A numerical simulation method for hydraulic control of water distribution hubs

Abstract

The present invention relates to a numerical simulation method for hydraulic control of a water distribution hub, including obtaining incremental correlation; conversion and integration; deriving the correlation between the flow increment and the water level increment of the inlet node of the water channel; back-substituting to solve the flow increment and water level increment of all nodes of the outlet channel; solving the flow increment and water level increment of the outlet node of the inlet channel and the inlet node of the outlet channel and the water level increment of the water distribution tank; back-substituting to solve the flow increment and water level increment of other nodes of each channel; calculating the flow and water level of each node. The present invention constructs the mass and energy conservation equations under the complex operating conditions of the inflow, outflow and overflow of the water distribution hub through theoretical analysis, provides the Newton-Simpson discrete algorithm, encoding method, Preissmann four-point implicit difference algorithm calculation program and other contents, and establishes a theoretical model of the water distribution hub; for the working conditions that may be encountered such as multiple inflows, multiple outflows, and pressure-free coupling, a numerical solution method is provided to provide a reference for the operation and scheduling of the project.

Description

Technical Field

[0001] The invention relates to a numerical simulation method for hydraulic control of a water distribution hub, an analysis method for a hydraulic project, and a method for analyzing a hydraulic project using a computer and assisting in operation and dispatching. Background Art

[0002] Water diversion 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 dispatch and operation are the key to ensuring the security of national water resources. The control and regulation of the project usually requires the calculation of hydraulic transition process, and the analysis of the flow, water level and pressure changes of pipes, culverts, channels and tunnels under certain operation strategies of gates, pumps, valves and other equipment to meet the requirements of specifications, pipeline pressure, equipment limits and so on.

[0003] A water distribution hub is a common hydraulic structure in water diversion projects and large and medium-sized irrigation areas. It is used to connect the pipe culverts, canals and tunnels of long-distance water diversion projects and subsequent water distribution projects, as well as the main water diversion canal (pipe) and the water distribution trunk and branch canals (pipes). If a reservoir is used as a water distribution hub, it can be regarded as a constant water level boundary, and the hydraulic transition process of the water diversion project and the water distribution project is calculated separately. Due to the constraints of factors such as the overall layout of the project, topographic and geological conditions, and immigration relocation, many projects do not have a suitable reservoir area as a water distribution hub, and artificial pools, canals and other facilities need to be built. Artificial construction must consider cost, environmental issues, and limit the pools and canals to 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, water level and pressure along the entire water transmission system will change accordingly. If the water distribution hub is regarded as a constant water level boundary, it cannot reflect the true hydraulic characteristics of the water diversion project and the water distribution project. What is the impact on the operation and scheduling of the project and the extent of the impact? It is necessary to couple the water diversion project, the water distribution hub and the water distribution project for simulation analysis. How to conduct simulation analysis on water distribution hubs to provide effective data and ensure water transmission safety is a problem that needs to be solved. Summary of the invention

[0004] 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 hub. The method studies the hydraulic characteristics of a finite volume water distribution hub under complex inflow and outflow conditions, provides control equations, discrete methods, solution procedures, etc., establishes a theoretical model of a water distribution hub, and simulates and analyzes the hydraulic control of a water distribution hub in combination with typical examples, solving the model problem of the joint solution of water diversion projects, water distribution hubs, and water distribution projects.

[0005] The object of the present invention is achieved as follows: a numerical simulation method for hydraulic control of a water distribution hub, wherein the water distribution hub analyzed by the method comprises: M water inlet channels, a water distribution tank with an overflow channel, and N water outlet channels; m+1 computing nodes are set along each water inlet channel, and the computing nodes of each water inlet channel are coded from upstream to downstream; n+1 computing nodes are set along the first water outlet channel, and the computing nodes of water outlet channel 1 are coded from upstream to downstream; n+1 computing nodes are set along the second to N water outlet channels, and the computing nodes of water outlet channels 2 to N are coded from downstream to upstream; that is, the terminal computing nodes of M water inlet channels, the terminal computing nodes of 2 to N water outlet channels, and the head end computing node of 1 water outlet channel are all computing nodes at the junction of the water inlet channel or the water outlet channel and the water distribution tank, and the method is characterized in that the steps of the method are as follows:

[0006] Step 1, obtain the incremental correlation: use the double-sweep method to eliminate the elements of the matrix B and column vector P of the inlet channel 1, inlet channel 2, ..., inlet channel M, outlet channel 2, ..., outlet channel N, and obtain the correlation between the flow increment and the water level increment of the end node. The number of equations is M+N-1; the calculation formula is as follows:

[0007] X=BX+P (1)

[0008] Where: ΔQ is the flow increment of the current iteration step, the subscript K represents the calculation node number, for the inlet channel, the subscript value is 0~m; for the outlet channel, the subscript value is 0~n; Δh is the water depth increment of the current iteration step; U, W, P are double sweep coefficients; the subscripts of U, W, P represent the change in the number of rows of the matrix, for the inlet channel, the number is 0~2m-1; for the outlet channel, the number is 0~2n-1;

[0009] Each element is calculated by the following recursive formula:

[0010]

[0011]

[0012]

[0013] Where: c, b, D, a, e are coefficients; j is the node number, j = 1, 2, ..., K-2;

[0014] For the inlet channel 1, the flow increment and water volume increment at the terminal calculation node satisfy the following relationship:

[0015] ΔQ In1,m =U In1,2m-2 Δy In1,m +P In1,2m-2(5.1)

[0016] For the inlet channels 2 to M, the flow increment and water volume increment at the terminal calculation node satisfy the following relationship:

[0017] ΔQ In2,m =U In2,2m-2 Δy In2,m +P In2,2m-2 (5.2)

[0018] …

[0019] ΔQ InM,m =U InM,2m-2 Δy InM,m +P InM,2m-2 (5.M)

[0020] For outlet channels 2 to N, the flow increment and water volume increment at the terminal calculation node satisfy the following relationship:

