CN114460981A - A Numerical Simulation Method for Hydraulic Control of Water Distribution Project - Google Patents

Abstract

The invention relates to a numerical simulation method for hydraulic control of a water distribution hub, which comprises the steps of obtaining an increment correlation relation; converting and integrating; deducing a correlation between the flow increment of the inlet node of the water channel and the water level increment; flow increment and water level increment of all nodes of the water outlet channel are solved by back substitution; solving the flow increment and the water level increment of the outlet node of the water inlet channel and the inlet node of the water outlet channel and the water level increment of the distribution pool; flow increment and water level increment of other nodes of each channel are solved in a retrospective mode; and calculating the flow and the water level of each node. The mass and energy conservation equation under the complex operating conditions of inflow, outflow and overflow of the water distribution hub is constructed through theoretical analysis, contents such as a Newton-Simpson discrete algorithm, a coding mode, a calculation program of a Preissmann four-point implicit difference algorithm and the like are provided, and a theoretical model of the water distribution hub is established; a numerical solving method is provided for the working conditions of multiple water inlets, multiple water outlets, pressure-free coupling and the like, and reference is provided for the operation scheduling of the engineering.

Description

A Numerical Simulation Method for Hydraulic Control of Water Distribution Project

technical field

The invention relates to a numerical simulation method for hydraulic control of a water distribution junction, which is an analysis method for hydraulic engineering, and a method that can apply computer to analyze hydraulic engineering and assist in operation scheduling.

Background technique

Water transfer projects and large and medium-sized irrigation areas are engineering measures to optimize the spatial allocation of water resources and alleviate the shortage of water resources in local areas or agricultural irrigation. Their safe scheduling and operation are the key to ensuring the security of national water resources. The control and adjustment of the project usually requires the calculation of the hydraulic transition process, and the analysis of the flow, water level and pressure changes of the pipe, culvert, tunnel and tunnel under a certain operation strategy of gates, pumps, valves and other equipment, so as to meet the specifications, pipeline pressure, equipment limits, etc. Require.

A water distribution hub is a common hydraulic structure in water transfer projects and large and medium-sized irrigation areas. It is used to connect pipes, culverts and tunnels of long-distance water diversion projects and subsequent water distribution projects, as well as main diversion channels (pipes) and water distribution trunks and branch channels ( Tube). If the reservoir is used as the water distribution hub, it can be regarded as the boundary of the constant water level, and the hydraulic transition process of the water diversion project and the water distribution project are calculated separately. Restricted by factors such as the overall layout of the project, topographic and geological conditions, and resettlement relocation, many projects do not have suitable reservoir areas as water distribution hubs, and facilities such as pools and canals need to be constructed manually. Manual construction must consider issues such as cost and environment, and limit the pool canal within a certain range. Due to the limited volume of the artificial pool of the water distribution hub, when the water supply flow or user demand changes, the flow rate, water level and pressure of the entire water delivery system will also change. Considering the water distribution hub as the boundary of the constant water level, it cannot reflect the water diversion project. , The real hydraulic characteristics of the water distribution project, how it affects the operation and scheduling of the project and the degree of its impact, it is necessary to couple the three parts of the water diversion project, the water distribution hub and the water distribution project for simulation analysis. How to simulate and analyze the water distribution hub in order to provide effective data and ensure the safety of water delivery is a problem that needs to be solved.

SUMMARY OF THE INVENTION

In order to overcome the problems of the prior art, the present invention proposes a numerical simulation method for hydraulic control of a water distribution project. The described method studies the hydraulic characteristics of the limited volume water distribution project under complex inflow and outflow conditions, gives the control equations, discrete methods, solving procedures, etc., establishes the theoretical model of the water distribution project, and combines the typical calculation examples to carry out the hydraulic control of the water distribution project. Simulation analysis to solve the model problem of simultaneous solution of water diversion project, water distribution hub and water distribution project.

The object of the present invention is achieved in this way: a numerical simulation method for hydraulic control of a water distribution junction, the water distribution junction analyzed by the method includes: M water inlet channels, a water distribution pool with overflow channels, and N water outlet channels channel; set m+1 computing nodes along each water inlet channel, and the computing nodes of each water inlet channel are coded from upstream to downstream; set n+1 computing nodes along the first water outlet channel, and the computing node of water outlet channel 1 Coding from upstream to downstream; n+1 calculation nodes are set along the 2nd to N water outlet channels, and the calculation nodes of water outlet channels 2 to N are coded from downstream to upstream; namely: the end calculation nodes of M water inlet channels, 2～N The end computing nodes of the N water outlet channels and the head end computing nodes of one water outlet channel are both computing nodes at the connection of the water inlet channel or the water outlet channel and the water distribution tank, and it is characterized in that, the steps of the method are as follows:

Step 1, obtain the incremental correlation: use the elimination process of the double sweep method to calculate the matrix B and The elements of the column vector P, to obtain the correlation between the flow increment at the end node and the water level increment, the number of equations is M+N-1; the calculation formula is as follows:

X=BX+P (1)

ΔQ为当前迭代步的流量增量，下角标K代表计算节点编号，对于进水渠道，下角标的数值为0～m；对于出水渠道，下角标的数值为0～n；Δh为当前迭代步的水深增量；U、W、P为双扫法系数；U、W、P的下角标代表矩阵的行数变化，对于进水渠道，编号为0～2m-1；对于出水渠道，编号为0～2n-1；where:

ΔQ is the flow increment of the current iteration step, and the subscript K represents the calculation node number. For the water inlet channel, the subscript value is 0～m; for the water outlet channel, the subscript value is 0～n; Δh is the water depth of the current iteration step Increment; U, W, and P are the coefficients of the double sweep method; the subscripts of U, W, and P represent the change of the number of rows in the matrix. For the water inlet channel, the number is 0～2m-1; 2n-1;

Each element is calculated by the following recursive formula:

In the formula: c, b, D, a, e are coefficients; j is the node number, j=1, 2, ..., K-2;

For water inlet channel 1, the flow increment and water volume increment of the end computing node satisfy the following relationship:

ΔQ _In1,m =U _In1,2m-2 Δy _In1,m +P _In1,2m-2 (5.1)

For water inlet channels 2 to M, the flow increment and water volume increment of the terminal computing node satisfy the following relational expressions:

ΔQ _In2,m =U _In2,2m-2 Δy _In2,m +P _In2,2m-2 (5.2)

...

