Laminar open-channel film — velocity profile: 16-Civ-A5 Hydraulic Engineering, December 2013, Q5

Question 5: Laminar open-channel film — velocity profile (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 flow of a Newtonian fluid down a plane inclined at angle $\theta$; total film depth $d$; the momentum balance $W\sin\theta=\tau\,\Delta s$ and Newton’s viscosity law $\tau=\mu\,du/dy$.

Find. The velocity profile $u(y)$ in closed form (and the surface/maximum velocity).

Figure 5. Elemental fluid volume on the incline: weight component $W\sin\theta$ balanced by boundary shear $\tau\,\Delta s$; $y$ measured up from the bed, $d$ the free-surface depth.

Approach. Apply the momentum balance to a control volume extending from the depth $y$ up to the free surface to get the shear stress at $y$; substitute Newton’s law and integrate with no-slip at the bed.

Shear stress at depth $y$. Take a slab from level $y$ to the surface $d$, of slope-length $\Delta s$ and unit width. Its weight component along the slope is $W\sin\theta=\rho g\,(d-y)\,\Delta s\,\sin\theta$, balanced by shear on its base $\tau(y)\,\Delta s$. Hence $$\tau(y)=\rho g\,(d-y)\sin\theta.$$
Introduce Newton’s viscosity law. With $\tau=\mu\,du/dy$, $$\mu\dfrac{du}{dy}=\rho g\sin\theta\,(d-y)\quad\Rightarrow\quad \dfrac{du}{dy}=\dfrac{\rho g\sin\theta}{\mu}\,(d-y).$$
Integrate over depth. $$u(y)=\dfrac{\rho g\sin\theta}{\mu}\left(d\,y-\dfrac{y^2}{2}\right)+C.$$
Apply no-slip at the bed ($u=0$ at $y=0$) ⇒ $C=0$, giving the closed-form profile: $$\boxed{\,u(y)=\dfrac{\rho g\sin\theta}{\mu}\left(d\,y-\dfrac{y^{2}}{2}\right)=\dfrac{\rho g\sin\theta}{2\mu}\,y\,(2d-y).\,}$$
Surface (maximum) velocity at $y=d$: $$u_{\max}=u(d)=\dfrac{\rho g\sin\theta\,d^{2}}{2\mu}.$$ The profile is a half-parabola, zero at the bed and maximal at the free surface where $\tau=0$.

As an illustrative check, a $d=5$ mm glycerin-like film ($\rho=1260$ kg/m³, $\mu=1.0$ Pa·s) on a $5^{\circ}$ slope gives $u_{\max}=\rho g\sin\theta\,d^{2}/2\mu\approx 0.013$ m/s — small and firmly laminar, consistent with the assumption.

Question 5 — results
Quantity	Expression
Velocity profile	$u(y)=\dfrac{\rho g\sin\theta}{\mu}\left(dy-\dfrac{y^2}{2}\right)$
Surface (max) velocity	$u_{\max}=\dfrac{\rho g\sin\theta\,d^2}{2\mu}$

Topic. Laminar gravity-driven film flow (Nusselt film) on an incline.

Key relations. Slope momentum balance $\tau(y)=\rho g\sin\theta\,(d-y)$; Newton’s law $\tau=\mu\,du/dy$; no-slip boundary condition $u(0)=0$.

Why this works. For steady uniform flow there is no acceleration, so along the slope the weight component is carried entirely by viscous shear. The shear must vanish at the stress-free surface and grow linearly toward the bed, which upon integration yields the parabolic velocity distribution — the open-channel analogue of laminar flow between plates.

Common pitfall. Measuring $y$ from the surface instead of the bed (flips the sign of the boundary condition), or dropping the $\sin\theta$ and using the full weight instead of its slope component.

Source. Crowe, Elger & Roberson, Engineering Fluid Mechanics (laminar flow on an inclined plane); White, Fluid Mechanics, film-flow example.

Question 5 of 6: Laminar open-channel film — velocity profile

Question 5: Laminar open-channel film — velocity profile (20 marks)