University of Warsaw Advanced Hydrodynamics
Faculty of Physics Selected Topics in Fluid Mechanics
Summer Semester 2019/20
Exercise Sheet 6
Lubrication theory: two spheres in squeeze flow
Questions, comments and corrections: e-mail togustavo.abade@fuw.edu.pl
Consider two spheres of radii a1and a2approaching each other along their line of centers at relative velocity V = −V ez as illustrated in Fig. 1. This relative motion generates a
“squeezing” flow. Our aim is to use the lubrication approximation do evaluate the axial hydrodynamic (lubrication) force Flubresisting the relative motion.
Figure 1: Squeezing flow produced by a sphere approaching a fixed sphere.
Show that the magnitude of the lubrication force is
Flub = 6πηa2V
h0 = 6πηaV −1, for 1 (1)
where h0 is the gap between the two spheres (defined as the separation between the closest points on the two particle surfaces), a is the reduced radius,
a ≡ 1 a1 + 1
a2
−1
,
and ≡ h0/a 1 is a small parameter. As the lubrication force is inversely proportional to the small separation distance, it diverges as the spheres approach one another at a fixed relative velocity.
1. Geometry approximation. Obtain an approximate expression for the surface separation h(r) for r a (by approximating the spherical surfaces by paraboloids for small r) and obtain the relevant lubrication length scale `ralong the radial direction.
Solution. Near the origin, the surfaces of the spheres are given by for r a1 : z1(r) = a1+ h0− (a21− r2)1/2∼ h0+ r2
2a1
, and
for r a2 : z2(r) = −a2+ (a22− r2)1/2∼ − r2 2a2, and the distance between the spherical surfaces is approximated by
h(r) ≡ z2− z1 ≈ h0+r2 2
1 a1 + 1
a2
= h0
1 + r2 2ah0
. (2)
For the curvature term [in the RHS of (2)] r2/2ah0be O(1), the radial length scale `rhas to be
`r ∼p
ah0 = 1/2a.
The axial length scale is `z= h0 = a.
2. Approximately unidirectional flow. Assuming that the symmetry in the flow near the origin suggests axisymmetric flow (in cylindrical coordinates),
u = ur(r, z)er+ uz(r, z)ez,
perform a scale analysis of the continuity equation ∇ · u = 0 and show that the flow is approximately unidirectional (i.e. mainly in the radial direction) for a very small gap (h0 a or 1).
Solution. Let `r and `z denote, respectively, the radial and axial characteristic length scales.
From the discussion above
`r ∼ 1/2a, `z ∼ a. (3)
Let Urand Uzdenote, respectively, the radial and axial characteristic velocity scales. The scale Uz is unambiguously given by V , the relative velocity particle velocity. The scale Ur may be obtained by using the continuity equation
1 r
∂
∂r(rur) +∂uz
∂z = 0. (4)
Hence Ur
`r ∼ Uz
`z, Ur∼ −1/2V,
Ur Uz, for 1, (5) and the flow is approximately unidirectional.
3. Velocity field. Use (3) and (5) to show that the equation of motion reduces to
∂2ur
∂z2 = 1 η
dp
dr (6)
Solve (6) with appropriate boundary conditions on the sphere surfaces and obtain the parabolic velocity profile
ur(r, z) = 1 2η
dp
dr[z − z1(r)][z − z2(r)]. (7)
Solution. The r-component of the momentum equation reads:
−∂p
∂r + η ∂
∂r
1 r
∂
∂r(rur)
+ η∂2ur
∂z2 = 0
∼ P
`r
∼ ηUr
`2r ∼ ηUr
`2z
For 1, ηUr/`2r ηUr/`2z, then the r-component of the momentum equation may be approximated by
−∂p
∂r + η∂2ur
∂z2 = 0, with ∂p
∂r ∼ ηUr
`2z = ηV
a2 −5/2. (8)
and the pressure scale is P ∼ Ur`r
`2z = ηaV h20 = ηV
a −2. (9)
Using the same procedure we may show that the largest term in the z-component of the mo- mentum equation,
−∂p
∂z + η 1 r
∂
∂r
r∂uz
∂r
+ η∂2uz
∂z2 = 0, (10)
is of order −2ηV /a2. Thus for 1, the axial pressure gradients is much smaller than the radial pressure gradients,
∂p
∂z ∼ ηV
a2 −2 ∂p
∂r ∼ ηV
a2 −5/2,
Thus, to a first approximation, the pressure is constant across the gap and p = p(r). The equation of motion then takes the approximate form given by Eq. (6).
Integrating (6) twice for uryields ur(r, z) = 1
2η dp
drz2+ A(r)z + B(r),
where A and B are obtained by applying the boundary conditions, ur[r, z1(r)] = ur[r, z2(r)] = 0,
yielding the parabolic profile (7).
4. Pressure field. Use Eq. (7) and requirement of mass conservation to obtain the follow- ing expression for the pressure
p(r) = 3ηaV
h0+ r2/2a2 = 3ηV /a
2(1 + r2/2a2)2. (11)
Solution. Let us first calculate the radial pressure gradient dp/dr from the solution (7) and mass conservation,
πr2V = 2πr Z z2
z1
ur(r, z)dz = −2πr h3 12η
dp dr.
The equation above states that the flux of fluid squeezed out over any radius must be equal to the volume displaced by the moving spheres. Then the pressure can be obtained by integration,
p(r) = −
Z r→∞
r
dp = 6ηV Z ∞
r
r
h3 dr = 6ηaV Z ∞
h
1 h3 dh, yielding
p(r) = 3ηaV
(h0+ r2/2a)2 = 3ηV /a
2(1 + r2/2a2)2.
This result shows that the pressure is concentrated near the axis of symmetry. The pressure is weak for r larger than the lubrication length scale `r∼ 1/2a.
In terms of the surface separation h, p(h) = 3ηaV /h2.
5. Lubrication force. Show that viscous tangential stresses at a rigid boundary are small compared with normal stresses for 1. Finally obtain the magnitude of the lubrica- tion force Flubby integrating the pressure.
Solution. It is straightforward to show that pressure dominates viscous stresses for 1, P ∼ ηV
a −2 η∂ur
∂z ∼ ηV
a −3/2.
The magnitude of the lubrication force can then be determined by integrating the pressure over the surface of either sphere,
Flub =
Z r→∞
r=0
p(r)2πr dr =
Z h→∞
h=h0
p(h)2πa dh
= 6πηa2 Z ∞
h0
1
h2 dh = 6πηa2V h0
= 6πηaV −1.
6. Assume that the particles a pushed together by constant applied forces of magnitude F in the axial direction. Show that the spheres will approach each other exponentially slowly,
h0(t) ∼ exp −F t/6πηa2 . (12)
Solution. Set V = −dh0/dt and the force is F = −6πηa2 1
h0 dh0
dt ,
which can be solved for h0(t), yielding the result (12).