ΔQ _InM,m =U _InM,2m-2 Δy _InM,m +P _InM,2m-2 (5.M)

For water outlet channels 2 to N, the flow increment and water volume increment of the terminal computing node satisfy the following relationship:

ΔQ _Out2,n =U _Out2,2n-2 Δy _Out2,n +P _Out2,2n-2 (6.2)

...

ΔQ _OutN,n =U _OutN,2n-2 Δy _OutN,n +P _OutN,2n-2 (6.N)

Thus, M+N-1 can be obtained about the flow increment and the water quantity increment, ΔQ _In1,m , Δy _In1,m , . . . , ΔQ _InM,m , Δy _InM,m , ΔQ _Out2,n , Δy _{Out2 ,n} , ..., ΔQ _OutN,n , Δy _OutN,n , the equation of the correlation; In in the subscript represents the water inlet channel, Out represents the water outlet channel, the following number represents the channel number, and the number after the comma represents the calculation node number ;

Step 2, conversion and integration: all the inflow, outflow and distribution pool water levels satisfy the energy conservation equation, and use the Newton-Simpson method to discretize, the number of equations is M+N; at the same time, the mass conservation equation is also satisfied, and the integration takes the second order Approximate, the number of equations is 1;

The energy conservation equation of the outlet node of inlet channel 1 and the distribution tank:

Where: y _In1,m is the piezometric head of the outlet node of the inlet channel 1; Q _In1,m is the flow rate of the outlet node of the inlet channel 1; g is the acceleration of gravity; A _In1,m is the outlet node of the inlet channel 1 y _s is the piezometric head of the distribution tank; ζ _In1 is the local head loss coefficient of the water from the inlet channel 1 entering the distribution tank;

The energy conservation equation of the outlet node 2～M of the inlet channel and the distribution pool:

In the formula: y _In2,m , ..., y _InM,m is the piezometric head of the outlet node of the water inlet channel 2～M; Q _In2,m , ..., Q _InM,m is the outlet of the water inlet channel 2～M The flow of the node; g is the acceleration of gravity; A _In2,m , ..., A _InM,m is the flow area of the outlet node of the water inlet channel 2～M; ζ _In2 , ... , ζ _InM is the water inlet channel 2～M The local head loss coefficient of the water entering the distribution tank;

The energy conservation equation of the inlet node of the distribution tank and outlet channel 1:

In the formula: y _Out1,0 is the piezometric head of the inlet node of the outlet channel 1; Q _Out1,0 is the flow rate of the inlet node of the outlet channel 1; A _Out1,0 is the flow area of the inlet node of the outlet channel 1; ζ _Out1 is local head loss coefficient of water entering outlet channel 1;

The energy conservation equation of the inlet node 2～N of the water distribution tank and the outlet channel:

...

In the formula: y _Out2,n ,...,y _OutN,n is the head of the piezometric pipe at the inlet node of the outlet channel 2～N; Q _Out2,n ,...,Q _OutN,n is the inlet node of the outlet channel 2～N Flow; A _Out2,n ,...,A _OutN,n is the flow area of the inlet node of the outlet channel 2～N; _ζOut2 ,..., _ζOutN is the local head loss coefficient of the water entering the outlet channel 2～N;

The mass conservation equation for the distribution tank is:

In the formula: A ₀ is the plane area of the water distribution tank; Q _w is the flow rate of the overflow channel, and the calculation formula is:

where μ is the flow coefficient; B _w is the overflow weir width; H _w is the overflow weir elevation;

Equation conversion:

For water inlet channel 1, it can be obtained from equation (7.1):

Using the Newton-Simpson method, equation (11) is transformed into:

F _In10 +e _In1 Δy _In1,m +a _In1 ΔQ _In1,m +e _s Δy _s =0 (12.1)

Δy_s为配水池水位的增量；where:

Δy _s is the increment of the water level in the distribution pool;

For water inlet channels 2 to M, you can get:

F _In20 +e _In2 Δy _In2,m +a _In2 ΔQ _In2,m +e _s Δy _s =0 (12.2)

...

F _InM0 +e _InM Δy _InM,m +a _InM ΔQ _InM,m +e _s Δy _s =0 (12.M)

……、where:

...,

……、

……、

...,

...,

Thereby, M equations for Δy _s , ΔQ _In1,m , Δy _In1,m , . . . , ΔQ _InM,m , Δy _InM,m are obtained;

For the outlet channel 1, it can be obtained by formula (8.1):

Using the Newton-Simpson method, equation (13) is transformed into:

F _Out10 +e _Out1 Δy _Out1,0 +a _Out1 ΔQ _Out1,0 +e _s Δy _s =0 (14.1)

where:

For outlet channels 2 to N, obtain

F _Out20 +e _Out2 Δy _Out2,n +a _Out2 ΔQ _Out2,n +e _s Δy _s =0 (14.2)

...

F _OutN0 +e _OutN Δy _OutN,n +a _OutN ΔQ _OutN,n +e _s Δy _s =0 (14.N)

……、where:

...,

……、

...,

……、

...,

For the mass conservation equation of the distribution tank, it can be obtained by integrating equation (9) and taking the second-order approximation:

In the formula: Δt is the time step; the suffix 0 represents the value of the physical quantity in the previous time step;

Using the Newton-Simpson method, equation (15) can be transformed into:

where:

In this way, one can be obtained about _{Δy s} , _{ΔQ In1,m , Δy In1,m ,} . Equations of _Out2,n ,...,ΔQ _OutN,n ,Δy _OutN,n ;

Step 3, deduce the correlation between the flow increment of the inlet node of water channel 1 and the water level increment: for the current iteration step, the unknowns include: water inlet channel 1, water inlet channel 2, ..., water inlet channel M, water outlet channel 1, Flow increment and water level increment of outlet channel 2,..., outlet channel N, ΔQ _In1,m , Δy _In1,m ,..., ΔQ _InM,m , Δy _InM,m , ΔQ _Out1,0 , Δy _{Out1, 0} , ΔQ _Out2,n , Δy _Out2,n ,..., ΔQ _OutN,n , Δy _OutN,n , the water level increment of the distribution tank, Δy _s , a total of 2(M+N)+1; the number of equations is 2( M+N), the correlation between the flow increment ΔQ _Out1,0 of the inlet node of water channel 1 and the water level increment Δy _Out1,0 can be obtained by deduction;

