# Another good one, 8.2



## Wolverine (Oct 20, 2006)

A two winding single-phase transformer is rated as 6.8kV : 115kV at 500kVA. A short circuit test at rated current on the high voltage side yields:

Ps.c. = 435 W

Vs.c. = 2500V

Is.c. = 4.35A

Determine the p.u. impedance.


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## Art (Oct 20, 2006)

> A two winding single-phase transformer is rated as 6.8kV : 115kV at 500kVA. A short circuit test at rated current on the high voltage side yieldss.c. = 435 W
> 
> Vs.c. = 2500V
> 
> ...


0.1345


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## Frontier05 (Oct 21, 2006)

> > A two winding single-phase transformer is rated as 6.8kV : 115kV at 500kVA.  A short circuit test at rated current on the high voltage side yieldss.c. = 435 W
> >
> > Vs.c. = 2500V
> >
> ...


What book shows how to obtain pu from these transformer test? The books I have just show how to get all element values on the transformer models from sc or oc conditions .. maybe it is there, but I don't see it yet. Willing to share the solution? Thanks!


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## Art (Oct 21, 2006)

> > > A two winding single-phase transformer is rated as 6.8kV : 115kV at 500kVA.? A short circuit test at rated current on the high voltage side yieldss.c. = 435 W
> > >
> > > Vs.c. = 2500V
> > >
> ...


you can only get the |Z| value with the data provided, ie, no power factor or phase shift given, only real power in W

basically Isc x Tr gives the current, you know the power 435W solve for |Z|

once you calc this value simply calc the base Z from the rated VA and voltage...

once you have this take the ratio of |Z| / Z base = pu

I didn't use a book, but most EE texts on power system design/analysis will have this info....


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## Frontier05 (Oct 22, 2006)

> basically Isc x Tr gives the current, you know the power 435W solve for |Z|


Art, I'm not really sure what Tr is? The Z base wasn't a problem and had that part, it's finding the actual (|Z|) with the given info. Pretty sure for short circuit test it's actually the r value, X value for open. Anyway.

I took 2 different approaches to this one, both a different answer though, but off by a factor of 1000.

1. Using Zpu = Vpu / I pu. Where Iact=4.35 (Isc), Ibase= 500MVA/115KV=4.34 and Vact=2500 (Vsc) and Vbase given as 115K.

so Ipu=4.35/4.34=1.01 and Vpu=2500/115000=.0217

then Zpu=.0217/1.01= .0217

The other I tried was this:

2.

Zbase=Vbase^2/VA=115^2/500=26.45

Zact=V/I=2500/4.35=574.7 (but -- using the Psc=(Isc)^2 x R would give a different answer?)

so zpu=574.7/26.45=21.7

Guess I'm off somewhere and probably something very simple - I'll be doing one of these :brick:

Wolverine: possible to post what the solution manual has?

Thanks!


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## Art (Oct 22, 2006)

> > basically Isc x Tr gives the current, you know the power 435W solve for |Z|
> 
> 
> Art, I'm not really sure what Tr is? The Z base wasn't a problem and had that part, it's finding the actual (|Z|) with the given info. Pretty sure for short circuit test it's actually the r value, X value for open. Anyway.
> ...


Tr = turns ratio = 6.8 / 115

Ip = 4.35 x 115/6.8 ~ 73.57 A

Z = 73.57^2 /435 ~ 12.44 ohm

Z base = V base/I base = 6800 / (500000/6800) = 92.48

or Z base = V base ^2 / VA base = 6800 ^2 / 500000 = 92.48

Zpu = 12.44 / 92.48 = 0.1345

be careful with your exponential components...you cancelled them, but the V one is squared...I always do things in scientific notataion...

Z base (s) = 115000 ^ 2 / 500000 = 26,450 ohms

Z base (p) = Z base (s) x 1/Tr ^ 2 = 92.48 ohms same as above...

if I could offer one word of advice, forget memorizing formulas...you know 90% of what you need to with KVL &amp; KCL and the power relationships V = PI and V = V^2/Z = I^2 Z, which can be derived from the Kirchoffs...it all comes down to basics


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## Frontier05 (Oct 22, 2006)

> Z = 73.57^2 /435 ~ 12.44 ohm|


Does this come from the power P=I^2 x R? ie - R=P/I^2? it appears flipped being 73.57^2/425 (I^2/P) or 1/R?



> you know 90% of what you need to with KVL &amp; KCL and the power relationships V = PI and V = V^2/Z = I^2 Z,


yes, I agree ...... P=IV and P=I^2R=V^2/R

ok, I did not see where the turns ratio came into affect being the problem statement said high side, and in this case, the high side is the secondary and value parameters are on the secondary side so there wouldn't be a need to reflect to the primary side (lower voltage side).