[0021] ΔQ Out2,n =U Out2,2n-2 Δy Out2,n +P Out2,2n-2 (6.2)

[0022] …

[0023] ΔQ OutN,n =U OutN,2n-2 Δy OutN,n +P OutN,2n-2 (6.N)

[0024] Thus, we can obtain M+N-1 flow increments and water volume increments, 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 relationship; the subscript In represents the water inlet channel, Out represents the water outlet channel, the following numbers represent the channel number, and the numbers after the comma represent the calculation node number;

[0025] Step 2, conversion and integration: All water levels of inlet, outlet and distribution tanks satisfy the energy conservation equation and are discretized using the Newton-Simpson method, with the number of equations being M+N; they also satisfy the mass conservation equation, which is integrated and approximated to the second order, with the number of equations being 1;

[0026] Energy conservation equation between the outlet node of inlet channel 1 and the water distribution tank:

[0027]

[0028] Where: y In1,m is the water head of the piezometric pipe at the outlet node of inlet channel 1; Q In1,m is the flow rate at the outlet node of inlet channel 1; g is the acceleration of gravity; A In1,m is the flow area of ​​the outlet node of inlet channel 1; y s is the water head of the pressure measuring pipe of the water distribution tank; In1 is the local head loss coefficient of water from inlet channel 1 entering the distribution tank;

[0029] Energy conservation equation of the outlet node 2~M of the water inlet channel and the water distribution tank:

[0030]

[0031] Where: y In2,m ,……,y InM,m Q is the water head of the pressure gauge at the outlet node of the water inlet channel 2~M; In2,m ,……,Q InM,m is the flow rate of the outlet node of the water inlet channel 2~M; g is the acceleration of gravity; A In2,m ,……,A InM,m is the flow area of ​​the outlet node of the inlet channel 2~M; In2 、……、ζ InM is the local head loss coefficient of water from inlet channel 2 to M entering the distribution tank;

[0032] Energy conservation equation for the inlet node of the water distribution tank and outlet channel 1:

[0033]

[0034] Where: y Out1,0 is the water head of the piezometric tube at the inlet node of outlet channel 1; Q Out1,0 A is the flow rate at the inlet node of outlet channel 1; Out1,0 is the flow area of ​​the inlet node of outlet channel 1; ζ Out1 is the local head loss coefficient of water entering the outlet channel 1;

[0035] Energy conservation equation for the water distribution tank and the inlet nodes 2 to N of the outlet channel:

[0036]

[0037] …

[0038]

[0039] Where: y Out2,n ,……,y OutN,n Q is the water head of the pressure gauge at the inlet node of the outlet channel 2~N; Out2,n ,……,QOutN,n A is the flow rate of the inlet node of the outlet channel 2~N; Out2,n ,……,A OutN,n is the flow area of ​​the inlet node of the outlet channel 2~N; Out2 、……、ζ OutN is the local head loss coefficient of water entering the outlet channel 2~N;

[0040] The mass conservation equation of the water distribution tank is:

[0041]

[0042] Where: A0 is the plane area of ​​the water distribution tank; Q w is the flow rate of the overflow channel, and the calculation formula is:

[0043]

[0044] Where: μ is the flow coefficient; B w is the overflow weir width; H w is the overflow weir elevation;

[0045] Equation conversion:

[0046] For the water inlet channel 1, formula (7.1) gives:

[0047]

[0048] Using the Newton-Simpson method, equation (11) is transformed into:

[0049] F In10 +e In1 Δy In1,m +a In1 ΔQ In1,m +e s Δy s =0 (12.1)

[0050] Where: Δy s is the increment of water level in the distribution tank;

[0051] For water inlet channels 2 to M, we can obtain:

[0052] F In20 +e In2 Δy In2,m +a In2 ΔQ In2,m +e s Δy s =0 (12.2)

[0053] …

[0054] F InM0 +e InM Δy InM,m +a InM ΔQ InM,m +e s Δy s =0 (12.M)

[0055] Where: ……、

[0056] ……、 ……、

[0057] Thus, we obtain M about Δy s , ΔQ In1,m , Δy In1,m , ..., ΔQ InM,m , Δy InM,m The equation of

[0058] For outlet channel 1, formula (8.1) gives:

[0059]

[0060] Using the Newton-Simpson method, equation (13) is transformed into:

[0061] F Out10 +e Out1 Δy Out1,0 +a Out1 ΔQ Out1,0 +e s Δy s =0 (14.1)

[0062] Where:

[0063]

[0064] For outlet channels 2 to N, we obtain

[0065] F Out20 +e Out2 Δy Out2,n +a Out2 ΔQ Out2,n +e s Δy s =0 (14.2)

[0066] …

[0067] F OutN0 +e OutN Δy OutN,n +a OutN ΔQOutN,n +e s Δy s =0 (14.N)

[0068] Where: ……、

[0069] ……、

[0070] ……、

[0071]

[0072] For the mass conservation equation of the water distribution tank, the integration of equation (9) and the second-order approximation are as follows:

[0073]

[0074]

[0075] Where: Δt is the time step; the suffix 0 represents the value of the physical quantity in the previous time step;

[0076] Using the Newton-Simpson method, equation (15) can be transformed into:

[0077]

[0078] Where:

[0079]

[0080]

[0081]

[0082] Thus, we can obtain a value about Δy s , 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 equation of

[0083] Step 3, derive the correlation between the flow increment and water level increment of the inlet node of water channel 1: For the current iteration step, the unknowns include: the flow increment and water level increment of inlet channel 1, inlet channel 2, ..., inlet channel M, outlet channel 1, outlet channel 2, ..., outlet channel N, ΔQ In1,m , ΔyIn1,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 , water level increment in the distribution tank, Δy s , a total of 2(M+N)+1; the number of equations is 2(M+N), and the flow increment ΔQ of the inlet node of water channel 1 can be obtained by derivation Out1,0 With water level increment Δy Out1,0 the relevant relationship;

