Question

A 3 m wide rectangular channel carries a flow of 6 m³/s. The depth of flow at a section P is 0.5 m.

A flat-topped hump is to be placed at the downstream of the section P. Assume negligible energy

loss between section P and hump, and consider g as 9.81 m/s². The maximum height of the hump

(expressed in m) which will not change the depth of flow at section P is _________

Answer 1

Concept:

The Froude number is given by,

\({F_r} = \frac{V}{{\sqrt {gD} }}\)

Where,

V = velocity of water, D = Y = depth of water

For critical flow,

F_r = 1, V = V_c, and D =Y_c

\({1} = \frac{V}{{\sqrt {gD} }}\)

\({{\bf{V}}_{\bf{C}}} = \sqrt {{\bf{g}}{{\bf{Y}}_{\bf{c}}}\;} \)

The specific energy equation for the channel

\({\rm{E}} = {\rm{Y}} + \frac{{{{\rm{V}}^2}}}{{2{\rm{g}}}}\)

Specific energy at a critical depth

\({\rm{E}} = {{\rm{Y}}_{\rm{c}}} + \frac{{{{\left( {{{\rm{V}}_{\rm{c}}}} \right)}^2}}}{{2{\rm{g}}}},{\rm{\;we\;know\;that\;\;}}{{\rm{V}}_{\rm{c}}} = \sqrt {{\rm{g}}{{\rm{Y}}_{\rm{c}}}} \)

\({\rm{E}} = {{\rm{Y}}_{\rm{c}}} + \frac{{{{\left( {\sqrt {{\rm{g}}{{\rm{Y}}_{\rm{c}}}} } \right)}^2}}}{{2{\rm{g}}}} = {Y_C} + \frac{1}{2} \times {Y_c} = \frac{3}{2} \times {Y_c}\)

\(\therefore {\rm{E}} = \frac{3}{2} \times {Y_c}\)

Calculation:

Given,

B = width of channel = 3 m

Q = 6 m³/sec, y₁ = 0.5 m

q = Discharge per unit width = Q/B = 6/3 = 2

 \({V_1} = \frac{Q}{{B{y_1}}} = \frac{6}{{3 \times 0.5}} = 4\ m/sec\)

Hump is to be provided.

\(\begin{array}{l} {E_1} = \frac{{V_1^2}}{{2g}} + {y_1}\\ {E_1} = \frac{{{4^2}}}{{2 \times 9.81}} + 0.5 = 1.315\ m \end{array}\)

\({E_2} = \frac{3}{2}{y_2}c = \frac{3}{2}{\left( {\frac{{{q^2}}}{g}} \right)^{\frac{1}{3}}} = 1.5 \times {\left( {\frac{{{2^2}}}{{9.81}}} \right)^{\frac{1}{3}}} = 1.112\ m\)

Relation between height of hump (ΔZ ) and specific energy (E)

E1 = E2 + Δz

Δz = E₁ – E₂ = 1.315 – 1.112 = 0.202 m