text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
= 0
′
′
1
Re
The normal stress condition assumes the dimensionless form:
′
−pd +
1
Fr
′
z +
2
Re
n ·
E ′
· n =
1
We
∇ ′
· n
(2.12)
(2.13)
The relative importance of surface tension to gravity is prescribed by the Bond number Bo, while that
of surface tension to viscous stresses by the capillary number Ca. In the high Re limit of interest, the
normal force balance requires that the dynamic pressure be balanced by either gravitational or curvature
stresses, the relative magnitudes of which are prescribed by the Bond number.
The nondimensionalization scheme will depend on the physical system of interest. Our purpose here
was simply to illustrate the manner in which the dimensionless groups arise in the theoretical formulation
of the problem. Moreover, we see that those involving surface tension enter exclusively through the
boundary conditions.
MIT OCW: 18.357 Interfacial Phenomena
8
Prof. John W. M. Bush
2.6. A few simple examples
Chapter 2. Definition and Scaling of Surface Tension
Figure 2.3: Surface tension may be measured by drawing a thin plate from a liquid bath.
2.6 A few simple examples
Measuring surface tension. Since σ is a tensile force per unit length, it is possible to infer its value by
slowly drawing a thin plate out | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
is a tensile force per unit length, it is possible to infer its value by
slowly drawing a thin plate out of a liquid bath and measure the resistive force (Fig. 2.3). The maximum
measured force yields the surface tension σ.
Curvature/ Laplace pressure: consider an oil drop in water (Fig. 2.4a). Work is required to increase
the radius from R to R + dR:
dW = −podVo − pwdVw + γowdA
(2.14)
where dVo = 4πR2dR = −dVw and dA = 8πRdR.
For mechanical equilibrium, we require
dW = −(p0 − pw)4πR2dR + γow8πRdR = 0 ⇒
ΔP = (po − pw) = 2γow/R.
mech. E
v
surf ace E
' v '
'
'
Figure 2.4: a) An oil drop in water b) When a soap bubble is penetrated by a cylindrical tube, air is
expelled from the bubble by the Laplace pressure.
MIT OCW: 18.357 Interfacial Phenomena
9
Prof. John W. M. Bush
2.6. A few simple examples
Chapter 2. Definition and Scaling of Surface Tension
Note:
1. Pressure inside a drop / bubble is higher than that outside ΔP ∼ 2γ/R ⇒ smaller bubbles have
higher Laplace pressure ⇒ champagne is louder than beer.
Champagne bubbles R ∼ 0.1mm, σ ∼ | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
champagne is louder than beer.
Champagne bubbles R ∼ 0.1mm, σ ∼ 50 dynes/cm, ΔP ∼ 10−2 atm.
2. For a soap bubble (2 interfaces) ΔP =
4σ
R
, so for R ∼ 5 cm, σ ∼ 35dynes/cm have ΔP ∼ 3×10−5atm.
More generally, we shall see that there is a pressure jump across any curved interface:
Laplace pressure Δp = σ∇ · n.
Examples:
1.
Soap bubble jet - Exit speed (Fig. 2.4b)
Force balance: Δp = 4σ/R
4σ
∼ ρairU 2 ⇒ U ∼ ρair R
(
1/2
∼
)
(
4×70dynes/cm
0.001g/cm3·3cm
)
∼ 300cm/s
2.
Ostwald Ripening: The coarsening of foams (or emulsions) owing to diffusion of gas across inter
faces, which is necessarily from small to large bubbles, from high to low Laplace pressure.
3.
Falling drops: Force balance M g ∼ ρairU 2a gives
fall speed U ∼ ρga/ρair.
drop integrity requires ρairU 2 ∼ ρga < σ/a
raindrop size a < ℓc =
If a drop is small relative to the capillary length, σ maintains it against the destabilizing influence
v
of aerodynamic stresses.
σ/ρg ≈ 2mm.
v
2
MIT OCW: 18.357 Interfac | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
.
σ/ρg ≈ 2mm.
v
2
MIT OCW: 18.357 Interfacial Phenomena
10
Prof. John W. M. Bush
3. Wetting
Puddles. What sets their size?
Knowing nothing of surface chemistry, one anticipates that Laplace pressure balances hydrostatic pressure
if σ/H ≥ ρgH ⇒ H < ℓc =
σ/ρg = capillary length.
Note:
J
1. Drops with R < ℓc remain heavily spherical
2. Large drops spread to depth H ∼ ℓc so that
Laplace + hydrostatic pressures balance at the
drop’s edge. A volume V will thus spread
to a radius R s.t. πR2ℓc = V , from which
R = (V /πℓc)
1/2
.
3. This is the case for H2O on most surfaces,
where a contact line exists.
Figure 3.1: Spreading of drops of increasing size.
Note: In general, surface chemistry can dominate and one need not have a contact line.
More generally, wetting occurs at fluid-solid contact. Two possibilities exist: partial wetting or total
wetting, depending on the surface energies of the 3 interfaces (γLV , γSV , γSL).
Now, just as σ = γLV is a surface energy per area or tensile force per length at a liquid-vapour surface,
γSL and γSV are analogous quantities at solid-liquid and solid-vapour interfaces.
The degree of wet | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
γSV are analogous quantities at solid-liquid and solid-vapour interfaces.
The degree of wetting determined by spreading parameters:
S = [Esubstrate]dry − [Esubstrate]wet = γSV − (γSL + γLV )
(3.1)
where only γLV can be easily measured.
Total Wetting: S > 0 , θe = 0 liquid spreads completely in order to minimize its surface energy. e.g.
silicon on glass, water on clean glass.
Note: Silicon oil is more likely to spread than H2O since σw ∼ 70 dyn/cm > σs.o. ∼ 20 dyn/cm. Final
result: a film of nanoscopic thickness resulting from competition between molecular and capillary forces.
Partial wetting: S < 0, θe > 0.
In absence of
g, forms a spherical cap meeting solid at a con
tact angle θe. A liquid is “wetting” on a particular
solid when θe < π/2, non-wetting or weakly wetting
when θe > π/2. For H2O, a surface is hydrophilic
if θe < π/2, hydrophobic if θe > π/2 and superhy
drophobic if θe > 5π/6.
Note: if g = 0, drops always take the form of a spherical cap ⇒ flattening indicates the effects of gravity.
Figure | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
of a spherical cap ⇒ flattening indicates the effects of gravity.
Figure 3.2: The same water drop on hydrophobic
and hydrophilic surfaces.
11
4. Young’s Law with Applications
Young’s Law: what is the equilibrium contact angle θe ? Horizontal force balance at contact line:
γLV cos θe = γSV − γSL
γSV − γSL
γLV
S
γLV
= 1 +
(Y oung 1805)
(4.1)
Note:
cos θe =
1. When S ≥ 0, cos θe ≥ 1 ⇒ θe undefined and
spreading results.
2. Vertical force balance not satisfied at contact
line ⇒ dimpling of soft surfaces.
E.g. bubbles in paint leave a circular rim.
3. The static contact angle need not take its equi
librium value ⇒ there is a finite range of pos
sible static contact angles.
Back to Puddles: Total energy:
Figure 4.1: Three interfaces meet at the contact line.
E = (γSL − γSV )A + γLV A +
ρgh2A
1
2
= −S + ρgV h
V
h
1
2
(4.2)
Minimize energy w.r.t. h:
dE
dh
surf ace energy
v
1
2 + ρgV = 0 when −S/h2 = ρg ⇒
= SV h
grav. pot. energy
v
1
2
1 | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
2 = ρg ⇒
= SV h
grav. pot. energy
v
1
2
1
2
h0 =
J
−2S
ρg = 2ℓc sin θ
e
2 gives puddle depth, where ℓc =
σ/ρg.
J
Capillary Adhesion: Two wetted surfaces can
stick together with great strength if θe < π/2, e.g.
Fig. 4.2.
Laplace Pressure:
ΔP = σ
)
i.e. low P inside film provided θe < π/2.
If H ≪ R, F = πR2 2σ
between the plates.
is the attractive force
1 − cos θe
H/2
R
≈ − 2σ
cos θe
H
cos θe
H
(
Figure 4.2: An oil drop forms a capillary bridge
between two glass plates.
E.g. for H2O, with R = 1 cm, H = 5 µm and θe = 0, one finds ΔP ∼ 1/3 atm and an adhesive force
F ∼ 10N , the weight of 1l of H2O.
Note: Such capillary adhesion is used by beetles in nature.
4.1 Formal Development of Interfacial Flow Problems
Governing Equations: Navier-Stokes. An incompressible, homogeneous fluid of density ρ and viscosity
µ = ρν (µ is dynamic and ν kinematic viscosity) acted upon by an external force per unit | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
µ = ρν (µ is dynamic and ν kinematic viscosity) acted upon by an external force per unit volume f evolves
according to
∇ · u = 0
∂u
∂t
ρ
(
+ u · ∇u = −∇p + f + µ∇2 u
)
(continuity)
(4.3)
(Linear momentum conservation)
(4.4)
12
4.1. Formal Development of Interfacial Flow Problems
Chapter 4. Young’s Law with Applications
This is a system of 4 equations in 4 unknowns (u1, u2, u3, p). These N-S equations must be solved subject
to appropriate BCs.
Fluid-Solid BCs: “No-slip”: u = Usolid.
E.g.1 Falling sphere: u = U on sphere surface, where U is the sphere velocity.
E.g.2 Convection in a box: u = 0 on the box surface.
But we are interested in flows dominated by interfacial effects. Here, in general, one must solve N-S
equations in 2 domains, and match solutions together at the interface with appropriate BCs. Difficulty:
These interfaces are free to move ⇒ Free boundary problems.
Figure 4.3: E.g.3 Drop motion within a fluid.
Figure 4.4: E.g.4 Water waves at an air-water in
terface.
Continuity of Velocity at an interface requires that u = uˆ.
And what about p ? We’ve seen Δp ∼ | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
requires that u = uˆ.
And what about p ? We’ve seen Δp ∼ σ/R for a static bubble/drop, but to answer this question in
general, we must develop stress conditions at a fluid-fluid interface.
Recall: Stress Tensor. The state of stress within an incompressible Newtonian fluid is described by
the stress tensor: T = −pI + 2µE where E =
is the deviatoric stress tensor. The
associated hydrodynamic force per unit volume within the fluid is ∇ · T .
One may thus write N-S eqns in the form: ρ Du = ∇ · T + f = −∇p + µ∇2u + f.
Now: Tij = force / area acting in the ej direction on a surface with a normal ei.
(∇u) + (∇u)T
1
2
Dt
[
]
Note:
1.
normal stresses (diagonals) T11, T22, T33 in
volve both p and ui
2. tangential
stresses (off-diagonals) T12, T13,
etc., involve only velocity gradients, i.e. vis
cous stresses
3.
Tij is symmetric (Newtonian fluids)
4.
t(n) = n·T = stress vector acting on a surface
with normal n
∂u
x
∂y
E.g. Shear flow. Stress in lower boundary is tan
gential. Force | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
E.g. Shear flow. Stress in lower boundary is tan
gential. Force / area on lower boundary:
Tyx = µ
y-surface in x-direction.
Note: the form of T in arbitrary curvilinear coordi- Figure 4.5: Shear flow above a rigid lower bound-
nates is given in the Appendix of Batchelor.
|y=0 = µk is the force/area that acts on
ary.
MIT OCW: 18.357 Interfacial Phenomena
13
Prof. John W. M. Bush
5. Stress Boundary Conditions
Today:
1. Derive stress conditions at a fluid-fluid inter
face. Requires knowledge of T = −pI + 2µE
2. Consider several examples of fluid statics
Recall: the curvature of a string under tension may
support a normal force. (see right)
Figure 5.1: String under tension and the influence
of gravity.
5.1 Stress conditions at a fluid-fluid interface
We proceed by deriving the normal and tangential stress boundary conditions appropriate at a fluid-fluid
interface characterized by an interfacial tension σ.
Figure 5.2: A surface S and bounding contour C on an interface between two fluids. Local unit vectors
are n, m and s.
Consider an interfacial surface S bounded by a closed contour C. One may | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
and s.
Consider an interfacial surface S bounded by a closed contour C. One may think of there being a force
per unit length of magnitude σ in the s-direction at every point along C that acts to flatten the surface S.
Perform a force balance on a volume element V enclosing the interfacial surface S defined by the contour
C:
Du
Dt
ρ
V
dV =
f dV +
V
S∗
[t(n) + tˆ(nˆ)] dS +
σs dℓ
C
(5.1)
Here ℓ indicates arc-length and so dℓ a length increment along the curve C.
t(n) = n · T is the stress vector, the force/area exerted by the upper (+) fluid on the interface.
The stress tensor is defined in terms of the local fluid pressure and velocity field as T = −pI+µ ∇u + (∇u)T
The stress exerted on the interface by the lower (-) fluid is ˆt(nˆ) = nˆ · Tˆ = −n · T
where Tˆ = −pˆI + ˆµ ∇uˆ + (∇uˆ)T
[
.
.
]
[
]
Physical interpretation of terms
ρ Du dV
f dV
body forces acting within V
:
:
V Dt
I
inertial force associated with acceleration of fluid in V
S
S
I
I
I
t(n) dS :
ˆt(nˆ) dS :
σs dℓ :
C
I
hydrodynamic force exerted by upper fluid
hydrodynamic force exert | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
dℓ :
C
I
hydrodynamic force exerted by upper fluid
hydrodynamic force exerted by lower fluid
surface tension force exerted on perimeter.
14
5.1. Stress conditions at a fluid-fluid interface
Chapter 5. Stress Boundary Conditions
Now if ǫ is the characteristic height of our volume V and R its characteristic radius, then the accel-
eration and body forces will scale as R2ǫ, while the surface forces will scale as R2. Thus, in the limit of
ǫ → 0, the latter must balance.
Now we have that
t(n) + tˆ(nˆ) dS +
σs dℓ = 0
Z
C
Z
S
t(n) = n · T ,
ˆ
ˆt(nˆ) = nˆ · T = −n · T
Moreover, the application of Stokes Theorem (see below) allows us to write
where the tangential (surface) gradient operator, defined
σs dℓ =
Z
S
Z
C
∇Sσ − σn (∇S · n) dS
∇
S = [I − nn] · = − n
∇ ∇
∂
∂n
(5.2)
(5.3)
(5.4)
(5.5)
appears because σ and n are only defined on the surface S. We proceed by dropping the subscript s on
∇, with this understanding. The surface force balance thus becomes
ˆ
n · T − n · T
(cid:17)
dS =
Z
S
Z (cid:16)
S
n (∇ · n) − ∇σ dS
σ
(5.6)
Now since the surface S is arbitrary, the integrand must vanish identically. One thus obtains the interfacial
stress balance equation, which is valid at every point on the interface:
Stress Balance Equation
n · T − n · T = σn (∇ · n) − ∇σ
ˆ
(5.7)
Interpretation of terms:
n · T stress (force/area) exerted by + on | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
− ∇σ
ˆ
(5.7)
Interpretation of terms:
n · T stress (force/area) exerted by + on - (will generally have both ⊥ and k components)
ˆn · T stress (force/area) exerted by - on + (will generally have both ⊥ and k components)
σn (∇ · n)
∇σ
normal curvature force per unit area associated with local curvature of interface, ∇ · n
tangential stress associated with gradients in σ
Normal stress balance Taking n·(5.7) yields the normal stress balance
ˆ
n · T · n − n · T · n = σ(∇ · n)
(5.8)
The jump in the normal stress across the interface is balanced by the curvature pressure.
Note: If ∇ · n = 0, there must be a normal stress jump there, which generally involves both pressure and
viscous terms.
