Question 2 of 5: Combined Couette–Gravity Flow Between Vertical Plates

Question 2: Combined Couette–Gravity Flow Between Vertical Plates (30 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. Two infinite vertical plates a distance $h$ apart; the left plate ($x=0$) is stationary and the right plate ($x=h$) moves vertically upward at constant $u_0$. The fluid ($\rho,\mu$) is in steady, fully developed, isothermal, incompressible flow; gravity $g$ acts vertically downward.

Find. (a) the velocity profile $u_z(x)$; (b) the mid-channel velocity $u_z(h/2)$; (c) the plate speed $u_0$ that makes the net mass flow zero.

Approach. Reduce the vertical ($z$) Navier–Stokes component to a one-dimensional ODE using the fully-developed assumption (no axial pressure gradient, gravity as the only body force), integrate with the no-slip conditions at the two plates, then impose zero net volumetric flow.

1. (a) Simplify the flow field. Fully developed flow between infinite plates has only a vertical component that varies across the gap: $\vec{u}=u_z(x)\,\hat{k}$ with $u_x=u_y=0$. Continuity is then automatically satisfied since $\partial u_z/\partial z=0$. Take $z$ vertically upward, so the body force is $g_z=-g$.
2. Reduce the $z$-momentum equation. Dropping the transient, inertial and cross-derivative terms and taking no imposed axial pressure gradient ($\partial P/\partial z=0$, both ends open), the Navier–Stokes $z$-component becomes $0=-\rho g + \mu\dfrac{d^2u_z}{dx^2}$, i.e. $\dfrac{d^2u_z}{dx^2}=\dfrac{\rho g}{\mu}$.
3. Integrate twice. $u_z=\dfrac{\rho g}{2\mu}x^2 + C_1 x + C_2$.
4. Apply the no-slip boundary conditions. $u_z(0)=0\Rightarrow C_2=0$; $u_z(h)=u_0\Rightarrow C_1=\dfrac{u_0}{h}-\dfrac{\rho g h}{2\mu}$.
5. Velocity distribution. Substituting the constants, $$u_z(x)=u_0\frac{x}{h}+\frac{\rho g}{2\mu}\left(x^2-hx\right),$$ the required result — a linear Couette term plus a downward gravity parabola ($x^2-hx\le0$ on $0\le x\le h$). Factoring $h^2$ out of the gravity term, $\dfrac{\rho g}{2\mu}(x^2-hx)=\dfrac{\rho g h^2}{2\mu}\left[\left(\dfrac{x}{h}\right)^2-\dfrac{x}{h}\right]$, which is exactly the printed form $u_z=\dfrac{\rho g h^2}{2\mu}\left[\left(\dfrac{x}{h}\right)^2-\left(\dfrac{x}{h}\right)\right]+u_0\left(\dfrac{x}{h}\right)$. $\boxed{\,u_z(x)=u_0\dfrac{x}{h}+\dfrac{\rho g}{2\mu}\left(x^2-hx\right)\,}$
6. (b) Mid-channel velocity. Setting $x=h/2$: $u_z(h/2)=\dfrac{u_0}{2}+\dfrac{\rho g}{2\mu}\!\left(\dfrac{h^2}{4}-\dfrac{h^2}{2}\right)=\boxed{\dfrac{u_0}{2}-\dfrac{\rho g h^2}{8\mu}}.$ The gravity term subtracts from the average drag, so the centre moves more slowly (or downward) than simple Couette flow.
7. (c) Impose zero net flow. With constant density, zero net mass flow is equivalent to zero net volumetric flow, $\displaystyle\int_0^h u_z\,dx=0$. Evaluating the two contributions, $\displaystyle\int_0^h u_0\frac{x}{h}\,dx=\frac{u_0 h}{2}$ and $\displaystyle\int_0^h \frac{\rho g}{2\mu}(x^2-hx)\,dx=\frac{\rho g}{2\mu}\!\left(\frac{h^3}{3}-\frac{h^3}{2}\right)=-\frac{\rho g h^3}{12\mu}.$
8. Solve for $u_0$. Setting the sum to zero, $\dfrac{u_0 h}{2}-\dfrac{\rho g h^3}{12\mu}=0\Rightarrow \boxed{\,u_0=\dfrac{\rho g h^2}{6\mu}\,}.$ At this speed the upward drag exactly balances the downward gravity drainage, so no fluid is netted through the channel.