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

[Phil Corso]

Gentlepeople… a reminder:

Here is the solution using my Gi.Fi.E.S. (Pronounced Jiffy’s) method:

Given:
o Hi-Leg distribution system: Vrw = 240<0⁰; Vbr = 240<-120⁰; Vwb = 240<+120⁰;
o Line-currents to load bus: Ib = 294A; Ir = 256A; Iw = 341A.
o Neutral-current: In = 0A, thus supply can be considered as typ 3-ph, Delta!
o Xfmr capacities: Tbr = 50kVA; Trw = 75kVA; Twb = 50kVA.

Not Given:
o Any info on Load-type.
o Any info on Load phase-currents.
o Any info on Load PF(s).
o Any info on Xfmr parameters like, %Imp; X/R; V-Reg.

Find:
o Which of the 3 Xfmrs is overloaded, meaning determine kVA generated, or produced in each.

Equations:

Solution:

Phil

 

[Smart $]

SmartA, InJunEar..

The only voltage any where are the supply voltages, i.e., 240 V@ 0, -120, +120 Deg!

Phil

:huh:... Stating something rhetorical for the 'el' of it?


Side note: those would be phase voltage vectors

 

[Smart $]

Gentlepeople… a reminder:

Here is the solution using my Gi.Fi.E.S. (Pronounced Jiffy’s) method:

Given:
o Hi-Leg distribution system: Vrw = 240<0⁰; Vbr = 240<-120⁰; Vwb = 240<+120⁰;
o Line-currents to load bus: Ib = 294A; Ir = 256A; Iw = 341A.
o Neutral-current: In = 0A, thus supply can be considered as typ 3-ph, Delta!
o Xfmr capacities: Tbr = 50kVA; Trw = 75kVA; Twb = 50kVA.

Not Given:
o Any info on Load-type.
o Any info on Load phase-currents.
o Any info on Load PF(s).
o Any info on Xfmr parameters like, %Imp; X/R; V-Reg.

Find:
o Which of the 3 Xfmrs is overloaded, meaning determine kVA generated, or produced in each.

Equations:

Solution: ????

Phil

My reply within quote. Should be obvious to discern which is which.

 

[LongCircuit]

Gentlepeople… a reminder:

Here is the solution using my Gi.Fi.E.S. (Pronounced Jiffy’s) method:

Given:
o Hi-Leg distribution system: Vrw = 240<0⁰; Vbr = 240<-120⁰; Vwb = 240<+120⁰;
o Line-currents to load bus: Ib = 294A; Ir = 256A; Iw = 341A.
o Neutral-current: In = 0A, thus supply can be considered as typ 3-ph, Delta!
o Xfmr capacities: Tbr = 50kVA; Trw = 75kVA; Twb = 50kVA.

Not Given:
o Any info on Load-type.
o Any info on Load phase-currents.
o Any info on Load PF(s).
o Any info on Xfmr parameters like, %Imp; X/R; V-Reg.

Find:
o Which of the 3 Xfmrs is overloaded, meaning determine kVA generated, or produced in each.

Equations:

Solution:

Phil

Imagine that the TF bank is feeding a street with residential customers. Therefore it would be impossible to tell what the load phase currents are and what the power factor is.

 

[Smart $]

Imagine that the TF bank is feeding a street with residential customers. Therefore it would be impossible to tell what the load phase currents are and what the power factor is.

If you knew the load phase currents, you likely would not need to know the supply transformer winding currents.

As to power factor, just give three reasonably real but random values, say 90, 85, and 87, respective of earlier-enumerated lines.

 

[Phil Corso]

LongCircuit... you are correct...

Even if one used the Law of Cosines to determine the relative angles between the Line-currents, there is NO Given data allowing one to reference Line-current to Phase-voltage.

But if you calculate Total kVA using the average, or even the RMS, or even using the largest of line-currents you'll find that it is quite less than the total capacity of the 3 transformers! Very interesting!

Phil

GadFly... are you following? And no, there is nothing proprietary about what the solution is!

 

[Phil Corso]

LongCircuit...

