Question

A short reach of a 2 m wide rectangular open channel has its bed level rising in the direction of flow at a slope of 1 in 10000. It carries a discharge of 4 m³/s and its Manning’s roughness coefficient is 0.01. The flow in this reach is gradually varying. At a certain section in this reach, the depth of flow was measured as 0.5 m. The rate of change of the water depth with distance, dy/dx, at this section is _______________ (use g = 10 m/s²).

Answer 1

Concept:

Equation of Gradually Varied Flow(G.V.F):

The rate of change of water depth with distance is given by,

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{{{\rm{S}}_{\rm{o}}} - {{\rm{S}}_{\rm{f}}}}}{{1 - {\rm{F}}_{\rm{r}}^2}}\)

Where,

S_o = slope of the channel bottom

S_f = slope of the Total Energy Line(T.E.L)

F_r = Froude's number

Froude number is given by:

\({{\rm{F}}_{\rm{r}}} = \frac{{\rm{V}}}{{\sqrt {{\rm{gD}}} {\rm{\;}}}}\)

Where,

D = Y = hydraulic depth

Calculation:

B = 2m, S = - 1/10000 (- ve as bed level is rising),

 Q = 4 m³/sec, n = 0.01, g = 10 m/sec²

\(\begin{array}{l} \frac{{dy}}{{dx}} = \frac{{{S_o} - {S_e}}}{{1 - F{r^2}}}\\ Fr = \frac{V}{{\sqrt {gy} }} = \frac{Q}{{2y\sqrt {gy} }} = \frac{4}{{2 \times 0.5\sqrt {10 \times 0.5} }} = 1.789\\ R = \frac{A}{P} = \frac{{2 \times 0.5}}{{2 + 2 \times 0.5}} = \frac{1}{3}\\ Q = \frac{1}{n}{R^{\frac{2}{3}}}{S^{\frac{1}{2}}}.A\\ 4 = \frac{1}{{0.01}} \times {\left( {\frac{1}{3}} \right)^{\frac{2}{3}}}\times S{_f^{\frac{1}{2}}} \times 1 \end{array}\)

S_f = 6.923 × 10^-3

\(\frac{{dy}}{{dx}} = \frac{{ - \frac{1}{{10000}} - 6.923 \times {{10}^{ - 3}}}}{{1 - {{\left( {1.789} \right)}^2}}} = 0.0031\)