MIT OCW: 18.357 Interfacial Phenomena
15
Prof. John W. M. Bush
6
5.2. Appendix A : Useful identity
Chapter 5. Stress Boundary Conditions
Tangential stress balance Taking d·(5.7), where d is any linear combination of s and m (any tangent
to S), yields the tangential stress balance at the interface:
ˆ
n · T · d − n · T · d = ∇σ · d
(5.9)
Physical Interpretation
• LHS represents the jump in tangential components of the hydrodynamic stress at the interface
• RHS represents the tangential stress (Marangoni stress) associated with gradients in σ, as may
result from gradients in temperature θ or chemical composition c at the interface since in general
σ = σ(θ, c)
• LHS contains only the non-diagonal terms of T - only the velocity gradients, not pressure; therefore
any non-zero ∇σ at a fluid interface must always drive motion.
5.2 Appendix A : Useful identity
Recall Stokes Theorem:
Z
Along the contour C, dℓ = m dℓ, so that we have
Z
C
F · dℓ =
F · m dℓ =
Z
C
n · (∇ ∧ F ) dS | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
we have
Z
C
F · dℓ =
F · m dℓ =
Z
C
n · (∇ ∧ F ) dS
S
Z
S
n · (∇ ∧ F ) dS
Now let F = f ∧ b, where b is an arbitrary constant vector. We thus have
(f ∧ b) · m dℓ =
Z
S
Z
C
n · (∇ ∧ (f ∧ b)) dS
Now use standard vector identities to see (f ∧ b) · m = −b · (f ∧ m) and
(5.10)
(5.11)
(5.12)
∇ ∧ (f ∧ b) = f (∇ · b) − b (∇ · f ) + b · ∇f − f · ∇b = −b (∇ · f ) + b · ∇f
(5.13)
since b is a constant vector. We thus have
b ·
Z
C
(f ∧ m) dℓ = b ·
Z
S
[n (∇ · f ) − (∇f ) · n] dS
Since b is arbitrary, we thus have
(f ∧ m) dℓ =
Z
S
Z
C
[n (∇ · f ) − (∇f ) · n] dS
(5.14)
(5.15)
We now choose f = σn, and recall that n ∧ m = −s. One thus obtains
−
[n∇ · (σn) − ∇ (σn) · n] dS = [n∇σ · n + σn (∇ · n) − ∇σ − σ (∇n) · n] dS.
σsdℓ =
S
C
We note that ∇σ · n = 0 since ∇σ must bRe tangent to the surface S and (
(1) = 0, and so obtain the desired result:
R
S
R
1 ∇2
∇n) · n = ∇2
1
(n · n) =
(5.16)
σs dℓ =
Z
S
Z
C
[∇σ − σn (∇ · n)] dS
MIT OCW: 18.357 | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
S
Z
C
[∇σ − σn (∇ · n)] dS
MIT OCW: 18.357 Interfacial Phenomena
16
Prof. John W. M. Bush
5.3. Fluid Statics
Chapter 5. Stress Boundary Conditions
5.3 Fluid Statics
We begin by considering static fluid configurations, for which the stress tensor reduces to the form T = −pI,
so that n · T · n = −p, and the normal stress balance equation (5.8) assumes the simple form:
pˆ − p = σ∇ · n
(5.17)
The pressure jump across a static interface is balanced by the curvature force at the interface. Now since
n · T · d = 0 for a static system, the tangential stress balance indicates that ∇σ = 0. This leads to
the following important conclusion: There cannot be a static system in the presence of surface tension
gradients. While pressure jumps can sustain normal stress jumps across a fluid interface, they do not
contribute to the tangential stress jump. Consequently, tangential surface (Marangoni) stresses can only
be balanced by viscous stresses associated with fluid motion. We proceed by applying equation (5.17) to
describe a number of static situations.
1. Stationary Bubble : We consider a spherical air bubble of radius R submerged in a static fluid.
What is the pressure drop across the bubble surface?
The divergence in spherical coordinates of F = (Fr, Fθ, Fφ) is given by
∇ · F =
Hence ∇ · n | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
F = (Fr, Fθ, Fφ) is given by
∇ · F =
Hence ∇ · n|S =
so the normal stress jump (5.17) indicates that
1
∂
r sin θ ∂θ
r2|r=R =
∂
r sin φ ∂φ
(sin θFθ) +
2
R
r2Fr +
1 ∂
)
r2 ∂r
1 ∂
r2 ∂r
Fφ.
1
(
ΔP = pˆ − p =
2σ
R
(5.18)
The pressure within the bubble is higher than that outside by an amount proportional to the surface
tension, and inversely proportional to the bubble size. As noted in Lec. 2, it is thus that small bubbles
are louder than large ones when they burst at a free surface: champagne is louder than beer. We note
that soap bubbles in air have two surfaces that define the inner and outer surfaces of the soap film;
consequently, the pressure differential is twice that across a single interface.
2. The static meniscus (θe < π/2)
Consider a situation where the pressure within a
static fluid varies owing to the presence of a gravi- meniscus (below).
tational field, p = p0 + ρgz, where p0 is the constant
ambient pressure, and g = −gzˆ is the grav. acceler
ation. The normal stress balance thus requires that
the interface satisfy the Young-Laplace | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
acceler
ation. The normal stress balance thus requires that
the interface satisfy the Young-Laplace Equation:
early with z. Such a situation arises in the static
ρgz = σ∇ · n
(5.19)
The vertical gradient in fluid pressure must be bal
anced by the curvature pressure; as the gradient is
constant, the curvature must likewise increase lin-
Figure 5.3: Static meniscus near a wall.
The shape of the meniscus is prescribed by two factors: the contact angle between the air-water
interface and the wall, and the balance between hydrostatic pressure and curvature pressure. We treat
the contact angle θe as given; noting that it depends in general on the surface energy. The normal
force balance is expressed by the Young-Laplace equation, where now ρ = ρw − ρair ≈ ρw is the density
difference between water and air. We define the free surface by z = η(x); equivalently, we define a
functional f (x, z) = z − η(x) that vanishes on the surface. The normal to the surface z = η(x) is thus
n =
∇f
|∇f |
=
zˆ − η′(x)xˆ
[1 + η′(x)2]
1/2
(5.20)
MIT OCW: 18.357 Interfacial Phenomena
17
Prof. John W. | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
MIT OCW: 18.357 Interfacial Phenomena
17
Prof. John W. M. Bush
5.3. Fluid Statics
Chapter 5. Stress Boundary Conditions
As deduced in Appendix B, the curvature of the free surface ∇ · nˆ , may be expressed as
∇ · nˆ =
−ηxx
(1 + η2)3/2
x
≈ −ηxx
(5.21)
Assuming that the slope of the meniscus remains sufficiently small, η2
x
(5.21), so that (5.19) assumes the form
≪ 1, allows one to linearize equation
ρgη = σηxx
(5.22)
Applying the boundary condition η(∞) = 0 and the contact condition ηx(0) = − cot θ, and solving (5.22)
thus yields
η(x) = ℓc cot θee
−x/ℓc
(5.23)
σ/ρg is the capillary length. The meniscus formed by an object floating in water is exponen-
where ℓc =
tial, decaying over a length scale ℓc. Note that this behaviour may be rationalized as follows: the system
arranges itself so that its total energy (grav. potential + surface) is minimized.
p
3. Floating Bodies
Floating bodies must be supported by some combination of buoyancy and curvature forces. Specifically,
since the fluid pressure beneath the interface is related to the atmospheric pressure p0 above the interface
by
p = p0 + ρgz + σ∇ · n ,
one may express the vertical force balance as
The buoyancy force
M g = z ·
Z
C
−pndℓ = Fb + F
c
.
buo
yancy
|{z}
c
urvature
|{z}
Fb = z ·
Z
C
ρgzn dℓ = ρgVb
is thus simply the weight of the fluid displaced above the object and inside the line of tangency | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
zn dℓ = ρgVb
is thus simply the weight of the fluid displaced above the object and inside the line of tangency (see figure
below). We note that it may be deduced by integrating the curvature pressure over the contact area C
using the first of the Frenet-Serret equations (see Appendix C).
F
c = z ·
Z
C
∇ ·
σ (
n) n dℓ = σz ·
dt
C dℓ
Z
dℓ = σz · (t1 − t2) = 2σ sin θ
(5.27)
At the interface, the buoyancy and curvature forces must balance precisely, so the Young-Laplace relation
is satisfied:
0 = ρgz + σ∇ · n
(5.28)
Integrating this equation over the meniscus and taking the vertical component yields the vertical force
balance:
where
b + F m
F m
c = 0
F m
b = z ·
Z
Cm
ρgzn dℓ = ρgVm
(5.29)
(5.30)
F m
c = z ·
Z
Cm
σ (∇ · n) n dℓ = σz ·
dt
Cm dℓ
Z
dℓ = σz · (t1 − t
2) = −2σ sin θ
(5.31)
where we have again used the Frenet-Serret equations to evaluate the curvature force.
MIT OCW: 18.357 Interfacial Phenomena
18
Prof. John W. M. Bush
(5.24)
(5.25)
(5.26)
5.3. Fluid Statics
Chapter 5. Stress Boundary Conditions
Figure 5.4: A floating non-wetting body is supported by a combination of buoyancy and curvature forces,
whose relative magnitude is prescribed by the ratio of displaced fluid volumes Vb and Vm.
Equations (5.27- | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
the ratio of displaced fluid volumes Vb and Vm.
Equations (5.27-5.31) thus indicate that the curvature force acting on the floating body is expressible in
terms of the fluid volume displaced outside the line of tangency:
Fc = ρgVm
(5.32)
The relative magnitude of the buoyancy and curvature forces supporting a floating, non-wetting body is
thus prescribed by the relative magnitudes of the volumes of the fluid displaced inside and outside the
line of tangency:
Fb
Fc
Vb =
Vm
(5.33)
For 2D bodies, we note that since the meniscus will have a length comparable to the capillary length,
ℓc = (σ/(ρg))
, the relative magnitudes of the buoyancy and curvature forces,
1/2
Fb
Fc
r ≈
ℓc
,
(5.34)
is prescribed by the relative magnitudes of the body size and capillary length. Very small floating objects
(r ≪ ℓc) are supported principally by curvature rather than buoyancy forces. This result has been
extended to three-dimensional floating objects by Keller 1998, Phys. Fluids, 10, 3009-3010.
MIT OCW: 18.357 Interfacial Phenomena
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Prof. John W. M. Bush
5.3. Fluid Statics
Chapter 5. Stress Boundary Conditions
Figure | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
M. Bush
5.3. Fluid Statics
Chapter 5. Stress Boundary Conditions
Figure 5.5: a) Water strider legs are covered with hair, rendering them effectively non-wetting. The
tarsal segment of its legs rest on the free surface. The free surface makes an angle θ with the horizontal,
resulting in an upward curvature force per unit length 2σ sin θ that bears the insect’s weight. b) The
relation between the maximum curvature force Fs = 2σP and body weight Fg = M g for 342 species of
water striders. P = 2(L1 + L2 + L3) is the combined length of the tarsal segments. From Hu, Chan &
Bush; Nature 424, 2003.
4. Water-walking Insects
Small objects such as paper clips, pins or insects may reside at rest on a free surface provided the curvature
force induced by their deflection of the free surface is sufficient to bear their weight (Fig. 5.5a). For example,
for a body of contact length L and total mass M , static equilibrium on the free surface requires that:
M g
2σL sin θ
< 1 ,
(5.35)
where θ is the angle of tangency of the floating body.
This simple criterion is an important geometric constraint on water-walking insects. Fig. 5.5b indicates
the dependence of contact length on body weight | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
-walking insects. Fig. 5.5b indicates
the dependence of contact length on body weight for over 300 species of water-striders, the most common
water walking insect. Note that the solid line corresponds to the requirement (5.35) for static equilibrium.
Smaller insects maintain a considerable margin of safety, while the larger striders live close to the edge.
The maximum size of water-walking insects is limited by the constraint (5.35).
If body proportions were independent of size L, one would expect the body weight to scale as L3 and
the curvature force as L. Isometry would thus suggest a dependence of the form Fc ∼ Fg
, represented
as the dashed line. The fact that the best fit line has a slope considerably larger than 1/3 indicates a
variance from isometry: the legs of large water striders are proportionally longer.
1/3
MIT OCW: 18.357 Interfacial Phenomena
20
Prof. John W. M. Bush
5.4. Appendix B : Computing curvatures
Chapter 5. Stress Boundary Conditions
5.4 Appendix B : Computing curvatures
We see the appearance of the divergence of the surface normal, ∇ · n, in the normal stress balance. We
proceed by briefly reviewing how to formulate this curvature term in two common geometries.
In cartesian coordinates (x, y, z), we consider a surface defined | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
two common geometries.
In cartesian coordinates (x, y, z), we consider a surface defined by z = h(x, y). We define a functional
f (x, y, z) = z − h(x, y) that necessarily vanishes on the surface. The normal to the surface is defined by
n =
∇f
|∇f |
=
and the local curvature may thus be computed:
zˆ − hxxˆ − hyyˆ
1/2
1 + h2 + h2
y
x
)
(
− (hxx + hyy) − hxxhy
∇ · n =
2 + hyyhx
3/2
2 + 2hxhyhxy
)
(
1 + h2 + h2
y
x
)
(
In the simple case of a two-dimensional interface, z = h(x), these results assume the simple forms:
n =
zˆ − hxxˆ
1/2
(1 + h2 )
x
, ∇ · n =
−hxx
(1 + h2 )
x
3/2
Note that n is dimensionless, while ∇ · n has the units of 1/L.
In 3D polar coordinates (r, θ, z), we consider a surface defined by z = h(r, θ). We define a functional
g(r, θ, z) = z − h(r, θ) that vanishes on the surface, and compute the normal:
n =
∇g
|∇g|
=
from which the local curvature is computed:
zˆ − hr ˆ
1 + h2 +r
r − 1 | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
local curvature is computed:
zˆ − hr ˆ
1 + h2 +r
r − 1 ˆhθθ
r
1/2
1 h2
r2 θ
,
(
)
−hθθ − h2hθθ + hrhtheta − rhr
r
∇ · n =
− 2 hrh2 − r2hrr − hrrh2
1/2
r
1 + h2 +r
θ
1
h2
r2 θ
θ + hrhθhrθ
r2
(
)
In the case of an axisymmetric interface, z = h(r), these reduce to:
n =
zˆ − hrrˆ
1/2
(1 + h2)
r
, ∇ · n =
−rhr − r2hrr
3/2
r2 (1 + h2)
r
5.5 Appendix C : Frenet-Serret Equations
(5.36)
(5.37)
(5.38)
(5.39)
(5.40)
(5.41)
ten useful in computing curvature forces on 2D in-
terfaces.
Differential geometry yields relations that are of- Note that the LHS of (5.42) is proportional to the
curvature pressure acting on an interface. Therefore
the net force acting on surface S as a result of cur
vature / Laplace pressures:
σ (∇ · n) n dℓ = σ
dℓ = σ (t2 − t1)
F =
and so the net force on an interface resulting from
curvature pressure can be deduced in terms of the
geometry of the end points.
− (∇ · n) t = | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
uced in terms of the
geometry of the end points.
− (∇ · n) t =
(∇ · n) n =
dt
C dℓ
I
dt
dℓ
dn
dℓ
(5.42)
(5.43)
C
I
MIT OCW: 18.357 Interfacial Phenomena
21
Prof. John W. M. Bush
6. More on Fluid statics
Last time, we saw that the balance of curvature and hydrostatic pressures requires
−ρgη = σ∇ · n = σ
We linearized, assuming ηx ≪ 1, to find η(x). Note: we can integrate directly
−ηxx
(1+η2 )3/2 .
x
ρgηηx = σ
1
2σ
ρgη2
=
Z
ηxηxx
(1 + η2
x)
∞
d
dx
x
ρg
⇒
d
dx
η2
(cid:18) 2 (cid:19)
= σ
d
dx
1
(1 + η2
x)
⇒
1/2
3/2
1
(1 + η2
x)
dx = 1
−
1/2
1
(1 + η2
x)
1/2
= 1
−
sin θ
1
σ sin θ + ρgη2 = σ
2
Figure 6.1: Calculating the shape and maximal rise height of a static meniscus.
