Water hammer air charging type pressure tank water transient simulation method

Abstract

The present application relates to a kind of water hammer air supply type pressure tank hydraulic transient simulation method.The described method includes: the functional relationship of gas temperature in pipe and gas pressure, so as to derive new air valve exhaust basic equation, new air valve hydraulic transient mathematical model and its solution algorithm are proposed, process is: the process of constructing mathematical model;Hydraulic transient analysis process;The process of solving nonlinear equation set.The present application uses the aerodynamics theory based on isentropic flow, establishes the functional relationship of gas temperature in pipe and gas pressure, so as to derive new air valve exhaust basic equation, considers the effect of air valve installation mode air valve-repair valve-connection pipe structure size, new air valve hydraulic transient mathematical model and simulation method are proposed.The described method further improves the existing transient simulation condition, makes model simulation more accurate, and verification shows that the simulation carried out by the method is closer to experimental data, provides more accurate design basis for the design of pressure tank.

Description

Technical Field

[0001] This invention relates to a hydraulic transient simulation method for a water-swing pressure tank, which is a simulation calculation method for hydraulic engineering research and a simulation calculation method for safety facilities of long-distance water transmission pipelines. Background Technology

[0002] Air valves are essential pressure regulating devices in water pipeline projects to prevent water hammer damage, and are typically installed every 500-1000 meters. Furthermore, long-distance water pipelines are generally buried underground. When the pipeline diameter is large, inspection holes need to be installed every 1-3 kilometers, with an inner diameter of 0.8-0.9 meters, to allow access for inspection of serious leaks. In some urban water pipelines with significant burial depth, the height of the inspection hole or the distance between its cover and the top of the pipeline (h) is... ori The height can reach 3-5 meters or more. To save space, air valves and access panels often share a basement. Furthermore, to allow for uninterrupted water supply maintenance of the air valves, each air valve is equipped with a manual access butterfly valve of the same specifications. This results in two main air valve installation methods: 1) Air valve-access valve-connecting pipe vertically installed on top of the water supply pipe; 2) Air valve-access valve-connecting elbow installed on the side wall of the access panel. To facilitate the installation and removal of the access panel cover, a certain distance is required between the connecting elbow and the cover. This allows some of the air entering the air valve to remain in the upper part of the access panel, creating a water hammer protection effect similar to an air cushion pressure tank.

[0003] To address the need for protection against water hammer hazards during pipeline filling and power outages at pumping stations, regulations require the use of computational simulation to determine the location, type, and orifice diameter of air valves. Currently, the mathematical model of renowned transient flow experts Wylie and Streeter is generally used to solve for the hydraulic transients of air valves. The basic assumptions of this model include: 1) the gas changes within the pipe follow isothermal laws; 2) the liquid surface height remains essentially constant, neglecting the influence of the height of the air valve, its associated maintenance valve, and connecting pipes. Although Wylie and Streeter's mathematical model for hydraulic transients of air valves is widely used in engineering calculations, including in existing commercial software, its basic assumptions 1 and 2 have problems both theoretically and practically. Summary of the Invention

[0004] To overcome the problems of existing technologies, this invention proposes a hydraulic transient simulation method for water-jet-assisted air-filling pressure tanks. The method establishes a functional relationship between air temperature and pressure within the pipe, thereby deriving new fundamental equations for air valve intake and exhaust, and proposing a new mathematical model for hydraulic transients in air valves and its solution algorithm.

[0005] The objective of this invention is achieved as follows: a hydraulic transient simulation method for a water hammer-type pressure tank, wherein the simulation method is based on a pump station pressurized water supply system comprising a vacuum breaker valve-type pressure tank or an air valve-type pressure tank installed on a water supply pipeline, and the method comprising the following steps:

[0006] The process of constructing a mathematical model:

[0007] assumed:

[0008] 1) The gas is an ideal gas and flows in and out of the air valve with isentropic properties, that is, the energy loss along the flow path and the heat exchange between the flow channel wall and the water body are ignored.

[0009] 2) Air entering the water supply pipe remains near the inspection hole where it can be discharged, so that the hydraulic transients of the water supply pipe can be solved using the conventional water hammer characteristic line method;

[0010] 3) The air entering the pressure tank will not be carried away by the water flow, but will rise above the water surface in the pressure tank;

[0011] I. The mathematical model for the air intake and exhaust of the air valve is as follows:

[0012] 1.1 Air enters at subsonic speeds:

[0013]

[0014] In the formula: C is the gas mass flow rate; in A is the air intake flow coefficient of the air valve; in p is the flow area of ​​the air inlet of the air valve; a The absolute pressure of the atmosphere; k is the polytropic index; R is the gas constant; T a p represents the absolute temperature of the air outside the valve. r The gas pressure ratio p inside the valve r =p / p a p is the absolute pressure of the gas inside the valve; rc,in =(2 / (k+1)) k / (k-1) The valve inlet critical pressure ratio;

[0015] 1.2 Air intake at the critical speed of sound:

[0016]

[0017] 1.3 Air is exhausted at subsonic speeds:

[0018]

[0019] In the formula: C out A is the air valve exhaust flow coefficient; out p is the flow area of ​​the air valve exhaust port;rc,out =1 / p rc,in This is the critical exhaust pressure ratio;

[0020] 1.4 Air is exhausted at the critical speed of sound:

[0021]

[0022] In the formula: ρ a Atmospheric density, kg / m³ 3 ; The speed of sound is denoted in m / s.

[0023] 1.5 Gas law under isentropic conditions:

[0024]

[0025] In the formula: V is the gas volume; M a For gas mass;

[0026] II. Mathematical Model of Hydraulic Transient in Air Valve-Type Pressure Tank:

[0027] Q s =C1-C2H P

[0028] In the formula: Q s H represents the flow rate through the impedance orifice at the bottom of the pressure tank. P The pressure head at the top of the water supply pipe; C1 and C2 are transition coefficients:

[0029]

[0030] In the formula: C P C M B P B M The coefficients are those of the characteristic line method.

