CAVITATION: A POSSIBLE SOLUTIONTSOGTGEREL GANTUMUR

Note: The chosen units might be a bit diﬀerent from the “oﬃcial” problem statement.
For instance, we might write 1 µm for 10−6 m. “Table 1” refers to the “notation table”
from the statement.
A1. By performing a simple dimensional analysis, estimate the collapse time τ of a purevapour bubble, in terms of bubble’s initial radius R0 , water density ρ, water pressure p∞ ,and the vapour pressure pv . Evaluate the formula when R0 = 1 mm and the quantities ρ,p∞ and pv take their typical values from Table 1. Assume no surface tension: σ = 0.
Solution. It is reasonable to expect that the bubble would not collapse if p∞ ≤ pv . Hence
we take p∞ > pv , and presume that the diﬀerence p∞ − pv will feature in the final formula.
The dimensions of the quantities are[R0 ] = m,

[ρ] = kg/m3 ,[p∞ − pv ] = N/m2 = kg/(m · s2 ).

The only combination of these that has the dimension of time is√ρτ ∼ R0≈ 0.1 s.
p∞ − pv(1)(2)

A2. Suppose that a micro-bubble consisting of air and vapour, with radius R0 = 10 µm,
is in equilibrium when the external pressure p∞ = 100 kPa. Find the partial pressure q0of air in the bubble. Now suppose that the external pressure p∞ was gradually decreased,
and that the air inside the bubble follows an isothermal process. Find the critical pressure
pc , defined by the condition that if p∞ < pc the bubble size grows without bound. Thequantities pv and σ take their typical values from Table 1.
Solution. The pressure equilibrium condition ispv + q = p∞ +2σ,R(3)which, under R = R0 and q = q0 , yields
2σ≈ 170 kPa.
R0
On the other hand, taking into account the isothermal law
q0 = p∞ − pv +q0 R03 = qR3 ,(4)(5)

we getq0 R03 2σ−.
(6)
R3R
The critical pressure corresponds to the minimum of p∞ as a function of R. The radiusat the minimum is easily found to be√
3q0 R0≈ 60 µm,(7)
Rc = R02σ
p∞ = pv +1

2

TSOGTGEREL GANTUMUR

and the corresponding pressure ispc = pv − 2q0( R0 )3≈ 700 Pa.
Rc(8)B1. Suppose that a single spherical bubble resides within water that fills space uniformly,
and that the bubble may evolve in size without distorting its spherical shape, due tochanges, e.g., in the external pressure p∞ . Derive an equation that relates the bubble
radius R(t) and its time derivatives R′ (t) and R′′ (t), surface tension σ, water density ρ,
the pressure far from the bubble p∞ , and the pressure inside the bubble p. Then split thepressure p into two terms, by assuming that the bubble has both vapour and air in it, andthat the air follows an adiabatic process with exponent γ. To give a reference point, thepartial air pressure must be q0 when the bubble size equals R0 . Assume that evaporation,condensation, or transfer of air between the bubble cavity and the surrounding water has
no eﬀect on the water volume.
Solution. Let v(r, t) denote the radial velocity of the fluid element at the distance r fromthe bubble center and at the time moment t. Let also u(t) = v(R, t), that is,u(t) = R′ (t).

(9)

Then the incompressibility condition yieldsv(r, t) =R2 u(t),r2(10)and hence∂v∂ R2 u2R 2 R2 ′==u + 2u.
∂t∂t r2r2r
The radial acceleration can now be computed as(11)R22R2R4∂v ∂v+v = 2 u′ + 2 u2 − 5 u2 .
(12)
∂t
∂rrrr
With p = p(r) denoting the pressure field (where possible time dependence is suppressedin the notation), Newton’s law readsa=ρa = −p′ ,or(13)R2 ′ ( 2R 2R4 ) 2p′u+−u=−.
(14)
r2r2r5ρ
We integrate it from r = R to r = ∞, to get
3p(R) − p∞Ru′ + u2 =.
(15)
2ρSince the pressure inside the bubble satisfies
2σp = p(R) +,(16)
R
we conclude that3
2σp − p∞RR′′ + (R′ )2 +=.
(17)
2
ρR
ρ
Finally, taking into account the fact that the bubble pressure p consists of the vapour
pressure pv and the partial air pressure q, as( R0 )3γ,(18)p = pv + q = pv + q0R

