Question
Download Solution PDFA rectangular channel of bed width 2 m is to be laid at a bed slope of 1 in 1000. Find the hydraulic radius of the canal cross-section for the maximum discharge condition? Take Chezy’s constant as 50
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Most efficient channel: A channel is said to be efficient if it carries the maximum discharge for the given cross-section which is achieved when the wetted perimeter is kept a minimum.
Rectangular Section:
Area of the flow, A = b × d
Wetted Perimeter, P = b + 2 × d
For the most efficient Rectangular channel, the two important conditions are
- b = 2 × d
- \(R = \frac{A}{P} = \frac{{b\times d}}{{b + 2 \times d}} = \frac{{2{d^2}}}{{4d}} = \frac{d}{2}= \frac{b}{4}\)
Calculation
Given: b = 2 m
\(R = \frac{2}{4}\)
⇒ R = 0.5
| Rectangular channel section | Trapezoidal channel section |
|
|
Where R = hydraulic radius, A = Area of flow, P = wetted perimeter, y = depth of flow, T = Top width
Last updated on May 28, 2025
-> SSC JE notification 2025 for Civil Engineering will be released on June 30.
-> Candidates can fill the SSC JE CE application from June 30 to July 21.
-> The selection process of the candidates for the SSC Junior Engineer post consists of Paper I, Paper II, Document Verification, and Medical Examination.
-> Candidates who will get selected will get a salary range between Rs. 35,400/- to Rs. 1,12,400/-.
-> Candidates must refer to the SSC JE Previous Year Papers and SSC JE Civil Mock Test, SSC JE Electrical Mock Test, and SSC JE Mechanical Mock Test to understand the type of questions coming in the examination.
-> The Staff Selection Commission conducts the SSC JE exam to recruit Junior Engineers in different disciplines under various departments of the Central Government.