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Bonetto1992-Simulation-TwoPhaseJet.pdf
AD-A259 326
A NUMERICAL SIMULATION OF TWO-PHASE JET
SPREADING USING AN EULER-LAGRANGIAN TECHNIQUE
F. Bonetto, D. Drew, R.T. Lahey, Jr.
Centerfor MultiphaseResearch
Rensselaer Polytechnic Institute
Troy, NY 12180-3590 USA
D I
ELECT E
JAN 12 19 9 3
C
74VIM
jJ'
The objective of this paper is to study the spreading of a submerged twophase jet. A plunging liquid jet impacting on a still liquid pool may
carryunder bubbles by entraining the surrounding gas. For low liquid
jet turbulence, the measured average bubble size was about 200 m
[Bonetto and Lahey, 1992]. For low void fraction, one may assume that
the liquid velocity field is not disturbed by the presence of the bubbles.
This allows the use of known solutions for the velocity field in a planar
liquid jet. That is the liquid flow may be solved using Eulerian
coordinates and neglecting the effect of the bubbles. The momentum
equation for the bubbles has been written in Lagrangian coordinates.
The resulting equations are solved for bubble trajectories in the known
liquid velocity field.
INTRODUCTION
1Z
'1+
Z
e,-(
The gas entrapment by a plunging liquid jet and the resultant two-phase
jet dispersion occur in many problems of practical interest. The
ecological balance in lakes and oceans is dependent on the amount of
dissolved oxygen. In addition, the air carryunder that occurs at most
hydraulic structures in rivers is primarily responsible for the large
air/water mass transfer that is associated with these structures [Avery
and Novak, 1978]. Also, the absorption of C02 by the oceans appears to
play an important role (as a sink term) in reducing the magnitude of the
"greenhouse effect". The absorption of greenhouse gases has been
hypothesized to be highly dependent upon the air carryunder that occurs
due to breaking waves.
=
I S..
93-00637'
III~niliillllVlVl
Plunging jets are also used for dissolving gas into liquid. Some
chemical reactors use a plunging jet to entrain bubbles, the reaction rate
is increased because of the increase in interfacial area and the
enhancement in the mass transfer characteristics produced by the
turbulence in the two-phase jet.
The problem under study is a planar liquid jet impacting on a
liquid pool surrounded by a still gas. Lezzi and Prosperetti [1991] and
Bonetto et al [1992] have recently proposed mechanisms for the gas
entrainement. In this work the dispersion of a two-phase jet is studied.
A dilute dispersion of bubbles in the liquid is assumed. The liquid
velocity field is assumed to not be affected by the presence of the
bubbles. The momentum balance equations are presented in the
Momentum Balance Section. Then, the Tolmien's solution for the mean
velocity of a turbulent liquid jet is derived. Finally, the integrated
bubble trajectories and other derived quantities of interest, such as the
maximum penetration depth of the bubble, the total time that the bubbles
are under water and their emergence position at the pool's surface are
presented.
MOMENTUM BALANCE
Newton's law for the gas phase takes the form
Dgu
p V -9
g g Dt
(1)
-
where g, Vb, Ug are the particle density, volume and velocity,
D Ug
respectively.
g
is the material derivative of the particle velocity
Dt
Ac*aae%.j
following the particle and F is the applied force.
N'
e & V
QtW
M~e IArp
It is convenient to divide the force, F, into three parts, one being the!
inviscid flow force (Re=oo) and the others corresponding to viscous flows
FV and the buoyancy force FB.
E=
+F +F()
-1I+
_V+_B
§1
(2)~'
2
For
"O
,,'.,
i) Inviscid flow force: The inviscid flow force can be further separated
into two parts, one corresponding to an irrotational flow and the other
corresponding to a rotational flow.
FI=Firrot) +(Frot)
(3)
+ (o
=
For spherical bubbles in inviscid irrotational flows the force EI. assuming
that the liquid phase pressure gradient is impressed on the bubbles, is
given by:
EI
F
DuMt
pjV b(1 + C )L
v"
Dt
D Ug
g
vm Dtj
_C
(4)
D~ue
Dt- is the material derivative following
the liquid and Cvm is the virtual mass coefficient (for spheres Cvm=1/2).
where ue is the liquid velocity,
aue.
