A rectangular open channel of 6 m width is carrying a discharge of 20 m³/s. Consider the acceleration due to gravity as 9.81 m/s² and assume water as incompressible and inviscid. The depth of flow in the channel at which the specific energy of the flowing water is minimum for the given discharge will then be
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For minimum specific energy in a rectangular open channel, use the relationship derived from energy equations to calculate the depth based on the discharge and channel width.
The specific energy of the flow in an open channel is given by:
\[
E = h + \frac{Q^2}{2gA^2},
\]
where:
- \( h \) is the flow depth,
- \( Q \) is the discharge,
- \( g \) is the acceleration due to gravity,
- \( A \) is the cross-sectional area of the flow.
To minimize the specific energy, we use the formula for the depth at minimum specific energy, derived from the above equation:
\[
h_{\text{min}} = \frac{Q^2}{2gA^2}.
\]
Given the discharge \( Q = 20 \, \text{m}^3/\text{s} \) and the channel width \( b = 6 \, \text{m} \), we can calculate the depth that minimizes the specific energy. Using the appropriate formula and calculations, we find that the depth at which the specific energy is minimized is \( 1.04 \, \text{m} \).
