Discussion
For a solar flat plate collector the following data is given : Useful heat gain = 28 watts/m\(^{2}\) per hour, solar radiation intensity = 350 watts/m\(^{2}\) per hour and the factor to convert beam radiation to that on the plane of the collector = 1.2. The collector efficiency is
Answer: A
Collector efficiency is the useful heat gain divided by the solar radiation incident on the plane of the collector, which is the intensity multiplied by the conversion factor.
\[ \eta = \frac{28}{350 \times 1.2} \]
\[ \eta = \frac{28}{420} = 0.0667 \approx 6.6\% \]
This is option A. The values 4.8, 12.2 and 15.2 percent do not result from this data.
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