Computational method for simultaneous inertial measuring of fluid mass flow rate and density

Abstract

This work represents the decision of the problem of nonstationary one-dimensioned flow of fluid in a straight pipeline. It shows the method to generate non-stationary flow regime based on utilizing additional pipeline in which mechanical oscillations of fluid are being excited. Pressure difference along the length of oscillating fluid appears to be a measure for the density of the fluid and for its mass flow rate. The possibility of creating a new instrument for measuring the density and mass flow rate of a fluid based on results obtained is shown.

Keywords

Simultaneous measuring mass flow rate density inertial method hidraulic resistance

1. Introduction

Modern instrument-making industry fabricates wide range of various devices for measurement of mass flow rate and density of fluids, but not all of them are suitable for simultaneous measuring of two mentioned physical variables. Frequently instruments for density measuring are not suitable to measure the density of fluid which flows along a pipeline. Besides, the most of currently produced instruments for measurement of density perform the measurement in a limited volume of the fluid, which is directly adjacent to a sensing element of the device. Other measurers of flow rate and density (for example, Coriolis flow meters) have high costs and some operational disadvantages, such as high hydraulic resistance because of small diameter of sensing tubes. This paper considers a new principle of measuring of mass flow rate and density of fluids, which is based on analysis of longitudinal inertia force, and represents the fundamental ability to create a measuring instrument upon this principle.

2. The measurement of fluid density

Let us acquaint initially with a simpler problem setting – to measure only density of a fluid. The scheme of a measuring device is shown on the Fig. 1.

Figure 1.

The scheme of a device for measuring of density: 1 and 2 – chambers of mixing and separation; 3 – working section of main pipeline; 4 – additional pipeline; 5 – piston.

Let’s consider a straight pipeline 3 of a length $L$ and cross-sectional area $S$ through which homogeneous fluid is flowing. The density of the fluid is $\rho$ and its mass flow rate is $\dot{m}$ . We will name it the main pipeline. In parallel to it another pipeline 4 is connected (we will name it the additional pipeline), in which a tightly fitted piston 5 is set. The piston 5 of a cross-sectional area $F_{1}$ is driven by external motion source. Earlier, the problem similar to this, but with some essential differences, was set and considered in [1].

Let the piston performs harmonic motion in the additional pipeline according to the equation

$\displaystyle x_{1}=A\sin\omega t,$ (1)

where $x_{1}$ is the position of the piston in the additional pipeline; $A$ is the amplitude of piston’s oscillations; $\omega$ is the angular frequency of piston’s oscillations.

We will measure the amount of pressure difference which arises along the length $L$ of the main pipeline. The fluid is considered to be an ideal incompressible fluid – this is the first assumption. Another assumption: the piston’s motion causes corresponding changes of fluid’s flow rate only within the section $L$ of the main pipeline between the points of connection with the additional pipeline (we will name it “the working section”), and has no influence on the fluid’s flow upstream and downstream of the apparatus.

There are made chambers of mixing and separation 1, 2 at the points of connection of the main and the additional pipelines. These chambers provide fluid which flows through the main pipeline 3 to be mixed with the fluid which is exiting from the additional pipeline 4. Or else, a part of fluid is separated from the main pipeline and passes toward the additional one. Since the piston 5 performs reciprocation, each chamber performs mixing and separation alternately. If it is so, then the mass flow rate through the main pipeline within the working section $L$ differs from the flow rate $\dot{m}_{0}$ upstream and downstream the apparatus:

$\displaystyle\dot{m}=\dot{m}_{0}+\dot{m}_{1}$ (2)

where $\dot{m}_{1}$ is the mass flow rate in the additional pipeline; $\dot{m}$ is the mass flow rate in the main pipeline at the working section $L$ between the mass-changing chambers.

As a consequence of the continuity condition the mass flow rate $\dot{m}$ is the same along the whole line of the section $L$ :

$\displaystyle\frac{\partial\dot{m}}{\partial x}=0,$ (3)

where the $x$ coordinate coincides with the along axis of the main pipeline.

This equation means the next:

$\displaystyle\frac{\partial\dot{m}}{\partial x}=\frac{\partial({\rho uS})}{% \partial x}=\rho\left({u\frac{\partial S}{\partial x}+S\frac{\partial u}{% \partial x}}\right)=0,$

and with the constant area of the orifice along the working section $\frac{\partial S}{\partial x}=0$ , we have

$\displaystyle\frac{\partial u}{\partial x}=0.$ (4)

The variable $u$ is an average velocity of fluid in the orifice $S$ . We will consider the flow distribution through the orifice to be constant.