[0031] 2.1 Functional relationship between air valve - maintenance valve - connecting pipe - pressure tank water level, gas volume, flow rate, and gas pressure:

[0032] H a p r =C1 / C2-Q s / C2+H a -C 31 -C 41 Q a -2C5|Q s0 |Q s +C5|Q s0 |Q s0 -2C 5a |Q a0 |Q a +C 5a|Q a0 |Q a0

[0033] In the formula: H a atmospheric pressure head H a =p a / γ;p r Q represents the gas pressure ratio inside the valve. s Q is the flow rate through the impedance orifice at the bottom of the pressure tank; s0 Q is the flow rate at the pressure vessel bottom impedance orifice at time t0; a The flow rate within the air valve-maintenance valve-connecting pipe; Q a0 C is the flow rate of the air valve or connecting pipe at time t0; C5 is the impedance coefficient of the pressure tank impedance orifice; C 5a C is the impedance coefficient at the outlet of the connecting pipe. 31 C 41 , where is the transition coefficient:

[0034]

[0035] Where: H s0 The water level in the air valve-maintenance valve-connecting pipe at time t0; Δt is the time step; A s For the corresponding water level H s The cross-sectional area of ​​the flow path; A a For the cross-sectional area of ​​the air valve-inspection valve-connecting pipe; A c Z represents the cross-sectional area of ​​the pressure tank; Z represents the water level in the water supply pipe; Z mt This is the highest water level inside the pressure tank;

[0036] 2.2 Functional relationship between pressure tank water level, gas volume, gas pressure and flow rate:

[0037] H a p cr =C1 / C2-Q s / C2+H a -C 32 -C 42 Q c -2C5|Q s0 |Q s +C5|Q s0 |Q s0

[0038] In the formula: C 32 C 42 Transition coefficient;

[0039]

[0040] Where: H c0 Q represents the water level in the pressure tank at time t0; c0The elevation Z of the pressure tank at time t0 ct The flow rate at the top;

[0041] The process of hydraulic transient analysis:

[0042] Hydraulic transients are divided into four stages:

[0043] Phase 1: There is no gas at the top of the pressure tank and the water level inside the pressure tank is at height H. c ≡Z mt Keeping unchanged, air valve intake and exhaust, air valve-inspection valve-water level H in connecting pipe s At elevation Z ct and Z at Changes between; H c H represents the water level inside the pressure tank. s For the water level in the air valve-maintenance valve-connecting pipe; Z at Z represents the air valve inlet elevation. ct Elevation of the connecting pipe; Z mt This is the highest water level inside the pressure tank;

[0044] Phase 2, water level H s =Z ct The air valve remains unchanged, and air rises above the water surface in the pressure tank, causing the water level H to rise. c From elevation Z mt Gradually decrease to Z ct ;

[0045] Phase 3, air valve intake and exhaust, water level H c =H s <Z ct ;

[0046] Phase 4, once the water level H c =H s ≥Z ct Then the gas is divided into two by the water body. One part is sealed in the upper part of the pressure tank. The gas volume will change with the rise and fall of the water pressure in the water pipe, and its function is equivalent to an air cushion. The other part of the gas is discharged from the air valve as the water pressure rises. When the water pressure drops, the air valve will re-intake air, repeating the similar hydraulic transient process of stages 2 to 4.

[0047] The process of solving a system of nonlinear equations: [The process involves] breaking down the system of nonlinear equations...

[0048]

[0049] Linearization is as follows:

[0050] a 51 Δp r +a 52 Δp cr =d5

[0051] a 61 Δp r +a 62 Δp cr =d6

[0052] Where: F1 and F2 are functions; d3, d4, d5, d6, a 31 a 32 a 41 a 42 a 51 a 52 a 61 a 62 M is the transition coefficient; a0 The mass of gas in the air valve-maintenance valve-connecting pipe at the initial time t0 of stage 4; p is the gas mass flow rate at time t0; C is the constant relating the gas pressure ratio and gas volume in the pressure tank; cr Δp represents the gas pressure ratio inside the pressure tank. r For p r The increment; Δp cr For p cr The increment;

[0053]

[0054]

[0055]

[0056]

[0057]

[0058]

[0059]

[0060] Where: δ is p r For a small increment, δ > 0;

[0061] Solving the system of linear equations yields:

[0062]

[0063] In the formula: D, D pr D pcr Transition coefficient;

[0064] D = a 51 a 62 -a 52 a 61 ;

[0065]

[0066]

[0067] Δp was calculated r and Δp cr season:

[0068]

[0069] p cr =p cr +Δp cr

[0070] In the formula: σ is the convergence factor, 0<σ≤1.

[0071] The advantages and beneficial effects of this invention are as follows: This invention utilizes aerodynamic theory based on isentropic flow to establish a functional relationship between air temperature and air pressure inside the pipe, thereby deriving new fundamental equations for air valve intake and exhaust. Considering the effects of the air valve installation method, the structural dimensions of the air valve, maintenance valve, and connecting pipe, a new mathematical model and simulation method for hydraulic transients of the air valve are proposed. The method described further improves the conditions of existing transient simulations, making the model simulation more accurate. Verification shows that the simulation performed by the method is closer to the experimental data, providing a more accurate design basis for pressure tank design. Attached Figure Description

[0072] The present invention will be further described below with reference to the accompanying drawings and embodiments.

[0073] Figure 1 This is a schematic diagram of the vacuum breaker valve-type pressure tank structure upon which the method described in the embodiments of the present invention is based;

[0074] Figure 2 This is a schematic diagram of the air valve-assisted pressure tank structure upon which the method described in the embodiments of the present invention is based;

[0075] Figure 3 The water level H in the air valve pressure regulating chamber of the engineering example in this invention embodiment is... s Curve showing the change of bottom water pressure H over time t (at station 2071m);

[0076] Figure 4 The water level H of the vacuum breaker valve-type pressure tank in the engineering example of this invention embodiment is shown. s A graph showing the change of bottom water pressure H over time t (at station 2071m). Detailed Implementation

[0077] Example:

[0078] This embodiment describes a hydraulic transient simulation method for a water-hammer-assisted air-filled pressure tank. The simulation method is based on a pump station pressurized water supply system, including a vacuum-rupture valve-assisted air-filled pressure tank or an air valve-assisted air-filled pressure tank installed on the water supply pipeline, such as... Figure 1 , 2 As shown.

[0079] Pressure tanks are commonly used safety devices in pumping station pressurized water supply systems to protect against water hammer hazards. Compared to unidirectional pressure regulating chambers, pressure tanks not only prevent negative pressure but also reduce maximum water hammer pressure. However, conventional pressure tanks require dedicated air compressors and automatic monitoring devices, while airbag-type pressure tanks have higher investment costs and require regular airbag replacement. Therefore, referencing air valve-inspection port devices, a new type of water hammer self-priming pressure tank device can be developed. Water hammer self-priming pressure tanks can be divided into two categories: vacuum breaker valve-priming pressure tanks, such as... Figure 1 As shown, an air valve-replenished pressure tank, such as Figure 2 As shown. Figure 2 In the middle, Z ct Z represents the inlet elevation of the connecting pipe. mt This refers to the elevation of the top of the pressure tank.

