# Complex Imaginary Test#2 problem 79



## TFT (Sep 30, 2011)

This is a fault current problem asking for the generator's contribution at the secondary of the transformer. The subtransient impedance of the generator is .39, rated power is 11MVA. The transformer is rated 5kV/25kV, 6MVA, and 6% . The generator is connected to the low side of the transformer, assume loss-less conductors.

Choosing 6MVA and 25kV as bases;

Zpu-gen = .39(5/25)2(6/11)

= .00851pu

Isc = 1/.00851

= 117.51pu

Ibase = 6MVA/25kVA

= 240

I = 117.51*240

= 28202.4 Amps

28202.4 was not an answer choice. Where did I go wrong? Should the Ibase equation be [6MVA/(25kVA/sqrt3)]


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## xd-data-ii (Sep 30, 2011)

I think you are right.

The z conversion is Zold x (Snew/Sold) x (Vold/Vnew)^2

which is what you have - specifically for the kVA being 6/11

They have 11/6 which is (Sold/Snew) - switching from 11MVA to 6MVA and from 5kV to 25kV

They say 11MVA to be converted to 6MVA but have calculated it the other way around - 11MVA is used and is used to convert PU back to actual.

Either just change the transformer in the question to be 11MVA and the Gen to 6MVA - then their solution is fine ..... or get messy with correcting the issues in the solution.


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## TFT (Sep 30, 2011)

xd-data-ii said:


> I think you are right.
> The z conversion is Zold x (Snew/Sold) x (Vold/Vnew)^2
> 
> which is what you have - specifically for the kVA being 6/11
> ...



What about the sqrt3 in the Ibase equation?


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## xd-data-ii (Sep 30, 2011)

Doesnt say how the generator or the transformer is connected.

But for per unit i dont think the sqrt(3) comes into it. think that is one of the points of per unit.

I think the dividing by 25kV is correct. (not 25kVA)


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## TFT (Oct 1, 2011)

xd-data-ii said:


> Doesnt say how the generator or the transformer is connected.
> But for per unit i dont think the sqrt(3) comes into it. think that is one of the points of per unit.
> 
> I think the dividing by 25kV is correct. (not 25kVA)



Yes 25kV, typo.

See the solution for #44, the sqrt3 is used in the Ibase equation. So which one is correct? Both questions are 3phase and both voltage values are given in line to line values. Delta or Wye connection should not matter, S3phase=srt3*I*V either way.

I believe Ibase = 6MVA/(sqrt3*25kV) = 138.564A

Therefore Isc-gen = 117.508*138.564 = 16282.4A


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## Insaf (Oct 4, 2011)

I did not see the question, but I feel you need to consider the transformer impedance as well (ref: NCEES 540).

Thanks


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## DK PE (Oct 4, 2011)

TFT said:


> This is a fault current problem asking for the generator's contribution at the secondary of the transformer.






Insaf said:


> I did not see the question, but I feel you need to consider the transformer impedance as well (ref: NCEES 540).
> Thanks


Be careful reading the question asked...


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## Wildsoldier PE (Oct 6, 2011)

TFT said:


> This is a fault current problem asking for the generator's contribution at the secondary of the transformer. The subtransient impedance of the generator is .39, rated power is 11MVA. The transformer is rated 5kV/25kV, 6MVA, and 6% . The generator is connected to the low side of the transformer, assume loss-less conductors.
> Choosing 6MVA and 25kV as bases;
> 
> Zpu-gen = .39(5/25)2(6/11)
> ...




Isn't Ibase = 6MVA/(25000V * square(3))=138.56A &lt;since the problem specify is a 3 phase system it needs to be multiply by square of 3 to get the line current.

then I = Ipu x Ibase= 117.51pu x 138.56=16,282 Amps


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## DK PE (Oct 6, 2011)

DK PE said:


> Be careful reading the question asked...


Apologies if my earlier comment confused the issue, without any diagram I thought maybe the generator and transformer were both tied to the bus and they were just trying to isolate the generator's contribution.

In this case, assuming the generator is tied to the primary of the txfmr and you are determining the fault at the secondary, the transformer's impedance in needed.


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## Wildsoldier PE (Oct 6, 2011)

DK PE said:


> DK PE said:
> 
> 
> > Be careful reading the question asked...
> ...


DK

do you think that the I base =6MVA/(25000V * square(3))?