CAVITATION: A POSSIBLE SOLUTION

3

we have3
2σ q0 R03γ
ρRR′′ + ρ(R′ )2 +− 3γ = pv − p∞ .
(19)
2RR
B2. A water tank under the external pressure p−∞ = 100 kPa, containing a nucleus ofradius R0 = 10 µm initially in equilibrium, was exposed to vacuum, so that the systemsuddenly has p∞ = 0. Estimate the terminal (asymptotic) value of the growth speed R′ ,as well as the time it reaches this terminal value.
Solution. Putting R′′ = 0 and R → ∞ in (19) yields√√
2(pv − p∞ )2pv′R (∞) ==≈ 1.24 m/s.
3ρ
3ρ

(20)

The initial acceleration can also be found from (19) as
q0 + pv − 2σ/R0p−= ∞ ≈ 107 m/s2 ,ρR0
ρR0
and hence the time for the speed to be stabilized can be estimated asR′ (∞)
t = ′′≈ 0.1 µs.
R (0)R′′ (0) =(21)(22)

B3. A water tank under the external pressure p−∞ = 1600 Pa, containing a gas bubbleof radius R0 = 10 µm initially in equilibrium, was suddenly exposed to the atmosphericpressure p∞ = 100 kPa. Estimate the minimum radius of the bubble before it rebounds.
Solution. Multiply (19) by 2R2 R′ to getρ[R3 (R′ )2 ]′ + 4σRR′ − 2q0 R03γ R2−3γ R′ = 2(pv − p∞ )R2 R′ ,

(23)

or

22q0 R03γ (R3−3γ )′= (pv − p∞ )(R3 )′ .
(24)
ρ[R (R ) ] + 2σ(R ) +3(γ − 1)
3This can easily be integrated, from R(0) = R0 and R′ (0) = 0 to R(t) = R and R′ (t) = R′ ,which yields
33′ 2

ρR (R ) + 2σ(R′ 2 ′22 ′2q0 R03γ (R3−3γ − R03−3γ )22− R0 ) += (pv − p∞ )(R3 − R03 ),3(γ − 1)or2(pv − p∞ ) 2σ 2R03
ρ(R ) =−+ 33RR
When R ≪ R0 , it shows first that′ 2((25)

3σp∞ − pv
q0
q0 R03γ−3++−R033(γ − 1) 3(γ − 1)R3γ−3R′ ∼ −R−3/2 ,and moreover that the rebound radius satisfies( R )3(γ−1) (γ − 1)(p − p + 3σ/R )
∞v
00
=+ 1.
Rq0
Taking into account thatq0 = p−∞ − pv + 2σ/R0 ≈ 10 Pa,). (26)

(27)

(28)

(29)

we conclude that the rebound radius isR ≈ 0.4 µm.

(30)

4

TSOGTGEREL GANTUMUR

B4. If there is no gas other than water vapour present in a bubble, the bubble completelycollapses in finite time. Determine the characteristic exponent α inR(t) ∼ (T − t)α ,(31)

where T is the collapse time.
Solution. We have found in the previous part thatR′ ∼ −R−3/2 .

(32)

Putting R(t) ∼ (T − t) into it we getα

(T − t)α−1 ∼ −(T − t)−3α/2 ,

(33)

or2
α= .
(34)
5
B5. Based on the equation derived in Part 3, find the natural frequency of the sphericaloscillation of a bubble of radius R0 = 0.1 mm.
Solution. Introducing the new variable x by R = R0 x, we write (19) as32σ
ρR02 xx′′ + ρR02 (x′ )2 +− q0 x−3γ = pv − p∞ .
2R0 x
Now put x = 1 + y and retain the terms up to linear in y, to have2σρR02 y ′′ +(1 − y) − q0 (1 − 3γy) = pv − p∞ .
R0
Taking into account the equilibrium condition
2σ− q0 = pv − p∞ ,R0
we infer
3γq0 − 2σ/R0y ′′ +y = 0.
ρR02
Thus the bubble is unstable if
3γq0 ≤ 2σ/R0 = q0 + pv − p∞ ,

(35)

(36)

(37)

(38)

(39)

or equivalently, if
p∞ ≤ pv − (3γ − 1)q0 .
(40)
On the other hand, if p∞ > pv − (3γ − 1)q0 then the bubble oscillates with the natural
frequency
√
f0 =1
2πR0

(3γ − 1)q0 + p∞ − pv
≈ 33 kHz.
ρ(41)B6. Suppose that the bubble described in the previous part is subjected to a standing
sound wave along the x-axis, whose pressure field is given by( 2πf)p(x, t) = p0 + A sin(x + a) sin(2πf t),(42)cwhere f is the frequency, and c is the speed of sound. The parameters p0 , A, and a areconstants, whose meanings may readily be deduced from the equation. Find the averageforce exerted upon the bubble. The bubble is situated at the origin of the xyz coordinate
system, and its size is much smaller than the wavelength of the sound.