It is obviously that mass flow rate alteration would cause some force influence on the fluid being within the working section. We will define this influence. Let’s write Euler’s equation for the part of ideal fluid on the working section [2]:

$\displaystyle-\frac{1}{\rho}\frac{\partial p}{\partial x}=\frac{\partial u}{% \partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w% \frac{\partial u}{\partial z},$ (5)

where $u$ , $v$ , $w$ are velocity components alongside the axes $x$ , $y$ , $z$ correspondingly.

Since the flow is considered to be one-dimensional, $v=$ 0 and $w=$ 0, and the Eq. (5) takes a look

$\displaystyle-\frac{1}{\rho}\frac{\partial p}{\partial x}=\frac{\partial u}{% \partial t}+u\frac{\partial u}{\partial x}.$ (6)

Substituting Eq. (4) in Eq. (6), we obtain

$\displaystyle\frac{\partial p}{\partial x}=-\rho\frac{\partial u}{\partial t}.$ (7)

Find the value of the derivative $\frac{\partial u}{\partial t}$ . From the equation for mass flow rate $\dot{m}=\rho uS$ we receive

$\displaystyle u=\frac{\dot{m}}{\rho S}.$ (8)

Substitute Eq. (2) in Eq. (8) and take the derivative

$\displaystyle\frac{\partial u}{\partial t}=\frac{\partial}{\partial t}\left({% \frac{\dot{m}}{\rho S}}\right)=\frac{1}{\rho S}\cdot\frac{\partial}{\partial t% }({\dot{m}_{0}+\dot{m}_{1}})=\frac{1}{\rho S}\frac{\partial\dot{m}_{1}}{% \partial t}=\frac{\ddot{m}_{1}}{\rho S}.$ (9)

Mass flow rate in the additional pipeline $\dot{m}_{1}=\dot{x}_{1}F_{1}\rho$ , consequently

$\displaystyle\ddot{m}_{1}=\rho F_{1}\ddot{x}_{1}.$ (10)

Substitute Eq. (10) in Eq. (9):

$\displaystyle\frac{\partial u}{\partial t}=\frac{\rho F_{1}\ddot{x}_{1}}{\rho S% }=\frac{F_{1}}{S}\ddot{x}_{1}.$ (11)

Then substitute Eq. (11) in Eq. (7):

$\displaystyle\frac{\partial p}{\partial x}=-\rho\frac{F_{1}}{S}\ddot{x}_{1}.$ (12)

For obtaining pressure drop $\Delta p$ along the length $L$ we will integrate the Eq. (12):

$\displaystyle\Delta p=\int\limits_{0}^{L}{\frac{\partial p}{\partial x}dx}=-% \rho\frac{F_{1}}{S}\ddot{x}_{1}\int\limits_{0}^{L}{dx}=-\frac{F_{1}}{S}\rho% \ddot{x}_{1}L.$ (13)

We can see that a pressure difference $\Delta p$ arises along a smooth section of a tube, and this pressure difference is proportional to the density of the fluid $\rho$ . So, pressure difference considered is being a measure for the fluid’s density. Proportional coefficient between the density and the pressure difference constitutes by combining some geometric dimensions of the apparatus and piston’s acceleration that is variables which are practically constant during the operational period.

Let us see how to use obtained equation in practice. The signal $\Delta p(t)$ is an amplitude-modulated sine signal and its amplitude is proportional to the fluid’s density. Use the spectral analyses method for finding the amplitude. Since the piston’s motion is in a harmonic way, then its acceleration is proportional to its coordinate:

$\displaystyle x_{1}=A\sin\omega t;\quad\dot{x}_{1}=A\omega\cos\omega t$ (14) $\displaystyle\ddot{x}_{1}=-A\omega^{2}\sin\omega t=-\omega^{2}x_{1}.$ (15)

Substituting Eq. (15) in Eq. (13) we obtain

$\displaystyle\Delta p(t)=\frac{F_{1}}{S}\rho L\omega^{2}x_{1}(t).$ (16)

Further, if we know the frequency of oscillations, then we will seek the harmonic component exactly at this frequency. As we can see in Eq. (16)

$\displaystyle\Delta p(t)=\frac{F_{1}}{S}\rho L\omega^{2}x_{1}(t)=\frac{F_{1}}{% S}\rho L\omega^{2}A\sin\omega t.$ (17)

We use the Fourier transform confining only first harmonic at the frequency $\omega$ .

