4.12. Dynamic air valve with water release

4.12.1. DYNAIRVA (class)

Fig. 4.12.1 Dynamic air valve

Supplier type

type label	description	active
Dynamic air valve	Detailed air valve model with automated level effect. Water is released from the air valve to dampen the surge during the transition from open to closed	No

4.12.1.1. Mathematical model

Fig. 4.12.2 shows the dynamic air valve which is modelled. Air can enter the system when the stem (yellow part) is in the open position (Left figure of Fig. 4.12.2). The stem will open when the pressure difference over the air valve (difference between atmospheric pressure and the pressure in the pipeline) is larger than the pressure required for opening the air valve (opening pressure). It will then start to move towards its full open position when this pressure difference increases. When the pressure difference is equal or greater than the full lift pressure the valve is fully open. In reality air can also enter via the piston at the top of the air valve (red part). This is neglected since the air flow through this part is small compared to the flow through the in/outlet.

Air is also released via the opening of the stem. The stem starts to move to its fully open position when the pressure difference over the air valve (difference between the pressure in the pipeline and atmospheric pressure) is larger than the opening pressure. When this pressure difference reaches its fully lift pressure the stem is in its full open position. Air is then expelled from the air valve and the water level rises. When the water level reaches the air valve, air and water will be released. When all air is expelled the stem will start to move towards its fully closed position. Table 4.12.1 shows an overview of the different stages of the air valve for different pressure differences. In the following sections the equation for the closed and open stage will be derived.

Fig. 4.12.2 Dynamic air valve

Table 4.12.1 Different modes of the air valve based on the pressure inside the system.

Situation	Result
\(P_{air} - P_{atm} < -P_{fyllyopen}\)	Stem fully open air inlet
\(-P_{fullyopen} < P_{air} - P_{atm} < -P_{lift}\)	Stem between fully open and closed, air inlet
\(-P_{lift} < P_{air} - P_{atm} < P_{lift}\)	Air valve closed
\(P_{lift} < P_{air} - P_{atm} < P_{fyllyopen}\)	Stem between fully open and closed, air outlet
\(P_{air} - P_{atm} > P_{fyllyopen}\)	Stem fully open air outlet and possible water outlet depending on the water level

Closed state

In the closed stage air will be trapped in the air valves, the amount will depend on the history of the system. This air can be compressed or expanded, depending on the pressure changes in the system. This is governed by the ideal gas law:

(4.12.1)\[P_{air} V_{air}^{k} = C\]

with

variable	Description	Units
\(P_{air}\)	Absolute air pressure on fluid level	Pa
\(V_{air}\)	Air volume	m3
\(k\)	Laplacer coefficient	-
\(C\)	Constant	J

This will result in a change of air volume and this change in volume will result in a discharge in the connected system. This is governed by:

(4.12.2)\[Q_1 - Q_2 + Q_{airce} = 0\]

with

variable	Description	Units
\(Q_{1}\)	Discharge at connection point 1	m3/s
\(Q_{2}\)	Discharge at connection point 2	m3/s
\(Q_{airce}\)	Air volume change due to compression and expansion	m3/s

During the closed state the change of volume of air is due to compression and expansion of the air. This is given by:

(4.12.3)\[Q_{airce} = \left( \left( \frac{P_{old}}{P_{new}} \right)^{1/\lambda} - 1 \right) \frac{V_{old}}{dt}\]

with

variable	Description	Units
\(P_{old}\)	Air pressure at previous time step	Pa
\(P_{new}\)	Air pressure at present time step	Pa
\(V_{old}\)	Air volume at previous time step	m3
\(dt\)	Time step	s

The air pressure is given by:

(4.12.4)\[P = P_{atm} + \rho g (H_i - w_i)\]

with

variable	Description	Units
\(P\)	Absolute air pressure on fluid level	Pa
\(P_{atm}\)	Atmospheric pressure	Pa
\(\rho\)	Denisty of the fluid	kg/m3
\(H_i\)	Head at the \(i^{th}\) connection point	m
\(w_i\)	Fluid level at the \(i^{th}\) connection point	m

The equations (4.12.1), (4.12.2), (4.12.3) and (4.12.4) govern the behavior of the air valve in closed state.