CAVITATION: A POSSIBLE SOLUTION

Solution. Small oscillation of the bubble is described by( 2πf )ρR02 (y ′′ + 4π 2 f02 y) = A sin
a sin(2πf t).
c
Looking for the solution in the formy(t) = B sin(2πf t),

5(43)

(44)

we findA sin(2πf a/c).
4π 2 ρR02 (f02 − f 2 )(45)

R(t) = R0 + R0 y(t),(46)B=Sincethe volume of the bubble is

4π
4π 3R(t)3 ≈R [1 + 3y(t)].
33 0Now recalling that the average force is∂p
F = −⟨V⟩,∂x
where the average is taken over time, and that the pressure gradient is( 2πf )∂p
2πf A=cosa sin(2πf t),∂xccwe conclude( 4πf a )f A2 R0F =−sin.
222ρc(f0 − f )cV (t) =(47)

(48)

(49)

(50)

C1. Suppose that a nucleus consisting of air and vapour, with radius R0 = 10 µm, isplaced in water-air solution, in which the dissolved air is in equilibrium with the atmospheric pressure above the water. The partial pressure of air in the bubble is q = 170 kPa,
and the vapour pressure can be neglected. Estimate the time required for the bubble tobe completely resorbed into water. The quantities p∞ , κ, δ and σ take their typical values
from Table 1. Assume that the region surrounding the bubble in which air diﬀusion takes
place immediately gets much larger than the bubble itself.
Solution. From Henry’s law, the initial concentration of dissolved air in the body ofwater isui = Hp∞ ≈ 0.024 kg/m3 ,(51)
and the initial concentration of dissolved air in the immediate vicinity of the bubble isu = Hq ≈ 0.041 kg/m3 .

(52)

Since u > ui , there will be diﬀusive flux directed away from the bubble, and the bubble
will start losing air. As a result, the bubble shrinks and the surface tension term 2σ/R
increases. Since the external pressure p∞ is constant, the partial air pressure q increases,
which leads to more diﬀusive flux. In the end, the bubble will get completely resorbed
into water.
To quantify the diﬀusion of air, let us consider the region enclosed by concentric spheresof radii r and r + ∆r with ∆r small. The rate of change of air mass in this region is∂udm= 4πr2 ∆r ,(53)dt∂t

6TSOGTGEREL GANTUMUR

Figure 1. (a) Initial configuration of the system (b) The geometry usedin the solution
where 4πr2 ∆r is the volume of the region and the air concentration u = u(r, t) is afunction of r and t. On the other hand, the same quantity can be computed as thediﬀerence between the diﬀusive fluxes through the concentric spheres:dm= 4π(r + ∆r)2 J(r + ∆r) − 4πr2 J(r)dt∂u∂u
= 4π(r + ∆r)2 κ (r + ∆r) − 4πr2 κ (r)∂r∂r2∂ u∂u≈ 4πr2 κ∆r 2 (r) + 8πrκ∆r (r),∂r∂r
where we have taken into account the directions of the fluxes, and the fact that
∂u∂u
∂ 2u
(r + ∆r) ≈(r) + ∆r 2 (r).
∂r∂r∂rComparing the two equations, we infer

(54)

(55)

∂u
∂ 2 u 2κ ∂u=κ 2 +.
(56)∂t∂rr ∂r
The dissolved air concentration in water at the initial time moment t = 0 is uniformly ui ,and assuming that the radius R of the bubble remains constant, the air concentration in
the immediate vicinity of the bubble should be equal to Hq:{u(r, 0) = ui
for r > R,(57)u(R, t) = Hq for t > 0.
Introducing a new variable v = r(u−qH), and a new time parameter τ = κt, the diﬀusion
equation (56) becomes∂ 2v∂v= 2,(58)∂τ∂rwith{v(r, 0) = r(ui − qH) for r > R,(59)v(R, τ ) = 0for τ > 0.

CAVITATION: A POSSIBLE SOLUTION

7We can go further by introducing
ξ =r−Randw(ξ, τ ) = v(R + ξ, τ ),(60)to write (56) as∂w
∂2w=,∂τ∂ξ 2with{w(ξ, 0) = (R + ξ)(ui − qH) for ξ > 0,w(0, τ ) = 0
for τ > 0.

