5% impedance

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
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.
Reply to
Quentin Grady
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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.)
Reply to
Fred Lotte
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.
Reply to
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 =2400V and the base current 1pu =4.17A and base power =10KVA The base impedance seen from the primary is 2400/4.17=576 ohms =1 pu the actual impedance is 29/576 = 0.05 pu or 5% From the secondary side the base voltage =240V, base current =41.7A 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 =12,500A (answer to (ii)) and the fault va would be 12500*400=5000000Va =5000KVA (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?
Reply to
Don Kelly
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voltage to get
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.
% comes from) is
the per unit or %
Level had me fooled. Fault current would have made sense.
rated voltage to
particular voltage.
capacity. At least
since I looked at
Special thanks,
Best wishes,
Reply to
Quentin Grady
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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.
I don't mind the wordiness. I'm looking for understanding and will do what it takes to get there.
Thanks and best wishes,
Reply to
Quentin Grady
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.
It sure is. Thank you,
Best wishes,
Reply to
Quentin Grady
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=250kVA/0.05
The available fault level is 20 times the transformer rated VA, because the transformer Z is 5% and 1/5%=20. 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)) =7216.878A. 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.
Reply to
operator jay
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G'day G'day Operator Jay,
I love your internet name, operator j.
Thanks for making a complex subject at least seem simple.
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,
Reply to
Quentin Grady
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.
There is also a wiki on the perunit 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
Reply to
Fred Lotte
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phase problems.
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.
When the system is
voltages are not
components is
system is just a
symmetrical components
of symmetrical
T&D reference
quotes) produces
Thank you. I spent part of this morning searching the tertiary institute library for any reference to P.U. impedance and found only one.
analysis is now
machine or
anymore so I
away, so I can't
symmetrical components is
sources. It is
the 'perunit'
example) in the
transformation is handled
perunit gets rid of
to phase fault or
are connected to
unbalance. In addition,
voltages are
have multiple
fault is between
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,
Reply to
Quentin Grady
----------- per unit expresses the value on a basis of 1 while % is on a basis of 100 a 5% discount on a $30 item is $30*5/100 =30*0.05 Similarly 5% impedance =0.05 per unit
The original question/ answer given did not specify whether this was 3 phase or not. If it was 3 phase the 5000MVA would be the total 3 phase bolted fault MVA rather than that of a single phase (i.e. each phase rated 250/3KVA). This is the usual way to express it.
If we have a transformer with 5% impedance and rated at 416V, 30 KVA 3 phase then we have, per phase, for a y connected transformer
Prated =10KVA Vrated 240V and using these as a per phase base Pbase =10KVA Vbase =240V and Zbase =Vbase^2/Pbase =240^2/10000=5.76 ohms and 5%of this is 0.288 ohms which is the ohmic impedance of each phase (Ibase =41.7A). On a 3 phase basis Pbase =30KVA, Vbase =416V so Zbase =416^2/30000 =5.76 ohms as before giving Zbase =0.288 ohms. Zbase is always the actual per phase impedance assuming a Y. (Ibase =30000/1.732*416 =41.7A but don't calculate Zbase from this.) The results using line quantities are the same as using per phase values for a Y. If one wanted to look at a Delta then use Pbase =10KVA, Vbase =416V so Zbase =17.3 ohms and I base =10000/416 =24A (corresponding to 41.7A line) Zbase/3 =5.76 ohms (familiar?). In typical one line analysis of a balanced system, any deltas are converted to Y equivalents as seen by the external system. (Zy =Zd/3) so using line or phase voltage and power leads to the same result. Only if you want to know what is going on internally in the transformer, or in the case of unbalanced faults with ground currents, do you actually need to "know" for analysis, that it is a Delta-delta, delta wye, Wye -wye-delta or whatever. (there are other concerns but those are another matter) -------------
------- Sure a transformer has resistance but transformers of this size will typically have R/X
------ 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.
Reply to
Don Kelly
"Quentin Grady" wrote in message
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.
Reply to
operator jay
The basic rule is divide impedance into 100 and multiply this times the full load amperes. 100/5 = 20 FLA =250000/400x1.732 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.
Reply to
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.
Reply to
Don Kelly
G'day G'day
Thank you,
Best wishes,
Reply to
Quentin Grady

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