TASKS

To perform calculations - graphic work

In the discipline "Hydraulics"

Topic: hydrostatics

Severodvinsk

BASIC THEORETICAL PROVISIONS

Hydraulics, or technical mechanics of fluids is the science of the laws of equilibrium and movement of fluids, of the methods of applying these laws to the solution of practical problems;

Liquid They call a substance in a state of aggregation that combines the features of a solid state (very low compressibility) and a gaseous state (fluidity). The laws of equilibrium and motion of droplet liquids, within certain limits, can also be applied to gases.

A liquid can be acted upon by forces distributed over its mass (volume), called massive, and along the surface, called superficial. The first includes the forces of gravity and inertia, the second - the forces of pressure and friction.

Pressure is called the ratio of the force normal to the surface to the area. With uniform distribution

Shear stress The ratio of the frictional force tangent to the surface to the area is called:

If the pressure R counted from absolute zero, then it is called absolute (p abs), and if from conditional zero (i.e., compared with atmospheric pressure r a, That redundant(r hut):

If R abs< Р а, то имеется vacuum, whose value:

Rvac = R a - R abs

The main physical characteristic of a liquid is densityρ (kg/m3), determined for a homogeneous liquid by the ratio of its mass m to volume V:

Density fresh water at temperature T = 4°C ρ = = 1000 kg/m 3. In hydraulics, the concept is often used specific gravity γ(N/m 3), i.e. weight G units of liquid volume:

Density and specific gravity are related to each other by the relation:

Where g- acceleration of gravity.

For fresh water γ water = 9810 N/m 3

The most important physical parameters of liquids that are used in hydraulic calculations are compressibility, thermal expansion, viscosity and volatility.

Compressibility liquids are characterized by a bulk modulus of elasticity TO, included in the generalized Hooke's law:

Where ΔV- increase (in this case decrease) of liquid volume V, caused by an increase in pressure by Δр. For example, for water K water ≈2. 10 3 MPa.

Temperature expansion is determined by the corresponding coefficient equal to the relative change in volume when the temperature changes by 1 °C:

Viscosity is the ability of a fluid to resist shear. There are dynamic (μ) and kinematic (ν) viscosity. The first is included in Newton's law of fluid friction, which expresses the tangential stress τ through the transverse velocity gradient dv/dt:

Kinematic viscosity associated with dynamic ratio

The unit of kinematic viscosity is m2/s.

Volatility liquids are characterized by saturated vapor pressure as a function of temperature.

Saturated vapor pressure can be considered the absolute pressure at which a liquid boils at a given temperature. Therefore, the minimum absolute pressure at which a substance is in a liquid state is equal to the saturated vapor pressure R n.p. .

The main parameters of some liquids, their SI units and non-system units temporarily allowed for use are given in Appendices 1...3.

HYDROSTATICS

The pressure in a stationary fluid is called hydrostatic and has the following two properties:

On the outer surface of the liquid it is always directed normally to the inside of the liquid volume;

At any point inside the liquid it is the same in all directions, i.e. it does not depend on the angle of inclination of the platform along which it acts.

Equation expressing hydrostatic pressure R at any point of a stationary fluid in the case when, among the mass forces, only one force of gravity acts on it, it is called the basic equation of hydrostatics:

Where p 0- pressure on any surface of the liquid level, for example on a free surface; h- the depth of the location of the point under consideration, measured from the surface with pressure p 0.

In cases where the point under consideration is located above the surface with pressure p 0, the second term in formula (1.1) is negative.

Another form of writing the same equation (1.1) has the form

(1.2)

Where z and z 0 - vertical coordinates of an arbitrary point and a free surface, measured from horizontal plane up; p/(pg)- piezometric height.

Hydrostatic pressure can be conventionally expressed by the height of the liquid column p/ρg.

In hydraulic engineering practice, external pressure is often equal to atmospheric: P 0 = P at

The pressure value P at = 1 kg/cm 2 = 9.81. 10 4 n/m g called technical atmosphere.

A pressure equal to one technical atmosphere is equivalent to the pressure of a column of water 10 meters high , i.e.

Hydrostatic pressure determined by equation (1.1) is called complete or absolute pressure . In what follows we will denote this pressure p abs or p’. Usually in hydraulic engineering calculations one is interested not in total pressure, but in the difference between total pressure and atmospheric pressure, i.e. the so-called gauge pressure

In the following presentation we will retain the notation R for gauge pressure.

Figure 1.1

The sum of the terms gives the value total hydrostatic head

Amount -- expresses hydrostatic head N excluding atmospheric pressure p at /ρg, i.e.

In Fig. 1.1, the plane of total hydrostatic head and the plane of hydrostatic head are shown for the case when the free surface is under atmospheric pressure p 0 = p at.

