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Stokes flow at low Reynolds number

Dropping inertia: the linear Stokes equations, drag on a sphere, reversibility and the scallop theorem.

When \(\mathrm{Re} \ll 1\) the inertial terms are negligible compared with viscosity, and the steady Navier-Stokes equations reduce to the Stokes equations \[ -\nabla p + \mu\Delta\mathbf{u} = 0, \qquad \nabla\cdot\mathbf{u} = 0 \] (with pressure measured again in true units, since we are balancing forces directly). They are linear, and they contain no time derivative: the flow at each instant is determined by the boundary motion at that instant. Taking the divergence gives \(\Delta p = 0\), so the pressure is harmonic; taking the curl gives \(\Delta\boldsymbol{\omega} = 0\).

The most famous solution is the flow past a sphere of radius \(a\) moving at speed \(U\), solved by Stokes in 1851. Its drag is \[ F = 6\pi\mu a U. \] It is linear in speed and in size, unlike high-Reynolds-number drag, which grows like \(U^2\). Balancing it against weight minus buoyancy gives the settling speed of a small particle, \[ 6\pi\mu aU = \tfrac43\pi a^3(\rho_s - \rho)g \quad\Longrightarrow\quad U = \frac{2a^2(\rho_s - \rho)g}{9\mu}. \] A grain of sand of radius 25 micrometres sinks through water at about 2 mm/s. Millikan used this law in his oil-drop measurement of the electron charge. Always check the answer is self-consistent: the Reynolds number \(2aU/\nu\) of the result must indeed be small.

Linearity plus the absence of time has striking consequences. Reverse the boundary motion and the whole flow reverses: G. I. Taylor's demonstration unmixes a blob of dye in syrup by turning the handle back. A swimmer whose stroke looks the same played backwards (a scallop opening and closing) makes no net progress, which is Purcell's scallop theorem (1977); bacteria swim with rotating corkscrew flagella instead. Energy methods also give uniqueness: if two Stokes flows share boundary values, their difference \(\mathbf{w}\) has \(\int\mu|\nabla\mathbf{w}|^2 = 0\), so \(\mathbf{w} = 0\). Helmholtz (1868) showed that the Stokes flow dissipates less energy than any other divergence-free field with the same boundary values.

The approximation has a subtle edge. In two dimensions there is no Stokes solution for flow past a cylinder that matches a uniform stream at infinity (the Stokes paradox), because far from the body inertia is never negligible. Oseen (1910) fixed this by keeping a linearised inertial term; matched asymptotic expansions (Kaplun, and Proudman and Pearson, 1957) later explained why this works.

Picture it: a teaspoon moved through honey leaves no swirl behind; stop the spoon and everything stops at once. Think it: the Stokes operator is minus the Leray projection of the Laplacian, acting on divergence-free fields, and Stokes flow is its elliptic boundary value problem; the time-dependent Navier-Stokes equations are that operator plus the nonlinear term.

Разобранный пример · 2*(0.000025)^2*(2650 - 1000)*9.81/(9*0.001)

Evaluate 2*(0.000025)^2*(2650 - 1000)*9.81/(9*0.001)

0.00224813

Шаг за шагом

  1. 0.002248125 = \frac{3597}{1600000}

    Multiply: 2(1/1600000000)·1650·9.81(1000/9) = 3597/1600000.

  2. \frac{3597}{1600000} = 0.00224812

    Simplify.

Показать ответ
0.00224812

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Как: Stokes flow at low Reynolds number

  1. Check Re is small with the expected speed; Stokes flow needs Re well below 1.
  2. Write the force balance: Stokes drag 6 pi mu a U against the net weight.
  3. Solve for the unknown (speed, radius or viscosity).
  4. Recompute Re with the answer to confirm the approximation was valid.

Вопросы, которые люди задают

Why is Stokes drag proportional to the radius and not the area?

Viscous stress scales like mu U / a and acts over an area of order a squared, so the force scales like mu U a. Area-proportional drag belongs to high Reynolds number, where the stress is dynamic pressure rho U squared.

How do microorganisms swim at all?

By moving non-reciprocally: a rotating helix or a whip-like flagellum beating in a wave traces a stroke that is not the same backwards, which the scallop theorem allows.

What do I need before starting fluid dynamics?

Multivariable calculus (divergence, curl, the divergence and Stokes theorems), linear algebra (symmetric matrices and eigenvalues) and partial differential equations (the heat and Laplace equations). The last lessons also use Sobolev spaces, which are introduced where they are needed.

What are the Navier-Stokes equations in one sentence?

Newton's second law for each particle of a viscous incompressible fluid: acceleration equals the pressure force plus viscous diffusion of momentum, with the constraint that the velocity field is divergence-free.

Why is two-dimensional flow easier than three-dimensional flow?

In 2D the vorticity is a scalar that is only carried and diffused, so its maximum never grows; in 3D vortex lines can be stretched, which amplifies vorticity, and no known bound rules out unlimited growth.

Is this course physics or mathematics?

Both, in order. The first half derives the equations from physical principles and solves classical flows; the second half treats the equations as mathematical objects and studies which of their properties can be proved.

Ещё в разделе Fluid Dynamics