Question 4 of 6: Combined Couette–Poiseuille Flow between Parallel Plates
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — December 2019, 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 as a study resource.
Given. Steady, incompressible, two-dimensional flow of a Newtonian fluid between horizontal parallel plates a distance $2b$ apart; the top plate moves at $U$ and a constant $dp/dx$ acts along $x$.
Find. (a) the two constituent classical flows; (b) the reduced $x$-momentum equation with justification; (c) the velocity profile $u(y)$.
Figure 4. Superposition of a linear Couette profile (moving top plate) and a parabolic Poiseuille profile (pressure gradient); $y$ is measured from the centreline, plates at $y=\pm b$.
(a) The two classical flows. (i) Plane Couette flow — shear-driven flow between two plates, one moving; with no pressure gradient the profile is linear, $u=U(1+y/b)/2$. (ii) Plane Poiseuille flow — pressure-driven flow between two stationary plates; the profile is parabolic and symmetric, $u=-\dfrac{1}{2\mu}\dfrac{dp}{dx}(b^2-y^2)$. The present flow is their linear superposition (both are solutions of the same linear equation).
(b) Reduced $x$-momentum equation. The full $x$-Navier–Stokes is
$$\rho\Big(\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}\Big)=-\frac{\partial p}{\partial x}+\mu\Big(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\Big).$$
Steady $\Rightarrow\partial u/\partial t=0$; fully developed $\Rightarrow\partial u/\partial x=0$, and continuity $\partial u/\partial x+\partial v/\partial y=0$ with $v=0$ at both walls then forces $v\equiv0$; hence $\partial^2u/\partial x^2=0$ and the convective terms vanish. What remains is a balance of pressure and viscous forces:
$$\boxed{0=-\frac{dp}{dx}+\mu\frac{d^2u}{dy^2}\quad\Longleftrightarrow\quad \mu\frac{d^2u}{dy^2}=\frac{dp}{dx}.}$$
(c) Integrate and apply boundary conditions. Because $dp/dx$ is constant, integrate twice:
$$u(y)=\frac{1}{2\mu}\frac{dp}{dx}\,y^2+C_1 y+C_2 .$$
With $y$ from the centreline, the no-slip conditions are $u(-b)=0$ (fixed lower plate) and $u(+b)=U$ (moving upper plate). Solving for $C_1,C_2$,
$$\boxed{u(y)=\frac{1}{2\mu}\frac{dp}{dx}\big(y^2-b^2\big)+\frac{U}{2}\Big(1+\frac{y}{b}\Big).}$$
The first term is the symmetric parabolic (Poiseuille) contribution — note it is negative for a favourable $dp/dx\lt0$, i.e. it adds forward flow — and the second is the linear (Couette) contribution from the moving plate.
