Discussion
A generator operating at 50 Hz delivers 1 pu power to an infinite bus through a transmission circuit in which resistance is ignored. A fault takes place reducing the maximum power transferable to 0.5 pu, whereas before the fault, this power was 2 pu.
What is the initial load angle?
Answer: A
The power transferred is \(P_e = P_{max}\sin\delta\). Before the fault \(P_{max} = 2\) pu and the load is 1 pu.
\[ \begin{aligned} \sin\delta_0 = \frac{1}{2} \Rightarrow \delta_0 \\ &= 30^\circ = \frac{\pi}{6} = 0.5236\ \text{rad} \end{aligned} \]
The fault reduces the limit to 0.5 pu, which is what sets the swing afterwards, but the initial angle only depends on the pre fault values. The other options are larger angles of 36, 42 and 48 degrees. The answer is 0.523 rad, option A.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development