h c= h d , (4.7)
where h c is the distance from the free surface of the liquid to the center of gravity, m;
h d is the distance from the free surface of the liquid to the center of pressure, m.
If some pressure also acts on the free surface of the liquid R , then the force of the total overpressure on a flat wall is equal to:
R = (R + ρ · g· h) F, (4.8)
Where R is the pressure acting on the free surface of the liquid, Pa.
The question of determining the pressure force of a liquid on flat walls is often encountered when calculating the strength of various tanks, pipes and other hydraulic structures.
Fluid pressure on a cylindrical surface.
Horizontal pressure force component on a cylindrical surface see fig. 4.5 is equal to the fluid pressure force on the vertical projection of this surface and is determined by the formula:
R x = ρ · g· h c F y , (4.9)
where R X is the horizontal component of the pressure force on the cylindrical surface, H;
Fy is the vertical projection of the surface, m 2.
vertical pressure force component is equal to the gravity of the fluid in the volume of the pressure body and is determined by the formula:
R y= ρ · g· V, (4.10)
where R at is the vertical component of the pressure force on the cylindrical surface, H;
V– total volume obtained as a result of summation of elementary volumes ΔV , m 3.
Volume V called pressure body and is the volume of liquid bounded from above by the level of the free surface of the liquid, from below by the considered curvilinear surface of the wall wetted by the liquid, and from the sides by vertical surfaces drawn through the boundaries of the wall.
Total fluid pressure force defined as the resultant force R x and RU according to the formula:
R = √P x 2 + P y 2 , (4.11)
where R is the total force of fluid pressure on a cylindrical surface, H.
Injection β , composed of the resultant with the horizon, is determined from the condition by the formula:
tgβ = R y / R x, (4.12)
where β is the angle formed by the resultant with the horizon, hail.
Fluid pressure on pipe walls.
Let's determine the force of pressure R liquid on the wall of a round pipe with a long l with inner diameter d .
Neglecting the mass of the liquid in the pipe, we compose the equilibrium equation:
p· l· d = P x= P y= P , (4.13)
where l· d is the area of the diametrical section of the pipe, m 2;
P is the desired force of fluid pressure on the pipe wall, H.
Required pipe wall thickness is determined by the formula:
δ = p· d / (2σ ), (4.14)
where σ is the allowable tensile stress of the wall material, Pa.
Obtained by the formula ( 4.14 ) the result is usually increased by α
δ = p· d / (2σ ) + α , (4.15)
where α - safety factor that takes into account possible corrosion, inaccuracy of the ebb, etc.
α = 3…7.
Work procedure
5.2. Familiarize yourself with pressure measuring instruments.
5.3. Convert the pressure dimensions of various technical systems into the pressure dimension of the international SI system - Pa:
740 mmHg Art.;
2300 mm w.c. Art.;
1.3 at;
2.4 bar;
0.6 kg/cm 2 ;
2500 N/cm2.
5.4. Solve problems:
5.4.1. The rectangular open tank is designed to store water. Determine the pressure forces on the walls and bottom of the tank, if the width a , length b , volume V . Take data from tab. 5.1 (odd options ).
Table 5.1
Data for odd variants (clause 5.4.1.)
| Options | Option | ||||||||
| V, m 3 | |||||||||
| a, m | |||||||||
| b, m | |||||||||
| Options | Option | ||||||||
| V, m 3 | |||||||||
| a, m | |||||||||
| b, m |
5.4.2. Determine the forces of liquid pressure on the bottom and side surface of a vertically located cylinder in which water is stored, if the diameter of the cylinder corresponds to the number of letters in the name (passport) in m, and the height of the cylinder is the number of letters in the surname in m (even options ).
5.5. Make a conclusion.
6.1. Draw diagrams of devices for measuring pressure: fig. 4.1 liquid barometers ( Var. 1…6; 19…24), rice. 4.2 pressure gauges and vacuum gauges ( Var. 7…12; 25…30) and fig. 4.3 differential pressure gauges ( Var. 13…18; 31…36). Apply positions and provide specifications. Provide a brief description of the scheme.
6.2. Write down the conversion of pressure dimensions of various technical systems into the pressure dimension of the international SI system - Pa (5.3.).