[0080] The vacuum breaker valve of the vacuum breaker type pressure tank should be located at the top of the pressure tank. Its characteristic is that it only allows air to enter the pressure tank and does not allow gas to exit the pressure tank.

[0081] The air valve connection pipe of an air-valve-replenished pressure tank needs to be located on the side wall of the pressure tank and inclined upwards. Its characteristic is that when the water level H... s Elevation Z below the inlet elevation of the connecting pipe ct At that time, gas is allowed to flow into and out of the pressure tank, while when H s >Z ct At this time, the gas inside the pressure tank will not be discharged by the air valve. The advantage is that it can meet the requirements of water filling of the water supply pipe and water hammer protection during power outages. Obviously, the air valve-inspection port device can be regarded as a special case of the air valve-replenished pressure tank.

[0082] The hydraulic transient of an air-valve-replenished pressure tank can be divided into four stages: Stage 1, there is no gas at the top of the pressure tank and the water level H in the pressure tank is... c ≡Z mt Keeping unchanged, air valve intake and exhaust, air valve-inspection valve-connecting pipe water level H s At elevation Z ct and Z at Changes between stages; Stage 2, water level H s =Z ct The air valve remains unchanged, and air rises above the water surface in the pressure tank, causing the water level H to rise. c From elevation Z mt Gradually decrease to Z ctStage 3, air valve intake and exhaust, water level H c =H s <Z ct Phase 4, once the water level H c =H s ≥Z ct The gas is then split in two by the water. One part is sealed in the upper part of the pressure tank, and its volume will change due to compression and expansion as the water pressure in the pipeline rises and falls, acting as an air cushion. The other part of the gas is discharged through the air valve as the water pressure rises, and when the water pressure drops, the air valve will re-intake gas, repeating a similar hydraulic transient process as in stages 2-4. Clearly, in stage 4, the air valve's intake and exhaust and the pressure changes in the pressure tank will affect each other. Furthermore, during the air valve's intake process, H2O may occur. c <Z ct And H s >Z ct In this phenomenon, the gas separates the water into two parts: a small portion is in the air valve-maintenance valve-connecting pipe, and the rest is in the pressure tank. It should be noted that in some cases, the hydraulic transient process does not follow the sequence of stages 1, 2, 3, 4. For example, in stage 2, when the air valve begins to release gas due to the renewed rise in water pressure in the supply pipe, the hydraulic transient process skips stage 3 and directly enters stage 4.

[0083] When gas is present at the top of the initial pressure tank, the hydraulic transient of the air-valve-replenished pressure tank will begin in stage 4, and then vary between stages 2 and 4. For Figure 1 The vacuum breaker valve-type pressure tank shown has only stages 1 and 3 in its hydraulic transition process.

[0084] In general, water-steam pressure tanks can be used instead of unidirectional pressure regulating chambers where they are suitable. Compared to unidirectional pressure regulating chambers, water-steam pressure tanks not only prevent liquid vaporization but also reduce the maximum water hammer pressure.

[0085] The main factors affecting the pressure tank of the water-hammer-assisted gas-filling type are the orifice diameter of the air valve or vacuum breaker valve, the volume and height of the pressure tank, the diameter of the impedance orifice, and the connection position between the outlet of the connecting pipe and the pressure tank. These can be determined through hydraulic transient calculations. Since the duration of hydraulic transients is generally very short, the release and dissolution of gas can be ignored in the calculation and analysis.

[0086] 1. Mathematical Model:

[0087] To simplify the problem, we first assume: 1) the gas is an ideal (perfect) gas and flows in and out of the air valve with isentropic flow, i.e., we ignore the influence of energy loss along the flow path and heat exchange between the flow channel wall and the water body within the air valve; 2) the air entering the water pipe remains near the inspection hole where it can be discharged, so that the hydraulic transients of the water pipe can be solved using the conventional water hammer characteristic line method; 3) the air entering the pressure tank (or inspection hole) will not be carried away by the water flow, but will float above the water surface in the pressure tank.

[0088] Since the hydraulic transient process of the vacuum breaker valve-replenished pressure tank is only a part of the air valve-replenished pressure tank, the latter will be the main research object below.

[0089] 1. Mathematical model of air valve intake and exhaust:

[0090] The mathematical model for air intake and exhaust of the air valve is as follows:

[0091] 1.1 Air enters at subsonic speeds:

[0092]

[0093] In the formula: C represents the gas mass flow rate, in kg / s; in The air intake flow coefficient for the air valve is typically taken as 0.7; A in The air inlet area of ​​the air valve is in meters. 2 ;p a The absolute pressure of the atmosphere is Pa; k is the polytropic index; R is the gas constant, typically taken as 287 J / kg·K; T a The absolute temperature of the air outside the valve, in K; p r =p / p a Gas pressure ratio inside the valve; p is the absolute pressure of the gas inside the valve, Pa; p rc,in =(2 / (k+1)) k / (k-1) This is the valve inlet critical pressure ratio, typically ranging from 0.53 to 0.57. For example, for adiabatic flow, k = 1.4, then p rc,in =0.53.

[0094] 1.2 Air intake at the critical speed of sound:

[0095]

[0096] 1.3 Air is exhausted at subsonic speeds:

[0097]

[0098] In the formula: C out The exhaust flow coefficient is typically taken as 0.7; A out The flow area of ​​the exhaust port is m. 2 ;prc,out =1 / p rc,in This is the critical exhaust pressure ratio, typically ranging from 1.75 to 1.89. For example, for adiabatic flow, k = 1.4, then p rc,out =1.89.

[0099] 1.4 Air is exhausted at the critical speed of sound:

[0100] The American Water Industry Association standard AWWA M51, "Micro-inlet / outlet valves, quick-inlet / outlet valves, and combination quick-inlet / outlet valves," states that when the exhaust pressure ratio is equal to or greater than the critical sonic exhaust pressure ratio p... r ≥p rc,out At that time, air will flow at the speed of sound, even as the pressure difference continues to increase. When p in equation (3) is given... r =p rc,out The mass flow rate of air exhausted at the critical speed of sound can be obtained as follows:

[0101]

[0102] In the formula: ρ a Atmospheric density, kg / m³ 3 ; The speed of sound is denoted in m / s.

