Question

A 100 MVA, 20 kV alternator has X 1 = X 2 = 0.20 pu and X 0 = 0.05 pu. The neutral of the generator is grounded through a reactor. Find the ohmic value of reactor to limit the current for a single line to ground fault equal to the current for a three-phase fault.

Name: a 100 mva 20 kv alternator has x 1 x 2 020 pu and x 0 005 pu the neutral of the generator is grounded through a reactor find the ohmic value of reactor to limit the current for a single line to ground
Uploaded: 2021-11-10T11:57:45-08:00
Duration: 1 min 35 s
Channel: Prem Bijarniya
Description: a 100 mva 20 kv alternator has x 1 x 2 020 pu and x 0 005 pu the neutral of the generator is grounded through a reactor find the ohmic value of reactor to limit the current for a single line to ground

          A 100 MVA, 20 kV alternator has X 1 = X 2 = 0.20 pu and X 0 =
0.05 pu. The neutral of the generator is grounded through a
reactor. Find the ohmic value of reactor to limit the
current for a single line to ground fault equal to the
current for a three-phase fault.

Added by Dennis B.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Prem Bijarniya

11/10/2021

Step 1

20 pu and X0 = 0.05 pu). The goal is to find the ohmic value of a reactor that is used to ground the neutral of the generator, such that the current for a single line to ground (SLG) fault equals the current for a three-phase fault. Show more…

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A 100 MVA, 20 kV alternator has X 1 = X 2 = 0.20 pu and X 0 = 0.05 pu. The neutral of the generator is grounded through a reactor. Find the ohmic value of reactor to limit the current for a single line to ground fault equal to the current for a three-phase fault.