Optimal shapes for best draining
J.D. Sherwood
14 October 2009
Department of Applied Mathematics and Theoretical Physics, University of
Cambridge, Wilberforce Road, CB3 0WA, UK
Abstract
The container shape that minimizes the volume of draining fluid remaining on the walls of the container after it has been emptied from its base is
determined. The film of draining fluid is assumed to wet the walls of the
container, and is sufficiently thin so that its curvature may be neglected.
Surface tension is ignored. The initial value problem for the thickness of a
film of Newtonian fluid is studied, and is shown to lead asymptotically to a
similarity solution. From this, and from equivalent solutions for power-law
fluids, the volume of the residual film is determined. The optimal container
shape is not far from hemispherical, to minimize the surface area, but has
a conical base to promote draining. The optimal shape for an axisymmetric
mixing vessel, with a hole at the centre of its base for draining, is also optimal when inverted in the manner of a washed wine glass inverted and left to
drain.
1
1
Introduction
The draining of liquid from a solid surface is of practical importance (e.g. for
withdrawal coating), and has been much studied. The surfaces considered
are often plane [1] or cylindrical [2]. The initial stages of draining from
containers have also been studied [3, 4, 5] but the time taken to empty a
container will be dominated by the slow draining of thin films during the
final stages of draining. Here we investigate the influence of container shape
upon the final draining process, in order to determine an optimal shape.
Such hydrodynamic optimization problems are frequently encountered by
designers of ships and aeroplanes [6]; examples at low Reynolds numbers are
rarer (e.g. [7, 8, 9]). In particular, we note an investigation of the effect of
container shape upon the draining of foams by Saint-Jalmes et al. [10].
We first study the final stages of draining of an axisymmetric vessel from
an opening at its base. The initial value problem for a Newtonian fluid leads
to a well-known similarity solution, discussed in Sec. 2. The corresponding,
more general class of similarity solutions for power-law fluids is obtained in
Sec. 3, and then is used in Sec. 4 to determine an optimal shape for most
rapid draining through a hole at the centre of the base of the vessel. The
analogous plane, 2-dimensional problem of a trough draining through a slit
along its base is considered in Sec. 5. In Sec. 6 it is shown that the shapes
determined in Sec. 4 are also optimal when inverted and drained in the
opposite direction, as occurs when a wine glass is washed and then inverted
to drain.
2
2
A Newtonian fluid in a conical container
We consider a Newtonian fluid of density ρ and viscosity µ. The fluid flows
as a thin film of thickness h over the surface of a plate inclined at an angle
φ to the horizontal; the acceleration due to gravity is g. We assume that
the film is already sufficiently thin that curvature of the vessel wall may be
neglected locally. The fluid is assumed to wet the walls of the container,
but surface tension effects are otherwise neglected. In particular, we ignore
any possibility that a sufficiently thin film breaks into a series of small fluid
drops. The volume flux Q (per unit length perpendicular to the direction of
flow) of the fluid draining under gravity is therefore
Q = ρgh3 sin φ/(3µ).
(1)
Now consider an axisymmetric vessel, the axis of which is aligned with
the z axis of cylindrical coordinates (r, z). The base of the vessel is at (r, z) =
(0, 0), and the vessel walls are at r = R(z) (or alternatively, at z = Z(r)), as
depicted in Fig. 1. The local volumetric flux of the draining fluid, Eq. (1)
becomes
Q = αh3 (1 + R02 )−1/2
(2)
where α = ρg/(3µ) and R0 = dR/dz. Volume conservation then gives the
change in film thickness with time as
¸
·
∂
2πRαh3
=
.
∂t
∂z (1 + R02 )1/2
02 1/2 ∂h
2πR(1 + R )
Setting
H3 =
and
Z
ξ=α
z
−1
Rh3
(1 + R02 )1/2
R2/3 (1 + R02 )2/3 dz 0 ,
0
3
(3)
(4)
(5)
the conservation equation (3) becomes
∂H
∂H 3
−
=0
∂t
∂ξ
(6)
with h = h0 (z), H = H0 (x) at t = 0. Transforming to coordinates (τ, η) such
that ξ = η on τ = 0, the solution of Eq. (6) can be found by the method of
characteristics:
H = H0 (η)
(7a)
t = τ
(7b)
ξ = −3H 2 τ + η.