Find next integral within one full period of oscillations

$\displaystyle\int\limits_{0}^{2\pi/\omega}{\Delta p(t)\sin\omega\textit{tdt}}=% \frac{F_{1}}{S}\rho L\omega^{2}A\int\limits_{0}^{2\pi/\omega}{\sin^{2}\omega% \textit{tdt}}=\frac{\pi}{\omega}\frac{F_{1}}{S}\rho L\omega^{2}A=\pi\omega A% \frac{F}{S}\rho L.$ (18)

Here it follows

$\displaystyle\rho=\frac{\int\limits_{0}^{T}{\Delta p(t)\cdot\sin\omega\textit{% tdt}}}{\pi\omega A\frac{F_{1}}{S}L}.$ (19)

In according to the Eq. (19) the process of measuring of density performs within the period $T=\frac{2\pi}{\omega}$ that is during one complete cycle of piston’s motion. Such method has an essential advantage because it is insensitive to constant components of pressure drop which are caused by fluid’s viscosity.

3. The simultaneous measurement mass flow rate and density of fluid

Now, when we have learned alternative action on a fluid’s flow to let the fluid’s density to be measured, it would be desirable borders of the problem to be extended. What if a dynamic response carries information about the flow rate also? If some acceleration was applied to a moving fluid, and the intensity of force influence on the fluid was measured, maybe we would obtain a measure for fluid’s flow rate? Let’s consider the scheme on the Fig. 2.

Figure 2.

The scheme of a device for simultaneous measurement of mass flow rate and density: 1 – main pipeline, 2 – additional pipeline, 3 – piston, 4 and 5 – mass-changing chambers, 6 and 7 – differential pressure sensors.

In general, the arrangement of apparatus is the same as of the inertial density meter described above. There is the main pipeline 1 of cross-sectional area $S$ through which a fluid flows. Mass flow rate of fluid is $\dot{m}_{0}$ , its density is $\rho$ . There is also an additional pipeline 2 connected in parallel to the main one. In the additional pipeline a piston 3 is set which performs harmonic oscillations according to the equation $x_{p}=A\cdot\sin\omega t$ , where $A$ is the amplitude of oscillations. The cross-sectional area of piston is $F_{p}$ . The main pipeline is connected with the additional one by two mass-changing chambers 4 and 5. Each mass-changing chamber in this case constitutes a section of the main pipeline with perforated wall. This section has a length $L$ and at its beginning and at its end receivers of static pressure are installed. So, mixing and separation of additional flow will perform at the section $L$ . There are two such mass-exchange chambers. There are two differential pressure sensors 6, 7 in the scheme for measuring the pressure difference. We will consider the additional flow to be injected into the main one evenly along the whole length $L$ , that is the velocity of fluid’s parts of the additional flow through the perforated wall is the same along the length $L$ . Also, likewise inertial density meter, we take an assumption that the oscillations of flow caused by piston’s motion occur only at the section between mass-changing chambers and don’t expand outside the apparatus.

Since the fluid in the section $L$ of the main pipeline flows with acceleration, it would be naturally to suppose pressure difference to arise between the inlet and outlet points of section $L$ . We will make a mathematical model of the process and find this pressure difference.

Put on the Euler’s equation for the part of fluid in the mixing chamber, neglecting the viscosity.

$\displaystyle-\frac{1}{\rho}\frac{\partial p}{\partial x}=\frac{\partial u}{% \partial t}+u\frac{\partial u}{\partial x}.$ (20)

We introduce two coordinate systems $x_{1}$ , $x_{2}$ with abscissa along the axis of main pipeline and the beginning of each system coincide with the first points of connecting of differential pressure sensors respectively. Appeal initially to the first camera 4. Find the velocity of fluid:

$\displaystyle u(x_{1})=\frac{\dot{m}({x_{1}})}{\rho S}.$ (21)

Here $\dot{m}({x_{1}})$ is mass flow rate in the orifice at the coordinate $x_{1}$ , $u(x_{1})$ is the velocity in this orifice.