[0103] 1.5 Gas law under isentropic conditions:

[0104]

[0105] In the formula: V is the gas volume, m 3 M a The mass of the gas is expressed in kg.

[0106] 2. Mathematical model of hydraulic transients in air valve-assisted pressure tank:

[0107] refer to Figure 2 The continuity equation for the air valve connecting pipe inlet is:

[0108] Q a =Q s -Q c (6)

[0109] In the formula: Q a For the flow rate of the air valve or connecting pipe, m 3 / s;Q s The flow rate at the impedance orifice at the bottom of the pressure tank, m 3 / s;Q c Elevation Z of the pressure tank ct The flow rate at the top, m 3 / s.

[0110] The continuity equation for node P at the bottom of the pressure tank is:

[0111] Q s =Q T -Q (7)

[0112] In the formula: Q T Let m be the flow rate of the water pipe flowing into node P. 3 / s; Q is the flow rate of the water pipe flowing out of node P, in meters. 3 / s.

[0113] Based on the compatibility of water hammer characteristic lines, we can conclude that:

[0114] C+:Q T =C P / B P -H P / B P (8)

[0115] C - :Q=-C M / B M +H P / B M (9)

[0116] C P C M B P B M For the characteristic variables of the method of characteristics, the parameter C at time t P C M B P B M It is a known quantity, determined by hydraulic transient calculations of the water pipeline system.

[0117] Substituting equations (8) and (9) into equation (7), we get:

[0118] Q s =C1-C2H P (10)

[0119] In the formula, C1 and C2 are transition coefficients:

[0120]

[0121] Under normal water supply conditions, the air valve is fully closed, Q a =Q c =Q s =0. When a hydraulic transient occurs in the pipeline system, if there is no gas in the water supply pipe... And the pressure head H P =C1 / C2>Z at If the air valve does not work, Q a =Q c =Q s=0; otherwise, the following mathematical model of hydraulic transients in the air valve-pressure tank should be used. It should be noted that... It is a common symbol used in hydraulics to represent the volume of air. The horizontal line in the middle does not mean it is crossed out, but rather indicates that it is empty.

[0122] 2.1 Functional relationship between air valve - maintenance valve - connecting pipe - pressure tank water level, gas volume, flow rate, and gas pressure:

[0123]

[0124]

[0125] In the formula: t is time, in seconds; H s For the air valve-maintenance valve-connecting pipe-pressure tank water level, in meters (m); A s For the corresponding H s The cross-sectional area of ​​the flow path, m 2 V a For H s Upper gas volume, m 3 Z represents the elevation of the top of the water pipeline, in meters. at The elevation of the air valve inlet is in meters (m).

[0126] Integrating equations (12) and (13) and taking a second-order approximation, we get:

[0127] H s =C 31 +C 41 Q a Z≤H s ≤Z at (14)

[0128]

[0129] In the formula: C 31 C 41 It is the transition coefficient; the subscript "0" indicates time t0, Δt = t - t0; V a0 H at time t0 s Upper gas volume;

[0130]

[0131] In the formula: A a The cross-sectional area of ​​the air valve-maintenance valve-connecting pipe is in meters. 2 A c The cross-sectional area of ​​the pressure tank is m. 2 C 31 =Z and C 41 =0 is equivalent to taking H s =Z, meaning it is assumed that the air entering the water pipe remains near the pressure tank from which it can be discharged; Hs0 Q represents the water level in the connecting pipe at time t0; ao The flow rate of the air valve or connecting pipe at time t0.

[0132] When the effects of water inertia and head loss along the pipeline are not considered, the absolute gas pressure p and the piezometric head H at the top of the water pipe are... P The relationship is:

[0133] H a p r =H P +H a -H s -C5|Q s |Q s -C 5a |Q a |Q a (17)

[0134] Where: H a =p a / γ is the atmospheric pressure head, in meters; γ is the specific gravity of water, typically taken as 9800 N / m. 2 C5 is the impedance coefficient of the pressure tank impedance orifice; C 5a The impedance coefficient at the outlet of the connecting pipe;

[0135] Substituting equations (10) and (14) into equation (17), eliminate H. P and H s We can approximate it as follows:

[0136] H a p r =C1 / C2-Q s / C2+H a -C 31 -C 41 Q a -2C5|Q s0 |Q s +C5|Q s0 |Q s0 -2C 5a |Q a0 |Q a +C 5a |Q a0 |Q a0 (18)

[0137] in:

[0138] 2.2 Functional relationship between pressure tank water level, gas volume, gas pressure and flow rate:

[0139]

[0140]

[0141] Where: H c The pressure tank water level is measured in meters (m); V c Let m be the gas volume of the pressure tank. 3 .

[0142] Integrating equations (19) and (20) and taking a second-order approximation, we get

[0143] H c =C 32 +C 42 Q c Z≤H c ≤Z mt (twenty one)

[0144]

[0145] In the formula: Q represents the gas volume in the pressure tank at time t0; c0 The elevation Z of the pressure tank at time t0 ct The flow rate at the top:

[0146]

[0147] Where: H c0 C represents the water level in the pressure tank at time t0. 32 =Z and C 42 =0 is equivalent to taking H c =Z, which means assuming that the air entering the water pipe remains near the pressure tank from which it can be discharged.

[0148] When the effects of inertial forces of the water in the pressure tank and head loss along the pressure vessel are not considered, the absolute pressure p of the gas in the pressure tank is... c Water head H at the top of the water supply pipe P The relationship is:

[0149] H a p cr =H P +H a -H c -C5|Q s |Q s (twenty four)

[0150] In the formula: p cr =p c / p a This refers to the gas pressure ratio inside the pressure tank.

[0151] Substituting equations (10) and (21) into equation (24), eliminate H. P and H c We can obtain:

[0152] Ha p cr =C1 / C2-Q s / C2+H a -C 32 -C 42 Q c -2C5|Q s0 |Q s +C5|Q s0 |Q s0 (25)

[0153] Among them: Q s0 Let t0 be the flow rate at the impedance orifice at the bottom of the pressure tank.

[0154] 3.3 Numerical solution for the hydraulic transients of an air-valve-replenished pressure tank:

[0155] Assuming there is no gas in the pressure tank at the initial moment, the numerical solution method for each stage of the hydraulic transient of the air valve-assisted pressure tank is studied.