Finally, Equations/Solutions:

Following are three solutions based on your original post. The first two are simplified methods assuming no neutral involvement.

Solution 1
Transform the "balanced" Delta-source into a "balanced" Wye-source. As you probably already know, the load can be a "black’ box, either Y or delta, having no impact on outcome! Each winding’s voltage-magnitude is equal to |Vpp|/Sqrt(3) or 138.6V. Then:

o kVAbn = (|Vbn| x |Ib|)/1,000 = (138.6 x 294)/1,000 ~ 41kVA!
o kVArn ~ 36kVA!
o kVAwn ~ 47kVA!
Note: none of the windings is overloaded! Furthermore, the total is close to that shown in my calculation yesterday which was based on averaging the 3 currents!

Solution 2
This method is similar to that of a Power Calculation, but instead of 2-Wattmeters substitute 3-theoretical VoltAmp-meters. The answer is identical to Solution 1.

Solution 3
This based on a true Hi-Leg source, i.e., the mid-point of winding Trw is grounded! But, because nothing is known about the load an iterative process such as the one promoted by InJunEar, must be used. It is very, very difficult because there are 6 unknowns, and not just 3!

Regards, Phil

 

[Ingenieur]

Solution 1
solve for each phase current (in essence all you did was divide the line current / sqrt3 x 240 = assumed phase current x V = kva for each xfmr, wye-delta transform)
kva / 240 ang
where ang = 0, -120, 120

this gives 3 phase currents
170/0 (294 line)
150/120 (256 line)
196/-120 (341 line)

adding the phase currents to solve for the line currents they do not equal 294, 256, 341 (magnitudes)???
hard to tell if the ang is correct without knowing the load

what is ac in my drawing (b, r or w?)
what is ba in my drawing (b, r or w?)
what is cb in my drawing (b, r or w?)
perhaps my references are incorrect

as far as I can tell iteration is the only way
6 variables
ac real + Img
ba real + Img
cb real + Img

3 knowns (assumed)
line a 256/0
line b 294/0
line c 341/0

[identity matrix] x [phase I Re Img] = [line Re Img]
easily derived.....x.....unknowns......= knowns
but the identity is singular so no inverse exists
therefore must use an iterative method

 

[Ingenieur]

Typo
should be
3 knowns (assumed)
line a 256/0
line b 294/-120
line c 341/120

got close in excel using manual iteration
may play with the excel 'solver'
may also try matlab

we all know a meter measurement is the way to go, lol
next weighted average
act line reading / avg line reading x act line reading x 1/1.732

 

[topgone]

Typo
should be
3 knowns (assumed)
line a 256/0
line b 294/-120
line c 341/120

got close in excel using manual iteration
may play with the excel 'solver'
may also try matlab

we all know a meter measurement is the way to go, lol
next weighted average
act line reading / avg line reading x act line reading x 1/1.732

When one uses 1.732, the assumption is that of a balanced current, which the OP's problem is not!

 

[Ingenieur]

When one uses 1.732, the assumption is that of a balanced current, which the OP's problem is not!

correct
that is why his solution is not valid (mine was a best approximation)
I tried several iterative methods: Gaussian, etc,
all said not solvable, non-convergent
the matrix is indeterminant

the matrix could have been 2x6 but I made it 6x6 to make it sparse, ie, better chance of a solution
I also mixed up the phase and Re/Im components to introduce a randomness

x1 = Iba Re
x2 = Iac Re
x3 = Iba Im
x4 = Iac Im
x5 = Icb Re
x6 = Icb Im
conventions from my previous sketch
details below
matrix
testing for compatibility
says 'has no solution' lol

imho can't be solved numerically (maybe a supercomputer, lol)
imho the best WAG is the weighted average I detailed in my previous post

line 256/phase 127/30.6 kva
line 294/phase 168/40.3 kva
line 341/phase 226/54.3 kva

or a thermal imaging camera

I surrender!!!!! lol

 

