Discussion
A plume from an effective height of \( 100 \) m is being modelled at a point where \( \sigma_z = 100 \) m. The vertical exponential term \( \exp(-H^2/2\sigma_z^2) \) equals
Answer: D
Correct answer: 0.607
With \( H = \sigma_z = 100 \) m the exponent simplifies:
\[ \exp\left(-\frac{H^2}{2\sigma_z^2}\right) = \exp\left(-\frac{100^2}{2 \times 100^2}\right) = \exp(-0.5) = 0.607 \]
This is the point at which the vertical spread has grown to equal the release height, so the ground is one standard deviation from the plume centreline and the term has fallen only to about \( 0.61 \) - a substantial fraction of the plume is now reaching the surface. The value \( 0.368 \) is \( \exp(-1) \) and \( 0.135 \) is \( \exp(-2) \), which correspond to larger \( H/\sigma_z \) ratios.
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