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

#### Ingenieur

##### Senior Member
But if there are more than one unknown you have at least n! (I believe) possible correct solutions. You need more information on phase/current or more boundary conditions that could fix one or more values.
:thumbsup:
imho we need load or line current phase information
or at least some way to derive it
we have neither

fea is still an iterative approximation method
without enough info no boundary, no convergence

ECE/ME 2646: Linear System Theory (3 Credits, Fall 2015)
Description:
Linear spaces and operators, mathematical descriptions of linear systems, controllability and observability, irreducible realization of rational transfer-function matrices, canonical forms, state feedback and state estimators, and stability.
Prerequisite:
Knowledge of linear algebra, differential equations, and feedback control systems (e.g. ECE 1673).

InJunEar...

Phil

#### GoldDigger

##### Moderator
Staff member
But if there are more than one unknown you have at least n! (I believe) possible correct solutions. You need more information on phase/current or more boundary conditions that could fix one or more values.
:thumbsup: and :thumbsdown:
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.

Since current is continuously variable you would have an infinite number of solutions, not n!

#### Phil Corso

##### Senior Member
JunkHound...

Par 3) of my solution was an "iteritive" process!

Phil

##### Senior Member
:thumbsup: and :thumbsdown:
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.

Since current is continuously variable you would have an infinite number of solutions, not n!
I see your point. I was envisioning a scenario where you had discrete variables with unique but unknown values. The way you describe it, the solution set blows up to infinity as soon as you have two unknowns more than you have equations.

#### GoldDigger

##### Moderator
Staff member
...
... The way you describe it, the solution set blows up to infinity as soon as you have two unknowns more than you have equations.
Way off!
If you have even one more unknowns than equations you have an infinite number of solutions. Each additional unknown multiplies by infinity again, leaving you with exactly the same infinity. :angel:

#### Smart \$

##### Esteemed Member
So what y'all are saying is that three transformers are smarter than y'all. :angel:

#### iwire

##### Moderator
Staff member
you have an infinite number of solutions.
With so many solutions you would think it would be solved by now. #### topgone

##### Senior Member
Approximations can help, I guess. The OP's figures, being measured values could have been the culprit! From a quick calc I did, the results tell me that the 50 kVA on "A" phase gets overloaded by 17% when the transformer on "C" phase is replaced with a bigger rated single-phase transformer (75 kVA, in OP's case).
The line current values I punched into my calc sheets:
Line 1 = 253.61/ 0.0 ⁰
Line 2 = 297.02/ 252.3 ⁰
Line 3 = 340.02/ 117.0 ⁰​
The transformer phase currents I got (in answer to the original poster's question):
"A" = 243.62/-18.90 ⁰ Amps --> overloaded by 17%
"B" = 60.73/ -103.63 ⁰ Amps --> 29% loaded
"C" = 236.48/ 71.27 ⁰ Amps --> 76% loaded​

##### Senior Member
Way off!
If you have even one more unknowns than equations you have an infinite number of solutions. Each additional unknown multiplies by infinity again, leaving you with exactly the same infinity. :angel:
I don't believe so. Using iterative numerical methods you will arrive at a unique solution.

#### GoldDigger

##### Moderator
Staff member
I don't believe so. Using iterative numerical methods you will arrive at a unique solution.
Trust me, it is not unique if you do not specify input current phase angles.
If you assume all known (given) currents are at PF=1 (or any known PF), then there is a unique solution.
It is possible for an iterative method to converge to a non-unique solution. Often in those cases a different "guess" will lead to a different solution.

As a very simple proof of my statement, take your converged solution and then rotate all input currents CW by 30 degrees. You should get a different iterated solution.

