Hydraulic transient calculation method considering air valve gas temperature variation and installation height

Abstract

The present application relates to a kind of hydraulic transient calculation method considering air valve gas temperature variation and installation height, comprising: the functional relationship of mass flow rate of inflow and outflow air valve and the pressure in valve;Gas pressure, volume and the functional relationship of water pressure of water conveyance pipeline;Iterative calculation.The present application proposes new air valve air intake and exhaust basic equation, air valve, and supporting repair valve and connecting pipe are equivalent with a pressure regulating chamber with automatic air intake and exhaust function, new air valve hydraulic transient mathematical model and its solution algorithm are proposed.Compared with existing air valve hydraulic transient mathematical model, under the condition of not considering air valve height, the maximum water pressure calculated by new model is greater, and the negative pressure is also lower, which means that there is a safety risk in using existing model calculation results as design basis.After considering the height of air valve, the maximum water pressure of water conveyance pipeline is reduced, but the negative pressure is more serious, because the water body in the equivalent pressure regulating chamber has a delay effect on air intake and exhaust.

Description

TECHNICAL FIELD

[0001] The present application relates to a hydraulic transient calculation method considering air valve gas temperature change and installation height, which is a design calculation method of water conservancy equipment. BACKGROUND

[0002] The air valve is an intelligent safety device commonly used in water pipeline systems to prevent water hammer damage. The air valve can not only be used for automatic air intake and exhaust during pipeline water filling and emptying to prevent water hammer damage, but also be used for automatic air intake and exhaust during pump station accident power failure hydraulic transient process to prevent liquid vaporization and liquid column closure water hammer damage. In order to meet the requirements of both, the layout position, type and aperture of the air valve generally need to be determined through hydraulic transient calculation and analysis. At present, the mathematical model and algorithm of Wylie and Streeter, international famous transient flow experts, are generally used to solve the air valve hydraulic transient. The basic assumptions for the mathematical model to be established are:

[0003] 1) Air flows into and out of the air valve isentropically, that is, the energy loss of the gas in the air valve and the heat exchange between the water body and the valve body are ignored;

[0004] 2) The change of the gas in the pipe follows the isothermal law, and the temperature of the gas is equal to the temperature of the liquid, considering that the mass of the air in the pipe is usually very small;

[0005] 3) The air entering the pipeline stays near the valve where it can be discharged, and the relationship between the air valve flow and the water pressure and flow of the water conveying pipe can be determined by using the conventional characteristic line method;

[0006] 4) The height of the liquid surface is basically unchanged, and the volume of the air is small compared with the volume of the liquid in the pipe section, that is, the influence of the height of the air valve and its supporting maintenance valve and connecting pipe is ignored.

[0007] Although the air valve hydraulic transient mathematical model of Wylie and Streeter is widely used in engineering calculation, the basic assumptions 1) and 2) of the model are contradictory. In fact, the water hammer process in the water conveying pipeline engineering belongs to a transient process, and the duration is very short. The gas temperature T cannot be equal to the water temperature, but changes with the change of the air pressure p. In addition, the basic assumption 4) does not consider the height of the air valve and its supporting maintenance valve and connecting pipe, which may cause the calculation results to deviate from the actual situation and cause serious operation safety problems. The deviation caused by the assumption that the gas temperature T is constant and the height of the air valve is ignored in the current air valve hydraulic transient mathematical model, as well as the height of the air valve and its supporting maintenance valve and connecting pipe in the actual engineering application, is a problem that should be solved. SUMMARY

[0008] In order to overcome the problems of the prior art, the present application provides a hydraulic transient calculation method considering air valve gas temperature change and installation height. The method proposes new air valve inlet and exhaust basic equations, and equivalent of the air valve, a matching repair valve and a connecting pipe with an automatic inlet and exhaust function pressure regulating chamber, and proposes a new air valve hydraulic transient mathematical model and its solving algorithm.

[0009] The purpose of the present application is achieved by a hydraulic transient calculation method considering air valve gas temperature change and installation height, characterized in that the steps of the method are as follows:

[0010] Step 1, establish the functional relationship between the mass flow rate flowing into and out of the air valve and the pressure in the valve:

[0011] Air valve inlet equation:

[0012]

[0013]

[0014] p rc,in =(2 / (k+1)) k / (k-1)

[0015] In the formula: is the gas mass flow rate; A in is the opening area of the valve, m 2 ; C in is the flow coefficient of the valve inlet; p a is the absolute pressure of the atmosphere; k is the polytropic index; R is the gas constant; T a is the absolute temperature of the atmosphere; p r is the pressure ratio, indicating the ratio of the air pressure in the air valve to the atmospheric pressure, p r =p / p a ; p is the absolute pressure of the gas in the valve; p rc,in is the gas pressure ratio of the valve inlet at the critical sound speed;

[0016] Air valve exhaust equation:

[0017]

[0018]

[0019] p rc,out =1 / p rc,in =(2 / (k+1)) -k / (k-1)

[0020] In the formula: A out is the valve orifice exhaust flow area; C out is the valve orifice exhaust flow coefficient; p rc,outThe pressure ratio at which the valve exhausts gas at the critical speed of sound; the gas state equation under isentropic conditions:

[0021]

[0022] or:

[0023]

[0024] In the formula: M is the gas volume; a The mass of the gas is represented by the subscript "0", which indicates time t0.