Step 4, back-substitution to solve the flow increments and water level increments of all nodes of the outlet channel 1: use the back-substitution process to solve the flow increments and water level increments of all nodes of the outlet channel 1, and the calculation formula is:

Step 5, solve the flow increment and water level increment of the outlet node of the water inlet channel, the inlet node of the outlet channel, and the water level increment of the distribution tank: solve the outlet of the inlet channel 1, the inlet channel 2, ..., the inlet channel M nodes, as well as the flow increment and water level increment of the inlet node of outlet channel 2, ..., outlet channel N, and the water level increment of the distribution tank; the flow increment ΔQ _Out1,0 of the inlet node of outlet channel 1 obtained in step 4 With the water level increment Δy _Out1,0 , into the equation established above, determine the outlet nodes of inlet 1, inlet 2, ..., inlet M, and the flow increments of inlet nodes of outlet 2, ..., outlet N, and Water level increment, ΔQ _In1,m , Δy _In1,m , …, ΔQ _InM,m , Δy _InM,m , ΔQ _Out2,n , Δy _Out2,n , …, ΔQ _OutN,n , Δy _OutN,n , The water level increment of the distribution tank, Δy _s ;

Step 6, back-substitution to solve the flow increment and water level increment of other nodes of each channel: Back-substitution to solve the water inlet channel 1, the water inlet channel 2, ..., the water inlet channel M, the water outlet channel 2, ..., the water outlet channel The flow increment and water level increment of all nodes of N are calculated as:

Step 7: Calculate the flow and water level of each node: Calculate the water inlet channel 1, the water inlet channel 2, ..., the water inlet channel M, the water outlet channel 1, the water outlet channel 2, ..., the water outlet channel N. All nodes are in the current iteration step The flow and water level of , and the water level of the distribution tank in the current iteration; the flow is equal to the value of the previous iteration plus the flow increment, and the water level is equal to the value of the previous iteration plus the water level increment.

Further, if the water inlet channel or the water outlet channel is a pressure flow, the narrow gap method is used to solve the problem, and the gap width is:

B=gA/a ² (19)

In the formula: B is the width of the gap; A is the cross-sectional area of the pressurized flow; a is the water hammer velocity.

The advantages and beneficial effects of the present invention are as follows: the present invention constructs the mass and energy conservation equations under complex operating conditions of water distribution hub inflow, outflow and overflow through theoretical analysis, and provides Newton-Simpson discrete algorithm, coding method, Preissmann The calculation procedure of the four-point implicit difference algorithm, etc., establishes the theoretical model of the water distribution hub; for the possible working conditions such as multiple inflows, multiple outflows, pressure and non-pressure coupling, etc., the numerical solution method is given, which can be used for The operation scheduling of the project provides a reference.

Description of drawings

The present invention will be further described below with reference to the accompanying drawings and embodiments.

1 is a schematic structural diagram of a water distribution hub analyzed by the method described in Embodiment 1 of the present invention;

2 is a flowchart of the method according to Embodiment 1 of the present invention;

3 is a schematic structural diagram of a water distribution hub with one inlet and two outlets analyzed in Embodiment 1 of the present invention;

4 is a schematic diagram of the water distribution system of the application example described in Embodiment 1 of the present invention;

Fig. 5 is the flow rate and water level process of the typical position of the water distribution system of the application example described in the first embodiment of the present invention, wherein (a) the flow rate process, (b) is the water level process;

6 is the hydraulic characteristics of the upstream tunnel of the application example according to the first embodiment of the present invention, wherein (a) flow process, (b) water level process.

Detailed ways

Example 1:

This embodiment is a numerical simulation method for hydraulic control of a water distribution hub. The water distribution hub analyzed by the method includes: M water inlet channels ①, water distribution pools with overflow channels ② ③, and N water outlet channels ④, as shown in FIG. 1 . The arrows in Figure 1 indicate the direction of water flow.

The water distribution hub described in this embodiment is a common hydraulic structure in water transfer projects and large and medium-sized irrigation areas. It is used to connect the pipes, culverts, tunnels, and main diversion channels (pipes) of long-distance water diversion projects and subsequent water distribution projects. And water distribution trunk, branch channel (pipe). If the reservoir is used as the water distribution hub, it can be regarded as the boundary of the constant water level, and the hydraulic transition process of the water diversion project and the water distribution project are calculated separately. Restricted by the overall layout of the project, topographic and geological conditions, resettlement and relocation and other factors, many projects do not have suitable reservoir areas as water distribution hubs and need to be constructed. When designing, there are generally two references for the selection of the volume of the water distribution hub: i. The underwater volume of the inlet pool can be determined by 30 to 50 times the design flow of the water pump sharing the inlet pool; ii. When the scale of water delivery is not large or the requirements are not high , the volume of the pool in the middle of the gravity water pipeline can be determined according to the maximum design water volume of not less than 5min. It can be seen that the volume of a water distribution hub is generally 30 to 300 times the design flow. Due to the small size of the water distribution hub, when the water supply flow or user demand changes, the flow, water level, and pressure of the entire water delivery system will change accordingly. Considering the water distribution hub as the boundary of the constant water level, it cannot reflect the water diversion project and water distribution. The true hydraulic properties of the project. Improper control, unable to deliver the required amount of water to users in a timely and appropriate amount, may cause engineering accidents such as flooding, burst pipes, and structural building damage. In this case, it is necessary to couple the three parts of the water diversion project, the water distribution project and the water distribution project for simulation analysis.

The purpose of this example is to study the hydraulic characteristics of a limited volume water distribution hub under complex inflow and outflow conditions, to give control equations, discrete methods, solving procedures, etc., to establish a theoretical model of the water distribution Control the simulation analysis to solve the model problem of the simultaneous solution of the water diversion project, the water distribution project and the water distribution project.

The water distribution hub described in this embodiment is mainly composed of three elements: the water inlet channel, the water outlet channel and the water distribution pool. The inlet and outlet channels can be open channels or unpressurized culverts, or pressurized pipes. The number of inlet channels and the number of outlet channels may or may not be equal, and in most cases, the number of inlet channels is usually less than the number of outlet channels. The water distribution pool is an artificially constructed pool with a design flow rate of 30 to 300 times.