Maximal rise height: At z = h we have θ = θe, so from (6.1) 1 ρgh2 = σ(1
2
− sin θe), from which
√
h =
2ℓc(1 − sin θe)1/2
where ℓc =
σ/ρg
Alternative perspective: Consider force balance on the meniscus.
Horizontal force balance:
σ sin θ
+
oj
horiz. pr
|
on of T1 | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
: Consider force balance on the meniscus.
Horizontal force balance:
σ sin θ
+
oj
horiz. pr
|
on of T1
ecti
}
{z
Vertical force balance:
ρgz2
= σ
1
2
hydrostatic suction
| {z }
∞
T2
|{z}
p
At x = 0, where θ = θe, gives σ cos θe = weight of fluid d|
isplaced above z = 0.
σ cos θ
=
ρgzdx
vert. proj. of T
| {z }
1
Z
x
wei
ght
id
of f lu
}
{z
(6.1)
(6.2)
(6.3)
(6.4)
Note: σ cos θe = weight of displaced fluid is +/− according to whether θe is smaller or larger than π .2
Floating Bodies Without considering interfacial effects, one anticipates that heavy things sink and light
things float. This doesn’t hold for objects small relative to the capillary length.
Recall: Archimedean force on a submerged body FA =
In general, the hydrodynamic force acting on a body in a fluid
Fh =
T · ndS, where T = pI + 2µE = pI for static fluid.
−
Here Fh = − pndS =
−ρg
of a body M g = ρBgV if ρF > ρB (fluid density larger than body density); otherwise, it sinks.
−ρg ∇z dV by divergence theorem. This is equal to
dV zˆ = −ρgV zˆ = weight of displaced fluRid. The archimedean force can thus support weight
pndS = ρgVB.
ρgzndS =
−
−
R
R
R
R
S
S
S
S
V
V
R
22
6.1. Capillary forces on floating bodies
Chapter 6. More on Fluid statics
Figure 6.2: | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
on floating bodies
Chapter 6. More on Fluid statics
Figure 6.2: A heavy body may be supported on a fluid surface by a combination of buoyancy and surface
tension.
6.1 Capillary forces on floating bodies
• arise owing to interaction of the menisci of floating bodies
• attractive or repulsive depending on whether the menisci are of the same or opposite sense
• explains the formation of bubble rafts on champagne
• explains the mutual attraction of Cheerios and their attraction to the walls
• utilized in technology for self-assembly on the microscale
Capillary attraction Want to calculate the attractive force between two floating bodies separated by
a distance R. Total energy of the system is given by
Etot = σ
f
dA(R) +
dx
1
0
1
−∞
ρgzdz
∞
h(x)
(6.5)
where the first term in (6.5) corresponds to the total surface energy when the two bodies are a distance
R apart, and the second term is the total gravitational potential energy of the fluid. Differentiating (6.5)
yields the force acting on each of the bodies:
Such capillary forces are exploited by certain water walking insects to climb menisci. By deforming the
free surface, they generate a lateral force that drives them up menisci (Hu & Bush 2005).
F (R) = −
dEtot(R)
dR
(6.6)
MIT | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
& Bush 2005).
F (R) = −
dEtot(R)
dR
(6.6)
MIT OCW: 18.357 Interfacial Phenomena
23
Prof. John W. M. Bush
7. Spinning, tumbling and rolling
drops
7.1 Rotating Drops
We want to find z = h(r) (see right). Normal stress
balance on S:
1
2
2
ΔP + ΔρΩ2 r = σ∇ · n
curvature
"
centrif ugal
"
Nondimensionalize:
Δp + 4B0
′
= ∇ · n,
, Σ =
r
a
(
=
2
′
aΔp
)
σ
3 a
ΔρΩ2
8σ
=
where Δp
= =
Rotational Bond number = const. Define surface
functional: f (r, θ) = z − h(r) ⇒ vanishes on the
surface. Thus
∇f
n = |∇|
and ∇ · n =
zˆ−hr (r)ˆr
(1+h2 (r))1/2
r
2
−rhr −r hrr
)3/2
r2(1+h2
r
centrif ugal
curvature
=
Figure 7.1: The radial profile of a rotating drop.
Brown + Scriven (1980) computed drop shapes and stability for B0 > 0:
1. for
Σ < 0.09, only axisymmetric solutions,
oblate ellipsoids
2. for 0.09 < Σ < 0.31, both axisymmetric and
lobed solutions | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
0.09 < Σ < 0.31, both axisymmetric and
lobed solutions possible, stable
3. for
Σ > 0.31 no stable solution, only lobed
forms
Tektites: centimetric metallic ejecta formed from
spinning cooling silica droplets generated by mete
orite impact.
σ
Δρg
Q1: Why are they so much bigger than raindrops?
From raindrop scaling, we expect ℓc ∼
but
both σ, Δρ higher by a factor of 10 ⇒ large tektite
size suggests they are not equilibrium forms, but
froze into shape during flight.
Q2: Why are their shapes so different from those of
raindrops? Owing to high ρ of tektites, the internal
dynamics (esp. rotation) dominates the aerodynam
ics ⇒ drop shape set by its rotation.
V
Figure 7.2: The ratio of the maximum radius to
the unperturbed radius is indicated as a function of
Σ. Stable shapes are denoted by the solid line, their
metastable counterparts by dashed lines. Predicted
3-dimensional forms are compared to photographs
of natural tektites. From Elkins-Tanton, Ausillous,
Bico, Qu´er´e and Bush (2003).
24
7.2. Rolling drops
Chapter 7. Spinning, tumbling and rolling drops
Light drops: For the case of Σ < 0, ∆ρ < 0, a spinning drop is | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
inning, tumbling and rolling drops
Light drops: For the case of Σ < 0, ∆ρ < 0, a spinning drop is stabilized on axis by centrifugal pressures.
For high |Σ|, it is well described by a cylinder with spherical caps. Drop energy:
E =
IΩ2
1
2
+
2πrLγ
at
ional
{z
|
}
Neglecting the end caps, we write volume V = πr2L and moment of inertia I = ∆mr = ∆ρ π Lr4.
K.E.
ot
R
2
2
2
S f
ur
ce energ
a
| {z }
y
Figure 7.3: A bubble or a drop suspended in a denser fluid, spinning with angular speed Ω.
The energy per unit drop volume is thus E = 1 ∆ρΩ2r2 +
Minimizing with respect to r:
V
4
2γ .
r
2
2
(cid:1)
= 1 ∆ρΩ2r − 2γ = 0, which occurs when r =
d
E
dr V
r
Vonnegut’s Formula: γ = 1 ∆ρΩ2 V
L
as it avoids difficulties associated with fluid-solid contact.
(cid:1)
Note: r grows with σ and decreases with Ω.
3/2
4π
(cid:16)
3/2
4γ
∆ρΩ2
(cid:17)
1
/
3
. Now r = V
πL
/2
1
=
1/3
⇒
4γ
∆ρΩ2
(cid:16)
(cid:17)
(cid:1)
allows inference of γ from L, useful technique for small γ
7.2 Rolling drops
Figure 7.4: A liquid drop rolling down an inclined plane.
(Aussillous and Quere 2003 ) Energetics:
dissipation= 2µ
speed. Stability characteristics different: bioconcave oblate ellipsoids now stable.
for steady descent at speed V, M gV sin θ =Rate of viscous
(∇u)2dV , where | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
oids now stable.
for steady descent at speed V, M gV sin θ =Rate of viscous
(∇u)2dV , where Vd is the dissipation zone, so this sets V ⇒ Ω = V /R is the angular
Vd
R
MIT OCW: 18.357 Interfacial Phenomena
25
Prof. John W. M. Bush
8. Capillary Rise
Capillary rise is one of the most well-known and vivid illustrations of capillarity. It is exploited in a number
of biological processes, including drinking strategies of insects, birds and bats and plays an important role
in a number of geophysical settings, including flow in porous media such as soil or sand.
Historical Notes:
• Leonar do da Vinci (1452 - 1519) recorded the effect in his notes and proposed that mountain streams
may result from capillary rise through a fine network of cracks
• Jacques Rohault (1620-1675): erroneously suggested that capillary rise is due to suppresion of air
circulation in narrow tube and creation of a vacuum
• Geovanni Borelli (1608-1675): demonstrated experimentally that h ∼ 1/r
• Geminiano Montanari (1633-87): attributed circulation in plants to capillary rise
• Francis Hauksbee (1700s): conducted an extensive series of capillary rise experiments reported by
Newton in his Opticks but was left unattributed
• James Jurin (1684-1750): an English physiologist who independently confirmed h | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
in (1684-1750): an English physiologist who independently confirmed h ∼ 1/r; hence
“Jurin’s Law”.
Consider capillary rise in a cylindrical tube of inner radius a (Fig. 8.2)
Recall:
Spreading parameter: S = γSV − (γSL + γLV ).
We now define Imbibition / Impregnation parame
ter:
I = γSV − γSL = γLV cos θ
via force balance at contact line.
Note: in capillary rise, I is the relevant parameter,
since motion of the contact line doesn’t change the
energy of the liquid-vapour interface.
Imbibition Condition: I > 0.
Note: since I = S + γLV , the imbibition condition
I > 0 is always more easily met than the spreading
condition, S > 0
⇒ most liquids soak sponges and other porous me- values of the imbibition parameter I:
dia, while complete spreading is far less common.
I > 0 (left) and I < 0 (right).
Figure 8.1: Capillary rise and fall in a tube for two
26
Chapter 8. Capillary Rise
We want to predict the dependence of rise height H on both tube radius a and wetting properties. We
do so by minimizing the total system energy, specifically the surface and gravitational potential energies.
The energy of the water | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
energy, specifically the surface and gravitational potential energies.
The energy of the water column:
E = (γSL − γSV ) 2πaH +
ρga2πH 2 = −2πaHI +
1
2
ρga2πH 2
1
2
surf ace energy
;
grav.P.E.
;
will be a minimum with respect to H when dE = 0
⇒ H =
, from which we deduce
dH
2 γSV −γSL
ρga
2 I
ρga
=
Jurin’s Law
H = 2
γLV cos θ
ρgr
(8.1)
Note:
1. describes both capillary rise and descent: sign
of H depends on whether θ > π/2 or θ < π/2
2. H increases as θ decreases. Hmax for θ = 0
3. we’ve implicitly assumed R ≪ H & R ≪ lC.
The same result may be deduced via pressure or
force arguments.
By Pressure Argument
Provided a ≪ ℓc, the meniscus will take the form
of a spherical cap with radius R =
. Therefore
cos θ
pA = pB − 2σ
a
2σ cos θ
⇒ H =
ρga
By Force Argument
The weight of the column supported by the tensile
force acting along the contact line:
ρπa2Hg = 2πa (γSV − γSL) = 2πaσ cos θ,
which Jurin’s Law again follows.
cos θ | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
− γSL) = 2πaσ cos θ,
which Jurin’s Law again follows.
cos θ
= p0 − 2σ
a
as previously.
a
cos θ
= p0 − ρgH
from
Figure 8.2: Deriving the height of capillary rise in
a tube via pressure arguments.
MIT OCW: 18.357 Interfacial Phenomena
27
Prof. John W. M. Bush
8.1. Dynamics
8.1 Dynamics
Chapter 8. Capillary Rise
The column rises due to capillary forces, its rise being resisted by a combination of gravity, viscosity, fluid
inertia and dynamic pressure. Conservation of momentum dictates d (m(t) ˙z(t)) = FT OT + ρvv · ndA,
where the second term on the right-hand side is the total momentum flux, which evaluates to πa2ρz˙ = m˙ z˙,
so the force balance on the column may be expressed as
dt
I
S
2
m + ma
Inertia
;
Added mass
;
z¨ = 2πaσ cos θ − mg −
1
πa2 ρz˙
2
2
− 2πaz · τv
(8.2)
capillary f orce
;
weight
;
dynamic pressure
;
viscous f orce
;
2
r
2
2
a
(
where m = πa2zρ. Now assume the flow in the tube is fully developed Poiseuille flow, which will | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
flow in the tube is fully developed Poiseuille flow, which will be
, and F = πa2z˙ is the flux along the
established after a diffusion time τ = ν . Thus, u(r) = 2 ˙z 1 − a
)
tube.
− 4µ
The stress along the outer wall: τν = µ
a
Finally, we need to estimate ma, which will dominate the dynamics at short time. We thus estimate the
change in kinetic energy as the column rises from z to z+Δz. ΔEk = Δ mU 2
, where m = mc+m0+m∞
(mass in the column, in the spherical cap, and all the other mass, respectively). In the column, mc = πa2zρ,
(
a3ρ, u = U . In the outer region, radial inflow extends to ∞, but
u = U . In the spherical cap, m0 =
u(r) decays.
Volume conservation requires: πa2U = 2πa2ur(a) ⇒ ur(a) = U/2.
Continuity thus gives: 2πa2ur(a) = 2πr2ur(r) ⇒ ur(r) = r2
Thus, the K.E. in the far field: 1 m
a
a
ur(a) = 2r
dm, where dm = ρ2πr2dr.
|r=a =
ef f U 2
(r)2
2 U .
1
2
2π
3
du
dr
∞
z˙.
=
)
2
2
ur
∞2
1 | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
1
2
2π
3
du
dr
∞
z˙.
=
)
2
2
ur
∞2
1
2 a
I
Hence
m
ef f
∞
1
U 2
=
1
a
= πρa4
∞
ρ
(
∞ 1
2r2
1
a
a2
2r2
U
2
)
2πr2dr =
1
dr = ρπa3
2
Now
ΔEk =
1
2
1
2
1
2
Δ (mc + m0 + m∞
) U 2
+ m2U ΔU =
1
2
1
= Δmc + mc + m0 + m
2
)
(
+ πa2ρz+ πa3ρ+ πa3ρ U ΔU
U 2
U 2
πa2ρΔz
2U ΔU =
ef f
∞
=
2
3
1
2
(
)
(
)
Figure 8.3: The dynamics of capillary rise.
Substituting for m = πa2zρ, ma = πa3ρ (added mass) and τv =
7
6
− 4µ
a
z˙ into (8.2) we arrive at
z + a z¨ =
7
6
)
(
2σ cos θ
ρa
1
− z˙2 −
2
8µzz˙
ρa2
− gz
(8.3)
The static balance clearly yields the rise height, i.e. Jurin’s Law. But how do we get there?
MIT OCW: 18.357 Interfacial Phenomena
28
Prof. John W. M. Bush
8.1. Dynamics
Inertial Regime | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
. John W. M. Bush
8.1. Dynamics
Inertial Regime
Chapter 8. Capillary Rise
1. the timescale of
flow is τ ∗ = 4a
ν
ary effects to diffuse across the tube
establishment of Poiseuille
, the time required for bound
2
2.
until this time, viscous effects are negligible
and the capillary rise is resisted primarily by Figure 8.4: The various scaling regimes of capillary
fluid inertia
rise.
˙z ∼ 0, so the force balance assumes the form 6
Initial Regime: z ∼ 0,
z(t) =
Once z ≥ a, one must also consider the column mass, and so solve z + a z¨ =
umn accelerates from z˙ = 0,
˙z becomes important, and the force balance becomes: 1 z˙ =
2
6 σ cos θ
7 ρa2
7
6
2σ cos θ
ρa
az¨ =
t2 .
7
6
)
(
2
2σ cos θ
ρa
7
We thus infer
. As the col
2σ cos θ
⇒
ρa
2
4σ cos θ
ρa
1/2
z˙ = U =
(
4σ cos θ
ρa
)
z =
(
is independent of g, µ.
1/2
)
t.
