NCEES 2009 Power question 530

[nuclear bus]

Hi Everybody,

I'm having an issue with this problem and it's driving me kind of nuts. I know I'm so close to the solution it's killing me but I just can't connect the dots. If anybody can help I'd greatly appreciate it!

A 1,000-kVA, 12.47-kV-480Y/277-V transformer with 4% impedance feeds a 480-V bus. The fault duty of the 12.47 kV system is 40 MVA. Assuming the transformer and the 12.47-kV system have the same X/R ration, the 3-phase short-circuit current (amperes) available at the 480-V bus is most nearly:

The solution is shown as Isc = 1.0 / 0.04 + 0.025 = 15.4 pu on a 1 MVA base. It looks like the numerator is the MVA base, and the denominator of this fraction is the combined per unit Z of the transformer and system, I get the 4% (0.04) impedance is in series with the system impedance, but I don't get how we get the 0.025 value. From there on I get it, finding the I base using the Base MVA / sqrt*3*480 = 1203 A and then Isc = 15.4 * 1203 = 18,526 A. That makes sense.

But..

0.025 is 1/40, but there's something I'm missing here. Is the impedance of the fault duty of a system is simply the inverse of the fault duty MVA rating? I can't find the term "fault duty" in any of the references I have. Since we know that the MVA of the system is 40 MVA, and the KV of the system is 12.47, then I first tried to find the current using those two (MVA / KV) and then find the Z using the result of that (KV / A) but that doesn't work here. I think the key is the 40 MVA fault duty but I can't figure out how it works. Please help! Thanks

 

[Flyer_PE]

Take a look at this thread and see if it helps any.

 

[nuclear bus]

Take a look at this thread and see if it helps any.

Thanks for your help! Sorry I didn't see that thread in the first place. I did try and search for question 530 before posting my question but didn't see it. The information in that thread is exactly what I needed to know. Thanks again for your support and pointing me in the right direction.

 

[chelapati Irvine Institute]

We are now offering PEEE (Power) License Seminars at Irvine Institute. 30 hours - 10 evenings (January 26, 2010-March 30, 2010). These are webcast and archived for future reference and can be seen by anyone worldwide. Your post is sent to Ms. Allie Auld, UCI and she will discuss these questions in her lectures. One needs to know the theoretical basis as well as solve the problem. I will let you know when these problems will be discussed in our seminars. Our seminars cover relevant background needed to solve the problems as well discuss the problems published in NCEES Sample Problems.

We have been offering programs to prepare for PE 'licensing' exams in many areas since 1973 and also published several manuals for these seminars. Two volumes are produced for PEEE Electrical Volume 1 and Volume 2. Volume 1 is for Power. TOC's are on our website.

We are trying to post our website www.irvine-institute.org on this forum and sent e-mail to the EngineringBoards.com. I or our staff will be answering the queries posted on this 'forum'

Chelapati - Irvine Institute

 

[yellowjacket03]

The solution posted in the NCEES book works, but is not very clear. I prefer to use the "MVA Method" to quickly calculate short circuit currents. Refer to the paper "Short circuits ABC - Learn it in an hour, Use it anywhere, Memorize no formula" by Moon H. Yuen. It is a great paper that was published in the IEEE Transactions on industry Applications, Vol 1A-10, No. 2, March/April 1974. It will help you breeze through problems like #530.

Hope this helps

 

[nuclear bus]

Thanks yellowjacket03, I'll look for that.

I did just get a copy of the EC&M's Electrical Calculations Handbook by Paschal, 2001, and there's a great example calculation in Chapter 5 (short circuit calculations) that covers this exact topic.

You determine the VA from the utility which is given in this case as 40 MVA, and the let-through of the transformer which is 1 MVA / 0.04 then add the inverse of the two and then inverse the result. This is kind of what is shown in the NCEES solutions manual, but using this method you first get 15.4 MVA as the answer, and then change it to 15.4 pu if using a 1 MVA base, and then the rest is pretty straight-forward standard calculations. In fact, I believe you can just got from 15.4 MVA directly to the SC-current by dividing 15.4 MVA by (sqrt 3 * 480)...

I see now that the 1/25 MVA (=0.04) and 1/40 MVA (=0.025) are also standard calculations, I just couldn't quite put it all together.

Thanks again for your and Flyer_PE's help.

This EC&M book seems like it's going to be very handy. It has lots of example calculations and discussions in pretty plain English on the subjects it covers.

 

[nmh0408]

Thanks yellowjacket03, I'll look for that.
I did just get a copy of the EC&M's Electrical Calculations Handbook by Paschal, 2001, and there's a great example calculation in Chapter 5 (short circuit calculations) that covers this exact topic.