[0025] Step 2, the functional relationship between gas pressure, volume, and water pressure in the water supply pipeline:

[0026] Calculate and solve p r The interlocking equations:

[0027] △p r =-F / F p

[0028]

[0029]

[0030]

[0031] In the formula: △p r This is the pressure ratio increment; C8, C9, M a0 Let be a known quantity at time t0; Δt = t - t0 is the time step; σ is the convergence factor, 0 < σ ≤ 1 is the convergence factor;

[0032] Due to the derivative of the subsonic intake and exhaust mass flow rate of the air valve That is, the mass flow rate of the air valve relative pressure ratio p r The derivative of p r =1 is discontinuous, and the following numerical difference method was used:

[0033]

[0034] In the formula: δ>0 represents p r A tiny increment, taking δ = 10 -7 ;

[0035] Step 3, the calculation of the hydraulic transient of the air valve at time t is as follows:

[0036] 1) Take p at time t0 r Value, i.e.: p r =p r0 ;

[0037] 2) Calculation F, F p and △p r ;

[0038] 3) Discrimination |△p r |≤10 -6 Does it hold true? If it does, then p r If it is a solution, then determine |△p. r If |≤σ holds true, then use p. r +△p r Replace p r If not true, then use p. r +σ△p r / |△p r |Replace p r Repeat steps 2) and 3) until |△p r |≤10 -6 .

[0039] The advantages and beneficial effects of this invention are as follows: This invention proposes a new basic equation for air valve intake and exhaust, equating the air valve, its associated maintenance valve, and connecting pipes with a pressure regulating chamber that has an automatic intake and exhaust function. A new hydraulic transient mathematical model for air valves and its solution algorithm are also proposed. Compared with existing hydraulic transient mathematical models for air valves, the new model calculates a higher maximum water pressure and a lower negative pressure without considering the air valve height. However, considering the air valve height, the calculated maximum water pressure in the water supply pipe decreases, but the negative pressure becomes more severe because the water in the equivalent pressure regulating chamber hinders and delays the intake and exhaust. This means that using the calculation results of existing models as the design basis carries safety risks. Furthermore, using the calculation results obtained by treating the air valve gas as adiabatic flow as the design basis is reasonable and leans towards safety, because the calculated maximum water pressure is higher and the minimum water pressure is lower under these conditions. Attached Figure Description

[0040] The present invention will be further described below with reference to the accompanying drawings and embodiments.

[0041] Figure 1 This is a schematic diagram of the air valve's intake and exhaust functions;

[0042] Figure 2 It is an air valve Relationship curve;

[0043] Figure 3 This is a schematic diagram of the working principle of the equivalent air valve pressure regulating chamber;

[0044] Figure 4 This does not consider the air valve height or the hydraulic transient curve of the air valve.

[0045] Figure 5The hydraulic transient curve of the air valve takes into account the influence of the air valve height. Detailed Implementation

[0046] Example:

[0047] The boundary conditions for air valves are quite complex, involving gas dynamics and hydraulic transients. The following section will first establish the boundary conditions for the inflow and outflow of air valves. Mass flow versus pressure in valve The functional relationship, and then establish Gas pressure, volume, and waterway pressure Air valve hydraulic transients The functional relationship was finally studied. Numerical calculations of Figure 1 method.

[0048] 1. Basic equations for air valve intake and exhaust:

[0049] To simplify the problem, let's first assume:

[0050] 1) The flow is quasi-steady state, the gas is an ideal (perfect) gas and is entropy, and the air valve flows in and out, that is, the energy loss along the flow path and the heat exchange between the flow channel wall and the water body are ignored.

[0051] 2) The air entering the pipeline remains near the valve from which it can be discharged, and the elevation difference is small, so the effect of gas gravity can be ignored. Under the above conditions, the effect of local energy loss between the air valve and the lower water supply pipe is handled by the equivalent resistance coefficient of the valve orifice.

[0052] The equation of state for an ideal gas is:

[0053] p=ρRT (1)

[0054] Right now:

[0055] p / p a =ρT / (ρ a T a (2)

[0056] In the formula: p is the absolute pressure of the gas inside the valve, Pa; ρ is the density of the gas inside the valve, kg / m³. 3 R is the gas constant, typically taken as 287 J / kg·K; T is the absolute temperature of the gas inside the valve, in K; p a ρ is the absolute pressure of the atmosphere, in Pa; a The density of the atmosphere, kg / m³ 3 ;T a Let K be the absolute temperature of the atmosphere.