6.3. Solve one problem given in p.p. 5.4.1 and 5.4.2 , according to the selected option, numerically corresponding to the student's serial number in the journal on the PAPP page.
6.4. Write a conclusion about the work done.
7 Security questions
7.1. In what units is pressure measured?
7.2. What is absolute and gauge pressure?
7.3. What is a vacuum, how to determine the absolute pressure in a vacuum?
7.4. What instruments are used to measure pressure and vacuum?
7.5. How is Pascal's law formulated? How is the pressing force of a hydraulic press determined?
7.6. How is the force of liquid pressure on vertical, horizontal and inclined flat walls determined? How is this force directed? Where is the point of its application?
Practice #5
The study of the device of the sump, its calculation
performance and area of deposition
Objective
1.1. The study of the device of various sedimentation tanks.
1.2. Instilling the skills to determine the productivity and area of sedimentation of the sump.
9. Determination of the pressure force of a fluid at rest on flat surfaces. Center of pressure
In order to determine the force of pressure, we will consider a fluid that is at rest relative to the Earth. If we choose an arbitrary horizontal area ω in the liquid, then, provided that p atm = p 0 acts on the free surface, excess pressure is exerted on ω:
R iz = ρghω. (one)
Since in (1) ρgh ω is nothing but mg, since h ω and ρV = m, the excess pressure is equal to the weight of the fluid contained in the volume h ω . The line of action of this force passes through the center of the area ω and is directed along the normal to the horizontal surface.
Formula (1) does not contain a single quantity that would characterize the shape of the vessel. Therefore, R izb does not depend on the shape of the vessel. Therefore, an extremely important conclusion follows from formula (1), the so-called hydraulic paradox- with different shapes of vessels, if the same p 0 appears on the free surface, then if the densities ρ, areas ω and heights h are equal, the pressure exerted on the horizontal bottom is the same.
When the bottom plane is inclined, wetting of the surface with area ω takes place. Therefore, unlike the previous case, when the bottom lay in a horizontal plane, it cannot be said that the pressure is constant.
To determine it, we divide the area ω into elementary areas dω, any of which is subject to pressure
By definition of pressure force,
where dP is directed along the normal to the area ω.
Now, if we determine the total force that acts on the area ω, then its value is:
Having determined the second term in (3), we find Р abs.
Pabs \u003d ω (p 0 + h c. e). (4)
We have obtained the desired expressions for determining the pressures acting on the horizontal and inclined
plane: R izb and R abs.
Consider one more point C, which belongs to the area ω, more precisely, the point of the center of gravity of the wetted area ω. At this point, the force P 0 = ρ 0 ω acts.
The force acts at any other point that does not coincide with point C.
Center of pressure the point at which the line of action of the resultant of the pressure forces of the environment (liquid, gas) applied to a resting or moving body intersects with some plane drawn in the body. For example, for an airplane wing ( rice.
) C. d. is defined as the point of intersection of the line of action of the aerodynamic force with the plane of the wing chords; for a body of revolution (body of a rocket, airship, mine, etc.) - as the point of intersection of the aerodynamic force with the plane of symmetry of the body, perpendicular to the plane passing through the axis of symmetry and the velocity vector of the center of gravity of the body. The position of the center of gravity depends on the shape of the body, and for a moving body it may also depend on the direction of motion and on the properties of the environment (its compressibility). Thus, at the wing of an aircraft, depending on the shape of its airfoil, the position of the central airfoil may change with a change in the angle of attack α, or it may remain unchanged (“a profile with a constant central airfoil”); in the latter case x cd ≈ 0,25b (rice.
). When moving at supersonic speed, the center of gravity shifts significantly towards the tail due to the influence of air compressibility. A change in the position of the central engine of moving objects (aircraft, rocket, mine, etc.) significantly affects the stability of their movement. In order for their movement to be stable in the event of a random change in the angle of attack a, the central air must shift so that the moment of aerodynamic force about the center of gravity causes the object to return to its original position (for example, with an increase in a, the central air must shift towards the tail). To ensure stability, the object is often equipped with an appropriate tail unit. Lit.: Loitsyansky L. G., Mechanics of liquid and gas, 3rd ed., M., 1970; Golubev V.V., Lectures on the theory of the wing, M. - L., 1949. The position of the center of flow pressure on the wing: b - chord; α - angle of attack; ν - flow velocity vector; x dc - distance of the center of pressure from the nose of the body.
