Free surface flow over a step
Abdelkrim Merzougui
Department of Mathematics, Faculty of Mathematics & Informatic M’Sila University 28000 Algeria,
Email [email protected]
ICM 2012, 11-14 March, Al Ain
ABSTRACT
A free surface flow over a step bottom is solved numerically by
series truncation method. The flow is assumed to be steady 2dimensional irrotational, meanwhile the fluid is inviscid and incompressible. The effect of surface tension is taken into account and the
effect of gravity is neglected. Numerical solutions are obtained via
series truncation procedure. The problem is solved numerically for
various values of the Weber number .
Keywords: surface tension, Schwartz-Christoffel transformation, series truncation methods.
1
INTRODUCTION
Free-surface flows over various types of bottom configurations
have been studied and received a great attention due to man's
interest in the flow of water in rivers and channels. That problem
has been considered as an important subject in the field of civil
hydraulic engineering, marine engineering, and provides at least
qualitative insight into the mechanism of wave generation by
submerged obstacles moving beneath a free surface as well as
description of the flow caused by a long body moving close to sea
bottom. The literature of the topic is rich and in particular we may
mention the work of Forbes [4], Forbes and Schwartz [5]. Recently, a considerable amount of work has been done by Boutros, Abdel-Malek [2], Abd-el-Malek and Masoud [1], King and Bloor [13],
and Abd-el-Malek and Hanna [14].
In 1961, Birkhoff gave a brief analysis of methods for computing
potential flows having fixed boundaries and in 1979, van der Zanden applied the conformal mapping method to solve some problems of free streamline potential flow theory. For a full survey of
the conformal mapping technique, one consults Birkhoff, and
Zarantonello [8], In this paper, we consider a steady two~
dimensional potential flow over a step of height and length H
(fig.1(b)). The fluid is assumed to be inviscid, incompressible and
the flow is irrotational. If we take the symmetry of the flow with
respect to the bottom wall, which is a streamline, we obtain a
symmetrical flow over a rectangular obstacle (fig. 1(b)). Flow over
polygonals obstacles were studied by many authors.
In the present work we neglect the effect of gravity but we take
into account the effect of surface tension. Far upstream the velocity
~
of the flow is a constant U .
When the effect of surface tension and gravity g are neglected,
the problem has an exact solution that can be computed via the
streamline method due to Kirchhoff or via the hodograph and
Schwartz-christoffel transform (see, for example [4]).
If the effect of surface tension or gravity is considered, the bounda-
ry condition on the free surface is nonlinear and the problem does
not have a known analytical solution. A series truncation procedure
is employed to calculate the flow over a step. This technique has
been used successfully by Birkhoff and Zarantonello [4] ,
Vanden-Broeck and Keller [11] , F. Dias and Vanden-Broeck [7]
, to calculate nonlinear free surface flow and bow flow.
As we shall see, the flow is characterized by the Weber number
 defined by:
𝛼=
̃2 𝐻
̃
̃𝑈
𝜌
𝑇̃
.
(1.1)
~
~
Here T is the surface tension and  is the density of the fluid.
The problem is formulated in section 2, the numerical procedure is
described in section 3 and the results are discussed and presented
in section 4.
1
A.Merzougui
2
over a step. We choose the Cartesian coordinates such that the
~
x  axis is along the bottom streamline and passes through the
y  axis is vertically upward through
stagnation point O and the ~
FORMULATION OF THE PROBLEM
x
Free surface
̃
𝑈
.
̃
𝐻
C
B
̃
2𝐻
y
O
A
B’
D
C’
D’
Fig. 1(a)
the point O (considered as the origin of the axes).
