Zach Stone P.E.
Well-known member
The MVA Method is used for worse case three phase symmetrical faults to solve problems in the NCEES exam spec IV.A.5. Fault Current Analysis.
While the Per Unit Method can also be used, it tends to be less intuitive and more error prone. It is still recommended to know both so that you can check your work and be familiar working with the per unit system.
For anyone needing a little extra help this weekend, in the below video I solve a more complicated fault current analysis problem using the MVA method with two parallel generators, a transformer, and line impedance feeding a 220kV faulted bus.
The same exact problem in this video is also solved with the Per Unit method along with 13 videos total on fault current analysis on the Electrical PE Review course website if you'd like to watch it to compare the MVA method with the Per Unit method.
Power Contributions from each circuit element in this problem are:
SfG1 = 982.14 MVA
SfG1 = 924.93 MVA
SfXfmr = 978.26 MVA
SfZ = 4,914.28 MVA
Remember that fault contributions in Volt-Amps will add opposite of resistors.
That means in series they add like parallel resistors using the reciprocating sums method, and in parallel, they add like resistors in series.
To find the total power at the fault on the bottom bus, find the total power in Volt-Amps by calculating the total fault power by combining the power contributions.
For this particular problem that means:
Sf = (SfG1 + SfG1)//SfXfmr//SfZ
Sf = 571.49 MVA
and
Fault Current I SC = 1.5KA, or, 1,500 Amps on the 220kV faulted bus.
Can you solve it?
While the Per Unit Method can also be used, it tends to be less intuitive and more error prone. It is still recommended to know both so that you can check your work and be familiar working with the per unit system.
For anyone needing a little extra help this weekend, in the below video I solve a more complicated fault current analysis problem using the MVA method with two parallel generators, a transformer, and line impedance feeding a 220kV faulted bus.
The same exact problem in this video is also solved with the Per Unit method along with 13 videos total on fault current analysis on the Electrical PE Review course website if you'd like to watch it to compare the MVA method with the Per Unit method.
Power Contributions from each circuit element in this problem are:
SfG1 = 982.14 MVA
SfG1 = 924.93 MVA
SfXfmr = 978.26 MVA
SfZ = 4,914.28 MVA
Remember that fault contributions in Volt-Amps will add opposite of resistors.
That means in series they add like parallel resistors using the reciprocating sums method, and in parallel, they add like resistors in series.
To find the total power at the fault on the bottom bus, find the total power in Volt-Amps by calculating the total fault power by combining the power contributions.
For this particular problem that means:
Sf = (SfG1 + SfG1)//SfXfmr//SfZ
Sf = 571.49 MVA
and
Fault Current I SC = 1.5KA, or, 1,500 Amps on the 220kV faulted bus.
Can you solve it?
