Stuck on Some Calculations

Did you mean to post a diagram or other information about the system containing "bus 4" ??

Look at Wikipedia or other web sources/search engines for information on "3 phase electrical fault" or "earth fault" or "power system fault calculations".

Simple, V=0.

Apologies, forgot to post diagrams I have been working on:

https://www.dropbox.com/s/zjubb5yupoknwzc/eracs.docx?dl=0

How many points is a correct answer worth?

? Have I missed something here -- is this not a forum to ask questions. I've taken the work as far as I can (see the attached doc), and am having issues with these. If someone has a suggestion for an alternative resource, do let me know...

It is certainly a forum to ask questions.

You asked a question and so did I.

So far, you have gotten two answers to your question and I'm sure you will get more, from many other people.

My question seems very simple, but only you know the answer. So far you have evaded answering it.

Maybe this will help:

Short Circuit Fault Calculations - Electrical Construction & Maintenance

Frequently asked questions on grounding and ground-fault current

I have looked at your diagram - at first sight, assuming generator at left is only power source, you need the voltage, impedance and MVA rating of that generator and the impedance, ratio and MVA rating of the the transfos bus1 to bus 2 and bus 4 to bus 5. This assumes impedance of 125 kV lines is negligible.

Look again, both generators show line current so both are generating. The thing that stands out the most is that the generator on Bus1 is connected to a transformer with Vector Group Dyn while the generator on Bus5 is connected to a Ynyn, and both feed the same bus! I do hope there's a phase shift in one of those transformers.

Until that little detail is cleared up, nothing can be said...

Two generators, you are right!

Two independent generators would adjust their frequency and phase to synchronise and meet the phase relations required by the loads and transformers (as do all parallel generators).

Is it a system grown organically from affordable parts or a theoretical example?

In my humble opinion, since the are two independent sources-it is not a single synchronized grid-and if the phase rotation is the same one could synchronized a second generator with an already energized grid by the first generator.

As it is already underlined above no equipment data is presented. The loads and short-circuit current levels only it does not permit any accurate calculation.

Generator rated MVA and X"d ,Transformer rated MVA, Zk1, Zo, line R, X1, Xo, PQ source data all these are required.

For calculations of this type I use "Inductance Calculations" by Frederick W. Grover

I agree, it is a very interesting book.

http://g3ynh.info/zdocs/magnetics/appendix/Grover46/Grover_errata.pdf

However, I am not sure it is connected to this required calculation [see o.p.]:“How does one go about calculating the short circuit level (in MVA) for a three-phase fault on Bus 4?”

From this book, one can calculate the energy stored in a bus system. Knowing this energy will permit calculation of the short circuit current under specified fault conditions. We do this when we design high current welding equipment.

You can calculate the 3 phase equivalent impedance from the fault data given, according to the (standard) ratio between fault types. I=E/Z

Is not question 2 answered on page 2?

Some algebra to separate the zero sequence impedance from the difference of the double line to ground & single line to ground equivalent fault impedance...

No algebra is required, any book on symmetrical components will have the sequence diagrams and the equations for all the possible types of faults; i.e., L-G, L-L, L-L-G, L-L-L, and L-L-L-G, in terms of line, phase, and sequence currents.

As presented by OP, there is insufficient information to accurately calculate the short circuit currents.

It's all there, look at all 3 diagrams

I did, the necessary parameters are not given on the diagram, and I'm certainly not going to back-calculate to find them. Perhaps you can tell me what they are, given the information on the diagrams.

I'm particularly interested in the time dependent machine parameters such as X"d, X'd, Xd, Τf_, etc.? How about the transmission line parameters like R+jX (that one is relatively easy), Transformer parameters like rated Voltage, rated MVA, per unit impedance, etc. Then there's the Loads on bus 2 and 4, they all seem to be quite inductive, probably large motors, but we are not told whether the machines are synchronous or asynchronous, nor any of the above parameters that describe them, or if they're just ordinary distribution loads.

You can't calculate the individual quantities, just the lumped effect at the point of the fault. The professor conveniently placed the fault in the same place for all cases, and gave you much more other information that you don't need.

And, to answer the question, you do need to back calculate, that's the whole point.

I agree. If the reported currents are correctly calculated;

If we shall neglect resistances and consider X2=X1

[and take the equivalent reactances X1,X2 and Xo]

Note:

IEC 60909-0 CH.4.2.2 Line-to-line short circuit states:

"During the initial stage of the short circuit, the negative impedance is approximately equal to the positive-sequence impedance, independent of whether the short circuit is a near-to-generator or a far-from-generator short circuit."

