G'day G'day Folks,
I have been working my way through some old advanced trades exam
papers and came across this question that frankly I don't understand.

A 250 kVA, 400 V 3-phase transformer has 5% impedance.
(i) Determine the fault level which could be produced by the
transformer.
The model answer is 250 kVA/0.05 = 5000 kVA.
OK, the calculation is easy to perform but I don't know what it means.
Clearly the 0.05 is 5% but what is being calculated and how did an
impedance get to have a % instead of Ohm units.
What the heck is fault level?
(ii) Determine the prospective short circuit current that would flow
if a short circuit of negligible impedance occurs across the
transformer output terminals.
Thanks,

It is a common practice to use "per-unit" values for voltage,current, power,
etc
A transformer may be rated 2400/240V, (4.17/41.7 A)10KVA and have an
impedance of 29 ohms as seen from the primary and 0.29 ohms as seen from the
secondary.
On the primary the base voltage =1 pu $00V and the base current 1pu
=4.17A and base power KVA
The base impedance seen from the primary is 2400/4.17W6 ohms =1 pu
the actual impedance is 29/576 = 0.05 pu or 5%
From the secondary side the base voltage $0V, base current A.7A and
power 10KVA
the per unit impedance becomes 0.05pu as seen from the secondary.
If you had rated power 1.0pf at 240V the current would be 1 pu and the
input voltage would be 1 +(j0.05)*1 =1.0012 (or 2403V) @ a phase angle of
2.9 degrees The voltage regulation would be 0.12% (impedance assumed
purely reactive)
In the per-unit notation, transformers essentially become 1:1 and are
representable by their pu impedances. Typical distribution transformers will
be about 0.05 pu impedance and station transformers will be about 0.1 pu.
Generator impedances fall into a narrow range-dependent on the type of
machine, etc. A complete power system can then be represented this way and
if one wants the actual voltage at a point -simply multiply the pu value by
the base voltage chosen at that point. Note that for a system, a single base
power is chosen and all pu impedances are converted to that base. The above
transformer, on a 100KVA base would have an impedance of 0.5pu reflecting
its smaller size.
Why do this? On a system basis , it makes calculations a lot easier and
relative quantities are easier to see.
Scaling would be necessary in case of a unit having a base voltage or power
which differs from that chosen for the system. This is straightforward.
There are corrections for changing the tap of a transformer without changing
the base voltage so that, in such cases, there is a turns ratio such as
1:1.03 but for computer modelling a pi circuit model can eliminate
this-keeping a 1:1 ratio. Also straightforward.
***
Fault level as can be seen from the calculation is the short circuit
current KVA with rated voltage applied, as limited only by the transformer
impedance. It is the corresponding short circuit current * the rated
voltage. Important for breaker ratings
In your example: the impedance is 5% of 400*400/250000=0.032 ohms
The short circuit current would be 400/0.032 ,500A (answer to (ii)) and
the fault va would be 12500*400P00000Va P00KVA (the answer to (i))
What is expected for (ii) is that you take 5000KVA and divide it by the
voltage (400V) to get the current .
Now isn't that a lot easier?

--

Don Kelly snipped-for-privacy@shawcross.ca
remove the X to answer

G'day G'day Don,
What I find remarkable is that there can be different yet equally
valid descriptions of impedances as percentages or P.U.
I notice you appear to be performing a single phase transformer
calculation. Should I modify this for a three phase transformer?

OK. Knowing which assumptions are valid is so important and to an
outsider like myself not immediately apparent. I'm happy with a
transformer being purely reactive though I must say I thought the
windings would have had resistance.

Cunning. It opens up a new word of thought for me.

OK. Let's check that I have that right.
So "level" is a kVA rating
= short circuit current x rated voltage.

That seems so simple.
What puzzles me here is that it seems OK to treat this as a single
phase calculation. Sometimes one can't. Frankly I don't have rule
for when it is OK and when it isn't.
Obviously it generates the correct answer to think in single phase
transformer terms. That puzzles me.

------
For 3 phase the Z is that per phase for a Y. The voltage and power are based
on line to line voltage and total 3 phase power. This gives the same result
as using per phase power and voltage. You are always safe using phase
voltage and power and phase impedance.
---------

--------
Balanced system. If unbalanced the usual approach is to use symmetrical
components and consider 3 systems- 2 balanced (positive and negative phase
rotation)and the third a set of equal single phase (ground related-0 phase
rotation) quantities. The analysis reduces to solution of 3 single phase
systems connected according to the fault conditions.
Get a good book on power systems - it's all covered there.