Open stage

When the air valve is open air can either enter or exit the system. The actual direction depends upon the pressure inside the system. Table 4.12.1 shows an overview of the different possibilities. The air in- or outflow can be defined by either coefficients or by a characteristic.

Vent capacity (defined by coefficients)

The following formuleas are based on [1]:

1. Subsonic air flow

(4.12.5)\[\begin{eqnarray} Q_{air} = C_{in} A_{in} \sqrt{7RT_0} \sqrt{ \left(\frac{P}{P_{atm}}\right)^{1.4286} - \left( \frac{P}{P_{atm}} \right)^{1.714}} & P_{atm} > P > 0.53 P_{atm} \end{eqnarray}\]

with

variable	Description	Units
\(A_{in}\)	Inlet area	m2
\(C_{in}\)	Inlet discharge coefficient	-
\(R\)	Gas constant	J/(kg K)
\(T_0\)	Ambient temperature	K

2. Critical flow in

(4.12.6)\[\begin{eqnarray} Q_{air} = C_{in} A_{in} \sqrt{7RT_0} 0.259 & \textrm{ when } P_{atm} > P > 0.53 P_{atm} \end{eqnarray}\]

3. Subsonic air flow out

(4.12.7)\[\begin{eqnarray} Q_{air} = -C_{out} A_{out} \sqrt{7RT_0} \left( \frac{P}{P_{atm}} \right)^{\frac{k+1}{2k}} \sqrt{ \left( \frac{P}{P_{atm}} \right)^{1.4286} - \left( \frac{P_{atm}}{P} \right)^{1.714} } & \mathrm{ when } \frac{P_{atm}}{0.53} > P > P_{atm} \end{eqnarray}\]

variable	Description	Units
\(A_{out}\)	Inlet area	m2
\(C_{out}\)	Inlet discharge coefficient	-

4. Critical flow out

(4.12.8)\[\begin{split}\begin{eqnarray} Q_{air} = -0.259 C_{out} A_{out} \sqrt{7RT_0} \left( \frac{P}{P_{atm}} \right)^{\frac{k+1}{2k}} & \textrm{ when } \frac{P_{atm}}{0.53} < P \\ \end{eqnarray}\end{split}\]

Vent capacity (defined by characteristic)

The capacity of the vent can also be defined by the air valve characteristic (see Fig. 4.12.3). In this case, you can define both the characteristic for inflow and outflow. The characteristic should be supplied as pressure against discharge.

Note: Often the characteristic supplied by the manufacturer will have the axes reversed from this definition. See also the example characteristic in this manual.

Fig. 4.12.3 Example of an air valave characteristic

Level effect

The air entering or leaving the pipeline will change the water level in the system. This is taken into account via the so called level effect. This is automatically included based upon the pipe profile of pipes directly connecting to the air valve. There are 4 possible cases:

1. No pipes are connected to the air valve, only the level effect for the stand pipe as specified by the user is taken into account. A warning is given when the water level drops below the bottom of the standpipe. The calculations will continue assuming that the standpipe continues with the same diameter below this point.

2. One pipe is connected to the air valve. For the side the pipe is connected the complete level area effect is taken into account. For the other side only the standpipe is taken into account.

3. Two pipes are connected to the air valve. For both sides the complete level area table is calculated based on the profile of the connected pipes.

4. More than 1 pipe is connected to one side of the air valve. In this case one pipe is selected and a message is given which pipe is selected.

A connection pipe (stand pipe) between the air valve and the pipeline can be included by specifying its diameter and height see Fig. 4.12.4. For the level effect an engineering approach is used to calculate the volume in bends. Since the most important result is the expelling of the last air, this approximation is acceptable.

The dynamic behaviour of the water levels in both legs depends on the actual water levels relative to the bottom elevation. If we denote the local pipe bottom elevation with b, then the following states are feasible:

Table 4.12.2 States of the pipe.