Last edited:

#### Ingenieur

##### Senior Member
I tried several iterative methods
none converged

you need either current or load phase informatiion

#### topgone

##### Senior Member
I tried several iterative methods
none converged

you need either current or load phase informatiion
You are partly correct by saying that. But, granting that the line current values posted by the OP are correct (only values of I2 and I3 are used in my formulations), one can arrive at a solution set. Please look at the following steps that I took:

Formulation of equations:
Ia.Za + Ib.Zb + Ic.Zc = 0​
Assume:
w = Zb/Za; or: Zb = w.Za
x = Zc/Za: or: Zc = x.Za​
Then the equation becomes:
Ia.Za + Ib.w.Za + Ic.x.Za = 0​
Finally, dividing the equation by Za:
(1) Ia + w.Ib + xIc = 0​
The other equations are derived from summation of currents at selected delta corners:
(2) I2 = Ib - Ic = 294/ 240⁰ amperes
(3) I3 = Ic - Ia = 341/120⁰ amperes​

The final equations in matrix form will be:
Code:
``````Ia      1     w     x          0
Ib      0     1     1   =     294/240⁰
Ic     -1     0     1         341/120V``````
Please do the inversion of the complex matrix and you'll get results!
Here's my results:
Phase "A" amps = 184.71/ -27.04⁰
Phase "B" amps = 159.00/ -164.20⁰
Phase "C" amps = 211.42/ 91.62⁰​
Re-computed the line amps:
I1 = 320.10/ -7.305416128⁰
I2 = 294.00/ -88.38⁰
I3 = 341.00/ 152.96⁰​

Take note that the computed line current of Line 1 does not jibe with the given figure by the OP (256). It's either those line currents are erroneous or readings were not taken simultaneously or what! You decide folks!
My two cents!

#### Ingenieur

##### Senior Member
You are partly correct by saying that. But, granting that the line current values posted by the OP are correct (only values of I2 and I3 are used in my formulations), one can arrive at a solution set. Please look at the following steps that I took:

Formulation of equations:
Ia.Za + Ib.Zb + Ic.Zc = 0​
Assume:
w = Zb/Za; or: Zb = w.Za
x = Zc/Za: or: Zc = x.Za​
Then the equation becomes:
Ia.Za + Ib.w.Za + Ic.x.Za = 0​
Finally, dividing the equation by Za:
(1) Ia + w.Ib + xIc = 0​
The other equations are derived from summation of currents at selected delta corners:
(2) I2 = Ib - Ic = 294/ 240⁰ amperes
(3) I3 = Ic - Ia = 341/120⁰ amperes​

The final equations in matrix form will be:
Code:
``````Ia      1     w     x          0
Ib      0     1     1   =     294/240⁰
Ic     -1     0     1         341/120V``````
Please do the inversion of the complex matrix and you'll get results!
Here's my results:
Phase "A" amps = 184.71/ -27.04⁰
Phase "B" amps = 159.00/ -164.20⁰
Phase "C" amps = 211.42/ 91.62⁰​
Re-computed the line amps:
I1 = 320.10/ -7.305416128⁰
I2 = 294.00/ -88.38⁰
I3 = 341.00/ 152.96⁰​

Take note that the computed line current of Line 1 does not jibe with the given figure by the OP (256). It's either those line currents are erroneous or readings were not taken simultaneously or what! You decide folks!
My two cents!
you are assuming the load is balanced phase wise
it's not
294/ 240⁰ amperes
341/120⁰ amperes
the reason you did not get the 256 is it's not in your equations
a solution is forced

no consideration is given to the load phase
assumed purely real

then to make it fit back in the line currents are now not 0/120/240
I1 = 320.10/ -7.305416128⁰
I2 = 294.00/ -88.38⁰
I3 = 341.00/ 152.96⁰

#### topgone

##### Senior Member
you are assuming the load is balanced phase wise
it's not
294/ 240⁰ amperes
341/120⁰ amperes
the reason you did not get the 256 is it's not in your equations
a solution is forced

no consideration is given to the load phase
assumed purely real

then to make it fit back in the line currents are now not 0/120/240
I1 = 320.10/ -7.305416128⁰
I2 = 294.00/ -88.38⁰
I3 = 341.00/ 152.96⁰
You are basically getting the results I posted just above your latest reply! Upon further analyses, I found out that the nearest true values of the line currents happens when one doesn't use I2 = 294 amps! I tried running my calcs again using I1 and I3, together with the impedance constraints and I got the following results:
Ia = 156.65/ -25.73⁰ amps
Ib = 133.51/ -149.37⁰ amps
Ic = 229.2/ 97.37⁰ amps
The recalculated line currents are:
I1 = 256/ 0⁰ amps
I2 = 307.44/-106.15⁰ amps, and
I3 = 341/ 120⁰ amps!