--

Don Kelly snipped-for-privacy@shawcross.ca
remove the X to answer

Welcome to the world of P.U.
When calculations and things had to be done by hand, it was often a lot
easier to convert things to some common, normalized base. And because
engineers sometimes are somewhat 'math-challenged', they didn't always like
using percentages. So they came up with 'per unit' (often abbreviated P.U.)
To calculate the P.U. rating of something, you simply divide by the 'rated
value' and expressed the results in decimal. This avoided having to
remember to always divide by 100 like you might do when using percentages.
It worked something like this. If the rated voltage of a unit was 4160V,
then if you applied 80% of nominal voltage, you don't calculate it as 4160V
* 80 = 332800V. You applied 0.8 P.U. and calculated 4160V * 0.8 = 3328V.
Yeah, I know it's just a matter of remembering what a percentage was, but
that's what P.U. started out as. It also made multiplying percentages
easier. You know of course that you can't just say 2.5% times 80% = 200%.
You have to remember that 2.5% equals 0.025, 80% equals 0.80 and 0.025*0.8 0.02 => 2%
Once things are in P.U., some calculations, like primary/secondary
calculations of a transformer can be easier. After all, if the primary is
supplied with 0.8 P.U. voltage, then the secondary will output 0.8 P.U.
voltage. The primary might be 4160V (0.8 of that is 3328V) and the
secondary 400V (0.8 of that is 350V).
Next, you have to figure out what the 'rated' units were. In the case of
generators and transformers, the one 'rated' value is the kVA or MVA rating.
So the base for your transformer is 250 kVA.
So the 'base' impedance for such a transformer is that impedance that will
allow 250 kVA apparent power when rated voltage is applied. If the voltage
is 400V, then 250kVA / 400 = 625 amps. But to get 625 amps at 400 volts, it
would mean the impedance of the connected load is 400V / 625A = 0.64 ohms.
So that is the P.U. impedance of the secondary. Not the *actual* impedance,
but just the 'base' for impedance ratings. If the *actual* impedance is
0.05 P.U. then it would be 0.64 ohms * 0.05 P.U. = 0.032 ohms.
With a 'bolted fault' on the secondary, you have 400V and an internal
impedance of 0.032 ohms, so current would be 400V / 0.032ohms = 12500 amps.
12500 amps, times the 400V would be 5000 kVA.
A lot of math. Or, just use 250 kVA / 0.05 = 5000 kVA. You can see that
using the P.U. calculations gets the same answer, and quicker. But it takes
a little 'leap of faith' and understanding of the whole idea of P.U.
ratings.
If you have several different pieces of equipment chained together, like a
generator, a step up transformer, and a transmission line, the calculations
can be quite hairy. But there are simple ways to re-define the P.U. ratings
of each into one common base. Then the P.U. method can be used to simply
add up the impedances and such and get the right answer pretty easy.
For example, an 800 MVA generator with a synchronous reactance of 0.02 P.U.,
combined with a 25kV/350kV step up transformer rated at 950 MVA with
impendance of 0.05 P.U. and a 350 kV / 1000MVA transmission line with an
impedance of 0.04 P.U. to the point of interest. If these are all converted
to the same 'base' (such as 350kV and 950 MVA), then the rest of the
calculations are easier.
But nowadays, with spread-sheets and computer software (not to mention
specialized programmable calculators), the use of P.U. calculations have
somewhat fallen from favor. Nevertheless, unit ratings of power equipment
are still given in P.U. or percentages. The internal impedance of
generators (synchronous reactance, transient reactance, etc...) are also
expressed this way.
A bit wordy, sorry about that. But that's some of the history behind this
practice.
daestrom

This post not CC'd by email
On Wed, 02 Aug 2006 22:48:02 GMT, "daestrom"

G'day G'day Daestrom,
Thank you for your extensive answer. It is much appreciated. I'm
bit puzzled by the last paragraph. The calculation seems to be for a
single phase transformer while the problem was for a three phase
transformer. I know when we think and type at the same time it is
easy to overlook such things.
Does the number of phases alter the calculation?
Is this one of those "equivalent" resistances that is used when a
three phase circuit is described in terms of a single line and an
"equivalent resistance?"

Now I am really puzzled. 5000 kVA is the model answer.

OK.