\(W_1\)	\(W_2\)	Description	Applied equations
\(>b\)	\(>b\)	both levels are identical \(w_1 = w_2\)	\(Q_1 - Q_2 = A_1(w_1) \frac{dw_1}{dt} + A_2(w_2)\frac{dw_2}{dt}\)
\(=b\)	\(<b\)	leg 1 fills leg 2, if \(Q_1 > 0\)	\(Q_1 - Q_2 = A_2(w_2)\frac{dw_2}{dt}\)
\(<b\)	\(=b\)	leg 2 fills leg 1, if \(Q_2 < 0\)	\(Q_1 - Q_2 = A_1(w_1)\frac{dw_1}{dt}\)
\(<b\)	\(<b\)	system is fully decoupled	\(Q_1 = A_1(w_1) \frac{dw_1}{dt}\) and \(Q_2= A_2(w_2)\frac{dw_2}{dt}\)

The fact that the water level remains equal to the local bottom elevation implies that the overflow is idealised.

Fig. 4.12.4 Example of a two-legged air valve with vertical stand pipe.

During the open stage the volume balance of (4.12.2) is extended with the air flow rate, this gives:

(4.12.9)\[Q_1 - Q_2 + Q_{airce} + Q_{airflow} - Q_{water} = 0\]

variable	Description	Units
\(Q_{airflow}\)	Flow of air into the system	m3/s
\(Q_{water}\)	Flow of air into the system	m3/s

4.12.1.2. Closing of the air valve

When the water level reaches the top of the air valve it will start to close. The speed of closure can be determined from a force balance over the stem. This gives:

(4.12.10)\[F_g + F_{p1} - F_{p2} - F_{p3} - F_f = 0\]

variable	Description	Units
\(F_{g}\)	Gravitational force	N
\(F_{i}\)	Force on surface \(i\) due to the pressure (Fig. 4.12.5)	N
\(F_f\)	Frictional force	N

Fig. 4.12.5 shows an overview of the different forces acting on the stem. Putting the correct formulas for the different forces gives:

(4.12.11)\[mg + PA_1 - PA_2 - P_{atm}A_3 - \frac{1}{2} \rho_l \xi A_1 \left( \frac{A_{ori} v_s}{A_1} \right)^2 = 0\]

variable	Description	Units
\(m\)	Mass of the stem	kg
\(A_i\)	Area of the air valve (Fig. 4.12.6)	m2
\(\xi\)	Loss coefficient of the orifice in the stem	-
\(v_s\)	velocity of the stem	m/s
\(A_{ori}\)	Area of the orifice	m2

The velocity of the stem can be rewritten as:

(4.12.12)\[v_s = \frac{A_1}{A_{ori}} \sqrt{\frac{2\left(mg + PA_1 - PA_2 - P_{atm}A_3 \right)}{\rho_l \xi A_1}}\]

This equation can be used to calculate the terminal velocity of the stem when it starts to close. Assuming that the stem reached this velocity almost instantaneously after all air has been released, this can be used to determine the closure time with:

(4.12.13)\[t_{closure} = \frac{x}{v_s}\]

This can then be used to include the discharge of water during closure. This discharge will gradually decrease due to the closure of the valve.

Fig. 4.12.5 Forces acting on the stem

Fig. 4.12.6 Area definitions for the air valve

4.12.1.3. Initial states

At first it needs to be selected if the air valve is open or closed. When the airvalve is closed the air flow is zero and the initial air volume is equal to the residual air volume. When the air valve is open the user needs to provide two out of the following 4 properties: 1. Initial air volume 2. Initial air pressure 3. Initial water level on side 1 4. Initial water level on side 2

These are then used as initial state. Please note that it is possible that the stem is in the closed phase, for example prior to air release. This can be modelled via the open state of the air valve.

Air transport Wanda is a single phase simulation tool, thus air transport through pipelines is not included. The air valve model can be used for all processes (such as slow filling) where air is expelled or taken in at the same air valve (air is not transported through the pipeline). To inform the user a warning is given when in reality air will be transported through the pipeline and the model is thus no longer applicable. Air is partially transported when the flow number is greater than 0.6: If the flow number is greater than 0.9 all air is transported. The flow number is given by:

(4.12.14)\[Fr = \frac{v}{\sqrt{gD}}\]

variable	Description	Units
\(Fr\)	Flow number	-
\(D\)	Pipe diameter	m
\(v\)	Flow velocity	m/s

A warning is given when one of these criteria is exceeded.

Note simulation results are invalid if the flow number exceeds 0.6. Background information can be found in [2, 3].