[0156] Stage 1: There is no gas at the top of the pressure tank and the water level in the pressure tank is H. c ≡Z mt Keeping unchanged, air valve intake and exhaust, air valve-inspection valve-connecting pipe water level Z ct ≤H s ≤Z at At this time, Q c ≡0 and Q a ≡Q s When H P =C1 / C2<Z at At that time, air enters through the air valve, and the water level H... s Decrease; when H P =C1 / C2≥Z at And V a0 When the value is greater than 0, the air valve releases air.

[0157] Q a ≡Q s Substituting into equation (18), we get:

[0158] Q s =C6-C7p r (26)

[0159] In the formula: C6 and C7 are transition coefficients.

[0160]

[0161]

[0162] In the formula: ω is the cross-sectional area of ​​the pressure tank impedance orifice, m 2 ;ζ and ζa These are the local resistance coefficients of the impedance orifice and the connecting pipe outlet, respectively, and represent a combination of sudden expansion or contraction and a 90° turn. C 5a The second item on the right is the water level H. s The variation in the air valve-maintenance valve-connecting pipe creates an additional resistance coefficient.

[0163] Substituting equation (26) into equation (15), we get:

[0164] V a =C8+C9p r (28)

[0165] In the formula: For H s Upper gas volume;

[0166]

[0167] In the formula: H at time t0 s Upper gas volume;

[0168] The equation of state for isentropic gas flow can be described as follows:

[0169]

[0170] Equation (28) Gas volume V a Substituting into equation (30), we get:

[0171]

[0172] In the formula: F is a function; M a For gas mass; M a0 Let be the mass of the gas at time t0;

[0173] When the parameters at time t0 are known, equation (31) only applies to the pressure ratio p. r It is an unknown function. The method is to solve it, and then calculate Q at time t using equations (26), (28), (14), (10), (8), and (9). s , H s H P Q T Q. However, when calculating H s ≥Z at If the air valve is completely closed, then let H... s =Z at Q s =0、 M a =0, and then H is determined by equations (10), (8), and (9). P QT Q.

[0174] Phase 2: Air intake via air valve, H s ≡Z ct and Z ct ≤H c ≤Z mt

[0175] Of the gas entering the pressure tank, some may be carried by the water flow into the water pipe, while the rest will rise to the top of the pressure tank, raising the water level H. c Descending to Z ct The ratio of gas carried into the water pipe to gas rising above the water surface in the pressure tank is related to the flow velocity in the pressure tank, V = Q. s / A c And the height of the pressure tank h = Z ct -Z is relevant. The larger V and the smaller h are, the more gas enters the water supply pipe, but the exact amount is currently poorly understood. To simplify the problem, we assume that the air entering the pressure tank will not be carried away by the water flow, but will instead rise above the water surface in the pressure tank.

[0176] In command:

[0177]

[0178] In the formula: g is the acceleration due to gravity.

[0179] Then p can be calculated and determined using the Stage 1 mathematical model. r Q s , H P Q T Q, until the following formula:

[0180]

[0181] Up to the establishment date, i.e., the elevation Z of the pressure tank. ct The space above is completely occupied by gas. (Right side of equation (33)) and These refer to the gas volume of the air valve-maintenance valve-connecting pipe and the gas volume of the upper part of the pressure tank.

[0182] In stage 2, the absolute gas pressure at the top of the pressure tank is:

[0183] p c / γ=pγ-(H c -Z ct (34)

[0184] The above formula shows that during stage 2, the pressure tank pressure p c The pressure is less than the air valve pressure p, which means that the relative air pressure at the top of the pressure tank is less than atmospheric pressure.

[0185] However, if p is in the calculation process r If ≥1, meaning the air valve starts to release air, then the hydraulic transient skips stage 3 and directly enters stage 4.

[0186] Phase 3: Air valve intake and exhaust, H s =H c ≤Z ct

[0187] In stage 3, under isentropic gas flow conditions, the pressure in the pressure tank is the same as the pressure in the air valve-maintenance valve-connecting pipe, i.e., p c =p, when:

[0188]

[0189] The hydraulic transients in stage 3 can still be calculated using the mathematical model from stage 1. The second term on the right-hand side of the pressure tank impedance coefficient C5. For water level H s Or H c The additional impedance coefficient generated by the change within the pressure vessel.

[0190] Stage 4, gas is present in the upper part of the pressure tank, H s >Z ct

[0191] I. Relationship between gas pressure ratio and volume:

[0192] The relationship between the gas pressure ratio and gas volume in a pressure vessel is:

[0193]

[0194] In the formula: C is the coefficient relating the gas pressure ratio and the gas volume in the pressure tank. p is a constant; cr1 The gas volume at the top of the pressure tank is The pressure ratio at the end of stage 3 is p. r0 Then p cr1 =p r0 .

[0195] The relationship between the gas pressure ratio and gas volume in the air valve-maintenance valve-connecting pipe is:

[0196]

[0197] Let the mass of the gas at the end of stage 3 be M. a30 The gas masses at the top of the pressure tank and the air valve-inspection valve-connecting pipe are respectively M ac0 and M a0 Then the following relationship exists:

[0198]

[0199] Solving for the given information yields:

[0200]

[0201] Where: M ac0 The mass of gas in the pressure tank at the initial time t0 of stage 4 is given in kg; M a0 The mass of gas in the air valve-maintenance valve-connecting pipe at the initial time t0 of stage 4 is expressed in kg.

