wiz-icon
MyQuestionIcon
MyQuestionIcon
4
You visited us 4 times! Enjoying our articles? Unlock Full Access!
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 (Xn) is connected between the neutral of the generator and ground. A double line to ground fault takes place involving phase 'b' and 'c' with a fault impendance of j0.1 p.u.. The value of xn(in p.u.) that will limit the positive sequence generator current to 4270 A is
  1. 1.07

Open in App
Solution

The correct option is A 1.07
Base current,

IB = 30×1033×13.8 = 1255.109 A

If = 4270 A

Ip.u. = 42701255.109 = 3.402 p.u.

Ig1 = EaX1+(X2||X0)

where, X0 = X0 + 3(Zn + Zf)

X0 = 3Zn + 0.35

3.402 = 1.00.15+0.15×(3Zn+0.35)0.15+3Zn+0.35

By solving the equation,

Zn = 1.07 p.u.

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Line - To - Line - To - Ground Fault (LLG Fault)
POWER SYSTEMS
Watch in App
Join BYJU'S Learning Program
CrossIcon