4.12.2. Dynamic air valve properties

4.12.2.1. Hydraulic specifications

Description	Input	unit	range	default	remarks
Height connection pipe	Real	m
Inner diameter connection pipe	Real	m
Diameter stem	Real	m
Inner diameter orifice	Real	m
Travel distance stem	Real	m
Mass of the stem	Real	kg
Kv Characteristic for water flow	Table				Kv values vs. stem position
Opening pressure	Real	Pa
Full open pressure	Real	Pa
Laplace coefficient	Real	-
Ambient air temperature	Real	oC
Air flow defined by	Coefficients Table			Coefs.
Inlet discharge coef.	Real	-			If “Discharge defined by” = Coefs
Outlet discharge coef.	Real	-			If “Discharge defined by” = Coefs
Air inflow characteristic	Table				If “Discharge defined by” = Table
Air outflow characteristic	Table				If “Discharge defined by” = Table
Normal air temp char	Real	oC			If “Discharge defined by” = Table
Residual air volume	Real	m3
Initial state	Closed Open			Closed
Open state defined by:	Air volume and Air pressure Air volume and WL1 Air volume and WL2 Air pressure and WL1 Air pressure and WL1 WL1 and WL2				If “Initial state” = Open
Initial air volume	Real	m3
Initial air pressure	Real	Pa.a
Initial upstream fluid level	Real	m
Initial downstream fluid level	Real	m

Remarks

1. Remark 1

Please be aware that for deriving the discharge coefficients Cin and Cout from a Kv-value (m3/h if = 1 bar), the manufacturer most probably specifies the capacity in atmospheric cubic metres per second. In general, the discharge coefficients describe the amount of contraction of the air flow through the orifices and will probably be in the range of 0.5 to 1.0.

To simulate an air valve, which only lets air into the system the outlet discharge coefficient can be set to 0, in the same way an outlet air valve can be modelled by setting the inlet discharge coefficient to zero.

2. Remark 2

Notice: In case of vents with large capacity it is well possible to have a very sensitive computation in which small pressure differences cause large air flows and volumetric effects. Although the program is optimised to find the most accurate solution, numerical oscillations may still occur (inertia is not taken into account). In that case it is up to the user to adjust (reduce) the time step and the convergence criteria to reduce the oscillations.

The table for air inflow or outflow consists of the pressure difference between the pressure in the pipeline at the location of the valve and the atmospheric pressure. The table should start at the opening pressure and should be monotonously increasing. The discharge should be supplied in atmospheric m3/h. For both characteristic, the values are given as positive values (see examples). When the pressure drop falls out of table, the corresponding air flow will be linearly extrapolated. Please note that this does not include choking flow or other phenomena. It is highly recommended to extend the characteristic if the warning “table out of range” is given.

The characteristic for in/outflow is sometimes measured at a temperature different from the actual ambient temperature. Therefore, the temperature used to define the characteristic has to be specified as well. If this unknown, you can use the same temperature as the ambient temperature.

4.12.2.2. Component specific output properties

Property	Messages
Air volume (m3)
Air flow (m3/s)
Air pressure (absolute) (N/m2.abs)
Fluid level 1 (m)
Fluid level 2 (m)
Flow number (-)

4.12.2.3. Component messages

Message	Type	Explanation
Air valve opens	Info
Air valve closes	Info
Air is transported	Warning
All is transported	Warning

4.12.3. Examples

4.12.3.1. Outflow table

Below you see an example of a outflow characteristic as supplied by the manufacturer, together with the corresponding input table for Wanda.

Fig. 4.12.7 Example of an outflow characteristic

Table 4.12.3 Outflow characteristic tabulated data.

Pressure difference (bar)	Air flow m3/h
0.0	0.0
0.1	7.2
0.4	15.5
0.8	23
1.2	28
1.5	30.6

4.12.3.2. Inflow table

Below you see an example of an inlow characteristic as supplied by the manufacturer, together with the corresponding input table for Wanda. Please note that the pressure difference and discharge are entered as positive values.

Fig. 4.12.8 Example of an inflow characteristic

Table 4.12.4 Inflow characteristic tabulated data.

Pressure difference (bar)	Air flow m3/h
0.0	0.0
0.01	360
0.02	540
0.03	684
0.04	792
0.05	864
0.06	936
0.07	1008
0.08	1080