[0202] II. Relationship between flow rate and gas pressure ratio in pressure tanks and air valves:

[0203] For the air valve-maintenance valve-connecting pipe, substituting equation (6) into equation (18) yields Q. s Q c With p r Linear algebraic equations,

[0204] a 11 Q s +a 12 Q c =d1-p r H c ≥Z ct H s ≥Z ct (39)

[0205] In the formula: a 11 a 12 d1 is the transition coefficient:

[0206] a 11 =(1 / C²+C) 41 +2C5|Q s0 |+2C 5a |Q s0 -Q c0 |) / H a a 12 =-(C 41 +2C 5a |Q s0 -Q c0 |) / H a d1=(C1 / C2+H a -C 31 +C5|Q s0 |Q s0 +C 5a |Q s0 -Q c0 |(Q s0 -Q c0 )) / H a (40)

[0207] For a pressure vessel, Q can be obtained from equation (25). s Q c With p cr Linear algebraic equations,

[0208] a 21 Q s +a 22 Q c =d2-p cr (41)

[0209] In the formula: a 21 a 22 d2 is the transition coefficient:

[0210] a 21 =(1 / C2+2C5|Q) s0 |) / H a a 22 =C 42 / H a d2=(C1 / C2+H a -C 32 +C5|Q s0 |Q s0 ) / H a (42)

[0211] In the formula: C 32 C 42 Transition coefficient;

[0212] When H c <Z ct H s >Z ct When the inertia of the water flow in the pressure tank and the head loss along the flow path are neglected, we can obtain:

[0213] H a p cr +Z ct =H a p r +H s +C 5a |Q a |Q a H c <Z ct H s >Z ct (43)

[0214] Substituting equations (6), (14), and (41) into equation (43) yields:

[0215] H a (d2-a 21 Q s -a 22Q c )+Z ct =H a p r +C 31 +(C 41 +2C 5a |Q s0 -Q c0 |)(Q s -Q c )-C 5a |Q s0 -Q c0 |(Q s0 -Q c0 )

[0216] Summarized as follows:

[0217] a 11 Q s +a 12 Q c =d1-p r H c <Z ct H s >Z ct (44)

[0218] In the formula:

[0219] a 11 =a 21 +(C 41 +2C 5a |Q s0 -Q c0 |) / H a a 12 =a 22 -(C 41 +2C 5a |Q s0 -Q c0 |) / H a d1=d2+(Z ct -C 31 +C 5a |Q s0 -Q c0 |(Q s0 -Q c0 )) / H a (45)

[0220] Solving the system of linear equations (39)(41) or (41)(44) simultaneously using Cramer's rule yields:

[0221]

[0222]

[0223] In the formula: Δ is an intermediate variable; For Q s The increment; For Q c The increment;

[0224]

[0225] three, With pressure ratio p r p cr Relationship:

[0226] Substitute equations (46) and (47) into equation (15) to eliminate Q. s and Q c We can obtain:

[0227]

[0228] In the formula: d3 is the transition coefficient.

[0229]

[0230] Substituting equation (47) into equation (22) to eliminate Q c We can obtain:

[0231]

[0232] In the formula: d4, a 41 a 42 Transition coefficient:

[0233]

[0234] IV. Pressure ratio p r p cr Numerical solution:

[0235] Substituting equations (49) and (51) into equations (37) and (36) respectively, we get:

[0236]

[0237]

[0238] In the formula: F1 is; F2 is;

[0239] At time t, the depressurization ratio p r and p rc Apart from the unknown quantity, all other parameters are known quantities.

[0240] Using the Newton-Raphson method, the nonlinear equation system of equation (53) can be linearized.

[0241] a 51 Δp r +a 52 Δp cr =d5 (54a)

[0242] a 61 Δp r +a 62 Δp cr =d6 (54b)

[0243] In the formula: Δp r For p r The increment; Δp cr For p rc The increments; d5, d6, a 51 a 52 a 61 a 62 Transition coefficient:

[0244]

[0245]

[0246]

[0247]

[0248] In the formula: Let δ be the gas mass flow rate at time t0; δ > 0 represents p. r For a small increment, we can take δ = 10. -7 .

[0249] Solving the system of linear equations (54) simultaneously yields:

[0250]

[0251]

[0252] In the formula: D, Transition coefficient:

[0253]

[0254] When Δp is calculated from equation (54) r and Δp cr season:

[0255]

[0256] p cr =p cr +Δp cr (55e)

[0257] In the formula: 0 < σ ≤ 1 is the convergence factor. When σ = 1, it is equivalent to calculating p in each iteration. r The change is limited to a range of 1, meaning the change in relative water pressure at the bottom of the air valve is limited to a range of 10m. Furthermore, if p occurs during the iterative calculation... r <0 will lead to If there are no real solutions, then we can let p r =0.1 (equivalent to a negative pressure head of -9m, the pressure required for liquid vaporization); similarly, if p appears during the iterative calculation... cr <0, p can be set cr =0.1, so that iterative calculations can continue to solve the problem.

[0258] The following analysis examines two special cases in phase 4:

[0259] Case 1: When calculating water level H c <Z ct and H s ≤Z ct When, then the pressure ratio p cr =p r From equation (36) + (37), we get:

[0260]

[0261] In the formula M ac0 It is determined by equation (38).

[0262] Assume that at time t0 in stage 4, condition H c <Z ct and H s ≤Z ct Established, at this time M a =M a0 It only needs to be taken from the computer program M a0 =M ac0 +M a0 And let H s0 =H c0 Q a0 =Q s0 Then equation (56) can be rewritten as:

[0263]

[0264] (28) Substituting into the above equation, we get:

[0265]

[0266] Then, the hydraulic transient calculation mathematical model of stage 3 is used to solve for the pressure ratio p at time t. r Q s , H s H P Q T Q.

[0267] Case 2: When the calculated H s ≥Z at If the air valve is completely closed, then let H... s =Z at Q c =Q s , M a =0, then the pressure tank parameter p can be determined using the following method. rc Q s =Q c V c H P Q T Q.

[0268] Due to Q c =Q s Equation (25) can be rewritten as:

[0269] Q s =C 10 -C 11 p cr (58)

[0270] In the formula: C 10 C 11 Transition coefficient;

[0271] C 10 =(C1 / C2+H a -C 32 +C5|Q s0 |Q s0 ) / (1 / C2+C 42 +2C5|Q s0 ), C 11 =H a / (1 / C2+C 42 +2C5|Q s0 |)(59)

[0272] Substituting equation (58a) into equation (22), we get:

[0273]

[0274] In the formula: C 12 C 13 Transition coefficient;

[0275]

[0276] Substituting equation (60) into equation (36), we get:

[0277]

[0278] p can be calculated from equation (62) using the conventional Newton-Raphson method. rc The solution is obtained, and then Q is calculated using equations (58), (60), (21), (10), (8), and (9). s , H c H P Q T Q.

[0279] special case:

[0280] Special Case 1: Gas is present in the initial pressure tank. In this case, the hydraulic transient of the air-valve-replenished pressure tank will begin at stage 4 and then vary between stages 2 and 4.

[0281] Special Case 2: The influence of the height of the air valve-maintenance valve-connecting pipe is not considered, i.e., the elevation Z is taken. at =Z ct In this case, the hydraulic transient process only exists in stages 2 to 4, and the hydraulic transient in stage 4 can be calculated using the model of stage 3. At this time, the water level H in the pressure tank... s =H c >Z ct M a =M ac0 It is a constant (the air valve is in the closed state).

