Question 6 of 7: Question 6 (Part B, Question B2): Thin Liquid Film Entrained by an Inclined Conveyor Belt

Question 6 (Part B, Question B2): Thin Liquid Film Entrained by an Inclined Conveyor Belt (26 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.

Find. The velocity profile u(y) with its boundary conditions, the shear stress transmitted to the belt, and the smallest belt speed for which the net flow is upward.

Approach. The film is fully developed, so the velocity depends only on the coordinate normal to the belt and the Navier–Stokes x-momentum equation collapses to a one-dimensional balance between the gravity component along the belt and the viscous stress gradient. Integrate twice, apply no-slip at the belt and zero shear at the free surface, then integrate the profile across the film for the flow rate.

1. Part (a) — reduce the momentum equation. Take x up the belt and y normal to it, with y = 0 at the belt surface. For a steady, fully developed film of constant thickness, $v=w=0$ and $u=u(y)$ only, so continuity is satisfied identically and every convective term in the aid-sheet $x$-momentum equation vanishes. With $\partial P/\partial x=0$ (the surface pressure is atmospheric and constant, and the thin film transmits it unchanged across its thickness) and the gravity component along the belt equal to $-g\sin\theta$, what remains is $$\begin{aligned} 0&=-\rho g\sin\theta+\mu\frac{d^{2}u}{dy^{2}}\\ \Longrightarrow\quad \frac{d^{2}u}{dy^{2}}&=\frac{\rho g\sin\theta}{\mu} \end{aligned}$$
2. State the two boundary conditions. They are the whole physical content of the problem:
   • No slip at the belt (the liquid wets the belt, which is non-porous): $u(0)=U_p$.
   • Zero shear at the free surface (the air exerts negligible stress): $\tau(\delta)=\mu\left.\dfrac{du}{dy}\right|_{y=\delta}=0$, i.e. $\left.\dfrac{du}{dy}\right|_{\delta}=0$.
   A third condition is implicit and worth naming: the surface is a streamline, which is what allows the film thickness to be treated as constant.
3. Integrate twice and apply the conditions. Integrating once gives $du/dy=(\rho g\sin\theta/\mu)y+C_1$, and the zero-shear condition at $y=\delta$ fixes $C_1=-\rho g\delta\sin\theta/\mu$. Integrating again and applying $u(0)=U_p$ fixes $C_2=U_p$, so $$\boxed{\ u(y)=U_p-\frac{\rho g\sin\theta}{\mu}\left(\delta y-\frac{y^{2}}{2}\right) =U_p+\frac{\rho g\sin\theta}{2\mu}\left(y^{2}-2\delta y\right)\ }$$ The profile is a downward-opening parabola in the frame of the belt: the belt drags the liquid up by viscosity, gravity drains it back down, and the two effects are equal and opposite in gradient exactly at the free surface. Its lowest value is at the surface, $u(\delta)=U_p-\rho g\delta^{2}\sin\theta/2\mu$, which may be positive or negative depending on how fast the belt runs.
4. Part (b) — evaluate the shear stress on the belt. The stress the fluid exerts on the belt is $\tau_{w}=\mu\left.du/dy\right|_{y=0}=\mu C_1$, so $$\boxed{\ \tau_{w}=-\rho g\,\delta\sin\theta\ }$$ i.e. a stress of magnitude $\rho g\delta\sin\theta$ acting down the slope, opposing the belt's motion. The result is independent of both $U_p$ and $\mu$, and a control-volume check shows why it must be: the film is not accelerating, no pressure gradient acts on it and the air supplies nothing, so the belt alone must carry the whole down-slope weight component of the film, $\rho g\delta\sin\theta$ per unit belt area. The power the belt spends against this is $U_p\rho g\delta\sin\theta$ per unit area.
5. Part (c) — integrate the profile for the volume flow rate. Per unit belt width, $$q=\int_0^{\delta}u\,dy=U_p\delta+\frac{\rho g\sin\theta}{2\mu} \left[\frac{\delta^{3}}{3}-\delta^{3}\right] =U_p\delta-\frac{\rho g\,\delta^{3}\sin\theta}{3\mu},$$ and the mass flow rate is $\dot m'=\rho q$. The first term is the drag the belt supplies; the second is the gravity drainage, and it grows as the cube of the film thickness, which is why thick films are so much harder to entrain.
6. Solve for the minimum belt speed. Net upward flow requires $q>0$, so $$\begin{aligned} U_p\delta\;&>\;\frac{\rho g\,\delta^{3}\sin\theta}{3\mu}\\ \Longrightarrow\quad \boxed{\ U_{p,\min}=\frac{\rho g\,\delta^{2}\sin\theta}{3\mu} =\frac{g\,\delta^{2}\sin\theta}{3\nu}\ }& \end{aligned}$$ Below this speed the belt still drags a layer up next to its surface, but the outer part of the film drains down faster and the net transport is downward. Note that at exactly $U_p=U_{p,\min}$ the surface velocity is $u(\delta)=U_{p,\min}-\rho g\delta^{2}\sin\theta/2\mu=-\rho g\delta^{2}\sin\theta/6\mu$, which is negative: the free surface is already running down the belt while the net flow is only just zero.
7. Put illustrative numbers to the three results. Take a coating line handling a glycerol-water mixture, $\rho=1200\ \text{kg/m}^{3}$, $\mu=0.50\ \text{Pa}\cdot\text{s}$, carried on a belt at $\theta=30^{\circ}$ with a film $\delta=1.5\ \text{mm}$ thick. Then the belt stress is $\tau_w=1200\times9.81\times0.0015\times0.5=8.83\ \text{Pa}$ and the threshold speed is $$U_{p,\min}=\frac{1200\times9.81\times(0.0015)^{2}\times0.5}{3\times0.50} =8.83\times10^{-3}\ \text{m/s}=8.8\ \text{mm/s}.$$ Running the belt at $U_p=20\ \text{mm/s}$, comfortably above the threshold, delivers $q=0.020\times0.0015-8.829\times10^{-3}\times0.0015=1.68\times10^{-5}\ \text{m}^{3}/\text{s}$ per metre of width, with the free surface creeping up at only $u(\delta)=6.76\ \text{mm/s}$ — about a third of the belt speed.