NCEES Power 513

[FatDirk]

I didn't see a thread regarding this question so I'll ask it.

Question:

for a 3-phase fault located at location A, the 12-kV short circuit current (amperes) is most nearly:

Distribution line length 20 miles

line impedance Z = j0.145 ohms/1000 ft

Transformer - 3ph 60/12kV, 7.5MVA with 7.5% impedance

------------------------------------

1. Why is 100 MVA used as the base power?

2. Why is the Ibase divided by the sum of the pu impedances?

Thanks in advance for any help...

 

[Flyer_PE]

For any PU problem the base MVA is somewhat arbitrary choice. Usually it will either be the largest transformer on the system or 10 MVA or 100 MVA. The two things you get to choose in PU calculations are the base MVA and the base voltage. Everything else is based off of those.

Their solution is abbreviated. There is a 1.0 Volt pu term included in the numerator that they don't bother showing. Basically, they are combining the following two equations in one line:

1. Ipu = 1.0Vpu/(Z in pu)

2. IActual=Ipu*IBase

 

[FatDirk]

For any PU problem the base MVA is somewhat arbitrary choice. Usually it will either be the largest transformer on the system or 10 MVA or 100 MVA. The two things you get to choose in PU calculations are the base MVA and the base voltage. Everything else is based off of those.
Their solution is abbreviated. There is a 1.0 Volt pu term included in the numerator that they don't bother showing. Basically, they are combining the following two equations in one line:

1. Ipu = 1.0Vpu/(Z in pu)

2. IActual=Ipu*IBase

Ahhhhhh... I see now. Thanks so much!!

 

[ElecPwrPEOct11]

I have a question on this problem too so thought I'd revive this thread.

The NCEES solution doesn't show the 'j' for the transmission line or transformer per-unit impedances. For this particular problem there aren't any resistances so it doesn't end up affecting the math, but shouldn't both the distribution line and transformer per-unit impedances be purely inductive, jX? If a similar problem had both a resistances & a reactance you'd have to add the per-unit values with vector math. Please let me know if I'm wrong on this.

 

[DK PE]

The poster of original problem had

- Distribution line length 20 miles

- line impedance Z = j0.145 ohms/1000 ft

Transformer - 3ph 60/12kV, 7.5MVA with 7.5% impedance

My opinion is they should have (and maybe did?) include the j with the transmission line parameter as depending on length and conductor size, both resistance and reactance can be a factor.

For the transformer, unless X/R ratio is mentioned, it is a reasonable assumption that 7.5% on its own base is same as j0.075 pu

 

[EEVA PE]

The poster of original problem had

- Distribution line length 20 miles

- line impedance Z = j0.145 ohms/1000 ft

Transformer - 3ph 60/12kV, 7.5MVA with 7.5% impedance

My opinion is they should have (and maybe did?) include the j with the transmission line parameter as depending on length and conductor size, both resistance and reactance can be a factor.

For the transformer, unless X/R ratio is mentioned, it is a reasonable assumption that 7.5% on its own base is same as j0.075 pu

The X/R ratio is a little fuzzy for me. I assume it is the impedance/resistance ratio. I have seen a comment in one of Camara's problem stating "the three phase system X/R ratio is the same for the generator and the transformer". Then the solution went ahead and equated S(MVA) of the generator in proportion with the transformer. A little fuzzy on the relationship of x/r and S(MVA). Can anyone add info of what to be concerned with the X/R ratio.

 

[DK PE]

No endorsement of the author but read here... http://www.brainfiller.com/documents/XRRatio.pdf

 

[ElecPwrPEOct11]

^ Good link DK, thanks. I hadn't learned about the XR ratio before.

Sounds like we agree that the transformer impedance and (in this case) line impedance should both have reactive impedance. My issue with the NCEES solution was that when they converted the actual impedance to pu values they didn't show the 'j's anymore. On a 100MVA base (per the NCEES solution), this would be

line impedance = j10.6 pu

xfmr impedance= j1.0 pu

 

[Kalika PE]

Sorry to revive this thread but can the MVA method be used to solve this problem?  This topic is still a bit confusing for me.

 

[nukem2k5]

Sorry to revive this thread but can the MVA method be used to solve this problem?  This topic is still a bit confusing for me.

Yes.  For future reference, if you just Google "NCEES power #" with the problem number, you can usually find one or several threads on EB answering the questions that you have.  

Have a look at this one, as well as the attachment.  They solve it using both PU method and MVA method.