(7c)
We now take as a simple example the conical vessel R = cz, with 0 <
z < Z0 . The internal half-angle of the cone is θ = tan−1 c. In this case, by
Eq. (5),
3
(8)
ξ = α−1 c2/3 (1 + c2 )2/3 z 5/3 .
5
The film thickness is assumed to be initially uniform over the wall of the
cone, apart from a small region Z0 − δ < z < Z0 near the rim, with
h = h0 ,
0 < z < Z0
= (Z0 + δ − z)h0 /δ,
Z 0 < z < Z0 + δ
= 0,
Z0 + δ < z
(9a)
(9b)
(9c)
where Eqs. (9b) and (9c) have been introduced to ensure that the initial film
thickness is defined and continuous for z > 0. We assume that the constant
δ ¿ Z0 is small, but do not need to specify it precisely.
We now find the initial film thickness H0 as a function of η. At t = τ = 0,
we find from Eqs. (7c) and (8)
3
η = αc2/3 (1 + c2 )2/3 z 5/3
5
4
(10)
and we can define η1 corresponding to z = Z0 at t = 0 and η2 corresponding
to z = Z0 + δ. The initial film thickness h = h0 is independent of z, but the
modified thickness H, defined by Eq. (4), is a function of z (or ξ), and at
t = 0 can be written as
µ
1/3
H0 (η) = h0 c
2 −3/10
(1 + c )
5α
3
¶1/5
η 1/5 ,
η < η1
(11)
with H(η2 ) = 0 and H continuous over η1 < η < η0 . At fixed ξ (i.e. fixed z),
we see that on the characteristic (7) η increases as τ increases. H0 increases
until η = η1 . H0 then decreases to zero as η increases from η1 to η2 . We have
assumed δ ¿ Z0 , so that η2 − η1 ¿ η1 during the final stages of draining
when η1 < η < η2 , and Eq. (7) can then be approximated as
i.e.
ξ = −3H 2 τ + η1
(12)
³
´
c2/3
5/3
(1 + c2 )2/3 Z0 − z 5/3 = H02 t
5α
(13)
so that
µ
h∼
1 + c2
5αt
¶1/2
5/3
(Z0
− z 5/3 )1/2
.
z 1/3
(14)
The t−1/2 dependence of the film thickness is well-known, though the nonuniform geometry has introduced a novel spatial variation in Eq. (14). This
similarity solution breaks down at the origin z = 0, where we have in any
case ignored the presence of the opening. However, the contribution of this
region to the total volume of fluid in the film, Vfilm , is integrable, and for a
vessel of height Z0 we have
Z Z0
Vfilm = 2π
R(1 + R02 )1/2 h dz
0
µ
2 1/2
= 2πc(1 + c )
1 + c2
5αt
¶1/2 Z
Z0
0
5/3
z 2/3 (Z0
− z 5/3 )1/2 dz
2
=
2πc(1 + c ) 5/2
Z0 I
(5αt)1/2
(15)
5
where
Z
1
I=
u2/3 (1 − u5/3 )1/2 du.
(16)
0
At any fixed time t we can make Vfilm as small as we wish by making the
container height Z0 sufficiently small. It is more sensible to consider a fixed
vessel size, and we assume that the vessel volume
Vvessel =
πc2 3
Z
3 0
(17)
is held constant. The volume of the film is then
µ
¶5/6
2πIc(1 + c2 ) 3Vvessel
Vfilm =
(5αt)1/2
πc2
(18)
and this is minimum when c = 2−1/2 which corresponds to a cone with
internal half angle θ = 35◦ . The internal surface area of the cone is
µ
2 1/2
Svessel = πc(1 + c )
3Vvessel
πc2
¶2/3
(19)
and this too is minimized when c = 2−1/2 .
3
A power-law fluid in a conical container
Note that the final similarity solution, Eq. (14), is independent of the initial
condition h0 (z). Thus the precise nature of the initial conditions is eventually
unimportant: any initial non-uniformities are convected away. We shall not
investigate the stability of our time-dependent flow: Benjamin [11] showed
that a non-vertical film draining under gravity is stable at sufficiently low
Reynolds numbers.
