Discussion
The exit area of the nozzle in cm2 is
Answer: D
The exit area is about 12.9 cm2.
Take R = 287 J/kg.K for air. The mass flow rate at inlet is:
\[ \begin{aligned} \rho_{1} &= \frac{300\,000}{287 \times 500} = 2.091\ \text{kg/m}^{3} \\ \dot{m} &= 2.091 \times 0.008 \times 10 = 0.1673\ \text{kg/s} \end{aligned} \]
At exit, with T2 = 484 K from the previous question:
\[ \begin{aligned} \rho_{2} &= \frac{100\,000}{287 \times 484} = 0.720\ \text{kg/m}^{3} \\ A_{2} &= \frac{0.1673}{0.720 \times 180} = 12.9\ \text{cm}^{2} \end{aligned} \]
4.4 cm2 is what results from ignoring the density change (80 × 10/180). 90.1 and 56.3 cm2 would make the exit larger than or close to the inlet, which an accelerating subsonic nozzle does not do.
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