PE Exam Question: Short circuit study and X/R

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Defenestrator

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Can someone explain how the X/R ratio is used to solve? I couldn't find the relationship between the generator's impedance and the X/R ratio.

See attached photos


IMG_20180903_172424022.jpg IMG_20180903_174304479.jpg
 

beanland

Senior Member
Location
Vancouver, WA
Same X/R

Same X/R

The issue is that you have to sum the impedances of the transformer and generator. Impedance is a vector (R + jX), not a scalar. As a result you must sum the real portions and separately sum the reactive portions. But, with the X/R of both being the same, you can treat them as scalars and sum the impedances as scalars rather than vectors.
 

Defenestrator

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Location
Denver, CO
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Electrical Engineer
The issue is that you have to sum the impedances of the transformer and generator. Impedance is a vector (R + jX), not a scalar. As a result you must sum the real portions and separately sum the reactive portions. But, with the X/R of both being the same, you can treat them as scalars and sum the impedances as scalars rather than vectors.


but how was he able to get 2MVA/20MVA = 0.1PU impedance of the generator?
 

beanland

Senior Member
Location
Vancouver, WA
Data is missing

Data is missing

I agree with you, I had not seen the first image. There is no data about the generator impedance other than the X/R ratio. The generator could be 1% impedance or 100% impedance and still have the same X/R ratio as the transformer. As a result, there is not enough data to answer the question.
 

Phil Corso

Senior Member
Beanland...

No, data is not missing! But, the given solution is dead wrong!

The significance of both electrical apparatus (or is it apparati) having per-unit impedances with equal X/R ratios, means rated MVA to Zpu ratios must be equal as well, as follows:

MVA(g)/Z(g) = MVA(t)/Z(t)!

So, Z(g) is not 0.10, but instead should be should 0.50! Try it! I assure you, the answer will not be 'B'! It is... (I'll let TMC finish.)

Regards, Phil Corso
 

beanland

Senior Member
Location
Vancouver, WA
Modeled

Modeled

Just because this has got my goat, I created a computer model in ASPEN Distriview. I used a 2MVA 5%IZ, X/R=10 transformer and a 20MVA, 50%IZ, X/R=10 generator. 460V fault current ~25kA. When I changed the generator to 5%IZ, 460V fault current is ~45kA. With a 100%IZ generator, 460V current is ~17kA. With a 0.1%IZ generator, 460V fault current is ~50kA.

In all cases, the model parameters meet the requirements of the problem. Without knowing the generator impedance, the problem has no one solution.
 

kingpb

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SE USA as far as you can go
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Intuitively I have to agree with Phil. But at the same time, I question myself; am I overlooking something?

The X/R being the same is where I am struggling with how they jump to the impedance being a ratio of 2MVA/20MVA; this is where I am scratching my head.

It certainly means there is no additional fault current that has to be accounted for (dc offset), from an equipment rating, but that wasn't the question.

Otherwise it's a simple MVA method calc.
 

kingpb

Senior Member
Location
SE USA as far as you can go
Occupation
Engineer, Registered
Just because this has got my goat, I created a computer model in ASPEN Distriview. I used a 2MVA 5%IZ, X/R=10 transformer and a 20MVA, 50%IZ, X/R=10 generator. 460V fault current ~25kA. When I changed the generator to 5%IZ, 460V fault current is ~45kA. With a 100%IZ generator, 460V current is ~17kA. With a 0.1%IZ generator, 460V fault current is ~50kA.

In all cases, the model parameters meet the requirements of the problem. Without knowing the generator impedance, the problem has no one solution.

I'm glad you did it, cause it was perplexing me to the point I was going to do it in ETAP. Thanks for saving me from that.
 

Phil Corso

Senior Member
Incredible...

Way too much over reach! Fellas, its a simple source plus two-series impedance calculation! All needed info is there! Use your heads, not the computer!

BTW, my comments are not meant to be critical of Elecs & Techs, who find this forum useful! But, I sadly cringe whenever I hear an EE say "But, the Computer said... !"

Remember the fatal Tesla car accidents? The Robot smiled and said, "It's the dead-driver's fault, and the "in the know crowd" all nodded their heads in agreement!

If you want the answer, then you have to wait until I return from our 60th Anniversary !

Regards, Phil
 

Defenestrator

Member
Location
Denver, CO
Occupation
Electrical Engineer
MVA(g)/Z(g) = MVA(t)/Z(t)!

So, Z(g) is not 0.10, but instead should be should 0.50! Try it! I assure you, the answer will not be 'B'! It is... (I'll let TMC finish.)

Regards, Phil Corso

so MVA/Z is equal to X/R?

Could you provide a link to source which details this? The only reason i ask is because i couldn't find this relationship in my copy of Engineer's Power Reference Manual by Camara.
 