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## DK PE (Oct 6, 2011)

TFT said:


> This is a fault current problem asking for the generator's contribution at the secondary of the transformer. The subtransient impedance of the generator is .39, rated power is 11MVA. The transformer is rated 5kV/25kV, 6MVA, and 6% . The generator is connected to the low side of the transformer, assume loss-less conductors.
> Choosing 6MVA and 25kV as bases;
> 
> Zpu-gen = .39(5/25)2(6/11)
> ...


I believe since you chose the base voltage as 25kV on the secondary side of transformer, the base voltage on the primary (and at the generator) is 5kV and doesn't need to be modified.

Seems like it should be Zpu-gen = .39*(6/11)

Then you need to add the p.u impedance of transformer and determine fault current in p.u. --&gt;then use base current at that point to get absolute current.


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## Wildsoldier PE (Oct 6, 2011)

DK PE said:


> TFT said:
> 
> 
> > This is a fault current problem asking for the generator's contribution at the secondary of the transformer. The subtransient impedance of the generator is .39, rated power is 11MVA. The transformer is rated 5kV/25kV, 6MVA, and 6% . The generator is connected to the low side of the transformer, assume loss-less conductors.
> ...



Dk

The problem ask for the generator contribution under fault conditions at the transformer secondary side. I think thats why you need to use voltages to change bases.


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## DK PE (Oct 6, 2011)

I agree with the 138.56A base current on the secondary of the transformer side you posted earlier. I wasn't reading the problem very well and assumed they are asking for fault current when the secondary is faulted. If they are asking for primary current (generator current) with a fault on Txfmr's secondary which should be about 3.6 pu which is ~ 2500 amps on that side.

I guess if they want the fault current on the secondary side that would be ~ 508 Amps (but I'm getting tired so could use some double checking)

I don't agree you need to use the voltages to change bases.


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## DK PE (Oct 6, 2011)

I guess I should state I am assuming the given subtransient impedance of the generator is 0.39, with a rated power is 11MVA and rated voltage of 5kV just for clarity.


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## DK PE (Oct 9, 2011)

I guess I left this hanging a bit earlier this week as was busy but if I understand the problem statement, the solution I would get is below. Don't know if agrees with the published solution or not. Again, I am assuming the voltage rating of the generator is 5kV which makes sense.

Fault current 1.pdf


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## Insaf (Oct 11, 2011)

This also can be solved this way:

Base S=11 MVA, Base kV: 5 KV (LT), 25 kV (HT)

Base Current (HT) = 11* 10^6 /(sqrt(3)* 25* 10^3) = 254 Amps

Z(Transformer) refer to 11 MVA= 0.06(11/6)=0.5 pu

I(fault)=1/0.5=2 pu

I(fault_actual)= 254*2= 508 A, this the generator contribution for bus fault on transformer HS bus.


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## DK PE (Oct 11, 2011)

Insaf said:


> This also can be solved this way:
> 
> Base S=11 MVA, Base kV: 5 KV (LT), 25 kV (HT)
> 
> ...


You lost me at this step: --&gt; Z(Transformer) refer to 11 MVA= 0.06(11/6)=0.5 pu ? seems too high?

and I think you are calculating the short circuit capability of the transformer with an "infinite source" which is ~9pu or 2300 Amps. Your solution doesn't take into account the generator finite source capability.

Maybe you meant to add the 0.39 from the generator to the 0.11 from the tsfmr and then you get 1/0.5 - 2pu ?


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## Insaf (Oct 11, 2011)

You are correct, Total impedance will be 0.39+ (0.6*(11/6))=0.5 pu. Z(transformer refer to 11 MVA is 0.11 (0.6*11/6) and then total system impedance is 0.5 pu. Good catch and thanks for your input.


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## Insaf (Oct 11, 2011)

Can anyone tell me answer from Complex Imaginary ( as I did not see the problem or solution)


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## xd-data-ii (Oct 11, 2011)

answer is 28,200A.


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## Insaf (Oct 11, 2011)

xd-data-ii said:


> answer is 28,200A.



The answer doesn't look very reasonable. Any explanation from complex imaginary?

