2. Orifice Flow Theory
Air flows through valves and orifices are traditionally modeled after an initial application of the mass, momentum and energy balances over a small region in space, as illustrated in
Figure 1, where a convergent–divergent nozzle is represented. As air flows from point 1 to point 2, there may or may not be a significant change in density. If the density does not change much, we may treat the flow as incompressible. However, this is not the general practice adopted in pneumatic valve design.
In what follows, we analyze the basis of compressible and incompressible flow modelling for the unidimensional flow represented in
Figure 1 and further apply the resulting equations to the typical situation where flow is discharged and charged from and into an air tank. We begin with the incompressible flow approach.
It can be shown [
3] that as long as the fluid speed remains below 0.3
M (
M is the Mach number), incompressible flow models can be applied and, in particular, the so-called “orifice equation” can be used; that is [
4]
where
Q is the volumetric flow;
and
are the pressures at points 1 and 2;
and
are the cross-sectional areas at points 1 and 2 (see
Figure 1) and
is the density of the fluid. Note that
K is constant in Equation (1).
The mass flow,
, can be obtained from (1) through the equation
. Since we are considering that
remains constant in space and in time, Equation (1) can be simplified to
where
. Here, the coefficient
can be experimentally adjusted to include pressure losses.
If fluid compressibility is considered, Equation (2) cannot be used. Still, an analytical expression can be developed if we assume that the flow is inviscid, the process is adiabatic and the flow regime has reached the steady-state. Considering
Figure 1 again, it can be shown that a momentum and a mass balance between points
and
produces the following equation [
3]
where
,
and
are the density, pressure and velocity, respectively. As
, in
Figure 1, becomes infinitesimal (
,
and
).
The assumption of an adiabatic flow, for which
(
is a constant and
is the ratio of specific heat capacities) allows us to integrate Equation (3) between points 1 and 2, resulting in
Equation (4) can be written in a more meaningful form if we write
. After some mathematical work, the following equation is obtained
Where R = 287 J/kgK is the gas constant for air and is the absolute temperature (K) inside the tank.
Of particular interest are the situations where air contained in a tank (a) discharges into the atmosphere through an orifice (or a valve), as illustrated in
Figure 2a,b, is pressurized by an incoming flow, as shown in
Figure 2b. Considering that air flows from point 1 to point 2, in Equation (4), we have that
,
,
and
for the tank discharge (
Figure 2a);
,
,
and
for the tank pressurization (
Figure 2b).
Substituting the values at points 1 and 2 for each case represented in
Figure 2 into Equation (5), and considering the speed of sound at the tank nozzle,
, and the corresponding Mach number,
, it is possible to arrive at the following equations for the pressure ratio
- 1.
The critical pressure ratio,
b, is defined as the ratio,
, for which sonic speed is attained at the nozzle. Its value is not the same for charging and discharging, being dependent on the ratio between nozzle and tank temperatures,
, at discharge (but not on charge). Although this fact was recognized long ago [
5], it has been ignored in most of the references on the topic (see, for example, references [
6,
7,
8]), where in both charge and discharge,
is obtained from the second equation in (6) by making
; that is
- 2.
In spite of the fact that Equation (6) makes no reference to the nozzle geometry, the Mach number at the nozzle,
, is directly related to the cross-sectional area,
A, and the way it changes along the
axis (
Figure 1). For instance, if pressure decreases along the flow direction, it can be shown that, for subsonic flows, the fluid velocity increases as
A is reduced. An opposite effect exists when the flow becomes supersonic. At sonic speed,
reaches a minimum value [
9];
- 3.
Considering a hypothetical scenario where the pressure tank discharges into the absolute vacuum ( and ), the first equation in (6) fails to represent the flow. To see that, we note that for a perfect gas, implies in . As a result, the first equation in (6) would result in , which is incongruent. The second equation in (6) also fails to represent the flow at . In addition, liquefaction of the gaseous components of air would not allow for using the perfect-gas hypothesis at very low temperatures. These considerations pose a limitation on the use of Equation (6) for adiabatic flows through orifices.
