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Control of flow in the side outflow channel
Iroda Babajanova1*, Orifjan Bazarov2, Sobir Eshev2, Yuldash Babajanov3, and Alisher
Isakov2
1Higher military aviation school of the republic of Uzbekistan, Karshi, Uzbekistan
2Karshi engineering-economics institute, 180100, Karshi, Uzbekistan
3Karshi State University, Karshi, Uzbekistan
Abstract. The article discusses the task of regulating the flow in
the channel with a lateral outflow. A form of retraction channel is
recommended to ensure a smooth flow. Solving the issues with
water distribution along the channel route and determining the
shape of these channels, providing non-cavitation and non-water
current, comes down to the problem of the flow of the ideal
incompressible liquid in the channels with a side drain.
1 Introduction
The task of fluid movement in the channel with a side drain is three-dimensional, and it is
impossible to solve it analytically. It offers a two-dimensional jet model of the current. The
flow of liquid in the main channel is much larger than the required amount of water in the
consuming object.
Let's take a look at the physical picture of the current. When entering from the main
channel into the drain (nozzles), the fluid jet is compressed, then expands and fills its
sections (Figure 1).
In the gaps between the compressed section and the walls of the nozzle, a vortex zone is
formed, where there is a process of erosion of the channel and deposition of particulate
matter sediments (for example, when the water from Amudarya to the Karshin main canal
these zones are filled with sediments, and the drainage channel is buried).[3-7]
The pressure losses in the water distribution channels are mainly from losses to sudden
expansion after the flow compression (in the nozzle). Therefore, the definition of the shape
of the non guide surface of the water distribution channel is of great importance in the
design of hydraulic facilities.
In a whirlpool stream, the free surface is indeed a guide surface. If the stagnation zone is
filled with solids, the current should remain unchanged, and the new hard surface should be
without a swirling [2, 8-11] (Figure 1 and 2).
In practice, this statement requires clarification. First, on the surface of the section of
the stagnant zone, friction is practically absent, and if the stagnation zone is filled with
solids, there will be surface friction (with the flow of real liquid), which slightly changes
the direction of the current. However, this influence is quite small, and the resulting change
* Corresponding author: babadjanova.i@mail.ru
E3S Web of Conferences 365, 03038 (2023) https://doi.org/10.1051/e3sconf/202336503038
CONMECHYDRO - 2022
© The Authors, published by EDP Sciences. This is an open access article distributed under the terms of the Creative
Commons Attribution License 4.0 (http://creativecommons.org/licenses/by/4.0/).
in the current should not cause vortex formation. In the current ideal incoherent liquid, this
friction does not exist.
Secondly, at the point where the flow is straightened in the desired direction (point M in
Figure 1), the guide plane must coincide with the tangent surface of the boundary of the
stagnant zone (so that there is no return current). Therefore, this point moves along the
boundary of the stagnant zone (i.e., along the guide) with a change of angle to the direction
of the main channel, under which you need to have a channel.
2 Research method
Let us consider the jet problem of the lateral outflow of liquid from a channel of width H
through a nozzle located at an angle to the direction of motion of the main flow (Fig. 1).
The flow is flat, potential; the liquid is incompressible. Let's introduce a coordinate
system (Fig. 1). As noted in the previous paragraph, at the entrance to the branch, a free
boundary DE appears, at which the fluid velocity is constant and equal to Vk.The
curvilinear section of the CM-channel is unknown; it is determined by the condition that the
flow rate at which is constant. When solving the problem, it can be assumed that the free
boundaries DE and CM are straightened (planed) by the planing plate ME.
Fig. 1. Scheme of a vortex-free flow of a liquid in a channel with a side outlet.
The given are the parameters,
,
1
H
H
,
k
V
U
,
1
H
is the channel width in section BB;
U is the speed in the channel at infinity (point A).
When solving the problem, it is necessary to determine the free boundary DE and the
curved section CM shapes, as well as the outflow coefficient [14, 15].
is the width of the jet at infinity (point E)
is the abscissa of point C
Let us solve the problem using the Zhukovsky method, for which we consider the
function.