In this article, we neglect the effect of gravity but we take into
account the effect of surface tension. If we neglect the effects of
surface tension and gravity the problem has an exact analytical
solution that can be computed via Schwartz-christoffel transform.
The purpose of this study is two-fold. Firstly, it aims to provide
analytical solution to the problem concerned. The second is to
establish the basis of numerical accuracy of the computer program,
in order to allow satisfactory computation of similar flows with
surface tension present.
Since the flow is irrotational and the fluid is incompressible, we
x  i~
y and the complex potential
define the complex variable ~z  ~
~ ~ ~
~
function f    i where  is the potential function and ~ is
~
the stream function. Since  and ~ are conjugate solutions of
~
Laplace's equation, f (~z ) is an analytic function of ~
z within the
flow region. The complex conjugate velocity is given by
x
̃
𝑑𝑓
𝜁̃ = = 𝑢̃(𝑥̃, 𝑦̃) − 𝑖𝑣̃(𝑥̃, 𝑦̃).
(2.1)
𝑑𝑧̃
~ and v~ are the horizontal and vertical components of
Where u
the fluid velocity, respectively, and may be expressed as
Free surface
̃
𝑈
ũ =
C
B
̃
𝐻
A
O
D
y
̃
∂ϕ
∂x̃
=
̃
∂ψ
∂y
̃
,
ṽ =
̃
∂ϕ
∂y
̃
=−
̃
∂ψ
∂x̃
.
(2.2)
Without loss of generality, we choose ~  0 on the streamline
~
~ ~
x, ~
y )  ( H , H ) ). The
AOBCD and   0 at the point C ( ( ~
~ ~
flow configuration in the complex potential plane f    i~ is
illustrated in Fig. 2.
Fig. 1(b)
𝜓
Figure. 1(a) sketch of flow and the coordinate. The flow is assumed to be symmetrical, so the dividing streamline is x’ox the xaxis is along the streamline AO and the y-axis is vertically upward
through the point O. 1(b) sketch of flow over a step, The free
̃
surface is CD, the velocity far downstream is 𝑈
Let us consider the motion of a two-dimensional potential flow
over a step. We assume that the fluid is inviscid, incompressible
and the flow is irrotational and steady. Since the flow is considered
to be potential the normal velocity vanishes on the horizontal
bottom and the vertical rigid boundaries of the step. Far upstream,
we assume that the flow is uniform so that the velocity approaches
~
a constant U . The flow is limited by the free streamline CD , the
horizontal bottom AO and the rigid boundaries of the step OB
and BC respectively. In the absence of gravity the main flow
extends to infinity in the direction of the bottom AO far downstream (fig. 1(a)). If we take the symmetry of the flow with respect
to the straight streamline AO, we obtain a symmetrical flow over a
rectangular obstacle (fig. 1(b)). Thus, the following formulation is
valid for the two problems. Our formulation is made for the flow
2
𝜙
A
O
B
C
D
~ ~
Fig.2 The complex potential f-plane, f    i~
On the free streamline (free surface) CD , the Bernoulli equation is
to be satisfied, that is
1 2
𝑝̃
𝑞̃ + ̃ = 𝐶 𝑜𝑛 𝜓̃ = 0, 0 ≤ 𝜙̃ ≤ +∞
(2.3)
2
𝜌
Where ~
p is the pressure of the fluid in a point on the free sur-
Free surface flow over a step
face CD , ~ is the density of the fluid and q~ 
speed of the fluid particle on the free surface. Let
u~ 2  v~ 2 is the
~
p0 be the pres~
sure outside the fluid just above the free surface. p0 is considered
to be a constant. Since far upstream the free surface is horizontal,
we have ~
p~
p0 . Thus, the constant C in equation ( 2.3) is evaluated far upstream and is given by
1 2 𝑝̃0
̃ + = 𝐶.
𝑈
2
𝜌̃
A relationship between ~p and ~
p0 is given by Laplace's capillary formula
̃.
𝑝̃ − 𝑝̃0 = −𝑇̃ 𝐾
(2.4)
~
~
Here K is the curvature of the free surface and T the surface
tension. If we substitute (2.4) into (2.3) we obtain :
1 2
𝑇̃
̃ = 1𝑈
̃2 .
𝑞̃ − ̃ 𝐾
(2.5)
2
𝜌
2
~
We introduce the dimensionless variables by taking H as the unit
~
length and U as the unit velocity. The dimensionless variables are
given by:
𝑥̃
𝑦̃
𝑞̃
̃ 𝑅̃
𝑥= ̃ , 𝑦= ̃ , 𝑞= ̃ , 𝐾=𝐻
(2.6)
𝐻
𝐻
𝑈
The dimensionless complex velocities are given by:
f    i ,
Hence u 

