Given that, S_{0} = slope of the channel bottom, S_{e }= slope of the energy line, F = Froude Number, the equation of gradually varied flow is expressed as:

This question was previously asked in

HPPSC AE Civil 2016 (HPPTCL) Official Paper

Option 2 : \(\frac{{dy}}{{dx}} = \frac{{{S_o} - {S_e}}}{{1 - {F^2}}}\)

__Explanation:__

In gradually varied flow, the total head H of the liquid at any cross-section can be written as

\(H = {z_b} + y + \frac{{{V^2}}}{{2g}}\).........(1)

Differentiating H with respect to x, we get

\(\frac{{dH}}{{dx}} = \frac{{d{z_b}}}{{dx}} + \frac{{dy}}{{dx}} + \frac{d}{{dx}}\left( {\frac{{{V^2}}}{{2g}}} \right)\)

\(\frac{{dH}}{{dx}}\) represent the energy slope (Sf). Since total energy always decreases in the direction of motion, therefore

\(\frac{{dH}}{{dx}} = - {S_f}\)

\(\frac{{d{z_b}}}{{dx}}\) denotes the bottom slope. It is common to consider the channel slope with bed elevations decreasing in the downstream direction, therefore

\(\frac{{d{z_b}}}{{dx}} = - {S_o}\)

\(\frac{{dy}}{{dx}}\) denotes water surface slope relative to the bottom of the channel.

\(\frac{d}{{dx}}\left( {\frac{{{V^2}}}{{2g}}} \right) = \frac{d}{{dy}}\left( {\frac{{{Q^2}}}{{2g{A^2}}}} \right)\frac{{dy}}{{dx}} = - \frac{{{Q^2}}}{{g{A^3}}}\frac{{dA}}{{dy}}\frac{{dy}}{{dx}} = - - \frac{{{Q^2}T}}{{g{A^3}}}\frac{{dy}}{{dx}}\)

Now, equation (1) will become,

\(- {S_f} = - {S_o} + \frac{{dy}}{{dx}} - \frac{{{Q^2}T}}{{g{A^3}}}\frac{{dy}}{{dx}}\)

\({F_r} = \frac{V}{{\sqrt {gy} }} = \frac{Q}{{A\sqrt {g\frac{A}{T}} }} = \frac{{Q\sqrt T }}{{\sqrt {g{A^3}} }}\)

So, the equation of gradually varied flow is expressed as

\(\frac{{dy}}{{dx}} = \frac{{{S_o} - {S_f}}}{{1 - \frac{{{Q^2}T}}{{g{A^3}}}}} = \frac{{{S_o} - {S_f}}}{{1 - F_r^2}}\)