Correct Answer - Option 3 : 2.5
Concept:
Critical flow analysis:
1) Specific energy is minimum for a given discharge.
2) Specific force is minimum for a given discharge.
3) Discharge is maximum for a given specific energy.
4) Discharge is maximum for a given specific force.
5) Froude number = Fr = 1
6) Velocity head is equal to half of hydraulic depth, \(\frac{{{{\rm{V}}^2}}}{{2{\rm{g}}}} = \frac{{\rm{D}}}{2}\)
Relations for critical flow:
General equation valid for critical flow of any shape of channel
\(\frac{{{{\rm{Q}}^2}{\rm{T}}}}{{{\rm{g}}{{\rm{A}}^3}}} = {\rm{F}}_{\rm{r}}^2\)
For critical flow in reactangular channel
\({{\rm{Y}}_{\rm{c}}} = {\rm{critical\;depth}} = {\left( {\frac{{{{\rm{Q}}^2}}}{{{\rm{g}}{{\rm{b}}^2}}}} \right)^{\frac{1}{3}}}\)
\({{\rm{E}}_{\rm{c}}} = {\rm{Specific\;energy}} = \frac{3}{2} × {{\rm{Y}}_{\rm{c}}}\)
Calculation:
Given,
b = Width of channel = 2.5 m
Q = Discharge = 4 m3/sec
g = 9.81 m/sec2
\({{\rm{Y}}_{\rm{c}}} = {\left( {\frac{{{{\rm{Q}}^2}}}{{{\rm{g}}{{\rm{b}}^2}}}} \right)^{\frac{1}{3}}}\)
\({{\rm{Y}}_{\rm{c}}} = {\left( {\frac{{{{\rm{4}}^2}}}{{{\rm{9.81× }}{{\rm{2.5}}^2}}}} \right)^{\frac{1}{3}}}\)
Yc = 0.64 m
Now,
Q = A × Vc
\({\rm{V_c}} = \frac{{\rm{Q}}}{{\rm{A}}} = \frac{4}{{{\rm{B}} \times {{\rm{Y}}_{\rm{c}}}}} = \frac{4}{{2.5 \times 0.64}} = 2.5{\rm{\;m}}/{\rm{sec}}\)
Hence the velocity of flow corresponding to critical depth = 2.5 m/sec
Important point:
For critical flow in parabolic channel
\({Y_c} = {\left( {\frac{{27{Q^2}}}{{8gB}}} \right)^{1/4}}\)
\({{\rm{E}}_{\rm{c}}} = {\rm{Specific\;energy}} = \frac{4}{3} × {{\rm{Y}}_{\rm{c}}}\)
For critical flow in triangular channel
\({{\rm{Y}}_{\rm{c}}} = {\rm{critical\;depth}} = {\left( {\frac{{{{\rm{2Q}}^2}}}{{{\rm{g}}{{\rm{m}}^2}}}} \right)^{\frac{1}{5}}}\)
\({{\rm{E}}_{\rm{c}}} = {\rm{Specific\;energy}} = \frac{5}{4} × {{\rm{Y}}_{\rm{c}}}\)