The term -'
a t is zero because in this study the liquid flow was steady,
thus the liquid acceleration is given by,
De 1 = (u
Dt
V)u j
e)~
A general expression has been derived for the lateral force on a sphere in
weakly rotational flows (Vu #0, Nu b<< 1). Auton et al [1988] and
Iur I
Drew & Lahey [1987,1990] show that,
-
(5)
Fi (rot) = CL (Ug - !1j) X V X _uJ
where CL is the lift coefficient which, for spheres is, CL=l/2.
ii) Viscous Flows (moderate Re): The effect of the viscous stress will now
be considered. A particle moving in a steady flow has a non-steady
component of the drag force due to vortex shedding. The viscous force is
then:
3
Fv(t) =FD +
Iu(t)
(6)
(7)
PICDAb lUrlUr
-2D=
Where ED is the steady drag force component parallel to the relative
velocity and E.L(t) is the non steady force component perpendicular to the
relative motion. For moderate Re numbers (Re =_100) the drag force is
caused by the pressure field asymmetry in the flow around the particle.
The vortex shedding produces an oscillating pressure distribution around
the bubble and a non-steady force component. Graham [1980] showed
that if the Strouhal number is S=Rb/(Iurl T) < 0.3, F_(t) is less than 10 %
of the value of FD, where T is the period of the vortex shedding. In our
simulation the previous condition is satisfied and hence EL(t) is neglected.
That is,
EV-ED
(8)
iii) Buoyancy force: The buoyancy force is due to the difference in
density between the bubbles and the liquid,
FB =_(P
(9)
- Pg)Vbg
Finally, we note that,
DgXg
Dg x-g
Dt
(10)
= Ug
where x is the position of the particle at time t. The momentum balance
equation is obtained by combining Eqs. (1) to (5), (8) and (9). It is
convenient to write Eq. (10) and the momentum balance in component
form for a two-dimensional steady-state flow:
U
Dt
(1 la)
gY
4
D z
g gg
Dt
PVDugg
PgVb
(1lb)
gz
Dt
PV(i0
j
)!
[0! 0_ )Ai
Dg9ugl
Dt
PVb(-UV)u-+CvrPVb (uOV)u-i
PjA
Iui Ug (1j-)+
(uI
(Pj -pg)Vbg
P£bD
b
- 9
U
+ CLPeVb ur X (V X uj)
(1 Ic)
TOLMIEN SOLUTION
Let us now consider the solution for a spreading turbulent single-phase
liquid jet. The jet has a uniform velocity profile where it impacts the
liquid pool at z=O. After the impact, a boundary layer develops at the
edge of the jet. Figure-1 shows the different jet regions. Region-1 is the
inviscid core. Inside this region the velocity is the inlet velocity, V.
Region-2 is where the jet is approximately self-similar. At larger
distances from the impacting point the jet is closer to a true self-similar
condition. Tolmien's calculation will be followed to obtain [Hinze, 1975]
the mean velocity values in the self-similar region. It is assumed that the
liquid emanates from an infinitely narrow cross section. Also it is
assumed that the pressure inside the jet is uniform and equal to the
Then the momentum crossing a plane
surrounding pressure.
perpendicular to the direction of the flow is constant:
T
(12)
2pu2dy = constant
Assuming the velocity profile is self-similar,
5
Uz
u0-
Y
2 dY
then
z= =constant
Q0
z
(13)
z
where z' is the axial distance measured from the infinitely narrow cross
section. Combining eqs. 12 and 13 we obtain the relationship between
the center-line axial velocity uz(y=O) and z' is as:
(14)
u0=u (y=O)=- Cz
where c is a constant. The velocity at an arbitrary point is given by:
U =--
f(1)
(15)
It is useful to define the stream function as:
z
uay
ýY
u y =--=a}z
(16)
then
V= JU dy=c-zF(T1)
,where,
(17)
F(TI) = j f(i1)dil
The momentum balance in the z direction for the control volume shown
in Figure-1 is:
u + ayUz2 dy + d2z ,2au
=0
.• z •2=0(8
UzUy+-z
z y
(18)
aPfzD
Co
where d is the mixing length constant and, according to mixing length
theory, the turbulent shear stress was assumed to be,
6
-Pd2z'2
(19)
Using Eqs. (16) and (17) we obtain:
2d2(F"
(20)
= F(71)F'(il)
-I))2
Introducing the additional change of variables,
Y=R
a
,where,
a=-2)
(21)
(F"(p))2 = F(p)F'(9)
(22)
yields,
The appropriate boundary conditions at the centerplane p=O are that the
lateral velocity Uy equals zero and the axial velocity Uz is uz(y=O), thus,
F(p=O)=O
, F'(9=0)=l
(23)
Equation (22) has been numerically integrated with the boundary
conditions Eq. (23). In order to compute the velocities the value of the
parameter has to be known. The value obtained experimentally by
Forthmann [Abramovich, 1963] was used, that is, a=0.11.