A graphical representation of the magnitude and direction of hydrostatic pressure acting on any point on the surface is called a hydrostatic pressure diagram. To construct a diagram, you need to plot the value of hydrostatic pressure for the point under consideration normal to the surface on which it acts. So, for example, a diagram of manometric pressure on a flat inclined shield AB(Fig. 1.2,a) will represent a triangle ABC, and the diagram of the total hydrostatic pressure is a trapezoid A"B"C"D"(Fig. 1.2, b).

Figure 1.2

Each segment of the diagram in Fig. 1.2,a (for example OK) will represent the gauge pressure at the point TO, i.e. p K = ρgh K , and in Fig. 1.2, b - total hydrostatic pressure

The force of fluid pressure on a flat wall is equal to the product of hydrostatic pressure ρ s at the center of gravity of the wall area by the wall area S, i.e.

Center of pressure(point of application of force F) located below the center of gravity of the area or coincides with the latter in the case of a horizontal wall.

The distance between the center of gravity of the area and the center of pressure in the direction of the normal to the line of intersection of the wall plane with the free surface of the liquid is equal to

where J 0 is the moment of inertia of the wall area relative to the axis passing through the center of gravity of the area and parallel to the line of intersection of the wall plane with the free surface: y s- coordinate of the center of gravity of the area.

The force of fluid pressure on a curved wall, symmetrical relative to the vertical plane, is the sum of the horizontal F G and vertical F B components:

Horizontal component F G equal to the force of fluid pressure on the vertical projection of a given wall:

Vertical component F B equal to the weight of the liquid in volume V, enclosed between this wall, the free surface of the liquid and a vertical projection surface drawn along the contour of the wall.

If overpressure p 0 on the free surface of the liquid is different from zero, then when calculating this surface should be mentally raised (or lowered) to a height (piezometric height) p 0 /(ρg)

Floating bodies and their stability. The condition for a body to float is expressed by the equality

G=P (1.6)

Where G- body weight;

R- the resulting force of fluid pressure on a body immersed in it - Archimedean force.

Force R can be found by the formula

P=ρgW (1.7)

Where ρg- specific gravity of the liquid;

W- the volume of fluid displaced by a body, or displacement.

Force R is directed upward and passes through the center of gravity of the displacement.

Draft body at is called the immersion depth of the lowest point of the wetted surface (Fig. 1.3,a). The axis of swimming is understood as a line passing through the center of gravity WITH and center of displacement D, corresponding to the normal position of the body in a state of equilibrium (Fig. 1.3, a )-

Waterline is called the line of intersection of the surface of a floating body with the free surface of the liquid (Fig. 1.3, b). Floating plane ABEF called the plane obtained from the intersection of the body with the free surface of the liquid, or, in other words, the plane limited by the waterline.

Figure 1.3

In addition to fulfilling the navigation conditions (1.5), the body (ship, barge, etc.) must satisfy stability conditions. A floating body will be stable if, during a roll, the weight force G and Archimedean force R create a moment that tends to destroy the roll and return the body to its original position.

Figure 1.4

When the body surfaces (Fig. 1.4), the center of displacement at small roll angles (α<15°) перемещается по некоторой дуге, проведенной из точки пересечения линии действия силы R with a floating axis. This point is called the metacenter (in Fig. 1.4 the point M). In the future, we will consider stability conditions only when the body floats on the surface at small roll angles.

If the center of gravity of body C lies below the center of displacement, then swimming will be unconditionally stable (Fig. 1.4, a).

If the center of gravity of body C lies above the center of displacement D, then swimming will be stable only if the following condition is met (Fig. 1-9, b):

Where ρ - metacentric radius, i.e. the distance between the center of displacement and the metacenter

δ - distance between the center of gravity of the body C and the center of displacement D. Metacentric radius ρ is found by the formula:

where J 0 is the moment of inertia of the floating plane or area limited by the waterline relative to the longitudinal axis (Fig. 1-8.6);

W- displacement.

If the center of gravity of body C is located above the center of displacement and metacenter, then the body is unstable; emerging force couple G And R tends to increase the roll (Fig. 1.4, V).

GUIDELINES FOR SOLVING PROBLEMS

When solving problems in hydrostatics, first of all, you need to thoroughly understand and not confuse concepts such as pressure R and strength F.

When solving problems of determining the pressure at a particular point of a stationary fluid, one should use the basic equation of hydrostatics (1.1). When applying this equation, you need to keep in mind that the second term on the right side of this equation can be either positive or negative. Obviously, as the depth increases, the pressure increases, and as the depth increases, it decreases.