[0282] Special Case 3: The air valve is a vacuum breaker valve and the pressure tank elevation Z mt =Z ct ,Right now Figure 1 Air valve-assisted pressure tank converted into Figure 2 Vacuum-ruptured valve-type pressure tank. In this case, the hydraulic transient process only exists in stages 1 and 3.

[0283] Special case 4: Figure 1 The vacuum breaker valve is replaced by an air valve, specifically an air valve pressure regulating chamber, which allows for both air intake and exhaust. In this case, the hydraulic transient process can also be calculated using the Stage 1 and Stage 3 procedures.

[0284] Engineering example:

[0285] Taking a pumping station pressurized water transmission project as an example, the water hammer protection effects of installing air valves, air valve pressure regulating chambers, and vacuum breaker valve-type pressure tanks are calculated and compared.

[0286] Calculation conditions: The air valve inlet and outlet diameters are 0.1m, the vacuum breaker valve inlet diameter is 0.1m, and the air valve locations are listed in Table 1. Initial conditions: 6 pumps are running at rated speed, and the butterfly valves are fully open. The two-stage closing rules of the hydraulic butterfly valve are: 0-2s, y = 1.0-0.1; 2-20s, y = 0.1-0.0.

[0287] Table 1. Location and corresponding elevation of air valves

[0288]

[0289]

[0290] Table 2 lists the characteristic parameters for hydraulic transient calculations under the condition of simultaneous power outage of 6 pumps and linear closure of two sections of the hydraulic control butterfly valve, where: h, D s d w These are the height, inner diameter, and impedance orifice diameter of the air valve-maintenance valve-connecting pipe, air valve pressure regulating chamber, and vacuum breaker valve replenishment pressure tank, respectively; n max This represents the relative value of the unit's maximum reverse speed; "-" indicates reverse rotation. H max H represents the maximum water pressure in the water supply pipe. min The minimum water pressure in the water supply pipe is indicated by "-", which represents negative pressure. Serial numbers 1 and 2 are the results with only air valves installed. Serial numbers 3 and 4 are the calculation results with air valves replaced by air valve pressure regulating chambers. Serial number 5 is the calculation result with air valves replaced by vacuum breaker valve air-replenishing pressure tanks.

[0291] It should be noted that in the calculation of the hydraulic transients of the air valve pressure regulating chamber and the vacuum breaker valve air-replenishing pressure tank, the influence of the corresponding air valve-maintenance valve-connecting pipe height was not considered. In Table 2, h and D... s These refer to the height and diameter of the pressure regulating chamber or pressure tank, respectively.

[0292] Figure 3 and Figure 4 The hydraulic transient curves of a typical operating condition of a pressure tank with an air valve regulating chamber and a vacuum breaker valve installed on the water pump outlet pipe are shown, including the transient curves of the water level in the regulating chamber and the pressure tank over time.

[0293] Table 2. Overview of Characteristic Parameters of Hydraulic Transition Process

[0294]

[0295] Based on Table 2, the following conclusions can be drawn:

[0296] 1) When the outlet pipe is only equipped with an air valve, the calculated H is as follows: air valve - maintenance valve - connecting pipe height h = 1.0m. min Much smaller than when h = 0.0m, at some locations H minReaching the water vaporization pressure (-9.0m) means that reducing the height of the connecting pipe helps to reduce the risk of liquid vaporization;

[0297] 2) When replacing the air valve on the outlet pipe with an air valve pressure regulating chamber, as the diameter D of the pressure regulating chamber... s and impedance aperture d w When it increases to a certain extent, not only H max Decrease, and H min Also increases, for example, when D is made s and d w When H is increased from 0.1m to 0.4m and 0.2m respectively, then max The diameter decreased from 486.7m to 476.0m, while H min Increased from -9.0m to -3.5m;

[0298] 3) When the air valve pressure regulating chamber is replaced with a vacuum breaker valve air-replenishing pressure tank, under the same conditions of height, diameter and impedance orifice diameter, the maximum water pressure of the vacuum breaker valve air-replenishing pressure tank is significantly lower than that of the air valve pressure regulating chamber, with a difference of up to 27m; while the difference in minimum water pressure between the two is small, about 0.1m.

[0299] 4) The air valve, air valve pressure regulating chamber, vacuum breaker valve, and air replenishment pressure tank have minimal impact on the unit's maximum reverse speed.

[0300] from Figure 3 and 4 It can be seen that when D s =0.4m, d w When the water level H is 0.2m, the water level H in the air valve pressure regulating chamber and pressure tank at station 2071m is... s The difference in elevation between the top of the air valve and the top (H) s -Z at ) min >-0.8m indicates the lowest water level H in the surge tank. smin Above the top elevation of the water pipe, gas will not enter the water pipe, which is very beneficial to the safety of water transmission.

[0301] This embodiment proposes a design method and a numerical simulation mathematical model for hydraulic transients for a water-hammer-replenished pressure tank device. Taking an actual pumping station pressurized water transmission project as an example, the water-hammer protection effects of setting air valves, air valve pressure regulating chambers, and vacuum breaker valve-replenished pressure tanks are calculated and compared. The results show that: 1) Reducing the height of the air valve connecting pipe helps reduce the risk of liquid vaporization; 2) When replacing the air valve with an air valve pressure regulating chamber, not only can the maximum water pressure be reduced, but the minimum water pressure can also be increased; 3) When the air valve pressure regulating chamber is replaced with a vacuum breaker valve-replenished pressure tank, under the conditions that the orifice diameter, flow coefficient, and structural parameters of the pressure regulating chamber and pressure tank are the same, the vacuum breaker valve-replenished pressure tank has the water-hammer protection effect; 4) Reasonably selecting the diameter and height of the water-hammer-replenished pressure tank can prevent gas from entering the water transmission pipe, which is very beneficial to the safety of water transmission.

[0302] Finally, it should be noted that the above is only used to illustrate the technical solution of the present invention and not to limit it. Although the present invention has been described in detail with reference to the preferred arrangement, those skilled in the art should understand that modifications or equivalent substitutions can be made to the technical solution of the present invention (such as the computing platform used in the simulation, the application of various formulas, the order of steps, etc.) without departing from the spirit and scope of the technical solution of the present invention.

