Discussion
A turboprop produces \( 1200 \ \mathrm{kW} \) of shaft power and \( 1.5 \ \mathrm{kN} \) of residual jet thrust at \( 100 \ \mathrm{m/s} \). Taking propeller efficiency as 0.85, the equivalent shaft power is approximately
Answer: C
Correct answer: 1376 kW
Equivalent shaft power credits the residual jet thrust by converting it to the shaft power that would have been needed to produce the same thrust through the propeller:
\[ \mathrm{ESHP} = \mathrm{SHP} + \frac{F_j V}{\eta_p} = 1200 + \frac{1500 \times 100}{0.85 \times 1000} \]\[ = 1200 + \frac{150}{0.85} = 1200 + 176 = 1376 \text{ kW} \]Adding the jet power of 150 kW without dividing by the propeller efficiency gives 1350 kW, which under-credits the jet: the propeller would have needed more than 150 kW of shaft power to deliver that thrust.
No comment is present. Be the first to comment.
Loading…
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