Complex Imaginary Test#2 problem 79

[Insaf]

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.

 

[Insaf]

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.

 

[xd-data-ii]

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. ------> this should be 6MVA/25kV

I(actual)=34.5 x 440 = 28200 A -------> 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.

 

[DK PE]

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

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???

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.

 

[ElecPwrPEOct11]

Wow, what a poorly crafted question by Complex Imaginary.

I agree with DK &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.