For the convenience of calculation, calculation nodes are set in the water inlet channel and the water outlet channel in this embodiment: m+1 calculation nodes are set along each water inlet channel, and the calculation nodes of each water inlet channel are coded from upstream to downstream. Set n+1 computing nodes along the first outlet channel, and the computing nodes of outlet channel 1 are coded from upstream to downstream; set n+1 computing nodes along the 2nd to N outlet channels, and the computing nodes of outlet channel 2～N Coding from downstream to upstream; that is, the end computing nodes of M inlet channels, the end computing nodes of 2 to N outlet channels, and the head computing node of 1 outlet channel are all inlet channels or outlet channels connected to the distribution tank The computing nodes at the location are shown in Figure 1.

Distribution hubs are usually open channel flows with free surfaces. The governing equations are Saint-Venant equations, including momentum equations and continuous equations. The main solving methods are the characteristic line method, the explicit difference method, and the implicit difference method. Among them, the implicit difference method has good numerical calculation stability. Therefore, the research adopts the Preissmann four-point implicit difference scheme which is widely used in engineering calculation.

(1) Energy conservation of the water inlet system:

The energy conservation equations of the inlet channel 1, the inlet channel 2, ..., the inlet channel M and the distribution pool are as follows.

The incoming water from the inlet channel 1 enters the distribution tank, and the energy conservation equation between the end node of the inlet channel 1 and the distribution tank can be obtained from the unsteady flow Bernoulli energy equation:

The energy conservation equation of the outlet node of inlet channel 1 and the distribution tank:

Where: y _In1,m is the piezometric head of the outlet node of the inlet channel 1; Q _In1,m is the flow rate of the outlet node of the inlet channel 1; g is the acceleration of gravity; A _In1,m is the outlet node of the inlet channel 1 y _s is the piezometric head of the distribution tank; ζ _In1 is the local head loss coefficient of the water from the inlet channel 1 entering the distribution tank; since the cross-sectional area of the distribution tank is generally much larger than that of the channel, the distribution The flow head of the pool is negligible.

The incoming water from the inlet channel 2 enters the distribution tank, and the energy conservation equation between the end node of the inlet channel 2 and the distribution tank can be obtained from the unsteady flow Bernoulli energy equation:

In the formula: y _In2,m is the piezometric head of the outlet node of the water inlet channel 2; is the flow rate of the outlet node of the water inlet channel 2; A _In2,m is the flow area of the outlet node of the water inlet channel 2; The local head loss coefficient of 2-M water entering the distribution tank.

...(the energy conservation equation from the end node of the water inlet channel 3 to the water inlet channel M-1 and the distribution tank is omitted in the middle)

The incoming water from the inlet channel M enters the distribution tank, and the energy conservation equation between the end node of the inlet channel M and the distribution tank can be obtained from the unsteady flow Bernoulli energy equation:

Where: y _InM,m is the piezometric head of the outlet node of the inlet channel M; Q _InM,m is the flow rate of the outlet node of the inlet channel M; A _InM,m is the flow area of the outlet node of the inlet channel M; ζ _InM is the local head loss coefficient of the water from the inlet channel M entering the distribution tank;

(2) Energy conservation of the effluent system:

The energy conservation equations of outlet channel 1, outlet channel 2, ..., outlet channel N are as follows:

The water of the distribution tank enters the outlet channel 1, and the energy conservation equation of the inlet node of the distribution tank and the outlet channel 1 can be obtained from the unsteady flow Bernoulli energy equation:

In the formula: y _Out1,0 is the piezometric head of the inlet node of the outlet channel 1; Q _Out1,0 is the flow rate of the inlet node of the outlet channel 1; A _Out1,0 is the flow area of the inlet node of the outlet channel 1; ζ _Out1 is Local head loss coefficient for water entering outlet channel 1.

The water in the distribution tank enters the outlet channel 2, and the energy conservation equation of the inlet node of the distribution tank and the outlet channel 2 can be obtained from the unsteady flow Bernoulli energy equation

In the formula: y _Out2,n is the piezometric head of the inlet node of the outlet channel 2; Q _Out2,n is the flow rate of the inlet node of the outlet channel 2; A _Out2,n is the flow area of the inlet node of the outlet channel 2; ζ _Out2 is Local head loss coefficient for water entering outlet channel 2. The compute nodes of outlet channel 2 are coded from downstream to upstream.

...(the energy conservation equation between the end node of the inlet and outlet channel 3 to the inlet channel M-1 and the distribution pool is omitted in the middle)

The water of the distribution tank enters the outlet channel N, and the energy conservation equation of the inlet node of the distribution tank and the outlet channel N can be obtained from the unsteady flow Bernoulli energy equation:

Where: y _OutN,n is the piezometric head of the inlet node of the outlet channel N; Q _OutN,n is the flow of the inlet node of the outlet channel N; A _OutN,n is the flow area of the inlet node N of the outlet channel; ζ _OutN is The local head loss coefficient of water entering the outlet channel N. The compute nodes of the outlet channel N are coded from downstream to upstream.

(3) Conservation of the quality of the distribution pool:

The increment of the water body in the distribution tank is equal to the sum of the inflow, outflow and overflow, and the mass conservation equation is:

In the formula: A ₀ is the plane area of the water distribution tank, m ² ; Q _w is the flow rate of the overflow channel, m ³ /s, and the calculation formula is

where μ is the flow coefficient; B _w is the overflow weir width, m; H _w is the overflow weir elevation, m.

(4) Equation conversion:

For water inlet channel 1, it can be obtained from equation (7.1):

Using the Newton-Simpson method, equation (11) can be transformed into:

F _In10 +e _In1 Δy _In1,m +a _In1 ΔQ _In1,m +e _s Δy _s =0 (12.1)

Δy_s为配水池水位的增量。where:

Δy _s is the increment of the water level in the distribution tank.

The equations for other influent channels can be transformed in a similar way.

From formula (8.1) we get:

Using the Newton-Simpson method, equation (13) can be transformed into:

F _Out10 +e _Out1 Δy _Out1,0 +a _Out1 ΔQ _Out1,0 +e _s Δy _s =0 (14.1)

where:

The equations for other outlet channels can be transformed in a similar way.

For the mass conservation equation of the distribution tank, it can be obtained by integrating equation (9) and taking the second-order approximation:

In the formula: Δt is the time step; the suffix 0 represents the value of the physical quantity in the previous time step.