[0057] The condition for isentropic gas flow is:

[0058] pρ -k =const (3)

[0059] Right now:

[0060] (ρ / ρ a ) k =p / p a (4)

[0061] In the formula: k is the polytropic index, and for an adiabatic process, k = 1.4.

[0062] According to equations (2) and (4), the functional relationship between the isentropic flow density, absolute temperature, and pressure of an ideal gas can be obtained.

[0063]

[0064] In the formula: p r =p / p a The pressure ratio, p, represents the ratio of the air pressure inside the air valve to the atmospheric pressure. The above equation shows that, under isentropic conditions, the air temperature T increases with the pressure ratio p. r It increases with the increase of p. r It decreases as it decreases.

[0065] According to gas dynamics, the energy of isentropic flow between any two cross sections is conserved; the influence of potential energy is negligible; and the sum of internal energy and kinetic energy is a constant. The energy conservation equation is:

[0066]

[0067] In the formula: the first term is the internal energy of a unit mass of gas; the second term is the kinetic energy of a unit mass of gas; and u is the average flow velocity of any cross section, in m / s.

[0068] like Figure 2 As shown, when gas enters the water pipe through the air valve, the flow velocity u in the water pipe can be ignored because the cross-sectional area of ​​the water pipe is often two orders of magnitude larger than that of the air valve. Similarly, since the space outside the valve is infinite, the flow velocity u can also be ignored. When the gas exists only inside the air valve, the kinetic energy term can be attributed to the energy loss at the valve inlet orifice.

[0069] For air intake through an air valve, the energy equations for the external atmosphere and any cross-section inside the valve can be written as follows:

[0070]

[0071] Summarized as follows:

[0072]

[0073] In the formula: This refers to the total local resistance coefficient of the air valve intake, including local resistances such as inlet / outlet and cross-sectional changes; u in The airflow velocity at the valve orifice.

[0074] When gas flows into the air valve at a critical velocity, that is, at the speed of sound If air flows into the air valve, then... Substituting into equation (7), we can obtain the critical pressure ratio:

[0075] p rc,in =(2 / (k+1)) k / (k-1) (8)

[0076] In the formula: p rc,in The critical gas pressure ratio at which the valve allows air to enter at the critical speed of sound is called the critical intake pressure ratio, typically ranging from 0.53 to 0.57. For example, for adiabatic flow, k = 1.4, then p rc,in =0.53; when k=1.2, then p rc,in =0.56. Aerodynamic studies show that sound is an infinitesimally small pressure wave. Therefore, as long as the flow at the air valve orifice (throat) reaches the speed of sound, the pressure disturbance cannot pass through the orifice, even if the pressure ratio further decreases, i.e., p r <p rc,in The flow velocity u flowing into the orifice in It will not follow p r The change occurs due to the decrease in [something].

[0077] Similarly, for air valve exhaust, we can obtain:

[0078]

[0079] In the formula: u is the total local resistance coefficient of the air valve exhaust. out The exhaust flow rate at the air valve orifice.

[0080] When gas is expelled from the air valve at the critical speed of sound, that is, at the speed of sound... When it is released into the atmosphere, it causes Substituting into equation (9), we can obtain the critical pressure ratio:

[0081] p rc,out =1 / p rc,in =(2 / (k+1)) -k / (k-1) (10)

[0082] In the formula: p rc,out The pressure ratio at which the valve exhausts gas at its critical speed of sound is called the exhaust critical pressure ratio, typically ranging from 1.75 to 1.89. For example, for adiabatic flow, k = 1.4, then p rc,out =1.89.

[0083] When both sides of equation (7) are multiplied by (A) in ρ) 2 ,because The mass flow rate of the gas flowing into the air valve at subsonic speed is:

[0084]

[0085] In the formula: Intake mass flow rate, kg / s; A in Let m be the valve's opening area. 2 ; This is the flow coefficient of the valve intake.

[0086] Using p in equation (8) rc,in In the substitution formula (11), p r The mass flow rate of the intake air at the critical speed of sound can be obtained as follows:

[0087]

[0088] When both sides of equation (9) are multiplied by (A) out ρ a ) 2 ,because The mass flow rate of the gas exiting the air valve at subsonic speed is:

[0089]

[0090] In the formula: The exhaust mass flow rate is kg / s; A out The exhaust flow area at the valve orifice is in meters. 2 ; This is the valve orifice exhaust flow coefficient. A negative value indicates that the air valve is venting.

