns (3x1)
Ia
Ib
Ic

C line currents, magnitude known, phase unknown (3x1)
256/?
294/?
341/?

B = (A^-1) C

solve it without assumptions

#### wwhitney

##### Senior Member
here's the equation that needs solved

A B = C where

A (3x3)
1 -1 0
0 1 -1
-1 0 1

B phase currents, ie, the unknowns (3x1)
Ia
Ib
Ic

C line currents, magnitude known, phase unknown (3x1)
256/?
294/?
341/?

B = (A^-1) C

solve it without assumptions
First, let me say I've only made through about the first 20 posts in this thread, plus the above post, so my apologies if what I say here has already been stated.

In the form above, the magnitudes of Ia, Ib, and Ic are unbounded. Let me restate the problem to be sure I've understood it correctly. Ia, Ib, and Ic are the coil currents, consider them to be complex numbers (in polar form, if you like, comprising a magnitude and a phase angle). The above equations are just:

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

Here | | is the magnitude operator on complex numbers. So all the equations are saying is that the three points Ia, Ib, and Ic form a triangle in the complex plane with sides of length 256, 294, 341. That triangle could be anywhere in the plane, so individually Ia, Ib, and Ic could have arbitrary magnitude.

Basically, given any solution, you could add an arbitrary (complex) value to Ia, Ib, and Ic, and you have another solution. If I understand correctly, that would correspond physically to saying that there could be an arbitrary circulating current in the delta coils that would never show up in the line currents as it cancels out of each pairwise difference.

If there is some physics that says this won't happen the problem may be tractable. For example, suppose the physics says further Ia + Ib + Ic = 0 (as complex numbers). Now the translational degrees of freedom have been removed, and the triangle must be centered about the origin. There is still a rotational degree of freedom present, but that is just a convention about phase angle, so we can arbitrarily set Ia to be purely real. This leaves just two possible solutions which are mirror images of each other. If we arbitrarily insist that Im Ib > 0, then there is only one solution, which we can compute.

Since I have no idea if the requirement that Ia + Ib + Ic = 0 is supported by the physics, I won't proceed with the computation.

Cheers, Wayne

#### Ingenieur

##### Senior Member
The coil currents can't be assumed to sum to zero
an imbalance will circulate
it can be assumed that the load is purely resistive so phase voltages can be of 0/120/240 phase
again not valid and forces an arbitrary solution

the line currents can be assumed to sum to zero if a wye w/o a neutral is the load

as you said any assumption about the phase will result in a solution
but there are infinite possible assumptions

#### topgone

##### Senior Member
First, let me say I've only made through about the first 20 posts in this thread, plus the above post, so my apologies if what I say here has already been stated.

In the form above, the magnitudes of Ia, Ib, and Ic are unbounded. Let me restate the problem to be sure I've understood it correctly. Ia, Ib, and Ic are the coil currents, consider them to be complex numbers (in polar form, if you like, comprising a magnitude and a phase angle). The above equations are just:

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

Here | | is the magnitude operator on complex numbers. So all the equations are saying is that the three points Ia, Ib, and Ic form a triangle in the complex plane with sides of length 256, 294, 341. That triangle could be anywhere in the plane, so individually Ia, Ib, and Ic could have arbitrary magnitude.

Basically, given any solution, you could add an arbitrary (complex) value to Ia, Ib, and Ic, and you have another solution. If I understand correctly, that would correspond physically to saying that there could be an arbitrary circulating current in the delta coils that would never show up in the line currents as it cancels out of each pairwise difference.

If there is some physics that says this won't happen the problem may be tractable. For example, suppose the physics says further Ia + Ib + Ic = 0 (as complex numbers). Now the translational degrees of freedom have been removed, and the triangle must be centered about the origin. There is still a rotational degree of freedom present, but that is just a convention about phase angle, so we can arbitrarily set Ia to be purely real. This leaves just two possible solutions which are mirror images of each other. If we arbitrarily insist that Im Ib > 0, then there is only one solution, which we can compute.

Since I have no idea if the requirement that Ia + Ib + Ic = 0 is supported by the physics, I won't proceed with the computation.

Cheers, Wayne
You missed a lot. If you only backread, the formulation of the second equation is not Ia + Ib + Ic =0. The equation is an offshoot of taking a voltage loop inside of the delta of the bank = 0 (Ia.Za + Ib.Zb + Ic.Zc = 0).

#### Ingenieur

##### Senior Member
You missed a lot. If you only backread, the formulation of the second equation is not Ia + Ib + Ic =0. The equation is an offshoot of taking a voltage loop inside of the delta of the bank = 0 (Ia.Za + Ib.Zb + Ic.Zc = 0).
only if the load is shorted
look at it this way
Za ~ 0.017 ohm on a 50kva / 240 vac base with 2% puZ

Vphase = Za x Ia which is summed per kvl around the loop
or 240 = 0.017 x Ia (or any ph coil)
Ia = 14.2 kA which is obviously not valid