Using the Newton-Simpson method, equation (15) can be transformed into:

where:

Calculation of other nodes:

The continuity equation and momentum equation of one-dimensional unsteady flow in an open channel are:

Where: A is the flow area, m ² ; t is the time variable, s; Q is the flow rate, m ³ /s; x is the spatial variable, m; q is the lateral flow per unit channel length, m ³ /s; β is the correction coefficient introduced by the uneven distribution of flow velocity in the section; g is the acceleration of gravity, m ² /s; _h is the water depth, m; S ₀ is the bottom slope of the riverbed;

S _f =Q|Q|/K ₂ (22)

n为渠道糙率；R为水力半径，m。利用Preissmann隐式格式，连续性和动量方程离散为：In the formula: K is the flow modulus,

n is the channel roughness; R is the hydraulic radius, m. Using the Preissmann implicit format, the continuity and momentum equations are discretized as:

In the formula: y is the water surface elevation; ψ _R is the spatial coefficient, which is not necessarily the same as ψ.

After substituting into the continuity equation (23), we get:

a _2j+1 Δh _j +b _2j+1 ΔQ _j +c _2j+1 Δh _j+1 +d _2j+1 ΔQ _j+1 =D _2j+1 (25)

b_2j+1＝-θ/Δx；

d_2j+1＝θ/Δx；where:

b _2j+1 =-θ/Δx;

d _2j+1 =θ/Δx;

After substituting into the momentum equation (24), we get:

e _2j+2 Δh _j +a _2j+2 ΔQ _j +b _2j+2 Δh _j+1 +c _2j+2 ΔQ _j+1 =D _2j+2 (26)

where:

Boundary conditions:

There are three common boundary conditions:

a.F=h-h(t)=0

b.F=Q-Q(t)=0

c.F=Q-f(h)=0

For type a boundaries, water depth is a function of time; for type b boundaries, flow is a function of time; for type c boundaries, flow is a function of water depth. For example, the downstream of the open channel is a wide-top weir, and the relationship between its flow and water level is:

In the formula: μ is the flow coefficient; B is the width of the top of the weir, m; h is the head of the weir, m.

Using the Newton-Simpson formula, the above three types of boundary conditions can be transformed into:

F _h Δh+F _Q ΔQ=-F ₀

For type a boundary conditions,

For type b boundary conditions,

For boundary conditions of type c,

The nodes in the channel are discretized by Preissmann four-point implicit difference, and the boundary conditions are linearized by the Newton-Simpson formula. The matrix form of the equation is:

AX=D (31)

系数b₀、c₀和D₀由渠道进口边界条件确定，b₀＝F_h、c₀＝F_Q和D₀＝-F₀；a_2K-1、b_2K-1和D_2K-1由渠道出口边界条件确定，a_2K-1＝F_h、b_2K-1＝F_Q和D_2K-1＝-F₀。where:

The coefficients b ₀ , c ₀ and D ₀ are determined by the channel inlet boundary conditions, b ₀ =F _h , c ₀ =F _Q and D ₀ =-F ₀ ; a _2K-1 , b _2K-1 and D _2K-1 are given by The channel outlet boundary conditions are determined, a _2K-1 =F _h , b _2K-1 =F _Q and D _2K-1 =-F ₀ .

The coefficient matrix A of the above nonlinear equation system is in the shape of a strip, and its non-zero elements are all located near the diagonal. The above equation system can be solved by the double sweep method. Assuming that the coefficient b ₀ ≠ 0, the above equation is transformed by the elimination method:

X=BX+P (1)

where:

Each element is calculated by the following recursive formula:

Since the matrix B of the above linear equation system is an upper triangular matrix with 0 elements on the diagonal, it is solved by recursive back-substitution process of the following formula:

The specific process of numerical solution can be described as (the process is shown in Figure 2):

Step 1, obtain the incremental correlation. Use the elimination process of the double sweep method to calculate the elements of the matrix B and column vector P of influent 1, influent 2, ..., influent M, effluent 2, ..., and effluent N, and obtain the flow increment at the end node and The correlation of the water level increment, the number of equations is M+(N-1).

Step 2, conversion and integration. All influent, effluent and distribution tank water levels satisfy the energy conservation equation, and are discretized by the Newton-Simpson method, and the number of equations is M+N; it also satisfies the mass conservation equation, and the second-order approximation is integrated and the number of equations is 1.

Step 3, deduce the correlation between the flow increment at the inlet node of water channel 1 and the water level increment. For an iterative step, the unknowns include: Influent 1, Influent 2, ..., Influent M, Outlet 1, Outlet 2, ..., the flow increment and water level increment of effluent N, and the water level increment of the distribution tank , a total of 2(M+N)+1; the number of equations is 2(M+N), the correlation between the flow increment at the inlet node of water channel 1 and the water level increment can be obtained by deduction.

Step 4, back-substitute to solve the flow increment and water level increment of all nodes of outlet channel 1.

Step 5, solve the flow increment and water level increment of the outlet node of the water inlet channel, the inlet node of the water outlet channel, and the water level increment of the water distribution tank.

Step 6, back-substitute to solve the flow increment and water level increment of other nodes of each channel.

Step 7: Calculate the flow and water level of each node.

When the flow increment and water level increment of the calculation node are less than the set error, the calculation of the current time step ends, the upstream and downstream boundaries are updated, and the simulation calculation of a new time step starts; otherwise, the new flow and water level are used to start a new time step. One iterative step until the accuracy requirement is met.

There is another special case in practical engineering, that is, some influent or effluent flows under pressure. At this time, the narrow slit method can be used to solve the problem, or other similar methods can be used to convert the pressure into the equivalent pressureless method.

Calculation example:

In order to further illustrate the method described in this embodiment, a water distribution hub of "one inlet, two outlets, and one overflow (discharge)" is used below, as shown in Figure 3, for a specific description. The outlet channels are named as channels 2 and 3, respectively. The incoming water from channel 1 enters the distribution pool, and is distributed to channel 2 and channel 3 after being controlled by the gate (valve). The overflow channel and the venting pipe are the boundary of the water level-flow relationship. In this embodiment, the overflow channel is used for analysis.

Governing equation:

The incoming water from channel 1 enters the distribution tank, and the energy conservation equation between the end node of channel 1 and the distribution tank can be obtained from the unsteady flow Bernoulli energy equation:

Where: y _1,m is the head of the piezometric pipe at the end node of channel 1, m; Q _1,m is the flow rate at the end node of channel 1, m ³ /s; g is the acceleration of gravity, m/s ² ; A _{1, m} is the flow area at the end node of channel 1, m ² ; y _s is the piezometric head of the distribution tank, m; ζ ₁ is the local head loss coefficient of the water entering the distribution tank, including section change, water outlet, control gate, etc. . Since the cross-sectional area of the distribution pool is generally much larger than that of the channel, the flow velocity and head of the distribution pool can be ignored.