ρa
Viscous Regime (t ≫ τ ∗) Here, inertial effects become negligible, so the force balance assumes the form:
2σ cos θ − | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
τ ∗) Here, inertial effects become negligible, so the force balance assumes the form:
2σ cos θ − 8µzz˙
− gz = 0. We thus infer H − z =
ρa2
∗
Nondimensionalizing: z = z/H, t∗ = t/τ , τ =
1
We thus have ˙z = z ∗ ⇒ dt∗ =
∗
−1
Note: at t ∗ → ∞, z → 1.
8µzz˙
ρga2
8µH
;
ρga2
dz∗ = (−1 − 1−z ∗ )dz∗
= −z − ln(1 − z ∗).
, where H =
2σ cos θ
ρga
∗
z
1−z ∗
⇒ t∗
, ˙z =
ρga
8µ
− 1
H
z
∗
1
)
(
∗
2
Early Viscous Regime: When z ≪ 1, we consider ln(z − 1) = −z ∗ − 1 z ∗2and so infer z = 2t∗ .
∗
∗
∗
2
√
Redimensionalizing thus yields Washburn’s Law: z =
Note that z˙ is independent of g.
[
1/2
σa cos θ
2µ
t
]
Late Viscous Regime: As z approaches H, z ≈ 1. Thus, we consider t∗ = [−z ∗ −ln(1−z ∗)] = ln(1−z ∗)
and so infer z = 1 − exp(−t ∗).
∗
Redimensionalizing yields z = H [1 − exp(−t/τ )], where H = | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
∗
Redimensionalizing yields z = H [1 − exp(−t/τ )], where H =
and τ =
∗
2σ cos θ
ρga
Note: if rise timescale ≪ τ =
overshoot arises, giving rise to oscillations of the water column about its equilibrium height H.
, inertia dominates, i.e. H ≪ Uintertialτ =
)
(
∗
∗
2
4a
ν
8µH
ρga2 .
4σ cos θ
ρa
1/2
2
4a
ν
⇒ inertial
Wicking In the viscous regime, we have 2σ
=
8µzz˙
ρa2 + ρg. What if the viscous stresses dominate
gravity? This may arise, for example, for predomi
nantly horizontal flow (Fig. 8.5).
Force balance:
z ⇒ z =
cos θ
ρa
2σa cos θ
8µ
= zz˙ =
1 d 2
2 dt
t (Washburn’s Law).
1/2 √
∼
(
Note: Front slows down, not due to g, but owing to
increasing viscous dissipation with increasing col-
umn length.
σa cos θ
2µ
t
)
Figure 8.5: Horizontal flow in a small tube.
MIT OCW: 18.357 Interfacial Phenomena
29
Prof. John W. M. Bush
9. Marangoni Flows
Marangoni flows are those driven by surface gradients. In general, surface tension σ depends on both the
temperature and chemical | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
Marangoni flows are those driven by surface gradients. In general, surface tension σ depends on both the
temperature and chemical composition at the interface; consequently, Marangoni flows may be generated
by gradients in either temperature or chemical composition at an interface. We previously derived the
tangential stress balance at a free surface:
n · T · t = −t · ∇σ ,
(9.1)
where n is the unit outward normal to the surface, and t is any unit tangent vector. The tangential
component of the hydrodynamic stress at the surface must balance the tangential stress associated with
gradients in σ. Such Marangoni stresses may result from gradients in temperature or chemical composition
at the interface. For a static system, since n · T · t = 0, the tangential stress balance equation indicates
that: 0 = ∇σ. This leads us to the following important conclusion:
There cannot be a static system in the presence of surface tension gradients.
While pressure jumps can arise in static systems characterized by a normal stress jump across a fluid
interface, they do not contribute to the tangential stress jump. Consequently, tangential surface stresses
can only be balanced by viscous stresses associated with fluid motion.
Thermocapillary flows: Marangoni flows induced by temperature gradients σ(T ).
Note that in general
surface is less energetically unfavourable; therefore, σ is lower.
Approach Through the interfacial BCs (and σ(T )’s appearance therein), N-S equations must be coupled
< 0 Why? A warmer gas phase has more liquid molecules, so | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
equations must be coupled
< 0 Why? A warmer gas phase has more liquid molecules, so the creation of
dσ
dT
Figure 9.1: Surface tension of a gas-liquid interface decreases with temperature since a warmer gas phase
contains more suspended liquid molecules. The energetic penalty of a liquid molecule moving to the
interface is thus decreased.
to the heat equation
∂T
∂t
+ u · ∇T = κ∇2T
(9.2)
30
Note:
1. the heat equation must be solved subject to appropriate BCs at the free surface. Doing so can be
complicated, especially if the fluid is evaporating.
Chapter 9. Marangoni Flows
2. Analysis may be simplified when the Peclet number Pe = ≪ 1. Nondimensionalize (9.2):
U a
κ
′
x = ax , t =
t ′ , u = U u to get
′
a
U
Pe
′
∂T
∂t′
(
+ u · ∇ ′ T = ∇2T
′
′
)
′
(9.3)
Note:
Pe = Re · Pr =
The Prandtl number Pr = O(1) for many common (e.g. aqeous) fluids.
E.g.1 Thermocapillary flow in a slot (Fig.9.2a)
Surface Tangential BCs τ =
· ≪ 1 if Re ≪ 1, so one has ∇2T = 0.
ν
κ
viscous stress U ∼ 1 H Δσ.
≈ µ
U a
ν
=
Δσ
L
dσ ΔT
dT L
µ L
U
H | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
≈ µ
U a
ν
=
Δσ
L
dσ ΔT
dT L
µ L
U
H
Figure 9.2: a) Thermocapillary flow in a slot b) Thermal convection in a plane layer c) Thermocapillary
drop motion.
E.g.2 Thermocapillary Drop Motion (Young, Goldstein & Block 1962)
can trap bubbles in gravitational field via thermocapillary forces. (Fig.9.2c).
E.g.3 Thermal Marangoni Convection in a Plane Layer (Fig.9.2b).
Consider a horizontal fluid layer heated from below. Such a layer may be subject to either buoyancy- or
Marangoni-induced convection.
Recall: Thermal buoyancy-driven convection (Rayleigh-Bernard) ρ(T ) = ρ0 (1 + α(T − T0)), where α is
the thermal expansivity. Consider a buoyant blob of characteristic scale d. Near the onset of convection,
one expects it to rise with a Stokes velocity U ∼ gΔρ d
. The blob will rise, and so convection
ν
will occur, provided its rise time τrise = =
is less than the time required for it to lose its heat
and buoyancy by diffusion, τdif f =
dν
gαΔT d2
gαΔT d
ν
=
d
U
ρ
2
2
2
d
.κ
Criterion for Instability:
τdif f
τrise
3
∼
gαΔT d
κν
≡ Ra > Rac ∼ 103, where Ra is the Rayleigh number.
Note: for Ra < Rac, heat is transported solely through diffusion, so the layer remains static. For
Ra > Rac, convection arises.
The subsequent behaviour depends on Ra and Pr. Generally, as Ra increases | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
static. For
Ra > Rac, convection arises.
The subsequent behaviour depends on Ra and Pr. Generally, as Ra increases, steady convection rolls ⇒
time-dependency ⇒ chaos ⇒ turbulence.
MIT OCW: 18.357 Interfacial Phenomena
31
Prof. John W. M. Bush
Chapter 9. Marangoni Flows
Thermal Marangoni Convection
Arises because of the dependence of σ on temperature: σ(T ) = σ0 − Γ(T − T0)
Mechanism:
• Imagine a warm spot on surface ⇒ prompts surface divergence ⇒ upwelling.
• Upwelling blob is warm, which reinforces the perturbation provided it rises before losing its heat via
diffusion.
• Balance Marangoni and viscous stress:
Δσ
d
∼
µU
d
• Rise time: ∼
d
U
µd
Δσ
• Diffusion time τdif f = d
κ
2
Criterion for instability:
Note:
τdif f ∼ ΓΔT d
τrise
µκ
≡ Ma > Mac, where Ma is the Marangoni number.
1. Since Ma ∼ d and Ra ∼ d3, thin layers are most unstable to Marangoni convection.
2. B´enard’s original experiments performed in millimetric layers of spermaceti were visualizing Marangoni
convection, but were misinterpreted by Rayleigh as being due to buoyancy ⇒ not recognized until
Block (Nature 1956).
3. Pearson (1958) performed stability analysis with flat surface ⇒ deduced Mac = 80 .
4. Scriven & Sternling (1964) considered a deformable interface, which renders the system unstable at
all Ma. Downwelling beneath peaks in Marangoni convection, upwelling between peaks in Rayleigh
B´enard convection (Fig | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
. Downwelling beneath peaks in Marangoni convection, upwelling between peaks in Rayleigh
B´enard convection (Fig. 9.3a).
5. Smith (1966) showed that the destabilizing influence of the surface may be mitigated by gravity.
< ρgd ⇒ thin layers prone to instability.
Stability Criterion:
dσ dT
dT dz
2
3
MIT OCW: 18.357 Interfacial Phenomena
32
Prof. John W. M. Bush
E.g.4 Marangoni Shear Layer (Fig. 9.3)
Lateral ∇θ leads to Marangoni stress ⇒ shear flow. The resulting T (x, y) may destabilize the layer to
Chapter 9. Marangoni Flows
Figure 9.3: a) Marangoni convection in a shear layer may lead to transverse surface waves or streamwise
rolls (c). Surface deflection may accompany both instabilities (b,d).
Marangoni convection.
Smith & Davis (1983ab) considered the case of flat free surface. System behaviour depends on Pr = ν/κ.
Low Pr: Hydrothermal waves propagate in direction of τ .
High Pr: Streamwise vortices (Fig. 9.3c).
Hosoi & Bush (2001) considered a deformable free surface (Fig. 9.3d)
E.g.5 Evaporatively-driven convection
e.g. for an alcohol-H2O solution, evaporation affects both the alcohol concentration c and temperature θ.
The density ρ(c, θ) and surface tension σ(c, θ) are such that ∂ρ < 0, ∂ρ < 0, dσ < 0, dσ < 0. Evaporation
results in surface cooling and so may | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
0, dσ < 0, dσ < 0. Evaporation
results in surface cooling and so may generate either Rayleigh-B´enard or Marangoni thermal convection.
Since it also induces a change in surface chemistry, it may likewise generate either Ra − B or Marangoni
chemical convection.
∂θ
∂c
dθ
dc
E.g.6 Coffee Drop
Marangoni flows are responsible for the ring-like stain left by a
coffee drop.
• coffee grounds stick to the surface
• evaporation leads to surface cooling, which is most pro
nounced near the edge, where surface area per volume ratio
is highest
• resulting thermal Marangoni stresses drive radial outflow
on surface ⇒ radial ring
Figure 9.4: Evaporation of water from
a coffee drop drives a Marangoni flow.
MIT OCW: 18.357 Interfacial Phenomena
33
Prof. John W. M. Bush
10. Marangoni Flows II
10.1 Tears of Wine
The first Marangoni flow considered was the tears of wine phenomenon (Thomson 1885 ), which actually
predates Marangoni’s first published work on the subject by a decade. The tears of wine phenomenon
is readily observed in a wine glass following the establishment of a thin layer of wine on the walls of the
glass.
An illustration of the tears of wine phenomenon is shown in Fig. 10.1. Evaporation of alcohol occurs
everywhere along the free surface. The alcohol concentration in the thin layer is thus reduced relative to
that in the bulk owing to the enhanced surface area to | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
in the thin layer is thus reduced relative to
that in the bulk owing to the enhanced surface area to volume ratio. As surface tension decreases with
alcohol concentration, the surface tension is higher in the thin film than the bulk; the associated Marangoni
stress drives upflow throughout the thin film. The wine climbs until reaching the top of the film, where
it accumulates in a band of fluid that thickens until eventually becoming gravitationally unstable and
releasing the tears of wine. The tears or “legs” roll back to replenish the bulk reservoir, but with fluid
that is depleted in alcohol.
The flow relies on the transfer of chemical potential energy to kinetic and ultimately gravitational
potential energy. The process continues until the fuel for the process, the alcohol is completely depleted.
For certain liquors (e.g. port), the climbing film, a Marangoni shear layer, goes unstable to streamwise
vortices and an associated radial corrugation - the “tear ducts of wine” (Hosoi & Bush, JFM 2001). When
the descending tears reach the bath, they appear to recoil in response to the abrupt change in σ. The
tears or legs of wine are taken by sommeliers to be an indicator of the quality of wine.
Figure 10.1: The tears of wine. Fluid is drawn from the bulk up the thin film adjoining the walls of the
glass by Marangoni stresses induced by evaporation of alcohol from the free surface.
34
10.2. Surfactants
Chapter 10. Marangoni Flows II
Figure 10.2: | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
10.2. Surfactants
Chapter 10. Marangoni Flows II
Figure 10.2: a) A typical molecular structure of surfactants. b) The typical dependence of σ on surfactant
concentration c.
10.2 Surfactants
Surfactants are molecules that have an affinity for interfaces; common examples include soap and oil.
Owing to their molecular structure (e.g. a hydrophilic head and hydrophobic tail, Fig. 10.2a), they
find it energetically favourable to reside at the free surface. Their presence reduces the surface tension;
consequently, gradients in surfactant concentration Γ result in surface tension gradients. Surfactants thus
generate a special class of Marangoni flows. There are many different types of surfactants, some of which
are insoluble (and so remain on the surface), others of which are soluble in the suspending fluid and
so diffuse into the bulk. For a wide range of common surfactants, surface tension is a monotonically
decreasing function of Γ until a critical micell concentration (CMC) is achieved, beyond CMC there is no
further dependence of σ on Γ (Fig. 10.2b).
Surfactant properties:
• Diffusivity prescribes the rate of diffusion, Ds (bulk diffusivity Db), of a surfactant along an
interface
• Solubility prescribes the ease with which surfactant passes from the surface to the bulk. An
insoluble surfactant cannot dissolve into the bulk, must remain on the surface.
• | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
surface to the bulk. An
insoluble surfactant cannot dissolve into the bulk, must remain on the surface.
• Volatility prescribes the ease with which surfactant sublimates.
Theoretical Approach: because of the dependence of σ on the surfactant concentration, and the ap
pearance of σ in the boundary conditions, N-S equations must be augmented by an equation governing
the evolution of Γ. In the bulk,
∂c
∂t
+ u · ∇c
= Db∇2
c
(10.1)
(10.2)
The concentration of surfactant Γ on a free surface evolves according to
∂Γ
∂t
+ ∇s · (Γus) + Γ (∇s · n) (u · n) = J (Γ, Cs) + Ds
∇2
s
Γ
where us is the surface velocity, ∇s is the surface gradient operator and Ds is the surface diffusivity of
the surfactant (Stone 1999). J is a surfactant source term associated with adsorption onto or desorption
from the surface, and depends on both the surface surfactant concentration Γ and the concentration in the
bulk Cs. Tracing the evolution of a contaminated free surface requires the use of Navier-Stokes equations,
relevant boundary conditions and the surfactant evolution equation (10.2). The dependence of surface
tension on surfactant concentration, σ(Γ), requires the coupling of the flow field and surfactant field. In
certain special cases, the system may be made more tractable. For example, for insoluble surfactants,
J = 0. Many surfactants have sufficiently small D | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
example, for insoluble surfactants,
J = 0. Many surfactants have sufficiently small Ds that surface diffusivity may be safely neglected.
MIT OCW: 18.357 Interfacial Phenomena
35
Prof. John W. M. Bush
10.2. Surfactants
Chapter 10. Marangoni Flows II
Figure 10.3: The footprint of a whale, caused by the whales sweeping biomaterial to the sea surface. The
biomaterial acts as a surfactant in locally suppressing the capillary waves evident elsewhere on the sea
surface. Observed in the wake of a whale on a Whale Watch from Boston Harbour.
Special case: expansion of a spherical surfactant-laden bubble.
∂Γ
dR
∂t
R dt
the surfactant is conserved.
