h c = h d , (4.7)
Where h c– distance from the free surface of the liquid to the center of gravity, m;
h d– 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 total excess pressure on a flat wall is equal to:
R = (R + ρ · g· h) F, (4.8)
Where R – pressure acting on the free surface of the liquid, Pa.
The issue of determining the force of liquid pressure 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 force of fluid pressure 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– horizontal component of the pressure force on a cylindrical surface, N;
Fy– vertical projection of the surface, m 2.
Vertical pressure force component is equal to the gravity of the liquid in the volume of the pressure body and is determined by the formula:
R y = ρ · g· V, (4.10)
Where R at– vertical component of the pressure force on a cylindrical surface, N;
V– total volume obtained as a result of summation of elementary volumes ΔV , m 3.
Volume V called body pressure and represents the volume of liquid limited from above by the level of the free surface of the liquid, from below by the considered curved 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 is defined as the resultant force R x And RU according to the formula:
R = √P x 2 + P y 2 , (4.11)
Where R – total force of fluid pressure on a cylindrical surface, N.
Corner β , composed of the resultant with the horizon, is determined from the condition using the formula:
tg β = R y/ R x, (4.12)
Where β – the angle made by the resultant with the horizon, hail.
Fluid pressure on pipe walls.
Let's determine the force of pressure R liquid onto the wall of a long round pipe l with inner diameter d .
Neglecting the mass of the liquid in the pipe, we create an equilibrium equation:
p· l· d = P x = P y= P , (4.13)
Where l· d – diametrical cross-sectional area of the pipe, m 2;
P– the required force of liquid pressure on the pipe wall, N.
Necessary pipe wall thickness determined by the formula:
δ = p· d / (2σ ), (4.14)
Where σ – permissible tensile strength 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 taking into account possible corrosion, inaccuracy of low tide, etc.
α = 3…7.
Work procedure
5.2. Familiarize yourself with instruments for measuring pressure.
5.3. Convert pressure dimensions of various technical systems into pressure dimensions international system SI – Pa:
740 mmHg Art.;
2300 mm water. Art.;
1.3 at;
2.4 bar;
0.6 kg/cm 2 ;
2500 N/cm2.
5.4. Solve problems:
5.4.1. A 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 table 5.1 (odd options ).
Table 5.1
Data for odd options (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 cylinder located vertically, 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. Draw 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). List the positions and provide specifications. Lead short description scheme.
6.2. Write down the transformation of the pressure dimensions of various technical systems into the pressure dimensions of the international SI system - Pa (clause 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 down a conclusion about the practical work done.
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 measure excess pressure and vacuum?
7.5. How is Pascal's law formulated? How is pressing force determined? hydraulic press?
7.6. How is the force of fluid pressure on vertical, horizontal and inclined flat walls determined? How is this force directed? Where is its point of application?
Practical lesson No. 5
Study of the settling tank design, its calculation
productivity and settling area
Goal of the work
1.1. Study of the design of various settling tanks.
1.2. Instilling skills in determining the productivity and settling area of a settling tank.
The point of application of the resulting force of fluid pressure on any surface is called the center of pressure.
In relation to Fig. 2.12 the center of pressure is the so-called D. Let's determine the coordinates of the center of pressure (x D ; z D) for any flat surface.
From theoretical mechanics it is known that the moment of the resultant force about an arbitrary axis is equal to the sum of the moments of the component forces about the same axis. In our case, we take the Ox axis as the axis (see Fig. 2.12), then
It is also known that is the moment of inertia of the area relative to the axis Ox
As a result we get
Let us substitute formula (2.9) into this expression for F and geometric ratio:
Let's move the axis of the moment of inertia to the center of gravity of the site. Let us denote the moment of inertia about an axis parallel to the axis Oh and passing through T.S., through . Moments of inertia about parallel axes are related by the relation
then we will finally get
The formula shows that the center of pressure is always located below the center of gravity of the platform, except for the case if the platform is horizontal and the center of pressure coincides with the center of gravity. For simple geometric figures, moments of inertia about an axis passing through the center of gravity and parallel to the axis Oh(Fig. 2.12) are determined by the following formulas:
for rectangle
Oh;
for an isosceles triangle
where the side of the base is parallel Oh;
for a circle
Coordinate for flat surfaces building structures most often determined by the coordinate of the location of the axis of symmetry geometric figure, limiting a flat surface. Since such figures (circle, square, rectangle, triangle) have an axis of symmetry parallel to the coordinate axis Oz, location of the axis of symmetry and determines the coordinate x D . For example, for a rectangular slab (Fig. 2.13), determining the coordinate xD clear from the drawing.