x


y
and
v
df
 u  iv
dz
  x .
 

y
( z  x  iy ,  is the dimensionless potential function and  is
the dimensionless stream function u is the component of the
dimensionless velocity in the direction of the x -axis, v is its
component in the direction of the y -axis). In the dimensionless
form, the Bernoulli boundary equation (2.5) reduces to
2
𝑞2 − 𝐾 = 1.
(2.7)
𝛼
Here  is the Weber number defined in (1.1).
The physical flow problem as formulated above can be formulated
as a boundary value problem in the potential function  ( x, y) .
2  0 in the flow domain

2)
=0 on the rigid boundary, (the normal velocity vanishes

on the rigid boundaries)
2
  2 K  1 on the free surface
3)

4)
 (1,1)  0 .
Solving the problem in this form is very difficult especially that the
nonlinear boundary condition is specified on an unknown boundary (the free surface). Instead of solving the problem in its partial
differential equation form in  , we take advantage of the property that for the bidimensional potential flow (as is in our problem)
and if the plane in which the flow is embedded is identified to the
complex plane, the complex velocity   u  iv and the complex
potential function f    i are analytical functions. Hence, we
use all the necessary properties of analytical functions of a complex variable: integral formulation, series formulation, conform
transformation etc...
We rewrite the dimensionless complex velocity in the new variables  and  as
𝜁 = 𝑢 − 𝑖𝑣 = 𝑒 𝜏−𝑖𝜃 .
(2.8)
where e  
and  is the angle the streamline makes with the
x-axis. We seek  and  as functions of the variables  and  .
In these variables (𝜙, 𝜓) the flow domain is the upper half plane
     and   0 .
Hence, the equation (2.7) becomes
𝜕𝜃
𝛼
| | = (𝑒 𝜏 − 𝑒 −𝜏 ) − ∞ ≤ 𝜙 ≤ +∞ and 𝜓 ≥ 0
(2.9)
𝜕𝜙
2
The kinematic condition on AO and OB yields
𝜃 = 0, 𝜓 = 0,
𝜙 ≥ −2 𝑜𝑛 𝐴𝑂
{
𝜃 = 0, 𝜓 = 0, −2 ≤ 𝜙 ≤ −1 𝑜𝑛 𝐴𝐵
(2.10)
We shall seek   i as an analytic function of f    i in
the region   0 .
We go a step further with conformal mapping. Using the schwartzchristoffel transformation, we map the flow domain the strip
  0 in the f -plane onto the first quadrant of the unit disk an
auxiliary t -plane by the transformation
2
2𝑡
𝜋
1+𝑡 2
O
is
𝑓 = log
The
stagnation
point
.
(2.11)
mapped
into
the
point
t  e  e  1 , the point B into t  e  e  1 , the points
at infinity A and D correspond to the points t  0 and t  i
respectively and the separation point C into t  1 . The solid
boundary maps onto the real diameter and the free surface onto the
circumference of the quadrant of the unit circle (fig. 3).

 /2
2

D
1)

A
O
B
C
Fig.3 The complex potential t-plane
In all the flow domain, the complex velocity  (t )  u  iv is
analytic except at the point O , the point B and the separation
point C which correspond to t  e  e2  1, t  e / 2  e 1
and t  1 . Hence, a close study in the neighborhood of these
points is to be done.
 At t  e  e2 1 , we have a flow inside an angle of measure 2 , this gives the local behavior of  (t ) as
2
𝜁(𝑡)~𝑂 (𝑡 2 − (𝑒 𝜋 − √𝑒 2𝜋 − 1) ) as 𝑡 ⟶ e  e2  1 (2.12)