The asymptotic solution can be obtained most easily by looking for a
similarity solution h = g(z)tγ , where g(Z0 ) = 0 and γ has to be determined
6
as part of the solution. The analysis is now sufficiently simple that it can
readily be extended to power-law fluids in which the shear stress τ is related
to the shear rate ∂u/∂s by a power law
µ ¶n
∂u
τ =K
∂s
(20)
where n = m−1 is the power law index. The local volume flux of draining
fluid is
µ
Q=h
2+m
ρg sin θ
K
¶m
n
.
2n + 1
(21)
The conservation equation (3) becomes
·
¸
Rh2+m
n ³ ρg ´m ∂
=
.
∂t
2n + 1 K
∂z (1 + R02 )m/2
02 1/2 ∂h
R(1 + R )
There is a similarity solution of the form
¾np
µ ¶p ½Z Z0
K
(n+1)q
02 (n+1)q
0
−np
02 q/2 −nq
R
(1 + R )
dz
h = t (1 + R ) R
ρg
z
(22)
(23)
with
q = (2n + 1)−1
,
p = (n + 1)−1 .
If we restrict our attention to conical containers, with R = cz, then
µ ¶p
K
−np
2 1/2
−nq
h = t (1 + c ) (cz)
cnq s−np (Z0s − z s )np
ρg
(24)
(25)
where s = (3n + 2)/(2n + 1), and the volume of fluid in the film is, by Eq.
(15),
µ
Vfilm
¶p
Z Z0
K
−np −np
s t
z 1−nq (Z0s − z s )np dz
= 2πc(1 + c )
ρg
0
µ ¶p µ
¶(3n+2)p/3 Z 1
2
2πc(1 + c ) K
3Vvessel
=
u1−nq (1 − us )np dz,
np
2
(st)
ρg
πc
0
(26)
2
which is minimum when
µ
c = tan
−1
θ=
7
3n + 1
3n + 5
¶1/2
.
(27)
As n decreases towards zero, the optimal value of c decreases towards 5−1/2 ,
and the internal half-angle of the cone decreases to θ = 24.1◦ . The internal
surface area, Eq. (19), of the optimal cone increases, but the steeper slope,
and correspondingly higher shear rate, tends to reduce the viscosity of the
fluid and enhances draining. As n → ∞, the optimal slope c → 1, corresponding to a cone with internal half-angle θ = 45◦ . The fluid shear-thickens
when n > 1, and the reduction in shear rate caused by the lower slope makes
up for the increase in internal surface area of the cone.
4
Optimal axisymmetric shapes
We return to the case of an arbitrary axisymmetric vessel R(z). The volume
of fluid in the film is, by Eq. (15),
µ ¶ p Z Z0
¾np
½ Z Z0
K
−np
(n+1)q
02 (n+1)q
0
02 (n+1)q (n+1)q
dz
Vfilm = 2πt
R
(1 + R )
dz
(1 + R )
R
ρg
z
0
µ ¶p
½ Z Z0
¾np+1
K
−np
(n+1)q
02 (n+1)q
0
= 2πt
.
(28)
(n + 1)q
R
(1 + R )
dz
ρg
0
The total capacity of the vessel is
Z
Z0
Vvessel = π
R2 dz.
(29)
0
Thus we seek the shape R(z) that minimizes Vfilm , subject to keeping Vvessel
constant, and so we seek the stationary points of
½ Z Z0
¾np+1
Z
(n+1)q
02 (n+1)q
M=
R
(1 + R )
dz
−β
0
Z0
R2 dz
(30)
0
where β is an as yet undetermined multiplier. In order to make further
progress, we change the independent variable from z to r, so that the vessel
shape is described by the curve z = Z(r) for 0 < r < R0 , and we set
µ ¶−1
dZ
dR
w=
=
.
dr
dz
8
(31)
M can now be written as
Z
np+1
M =L
R0
−β
r2 w dr
(32)
0
where
Z
L=
Z
R0
r
(n+1)q
−2 (n+1)q
(1 + w )
R0
w dr =
0
r(n+1)q (1 + w2 )(n+1)q w−q dr. (33)
0
We require the variational derivative of M with respect to changes in w, and
consider a small perturbation in which Z(r) is perturbed to Z + δZ, with
w = Z 0 perturbed to w + δw so that
Z R0
£
¤
np
r(n+1)q q (n + 1)(1 + w2 )−nq 2w1−q − (1 + w2 )(n+1)q w−q−1 δw dr
δM = (np + 1)L
0
Z R0
−β
r2 δw dr.