I don't mind the wordiness.
I'm looking for understanding and will do what it takes to get there.

The 5% means that if you short circuit the secondary and apply enough primary
voltage to get
*rated* current you will find that the primary voltage is 5% of rated.
I don't know a short answer for why the 'per unit' system (which is where your %
comes from) is
used, but I do know that if you have a lot of similarly constructed equipment,
the per unit or %
impedance will be the about the same over a wide range of equipment ratings.

Short circuits are called 'faults'.
5000kVA would be (about) 7200 amps at 400V. (5000/(sqrt(3)*400)
This is the current that you would get if you shorted the secondary and applied
rated voltage to
the primary. It's also 1/0.05 * rated current.... 20 times rated current.
Breakers (at higher voltages) are rated in MVA of interrupting capacity at a
particular voltage.
I would expect a 400V system to have breakers rated in interrupting current
capacity. At least
that's the way 480V breakers are (were?) rated in the US. (It's been 20 years
since I looked at
a low voltage system calc.)

--
Fred Lotte
snipped-for-privacy@nospam.stratos.net

This post not CC'd by email
On Wed, 02 Aug 2006 17:30:18 -0400, Fred Lotte

G'day G'day Fred,
Thanks. That makes sense now. I was trying to relate it to input
regulation. change in Vout/change in Vin. It wasn't working for this
situation.

Level had me fooled. Fault current would have made sense.

Available fault level can refer to current or to VA. They are directly
dependent, so either value gives the same information ultimately. Though
each (Amps and VA) have some different uses, e.g. the breaker ratings
someone mentioned.
The short circuit VA at a three phase transformer secondary with an
"infinite primary" (no impedance in the primary - a worst case assumption
used in certain very common calculations) is
VAsc = VAtxf / Zpu_txf
The short circuit VA equals the transformer rated VA divided by the
impedance in per unit (5%Z -> 0.05pu)
In your example,
5000000 = 250000/0.05
5MVA%0kVA/0.05
The available fault level is 20 times the transformer rated VA, because the
transformer Z is 5% and 1/5% . 20=1/0.05.
What happens there is a bit of a trick. You prolly know that V/Z=I. You
prolly know that S=V^2/Z. What happens when V=1, as in 1pu? That equation
becomes S=1/Z. 20=1/0.05. The 20 is 20pu VA, the 1 is pu voltage, the 0.05
is pu Z. To get from 20pu VA to the actual VA you multiply by the base,
which here is 250000VA. You get 5000000VA. 5MVA.
The current is 5000000VA/(400*sqrt(3)) r16.878A. You are right, you do
have to be careful about your three phase calcs. You have to be more
careful with pu calcs. To do the 'routine' of
SCkVA=txfkVA/txfZpu
(that 'shortcut' combines the pu calc and the conversion back to actual kVA
from pukVA) then
SCI=SCkVA/(Vll*sqrt(3))
is fine but don't try extending the calculation to consider other equipment,
connections, impedances, etc. without knowing your pu. As people have
pointed out it gets complicated with the different 'base' values that are
used and often need to be converted.
The factor to change between fault level given in Amps versus fault level
given in VA is Vll*sqrt(3). Watch out for kVA and MVA values for fault
level, remember to factor the 1000 or 1000000.
j

G'day G'day Operator Jay,
I love your internet name, operator j.
Thanks for making a complex subject at least seem simple. <grin>
Thank you all, who have replied. I've spent a couple of decades
teaching electrical theory but not in a trades related field and the
terminology threw me.
Over a year ago I was diagnosed with an incurable cancer and treatment
to control it failed. At the moment I scratch up something of a
living being a support tutor. I get given the students whose
technical subjects fit into the other tutor's too hard basket.
The mental stimulation of learning about topics like P.U impedance
helps keep me alive, so may I take the opportunity to thank everyone
who has generously given of their expert knowledge.
You have helped in ways you could not have imagined.
I've still to sort out when I should use single phase formulae and
when to use three phase formulae so expect some more questions. Your
extensive answers have been much appreciated.
Best wishes,