Great Soviet Encyclopedia. - M.: Soviet Encyclopedia. 1969-1978 .
See what the "Center of Pressure" is in other dictionaries:
This is the point of the body at which they intersect: the line of action of the resultant forces of pressure on the body of the environment and some plane drawn in the body. The position of this point depends on the shape of the body, and for a moving body it also depends on the properties of the environment ... ... Wikipedia
A point at which the line of action of the resultant of the pressure forces of the environment (liquid, gas) applied to a body at rest or moving intersects with a certain plane drawn in the body. For example, for an airplane wing (Fig.) C. d. determine ... ... Physical Encyclopedia
The conditional point of application of the resultant aerodynamic forces acting in flight on an aircraft, projectile, etc. The position of the center of pressure depends mainly on the direction and speed of the oncoming air flow, as well as on the external ... ... Marine Dictionary
In hydroaeromechanics, the point of application of the resultant forces acting on a body moving or at rest in a liquid or gas. * * * CENTER OF PRESSURE CENTER OF PRESSURE, in hydroaeromechanics, the point of application of the resultant forces acting on the body, ... ... encyclopedic Dictionary
center of pressure- The point at which the resultant of pressure forces is applied, acting from the side of a liquid or gas on a body moving or resting in them. Engineering topics in general… Technical Translator's Handbook
In hydroaeromechanics, the point of application of the resultant forces acting on a body moving or at rest in a liquid or gas ... Big Encyclopedic Dictionary
The point of application of the resultant aerodynamic forces. The concept of C. D. is applicable to the profile, wing, aircraft. In the case of a flat system, when the lateral force (Z), transverse (Mx) and track (My) moments can be neglected (see Aerodynamic forces and ... ... Encyclopedia of technology
center of pressure- slėgimo centras statusas T sritis automatika atitikmenys: angl. center of pressure vok. Angriffsmittelpunkt, m; Druckmittelpunkt, m; Druckpunkt, m rus. center of pressure, m pranc. center de poussee, m … Automatikos terminų žodynas
center of pressure- slėgio centras statusas T sritis fizika atitikmenys: engl. center of pressure vok. Druckmittelpunkt, m rus. center of pressure, m pranc. center de pression, m … Fizikos terminų žodynas
center of pressure Encyclopedia "Aviation"
center of pressure- center of pressure - the point of application of the resultant aerodynamic forces. The concept of C. D. is applicable to the profile, wing, and aircraft. In the case of a flat system, when the lateral force (Z), transverse (Mx) and track (My) can be neglected ... ... Encyclopedia "Aviation"
Books
- Historians of the Iron Age, Gordon Alexander Vladimirovich. The book examines the contribution of Soviet scientists to the development of historical science. The author seeks to restore the connection of times. He believes that the history of historians deserves not ...
n1.doc
Center of pressure
T.K.r 0 is transmitted to all points of area A equally, then its resultant F 0 will be applied at the center of mass of area A. To find the point of application of the pressure force F W from the weight of the liquid (t.D), we apply the theorem of mechanics according to which: the moment of the resultant force about the x-axis is equal to the sum of the moments of the component forces.Y d - coordinate of the point of application of the force F w.
We express the forces F w through the coordinates y c and y and then we get
- the moment of inertia of the area A about the x-axis.
then 
(1)
J x0 - moment of force of area A relative to the central axis parallel to x 0. thus, the point of application of the force F W located below the center of mass of the wall, the distance between them is determined by the expression
(2)
If the pressure p 0 is equal to atmospheric pressure, then the center of pressure.
At p 0 > p atm, the center of pressure is located as the point of application of the resultant 2x forces F 0 and F well. The larger F 0 compared to F W, the closer the center of pressure is to the center of mass of area A.
