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What does Bernoulli's law have to do with aviation? It turns out the most direct. With its help, it is possible to explain the emergence of the lifting force of the aircraft wing and other aerodynamic forces.

Bernoulli's law

The author of this law Swiss universal physicist, mechanic and mathematician. Daniel Bernoulli is the son of the famous Swiss mathematician Johann Bernoulli. AT 1838 he published fundamental scientific work "Hydrodynamics", in which he derived his famous law.

It should be said that in those days aerodynamics as a science did not yet exist. And Bernoulli's law described the dependence of the flow rate of an ideal fluid on pressure. But at the beginning of the twentieth century, aviation began to emerge. This is where Bernoulli's law came in handy. After all, if we consider the air flow as an incompressible liquid, then this law is also valid for air flows. With its help, they were able to understand how to lift an aircraft heavier than air into the air. This is the most important law of aerodynamics, since it establishes a relationship between the speed of air movement and the pressure acting in it, which helps to make calculations of the forces acting on the aircraft.

Bernoulli's law is a consequence of the law of conservation of energy for stationary flow of an ideal and incompressible fluid .

In aerodynamics, air is considered as incompressible fluid , that is, a medium whose density does not change with pressure. BUT stationary a flow is considered in which particles move along trajectories that are constant in time, which are called streamlines. Vortices do not form in such flows.

To understand the essence of Bernoulli's law, let's get acquainted with the jet continuity equation.

Jet continuity equation

From it it is clear that the higher the fluid flow rate (and in aerodynamics, the air flow rate), the lower the pressure, and vice versa.

The Bernoulli effect can be observed sitting by the fireplace. During strong gusts of wind, the airflow speed increases and the pressure drops. The air pressure in the room is higher. And the flames rush up into the chimney.

Bernoulli's law and aviation

Using this law, it is very easy to explain how the lifting force for an aircraft heavier than air arises.

During the flight, the wing of the aircraft, as it were, cuts the air flow into two parts. One part flows around the upper surface of the wing, and the other the lower one. The shape of the wing is such that the upper stream must overcome a greater distance in order to connect with the lower one at one point. This means he is moving faster. And since the speed is greater, then the pressure above the upper surface of the wing is less than under the lower one. Due to the difference in these pressures, the lifting force of the wing arises.

As the aircraft climbs, the pressure difference increases, which means that the lift force also increases, which allows the aircraft to rise.

We will immediately clarify that the laws described above are valid if the speed of the air flow does not exceed the speed of sound (up to 340 m / s). After all, we considered air as an incompressible fluid. But it turns out that at speeds above the speed of sound, the airflow behaves differently. The compressibility of air can no longer be neglected. And the air under these conditions, like any gas, tries to expand and occupy a larger volume. Significant pressure drops or shock waves occur. And the air flow itself does not narrow, but, on the contrary, expands. The solution of problems on the movement of air flows with velocities close to or exceeding the speed of sound is gas dynamics , which arose as a continuation of aerodynamics.

Using aerodynamic laws, theoretical aerodynamics allows you to make calculations of the aerodynamic forces acting on the aircraft. And the correctness of these calculations is checked by testing the constructed model on special experimental facilities, which are called wind tunnels . These installations allow you to measure the magnitude of forces with special instruments.

In addition to studying the forces acting on aerodynamic models, aerodynamic measurements are used to study the distribution of the values of velocity, density and temperature of the air flowing around the model.

