Additional problem

University of Warsaw Advanced Hydrodynamics

Faculty of Physics Selected Topics in Fluid Mechanics

Summer Semester 2019/20

Exercise Sheet 5

An application of the Lorentz reciprocal theorem: the Stokes resistance of an arbitrary particle Questions, comments and corrections: e-mail togustavo.abade@fuw.edu.pl

Consider a rigid particle of an arbitrary shape translating with steady velocity U through an unbounded fluid otherwise at rest [Fig.1.a]. The instantaneous flow velocity and its asso- ciated stress field are u_tand σ_t, respectively. Denote by

f(r) ≡ σt(r) · n(r), for r on∂V, (1)

the stress vector (force density) on the surface of the translating particle. From linearity of the Stokes equations and boundary conditions, one may write

f(r) = −Z(r) · U, for r on∂V, (2)

where Z(r) is a symmetric resistance tensor. It depends solely on the shape of the immersed particle and is independent of U.

∂V ∂V

U F^h

u0

(a) (b)

Figure 1: Rigid particle of arbitrary shape (a) translating with steady velocity U, and (b) immersed in an ambient flow u0.

1. Consider the same particle immersed in an arbitrary ambient flow u₀ [Fig. 1.b]. Use the Lorentz reciprocal theorem to show that the hydrodynamic drag F^h on the particle when stationary is

F^h = Z

∂V

Z(r) · u0(r) dS(r). (3)

Solution. The flow velocity may be decomposed as

u(r) = u₀(r) + u⁰(r), (4)

where u⁰ is the flow “perturbation” (due to the presence of the particle) with associated stress field σ⁰.

We invoke the Lorentz reciprocal theorem for the pair (ut, σ_t) and (u⁰, σ⁰), Z

∂V

ut· (σ⁰· n) dS = Z

∂V

u⁰· (σ_t· n) dS, (5)

where ∂V is the surface of the immersed particle.

The boundary conditions for the two considered Stokes flow problems read

u_t(r) = U, u⁰(r) = −u0(r), for r on ∂V . (6)

Then Eq. (5) may be written as U ·

Z

∂V

σ⁰· n dS = − Z

∂V

u₀· (σ_t· n) dS = U · Z

∂V

Z(r) · u₀(r) dS(r), (7) where Eqs. (1) and (2) where used in the last equality.

The stress field σ0associated with the ambient flow u0 is defined inside the volume V of the particle enclosed by ∂V . Also σ0satisfies the Stokes momentum equation, ∇ · σ0 = 0. Then

Z

V

∇ · σ₀dV = Z

∂V

σ₀· n dS = 0, (8)

and Eq. (7) may be written as U · F^h= U ·

Z

∂V

Z(r) · u₀(r) dS(r), (9)

where F^h =

Z

∂V

(σ₀+ σ⁰) · n dS, (10)

is the hydrodynamic drag on the immersed particle. The result (3) holds for arbitrary U.

2. Show that the hydrodynamic drag is F^h =

Z

∂V

Z(r) · [u0(r) − V] dS(r), (11)

for a particle translating with velocity V.

Solution. Follow the same procedure above using the appropriate boundary conditions, ut(r) = U, u⁰(r) = −[u0(r) − V], for r on ∂V ,

in place of (6).

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Additional problem

Using the development described above, derive a similar relation between the hydro- dynamic torque N^hon the particle in an arbitrary ambient field u0and the solution for the particle rotating with steady angular velocityΩ.

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