Help with Pu conversion for sample PE 3-Phase Fault question

[wfg42438]

Hello,

In the question below we are after the 3Phase fault current on the secondary of a 12.47/0.480 kV XFMR.

Now in addition to the XFMR parameters we are told the "Fault Duty" on of the 12.47kV side of the system.

My interpretation is that this represents the equivalent MVAsc of the system above the XFMR.




Based on this info I did the following:

Isc above the XFMR is Isc_12.47= MVAsc/ (12,470 * sqrt (3)) =1851.965 amps
Therefore Z_12.47kV= V/Isc= 12470/1851.965=6.73 ohms

Since the XFMR imp is already in pu i decided to convert Z_1247kV to pu as shown below:
Z_base= (0.480^2)/(10^6)
Z_12.47kV ref to secondary= Z_12.47kV/ (12.47/0.480)^2=0.009 ohms
Z_12.47kV_pu= Z_12.47kV ref to secondary/Z_base=0.0433 pu

When i looked at the solution what i saw is that the pu value NCESS gets is 0.025 pu

Does anyone see where i went wrong with my pu conversion?

 

[main197]

Can you upload the diagram?

 

[wfg42438]

Hello,

In the question below we are after the 3Phase fault current on the secondary of a 12.47/0.480 kV XFMR.

Now in addition to the XFMR parameters we are told the "Fault Duty" on of the 12.47kV side of the system.

My interpretation is that this represents the equivalent MVAsc of the system above the XFMR.
View attachment 27123

View attachment 27122

Based on this info I did the following:

Isc above the XFMR is Isc_12.47= MVAsc/ (12,470 * sqrt (3)) =1851.965 amps
Therefore Z_12.47kV= V/Isc= 12470/1851.965=6.73 ohms

Since the XFMR imp is already in pu i decided to convert Z_1247kV to pu as shown below:
Z_base= (0.480^2)/(10^6)
Z_12.47kV ref to secondary= Z_12.47kV/ (12.47/0.480)^2=0.009 ohms
Z_12.47kV_pu= Z_12.47kV ref to secondary/Z_base=0.0433 pu

When i looked at the solution what i saw is that the pu value NCESS gets is 0.025 pu

Does anyone see where i went wrong with my pu conversion?

For this question the sample Exam does not provide any diagrams, we only have the problem description.

 

[main197]

See this: NCEES Problem #70

 

[EE_transformer]

For this question the sample Exam does not provide any diagrams, we only have the problem description.

Someone else may want to chime in here, but I believe since they give the fault duty of the transformer as 40MVA and the problem states "the transformer and system have the same X/R ratio" you just need to take 1000KVA/40MVA = 0.025. Then add that to the 4% of the transformer (1/(.04+.025) = 15.4).

 

[wfg42438]

See this: NCEES Problem #70

Thank you for sharing the solution!

Im used to applying the pu method so to avoid trying to learn a method i will stick to that.

However when determining the pu impedance for the given MVAsc i realized a mistake on my end.

For example what i did was MVAsc= srt(3)*Vnom*Isc (Based on the NCEES hand book)


So What i did was Isc= MVAsc/ (sqrt (3)*Vnom)

Then Z_ohm= Vnom/Isc

however this leads to the wrong impedance

I noticed that if instead use MVAsc / kVnom then i get the correct impedance

Can anyone comment on why the sqrt(3) factor is not applicable?

Did i misinterpret the line voltage as Line-to-Line when it actually is the Line to Neutral voltage?

I realize kV^2/MVAsc is much easier but id like a clarification just in case i forget this Shortcut on the exam

Thanks in Advance!

 

[main197]

So What i did was Isc= MVAsc/ (sqrt (3)*Vnom)

Do not use that equation to calculate the pu value. That is the actual value of short circuit current.

I_base= VA_base/sqrt(3)*V_base
and Isc_actual= Isc_pu*I_base

Use this equation to find the impedance, S=V^2/Z,
so Z=V^2/S.

To make it easier to memorize, remember the Ohm's law (V=IR). P=IV, and I=V/R, so P=V^2/R.

 

[wfg42438]

Do not use that equation to calculate the pu value. That is the actual value of short circuit current.

I_base= VA_base/sqrt(3)*V_base
and Isc_actual= Isc_pu*I_base

Use this equation to find the impedance, S=V^2/Z,
so Z=V^2/S.

To make it easier to memorize, remember the Ohm's law (V=IR). P=IV, and I=V/R, so P=V^2/R.

View attachment 27253

I see so by using Isc= MVAsc/ (sqrt (3)*Vnom) what im doing is finding the ISc from the utility with a 40MVAsc assuming i want the ISC contribution to the node directly below it. this is not a relationship i can use to find the impedance

However what we really want is the impedance which is easiest to derive using V^2/MVA
Ill stick this formula given its simplicity and how quick it is\

Thanks for the help!