[0084] Step 4, back-substitution to solve the flow increment and water level increment of all nodes of outlet channel 1: Use the back-substitution process to solve the flow increment and water level increment of all nodes of outlet channel 1. The calculation formula is:

[0085]

[0086] Step 5, solve the flow increment and water level increment of the outlet node of the inlet channel and the inlet node of the outlet channel, as well as the water level increment of the distribution tank: solve the outlet nodes of the inlet channel 1, the inlet channel 2, ..., the inlet channel M, and the flow increment and water level increment of the inlet node of the outlet channel 2, ..., the outlet channel N, as well as the water level increment of the distribution tank; use the flow increment ΔQ of the inlet node of the outlet channel 1 obtained in step 4 Out1,0 With water level increment Δy Out1,0 , Substitute into the equation established above to determine the flow increment and water level increment of the outlet nodes of inlet 1, inlet 2, ..., inlet M, and the inlet nodes of outlet 2, ..., outlet N, ΔQ In1,m , Δy In1,m , ..., ΔQ InM,m , Δy InM,m , ΔQ Out2,n , Δy Out2,n , ..., ΔQ OutN,n , Δy OutN,n , water level increment in the distribution tank, Δy s ;

[0087] Step 6, back-substitute to solve the flow increment and water level increment of other nodes of each channel: back-substitute to solve the flow increment and water level increment of all nodes of inlet channel 1, inlet channel 2, ..., inlet channel M, outlet channel 2, ..., outlet channel N, respectively. The calculation formula is:

[0088]

[0089] Step 7, calculate the flow and water level of each node: calculate the flow and water level of all nodes of inlet channel 1, inlet channel 2, ..., inlet channel M, outlet channel 1, outlet channel 2, ..., outlet channel N in the current iteration step, and the water level of the distribution tank in the current iteration step; the flow is equal to the value of the previous iteration step plus the flow increment, and the water level is equal to the value of the previous iteration step plus the water level increment.

[0090] Furthermore, if the water inlet channel or the water outlet channel is under pressure, the narrow slit method is used for solution, and the slit width is:

[0091] B=gA / a 2 (19)

[0092] Where: B is the gap width; A is the cross-sectional area of ​​pressurized flow; a is the water hammer wave velocity.

[0093] The advantages and beneficial effects of the present invention are as follows: the present invention constructs the mass and energy conservation equations under the complex operating conditions of inflow, outflow and overflow of the water distribution hub through theoretical analysis, provides the Newton-Simpson discrete algorithm, encoding method, calculation procedures of the Preissmann four-point implicit difference algorithm and other contents, and establishes a theoretical model of the water distribution hub; for the operating conditions that may be encountered such as multiple water inlets, multiple water outlets, and pressure-free coupling, a numerical solution method is provided, which can provide a reference for the operation and scheduling of the project. BRIEF DESCRIPTION OF THE DRAWINGS

[0094] The present invention will be further described below in conjunction with the accompanying drawings and embodiments.

[0095] Figure 1 is a schematic diagram of the structure of the water distribution hub analyzed by the method described in the first embodiment of the present invention;

[0096] Figure 2 is a flow chart of the method described in Embodiment 1 of the present invention;

[0097] Figure 3 This is a schematic diagram of the structure of a one-inlet and two-outlet water distribution hub analyzed in Example 1 of the present invention;

[0098] Figure 4 This is a schematic diagram of a water distribution system in an application example according to the first embodiment of the present invention;

[0099] Figure 5 It is the flow rate and water level process at a typical position of the water distribution system of the application example described in the first embodiment of the present invention, wherein (a) is the flow rate process and (b) is the water level process;

[0100] Figure 6 It is the hydraulic characteristics of the upstream tunnel of the application example described in the first embodiment of the present invention, wherein (a) is the flow process and (b) is the water level process. DETAILED DESCRIPTION

[0101] Embodiment 1:

[0102] 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, a water distribution tank ③ with an overflow channel ②, and N water outlet channels ④, such as Figure 1 shown. Figure 1 The arrows in the figure indicate the direction of water flow.

[0103] The water distribution hub described in this embodiment is a common hydraulic structure in water diversion projects and large and medium-sized irrigation areas. It is used to connect the pipe culvert tunnels of long-distance water diversion projects and subsequent water distribution projects, as well as the main water diversion channel (pipe) and the water distribution trunk and branch channels (pipes). If a reservoir is used as a water distribution hub, it can be regarded as a normal water level boundary, and the hydraulic transition process of the water diversion project and the water distribution project is calculated separately. Restricted by factors such as the overall layout of the project, topographic and geological conditions, and immigration relocation, many projects do not have a suitable reservoir area as a water distribution hub and need to be built. During the design, the volume selection of the water distribution hub generally has two references: i. The underwater volume of the water inlet pool can be determined according to 30 to 50 times the design flow rate of the water pump sharing the water inlet pool; ii. When the water transfer scale is not large or the requirements are not high, the volume of the water pool in the middle of the gravity water transfer pipeline can be determined according to the maximum design water volume of not less than 5min. It can be seen that the volume of the water distribution hub is generally 30 to 300 times the design flow rate. 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 along the entire water delivery system will change accordingly. If the water distribution hub is regarded as a constant water level boundary, it will not be able to reflect the true hydraulic characteristics of the water diversion project and the water distribution project. Improper control will not be able to deliver the required water to users in a timely and appropriate manner, which may cause engineering accidents such as overbank overflow, pipe burst, and structural damage. In this case, it is necessary to couple the water diversion project, water distribution hub and water distribution project for simulation analysis.

[0104] The purpose of this embodiment is to study the hydraulic characteristics of a finite volume water distribution hub under complex inflow and outflow conditions, provide control equations, discrete methods, solution procedures, etc., establish a theoretical model of the water distribution hub, and simulate and analyze the hydraulic control of the water distribution hub in combination with typical examples to solve the model problem of the joint solution of water diversion projects, water distribution hubs and water distribution projects.

