Question 4 of 6: Combined Gravity–Couette Flow Between Inclined Plates

Question 4: Combined Gravity–Couette Flow Between Inclined Plates (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. Incompressible Newtonian fluid ($\mu$, $\rho$) in a gap $h$ between two long inclined parallel plates at angle $\theta$ to the horizontal. Lower plate stationary; upper plate translates up-slope at $V_{top}$. Coordinate $x$ up the incline, $y$ normal to the plates from the lower wall.

Find. The modelling assumptions, the governing Navier–Stokes equations with boundary conditions, their reduction to a single ODE, and the velocity profile $u(y)$.

(a) Assumptions. Steady flow ($\partial/\partial t=0$); incompressible, constant $\mu$ and $\rho$; two-dimensional ($w=0$, $\partial/\partial z=0$); very long, wide plates so the flow is fully developed and unidirectional, $u=u(y)$ only with $v=0$ (continuity then forces $\partial u/\partial x=0$); laminar Newtonian flow; no imposed streamwise pressure gradient ($\partial p/\partial x=0$, the gap being open so the flow is driven only by the moving wall and gravity); gravity acts vertically down, with component $-g\sin\theta$ along $+x$ and $-g\cos\theta$ along $+y$; no-slip at both walls.

(b) Navier–Stokes and boundary conditions. The incompressible Navier–Stokes momentum equations are $$\rho\frac{D\mathbf{V}}{Dt}=-\nabla p+\mu\nabla^2\mathbf{V}+\rho\mathbf{g}.$$ With $\mathbf{V}=(u,v,0)$ and $\mathbf{g}=(-g\sin\theta,\,-g\cos\theta,\,0)$, the two in-plane components are $$x:\ \rho\!\left(u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}\right)=-\frac{\partial p}{\partial x}+\mu\!\left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right)-\rho g\sin\theta,$$ $$y:\ \rho\!\left(u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}\right)=-\frac{\partial p}{\partial y}+\mu\nabla^2 v-\rho g\cos\theta.$$ Boundary conditions (no slip): $u(0)=0$ at the lower wall and $u(h)=V_{top}$ at the upper wall.

(c) Reduction to an ODE. Continuity $\partial u/\partial x+\partial v/\partial y=0$ with $v=0$ gives $\partial u/\partial x=0$, so $u=u(y)$ and the convective terms vanish. Then $\partial^2u/\partial x^2=0$. The $y$-equation reduces to $\partial p/\partial y=-\rho g\cos\theta$ (a hydrostatic pressure variation across the gap that does not affect $u$). With $\partial p/\partial x=0$, the $x$-equation collapses to the second-order ODE $$\boxed{\mu\frac{d^2u}{dy^2}=\rho g\sin\theta\quad\Longleftrightarrow\quad \frac{d^2u}{dy^2}=\frac{\rho g\sin\theta}{\mu}.}$$ The right-hand side is a positive constant: gravity acts like a favourable-to-downslope body force that curves the profile.

(d) Solution. Integrating twice, $$u(y)=\frac{\rho g\sin\theta}{2\mu}\,y^2+C_1 y+C_2.$$ Applying $u(0)=0\Rightarrow C_2=0$, and $u(h)=V_{top}\Rightarrow C_1=\dfrac{V_{top}}{h}-\dfrac{\rho g\sin\theta}{2\mu}h$. Hence $$\boxed{u(y)=\frac{V_{top}}{h}\,y-\frac{\rho g\sin\theta}{2\mu}\,y\,(h-y).}$$ The first term is the linear Couette flow dragged up-slope by the moving wall; the second is the parabolic gravity contribution that pulls the fluid back down the incline. If $V_{top}$ is small enough, the near-lower-wall fluid can actually flow downhill (net back-flow), the classic signature of opposed shear- and gravity-driving.