A three phase transmission line has a self-reactance of 0.2 pu and mutual reactance of 0.05 pu. The sum of positive sequence reactance, negative sequence reactance and zero sequence reactance is __ (in pu)Correct answer is '0.6'. Can you explain this answer?

Question

Answer

Solution:

Given, self-reactance (Xs) = 0.2 pu and mutual reactance (Xm) = 0.05 pu

Positive sequence reactance (X1):

In three-phase transmission lines, the positive sequence reactance is given by the self-reactance of the line (Xs).

Therefore, X1 = Xs = 0.2 pu

Negative sequence reactance (X2):

The negative sequence reactance is given by:

X2 = Xs + 2Xm/3

Substituting the given values, we get:

X2 = 0.2 + 2*0.05/3 = 0.2667 pu

Zero sequence reactance (X0):

The zero sequence reactance is given by:

X0 = Xs + Xm

Substituting the given values, we get:

X0 = 0.2 + 0.05 = 0.25 pu

Sum of positive sequence, negative sequence and zero sequence reactances:

The sum of positive sequence, negative sequence and zero sequence reactances is given by:

Xsum = X1 + X2 + X0

Substituting the values, we get:

Xsum = 0.2 + 0.2667 + 0.25 = 0.7167 pu

Therefore, the sum of positive sequence reactance, negative sequence reactance and zero sequence reactance is 0.7167 pu.

However, the question asks for the sum in pu (per unit), which means we need to divide the sum by the base reactance (Xbase) of the transmission line. Since Xbase is not given in the question, we cannot convert the sum to pu. Therefore, the answer is incorrect and cannot be 0.6 pu.

Similar GATE Doubts

A 15 MVA, 11 kV, alternator has positive, negative and zero sequence reactance are 0.3 p.u., 0.2 p.u. and 0.05 p.u. respectively. Find the value of resistance to be added in the generator neutral so that the fault current of LG fault does not exceed 1.5 times of rated line current.

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