Note that this calculation corresponds to a jet that is self-similar
everywhere and that it emerges from an infinitely narrow slit. Actually,
the jet emerges from a finite slit with half width equal to h and constant
velocity V. In order to reconcile these differences an effective origin is
computed for the Tolmien jet, and it is required that its momentum agree
with the actual jet momentum at impact. The momentum flux
conservation condition is:
7
(24)
Pto
= pV 2 h
PUf2zdy
The distance from the infinitely narrow slit, z', and the distance from the
finite slit, z, are related by,
z'=z+s
,where, s--0.41h/a
A final modification has to be made to the self-similar solution to account
for the inviscid core region. For the infinitely narrow solution the
velocities goes to infinity we get closer to the slit. In the finite slit, the
velocity can not exceed the inlet velocity V, thus wherever the velocity
exceeds V, it is reset to V. Figs. 2 and 3 show the lateral and the axial
velocity Uy and Uz as a function of the lateral coordinate y for different
axial positions z. Note that at the inlet the velocity profile is uniform and
the jet is spreading as we increase the axial position z. Figs. 4 to 8 show
the various spatial derivatives.
DRAG LAW
To this point, the analysis is valid for both bubbles and solid particles.
However, there is a difference between the drag that the liquid exerts on a
bubble compared to that for a solid particle.
The so-called "dirty water" Wallis drag correlation as been used to
compute the drag on bubbles:
6.3
CD - Re0 .385
(25)
In contrast the Clift, Grace and Weber correlation is generally used to
compete the drag in liquid/solid flows
C = 24 (1 +0.15 Re 0 .6 87 ) +
CD=e10[1
0.42
+ 4.25 104 Re- 1 .16 ]
8
(26)
where Re=2 Rs lurl/10t
RESULTS
Before presenting the results is convenient to make the following change
of variables:
x1 = y cos(0)+ z sin(0)
x2 = y sin(0)- z cos(0)
Figure 9 shows the trajectories of three bubbles with initial positions (a)
(y=-h/2,z=Rb), (b) (y=O,z=Rb) and (c) (y=h/2,z=Rb). Bubble (a) is
initially at the center plane, while the other two are initially at one side or
the other of the jet centerplane halfway to the edge of the jet. For the
case shown in Figure 9, the liquid planar jet is impacting the pool at an
angle of 0=45 degrees with respect to the vertical. The uniform velocity
of the jet at the impacting point is V=5m/s, the jet width is 2h=4.16 mm
and the radius of the bubble is Rb=2mm. For the base case we used the
coefficients of the lift force and the virtual mass force computed
analytically assuming that the liquid was inviscid, that is, Cvm=l/2 and
CL=1/2.
A qualitative description of the flow follows. We consider first the
bubble that is initially at the center of the jet. In the inviscid core region,
all derivatives of the liquid velocity field are zero, then the only forces
acting acting on the bubbles are drag and buoyancy. The drag force acts
on the bubble in the same direction that the liquid flow. In the inviscid
core the lateral velocity, uy, is zero and the axial velocity, uz, is equal to
V. Thus, the drag force in the inviscid core tends to move the bubble
parallel to the jet and at a gas velocity near V. The buoyancy force acts in
the upward vertical direction. That is, buoyancy tends to cause the bubble
to move towards the free surface. In the self-similar region, the lift force
keeps the bubble near the centerplane, the virtual mass reduces the
acceleration that the bubble experiences for a given force.
We take the previous solution as the base solution in parameter space.
Then the effect of the different parameters are explored one at a time and
9
compared with the base solution.