Claims

1. A hydraulic transient simulation method for a water-hammer-assisted air-filled pressure tank, wherein the simulation method is based on a pump station pressurized water supply system comprising: a vacuum-rupture valve-assisted air-filled pressure tank or an air valve-assisted air-filled pressure tank installed on a water supply pipeline, characterized in that the method comprises the following processes: The process of constructing a mathematical model: assumed: 1) The gas is an ideal gas and flows in and out of the air valve with isentropic properties, that is, the energy loss along the flow path and the heat exchange between the flow channel wall and the water body are ignored. 2) Air entering the water supply pipe remains near the inspection hole where it can be discharged, so that the hydraulic transients of the water supply pipe can be solved using the conventional water hammer characteristic line method; 3) The air entering the pressure tank will not be carried away by the water flow, but will rise above the water surface in the pressure tank; I. The mathematical model for the air intake and exhaust of the air valve is as follows: 1.1 Air enters at subsonic speeds: In the formula: C is the gas mass flow rate; in A is the air intake flow coefficient of the air valve; in p is the flow area of ​​the air inlet of the air valve; a The absolute pressure of the atmosphere; k is the polytropic index; R is the gas constant; T a p represents the absolute temperature of the air outside the valve. r The gas pressure ratio p inside the valve r =p / p a p is the absolute pressure of the gas inside the valve; rc,in =(2 / (k+1)) k / (k-1) The valve inlet critical pressure ratio; 1.2 Air intake at critical speed of sound: 1.3 Air is exhausted at subsonic speeds: In the formula: C out A is the air valve exhaust flow coefficient; out p is the flow area of ​​the air valve exhaust port; rc,out =1 / p rc,in This is the exhaust critical pressure ratio; 1.4 Air is exhausted at the critical speed of sound: In the formula: ρ a Atmospheric density, kg / m³ 3 ; The speed of sound is in m / s; 1.5 Gas law under isentropic conditions: In the formula: M represents the gas volume; a For gas mass; II. Mathematical Model of Hydraulic Transient in Air Valve-Type Pressure Tank: Q s =C1-C2H P In the formula: Q s H represents the flow rate through the impedance orifice at the bottom of the pressure tank. P The pressure head at the top of the water supply pipe; C1 and C2 are transition coefficients: In the formula: C P C M B P B M The coefficients are those of the characteristic line method. 2.1 Functional relationship between air valve - maintenance valve - connecting pipe - pressure tank water level, gas volume, flow rate, and gas pressure: H a p r =C1 / C2-Q s / C2+H a -C 31 -C 41 Q a -2C5|Q s0 |Q s +C5|Q s0 |Q s0 -2C 5a |Q a0 |Q a +C 5a |Q a0 |Q a0 In the formula: H a atmospheric pressure head H a =p a / γ;p r Q represents the gas pressure ratio inside the valve. s Q is the flow rate through the impedance orifice at the bottom of the pressure tank; s0 Q is the flow rate at the pressure vessel bottom impedance orifice at time t0; a The flow rate within the air valve-maintenance valve-connecting pipe; Q a0 C is the flow rate of the air valve or connecting pipe at time t0; C5 is the impedance coefficient of the pressure tank impedance orifice; C 5a C is the impedance coefficient at the outlet of the connecting pipe. 31 C 41 , where is the transition coefficient: In the formula: H s0 The water level in the air valve-maintenance valve-connecting pipe at time t0; Δt is the time step; A s For the corresponding water level H s The cross-sectional area of ​​the flow path; A a For the cross-sectional area of ​​the air valve-inspection valve-connecting pipe; A c Z represents the cross-sectional area of ​​the pressure tank; Z represents the water level in the water supply pipe; Z mt This is the highest water level inside the pressure tank; 2.2 Functional relationship between pressure tank water level, gas volume, gas pressure and flow rate: H a p cr =C1 / C2-Q s / C2+H a -C 32 -C 42 Q c -2C5|Q s0 |Q s +C5|Q s0 |Q s0 In the formula: C 32 C 42 Transition coefficient; In the formula: H c0 Q represents the water level in the pressure tank at time t0; c0 The elevation Z of the pressure tank at time t0 ct The flow rate at the top; The process of hydraulic transient analysis: Hydraulic transients are divided into four stages: Phase 1: There is no gas at the top of the pressure tank and the water level inside the pressure tank is at height H. c ≡Z mt Keeping unchanged, air valve intake and exhaust, air valve-inspection valve-water level H in connecting pipe s At elevation Z ct and Z at Changes between; H c H represents the water level inside the pressure tank. s For the water level in the air valve-maintenance valve-connecting pipe; Z at Z represents the air valve inlet elevation. ct Elevation of the connecting pipe; Z mt This is the highest water level inside the pressure tank; Phase 2, water level H s =Z ct The air valve remains unchanged, and air rises above the water surface in the pressure tank, causing the water level H to rise. c From elevation Z mt Gradually decrease to Z ct ; Phase 3, air valve intake and exhaust, water level H c =H s <Z ct ; Phase 4, once the water level H c =H s ≥Z ct Then the gas is divided into two by the water body. One part is sealed in the upper part of the pressure tank. The gas volume will change due to compression and expansion as the water pressure in the water pipe rises and falls, which is equivalent to an air cushion. The other part of the gas is discharged from the air valve as the water pressure rises. When the water pressure drops, the air valve will re-intake air, repeating the similar hydraulic transient process of stages 2 to 4. The process of solving a system of nonlinear equations: [The process involves] breaking down the system of nonlinear equations... Linearization is as follows: a 51 Δp r +a 52 Δp cr =d5 a 61 Δp r +a 62 Δp cr = d6 in: F1 and F2 are functions; d3, d4, d5, d6, a 31 a 32 a 41 a 42 a 51 a 52 a 61 a 62 M is the transition coefficient; a0 The mass of gas in the air valve-maintenance valve-connecting pipe at the initial time t0 of stage 4; p is the gas mass flow rate at time t0; C is the constant relating the gas pressure ratio and gas volume in the pressure tank; cr Δp represents the gas pressure ratio inside the pressure tank. r For p r The increment; Δp cr For p cr The increment; Where: δ is p r For a small increment, δ > 0; Solving the system of linear equations yields: In the formula: D, Transition coefficient; D=a 51 a 62 -a 52 a 61 ; Δp was calculated r and Δp cr season: p cr =p cr +Δp cr In the formula: σ is the convergence factor, 0<σ≤1.

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