Rice. 2.13. Diagram of the location of the center of pressure for a rectangular surface
Hydrostatic paradox. Let us consider the force of liquid pressure on the bottom of the vessels shown in Fig. 2.14.
Let there be a figure of arbitrary shape with area co in the plane Ol , inclined to the horizon at an angle α (Fig. 3.17).
For the convenience of deriving the formula for the force of fluid pressure on the figure under consideration, let us rotate the plane of the wall 90° around the axis 01 and combine it with the drawing plane. Let us highlight on the flat figure under consideration at 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 flat figure
Considering that , we get
The last integral is equal to the static moment of the platform c relative to the axis OU, those.
Where l WITH – distance from axis 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 eccentricity value is set out in paragraph 3.13.
In the special case of a vertical rectangular wall we obtain (Fig. 3.18)
Rice. 3.18.
In the case of a horizontal rectangular wall we will have
Hydrostatic paradox
The formula for the force of pressure on a horizontal wall (3.31) shows that the total pressure on a flat figure is determined only by the depth of immersion 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 you 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 caused in this case by the force of 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? This apparent contradiction is what is called hydrostatic paradox. The revelation 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 towards the top, it is enough to remember that the weight of the body of pressure G in this case it 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. Let's determine the coordinates of the center of pressure l d and y d (Fig. 3.20). As is known from theoretical mechanics, in equilibrium, the moment of the resultant force F relative to a certain axis is equal to the sum of the moments of the component forces dF about the same axis.
Rice. 3.20.
Let's create an equation for moments of force F and dF relative to the axis OU:
Powers F And dF determine by formulas
- Introductory lesson for free;
- Big number experienced teachers (native and Russian-speaking);
- Courses are NOT for a specific period (month, six months, year), but for a specific number of lessons (5, 10, 20, 50);
- More than 10,000 satisfied customers.
- The cost of one lesson with a Russian-speaking teacher is from 600 rubles, with a native speaker - from 1500 rubles
Center of pressure atmospheric pressure forces p0S will be located at the center of gravity of the site, since atmospheric pressure is transmitted equally to all points of the liquid. The center of pressure of the fluid itself on the platform can be determined from the theorem on the moment of the resultant force. Resultant moment
forces about the axis OH will be equal to the sum of the moments of the component forces relative to the same axis.
Where 
where: - position of the center of excess pressure on the vertical axis, - moment of inertia of the platform S relative to the axis OH.
The center of pressure (the point of application of the resultant force of excess pressure) is always located below the center of gravity of the site. In cases where the external force acting on the free surface of the liquid is the force of atmospheric pressure, then two forces equal in magnitude and opposite in direction will simultaneously act on the wall of the vessel atmospheric pressure(on the inside and outside of the wall). For this reason, the real unbalanced force remains the force of excess pressure.
| Previous materials: |
n1.doc
Center of pressure
Since T.K.p 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 from the weight of the liquid (t.D), we apply the theorem of mechanics according to which: the moment of the resultant force relative to the x axis is equal to the sum of the moments of the component forces.Y d - coordinate of the point of application of force F.
Let's express the forces F through the coordinates y c and y and then we get
- moment of inertia of area A relative to 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 force F 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.
When p 0 > p atm, the center of pressure is located as the point of application of the resultant 2 forces F 0 and F l. The greater 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 pressure centers are taken conditionally.
from silt pressure on curved walls
Let us consider a cylindrical surface AB with a generatrix perpendicular to the plane of the drawing and determine the force of pressure on this surface AB. Let us select the volume of liquid limited by the surface AB. Vertical planes drawn through the boundaries of this area and the free surface of the liquid, i.e. volume ABCD and consider the conditions of its equilibrium in the vertical and horizontal. directions.
If the liquid acts on the wall with a force F, then the walls AB act with a force F directed in the opposite direction (reaction force). Let us decompose the reaction force into 2 components, horizontal and vertical. Equilibrium condition in the vertical direction:
(1)
G - weight of the allocated volume of liquid
A g is the area of the horizontal projection of area AB.
The equilibrium condition in the horizontal direction is written taking into account the fact that the fluid pressure forces on the surfaces EC and AD are mutually balanced. All that remains is the pressure force on BE, then
h c - depth of location of the center of mass of area BE.
