Complex Imaginary Test#2 problem 79

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)]

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.

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

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.

Click to expand...

What about the sqrt3 in the Ibase equation?

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)

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)

Click to expand...

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

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

Thanks

TFT said:

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

Click to expand...

Insaf said:

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

Click to expand...

Be careful reading the question asked...

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)

= .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)]

Click to expand...

Isn't Ibase = 6MVA/(25000V * square(3))=138.56A <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

DK PE said:

Be careful reading the question asked...

Click to expand...

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.

DK PE said:

DK PE said:

Be careful reading the question asked...

Click to expand...

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.

You have calculated the Gen Z in pu, add the transformer and determine the fault current in p.u. then convert to absolute using base I on secondary of Txfmr.

Click to expand...

DK

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

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)

= .00851pu

Where did I go wrong?

Click to expand...

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. -->then use base current at that point to get absolute current.

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.
Choosing 6MVA and 25kV as bases;

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

= .00851pu

Where did I go wrong?

Click to expand...

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. -->then use base current at that point to get absolute current.

Click to expand...

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.

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.

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.

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

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.

Insaf said:

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.

Click to expand...

You lost me at this step: --> 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 ?

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.

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

answer is 28,200A.