(34)
0
For M to be stationary, we require δM = 0, and hence
£
¤
(1 + w2 )−nq w−q−1 (2n + 1)w2 − 1 = γr(3n+1)q
(35)
γ = (n + 1)βL−np .
(36)
where
If, in addition, the endpoint R0 is a parameter at our disposal, then the
variation of M with respect to a change δR0 in R0 is
h
i
(n+1)q
δM = (np + 1)Lnp R0
(1 + w2 )(n+1)q w−q − βR02 w
r=R0
δR0 .
(37)
Stationarity of M then requires that the right-hand side of Eq. (37) be zero,
i.e.
£
(1 + w2 )(n+1)q w−2(n+1)q
¤
(3n+1)q
r=R0
= γqR0
.
(38)
Equations (35) (evaluated at r = R0 ) and (38) together imply that γ =
−(3n+1)q
(2n + 1)R0
, and
µ
w∼
n+1
3n + 1
¶1/2 µ
r
1−
R0
9
¶−1/2
,
r → R0 .
(39)
The vessel walls are therefore vertical at the outer edge of the vessel, in order
to promote draining. At the base of the container, w → (2n + 1)−1/2 as
r → 0, and hence the base of the vessel is conical with an interior half-angle
θ = tan−1 (2n + 1)1/2 .
(40)
We now scale lengths by R0 and volumes by R03 , and denote non-dimensional
quantities by a caret. Equation (35) becomes
£
¤
(1 + w2 )−nq w−q−1 (2n + 1)w2 − 1 = (2n + 1)r̂(3n+1)q .
(41)
When n = 0 we can integrate Eq. (41) to obtain
Ẑ = 2 − 2 (1 − r̂)1/2 ,
(42)
so that V̂vessel = 16π/15. In the limit n → ∞, we find w ∼ r̂3/2 (1 − r̂3 )−1/2 ,
except in a region r̂ ∼ (2n + 1)−1/3 around the origin, and V̂vessel → π 2 /6.
More generally, Eq. (41) can be solved numerically to give w(r̂), which
can then be integrated to give Ẑ(r̂). The volume V̂vessel of the optimal vessel
is shown in Fig. 2. Vessel shapes for various n are shown in Fig. 3: the
1/3
shapes have been further scaled by V̂vessel so that all the containers in the
figure have the same volume. Although a hemisphere would minimize the
area of the film, drainage would be slow at the horizontal base: the optimal
vessel therefore has a conical base to enhance draining in this region. As
with the cones discussed in Sec. 3, the slope of the wall increases closer to
the vertical when n decreases.
The slope of the wall of the optimal axisymmetric shape varies from
w = (2n + 1)−1/2 at the base to infinite at the rim z = Z0 . The slope of
the optimal cone is uniform, and represents a compromise between the best
slope at the base and that at the rim. In consequence, the optimal cone halfangle,given by Eq. (27), is smaller than that for the arbitrary axisymmetric
vessel, Eq. (40). Both angles are plotted in Fig. 4.
10
5
Plane, two-dimensional results
The equivalent problem for a 2-dimensional trough is similar to the axisymmetric case. We now describe the vessel shape as y = Y (x), where (x, y) are
Cartesian coordinates with y along the axis of symmetry (Fig. 5). The base
of the trough is at the origin, and its extremity is at (X0 , Y0 ).
As before, the flux of fluid in the draining film is given by Eq. (21);
conservation of volume in the film now requires
³ ρg ´m ∂ · h2+m v m ¸
2 1/2 ∂h
(1 + v )
= nq
∂t
K
∂x (1 + v 2 )m/2
where v = Y 0 . There is a similarity soution of the form
µ ¶p
·Z X0
¸np
K
−np
−q
2 q/2
−q
2 (n+1)q
0
h=t
.
v (1 + v )
v (1 + v )
dx
ρg
x
(43)
(44)
The volume of fluid in the film (per unit length of the trough in the z direction) is
Z
2D
Vfilm
X0
h(1 + v 2 )1/2 dx
0
·Z X 0
¸np
µ ¶p Z X0
K
−q
2 (n+1)q
−q
2 (n+1)q
0
−np
= 2t
v (1 + v )
v (1 + v )
dx
dx
ρg
0
x
µ ¶p ·Z X0
¸np+1
2(n + 1) −np K
−q
2 (n+1)q
0
=
t
v (1 + v )
dx
.