It is necessary to firmly distinguish between absolute, excess and vacuum pressures and be sure to know the relationship between pressure, specific gravity and the height corresponding to this pressure (piezometric height).

When solving problems in which pistons or systems of pistons are given, one should write an equilibrium equation, i.e., the sum of all forces acting on the piston (piston system) is equal to zero.

Problems should be solved in the international system of measurement units SI.

The solution to the problem must be accompanied by the necessary explanations, drawings (if necessary), a listing of the initial quantities (the “given” column), and conversion of units to the SI system.

EXAMPLES OF SOLVING PROBLEMS IN HYDROSTATICS

Task 1. Determine the total hydrostatic pressure at the bottom of a vessel filled with water. The vessel is open at the top, the pressure on the free surface is atmospheric. Depth of water in the vessel h = 0,60 m.

Solution:

In this case, we have p 0 = p at and therefore apply formula (1.1) in the form

p"=9.81.10 4 +9810. 0.6 = 103986 Pa

Answer p’=103986 Pa

Task 2. Determine the height of the water column in the piezometer above the liquid level in a closed vessel. The water in the vessel is under absolute pressure p" 1 = 1.06 at(drawing for task 2).

Solution.

Let's create equilibrium conditions for a common point A(see picture ). Point pressure A left:

Right pressure:

Equating the right-hand sides of the equations and reducing by γg we get:

The indicated equation can also be obtained by creating an equilibrium condition for points located in any horizontal plane, for example in the plane OO(see picture). Let us take the plane as the beginning of the piezometer reading scale OO and from the resulting equation we find the height of the water column in the piezometer h.

Height h is equal to:

=0.6 meters

A piezometer measures the magnitude of gauge pressure expressed by the height of a liquid column.

Answer: h = 0.6 meters

Task 3. Determine the height to which water rises in a vacuum gauge if the absolute air pressure inside the cylinder p’ = 0.95 at(Figure 1-11). State what pressure a vacuum gauge measures.

Solution:

Let's create an equilibrium condition relative to the horizontal plane O-O:

hydrostatic pressure acting from the inside:

Hydrostatic pressure in plane ABOUT-ABOUT, acting from the outside,

Since the system is in equilibrium, then

Task 4. Determine the gauge pressure at the point A pipeline, if the height of the mercury column according to the piezometer is h 2 = 25 cm. The center of the pipeline is located at h 1 = 40 cm below the dividing line between water and mercury (drawing for the problem).

Solution: Find the pressure at point B: p" B = p" A-γ h 1, since point IN located above the point A by the amount h 1. At point C the pressure will be the same as at point IN, since the pressure of the water column h mutually balanced, i.e.

hence the gauge pressure:

Substituting numeric values , we get:

r "A -r atm=37278 Pa

Answer: p" A -r atm=37278 Pa

TASKS

Task 1.1. A canister filled with gasoline and containing no air was heated in the sun to a temperature of 50 °C. How much would the pressure of gasoline inside the canister increase if it were absolutely rigid? The initial temperature of gasoline is 20 0 C. The bulk modulus of gasoline is taken equal to K = 1300 MPa, the coefficient of thermal expansion β = 8. 10 -4 1/deg.

Problem 1.2. Determine the excess pressure at the bottom of the ocean, the depth of which is h = 10 km, taking the density of sea water ρ = 1030 kg/m 3 and considering it incompressible. Determine the density of water at the same depth, taking into account compressibility and taking the modulus of bulk elasticity K = 2. 10 3 MPa.

Problem 1.3. Find the law of pressure change R atmospheric air at height z , considering the dependence of its density on pressure isothermal. In fact, up to a height of z = 11 km, the air temperature drops according to a linear law, i.e. T=T 0 -β z , where β = 6.5 deg/km. Define dependency p = f(z) taking into account the actual change in air temperature with altitude.

Problem 1.4. Determine excess water pressure in the pipe IN, if the pressure gauge reading is p m = 0.025 MPa. The connecting tube is filled with water and air, as shown in the diagram, with H 1 = 0.5 m; H 2 = 3 m.

How will the pressure gauge reading change if, at the same pressure in the pipe, the entire connecting tube is filled with water (the air is released through tap K)? Height H 3 = 5 m.

Problem 1.5. The U-shaped tube is filled with water and gasoline. Determine the density of gasoline if h b = 500 mm; h in = = 350 mm. Ignore the capillary effect.

Problem 1.6. Water and gasoline are poured into a cylindrical tank with a diameter of D = 2 m to a level of H = 1.5 m. The water level in the piezometer is h = 300 mm lower than the gasoline level. Determine the amount of gasoline in the tank if ρ b = 700 kg/m 3 .

