Bernoulli equation in general form. Bernoulli differential equation

A differential equation of the form , where , is called Bernoulli's equation.

Assuming that , we divide both sides of the Bernoulli equation by . As a result, we obtain: (8.1) Let us introduce a new function . Then . Let us multiply equation (8.1) by and pass to the function z(x): , i.e. for function z(x) obtained a linear inhomogeneous equation of the 1st order. This equation is solved using the methods discussed in the previous paragraph. Let us substitute into its general solution instead z(x) expression, we obtain the general integral of the Bernoulli equation, which is easily resolved with respect to y. When a solution is added y(x)=0. Bernoulli's equation can also be solved without making the transition to a linear equation by substitution, but using Bernoulli's method.

Differential equations in total differentials.

Definition. If in Eq. M(x,y)dx+N(x,y)dy=0(9.1) the left side is the total differential of some function U(x,y), then it is called a total differential equation. This equation can be rewritten as du(x,y)=0, therefore, its general integral is u(x,y)=c.

For example, the equation xdy+ydx=0 there is an equation in total differentials, since it can be rewritten in the form d(xy)=0. The general integral will be xy=c.

Theorem. Let's assume that the functions M And N defined and continuous in some simply connected domain D and have continuous partial derivatives in it, respectively, with respect to y and by x. Then, for equation (9.1) to be a total differential equation, it is necessary and sufficient that identity (9.2) holds.

Proof. The proof of the necessity of this condition is obvious. Therefore, we prove the sufficiency of condition (9.2). Let us show that such a function can be found u(x,y), that and .

Indeed, since , then (9.3) , where is an arbitrary differentiable function. Let us differentiate (9.3) with respect to y: . But, therefore, let us assume and then .So, the function is constructed , for which , a .

Integrating factor.

If the equation M(x,y)dx + N(x,y)dy = 0 is not a total differential equation and there is a function µ = µ(x,y), such that after multiplying both sides of the equation by it, we get the equation

µ(Mdx + Ndy) = 0 in total differentials, i.e. µ(Mdx + Ndy)du, then the function µ(x,y) is called the integrating factor of the equation. In the case where the equation is already an equation in total differentials, we assume µ = 1.

If the integrating factor is found µ , then the integration of this equation is reduced to multiplying both its sides by µ and finding the general integral of the resulting equation in total differentials.

If µ is a continuously differentiable function of x And y, That .

It follows that the integrating factor µ satisfies the following 1st order partial differential equation: (10.1). If it is known in advance that µ= µ(ω) , Where ω – given function from x And y, then equation (10.1) reduces to an ordinary (and, moreover, linear) equation with an unknown function µ on independent variable ω : (10.2), where , i.e. the fraction is a function only of ω .

Solving equation (10.2), we find the integrating factor, With= 1. In particular, the equation M(x,y)dx + N(x,y)dy = 0 has an integrating factor that depends only on x(ω = x) or only from y(ω = y), if the following conditions are met, respectively: , or , .

10. Properties of solutions of second-order LDEs (with proof). The 2nd order linear differential equation (LDE) has the following form: , (2.1)

where , , and are given functions that are continuous on the interval on which the solution is sought. Assuming that a 0 (x) ≠ 0, we divide (2.1) by and, after introducing new notations for the coefficients, we write the equation in the form: (2.2)

Let us accept without proof that (2.2) has a unique solution on some interval that satisfies any initial conditions , , if on the interval under consideration the functions , and are continuous. If , then equation (2.2) is called homogeneous, and equation (2.2) is called inhomogeneous otherwise. Let us consider the properties of solutions to the 2nd order lode.

Definition. A linear combination of functions is the expression , where are arbitrary numbers.

Theorem. If and is a solution to Lod, (2.3) then their linear combination will also be a solution to this equation.

When a real fluid moves, due to its viscosity, there are hydraulic resistances, which require energy to overcome. This energy turns into heat and is further dissipated by the moving fluid.

Bernoulli's equation for a stream of real fluid has the form

Where ─ pressure loss over a section length along the axis of the stream between two sections.

Bernoulli's equation for real fluid flow is:

(3.9)

Where
─ Coriolis coefficients, taking into account the difference in velocities at different cross-sectional points of a real fluid flow.

On practice
: for laminar fluid flow in round pipes
; for turbulent mode
.

Using the Bernoulli equation, most problems of practical hydraulics are solved. To do this, select two sections along the length of the flow, so that for one of them the values ​​of
, and for the other section one or values ​​were to be determined. With two unknowns for the second section, the equation of constant fluid flow is used υ 1 ω 1 = υ 2 ω 2 .

Hydraulic resistance

A moving fluid flow along its path overcomes the frictional forces of the fluid against the walls of a pipe or channel and various local resistances, as a result of which specific energy losses occur. There are two types of pressure losses:

Losses along the flow length ;

Losses to overcome local resistances
.

Total pressure losses are equal to the sum of all losses

(3.10)

Head loss along length

With uniform movement in pipes, pressure loss along the length, both during turbulent and laminar movement, is determined for round pipes using the Darcy formula

(3.11)

and for pipes of any other cross-sectional shape according to the formula

(3.12)

In some cases the formula is also used

(3.13)

Pressure loss due to friction along the length
, Pa, are determined by the formula

(3.14)

Where ─ length of the pipe or channel section, m;

─equivalent diameter, m;

─average current speed, m/s;

─hydraulic radius of the pipe, m;

─hydraulic friction coefficient;

─Chezy coefficient, related to the coefficient of hydraulic friction by dependencies

;

Depending on the driving mode, different formulas are used to determine the coefficient of hydraulic friction.

During laminar movement through round pipes, the coefficient of hydraulic friction is determined by the formula

(3.15)

and for pipes of any cross-sectional shape

(3.16)

Where A─ coefficient, the numerical value of which depends on the cross-sectional shape of the pipe.

Then the formula for determining the pressure loss along the length in laminar mode takes the form

(3.17)

For the first time, the most comprehensive works on the definition were given to I.I. Nikuradze, who, based on experimental data, constructed a dependence graph
from
for a range of values
. Nikuradze's experiments were carried out on pipes with an artificially specified roughness, obtained by gluing sand grains of a certain size to the inner walls of the pipeline. The results of these studies are presented in Figure 3.5, where the dependencies are plotted
from
for a range of values
.

Straight line I corresponds to the laminar mode of fluid motion in accordance with expression (3.15).

In turbulent mode, three areas of hydraulic resistance are distinguished, established as a result of experiments conducted by Nikuradze (see Figure 3.5)

Figure 3.5 ─ Nikuradze graph

The first area is the area of ​​small
And
, where the coefficient does not depend on roughness, but is determined only by the number
(marked in Figure 3.5 as straight II).

This area of ​​hydraulically smooth pipes. If the Reynolds number is in the range coefficient determined by the semi-empirical Blasius formula

. (3.18)

Topic 7

Analysis and application of Bernoulli's equation

1. Continuity equation in hydraulics. Consumption.

2. Analysis of the Bernoulli equation.

3. Energy meaning of the Bernoulli equation.

4. Limit of applicability of the Bernulia equation.

5. Examples of application of the Bernoulli equation.

5.1. Venturi flow meter.

5.2. Velocity measurement (Pitot tube).

5.3. Cavitation.

5.4. Toricelli's formula.

6. Continuity equation in hydraulics. Consumption.

7.1. Consumption. Continuity equation in hydraulics

Let us consider the steady flow between living sections 1,2 (Fig. 26).

where is the living cross-sectional area, is the average speed in the cross-section.

During this time, a volume of liquid flows out through the living section 2

where is the area of ​​the live section 2, is the average speed in section 2.