Pressure force 
9. Model ideal liquid tee. Bernoulli's equation
By ideal we mean a liquid that is absolutely incompressible and non-expandable, incapable of resisting stretching and shear, and also lacking 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 - compression stress (p ).
The basic equations that allow solving the simplest problems of the movement of an ideal fluid are the flow equation and the Bernoulli equation.
Bernoulli's equation for an ideal fluid flow expresses the law of conservation of specific energy of the fluid along the flow. Specific energy is understood as energy per unit weight, volume or mass of a liquid. If we relate energy to a unit of weight, then in this case the Bernoulli equation, written for an ideal fluid flow, has the form
where z is the vertical coordinates of the centers of gravity of the sections;
- piezometric height, or specific energy pressure; - pressure, or specific kinetic energy; N- total pressure, or total specific energy of the fluid.
If the energy of a liquid is related to a unit of its volume, the equation takes the form:
E 
If the energy of the liquid is related to a unit of mass, then we can obtain the 3rd formula:
10.Bernoulli equation for real fluid flow.
When a real (viscous) liquid moves in a tube, the flow is slowed down due to the influence of viscosity, as well as due to the action of molecular adhesion forces between the liquid and the walls, therefore highest value reaches speeds in the central part of the flow, and as they approach the wall they decrease to almost zero. The result is a speed distribution:
In addition, the movement of a viscous fluid is accompanied by rotation of particles, vortex formation and mixing. All this requires energy expenditure, 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 resistance and therefore decreases along the flow. Thus, when transitioning from an elementary stream of an ideal liquid to a flow of a real (viscous) liquid, it is necessary to take into account: 1) the unevenness of velocities across the cross section of the flow; 2) loss of energy (pressure). Taking into account these features, the movement of a viscous fluid Bernoulli’s equation has the form:
(1) .
- the total loss of total pressure 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 across sections and is equal to the ratio of the actual kinetic energy of the flow to the kinetic energy of the same flow at uniform
11 Bernoulli's equation for relative motion
Bernoulli's equation in formulas is valid in those cases of steady fluid flow when, of the mass forces, only gravity acts on the fluid. However, sometimes it is necessary to consider such flows, when calculating which, in addition to the force of gravity, the inertial forces of the portable motion should be taken into account. If the inertial force is constant in time, then the fluid flow relative to the channel walls can be steady, and for it the Bernoulli equation can be derived
They did and... To the left side of the equation, to the work of the forces of pressure and gravity, we should add the work of the inertial force acting on the stream element with a weight dG when it moves from the section 1 -1 in cross section 2 -2 . Then we divide this work, like other terms of the equation, by dG, i.e., we relate it to a unit of weight, and, having received some pressure, we transfer it to the right side of the equation. We obtain the Bernoulli equation for relative motion, which in the case of a real flow takes the form
Where ? Nin - the so-called inertial pressure, which represents the work of inertia force per unit weight and taken with the opposite sign (the opposite sign is due to the fact that this work is transferred from the left side of the equation to the right).
Straightforward uniformly accelerated motion beds. What if the channel along which the fluid flows moves in a straight line with constant acceleration? (Fig. 1.30, a), then all particles of the liquid are affected by the same and time-constant force of inertia of portable motion, which can promote or impede flow. If this force is per unit mass, will it be equal to the corresponding acceleration? and is directed in the direction opposite to it, and each unit of liquid weight will be acted upon by the force of inertia alg. The work done by this force when moving fluid from the section 1- 1 in cross section 2-2 (just like the work of gravity) does not depend on the shape of the path, but is determined only by the difference in coordinates measured in the direction of acceleration and, therefore,
Where 1 A - projection of the channel section under consideration onto the direction of acceleration a.
If acceleration? directed from the section 1-1 to section 2-2, and the inertial force is the opposite, then this force impedes the flow of fluid, and the inertial pressure must have a plus sign. In this case, the inertial pressure reduces the pressure in the section
2-2 compared to the pressure in the section 1-1 and therefore similar to hydraulic losses? h a , which always appear on the right side of the Bernoulli equation with a plus sign. What if acceleration? directed from section 2- 2 to the section 1 -1, then the inertial force contributes to the flow and the inertial pressure must have a minus sign. In this case, the inertial pressure will increase the pressure in section 2-2, i.e., it will, as it were, reduce hydraulic losses.
2. Rotation of the channel around a 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 fluid is acted upon by the force of inertia of rotational motion, which is a function of the radius. Therefore, to calculate the work done by this force or the change in potential energy caused by its action, it is necessary to apply integration. 
