# Power question 513



## maxxpower71 (Oct 19, 2009)

For a 3PH fault at Location A, the 12kV short circuit current is most nearly:

Iactual (3Phase) = Ibase / (ZT1 (pu) + ZL (pu))

Ibase = 100MVA / 1.73 x 12kV

Iactual = 4,811/(1 p.u. + 10.6 p.u.) = 415A

A few questions for this solution,

1. Wwhere did they get the first formula, I am not following it? I know Iactual = Ipu / Ibase. Why is the impedance in the equation?

2. How did they come up with Sbase = 100MVA? I'm guessing they chose an arbitray value.

3. How did they come up with the ZT1 pu and ZL pu values? Again assuming the chosen Sbase and Vbase gave me those values but if anyone can help me visualize how these numbers were calculated I would really appreciate it.

Thanks.


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## Flyer_PE (Oct 19, 2009)

maxxpower71 said:


> 1. Wwhere did they get the first formula, I am not following it? I know Iactual = Ipu / Ibase. Why is the impedance in the equation?


They are doing a couple of steps in one line. The current is actually voltage divided by impedance. In p.u. the base voltage is 12kV=1p.u.

Iactual=Ibase*Ipu

Expanded they are stating that Iactual=Ibase*Ipu where Ipu=Vpu/Zpu



maxxpower71 said:


> 2. How did they come up with Sbase = 100MVA? I'm guessing they chose an arbitray value.


The 100MVA base is an arbitrary choice in that any chosen MVA base will result in the same solution. In this instance, the 100MVA base is chosen since the 7.5% transformer impedance on a 7.5 MVA base translates to a 1 p.u. impedance on a 100 MVA base.



maxxpower71 said:


> 3. How did they come up with the ZT1 pu and ZL pu values? Again assuming the chosen Sbase and Vbase gave me those values but if anyone can help me visualize how these numbers were calculated I would really appreciate it.


ZT1 is a simple base conversion. Znew=Zold*MVAnew/MVAold

0.075 * 100/7.5 = 1 p.u.

The transmission line impedance is a little more involved since they give you the value in ohms. For that you have to determine the base impedance. Since the base is already chosen at 100 MVA and the fault is on the 12 kV bus,

Zbase = kVbase^2/MVAbase = 12^2/100 = 1.44 Ohms

Zactual = (j0.145 ohms/1000ft)(5280 ft/1 mile)(20 miles) = 15.31 Ohms

Zpu = Zactual/Zbase = 15.31/1.44 = 10.63 pu


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## maxxpower71 (Oct 19, 2009)

Thanks for the quick and detailed response!!


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## z06dustin (Oct 20, 2009)

hey flyer, why is the root(3) there? i missed this problem too. i mean i realize it's a line-line to line-ground cconversion, but shouldn't it be 12/root(3) instead of root(3)*12?


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## Flyer_PE (Oct 20, 2009)

^The are solving the 3-phase power equation for I:

SMVA = sqrt3 * VLine * ILine

ILine = SMVA / (sqrt3 * VLine)


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## z06dustin (Oct 20, 2009)

doh ok.


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## maxxpower71 (Oct 20, 2009)

Flyer, you really seem to know the stuff on the PE. I wanted to ask you if you know where I can find material about simple full wave bridge rectifier, inverter schematic, thyristor bridge, half wave rectifier, 3 phase battery charger circuit, ac motor drive circuit. Those electronic power questions.

There are a couple of questions that ask you about the wave form of a circuit and to be honest, I do not have a clue.

Anyone who was taken the previous power exam, did you guys see a lot of these?


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## Flyer_PE (Oct 20, 2009)

^Unfortunately, most of the stuff I know about power electronics comes from several years spent troubleshooting battery chargers and inverters.

The only advice I can give for that stuff is to take the three phase circuit they give you and draw the single phase equivalent. The trick is keeping track of which diodes/SCRs are conducting at any given time.


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## reqex78 (Jan 5, 2010)

I would recommend Power Electronics: Converters, Applications And Design, Media Enhanced, 3rd Ed by Mohan/Robbins/Undeland

Flipkart



maxxpower71 said:


> Flyer, you really seem to know the stuff on the PE. I wanted to ask you if you know where I can find material about simple full wave bridge rectifier, inverter schematic, thyristor bridge, half wave rectifier, 3 phase battery charger circuit, ac motor drive circuit. Those electronic power questions.
> There are a couple of questions that ask you about the wave form of a circuit and to be honest, I do not have a clue.
> 
> Anyone who was taken the previous power exam, did you guys see a lot of these?