The velocity and the mass flow at the nozzle,
and
, can be obtained from Equation (5) for both situations shown in
Figure 2. Therefore,
is given by
To obtain the mass flow, we make
The
discharge equation for
has been extensively used in pneumatic circuits to simulate flow through valves and is usually represented in a different way, following the work of St. Venant and Wantzel in 1839 [
6]. To modify the discharge equation, we first write
. Then, we obtain
from this last expression and substitute it into the first equation in (9). Finally, we write
and, after some algebraic work, arrive at the following expression for
where
, known as the “flow function”, is given by
The following expression is obtained for
(remember that
)
Equation (12) can be solved for
, which yields the maximum value of the flow function,
, in Equation (11). The solution of Equation (12) is
which is the critical pressure ratio, already obtained in Equation (7) for the case where the air tank is being charged. Substituting
, given by Equation (13), into the first equation in (8), we obtain the maximum value of the air velocity when the tank is being discharged,
, as
where
is the speed of sound at temperature
T. Note that
is close to sonic speed (for
,
).
The ratio
corresponding to the maximum value of
has been used as a divider in compressible flow discharge. Flows where
are termed as
critical flows while flows where
are denominated
subcritical flows [
10].
Figure 3 illustrates this division in a graphical manner. Observe that the function
resembles half of an ellipse at the interval
.
Figure 3 shows the ellipse as a dashed curve and its corresponding equation.
We may write the ellipse equation in
Figure 3 as
, as follows
where
is obtained by substituting
rp, given by Equation (13) into Equation (11), resulting in the following expression
The theory presented so far led us to two equations that describe the steady-state behavior of incompressible and compressible air flows as they discharge through a nozzle; that is, Equations (2) and (10), plus the elliptical flow function approximation (15). Although pressure losses due to viscosity effects and the influence of nozzle geometry can be factored into the coefficient
, in Equation (2), the compressible flow model, given by Equation (10), assumes that the flow is inviscid and makes no particular reference to the nozzle geometry. One simple adjustment aiming to correct the mass flow to include these effects consists in adding a discharge coefficient,
, to Equation (10), as follows
Like the fluid velocity,
, the mass flow,
, has a maximum at
in the case where the upstream pressure,
p, is kept constant. This is very important, because in the situation depicted in
Figure 2a, where no flow enters the air tank, such assumption is not correct. The fact that
achieves a maximum when
is constant can be concluded by a simple observation of Equation (17), where all the terms on the right-hand side, except the flow function
, become constant, so that
has the same behaviour as
(see
Figure 3). When the upstream pressure,
p, is not constant, as is the case in
Figure 2a, we can write
in Equation (17) so that the mass flow can then be written as
, as follows
where
is given by
Assuming that
is constant, the mass flow,
, can be plotted against the pressure ratio,
, for a vessel discharging into the atmosphere, as in
Figure 2a. Here, we have assumed the typical industrial range for the pressure ratio, 0.1
.
Figure 4 shows the curve
.
The problem in using Equation (17) in the whole range of the pressure ratio was addressed by Lord Rayleigh in 1916 [
4], who experimentally proved the non-validity of Equation (18) for the interval
, where the mass flow would, in fact, be constant for a constant value of
. His results contradicted a previous statement made by Osborne Reynolds, in 1885. According to Hartshorn [
11], Reynolds wrote that the mass flow would remain constant within the interval
. Hartshorn concluded that neither Rayleigh nor Reynolds were 100% correct, for experiments showed that the interval limit where the mass flow is constant depends on the type of nozzle that is used. According to his experiments, the mass flow would become constant whenever
, with
L varying from 0.2 to 0.8, depending on the shape of the discharge nozzle.
Figure 5 has been based on the experimental curves published by Hartshorn and illustrates the experimental results that he obtained.
In summary, the following equations can be used for adiabatic discharges through orifices and valves
only for the case where is constantAttempts to find the experimental curves for
have followed ever since. For instance, in 1955, Druett obtained experimental curves for divergent conic-shaped orifices [
12]. However, a somewhat common practice is to assume
, corresponding to the critical pressure ratio,
, for
(see, for example, [
2,
13,
14,
15,
16,
17,
18,
19,
20,
21]).
