Work in the frame translating with the bubble. The bubble is the fixed circle

r = a

, while the velocity tends to

- U e_{x}

at infinity. The potential flow around a circular cylinder has velocity potential

ϕ = - U (r + \frac{a ^{2}}{r}) cos θ .

(1)

It is harmonic for

r > a

and gives

u_{r} = - U (1 - \frac{a ^{2}}{r ^{2}}) cos θ, u_{θ} = U (1 + \frac{a ^{2}}{r ^{2}}) sin θ .

(2)

Thus $u_{r} = 0$ on $r = a$ , satisfying the kinematic boundary condition, and $u \to - U e_{x}$ as $r \to \infty$ . The laboratory-frame flow is obtained by adding $U e_{x}$ .

Solution

 0  0

For the outer flow from part (i), the rate-of-strain tensor in polar coordinates has components

e_{rr} = - \frac{2 U a ^{2}}{r ^{3}} cos θ, e_{θθ} = \frac{2 U a ^{2}}{r ^{3}} cos θ, e_{r θ} = - \frac{2 U a ^{2}}{r ^{3}} sin θ .

(1)

Therefore

e : e = e_{rr}^{2} + e_{θθ}^{2} + 2 e_{r θ}^{2} = \frac{8 U ^{2} a ^{4}}{r ^{6}} .

(2)

The leading dissipation per unit axial length is

D = 2 μ \int_{a}^{\infty} \int_{0}^{2 π} e : e r d θ d r = 16 μ U^{2} a^{4} (2 π) \int_{a}^{\infty} r^{- 5} d r = 8 π μ U^{2} .

(3)

Balancing this with drag power

D U

gives the drag on a two-dimensional circular bubble from outer-flow dissipation

D = 8 π μU

(4)

per unit axial length, directed opposite to the bubble velocity.

Solution

 0  0

A rigid body obeys the no-slip boundary condition. Its outer potential flow generally has an

O (U)

tangential slip velocity at the surface, so the boundary layer must make an

O (U)

velocity correction across thickness

δ

. Its strain is therefore

O (U / δ)

, and in two dimensions its dissipation per unit axial length is

D_{layer} \sim μ (\frac{U}{δ})^{2} a δ = μ U^{2} \frac{a}{δ},

(1)

which exceeds the

O (μ U^{2})

outer dissipation by

a / δ \sim Re^{1/2}

The dominant drag consequently depends on the detailed boundary-layer velocity field and possibly on boundary-layer separation. It cannot be obtained by integrating the outer irrotational strain alone.