My understanding that for SC calculation, one has to account both generator and transformer impedance. As transformer 6% impedance not based on primary or secondary voltage rather it indicates full load current will flow to short-circuited secondary (Secondary short) when 6% of primary rated voltage have been applied. So to get pu impedance of generator, sub-transient impedance should be converted by using base MVA of transformer (6 MVA, if someone choose as base) and gen MVA (11 MVA), voltage will not come to account. In that case, no way to get the 28,200 A.

For this problem, it will be more convenient to choose generator MVA as base MVA, KV base 5 (LS), 25 (HS) and then converting transformer impedance based on generator MVA (0.06*(11/6)). Then system imp will be 0.11+0.39= 0.5 pu.


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## Insaf (Oct 11, 2011)

Dear DK PE,

You are correct, total impedance will be 0.39+ (0.06*(11/6))=0.5 pu. Z(transformer refer to 11 MVA) is 0.11 pu (0.06*11/6=0.11) and total system impedance is 0.5 pu (0.11+0.39=0.5). Good catch and thanks for your input.


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## xd-data-ii (Oct 12, 2011)

Their corrections of the solutions are still full of massive errors. It's ridiculous. Their solution looks a mess at this point.

Qtn:

3phase 11MVA Gen (Xd"=0.39j pu) serves 6MVA 5/25kV transformer (6% impedance). What is The MVA contribution at the generator with a fault at the transformer secondary terminals?

- question asks for MVA but the answers are all Amps so I guess it should be the Amps on the secondary

Their solution:

Convert to 6MVA base and 25kV

Z(new) = 0.0085j pu

I=1.0/0.0085=117.5 pu

I(base)=11MVA/25kV=240A. ------&gt; this should be 6MVA/25kV

I(actual)=34.5 x 440 = 28200 A -------&gt; this should be 117.5 x 240 = 28200 A

The solution should have

I(base)= 6MVA/(sqrt(3)x25kV) = 138.56

I(actual) = 117.5 x 138.56 = 16281.28 A ......... as someone else above pointed out

So they converted the generator Xd" to 6MVA base and 5kV to 25kV.

They left out the transformer impedance. ......... So the 6% is on the 6MVA referred to primary side 5kV. This is pu so is same on primary and on secondary (I think).

Looks like similar question in NCEES sample exam 540

So solution looks like:

Convert Tx to Gen base - 6MVA to 11MVA base

Z(Tx)=0.06(11/6)=0.11

S(s/c)=1.0/(0.39+0.11)=2 pu

S(actual)=2 x 11MVA = 22 MVA. ....... they asked for MVA in the question

I(s/c) = 22MVA/(sqrt(3)x5kV) = 2540.3 A

.... Is this correct or am I way off???

This seems to agree with the posts above this, however I think in your alternative solution, Insaf, you seemed to have assumed the transformer is a step down transformer (its step up) or that the voltage of the Gen is 25kV. --- did not calculate based on 5kV generator voltage.


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## DK PE (Oct 12, 2011)

xd-data-ii said:


> Their corrections of the solutions are still full of massive errors. It's ridiculous. Their solution looks a mess at this point.
> 
> Qtn:
> 
> ...





DK PE said:


> I agree with the 138.56A base current on the secondary of the transformer side you posted earlier. I wasn't reading the problem very well and assumed they are asking for fault current when the secondary is faulted. If they are asking for primary current (generator current) with a fault on Txfmr's secondary which should be about 3.6 pu which is ~ 2500 amps on that side.


Well at least for me this question has come full circle back to what I thought... about 10 posts earlier I said ~2500 Amps at generator, then I thought they were asking at the fault... now we have the real question. Earlier posts have stated 508 at txfmr secondary which when multiplied *5 would agree with the 2540A on primary. I didn't even realize they were asking for MVA and giving answers in Amps (that's some bad proofreading).

Going back to the pdf diagram I posted earlier, the base current (using the bases I chose) on the generator side of txfmr is 6000/(sqrt3 *5) = 692.5A. Multiply this times the 3.67pu I showed and you get 2541Amps which agree with you.

That would give the ~22MVA at generator you showed. I believe this is correct.


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## ElecPwrPEOct11 (Oct 24, 2011)

Wow, what a poorly crafted question by Complex Imaginary.

I agree with DK &amp;xd-data above that Ssc = 22MVA. Since this is a step-up transformer the current on the secondary will be 22MVA / (sqrt(3) * 25kV) = 508A. This also aligns with the solution to NCEES #540, which is exactly the same type of problem.


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