[0105] The water distribution hub described in this embodiment mainly consists of three elements: an inlet channel, an outlet channel and a water distribution pool. The inlet channel and the outlet channel can be open channels or non-pressure culverts, or pressurized pipes. The number of inlet channels and the number of outlet channels can be equal or unequal. 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.

[0106] To facilitate calculation, this embodiment sets computing nodes in the water inlet channel and the water outlet channel: m+1 computing nodes are set along each water inlet channel, and the computing nodes of each water inlet channel are coded from upstream to downstream. n+1 computing nodes are set along the first water outlet channel, and the computing nodes of water outlet channel 1 are coded from upstream to downstream; n+1 computing nodes are set along the second to N water outlet channels, and the computing nodes of water outlet channels 2 to N are coded from downstream to upstream; that is, the terminal computing nodes of M water inlet channels, the terminal computing nodes of 2 to N water outlet channels, and the head end computing node of 1 water outlet channel are all computing nodes at the connection between the water inlet channel or the water outlet channel and the water distribution tank, see Figure 1 .

[0107] The water distribution hub is usually an open channel flow with a free surface. The control equation is the Saint-Venant equation, including the momentum equation and the continuity equation. The main solution methods include the characteristic line method, the explicit difference method, the implicit difference method, etc. Among them, the implicit difference method has good numerical stability. Therefore, the study adopted the Preissmann four-point implicit difference format, which is widely used in engineering calculations.

[0108] (I) Energy conservation of water inlet system:

[0109] The energy conservation equations of inlet channel 1, inlet channel 2, ..., inlet channel M and the distribution tank are as follows.

[0110] The water from inlet channel 1 enters the distribution tank. The energy conservation equation between the end node of inlet channel 1 and the distribution tank can be obtained from the Bernoulli energy equation for unsteady flow:

[0111] Energy conservation equation between the outlet node of inlet channel 1 and the water distribution tank:

[0112]

[0113] Where: y In1,m is the water head of the piezometric pipe at the outlet node of inlet channel 1; Q In1,m is the flow rate at the outlet node of inlet channel 1; g is the acceleration of gravity; A In1,m is the flow area of ​​the outlet node of inlet channel 1; y s is the water head of the pressure measuring pipe of the water distribution tank; In1 is the local head loss coefficient of water from inlet channel 1 entering the distribution tank. Since the cross-sectional area of ​​the distribution tank is generally much larger than the cross-sectional area of ​​the channel, the flow velocity head of the distribution tank can be ignored.

[0114] The water from inlet channel 2 enters the distribution tank. The energy conservation equation between the end node of inlet channel 2 and the distribution tank can be obtained from the Bernoulli energy equation for unsteady flow:

[0115]

[0116] Where: y In2,m is the water head of the piezometer at the outlet node of inlet channel 2; is the flow rate at the outlet node of inlet channel 2; A In2,m is the flow area of ​​the outlet node of inlet channel 2; is the local head loss coefficient of water from inlet channel 2~M entering the distribution tank.

[0117] ...(The energy conservation equation of the water inlet channel 3 to the end node of the water inlet channel M-1 and the water distribution tank is omitted)

[0118] The water from the inlet channel M enters the distribution tank. The energy conservation equation for the end node of the inlet channel M and the distribution tank can be obtained from the unsteady flow Bernoulli energy equation:

[0119]

[0120] Where: y InM,m is the water head of the piezometric pipe at the outlet node of the inlet channel M; Q InM,m A is the flow rate at the outlet node of the inlet channel M; InM,m is the flow area of ​​the outlet node of the inlet channel M; InM is the local head loss coefficient of water from the inlet channel M entering the distribution tank;

[0121] (II) Energy conservation of water outlet system:

[0122] The energy conservation equations of outlet channel 1, outlet channel 2, ..., outlet channel N are as follows:

[0123] The water in the distribution tank enters the outlet channel 1. 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:

[0124]

[0125] Where: y Out1,0 is the water head of the piezometric tube at the inlet node of outlet channel 1; Q Out1,0 A is the flow rate at the inlet node of outlet channel 1; Out1,0 is the flow area of ​​the inlet node of outlet channel 1; ζ Out1 is the local head loss coefficient of water entering the outlet channel 1.

[0126] The water in the distribution tank enters the outlet channel 2. The energy conservation equation of the inlet node of the distribution tank and the outlet channel 2 can be obtained from the Bernoulli energy equation of unsteady flow.

[0127]

[0128] Where: y Out2,n is the water head of the piezometric tube at the inlet node of outlet channel 2; Q Out2,n A is the flow rate at the inlet node of outlet channel 2;Out2,n is the flow area of ​​the inlet node of outlet channel 2; Out2 is the local head loss coefficient of water entering outlet channel 2. The computation nodes of outlet channel 2 are coded from downstream to upstream.

[0129] ...(The energy conservation equation of the water distribution tank and the terminal node from the inlet and outlet channel 3 to the inlet channel M-1 is omitted)

[0130] The water in the distribution tank enters the outlet channel N. The energy conservation equation for the inlet node of the distribution tank and the outlet channel N can be obtained from the unsteady flow Bernoulli energy equation:

[0131]

[0132] Where: y OutN,n is the water head of the piezometric tube at the inlet node N of the outlet channel; Q OutN,n A is the flow rate at the inlet node of the outlet channel N; OutN,n is the flow area of ​​the inlet node N of the outlet channel; OutN is the local head loss coefficient of water entering the outlet channel N. The computation nodes of the outlet channel N are coded from downstream to upstream.

[0133] (III) Conservation of mass of water distribution tank:

[0134] The increment of water in the distribution tank is equal to the sum of inflow, outflow and overflow, and the mass conservation equation is:

[0135]

[0136] Where: A0 is the plane area of ​​the water distribution tank, m 2 ;Q w is the flow rate of the overflow channel, m 3 / s, the calculation formula is

[0137]

[0138] Where: μ is the flow coefficient; B w is the overflow weir width, m; H w is the overflow weir elevation, m.