[LongCircuit]

correct
that is why his solution is not valid (mine was a best approximation)
I tried several iterative methods: Gaussian, etc,
all said not solvable, non-convergent
the matrix is indeterminant

the matrix could have been 2x6 but I made it 6x6 to make it sparse, ie, better chance of a solution
I also mixed up the phase and Re/Im components to introduce a randomness

x1 = Iba Re
x2 = Iac Re
x3 = Iba Im
x4 = Iac Im
x5 = Icb Re
x6 = Icb Im
conventions from my previous sketch
details below
matrix
testing for compatibility
says 'has no solution' lol

imho can't be solved numerically (maybe a supercomputer, lol)
imho the best WAG is the weighted average I detailed in my previous post

line 256/phase 127/30.6 kva
line 294/phase 168/40.3 kva
line 341/phase 226/54.3 kva

or a thermal imaging camera

I surrender!!!!! lol

Thank you for your effort. I suspect that this problem really is unsolvable with the limited data in this scenario.

 

[Ingenieur]

Thank you for your effort. I suspect that this problem really is unsolvable with the limited data in this scenario.

you're welcome but no problem
I like this stuff lol

I agree
we have 6 unknowns 3 phase x mag/ang
and 6 knowns 3 line currents and can assume 3 phase voltages
but the link between them is missing, phase relationship

 

[topgone]

Anyone can formulate a 3x3 complex matrix out of the given data plus the knowledge of the complex percent impedance of the transformers. I did my calcs using these additional data from typical transformers:

50 kVA: %IR = 1.1%; %IX = 1.7% or a %Z = 2.02%
75 kVA: %IR = 0.9%; %IX = 1.6% or a %Z = 1.84%​

The additional ratio used to reduce the unknowns are:

w = Zb/Za
x = Zc/Za​

The equations are:
KCL inside the delta = 0

Ia + w(Ib) + x(Ic) = 0 (eq. 1)​

At delta corner where current I2 leaves:

Ia - Ib = I2 (eq. 2)​

At delta corner where I3 leaves:

Ib - Ic = I3 (eq. 3)​

The voltages were assumed to be 240V and 120 degrees apart.
The resulting matrices are complex in nature! Inverting the complex matrices was a bit challenging and multipying, the results I got are:
Ia = 205.2 @ angle -145.31; TR A (50kVA) 98% loaded!
Ib = 227.28 @ angle 42.46; TR B (50kVA) 109% loaded!
Ic = 321.79 @ angle 19.03; TR C (75kVA) 103% loaded!

I tried changing the load power factors but the load sharing doesn't change! If the load power factors are different on each phase, the load sharing of the three single-phase units changes!

 

[Ingenieur]

Anyone can formulate a 3x3 complex matrix out of the given data plus the knowledge of the complex percent impedance of the transformers. I did my calcs using these additional data from typical transformers:

50 kVA: %IR = 1.1%; %IX = 1.7% or a %Z = 2.02%
75 kVA: %IR = 0.9%; %IX = 1.6% or a %Z = 1.84%​

The additional ratio used to reduce the unknowns are:

w = Zb/Za
x = Zc/Za​

The equations are:
KCL inside the delta = 0

Ia + w(Ib) + x(Ic) = 0 (eq. 1)​

At delta corner where current I2 leaves:

Ia - Ib = I2 (eq. 2)​

At delta corner where I3 leaves:

Ib - Ic = I3 (eq. 3)​

The voltages were assumed to be 240V and 120 degrees apart.
The resulting matrices are complex in nature! Inverting the complex matrices was a bit challenging and multipying, the results I got are:
Ia = 205.2 @ angle -145.31; TR A (50kVA) 98% loaded!
Ib = 227.28 @ angle 42.46; TR B (50kVA) 109% loaded!
Ic = 321.79 @ angle 19.03; TR C (75kVA) 103% loaded!

