Fault Current Analysis (PU method)

[supra33202]

@Zach Stone, P.E.

I am working on Fault Current Analysis quiz question 9 at electricalpereview.com.

And I am trying to use PU method to solve it.

Set Vb=4160V, Sb=1.5MVA

Zg, pu (Zpu at generator) = 0.079 pu * (1.5MVA/50MVA)=0.00237pu

Zx, pu (Zpu at transformer) = 0.075pu

Since Zg, pu is parallel with Zx, pu, so Ztotal, pu = (Zg, pu*Zx, pu) / (Zg, pu+Zx, pu) = 0.0028 pu

Isc, pu = Vpu/Ztotal, pu = 1pu/0.0028pu = 438.6pu

Ib=Sb/(sqrt(3)*Vb= 1804.2A

Isc = Isc, pu * Ib = 791316A

The correct answer should be 91.5KA.

What did I do wrong with the PU method?

Thanks!

 

[Zach Stone P.E.]

18 hours ago, supra33202 said:
@Zach Stone, P.E.

I am working on Fault Current Analysis quiz question 9 at electricalpereview.com.

And I am trying to use PU method to solve it.

Set Vb=4160V, Sb=1.5MVA

Zg, pu (Zpu at generator) = 0.079 pu * (1.5MVA/50MVA)=0.00237pu

Zx, pu (Zpu at transformer) = 0.075pu

Since Zg, pu is parallel with Zx, pu, so Ztotal, pu = (Zg, pu*Zx, pu) / (Zg, pu+Zx, pu) = 0.0028 pu

Isc, pu = Vpu/Ztotal, pu = 1pu/0.0028pu = 438.6pu

Ib=Sb/(sqrt(3)*Vb= 1804.2A

Isc = Isc, pu * Ib = 791316A

The correct answer should be 91.5KA.

What did I do wrong with the PU method?

Thanks!


 

Hi Tommy,

Your calculation for the equivalent per unit impedance is off. 



Set Vb=4160V, Sb=1.5MVA

Zg, pu (Zpu at generator) = 0.079 pu * (1.5MVA/50MVA)=0.00237pu

Zx, pu (Zpu at transformer) = 0.075pu

Since Zg, pu is parallel with Zx, pu, so Ztotal, pu = (Zg, pu*Zx, pu) / (Zg, pu+Zx, pu) = 0.0028 pu

     Try this:

      Ztotal pu = 0.00237pu //0.075pu

      Ztotal pu = (0.00237pu•0.075pu )/(0.00237pu+0.075pu )

      or:

      Ztotal pu =  (0.00237pu^-1 + 0.075pu ^-1)^-1

      Ztotal pu =  0.002275pu

Isc, pu = Vpu/Ztotal, pu = 1pu/0.0028pu = 438.6pu

      Isc, pu = 1/0.002275pu

      Isc, pu = 439.56pu

Ib=Sb/(sqrt(3)*Vb= 1804.2A

      Not sure what numbers you plugged in, but try this using the same base values:

      Ib=Sb/(sqrt(3)*Vb= 1804.2A

       Ib=1.5MVA/(sqrt(3)*4,160V

       Ib= 208.18A

Isc = Isc, pu * Ib = 791316A

     Now plug in the correct values for the per unit fault current and base current:

     Isc = Isc, pu * Ib

     Isc = (439.56pu)(208.18A)

     Isc = 91,507.6A

     Isc = 91.5kA

If anyone else would like to try this, the entire fault current analysis chapter including the quiz with 11 practice exam problems with full-length solutions available in the free trial. All you need is an email address to sign up. Click on the link and scroll down a little bit until you see "enroll in the free course":

Electrical PE Review - Online Review Course for the Electrical PE Exam

 

[supra33202]

@Zach Stone, P.E.

Thanks for the help!