Engineering Pro Guides Question 54

So this is a phase to phase fault question. I thought that IA = 0 in phase to phase faults?

eatsleep said:

So this is a phase to phase fault question. I thought that IA = 0 in phase to phase faults?

View attachment 12636

Click to expand...

It depends on which phase to phase fault occurred. IA=0 if there is Phase B to Phase C fault. 

Here in this problem phase A to phase B fault occurs so condition will be IC=0.

I think the answer options are incorrect or problem is not worded correctly. In case of line to line fault, Ia1 = -Ia2 since there are no zero sequence components in line to line faults, the current in IA = i0+i1+i2 = 0

Feel free to correct if my understanding is not correct, I am going by the formula in graffeo. 

@roy167 The equation you are using is for a fault between phase B & phase C.  I think that is the confusion.  The graffeo book has V_b = V_c, meaning that the fault is between phases B & C.  For a fault between phase A and B, V_a = V_b and Ic1 = -Ic2; so the current in IC = i0 + i1 + i2 = 0. 

I have been working on these cheat sheets for Engineering Pro Guides and it is only a draft.  This one is for fault current analysis, hopefully it clears up the confusion on the problem.  I appreciate any feedback. 

View attachment Fault Current Analysis - Draft Engineering Pro Guides.pdf

Justin,

Cheat sheet looks good. hope you can have a final version in a few weeks before the april 2019 exam. Can you please update it here when the final version is out? 

roy167 said:

Justin,

Cheat sheet looks good. hope you can have a final version in a few weeks before the april 2019 exam. Can you please update it here when the final version is out? 

Click to expand...

Yes, will do. 

@justin-hawaii

For those that purchased the Engproguides, will an e-mail be sent out to download the updated version and if so, how long until release?

I'm sure I'm just missing something simple here. This a part of the solution where the sequence current is converted into phase current.



That equation does not match the equation used in the EPG study guide. The study guide has 𝐼𝐵 = 𝐼𝐴1 ∗ (𝑎^2 − 𝑎) and i understand that IA=-IB. Can anyone explain this part of the solution (just how this formula was determined)?

chendon said:

@justin-hawaii

For those that purchased the Engproguides, will an e-mail be sent out to download the updated version and if so, how long until release?

Click to expand...

I will send an email out to those that purchased a engproguides product.  I am looking at a couple of weeks from today. 

eatsleep said:

I'm sure I'm just missing something simple here. This a part of the solution where the sequence current is converted into phase current.

View attachment 12664

That equation does not match the equation used in the EPG study guide. The study guide has 𝐼𝐵 = 𝐼𝐴1 ∗ (𝑎^2 − 𝑎) and i understand that IA=-IB. Can anyone explain this part of the solution (just how this formula was determined)?

Click to expand...

Sorry about the confusion with the two different formulas.  You will get the same magnitude with both formulas, it is just the angle of currents will be 90 and -90 vs. 30 and -30.  Since the angles are relative, both will satisfy the solution.  The derivation of the a^2 - a term can be seen below.

IB = I0 +a^2 * I1 + a * I2

IC = I0 +a* I1 + a^2 * I2

but you know that I1 = -I2 and I0 = 0

IB = a^2 * I1 - a * I1

IB = (a^2 - a)* I1

and

IC = I0 +a* I1 - a^2 * I1

IC = (a-a^2)* I1

Sym components are cool, but sometimes there are simpler methods.

find Z base using 1 mva and 1 kv = 1 ohm and use pos seq values since pos=neg and convert PU values, consider we only need magnitude, no phase or Vdrop

Gen Z = 0.15 Ohm and xfmr Z = 0.08 Ohm

2 lines x sum = 0.46 Ohm

1000 v / 0.46 Ohm = 2173.9 A