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

#### Ingenieur

##### Senior Member
Ia + w(Ib) + x(Ic) = 0

w = x = 2.02/1.84 ~ 1.1

so
Ia + w(Ib) + x(Ic) = 348 + 167j = 386/65 using your Iabc, and not 0

#### Phil Corso

##### Senior Member
Gentlepeople...

InJunEar's Prof is correct... not Complex-Numbers, not Matrix-Algebra, not Symmetrical-Components, and certainly not Xfmr-Impedance, will yield an answer to the OP's question... unless there is something known about the load!

The good news is that bank kVA, can be accurately determined! And, it's 110 kVA... close to the alternative calculations provided earlier. Thus, the probability one Xfmr is over-loaded is low, because total kVA is only 63% of the bank's capacity! Following is the procedure I used:

1) The LAW of COSINES was used to determine line-current vectors with respect to each other and the source voltage! For example, Vab, and Ia, were referenced @ zero degrees!

2) The TWO-WATTMETER METHOD was used to determine total kW and kVA!

3) Assuming Ia lagged Vab, I incremented the angle between them until kVA reached a maximum!

Regards, Phil Corso

#### topgone

##### Senior Member
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
Thank you for taking time to investigate the "erroneous" results I got in my previous post. I should posted "voltage loop inside the banked transformer delta" there.

In my haste to post the results when I got a convergence, I missed a lot of things including that of expressing the impedances with their respective base impedances and formulating the first equation wrongly.
BTW, to make up for my previous bo-bo, here's the latest figures I got (now back-checked wrt the OP line currents I1=256A, I2=294A, and I3=341A).
The set of equations involved are:
(1) Voltage loop inside the transformer bank delta:
Za.Ia + Zb.Ib + Zc.Ic = 0
(2) At the point where I2 leaves:
Ia - Ib = I2​
(3) At the point where I3 leaves:
Ib - Ic = I3
The results I got with the new corrected equations are:
IA = -130.08 -j 88.35 = 157.25 @ -145.82 angle ---> TR A 75% loaded (157.25/208)
IB = 16.92 +j166.26 = 167.12 @ 84.19 angle ---> TR C 80% loaded (167.12/208)
IC = 187.42 -j129.05 = 227.55 @ -34.55 angle ---> TR C 73% loaded (227.55/312)​

#### topgone

##### Senior Member
Typo error in the previous post-> the second line of the results should read "TR B 80% loaded" not "TR C"

#### Phil Corso

##### Senior Member
TopGone...

Several questions::

1) OP said 256, 294, 341 for Ia, Ib, Ic, so why did you ignore those values?

2) Op said source was hi-leg, so how did you come up with 3-ph, 208?

3) Assuming ph sequence is A-B-C, then (usually) Ia is the vectoral sum of Iab + Ica => Vab/Za + Vca/Zb, then how did you derive Za anc Zc?

Phil

#### topgone

##### Senior Member
TopGone...

Several questions::

1) OP said 256, 294, 341 for Ia, Ib, Ic, so why did you ignore those values?

2) Op said source was hi-leg, so how did you come up with 3-ph, 208?

3) Assuming ph sequence is A-B-C, then (usually) Ia is the vectoral sum of Iab + Ica => Vab/Za + Vca/Zb, then how did you derive Za anc Zc?

Phil
Ans.1 : In fact, I did consider those currents! Those values were used, multiplied with the inverted matrix to get the phase currents. Please see the vectorial addition of the resulting phase currents in relation to the line current: I've cut and pasted from the worksheet that I did, here:

CHECKING:
1 ZA.Ia + ZB.Ib + ZCIc = 0
ZA.Ia = 0.0817907627694643 -3.66702831201562i
ZB.Ib = -3.04173624049858 +2.43825621951076i
ZC.Ic = 2.88120218770385 +1.41101608222643i
-0.0787432900252658+0.18224398972157i = 0.198528027 --> small error as this should have been zero!​
2
IA = -130.08 -88.35i
-IB = -16.92 -166.2651052i
I_2? = -147.00 -254.61i
I_2 = 294.00​
3
Ib - Ic = I3
IB = 16.92 +166.27i
IC = -187.42 129.0495575 187.42 -129.05
I_3? = -170.50 295.31
I_3 = 341.00​
Ans.2: Also, the figure 208A (208.33 = 50,000VA/240V) is the full load amps of a 50 kVA, 240V singlephase transformer!

Ans. 3: Za, Zb, and Zc are transformer impedances!
Both Za and Zb were 50kVA, and the percent impedance used are typical values that I found on manufacturer data; 1.1%IR, 1.7%IX or a %Z = 2.02%
The third transformer was a 75kVA of which the impedance used was: 0.9%IR and 1.6%IX; or a %Z = 1.84%.

