Fault Calculations

Professional Engineer & PE Exam Forum

Help Support Professional Engineer & PE Exam Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

eng.dork

Active member
Joined
Jan 30, 2007
Messages
39
Reaction score
0
Okay...I am posting a couple problems that I can't seem to get the answers that the book porvides. The two problems that I am struggling with are 6.24 and 6.25. Can someone work these and make sure that the answer in the book is correct. I know for a delta-connected load that the zero-sequence component is 0. I can't get the positive and negative-sequences (6.24). On 6.25 I am stuggling with getting the correct angle. The problems are attached.

 
I posted a link time ago about a presentation for Fault Calculations. I did not more than that to prepare to the test and actually used it during the test on one problem. I am going to try to find it. Has to be somewhere on in the Electrical forum.

 
Did anybody get the solution for these problems. I couldn't get it either. I will try to post my solution(?) :dunno: later.

Thanks,

Ilan.

 
I think there is an error in the problem statement for 6.24. The zero sequence current should not only be logically 0, the math should get you there also. In this case, the sum of the three currents is not zero. In rectangular form:

Ia = 5 + j0

Ib = -6.58 - j2.39

Ic = 0 + j5

As a quick test, I replaced Ib with -5 - j5 and got their answer for the positive sequence current. I didn't get the negative sequence to work out but that's probably to be expected since I adjusted the input values.

I haven't played with 6.25 yet.

Jim

Edit: Negative sequence answer to 6.24 is wrong. The maximum possible magnitude for I2 is (5+5+7)/3 = 5.67.

 
Last edited by a moderator:
Despite an error in this posted question, would you tackle the sequence currents for phase A as follows? :f_115m_e45d7af:

zero-sequence Ia_0 = 1/3 * (5/_0 + 7/_200 + 5/_90)

positive-sequence Ia_1 = 1/3 * (5/_0 + 7/_(200+120) + 5/_(90+240))

negative-sequence Ia_2 = 1/3 * (5/_0 + 7/_(200+240) + 5/_(90+120))

Also, do we always use line currents in symmetrical components analysis?

Does the above set of equations hold true for a wye-connected load as well?

BTW eng_dork, I have yet to get their posted answers as well. I wonder what we might be doing wrong in our calcs?

 
Any takers on my question? I know most have said that we can expect a qualitative question on the exam regarding symmetrical components, but I just want to be prepared in case a quantitative question is also asked.

 
I think your calculation for the phase A current is correct.

The purpose of symmetrical components is to analyze unbalanced faults/loads. The only bearing the configuration of the load has is whether or not you expect a zero sequence (neutral) current.

Yes the sequence currents are valid for both delta and wye connected systems.

Jim

 
Thanks for clearing that up, Jim.

I believe there is an error in EERM on p. 35-6 and it's not on the errata list.

In that section, it refers to a single line-to-ground fault as a balanced three-phase symmetrical fault in Figure 35.7(a)

Correct me if I am wrong but I understand the classification as follows:

Balanced Symmetrical fault occur as (only one type?):

- three phase fault

Unbalanced Unsymmetrical faults occur as:

- single line-to-ground

- 2-phase line-to-line

- double line-to-ground

- triple line-to-ground (or does this fall under symmetrical fault classification??)

- single low-impedance-to-ground fault

 
I'm not sure about the error in the EERM. I didn't like the way this stuff was presented in there and used my college text instead. If somebody else hasn't already answered, I'll take a look at the EERM tonight when I get back from Chuckee Cheese Hell.

Your understanding of the fault types agrees with my understanding. A triple-line to ground fault is a balanced symmetrical fault.

Jim

 
Balanced Symmetrical fault occur as (only one type?):- three phase fault

Unbalanced Unsymmetrical faults occur as:

- single line-to-ground

- 2-phase line-to-line

- double line-to-ground

- triple line-to-ground (or does this fall under symmetrical fault classification??)

- single low-impedance-to-ground fault
The only thing I can think of for the triple-line-to-ground is they are implying there is a different impedance to ground in each phase. I've never heard of such a fault, but I suppose it's possible--say a crane contacts a line and one phase falls on the crane (low impedance), one phase falls on a nearby tree (high impedance) and the other falls on a squirrel (medium impedance).

 
Last edited by a moderator:
And, BTW, chicago, I was never able to get their answers either. It looks like you are doing the calculations correctly.

 
Back
Top