+ Γ (∇ · n) ur = 0. Here ∇ · n = 2/R, ur =
dΓ
so + Γ 2
dt
dR
dt
= 0 ⇒ dΓ
Γ
= −2 dR
R
4πR2Γ =const., so
Marangoni Elasticity The principal dynamical influence of surfactants is to impart an effective
elasticity to the interface. One can think of a clean interface as a “slippery trampoline” that resists
deformation through generation of normal curvature pressures. However, such a surface cannot generate
traction on the interface. However, a surface-laden interface, like a trampoline, resists surface deformation
as does a clean interface, but can also support tangential stresses via Marangoni elasticity. Specifically,
the presence of surfactants will serve not only to alter the normal stress | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
fically,
the presence of surfactants will serve not only to alter the normal stress balance (through the reduction
of σ), but also the tangential stress balance through the generation of Marangoni stresses.
The presence of surfactants will act to suppress any fluid motion characterized by non-zero surface
divergence. For example, consider a fluid motion characterized by a radially divergent surface motion.
The presence of surfactants results in the redistribution of surfactants: Γ is reduced near the point of
divergence. The resulting Marangoni stresses act to suppress the surface motion, resisting it through an
effective surface elasticity. Similarly, if the flow is characterized by a radial convergence, the resulting
accumulation of surfactant in the region of convergence will result in Marangoni stresses that serve to
resist it. It is this effective elasticity that gives soap films their longevity: the divergent motions that
would cause a pure liquid film to rupture are suppressed by the surfactant layer on the soap film surface.
The ability of surfactant to suppress flows with non-zero surface divergence is evident throughout
the natural world. It was first remarked upon by Pliny the Elder, who rationalized that the absence of
capillary waves in the wake of ships is due to the ships stirring up surfactant. This phenomenon was also
widely known to spear-fishermen, who poured oil on the water to increase their ability | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
ely known to spear-fishermen, who poured oil on the water to increase their ability to see their prey,
and by sailors, who would do similarly in an attempt to calm troubled seas. Finally, the suppression of
capillary waves by surfactant is at least partially responsible for the ‘footprints of whales’ (see Fig. 10.3).
In the wake of whales, even in turbulent roiling seas, one seas patches on the sea surface (of characteristic
width 5-10m) that are perfectly flat. These are generally acknowledged to result from the whales sweeping
biomaterial to the surface with their tails, this biomaterial serving as a surfactant that suppresses capillary
waves.
MIT OCW: 18.357 Interfacial Phenomena
36
Prof. John W. M. Bush
10.2. Surfactants
Chapter 10. Marangoni Flows II
Surfactants and a murder mystery. From Nature, 22, 1880 :
“In the autumn of 1878 a man committed a terrible crime in Somerset, which was for some time involved
in deep mystery. His wife, a handsome and decent mulatto woman, disappeared suddenly and entirely from
sight, after going home from church on Sunday, October 20. Suspicion immediately fell upon the husband,
a clever young fellow of about thirty, but no trace of the missing woman was left behind, and there seemed
a strong probability that the crime would remain undetected. On Sunday, however, October 27, a week
after the woman had disappeared, some Som | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
Sunday, however, October 27, a week
after the woman had disappeared, some Somerville boatmen looking out towards the sea, as is their custom,
were struck by observing in the Long Bay Channel, the surface of which was ruffled by a slight breeze, a
streak of calm such as, to use their own illustration, a cask of oil usually diffuses around it
when in the water. The feverish anxiety about the missing woman suggested some strange connection
between this singular calm and the mode of her disappearance. Two or three days after - why not sooner
I cannot tell you - her brother and three other men went out to the spot where it was observed, and from
which it had not disappeared since Sunday, and with a series of fish-hooks ranged along a long line dragged
the bottom of the channel, but at first without success. Shifting the position of the boat, they dragged a
little further to windward, and presently the line was caught. With water glasses the men discovered that
it had caught in a skeleton which was held down by some heavy weight. They pulled on the line; something
suddenly gave was, and up came the skeleton of the trunk, pelvis, and legs of a human body, from which
almost every vestige of flesh had disappeared, but which | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
human body, from which
almost every vestige of flesh had disappeared, but which, from the minute fragments remaining, and the
terrible stench, had evidently not lain long in the water. The husband was a fisherman, and Long Bay
Channel was a favourite fishing ground, and he calculated, truly enough, that the fish would very soon
destroy all means of identification; but it never entered into his head that as they did so their ravages,
combined with the process of decomposition, set free the matter which was to write the traces of
his crime on the surface of the water. The case seems to be an exceedingly interesting one; the calm
is not mentioned in any book on medical jurisprudence that I have, and the doctors seem not to have had
experience of such an occurrence. A diver went down and found a stone with a rope attached, by which
the body had been held down, and also portions of the scalp and of the skin of the sole of the foot, and of
clothing, by means of which the body was identified. The husband was found guilty and executed.”
MIT OCW: 18.357 Interfacial Phenomena
37
Prof. John W. M. Bush
10.3. Surfactant-induced Marangoni flows
Chapter 10. Marangoni Flows II
Figure 10.4: The soap boat. | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
Marangoni flows
Chapter 10. Marangoni Flows II
Figure 10.4: The soap boat. A floating body (length 2.5cm) contains a small volume of soap, which serves
as its fuel in propelling it across the free surface. The soap exits the rear of the boat, decreasing the local
surface tension. The resulting fore-to-aft surface tension gradient propels the boat forward. The water
surface is covered with Thymol blue, which parts owing to the presence of soap, which is thus visible as a
white streak.
10.3 Surfactant-induced Marangoni flows
1. Marangoni propulsion
Consider a floating body with perimeter C in contact with the free surface, which we assume for the sake
of simplicity to be flat. Recalling that σ may be though of as a force per unit length in a direction tangent
to the surface, we see that the total surface tension force acting on the body is:
Fc =
�
C
σsdℓ
(10.3)
where s is the unit vector tangent to the free surface and normal to C, and dℓ is an increment of arc
length along C. If σ is everywhere constant, then this line integral vanishes identically by the divergence
theorem. However, if σ = σ(x), then it may result in a net propulsive force. The ‘soap boat’ may be
simply generated by coating one end of a toothpick with soap, which acts to reduce surface tension (see
right). The concomitant gradient in surface tension results in a net propulsive force that drives the boat
away from the soap. We note that an analogous Marangoni propulsion technique arises in the natural
world: certain | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
We note that an analogous Marangoni propulsion technique arises in the natural
world: certain water-walking insects eject surfactant and use the resulting surface tension gradients as
an emergency mechanism for avoiding predation. Moreover, when a pine needle falls into a lake or pond,
it is propelled across the surface in an analogous fashion owing to the influence of the resin at its base
decreasing the local surface tension.
2. Soap film stability
Pinching a film increases the surface area, decreases
Γ and so increases σ. Fluid is thus drawn in and
the film is stabilized by the Marangoni elasticity.
Figure 10.5: Fluid is drawn to a pinched area of a
soap film through induced Marangoni stresses.
MIT OCW: 18.357 Interfacial Phenomena
38
Prof. John W. M. Bush
10.4. Bubble motion
Chapter 10. Marangoni Flows II
3. Vertical Soap Film
• Vertical force balance: ρgh(z) = dσ . The weight of a soap
dz
film is supported by Marangoni stress.
• Internal dynamics: note that film is dynamic (as are all
Marangoni flows), if it were static, its max height would
be ℓc. It is constantly drying due to the influence of gravity.
• On the surface:
cous stresses.
dσ
dz
∼ µ
du
dx
balance of Marangoni and vis
2 u
• Inside: ρg ∼ µ dx2 Gravity-viscous.
d
10.4 Bubble motion
Figure 10.6: The | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
� µ dx2 Gravity-viscous.
d
10.4 Bubble motion
Figure 10.6: The weight of a vertical
soap film is supported by Marangoni
stresses on its surface.
Theoretical predictions for the rise speed of small drops or bub
bles do not adequately describe observations. Specifically, air bubbles rise at low Reynolds number at
rates appropriate for rigid spheres with equivalent buoyancy in all but the most carefully cleaned fluids.
This discrepancy may be rationalized through consideration of the influence of surfactants on the surface
dynamics. The flow generated by a clean spherical bubble or radius a rising at low Re = U a/ν is in
tuitively obvious. The interior flow is toroidal, while the surface motion is characterized by divergence
and convergence at, respectively, the leading and trailing surfaces. The presence of surface contamination
changes the flow qualitatively.
The effective surface elasticity imparted by the surfactants acts to suppress the surface motion. Sur
factant is generaly swept to the trailing edge of the bubble, where it accumulates, giving rise to a local
decrease in surface tension. The resulting for-to-aft surface tension gradient results in a Marangoni stress
that resists surface motion, and so rigidifies the bubble surface. The air bubble thus moves as if its surface
were stagnant, and it is thus that its rise speed is commensurate with that predicted for a rigid sphere: the
no-slip boundary condition is more appropriate than the free-slip. Finally, we note that the characteristic
Marangoni stress Δσ/a is most pronounced for small bubbles. It is thus that | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
-slip. Finally, we note that the characteristic
Marangoni stress Δσ/a is most pronounced for small bubbles. It is thus that the influence of surfactants
is greatest on small bubbles.
Figure 10.7: A rising drop or bubble (left) is marked by internal circulation in a clean system that is
absent in a contaminated, surfactant-laden fluid (right). Surfactant sticks to the surface, and the induced
Marangoni stress rigidifies the drop surface.
MIT OCW: 18.357 Interfacial Phenomena
39
Prof. John W. M. Bush
11. Fluid Jets
11.1 The shape of a falling fluid jet
Consider a circular orifice of a radius a ejecting a flux Q of fluid density ρ and kinematic viscosity ν
(see Fig. 11.1). The resulting jet accelerates under the influence of gravity −gzˆ. We assume that the jet
Reynolds number Re = Q/(aν) is sufficiently high that the influence of viscosity is negligible; furthermore,
we assume that the jet speed is independent of radius, and so adequately described by U (z). We proceed
by deducing the shape r(z) and speed U (z) of the evolving jet.
Applying Bernoulli’s Theorem at points A and B:
ρU 2 + ρgz + PA =
0
1
2
ρU 2(z) + PB
1
2
(11.1)
The local curvature of slender threads may be expressed in terms of
the two principal radii of curvature, R1 and R2 | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
be expressed in terms of
the two principal radii of curvature, R1 and R2:
∇ · n =
1
1
+
R1 R2
≈
1
r
(11.2)
Thus, the fluid pressures within the jet at points A and B may be
simply related to that of the ambient, P0:
PA ≈ P0 +
σ
a
,
PB ≈ P0 +
σ
r
Substituting into (11.1) thus yields
1
2
σ
ρU 2 + ρgz + P0 + = ρU 2(z) + P0 +
a
1
2
0
σ
r
from which one finds
1/2
U (z)
U0
1 +
=
2 z
Fr a
acc. due to g
+
1 −
2
a
We
r
dec. due to σ
(
)
�
where we define the dimensionless groups
�
(11.3)
(11.4)
(11.5)
Fr =
We =
=
U 2
0
ga
2a
ρU0
σ
IN ERT IA
GRAV IT Y
=
IN ERT IA
CU RV AT U RE
Now flux conservation requires that
= F roude N umber
(11.6)
= W eber N umber
(11.7)
r
Q = 2π
0
from which one obtains
U (z)r(z)dr = πa2U0 = πr2U (z)
(11.8) | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
(z)r(z)dr = πa2U0 = πr2U (z)
(11.8)
r(z)
a
U0
U (z)
)
=
(
1/2
= 1 +
[
2 z
2
+
Fr a We
1 −
(
a
r
)]
Figure 11.1: A fluid jet extruded
from an orifice of radius a ac
celerates under the influence of
Its shape is influenced
gravity.
both by the gravitational acceler
ation g and the surface tension σ.
Note that σ gives rise to a gradi-
ent in curvature pressure within
the jet, σ/r(z), that opposes the
acceleration due to g.
−1/4
(11.9)
This may be solved algebraically to yield the thread shape r(z)/a, then this result substituted into (11.5)
to deduce the velocity profile U (z). In the limit of We → ∞, one obtains
r
a
1 +
=
(
2gz
2
U0
)
−1/4
,
U (z)
U0
1 +
=
(
2gz
2
U0
)
1/2
(11.10)
40
11.2. The Plateau-Rayleigh Instability
Chapter 11. Fluid Jets
11.2 The Plateau-Rayleigh Instability
We here summarize the work of Plateau and Rayleigh on the instability
of cylindrical fluid jets bound by surface tension. It is precisely this
Rayleigh-Plateau instability that is responsible for the pinch-off of thin
water jets emerging from kitchen taps (see Fig. 11.2).
The | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
that is responsible for the pinch-off of thin
water jets emerging from kitchen taps (see Fig. 11.2).
The equilibrium base state consists of an infinitely long quiescent
cylindrical inviscid fluid column of radius R0, density ρ and surface
tension σ (see Fig. 11.3). The influence of gravity is neglected. The
pressure p0 is constant inside the column and may be calculated by
balancing the normal stresses with surface tension at the boundary.
Assuming zero external pressure yields
p0 = σ∇ · n ⇒ p0 =
σ
R0
.
(11.11)
We consider the evolution of infinitesimal varicose perturbations on
the interface, which enables us to linearize the governing equations.
The perturbed columnar surface takes the form:
,
˜ R0 + ǫeωt+ikz
R =
(11.12)
where the perturbation amplitude ǫ ≪ R0, ω is the growth rate of
the instability and k is the wave number of the disturbance in the z-
direction. The corresponding wavelength of the varicose perturbations
is necessarily 2π/k. We denote by ˜ur the radial component of the
perturbation velocity, ˜uy the axial component, and ˜p the perturbation
pressure. Substituing these perturbation fields into the N-S equations
and retaining terms only to order ǫ yields:
(11. | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
into the N-S equations
and retaining terms only to order ǫ yields:
(11.13) Figure 11.2: The capillary-driven
thread
instability of a water
falling under
influence of
the
gravity. The initial jet diameter
is approximately 3 mm.
(11.14)
∂u˜r
∂t
∂u˜z
∂t
= −
= −
1 ∂p˜
ρ ∂r
1 ∂p˜
ρ ∂z
The linearized continuity equation becomes:
+
+
= 0
u˜r
r
∂u˜r
∂r
∂u˜z
∂z
We anticipate that the disturbances in velocity and
pressure will have the same form as the surface dis
turbance (11.12), and so write the perturbation ve
locities and pressure as:
(11.15)
.
(˜ur, u˜z, p˜) = R(r), Z(r), P (r) e
)
(
ωt+ikz
.
(11.16)
Substituting (11.16) into equations (11.13-11.15)
yields the linearized equations governing the per
turbation fields:
Momentum equations : ωR = −
1 dP
ρ dr
ik
ρ
P
ωZ = −
Continuity:
dR R
r
dr
+ + ikZ = 0
.
(11.17)
(11.18) Figure 11.3: A cylindrical column of initial radius
R0 comprised of an inviscid fluid of density ρ, | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
of initial radius
R0 comprised of an inviscid fluid of density ρ, bound
by surface tension σ.
(11.19)
MIT OCW: 18.357 Interfacial Phenomena
41
Prof. John W. M. Bush
11.2. The Plateau-Rayleigh Instability
Chapter 11. Fluid Jets
Eliminating Z(r) and P (r) yields a differential equation for R(r):
2 d2R
dr2
r
+ r
dR
dr
− 1 + (kr)2
R = 0
.