In a liquid, only force distributions are possible, so the centers of pressure are taken conditionally.
with silts of pressure on curved walls
Consider a cylindrical surface AB with a generatrix perpendicular to the square of the drawing and determine the pressure force on this surface AB. Let us single out the volume of the liquid by the bounded surface AB. Vertical planes drawn through the boundaries of this section and the free surface of the liquid i.e. the volume of ABSD and consider the conditions for its equilibrium in the vertical and horizon. directions.
If the fluid acts on the wall with a force F, then the walls AB act with a force F directed in the opposite direction (reaction force). We decompose the reaction force into 2 components, horizon and vertical. Equilibrium condition in the vertical direction:
(1)
G is the weight of the allocated liquid volume
And g - the area of the horizontal projection of the line AB.
The condition of equilibrium in the horizontal direction is written taking into account the fact that the forces of fluid pressure on the surfaces of the EU and AD are mutually balanced. Only the force of pressure on BE remains, then
h c - depth of location of the center of mass of the area BE.
pressure force 
9. Model of an ideal liquid. Bernoulli equation
An ideal liquid is understood as a liquid that is absolutely incompressible and non-expandable, unable to resist stretching and shear, and also devoid of the property of evaporation. The main difference from a real liquid is its lack of viscosity, i.e. ( 
=0).
Consequently, in a moving ideal fluid, only one type of stress is possible - the compressive stress (p ).
The basic equations that allow solving the simplest problems of the motion of an ideal fluid are the flow equation and the Bernoulli equation.
The Bernoulli equation for the flow of an ideal fluid expresses the law of conservation of the specific energy of the fluid along the flow. Under the specific understand the energy related to the unit weight, volume or mass of the liquid. If we relate the energy to a unit of weight, then in this case the Bernoulli equation, written for the flow of an ideal fluid, has the form
where z - vertical coordinates of the centers of gravity of the sections;
- piezometric height, or specific pressure energy; - pressure, or specific kinetic energy; H is the total head, or the total specific energy of the fluid.
If the energy of the liquid is related to a unit of its volume, the equation takes the form:
E 
If the energy of the liquid is attributed to a unit of mass, then the 3rd formula can be obtained:
10. Bernoulli equation for real fluid flow.
When a real (viscous) fluid moves in a tube, the flow is decelerated due to the influence of viscosity, and also due to the action of molecular cohesion forces between the fluid and the walls, therefore, the velocity reaches its highest value in the central part of the flow, and as it approaches the wall, they decrease practically. down to zero. The result is a velocity distribution:
In addition, the movement of a viscous fluid is accompanied by particle rotation, vortex formation, and mixing. All this requires an expenditure of energy, and therefore the specific energy of a moving viscous fluid does not remain constant, as in the case of an ideal fluid, but is gradually spent on overcoming resistances and, consequently, decreases along the flow. Thus, when passing from an elementary stream of an ideal liquid to a flow of a real (viscous) liquid, it is necessary to take into account: 1) uneven velocities along the flow cross section; 2) energy loss (pressure). Taking into account these features, the motion of a viscous fluid, the Bernoulli equation has the form:
(1) .
- total loss of total head between the considered sections 1-1 and 2-2 due to the viscosity of the liquid; 
- Coriolis coefficient, takes into account the uneven distribution of V over the cross sections and is equal to the ratio of the actual kinetic energy of the flow of the kinetic energy of the same flow at a uniform
11 Bernoulli's equation for relative motion
The Bernoulli equation in the formulas and is valid in those cases of a steady flow of a liquid, when only gravity acts on the liquid from the body forces. However, sometimes it is necessary to consider such flows, in the calculation of which, in addition to the force of gravity, one should take into account the inertia forces of the portable motion. If the inertial force is constant in time, then the fluid flow relative to the channel walls can be steady-state, and the Bernoulli equation can be derived for it
Did and. To the left side of the equation, to the work of the pressure and gravity forces, one should add the work of the inertial force acting on the jet element with the weight dG when moving from the section 1 -1 in section 2 -2 . Then we divide this work, as well as other terms of the equation, by dG, i.e., we refer to the unit of weight, and, having received some pressure, we transfer it to the right side of the equation. We obtain the Bernoulli equation for the relative motion, which in the case of a real flow takes the form
Where ? Ning - the so-called inertial force, which is the work of the inertia force, related to the unit of weight and taken with the opposite sign (the reverse sign is due to the fact that this work is transferred from the left side of the equation to the right).