MAXIMUM DEPTH
One of the important parameters is the maximum depth that a given
bubble reaches. Figure 10 shows the maximum depth as a function of the
inlet liquid velocity V for three different inlet positions. The increase of
the maximum depth with the velocity is easily seen. This is expected
because when we increase the velocity by a factor AV, the lift force that
keeps the bubble near the center plane increases a factor (AV) 2 while the
buoyancy force that tends to move the bubble toward the free surface
remains the same. Notice that the depth reached by bubble with initial
conditions in the lower half of the jet have similar trajectories compared
to the bubbles initially at the center plane. This is again due to the fact
that the bubbles that remain near the centerplane longer are the ones that
reach greater depths. A typical trajectory of a bubble that is at the lower
half of the jet initially is as follows: buoyancy tends to move it up (i.e.
toward the center plane), however as the bubble gets closer to the center
plane, the higher liquid velocities there produce a larger drag and the
bubble flows toward greater depths.
EMERGENCE DISTANCE
Another quantity of importance is the distance from the jet impact
position to the place where the bubble reemerges at the free surface.
Figure 11 shows the distance of emergence, s, as a function of the jet inlet
velocity V for three different initial bubble positions. Again thc
"
ulcreases with the inlet velocity V because of the effect of the lift force.
It is clearly seen in Figure 11 that the bubbles that are initially in the
lower half of the jet (yi<O) have a longer emergence distance s than the
bubbles initially in the upper half of the jet (y>O). Therefore s increases
as a function of the initial bubble position yo (-h<yo<h). Thus, the larger
the initial position yo (-h<y0 <h), the larger the emergence distance s,
except for yo very near -h. For yo values near -h, an interesting
phenomenon occurs. Figure 9 shows two typical trajectories with y0 1 >Y02
both in the neighborhood of y0 =-h. The buoyancy force acts over the
bubble with initial position y01 producing a lateral velocity uey >0 which
10
is greater than the lateral velocity of bubble number 2, u y
>U1y
2
>0.
Then, bubble-1 crosses the jet centerplane sooner than the bubble-2.
Thus, bubble 2 has a larger emergence distance s.
Notice that the two trajectories cross each other . An Eulerian approach
would give some kind of average behavior at this point. This result does
not contradict the fact that a set of Ordinary Differential Equations
(ODE) has an unique solution. The trajectories are non-crossing in full
phase space. The dimension of the space is equal to the number of
independent equations (i.e., four). Thus the coordinates in phase space
are y,zuyuz. The trajectories presented in the y-z plane can be seen as
projections of the trajectories in full phase space on the y-z plane. Note
that the x1-x2 plane is identical to the y-z plane. This effect can not occur
using partial differential equations resulting from the Eulerian approach.
The corresponding solution would effectively average the different
velocities at trajectory crossings, and would result in a single velocity at
each point.
SENSITIVITY
PARAMETERS
OF THE
SOLUTION
TO
THE
PROBLEM
We wish to study the effect of the various parameters on the resulting
bubble trajectories. What follows is a parametric study of the solution of
eq (11). In the following figures the dashed lines correspond to the base
solution trajectories from the previous section. The solid lines show the
effect of changing the corresponding parameters. Again curves (a), (b)
and (c) correspond to the initial conditions described in Fig. 9.
In Figs. 13 and 14, the lift coefficient was varied, with CL=0.25 (for
a base case solution of CL=0. 5 ). For the trajectories (a) and (b) the
reduction in the lift coefficient produces a reduction in the ability of the
bubble to remain near the center plane. The y component of the lift force
is:
(
F(rot)y
ly
=CL(uzu)
L
au
z
y <0,
azJ 0 '
if, y>0
fa
>1
Figure 8 shows the derivatives as a function of y. This factor is always
negative for y>0. Also the axial component of the relative velocity is
positive. Thus for y>O the lift force tries to push the bubblea to the
centerline. At the center plane the axial liquid velocity has a maximum
and therefore the drag acts to move the bubble to greater depths.
Obviously, there is a strong dependence of the emergence point, s, on the
vaLae of the lift coefficient.
There is also a lift force component in the axial direction:
- U )
F(rot) = C (u
Iz
L
y
•yi
ay
z
a
For bubbles in the upper half of the jet, the lateral bubble velocity
component is in general greater than the lateral liquid velocity
component. Thus, the lift force component in the axial direction is
negative. For trajectory (c) the negative lift force component in the z
direction is reduced and the emergence position, s, is increased. In Fig 14
the lift force has been completely eliminated (CL=O) and we find that the
emergence distance of the trajectories, s, are fairly close together. Hence
we find that a larger lift coefficient results in a spreading of the
emergence distance for the various trajectories.
