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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

E3S Web of Conferences 365, 03038 (2023) https://doi.org/10.1051/e3sconf/202336503038

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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

CONMECHYDRO - 2022

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.

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