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# 3.8 A 138 kV/13.8 kV, 20 MVA, three-phase transformer has

3.8
A 138 kV/13.8 kV, 20 MVA, three-phase transformer has a reactance of 10% and a resistance of 0.4%. Calculate a reasonably close approximation the peak fault current in the low voltage winding under the worst conditions, if a three-phase fault occurs at the low voltage terminals. For the purpose of the calculation the impedance of the source can be considered negligible.
If the fault current is interrupted by a circuit breaker on the high voltage side of the transformer, and if the capacitance per phase of this winding is 5200 pF, determine the frequency of the transient recovery voltage seen by the breaker.

Then;
3.9
Refer again to Problem 3.8. Let us now assume that the impedance of the 138 kV source is not negligible and that it has the same X/R ration as the transformer itself. Let us further state that a symmetrical fault on the 138 kV bus would develop a fault current of 18 kA rms.
Determine the frequency and relative magnitude of the transient recovery voltage of the circuit breaker when it interrupts the fault described in Problem 3.2.
Assume that the stray capacitance on the source side of the breaker is 12,000 pF.
Kindly indicate time in hours by which you would like to have answers for these questions.
Customer: replied 5 years ago.

I'm trying to finish and need within 5 hours

I will try to solve by that time and upload the answers. Thanks.
Customer: replied 5 years ago.

Ok thanks, XXXXX XXXXX graduate level Power Systems Transients course problems and the use of laplace transforms is requested as needed by the prof, as much detail as possible would be helpful. Thanks

Where ever use of Laplace transform is required, I will try to use. For calculating transient recovery voltage, formulae are available and I will use them.
What is frequency as it is required to calculate frequency of recovery voltage? Kindly inform.
Customer: replied 5 years ago.

It is not stated in the problem so assume 60Hz

OK. Thanks.
Customer: replied 5 years ago.

I just noticed it requested information from question 3.2 so i'm uploading the question for reference.

I am working. If required, I will use it.
It seems that transformers in 3.8 and 3.9 are different. When we are assuming source impedance zero, fault current is coming 8.36kA.
In 3.9, it is stated that fault current is 18kA considering source impedance. It can not be with same same transformer as maximum fault current occurs with zero source impedance.
Kindly look into this aspect and reply. I have completed 3.8.
Customer: replied 5 years ago.

I have reviewed the problem and it is verbatim from the book.

OK. I will solve it considering 18kA fault current on HV side itself.
Customer: replied 5 years ago.

Ok, thank you, XXXXX XXXXX is much appreciated.

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Customer: replied 5 years ago.

thank you, I have another 3 questions i've posted under engineering, if you are interrested in looking at them for me. they are simpler than the other 2

Thanks for rating excellent.
Kindly post those questions under home work category as I am not authorized to reply in Engineering category.
Customer: replied 5 years ago.

* it keeps erroring out when i try to post a new question. Is it possible to increase pay for this problem?

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

Kindly indicate time in hours by which you would like to have answers.
Customer: replied 5 years ago.

Soon, today preferably, I'm a web student and going to submit my assignment late but i'd rather late than never. I'm hoping by getting help i'll be able to drill down the funamentals of the problems more accurately.

Customer: replied 5 years ago.

I was finally able to resubmit the homework question if you would like to pick it up from there.

Here it is night time and I would like to work up to 1pm or so. If it is possible to finish by that time, I will upload the answers; otherwise I will wok tomorrow and will try to upload the answers in 12 hrs or so.
I will look in home work category for these questions.