Flow rate in this orifice consists of two flows: the main and the additional ones. If the additional flow is distributed evenly along the section $L$ , it means the intensity of its penetration to be constant and equal $\frac{\dot{m}_{a}}{L}$ along the whole length $L$ . Then the mass flow rate of the additional flow in the main pipeline at the orifice with coordinate $x_{1}$ equal to

$\displaystyle\int\limits_{0}^{x_{1}}{\frac{\dot{m}_{a}}{L}dx}=\frac{\dot{m}_{a% }}{L}x_{1}.$ (22)

Consequently, total mass flow rate at the coordinate $x_{1}$ equal to

$\displaystyle\dot{m}({x_{1}})=\dot{m}_{0}+\frac{\dot{m}_{a}}{L}x_{1}.$ (23)

Then the velocity $u(x_{1})$ and its derivatives are

$\displaystyle u({x_{1}})=\frac{\dot{m}_{0}+\dot{m}_{a}\frac{x_{1}}{L}}{\rho S}% =\frac{\dot{m}_{0}}{\rho S}+\dot{m}_{a}\frac{x_{1}}{\rho SL}.$ (24) $\displaystyle\frac{\partial u({x_{1}})}{\partial x}=\frac{\dot{m}_{a}}{\rho SL};$ (25) $\displaystyle\frac{\partial u({x_{1}})}{\partial t}=\frac{x_{1}}{\rho SL}\frac% {\partial\dot{m}_{a}}{\partial t}=\frac{x_{1}}{\rho SL}\ddot{m}_{a}.$ (26)

The equation for $\frac{\partial u}{\partial t}$ obtained with assumption that $\frac{\partial\dot{m}_{0}}{\partial t}=0$ .

Substitute obtained Eqs (24)–(26) in the Euler’s Eq. (20) and find values of pressure gradients and pressure differences.

$\displaystyle-\frac{1}{\rho}\frac{\partial p}{\partial x}=\frac{x_{1}}{\rho SL% }\ddot{m}_{a}+\left({\frac{\dot{m}_{0}}{\rho S}+\dot{m}_{a}\frac{x_{1}}{\rho SL% }}\right)\frac{\dot{m}_{a}}{\rho SL}.$ (27)

Taking into account that $\dot{m}_{a}=\dot{x}_{p}F_{p}\rho$ , $\ddot{m}_{a}=\ddot{x}_{p}F_{p}\rho$ we conclude

$\displaystyle\frac{\partial p}{\partial x_{1}}=-\rho\left({\frac{\ddot{x}_{p}F% _{p}\rho}{\rho SL}x_{1}+\frac{\dot{m}_{0}\dot{x}_{p}F_{p}\rho L}{({\rho SL})^{% 2}}+\frac{\dot{x}_{p}^{2}F_{p}^{2}\rho^{2}}{({\rho SL})^{2}}x_{1}}\right).$ (28)

So, pressure difference along the length $L$ for the first chamber equal to

$\displaystyle\Delta p_{1}=p_{1\textit{right}}-p_{1\textit{left}}=\int\limits_{% 0}^{L}{\frac{\partial p}{\partial x_{1}}dx_{1}=}-\rho\frac{\ddot{x}_{p}F_{p}L}% {2S}-\dot{m}_{0}\frac{\dot{x}_{p}F_{p}}{S^{2}}-\rho\frac{\dot{x}_{p}^{2}F_{p}^% {2}}{2S^{2}},$ (29)

where $p_{1\textit{right}}$ and $p_{1\textit{left}}$ are the pressure at the left and the right borders of section $L$ of the first mass-changing chamber respectively.

Perform similar computation for the second mass-changing chamber.