(45)
2n + 1
ρg
0
= 2
The volume of the trough (per unit length of the trough in the z direction)
is
Z
X0
Vtrough = 2
xv dx.
(46)
0
If we restrict our attention first to the case of a wedge y = vx, with v
constant, then
Vtrough = vX02
11
(47)
and
2D
Vfilm
µ ¶p
2(n + 1) −np K
=
t
v −p (1 + v 2 )X0np+1
2n + 1
ρg
µ ¶p
2(n + 1) −np K
(np+1)/2
=
t
v −(2n+3)p/2 (1 + v 2 )Vtrough ,
2n + 1
ρg
(48)
which is minimum when v 2 = (2n + 3)/(2n + 1), implying an internal half
angle θ = 30◦ for n = 0, θ = 37.8◦ for n = 1 and θ → 45◦ as n → ∞.
Reverting to the case of an arbitrary, plane 2-dimensional shape, with
film volume (Eq. 45) and vessel volume (Eq. 46), we seek to minimize
·Z
¸np+1
X0
−q
2 (n+1)q
v (1 + v )
dx
0
Z
X0
−β
0
xv dx,
(49)
0
where β is an undetermined multiplier. As in Sec. 4 we use the calculus of
variations: the optimal v satisfies
£
¤
v −2(n+1)q (1 + v 2 )−nq (2n + 1)v 2 − 1 = κx
where
·Z
X0
¸−np
−q
2 (n+1)q
v (1 + v )
κ = β(n + 1)
(50)
dx
0
.
(51)
0
If X0 is at our disposal, then stationarity of expression (49) with respect to
changes in X0 requires
£
v −2(n+1)q (1 + v 2 )(n+1)q
¤
X0
= κqX0
(52)
which, when combined with Eq. (50) (evaluated at x = X0 ) tells us that
κ = (2n + 1)X0−1 and that
µ
v∼
n+1
2n + 1
¶1/2 µ
x
1−
X0
¶−1/2
,
x → X0
(53)
so that the slope v is infinite at x = X0 , as in the axisymmetric case. At
the base of the vessel, v → (2n + 1)−1/2 as x → 0, and so the half-angle
at the base of the plane, 2-dimensional shape is the same as that at the
12
base of the axisymmetric vessel of Sec. 4, shown as curve (a) of Fig. 4.
We scale lengths by X0 and volumes (per unit length of the trough in the z
direction) by X02 , and denote non-dimensional quantities by a tilde. If n = 0,
then Ỹ = 2 − 2(1 − x̃)1/2 and the cross-section is the same as that for an
axisymmetric container (Eq. 42), but this is not true for n > 0. Figure
5 shows the optimal shapes for various power-law indices n, with lengths
1/2
further scaled by Vtrough so that all the containers in the figure have the same
volume.
6
Draining on the underside of the vessel
We have not considered the precise nature of the flow around the central exit
hole. Away from this exit, we do not need to specify whether the flow is
on the upper or lower side of the container. Thus the same analysis holds
for draining of the underside of a spoon dipped in fluid. We neglect surface
tension, which eventually plays an important role when fluid leaves the base
of the spoon in discrete drops.
Suppose that we invert the container, as shown in Fig. 6. Fluid now
drains from the centre of what was previously the base of the container,
towards the extremities. The conservation equation for the axisymmetric
problem is still given by Eq. (22), but with the sign of the right-hand side
reversed. The previous similarity solution, Eq. (23), is modified, and the
film thickness is now found to be
µ
¶p ½Z z
¾np
K
02 q/2 −nq
(n+1)q
02 (n+1)q
0
h = (1 + R ) R
.
