GATE EE 2016 Set 1 — Question 51
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Power Systems → Fault Analysis → Symmetrical Components
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Question
A 30 MVA, 3-phase, 50 Hz, 13.8 kV, star-connected synchronous generator has positive, negative and zero sequence reactances, 15%, 15% and 5% respectively. A reactance () is connected between the neutral of the generator and ground. A double line to ground fault takes place involving phases ‘b’ and ‘c’, with a fault impedance of p.u. The value of (in p.u.) that will limit the positive sequence generator current to 4270 A is _________.
Correct answer
1.05 to 1.15
Solution
1.Calculate the base current:
A.2.Convert the positive sequence current limit to per-unit:
p.u.3.For a double line to ground (LLG) fault, the positive sequence current is given by:
4.Given sequence reactances and fault impedance in p.u.:
p.u.p.u.
p.u.
p.u. (where is the neutral grounding reactance)
p.u.
5.Calculate the equivalent zero sequence impedance including fault impedance:
6.Substitute the values into the formula (ignoring as all components are purely reactive):
7.Solve for :
p.u.The value of is approximately 1.06 p.u., which falls within the range [1.05, 1.15].
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