The Bernoulli equation is considered one of the basic laws of hydromechanics; it establishes a relationship between the pressure in a fluid flow and its velocity in hydraulic systems: with an increase in the velocity of the flow, the pressure in it must fall. It explains many hydrodynamic effects. Let's take a look at some of the well-known ones. The rise and spraying of the liquid in the spray gun (Fig. 1) occurs due to the reduced pressure in the air stream passing at high speed over the tube lowered into the vessel with the liquid. The liquid rises upwards due to atmospheric pressure, which is greater than the pressure in the air stream.
A ping-pong ball (Fig. 2) hovers steadily in a vertical jet of air, since the pressure in the jet is less than atmospheric pressure, which presses the ball against the jet, preventing it from falling.
Ships traveling in parallel (Fig. 3) are attracted to each other, which is the cause of many maritime disasters. This is due to a decrease in pressure between ships due to the greater speed of water in the narrowed space between them.
The lifting force of the wing (Fig. 4) is due to the presence of a pressure difference p1 and p2 because of the speed difference V1 and V2, when V1 less V2, since air particles located above the wing, before meeting at the end of the wing, travel a longer path than particles located below.
If you blow between two sheets of paper touching each other (Fig. 5), then they will not separate, as it would seem, should happen, but, on the contrary, they will press against each other.
Thus, we see that the Bernoulli equation has a wide range of applications for explaining many hydrodynamic phenomena. Daniel Bernoulli published it in 1738 after many years of reflection and research, search and doubt. He was absolutely sure of the correctness of the law he discovered, which relates the static pressure in a liquid to the speed of its movement.
Consider the derivation of this equation for an elementary fluid stream (streamline), as it is given in all textbooks, for a stationary laminar flow of an ideal incompressible fluid. To eliminate the effect of gravity on the movement of the liquid, we take a horizontal section of the pipe (Fig. 6), we also place the elementary stream horizontally.
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- the cross-sectional area on the left side of the selected section of the liquid, and the resistance force determined by the static pressure p2:
, (2)
where S2- cross-sectional area on the right side of the plot.
The pressure acting on the side surface of the fluid element, according to the authors, is perpendicular to the displacements and will not perform work.
Under the action of these two forces, the selected part of the liquid will move from left to right. Let us suppose that it moves some small distance and takes up 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 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 is the mass of the selected liquid element, and are 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 introduced into equation (3) by adding and subtracting it on the right side:
(4)
where mtot- mass of the common part, - speed of the center of mass of the common part.
The expressions in parentheses are the kinetic energies of the shaded sections 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 are the masses of the shaded areas of the liquid.
Due to the continuity of the fluid flow, the volumes and masses of the shaded parts will be equal:
, (6)
where r is the density of the liquid.
Dividing expression (5) by S1Dl1=S2Dl2, let's transform it to the form:
(7)
After rearranging the terms, the equation will take the form:
(8)
This is the Bernoulli equation. Since the 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 in any place of 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. Take, for example, a Venturi tube (Fig. 7). The liquid levels in the gauge tubes clearly show that the static pressure is less at the constriction where the flow velocity is greater. In addition, confirmation of what has been said can be 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 narrow one, its speed increases and the acceleration is directed in the direction of movement. And since acceleration is determined by the difference in pressure acting on the fluid element from the left and right, the pressure in the wide part of the tube must be greater than in the narrow part. True, here one can notice that acceleration is determined not by pressure, but by force, while the force depends not only on pressure, but also on the cross-sectional area. Therefore, more force can be obtained with less pressure, so the argument is not convincing.
So, everything seems to be logical in the above reasoning. However, there is a possibility 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 in a completely different way.
Consider what will happen to a viscous fluid flowing through a pipe (Fig. 8). Due to the presence of friction between the fluid flow and the pipe walls, as well as between the layers of the fluid itself, the velocity of the fluid particles will be different at different points in the same flow section: in the center of the pipe it will be maximum, near the walls it will be equal to zero. As a result, the velocity field in the liquid flow section will be determined by the expression:
, (10)
where V is the speed at the center of the stream, r- current radius, R is the radius of the pipe, and will have the form shown in Figure 8. The scalar field of kinetic energy is inextricably linked with the velocity field, which is characterized by the expression:
, (11)
where Edm- kinetic energy of the allocated elementary mass dm, which is determined by the expression:
(12)
Here: dl- elementary length in the axial direction, r- liquid density.
Since the field of kinetic energy is non-uniform, a force directed towards the center of the flow will act on the elementary particle of the liquid:
(13)
This force, related to the cylindrical part of the particle surface dS, normally located to the force:
, (14)
determines the pressure arising at a given point in the flow under the action of a given force:
(15)
This pressure depends only on the elemental 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. Since all forces dF have a radial direction and are directed towards the center of the flow, the total pressure at the point will be determined by the forces lying on the same radius and located on the side external to the point under consideration. 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).
An amazing result was obtained, since this expression is similar to the expression for the kinetic energy (11), referred to the volume of the elementary mass dm:
, (17)
those. the total pressure is the density of kinetic energy in some elementary volume in the vicinity of the considered point.
It follows from expression (16) that on the flow axis (at r=0), the pressure will be maximum, and at its boundary (at r=R) will be equal to zero.
Under the action of radial forces, the flow will be compressed in the direction towards its axis, as a result of which the pressure on the pipe walls will decrease, i.e. a kind of negative pressure will appear, the value of which can be found as the average over the radius of expression (16). To do this, we integrate it in 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 axial line of the pipe, for which this expression, taking into account expression (12), must be integrated within the range from 0 to R:
(19)
Dividing this force by the value of the elementary area:
, (20)
we obtain the value of the 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 given by:
(21)
Since the magnitude of the negative pressure depends on the square of the velocity, it is quite natural that its magnitude will be much greater in the narrow part of the flow than in the wide part. Therefore, in the Venturi tube in its narrow part, the pressure gauges will show less pressure than in the wide part. The dependence of the value of the 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 jet sucks in the liquid in the vessel (see Fig. 1). It is believed that the liquid is sucked in due to the fact that the pressure in the gas jet becomes lower than atmospheric pressure due to its speed, which squeezes the liquid out of the vessel, and the gas jet drags it along. However, the same effect will also be caused by the presence of negative pressure, due to the presence of an inhomogeneous field of kinetic energy in the flow of a gas jet emitted from the nozzle of the atomizer. In addition, the jet will drag particles of the surrounding air along with it, which will lead to the appearance of its own kinetic energy field in it, 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 atomizer may not be a decrease in pressure in the flow of a moving liquid or gas, then how to understand the essence of the Bernoulli equation? After all, the velocity of the liquid in the narrowed part of the flow really increases, and this, it seems, is possible only with a decrease in counteraction, and experiments show that the pressure in the flow can be lower than atmospheric pressure, since in the manometric tube the liquid rises above the level corresponding to atmospheric pressure (Fig. . ten). But on the other hand, it is also undoubted that the narrowing of the flow should increase the resistance to movement, and hence increase the pressure inside the fluid flow. In this case, an increase in the flow velocity can occur only due to an increase in the driving force, i.e. pressure on the left side of the selected flow element. Indeed, a similar conclusion can be drawn if we turn to equation (7):