(61)

(62)

The solution to this problem is∫)ui − qH ∞ ( −(ξ−η)2 /(4τ )2e− e−(ξ+η) /(4τ ) (η + R)dη.
(63)
w(ξ, τ ) = √4πτ 0Sincev(r, κt)w(r − R, κt)u(r, t) = qH += qH +,(64)rrwe have
1 ∂ww∂u=− 2,(65)∂rr ∂ξrand hence(1∂u1 )
1 ∂w
+√.
(66)
== (ui − qH)∂r r=R R ∂ξ ξ=0Rπκt
The second term corresponds to the width of the diﬀusion layer surrounding the bubble,
which we assume to be much larger than the bubble itself. Hence the rate of change ofthe mass of the bubble can be estimated asdm
∂u= 4πR2 κ≈ 4πκR(ui − qH).
(67)
dt
∂r r=ROn the other hand, the mass of the bubble is related to the air density, which in turn is
proportional to the pressure:
4π 3 δ0 q
4π 3R δ=R ·,(68)
m=33p∞
where δ0 = 1.2 kg/m3 is the air density at the atmospheric pressure p∞ = 105 Pa. Furthermore, neglecting vapour pressure, we have the mechanical equilibrium condition
2σq = p∞ +,(69)R
leading toε4π 3
R · (δ0 + ),(70)m=3R
with2σδ0ε=≈ 1.73 · 10−6 kg/m2 .
(71)
p∞After taking the derivative of (70) with respect to t, we equate it to (67), and get
2σHdR 8π dR
4πκR(ui − p∞ H −) = 4πδ0 R2+εR .
(72)
Rdt
3dt
Since ui − p∞ H = 0, we have2(73)−2σκHdt = δ0 R2 dR + εRdR,38TSOGTGEREL GANTUMUR

and a direct integration yields
δ0 R03 4εR02+,
33(74)(δ0 R0 + 4ε)R02t=≈ 9 s.
6σκHThus the nucleus collapses in a matter of seconds.

(75)

2σκHt =
finally giving

C2. Consider a conical crevice in the wall of a water container, with an aperture angle α.
A small amount of air and vapour is trapped within the cone. Write down the conditionof mechanical and diﬀusive equilibrium. Determine when the pocket of air stays in thecrevice without disappearing.

Figure 2. Conical crevice
Solution. The equilibrium conditions are
2σpv + q = p∞ ±andu = Hq.
(76)
RThe plus sign corresponds to the situation where the water surface is concave, and thesurface tension tends to squeeze the air out of the crevice through diﬀusion. When thewater surface is convex, we have the opposing sign. Supposing that initially the partial
air pressure q is large, q will decrease as the region of trapped air shrinks due to diﬀusion.
At some point, a diﬀusive equilibrium will be found. Thus the pocket of air does notdisappear when the water surface is convex, meaning thatπ + α < 2θ.

(77)

Theoretical Problem 3: Cavitation
Q3 - Marking Scheme
Question part
A1Total marks

Partial marks
0.5Explanation

0.4 Derivation of the formula
0.1 Numerical answer

A210.3 Pressure equilibrium condition
0.1 Numerical answer for partial air pressure
0.1 Equilibrium condition with isothermal law
0.4 Formula for critical pressure
0.1 Numerical answer for critical pressure

B1

1.5

0.3 Incompressibility condition
0.5 Computation of radial acceleration
0.3 Taking into account Newton’s second law
0.2 Final equation with p, where surface tension enters
0.2 Final equation with p split and adiabatic law taken intoaccountB2

10.4 Formula for terminal speed
0.1 Numerical answer for terminal speed
0.4 Formula for time or for initial acceleration
0.1 Numerical answer for timeB3

10.5 Formula for speed R’
0.4 Formula for rebound radius
0.1 Numerical answer for rebound radius

B4

0.5

0.2 Writing down the asymptotic equation for R’
0.2 Substitution of the form into the equation

Question part

Total marks

Partial marks

Explanation

0.1 Numerical answer for alphaB510.4 Writing down the linear second order equation
0.2 Recognition of the stability condition
0.3 Formula for the natural frequency
0.1 Numerical answer for the natural frequency

B61

0.2 Writing down the forced oscillator equation
0.3 Solution of the equation
0.1 Computation of volume
0.1 Computation of pressure gradient
0.3 Final answer

C1

20.1 Initial concentration of air in water
0.1 Initial concentration of air near bubble
0.1 Mass change rate in a shell in terms of concentration
change rate
0.1 Mass change rate in a shell in terms of flux
0.1 Use of Fick’s law
0.2 Diffusion equation in terms of r and t
0.2 Initial and boundary conditions
0.2 Reformulation and solution
0.2 Computing the air flux at the bubble wall
0.1 Simplification due to smallness of the bubble
0.2 Mass of the bubble in terms of pressure with ideal gas
law taken into account
0.1 Mass balance
0.2 Final formula for resorption timeQuestion part

Total marks

Partial marks

Explanation

0.1 Numerical answer for resorption timeC2

0.5

0.2 Condition of equilibrium
0.3 Final condition on angles