 At t  e 2  e  1 , we have a flow over an angle of measure
3
The appropriate singularity is
2 .
3
A.Merzougui
2 −1
𝜋
𝜁(𝑡)~𝑂 (𝑡 2 − (𝑒 2 − √𝑒 𝜋 − 1) )

as 𝑡 ⟶ e 2  e  1 (2.13)
 At the separation point which correspond to t  1 the local
behavior of  (t ) is
𝛾
2(1− )
𝜋
𝜁(𝑡) = 𝑂(𝑡 − 1)
𝑎𝑠 𝑡 ⟶ 1.
(2.14)
Where  is the angle between the free surface and the rigid
boundary of the step (    if the surface tension T  0 and
   if T  0 )
Now that we know the local behavior of  (t ) near the singularity, we seek the function  (t ) as a series of the form
2 −1
− 1) )
𝛾
𝜋
2(1− )
× (𝑡 − 1)
×
2𝑘
𝑒𝑥𝑝(∑∞
𝑘=0 𝑎𝑘 𝑡 )
(2.15)
The coefficients a k are to be determined. Since (2.15) satisfies
(2.12), (2.13) and (2.14) we expect the series to converge in the
quadrant of disk in the t -plane. The coefficients a k are chosen to
be real, so that the boundary conditions (2.10) are satisfied i.e.
u  0 on OB and v  0 on AO .
We use the notation t  t e
i
so that the points on CD are given
𝜋
i
by t  e
and 0 < 𝜎 < . Using (2.11) the expression (2.9) is
2
rewritten as
𝜋
𝜕𝜃
𝑒 2𝜏 − 𝑐𝑜𝑡𝜎 | | 𝑒 𝜏 = 1
(2.16)
𝛼
Here  ( )
𝜕𝜙
and  ( )
denote the values of  and  on
𝜋
i
i
and 0 < 𝜎 <
Setting t  e and  (t )  e
2
(2.16) yields an equation to determine the unknowns a k .
te
 i
𝜎𝐼 =
2𝑁
1
(𝐼 − ) ,
𝐼 = 1,2, … , 𝑁
2
substituting these expressions into (2.15) we obtain N  1 non-
linear algebraic equations for the N unknowns an n1 and  .
N
That is, for each  ( j ), j  1,, N

exp( j ) cot( j )  j  1.



|  j and
2
𝜁(𝑡𝑗 ) = 𝑒 𝜏𝑗 −𝑖𝜃𝑗 = (𝑡𝑗 2 − (𝑒 𝜋 − √𝑒 2𝜋 − 1) )
2 −1
𝜋
× (𝑡𝑗 2 − (𝑒 2 − √𝑒 𝜋 − 1) )
× (𝑡𝑗 − 1)
𝛾
𝜋
2(1− )
𝜕𝜙
𝜁
(3.2)
𝑢−𝑖𝑣
In the new variables  and  , (3.2) rewrites
𝜕𝑥
{𝜕𝜎
𝜕𝑦
𝜕𝜎
2
= exp(−𝜏) cos(𝜃) tan(𝜎)
𝜋
2
(3.3)
= exp(−𝜏) sin(𝜃) tan(𝜎)
𝜋
To obtain the form of the free surface CD , we integrate numerically the expression (3.3) in the interval 0     , with the
2
initial condition x (0)  1 , y (0)  1. The Euler method was used
to integrate numerically the relation (3.3).
4 RESULTS AND DISCUSSION
We use the numerical procedure described in section 3 to compute
solutions of the problem for various values of the Weber number
 . For fixed values of  (0    ) the coefficients a n were
found to decrease very rapidly as n increases (table 2).
For values of  very large, (    ) (table 1), all the coefficient of the series (2.15) are zeros ( ai  0 for all i  1) . This
gives the exact solution for (  )
2
 