We must not forget that this equation refers to the entire volume of liquid that we have allocated, which we consider as a single whole. Therefore, it is impossible to separate it, as it was done in expression (9). This is very important to remember. From expression (7) it follows that with increasing speed V2 at constant speed V1 pressure difference will increase. p1 and p2. This increase may be due to a decrease p2, as well as by increasing p1. When analyzing the Bernoulli equation, they prefer to talk about a decrease in pressure p2. But what is the pressure p2? This is the pressure that prevents the movement of a liquid or gas. How is it defined? Take for example a conical nozzle for a pipeline (Fig. 11). It is clear that the counterpressure p2 cannot be less than atmospheric pressure, otherwise the liquid will not flow out of the nozzle. If we want to increase the fluid outflow rate with a given nozzle, then we, in accordance with equation (7), must increase the pressure p1. But that's not all. Since the speed V1 and V2 interdependent, with increasing speed V2 will increase the speed V1, and then the pressure difference p1 and p2 should decrease, which corresponds to an increase in pressure p2 at constant pressure p1.
Thus, the analysis of the Bernoulli equation reveals the presence of a problem in understanding its essence. In order to better understand this problem, we apply equation (7) to the study of the movement of fluid in a conical nozzle (see Fig. 11). It follows from the flow continuity condition 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 with respect to the speed V2, we get:
(23)
Let's analyze this expression. Let's take the limit ratios R2/R1. At R2/R1=0 speed V2 will be equal to:
, (24)
while it is quite clear that it must be equal to zero. True, common sense dictates that the pressure p1 and p2 according to Pascal's law 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, even here one can find a way out by declaring that the pressure p1 and p2 will also be equal, since the magnitude of the speed must be constant. However, we cannot find the value of the speed V2, since it will be determined by the ratio of zeros.
But what about intermediate values of the ratio R2/R1? Can't the pressure difference p1 and p2 be zero all the time. How will this difference change? There are no answers to these questions. Only one thing becomes clear: the Bernoulli equation, even for an ideal fluid, is not accurate and cannot be used to calculate velocities or pressures, something is missing in it. This question should be dealt with, moreover, with digital calculations.
Such calculations, although approximate, exist for the outflow of liquid from the 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 \u003d 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 because V1<<V2, then from expression (26) it turns out:
, (27)
where p2 equal to atmospheric pressure.
If a p1 will be equal to p2, then formula (27) will take an even simpler form:
, (28)
whence it follows that the velocity of the outflow of the liquid is equal to the velocity of the free fall of the rigid body from a height H.
This expression was obtained by Toricelli 100 years before Bernoulli and is therefore called Toricelli's formula.
However, even here, despite the obviousness of the derivation of this equation, questions arise that have no answer: whether, for example, the fluid outflow rate will depend on the size of the hole or on the size of a conical nozzle that can be attached to the tank (see Fig. 12,b )? Can the outflow of a liquid through a small hole be similar to its free fall? This, of course, is highly doubtful even for an approximate determination of speed.
To simplify the analysis of this problem, let us take a vertically located tank of a conical shape (Fig. 13), into which liquid enters and flows out, moreover, so that its level remains the same all the time. Taking into account relation (22), from the Bernoulli equation we obtain:
(29)
It follows from this expression that when R2/R1=0 speed V2 will be equal to zero only if:
, (30)
from where follows:
, (31)
which does not follow from the condition 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: p2=p0, and the rate of fall must have a very specific value.
Thus, we have established that the pressure p2 in the 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 dependence, let us first consider a closed conical vessel in which a 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 pressure of the gas at all points of the vessel will be the same. We 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, strength F2 there will be less power F1. It is quite obvious that the difference between these forces:
(35)
will be compensated by the resistance from the side walls of the vessel.
Thus, the constriction of the vessel provides additional resistance to the force F1, as a result of which a smaller 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. Obviously, in this case, the ratio between the pressure in the first and second sections must 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 it only remains to assume 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 the pressure of the gas, due to the 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 the pressure forces was transferred into the 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 value of the volumetric densities of kinetic energies in them:
; (36)
, (37)
because p2=p1.
It can be seen from these relations that we establish a relationship 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 and are interdependent.
If we add to the pressures and the losses of potential energy converted into the kinetic energy of motion and , then the static pressure in the first and second sections will be equal to each other and equal to p1, i.e.:
, (38)
which is analogous to the Bernoulli equation.
Thus, we have obtained the Bernoulli equation, based on the law of conservation of energy for a steady flow of an ideal fluid. In fact, we have expanded the scope of Pascal's law by transferring 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 reaction force D.F. Defining it as the difference of forces and , we find:
, (41)
whence it follows that the resistance force from the side of the walls increases.
From the considered example and the assumptions that we made, we can draw the following conclusions.
Firstly, any narrowing of the channel through which a liquid or gas moves, resists this movement, the value 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 resistance value 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 velocity of the fluid flow, decreasing by . It follows from this that pressure is not an external resistance in relation to the selected element of the liquid, it is an internal property of the considered part of the liquid. And this, in fact, is the pressure that the selected element of the liquid exerts on the subsequent, discarded part of the liquid, i.e. creates a force that causes the movement of subsequent sections of the fluid. And what is very important, this pressure will not directly depend on the pressure external to the selected element of the liquid from the side of the discarded subsequent part of the liquid, which we denote by . Here the dependence will be indirect: the velocities will depend on the pressure V1 and V2, but already on the speed V2 pressure will depend. It should be noted that one of the components of pressure will generally be the ambient pressure, in particular atmospheric pressure. This directly implies the fact that the pressure in the fluid flow cannot be less than atmospheric pressure. Thus, from all of the above, it follows that when deriving the Bernoulli equation, pressure should not be taken into account as the cause of the appearance of a resistance force - the resistance force will be created only by pressure.
Thirdly, 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 and directly counteracts the force , i.e. we can assume that it is applied in the first section. Since force is determined by pressure, depending on pressure p1, then the counteracting force D F also depends on pressure p1 and, therefore, is, as it were, the force of self-braking 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, it must be multiplied by the displacement of the left end of the fluid 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 corresponds to the task at hand.
Now let's return to the consideration of the problem of the outflow of water from a conical tank (see Fig. 13). In a tank with liquid, the pressure in the second section, which will be used to determine the reaction force D.F. except pressure p1 will also be determined by the pressure pH created by the weight of the liquid:
, (42)
where H- the height of the liquid column, counted from its upper level, in connection with which expressions (36) and (37) will take the form:
; (43)
(44)
In connection with the foregoing, 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 compiling 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 the fluid in the first section - D l1. It remains to find out the question of how to deal with the resistance force: by what displacement D l it must be multiplied - by D l1 or D l2? To solve this problem, we combine the forces of D F and :
(49)
whence we obtain that the second expression in brackets is the excess pressure of the liquid in the second section with respect to the pressure:
(50)
It follows from this that the work of a force must be determined by multiplying it by the displacement Dl1.
Thus, the equation of motion in the form of the law of change of 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 converted to the form:
(52)
which, after dividing 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) for the 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 the corresponding value. The results of calculating this speed for === n/m2 and for H= 10.2 m are shown in Figure 15. As one might expect with an increase in the ratio R2/R1 the speed gradually increases from zero to the maximum value corresponding to free fall. In addition, using formula (44), one can find the pressure in the liquid jet flowing out of the conical tank. An analysis of this formula shows that when V2=0 pressure in the liquid will be equal to:

and at , which corresponds to free fall, =. The calculated curve for pressure =+= is shown in Figure 15, from which it can be seen that the pressure in the outflowing jet will be greater than atmospheric pressure for all ratios of radii R2/R1, unless these pressures are equal.
For greater persuasiveness of the above, we will give another derivation of the equation of motion, which takes 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 under consideration will be in equilibrium.
To determine the force of inertia, consider a part of the conical channel through which the fluid moves (Fig. 16). Let us single out the elementary volume of the liquid dm, which will move from the first position to the second, while changing the speed of its center of mass from the value to the value . The elementary force of inertia arising in this case can be determined by the formula:
, (56)
where
, (57)
and the minus sign shows the direction of the inertial force.
The ratio between the velocities in the two considered positions of the elementary mass dm is defined by the expression:
, (58)
where
(59)
Using this ratio we get:
(60)
Raising the binomial to the fourth power, dividing each term by D ls and then accepting D ls equal to zero, we find the expression for the elementary force of inertia:
(61)
Let's 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 look like:
; (62)

(63)
Substituting these values of speed and radius into expression (61), we obtain:
(64)
Now it is necessary to sum the elementary forces of inertia over the entire allocated volume of the moving fluid, i.e. by lenght l. Substituting into expression (64) the value of the mass dm:
(65)
and taking the integral of expression (64) in the range from 0 to L, we find the force of inertia acting from the side of the entire moving mass of fluid on the first section, where the driving force is applied F1:
(66)
where .
It follows from expression (66) that the force of inertia is indeed applied to the first section, since the difference between the energy densities in the second and first sections (expression in brackets) is multiplied by the area of the first section.
Thus, the following forces will act on the selected volume of liquid:
;
;
;
, (67)
under the influence of which this volume of liquid, considered by us as a single body, will be in equilibrium in accordance with the laws of mechanics, 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 pressures are correct.
Thus, the problem of finding the fluid flow velocity seems to be solved by us. However, if we comprehend the situation from the point of view of the laws of mechanics, there are doubts about the validity of the obtained formulas. Indeed, if, as an example, we look at a vertically falling fluid flow flowing out of a pipe of constant cross section (Fig. 17), then we can immediately notice that the fluid flow, even outside the pipe, moves as a single body with the fluid in the pipe and, therefore, in all its points must have the same speed. If this is not the case, then the flow will break, since when falling under the action of gravity, the speed must continuously increase. However, in practice, such a gap is not observed. This circumstance is due to the presence of cohesive forces (cohesion) between the molecules of the liquid, 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 also exist in an ideal fluid. Therefore, when moving any fluid element r m on it except for gravity Ftyazh there will be a resistance force Fresist from the upper parts of the liquid and the driving force Fmot from the bottom side. As a result of the free fall of the fluid element r m will not, and the element itself, under the action of forces applied to it, will experience tensile deformations, due to which it will be compressed in the transverse direction, and the entire flow will narrow (in Figure 17, the flow narrowing is shown by dash-dotted lines). Due to this narrowing, the speed of the element dm should change as it falls, and neither the speed V1, nor speed V2 we do not know, and, as follows from our reasoning, cannot be found by the above formulas.
In order to somehow get out of the situation, we take into account, at least approximately, the effect of the outflowing part of the flow external to the pipe on the liquid in the pipe. This external influence will be pulling, i.e. it will create some additional pressure rd in the flow, contributing to 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 is the mass of the liquid column, R2 and Rh are the 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 value of the height of the flow fall, for example, into some vessel, or the loss of cohesion between the particles of the liquid during its thinning, when the flow begins to break up into separate drops. We will set the value h arbitrarily, without considering critical situations in relation to the disintegration of the jet, since this issue requires a special study.
To find the weight of a liquid column, it is necessary with a known radius R2 find the radius Rh, corresponding to the height of the fall h. For an approximate determination of this radius, consider the fall of some liquid element with a mass Dm from high h under the action of only its own weight, although it will be affected by cohesive 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 fall velocity Vh from the coordinate h:
(76)
Using the flow continuity condition:
, (77)
we get:
(78)
On fig. 19 shows the shapes of the liquid jets obtained as a result of calculations of the ratio Rh/R2 according to formula (78) for the outflow velocities V2, equal to 0.1 m/s and 0.5 m/s, depending on the height of the fall h. It can be seen from the figures that, at a low outflow velocity, the narrowing of the jet will be sharper.
To take into account the influence of the additional driving force on the flow velocity and pressure inside it, it must be taken into account in the equations we have obtained. This can be done by referring it to the first section, where the driving force is acting, determined by the pressure p1 and cross-sectional area S1. Then the pressure created by this additional force will be equal to:
(79)
It is more convenient to represent this expression in the form:
, (80)
because then the relation 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 with ==:
, (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 10.32875 m and 1 m of the conical vessel from which the liquid flows. The first height corresponds to atmospheric pressure. In both cases, the height h was taken equal H and H/R1=10, =.
As can be seen from the curves, the flow velocity can increase significantly due to the drop 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) is compensated by the added pressure due to an increase in the flow velocity. However, in the free fall of a liquid at R2/R1=1 pressure in both cases becomes equal to atmospheric.
Thus, the formulas obtained by us can be used to approximately determine the flow velocities in its various sections, and these velocities will largely depend on the value h(see Fig. 22, a and b).