You determine the VA from the utility which is given in this case as 40 MVA, and the let-through of the transformer which is 1 MVA / 0.04 then add the inverse of the two and then inverse the result. This is kind of what is shown in the NCEES solutions manual, but using this method you first get 15.4 MVA as the answer, and then change it to 15.4 pu if using a 1 MVA base, and then the rest is pretty straight-forward standard calculations. In fact, I believe you can just got from 15.4 MVA directly to the SC-current by dividing 15.4 MVA by (sqrt 3 * 480)...

I see now that the 1/25 MVA (=0.04) and 1/40 MVA (=0.025) are also standard calculations, I just couldn't quite put it all together.

Thanks again for your and Flyer_PE's help.

This EC&M book seems like it's going to be very handy. It has lots of example calculations and discussions in pretty plain English on the subjects it covers.

I do agree with you on Paschal, as it offers very clear and simple way of finding the short circuit current, but this method applies when you have MVA(KVA) ratings for every component in a system, I don't think it can be used when you only given the KV ratings? Does it?

 

[nmh0408]

The solution posted in the NCEES book works, but is not very clear. I prefer to use the "MVA Method" to quickly calculate short circuit currents. Refer to the paper "Short circuits ABC - Learn it in an hour, Use it anywhere, Memorize no formula" by Moon H. Yuen. It is a great paper that was published in the IEEE Transactions on industry Applications, Vol 1A-10, No. 2, March/April 1974. It will help you breeze through problems like #530.
Hope this helps

Follow this link for the above paper

http://www.arcadvisor.com/pdf/ShortCircuitABC.pdf

 

[DK PE]

Maybe this will be too much information but here is a link to a paper from Jim Phillips website that is pretty good. It shows how you add the contribution from the utility in the denominator in place of assuming "infinite source"

Go here http://www.brainfiller.com/newsletterArchive.php

Then down at bottom choose --> Short Circuit Calculations – Transformer and Source Impedance

For some reason the I cannot get the direct link to work.

 

[Flyer_PE]

Maybe this will be too much information but here is a link to a paper from Jim Phillips website that is pretty good. It shows how you add the contribution from the utility in the denominator in place of assuming "infinite source"
Go here http://www.brainfiller.com/newsletterArchive.php

Then down at bottom choose --> Short Circuit Calculations – Transformer and Source Impedance

For some reason the I cannot get the direct link to work.

Direct Link

I took one of the arc flash calculation courses from him a few years ago. He's pretty good as far as those classes go.

 

[cableguy]

Follow this link for the above paper
http://www.arcadvisor.com/pdf/ShortCircuitABC.pdf

I cannot say enough good things about this method. I'd never heard of it before, I got the paper, I have given it a try, and dang if it doesn't work slicker than snot. It is FAST. It is SIMPLE. It is great!

 

[tesla314]

The solution posted in the NCEES book works, but is not very clear. I prefer to use the "MVA Method" to quickly calculate short circuit currents. Refer to the paper "Short circuits ABC - Learn it in an hour, Use it anywhere, Memorize no formula" by Moon H. Yuen. It is a great paper that was published in the IEEE Transactions on industry Applications, Vol 1A-10, No. 2, March/April 1974. It will help you breeze through problems like #530.
Hope this helps

Follow this link for the above paper

http://www.arcadvisor.com/pdf/ShortCircuitABC.pdf

Thanks, yellowjacket03 and nmh0408, this is super useful.

 

[pbo064]

Has anyone used the MVA method to solve it? If so, what MVAs did you find for the system and the transformer? I have been doing 40/0.01=4000 MVA for the system and 1/4%=25 for the transformer and did not get the right answer. This is the first time I've had trouble using the MVA method.

 

[BigBang]

pbo064:

In the MVA method, the bases must be unified first. Don't confuse it with converting everything to MVA. Rather get impedances of every component and that will help you construct the reactance diagram to calculate fault problems. The 40 MVA is not the system rating, rather its the available SCD contribution to downstream faults, so you cant divide it by the impedance. This value is determined in a similar way to the "let through" fault current of the distribution transformer (MVA rating / Z%). As an example, if the 40 MVA limit was not specified, the the "let through" SCD of the 1000 KVA 480 V transformer with a 4% impedance (1 MVA / 0.04) = 25 MVA = 30,071 Amps of fault current on the 480 V side. But including system SCD is a lot like adding another impedance in series with the transformer impedance, consequently limiting the fault current. All you have to do is calculate this impedance and add to to the transformer impedance (@ the new base MVA). The system impedance in per unit is determined by the equation: X(system) = (MVA base / MVA scd). In #530, the MVA was chosen to be the transformer rating (1000 = 1MVA). So X system = 1/40 = 0.025 pu.