Quentin Grady said:
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.
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