The water in the distribution pool enters channel 2, and the energy conservation equation of the first node of the distribution pool and channel 2 can be obtained from the unsteady flow Bernoulli energy equation:

In the formula: y _2,n is the head of the piezometric pipe at the first node of channel 2, m; Q _2,n is the flow rate of the first node of channel 2, m ³ /s; A _2,n is the flow area of the first node of channel 2 , m ² ; ζ ₂ is the local head loss coefficient of water entering channel 2, including section change, water inlet, control gate, etc. The compute nodes of channel 2 are encoded from downstream to upstream.

The water in the distribution tank enters channel 3, and the energy conservation equation of the first node of the distribution tank and channel 3 can be obtained from the unsteady flow Bernoulli energy equation:

In the formula: y _3,0 is the head of the piezometric pipe at the first node of the channel 3, m; Q _3,0 is the flow rate of the first node of the channel 3, m ³ /s; A _3,0 is the flow area of the first node of the channel 3 , m ² ; ζ ₃ is the local head loss coefficient of water entering the channel 3, including section change, water outlet, control gate, etc.

The increment of the water body in the distribution tank is equal to the sum of the inflow, outflow and overflow, and the mass conservation equation is:

In the formula: A ₀ is the plane area of the water distribution tank, m ² ; Q _w is the flow rate of the overflow channel, m ³ /s, and the calculation formula is:

where μ is the flow coefficient; B _w is the overflow weir width, m; H _w is the overflow weir elevation, m.

Solving algorithm:

From formula (1a) we get:

Using the Newton-Simpson method, equation (6a) can be transformed into:

F ₁₀ +e ₁ Δy _1,m +a ₁ ΔQ _1,m +e _s Δy _s =0(7a)

Δ表示相应物理量在某一迭代步的增量；

where:

Δ represents the increment of the corresponding physical quantity in an iterative step;

From formula (2a), we get:

Using the Newton-Simpson method, equation (8a) can be transformed into:

F ₂₀ +e ₂ Δy _2,n +a ₂ ΔQ _2,n +e _s Δy _s =0 (9a)

where:

From formula (3a), we get:

Using the Newton-Simpson method, equation (10a) can be transformed into:

F ₃₀ +e ₃ Δy _3,0 +a ₃ ΔQ _3,0 +e _s Δy _s =0 (11a)

where:

By integrating equation (4a) and taking the second-order approximation, we get:

Arranged:

In the formula: Δt is the time step, s; the subscript 0 represents the value of the physical quantity in the previous time step.

Using the Newton-Simpson method, equation (13a) can be transformed into:

F ₄₀ +F _4,ys Δy _s +F _4,Q1,m ΔQ _1,m +F _4,Q2,n ΔQ _2,n +F _4,Q3,0 ΔQ _3,0 =0 (14a)

where:

The inlet of channel 1 is usually the flow boundary, and the elimination process of the double sweep method can obtain that the end node satisfies:

ΔQ _1,m =U _1,2m-2 Δy _1,m +P _1,2m-2 (15a)

In the formula: U _i,j and P _i,j are the coefficients of the double sweep method.

When the outlet of channel 2 is the water level boundary or the water level-flow relationship, the elimination from the outlet to the inlet can be obtained:

Δy _2,n =U _2,2n-2 ΔQ _2,n +P _2,2n-2 (16a)

From formula (7a), we get:

Substitute equation (17a) into equation (15a) to get:

From formula (9a), we get:

Substitute equation (19a) into equation (16a) to get:

Substitute equations (18a) and (20a) into equation (14a) to get:

where:

Substitute equation (21a) into equation (11a):

b _3,0 Δy _3,0 +c _3,0 ΔQ _3,0 =D _3,0 (22a)

In the formula: b _3,0 =e ₃ ;

The procedure for numerically solving the entire water delivery system is:

(1) Calculate the elements U _1,i , W _1,i , P _1,i , U _2,j , W ₂ of the matrix B and column vector P of the inlet channel and channel outlet 1 by the elimination process of the double sweep method _,j and P _2,j (i=0,1,...,2m-2,j=0,1,...,2n-2);

(2) Use formula (22a) to determine the double sweep method coefficients of the inlet boundary of channel 1, b _3,0 , c _3,0 and D _3,0 ;

(3) Use the double sweep method to solve the water depth and flow increment of channel 3;

(4) Solve Δy _s , ΔQ _2,n , Δy _2,n , ΔQ _1,m and Δy _1,m sequentially by using equations (21a), (20a), (19a), (18a) and (17a);

(5) Use the back-substitution process of the double sweep method to solve the water depth and flow increments of channel 1 and channel 2.

Applications:

A water transfer project transfers water to another area through aqueduct tunnel 1. As shown in Figure 4, the water distribution pool in the middle distributes water to 2 aqueduct tunnels, and then transports it to the water receiving area.

The water discharge from the water conservancy project enters the aqueduct tunnel 1, and then enters the distribution tank, with a design flow of 70m ³ /s. The water distribution tank is a rectangular tank with a length of 80m, a width of 46m, and a volume of 28,000 m ³ ; the water in the distribution tank enters the aqueduct tunnel 2 and aqueduct tunnel 3 after regulation, and discharges through the overflow weir when the water level is high, as shown in the figure 4 shown.

There are two cases of water distribution flow of the aqueduct tunnel 2 and the aqueduct tunnel 3 of 47m ³ /s+23m ³ /s and 40m ³ /s+30m ³ /s. When the water distribution flow is switched from 47m ³ /s+23m ³ /s to 40m ³ /s+30m ³ /s, the gate of the canal tunnel 2 is reduced from fully open to 2.1m, and the opening of the gate of the canal tunnel 3 is changed from 1.7m to fully open , the action rate is 0.2m/min, and the hydraulic transition process characteristics of the water distribution tank and the water distribution project are shown in Figure 5. At 5km and 10km of canal tunnel 2, the target flow rate is 40m ³ /s after the gate actuated for 103min and 136min respectively; the water storage tank reaches the target flow rate of 30m ³ /s after the gate actuation for 15min. The water level in the distribution pool decreased steadily from 258.88m to 258.74m, the water level behind the canal and tunnel 2 gates decreased from 258.33m to 257.78m, and the water level in the storage tank increased from 256.79m to 258.10m. The characteristics of the hydraulic transition process of the upstream channel-tunnel 1 of the distribution tank are shown in Figure 6. In the process of water distribution flow adjustment, the flow and water level of the canal and tunnel 1 within 15km upstream of the water distribution tank change accordingly. The flow at 5km upstream first increased to 70.14m ³ /s, and then slowly recovered to 70.00m ³ /s. The water level decreased from 260.84m to 260.79m, and the changes of flow and water level decreased with the increase of distance. During the adjustment of water distribution flow, the flow of canal-tunnel 1, canal-tunnel 2, and canal-tunnel 3 transitioned smoothly, the open-flow tunnel did not overflow, and the water distribution pool did not overflow.