2

1
. This
result was compared with the exact solution found via the hodograph transform due to kirchhoff and were found to agree exactly. In figure 1, the exact solutions via the hodograph transform
{
𝑥=
1
√ϕ+1−√ϕ
√2+log(√2−1) 2
1
√ϕ+1−√ϕ
[ log (
+ √ϕ(ϕ + 1))]
(4.1)
𝑦=1
was
compared
with

2
the
 
solution
2

 (t )   t 2   e  e2  1     t 2   e / 2  e  1  

  

 

1
.
The
two solutions were found to be identical.
With this numerical procedure we could compute solution for the
Weber number  very small. As an example we could compute
the solution for all   0.2 . There exists a critical value  
(   very small) of  such that for     the numerical
scheme ceases to converge . The free surface was smooth with no
capillary waves even for small  . As the Weber number decreases from  to   , the shape of the free surface tends to a
straight vertical line (fig.2).
1
Figure 3 shows the variation of the angle of separation 𝛾 versus .
𝛼
From the above numerical results, we conclude there exist a unique
solution with a smooth free surface for all     .
The results presented here are obtained with N  50 .
∞
× 𝑒𝑥𝑝 (∑ 𝑎𝑘 𝑡𝑗 2𝑘 )
𝑘=0
4
𝜕𝜙
(3.1)
by collocation. Using (2.16) we obtain  ( ), ( ) and  in

terms of coefficients a n and the separation angle  . Upon
Where  j =  ( j ) ,  j 
Newton method for given values of  .
To draw the free surface we use the identity
𝜕𝑥
𝜕𝑦
1
1
+𝑖 = =
= 𝑒 −𝜏+𝑖𝜃
 (t )   t 2   e  e2  1     t 2   e / 2  e  1  

  

 

The N coefficients an and the separation angle  are found
exp( 2 j ) 
The Weber number  is a parameter. We solve this system by

in
3 NUMERICAL PROCEDURE
We solve the problem by truncating the infinite series in (2.15)
after N terms. Introducing the N mesh points
𝜋
i j
𝜋
2
𝜁 = 𝑢 − 𝑖𝑣 = (𝑡 2 − (𝑒 𝜋 − √𝑒 2𝜋 − 1) ) × (𝑡 2 − (𝑒 2 −
√𝑒 𝜋
with t j  e
𝛾
𝑎1
𝑎2
𝑎5
𝑎10
𝑎20
𝑎30
𝑎40
𝑎49
Free surface flow over a step
31459
6 . 0 8 5
5
4
10
10
2 .5 7 5
8 .1 6
10
11
10
10
1 . 7 6 5
13
10
2 .5 9 9
 1.2 1
11
14
10
10
3 . 1 8 3
12
10
5 .3 1 9
1 014
Table 1.Some values of the coefficients 𝑎𝑛 and value of the angle
of separation 𝛾 of the series (2.13) for 𝛼 ⟶ ∞

10
3
100
50
30
2
0.88
0.2

a5
a 10
a 30
a 40
31446
10 4
31912
10 4
34814
10 4
44153
10 4
47
10 1
471
10 2
474
10 2
3.192
10 4
5.071
10 3
3.475
10 2
1.165
10 1
5.731
10 4
7.917
10 3
8.395
10 2
1.055
10 4
1.673
10 3
1.141
10 2
3.553
10 2
3.309
10 4
9.049
10 4
9.584
10 3
5.273
10 6
7.277
10 5
4.991
10 4
1.261
10 3
6.582
10 6
1.89
10 5
1.848
10 4
9.511
10 7
8.265
10 6
5.942
10 5
1.437
10 4
7.014
10 7
2.016
10 6
5.645
10 4
Table 2.Some values of the coefficients 𝑎𝑛 and values of the angle
of separation 𝛾 of the series (2.13) for different values of the
Weber number 𝛼.

3.14159
1

0
3.16
3.1912
3.4814
4.4153
4.638
5  10 3
10 2
2  10 2
3  10 2
5  10 2
Table 3.Some values of the angle of separation 𝛾 versus
4.69
0.076
1
𝛼
5
A.Merzougui
REFERENCES
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