[0139] (IV) Equation conversion:

[0140] For the water inlet channel 1, formula (7.1) gives:

[0141]

[0142] Using the Newton-Simpson method, equation (11) can be transformed into:

[0143] F In10 +e In1 ΔyIn1,m +a In1 ΔQ In1,m +e s Δy s =0 (12.1)

[0144] Where: Δy s is the increment of water level in the distribution tank.

[0145] The equations for other inlet channels can be converted in a similar way.

[0146] From formula (8.1), we get:

[0147]

[0148] Using the Newton-Simpson method, equation (13) can be transformed into:

[0149] F Out10 +e Out1 Δy Out1,0 +a Out1 ΔQ Out1,0 +e s Δy s =0 (14.1)

[0150] Where:

[0151]

[0152] The equations for other outlet channels can be converted in a similar way.

[0153] For the mass conservation equation of the water distribution tank, the integration of equation (9) and the second-order approximation are as follows:

[0154]

[0155]

[0156] Where: Δt is the time step; the suffix 0 represents the value of the physical quantity in the previous time step.

[0157] Using the Newton-Simpson method, equation (15) can be transformed into:

[0158]

[0159] Where:

[0160]

[0161]

[0162]

[0163] Calculation of other nodes:

[0164] The continuity equation and momentum equation for one-dimensional unsteady flow in an open channel are:

[0165]

[0166]

[0167] 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 uneven distribution of cross-sectional flow velocity; g is the gravitational acceleration, m 2 / s; h is the water depth, m; S0 is the riverbed slope; S f is the friction ratio drop, and the calculation formula is:

[0168] S f =Q|Q| / K2 (22)

[0169] Where: 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:

[0170]

[0171]

[0172] Where: y is the water surface elevation; ψ R is the spatial coefficient, which is not necessarily the same as ψ.

[0173] Substituting into the continuity equation (23), we get:

[0174] 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)

[0175] Where: b 2j+1 = -θ / Δx; d 2j+1 =θ / Δx;

[0176]

[0177] Substituting into the momentum equation (24), we get:

[0178] 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)

[0179] Where:

[0180]

[0181]

[0182]

[0183]

[0184] Boundary conditions:

[0185] There are three common boundary conditions:

[0186] aF=hh(t)=0

[0187] bF=QQ(t)=0

[0188] cF=Qf(h)=0

[0189] For a type of boundary, water depth is a function of time; for a type of boundary, flow is a function of time; for a type of boundary, flow is a function of water depth. For example, if the downstream of an open channel is a wide crested weir, the relationship between flow and water level is:

[0190]

[0191] Where: μ is the discharge coefficient; B is the weir top width, m; h is the water head above the weir, m.

[0192] Using the Newton-Simpson formula, the above three types of boundary conditions can be transformed into:

[0193] F h Δh+F Q ΔQ=-F0

[0194] For type a boundary conditions,

[0195]

[0196] For type b boundary conditions,

[0197]

[0198] For type c boundary conditions,

[0199]

[0200] The nodes in the channel are discretized by using the Preissmann four-point implicit difference, and the boundary conditions are linearized using the Newton-Simpson formula. The matrix form of the equation is:

[0201] AX=D (31)

[0202] Where: The coefficients b0, c0 and D0 are determined by the channel inlet boundary conditions, b0 = F h 、c0=F Q and D0 = -F0; a 2K-1 、b 2K-1 and D 2K-1 Determined by the channel outlet boundary conditions, a 2K-1 =F h 、b 2K-1 =F Q and D 2K-1 =-F0.

[0203] The coefficient matrix A of the above nonlinear equations is in a band shape, and its non-zero elements are all located near the diagonal. The double sweep method can be used to solve the above equations. Assuming that the coefficient b0≠0, the elimination method is used to transform the above equations:

[0204] X=BX+P (1)

[0205] Where:

[0206] Each element is calculated by the following recursive formula:

[0207]

[0208]

[0209]

[0210]

[0211] Since the matrix B of the above linear equations is an upper triangular matrix with 0 elements on the diagonal, the following back substitution process is used to recursively solve it:

[0212]

[0213] The specific process of numerical solution can be described as (the process is as follows Figure 2 shown):

[0214] Step 1, obtain the incremental correlation. Use the double-sweep method to calculate the elements of the matrix B and column vector P of inlet 1, inlet 2, ..., inlet M, outlet 2, ..., outlet N, and obtain the correlation between the flow increment and the water level increment at the end node. The number of equations is M+(N-1).

[0215] Step 2: Transformation and integration. All water levels of inlet, outlet and distribution tanks satisfy the energy conservation equation and are discretized using the Newton-Simpson method. The number of equations is M+N. At the same time, they also satisfy the mass conservation equation. The integration and second-order approximation are taken. The number of equations is 1.

[0216] Step 3, derive the correlation between the flow increment and the water level increment at the inlet node of water channel 1. For a certain iteration step, the unknowns include: the flow increment and water level increment of inlet 1, inlet 2, ..., inlet M, outlet 1, outlet 2, ..., outlet 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), and the correlation between the flow increment and the water level increment at the inlet node of water channel 1 can be derived.

[0217] Step 4: Back-substitute to solve the flow increment and water level increment of all nodes in outlet channel 1.

[0218] Step 5, solve the flow increment and water level increment of the outlet node of the inlet channel, the inlet node of the outlet channel, and the water level increment of the distribution tank.

[0219] Step 6: Back-substitute to solve the flow increment and water level increment of other nodes in each channel.

[0220] Step 7: Calculate the flow and water level of each node.

[0221] 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 begins; otherwise, a new iteration step is started using the new flow and water level until the accuracy requirements are met.

