NCEES Problem #70

[lost4ever P.E.]

Hi to all,
Can anyone help me with the impedance calculation.
base kva = 1000
base kV = 0.48
I got X p.u for transformer is 0.04
and for system is 0.67
I don't know what I am doing wrong?

 

[akyip]

Since the system and transformer have different KVA ratings, did you use the Zpu new = Zpu old * (Sbase new / Sbase old) * (Vbase old /Vbase new)^2 formula to convert one of the base impedances to the correct impedance value per the new system base?

 

[lost4ever P.E.]

Yes I used the similar formula
Transformer : 0.04 (1000/1000)*(0.48/0.48)^2 = 0.04 p.u
System: 0.04 (1000/40000)(12.47/0.48)^2 = 0.67 p.u

 

[RedRaider2020]

Your KV pu is not correct.

12.47 System - Vbase=12.47 V=1 pu, MVAbase=1MVA MVA= 40 pu Z= (1)^2/40=.025 pu
12.47 side Transformer = Vbase= 12.47V V=1pu, MVAbase=1MVA MVA=1pu, Z= .04pu
480 side Transformer = Vbase=480V V=1pu, MVAbase=1MVA MVA = 1pu, Z= .04pu

I=V/Z Z=Ztransformer + Z system = .04pu + .025pu V=1pu V/Z=1/(.04+.025)=15.4

 

[lost4ever P.E.]

@RedRaider2020, how did you get Z = 0.04p.u?

 

[RedRaider2020]

It's given in the problem "transformer with 4% impedence"
For transformers the PU impendence quantity is always given for the rated voltage and MVA.
If you're operating the transformer at a different voltage then you have to change the impendence but if you're operating it at the rated voltage and MVA and you've chosen those rated values as your base then you can just use the given PU impendence.

 

[lost4ever P.E.]

@RedRaider2020, thanks now its clear.

 

[giginubs]

@RedRaider2020, how did you get Z = 0.04p.u?
View attachment 20788

Hi @RedRaider2020, I'm just wondering how did you get the PU values for the 12.47 system.

 

[RedRaider2020]

Voltage is the same as the base so it is 1. The MVA of the system =40MVA/The base MVA = 1 so 40/1 = 40pu. Z=V^2/S so 1^2/40=.025 pu

 

[giginubs]

Voltage is the same as the base so it is 1. The MVA of the system =40MVA/The base MVA = 1 so 40/1 = 40pu. Z=V^2/S so 1^2/40=.025 pu

I'm assuming we are using the 1 MVA as a base since the Transformer Base is 1000KVA=1 MVA. You did this to make the math easier if I understand that correctly.

Also for the 40 MVA, is that considered the actual MVA value because I know that S_pu = S_actual/S_base so in that case would S_actual = 40 MVA and S_base=1MVA therefore S_pu is 40pu as you stated?

 

[RedRaider2020]

I'm assuming we are using the 1 MVA as a base since the Transformer Base is 1000KVA=1 MVA. You did this to make the math easier if I understand that correctly.

Also for the 40 MVA, is that considered the actual MVA value because I know that S_pu = S_actual/S_base so in that case would S_actual = 40 MVA and S_base=1MVA therefore S_pu is 40pu as you stated?

That's correct. You have to pick one base for the whole system so if you can pick one where all the values are 1 the math will be a lot easier. Yes 40MVA is the actual and divided by 1 becomes 40 pu. The values with units like MVA, KVA, KW, V, A will always be the actual value and the base is whatever base you choose it to be if they don't tell you to use a specific one in the question.

 

[giginubs]

That's correct. You have to pick one base for the whole system so if you can pick one where all the values are 1 the math will be a lot easier. Yes 40MVA is the actual and divided by 1 becomes 40 pu. The values with units like MVA, KVA, KW, V, A will always be the actual value and the base is whatever base you choose it to be if they don't tell you to use a specific one in the question.

Thank you for the clarification!

 

[CeeYem]

I thought a pictorial representation may be helpful to understand each step involved in this calculation.
I

 

[Sharmaa]

I thought a pictorial representation may be helpful to understand each step involved in this calculation.
View attachment 21388I

z in p.u = actual value/ base Value, then if you assume base as 1 MVA then how you did 1/40 should not it be 40/1, another question how MVA/MVA will be Z it will be MVA in p.u. Correct me if am wrong.

 

[DLD PE]

This problem can be solved using per-unit or, since this is a 3-phase fault, the MVA method.

For MVA method, you just take the MVA contribution of the transformer (1MVA/0.04)=25MVA, and the MVA contribution of the system (40MVA, which is given in the problem), and add them in series (MVAtotal = (25x40)/(25+40) = 15.38MVA, then solve for the short-circuit current, Isc=15.38MVA/(sqrt 3 x 480V) = 18505A.

For any NCEES practice exam problem, type in "NCEES #____" in the search bar on the top right. I think every problem has been discusses at one point or another. For this problem, either search "NCEES #70, or NCEES #530", since the older NCEES practice exams used 101-140 for the morning session and 501-540 for the afternoon session.

 

[kris7o2]

@RedRaider2020, how did you get Z = 0.04p.u?
View attachment 20788

 

[akyip]

I just posted this in another topic, attached are 2 different methods I used to solve NCEES problem 530 (which became question 70 in the updated practice exam).

 

[yaoyaodes]

This problem can be solved using per-unit or, since this is a 3-phase fault, the MVA method.

For MVA method, you just take the MVA contribution of the transformer (1MVA/0.04)=25MVA, and the MVA contribution of the system (40MVA, which is given in the problem), and add them in series (MVAtotal = (25x40)/(25+40) = 15.38MVA, then solve for the short-circuit current, Isc=15.38MVA/(sqrt 3 x 480V) = 18505A.

For any NCEES practice exam problem, type in "NCEES #____" in the search bar on the top right. I think every problem has been discusses at one point or another. For this problem, either search "NCEES #70, or NCEES #530", since the older NCEES practice exams used 101-140 for the morning session and 501-540 for the afternoon session.

Can you use MVA method for three phase short?

 

[Zach Stone P.E.]

Can you use MVA method for three phase short?

Yes. The MVA method is for worse case bolted three-phase faults (three-phase shorts).