If the depth of flow in a channel changes gradually over a long length of the channel the flow is said to be gradually, varied flow and is denoted by G.V.F.

Equation for Gradually varied flow:

Before deriving an equation, following assumptions are made:

1] Bed slope of the channel is small.

2] Flow is steady and hence Q is constant

3] Accelerative effect is negligible.

4] Energy correction factor 'a' is unity

5] Roughness co-efficient is constant for the length of the channel.

6] Chezy's formula, manning formula are applicable to G.V.F to find slope.

7] Channel is prismatic.

Consider a rectangular channel having gradually varied flow,

Let, Z = height of bottom of channel

h = depth of flow

v = mean velocity of flow

$i_b$ = slope of the channel bed

$i_e$ = slope of energy line

b = width of channel

Q = discharge through channel

Energy equation at any section is given by Bernoulli's equation,

$E = z + h + \frac{v^2}{2g}$ - - - - (1)

Differentiating w.r. to 'x' where 'x' is measured along the bottom of the channel in direction of flow,

$\frac{dE}{dx} = \frac{d_2}{dx} + \frac{dh}{dx} + \frac{d}{dx} (\frac{v^2}{2g})$ - - - - (2)

Now, $\frac{d}{dx} (\frac{v^2}{2g}) = \frac{d}{dx} (\frac{q^2}{A \times 2g})$

$v = \frac{Q}{A}$

$= \frac{d}{dx} (\frac{Q^2}{b^2h^2 \times 2g})$

$= \frac{Q^2}{b^2 \times 2g} \frac{d}{dx} (\frac{1}{h^2})$

($\because$ Q , b, g are) constant

$= \frac{Q^2}{b^2 \times 2g} \frac{d}{dh} [\frac{1}{h^2}) \frac{dh}{dx}$

$= \frac{Q^2}{b^2 \times 2g} [\frac{-2}{h^3}] \frac{dh}{dx}$

$= \frac{- 2 q^2}{b^2 \times 2gh^3} \frac{dh}{dx}$

$= \frac{Q^2}{b^2 h^2 \times gh} \frac{dh}{dx}$

$= \frac{-v^2}{gh}$

$[ \because \frac{Q}{bh} = v]$

Substituting the value of $\frac{d}{dx} (\frac {v^2}{2g})$

in equation (2), we get,

$\frac{dE}{dx} = \frac{dz}{dx} + \frac{dh}{dx} - \frac{v^2}{gh} \frac{dh}{dx}$

$= \frac{dz}{dx} + \frac{dh}{dx} [ 1 - \frac{v^2}{gh}]$

Now, $\frac{dE}{dx}$ = slope of energy line = - $i_e$.

$\frac{dz}{dx}$ = slope of the bed of the channel = - $i_b$

negative sign because as with the increase of x, the value of E and z decreases.

substituting the value of $\frac{dE}{dx}$ and $\frac{dz}{dx}$

in equation (3), we get,

$-$i_e$= -$i_b$+ \frac{dh}{dx} [ 1 - \frac{v^2}{gh}]$

$i_b - i_e = \frac{dh}{dx} [ 1 - \frac{v^2}{gh}]$

OR $\frac{dh}{dx} = \frac{(i_b - i_e)}{[ 1 - \frac{v^2}{gh}]}$

$\frac{dh}{dx} = \frac{(i_b - i_e)}{[1 - (fe)^2]}$

$[ \therefore \frac{v}{\sqrt gh} = fe]$