Math 2.0

Fluid mechanics Incompressible viscous flow Example

Couette flow

Builds on Navier–Stokes equationsNo-slip conditionViscosity

Example

Consider a fluid of density ρ>0\rho > 0 and dynamic viscosity μ>0\mu > 0 between two parallel plates y=0y = 0 and y=hy = h, with h>0h > 0. The lower plate is at rest and the upper plate moves in the xx-direction with constant speed UU. Plane Couette flow is u=(Uyh,0,0),p=p0,u = \Big( \frac{U y}{h}, 0, 0 \Big), \qquad p = p_0, with p0p_0 a constant.

Check: uu depends only on yy and points in the xx-direction, so ∇⋅u=0\nabla \cdot u = 0 and (u⋅∇)u=(Uy/h) ∂xu=0(u \cdot \nabla) u = (U y / h) \, \partial_x u = 0. Also Δu=0\Delta u = 0 since uu is linear, and ∇p=0\nabla p = 0. So the steady Navier–Stokes equations ρ(u⋅∇)u=−∇p+μΔu\rho (u \cdot \nabla) u = -\nabla p + \mu \Delta u hold. At y=0y = 0 the velocity is 00 and at y=hy = h it is (U,0,0)(U, 0, 0), so the no-slip condition holds on both plates.

Remarks

The shear stress μ ∂yu1=μU/h\mu \, \partial_y u_1 = \mu U / h is the same at every height. Measuring the force needed to drag the upper plate gives μ\mu, which is the principle of a Couette viscometer. With an added constant pressure gradient ∂xp=−G\partial_x p = -G the flow becomes u1=Uyh+G2μy(h−y)u_1 = \frac{U y}{h} + \frac{G}{2 \mu} y (h - y), a sum of Couette and Poiseuille flow. The rotating version between two coaxial cylinders is Taylor–Couette flow.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Couette_flow
About this page
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Kindexample
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MSC76D05, 76D07
IllustratesNavier–Stokes equations
UsesNavier–Stokes equations, No-slip condition, Viscosity
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Changed3 November 2026

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