From the calculation example, in the process of water distribution flow adjustment, the flow and water level along the water transfer project will change accordingly; similarly, the adjustment of the water transfer project will also affect the user's water inflow process, available water and other content. If the limited volume of the water distribution hub is used as the boundary of the constant water level, it is impossible to analyze the hydraulic transition process of the entire water delivery system, and it is difficult to quantify the real flow, water level, and pressure variation characteristics. Specific problems and specific analysis are needed to determine the degree of impact on project operation safety and scheduling scheme preparation.

Embodiment 2:

This embodiment is an improvement of the above-mentioned embodiment, and is the refinement of the above-mentioned embodiment about the water inlet pipe and the water outlet pipe. The water inlet channel or the water outlet channel described in this embodiment is a pressure flow, and the narrow slit method is used to solve it. The gap width is:

B=gA/a ² (18)

In the formula: B is the width of the gap; A is the cross-sectional area of the pressurized flow; a is the water hammer velocity.

The narrow slit method described in this embodiment is used to solve the problem, that is, it is assumed that there is a very narrow slit at the top of the pipeline, neither the cross-sectional area of the pipeline nor the hydraulic radius is increased, the method is extremely simple, and very accurate calculation can be achieved.

Finally, it should be noted that the above is only used to illustrate the technical solutions of the present invention and not to limit it. Although the present invention has been described in detail with reference to the preferred arrangement solutions, those of ordinary skill in the art should understand that the technical solutions of the present invention (such as The form of the water distribution system, the use of various formulas, the sequence of steps, etc.) can be modified or equivalently replaced without departing from the spirit and scope of the technical solution of the present invention.

Claims (2)

1. A numerical simulation method for hydraulic control of a water distribution junction, the analyzed water distribution junction of the method comprises: M water inlet channels, a water distribution pool provided with overflow channels, and N water outlet channels; The channel is set with m+1 computing nodes, and the computing nodes of each inlet channel are coded from upstream to downstream; n+1 computing nodes are set along the first outlet channel, and the computing nodes of outlet channel 1 are coded from upstream to downstream; There are n+1 calculation nodes for the 2nd to N water outlet channels, and the calculation nodes of the 2nd to N outlet channels are coded from downstream to upstream; that is: the end calculation nodes of M water inlet channels, the end calculation nodes of 2 to N water outlet channels The node and the computing node at the head end of one outlet channel are both computing nodes at the connection between the inlet channel or the outlet channel and the water distribution tank, and it is characterized in that the steps of the method are as follows: Step 1, obtain the incremental correlation: use the elimination process of the double sweep method to calculate the matrix B and The elements of the column vector P, to obtain the correlation between the flow increment of the end node and the water level increment, the number of equations is M+N-1; the calculation formula is as follows: X=BX+P (1) where:

ΔQ is the flow increment of the current iteration step, and the subscript K represents the calculation node number. For the water inlet channel, the subscript number is 0～m; for the water outlet channel, the subscript number is 0～n; Δh is the water depth of the current iteration step Increment; U, W, and P are the coefficients of the double sweep method; the subscripts of U, W, and P represent the change of the number of rows in the matrix. For the water inlet channel, the number is 0～2m-1; 2n-1; Each element is calculated by the following recursive formula:

In the formula: c, b, D, a, e are coefficients; j is the node number, j=1, 2, ..., K-2; For water inlet channel 1, the flow increment and water volume increment of the end computing node satisfy the following relational expressions: ΔQ _In1,m =U _In1,2m-2 Δy _In1,m +P _In1,2m-2 (5.1) For water inlet channels 2 to M, the flow increment and water volume increment of the terminal computing node satisfy the following relational expressions: ΔQ _In2,m =U _In2,2m-2 Δy _In2,m +P _In2,2m-2 (7.2) ... ΔQ _InM,m =U _InM,2m-2 Δy _InM,m +P _InM,2m-2 (5.M) For water outlet channels 2 to N, the flow increment and water volume increment of the terminal computing node satisfy the following relational expressions: ΔQ _Out2,n =U _Out2,2n-2 Δy _Out2,n +P _Out2,2n-2 (6.2) ... ΔQ _OutN,n =U _OutN,2n-2 Δy _OutN,n +P _OutN,2n-2 (6.N) Thus, M+N-1 can be obtained about the flow increment and the water quantity increment, ΔQ _In1,m , Δy _In1,m , . . . , ΔQ _InM,m , Δy _InM,m , ΔQ _Out2,n , Δy _{Out2 ,n} , ..., ΔQ _OutN,n , Δy _OutN,n , the equation of the correlation; In in the subscript represents the water inlet channel, Out represents the water outlet channel, the following number represents the channel number, and the number after the comma represents the calculation node number ; Step 2, conversion and integration: all the inflow, outflow and distribution pool water levels satisfy the energy conservation equation, and use the Newton-Simpson method to discretize, the number of equations is M+N; at the same time, the mass conservation equation is also satisfied, and the integration takes the second order Approximate, the number of equations is 1; The energy conservation equation of the outlet node of inlet channel 1 and the distribution tank:

Where: y _In1,m is the piezometric head of the outlet node of the inlet channel 1; Q _In1,m is the flow rate of the outlet node of the inlet channel 1; g is the acceleration of gravity; A _In1,m is the outlet node of the inlet channel 1 y _s is the piezometric head of the distribution tank; ζ _In1 is the local head loss coefficient of the water from the inlet channel 1 entering the distribution tank; The energy conservation equation of the outlet node 2～M of the inlet channel and the distribution pool:

...

