Experts are full of valuable knowledge and are ready to help with any question. Credentials confirmed by a Fortune 500 verification firm.

Get a Professional Answer

Via email, text message, or notification as you wait on our site. Ask follow up questions if you need to.

100% Satisfaction Guarantee

Rate the answer you receive.

Ask akch2002 Your Own Question

akch2002, Engineer

Category: Homework

Satisfied Customers: 3120

Experience: Home work expert

46493125

Type Your Homework Question Here...

akch2002 is online now

Power System Transients Homework Problems

This answer was rated:

★★★★★

* Note - All questions should be answered with Laplace Transforms when possible.

3.3

Figure 3P.2 shows two capacitor banks, C1 and C2, in a substation. C1 is energized, but C2 is discharged. The three-phase, 60 Hz rating of the banks are: C1, 5 MVA; C2, 3 MVA, on a 13.8 kV base. The source has a short circuit rating of 20kA rms at 13.8 kV. The inductance of the loop between C1 and C2, represented by L2 is 30μH.

Calculate the peak transient voltage that will appear on C2 and the peak transient current that will flow in L1, if the switch S is closed at the peak of the voltage cycle. Point out any assumptions you make.

3.5

A line to ground fault occurs as indicated in Fig. 3P.4, close to the secondary terminals of a 230/34.5 kV transformer. The transformer has a three-phase rating of 100 MVA; it has 0.1 pu reactance on this base.

Calculate:

a.The fault current

b.The time to peak of the transient recovery voltage when the circuit breaker opens to interrupt the fault current. A value of 12.7nF can be assumed for the effective capacitance per phase of the transformer secondary winding.

3.12

C1=100μF; C2=40μF; L1=10mH; L2=20mH; C1 and L1 in Fig. 3P.8. are oscillating; C2 and L2 are de-energized. The switch is closed when V(C1 )=0 and I2=500A. Calculate: a.Peak voltage reached on C2 b.Peak current reached in L2 c.Frequency of the current in L1 d.Maximum voltage on C1 prior to closing the switch

Thank you, XXXXX XXXXX to spend some time now reviewing your other answers in more depth to better my understanding, I may ask a couple of questions later.

It doesn't state explicitly, and I cannot say I understand the concept as well as I would like, but a simple L-G fault would be unsymmetrical, and would only be symmetrical if it were L-L-L-G. L -Line and G -Ground

Had this been asymmetrical fault, then zero sequence of the transformer should also have been given; but zero sequence impedance is not given. Hence, I conclude it to be symmetrical fault. The figure also represents symmetrical fault(LLLG). I am almost sure but I just wanted to confirm as language is confusing.

After reviewing some of the answers you provided, several do not mesh with solutions i found from another professor. Of which I found the final answers for 3.3 and 3.12

I can look into these problems tomorrow only. Should I opt out so that some expert may try. In3.12, frequency should be 159Hz. I a not able to follow how frequency will be 164.74Hz.

Not yet, I'm going to look at it a bit further and see if I can find any discrepencies. I feel i have enough information amoung the problems you've answered to really get a fell for the questions. I appreciate your work on this very much.

Actually, there are assumptions in my solutions. I have assumed that direct voltage is applied across L2 and C2 in 3.3 and solved the problem. Actually voltage will vary with time and this assumption may be leading to differences in answers. In 3.12, L1 and C1 are known, we can calculate the frequency. Had frequency been incorrect, I could not have got correct answer in(d).Hence, frequency as calculated seems to be correct. Anyway look into the solutions. Probably we may gave to write loop differential equations and solve them without making any assumptions.

I think you are on track, it appears the time variance is likely the issue in 3.3, I've been able to validate your c1 and c2 calculations for positive, but still working on the rest a bit. Thanks for staying with the problem for me.

,I was looking to mark the last question I asked you as satisfactory, but cannot find a link. The homework came back today finally as correct. Sorry, my comfort level for getting help online is not very strong. I do appreaciate your help and hopefully could utilize you again sometime.