Yes, I see the Zbase(P) and Zbase(s) - Thanks Art!

I believe it doesn't matter though, either way the same Z should result. 13.45% seems high for a transformer, but then again I realize these are exsercise problems.


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## Art (Oct 22, 2006)

> > Z = 73.57^2 /435 ~ 12.44 ohm
> >
> > |
> 
> ...


yep, just solve P = I^2 Z for Z, after finding the primary current...

typos 

well, it's not really 13% it's 0.1345 pu

it's basically 'unitized' for mathematical convenience...

if we use the Z base as reference, an 'average' xfmr would be 1 pu, so 0.1345 is pretty good...


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## Art (Oct 22, 2006)

I dug out the text book:

Z = R + j X

R = 12.44 ohm from before

500000 VA/6800 V = rated current

|Z| = 2500 V / (500000/6800) A = 2500/73.53 = 34 ohm

X =(Z^2 - R^2)^1/2 = 31.64 ohm

Z = 12.44 + j 31.64 or 34 at 68.54 deg

Z base same as before 92.48

Z pu = 0.3676 at 68.54 deg or 0.1345 + j 0.3421

I was only considering the real component, not the reactance


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## Frontier05 (Oct 22, 2006)

> I dug out the text book:
> Z = R + j X
> 
> R = 12.44 ohm from before
> ...


In my other post above:

It appears Z to be reversed unless I'm not seeing it correctly.

ok, you have listed above

Ip = 4.35 x 115/6.8 ~ 73.57 A

Z = 73.57^2 /435 ~ 12.44 ohm

But that Z is for Z=I^2/P (it's reversed from what it should be)? From the equation we use often, P=I^2 x Z, so this means Z=P/I^2

This would make Z= 435 / (73.57)^2 = .0804 , not 12.44 ohm - unless I'm not seeing something here, which may be the case.

I'm still not quite sure why the turns ratio is needed on this being all the parameters are given on the high side (115KV) per the original problem statement.

Curious how the solution is written out on this one.


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## Wolverine (Oct 23, 2006)

Sorry for the delay in posting the solution - I've been in the cave studying, then went back to redo this problem and couldn't get it right the second time around. Then couldn't make sense of the solution and had Op-Amps to study anyway, so I left it. Will try to post the solution tonight.

:drunk:

_Wake me up when October ends_


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## Frontier05 (Oct 23, 2006)

Sometimes I think I take way to many breakes between studying .......

Sounds good Wolverine, thanks.


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## Wolverine (Oct 24, 2006)

l Zsc l = 2500/4.35 = 574.7 ohms

Rsc = P/I^2 = 435/4.35^2 = 22.99 ohms

x^2 = l Zsc l^2 - Rsc^2 = 330272 , so X = 574.7ohms

Z = 22.99 + j 574.7

Zbase = (115^2 *10^6)/5*10^5 = 26450 ohms

Zpu = (23 + j575) / 26450 = .00087 +j.0217

The first time I did this problem, I got the right answer and moved on. Now I look at the solution and can hardly make sense of it. Oh fudge, I'm regressing! :brick:

WAIT A MINUTE! I'VE GOT IT!

The solution had a typo in the Zbase calculation using 440 instead of 115 (since corrected above)! I GET IT! I GET IT! WOO HOO, I MIGHT PASS! AAAA- HAHAHAHAHA!!!! HAHAHAHAHA! HAHAhaha! . . . . ha.ha. . . ha. . . . okay, I'm losing it.


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## Frontier05 (Oct 25, 2006)

I continued this problem a slightly different way. As seen in my previous post up there:

" 1. Using Zpu = Vpu / I pu. Where Iact=4.35 (Isc), Ibase= 500MVA/115KV=4.34 and Vact=2500 (Vsc) and Vbase given as 115K.

so Ipu=4.35/4.34=1.01 and Vpu=2500/115000=.0217

then Zpu=.0217/1.01= .0217 "

Then finding the real portion, find the power factor using the short circuit data:

Cos (theta)=Ppu/(Vpu x Ipu)

Cos (theta) = (435/500,000)/ (.0217)(1.01)

Cos (theta) =.0396

so Re = Zpu x Cos (theta) = (.0217)(.0396) = .00087

I looked at it like this: Since the test data was provided on the "high" side, readings were taken on the secondary "high" side, thus, the primary side was shorted out. Obviously things would be a little different if this were a step down xfmr, ie - the high side being the primary.

Anyways, I think the sloution manual was a little more direct with the process. That is a good problem.


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## Dolphin P.E. (Jan 18, 2011)

At the short circuit test, they usually short the low voltage side.


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