It is also of interest to consider the problem of upward motion 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 outflowing liquid column has a cylindrical shape).
This pressure will be included as a component in the pressure, which is included in the calculation formulas. Then the pressure will be determined by the expression:
(88)
It is clear that the speed V2 while decreasing. However, in order to calculate V2 you need to know the lift height h, which, in turn, depends on the velocity of the outflow V2. That's why h must somehow be expressed in terms of speed V2. We will argue as follows. Flow element r m in section 2 has some kinetic energy, which in the upper part of the flow turns into potential. Therefore, the following relation must hold:
, (89)
from where we get:
(90)
Then the pressure will take the form:
(91)

This pressure value should be substituted into the initial 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 at p1=p0 we get:
(94)
Substituting this speed value into expression (90), we find:
(95)
Thus, the height of the rise of the liquid will be half the difference in its levels H. Once again, we note that these will be approximate values for the speed V2 and lift height h, since the cross section of the external flow should not remain constant: it should increase with distance from the outlet due to a 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 falling part of the flow, which will create a pulling force that increases the speed of the flow.
Estimated speeds V2, pressure and altitude h water rise are shown in figures 20 and 21 for two cases when H=10.32875m and H=1m. The pressure in this case is determined by the usual formula:

Since the outflow velocity in this case will be less due to the presence of additional resistance of the water column, the pressure will also be greater than when the liquid flows down, 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 deceleration of the liquid layers against the pipe walls and between themselves leads to a decrease in the velocity of the 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 set 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 find the volumetric flow using expression (96) for elementary annular sections, the area of which is determined by the expression:
, (99)
where dr- the width of the ring. In accordance with this, the elementary volume flow will be equal to:
(100)
By integrating this expression in the range from 0 to R, we obtain the total volumetric flow rate of the liquid in the cross section l:
(101)
Using formula (98), we find the average flow velocity in the section l:
(102)
The kinetic energy of the flow in some section D l in this case will be equal to:
, (103)
where D m-corresponding to the length D l the mass of the fluid.
The equation of motion of the allocated volume of liquid in the form of the sum of forces, taking into account the friction force Ftr is defined by the expression:
(104)
This expression takes into account the cross-sectional flow velocities in sections 1 and 2. The friction force must be determined from existing experimental data.
Having made the necessary transformations, we bring expression (104) to the form:
(105)
where do we find the speed V2:
, (106)
where
(107)
pressure loss in length L=H(pressure decreases by this amount p1 in section 2).
An analysis of this expression shows that when R2/R1=0 speed V2 will be equal to zero, and R2/R1=1 expression (107) will take the form:
(108)
The average value of the flow velocity in the second section will be half as much.
The pressure value in the second section will decrease due to the energy loss to overcome the friction forces and will be determined by the expression:
(109)
When the liquid moves down, it is necessary to take into account intermolecular cohesion. Then the speed V2 is defined by the expression:
(110)
When a liquid flows vertically upwards, the pressure, as shown above, can be represented by the expression:
(111)
Then the expression for the speed V2 will take the form:
(112)
The pressure inside the liquid as it moves up and down will be determined by expression (109), only the velocities V2 they will, of course, be different. So the pressure will be different.
The pressure inside the liquid, taking into account its compression, will, in accordance with formula (18), be greater by the value of the average negative pressure:
,
the near-wall pressure is less by this value, i.e.:
; 113)
(114)
To calculate the fluid flow rates and 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 the laminar flow regime:
, (115)
where Q- fluid flow rate in m3/s, p1-p2- pressure drop in the fluid flow in a section of a cylindrical pipe with a length L in N/m2, m is the dynamic viscosity of the fluid in kg/ms, d- pipe diameter in m.
Using this expression, you can find the average speed over the pipe section:
, (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 speed in the section located at a distance from the first section l, Rl is the 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 flow continuity condition:
,
from where we find:
, (119)
where
(120)
Using these expressions, we get:
(121)
By integrating the resulting expression over l ranging from 0 to L, we find the pressure loss along the entire length L:
(122)
Since the expression in parentheses is:
, (123)
and tg a is defined 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 bring expression (125) to the form:
(127)
According to the obtained formulas, three variants of calculations were made for the following dimensions of the conical pipe:
1) H=L=10.32875 m (which corresponds to atmospheric pressure);
2) H=L=1.0 m;
3) H=L=0.1m
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 equals 0.001 kg/ms. Calculations showed that for the selected pipe sizes, the average water flow rate in the presence of viscosity practically did not differ from the ideal fluid velocity shown in 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 kinetic energy field gradient, 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 the 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 down without taking into account the link
between liquid molecules (curves 1), when the flow moves down, 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 large pipe sizes and therefore can be easily observed.
Thus, we have considered how the flow velocity 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 the atmospheric pressure for some time even when the liquid moves outside the pipe. Let's consider this question in more detail.
If the pressure in the liquid at the exit from the hole is greater than the 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. 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 had time to equalize. The force determined by the gradient will continue to compress the flow. Secondly, the fluid flow will be compressed by the force arising from the movement of air entrained by the fluid jet. In this case, a kinetic energy field will also appear in the air jet, the gradient of which will determine the acting force.
Determine the pressure with which air compresses a jet 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: rv- air density.
The derivative of this expression will determine the elemental force dFv:
,(131)
towards the center of the stream.
The ratio of this force to the elementary surface dS=rdjdh, corresponding to the elementary mass, will determine the differential pressure dpw:
(132)
(omit the minus sign).
The total pressure acting on the elementary mass from all air particles external to it is determined by the integral of expression (132), taken over r in the range from r to :
(133)
On the jet surface ( r=Rh) the air pressure will be:
(134)
Thirdly, the jet will be compressed due to the presence of tensile forces due to the adhesion between the liquid molecules, and also, as noted above, due to an increase in the falling velocity under the action of gravity.
Fourth, the jet will compress due to the presence of surface tension.
Thus, several forces will act on a jet of liquid flowing out of a pipe, the combination of which will determine both its shape and pressure in it, and the influence of which is difficult to take into account mathematically.
Let us try, however, to do this at least approximately. Since the jet has a well-defined conical shape, it can be assumed that the movement of the fluid in the jet will be similar to the movement in a narrowing channel (pipe), and we will know the speeds at the beginning and at the end of the movement V2 and Vh, as well as the pressure at the outlet of the jet from the pipe. Speed Vh due to motion under the action of gravity, as we have shown above, is determined by the approximate expression:

To solve the problem, we assume that the increase in velocity 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 fluid under the action of gravity is hampered by forces due to the adhesion between its particles (molecules), i.e. cohesion 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 the Bernoulli equation in its pure form:
, (135)
where can i find the pressure ph:
(136)
Using Expression for Speed Vh, we transform equation (136) to the form:
(137)
The resulting expression can be used to determine the height of the fall of the flow h, at which the pressure ph will be equal to atmospheric:
(138)
For the three examples we have considered, when H»10 m, H=1m and H=0.1m 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 the atmospheric pressure, turned out to be approximately 4 times greater than the height h=H. Of course, as already noted, these will be 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 liquids cannot be lower than atmospheric pressure. However, the near-wall pressure can be much less, which is manifested when using manometric tubes. Using expression (114), it is possible to determine the pressure in the fluid flow from the pressure found using a manometric tube:
(139)
The second term in this expression, in fact, is a methodological measurement error, since this is not an instrument error and not some random error, but an error associated with the measurement method itself.
Formula (114) can be used to determine the velocity of fluid 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)
Consider two cases of pressure measurement shown in Figures 7 and 10. The pressures shown by manometric tubes in sections 1 and 2 in the first case (Fig. 7) will differ by the value 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)
so their difference is given by:
(143)
Using relation (22), from expression (143) we find the speed V1:
(144)
For the second case (Fig. 10), we establish the relationship between the near-wall and atmospheric pressure in a narrow section in the form of a relation:
, (145)
where rm is the density of the liquid in the gauge tube, h- the height of the rise of the liquid in the tube above the level of the liquid in the vessel under atmospheric pressure. From expression (145) we find the fluid flow rate V:
(146)
Let us now find the error in measuring the pressure inside the fluid flow using a probe (Fig. 29). 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 movement of the flow, 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 fluid flow. The velocity field can be divided into two parts in relation to the maximum value of the flow velocity Vm: The first part is from the tube probe radius r3 up to radius rm corresponding to the maximum speed, and the second part - from rm to the pipe wall, i.e. up to radius R.
We accept that the velocity field in these sections will be determined by the expressions:
; (147)
(148)
From these expressions, it follows that r=rm speed and will have the same value Vm, and when r=r3 and r=R they will be zero.
The presence of the 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 a 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 sections will be determined, as shown above, by the average density of kinetic energies:
(149)
This pressure will increase with an increase in the diameter of the probe tube, since the flow rate will increase, the value of which can be found from the flow continuity condition:
, (150)
where V is the velocity of the fluid flow unperturbed by the probe. From expression (150) we find:
(151)
Thus, it turns out that existing measuring instruments cannot accurately measure the pressure inside the fluid flow. This circumstance, as we see, is due to the method of pressure measurement itself.
Our analysis of the problem of determining the fluid flow rate and 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 body, easily changes its shape due to much less adhesion between its particles. Nevertheless, the adhesion forces are enough to influence the movement of the entire volume of fluid, both in the hydraulic system itself and outside it. So, for example, with an expanding conical nozzle, the flow rate of the liquid 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 additional pressure. Therefore, the fluid in the hydraulic system and outside it should be considered as a single body subjected to different deformations in different parts of the system.
In the light of the foregoing, 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. It follows from this equation that the sum of static and dynamic pressures, i.e. 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 have shown above, should be understood not as the counter pressure from the side of the discarded part of the liquid, which we denoted as , but the pressure in the flow of the considered section of the liquid. In Bernoulli's law, this condition is not specified and is not even implied.
The essence of Bernoulli's law can be commented on in another way. Static pressure, in accordance with the law of conservation of energy during fluid movement, must 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 some part of it is decelerated . In fact, the expression is the volumetric density of kinetic energy, i.e. the amount of kinetic energy per unit volume of the moving fluid. In fact, this expression represents the loss of static pressure due to its conversion into the energy of motion. So if we to the static pressure R add the pressure loss , then we return to the original static pressure, which would have occurred in the absence of fluid movement. So the pressure p1 the Bernoulli equation actually has 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 to the pressures and we add the corresponding pressure losses due to fluid movement, then on the basis of equation (8) we can say that the initial static pressure in both sections in the absence of fluid movement was the same . In fact, 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 is the volumetric density of the kinetic energy of the moving fluid. Obviously, the same can be said about static pressure. R, which can also be considered the energy density, but not kinetic, but potential. As for 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, the analysis of Bernoulli's law shows that it has a completely rigorous physical meaning associated with the law of conservation of energy. However, the Bernoulli 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 constriction area. When deriving this equation, the work of forces was found incorrectly, since all of them had to be reduced to the first section and therefore multiplied by the displacement Dl1. Using the Bernoulli equation to determine velocities or pressures leads to significant errors. The use of Toricelli's formula to determine the velocity of fluid outflow from an arbitrary hole is also illegal, since there is no need to talk about any free fall in this case.
Consequently, for the entire time of its existence, Bernoulli's law was misunderstood, this 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, such an opportunity arose due to the mistakes that were made in the derivation of 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, in fact, consists in the transition of potential energy into kinetic energy and, as a result, 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 a kinetic energy field in a moving fluid and the accompanying gradient.
In conclusion, it should be recalled that the formulas obtained by us can only be used for an approximate calculation of the velocities and pressures inside the fluid flow, since the external pressure cannot be found exactly due to the action of cohesive forces on the fluid particles.