E3S Web of Conferences 365, 03038 (2023) https://doi.org/10.1051/e3sconf/202336503038
CONMECHYDRO - 2022
2
i
V
V
dw
dzV
kk
00
lnln
(1)
0
V
is the liquid velocity module;
is the angle of the velocity with the axis Along
the free boundary DE we have
,
0k
VV
where is
0ln
0
V
V
k
Сthe other side,
is
along DE changes from zero to
, so it goes from zero to
i
(fig. 1). Along DA,
AB, BK the imaginary part is constant and equal to zero. Along the KC and ME the
imaginary 'part
constant and equal
and
correspondingly.
On CM,
the real part is
,ln
c
k
V
V
and the imaginary changes from
before
.When passing through point K, at which the real part
increases to
, the
quantity
changes abruptly. So the area of change
is a pentagon, the vertex K of
which is removed to infinity.
With the indicated in Fig. 1. and 2 correspondences of points, formula (1) gives:
2
0
1)()( C
k
d
C
m
C
u
(2)
With the indicated in Fig. 1. and 2 correspondences of points, formula (1) gives:
Let's choose constantC1and C2, based on the following
Fig. 2. The area of change in the Zhukovsky function
At point D, we have
0)0(
.From here
0
2
C
and
)()(
0
1
k
d
C
m
C
u
(3)
Fig 1 and 2 it is seenthat at points ω(∞) =i
ߙ
π Then from formula (3) we find Ci
)(
1
J
C
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where
)()(
)(
0k
d
C
m
uJ
u
Means,
)()()(
)(
0
k
d
C
m
J
u
u
(4)
In the transition from DK to KC, the jump is, therefore (see Figure 2)
)(
)( kck
km
J
It follows from (1) and (4) that
)()()(
1
0
k
d
C
m
J
EXP
dz
dw
V
u
k
(5)
From formula (5), we obtain the corresponding expressions for the velocities
1
0
)()()(
k
d
C
m
J
EXP
V
U
k
(6)
in section AA and on the curvilinear section of the CM channel: Means,
C
k
c
k
d
C
m
J
EXP
V
V
0
)()()(
(7)
Function
00
)(
izw
conformally displays areas per strip
Q
0
with a
horizontal incision, the top of which corresponds to the point of splitting the flow - the
flow of liquid in the section of the AA). Conformal display of the upper semi-flatness
on the strip is carried out by an analytical function [12, 13].
B
E
iq
bk
bu
bk
k
u
k
b
q
uw
ln)(
1
1
ln)1(
)1(
)(
(8)
Where
q and
B
kE
Vq
are fluid costs in the ee EE and BB sections,
respectively. Differentiating
)(u
u, from (8) get
E3S Web of Conferences 365, 03038 (2023) https://doi.org/10.1051/e3sconf/202336503038
CONMECHYDRO - 2022
4
))(1(
)(
buu
ku
q
du
dw
E
(9)
Fluid consumption, in part BB, is defined as the increment of the imaginary part of the
function
)(uw
bypassing the point
:bu
Bu
B
du
du
dw
q
0
or
1
b
bk
q
q
E
B
(10)
From Figure 1. You can see that
EB
qqQ
(11)
With the help of formula (1), we will make the transition to the physical plane of
the current
du
du
d
e
V
uZ
R
0
1
)(
(12)
Substituting (4) and (2), we get
du
k
d
C
m
J
EXP
buu
ku
V
q
uZ
uu
k
E
00
)()()())(1(
)(
)(
(13)
Highlighting the actual and imaginary parts of the expression (13) on the border
)0( u
, find a form of free DE border:
duuJ
J
Sin
buu
ku
V
q
uy
duuJ
J
Cos
buu
ku
V
q
ux
u
C
K
E
u
k
E
)(
)())(1(
)(
)(
)(
)())(1(
)(
)(
0
(14)
3 Results and Discussions
The resulting ratios completely close the solution to the problem. This system is
solved by Newton's numerical method at the specified geometric values of the channel
and the different speeds of the incoming flow. Figures 3 and 4 show the shapes of the
E3S Web of Conferences 365, 03038 (2023) https://doi.org/10.1051/e3sconf/202336503038
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5
drainage channel, providing a non-vortex current in different ways
k
V
U
and at two
values
:
,
2
1
,
4
1
under which the guide plate is located.
4 Conclusions
Passage through the openings of the main channel, the jet of liquid, as shown by numerous
experiments, shrinks and, at some distance from the hole of the main channel, acquires the
smallest area of the section
. This section also depends on the direction of the diverted
channel. The compression of the jet is because the particles of the liquid, moving along the
walls of the AD and KC of the main channel, reach the edge of the hole and continue to
move in the same direction, licking gradually, deviating from it.
