Question
Download Solution PDFA channel is said to be the most economical section when, it gives maximum discharge for a given area and slope. What is the relation between hydraulic radius and depth for the most economical trapezoidal section?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
Conditions for the most economical or efficient section:
- From the equation of continuity, it can be seen that for a constant area of cross-section the discharge Q will be maximum for the minimum value of flow velocity.
- From Chezys and Mannings equation that the velocity of flow will be maximum when hydraulic radius R will be minimum for factors like the slope of bed, and surface roughness being constant.
- Hydraulic radius is maximum if the wetted perimeter is minimum.
Most economical rectangular section:
- b = 2d, i.e., the width of the channel is twice the depth of flow.
- R = \(d\over 2\), i.e., the hydraulic radius is half of the depth of flow.
Most economical trapezoidal section:
- Half of the top width must be equal to the sloping side. OR
- The hydraulic radius must be half the depth of water.
Most economical triangular section:
- The side slope has to be inclined at 45∘C. OR
- Hydraulic radius has to be \(1\over 2\sqrt2\) times the depth of flow.
Most economical circular section:
- The depth of flow for maximum velocity is 0.81D
- The hydraulic radius for maximum velocity is 0.3 D
- The depth of flow for maximum discharge is 0.95D
Where D = Diameter of channel
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