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## Q. 3.17

If the velocity distribution of a turbulent flow in an open channel is given by a power law
$\frac{v}{v_{max}}=\left(\frac{y}{y_{0}} \right)^{1/7}$
where v is the velocity at a distance y from the bed and $v_{max}$ is the maximum velocity in the channel with a flow depth of $y_{0}$, determine the average velocity and the energy (α) and momentum (β) correction factors. Assume the flow to be two-dimensional.

## Verified Solution

If the mean velocity of flow is V, the discharge per unit width of the channel is
$y_{0}V=q=\int_{0}^{y_{0}}{v dy}$
which gives q = (7/8)$v_{max}y_{0}$.
Therefore
$V=\frac{q}{y_{0}}=\left(\frac{7}{8} \right)v_{max}$

The kinetic energy correction factor α, given by Equation 3.11, can be written as

$\alpha =\frac{1}{A}\int\limits_{A}{\left(\frac{v}{\bar{V}} \right)^{3}}dA$       (3.11)

$\alpha =\frac{1}{A}\int\limits_{A}{\left(\frac{v}{V} \right)^{3}}dA=\frac{1}{y_{0}V^{3}}\int_{0}^{y_{0}}{\left[v_{max}\left(\frac{y}{y_{0}} \right)^{1/7} \right]}^{3}dy$

Replacing $v_{max}$ [=(8/7)V] and integrating, we obtain α = 1.045. The momentum correction factor β given by Equation 3.14 can be written as

$\beta =\frac{1}{A}\int\limits_{A}{\left(\frac{v}{\bar{V}} \right)^{2}}dA$          (3.14)

$\beta =\frac{1}{A}\int\limits_{A}{\left(\frac{v}{V} \right)^{2}}dA=\frac{1}{y_{0}V^{3}}\int_{0}^{y_{0}}{\left[v_{max}\left(\frac{y}{y_{0}} \right)^{1/7} \right]}^{2}dy$

Again replacing $v_{max}$ and integrating, we obtain β = 1.016.
Note: The energy and momentum correction factors α and β for open channels may be computed by the equations
α = 1 + 3ε² − 2ε³      and      β = 1 + ε²      where $\varepsilon =\frac{v_{max}}{V}-1$

If the velocity distributions are not described by any equation and if the measured data are available, α and β values may be computed by graphical methods; plots of ∫ v dy, ∫ v³ dy and ∫ v² dy will help to give V, α and β, respectively.