Short Circuit Study

A good approach.

If you take out the 1.1 overvoltage factor S/C becomes 28.5/1.1 = 25.91 kA.

That is 25.91/24.4 = 1.0618 more than the manual calculation.

This is presumably the "6%" you discard by changing the transfo impedance tolerance.

A real system would not deliver 33 kV from 220 kV with a transfo ratio of 220/33. The regulation of the transfo at a normal power factor & load would have to be taken into account.

A trial with various values indicates regulation of 13% X, 0.5%R transfo at 0.9 power factor full load is 6.2%.

This gives rise to a question about what exactly this ETAP program automates for one - does it include setting up a typical load flow solution & assumption of transfo ratios? Or did the writer of the original question tell the program the initial condition at 33 kV was zero load & transfo ratio is 220kV:33 kV?

One assumption is that the 3 transfos are paralleled at 33 kV.

An exact numerical solution is an asset if it is understood what procedure it is following, but does not facilitate understanding if it has hidden assumptions.

I agree the c factor i have consider max i.e. 1.1.

But tolerance on impedance consider is 7.5 % as per Is. Obviously by default Etap will also calculate the SC on 7.5% negative tolerance.

1. Why it will consider only 6 % as SC should be done for worst possible case i.e. 7.5% negative tolerance.

Do i have to consider no load voltage of transformer to 34.5 instead of 33 kV?

If i m taking c factor 1.05 & no load votage 34.5 kV

Then i am getting the same result of 24.4kA.

But in manual calculation these factors i hv not consider as i mentioned in my original query

In reply to posts #4 & #5 by "Pra2 4"....

It was post #1 reply which suggested 6%, bending tolerance to fit the numbers - I suggested a different reason for a discrepancy of about 6%.

You need to think what a "220/33 kV, 65 MVA" transformer means. Since it is a distribution transformer, the user expects it to deliver continuously 65 MVA at 33 kV output. That means an output current of 65000/(33*√3) = 1137 amps/phase.

What does Z=14% mean? It means that if you apply a sinusoidal 3 phase, symmetrical 50.00Hz (you never stated frequency, but 220 & 33kV suggest a 50Hz standard) source to the HV terminals and raise voltage until secondary current, with zero ohms loadimpedance, is "continuous rated" value (1137 amps r.m.s., for example), the applied voltage is 14% of 220kV [line to line] = 30.8 kV r.m.s. (+/- the permitted tolerances).

Note: real transformers are not symmetrical, for the usual 50Hz construction, the centre limb has a different flux to the outer ones, so impedances are not the same for each phase - you would have to look at the transformer standard to see if that 1137 amps is the mean value of 3 phases or ??

Getting 33 kV out on Low Voltage winding means more volts at HV to allow for regulation according to usual "per unit" equation, using the resistive and reactive components of the impedance voltage...

E = E_r*cos∅ + E_x*sin∅

Which was what I tried several values with, to guess a regulation value of about 6% [with 14% nominal impedance] corresponds to about 0.9 power factor lagging at rated current & voltage on LV winding.

Which all suggests that the computer program you are using has a different value for e.m.f. than 33 kV and transfo ratio than you assume with your simplified check calculation.

You really need to look at what values the program is using - it appears it will insert some realistic values unless you give it some system specific values.

Welcome, Pra2_4,

I am far from being an expert in this field, but I generally would trust the calculations from etap and similar software. I wonder if the calculations in either of your methods includes any contribution to the current from the collapsing magnetic fields in two of the three transformers when there is a sort on the third. Perhaps someone with more knowledge can educate me.

--JMM

jmueller: do you mean the 1st configuration:

Here the no HV two transformers are supplied from 33 kV -like two no-load step-up

transformers.

or the 2nd:

where the two other transformers are supplied from the first secondary and HV side of both is short-circuited. However the resulted impedance is more than if the short-circuit was at first transformer secondary terminals then the current will be less.

7anoter4,

in my limited experience, the fault energy is generally higher on the secondary (low voltage) side of a transformer than on the primary side, because the overcurrent protection on the primary side will respond more slowly to a secondary fault due to the impedance of the transformer. Now, if the transformer is being back-fed as a step-up transformer, then the available current and clearing time will result in comparatively low arc energies; but this configuration usually requires that the transformer be approved for this use.

Since the purpose of arc flash calculations is to determine the worst case energy a person could be exposed to, so proper steps for safety can be taken, I would do the calculations for all fault configurations and then post the results for the worst of them. At a transformer, unless the primary and secondary are physically separated by an suitable robust barrier capable of keeping the released energy and the arc itself from migrating from one side to the other, the relevant information that should be considered is the highest calculated fault energy and the highest voltage (these typically are from opposite sides of a transformer).

In regard to the two configurations you posted, I do not otherwise know how to respond to your question, other than to repeat, that all fault locations/types should be considered when doing the study.

--JMM

what exactly do you mean "Z = 14% with 7.5%"

This statement makes no sense

Z is either 14% or 7.5%

Which is it.

assuming Z = 14%

I get a 24.9KA

but that is assuming all idealizes.

What is your sentence meaning when you say 14 @ 7.5?

I am not familiar with this terminology

Pra2 4 meant to type "with 7.5% tolerance", which becomes clear in original post when he writes later....

"Zt=14%, with negative tolerance zt=12.95%"

14% x 0.925 = 12.95%

thank you for clarifying

His post is also very unclear to me what kind of fault

3P or L-L

Where does it occur and when?

Would you perhaps explain when a 7.5% tolerance means the Z goes to 12.95%?

Does that mean that the Z is kinda unknown by the manufacture and it could be plus or minus 7.5%?

I got 24.8KA with a 14% value doing it by hand

It would certainly be 3 phase. It involves the most power and disturbance to the system. In any case, there is no information on earthing or any other factors which could make looking at earth or line-line faults of benefit.

The calculation of the effect of the tolerance requires calculating the lowest value of impedance, which gives the highest and most destructive current.

7.5% tolerance means a maximum variation of +/- 7.5 parts in 100 from the average (design) value. That is actually assuming that the 7.5% is a maximum (every one is tested and wider values are rejected) rather than than a proportion of a sample of 100 from a crate of 10000 screws which is a reject level on some dimension.

That means a minimum of (100- 7.5)/100 as a proportion of the design value i.e. 92.5/100 = 0.925 - which is the value I wrote.

Since the design value is 14%, the value with 7.5% tolerance is then 0.925 x 14 = 12.95%, as appears in the original post.

The manufacturer designs for a particular value, but manufacturing variations in the size and position of the coils in the transformer and air gaps in assembly of the cores, tolerances of lamination thickness etc all have an effect.

According to my memory, tolerance in standards is +/- 15%, but the standard was set a lifetime ago and covers a very wide range of transfo size and voltage. I expect manufacturing tolerances have improved over time and typical test limits for Grid size/voltage power transformers have proved such that an expert program can assign 7.5% tolerance.

The definitive test for S/C impedance is a short circuit test but this requires test gear of sufficient power and accuracy which is expensive to build and use and probably an independent organisation from manufacturers. A customer does not want to pay for this unless it is necessary, nor does he want a stress which will probably occur only once in the life of plant applied more often than necessary. The standards usually have a proviso that any repeat of test should be at 80% of nominal voltage [mechanical stress is proportional to current x current so that is really 0.8 x 0.8 = 0.64 of stress], to avoid tests themselves becoming a cause of in-service failure.

The impedance test in-factory is usually done at rated current, not at the S/C current - there will be differences due to more saturation at high currents, and should really be compared with the prototype test at the same current.