(11.20)
(
This corresponds to the modified Bessel Equation of order 1, whose solutions may be written in terms of
the modified Bessel functions of the first and second kind, respectively, I1(kr) and K1(kr). We note that
K1(kr) → ∞ as r → 0; therefore, the well-behavedness of our solution requires that R(r) take the form
)
R(r) = CI1(kr)
,
(11.21)
where C is an as yet unspecified constant to be determined later by application of appropriate boundary
conditions. The pressure may be obtained from (11.21) and (11.17), and by using the Bessel function
identity I 0(ξ) = I1(ξ):
′
P (r) = −
ωρC
k
I0(kr) and Z(r) = − P (r).
ik | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
= −
ωρC
k
I0(kr) and Z(r) = − P (r).
ik
ωρ
(11.22)
We proceed by applying appropriate boundary conditions. The first is the kinematic condition on the free
surface:
∂R˜
∂t
= u˜ · n ≈ u˜r
.
Substitution of (11.21) into this condition yields
C =
ǫω
I1(kR0)
.
Second, we require a normal stress balance on the free surface:
p0 + ˜p = σ∇ · n
We write the curvature as σ∇ · n =
the jet surface:
1
R1
(
=
1
R1
1+
R2
, where R1 and R2 are the principal radii of curvature of
)
1
R0 + ǫeωt+ikz
1
R2
= ǫk2 e
≈
1
R0
−
ǫ ωt+ikz
R2
0
e
ωt+ikz
.
Substitution of (11.26) and (11.27) into equation (11.25) yields:
p0 + ˜p =
σ
R0
−
ǫσ
R2 1 − k2R0
0
(
ωt+ikz
2 e
)
Cancellation via (11.11) yields the equation for p˜ accurate to order ǫ:
p˜ = −
ǫσ
R2
0
1 − k2R2 e
(
)
0
ωtikz
.
Combining (11.22), (11.24) and (11.29) yields the dispersion relation, that indicates the dependence of
the growth rate ω on the wavenumber k:
ω2
= | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
relation, that indicates the dependence of
the growth rate ω on the wavenumber k:
ω2
=
σ
ρR3
0
kR0
I1(kR0)
I0(kR0)
We first note that unstable modes are only possible when
kR0 < 1
1 − k2R2
0
(
)
(11.30)
(11.31)
MIT OCW: 18.357 Interfacial Phenomena
42
Prof. John W. M. Bush
(11.23)
(11.24)
(11.25)
(11.26)
(11.27)
(11.28)
(11.29)
11.3. Fluid Pipes
Chapter 11. Fluid Jets
The column is thus unstable to disturbances whose
wavelengths exceed the circumference of the cylinder.
A plot of the dependence of the growth rate ω on the
is
for the Rayleigh-Plateau
wavenumber k
shown in Fig. 11.4.
The fastest growing mode occurs for kR0 = 0.697, i.e.
when the wavelength of the disturbance is
instability
λmax ≈ 9.02R0
(11.32)
By inverting the maximum growth rate ωmax one may
estimate the characteristic break-up time:
tbreakup ≈ 2.91
�
3
ρR0
σ
(11.33)
Figure 11.4: The dependence of the growth
rate ω on the wavenumber k for the | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
.4: The dependence of the growth
rate ω on the wavenumber k for the Rayleigh
Plateau instability.
Note: In general, pinch-off depends on Oh =
σR . µν
1/2
2
(
)
, λ = 9.02R.
At low Oh, we have seen that τpinch ∼ ρR
σ
At high Oh, when viscosity is important, τpinch ∼ µR , λ increases with µ.
σ
A water jet of diameter 1cm has a characteristic break-up time of about 1/8s, which is consistent with
casual observation of jet break-up in a kitchen sink.
Related Phenomena: Waves on jets
When a vertical water jet impinges on a horizontal reservoir of water, a field of standing waves may be
excited on the base of the jet (see Fig. 11.5). The wavelength is determined by the requirement that the
wave speed correspond to the local jet speed: U = −ω/k. Using our dispersion relation (11.30) thus yields
U 2 =
ω2
k2
=
σ I1(kR0)
ρkR2 I0(kR0)
0
1 − k2R2
0
(
)
(11.34)
Provided the jet speed U is known, this equation may be solved in order to deduce the wavelength of the
waves that will travel at U and so appear to be stationary in the lab frame. For jets falling from a nozzle,
the result (11.5) may be used to deduce the local jet speed.
11.3 Fluid Pipes
The following | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
.5) may be used to deduce the local jet speed.
11.3 Fluid Pipes
The following system may be readily observed in a kitchen sink. When the volume flux exiting the tap
is such that the falling stream has a diameter of 2 − 3mm, obstructing the stream with a finger at a
distance of several centimeters from the tap gives rise to a stationary field of varicose capillary waves
upstream of the finger. If the finger is dipped in liquid detergent (soap) before insertion into the stream,
the capillary waves begin at some critical distance above the finger, below which the stream is cylindrical.
Closer inspection reveal that the surface of the jet’s cylindrical base is quiescent.
An analogous phenomenon arises when a vertical fluid jet impinges on a deep water reservoir (see
Fig. 11.5). When the reservoir is contaminated by surfactant, the surface tension of the reservoir is
diminished relative to that of the jet. The associated surface tension gradient draws surfactant a finite
distance up the jet, prompting two salient alterations in the jet surface. First, the surfactant suppresses
surface waves, so that the base of the jet surface assumes a cylindrical form (Fig. 11.5b). Second, the jet
surface at its base becomes stagnant: the Marangoni stresses associated with the surfactant gradient are
balanced by the viscous stresses generated within the jet. The quiescence of the jet surface may be simply
demonstrated by sprinkling a small amount of talc | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
jet. The quiescence of the jet surface may be simply
demonstrated by sprinkling a small amount of talc or lycopodium powder onto the jet. The fluid jet thus
enters a contaminated reservoir as if through a rigid pipe.
A detailed theoretical description of the fluid pipe is given in Hancock & Bush (JFM, 466, 285-304).
We here present a simple scaling that yields the dependence of the vertical extent H of the fluid pipe on
MIT OCW: 18.357 Interfacial Phenomena
43
Prof. John W. M. Bush
12.3. Rupture of a Soap Film (Culick 1960, Taylor 1960)
Chapter 12. Instability Dynamics
Note: Surface area of rim/ length: p = 2πR where m = ρhx = πρR2 ⇒ R =
radius. Therefore the rim surface energy is σP = σ2π
is σ 2x + 2(πhx)1/2 .
]
[
hxπ
Scale:
2x
The rim surface area is thus safely neglected once the sheet has retracted a distance comparable to its
thickness.
Some final comments on soap film rupture.
SArim ∼ 2
SAsheet
≪ 1 for x ≫ h.
hπ
x
∼
1/2
J
√
)
(
√
J
= 2σ hxπ. Total surface energy of the system
hx
π
hx
π
where R is the rim
1 | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
Total surface energy of the system
hx
π
hx
π
where R is the rim
1. for dependence on geometry and influence of µ, see
Savva & Bush (JFM 2009).
√
2. form of sheet depends on Oh =
√
µ
2hρσ
.
3. The growing rim at low Oh is subject to Ra-Plateau
⇒ scalloping of the retracting rim ⇒ rim pinches off
into drops
4. At very high speed, air-induced shear
stress leads
to flapping. The sheet thus behaves like a flapping
flag, but with Marangoni elasticity.
Figure 12.5: The different shapes of a retract
ing sheet and rim depend on the value of Oh.
Figure 12.6: The typical evolution of a retracting sheet. As the rim retracts and engulfs fluid, it eventually
becomes Rayleigh-Plateau unstable. Thus, it develops variations in radius along its length, and the
retreating rim becomes scalloped. Filaments are eventually left by the retracting rim, and pinch off
through a Rayleigh-Plateau instability, the result being droplets.
MIT OCW: 18.357 Interfacial Phenomena
48
Prof. John W. M. Bush
12. Instability Dynamics
12.1
Capillary Instability of a Fluid Coating on a Fiber
We proceed by considering the surface tension-induced
instability of a � | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
Coating on a Fiber
We proceed by considering the surface tension-induced
instability of a fluid coating on a cylindrical fiber.
Define mean thickness
∗ =
h
λ
1
λ 0
h(x)dx
(12.1)
Local interfacial thickness
h(x) = h
∗ + ǫ cos kx
(12.2)
Volume conservation requires:
λ
0
π(r + h)2dx =
λ
0
π(r + h0)2dx ⇒ (r + h
∗ + ǫ cos kx)2dx = (r + h0)2λ ⇒
λ
(r + h
∗
)2λ + ǫ2 λ
2
= (r + h0)2λ ⇒ (r + h
∗
0
ǫ2
)2 = (r + h0)2 − = (r + h0)2 1 −
2
�
ǫ2
1
2 (r + h0)2
�
Figure 12.1:
a cylindrical fiber.
Instability of a fluid coating on
which implies
∗ = h0 −
h
ǫ2
1
4 r + h0
Note:
h∗ < h0 which suggests instability.
λ
So, when does perturbation reduce surface energy? i.e. when is
0
f
2 kx
ǫ2k2 sin
Note: ds2 = dh2 + dx2 ⇒ ds = dx 1 + dx
2 kx)1/2dx = (r + h∗)λ +
≈ dx 1 +
)
[
1
2
1 (r + h∗)ǫ2k2λ < (r + h0)λ.
So | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
1
2
1 (r + h∗)ǫ2k2λ < (r + h0)λ.
So the inequality holds provided (r + h∗)λ + 4
Substitute for h∗ from (12.3):
J (
(r + h∗ + ǫ cos kx)(1 + ǫ2k2 sin
(r + h)ds =
1/2
λ
0
λ
0
dh
1
2
f
f
]
2π(r + h)ds < 2π(r + h0)λ?
1 (r + h∗)ǫ2k2λ.
4
ǫ2
1
−
4 r + h0
+
1
4
(r + h
∗
)ǫ2k2
< 0
We note that the result is independent of ǫ:
k2 < (r + h0)
−1(r + h
∗
−1 ≈
)
1
(r + h0)2
i.e. unstable wavelengths are prescribed by
λ =
2π
k
> 2π(r + h0)
(12.3)
(12.4)
(12.5)
(12.6)
as in standard inviscid Ra-P. All long wavelength disturbances will grow. Which grows the fastest? That
is determined by the dynamics (not just geometry). We proceed by considering the dynamics in the thin
film limit, h0 ≪ r, for which we obtain the lubrication limit.
45
12.2. Dynamics of Instability (Rayleigh 1879)
Chapter 12. Instability Dynamics
12.2 Dynamics of Instability (Rayleigh 1879)
Physical picture: Curvature pressure induced by pert | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
2. Instability Dynamics
12.2 Dynamics of Instability (Rayleigh 1879)
Physical picture: Curvature pressure induced by perturbation drives Couette flow that is resisted by
viscosity
d2v
dy
η
−
dp
dx
= 0
(12.7)
where dp is the gradient in curvature pressure, which is independent of y ( a generic feature of lubrication
problems), so we can integrate the above equation to obtain
dx
v(y) =
1 dp
µ dx
y2
2
(cid:18)
− hy
(cid:19)
Flux per unit length:
h
Z
Conservation of volume in lubrication problems requires that Q(x + dx) − Q(x) = − ∂h dx∂t
0
Q =
v(y)dy =
−
1 dp
3µ dx
h3
dQ
dx
=
h3 d2
p
− 0
3µ dx2 −
=
∂h
∂t
Curvature pressure
Substitute (12.11) into (12.10):
p(x) = σ
1
R
1
(cid:18)
+
1
R2 (cid:19)
=
σ
1
r + h
(cid:18)
− hxx(cid:19)
∂h
∂t
=
σh3
0 ∂2
µ
3
∂x2 (cid:20)
1
r + h(x)
−
σhxx
(cid:21)
Now h(x, t) = h∗ + ǫ(t) cos kx ⇒ hx = −ǫk sin kx, hxx = −ǫ2k cos kx, ht = dǫ cos kx
k4
σh
⇒
So cos kx dǫ
= 0
3µ
dt
i
= 0 = 2k
(r+h0)2 − 4k∗3 so
Fastest growing mode when dβ
dk
σh
dǫ = βǫ where β = 0
3µ
dt
(r+h0)2 −
(r+h)2 −
ǫ cos kx
k4
dt
h | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
0
3µ
dt
(r+h0)2 −
(r+h)2 −
ǫ cos kx
k4
dt
h
i
h
k
k
2
3
2
3
8
√
= 2
2π (r + h0)
∗
λ
is the most unstable wavelength for the viscous mode.
Note:
• Recall that for classic Ra-P on a cylindrical fluid thread
λ∗ ∼ 9R.
• We see here the timescale of instability: τ ∗ =
12µ(r+h )4
3
σh0
0
.
• Scaling Argument for Pinch-off time.
When h ≪ r, ∇
p ∼ σh0 1 ∼ µ v
2 ⇒ v ∼ r
τ
r
h0
r
2
∼
h
3
0 σh0
µ r3 ⇒
⇒
(12.8)
(12.9)
(12.10)
(12.11)
(12.12)
(12.13)
τpinch ∼
µr4
σh3
0
(12.14)
Figure 12.2: Growth rate β as a func-
tion of wavenumber k for the system de-
picted in Fig. 12.1.
MIT OCW: 18.357 Interfacial Phenomena
46
Prof. John W. M. Bush
12.3. Rupture of a Soap Film (Culick 1960, Taylor 1960)
Chapter 12. Instability Dynamics
12.3 Rupture of a Soap Film (Culick
1960, Taylor 1960)
h
We assume O = µν
≪ 1, so that viscous effects are negligible.
σR
The driving curvature force is thus resisted principally by fluid
inertia. Assume dynamics is largely 2D (true for a planar film,
or for bubble burst for r(t) ≫ h).
Retraction of a Planar Sheet
Note: Force/ length acting on the rim | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
��lm,
or for bubble burst for r(t) ≫ h).
Retraction of a Planar Sheet
Note: Force/ length acting on the rim may be calculated exactly
via Frenet-Serret
F C =
Z
C
σ (∇ · n) ndl
(12.15) Figure 12.3: Rupture of a soap film of
thickness h.
where (∇ · n) n = dt
dl
⇒
F C =
dt
σ
C dl
Z
dl = σ (t1
− t2) = 2σxˆ
(12.16)
At time t = 0, planar sheet of thickness h punctured at x = 0, and retracts in xˆ direction owing to F c.
Observation: The rim engulfs the film, and there is no upstream disturbance.
Figure 12.4: Surface-tension-induced retraction of a planar sheet of uniform thickness h released at time
t = 0.
Rim mass: m(x) = ρhx and speed v = dx .dt
Since the inertial force on the rim is equal to the rate of change of rim momentum
FI = (mv) = v mv = v
d
dx
2 dm
dx
+ mv
dv
Dx
1 2 dm 1 d
= v
2 dx
2
dx
+
d
dt
The force balance us between the curvature force and the inertial force
2σ =
( mv2) + ρhv2
d 1
dx 2
1
2
Integrate from 0 to x:
1
2σx = ρhxv2
2
1
+ ρh
2
Z
0
x
v2dx
(mv2) .
(12.17)
(12.18)
(12.19)
The first term is the surface energy released per unit length, the 2nd term the K.E. of the rim, and the
3rd term the energy required to accelerate the rim. Now we assume v is independent of x (as observed in
experiments), thus
v2dx = xv2 and the force balance becomes | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
rim. Now we assume v is independent of x (as observed in
experiments), thus
v2dx = xv2 and the force balance becomes 2σx = ρhxv2
⇒
x
0
R
1/2
v =
2σ
ρh
(cid:18) (cid:19)
is the retraction speed (Taylor-Culick speed)
(12.20)
E.g. for water-soap film, h ∼ 150µm ⇒ v ∼ 102cm/s.
MIT OCW: 18.357 Interfacial Phenomena
47
Prof. John W. M. Bush
12.3. Rupture of a Soap Film (Culick 1960, Taylor 1960)
Chapter 12. Instability Dynamics
Note: Surface area of rim/ length: p = 2πR where m = ρhx = πρR2 ⇒ R =
radius. Therefore the rim surface energy is σP = σ2π
is σ 2x + 2(πhx)1/2 .