Figs. 14 and 15 correspond to varying the virtual mass coefficient
from the base case value of Cvm = 0.5, to Cvm =0.4 and Cvm =1.0. We
note that a larger virtual mass coefficient results in less acceleration of the
bubble, so that an increase in the virtual mass coefficient produces an
increase in the ability of the bubble to follow the liquid flow.
Figs. 16 and 17 corresponds to the effect of the jet inclination angle,
0. Fig. 16 shows a jet impacting at an argle of 40 degrees (base case
=45'). The maximum depth dm is seen to be sensitive to the angle. On
the contrary, the location of the emergent bubble is seen to be somewhat
insensitive to the angle . Fig. 18 shows a jet impacting at an angle =60
degrees. For this case both the maximum depth, dm, and the distance of
emergence, s, are seen to be strongly affected by the impacting angle.
12
SUMMARY AND CONCLUSIONS
A dilute bubbly jet has been analyzed using a Eulerian-Lagrangian
approach. It is assumed that the presence of the bubbles does not affect
the liquid flow. Under this assumption the liquid flow is decoupled from
the bubble motion, and is given by the Tollmien self-similar jet solution
modified to account for the inviscid core. That is, it was not necessary to
solve for the liquid flow field. The momentum balance for the gas phase
has been rewritten as a system of four ODEs. The independent variable is
time and the four dependent variables are the bubble position (y, z) and
the bubble velocity components (uy, uz). The system of ODEs was
integrated using a fifth order Runge Kutta scheme. The initial conditions
were: y(t--O)=yo, z(t=O)=O,uy(t=O)=O,uy(t=O)=V.
Based on the computed trajectories, three quantities of practical
interest were estimated: The maximum depth dm, the bubble emergence
distance, s, and the total time that the bubble is immersed in the pool, t.
One may be interested in knowing the aeration produced by a breaking
wave in the sea. The maximum depth provides information about the
amount of the aerated layer. The time under water gives information
about the amount of gas that is dissolved. However, in order to compute
the amount of dissolved gas, an additional model for the diffusion process
needs to be included. Finally, the distance of emergence gives
information about the horizontal extent of the aeration produced by the
breaking of a single wave. The results show that dm, t and s increase with
the jet velocity, with faster increase for low values of V. Also the bubbles
initially at the lower half of the jet have larger dm, t and s than the
bubbles initially at the upper half of the jet for the same velocity.
The relative velocity in the axial direction, ugz-utz is negative inside
the jet. Thus the lateral component of the lift force tends to move the
bubble toward the jet centerplane. The axial component of the lift force
is i. ,•ative if the relative velocity is positive and this force tends to retard
the bubble motion.
The effect of virtual mass force is not as straightforward to analyze.
13
The reason is that the terms 0 Iz and
W
uy change sign inside the liquid
az
jet. This can be easily seen in Figs. 4 and 7.
We conclude that both the virtual mass and the lift forces play an
important role in the spreading of a two-phase jet.
ACKNOWLEDGMENT
The authors wish to acknowledge the funding provided for this
research by the Office of Naval Research (ONR), Grant No N-0001491-J1271, Fluid Dynamics Program.
14
REFERENCES
Abramovich, G. "The Theory of Turbulent Jets", MIT Press, (1963).
Avery, S. & Novak, P., "Oxygen Transfer at Hydraulic Structures",
Journal of Hyd. Div., ASCE, vol 104 (HYll), pp. 1521-1540, (1978).
Auton, T.R.; Hunt, J.C.R. & Prudhomme, M., "The Force Exerted On A
Body In Inviscid Unsteady Non-Uniform Rotational Flow". J. Fluid
Mech., 197, pp. 241-257, (1988).
Bonetto, F. and Lahey, R.T.,"An Experimental Study On Air Carryunder
Due To A Plunging Liquid Jet". To be published in Int. Journal of
Multiphase Flow, (1992).
Drew, D.A. & Lahey, R.T., "The Virtual Mass And Lift Force On A
Sphere In Rotating And Straining Inviscid Flow", Int. J. Multiphase Flow,
13, #1, pp.113-121, (1987).
