Discussion
Air enters a diffuser at \( 200 \ \mathrm{m/s} \) and leaves at \( 80 \ \mathrm{m/s} \). With \( c_p = 1005 \ \mathrm{J/(kg\,K)} \), the ideal static temperature rise is approximately
Answer: C
Correct answer: 16.7 K
The kinetic energy given up appears as static temperature: \[ \Delta T = \frac{V_1^2 - V_2^2}{2c_p} = \frac{200^2 - 80^2}{2 \times 1005} = \frac{33{,}600}{2010} = 16.7\ \mathrm{K}. \] Note it is the difference of the squares, not the square of the difference - the latter would give 7.2 K.
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