- Thread starter LongCircuit
- Start date

- Location
- Boca Raton, Fl, USA

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

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.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

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)

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)

- Location
- Boca Raton, Fl, USA

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:

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

CHECKING:

1 ZA.Ia + ZB.Ib + ZCIc = 0

Ans.2: Also, the figure 208A (208.33 = 50,000VA/240V) is the full load amps of a 50 kVA, 240V singlephase transformer!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 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!

IA = -130.08 -88.35i

-IB = -16.92 -166.2651052i

I_2? = -147.00 -254.61i

I_2 = 294.00

3 -IB = -16.92 -166.2651052i

I_2? = -147.00 -254.61i

I_2 = 294.00

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

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. 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%.

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?**

w**hat 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

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

w

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

Last edited:

I rounded to simplify the math (done by hand)

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

???

???

- Location
- Boca Raton, Fl, USA

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

You ignored advice from your prof, (50 yrs, industry/Academic) and now me, (63 yrs, Industry/Gov/Academic)!

Phil

his answer looked convincing since the corner currents summed properly

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

You ignored advice from your prof, (50 yrs, industry/Academic) and now me, (63 yrs, Industry/Gov/Academic)!

Phil

I can't replicate his results though

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

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!

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

You ignored advice from your prof, (50 yrs, industry/Academic) and now me, (63 yrs, Industry/Gov/Academic)!

Phil

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

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

Last edited:

- Location
- Boca Raton, Fl, USA

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'

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

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 +/-

load is the same (obviously)

iirc you offered solutions using calculations, imaginary meters and such

Last edited:

- Location
- Renton, WA

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;

http://www.orcad.com/products/orcad-lite-overview?gclid=COG-5_yY7s0CFYVrfgod8eUPmA

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! lolNo 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;

http://www.orcad.com/products/orcad-lite-overview?gclid=COG-5_yY7s0CFYVrfgod8eUPmA

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

phase voltages Vabc = 240 at b0/-120/120 deg

line currents I123 = 256, 294, 341 (phasing not known)

- Location
- New Jersey

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.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;

http://www.orcad.com/products/orcad-lite-overview?gclid=COG-5_yY7s0CFYVrfgod8eUPmA

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