Rectilinear uniformly accelerated movement of the channel. If the channel along which the fluid flows moves in a straight line with constant acceleration? (Fig. 1.30, a), then all fluid particles are affected by the same and time-constant force of inertia of portable motion, which can promote or hinder the flow. If this force is attributed to a unit of mass, then it will be equal to the corresponding acceleration? and is directed in the direction opposite to it, and the force of inertia will act on each unit of fluid weight alg. The work of this force when moving the liquid from the section 1- 1 in section 2-2 (as well as the work of gravity) does not depend on the shape of the path, but is determined only by the difference in coordinates counted in the direction of acceleration and, therefore,
Where 1 a - projection of the section of the channel under consideration on the direction of acceleration a.
If acceleration? directed away from the section 1-1 to section 2-2, and the force of inertia is vice versa, then this force impedes the flow of the liquid, and the inertial head must have a plus sign. In this case, the inertial head reduces the head in the section
2-2 compared to the head in the section 1-1 and therefore similar to hydraulic losses? h a , which always enter the right side of the Bernoulli equation with a plus sign. What if acceleration? directed from section 2- 2 to section 1 -1, then the force of inertia contributes to the flow and the inertial pressure must have a minus sign. In this case, the inertial head will increase the head in section 2-2, i.e., it will, as it were, reduce hydraulic losses.
2. Rotation of the channel around the vertical axis. Let the channel along which the fluid moves rotate around a vertical axis with a constant angular velocity? (Fig. 1.30, b). Then the force of inertia of rotational motion, which is a function of the radius, acts on the liquid. Therefore, to calculate the work of this force or the change in potential energy due to its action, it is necessary to apply integration. 
12. Similarity of hydromechanical processes
There are 2 stages in the study of real liquids.
Stage 1 - the selection of those factors that are decisive for the process under study.
Stage 2 of the study is to establish the dependence of the quantity of interest on the system of selected determining factors. This stage can be carried out in two ways: analytical, based on the laws of mechanics and physics, and experimental.
Problems can be solved by theory hydrodyne mimic likeness (similarity of incompressible fluid flows). Hydrodynamic similarity consists of three components; geometric similarity, kinematic and dynamic.
Geometric similarity - understand the similarity of those surfaces that limit flows, i.e. sections of channels, as well as sections that are located immediately in front of them and behind them and which affect the nature of the flow in the sections under consideration.
The ratio of two similar sizes of similar channels will be called a linear scale and denoted by 
.This value is the same for similar channels a and b:
Kinematics to oh similarity- means the proportionality of local velocities at similar points and the equality of the angles characterizing the direction of these 
speeds:
Where k is the speed scale, which is the same for kinematic similarity.
As 
(where T- time, 
- time scale).
Dynamic similarity is the proportionality of the forces acting on similar volumes in kinematically similar flows and the equality of the angles characterizing the direction of these forces.
Different forces usually act in fluid flows: pressure forces, viscosity (friction), gravity, etc. Compliance with their proportionality means complete hydrodynamic similarity. We take the forces of inertia as a basis and we will compare other forces acting on the liquid with the inertial ones the general form of the law of hydrodynamic similarity, Newton's number (Ne):
Here under R the main force is implied: the force of pressure, viscosity, gravity, etc.
Criterion 1. Euler number. Only pressure and inertia forces act on the liquid. Then 
and the general law is: 
Consequently, the condition for hydrodynamic similarity of geometrically similar flows in this case is the equality of their Euler numbers.
Criterion 2. Reynolds number. The fluid is affected by the forces of viscosity, pressure and inertia. Then
And the condition after dividing the last expression by pv 2 L 2 will take the form 
Consequently, the condition for hydrodynamic similarity of geometrically similar flows in the case under consideration is the equality of the Reynolds numbers calculated for similar flow sections.