[0222] There is another special case in actual engineering, that is, some water inlet or outlet is pressurized flow. In this case, the narrow slit method can be used to solve it, or other similar methods that convert pressurized water into equivalent pressure-free water can be used to solve it.

[0223] Calculation example:

[0224] To further illustrate the method described in this embodiment, a water distribution hub with "one inlet, two outlets, and one overflow (drainage)" is used below. Figure 3, for the convenience of explanation, the water inlet channel is named channel 1, and the two water outlet channels are named channels 2 and 3. The water from channel 1 enters the water distribution tank and is distributed to channels 2 and 3 after being regulated by the gate (valve). In order to avoid full flow over the embankment or emptying the tank, an overflow channel or a drain pipe is provided. The overflow channel and the drain pipe are both water level-flow relationship boundaries. This embodiment is analyzed based on the overflow channel.

[0225] Governing equations:

[0226] The water from channel 1 enters the water distribution tank. The energy conservation equation between the end node of channel 1 and the water distribution tank can be obtained from the Bernoulli energy equation for unsteady flow:

[0227]

[0228] Where: y 1,m is the water 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 due to gravity, m / s 2 ; A 1,m is the flow area of ​​the last node of channel 1, m 2 ;y s is the pressure pipe head of the distribution tank, m; ζ1 is the local head loss coefficient of water entering the distribution tank, including cross-sectional changes, outlets, control gates, etc. Since the cross-sectional area of ​​the distribution tank is generally much larger than the cross-sectional area of ​​the channel, the velocity head of the distribution tank can be ignored.

[0229] The water in the distribution tank enters channel 2. The energy conservation equation between the distribution tank and the first node of channel 2 can be obtained from the Bernoulli energy equation for unsteady flow:

[0230]

[0231] Where: y 2,n is the water 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 ζ2 is the local head loss coefficient of water entering channel 2, including cross-sectional changes, water inlets, control gates, etc. The computational nodes of channel 2 are coded from downstream to upstream.

[0232] The water in the distribution tank enters channel 3. The energy conservation equation of the distribution tank and the first node of channel 3 can be obtained from the Bernoulli energy equation of unsteady flow:

[0233]

[0234] Where: y 3,0is the water head of the piezometric pipe at the first node of the channel, m; Q 3,0 is the flow of the first node of channel 3, m 3 / s; A 3,0 is the flow area of ​​the first node of channel 3, m 2 ; ζ3 is the local head loss coefficient of water entering channel 3, including cross-sectional changes, water outlet, control gate, etc.

[0235] The increment of water in the distribution tank is equal to the sum of inflow, outflow and overflow, and the mass conservation equation is:

[0236]

[0237] Where: A0 is the plane area of ​​the water distribution tank, m 2 ;Q w is the flow rate of the overflow channel, m 3 / s, the calculation formula is:

[0238]

[0239] Where: μ is the flow coefficient; B w is the overflow weir width, m; H w is the overflow weir elevation, m.

[0240] Solving algorithm:

[0241] From formula (1a), we get:

[0242]

[0243] Using the Newton-Simpson method, equation (6a) can be transformed into:

[0244] F 10 +e1Δy 1,m +a1ΔQ 1,m +e s Δy s =0(7a)

[0245] Where: Δ represents the increment of the corresponding physical quantity in a certain iteration step;

[0246] From formula (2a), we get:

[0247]

[0248] Using the Newton-Simpson method, equation (8a) can be transformed into:

[0249] F 20 +e2Δy 2,n +a2ΔQ 2,n +e s Δys =0 (9a)

[0250] Where:

[0251]

[0252] From formula (3a), we get:

[0253]

[0254] Using the Newton-Simpson method, equation (10a) can be transformed into:

[0255] F 30 +e3Δy 3,0 +a3ΔQ 3,0 +e s Δy s =0 (11a)

[0256] Where:

[0257]

[0258] By integrating formula (4a) and taking the second-order approximation, we can get:

[0259]

[0260] Arranged:

[0261]

[0262] Where: Δt is the time step, s; the subscript 0 represents the value of the physical quantity in the previous time step.

[0263] Using the Newton-Simpson method, equation (13a) can be transformed into:

[0264] 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)

[0265] Where:

[0266]

[0267]

[0268] The inlet of channel 1 is usually the flow boundary. The elimination process of the double sweep method can be used to obtain the final node satisfying:

[0269] ΔQ 1,m =U 1,2m-2 Δy 1,m +P 1,2m-2 (15a)

[0270] Where: U i,j , P i,j is the double sweep coefficient.

[0271] When the outlet of channel 2 is a water level boundary or a water level-flow relationship, eliminating variables from the outlet to the inlet gives:

[0272] Δy 2,n =U 2,2n-2 ΔQ 2,n +P 2,2n-2 (16a)

[0273] From formula (7a), we get:

[0274]

[0275] Substituting formula (17a) into formula (15a) yields:

[0276]

[0277] From formula (9a), we get:

[0278]

[0279] Substituting formula (19a) into formula (16a) yields:

[0280]

[0281] Substituting equations (18a) and (20a) into equation (14a) yields:

[0282]

[0283] Where:

[0284]

[0285] Substitute formula (21a) into formula (11a):

[0286] b 3,0 Δy 3,0 +c 3,0 ΔQ 3,0 =D 3,0 (22a)

[0287] Where: b3,0 =e3;

[0288] The procedure for numerically solving the entire water delivery system is:

[0289] (1) Use the double sweep method to calculate the matrix B and the element U of the column vector P of the water inlet channel and the water outlet channel 1. 1,i , W 1,i , P 1,i , U 2,j , W 2,j and P 2,j (i=0,1,...,2m-2, j=0,1,...,2n-2);

[0290] (2) Using equation (22a), determine the double sweep coefficient of the inlet boundary of channel 1, b 3,0 、c 3,0 and D 3,0 ;

[0291] (3) Use the double-sweep method to solve the water depth and flow increment of channel 3;

[0292] (4) Solve Δy using equations (21a), (20a), (19a), (18a) and (17a) in sequence s , ΔQ 2,n , Δy 2,n , ΔQ 1,m and Δy 1,m ;

[0293] (5) Use the back substitution process of the double-sweep method to solve the water depth and flow increment of Channel 1 and Channel 2.