12. Similarity of hydromechanical processes
There are 2 stages of studying real liquids.
Stage 1 - selection of those factors that are decisive for the process being studied.
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 performed in two ways: analytical, based on the laws of mechanics and physics, and experimental.
Theory allows you to solve problems hydrodyne mic resemblance (similar to 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 directly in front of them and behind them and that influence 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 angles characterizing the direction of these 
speeds:
Where k is the velocity scale, which is the same for kinematic similarity.
Because 
(Where T- time, 
- time scale).
Dynamic similarity – this 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.
Various forces usually act in fluid flows: pressure forces, viscosity (friction), gravity, etc. Compliance with their proportionality means complete hydrodynamic similarity. Let us take inertial forces as a basis and compare other forces acting on the fluid with inertial ones general form law of hydrodynamic similarity, Newton's number (Ne):
Here under R the main force is implied: the force of pressure, viscosity, gravity, or others.
Criterion 1. Euler's number. Only pressure and inertia forces act on the fluid. 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 forces of viscosity, pressure and inertia act on the fluid. 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's number The forces of gravity, pressure and inertia act on a liquid. 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 sections of flows.
Criterion 4: Weber number. When considering flows associated with surface tension (fuel atomization in engines), it is equal to the ratio of surface tension forces to inertial forces. For this case, the general GP law takes the form:
Criterion 5. Strouhal number. When considering unsteady (unsteady) periodic flows with a period T(for example, flows in a pipeline connected to a piston pump), takes into account inertia 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 law of GP takes the form
Criterion 6. Mach number. When considering the movements of a liquid 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 volumetric modulus of elasticity K =
.
Therefore, the elastic forces are proportional
13. Hydraulic resistance
There are two types of hydraulic head 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 changes, where the flow is deformed in one way or another - it 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 local resistance (during expansion) or behind it (during narrowing) and in cases where pressure losses in hydraulic fittings for various purposes are considered; ? m- dimensionless coefficient of local resistance. Numerical value of the coefficient ? is mainly determined by the shape of the local resistance, its geometric parameters, but sometimes the Reynolds number also influences. We can assume that in a turbulent regime the coefficients of local resistance ? 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 speed (quadratic resistance mode). In laminar mode, it is believed that
(2)
Where A- number determined by the form of local resistance; ? kv - local resistance coefficient in quadratic resistance mode, i.e. at Re??.
Head loss due to friction along the length l are determined by the general Darcy formula
(3)
Where is the dimensionless friction drag coefficient ? determined depending on the flow regime:
In laminar mode ? l The Reynolds number is uniquely determined, i.e.
In turbulent conditions ? t, in addition to the Reynolds number, also depends on the relative roughness?/d, i.e.
14 Resistance along the length.
Friction losses along the length are energy losses that occur in their 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 caused by internal friction in the liquid, and therefore occur not only in rough, but also in smooth pipes. The loss of pressure due to friction can be expressed using the general formula for hydraulic losses, i.e.
h Tp = Ј Tp 
2 /(2g), or in pressure units 
Dimensionless coefficient is denominated 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 pressure.
P 
In a turbulent flow, local pressure losses can be considered proportional to the speed (flow rate) to the second power, and the loss coefficients Ј are determined mainly by the form of local resistance and are practically independent of Re, while in a laminar flow, the pressure loss should be considered as the sum 
,
Where 
- loss of pressure caused by the direct action of friction forces (viscosity) in a given local resistance and proportional to the fluid viscosity and speed to the first power 
- loss associated with flow separation and vortex formation in the local resistance itself or behind it, proportional to the speed to the second power.
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 kinetic energy of the liquid into pressure energy. Particles of a moving fluid overcome increasing pressure due to their kinetic energy, which decreases along the diffuser and, most importantly, in the direction from the axis to the wall. The layers of liquid adjacent to the stools have such low kinetic energy that sometimes they are unable to overcome the increased pressure; they stop or even begin to move back. The reverse movement (countercurrent) causes separation of the main flow from the wall and vortex formation. The intensity of these phenomena increases e increasing the angle of expansion of the diffuser and at the same time the losses due to vortex formation increase. The total pressure loss in the diffuser is conventionally 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 friction of the flow at the entrance to a narrow pipe and, secondly, to losses due to vortex formation. The latter are caused by the fact that the flow does not flow around the inlet corner, but breaks away from it and narrows; the annular space around the narrowed part of the flow is filled with swirling liquid.