Bernoulli's equation for the flow of a real fluid, its physical meaning.

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

Here is the density of the liquid, is the flow velocity, is the height at which the element of the liquid under consideration is located, is the pressure at the point in space where the center of mass of the considered element of the liquid is located, is the acceleration of free fall.

In real fluid flows, there are forces of viscous friction. As a result, the fluid layers rub against each other as they move. Part of the flow energy is expended on this friction. For this reason, in the process of movement, energy losses are inevitable. This energy, as with any friction, is converted into thermal energy. Due to these losses, the energy of the fluid flow along the length of the flow, and in its direction, is constantly decreasing.

It follows from Bernoulli's law that as the flow cross section decreases, due to an increase in velocity, 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. Bernoulli's law is valid in its pure form only for liquids whose viscosity is zero. To describe the flows of real fluids in technical hydromechanics (hydraulics), the Bernoulli integral is used with the addition of terms that take into account losses in local and distributed resistances.

Bernoulli's equation for real fluid flow

Speed distribution:

What is a Pitot tube and what is it used for?

A pitot tube is a device for measuring velocity at points in a flow. to measure the dynamic head of a flowing liquid or gas. It is an L-shaped tube. The excess pressure established in the tube is approximately equal to: , where p is the density of the moving (incoming) medium; V? - the speed of the oncoming flow; ξ is a coefficient.

Pitot pressure tube is connected to special instruments and devices. It is used to determine the relative velocity and volume flow in gas ducts and ventilation systems, complete with differential pressure gauges.

It is used as an integral part of the Prandtl tube in aviation air pressure receivers for the possibility of simultaneously determining the speed and flight altitude.

How to convert the Bernoulli equation from the dimension of lengths to the dimension of pressures?

Bernoulli's equation in the form of heads, m

Bernoulli's equation in the form of pressures, Pa

Pressure loss from the first section to the second.

What are the flow regimes and how are the boundaries of the existence of these regimes determined?

1. Laminar mode of motion. Features - layered nature of the fluid flow, lack of mixing, invariance of pressure and velocity over time.

2. Transition mode.

3. Turbulent flow regime. Visible: vortex formation, rotational movement of the liquid, continuous pressure and velocity pulsations in the water flow.

1. Laminar is a layered flow without mixing of fluid particles and without pulsation of velocity and pressure. In a laminar fluid flow in a straight pipe of constant cross section, all streamlines are directed parallel to the pipe axis, and there are no transverse movements of fluid particles.

2. A flow is called turbulent, accompanied by intense mixing of the liquid with pulsations of velocities and pressures. Along with the main longitudinal movement of the liquid, transverse movements and rotational movements of individual volumes of the liquid are observed. 3. The transition from a laminar regime to a turbulent one is observed at a certain fluid velocity. This speed is called critical ( Vcr=kv/d).

The value of this velocity is directly proportional to the kinematic viscosity of the fluid v and inversely proportional to the pipe diameter d.

4. Dimensionless coefficient included in this formula k the same for all liquids and gases, as well as for all pipe diameters. This coefficient is called the critical Reynolds number. Recr and is defined as follows:

Recr = Vcrd/v = pVcrd/μ ≈ 2300-2320

How is the Reynolds number calculated?

The Reynolds similarity criterion (Reynolds number) makes it possible to judge the mode of fluid flow in the pipe. Reynolds number (criterion) Re - a measure of the ratio of inertia force to friction force

Re = Vd/v = pVd/μ, where μ is the dynamic viscosity coefficient, v = μ/p,

At Re< Reкр = 2320 течение является ламинарным;

Re > 3800-4200 the flow is turbulent.

Dependencies are valid only for round pipes.

As the speed increases, the force of inertia increases. In this case, the friction forces are greater than the inertial forces and up to some extent straighten the trajectories of the jets

At some speed vcr:

Inertia force Fand > friction force Ffr, the flow becomes turbulent

Bernoulli's equation for the steady motion of an ideal fluid, its physical meaning.

We reduce the Euler equations to a form convenient for integration by multiplying by dx, dy, respectively,

dz and adding:

We get

Considering that

- full differential pressure

Final expression:

If the fluid is only under the influence of gravity and its density is unchanged, then

Finally

Bernoulli's equation for a trickle of an ideal liquid

Bernoulli's equation for the steady motion of a viscous fluid.

Speed distribution:

1 - elementary trickle; ideal liquid;

2 - real (viscous) liquid

When a real viscous fluid moves, friction forces and vortices arise, to overcome which the fluid expends energy.

As a result, the total specific energy of the liquid in section 1-1 will be greater than the total specific energy in section 2-2 by the value of the lost energy

Here

V 1.2- average flow velocity in sections 1.2;

hW1,2 = hpot 1-2- lost pressure head loss between sections 1-2;

α1,2- dimensionless Coriolis coefficient - the ratio of the actual kinetic energy of the flow in a given section to the kinetic energy of the flow in the same section with a uniform distribution of velocities.