Since the shape of volume 1-2 does not change over time, the liquid is incompressible, the volume of the liquid must be equal to the volume flowing out.

Therefore we can write

This equation is called continuity equation.

From the continuity equation it follows that

Average speeds are inversely proportional to the areas of the corresponding sections.

7.2. Analysis of Bernoulli's equation

Let us write the Bernoulli equation for the steady motion of an ideal compressible fluid under the condition of its barotropy () in the field of mass forces

,

having integrated we have

.

For potential flow, the Bernoulli equation constant is constant over the entire flow region. In the vortex motion of an ideal fluid, the constant WITH in the Bernoulli integral retains a constant value only for a given vortex line, and not for the entire space, as in the case of irrotational flow.

Bernoulli's equation is one of the main ones in fluid dynamics, as it determines the change in the main parameters of the flow - pressure, velocity and height of the fluid.

Let's integrate Bernoulli's differential equation for the final section of the stream 1-2

.

The integral expresses the work of pressure forces to move a kilogram of liquid from region 1 with pressure R 1 to area 2 with pressure R 2 .

The value of the integral varies depending on the type of process (thermodynamic) that the liquid performs, that is, on the type of dependence.

Let's consider the isobaric process (Fig. 27)

In an isochoric process

For an incompressible fluid flowing without exchange of mechanical work with the external environment, we obtain, with from the Bernoulli equation

,

or multiplying by r

,

or dividing by rg

,

where the constants have the following physical meaning:

WITH- total mechanical energy of a kilogram of liquid or full pressure, ,

Total mechanical energy of a mass of liquid with a volume of cubic meter or full pressure, or Pa. ,

- total mechanical energy or full pressure in meters of column of a given liquid.

All three quantities have the same physical meaning; any of them is given a name full head.

The components of the total mechanical energy of a liquid are most clearly depicted and measured in meters of the liquid column,

g z,rgz,z- potential energy of the fluid position, measured from an arbitrarily selected horizontal leveling plane, or geometric head, ,

Potential energy of fluid pressure or piezometric head,,

-potential energy of the liquid or hydrostatic head,,

- kinetic energy of the liquid or express pressure, .

Piezometric head R can be measured from full vacuum p=0 or, for example, from environmental pressure. Absolute or excess pressure must be substituted in both sides of the equations.

The starting point for energy is arbitrary, but must be the same for both sides of the equations.

7.3. Energy meaning of Bernoulli's equation

Consists in establishing the law of conservation of total mechanical energy per unit mass of incompressible fluid

a) with potential flow for any point in space,

b) with a vortex - only along the vortex streamline and the elementary

This law is sometimes formulated as the three-height theorem.

Under the given conditions, the sum of three heights - geometric, piezometric and dynamic - remains unchanged.

In this case, the components of the total energy can be interconverted.

It should be borne in mind that the change in the kinetic energy of an incompressible fluid along an elementary stream cannot be specified arbitrarily: in accordance with the continuity equation, this change is uniquely determined by the change in the cross-sectional area of ​​the channel

Flow in a horizontal jet is of great practical importance; it is realized in engine nozzles. Let us write the Bernoulli equation at z= const

.

So, an increase in the speed of an incompressible fluid in a horizontal elementary stream is always accompanied by a decrease in pressure, and a decrease in speed is always accompanied by an increase in pressure up to v= 0. Therefore, high-velocity pressure is widely used, for example, for supplying water to the cooling system, breaking rocks, etc.

Due to the fact that the speed of an incompressible fluid can decrease only due to a change in cross-sectional area, we come to the important conclusion that the pattern of streamlines during the flow of an incompressible fluid uniquely determines not only the change in speed, but also the static pressure: when the streamlines become denser, the pressure decreases , with expansion it increases. This rule is widely used in analyzing the movement of fluid and its interaction with bodies.

7.4. Limit of applicability of the continuity and Bernoulli equations

When fluid flows through a channel at constant , and at arbitrarily variable area 2. It would seem that

.

However, according to the Bernoulli equation at

,

pressure would have to take the value minus infinity, which makes no sense: absolute pressure cannot be less than zero.

Thus, the continuity and Bernoulli equations are valid only as long as the minimum pressure in the flow remains greater than zero.