[0091] Using p in equation (10) rc,out In the substitution formula (13), p r The fundamental equation for the critical sonic velocity exhaust of the air valve can be obtained as follows:

[0092]

[0093] Multiplying both sides of equation (1) by the gas volume, we obtain the ideal gas law expressed in terms of gas mass:

[0094] pV = M a RT (15)

[0095] In the formula: V is the gas volume, m 3 M a The mass of the gas is expressed in kg.

[0096] Assuming the isentropic flow of the air valve is adiabatic, i.e., taking the polytropic index k = 1.4, then equations (11) to (15) are the basic equations for air valve intake and exhaust used by Wylie and Streeter. When the term k / (k-1) = 1 in equations (11) and (13), then equations (11) to (15) are the mathematical model of the air valve used by Lee et al. Since the isentropic process gas lies between the isothermal process k = 1 and the adiabatic process k = 1.4, the missing term k / (k-1) in Lee et al.'s model may be a typographical error. This is pointed out because some papers have adopted this model.

[0097] It should be noted that the aforementioned basic equations for air valve intake and exhaust are based on the isentropic flow theory of rocket engine nozzle gases. For rocket engines, the gas inside the nozzle comes from the combustion chamber, and the air temperature T is controlled by the combustion chamber temperature, reaching temperatures of several thousand K. Therefore, T is the primary control variable for rocket engines and can be considered a known constant during stable rocket flight. Unlike rocket engines, in water pipeline engineering, the air valve only operates during the transient water hammer process. The gas comes from the atmosphere, not the water body, and the intake and exhaust duration is very short. Furthermore, T changes instantaneously with changes in air pressure p.

[0098] The existing mathematical model for air valve exhaust completely adopts the assumption of constant air temperature T from the steady-state model of rocket engines, and it is obviously unreasonable to take T equal to water temperature. If T is constant, it means that the gas inside the air valve is isothermal, that is, the polytropic index k≡1, which contradicts the assumption of isentropic flow. If k=1 is substituted into equations (11)(13), the right side of the formula is 0 / 0. In fact, as the diameter of the water conveyance pipe increases, the size of the air valve also increases proportionally, and the air intake during the hydraulic transient process is quite large. Taking the Huinanzhuang Pumping Station of the South-to-North Water Diversion Project in Beijing as an example, the diameter of the water conveyance pipe is 4m, and the normal water conveyance flow of a single pipe is 30m3 / s. Once the pumping station experiences a power outage and a hydraulic transient occurs, hundreds or even thousands of cubic meters of gas will quickly enter the water conveyance pipe through the air valve. Obviously, it is impossible to make the air temperature equal to the water temperature and keep it constant. The following will transform the air temperature T into the pressure ratio p based on the ideal gas law and the isentropic condition. r and atmospheric temperature T a The function.

[0099] To ensure that the unknown quantity T is not included in the air valve exhaust equation, we first use equation (5) to obtain... and p = p a p r Then, substituting these equations into equations (13), (14), and (15) respectively, we obtain the following new air valve exhaust equations:

[0100]

[0101]

[0102] And the equation of state for a gas under isentropic conditions:

[0103]

[0104] Equations (11)(12)(16)(17)(18) are the basic equations for the intake and exhaust of the new air valve under isentropic conditions derived in this embodiment, where the parameter p a T a ,k,R,A in A out C in C out The data is the original data (known), and the only unknown quantity is the pressure ratio p. r =p / p a Gas volume V and gas mass M a These can only be determined by simultaneously solving for the hydraulic transients of the water supply pipeline. The American Water Industry Association standard AWWA M51, "Micro-air Inlet / Outlet Valves, Quick-Air Inlet / Outlet Valves, and Combination Quick-Air Inlet / Outlet Valves," recommends using C. in =C out =0.7.

[0105] To understand the differences in gas mass flow rate calculations between the Wylie and Streeter air valve intake / exhaust model and the new model, Figure 3 A typical example is given. The relationship curve, where: atmospheric temperature is 10℃, i.e., absolute temperature T. a =283K; p a Let p be the atmospheric pressure at sea level; the water temperature in the Wylie and Streeter models is taken as 4°C, i.e., the absolute air temperature T = 277K. Clearly, both models have an air inlet valve, i.e., a pressure ratio p. r <1.0 The values ​​are exactly the same, and then with the air valve exhaust pressure ratio p r The increase of both The increasing difference in values ​​will have a significant impact on the impact water pressure of the liquid column in the water delivery pipe at the end of the exhaust.

[0106] 2. Digital model of hydraulic transients in air valve:

[0107] Currently, none of the mathematical models for calculating hydraulic transients involving air valves consider the elevation difference between the air valve inlet and the water supply pipe. As the diameter of the water supply pipe increases, the size of the air valve also increases proportionally. Considering the need for air valve maintenance under normal water supply conditions, maintenance valves and connecting pipes are often installed between the air valve and the water supply pipeline. Currently, in water supply projects, the nominal orifice diameter D of the air valve is... a ≤0.4m, the maximum sum of the heights of the air valve, maintenance butterfly valve and connecting pipe exceeds 2m.