Phil Corso

Senior Member
TSM, Beanland, et al...

Boy, is my face red! I erred as well! Yes, it is ‘B’. My error was accepting the X/R-rule given in the PE Exam solution… that is, per-unit impedances were proportional to power… instead, the ratio must be in per-unit! (See Note 2 after the solution!)

Now, here is the correct solution using my Gi.Fi.E.S. method (pronounced Jiffy’s), where:
o Gi means ‘Given’
o Fi “ ‘Find’
o E “ ‘Equation’
o S “ ‘Solution’ The procedure follows:
Find:
Short-circuit current, ISC, at transformer secondary bus!
Equation(s):
If Capacity and Impedance were known for both the Gen’r and Xfmr, then the problem is easily solved, using the MVA, Ohm, or per-unit method! Unfortunately, for this problem, only ZT is known. But, ZG can be found using a little-known relationship that I call the X/R-Rule (see Note 2)! It is properly stated as Eq 1:
o Eq 1: Zg = Sb / Sg
o Eq 2, Ssc = Sbase / (Zg + Zt)
Then, using Per-Unit method,
o Eq 3, Isc = Ssc / ( Sqrt(3) x Vbase )
Solution:
o Zg = 2.0/20 = 1/10 or 0.10 (pu)
o Ssc = 2.0 x 10⁶ / (0.10 +0.05) = 13 x 10⁶
o Isc = 13.3 x 10E6 / Sqrt(3) x 460 = 16,729 A. Closest is B, 20 kA!

Note 1.

Since 460 was the only given voltage, I called it Vbase! Why? Because I perceived the series-circuit to be a generator connected to a 1:1 Xfmr or reactor!

Note 2
:
To TheSmoothestCriminal… you asked for some reference on the X/R rule! No, I couldn’t find any! But, a search of my archives revealed where it was! I did numerous fault-calcs, mostly at client-sites, using a hand-held HP-35 calculator! The X/R rule was pre-programmed! But, I’d forgotten the restrictions involved! It works, because the variables must be converted to base figures, i.e., base volts and base-MVA! In addition, having equal X/R ratio’s insure their vectors have a magnitude, and their angles are the same! BTW, it can also be applied to problems having reactance as the magnitude, and the angle is 90º!

(Caveat… because of recent events, your moniker might get you into trouble in the future. LOL!)
 

Phil Corso

Senior Member
The following definitions were omitted in the earlier thread:

Gi
ven: (System Parameters)
o Let Zg(pu) and Zt(pu) = Zg and Zt, respectively.
o Let MVAg, MVAt, MVAbase MVAsc = Sg, St, Sbase, Ssc, respectively. Then,
 

Defenestrator

Member
Location
Denver, CO
Occupation
Electrical Engineer
TSM, Beanland, et al...

Note 1. [/U]
Since 460 was the only given voltage, I called it Vbase! Why? Because I perceived the series-circuit to be a generator connected to a 1:1 Xfmr or reactor!

Note 2
:
To TheSmoothestCriminal… you asked for some reference on the X/R rule! No, I couldn’t find any! But, a search of my archives revealed where it was! I did numerous fault-calcs, mostly at client-sites, using a hand-held HP-35 calculator! The X/R rule was pre-programmed! But, I’d forgotten the restrictions involved! It works, because the variables must be converted to base figures, i.e., base volts and base-MVA! In addition, having equal X/R ratio’s insure their vectors have a magnitude, and their angles are the same! BTW, it can also be applied to problems having reactance as the magnitude, and the angle is 90º!

(Caveat… because of recent events, your moniker might get you into trouble in the future. LOL!)

Thank you for the long explanation. I'll need to chew on it for a while tonight before it makes sense to me, but all the info I need is given.

I swear I don't feel smart enough to be an engineer. If egr doesn't work out then I'll be a drafter lol.
 

jumper

Senior Member
Thank you for the long explanation. I'll need to chew on it for a while tonight before it makes sense to me, but all the info I need is given.

I swear I don't feel smart enough to be an engineer. If egr doesn't work out then I'll be a drafter lol.

Since you are smart enough to come here, ask questions, listen to the answers, and continue on - then you are smart enough to be an engineer IMO.
 

Phil Corso

Senior Member
Gentlemen… Following is an example illustrating the importance of the X/R caveat:

Multiply 2 complex impedances, Z1 & Z2. Their complex impedances are 3+j4, and 4+j3, respectively. The magnitude of each is 5, their X/R ratios = 3/4, and, 1.33, respectively, and the product of their magnitudes is 25!

Now add their x & y components. The resultant complex number is 7+j7! X/R ratio is 1.0! Magnitude is Sqrt(2)x7 or 9.89! And, their product is now 98.0 instead of 25!

Phil
 
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