[0294] Application examples:

[0295] A water diversion project transfers water to another area through a water diversion tunnel1, such as Figure 4 As shown, the water distribution tank in the middle distributes water to two water transmission tunnels, which are then transported to the receiving areas.

[0296] The water from the water conservancy project flows into the water channel tunnel 1 and then into the water distribution tank. The designed flow rate is 70m 3 / s. The water distribution tank is a rectangular tank, 80m long, 46m wide, and has a capacity of 28,000 m 3 ; The water in the water distribution tank is regulated and then flows into the water delivery tunnel 2 and the water delivery tunnel 3. When the water level is high, the water is discharged through the overflow weir. Figure 4 shown.

[0297] The water distribution flow of water channel tunnel 2 and water channel tunnel 3 is 47m 3 / s+23m 3 / s and 40m 3 / s+30m3 / s. The water distribution flow rate is 47m 3 / s+23m 3 / s switch to 40m 3 / s+30m 3 / s, the gate of channel tunnel 2 is reduced from full opening to 2.1m, and the gate opening of channel tunnel 3 is reduced from 1.7m to full opening, and the action rate is 0.2m / min. The hydraulic transition process characteristics of the water distribution tank and water distribution project are as follows: Figure 5 The target flow rate of 40m3 was reached at 5km and 10km of Tunnel 2 103min and 136min after the gate was actuated respectively. 3 / s; 15 minutes after the gate is activated, the water tank reaches the target flow rate of 30m 3 / s. The water level in the water distribution tank dropped steadily from 258.88m to 258.74m, the water level behind the gate of channel tunnel 2 dropped from 258.33m to 257.78m, and the water level in the water storage tank increased from 256.79m to 258.10m. The hydraulic transient characteristics of channel tunnel 1 upstream of the water distribution tank are shown in Figure 2. Figure 6 As shown in the figure. During the water distribution flow regulation process, the flow and water level of channel tunnel 1 within 15 km upstream of the water distribution tank change accordingly. The flow at 5 km upstream first increases to 70.14m 3 / s, then slowly recover to 70.00m 3 / s, the water level dropped from 260.84m to 260.79m, and the change range of flow and water level decreased with the increase of distance. During the water distribution flow regulation process, the flow of channel tunnel 1, channel tunnel 2, and channel tunnel 3 transitioned smoothly, the open flow tunnel did not overflow, and the water distribution tank did not overflow.

[0298] From the calculation example, we know that during the water distribution flow regulation process, the flow and water level along the water diversion project will change accordingly; similarly, the regulation of the water diversion project will also affect the water supply process and available water volume of users. If the water distribution hub with limited volume is used as the constant water level boundary, it is impossible to analyze the hydraulic transition process of the entire water transmission system, and it is difficult to quantify the actual flow, water level, and pressure change characteristics. The degree of impact on the safety of project operation and the preparation of scheduling plans requires specific problems and specific analysis.

[0299] Embodiment 2:

[0300] This embodiment is an improvement of the above embodiment, and is a refinement of the above embodiment regarding the water inlet pipe and the water outlet pipe. The water inlet channel or the water outlet channel described in this embodiment is a pressurized flow, and the narrow slit method is used for solution, and the slit width is:

[0301] B=gA / a 2 (18)

[0302] Where: B is the gap width; A is the cross-sectional area of ​​pressurized flow; a is the water hammer wave velocity.

[0303] The narrow gap method described in this embodiment is used for solving the problem, that is, assuming that there is a very narrow gap on the top of the pipe, which neither increases the cross-sectional area of ​​the pipe nor the hydraulic radius. The method is extremely simple and can achieve very accurate calculations.

[0304] Finally, it should be noted that the above is only used to illustrate the technical solution of the present invention rather than to limit it. Although the present invention has been described in detail with reference to the preferred arrangement scheme, a person skilled in the art should understand that the technical solution of the present invention (such as the form of the water distribution system, the application of various formulas, the sequence of steps, etc.) can be modified or replaced by equivalents without departing from the spirit and scope of the technical solution of the present invention.