[0108] Under normal circumstances, the diameter of the connecting pipe, the diameter of the butterfly valve orifice, and the diameter of the air valve orifice D are... a The air valve and its connecting pipe are similar in function, and the net cross-sectional area of ​​most of the air valve is not much different from that of the inlet and outlet ports. Therefore, the functions of the air valve, the maintenance butterfly valve, and the connecting pipe can be equivalent to a pressure regulating chamber with automatic air intake and exhaust functions. Figure 4 As shown, where: Z at Z is the elevation of the equivalent pressure regulating chamber top, i.e., the elevation of the air valve top cover; Z is the elevation of the water supply pipe top, and H is the elevation of the equivalent pressure regulating chamber top. s H represents the water level in the surge tank. p The pressure head of the pressure measuring pipe at the top of the water supply pipe.

[0109] To ensure the generality of the mathematical model established below, it is assumed that there is an impedance orifice between the pressure regulating chamber and the water supply pipe, where the diameter of the equivalent pressure regulating chamber can be approximated by the orifice diameter D of the air valve. a Alternatively, the cross-sectional area of ​​the surge tank can be described as the water level H, as needed. s The function.

[0110] The working principle of the equivalent pressure regulating chamber of the air valve is as follows: Under the condition of hydraulic transient change in the water pipeline, as the water head H in the pressure measuring tube increases... p Lower to the top elevation Z of the pressure regulating chamber at The following is H p <Z at The air valve starts to allow air in, and the water level H... s The gas volume in the pressure regulating chamber decreases. Increase, and the air pressure becomes negative; if H s When Z ≤ Z, air will enter the water supply pipe; then, as H... p As the pressure rises, the gas entering the water pipe is first pressurized and discharged into the pressure regulating chamber, V decreases, and the water level H rises. s As the gas rises, it becomes pressurized, and the air valve begins to release air until H... s =Z at That is, until the air valve is completely closed. If the pressure head H in the pressure measuring tube... p Lower back to the top elevation Z of the pressure regulating chamber at The aforementioned process will then repeat inside the pressure regulating chamber. Because the density of liquid is two orders of magnitude greater than that of gas, the flow into and out of the pressure regulating chamber will significantly slow down the air intake and exhaust of the air valve, thus greatly affecting the changes in water hammer pressure in the water supply pipe.

[0111] For ease of calculation, the gas pressure ratio p is established. r Gas volume V and mass M a Before considering the hydraulic transients of the water supply pipe, we first assume that the volume of gas entering the pipe is very small compared to the volume of liquid in the pipe section. The conventional characteristic line method can be used to solve for the hydraulic transients of the pipe.

[0112] The continuity equation for the bottom node P of the pressure regulating chamber is:

[0113] Q s =Q T -Q (19)

[0114] In the formula: Q s The flow rate in or out of the pressure regulating chamber, m 3 / s;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.

[0115] According to the equation of compatibility of characteristic lines, we can obtain:

[0116] C + :Q T =C P / B P -H P / B P (20)

[0117] C - :Q=-C M / B M +H P / B M (twenty one)

[0118] Substituting equations (20) and (21) into equation (19), we get:

[0119] Q s =C1-C2H P (twenty two)

[0120] In the formula:

[0121]

[0122] At time t, the coefficients C1 and C2 are known quantities.

[0123] Under normal water supply conditions, the air valve is fully closed, Q s ≡0. If there is no gas in the water pipe during a hydraulic transient in the pipeline system. And the pressure head H P =C1 / C2>Z at If the air valve does not work, Q s =0; otherwise, the following method is needed to establish the surge tank water level H. s Gas volume and traffic Q s The functional relationship,

[0124]

[0125]

[0126] In the formula: t is time, in seconds; A s The cross-sectional area of ​​the pressure regulating chamber is m. 2 .

[0127] Integrating equations (24) and (25) and taking a second-order approximation, we get

[0128] H s =C3+C4Q s (26)

[0129]

[0130] In the formula: the subscript "0" represents time t0; Δt = t - t0;

[0131]

[0132] When the inertial force of the water in the pressure regulating chamber, the head loss along the pipe, and the elasticity of the chamber wall are not considered, the gas pressure p (absolute pressure) and the piezometric head H at the top of the water supply pipe are... P Water level H s and traffic Q s The relationship is:

[0133] p / γ=H P +H a -H s -C5|Q s |Q s =H P +H a -H s -2C5|Q s0 |Q s +C5|Q s0 |Q s0 (29)

[0134] In the formula: γ is the specific gravity of water, taken as 9807 kg / m³. 3 H a =p a / γ represents the local atmospheric pressure head:

[0135]

[0136] In the formula: ω is the area of ​​the impedance hole at the bottom of the pressure regulating chamber, in meters. 2 ; Let ω be the local resistance coefficient of the impedance orifice at the bottom of the pressure regulating chamber. When the pressure regulating chamber is simply a combination of an air valve, a maintenance valve, and connecting pipes, then ω = A. s At this point, the impedance orifice at the bottom of the pressure regulating chamber can be considered as a combination of sudden expansion or contraction and a 90° turn. Preliminary calculations can take...