Criterion 3. Froude number Fluid is affected by gravity, pressure and inertia. Then 
And the general GP law has the form: 
whether 
Consequently, the condition for hydrodynamic similarity of geometrically similar flows in the case under consideration is the equality of the Froude numbers calculated for similar flow sections.
Criterion 4: Weber number. When considering flows associated with surface tension (spraying fuel in engines), it is equal to the ratio of surface tension forces to inertia forces. For this case, the general GP law takes the form:
Criterion 5. Strouhal number. When considering unsteady (nonstationary) periodic flows with a period T(for example, flows in a pipeline connected to a piston pump), takes into account the inertial forces from unsteadiness, called local. The latter are proportional to the mass (RL 3
)
and acceleration which, in turn, is proportional to 
.Consequently, the general GP law takes the form
Criterion 6. Mach number. When considering the movements of a fluid, taking into account its compressibility (for example, the movements of emulsions). Takes into account elastic forces. The latter are proportional to the area (L 2
)
and bulk modulus of elasticity K =
.
Therefore, the elastic forces are proportional
13. Hydraulic resistance
There are two types of hydraulic pressure losses: local losses and friction losses along the length. Local pressure losses occur in the so-called local hydraulic resistance, i.e., in places where the shape and size of the channel change, where the flow is somehow deformed - expands, narrows, bends - or a more complex deformation takes place. Local losses are expressed by the Weisbach formula
(1)
Where ? - the average flow velocity in the section in front of the local resistance (during expansion) or behind it (during narrowing) and in those cases when pressure losses in hydraulic fittings for various purposes are considered; ? m- dimensionless coefficient of local resistance. The numerical value of the coefficient ? is mainly determined by the form of local resistance, its geometric parameters, but sometimes the Reynolds number also affects. It can be assumed that in the turbulent regime the coefficients of local resistances ? do not depend on the Reynolds number and, therefore, as can be seen from formula (1), the pressure loss is proportional to the square of the velocity (quadratic resistance mode). In the laminar regime, it is assumed that
(2)
Where BUT- number determined by the form of local resistance; ? kv - coefficient of local resistance in the quadratic resistance mode, i.e. at Re??.
Loss of pressure due to friction along the length l are determined by the general Darcy formula
(3)
Where is the dimensionless friction drag coefficient ? is determined depending on the flow regime:
in laminar mode ? l the Reynolds number is uniquely determined, i.e.
In turbulent conditions ? m, in addition to the Reynolds number, also depends on the relative roughness?/d, i.e.
14 Length resistance.
Friction loss along the length, these are the energy losses that occur in pure form in straight pipes of constant cross section, i.e. with uniform flow, and increase in proportion to the length of the pipe. The losses considered are due to internal friction in the liquid, and therefore take place not only in rough, but also in smooth pipes. The loss of pressure due to friction can be expressed by the general formula for hydraulic losses, i.e.
h Tp = J Tp 
2 /(2g), or in units of pressure 
Dimensionless coefficient of kneading loss factorfor friction along the length, or the Daren coefficient. It can be considered as a coefficient of proportionality between the loss of pressure due to friction, and the product of the relative length of the pipe and the velocity head.
P 
In turbulent flow, local head losses can be considered proportional to the velocity (flow rate) to the second degree, and the loss coefficients J are determined mainly by the form of local resistance and practically do not depend on Re, then in laminar flow, the head loss should be considered as the sum 
,
Where 
- loss of pressure due to the direct action of friction forces (viscosity) in a given local resistance and proportional to the viscosity of the fluid and speed to the first degree 
- loss associated with flow separation and vortex formation in the local resistance itself or behind it is proportional to the speed to the second degree.
The gradually expanding pipe is called a diffuser. The flow of liquid in the diffuser is accompanied by a decrease in speed and an increase in pressure, and, consequently, the conversion of the kinetic energy of the liquid into pressure energy. The particles of the moving liquid overcome the increasing pressure due to their kinetic energy, which decreases along the diffuser and, what is especially important, in the direction from the axis to the wall. The layers of liquid adjacent to the pillars have such a low kinetic energy that sometimes they are unable to overcome the increased pressure, they stop or even begin to move back. The reverse movement (counterflow) causes the main flow to separate from the wall and vortex formation. an increase in the expansion angle of the diffuser, and along with this, the losses due to vortex formation also increase.The total pressure loss in the diffuser is conditionally considered as the sum of two terms 
A sudden narrowing of a channel (pipe) always causes less energy loss than a sudden expansion with the same area ratio. In this case, the loss is due, firstly, to the friction of the flow at the entrance to the narrow pipe and, secondly, to the losses due to vortex formation. The latter are caused by the fact that the flow does not flow around the input corner, but breaks off from it and narrows; the annular space around the narrowed part of the flow is filled with swirling fluid.