In the formula: y _In2,m , ..., y _InM,m is the piezometric head of the outlet node of the water inlet channel 2～M; Q _In2,m , ..., Q _InM,m is the outlet of the water inlet channel 2～M The flow of the node; g is the acceleration of gravity; A _In2,m , ..., A _InM,m is the flow area of the outlet node of the water inlet channel 2～M; ζ _In2 , ... , ζ _InM is the water inlet channel 2～M The local head loss coefficient of the water entering the distribution tank; The energy conservation equation of the inlet node of the distribution tank and outlet channel 1:

In the formula: y _Out1,0 is the piezometric head of the inlet node of the outlet channel 1; Q _Out1,0 is the flow rate of the inlet node of the outlet channel 1; A _Out1,0 is the flow area of the inlet node of the outlet channel 1; ζ _Out1 is local head loss coefficient of water entering outlet channel 1; The energy conservation equation of the inlet node 2～N of the water distribution tank and the outlet channel:

...

In the formula: y _Out2,n ,...,y _OutN,n is the piezometric head of the inlet node of the outlet channel 2～N; Q _Out2 ,n,...,Q _OutN,n is the inlet node of the outlet channel 2～N Flow; A _Out2,n ,...,A _OutN,n is the flow area of the inlet node of the outlet channel 2～N; _ζOut2 ,..., _ζOutN is the local head loss coefficient of the water entering the outlet channel 2～N; The mass conservation equation for the distribution tank is:

In the formula: A ₀ is the plane area of the water distribution tank; Q _w is the flow rate of the overflow channel, and the calculation formula is:

where μ is the flow coefficient; B _w is the overflow weir width; H _w is the overflow weir elevation; Equation conversion: For water inlet channel 1, it can be obtained from equation (7.1):

Using the Newton-Simpson method, equation (11) is transformed into: F _In10 +e _In1 Δy _In1,m +a _In1 ΔQ _In1,m +e _s Δy _s =0 (12.1) where:

Δy _s is the increment of the water level in the distribution pool; For water inlet channels 2 to M, you can get: F _In20 +e _In2 Δy _In2,m +a _In2 ΔQ _In2,m +e _s Δy _s =0 (12.2) ... F _InM0 +e _InM Δy _InM,m +a _InM ΔQ _InM,m +e _s Δy _s =0 (12.M) where:

Thereby, M equations for Δy _s , ΔQ _In1,m , Δy _In1,m , . . . , ΔQ _InM,m , Δy _InM,m are obtained; For the outlet channel 1, it can be obtained by formula (8.1):

Using the Newton-Simpson method, equation (13) is transformed into: F _Out10 +e _Out1 Δy _Out1,0 +a _Out1 ΔQ _Out1,0 +e _s Δy _s =0 (14.1) where:

For outlet channels 2 to N, obtain F _Out20 +e _Out2 Δy _Out2,n +a _Out2 ΔQ _Out2,n +e _s Δy _s =0 (14.1) ... F _OutN0 +e _OutN Δy _OutN,n +a _OutN ΔQ _OutN,n +e _s Δy _s =0 (14.N) where:

For the mass conservation equation of the distribution tank, it can be obtained by integrating equation (9) and taking the second-order approximation:

In the formula: Δt is the time step; the suffix 0 represents the value of the physical quantity in the previous time step; Using the Newton-Simpson method, equation (15) can be transformed into:

where:

In this way, one can be obtained about _{Δy s} , _{ΔQ In1,m , Δy In1,m ,} . Equations of _Out2,n ,...,ΔQ _OutN,n ,Δy _OutN,n ; Step 3, deduce the correlation between the flow increment of the inlet node of water channel 1 and the water level increment: for the current iteration step, the unknowns include: water inlet channel 1, water inlet channel 2, ..., water inlet channel M, water outlet channel 1, Flow increment and water level increment of outlet channel 2,..., outlet channel N, ΔQ _In1,m , Δy _In1,m ,..., ΔQ _InM,m , Δy _InM,m , ΔQ _Out1,0 , Δy _{Out1, 0} , ΔQ _Out2,n , Δy _Out2,n ,..., ΔQ _OutN,n , Δy _OutN,n , the water level increment of the distribution tank, Δy _s , a total of 2(M+N)+1; the number of equations is 2( M+N), the correlation between the flow increment ΔQ _Out1,0 of the inlet node of water channel 1 and the water level increment Δy _Out1,0 can be obtained by deduction; Step 4, back-substitution to solve the flow increments and water level increments of all nodes of the outlet channel 1: use the back-substitution process to solve the flow increments and water level increments of all nodes of the outlet channel 1, and the calculation formula is:

Step 5, solve the flow increment and water level increment of the outlet node of the water inlet channel, the inlet node of the outlet channel, and the water level increment of the distribution tank: solve the outlet of the inlet channel 1, the inlet channel 2, ..., the inlet channel M nodes, as well as the flow increment and water level increment of the inlet node of outlet channel 2, ..., outlet channel N, and the water level increment of the distribution tank; the flow increment ΔQ _Out1,0 of the inlet node of outlet channel 1 obtained in step 4 With the water level increment Δy _Out1,0 , into the equation established above, determine the outlet nodes of inlet 1, inlet 2, ..., inlet M, and the flow increments of inlet nodes of outlet 2, ..., outlet N, and Water level increment, ΔQ _In1,m , Δy _In1,m , …, ΔQ _InM,m , Δy _InM,m , ΔQ _Out2,n , Δy _Out2,n , …, ΔQ _OutN,n , Δy _OutN,n , The water level increment of the distribution tank, Δy _s ; Step 6, back-substitution to solve the flow increment and water level increment of other nodes of each channel: Back-substitution to solve the water inlet channel 1, the water inlet channel 2, ..., the water inlet channel M, the water outlet channel 2, ..., the water outlet channel The flow increment and water level increment of all nodes of N are calculated as:

Step 7: Calculate the flow and water level of each node: Calculate the water inlet channel 1, the water inlet channel 2, ..., the water inlet channel M, the water outlet channel 1, the water outlet channel 2, ..., the water outlet channel N. All nodes are in the current iteration step The flow and water level of , and the water level of the distribution tank in the current iteration; the flow is equal to the value of the previous iteration plus the flow increment, and the water level is equal to the value of the previous iteration plus the water level increment. 2. method according to claim 1, is characterized in that, described water inlet channel or water outlet channel is pressure flow, then adopt narrow slit method to solve, and slit width is: B=gA/a ² (19) In the formula: B is the width of the gap; A is the cross-sectional area of the pressurized flow; a is the water hammer velocity.

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