[0137] Substituting equations (22) and (26) into equation (29), we obtain the gas flow rate Q. s With pressure ratio p r Relationship,

[0138] p a p r / γ=(C1 / C2-Q s / C2)+H a -(C3+C4Q s )-(2C5|Q s0 |Q s -C5|Q s0 |Q s0 )

[0139] Summarized as follows:

[0140] Q s =C6-C7p r (31)

[0141] In the formula:

[0142]

[0143] Substituting equation (31) into equation (27) yields the gas volume V and pressure ratio p. r Relationship:

[0144]

[0145] In the formula:

[0146]

[0147] Substituting the gas volume V from equation (33) into the isentropic gas state equation (18), and considering the gas mass M... a Taking the second-order approximation, we get:

[0148]

[0149] Since the polytropic exponent k≠1, the function F is not the pressure ratio p. r The equation of the parabola cannot be solved using the numerical algorithms of Wylie and Streeter.

[0150] Since the function F in equation (35) only contains the pressure ratio p r Since it is an unknown quantity, the following is obtained using the Newton-Rayvosen numerical calculation method:

[0151]

[0152] The following iterative calculations are used to solve p. r The interlocking equations:

[0153] △pr =-F / F p (36a)

[0154] In the formula:

[0155]

[0156]

[0157]

[0158] 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 1 atmosphere. Furthermore, if p occurs during the iterative calculation... r If <0, it 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, which is generally the pressure required for liquid vaporization), so that iterative calculations can be continued to solve the problem.

[0159] Due to the derivative of the subsonic intake and exhaust mass flow rate of the air valve That is, the mass flow rate of the air valve relative pressure ratio p r The derivative of p r =1 is discontinuous. To overcome this difficulty, the following numerical difference method is recommended:

[0160]

[0161] In the formula: δ>0 represents p r For a small increment, we can take δ = 10. -7 .

[0162] When the effect of air valve height is not considered, simply let Z at If Z = , then the above mathematical model for calculating hydraulic transients of the air valve is also applicable.

[0163] The calculation starts from t0 = 0, with the initial conditions being:

[0164]

[0165] When the air valve has completely released the gas:

[0166]

[0167] When V=0, Q s =0, H s =Z at At that time, the boundary condition at the air valve connector is H. P QT General in-section solution for Q:

[0168] QP = QT = Q = (C P -C M ) / (B P +B M ), H P =C P -B p Q T (39)

[0169] Using the above calculation procedure, the calculation will always converge as long as the number of iterations is sufficient. Under normal circumstances, a few iterations are enough to meet the calculation accuracy requirements. However, in some cases, when the exhaust orifice diameter of the air valve is designed to be too small, in order to ensure calculation convergence, it is only necessary to adjust the size of the convergence factor σ. Generally, a smaller σ is chosen, such as σ = 0.05 or even smaller, while increasing the control value of the number of iterations.

[0170] 3. The calculation of the hydraulic transient of the air valve at time t is as follows:

[0171] 1) Take p at time t0 r Value, i.e.: p r =p r0 ;

[0172] 2) Calculation F, F p and △p r ;

[0173] 3) Discrimination |△p r |≤10 -6 Does it hold true? If it does, then p r If it is a solution, then determine |△p. r If |≤σ holds true, then use p. r +△p r Replace p r If not true, then use p. r +σ△p r / |△p r |Replace p r Repeat steps 2) and 3) until |△p r |≤10 -6 .

[0174] Using the calculated p r Other relevant hydraulic transient parameters, such as flow rate, water level, and water pressure, can be calculated.

[0175] The following example, a typical hydraulic transient of an air valve in a water pipe from Wylie and Streeter's monograph, is used to compare the differences between Wylie and Streeter's mathematical model of hydraulic transients in air valves and the new model presented in this paper. The water pipe is 1219.2m long, divided into two sections, with Δx = 609.6m; the pipe diameter D = 0.6096m, corresponding to a cross-sectional area A = 0.292m². 2 Water hammer wave velocity a = 1219.2 m / s; Darcy-Wiesbach friction coefficient f = 0.02; air valve inlet and outlet flow area A in =A out =0.001858m 2 Corresponding to air valve orifice diameter D a =0.049m; the air valve is installed in the middle of the water pipe, and the top elevation of the pipe is Z = 10.36m; the atmospheric absolute pressure head H a =p a / γ=10.36mH2O; Downstream of the water pipeline is a reservoir as a boundary condition, with a water level Z0=9.75m; The forced flow rate Q at the inlet of the water pipeline. P (1)=Q0-△Qsin(OM×t),Q0=0.3398m 3 / s, ΔQ=0.1133m 3 / s, OM = 0.3rad. The air valve flow coefficient C is taken in the calculation. in =C out =0.7.