Bernoulli's equation is considered one of the basic laws of fluid mechanics; it establishes a connection between the pressure in a fluid flow and the speed of its movement in hydraulic systems: with an increase in the speed of the flow, the pressure in it must fall. It helps explain many hydrodynamic effects. Let's look at some well-known ones. The lifting and spraying of liquid in a spray bottle (Fig. 1) occurs due to the reduced pressure in a stream of air passing at high speed over a tube lowered into a vessel with liquid. The liquid rises upward due to atmospheric pressure, which is greater than the pressure in the air stream.
A ping-pong ball (Fig. 2) floats steadily in a vertical stream of air, since the pressure in the stream is less than atmospheric pressure, which presses the ball against the stream, preventing it from falling.
Ships traveling on a parallel course (Fig. 3) are attracted to each other, which is the cause of many maritime disasters. This is explained by a decrease in pressure between ships due to the higher speed of water in the narrowed space between them.
The lift of the wing (Fig. 4) is due to the presence of a pressure difference p1 And p2 due to speed difference V1 And V2, When V1 less V2, since air particles located above the wing travel a longer distance before meeting at the end of the wing than particles located below.
If you blow between two sheets of paper touching each other (Fig. 5), they will not separate, as it would seem that should happen, but, on the contrary, will press against each other.
Thus, we see that Bernoulli's equation has a wide range of applications to explain many hydrodynamic phenomena. Daniel Bernoulli published it in 1738 after many years of thought and research, search and doubt. He was absolutely confident in the correctness of the law he discovered, connecting the static pressure in a liquid with the speed of its movement.
Let us consider the derivation of this equation for an elementary stream of liquid (streamline), as it is given in all textbooks, for a stationary laminar flow of an ideal incompressible fluid. To eliminate the influence of gravity on the movement of liquid, we take a horizontal section of the pipe (Fig. 6), and also place the elementary stream horizontally.
Let us consider the motion of a fluid element determined by the length l1. The selected part of the liquid will be affected by the driving force created by the static pressure p1:
, (1)
Where S1- cross-sectional area on the left side of the selected section of liquid, and the resistance force determined by the static pressure p2:
, (2)
Where S2- cross-sectional area on the right side of the site.
The pressure acting on the lateral surface of the fluid element, according to the authors, is perpendicular to the displacements and will not do any work.
Under the influence of these two forces, the released part of the liquid will move from left to right. Let us assume that it moves some short distance and takes a position determined by the length l2, while the left end of the fluid element will move by the amount D l1, and the right one by the value D l2.
In accordance with the laws of mechanics, the movement of a fluid element will be characterized by the fact that the change in its kinetic energy will be equal to the work of all forces acting on it:
, (3)
Where m- the mass of the selected fluid element, and - the final and initial velocities of its center of mass.
The right side of expression (3) can be transformed if we pay attention to the fact that in both positions of the selected element there is a common part (not shaded in Fig. 6), which will have the same kinetic energy. This part of the energy can be entered into equation (3) by adding and subtracting it on the right side:
(4)
Where mtotal- mass of the common part, - speed of the center of mass of the common part.
The expressions in parentheses represent the kinetic energies of the shaded areas of length D l1 and D l2, moving due to their small extent with constant velocities for all points V1 And V2. Therefore, equation (4) will take the form:
, (5)
Where Dm1 And Dm2- masses of shaded areas of liquid.
Due to the continuity of fluid flow, the volumes and masses of the shaded parts will be equal:
, (6)
Where r- liquid density.
Dividing expression (5) by S1Dl1=S2Dl2, transform it to the form:
(7)
After rearranging the terms, the equation will take the form:
(8)
This is Bernoulli's equation. Since a fluid element can be taken anywhere in the flow and of any length, Bernoulli's equation can be written as follows:
, (9)
where p and V are the static pressure and the speed of movement at any place in the elementary stream of liquid. Expression rV2/ 2 is called dynamic pressure.
From equation (9) it follows that at those points where the speed is greater, the static pressure will be less and vice versa. That this is indeed the case is confirmed by experience. Let's take a Venturi tube as an example (Fig. 7). The fluid levels in the gauge tubes clearly show that the static pressure is less at the constriction end where the flow rate is greater. In addition, this can be confirmed by the fact that the result obtained, as stated in the work, is a direct consequence of Newton’s second law. Indeed, when a fluid moves from a wide part to a narrowed part, its speed increases and the acceleration is directed in the direction of movement. And since the acceleration is determined by the difference in pressure acting on the fluid element on the left and right, the pressure in the wide part of the tube should be greater than in the narrow part. True, here you can notice that acceleration is determined not by pressure, but by force, and force depends not only on pressure, but also on the cross-sectional area. Therefore, greater force can be obtained with less pressure, so the argument presented is not convincing.
So, everything seems to be logical in the above reasoning. However, it is possible to explain all hydrodynamic effects differently. The fact is that we are always dealing not with an ideal, but with a viscous liquid that behaves completely differently.
Let's consider what will happen to a viscous liquid flowing through a pipe (Fig. 8). Due to the presence of friction between the liquid flow and the walls of the pipe, as well as between the layers of the liquid itself, the speed of liquid particles will be different at different points in the same section of the flow: in the center of the pipe it will be maximum, near the walls it will be zero. As a result, the velocity field in the cross section of the fluid flow will be determined by the expression:
, (10)
Where V- speed at the center of the flow, r- current radius, R is the radius of the pipe, and will have the form shown in Figure 8. Inextricably linked with the velocity field is the scalar field of kinetic energy, which is characterized by the expression:
, (11)
Where Edm- kinetic energy of the released elementary mass dm, which is determined by the expression:
(12)
Here: dl- elementary length in the axial direction, r- liquid density.
Since the kinetic energy field is non-uniform, an elementary particle of the liquid will be acted upon by a force directed towards the center of the flow:
(13)
This force, related to the cylindrical part of the particle surface dS, normally located to the force:
, (14)
will determine the pressure arising at a given point in the flow under the influence of a given force:
(15)
This pressure depends only on the elementary force dF, so it can be called differential pressure. The total pressure at a given point in the liquid will depend on the elementary forces of inertia acting on other particles of the liquid. Because all the strength dF have a radial direction and are directed towards the center of the flow, the total pressure at the point will be determined by forces lying on the same radius and located on the side external to the point in question. Therefore, the total pressure can be found by integrating expression (15) over r ranging from r before R:
(16)
Here the minus sign indicates the direction of compression (toward the center of the section).
The result was surprising, since this expression is similar to the expression for kinetic energy (11), related to the volume of the elementary mass dm:
, (17)
those. the total pressure is the density of kinetic energy in a certain elementary volume in the vicinity of the point in question.
From expression (16) it follows that on the flow axis (at r=0) the pressure will be maximum, and at its boundary (at r=R) it will be equal to zero.
Under the action of radial forces, the flow will be compressed towards its axis, as a result of which the pressure on the pipe walls will decrease, i.e. a negative pressure will appear, the value of which can be found as the radial average of expression (16). To do this, we integrate it over the range from 0 to R and divide by R:
. (18)
The same result will be obtained if, using expression (13), we find the force acting on the elementary area of ​​the surface of the pipe itself and directed to the center line of the pipe, for which this expression, taking into account expression (12), must be integrated in the range from 0 to R:
(19)
Dividing this force by the size of the elementary area:
, (20)
we obtain the value of negative pressure on the inner surface of the pipe:
.
Due to this pressure, the static pressure near the pipe walls will decrease. The resulting static pressure is determined by the expression:
(21)
Since the magnitude of the negative pressure depends on the square of the velocity, it is quite natural that its value will be significantly greater in the narrow part of the flow than in the wide one. Therefore, in a Venturi tube in its narrow part, pressure gauges will show lower pressure than in the wide part. The dependence of the magnitude of negative pressure at the pipe walls on the speed of movement for water is shown in Figure 9.
As another example, we can consider the principle of operation of a spray gun, when a gas stream sucks in the liquid in a vessel (see Fig. 1). It is believed that the liquid is sucked in due to the fact that the pressure in the gas stream due to its speed becomes lower than atmospheric pressure, which squeezes the liquid out of the vessel, and the gas stream carries it along with it. However, the same effect will be caused by the presence of negative pressure, caused by the presence of a non-uniform field of kinetic energy in the flow of a gas stream escaping from the spray nozzle. In addition, the jet will carry along particles of the surrounding air, which will lead to the appearance of its own kinetic energy field, the gradient of which will be the reason for the absorption of liquid from the vessel.
Then the question arises: if the reason for the decrease in pressure in the Venturi tube and suction in the spray gun may not be a decrease in pressure in the flow of moving liquid or gas, then how can we understand the essence of Bernoulli’s equation? After all, the speed of the liquid in the narrowed part of the flow actually increases, and this, it seems, is possible only with a decrease in the counteraction, and experiments show that the pressure in the flow can be lower than atmospheric, since in the manometric tube the liquid rises above the level corresponding to atmospheric pressure (Fig. . 10). But on the other hand, it is also undeniable that narrowing the flow should increase the resistance to movement, and therefore increase the pressure inside the fluid flow. In this case, an increase in flow speed can only occur due to an increase in the driving force, i.e. pressure on the left side of the highlighted flow element. Indeed, a similar conclusion can be made if we turn to equation (7):