#### Ingenieur

##### Senior Member
that looks like it
good work
The way you got around the indeterminent matrix is clever
rather than using current for eq 1 you used the voltage loop
after that just crunching numbers

the weighted average yielded
128 (vs 157)
168 (vs 167)
226 (vs 227)
close except for phase A, wonder why?

couple of questions
did you assume 0/-120/120 for the line currents?
what ohmic values for Za, b, c (0.0175, 0.0175, 0.024 ohm ???)
thanks

A B = C

A
Za Zb Zc
1 -1 0
0 1 -1
Real, non-complex

B
Ia
Ib
Ic

C
0
294/?
341/?
complex?

A^-1 C = B

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#### Ingenieur

##### Senior Member
show me the error of my ways lol
I rounded to simplify the math (done by hand)

did you assign 0/-120/120 to row 1 matrix A???
basically the phase voltages, made the xfmr Z complex?

A
0.0175 0.0175 0.024
1 -1 0
0 1 -1

A^-1
16.95 0.7 0.41
16.95 -0.3 0.41
16.95 -0.3 -0.59

C
0
294/0 or (294 + 0j)
341/120 or (-170 + 295j)
(rotation should not matter as long as consistent?)

A^-1 C = B
Ia 182/42
Ib 199/142
Ic 174/-86
???

???

#### Ingenieur

##### Senior Member
calculated it 2 ways using topgone's equations

first using Z as complex (phased) offset by 0/-120/120

second using line currents as complex (phased) by 0/-120/120

#### Phil Corso

##### Senior Member
InJunEar,

So... your one of those "Damn the torpedoes, full speed ahead" guys!

Phil

#### Ingenieur

##### Senior Member
made a data entry error on the previous run using Z 0/-120/120 as the phased known
for Zc used -0.012 + 0.21j instead of the correct -0.012 + 0.021j
still no go

Iabc all magnitudes > 1300
all angs between 15-25 deg

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#### Ingenieur

##### Senior Member
InJunEar,

So... your one of those "Damn the torpedoes, full speed ahead" guys!

Phil
his answer looked convincing since the corner currents summed properly
I can't replicate his results though

trust me, smart guys can be wrong on occasion too lol

#### topgone

##### Senior Member
InJunEar,

So... your one of those "Damn the torpedoes, full speed ahead" guys!

Phil
You could be right there, Phil. I am still unconvinced with the results i got. The thing that the eq. 1 doesn't zero out keeps me thinking. There must be something wrong in the equation!

#### Ingenieur

##### Senior Member
I'm not sure kvl applies
this is a loop of sources (no drops or loads)
the net sum will never be zero

although if you add them up assuming the same polarity = 0

what is the sum of Vabc 240 at 0/-120/120 ?
maybe that should be the first element of C?
have to be careful with polarity conventions

although I still have doubts it can be solved without load data

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#### Phil Corso

##### Senior Member
TopGone, InJunEar...

Perhaps the following examples will show you the error of your way(s):

1) Given: 1-phase circuit; V= 117V @ 127 Deg; Line current = 56.3A! What is the magnitude and nature of the load?

2) Given: All of the above, but an installed Wattmeter indicates 1812W? What is the magnitude and nature of the load?

Regards, Phil Corso

#### Ingenieur

##### Senior Member
TopGone, InJunEar...

Perhaps the following examples will show you the error of your way(s):

1) Given: 1-phase circuit; V= 117V @ 127 Deg; Line current = 56.3A! What is the magnitude and nature of the load?

2) Given: All of the above, but an installed Wattmeter indicates 1812W? What is the magnitude and nature of the load?

Regards, Phil Corso
there is no 'error in our ways'
I concluded, even before speaking to my prof, that is was indeterminent

but topgone put effort into his work that deserved respect and careful consideration
especially in light of the delta corner currents summing to the respective line (in his calculation)

1 (the Vdrop must be the opposite of the source 117/-53)
pf = cos(127 - (53)) = 0.2756
Z mag = 1/(0.2756) x 117/56.3 = 7.555 ang = -106

2 P = 127 x 56.3 x 0.2756 = 1815 +/-

iirc you offered solutions using calculations, imaginary meters and such

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#### junkhound

##### Senior Member
fea

fea

No one familiar with FEA? (finite element analysis)

Easy, just draw the circuit and insert values, answer spits out for even the most complicated VFD circuit.

'student/eval version' good for simple power problems like this;

Have used a similar type FEA since 1988, absolutely indispensable for circuit design, esp power electronic circuits such as vfd design.

#### Ingenieur

##### Senior Member
correction
Z = 2.08/180

S = (1812/0.275)/(117-53) = 6590/74

Z = V^2/S = (117^2/254)/(6590/74) = 2.08/180

#### Ingenieur

##### Senior Member
No one familiar with FEA? (finite element analysis)

Easy, just draw the circuit and insert values, answer spits out for even the most complicated VFD circuit.

'student/eval version' good for simple power problems like this;

Have used a similar type FEA since 1988, absolutely indispensable for circuit design, esp power electronic circuits such as vfd design.
well, plug it in! lol

phase voltages Vabc = 240 at b0/-120/120 deg
line currents I123 = 256, 294, 341 (phasing not known)