Drew, D.A. & Lahey, R.T., "Some Supplemental Analysis Concerning
The Virtual Mass And Lift Force On A Sphere In A Rotating And
Straining Flow", Int. J. Multiphase Flow, 16, #6, pp. 1127-1130, (1990).
Graham, J.M.R., "The Force On Sharp-Edged Cylinders In Oscillatory
Flow At Low Keulegan-Carpenter Numbers", J. Fluid Mech., 20, pp.
331-346, (1980).
Hinze, J.O., "Turbulence", McGraw-Hill, (1975).
15
a.)
Cu
1
--
00a)
4))c
.0
N
I03
ca
C"'
(a) z=O.lm
(b) z=-O.2m
(c) z=O.3m
0.20 •(d)
z=O.4m
(a)
"(b)
0.10
-(d)
u, (m•s)
(c)
-0. 0
-0.10
-0.20
-0.10
-0.05
-0.00
0.05
0.10
y(m)
Figure 2: Lateral velocity uIy as a function of the lateral position y for
different axial positions z
(a) z--O.lm
(b) z=O.2m
(c) z--O.3m
(d) z--O.4m
3.00
2.50
(a)
2.00
U1z(m/s)
()
1.50
(d)
1.00
0.50
0
-0.10
-0.05
-0.00
0.05
0.10
y(m)
Figure 3: Axial liquid velocity Ulz as a function of the lateral position y for
different axial positions z
(a) z=O.lm
(b) z=O.2m
(c) z--O.3m
(d) z=0.4m
20 -'
10
0'
-t-(l/s)
dy
(C.d) d
-10
(a)
-20-
-0.10
-0.05
-0.00
0.05
0.10
y(m)
Figure 4:
positions z
-uty
dy
as a function of the lateral position y for different axial
(a) z--O.lm
(b) z---0.2m
5-
(c) z=0.3m
(d) z=0.4m
3
dlut
Ž0y (l/s)
y(d)
(c)
-i..
-
(b)
(a)
-0.10
-0.05
-0.00
0.05
0.10
y(m)
Figure 5: 8Uty as a function of the lateral position y for different axial
dz
z
positions
(a) z--O.lm
(b) z=0.2m
(c) z--0.3m
(d) z--0.4m
15"0
(a)
100_
50O 50
°uJL(l/s)
du
0-
(b)
(C)
(d)
dy
-50
-100
- 15 0 - ..
-0.10
, ,
..
-0.05
,
...
-0.00
y(m)
..
0.05
0.10
Figure 6: duz as a function of the lateral position y for different axial
dy
positions z
(a) z=O.lm
(b) z=O.2m
z=O.3m
(d) z--O.4m
201
'1(c)
10-
dz
(a)
- 20
-0.10
, , , '
-0.05
. .
- ,.
-0.00
.
., .
0.05
. .
0.10
y(m)
du
Figure 7: ----- as a function of the lateral position y for different axial
positions z
(a) z--O.lm
(b) z--O.2m
(c) z--0.3m
(d) z--0.4m
150
100
50
dul
dy
cl~u
(d)
z
(c)
-50
'(b)
-100
(a)
-0.10
-0.05
-0.00
0.05
0.10
y(m)
Figure 8: Multiplier for the lift coefficient as a function of the lateral position
y for different axial positions z
x2 (r)
0.05
o'
I
i(c)
-0.05
'--'
(b)
-0.12
-0.15
-'..
,
-0.2
"-S
"".
(a)
-0.25-0.3
-0.35
-0.41
0
0.1
0.2
0.3
0.4
0.5
x, (M)
CL = 0.5
Cvm = 0.5
0=45 0
V=5m/s
Rb--0.002m
Figure 9: Bubble trajectories for the base solution and different initial
positions.
(a) y(t=O)=-h/2
(b) y(t=O)=O
(c) y(t=O)=+h/2
(a)
0.20
(b)
0
C1
0.15
0.10
(c)
00
C:C
0.05
0
0.00
i
0
I
1'
IIII
111
2
111
li
3...