I tried changing the load power factors but the load sharing doesn't change! If the load power factors are different on each phase, the load sharing of the three single-phase units changes!

matrix
[Ia] = [Ix]


1 w x
1 -1 0
0 1 -1

[Ia]
Ia
Ib
Ic

[I2]
0
I2
I3

what actual values did you use for w, x
not sure it matters, the ratio xfmr Z << load
xfmr <0.02 ohm, load ~240/250 ~ 1 ohm
need to calc actual Z = Zpu%/100 x kva/V^2
base = 0.87 for the 50 kva
if we use 2% act Z 0.017 ohm <<<1ohm
xfmr Z is moot compared to load

what values for I2 and I3?
are Iabc the phase currents?
and I23 the line?
where do voltages come into play?
if you invert I x I23 that gives Iabc directly?
???

KCL
at each node will be zero but around the loop a net unbalanced circulating current will result

thanks

 

[Ingenieur]

The reason I ask is I discussed this with my Proff and the below books author
50 year industry academic experience
PhD CMU
numerous text, papers and patents on power
wrote some of the first power flow control sw
the book is about implementing matrix solutions for grid control
I used the sw that came with it (among others)
took the course last term A

he indicated it could not be solved numerically (either direct calc or iterative estimation) without basic load info since it DETERMINES the imbalance

 

[Ingenieur]

Using 1for xfmr Z ratios
should be close
and some close line currents
this is the result

I used only magnitues
inverse x line = phase

 

[topgone]

matrix
[Ia] = [Ix]


1 w x
1 -1 0
0 1 -1

[Ia]
Ia
Ib
Ic

[I2]
0
I2
I3

what actual values did you use for w, x
not sure it matters, the ratio xfmr Z << load
xfmr <0.02 ohm, load ~240/250 ~ 1 ohm
need to calc actual Z = Zpu%/100 x kva/V^2
base = 0.87 for the 50 kva
if we use 2% act Z 0.017 ohm <<<1ohm
xfmr Z is moot compared to load

what values for I2 and I3?
are Iabc the phase currents?
and I23 the line?
where do voltages come into play?
if you invert I x I23 that gives Iabc directly?
???

KCL
at each node will be zero but around the loop a net unbalanced circulating current will result

thanks

A: It's all in the post above. You are given the percent impedances of the transformers I used to calculate into the matrices. Punch in the ohmic values of the transformers and get the needed numbers, w and x.

A: The general form of the matrices are [Ia,b,c][Matrix A] = [I1,2,3]. If you can invert the matrix A and then multiply with the constant matrix [I1,2,3]-->the measured line currents from the OP, you'll get the phase currents of the banked transformers [Ia,b,c].

Please digest what's written in this paper. LINK
Better still, here's a cut-n-paste from the other forum posted by cuky2000 way back 2003:

"The calculation of the inverse complex matrix can be determined solely based upon the inverses of decomposed real matrices.
Z = A + i.B Z^-1 = C + i.D

Where:
Z is a complex matrix Z^-1= inverse complex matrix
A & B are two real matrices C & D real matrices
I= Identity matrix (all elements=1) i = Imaginary unit = SQRT(-1)

Matrices C &D could be calculated in Excel as function of two given real matrices A & B as follow:

? For A nonsingular , det(A) ?0. [For example, for 3x3 matrix with data from A1 to C3 in Excel, select an empty cell MDETERM(A1:C3)]

C = [A + BA^-1 B]^-1 & D= -CBA^-1

? If A is singular (det(A)= 0):, but B is nonsingular, ( det(B) ?0):

C = B^-1 A [B + A B^-1A]^-1 & D = [CA – I]B^-1 "​

 

[Ingenieur]

Your assumption about the sum of the delta currents being zero in incorrect
there is a current
KCL is for a node not a loop


the xmfr Z is moot and not a factor, unless you assume a balanced load, which it is not
the Z is unknown and is what determines line current
in adfition the Z %pu has no phase or complex component

there are 6 unknowns
3 x phase i mag
3 x phase i ang

only 3 knowns
3 x line i

not sure the assumtion that the phase v's are balanced is valid in this caee

 

[Ingenieur]

Ia = 205/-145 = -168 -118j
Ib = 227/43 = 166 + 155j
Ic = 322/19 = 304 + 105j

Ia - Ib = I2 = -169 - 118j - (166 + 155j) = -334 -273j = 431/-130

op's currents
256
294
341

Ib - Ic = I3 = 145/-72

does not match