Fig. 3 Form of the drain channel, providing a swirlless current at
4
1
;3.0
H
and in different ways
:
k
V
U
;2.01
k
V
U
;43.02
k
V
U
66.03
k
V
U
.
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Fig. 4. Form of the drainage channel, which provides a swirlless current
2
1
;3.0
H
and in
different ways
:
k
V
U
;2.01
k
V
U
;43.02
k
V
U
66.03
k
V
U
Described physical processes and the pattern of the current are well illustrated in
the pic. 3 and 4. Calculations show that when the speed of the incoming flow
increases, the point M(where the guide plate is located) approaches point C, moving
along the boundary of the congested zone.
References
1. Ramamurthy A.S. Restangular lateral orifices in open channels. H.Envir. Engrg.,
ASCE., 112(2), pp.292-299. (1988)
2. Pouse H., Abdul –Fetouh A.H. Characteristies of irrotationall flow through Axially
symmetric orifices. Trans. ASME.J. Appl. Mech. 17, (1960)
3. Mahammad A., Gill M. Flow through side slots. Journal of Environmental Engineering.
113(5), pp. 1047-1057 (1987)
4. Batchelor O.K., Gill A.E. Analysis of the statility of axisymmetric jets. J. Fluid Mech.
14(4), pp.529-551 (1962)
5. Vigander S., Elder. R., Brooks. N. Internal Hydraulics of Thermat Discharpe Diffusers.
Journal Hydraulics Div., ASCE. 96(2). pp. 509-515 (1970)
6. Rawn A.M., Bowerman F.R. and Brooks N.H. Diffusers for Disposal of sewage in
Seawater. Journall of the Sanitary Engineering Division. ASCE. 86(2). pp. 65-74 (1960)
7. Bogomolov Ye.N., Bezvikhrevoye istecheniye zhidkosti iz potoko cherez bokovoye
otverstiye konechnoy glubiny. Izv RAN. Mekhanika zhidkosti i gaza. 4, pp. 162-164
(1994)
8. Kuznetsov A.V. Troyepol'skaya O.V., Ob odnoy zadache nestatsionarnogo istecheniya
zhidkosti iz kanala. Trudy seminara po krayevym zadacham. Kazan': Izd-vo KGU. 19,
pp. 97-103 (1982)
E3S Web of Conferences 365, 03038 (2023) https://doi.org/10.1051/e3sconf/202336503038
CONMECHYDRO - 2022
7
9. A. Yangiev, S. Eshev, Sh. Panjiev. Calculation of sediment flow in channels taking
into account passing and counter wind waves. In IOP Conf. Series: Materials Science
and Engineering .883, (2020).doi:10.1088/1757-899X/883/1/012036
10. S.Eshev, A. Rakhimov, I. Gayimnazarov, A.Isakov, B.Shodiev, F.Bobomurodov.
Dynamically stable sections of large soil canals taking into account wind waves. In
IOP Conf. Series: Materials Science and Engineering 1030, (2021) IOP Publishing.
doi:10.1088/1757-899X/1030/1/012131
11. S.Eshev, A.Khazratov, A. Rakhimov, Sh. A. Latipov. Influence of wind waves on the
flow in flowing reservoirs. IIUM Engineering Journal, 21(2), (2020) pp.125-132.
doi: https://doi.org/10/31436/iiumej.v21i2.1329
12. S. Eshev , Sh. Latipov, A. Qurbonov, J. Sagdiyev, М. Berdiev, N. Mamatov. Non-
eroding speed of water flow of channels running in cohesive soils. In IOP Conf. Series:
Materials Science and Engineering 1030, (2021). doi:10.1088/1757-
899X/1030/1/012131
13. Eshev S.S. Raschet deformatsiy bol'shikh zemlyanykh kanalov v usloviyakh
statsionarnosti vodnogo potoka. Monografiya. pp. 101-108, Tashkent (2017)
14. Hager W. H. Lateral outflow over side weirs. Journal of Hydraulic Engineering, 113(4),
491-504 (1987)
15. Gregg M. C., and Özsoy E. Flow, water mass changes, and hydraulics in the Bosphorus.
Journal of Geophysical Research: Oceans, 107(C3), 2-1. (2002)
E3S Web of Conferences 365, 03038 (2023) https://doi.org/10.1051/e3sconf/202336503038
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