MIT Department of Mechanical Engineering
2.25 Advanced Fluid Mechanics
Problem 2.02
This problem is from “Advanced Fluid Mechanics Problems” by A.H. Shapiro and A.A. Sonin
y
g
α
h
ys (x)
x
A liquid of density ρ and surface tension σ has been spilled on a horizontal plate so that it forms a very large
puddle whose depth (in the central parts) is h. Consider the region near the edge of the puddle, which can
be viewed to good approximation as two-dimensional. If the contact angle is α, derive an expression for the
shape of the liquid surface ys (x).
Assume for simplicity that α is small, so that the radius of curvature of the surface is large compared with
h and can be approximated by
1
R= 2
d ys
dx2
ans:
ys = h 1 − exp −
h = tan α
2.25 Advanced Fluid Mechanics
ρg/σx
σ/ρg
1
c 2010, MIT
Copyright @
Surface Tension
A.H. Shapiro and A.A. Sonin 2.02
Solution:
y
g
h « Rpuddle
⇓
treat as a 2-dimensional problem
α
Po
Pi
h
ys (x)
x
Given:
Rpuddle
σ, α, ρ
1
since α is small.
d ys
dx2
Radius of curvature, R1 ≈
2
Unknown: ys , h
Find Po − Pi :
1
1
+ ll
R1 lR2
Po − Pi = −σ
since R2 = Rpuddle , which is assumed to be very large
flat
Pa
Pa
ys
Pi
h
Pi = P a at ys = h since the surface is flat
(no curvature) ⇒ no surface tension!
Pj
Pj = Pa + ρg(h − y)
For this side of the puddle
d2 ys
<0
dx2
⇒
R1 = −
For y = ys (x): Pi = Pj = Po + ρg(h − ys )
ρg
ρg
d 2 ys
− ys = − h
σ
σ
dx2
⇒
1
d 2 ys
dx2
⇒
Po − Pi = σ
d2 y s
dx2
2
d y
−ρg(h − ys ) = σ dx
2
2nd order ODE w/ B.C.
ys = 0
ys = h
at x = 0
as x → ∞
(2.02a)
Solve Eq. (2.02a) by assuming the form of the solution to be: ys = AeBx + C
⇒
d 2 ys
= AB 2 eBx
dx2
∴C=h
∴B=±
→
ρg
→B=−
σ
AB 2 eBx −
ρg
σ
(so that ys = h at x → ∞)
⇒ ys = A exp −
2.25 Advanced Fluid Mechanics
ρg Bx ρg
ρg
Ae − C = − h
σ
σ
σ
2
ρg
x +h
σ
c 2010, MIT
Copyright @
Surface Tension
A.H. Shapiro and A.A. Sonin 2.02
Solve for A by invoking the first boundary condition:
⇒
A = −h
ρg
x
⇒ ys = h 1 − exp −
σ
ys = 0 = A + h
Find h by letting
dys
dx
= tan α at x = 0:
ρg
ρg
dys
=h
exp −
x
dx
σ
σ
dys ρg
=h
= tan α
dx x=0
σ
⇒
⇒ h=
(2.02b)
σ
tan α
ρg
Alternatively, starting from Young-Laplace, (assuming that α is small) such that curvature is equal to
2
1
d2 ys
dx2
⇒ ΔP = −σ ddxy2s , ⇒
d 2 ys
,
(2.02c)
dx2
besides, Pi (x, y) = Pi (x, ys ) + ρg(ys − y), from hydrostatics. Then, combining the information from hydro­
statics and Young Laplace,
d2 y s
(2.02d)
⇒ Pi (x, y) = −σ 2 + Po + ρg(ys − y).
dx
Now, since it’s hard to work with so many functions, let’s use one more property from hydrostatics, dP
dx = 0,
at any y. Then, differentiating all the expression,
Pi (x, ys (x)) − Po = −σ
0 = −σ
σ
ρg ,
and setting Lc =
dys
d 3 ys
+ ρg
,
dx
dx3
(2.02e)
⇒ for 0 < x < ∞ (large puddle),
3
0 = L2c ddxy3s −
dys
dx ,
(2.02f)
which is a 3rd order differential equation with the boundary conditions
ys (0)
dys (0)
dx
dys
→0
dx
Solving, we first introduce G =
Lc
dys
dx ,
d2 G
− G = 0,
dx2
=
0,
(2.02g)
=
tan α,
(2.02h)
x → ∞.
as
(2.02i)
so
G = C1 e− Lc +
x
⇒
x
C2 e Lc
(2.02j)
C2 =0, unbounded term
then,
ys = C1q e− Lc + C3 .
x
From the boundary condition ys (0) = 0, ⇒ C1q = C3 , ⇒ ys = C3 (1 − e
From the boundary condition
Finally,
dys (0)
dx
= tan α, ⇒
C3
Lc
(2.02k)
−x
Lc
).
−x
= tan α, ⇒ ys = Lc tan α(1 − e Lc ).
−x
ys (x) = Lc tan α(1 − e Lc ),
(2.02l)
from which we notice that Lc measures the extent of the effect of interface tension on the surface profile.
2.25 Advanced Fluid Mechanics
3
c 2010, MIT
Copyright @
Surface Tension
A.H. Shapiro and A.A. Sonin 2.02
Note that we can deduce h (for all x) by a force balance as done in class:
ρgh2
+ σ cos α = σ,
2
then, as α → 0
⇒
h2 =
2σ(1 − cos α)
ρg
2σ(1 − cos α)
1
σα2
≈ 2σ(1 − (1 − α2 + ...))
.
=
ρg
ρg
ρg
(2.02m)
(2.02n)
D
Problem Solution by Sungyon Lee, MC (Updated), Fall 2008
2.25 Advanced Fluid Mechanics
4
c 2010, MIT
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2.25 Advanced Fluid Mechanics
Fall 2013
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