Claims

1. A numerical simulation method for hydraulic control of a water distribution hub, wherein the water distribution hub analyzed in the method comprises: M inlet channels, a distribution tank with an overflow, and N outlet channels; m+1 computing nodes are set along each water inlet channel, and the computing nodes of each water 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; n+1 computing nodes are set along the 2nd to Nth water outlet channels, and the computing nodes of the water outlet channels 2 to N are coded from downstream to upstream; that is, the terminal computing nodes of the M water inlet channels, the terminal computing nodes of the 2nd to N water outlet channels and the head end computing node of the 1 water outlet channel are all computing nodes at the connection between the water inlet channel or the water outlet channel and the water distribution tank, characterized in that the steps of the method are as follows: Step 1, obtain the incremental correlation: use the double-sweep method to eliminate the elements of the matrix B and column vector P of the inlet channel 1, inlet channel 2, ..., inlet channel M, outlet channel 2, ..., outlet channel N, and obtain the correlation between the flow increment and the water level increment of the end node. 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, the subscript K represents the calculation node number, for the inlet channel, the subscript number is 0~m; for the outlet channel, the subscript number is 0~n; Δh is the water depth increment of the current iteration step; U, W, P are double sweep coefficients; the subscripts of U, W, P represent the change in the number of rows of the matrix, for the inlet channel, the number is 0~2m-1; for the outlet channel, the number is 0~2n-1; Each element is calculated by the following recursive formula: Where: c, b, D, a, e are coefficients; j is the node number, j = 1, 2, ..., K-2; For the inlet channel 1, the flow increment and water volume increment at the terminal calculation node satisfy the following relationship: ΔQ In1,m =U In1,2m-2 Δy In1,m +P In1,2m-2 (5.1) For the inlet channels 2 to M, the flow increment and water volume increment at the terminal calculation node satisfy the following relationship: 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 outlet channels 2 to N, the flow increment and water volume increment at the terminal calculation 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, we can obtain M+N-1 flow increments and water volume increments, 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 relationship; the subscript In represents the water inlet channel, Out represents the water outlet channel, the following numbers represent the channel number, and the numbers after the comma represent the calculation node number; Step 2, conversion and integration: All water levels of inlet, outlet and distribution tanks satisfy the energy conservation equation and are discretized using the Newton-Simpson method, with the number of equations being M+N; they also satisfy the mass conservation equation, which is integrated and approximated to the second order, with the number of equations being 1; Energy conservation equation between the outlet node of inlet channel 1 and the water distribution tank: Where: y In1,m is the water head of the piezometric pipe at the outlet node of inlet channel 1; Q In1,m is the flow rate at the outlet node of inlet channel 1; g is the acceleration of gravity; A In1,m is the flow area of ​​the outlet node of inlet channel 1; y s is the water head of the pressure measuring pipe of the water distribution tank; In1 is the local head loss coefficient of water from inlet channel 1 entering the distribution tank; Energy conservation equation of the outlet node 2~M of the water inlet channel and the water distribution tank: …… Where: y In2,m ,……,y InM,m Q is the water head of the pressure gauge at the outlet node of the water inlet channel 2~M; In2,m ,……,Q InM,m is the flow rate of the outlet node of the water inlet channel 2~M; g is the acceleration of gravity; A In2,m ,……,A InM,m is the flow area of ​​the outlet node of the inlet channel 2~M; In2 、……、ζ InM is the local head loss coefficient of water from inlet channel 2 to M entering the distribution tank; Energy conservation equation for the inlet node of the water distribution tank and outlet channel 1: Where: y Out1,0 is the water head of the piezometric tube at the inlet node of outlet channel 1; Q Out1,0 A is the flow rate at the inlet node of outlet channel 1; Out1,0 is the flow area of ​​the inlet node of outlet channel 1; ζ Out1 is the local head loss coefficient of water entering the outlet channel 1; Energy conservation equation for the water distribution tank and the inlet nodes 2 to N of the outlet channel: …… Where: y Out2,n ,……,y OutN,n Q is the water head of the pressure gauge at the inlet node of the outlet channel 2~N; Out2 ,n,……,Q OutN,n A is the flow rate of the inlet node of the outlet channel 2~N; Out2,n ,……,A OutN,n is the flow area of ​​the inlet node of the outlet channel 2~N; Out2 、……、ζ OutN is the local head loss coefficient of water entering the outlet channel 2~N; The mass conservation equation of the water distribution tank is: Where: A0 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 the water inlet channel 1, formula (7.1) gives: 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 water level in the distribution tank; For water inlet channels 2 to M, we can obtain: 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: Thus, we obtain M about Δy s , ΔQ In1,m , Δy In1,m , ..., ΔQ InM,m , Δy InM,m The equation of For outlet channel 1, formula (8.1) gives: 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, we 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 water distribution tank, the integration of equation (9) and the second-order approximation are as follows: Where: 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: Thus, we can obtain a s , 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 equation of Step 3, derive the correlation between the flow increment and water level increment of the inlet node of water channel 1: For the current iteration step, the unknowns include: the flow increment and water level increment of inlet channel 1, inlet channel 2, ..., inlet channel M, outlet channel 1, 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 , water level increment in the distribution tank, Δy s , a total of 2(M+N)+1; the number of equations is 2(M+N), and the flow increment ΔQ of the inlet node of water channel 1 can be obtained by derivation Out1,0 With water level increment Δy Out1,0 the relevant relationship; Step 4, back-substitution to solve the flow increment and water level increment of all nodes of outlet channel 1: Use the back-substitution process to solve the flow increment and water level increment of all nodes of outlet channel 1. The calculation formula is: Step 5, solve the flow increment and water level increment of the outlet node of the inlet channel and the inlet node of the outlet channel, as well as the water level increment of the distribution tank: solve the outlet nodes of the inlet channel 1, the inlet channel 2, ..., the inlet channel M, and the flow increment and water level increment of the inlet node of the outlet channel 2, ..., the outlet channel N, as well as the water level increment of the distribution tank; use the flow increment ΔQ of the inlet node of the outlet channel 1 obtained in step 4 Out1,0 With water level increment Δy Out1,0 , Substitute into the equation established above to determine the flow increment and water level increment of the outlet nodes of inlet 1, inlet 2, ..., inlet M, and the inlet nodes of outlet 2, ..., outlet N, ΔQ In1,m , Δy In1,m , ..., ΔQ InM,m , Δy InM,m , ΔQ Out2,n , Δy Out2,n , ..., ΔQ OutN,n , Δy OutN,n , water level increment in the distribution tank, Δy s ; Step 6, back-substitute to solve the flow increment and water level increment of other nodes of each channel: back-substitute to solve the flow increment and water level increment of all nodes of inlet channel 1, inlet channel 2, ..., inlet channel M, outlet channel 2, ..., outlet channel N, respectively. The calculation formula is: Step 7, calculate the flow and water level of each node: calculate the flow and water level of all nodes of inlet channel 1, inlet channel 2, ..., inlet channel M, outlet channel 1, outlet channel 2, ..., outlet channel N in the current iteration step, and the water level of the distribution tank in the current iteration step; the flow is equal to the value of the previous iteration step plus the flow increment, and the water level is equal to the value of the previous iteration step plus the water level increment.

2. The method according to claim 1, characterized in that If the water inlet channel or water outlet channel is under pressure flow, the narrow slit method is used for solution, and the slit width is: B = gA / a 2 (19) Where: B is the gap width; A is the cross-sectional area of ​​pressurized flow; a is the water hammer wave velocity.

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