15. Laminar mode of fluid movement
This mode is parallel to the jet concentrated motion of particles. All the main patterns of this flow are derived analytically.
R 
distribution of velocities and tangential stresses along the section. Let us consider the steady laminar flow of fluid in a pipe of circular cross-section with radius r. Let the pressure in the section be 1-1 P 1, and in the section 2-2 P 2. Taking into account that Z 1 = Z 2, we write the Bernoulli equation:
Р 1 /?Чg = Р 2 /?Чg + htr. (htr – pressure loss along the length)
Htr=(P 1 - P 2)/ ?Chg= P TR /?Chg.
Let's select a cylinder in the flow. Volume W, radius y and length ℓ. For this volume we write the equation uniform motion, i.e. equality 0 of the sum of pressure forces and resistance forces:
RtrCh?Chu 2 – 2Ch?ChuChℓCh?=0 (1)
?
– tangential stresses on the side surfaces of the cylinder.
Flow rate and average flow rate
In the cross section of the flow, we select an elementary section of the annular section with radius y and width dу. Elementary flow rate through the platform dA: dQ=VЧdA (1)
Knowing: dA=2H?HyHdy and Vtr=Ptr/4H?Hℓ we express:
DQ=(Ptr/4H?Hℓ)H(r 2 -y 2)H2H?HyHdy= =(?Ptr/2H?Hℓ)H(r 2 -y 2) ChyHdy (2)
Let's integrate (2) over the cross-sectional area of the pipe (from y=0 to y=r):
Q=(?Ptr/2H?Hℓ) 
(r 2 -y 2)Chydy=(?Ptr/8?ℓ)Chr 4 (3)
Let's substitute r=d/2 into (3): Q=(?d 4 /128?ℓ)Трtr (4)
Average speed over the cross section: Vav=Q/?r 2 (5). Let's substitute (3) into (5) then the average speed of the laminar section in the pipe: Vav = (r 2 /8?ℓ) CHRtr. The average speed of laminar flow in a round pipe is 2 times less than max, i.e. Vav=0.5Vmax.
Pressure loss during laminar fluid movement
Friction pressure loss Ptr is found from the formula for flow rate:
Q=(?ChPtr/8?ℓ) Ch r 4, Рtr=(8Q?ℓ/?Chr 4) (1) Divide by?g and replace?=?Ch?, the pressure drop will be expressed in terms of friction pressure:
Рtr=?ghtr, replace r=d/2, then htr=Рtr/?g=(128?ℓ/?gd 4)ЧQ (2)
Z.-n resistance (2) shows that frictional head loss in a round pipe is proportional to the flow rate and viscosity to the 1st power is inversely proportional to the diameter to the 4th power.
Z. Mr. Poiselle used for calculations with laminar flow. Let us replace the flow rate Q=(?d 2 /4)ХVср and then divide the resulting expression by Vср and multiply by Vср:
Htr=(128?ℓ/?gd 4)Ч(?d 2 /4)ЧVср=
=(64?/Vcрd)Ч(ℓ/d)Ч(V 2 ср/2g)=
=(64/Re)Х(ℓ/d)Ч (V 2 ср/2g)=?Ч(V 2 срЧℓ/2gЧd). ?
F.-la Weisbon-Darcy.
Weisbon-Darcy coefficient – friction loss coefficient for laminar flow: ?=64/Re.
16.Turbulent (TRB) mode of fluid movement
For TRB flow, the pressure, the phenomenon of pulsation, speed, i.e. different changes in pressure and speed at a given point in time in magnitude and direction. If in laminar mode energy is spent only on overcoming the forces of internal friction between layers of fluid, then in TRB mode energy is also spent on the process of chaotic mixing of fluid, which causes additional losses.
With TRB, a very thin laminar sublayer is formed near the pipe walls. significantly affects the velocity distribution across the flow cross section. The more intense the flow mixing and the greater the velocity equalization across the cross section, the smaller the laminar sublayer. The speed distribution in TRB mode is more uniform. Velocity plot:
ABOUT 
attitude cf. speed to max for TRB flow: Vav/Vmax=0.75…0.90? tends to the limit of 1 for large numbers.
The basic calculation formula for pressure losses during 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 losses
along the length are energy losses that occur in their 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 caused by internal friction in the liquid, and therefore occur not only in rough, but also in smooth pipes.
The friction head loss can be expressed using the general formula for hydraulic losses
.
However, a more convenient coefficient 
relate to the relative pipe length l/d.
; 
Or in pressure units 