How to calculate if one transformer of a three phase Star/Delta bank is overloaded?
- Thread starter LongCircuit
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Phil Corso
Senior Member
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- Boca Raton, Fl, USA
InJunEar...
See, others are learning! Here's the rest!
Part 1, only V and A are known, so S (Apparent Power) = V x A = 126 x 56.3 = 7,150 VA!
Part 2, since P = 1,182 W (Real Power) is now known, then Q (Reactive Power) = Sqrt(S^2 - P^2) = 7,052 VAr!
I hate to ask the following of you, but since you didn't participate, how about revealing the load's power-factor... lagging or leading (decimal or %'s acceptable)!
Phil
See, others are learning! Here's the rest!
Part 1, only V and A are known, so S (Apparent Power) = V x A = 126 x 56.3 = 7,150 VA!
Part 2, since P = 1,182 W (Real Power) is now known, then Q (Reactive Power) = Sqrt(S^2 - P^2) = 7,052 VAr!
I hate to ask the following of you, but since you didn't participate, how about revealing the load's power-factor... lagging or leading (decimal or %'s acceptable)!
Phil
Whoa! Whoa!
We know S, and not P or Q, is what I said. How did we jump to knowing P in Part 2?
Ingenieur cannot tell you the load power factor because it has not been stated or otherwise established. For his calculations, he in effect stated the phi as an assumed 0°/120°/240° respectively. However, in this case, phi and load power factor (or rather arccos thereof) are not necessarily, likely not the same thing.
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- Placerville, CA, USA
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- Retired PV System Designer
As written, only the magnitude of S is known.
If the phase offset on one L-L current is different than that of another connected to the same line it affects the magnitude of the corresponding L-N current and vice versa.
I am aware of that. Please keep the response simple. S* [scalar value; magnitude, if you will] is known, correct? P* and Q* are not, correct?
*If we knew P or Q we could establish S vector.
How is knowing P and /or Q relevant?
What would knowing P and/or Q change?
- Location
- Placerville, CA, USA
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- Retired PV System Designer
It is not relevant as long as the phase angle of all three loads is the same.
I assume you mean the same phi (difference of current to voltage angle).
You do realize the probability of that being the case for unbalanced current magnitudes is quite, quite low, right there next to improbable, right?
Phil Corso
Senior Member
- Location
- Boca Raton, Fl, USA
Fellas...
I'm not going to play your game of Obfuscation (typical of most engineers)! If anyone is interested in what I call "Vectors from Magnitudes" contact me off-forum!
Regards, Phil Corso
I'm not going to play your game of Obfuscation (typical of most engineers)! If anyone is interested in what I call "Vectors from Magnitudes" contact me off-forum!
Regards, Phil Corso
You want only phase current magnitudes. Kirchoff current law is applicable to both dc and ac circuits. So it should be applicable in an ac circuit with effective dc values only. So by considering line currents as dc values and replacing transformers by ideal dc sources (as a first approximation ) and by applying Kirchoff current law, equivalent dc phase current and so solution to your problem may be found.
Sorry. I committed blunder: Kirchoff law cannot be appled that way..................Well another try. Suppose phase sequence of line currents is known. Then line currents can be represented in trigonometric form as functions of time. Each line current may also be represented as difference between respective phase currents also in trigonometric forms by applying kirchoff law. The resulting equations may be solved for an arbitrary instant of time to find intantaneous values of phase currents. From that it is easy to find RMS values.
You are correct
but with the given info P or Q can't be calculated
either one will give power factor
his example is easily calculated
1812 = 117 x 56.3 x pf
pf = 0.2751
ang = 74 deg
pf delta ang v - i
since v = 127 i = 53
Z = v/i = 2.08/74 Ohm
and
P 1812 given
Q 6332 j
S 6603/74
but the op NEVER gave P , Q or V phase, only V and I which yields a per ph kva
assume load is a ungnded wye
S/ph = 240/1.732 x Ix
frmm the kva a per phase |Z| can be calculated
|Z| = ph kva/V^2
op kva
35.5
40.7
47.25
Z
1.85
2.12
2.46
Iph
148
170
197
no phase info can be determined
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AHA! I traced his "example" back to post #75.
Simple answers are best. It seems no one knows how to do that.
So it's just another "if" scenario that leads us nowhere, right?
I've often said people have a tendency to tell on themselves:
I'll buy anyone a beer (craft import to boot!) who can get mr phil to answer my question
:roll:
If a tree falls in the forest what is the power factor?
You have to be in the forest when the tree falls to hear his answer. :happyyes:
if that is the case I may jump into the trees fall path
I could not bear another imaginary watt meter postulation
Phil Corso
Senior Member
- Location
- Boca Raton, Fl, USA
InJunEar,
Just contact me off-forum!
Phil
Just contact me off-forum!
Phil
Hi folks! If each line current is known in trigonometric or vector form i.e if its intantaneous value as a function of time is known or if at least sufficient number of intantaneous values of each line current are known,then the phase currents may be found out.
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- Location
- Massachusetts
Phil this is lame.
You keep telling posters they are wrong but you won't go into the reasons why or show how you feel it can be done.
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