I hope the currents are calculated “at initial stage” [subtransient].

I"k2EL2=1.1*VL_L*SQRT((Xo-0.5*X2)^2+3/2X2^2)/(X1*X2+X1*Xo+X2*Xo)

Ik1=1.1*sqrt(3)*VL_L/(X1+X2+Xo)

Put X2=X1

Calculate X1[algebraically] and

Ik3=1.1VLL/sqrt(3)/X1

Sorry. In I"k2EL2 formula it is an error.

Actually the IEC 60909-0 formula(46) [in complex] is:

I"k2EL2=-j*c*Un*[Z(o)-a*Z(2)]/[Z(1)*Z(2)+Z(1)*Z(o)+Z(2)*Z(o)]

where : j=i=sqrt(-1) ; a=-0.5+i.sqrt(3)/2

If we neglect R the formula remains:

I"k2EL2=-i*c*Un*{iXo-[-0.5+i.sqrt(3)/2)]*X2}/(-X1*X2-X1*Xo-X2*Xo)

I"k2EL2=c*Un*{[Xo-i0.5*X2-sqrt(3)/2*X2]/(-X1*X2-X1*Xo-X2*Xo)

|I"k2EL2|=c*Un*sqrt[(Xo-sqrt(3)/2*X2)^2+0.25*X2^2)/(X1*X2+X1*Xo+X2*Xo)

If X2=X1 then:

|I"k2EL2|=1.1*VL_L*sqrt[(Xo-sqrt(3)/2*X1)^2+0.25*X1^2)/(X1^2+2*X1*Xo)

I'll would solve it by iteration. At first put Xo=0

I"k2EL2=1.1*VL_L*X1*sqrt[(3/4+0.25)/(X1^2) sqrt(3/4+0.25)=1

I"k2EL2=1.1*VL_L/X1 X1=1.1*VL_L/I"k2EL2

calculate Xo from I"k1:

Xo+2*X1=1.1*VL_L/I"k1 Xo=1.1*VL_L/I"k1-2*X1

Recalculate |I"k2EL2|

Suitable adjust X1 in order to fit |I"k2EL2|.

Recalculate Xo etc.

Now another mistake:

Xo+2*X1=1.1*VL_L/I"k1 it has to be: Xo+2*X1=1.1*sqrt(3)*VL_L/I"k1

You also deserve a star for your initiative!

I think X1=47.8 and Xo=8.87 for a less than 1% error.

Nicely done!

I think the purely algebraic solution would include the current X/R lumped, probably not at sub-transient conditions, and your results would be at the same time and condition as reported for the original two faults. You could assume fully symmetrical conditions, but at whatever unknown condition the original fault current was reported, your solution would be at those same conditions.

My guess is the 'professor' wanted his students to show their knowledge of the relationship between the driving voltages behind various types of faults, and picked a sneaky way to make them think.

O.K.I could live with this.

However something is still not so clear.

I don't know what it is 3F [Diagram 1]. I cannot reach more than 367.36 MVA- neglecting resistances.

Vrated=Vactual/pV=122.831/0.931=131.93 kV?

Only in the case the title of the Diagram 2 [ "Solid earth fault;phase to phase"] it has to be only" phase to phase"-then Ik3=Ik2*2/sqrt(3) Ik3=1.728*2/sqrt(3)=1.995 kA and MVA=SQRT(3)* 1.995*120=414.72 [close!] 120 kV?

I don't know why the 3 phase fault current is shown on a load flow diagram, but a quick check on the double line to ground at 365MVA puts it at the right magnitude, good detective work by Mr. RAM

You're right, it takes a bit of sleuthing to decode these diagrams, especially since 3 phase and single phase quantities are used without clear delineation.

Let's use Bus 1 as an example. The flow into it is listed as 50MW and 57.7MVAR, we can calculate its apparent power (S) as √(502 +58²⁾ = 77MVA, the same exercise on Bus 5 yields S=64. Adding them together yields 141, a number that doesn't exist on the diagram; but 3 x 141 = 423MVA which is in the ballpark of those "3F" numbers does, so that must be the 3phase fault current. For simplicity (and ease of typing) it is important to note that I used algebra when in fact these are vector calculations, but since the angles are all relatively small, so the error isn't that great.

NB- a three phase fault (grounded or not) is the largest and easiest to calculate (except under some very special circumstances) since it only involves positive sequence impedances. It's basically just combining the series impedances from the left generator to Bus 4, in parallel with the series impedances from the right generator to bus 4, driven by an infinite bus with a fixed voltage of 1pu.

Too bad the OP has not returned.