Please take note that the error between the OP's I2 = 294 versus the recomputed current here of 307.44 amps is just around 4.4%. I've compared the differences using the other two line currents as larger compared to not using the value 294A! I may have to say, the erroneous reading is this 294A thing!

##### Senior Member
Trust me, it is not unique if you do not specify input current phase angles.
If you assume all known (given) currents are at PF=1 (or any known PF), then there is a unique solution.
It is possible for an iterative method to converge to a non-unique solution. Often in those cases a different "guess" will lead to a different solution.

As a very simple proof of my statement, take your converged solution and then rotate all input currents CW by 30 degrees. You should get a different iterated solution.
Then you have 3 unknowns. You prove my point. I said if there was only one more unknown than equations you could use numerical methods to get to a unique solution.

#### Ingenieur

##### Senior Member
You are basically getting the results I posted just above your latest reply! Upon further analyses, I found out that the nearest true values of the line currents happens when one doesn't use I2 = 294 amps! I tried running my calcs again using I1 and I3, together with the impedance constraints and I got the following results:
Ia = 156.65/ -25.73⁰ amps
Ib = 133.51/ -149.37⁰ amps
Ic = 229.2/ 97.37⁰ amps
The recalculated line currents are:
I1 = 256/ 0⁰ amps
I2 = 307.44/-106.15⁰ amps, and
I3 = 341/ 120⁰ amps!

Please take note that the error between the OP's I2 = 294 versus the recomputed current here of 307.44 amps is just around 4.4%. I've compared the differences using the other two line currents as larger compared to not using the value 294A! I may have to say, the erroneous reading is this 294A thing!
I was illustrating the inconsistent in the currents
one is 0/120/240

you still are assuming line currents to be balanced at 0/120/240
they are not
you are forcing a solution to meet your initial conditions which are not valid
the sum of voltages are not 0, they are balanced in magnitude but not phase

the problem is no solvable with the givens

#### GoldDigger

##### Moderator
Staff member
Then you have 3 unknowns. You prove my point. I said if there was only one more unknown than equations you could use numerical methods to get to a unique solution.
Using normal mathematical terminology a set of equations has a unique solution only if the number of scalar unknowns and scalar equations is equal.
Consider x + y = 5. In your terms, two unknowns and only one equation.
You can find any number of solutions by trial and error or by arbitrarily choosing a value for y. But the set of solutions is the line represented by that equation, which has a infinite number of points.

#### Ingenieur

##### Senior Member
from you identities
(2) I2 = Ib - Ic = 294/ 240⁰ amperes

Ib = 133.51/ -149.37⁰ amps
Ic = 229.2/ 97.37⁰ amps

check (rounded)
Ib = 134/-149 = -115 -68j
Ic = 229/97 = -29 + 227j
Ib - Ic = -115 - (-28) -68j - 227j = -87 - 295j = 308/-106 NE to 294/ 240⁰
???

the measurements are not the issue
the problem is unsolvable without load or line phase info

and by doing so contriving the solution

#### topgone

##### Senior Member
from you identities
(2) I2 = Ib - Ic = 294/ 240⁰ amperes

Ib = 133.51/ -149.37⁰ amps
Ic = 229.2/ 97.37⁰ amps

check (rounded)
Ib = 134/-149 = -115 -68j
Ic = 229/97 = -29 + 227j
Ib - Ic = -115 - (-28) -68j - 227j = -87 - 295j = 308/-106 NE to 294/ 240⁰
???

the measurements are not the issue
the problem is unsolvable without load or line phase info

Anyway, I think this is a dead horse already. We might as well stop beating it. 