3. The ISO 6358 Equations
As is usually the case, the need to find a relation between mass flow and pressure has led to attempts to standardize the mathematical model for orifice and valve flows in pneumatic circuits. One known standard that has been extensively used is the ISO 6358, where the following equations are recommended for modelling pneumatic valve flows [
6]
where
and
are the density and absolute temperature measured at a pre-established rated condition, defined by ISO as
,
,
for a relative humidity of 65% and a corresponding gas constant
. The coefficient
is denominated “sonic conductance” and, together with the “sonic pressure ratio”,
, must be experimentally obtained.
The ISO standard uses the elliptical approximation,
, given by Equation (15), as can be seen in second radicand of the first equation in (21). In fact, the first equation in (21) can be obtained by first substituting
with
, in Equation (17), as follows
We can then use the perfect gas equation at a given reference state,
, to eliminate the constant
from Equation (22)
where
Equation (21) has been widely used to this date for modelling pneumatic valve flows. The usual procedure, however, is to skip the experimental determination of the sonic pressure ratio,
L, and use
L = 0.528, which, as we have seen in the previous section, is not a safe assumption (
Figure 5). On the other hand, the ISO 6358 Standard is highly demanding, requiring the experimental determination of two coefficients (
L and
C). The difficulties involved in determining the exact nature of the gas compression/expansion, being generally polytropic can move
away from the adiabatic index (
), thus changing the value of
in Equation (13). For an assumed range
, we would have
[
22]. In addition, as mentioned in the previous section, values of
L as low as 0.2 were experimentally obtained. In fact, due to the difficulty in obtaining
L, it is tempting to attribute a random value to
L, perhaps based on some intuitive criteria, as can be concluded by reading some of the several works published on pneumatic circuits.
One question to be asked here is whether a simpler model such as the one given by Equation (2) could be used for modelling mass flows through valves. Bobrow and McDonell [
2], for example, stated that “…nearly all previous results on pneumatic control incorrectly assume (14) is true”, where (14) is a reference to the ISO equations in their paper. They obtained better results with the following equations, used in the context of a proportional valve connected to a cylinder
where the coefficients
and
depend on the electric current that is applied to the proportional valve, according to the original paper [
2].
We would like to investigate whether Equation (25) can still be used to model air flow through valves and orifices when
and
are constants. In fact, the first equation in (25) is, precisely, the incompressible flow model (2), with
. The linear relation indicated by the second equation, on the other hand, characterizes a laminar flow [
4]. If these equations are able to represent flows through actual valves, the difficulties involved in determining
L in Equation (21) are bypassed. We deal with this matter in the following section.
4. Numerical Modelling and Simulation
Consider the pressurization and subsequent decompression of an air tank, as shown in
Figure 6. The schematics shown in the figure include elements that will be referred to, when we describe the test rig used in the experimental procedure.
To pressurize the tank 1, in
Figure 6a, the unidirectional flow–control valve 2 can be adjusted between totally closed and totally open. The manually activated directional valve 3 provides a constant pressure,
p, at the input of valve 2 while the tank pressure,
, changes from 0 to 0.6 MPa (gauge). The pressure evolution in time can be visually seen with the help of gauge 5 and measured in a timely manner by transducer 4, which sends data to an analog–digital converter connected to a computer. Depressurization (
Figure 6b) is performed by inverting the position of valve 2 and closing it with the air tank pressure at 0.6 MPa, while air escapes to the atmosphere through valve 6. Note that
p, at the valve 6 input, is variable this time, while
remains constant.
Mass conservation can be applied to both situations in
Figure 6. Considering that the air tank volume is
V and that it remains at a constant temperature,
T, the following equation is obtained
where
is negative when flowing into the air tank, as in
Figure 6a, and positive otherwise (
Figure 6b). If we write
, Equation (26) can be written in a more convenient form as
To circumvent the complexities of the analytical solutions for (27), when different expressions for
are used, we have solved this equation numerically.
Table 1 summarizes the equations for
we have seen so far, for the situation where the air tank discharges into the atmosphere. In the table, we have grouped different factors as constants
through
. Since the great majority of scientific and engineering works make use of only one model regardless of the direction of the air flow, only the equations corresponding to the tank discharge are being considered in
Table 1. Incidentally, both Equation (25) are represented, given that the first equation in (25) corresponds exactly to the incompressible flow model (2).