'
4
l
ii
11
5
6
V(m/s)
Figure 10: Maximum depth dm as a function of the jet velocity for
different initial positions:
(a) y(t=O)=-h/2
(b) y(t=-O)=0O
(c) y(t=-O)=+h/2
0.4
(a)
0.3
0
C
B 8
s(m)
8 n °
° (b)
n
(c)
H
0.2
0
C3
00
C
0.1
0.0
0
1
2
3
4
6
5
V(m/s)
Figure 11: Emergence distance s as a function of the jet velocity for
different initial positions:
(a) y(t=O)=-h/2
(b) y(t=O)=O
(c) y(t=O)=+h/2
1.4 !(a)
1.2
0
1.0
0]
0
(b)
0
0.80
8
t(s)
0.6
0
0.4
[
o
[](c)
0
0.2
0 .0
,,''''''
0
'I'
1
I',''
'III
'''I
2
''II'''''''''1
3
I
''''11
,,,,'''''I'''''
4
5
6
V(m/s)
Figure 12: Emergence time t as a function of the jet velocity for different
initial positions:
(a) y(t=O)=-h12
(b) y(t--O)=O
(c) y(t--O)=+h/2
9.05
(c)
X2 (M)
(b) (a)
0I
-0.05-I
-0.1
-
-0.15
-0 .2
""
-0.25
.
-0.3
-0.35
-0.40
0.1
0.2
0.3
0.4
0.5
x, (m)
CL = 0.5
Cvm = 0.5
0=450
V=5m/s
Rb--0.002m
Figure 13: Bubble trajectories for CL=0.25 and different initial positions
(a) y(t=0)=-h/2
(b) y(t--0)--0
(c) y(t--O)=+h/2
The dashed lines are the base solutions
x2 (M)
0.05
(b)
0o
!(c)
(a)
-0.05-0.1-0.15
-..
-0.2-0.25
-
-0.3
-0.35
-0.4 1
0
0.1
0.2
0.3
,
0.4
0.5
x, (M)
CL = 0.5
Cvm = 0.5
0-_450
V=5n/s
Rb--0.002m
Figure 14: Bubble trajectories for CL--O and different initial positions
(a) y(t=0)=-h/2
(b) y(t--O)=O
(c) y(t--O)=+h/2
The dashed lines are the base solutions
0.05
X2 (M)
(c)
(a)
'(b)
-0.051
-0.15
I
-01--
-0.2-0.25 -
-0.3
-0.35
-0.4
0
0.1
0.2
0.3
0.5
0.4
x, (M)
CL = 0.5
Cvm = 0.4
0--450
V=5rn/s
Rb=0.002m
Figure 15: Bubble trajectories for Cvm=0.4 and different initial positions
(a) y(t=0)=-h/2
(b) y(t=O)---O
(c) y(t=O)=+h/2
The dashed lines are the base solutions (Cvm = 0.5)
x 2 (in)
0.05
0-
(c)
-0.(c)
(a)
!(b)
(b
I
-0.1
05' -
-0.15....
-0.2
-0.25-
-0.3
-0.35
-0.4 1
0
0.1
0.2
0.3
0.5
0.4
x, (M)
CL = 0.5
Cvm = 1.0
0=450
V=5m/s
Rb--0.002m
Figure 16: Bubble trajectories for Cvm=l.0 and different initial positions
(a) y(t=O)=-h/2
(b) y(t=O)--0
(c) y(t--O)=+h/2
The dashed lines are the base solutions (Cvm = 0.5)
0.05
(c)
(a)
(b)
-0.05-0.15-0.152-
I
-0.25-
-0.25
-0.3
-0.35
-0.41
0
0.1
0.2
0.3
0.4
0.5
x, (M)
CL = 0.5
Cvm = 0.5
0=400
V=5m/s
Rb=0.002m
Figure 17: Bubble trajectories for 0=40' and different initial positions
(a) y(t=0)=-h/2
(b) y(t=0)--0
(c) y(t--O)=+h/2
The dashed lines are the base solutions (0--45*)
0.05
0.3
(b)
(a
-0.05-
-0.2(in)"
-0.1
.5
-0.15-
0.
-0.2-=
0.
.
-0.25-0.3-0.35-0.4
0
0.1
0.2
0.3
0.4
0.5
x, (in)
CL =0.5
Cvm = 0.5
0=600
V=5m/s
Rb--0.002m
Figure 18: Bubble trajectories for 0=600 and different initial positions
(a) y(t--O)=-h/2
(b) y(t--0)=0
(c) y(t--0)=+h/2
The dashed lines are the base solutions (0=450)
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