Question 12.4: Show that the vertical distribution of velocity is parabolic...
Show that the vertical distribution of velocity is parabolic for a uniform laminar flow in a wide open channel with constant slope and depth of flow.
Let the depth of flow be h (Fig. 12.10). A control volume abcda of length dL and of width B (the width of the channel) is taken as shown in the figure. Now we have to apply the momentum theorem to this control volume.
The forces acting on the surfaces ab and cd are the hydrostatic pressure forces as shown in the figure.
Let F_{1} and F_{2} be the hydrostatic pressure forces on these two surfaces ab and cd respectively.
Therefore the net force acting on the control volume in the direction of flow can be written as
F_{x}=F_{1}-F_{2}+\rho g(h-y) \mathrm{d} L B \sin \alpha-\tau \mathrm{d} L BSince F_{1}=F_{2}
F_{x}=\rho g(h-y) \mathrm{d} L B \sin \alpha-\tau \mathrm{d} L BFor a steady uniform flow, the momentum coming into the control volume across the face ab is equal to that leaving from the control volume across the face cd.
Therefore the net rate of momentum efflux from the control volume is zero.
Hence, we can write, from the momentum theorem applied to the control volume abcda,
F_{x}=\rho g(h-y) \mathrm{d} L B \sin \alpha-\tau \mathrm{d} L B=0which gives,
\tau=\rho g(h-y) \sin \alpha (12.12)
For a laminar flow,
\tau=\mu \frac{\mathrm{d} u}{\mathrm{~d} y}Substituting the expression of \tau in the Eq. (12.12), we get
\mathrm{d} u=\frac{\rho g}{\mu}(h-y) \sin \alpha \mathrm{d} yOr u=\frac{\rho g \sin \alpha}{\mu}\left(h y-\frac{y^{2}}{2}\right)+C_{1} (12.13)
For small values of a, sin a = tan a = S (slope of the channel). The constant of integration C_{1} in Eq. (12.13) can be obtained from the boundary condition that at y = 0, u = 0, which gives C_{1} = 0 . Hence, Eq. (12.13) becomes
u=\frac{\rho g \sin \alpha}{\mu}\left[(y / h)-\frac{1}{2}(y / h)^{2}\right] (12.14)
Equation (12.14) is the required velocity distribution which is parabolic in nature.