We must not forget that this equation applies to the entire volume of liquid we have isolated, which we consider as a whole. Therefore, it is impossible to separate it, as is done in expression (9). This is very important to remember. From expression (7) it follows that with increasing speed V2 at constant speed V1 the pressure difference will increase p1 And p2. This increase can occur either due to a decrease p2, and due to an increase p1. When analyzing the Bernoulli equation, they prefer to talk about a decrease in pressure p2. But what is pressure p2? This is the pressure that prevents the movement of a liquid or gas. How is it determined? Let's take as an example a conical nozzle for a pipeline (Fig. 11). It is clear that the back pressure p2 The pressure cannot be less than atmospheric pressure, otherwise the liquid will not flow out of the nozzle. If we want to increase the flow rate of liquid at a given nozzle, then, in accordance with equation (7), we must increase the pressure p1. But that is not all. Since the speed V1 And V2 interdependent, with increasing speed V2 the speed will also increase V1, and then the pressure difference p1 And p2 should decrease, which corresponds to an increase in pressure p2 at constant pressure p1.
Thus, analysis of the Bernoulli equation reveals a problem in understanding its essence. In order to better understand this problem, let us apply equation (7) to study the movement of liquid in a conical nozzle (see Fig. 11). From the condition of flow continuity it follows that the velocities in sections 1 and 2 are related by the relation:
, (22)
Where R1 And R2- radii of cross sections in sections 1 and 2.
Substituting this speed value into expression (7) and solving it for speed V2, we get:
(23)
Let's analyze this expression. Let us take the limiting relations R2/R1. At R2/R1=0 speed V2 will be equal to:
, (24)
whereas it is absolutely clear that it should be equal to zero. True, common sense dictates that pressure p1 And p2 in accordance with Pascal's law, they must be equal, and their difference must be equal to zero. However, this circumstance does not follow from expression (24).
At R2/R1=1 speed V2 will be equal to infinity:
, (25)
which, of course, cannot be true. However, here too you can find a way out by declaring that the pressure p1 And p2 will also be equal, since the speed must be constant. However, we will not be able to find the magnitude of the speed V2, since it will be determined by the ratio of zeros.
But what about intermediate values ​​of the ratio? R2/R1? The pressure difference can't p1 And p2 be equal to zero all the time. How will this difference change? There is no answer to these questions. Only one thing becomes clear: Bernoulli's equation, even for an ideal fluid, is not accurate and cannot be used to calculate velocities or pressures; there is something missing in it. This is the question that needs to be dealt with, and with digital calculations.
Such calculations, although approximate, exist for the outflow of liquid from a tank (Fig. 12). The Bernoulli equation in this case, taking into account the potential energy from the weight of the liquid, has the form:
(26)
where g=9.81 m/s2 is the acceleration of gravity, and the coordinates z 1 and z 2 are counted from some arbitrary level, since when solving the problem, only their difference is needed: H=z 1 - z 2 . It is accepted that V1=0, since V1<<V2, then from expression (26) it turns out:
, (27)
Where p2 equal to atmospheric pressure.
If p1 will be equal p2, then formula (27) will take an even simpler form:
, (28)
whence it follows that the speed of liquid outflow is equal to the speed of free fall of a solid body from a height H.
This expression was obtained by Toricelli 100 years before Bernoulli and is therefore called the Toricelli formula.
However, even here, despite the obviousness of the derivation of this equation, questions arise that have no answer: will, for example, the speed of liquid flow depend on the size of the hole or on the size of the conical nozzle that can be attached to the tank (see Fig. 12,b )? Could the flow of liquid through a small hole be similar to its free fall? This, of course, is very doubtful even for an approximate determination of speed.
To simplify the analysis of this problem, let us take a vertically located conical tank (Fig. 13), into which liquid flows and flows out, and so that its level remains the same all the time. Taking into account relation (22) from the Bernoulli equation we obtain:
(29)
From this expression it follows that when R2/R1=0 speed V2 will be equal to zero only if:
, (30)
from which follows:
, (31)
which does not follow at all from the conditions of the problem.
At R2/R1=1 V2=¥ , although it is quite obvious that the liquid will fall when counteracted by external pressure, which will be equal to atmospheric pressure: p2=p0, and the rate of fall should have a very specific value.
Thus, we have established that the pressure p2 in a fluid flow should vary depending on the ratio R2/R1 within:
, (32)
the law of change of which we do not know.
To establish this relationship, let us first consider a closed conical vessel in which the gas is under some pressure (Fig. 14). In this case, the weight of the gas, due to its smallness, can be ignored. In accordance with Pascal's law, the gas pressure at all points of the vessel will be the same. We will assume that the pressure in the vessel is created from the side of the first section by the force F1, the value of which will be equal to:
, (33)
Where S1- cross-sectional area in the first section. In the second section, the gas will act on the bottom with a force F2, equal to:
, (34)
Where p2=p1, S2- bottom area.
Since the area S2 less area S1, force F2 there will be less power F1. It is quite obvious that the difference between these forces:
(35)
will be compensated by resistance from the side walls of the vessel.
Thus, the narrowing of the vessel provides additional resistance to force F1, as a result of which less force will act on the bottom.
Now let's remove the bottom of the vessel. Since the gas in the vessel will be under pressure greater than atmospheric pressure, it will begin to flow out of the vessel at a certain speed. This movement can only occur due to a decrease in gas pressure, since the kinetic energy of gas movement can only appear due to the potential energy of its pressure. It is obvious that in this case the relationship between the pressure in the first and second sections should change, since the speeds of movement of gas particles in them will be different and therefore the amount of potential energy (pressure) converted into kinetic energy of motion will also be different.
Now all that remains is to guess how the pressures in both sections will change if the gas velocities in them are respectively V1 And V2, and the static pressure p1 will be maintained at a constant level. Since the source of motion is only gas pressure, due to a decrease in the potential energy of which the energy of motion appears, it is quite reasonable to use the law of conservation of energy, assuming that there are no energy losses. By the way, when deriving his equation, Bernoulli also used this law, since all the work of pressure forces turned into kinetic energy of motion.
In accordance with the law of conservation of energy, the static pressures in the first and second sections will become less than the initial ones by the amount of volumetric kinetic energy densities in them:
; (36)
, (37)
because p2=p1.
From these relations it is clear that we are establishing a connection between pressures and velocities in both sections, and the pressure in the second section will depend on the pressure in the first section. Speeds V1 And V2 are also interdependent. So it can be argued that the pressures are interdependent.
If we add to the pressures the losses of potential energy, converted into kinetic energy of motion and , then the static pressure in the first and second sections will be equal to each other and equal p1, i.e.:
, (38)
which is an analogue of Bernoulli's equation.
Thus, we have obtained Bernoulli's equation based on the law of conservation of energy for a steady flow of an ideal fluid. In essence, we have expanded the scope of Pascal's law by extending it to a moving fluid.
Due to the change in pressure in the first and second sections, the forces acting in them also change. In accordance with expressions (36) and (37), the magnitude of these forces will be equal to:
; (39)
(40)
Let's see what happens with the counterforce D.F. Defining it as the difference in forces and , we find:
, (41)
from which it follows that the counteraction force from the walls increases.
From the example considered and the assumptions we made, the following conclusions can be drawn.
Firstly, any narrowing of the channel through which a liquid or gas moves exhibits resistance to this movement, the magnitude of which depends on the degree of narrowing, i.e. the greater the narrowing, the greater the resistance. And the presence of this resistance will not depend on which channel the liquid flows through - through a wide pipe or in an elementary stream. The amount of resistance will also depend on the ratio of flow velocities in different sections, as follows from formula (41). When deriving the Bernoulli equation, this resistance is not taken into account.
Secondly, the pressure in the second section depends on the pressure in the first section, equal to:

The pressure in the second section will also depend on the speed of the fluid flow, decreasing by an amount. It follows from this that pressure is not an external resistance in relation to a selected element of the liquid, it is an internal property of the part of the liquid in question. And this, in essence, is the pressure that the released element of the liquid exerts on the subsequent, discarded part of the liquid, i.e. creates a force that causes movement of subsequent sections of liquid. And what is very important, this pressure will not directly depend on the pressure external to the selected liquid element from the side of the discarded subsequent part of the liquid, which we denote by . Here the dependence will be indirect: speeds will depend on pressure V1 And V2, and already from the speed V2 pressure will depend. It should be noted that one of the components of pressure will generally be environmental pressure, in particular atmospheric pressure. This directly follows from the fact that the pressure in a fluid flow cannot be less than atmospheric. Thus, from all of the above it follows that when deriving the Bernoulli equation, pressure should not be taken into account as the reason for the appearance of the resistance force - the resistance force will be created only by pressure.
Third, the drag force D F, arising due to the narrowing of the channel, is determined only by the difference in forces in the first and second sections and directly counteracts the force, i.e. we can assume that it is applied in the first section. Because force is determined by pressure, dependent on pressure p1, then the opposing force D F also depends on pressure p1 and, therefore, is, as it were, a self-braking force of the fluid flow when it moves in the narrowed part. Therefore, when deriving the Bernoulli equation, the force D F, firstly, must be taken into account, and secondly, to determine its work must be multiplied by the movement of the left end of the liquid D l1.
In conclusion, it should be said that all the conclusions we made became possible because we considered the movement of the selected fluid element as a single whole body, and not two small sections located at its ends. It is quite obvious that this approach most accurately meets the task at hand.
Now let's return to considering the problem of the outflow of water from a conical tank (see Fig. 13). In a tank with liquid, there is pressure in the second section, by which the reaction force will be determined DF except pressure p1 will also be determined by pressure rn created by the weight of the liquid:
, (42)
Where N- the height of the liquid column, measured from its upper level, in connection with which expressions (36) and (37) will take the form:
; (43)
(44)
In connection with the above, it is possible to determine the forces acting on the selected fluid element:
; (45)
; (46)
(47)
In addition, we must take into account the resistance force from the discarded subsequent part of the liquid:
, (48)
where in this case will be equal to atmospheric pressure ro.
When composing the equation of motion for the volume of liquid under consideration, we must take into account only the forces and , since it was shown above that the force is not a resistance force. It was also shown that when finding the work of forces and D F they must be multiplied by the movement of fluid in the first section - D l1. It remains to clarify the question of how to deal with the resistance force: what displacement D l it must be multiplied by D l1 or D l2? To solve this problem, let's combine forces D F And :
(49)
from which we obtain that the second expression in brackets represents the excess fluid pressure in relation to the pressure in the second section:
(50)
It follows that the work of a force should be determined by multiplying it by displacement Dl1.
Thus, the equation of motion in the form of the law of change in kinetic energy for this problem is determined by the expression:
(51)
After substituting the corresponding values ​​of the forces determined by expressions (45) and (49), expression (51) is transformed to the form:
(52)
which after division by the product S1 D l1 and the corresponding transformations will take the form:
(53)
Expressing speed V1 through speed V2 in accordance with expression (22) and solving equation (53) regarding speed V2, we get the calculation formula:
(54)
Let's analyze this formula. At R2/R1=0 speed V2 will be equal to zero, since the numerator will be equal to zero and the denominator to one. At R2/R1=1 speed V2 will be equal to:
, (55)
which coincides with expression (27). And this expression will in this case really correspond to the free fall of the liquid, since R2=R1. At intermediate values ​​of the ratio R2/R1 speed V2 will have a meaning corresponding to this relationship. The results of calculating this speed at values ​​=== n/m2 and at N=10.2 m are presented in Figure 15. As one might expect, with increasing ratio R2/R1 the speed smoothly increases from zero to the maximum value corresponding to free fall. In addition, using formula (44) one can find the pressure in a stream of liquid flowing from a conical tank. Analysis of this formula shows that when V2=0 the pressure in the liquid will be equal to:

and at , which corresponds to free fall, =. The calculated curve for pressure =+= is presented in Figure 15, from which it can be seen that the pressure in the outflowing jet will be greater than atmospheric pressure for all radius ratios R2/R1, except for the case when these pressures are equal.
To make everything stated more convincing, we will give another derivation of the equation of motion, taking into account the inertial forces acting on the selected element of an ideal fluid. In this case, based on the laws of mechanics, the forces acting on the fluid element in question will be in equilibrium.
To determine the inertial force, consider part of the conical channel through which the liquid moves (Fig. 16). Let us select the elementary volume of liquid dm, which will move from the first position to the second, changing the speed of its center of mass from value to value . The resulting elementary inertial force can be determined by the formula:
, (56)
Where
, (57)
and the minus sign shows the direction of the inertial force.
The relationship between the velocities in the two considered positions of the elementary mass dm is determined by the expression:
, (58)
Where
(59)
Using this relation we get:
(60)
Raising the binomial to the fourth power and dividing each term by D ls and then accepting D ls equal to zero, we find an expression for the elementary force of inertia:
(61)
Let us assume that the point Si is at a distance l from the first section, then the ratio of the velocities and radii of the sections at these points will have the form:
; (62)

(63)
Substituting these values ​​of speed and radius into expression (61), we obtain:
(64)
Now it is necessary to sum up the elementary forces of inertia over the entire selected volume of moving fluid, i.e. by lenght l. Substituting the mass value into expression (64) dm:
(65)
and taking the integral of expression (64) ranging from 0 to L, let us find the inertial force acting from the entire moving mass of fluid on the first section where the driving force is applied F1:
(66)
Where .
From expression (66) it follows that the inertial force is actually applied to the first section, since the difference in energy densities in the second and first sections (the expression in parentheses) is multiplied by the area of ​​the first section.
Thus, the following forces will act on the released volume of liquid:
;
;
;
, (67)
under the influence of which this volume of liquid, considered by us as a single body, in accordance with the laws of mechanics, will be in equilibrium, i.e. the following condition will be met:
, (68)
which, after substituting the values ​​of all forces, is transformed to the form:
(69)
After reducing terms and dividing by S1 expression (69) will take the form:
,
which completely coincides with the previously obtained expression (53). Therefore, our reasoning was fair, and the resulting formulas for determining the speed V2 and the pressures are correct.
Thus, it would seem that we have solved the problem of finding the speed of fluid flow. However, if we comprehend the situation from the point of view of the laws of mechanics, doubts arise about the validity of the resulting formulas. Indeed, if, as an example, we look at a vertically falling flow of liquid flowing out of a pipe of constant cross-section (Fig. 17), then we can immediately notice that the flow of liquid, even outside the pipe, moves as a single body with the liquid in the pipe and, therefore, in all its points must have the same speed. If this does not happen, the flow will break, since when falling under the influence of gravity, the speed must continuously increase. However, in practice such a gap is not observed. This circumstance is due to the presence of adhesion forces (cohesion) between liquid molecules, and these forces can be quite large. So for pure water without impurities, its tensile strength reaches 3107 N/m2, which corresponds to 300 atm or a water column of 3000 m. It is quite obvious that cohesive forces must exist in an ideal liquid. Therefore, when any fluid element moves r m on it except gravity Fstrand the resistance force will also act Fresistance from the upper parts of the liquid and the driving force Fdv from the lower side. As a result of the free fall of the fluid element r m will not, and the element itself, under the influence of forces applied to it, will experience tensile deformations, due to which it will be compressed in the transverse direction, and the entire flow as a whole will narrow (in Figure 17, the narrowing of the flow is shown by dash-dotted lines). Due to this narrowing, the speed of the element dm as it falls it should change, and neither the speed V1, nor speed V2 are not known to us, and, as follows from our reasoning, cannot be found using the above formulas.
In order to somehow get out of this situation, let us take into account, at least approximately, the effect of the outflowing part of the flow external to the pipe on the liquid located in the pipe. This external influence will be pulling, i.e. it will create some additional pressure rd in the flow, facilitating its movement. The magnitude of the external pulling force will be determined by the weight of the liquid column located outside the pipe. Since the flow narrows as it falls, the weight of the liquid column will be equal to the weight of the water cone (Fig. 18):
, (70)
Where mh- mass of the liquid column, R2 And Rh- radii of the column at the beginning and at the end of the considered part of the flow. Pole height h, obviously depends on the given height of the flow fall, for example, into some vessel, or the loss of adhesion between the particles of the liquid when it thins out, when the flow begins to disintegrate into individual drops. We will be given the value h arbitrarily, without considering situations critical to the decay of the jet, since this issue requires special research.
To find the weight of a liquid column, it is necessary, with a known radius R2 find the radius Rh, corresponding to the fall height h. To approximately determine this radius, consider the fall of some liquid element with a mass Dm from high h under the influence of its own weight only, although it will be subject to adhesion forces from both the upper and lower sides, the ratio between which will change as the selected element falls.
In accordance with Newton's second law, we will have:
(71)
We solve this equation with initial conditions:
(72)
As a result we get:
; (73)
(74)
From expression (74) we find the fall time t:
(75)
Substituting this value t into expression (73), we obtain the dependence of the rate of fall Vh from coordinate h:
(76)
Using the flow continuity condition:
, (77)
we get:
(78)
In Fig. Figure 19 shows the shapes of liquid jets obtained as a result of calculations of the ratio Rh/R2 according to formula (78) for exhaust velocities V2 equal to 0.1 m/s and 0.5 m/s, depending on the height of the fall h. From the figures it is clear that at a low outflow velocity the narrowing of the jet will be more sharp.
To take into account the influence of the additional driving force on the speed of the flow and the pressure inside it, it must be taken into account in the equations we obtained. This can be done by assigning it to the first section, where the driving force determined by pressure acts p1 and cross-sectional area S1. Then the pressure created by this additional force will be equal to:
(79)
It is more convenient to present this expression in the form:
, (80)
because then the attitude Gh/S2 will take a simple form:
, (81)
and expression (80) is transformed to the form:
(82)
Then the calculation formulas for velocities and pressures in the second section, taking into account adhesion, will be determined in accordance with the formulas we obtained earlier by the following expression:
; (83)
(84)
At R2/R1=1 formula (83) will take the form:
, (85)
and when ==:
, (86)
Figures 20 and 21 show the results of calculations of velocities and pressures without taking into account and taking into account adhesion inside the liquid at a height of the conical vessel from which the liquid flows at 10.32875 m and 1 m. The first height corresponds to atmospheric pressure. In both cases the height h was taken equal N And N/R1=10, =.
As can be seen from the curves, the flow rate can increase significantly due to the fall height h. This will bring the value of the outflow velocity closer to the result determined by the Toricelli formula. The pressure inside the jet will increase, since part of the lost pressure (potential energy) due to an increase in flow speed is compensated by added pressure. However, with free fall of liquid at R2/R1=1 pressure in both cases becomes equal to atmospheric pressure.
Thus, the formulas we obtained can be used to approximately determine the flow velocities in its various sections, and these velocities will largely depend on the magnitude h(see Fig. 22, a and b).