R
(1 + R )
dz
ρgtn
0
(54)
The draining film now has zero thickness at the central, highest part of the
container at z = 0, and is thickest at the rim (R0 , Z0 ). The volume of fluid
13
in the film is now
µ ¶ p Z Z0
½Z z
¾np
K
up
−np
02 (n+1)q (n+1)q
(n+1)q
02 (n+1)q
0
Vfilm = 2πt
(1 + R )
R
R
(1 + R )
dz
dz
ρg
0
0
= Vfilm
(55)
where Vfilm is given by Eq. (28). Thus the volume of the film is again given
by Eq. (28) of Sec. 4, and the optimal shapes derived in Sec. 4 are also
optimal when inverted.
The film thickness in this case, Eq. (54), is greatest at the outer rim
of the vessel, where r = R0 is maximum, so that the contribution of this
region to the total volume of the film is large. Conversely, in the analysis of
Sec. 4, the film thickness, Eq. (23), was zero at the outer rim of the vessel.
It is therefore surprising that the total volumes of the two films are equal.
However, we note that in the case of the container draining from its base,
the film thickness h, Eq. (23), is singular (but integrable) at the base, and
the contribution of this portion of the film to the total film volume cannot
be neglected.
7
Vessels with fixed rim diameter 2R0
The analysis of Secs. 3–6 assumed that R0 can be varied. We now consider a
more restricted problem in which R0 is fixed. In this case we may solve Eq.
(35) for w(r, γ), and then choose γ in order to give the desired volume. We
again non-dimensionalize lengths by R0 , so that the rim of the container is
at r̂ = R̂0 = 1, and the governing equation (35) becomes
£
¤
(1 + w2 )−nq w−q−1 (2n + 1)w2 − 1 = γ̂r̂(3n+1)q ,
(3n+1)q
where γ̂ = γR0
(56)
. We assume for the moment γ̂ > 0. By Eq. (56),
(3n+1)q
w → ∞ when r̂ = r̂1 , where r̂1
= (2n + 1)/γ̂ > 0, in the neighbourhood
14
of which
µ
w∼
n+1
3n + 1
¶1/2 µ
r
1−
r1
¶−1/2
,
r → r1 .
(57)
Optimal shapes that start from the base at r̂ = ẑ = 0 and reach r̂ = 1 are
therefore possible if r̂1 ≥ 1, i.e. if γ̂ ≤ 2n + 1. If γ̂ = 2n + 1 we recover the
solution of Sec. 4 (Fig. 3), but in general there are two possible solutions
for each γ̂. If γ̂ ¿ 1, one solution is a cone, with w = (2n + 1)−1/2 . The
other solution deviates from conical when r is sufficiently large, reaches a
maximum radius r̂1 À 1, and then closes back on itself to return to r̂ = 1.
In this second solution, liquid hangs from the underside of the upper portion
of the container wall, as discussed in Sec. 6. In this case Z(r) is not singlevalued. As a result, although the film volume of Eq. (28), given as an integral
over z, is still correct, minor modifications have to be made to the subsequent
analysis, Eqs. (32–33).
If we scale both ẑ and r̂ by r̂1 , the governing equation in the new scaled
variables reverts to Eq. (41). The upper and lower portions of the vessel are
simply scaled portions of the optimal shapes found in Sec. 4 (Fig. 3). Fig.
7 shows optimal vessel shapes for a Newtonian fluid (n = 1), for several r̂1 .
If γ > 0, as assumed above, the optimal shapes that satisfy Eq. (56)
have volume greater than πR03 (2n + 1)−1/2 /3, the volume of a cone with wall
slope w = (2n + 1)−1/2 . Solutions of Eq. (56) with smaller volume can be
obtained by taking γ < 0, so that w ≤ (2n + 1)−1/2 : these solutions are not
portions of the optimal shapes of Sec. 4.
I thank Professor E.J. Hinch and the Department of Applied Mathematics, University of Cambridge, for welcome hospitality.
15
References
[1] J.H. Snoeijer, J. Ziegler, B. Andreotti, M. Fermigier, and J. Eggers,
“Thick films of viscous fluid coating a plate withdrawn from a liquid
reservoir,” Phys. Rev. Lett. 100 (2008) 244502.
[2] D. Quéré, “Fluid coating on a fibre,” Annual Rev. Fluid Mech. 31, 347
(1999).
[3] B.T. Lubin, and G.S. Springer, “The formation of a dip on the surface
of a liquid draining from a tank,” J. Fluid Mech. 29, 385 (1967).