[0176] Example 1: When the height of the air valve is not considered, including the height of the maintenance valve and connecting pipe, i.e., Z at =H s =Z, and take the atmospheric temperature as 10℃ and the water temperature as 4℃, i.e., T a =273+10=283K and T=273+4=277K, then under the condition that the gas is in adiabatic flow and k=1.4, we get as follows Figure 4 The calculation results for the hydraulic transients of the air valve are shown in the figure. In the figure, the solid and dashed lines represent the calculation results for the hydraulic transients of the new model and the Wylie and Streeter air valve intake / exhaust models, respectively; H = H P -Z represents the pressure head at the bottom of the air valve.

[0177] from Figure 4 It is evident that the air valve was not venting properly. This can generate intense hydraulic jet pressure, with sudden increases and decreases in water pressure. The maximum water pressure H calculated using the new model versus the Wylie and Streeter model is compared between the two models. max and minimum water pressure H min The differences are obvious; the new model H max1 =39.26mH2O and H min1= -2.69mH2O, while the Wylie and Streeter models H max2 =36.04mH2O and H min2 = -1.58mH2O, the relative deviation between the two (H max1 -H max2 ) / H max2 =9% and (H) min1 -H min2 ) / H min2 =70%. This result indicates that the new model calculates a higher maximum water pressure and a lower minimum negative pressure. This means that engineering designs based on the new model's calculations are more reasonable and lean towards safety. Figure 5 The diagram also shows the change in air temperature inside the air valve over time: as the air intake V- increases, the air temperature T increases with the pressure ratio p. r It decreases as p decreases, and then decreases as p decreases. r The temperature increases with the increase of the absolute temperature T. max =293.2K, lowest absolute temperature T min =259.7K, or a maximum temperature of 20.2℃ and a minimum temperature of -13.3℃.

[0178] For further comparison, the new model was also used to calculate the hydraulic transients in the pipeline when the polyvariance index k = 1.2, and the maximum water pressure H. max = 37.97mH2O, while the minimum water pressure H min = -2.60mH2O. Compared with the results for adiabatic flow k=1.4, an important conclusion can be drawn: the maximum water pressure H in the water supply pipe... max It increases with the increase of the polytropic index k, but the minimum water pressure H min The value decreases as k increases, so it is reasonable and safe to use the calculation results assuming that the gas in the air valve is adiabatic as the design basis.

[0179] Example 2: When considering the height of the air valve, taking a typical air valve as an example, the orifice diameter D a The air valve height h is 0.049m. a ≈0.327m. When using a butterfly valve of the same nominal diameter as a maintenance valve, the butterfly valve height L = 0.108m. In this case, if the equivalent pressure regulating chamber height h = Z is taken... at When -Z=0.44m, then H s0 =Z at =Z + 0.44m. Under these conditions, if other conditions are the same as in Example 1, then the new model yields the following result: Figure 5 The calculation results are shown in the figure: the solid line and the dashed line represent the results without considering the air valve height (h=0.0) and with the air valve height considered, respectively.

[0180] observe ​It can be seen that, when considering the influence of the air valve height, H max3 =37.92mH2O and H min3 = -2.80mH2O; and without considering the effect of air valve height, H max1 =39.26mH2O and H min1 = -2.69mH2O; relative deviation (H max1 -H max3 ) / H max3 =3.5% and (H) min1 -H min3 ) / H min3 = -4%. This result shows that even though the air valve height in this example is small, only 0.44m, it has a significant impact on the maximum and minimum water pressure in the water supply pipe. Taking the air valve height into account reduces the calculated maximum water pressure, which is advantageous; however, the minimum water pressure also decreases, leading to a more severe negative pressure in the water supply pipe, which is disadvantageous. The main reason for this phenomenon is that the water in the equivalent pressure regulating chamber of the air valve has a hindering and delaying effect on air intake and exhaust.

[0181] It should be noted that the length of the connecting pipe for air valves varies considerably in actual engineering projects. Different projects and different locations of the air valves have different requirements for the type and size of the connecting pipe, and some requirements may even be nonexistent. Currently, the requirements and application restrictions for air valves are becoming increasingly stringent. If the height and diameter of the connecting pipe are designed appropriately to confine air within the connecting pipe and prevent it from entering the water supply pipe, the safety of water supply can be significantly improved. For example, using the air valve and connecting pipe to form an air valve pressure regulating chamber can greatly enhance the water hammer protection capability of water supply projects.

