Example · Couette Flow
Problem statement
A viscous fluid fills the gap \(h\) between two parallel plates. The lower plate is stationary; the upper plate slides at speed \(U\). There is no pressure gradient. Find the steady velocity profile \(u(y)\).
Given information
- Gap \(h\), plate speed \(U\), dynamic viscosity \(\mu\)
- Steady, incompressible, fully developed (\(\partial/\partial x = 0\)), no \(dp/dx\)
Solution strategy
Symmetry slaughters the Navier–Stokes equation: one velocity component, one coordinate, no time. Integrate twice, apply no-slip at both walls.
Step-by-step solution
- With \(\mathbf{v} = (u(y), 0, 0)\), steady, and no pressure gradient, Navier–Stokes reduces to $\(\frac{d^2 u}{dy^2} = 0\)$
- Integrate twice: \(u(y) = C_1 y + C_2\).
- Boundary conditions (no-slip): \(u(0) = 0 \Rightarrow C_2 = 0\); \(u(h) = U \Rightarrow C_1 = U/h\).
Final answer
\[\boxed{\,u(y) = \frac{U}{h}\,y\,}\]
A perfectly linear profile. The shear stress \(\tau = \mu\,du/dy = \mu U/h\) is uniform across the gap — every layer drags its neighbor equally.
Key takeaways
- Couette flow is the physical realization of Newton's law of viscosity; rotational viscometers measure \(\mu\) by building exactly this flow in an annulus.
- Momentum enters at the moving wall, diffuses across, and exits at the fixed wall — viscosity as momentum diffusion, in its purest form.
- Adding a pressure gradient superimposes a parabola on the line (Couette–Poiseuille flow) — linearity of the reduced equation makes superposition legal here.