Question

When a 50 MVA, 11 kV, 3-phase generator is subjected to a 3-phase fault, the fault current is – j5 pu (per unit). When it is subjected to a line-to-line fault, the positive sequence current is – j4 pu. The positive and negative sequence reactance are respectively​

Answer 1

Answer:

  j0.2 and 0.05 pu

Answer 2

Answer:  j0.2 p.u. and j0.05 p.u.

Concept: For a three-phase fault

                  If(3-Φ) = Ef/X1 = 1/X1

                Line to line fault in b and c phases,

                  I(a1) = Ef/(X1+X2)

Given: The fault current [If(3-Φ)] = –j5 p.u.

The positive sequence current [I(a1)] = –j4 p.u.

To find: The positive and negative sequence reactance’s(X1,X2)

Step by step explanation:

For a three-phase fault

If(3-Φ) = Ef/X1 = 1/X1

-j5 = 1/X1

X1 = -1/j5

X1 = j0.2 p.u.

Line to line fault on b and c phases,

I(a1) = Ef/(X1+X2)

-j4 = 1/(j0.2 + X2)

j0.2 + X2 = -1/j4

X2 = j0.25 - j0.2

X2= j0.05 p.u.

Answer: The positive sequence reactance is (X1) = j0.2 p.u.

The negative sequence reactance is (X2) = j0.05 p.u.

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