]
[
hxπ
Scale:
2x
The rim surface area is thus safely neglected once the sheet has retracted a distance comparable to its
thickness.
Some final comments on soap film rupture.
SArim ∼ 2
SAsheet
≪ 1 for x ≫ h.
hπ
x
∼
1/2
J
√
(
)
√
J
= 2σ hxπ. Total surface energy of the system
hx
π
hx
π
where R is the rim
1. for dependence on geometry and influence of µ, see
Savva & Bush (JFM 2009).
√
2. form of sheet depends on Oh | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
Savva & Bush (JFM 2009).
√
2. form of sheet depends on Oh =
√
µ
2hρσ
.
3. The growing rim at low Oh is subject to Ra-Plateau
⇒ scalloping of the retracting rim ⇒ rim pinches off
into drops
4. At very high speed, air-induced shear
stress leads
to flapping. The sheet thus behaves like a flapping
flag, but with Marangoni elasticity.
Figure 12.5: The different shapes of a retract
ing sheet and rim depend on the value of Oh.
Figure 12.6: The typical evolution of a retracting sheet. As the rim retracts and engulfs fluid, it eventually
becomes Rayleigh-Plateau unstable. Thus, it develops variations in radius along its length, and the
retreating rim becomes scalloped. Filaments are eventually left by the retracting rim, and pinch off
through a Rayleigh-Plateau instability, the result being droplets.
MIT OCW: 18.357 Interfacial Phenomena
48
Prof. John W. M. Bush
13. Fluid Sheets
13.1 Fluid Sheets: shape and stability
The dynamics of high-speed fluid sheets was first considered by Savart (1833) after his early work on
electromagnetism with Biot, and was subsequently examined by Rayleigh (1879), then in a series of
papers by | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
was subsequently examined by Rayleigh (1879), then in a series of
papers by Taylor (Proc. Roy. Soc., 1959 ). They have recently received a great deal of attention owing to
their relevance in a number of spray atomization processes. Such sheets may be generated from a variety
of source conditions, for example, the collision of jets on rigid impactors, and jet-jet collisions. There
is generally a curvature force acting on the sheet edge which acts to contain the fluid sheet. For a 2D
(planar) sheet, the magnitude of this curvature force is given by
Using the first Frenet-Serret equation (Lecture 2, Appendix B),
F c =
1
C
σ (∇ · n) ndl
thus yields
(∇ · n) n =
dt
dl
F c =
σ
1
C
dt
dl
dl = σ (t1 − t2) = 2σx
(13.1)
(13.2)
(13.3)
There is thus an effective force per unit length 2σ along the length of the sheet rim acting to contain the
rim. We now consider how this result may be applied to compute sheet shapes for three distinct cases: i)
a circular sheet, ii) a lenticular sheet with unstable rims, and iii) a lenticular sheet with stable rims.
49
13.2. Circular Sheet
Chapter 13. Fluid Sheets
Figure | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
stable rims.
49
13.2. Circular Sheet
Chapter 13. Fluid Sheets
Figure 13.1: A circular fluid sheet generated by the impact of a water jet on a circular impactor. The
impacting circle has a diameter of 1 cm.
13.2 Circular Sheet
2
We consider the geometry considered in Savart’s original experiment. A vertical fluid jet strikes a small
horizontal circular impactor. If the flow rate is sufficiently high that gravity does not influence the sheet
shape, the fluid is ejected radially, giving rise to a circular free fluid sheet (Fig. 13.1).
For We =
We < 1000, for which the sheet is stable. Scaling: Re =
so inertia dominates gravity.
The sheet radius is prescribed by a balance of radial forces; specifically, the inertial force must balance
the curvature force:
> 1000, the circular sheet is subject to the flapping instability. We thus consider
U R ∼ 30·10
∼ 30 ∼ 0.1
0.01
ν
∼ 3·104 ≫ 1. Fr =
2
103 ·10
ρU
σ
U
gR
D
2
ρu2h = 2σ
Continuity requires that the sheet thickness h depend on the speed u, jet flux Q and radius r as
h(r) =
Q
2πru
∼
1
r
(13.4)
(13.5)
Experiments (specifically, tracking of particles suspended within the | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
r
(13.4)
(13.5)
Experiments (specifically, tracking of particles suspended within the sheet) indicate that the sheet speed u
is independent of radius; consequently, the sheet thickness decreases as 1/r. Substituting the form (13.5)
for h into the force balance (13.4) yields the sheet radius, or so-called Taylor radius:
RT =
ρQu
4πσ
(13.6)
The sheet radius increases with source flux and sheet speed, but decreases with surface tension. We note
that the fluid proceeds radially to the sheet edge, where it accumulates until going unstable via a modified
Rayleigh-Plateau instability, often referred to as the Rayleigh-Plateau-Savart instability, as it was first
observed on a sheet edge by Savart.
MIT OCW: 18.357 Interfacial Phenomena
50
Prof. John W. M. Bush
13.3. Lenticular sheets with unstable rims (Taylor 1960)
Chapter 13. Fluid Sheets
13.3 Lenticular sheets with unstable rims (Taylor 1960)
Figure 13.2: A sheet generated by the collision of water jets at left. The fluid streams radially outward in
a thinning sheet; once the fluid reaches the sheet rim, it is ejected radially in the form of droplets. From
G.I. Taylor (1960).
We now consider the non-axisymmetric fluid , such as may be
formed by the oblique collision of water jets (see Fig. 13 | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
��uid , such as may be
formed by the oblique collision of water jets (see Fig. 13.2), a ge
ometry originally considered by Taylor (1960). Fluid is ejected
radially from the origin into a sheet with flux distribution given
by Q(θ), so that the volume flux flowing into the sector between
θ and θ + dθ is Q(θ)dθ. As in the previous case of the circular
sheet, the sheet rims are unstable, and fluid drops are contin
uously ejected therefrom. The sheet shape is computed in a
similar manner, but now depends explicitly on the flux distri
bution within the sheet, Q(θ). The normal force balance on the
sheet edge now depends on the normal component of the sheet
speed, un:
ρu2 (θ)h(θ) = 2σ
n
(13.7)
The sheet thickness is again prescribed by (13.5), but now Q =
Q(θ), so the sheet radius R(θ) is given by the Taylor radius
R(θ) =
ρu2 Q(θ)
n
4πσu
(13.8)
Computing sheet shapes thus relies on either experimental mea
surement or theoretical prediction of the flux distribution Q(θ)
within the sheet.
13.4 Lenticular sheets with stable rims
Figure 13.3:
The “Fluid fishbone”
formed by the collision of two jets of
a | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
.3:
The “Fluid fishbone”
formed by the collision of two jets of
a glycerine-water solution. Bush &
Hasha (2004).
In a certain region of parameter space,
fluids more viscous than water, one may encounter fluid sheets with stable rims (see www
math.mit.edu/∼bush/bones.html). The force balance describing the sheet shape must change accordingly.
When rims are stable, fluid entering the rim proceeds along the rim. As a result, there is a centripetal
force normal to the fluid rim associated with flow along the curved rim that must be considered in order
to correctly predict the sheet shapes.
specifically, with
MIT OCW: 18.357 Interfacial Phenomena
51
Prof. John W. M. Bush
13.4. Lenticular sheets with stable rims
Chapter 13. Fluid Sheets
The relevant geometry is presented in Fig. 13.4.
r(θ) is defined to be the distance from the origin
to the rim centreline, and un(θ) and ut(θ) the nor
mal and tangential components of the fluid velocity
in the sheet where it contacts the rim. v(θ) is de
fined to be the velocity of flow in the rim, R(θ)
the rim radius, and Ψ(θ) the angle between the po
sition vector r and | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
the rim radius, and Ψ(θ) the angle between the po
sition vector r and the local tangent to the rim
centreline. Finally, rc(θ) is defined to be the ra
dius of curvature of the rim centreline, and s the
arc length along the rim centreline. The differential
equations governing the shape of a stable fluid rim
bounding a fluid sheet may be deduced by consid
eration of conservation of mass in the rim and the
local normal and tangential force balances at the
rim.
For a steady sheet shape, continuity requires
that the volume flux from the sheet balance the
tangential gradient in volume flux along the rim:
Figure 13.4: A schematic illustration of a fluid sheet
bound by stable rims.
0 = unh −
∂
∂s
vπR2
(
)
(13.9)
The normal force balance requires that the curvature force associated with the rim’s surface tension
balance the force resulting from the normal flow into the rim from the fluid sheet and the centripetal force
resulting from the flow along the curved rim:
ρu2 h +
n
2
ρπR2v
rc
= 2σ
(13.10)
Note that the force balance (13.7) appropriate for sheets with unstable rims is here augmented by the
centripetal force. The tangential force | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
sheets with unstable rims is here augmented by the
centripetal force. The tangential force balance at the rim requires a balance between tangential gradients
in tangential momentum flux, tangential gradients in curvature pressure, viscous resistance to stretching
of the rim, and the tangential momentum flux arriving from the sheet. For most applications involving
high-speed sheets, the Reynolds number characterizing the rim dynamics is sufficiently large that viscous
resistance may be safely neglected. Moreover, the curvature term ∇ · nˆ generally depends on θ; however,
accurate to O(R/rc), we may use ∇ · nˆ = 1/R. One thus obtains:
∂
∂s
(
πR2 v = hutun −
2
)
πR2σ ∂
ρ
1
∂s R
(
.
)
Equations (13.9)-(13.11) must be supplemented by the continuity relation,
h(r, θ) =
Q(θ)
u0r
in addition to a number of relations that follow directly from the system geometry:
un = u0 sin Ψ
,
uT = u0 cos Ψ
,
1
rc
=
sin Ψ
r
∂Ψ
∂θ
(
+ 1
)
(13.11)
(13.12)
(13.13)
The system of equations (13.9-13.13) may be nondimensionalized, and reduce to a set of coupled ordinary
equations in the four variables r(θ), v(θ), R(θ) and Ψ(θ). Given a | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
in the four variables r(θ), v(θ), R(θ) and Ψ(θ). Given a flux distribution, Q(θ), the system may
be integrated to deduce the sheet shape.
MIT OCW: 18.357 Interfacial Phenomena
52
Prof. John W. M. Bush
13.5. Water Bells
Chapter 13. Fluid Sheets
13.5 Water Bells
All of the fluid sheets considered thus far have been confined to a plane. In §13.1, we considered circular
sheets generated from a vertical jet striking a horizontal impactor. The sheet remains planar only if
the flow is sufficiently fast that the fluid reaches its Taylor radius before sagging substantially under the
influence of gravity. Decreasing the flow rate will make this sagging more pronounced, and the sheet will
no longer be planar. While one might expect the sheet to fall along a parabolic trajectory, the toroidal
curvature of the bell induces curvature pressures that act to close the sheet. Consequently, the sheet may
close upon itself, giving rise to a water bell, as illustrated in Fig. 13.5. A recent review of the dynamics
of water bells has been written by Clanet (Ann.Rev.). We proceed by outlining the theory required to
compute the shapes of water bells.
We consider a fluid sheet extr | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
required to
compute the shapes of water bells.
We consider a fluid sheet extruded radially at a speed u0 and subsequently sagging under the influence
of a gravitational field g = −gzˆ. The inner and outer sheet surfaces are characterized by a constant
surface tension σ. The sheet has constant density ρ and thickness t(r, z). Q is the total volume flux in
the sheet. The characteristic Re is assumed to be sufficiently high so that the influence of viscosity is
negligible.
We define the origin to be the center of the impact plate; r and z are, respectively, the radial and
vertical distances from the origin. u is the sheet speed, and φ the angle made between the sheet and the
vertical. rc is the local radius of curvature of a meridional line, and s the arc length along a meridional
line measured from the origin. Finally, ΔP is the pressure difference between the outside and inside of
the bell as may be altered experimentally.
Flux conservation requires that
Q = 2πrtu
(13.14)
while Bernoulli’s Theorem indicates that
2
2
= u + 2gz
0
u
(13.15)
The total curvature force acting normal to the bell surface
is given by
2σ∇ · n = 2σ
1
rc
(
+
cos | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
normal to the bell surface
is given by
2σ∇ · n = 2σ
1
rc
(
+
cos φ
r
)
(13.16)
Note that the factor of two results from there being two Figure 13.5: A water bell produced by the
free surfaces. The force balance normal to the sheet thus
impact of a descending water jet on a solid
takes the form:
impactor. The impactor radius is 1 cm. Fluid
is splayed radially by the impact, then sags
(13.17) under the influence of gravity. The sheet may
close on itself owing to the azimuthal curva
ture of the bell.
− ΔP + ρgt sin φ −
2σ cos φ
r
ρtu2
rc
2σ
rc
= 0
+
Equations (13.14), (13.15) and (13.17) may be appropri
ately nondimensionalized and integrated to determine the
shape of the bell.
Note:
• the bell closes due to the out-of-plane curvature
• the influence of g is reflected in top-bottom asymmetry. Note that g is not significant in Fig. 13.5.
The relevant control parameter is Fr = INERTIA/GRAVITY = U 2/gL ∼ 1
• if deflected upwards by the impactor, the bell with also close due to σ
MIT OCW: 18.357 Interfacial Phenomena
53 | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
due to σ
MIT OCW: 18.357 Interfacial Phenomena
53
Prof. John W. M. Bush
13.6. Swirling Water Bell
Chapter 13. Fluid Sheets
13.6 Swirling Water Bell
Consider the water bell formed with a swirling jet (Bark et al. 1979 ).
Observation: Swirling bells don’t close. Why not?
Conservation of angular momentum: as r decreases, v increases as does FC ∼ v2/r.
Sheet velocity:
v = ueˆs + veˆθ
in plane
'-v"
Continuity:
swirl
'-v"
Q = 2πrhu
Conservation of Angular Momentum:
vr = v0r0
Energy conservation:
2
2
u + v = 2gz + u + v0z
Normal force balance:
2σ
R
+
2σ cos φ
r
+ ρgh sin φ = ΔP +
2
0
ρhu2
R
+
ρhv2 cos φ
r
(13.18)
(13.19)
(13.20)
(13.21)
(13.22)
Evidently, the bell fails to close owing to the influence of the centripetal forces induced by the swirl.
Figure 13.6: Swirling water bells extruded from a rotating nozzle.
MIT OCW: 18.357 Interfacial Phenomena
54
Prof. John W. M. Bush
14. Instability of Superposed Fluids
Figure 14.1: | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
Bush
14. Instability of Superposed Fluids
Figure 14.1: Wind over water: A layer of fluid of density ρ+ moving with relative velocity V over a layer
of fluid of density ρ− .
Define interface: h(x, y, z) = z − η(x, y) = 0 so that ∇h = (−ηx, −ηy, 1).
The unit normal is given by
nˆ =
∇h
|∇h|
=
(−ηx, −ηy, 1)
2 + ηy
ηx
2 + 1
)
(
1/2
Describe the fluid as inviscid and irrotational, as is generally appropriate at high Re.
∓ 1
Basic state: η = 0 , u = ∇φ , φ =
2 Vx for z±.
Perturbed state: φ = ∓ 1 Vx + φ± in z±, where φ± is the perturbation field.
Solve
2
∇ · u = ∇2φ± = 0
subject to BCs:
1.
φ± → 0 as z → ±∞
2.
Kinematic BC:
where
∂η
∂t
= u · n,
u = ∇ ∓ Vx + φ± = ∓ V xˆ +
1
2
)
∂φ±
∂x
xˆ +
∂φ±
∂y
yˆ +
∂φ±
∂z
zˆ
1
2
(
from which
∂η
∂t
1
= ∓ V +
2
(
∂ | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
zˆ
1
2
(
from which
∂η
∂t
1
= ∓ V +
2
(
∂φ±
∂x
)
(−ηx) +
∂φ±
∂y
(−ηy) +
∂φ±
∂z
(14.1)
(14.2)
(14.3)
(14.4)
Linearize: assume perturbation fields η, φ± and their derivatives are small and therefore can neglect
their products.
