22-Mec-A6 Fluid Machinery · December 2018

Question 4 of 6: Gravity–Couette Flow between Inclined Plates

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams — December 2018, 16-Mec-A6 Advanced Fluid Mechanics. Open book, 3 hours. Six questions of equal value (20 marks); any five constitute a complete paper. All six are solved here.

Reference texts. J.D. Anderson, Modern Compressible Flow, 3rd ed. (Q1); F.M. White, Fluid Mechanics, 8th ed. and Kundu, Cohen & Dowling, Fluid Mechanics, 6th ed. (Q2, Q5); F.M. White, Viscous Fluid Flow, 3rd ed. (Q3, Q4, Q6); Schlichting & Gersten, Boundary-Layer Theory, 8th ed. (Q6); Fox & McDonald, Introduction to Fluid Mechanics, 10th ed. (general).

Question 4: 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. A thin fluid layer between long inclined plates at angle $\theta$ to the horizontal; $x$ up the incline along the lower plate, $y$ normal to it. Lower plate fixed, upper plate translating at $V_{top}$ up the incline; no imposed streamwise pressure gradient (both ends open).

Given data (symbolic)
Gap	$h$ (small)
Incline angle	$\theta$ from horizontal
Upper-plate velocity	$V_{top}$, up-incline ($+x$)
Fluid	Newtonian, viscosity $\mu$, density $\rho$

Find. (a) modelling assumptions; (b) full Navier–Stokes + BCs; (c) the reduced 2nd-order ODE; (d) the velocity profile $u(y)$.

Figure 4. Inclined Couette layer: the moving upper plate drags fluid up the slope while the gravity component along $x$ opposes it.

Approach. Argue the flow is unidirectional and fully developed, strip the Navier–Stokes equations to a single balance between viscous shear and the streamwise gravity component, and integrate twice with the no-slip conditions at the two plates.

(a) Assumptions. Steady; incompressible, constant $\mu,\rho$; Newtonian; laminar; two-dimensional; unidirectional $\mathbf{u}=(u(y),0,0)$ with $v=w=0$; fully developed (very long plates $\Rightarrow \partial u/\partial x=0$, no entrance effects); no imposed streamwise pressure gradient ($\partial p/\partial x=0$, both ends open to the same conditions); gravity acts vertically down, with component $-g\sin\theta$ along $x$ and $-g\cos\theta$ along $y$.
(b) Full Navier–Stokes + BCs. Continuity $\partial u/\partial x+\partial v/\partial y=0$. The $x$- and $y$-momentum equations are $$\rho\Big(\tfrac{\partial u}{\partial t}+u\tfrac{\partial u}{\partial x}+v\tfrac{\partial u}{\partial y}\Big)=-\tfrac{\partial p}{\partial x}+\mu\Big(\tfrac{\partial^2u}{\partial x^2}+\tfrac{\partial^2u}{\partial y^2}\Big)-\rho g\sin\theta,$$ $$\rho\Big(\tfrac{\partial v}{\partial t}+u\tfrac{\partial v}{\partial x}+v\tfrac{\partial v}{\partial y}\Big)=-\tfrac{\partial p}{\partial y}+\mu\Big(\tfrac{\partial^2v}{\partial x^2}+\tfrac{\partial^2v}{\partial y^2}\Big)-\rho g\cos\theta.$$ Boundary conditions (no-slip): $u(0)=0$ (lower plate) and $u(h)=V_{top}$ (upper plate).
(c) Reduction to an ODE. Steady $(\partial_t=0)$; continuity with $v=0$ gives $\partial u/\partial x=0$, so $u=u(y)$ and the convective and $\partial^2u/\partial x^2$ terms vanish; with $\partial p/\partial x=0$ the $x$-momentum equation collapses to $$\boxed{\mu\frac{d^{2}u}{dy^{2}}=\rho g\sin\theta ,}$$ a second-order ODE. The $y$-momentum equation reduces to $\partial p/\partial y=-\rho g\cos\theta$ (hydrostatic across the gap), consistent with $\partial p/\partial x=0$.
(d) Solution. Writing $C=\rho g\sin\theta/\mu$ and integrating twice, $u=\tfrac{C}{2}y^2+Ay+B$. Applying $u(0)=0\Rightarrow B=0$ and $u(h)=V_{top}\Rightarrow A=V_{top}/h-\tfrac{C h}{2}$: $$\boxed{u(y)=\frac{V_{top}}{h}\,y-\frac{\rho g\sin\theta}{2\mu}\,y\,(h-y).}$$ The first term is the linear Couette drag from the upper plate; the second is the parabolic gravity contribution, which retards the up-slope flow and can produce near-wall back-flow at the lower plate when $V_{top}$ is small.

Question 4 — results
Item	Result
Governing ODE	$\mu\,d^2u/dy^2=\rho g\sin\theta$
Cross-gap pressure	$\partial p/\partial y=-\rho g\cos\theta$ (hydrostatic)
Boundary conditions	$u(0)=0,\ u(h)=V_{top}$
Velocity profile	$u(y)=\dfrac{V_{top}}{h}y-\dfrac{\rho g\sin\theta}{2\mu}y(h-y)$

Topic. Exact solution of the Navier–Stokes equations for unidirectional, fully-developed viscous flow — a body-force-driven Couette layer.

Key relations. Unidirectional flow $\Rightarrow$ convective terms vanish; force balance $\mu\,u_{yy}=\rho g\sin\theta$; superposition of a linear Couette part and a parabolic gravity part; no-slip at both walls.

Why this works. For a thin, long gap the flow forgets $x$, so every term carrying an $x$-derivative or a cross-velocity drops out and the vector Navier–Stokes equations reduce to a single ordinary differential equation in $y$. The two integration constants are set precisely by the two no-slip conditions, giving a closed-form profile. Because the equation is linear, the plate-drag and gravity effects simply add.

Common pitfall. Forgetting the streamwise gravity component (using only $-g$ vertically), or assuming a pressure gradient. With open ends there is no $\partial p/\partial x$; the driving force along the plates is purely $-\rho g\sin\theta$, while $\cos\theta$ only sets the hydrostatic pressure across the gap.

Source. White, Viscous Fluid Flow, 3rd ed., Ch. 3 (exact solutions of the Navier–Stokes equations; Couette and gravity-driven films).

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