Velocity profile in a laminar open-channel film: 16-Civ-A5 Hydraulic Engineering, May 2016, Q5

Question 5: Velocity profile in a laminar open-channel film (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Steady, uniform, laminar sheet flow of depth $d$ on a bed at slope angle $\theta$; Newtonian fluid $\tau=\mu\,du/dy$; momentum balance $W\sin\theta=\tau\,\Delta s$ on an element between the free surface and depth-coordinate $y$ (measured up from the bed).

Find. The velocity profile $u(y)$ in closed form.

Figure 5. Free body of an elemental volume: self-weight component W sin θ balanced by wall shear τΔs.

Approach. Write the weight of the fluid above height $y$ as the shear-resisting force at that level, giving $\tau(y)$; substitute Newton’s law of viscosity and integrate once, applying no-slip at the bed.

Shear-stress distribution. For a slab of unit plan area between level $y$ and the free surface $d$, the down-slope weight component is $\rho g (d-y)\sin\theta$, balanced by the shear at $y$: $$\tau(y)=\rho g\,(d-y)\sin\theta.$$
Introduce Newton’s law of viscosity. With $\tau=\mu\,\dfrac{du}{dy}$, $$\mu\frac{du}{dy}=\rho g\sin\theta\,(d-y)\;\Rightarrow\;\frac{du}{dy}=\frac{\rho g\sin\theta}{\mu}(d-y).$$
Integrate with no-slip at the bed. Integrating from the bed and applying $u(0)=0$: $$\boxed{\,u(y)=\frac{\rho g\sin\theta}{\mu}\left(d\,y-\frac{y^{2}}{2}\right)\,}.$$
Consequences. The profile is parabolic with maximum at the free surface, $u_{max}=u(d)=\dfrac{\rho g\sin\theta\,d^{2}}{2\mu}$; the depth-averaged velocity is $\bar u=\tfrac{2}{3}u_{max}$; and the unit-width discharge is $q=\displaystyle\int_0^d u\,dy=\dfrac{\rho g\sin\theta\,d^{3}}{3\mu}$.

Final results — Question 5
Quantity	Expression
Velocity profile	$u(y)=\dfrac{\rho g\sin\theta}{\mu}\left(dy-\tfrac{y^2}{2}\right)$
Surface (maximum) velocity	$u_{max}=\rho g\sin\theta\,d^2/(2\mu)$
Mean velocity	$\bar u=\tfrac{2}{3}u_{max}$
Unit discharge	$q=\rho g\sin\theta\,d^3/(3\mu)$

Topic. Laminar gravity-driven sheet (Nusselt) film flow.

Key relations. Shear balance $\tau(y)=\rho g\sin\theta(d-y)$; Newton $\tau=\mu\,du/dy$; no-slip $u(0)=0$; parabolic profile with $\bar u=\tfrac23 u_{max}$.

Why this works. In steady uniform flow there is no acceleration, so gravity along the slope is resisted entirely by internal shear. The shear must vanish at the free surface (no air drag) and grow linearly toward the bed, which — through Newton’s linear law — integrates to a parabola pinned at zero on the bed. The factor $\tfrac23$ between mean and surface velocity is the signature of a parabolic laminar profile.

Common pitfall. Applying no-slip at the wrong boundary (the free surface has zero shear, not zero velocity), or measuring $y$ from the surface and dropping a sign, which flips the parabola.

Source. Crowe, Elger & Roberson, Engineering Fluid Mechanics, ch. 10 (laminar flow between/over surfaces; falling-film derivation).

Question 5 of 6: Velocity profile in a laminar open-channel film

Question 5: Velocity profile in a laminar open-channel film (20 marks)