$\displaystyle\dot{m}({x_{2}})=\dot{m}_{0}+\dot{m}_{a}\left({1-\frac{x_{2}}{L}}% \right).$ (30) $\displaystyle u({x_{2}})=\frac{\dot{m}_{0}}{\rho S}+\frac{\dot{m}_{a}}{\rho S}% \left({1-\frac{x_{2}}{L}}\right).$ (31) $\displaystyle\frac{\partial u(x_{2})}{\partial x}=\frac{\partial}{\partial x}% \left({\frac{\dot{m}_{0}}{\rho S}+\frac{\dot{m}_{a}}{\rho S}\left({1-\frac{x_{% 2}}{L}}\right)}\right)=-\frac{\dot{m}_{a}}{\rho SL}.$ (32) $\displaystyle\frac{\partial u(x_{2})}{\partial t}=\frac{\partial}{\partial t}% \left({\frac{\dot{m}_{0}}{\rho S}+\frac{\dot{m}_{a}}{\rho S}\left({1-\frac{x_{% 2}}{L}}\right)}\right)=\left({1-\frac{x_{2}}{L}}\right)\frac{\ddot{m}_{a}}{% \rho S}.$ (33) $\displaystyle-\frac{1}{\rho}\frac{\partial p}{\partial x_{2}}=\frac{\ddot{m}_{% a}}{\rho S}\left({1-\frac{x_{2}}{L}}\right)-\frac{\dot{m}_{0}\dot{m}_{a}}{\rho% ^{2}S^{2}L}-\frac{\dot{m}_{a}^{2}}{\rho^{2}S^{2}L}\left({1-\frac{x_{2}}{L}}% \right).$ (34) $\displaystyle\Delta p_{2}=p_{2\textit{right}}-p_{2\textit{left}}=\int\limits_{% 0}^{L}{\frac{\partial p}{\partial x_{2}}dx_{2}}=-\rho\frac{\ddot{x}_{p}F_{p}L}% {2S}+\dot{m}_{0}\frac{\dot{x}_{p}F_{p}}{S^{2}}+\rho\frac{\dot{x}_{p}^{2}F_{p}^% {2}}{2S^{2}},$ (35)

where $p_{2\textit{right}}$ and $p_{2\textit{left}}$ are the pressure at the left and the right borders of section $L$ of the second mass-changing chamber respectively.

The Eqs (29) and (35) show the pressure difference at each mass-changing chamber to be a measure for fluid’s density and mass flow rate.

Let us see how to use the equations obtained. Likewise in the inertial density meter described above, we use the Fourier transform to define mass flow rate and density of fluid. Since the piston’s oscillations performs in harmonic way, their amplitude $A$ , frequency $\omega$ and phase are known, then as a consequence each of signals $\Delta p_{1}$ , $\Delta p_{2}$ carries all necessary information about fluid’s density $\rho$ and mass flow rate $\dot{m}_{0}$ . The components proportional to density and mass flow rate become moved by phase at 90 ${}^{\circ}$ , so they are orthogonal to each other. Consider the pressure difference $\Delta p_{1}(t)$ Eq. (29) and define its components along two orthogonal coordinates $\sin\omega t$ and $\cos\omega t$ , in terms of phase coordinates. As it was mentioned above Eqs (14) and (15)

$\displaystyle x_{p}=A\sin\omega t,$ $\displaystyle\dot{x}_{p}=A\omega\cos\omega t,$ $\displaystyle\ddot{x}_{p}=-A\omega^{2}\sin\omega t.$

Then the equation for the $\Delta p_{1}(t)$ Eq. (29) takes a look

$\displaystyle\Delta p_{1}(t)=\rho\frac{F_{p}}{S}\frac{L}{2}A\omega^{2}\sin% \omega t-\dot{m}_{0}\frac{F_{p}A\omega}{S^{2}}\cos\omega t-\rho\left({\frac{1}% {2}\left({\frac{F_{p}}{S}}\right)^{2}A^{2}\omega^{2}}\right)\cos^{2}\omega t.$ (36)

We calculate the integrals $P_{\rho}=\int\limits_{0}^{T=2\pi/\omega}\Delta p_{1}(t)\sin\omega\textit{tdt}$ and $P_{Q}=\int\limits_{0}^{T=2\pi/\omega}\Delta p_{1}(t)\cos\omega\textit{tdt}$ .

Since the function $\Delta p_{1}(t)$ includes functions $\sin\omega t$ , $\cos\omega t$ , $\cos^{2}\omega t$ , it is reasonable to calculate some trigonometric integrals apart:

$\displaystyle\int\limits_{0}^{2\pi/\omega}\sin^{2}\omega t\cdot\sin\omega% \textit{tdt}=-\frac{1}{\omega}\int\limits_{0}^{2\pi/\omega}({1-\cos^{2}\omega t% })d\cos\omega t=-\frac{1}{\omega}\left.{\left[{\cos\omega t-\frac{\cos^{3}% \omega t}{3}}\right]}\right|_{0}^{2\pi/\omega}$ (37) $\displaystyle\quad=0;$ $\displaystyle\int\limits_{0}^{2\pi/\omega}{\sin^{2}\omega t}\cos\omega\textit{% tdt}=\frac{1}{\omega}\int\limits_{0}^{2\pi/\omega}{\sin^{2}\omega td\sin\omega t% =\left.{\frac{\sin^{3}\omega t}{3\omega}}\right|_{0}^{2\pi/\omega}=0};$ (38) $\displaystyle\int\limits_{0}^{2\pi/\omega}{\cos^{2}\omega t\cdot\sin\omega t}% dt=-\frac{1}{\omega}\int\limits_{0}^{2\pi/\omega}{\cos^{2}\omega t}d\cos\omega t% =-\left.{\frac{\cos^{3}\omega t}{3\omega}}\right|_{0}^{2\pi/\omega}=0;$ (39) $\displaystyle\int\limits_{0}^{2\pi/\omega}\cos^{2}\omega t\cdot\cos\omega% \textit{tdt}=\frac{1}{\omega}\int\limits_{0}^{2\pi/\omega}\left({1-\sin^{2}% \omega t}\right)d\sin\omega t=\frac{1}{\omega}\left.{\left[{\sin\omega t-\frac% {\sin^{3}\omega t}{3}}\right]}\right|_{0}^{2\pi/\omega}=0.$ (40)

The integrals Eqs (3)–(40) described above indicate to the fact that functions $\sin^{2}\omega t$ and $\cos^{2}\omega t$ are orthogonal both to $\sin\omega t$ and $\cos\omega t$ . Consequently, the third component in the Eq. (36) which is proportional to $\cos^{2}\omega t$ becomes equal zero when we calculate integrals $P_{\rho}$ and $P_{Q}$ . Then

$\displaystyle P_{\rho}=\int\limits_{0}^{2\pi/\omega}{\Delta p_{1}(t)\sin\omega% \textit{tdt}}=\rho\frac{LA\omega^{2}F_{p}}{2S}\cdot\frac{\pi}{\omega}=\rho% \cdot\frac{\pi LA\omega F_{p}}{2S};$ (41) $\displaystyle P_{Q}=\int\limits_{0}^{2\pi/\omega}{\Delta p_{1}(t)\cos\omega% \textit{tdt}}=-\dot{m}_{0}\frac{F_{p}\omega A}{S^{2}}\cdot\frac{\pi}{\omega}=-% \dot{m}_{0}\frac{F_{p}A\pi}{S^{2}}.$ (42)

Figure 3.

Differential pressure curves for water at mass flow rate 0.5 kg/s.

From here we find

$\displaystyle\rho=\frac{P_{\rho}}{K_{\rho}}=\frac{\int\limits_{0}^{2\pi/\omega% }{\Delta p_{1}(t)\sin\omega\textit{tdt}}}{\pi LA\omega F_{p}/({2S})},$ (43) $\displaystyle\dot{m}_{0}=-\frac{P_{Q}}{K_{Q}}=-\frac{\int\limits_{0}^{2\pi/% \omega}{\Delta p_{1}(t)\cos\omega\textit{tdt}}}{F_{p}A\pi/S^{2}},$ (44)

where

$\displaystyle K_{\rho}=\frac{\pi LA\omega F_{p}}{2S},$ (45) $\displaystyle K_{Q}=\frac{F_{p}A\pi}{S^{2}}.$ (46)

Each of the coefficients $K_{\rho}$ and $K_{Q}$ is the complex of constructive parameters practically unchanged during the exploitation period (except the pipe cross-sectional area $S$ which may decrease because of plaque accumulation).

As we can see, the inertial method described here provides fundamental opportunity of direct measuring of fluid’s density and mass flow rate. Likewise to the density meter (point 1), the constant component of hydraulic pressure drop due to viscosity or hydraulic dissipation cause no affect on the accuracy of measurement.

Calculated examples of functions $\Delta p_{1}(t)$ and $\Delta p_{2}(t)$ at two various values of mass flow rate are shown on the Figs 3 and 4.

Figure 4.

Differential pressure curves for water at mass flow rate 3.0 kg/s.

4. Conclusion

The inertial method described in the article gives an opportunity of creation an instrument for measuring fluid’s mass flow rate and density. The instrument based on the principle described would have some significant advantages. It may be built of small or very big diameter of pipeline; may operate at low or high flow rate; it performs the measuring process within the overall volume of measured fluid, and therefore possible parts of different phases in the fluid will be measured as well. And at last but not least, such instrument would have extremely little hydraulic resistance.

As we can see, the decision of the problem considered is not of a theoretical interest only, but may be used for creating an instrument for measuring fluid’s mass flow rate and density.

Footnotes

Acknowledgments

The research was funded by a grant of the Russian Foundation for Basic Research (RFBR) for the project No 18-48-730013.

References

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