[0182] Existing hydraulic transient models for air valves suffer from two major problems: contradictory fundamental assumptions and failure to consider the influence of the air valve's installation height. Based on the fundamental theory of isentropic flow aerodynamics in air valves, and considering the variation of gas absolute temperature with gas pressure, a new fundamental equation for air valve inlet and outlet is proposed. Under the condition that the air valve, maintenance butterfly valve, and connecting pipes are equivalently represented by a pressure regulating chamber with automatic inlet and outlet functions, a new mathematical model for the hydraulic transient of air valves is proposed. In the numerical solution process, finite differences are used instead of differentials to solve for the air valve's mass flow rate. relative pressure ratio p r The problem of derivative discontinuity is addressed, and a computer program is developed to solve the hydraulic transients of air valves using the Newton-Rayvosen method. This program includes adjusting the convergence factor σ to ensure the convergence of the numerical iterative calculations. For air valves with small orifice diameters, a smaller σ is typically chosen, such as σ = 0.05 or even smaller, while simultaneously increasing the number of iterations.

[0183] Computational studies using typical examples show that:

[0184] 1) When the air intake V of the air valve increases, the air temperature T increases with the pressure ratio p. rIt decreases as p decreases, and then decreases as p decreases. r The temperature will rise as the temperature increases, with the lowest temperature possibly dropping below 0°C.

[0185] 2) The calculation results using the new air valve hydraulic transient model and the Wylie and Streeter models show significant differences. The maximum relative deviation of water pressure can reach over 9%, while the relative deviation of minimum water pressure can reach over 70%. In other words, neglecting the influence of temperature changes during the air valve exhaust process can lead to safety risks in the calculation results.

[0186] 3) The height h of the air valve and its matching maintenance valve and connecting pipe will have a significant impact on the maximum and minimum water pressure of the water supply pipe, even if h is small;

[0187] 4) Reasonably designing the height and diameter of the connecting pipe to confine air inside the connecting pipe and prevent it from entering the water supply pipe can significantly improve the safety of water supply. For example, using an air valve and connecting pipe to form an air valve pressure regulating chamber can greatly improve the water hammer protection capability of the water supply project.

[0188] 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 preferred arrangements, 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 structure of the air valve involved, the structure of the water pipeline, 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 method for calculating hydraulic transients considering gas temperature changes and installation height in air valves, characterized in that, The steps of the method are as follows: Step 1: Establish the functional relationship between the mass flow rate of the air flowing into and out of the valve and the pressure inside the valve: Air valve intake equation: p rc,in =(2 / (k+1)) k / (k-1) In the formula: A is the gas mass flow rate; in Let m be the valve's opening area. 2 C in p is the flow coefficient of the valve intake air; 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 atmosphere. r The pressure ratio, p, represents the ratio of the air pressure inside the air valve to the atmospheric pressure. r =p / p a p is the absolute pressure of the gas inside the valve; rc,in The gas pressure ratio at which the valve allows air to enter at the critical speed of sound; Air valve exhaust equation: p rc,out =1 / p rc,in =(2 / (k+1)) -k / (k-1) In the formula: A out C is the exhaust flow area at the valve orifice. out p is the valve orifice exhaust flow coefficient. rc,out The pressure ratio at which the valve exhausts gas at the critical speed of sound; Gas law under isentropic conditions: or: In the formula: V is the gas volume; M a The mass of the gas is represented by the subscript "0", which indicates time t0. Step 2, the functional relationship between gas pressure, volume, and water pressure in the water supply pipeline: Calculate and solve p r The interlocking equations: △p r =-F / F p In the formula: △p r This is the pressure ratio increment; C8, C9, M a0 Let be a known quantity at time t0; Δt = t - t0 is the time step; σ is the convergence factor, 0 < σ ≤ 1 is the convergence factor; Due to the derivative of the subsonic intake and exhaust mass flow rate of the air valve That is, the mass flow rate of the air valve relative pressure ratio p r The derivative of p r =1 is discontinuous, and the following numerical difference method was used: In the formula: δ>0 represents p r A tiny increment, taking δ = 10 -7 ; Step 3, the calculation of the hydraulic transient of the air valve at time t is as follows: 1) Take p at time t0 r Value, i.e.: p r =p r0 ; 2) Calculation F, F p and △p r ; 3) Discrimination |△p r |≤10 -6 Does it hold true? If it does, then p r If it is a solution, then determine |△p. r If |≤σ holds true, then use p. r +△p r Replace p r If not true, then use p. r +σ△p r / |△p r |Replace p r Repeat steps 2) and 3) until |△p r |≤10 -6 .

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