[4] C.Y. Chow, and W.M. Lai, “Unsteady axisymmetric flows of a liquid
draining from a circular tank,” A.I.A.A.J. 10, 1032 (1972).
[5] Q.N. Zhou, and W.P. Graebel, “Axisymmetrical draining of a cylindrical
tank with a free surface,” J. Fluid Mech. 221, 511 (1990).
[6] B. Mohammadi, and O. Pironneau, “Shape optimization in fluid mechanics,” Annual Rev. Fluid Mech. 36, 255 (2004).
[7] J.M. Bourot, “On the numerical computation of the optimum profile in
Stokes flow,” J. Fluid Mech. 65, 513 (1974).
[8] O. Pironneau, “Optimum design in fluid mechanics,” J. Fluid Mech. 64,
97 (1974).
[9] M. Roper, T.M. Squires, and M.P. Brenner, “Symmetry unbreaking in
the shapes of perfect projectiles,” Phys. Fluids 20, 093606 (2008).
[10] A. Saint-Jalmes, M.U. Vera, and D.J. Durian, “Free drainage of aqueous
foams: container shape effects on capillarity and vertical gradients,”
Europhys. Lett. 50, 695 (2000).
16
[11] T.B. Benjamin, “Wave formation in laminar flow down an inclined
plane,” J. Fluid Mech. 2, 554 (1957).
17
Figure captions
Fig. 1. The vessel, with liquid film draining towards its base, where fluid
is allowed to leave via a hole (not shown).
Fig. 2. The non-dimensional volume V̂vessel of the optimal container, as
a function of the power-law index n. The broken line shows the asymptote
π 2 /6 for n → ∞.
Fig. 3. Optimal shapes for the axisymmetric container. All shapes have
1/3
been scaled by Vvessel so that they have the same volume.
Fig. 4. The half-angle θ at the base of the optimal container, as a function
of power-law index n, (a) for arbitary axisymmetric shapes, Eq. (40) and for
the troughs of Sec. 5; (b) for conical containers, Eq. (27).
Fig. 5. Optimal shapes for the plane 2D trough. All shapes have been
1/2
scaled by Vtrough so that they have the same volume.
Fig. 6. The inverted container, with liquid film draining towards the
outer rim.
Fig. 7. The optimal container with fixed rim R̂0 = 1 at ẑ = 0. There are
two solutions for each value of r̂1 > 1.
18
z
(R0 , Z0)
(R, Z)
liquid film
h
container
r
Figure 1: The vessel, with liquid film draining towards its base, where fluid
is allowed to leave via a hole (not shown).
‘
19
4
3.5
V̂vessel
3
2.5
2
1.5
1
0
1
2
n
3
4
5
‘
Figure
2: The non-dimensional volume V̂vessel of the optimal container, as
a function of the power-law index n. The broken line shows the asymptote
π 2 /6 for n → ∞.
1.5
z
n=0
0.2
0.5
1.0
2.0
5.0
1
0.5
r
-1
-0.5
0
0.5
1
Figure 3: Optimal shapes for the axisymmetric container. All shapes have
1/3
been scaled by Vvessel so that they have the same volume.
20
80
70
θ/ degree
a
60
50
40
b
30
20
0
1
2
n
3
4
5
Figure 4: The half-angle θ at the base of the optimal container, as a function
of power-law index n, (a) for arbitary axisymmetric shapes, Eq. (40) and for
the troughs of Sec. 5; (b) for conical containers, Eq. (27).
1.5
y
n=0
0.2
0.5
1.0
2.0
5.0
1
0.5
x
-1
-0.5
0
0.5
1
Figure 5: Optimal shapes for the plane 2D trough. All shapes have been
1/2
scaled by Vtrough so that they have the same volume.
21
r
container
(R, Z)
liquid film
h
(R0 , Z0)
z
Figure 6: The inverted container, with liquid film draining towards the outer
rim.
22
-2
0
-1
1
r
2
2.0
1.5
1
1.0
2
1.5
3
4
r1 = 2.0
z
5
Figure 7: The optimal container with fixed rim R̂0 = 1 at ẑ = 0. There are
two solutions for each value of r̂1 > 1.
23