It also seems interesting to consider the problem of the upward movement of a fluid flow at the outlet of a pipe (Fig. 23). In this case, in section 2-2, an additional resistance force will act on the flow, equal to the weight of the outer part of the liquid flow with a height h. This force will create additional pressure in the second section, the value of which will be approximately equal to:
(87)
(we assume that the flowing column of liquid has a cylindrical shape).
This pressure will be included as a component in pressure, which is included in the calculation formulas. Then the pressure will be determined by the expression:
(88)
It is quite obvious that the speed V2 it will decrease. However, to calculate V2 need to know the lift height h, which, in turn, depends on the exhaust speed V2. That's why h should somehow be expressed in terms of speed V2. We will reason as follows. Flow element r m in section 2 it has some kind of kinetic energy, which in the upper part of the flow turns into potential. Therefore, the following relationship must be satisfied:
, (89)
from where we get:
(90)
Then the pressure will look like:
(91)

This pressure value should be substituted into the original equation (53), which, after solving it with respect to V2 will give the following expression:

(92)
For a pipe of constant cross-section, i.e. at R2/R1=1, this expression will take the form:
, (93)
and when p1=p0 we get:
(94)
Substituting this speed value into expression (90), we find:
(95)
Thus, the height of the liquid rise will be two times less than the difference in its levels H. Note again that these will be approximate values ​​for speed V2 and lifting heights h, since the cross-section of the external flow should not remain constant: it should increase with distance from the outlet due to the drop in speed and the condition of continuity of its flow. In addition, the value of the cross section of the flow will be influenced by the downward part of the flow, which will create a pulling force that increases the speed of the flow.
Estimated speed values V2, pressure and altitude h water rise are presented in Figures 20 and 21 for two cases when N=10.32875m and N=1m. The pressure in this case is determined by the usual formula:

Since the flow rate in this case will be less due to the presence of additional resistance of the water column, the pressure will be greater than when the liquid flows downwards, if we do not take into account the presence of additional force due to the adhesion of liquid particles.
Let us now consider the motion of not an ideal, but a real viscous fluid. The braking of liquid layers against the pipe walls and among themselves leads to a decrease in the speed of movement of liquid particles and, consequently, to the loss of some part of the kinetic energy of the flow. To determine the kinetic energy of the flow, we define the law of velocity change along the radius of an arbitrary section in the form:
, (96)
Where Vl And Rl- respectively, the fluid velocity on the flow axis and the cross-sectional radius at a distance l from the first section. Kinetic energy should be determined from the average flow rate, which can be found from the volumetric flow rate of the liquid Q:
, (97)
Where Sl- cross-sectional area at a distance l. From expression (97) we have:
(98)
We will find the volumetric flow rate using expression (96) for elementary annular sections, the area of ​​which is determined by the expression:
, (99)
Where dr- ring width. In accordance with this, the elementary volumetric flow rate will be equal to:
(100)
By integrating this expression from 0 to R, we obtain the total volumetric flow rate of the liquid in the section l:
(101)
Using formula (98), we find the average flow velocity in the cross section l:
(102)
Kinetic energy of the flow in a certain area D l in this case it will be equal to:
, (103)
where D m-corresponding to length D l mass of a section of liquid.
Equation of motion of a selected volume of liquid in the form of a sum of forces taking into account the friction force Ftr will be determined by the expression:
(104)
This expression takes into account the cross-sectional average flow velocities in sections 1 and 2. The friction force must be determined from existing experimental data.
Having made the necessary transformations, we reduce expression (104) to the form:
(105)
where do we find the speed? V2:
, (106)
Where
(107)
pressure loss along the length L=H(pressure decreases by this amount p1 in section 2).
Analysis of this expression shows that when R2/R1=0 speed V2 will be equal to zero, and when R2/R1=1 expression (107) will take the form:
(108)
The average flow velocity in the second section will be two times less.
The pressure value in the second section will decrease due to the loss of energy to overcome friction forces and will be determined by the expression:
(109)
When a liquid moves downwards, intermolecular cohesion must be taken into account. Then the speed V2 will be determined by the expression:
(110)
When liquid flows vertically upward, the pressure, as shown above, can be represented by the expression:
(111)
Then the expression for speed V2 will take the form:
(112)
The pressure inside the liquid as it moves down and up will be determined by expression (109), only the speed V2 they will naturally be different. This means the pressures will be different.
The pressure inside the liquid, taking into account its compression, will, in accordance with formula (18), be greater by the amount of average negative pressure:
,
the near-wall pressure is less by this amount, i.e.:
; 113)
(114)
To calculate the speed of fluid flow and the pressure inside it, taking into account the friction force, it is necessary to determine the friction force. To do this, we use the Poiseuille formula, which determines the fluid flow rate in a laminar flow regime:
, (115)
Where Q- liquid flow rate in m3/s, p1-p2- pressure drop in the fluid flow over a section of a cylindrical pipe of length L in N/m2, m- dynamic viscosity of the liquid in kg/ms, d- pipe diameter in m.
Using this expression, you can find the average speed over the cross section of the pipe:
, (116)
where, as noted above, the average speed is equal to half the maximum axial speed V.
Using expression (116), we find the pressure loss due to friction along the length L:
(117)
Since we are considering a vessel (pipe) of variable cross-section, we write expression (117) in differential form:
, (118)
Where Vl- axial velocity in a section located at a distance from the first section l, Rl- radius of this section, dl- elementary length of the section corresponding to the elementary pressure loss dp(Fig. 24).
For further transformations we use the condition of flow continuity:
,
where we find:
, (119)
Where
(120)
Using these expressions, we get:
(121)
By integrating the resulting expression over l ranging from 0 to L, let's find the pressure loss along the entire length L:
(122)
Since the expression in parentheses is:
, (123)
a tg a is determined by the expression:
, 124)
formula (122) is transformed to the form:
(125)
Let's express the speed V1 through speed V2, using the flow continuity condition:
(126)
and reduce expression (125) to the form:
(127)
Using the formulas obtained, three calculation options were made for the following sizes of conical pipe:
1) H=L=10.32875 m (which corresponds to atmospheric pressure);
2) H=L=1.0 m;
3) H=L=0.1 m
In all cases the ratio H/R1 was taken equal to 10, h=H, water was taken as a liquid, for which the coefficient of dynamic viscosity m equal to 0.001 kg/ms. Calculations showed that for the selected pipe sizes, the average water flow speed in the presence of viscosity was practically no different from the speed of an ideal liquid, presented by the graph in Figure 15. This is due to the small value of the coefficient m. The pressure in the jet, without taking into account the adhesion between molecules and its compression, due to the presence of a gradient of the kinetic energy field, will also be the same as for an ideal liquid. If these factors are taken into account, then the pressure inside the jet can increase significantly, and the near-wall pressure can decrease, becoming less than atmospheric and even negative. The calculation results for three options are presented in Figures 25-27. The figures show curves characterizing the change in pressure and in
relation functions R2/R1, when the flow moves downward without taking into account the clutch
interactions between liquid molecules (curves 1), when the flow moves downwards taking into account molecular cohesion (curves 2) and when the flow moves upwards (curves 3). It can be seen from the curves that pressure changes are most significant for larger pipe sizes and can therefore be easily observed.
Thus, we examined how the flow speed and pressure inside it change when a liquid flows through a pipe of variable cross-section. Calculations show that the pressure in the viscous liquid at the outlet of the pipe will be greater than atmospheric pressure. Obviously, this pressure will be greater than atmospheric pressure for some time even when the liquid moves outside the pipe. Let's take a closer look at this issue.
If the pressure in the liquid at the exit from the hole is greater than atmospheric pressure, then the jet should immediately expand at the exit, but this, however, does not happen; the jet even contracts. The reason for this has already been discussed by us. Firstly, this is explained by the conservation of the gradient of the kinetic energy field, due to the difference in velocities in the center and along the edges of the flow, which have not yet leveled off. The force determined by the gradient will continue to compress the flow. Secondly, the fluid flow will be compressed by the force generated by the movement of air carried along by the fluid stream. In this case, a kinetic energy field will also appear in the air stream, the gradient of which will determine the acting force.
Let us determine the pressure with which air compresses a stream of liquid. Figure 28 shows the pattern of the velocity field in the air, which can be characterized by the expression:
, (128)
Where r- distance from the center of the jet.
Then the kinetic energy of some elementary mass dm will be equal to:
, (129)
Where
(130)
Here: rв- air density.
The derivative of this expression will determine the elementary force dFв:
,(131)
directed towards the center of the flow.
The ratio of this force to the elementary surface dS=rdjdh, corresponding to the elementary mass, will determine the differential pressure dpv:
(132)
(we omit the minus sign).
The total pressure acting on the elementary mass from all air particles external to it will be determined by the integral of expression (132), taken over r in the range from r to:
(133)
On the surface of the jet ( r=Rh) the air pressure will be equal to:
(134)
Thirdly, the jet will be compressed due to the presence of tensile forces caused by the adhesion between liquid molecules, as well as, as noted above, by an increase in the falling speed under the influence of gravity.
Fourthly, the jet will compress due to the presence of surface tension.
Thus, several forces will act on a stream of liquid flowing from a pipe, the combination of which will determine both its shape and the pressure in it, and the influence of which is difficult to take into account mathematically.
Let's try, however, to do this at least approximately. Since the jet has a well-defined conical shape, we can assume that the movement of the liquid in the jet will be similar to the movement in a tapering channel (pipe), and we will know the speeds at the beginning and end of the movement V2 And Vh, as well as the pressure at the outlet of the jet from the pipe. Speed Vh caused by movement under the influence of gravity, as we showed above, is determined by the approximate expression:

To solve the problem, we assume that the increase in speed occurs only due to the use of the potential energy of the jet, i.e. by reducing its internal pressure. Such an assumption is to some extent possible if we remember that the movement of a liquid under the influence of gravity is prevented by forces caused by the adhesion between its particles (molecules), i.e. cohesive forces.
Since the flow movement is not formed by any channel and the weight of the jet does not take part in creating additional pressure, we use Bernoulli’s equation in its pure form:
, (135)
where can you find the pressure ph:
(136)
Using the speed expression Vh, we transform equation (136) to the form:
(137)
The resulting expression can be used to determine the height of the flow drop h, at which the pressure ph will be equal to atmospheric:
(138)
For the three examples we considered, when H»10 m, H=1m and H=0.1 m the values ​​will be respectively equal to:
1) m
2) m
3) m
In all three cases, the height of the jet fall, at which the internal pressure in it will be equal to atmospheric pressure, turned out to be approximately 4 times greater than the height h=H. Of course, these will be, as already noted, approximate values ​​that need to be verified experimentally.
All the examples we have considered convincingly show that the pressure inside the jet of both ideal and real liquid cannot be lower than atmospheric pressure. However, the wall pressure can be significantly lower, which is manifested when using pressure tubes. Using expression (114), you can use the pressure found using a manometric tube to determine the pressure in the liquid flow:
(139)
The second term in this expression, in fact, is a methodological measurement error, since it is not a device error or some random error, but an error associated with the measurement method itself.
Formula (114) can be used to determine the speed of fluid movement in a pipeline at a known wall pressure found experimentally. To do this, it must be presented in expanded form, taking into account expressions (109) and (107):
(140)
Let's consider two cases of pressure measurement, presented in Figures 7 and 10. The pressures shown by manometric tubes in sections 1 and 2 in the first case (Figure 7) will differ by the amount h due to the difference in fluid velocities in these sections. The wall pressures themselves for a horizontal pipe, in accordance with formula (140), will be equal to:
; (141)
, (142)
therefore their difference is determined by the expression:
(143)
Using relation (22), from expression (143) we find the speed V1:
(144)
For the second case (Fig. 10), we establish a relationship between wall and atmospheric pressure in a narrow section in the form of a relationship:
, (145)
Where rm- density of the liquid in the manometric tube, h- the height of the liquid in the tube above the liquid level in the vessel under atmospheric pressure. From expression (145) we find the fluid flow rate V:
(146)
Let us now find the error when measuring the pressure inside the liquid flow using a probe (Fig. 29). Let us consider the case when the probe tube is located along the flow axis. The presence of a tube will lead to a change in the nature of the flow movement, to a change in the pattern of the velocity field in it (Fig. 30), since the tube, like the walls of the pipe, will slow down the flow of liquid. The velocity field can be divided into two parts with respect to the maximum value of the flow velocity Vm: first part - from the radius probe tube r3 to radius rm, corresponding to the maximum speed, and the second part - from rm to the pipe wall, i.e. to radius R.
Let us assume that the velocity field in these sections will be determined by the expressions:
; (147)
(148)
From these expressions it follows that when r=rm speeds and will have the same value Vm, and when r=r3 And r=R they will be equal to zero.
The presence of corresponding kinetic energy fields leads to the appearance of radial inertia forces directed from the probe tube and from the tube wall to the middle of the flow. These forces will compress the flow and create negative pressure on the pipe wall and on the surface of the probe tube. This pressure will reduce the static pressure measured by the probe. The magnitude of the negative pressure in both areas will be determined, as shown above, by the average kinetic energy density:
(149)
This pressure will increase with increasing diameter of the probe tube, since the flow speed will increase, the value of which can be found from the condition of flow continuity:
, (150)
Where V- the speed of the liquid flow undisturbed by the probe. From expression (150) we find:
(151)
Thus, it turns out that existing measuring instruments cannot accurately measure the pressure inside a fluid stream. This circumstance, as we see, is due to the pressure measurement technique itself.
Our analysis of the problem of determining the speed of fluid flow and the pressure inside it shows that this problem does not have a fairly simple solution. This is due, first of all, to the fact that a liquid, unlike a solid, easily changes its shape due to significantly less adhesion between its particles. And yet, the adhesion forces are enough to influence the movement of the entire volume of fluid located both in the hydraulic system itself and outside it. So, for example, with an expanding conical nozzle, the liquid flow increases, i.e. the speed of its outflow from the vessel increases. This phenomenon can only be explained by an increase in the mass of the falling liquid and, consequently, an increase in the additional pressure. Therefore, the fluid in and outside the hydraulic system should be considered as a single body subject to different deformations in different parts of the system.
In light of all of the above, the question arises about the physical essence of the equation obtained by Daniel Bernoulli himself.
To clarify its essence, let us turn to this equation in the form of expression (8). Here p1 And p2 static and and are dynamic pressures. From this equation it follows that the sum of static and dynamic pressures, i.e. the total pressure is a constant value for an elementary current tube along its entire length. However, this statement will be true only under one condition - under pressure p2, as we showed above, should not be understood as the counter pressure from the rejected part of the liquid, which we denoted as , but the pressure in the flow of the section of liquid under consideration. In Bernoulli's law this condition is not specified or even implied.
The essence of Bernoulli's law can be commented in another way. Static pressure, in accordance with the law of conservation of energy, when a fluid moves, should decrease by the amount of dynamic pressure, although in fact, there is no dynamic pressure in the fluid flow, since the expression manifests itself as real pressure only when the entire flow or any part of it slows down . In fact, the expression is the volumetric kinetic energy density, i.e. the amount of kinetic energy per unit volume of moving fluid. In fact, this expression represents the loss of static pressure due to its conversion into motion energy. Therefore, if we go to static pressure R we add pressure loss, then we return to the original static pressure, which would occur in the absence of fluid movement. So the pressure p1 in Bernoulli's equation there is actually a pressure less than the original pressure p1. The same can be said about the pressures in the second section. However, this circumstance is also not specified when deriving the equation. Thus, if in the first and second sections of the flow we add the corresponding pressure losses due to fluid movement to the pressures, then based on equation (8) we can say that the initial static pressure in both sections in the absence of fluid movement was the same . In essence, this is the law of constancy of the initial hydrostatic pressure, i.e. this is an analogue of Pascal's law for a moving fluid.
There is another way to explain the physical essence of Bernoulli's law. We have already noted that the expression represents the volumetric density of the kinetic energy of a moving fluid. Obviously, the same can be said about static pressure R, which can also be considered energy density, but not kinetic, but potential. Regarding weight pressure rgH, then it can also be considered the potential energy density of the weight of the liquid. Therefore, Bernoulli’s law can also be interpreted as the law of conservation of volumetric energy density, i.e. law of conservation of energy per unit volume of liquid.
Thus, an analysis of Bernoulli's law shows that it has a very strict physical meaning associated with the law of conservation of energy. However, Bernoulli's equation cannot be used to directly find fluid flow rates from known pressures, or vice versa, even for an ideal fluid, since it does not take into account external resistance and resistance in the flow narrowing section. When deriving this equation, the work of forces was incorrectly calculated, since all of them had to be reduced to the first section and therefore multiplied by the displacement Dl1. Using Bernoulli's equation to determine velocities or pressures leads to significant errors. Using Toricelli's formula to determine the speed of fluid flow from an arbitrary hole is also illegal, since in this case there is no talk of any free fall.
Consequently, throughout its existence, Bernoulli's law has been misunderstood; it is, in fact, one of the myths of mechanics, however, with its help it turned out to be possible to explain almost all hydrodynamic phenomena (effects) in a moving fluid. And, surprisingly, this opportunity arose due to the mistakes that were made in deriving this equation. It just so happened that when deriving the equation, all the work from the pressure forces was spent on changing only the kinetic energy of equal volumes of liquid, mass r m, as a result of which a physically meaningful result was obtained, which essentially consists in the transition of potential energy into kinetic energy and, as a consequence, the constancy of the sum of these energies in all sections of the fluid flow.
The misunderstanding of Bernoulli's law was also facilitated by the absence of the concept of the kinetic energy field in a moving fluid and the accompanying gradient.
In conclusion, it is necessary to recall that the formulas we obtained can only be used for an approximate calculation of velocities and pressures inside a fluid flow, since external pressure cannot be found accurately due to the action of cohesive forces on fluid particles.

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Daniel Bernoulli (January 29 (February 8) 1700 - March 17, 1782), Swiss universal physicist, mechanic and mathematician, one of the creators of the kinetic theory of gases, hydrodynamics and mathematical physics. Academician and foreign honorary member (1733) of the St. Petersburg Academy of Sciences, member of the Academies: Bologna (1724), Berlin (1747), Paris (1748), Royal Society of London (1750). Son of Johann Bernoulli.

Bernoulli's law (equation) is (in the simplest cases) a consequence of the law of conservation of energy for a stationary flow of an ideal (that is, without internal friction) incompressible fluid:

Here

Bernoulli's equation can also be derived as a consequence of Euler's equation, which expresses the momentum balance for a moving fluid.

In scientific literature, Bernoulli's law is usually called Bernoulli's equation(not to be confused with Bernoulli's differential equation), Bernoulli's theorem or Bernoulli integral.

The constant on the right side is often called full pressure and depends, in general, on the streamline.

The dimension of all terms is the unit of energy per unit volume of liquid. The first and second terms in the Bernoulli integral have the meaning of kinetic and potential energy per unit volume of liquid. It should be noted that the third term in its origin is the work of pressure forces and does not represent a reserve of any special type of energy (“pressure energy”).

A relationship close to the one given above was obtained in 1738 by Daniel Bernoulli, whose name is usually associated Bernoulli integral. The integral in its modern form was obtained by Johann Bernoulli around 1740.

For a horizontal pipe, the height is constant and Bernoulli’s equation takes the form: .

This form of Bernoulli's equation can be obtained by integrating Euler's equation for stationary one-dimensional fluid flow, at constant density: .

According to Bernoulli's law, the total pressure in a steady flow of fluid remains constant along that flow.

Total pressure consists of weight, static and dynamic pressure.

From Bernoulli's law it follows that as the flow cross-section decreases, due to an increase in speed, that is, dynamic pressure, the static pressure decreases. This is the main reason for the Magnus effect. Bernoulli's law is also valid for laminar gas flows. The phenomenon of a decrease in pressure with an increase in flow rate underlies the operation of various types of flow meters (for example, a Venturi tube), water and steam jet pumps. And the consistent application of Bernoulli's law led to the emergence of a technical hydromechanical discipline - hydraulics.

Bernoulli's law is valid in its pure form only for liquids whose viscosity is zero. To approximate the flow of real fluids in technical fluid mechanics (hydraulics), the Bernoulli integral is used with the addition of terms that take into account losses due to local and distributed resistances.

Generalizations of the Bernoulli integral are known for certain classes of viscous fluid flows (for example, for plane-parallel flows), in magnetohydrodynamics, and ferrohydrodynamics.