Thus, the level of initial energy that the liquid has in the first section for the second section will be the sum of four components: geometric height, piezometric height, velocity height and lost head between sections 1-1 and 2-2
The velocity of a viscous fluid in a long tube: v = (ΔP / η) R 2 / (8 l), where ∆P is the pressure difference at the ends of the tube, η - viscosity of a liquid or gas (strongly depends on temperature), R is the inner radius of the tube, l- its length, l >> R.

Coriolis coefficients. The value of the coefficients for laminar and turbulent flow regimes.

The Coriolis coefficient is the ratio of the actual kinetic energy of the flow in a given section to the kinetic energy of the flow in the same section with a uniform distribution of velocities.

Power of an elementary stream:

For the flow

Dividing the resulting expression by and taking into account that (specific power per 1 N

liquid weight = average head in section Nsr) we get:

Here ? - Coriolis coefficient.

With a uniform distribution of speeds α =1 (elementary trickle/ideal liquid),

for non-uniform α>1. V- average speed in live section .

- Coriolis coefficient for laminar flow.

- Coriolis coefficient for turbulent regime (tends to 1.0 with increasing Re)

Rational choice of sections for solving the Bernoulli equation.

Sections are selected always perpendicular to the direction of fluid flow and must be located on straight sections of the flow

One of design sections must be taken where it is necessary to determine the pressure R, height z or speed V, the second, where the quantities R, z, and V known

number design sections should be so that the liquid moves from the section 1-1 to section 2-2

Comparison plane 0-0 - any horizontal plane. For convenience, it is carried out through the center of gravity of one of the sections

Practical application of the Bernoulli equation: Pitot tube.

A pitot tube is a device for measuring velocity at points in a flow.

Composing the Bernoulli equation for the sections a-a and b-b, we get

From here

Practical application of the Bernoulli equation: Venturi flowmeter.

a) Neglecting head losses and assuming z1 = z2, we write the Bernoulli equation for sections 1-1 and 2-2:

b) From the continuity equation

c) From the piezometer equation

Solving together, we get:

Energy interpretation of the Bernoulli equation.

Energy characteristics of the liquid. The total energy characteristic of a fluid is its hydrodynamic head.

From a physical point of view, this is the ratio of the magnitude of mechanical energy to the magnitude of the weight of the fluid that possesses this energy. Thus, the hydrodynamic head must be understood as the energy per unit weight of the fluid. And for an ideal fluid, this value is constant along the length. Thus, the physical meaning of the Bernoulli equation is law of conservation of energy for a moving fluid .

Here, from an energy point of view (in units of energy, J / kg) gz — specific potential energy of position; rР/ — specific potential energy of pressure; gz + rР/ — specific potential energy; u 2 /2 — specific kinetic energy; and — the speed of an elementary stream of an ideal liquid.

Multiplying all terms of the equation by the specific gravity of the liquid g , we get:

g z - weight pressure, Pa; P — hydrodynamic pressure, Pa; ir 2 /2 — dynamic pressure Pa; hg — total pressure, Pa

Geometric interpretation of the Bernoulli equation.

The position of any fluid particle relative to some arbitrary zero level line 0-0 determined by the vertical coordinate Z . For real hydraulic systems, this may be the level below which liquid cannot flow out of a given hydraulic system. For example, the floor level of a workshop for a machine tool or the basement level of a house for domestic plumbing.

All terms of the Bernoulli equation have the dimension of length and can be represented graphically.

Values - leveling, piezometric and speed heights can be determined for each section of an elementary stream of liquid. The locus of points whose heights are equal is called piezometric line . If we add speed heights equal to these heights, we get another line, which is called hydrodynamic or pressure line .

It follows from the Bernoulli equation for a trickle of a non-viscous liquid (and the graph) that the hydrodynamic head along the length of the trickle is constant.

Full pressure line and its construction.

The physical meaning of the Bernoulli equation.

It follows from Bernoulli's law that as the flow cross section decreases, due to an increase in velocity, 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 the 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, that is, liquids that do not stick to the surface of the pipe. In fact, it has been experimentally established that the velocity of a liquid on the surface of a solid body is almost always exactly zero (except in cases of jet separation under certain rare conditions).

Bernoulli's law explains the effect of attraction between bodies located at the boundary of the flow of a moving fluid (gas). Sometimes this attraction can create a security risk. For example, when the Sapsan high-speed train (moving speed over 200 km/h) is moving, there is a danger for people on the platforms to be thrown under the train. Similarly, a “pulling force” occurs when ships move in a parallel course: for example, similar incidents occurred with the Olympic liner .

Influence of the diagram of velocities in the channel on the specific kinetic energy of the flow. Its account in the Bernoulli equation.

Cavitation, causes, conditions of occurrence, measures to combat cavitation. Determination of the possibility of cavitation using the Bernoulli equation.

Cavitation is a phenomenon that occurs in a liquid at high fluid velocities, i.e. at low pressures. Cavitation is a violation of the continuity of a liquid with the formation of steam and gas bubbles (caverns), caused by a drop in the static pressure of the liquid below the saturated vapor pressure of this liquid at a given temperature.

p2 = pnp = f(t) - the condition for the occurrence of cavitation

Measures to combat cavitation:

Reduced fluid velocity in the pipeline;

Reducing differences in pipeline diameters;

Increasing the working pressure in hydraulic systems (pressurization of tanks with compressed gas);

Installation of the suction port of the pump is not higher than the permissible suction height (from the pump passport);

Application of cavitation-resistant materials.