Thus ηˆ ≈ (−ηx, −ηy, 1) and ∂η = ± 1 V ηx +
⇒
2
∂φ±
∂z
∂t
∂φ±
∂z
=
∂η
∂t
1
∂η
∓ V
2 ∂x
on z = 0
(14.5)
3.
Normal Stress Balance: p− − p+ = σ∇ · n on z = η.
Linearize: p− − p+ = −σ (ηxx + ηyy) on z = 0.
55
Chapter 14. Instability of Superposed Fluids
We now deduce p± from time-dependent Bernoulli:
ρ
∂φ
∂t
+
1
2
ρu2 + p + ρgz = f (t)
(14.6)
1
2
where u = 4 V 2 ∓ V ∂x + H.O.T.
Linearize:
∂φ±
ρ±
∂φ±
∂t
1
2
+ ρ± ∓V
(
∂φ±
∂x
)
+ p± + ρ±gη = G(t)
(14.7)
so
p− − p+ = ( | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
± + ρ±gη = G(t)
(14.7)
so
p− − p+ = (ρ+ − ρ−)gη + (ρ+
∂φ±
∂t
− ρ−
∂φ−
∂t
) +
V
2
(ρ−
∂φ−
∂x
+ ρ+
∂φ+
∂x
) = −σ(ηxx + ηyy)
(14.8)
is the linearized normal stress BC. Seek normal mode (wave) solutions of the form
η = η0e
iαx+iβy+ωt
φ± = φ0±e
∓kz iαx+iβy+ωt
e
where ∇2φ± = 0 requires k2 = α2 + β2 .
V ∂η
Apply kinematic BC:
∂x
∂φ±
∂z
∓ 1
2
∂η
∂t
=
at z = 0 ⇒
∓kφ0± = ωη0 ∓
1
2
iαV η0
(14.9)
(14.10)
(14.11)
Normal stress BC:
k2ση0 = −g(ρ− − ρ+)η0 + ω(ρ+φ0+ − ρ−φ0−) +
1
2
iαV (ρ+φ0+ + ρ−φ0−)
(14.12)
Substitute for φ0± from (14.11):
−k3σ = ω ρ+(ω −
[
1
2
iαV ) + ρ−(ω +
1
2
iαV ) + gk(ρ− − ρ+) +
]
1
2
iαV ρ+(ω −
[ | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
V ) + gk(ρ− − ρ+) +
]
1
2
iαV ρ+(ω −
[
1
2
iαV ) + ρ−(ω +
1
2
iαV )
]
so
ω2 + iαV
ρ− − ρ+
ρ− + ρ+
(
)
ω − α2V + k2C0
2
2 = 0
1
4
where C 2 ≡
k.
Dispersion relation: we now have the relation between ω and k
σ
ρ−+ρ+
ρ−−ρ+
ρ−+ρ+
+
g
k
(
)
0
ω =
1
2
i
(
ρ+ − ρ−
ρ− + ρ+
)
k · V ±
[
ρ−ρ+
(ρ− + ρ+)2
2
(k · V ) − k2C 2
0
1/2
]
where k = (α, β), k2 = α2 + β2 .
The system is UNSTABLE if Re (ω) > 0, i.e. if
ρ+ρ−
ρ− + ρ+
2
(k · V ) > k2
C 2
0
(14.13)
(14.14)
(14.15)
Squires Theorem:
Disturbances with wave vector k = (α, β) parallel to V are most unstable. This is a general property of
shear flows.
We proceed by considering two important special cases, Rayleigh-Taylor and Kelvin-Helmholtz instability.
MIT OCW: 18.357 Interfacial Phenomena
56
Prof. John W. M. Bush
14.1. | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
56
Prof. John W. M. Bush
14.1. Rayleigh-Taylor Instability
Chapter 14. Instability of Superposed Fluids
14.1 Rayleigh-Taylor Instability
We consider an initially static system in which heavy fluid overlies light fluid: ρ+ > ρ−, V = 0. Via
(14.15), the system is unstable if
C 2
0 =
ρ− − ρ+ g
ρ+ρ− k
+
σ
ρ− + ρ+
k < 0
(14.16)
i.e. if ρ+ − ρ− >
2
4π
σ
σk
g = gλ2 .
2
Thus, for instability, we require: λ > 2πλc where λc =
σ
Δρg
J
is the capillary length.
Heuristic Argument:
Change in Surface Energy:
ΔES = σ · Δl
= σ
λ
0 ds − λ = 4 σǫ2k2λ.
1
]
[f
arc length
�
0
2
f
=
λ
0
Change in gravitational potential energy:
− 1 ρg h2 − h2 dx = − 1 ρgǫ2λ.
ΔEG
4
When is the total energy decreased?
)
(
When ΔEtotal = ΔES + ΔEG < 0, i.e. when ρg > σk2 ,
so λ > 2πlc.
The system is thus unstable to long λ.
Note:
Figure 14.2: The base state and | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
is thus unstable to long λ.
Note:
Figure 14.2: The base state and the per-
turbed state of the Rayleigh-Taylor system,
heavy fluid over light.
1. The system is stabilized to small λ disturbances by
σ
2. The system is always unstable for suff. large λ
3. In a finite container with width smaller than 2πλc,
the system may be stabilized by σ.
4. System may be stabilized by temperature gradients
since Marangoni flow acts to resist surface defor
mation. E.g. a fluid layer on the ceiling may be
stabilized by heating the ceiling.
Figure 14.3: Rayleigh-Taylor instability may
be stabilized by a vertical temperature gradi
ent.
MIT OCW: 18.357 Interfacial Phenomena
57
Prof. John W. M. Bush
14.2. Kelvin-Helmholtz Instability
Chapter 14. Instability of Superposed Fluids
14.2 Kelvin-Helmholtz Instability
We consider shear-driven instability of a gravitationally stable base state. Specifically, ρ− ≥ ρ+ so the
system is gravitationally stable, but destabilized by the shear.
Take k parallel
criterion becomes:
instability
(V · k)
to V ,
k2V
and
the
so
=
2
2
ρ−ρ+V > (ρ− − ρ+)
2 | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
k2V
and
the
so
=
2
2
ρ−ρ+V > (ρ− − ρ+)
2
g
k
+ σk
(14.17)
Equivalently,
ρ−ρ+V > (ρ− − ρ+) g
2
λ
2π
+ σ
2π
λ
(14.18)
Note:
1. System stabilized to short λ disturbances by
surface tension and to long λ by gravity.
2. For any given λ (or k), one can find a critical
V that destabilizes the system.
Marginal Stability Curve:
Figure 14.4: Kelvin-Helmholtz instability: a gravi
tationally stable base state is destabilized by shear.
V (k) =
ρ− − ρ+ g
ρ−ρ+ k
(
+
1
ρ−ρ+
σk
)
1/2
(14.19)
dV
dk
= 0,
implies − Δρ + σ = 0 ⇒ kc =
V (k) has a minimum where
0.
This
1
lcap
The corresponding Vc = V (kc) = ρ−ρ+
imal speed necessary for waves.
√2
k2
.
i.e.
2
V =
d
dk
Δρg =
σ
J
Δρgσ is the min
E.g. Air blowing over water: (cgs)
2
V =
c
mum wind speed required to generate waves.
2
1.2·10−3
1 · 103 · 70 ⇒ Vc ∼ 650cm/s is the mini-
√
Figure 14 | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
70 ⇒ Vc ∼ 650cm/s is the mini-
√
Figure 14.5: Fluid speed V (k) required for
the growth of a wave with wavenumber k.
These waves have wavenumber kc =
waves.
J
1·103
70
≈ 3.8 cm , so λc = 1.6cm. They thus correspond to capillary
−1
MIT OCW: 18.357 Interfacial Phenomena
58
Prof. John W. M. Bush
15. Contact angle hysteresis, Wetting
of textured solids
Recall: In Lecture 3, we defined the equilibrium contact angle θe, which is prescribed by Young’s Law:
cos θe = (γSV − γSL) /γ as deduced from the horizontal force balance at the contact line.
Work done by a contact line moving a distance dx:
Figure 15.1: Calculating the work done by moving a contact line a distance dx.
dW = (γSV − γSL) dx −
γ cos θedx
(15.1)
contact line motion
f rom creating new interf ace
In equilibrium: dW = 0, which yields Young’s Law. It would be convenient if wetting could be simply
characterized in terms of this single number θe. Alas, there is:
15.1 Contact Angle Hysteresis
For a given solid wetting a given liquid, there is a range of | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
Hysteresis
For a given solid wetting a given liquid, there is a range of possible contact angles: θr < θ < θa, i.e. the
contact angle lies between the retreating and advancing contact angles; θr and θa, respectively. That is,
many θ values may arise, depending on surface, liquid, roughness and history.
Filling a drop
Draining a drop
• begin with a drop in equilibrium with θ = θe
• fill drop slowly with a syringe
• θ increases progressively until attaining θa, at
which point the contact line advances
• begin with a drop in equilibrium with θ = θe
• drain drop slowly with a syringe
• θ decreases progressively until attaining θr, at
which point the contact line retreats
Origins: Contact line pinning results from surface heterogeneities (either chemical or textural), that
present an energetic impediment to contact line motion.
The pinning of a contact line on impurities leads to increased interfacial area, and so is energetically
costly. Contact line motion is thus resisted.
Contact Line Pinning at Corners
A finite range of contact angles can arise at a corner θ1 < θ < π − φ + θ1; thus, an advancing contact line
will generally be pinned at corners. Hence surface texture increases contact angle | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
an advancing contact line
will generally be pinned at corners. Hence surface texture increases contact angle hysteresis.
59
15.1. Contact Angle Hysteresis
Chapter 15. Contact angle hysteresis, Wetting of textured solids
Figure 15.2: Pinning of a contact line retreating from left to right due to surface impurities.
Figure 15.3: A range of contact angles is possible at a corner.
Manifestations of Contact Angle Hysteresis
I. Liquid column trapped in a capillary tube.
θ2 can be as large as θa; θ1 can be as small as
θr.
so there is a net cap-
illary force available to support the weight of the
slug.
In general θ2 > θ1,
2πRσ(cos θ −1
cos θ2) = ρgπR2H
(15.2)
max contact force
{z
|
Force balance requires:
}
weight
{z
|
}
2σ
R
(cos θ −1
cos θ2) = ρgH
(15.3)
Thus, an equilibrium is possible only if 2σ (cos θ −r
ρgH.
R
cos θa
) >
Note: if θa = θr (no hysteresis), there can be no equilib- of gravity.
rium.
Figure 15.4: A heavy liquid column may be
trapped in a capillary tube despite the effects
MIT OCW: 18.357 Interfacial Phenomena
60
Prof. John W. M. Bush
15.2. Wetting of a Rough Surface
Chapter 15. Contact angle hysteresis, Wetting of textured solids
II. Raindrops on window panes (Dussan+Chow 1985)
If θ1 = θ2 then the drop will
fall due to unbalanced | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
85)
If θ1 = θ2 then the drop will
fall due to unbalanced
gravitational force.
large as θa, θ1 as small
as θr. Thus, the drop weight may be supported by the
capillary
force associated with the contact angle hystere
sis.
θ2 can be as
Note: Fg ∼ ρR3g, Fc ∼ 2πRσ(cos θ1 − cos θ2) which implies
that FG ∼
In general, drops on a window pane
FC
will increase in size by accretion until Bo > 1 and will then roll
downwards.
≡ Bo.
ρgR
σ
2
15.2 Wetting of a Rough Surface
Consider a fluid placed on a rough surface.
Define: roughness parameters
Figure 15.5: A raindrop may be pinned
on a window pane.
r =
Total Surface Area
Projected Surf. Area
> 1
φS =
Area of islands
Projected Area
< 1
The change in surface energy associated with the fluid front advancing a distance dz:
dE = (γSL − γSV ) (r − φS)dz + γ(1 − φS)ds
(15.4)
(15.5)
Spontaneous Wetting (demi-wicking) arises when dE < 0
i.e. cos θe =
≡ cos θc, i.e. when θe < θC. Note:
>
γSV −γSL
γ
1−φS
r−φS | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
θe < θC. Note:
>
γSV −γSL
γ
1−φS
r−φS
1.
can control θe with chemistry, r and φS with geometry, so can prescribe wettability of a solid.
2. if r ≫ 1, θC = 2 , so one expects spontaneous wicking when θe < π/2
3. for a flat surface, r ∼ 1, θc = 0: wicking requires cos θe > 1 which never happens.
π
4. most solids are rough (except for glass which is smooth down to ∼ 5˚A).
Wetting of Rough Solids with Drops
Consider a drop placed on a rough solid. Define: Effective contact angle θ∗ is the contact angle apparent
on a rough solid, which need not correspond to θe. Observation:
θ∗ < θe when θe < π/2 (hydrophilic)
θ∗ > θe when θe > π/2 (hydrophobic).
The intrinsic hydrophobicity or hydrophilicity of a solid, as prescribed by θe, is enhanced by surface
roughening.
Figure 15.6: A drop wetting a rough solid has an effective contact angle θ∗ that is generally different
from its equilibrium value θe.
MIT OCW: 18.357 Interfacial Phenomena
61
Prof. John W. M. Bush
15.3. Wenzel State (1 | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
Prof. John W. M. Bush
15.3. Wenzel State (1936)
Chapter 15. Contact angle hysteresis, Wetting of textured solids
15.3 Wenzel State (1936)
A Wenzel state arises when the fluid impregnates the rough solid. The change in wetting energy associated
with a fluid front advancing a distance dx (see Fig. 15.7) is
dEW = r(γSL − γSV )dx + γ cos θ
∗
dx
(15.6)
If r = 1 (smooth surface), Young’s Law emerges.
If r > 1: cos θ∗ = r cos θe
Note:
1.
wetting tendencies are amplified by roughening,
e.g. for hydrophobic solid (θe > π/2, cos θe < 0 ⇒ θe ≫
π/2 for large r )
2. for θe < θc (depends on surface texture) ⇒ demi-wicking
/ complete wetting
3. Wenzel state breaks down at large r ⇒ air trapped within
the surface roughness ⇒ Cassie State
15.4 Cassie-Baxter State
Figure 15.7: The wetting of a rough solid in
a Wenzel state.
In a Cassie state, the fluid does not impregnate the rough
solid, leaving a trapped vapour layer. A fluid placed on
the rough surface thus sits on roughness elements (e.g.
pillars or | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
uid placed on
the rough surface thus sits on roughness elements (e.g.
pillars or islands), and the change of energy associated
is
with
dx
advancing
∗
dEc = φS (γSL − γSV ) dx + (1 − φS) γdx + γ cos θ
distance
front
its
a
dx
(15.7)
For equilibrium (dEc/dx = 0), we require:
cos θ
∗ = −1 + φS + φS cos θe
(15.8)
Note:
1.
as pillar density φS → 0, cos θ∗ → −1, i.e. θ∗ → π
2. drops in a Cassie State are said to be in a “fakir
state”.
3.
contact angle hysteresis is greatly increased in the
Wenzel state, decreased in the Cassie.
4. the maintenance of a Cassie state is key to water
repellency.
Crossover between Wenzel and Cassie states:
Figure 15.8: The wetting of a rough solid in
a Cassie-Baxter state.
−1+φS
For dEW > dEc, we require −r cos θe + cos θ∗ > −φs cos θe + (1 − φs) + cos θ∗, i.e. cos θe < r−φS
= cos θc,
i.e. one expects a Cassie state to emerge for cos θe > cos θc. Therefore, the criterion for a Wenzel State | https://ocw.mit.edu/courses/18-357-interfacial-phenomena-fall-2010/04759cfc15c79928ca9f72a18b3864dd_MIT18_357F10_lec_all.pdf |
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