Originally posted by: icoggan
I have used the MVA Method for verifying some per unit Fault Level Calculations.
Both answers have matched well for my calculations in networks without parallel branches, my question regards a main switchboard fed from a generator, with bus-tie closed, feeding two transformers in parallel both incoming into another switchboard with the bus-tie closed.
The fault level I am calculating is on the switchboard fed from the transformers.
The MVA equivalent circuit has a single branch at the top from the generator, leading to 2 parallel branches with the cables / transformers, these branches recombine and the motor contributing MVA level is at the bottom.
How do I reduce this MVA equivalent circuit?
The MVA method can be think of as analogous to dealing with admittances. When the MVA elements are in series, the operation is to sum them as (1/MVA1 + 1/MVA2 +...) and evaluate the inverse of the result. When the MVA elements are in parallel, the result is to sum them up as (MVA1 + MVA2 + ...).
In your case, it will be finding the inverse of ((1/MVA-genny)+(1/(MVA-branch_a+MVA-branch_b))+(1/MVA-motor)).
Hope it will give you the same result as the p.u. method. Alternatively, you can do one branch as a time and sum the results using superposition theorem.