Control of flow in the side outflow channel

. 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.


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][4][5][6][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][9][10][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 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.

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 V k .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.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) D is the abscissa of point C Let us solve the problem using the Zhukovsky method, for which we consider the function.DS .When passing through point K, at which the real part Z increases to f , the quantity T changes abruptly.So the area of change Z 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: With the indicated in Fig. 1. and 2 correspondences of points, formula (1) gives: Let's choose constantC 1 and C 2 , based on the following In the transition from DK to KC, the jump is, therefore (see Figure 2) 1) and ( 4) that From formula (5), we obtain the corresponding expressions for the velocities in section AA and on the curvilinear section of the CM channel: Means, conformally displays areas per strip ^Q d d\ 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].
Where q and B k E V q G are fluid costs in the ee EE and BB sections, respectively.Differentiating Fluid consumption, in part BB, is defined as the increment of the imaginary part of the function From Figure 1.You can see that With the help of formula (1), we will make the transition to the physical plane of the current Substituting ( 4) and (2), we get Highlighting the actual and imaginary parts of the expression (13)

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

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 G .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. 1 .
Fig. 1.Scheme of a vortex-free flow of a liquid in a channel with a side outlet.

С
the liquid velocity module; T is the angle of the velocity with the axis Along the free boundary DE we have , the other side, T is along DE changes from zero to DS , so it goes from zero to DS i (fig.1).Along DA, AB, BK the imaginary part is constant and equal to zero.Along the KC and ME the imaginary 'part Z constant and equal S and DS correspondingly.On CM, Z the real part is changes from S before

Fig. 2 .Fig 1
Fig. 2. The area of change in the Zhukovsky function At point D, we have 0 ) 0 ( Z show the shapes of the drainage channel, providing a non-vortex current in different ways guide plate is located.

Fig. 3 Fig. 4 .U
Fig. 3 Form of the drain channel, providing a swirlless current at