15. Laminar regime of fluid motion
This mode is x-Xia parallel to the jet concentrated motion of particles. All the main regularities of this flow are derived analytically.
R 
distribution of velocities and shear stresses over the section. Consider a steady laminar flow W in a pipe with a circular cross section of radius r. Let the pressure in the section 1-1 Р 1, and in the section 2-2 Р 2. Given that Z 1 \u003d Z 2, we write the Bernoulli equation:
P 1 /? Chg \u003d P 2 /? Chg + htr. (htr - head loss along the length)
Htr \u003d (P 1 - P 2) /? Chg \u003d P TR /? Chg.
Let's select a cylinder in the flow. Volume W, radius y and length ℓ. For this volume, we write down the equation of uniform motion, i.e. equality 0 of the sum of pressure forces and resistance forces:
RtrCh?Chu 2 – 2H?ChuChℓCh?=0 (1)
?
are shear stresses on the side surfaces of the cylinder.
Flow rate and average flow rate
In the cross section of the flow, we single out an elementary section of the annular section with a radius y and a width dу. Elementary flow through the site dA: dQ=VЧdA (1)
Knowing: dA=2H?ChyChdy and Vtr=Ptr/4Ch?Chℓ we express:
DQ \u003d (Ptr / 4H? Hℓ) H (r 2 -y 2) H2H? ChyChdy = \u003d (? ChPtr / 2H? Hℓ) H (r 2 -y 2) ChyChdy (2)
We integrate (2) over the cross-sectional area of the pipe (from y=0 to y=r):
Q \u003d (? Ptr / 2H? Hℓ) 
(r 2 -y 2)Chydy \u003d (? Ptr / 8? ℓ) Chr 4 (3)
Substitute in (3) r=d/2: Q=(?d 4 /128?ℓ)ChPtr (4)
Average speed over the section: Vav=Q/?r 2 (5). Let us substitute (3) into (5) then the average velocity of the laminar section in the pipe: Vav = (r 2 /8?ℓ)ChRtr. The average laminar flow velocity in a round pipe is 2 times less than max, i.e. Vav=0.5Vmax.
Head loss in laminar fluid flow
The friction head loss Ptr is found from the formula for the flow rate:
Q=(?ChPtr/8?ℓ) Ch r 4 , Рtr=(8Q?ℓ/?Chr 4) (1) Divide by?g and replace?=?Ch?
Рtr=?ghtr, replace r=d/2, then htr=Рtr/?g=(128?ℓ/?gd 4)ЧQ (2)
Z.-n resistance (2) shows that the frictional head loss in a round pipe is proportional to the flow rate and viscosity to the 1st power and inversely proportional to the diameter to the 4th power.
Z.-n Poiselle is used for calculations in laminar motion. Let's replace the flow rate Q=(?d 2 /4) HVavg and then divide the resulting expression by Vcp and multiply by Vcp:
Htr \u003d (128? ℓ /? gd 4) H (? d 2 / 4) H Vcr \u003d
\u003d (64? / Vcrd) H (ℓ / d) H (V 2 cp / 2g) \u003d
\u003d (64 / Re) H (ℓ / d) H (V 2 cp / 2g) \u003d? H (V 2 cf ℓ / 2gCh d). ?
F.-la Weisbon-Darcy.
Coefficient-t of Weisbon-Darcy - coefficient-t of friction losses for laminar flow: ?=64/Re.
16. Turbulent (TRB) mode of fluid movement
For flow TRBs, but the pressure, the phenomenon of pulsation, speed, i.e. different changes in pressure and velocity at a given point in time in magnitude and direction. If, in the laminar regime, energy is spent only to overcome the forces of internal friction between the W layers, then in the TRB mode, in addition, energy is expended on the process of chaotic mixing of the W, which causes additional losses.