Part of the perunit paradigm is using 'single' phase methods to solve multi
phase problems. May
be better stated as reducing multiphase problems to a single phase problem. When
the system is
not 'balanced', that is each phase is not identical to the next or currents or
voltages are not
displace by 120d exactly from one phase to the next, the method of symmetrical
components is
used to convert the multi phase problem to a single phase problem. The balanced
system is just a
simple special case. Actually, the perunit system may be part of the symmetrical
components
paradigm.
It is very far beyond the scope of a newsgroup posting to describe the method of
symmetrical
components. The only book I can recommend is the long out of print Westinghouse
T&D reference
book. On line, a google search on the phrase "symmetrical components" (with the
quotes) produces
a credible number of hits including a wikipedia entry.
<http://en.wikipedia.org/wiki/Symmetrical_components
There is also a wiki on the perunit system.
<http://en.wikipedia.org/wiki/Per-unit_system
(I suspect that the article is wrong when it states that "Since power system
analysis is now
done by computer the use of this system is rare except for analysis of a single
machine or
manual analysis of small (3-4 node) power systems." but I'm not in the industry
anymore so I
don't have first hand knowledge.)
My copy of the second edition of the Stevenson book that is mention is packed
away, so I can't
look to see what the treatment is. I note from the google search that
symmetrical components is
taught at the senior or graduate level.
When everything is 'balanced' only the 'positive' sequence has non-zero voltage
sources. It is
simply a single phase representation of the power system. Values are given in
the 'perunit'
system which incorporates necessary factors (sqrt(3) and transformer ratios for
example) in the
development of the 'base' values. The phase shift 'thru' a wye-delta
transformation is handled
as a phase shift multiplier in the + and - sequence circuits. Converting to
perunit gets rid of
those pesky sqrt(3) and transformer ratios considerations.
When the system becomes 'unbalanced' such as a phase to ground fault or phase to
phase fault or
a significant single phase load, the 'negative' and 'zero' sequence circuits are
connected to
the positive sequence circuit in special ways depending on the type of
unbalance. In addition,
the negative sequence circuit will have non-zero voltage sources if the phase
voltages are
unbalanced in any way. Now you get to solve a more complex circuit that may have
multiple
parallel paths and loops, but is still single phase. In addition to the perunit
system
quantities, only transformer tap settings enter into the calculations and wye or
delta
connections are important only if ground or neutral current is involved (or the
fault is between
voltage levels).
Best wishes and Good luck

--
Fred Lotte
snipped-for-privacy@nospam.stratos.net

This post not CC'd by email
On Thu, 03 Aug 2006 18:07:51 -0400, Fred Lotte

G'day G'day Fred,
That makes things clearer. I had wondered whether some poeters were
making an oversight and canceling errors let the correct answer emerge
or whether it was somehow valid to use a single phase model for
balanced transformer loads.

Thank you. I spent part of this morning searching the tertiary
institute library for any reference to P.U. impedance and found only
one.

Thanks, Fred,
While I have taught transformer efficiency for decades I had no idea
this P.U system existed. Looking through the library gave no
indication that there was so much literature on the subject.
Best wishes,

Internet name?
Just kidding. I didn't make it up it is from a long winded joke I read once
in an email. People submitting empty envelopes ("imaginary cheques") to the
utility to pay for the imaginary power they'd billed for and so on. I'm not
convinced j is an operator (more of a "number" than an "operator"?), but it
sure does some operator like things, and I've seen lots of textbooks written
by people a lot smarter than me refer to it that way, so good enough.
j

The basic rule is divide impedance into 100 and multiply this times the
full load amperes.
100/5 = 20
FLA %0000/400x1.732 amperes
FLA = 361 amperes
20 x 360 = 7220 amperes
If a phase to phase short circuit were to occur at the transformer
output terminals the instantaneous fault current would be approximately
7220 amperes. This current would become less at the downstream
equipment as as conductors become longer and smaller.

I agree with your calculation for a 3 phase bolted fault at the secondary of
the transformer. I also agree that the current will be lower for a fault at
some distance "downstream". This "rule" is applicable and useful. However I
do have some boringly pedantic nit-picking.
These calculations deal with the nominal "steady state rms" current that
would occur -not the instantaneous current. There is no information given
that allows the determination of instantaneous currents as that depends on
the point in the cycle that the fault is applied and the related transient.
An RMS value is never "instantaneous"-by definition.
I also assume that what you are calling "phase to phase" is a 3 phase fault.
For what is normally called a "phase to phase" fault the 3rd leg is open
and the fault current between the other phases is not 20 times but 17.3
times (in this case) the rated current.
My problem is that certain expressions such as "instantaneous" and "phase
to phase" (as opposed to "3 phase" have well established usage which is not
how you are using these terms.

--

Don Kelly snipped-for-privacy@shawcross.ca
remove the X to answer

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