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## nuclear bus (Jan 24, 2010)

I have trouble when it comes to PU and fault calculations. I understand the concepts but the only way to be able to get through the problems with consistent success is to do as many problems as you can right? As bad as the reviews are for it, I found the Schaum's Oulines for Electric Power Systems has quite a few example problems and a nice discussion on how this works. A lot of the same stuff is also in the Camara EERM. However, I found that the material in the EERM regarding this made a lot more sense after going through the examples in the Shaum's book. Just a suggestion. It's only about $14 on Amazon so you can't really go wrong with it I think.

That being said, I would like to still find any other books that have lots of sample problems on these types of exercises. So please suggest away if anybody has any ideas.

I thought the Schaum's book did a better job than the EERM on symmetrical and unsymmetrical fault components. It's amazing how just adding one or two more sentences can make a world of difference to me. Although both could use a lot more sample problems on the subject and both discussions were just way too brief, in my opinion.


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## z06dustin (Jan 25, 2010)

nuclear bus said:


> I have trouble when it comes to PU and fault calculations. I understand the concepts but the only way to be able to get through the problems with consistent success is to do as many problems as you can right? As bad as the reviews are for it, I found the Schaum's Oulines for Electric Power Systems has quite a few example problems and a nice discussion on how this works. A lot of the same stuff is also in the Camara EERM. However, I found that the material in the EERM regarding this made a lot more sense after going through the examples in the Shaum's book. Just a suggestion. It's only about $14 on Amazon so you can't really go wrong with it I think.
> That being said, I would like to still find any other books that have lots of sample problems on these types of exercises. So please suggest away if anybody has any ideas.
> 
> I thought the Schaum's book did a better job than the EERM on symmetrical and unsymmetrical fault components. It's amazing how just adding one or two more sentences can make a world of difference to me. Although both could use a lot more sample problems on the subject and both discussions were just way too brief, in my opinion.


I think each person is different, but when I was studying this stuff I leaned heavily on my college texts and notes. So I would take a look at them if you have them, if not the ones I liked were the ones you mentioned above, and the texts:

Power System Analysis and Design (Glover Sarma)

Electrical Transients in Power Systems (Greenwood) (there are good practice problems in here for transients, especially sub-cycle and unsymmetrical)

Protective Relaying Principles and Appplications (Blackburn Domin)

YMMV.


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## z06dustin (Jan 25, 2010)

duplicate, mods feel free to delete.


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## pelaw (Apr 4, 2010)

Thank you for the explanation Flyer. I struggled with this problem. I also want to share my way of solving the problem.

Step 1: The important thing to look at here is % value of transformer impedance.

Step 2: calculate transformer impedance - S = V^2/Z -&gt; Z = V^2/S = 12000 V ^2 / 7,500,000 VA = 19.2 ohms

Step 3: The Z tx = % * Z = 0.075 * 19.2 ohms= 1.44 ohms

Step 4: Line Impedance Zl = 0.145 * 20 miles * 5280 ft/mile / 1000 = 15.3 ohms

Step 5: Z total = Z tx + Z l = 1.44 + 15.3 = 16.74 ohms

Step 6: I (3phase) = V / Z total = 12000 V / 16.74 ohms = 717 A

Step 7: Isc (per phase/line) = I (3 phase) / sqrt 3 = 414 A


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## Flyer_PE (Apr 4, 2010)

^Step 6 is not quite correct. The current in any given conductor will be the phase voltage divided by the impedance (VLine/sqrt3)/Z. That would have given you the correct answer right there.

For step 7, the only time you will reduce a current by the square root of 3 is when dealing with the internals of a delta-connected piece of equipment. Once you're outside the equipment and looking at power conductors, ILine=IPhase.

You arrived at the right answer but not quite for the right reason.


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## bacchi (Apr 12, 2010)

Flyer_PE said:


> ^Step 6 is not quite correct. The current in any given conductor will be the phase voltage divided by the impedance (VLine/sqrt3)/Z. That would have given you the correct answer right there.
> For step 7, the only time you will reduce a current by the square root of 3 is when dealing with the internals of a delta-connected piece of equipment. Once you're outside the equipment and looking at power conductors, ILine=IPhase.
> 
> You arrived at the right answer but not quite for the right reason.


There is an easier way to solve this using MVA method as follows:

Transformer SCMVA = 7.5/.075=100 MVA

Transmission Line SCMVA = kV^2/Z=144/15.312 = 9.40 MVA

Two impedance sources are in series.

Hence SCMVA at fault = 9.40X100/9.40+100 = 8.59 MVA

Ifault at Fault location I = 8.59x1000/1.732X12 = 413.3 A


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## phowardtx (Apr 14, 2010)

maxxpower71 said:


> For a 3PH fault at Location A, the 12kV short circuit current is most nearly:
> Iactual (3Phase) = Ibase / (ZT1 (pu) + ZL (pu))
> 
> Ibase = 100MVA / 1.73 x 12kV
> ...



NM


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