MIT Department of Mechanical Engineering
2.25 Advanced Fluid Mechanics
Problem 8.13
This problem is from “Advanced Fluid Mechanics Problems” by A.H. Shapiro and A.A. Sonin
Consider a gas bubble of fixed mass and radius R(t) which is expanding or contracting in an infinite sea of
incompressible liquid. The speed of the interface is dR/dt. The local Eulerian coordinate in the liquid is r.
Let pR , p, and p∞ be, respectively the pressure at r = R (on the liquid side of the interface), at r = r, and
at r = ∞.
(a) Determine the viscous contribution to the normal stress τrr in the liquid.
(b) Show that the dimensionless overpressure, (pR − p∞ )/ρ(dR/dt)2 , is independent of whether the fluid
is viscous or inviscid.
2.25 Advanced Fluid Mechanics
1
c 2010, MIT
Copyright ©
Viscous and Inviscid Flows
A.H. Shapiro and A.A. Sonin 8.13
Solution:
τ φφ
τ θθ
τrr
τrr
τ θθ
τ φφ
(a) First, we must determine the velocity field in the liquid at any point in time. We choose a control
volume taking the shape of a hollow sphere with inner control surface at radius R(t), which moves
outward at exactly the rate of expansion of the bubble dR/dt, and outer surface at an arbitrary radius
r, such that its volume is V = 43 π(r3 −R(t)3 ). Using Form A of the integral mass conservation equation,
d
dt
ρ(u − uCS ) · ndA
= 0
ˆ
ρdV +
CV
(8.13a)
CS
we solve for the radial velocity ur at any position r
ur =
R2 dR
r2 dt
(8.13b)
Using Eq. (8.13b) we can determine the average rate of strain from the following equations
∂ur
γ̇rr = 2
∂r
(8.13c)
and
ur
1 ∂uθ
+
r ∂θ
r
(8.13d)
ur
uθ cot θ
1 ∂uφ
+
+
r sin θ ∂φ
r
r
(8.13e)
γ̇θθ = 2
and
γ̇φφ = 2
There is no azimuthal or polar velocity in this flow, uθ = uφ = 0, and hence
γ̇rr = −4
2.25 Advanced Fluid Mechanics
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R2 dR
r3 dt
(8.13f)
c 2010, MIT
Copyright ©
Viscous and Inviscid Flows
A.H. Shapiro and A.A. Sonin 8.13
and
γ̇θθ = 2
R2 dR
r3 dt
(8.13g)
γ̇φφ = 2
R2 dR
r3 dt
(8.13h)
and
For a Newtonian fluid, τij = µγ̇ij , where µ is the dynamic viscosity. Accordingly the normal stresses
for this flow are
τrr = −4µ
R2 dR
r3 dt
(8.13i)
and
τθθ = 2µ
R2 dR
r3 dt
(8.13j)
τφφ = 2µ
R2 dR
r3 dt
(8.13k)
and
(b) The complete equation of motion in the radial direction for spherical coordinates is
!
u2θ + u2φ
∂ur
∂ur
uθ ∂ur
uφ ∂ur
ρ
+ ur
+
+
−
(8.13l)
∂t
∂r
r ∂θ
r sin θ ∂φ
r
"
#
1 ∂ 2
1
∂
1 ∂τφr
τθθ + τφφ
∂p
= 2 (r τrr ) +
(τθr sin θ) +
−
−
+ ρgr (8.13m)
r ∂r
r sin θ ∂θ
r sin θ ∂φ
r
∂r
Neglecting gravity and retaining only the non-zero terms, we have
∂ur
∂ur
ρ
+ ur
∂t
∂r
!
"
#
1 ∂ 2
τθθ + τφφ
∂p
= 2 (r τrr ) +
−
r ∂r
r
∂r
(8.13n)
Substituting Eq. (8.13i), (8.13j) and (8.13k) into Eq. (8.13n) to evaluate the net contribution of viscous
stresses acting on a fluid element, we obtain
∂ur
∂ur
ρ
+ ur
∂t
∂r
!
"
=
4µ ∂
− 2
r ∂r
R2 dR
r dt
!
#
R2 dR
∂p
− 4µ 4
−
r dt
∂r
(8.13o)
When we differentiate this term, we find that the net contribution of viscous stresses acting radially
is exactly zero and hence there is no net dissipation associated with this flow. Consequently, the
governing equation is
∂ur
∂ur
ρ
+ ur
∂t
∂r
2.25 Advanced Fluid Mechanics
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!
+
∂p
=0
∂r
(8.13p)
c 2010, MIT
Copyright Viscous and Inviscid Flows
A.H. Shapiro and A.A. Sonin 8.13
Integrating Eq. (8.13p) along dr, which is indeed the streamline coordinate, we obtain the unsteady
Bernoulli equation
r=∞
Z
ρ
r=R
∂ur 0
dr + ρ
∂t
Z
ur,∞ =0
ur = dR
dt
u0r du0r +
Z
p∞
dp0 = 0
(8.13q)
pR
which is
Z
r=∞
ρ
r=R
R 2 d2 R 0 1
dR
dr − ρ
r02 dt2
2
dt
!2
+ p∞ − pR = 0
(8.13r)
which gives
d2 R 1
dR
ρR 2 − ρ
dr
2
dt
!2
+ p∞ − pR = 0
(8.13s)
which at last yields the final result
2
2R ddtR
pR − p∞
2
2 = 2 − 1
1
dR
dR
2 ρ dt
dt
(8.13t)
So we have shown that the dimensionless overpressure is indeed independent of whether the fluid is
viscous or inviscid. Note that the dimensionless overpressure can be positive or negative depending on
the rate of change of the surface velocity of the gas bubble.
Problem Solution by TJO, Fall 2010
2.25 Advanced Fluid Mechanics
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2.25 Advanced Fluid Mechanics
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