With TRB, a very thin laminar sublayer is formed near the pipe walls, a cat. significantly affects the velocity distribution over the flow cross section. The more intense the mixing of the flow and the greater the equalization of the velocity over the cross section, the smaller the laminar sublayer. The distribution of speeds in the TRB mode is more even. Plot of speed:
O 
ratio cf. speed to max for flow TRB: Vav/Vmax=0.75…0.90 ? tends to the limit up to 1 for large numbers.
The main calculation formula for head loss in turbulent flow in round pipes is the formula called the Weisbach-Darcy formula:
Where 
- friction loss coefficient in turbulent flow, or Darcy coefficient.
17. Summary of the most commonly used formulas for the hydraulic coefficient of friction.
Friction loss
along the length, these are the energy losses that occur in pure form in straight pipes of constant cross section, i.e. with uniform flow, and increase in proportion to the length of the pipe. The losses under consideration are due to internal friction in the liquid, and therefore take place not only in rough, but also in smooth pipes.
The loss of pressure due to friction can be expressed by the general formula for hydraulic losses
.
However, a more convenient coefficient 
relate to the relative length of the pipe l/d.
; 
Or in units of pressure 
Let there be a figure of arbitrary shape with area ω in the plane Ol , inclined to the horizon at an angle α (Fig. 3.17).
For the convenience of deriving a formula for the fluid pressure force on the figure under consideration, we rotate the wall plane by 90 ° around the axis 01 and align it with the drawing plane. On the plane figure under consideration, we single out at a depth h from the free surface of the liquid to an elementary area d ω . Then the elementary force acting on the area d ω , will
Rice. 3.17.
Integrating the last relation, we obtain the total force of fluid pressure on a flat figure
Considering that , we get
The last integral is equal to the static moment of the platform with respect to the axis OU, those.
where l With – axle distance OU to the center of gravity of the figure. Then
Since then
those. the total force of pressure on a flat figure is equal to the product of the area of the figure and the hydrostatic pressure at its center of gravity.
The point of application of the total pressure force (point d , see fig. 3.17) is called center of pressure. The center of pressure is below the center of gravity of a flat figure by an amount e. The sequence for determining the coordinates of the center of pressure and the magnitude of the eccentricity is described in paragraph 3.13.
In the particular case of a vertical rectangular wall, we get (Fig. 3.18)
Rice. 3.18.
In the case of a horizontal rectangular wall, we will have
hydrostatic paradox
The formula for the pressure force on a horizontal wall (3.31) shows that the total pressure on a flat figure is determined only by the depth of the center of gravity and the area of the figure itself, but does not depend on the shape of the vessel in which the liquid is located. Therefore, if we take a number of vessels, different in shape, but having the same bottom area ω g and equal liquid levels H , then in all these vessels the total pressure on the bottom will be the same (Fig. 3.19). Hydrostatic pressure is due in this case to gravity, but the weight of the liquid in the vessels is different.
Rice. 3.19.
The question arises: how can different weights create the same pressure on the bottom? It is this seeming contradiction that constitutes the so-called hydrostatic paradox. The disclosure of the paradox lies in the fact that the force of the weight of the liquid actually acts not only on the bottom, but also on other walls of the vessel.
In the case of a vessel expanding upward, it is obvious that the weight of the liquid is greater than the force acting on the bottom. However, in this case, part of the weight force acts on the inclined walls. This part is the weight of the pressure body.
In the case of a vessel tapering to the top, it suffices to recall that the weight of the pressure body G in this case is negative and acts upward on the vessel.
Center of pressure and determination of its coordinates
The point of application of the total pressure force is called the center of pressure. Determine the coordinates of the center of pressure l d and y d (Fig. 3.20). As is known from theoretical mechanics, at equilibrium, the moment of the resultant force F about some axis is equal to the sum of the moments of the constituent forces dF about the same axis.
Rice. 3.20.
Let's make the equation of the moments of forces F and dF about the axis OU:
Forces F and dF define by formulas