Each equation in
Table 1 demands the knowledge of a constant, which must be determined through experimental data fitting. An easy way to make any of these models describe the actual air flow, however, is to change these constants into coefficients. For instance, considering a general coefficient
, we might think of a polynomial expansion of degree
,
(
), as proposed in [
13] for the ISO Equation (21). We do not favor this approach, given that, virtually, any model in
Table 1 can be adjusted to fit experimental results by correctly choosing the new constants of the coefficient
. Moreover, once the polynomial equation for
is introduced back into its corresponding mathematical model, we end up with a totally strange equation, for which no physical background can be traced.
In
Table 1, the constants
,
and
are given by:
The combination of the equations in
Table 1 and the differential Equation (27) results in three mathematical models for the pressure inside the air tank in
Figure 6, as listed in
Table 2.
We have solved each equation in
Table 2 for
, adjusting constants
through
so that
and
, simulating the pressurization of the air tank in
Figure 6a. In a reverse order, we also simulated the decompression of the air tank for
and
. The time setting for the simulation has been based on experimental data to be introduced in the following section. Considering an air tank volume
and assuming that the air temperature is kept at
, the constant
can be calculated as
We have assumed
and
. Simulation results, obtained by using the 4th order Runge–Kutta method to solve the equations in
Table 2, are shown in
Figure 7.
Figure 7a,b show the evolution in time of the pressure inside the air tank. In both cases, there is a closer match between results given by models 1 and 2. This is due to the much higher mass flow at the beginning when the linear model 3 is used, as can be seen in
Figure 7c,d. Note the constant mass flow during
at pressurization when model 2 is used. This is justified by the fact that model 2 is based on the theoretical–experimental approach explained in the previous section where the mass flow becomes constant when the pressure ratio drops below a given value of
(see
Figure 5).
It is very important to observe the values of the constants
,
and
change significantly between decompression and pressurization. As a matter of fact, they also change with the flow–control valve adjustment, as will be seen in the following section, where we present experimental results that will help us better compare the models in
Table 2.
5. Comparison with Experimental Results
Figure 8 shows the pneumatic test rig used to obtain the curve
for the two situations displayed in
Figure 6. The technical specifications for each component are given in
Table 3. All components are interconnected through plastic tubing whose external/internal diameters are 4 mm and 2.5 mm, respectively.
A comparison between experimental results and the theoretical curves is shown in
Figure 9 for two opening states of the flow–control valves. First, the valve was approximately 5% open. The curves corresponding to the decompression and pressurization of the air tank for this first setting are shown in
Figure 9a,b, respectively. The second set of experimental data was obtained for a 100% opening of the flow control valves, for which corresponding curves are shown in
Figure 9c,d, for decompression and pressurization, respectively. The curves were fitted so that they would match experimental data at
and
(gauge).
At a first glance, we conclude that the third model produces a far better approximation to experimental data when the air tank is discharging into the atmosphere (
Figure 9a,c). On the other hand, both the incompressible flow model 1 and the ISO model 2 are better when the tank is being charged (
Figure 9b,d). It is, therefore, reasonable, to prefer the one-constant modelling Equation (25) instead of the two-constant ISO Equation (21).
The reason why the linear dependence between flow and pressure provides a better approximation during discharge is a challenging one. As we already mentioned, the assumption of a laminar flow on discharge would help to explain this fact. We believe that a more detailed analysis would require the numerical solution of the Navier–Stokes equations. This is certainly one possibility for further developments to be pursued in a future work.
One might wonder whether the choice for
in model 2 would be the reason why the ISO model does not provide the best fit during tank discharge. To verify this, we simulated the same case shown in
Figure 9c using the two extreme values of
, 0 and 1. The results are presented in
Figure 10. Even with these changes we could not obtain a better fitting, which suggests that Equation (21) is not the best option for the case where the tank is depressurized. Moreover, the simpler incompressible model 1 also provides a good approximation for the case where the tank is being charged, with only one constant to be adjusted. We thus come to the same conclusion found in reference [
2] and agree that Equation (25) best fits our experimental data.