Problem 1.7. Determine the absolute air pressure in the vessel if the reading of the mercury device is h = 368 mm, height H = 1 m. Density of mercury ρ = 13600 kg/m 3. Atmospheric pressure 736 mm Hg. Art.

Problem 1.8. Determine the excess pressure p 0 of air in the pressure tank according to the reading of a pressure gauge made up of two U-shaped tubes with mercury. The connecting tubes are filled with water. Level marks are given in meters. What height N there must be a piezometer to measure the same pressure p 0 Density of mercury ρ = 13600 kg/m 3.

Problem 1.9. Determine the force of liquid (water) pressure on a manhole cover with a diameter of D=l m in the following two cases:

1) pressure gauge reading p m = 0.08 MPa; H 0 =1.5 m;

2) reading of a mercury vacuum gauge h= 73.5 mm at a= 1m; ρ RT = 13600 kg/m 3 ; H 0 =1.5 m.

Problem 1.10. Determine the volumetric modulus of elasticity of the liquid if under the influence of a load A with a mass of 250 kg, the piston travels a distance Δh = 5 mm. Initial height of the piston position (without load) H = 1.5 m, piston diameters d = 80 mm N tank D= 300 mm, tank height h = 1.3 m. Neglect the weight of the piston. The reservoir is considered absolutely rigid.

Problem 1.11. To pressurize an underground pipeline with water (check for leaks), a manual piston pump is used. Determine the volume of water (modulus of elasticity TO= 2000 MPa), which must be pumped into the pipeline to increase the excess pressure in it from 0 to 1.0 MPa. Consider the pipeline absolutely rigid. Pipeline dimensions: length L = 500 m, diameter d = 100 mm. What is the force on the pump handle at the last moment of crimping, if the diameter of the pump piston d n = 40 mm, and the ratio of the arms of the lever mechanism a/v= 5?

Problem 1.12. Determine the absolute air pressure in the tank p 1, if at atmospheric pressure corresponding to h a = 760 mm Hg. Art., reading of the mercury vacuum gauge h RT = 0.2 m, height h = 1.5 m. What is the reading of the spring vacuum gauge? The density of mercury is ρ=13600 kg/m3.

Problem 1.13. When the pipeline tap is closed TO determine the absolute pressure in a tank buried at a depth of H = 5 m, if the reading of a vacuum gauge installed at a height of h = 1.7 m is equal to pvac = 0.02 MPa. Atmospheric pressure corresponds to p a = 740 mm Hg. Art. Density of gasoline ρ b = 700 kg/m 3.

Problem 1.14. Determine pressure p' 1, if the piezometer reading h =0.4 m. What is gauge pressure?

Problem 1.15. Define vacuum rvac and absolute pressure inside the cylinder p" in(Fig. 1-11), if the vacuum gauge reading h =0.7 m aq. Art.

1) in the cylinder and in the left tube - water , and in the right tube - mercury (ρ = 13600 kg/m 3 );

2) in the cylinder and the left tube - air , and in the right tube there is water.

Determine what percentage is the pressure of the air column in the tube from the gauge pressure calculated in the second case?

When solving a problem, take h 1 = 70 cm,h 2 = = 50 cm.

Problem 1.17. What will be the height of the mercury column h 2 (Fig. for Problem 1.16) if the gauge pressure of the oil in the cylinder A p a = 0.5 at, and the height of the oil column (ρ=800 kg/m 3) h 1 =55 cm?

Problem 1.18. Determine the height of the mercury column h 2, (figure), if the location of the pipeline center A will increase compared to that indicated in the figure and will become h 1 = 40 cm above the dividing line between water and mercury. Take the gauge pressure in the pipe to be 37,278 Pa .

Problem 1.19. Determine at what height z the mercury level in the piezometer will be established if, at gauge pressure in the pipe R A =39240 Pa and reading h=24 cm the system is in equilibrium (see figure).

Problem 1.20. Determine the specific gravity of a beam having the following dimensions: width b=30 cm, height h=20 cm and length l = 100 cm if its draft y=16 cm

Problem 1.21. A piece of granite weighs 14.72 N in air and 10.01 N in a liquid having a relative specific gravity of 0.8. Determine the volume of a piece of granite, its density and specific gravity.

Problem 1.22 A wooden beam measuring 5.0 x 0.30 m and 0.30 m high is lowered into the water. To what depth will it sink if the relative weight of the beam is 0.7? Determine how many people can stand on the beam so that the upper surface of the beam is flush with the free surface of the water, assuming that each person has an average mass of 67.5 kg.

Problem 1.23 A rectangular metal barge 60 m long, 8 m wide, 3.5 m high, loaded with sand, weighs 14126 kN. Determine the draft of the barge. What volume of sand V p needs to be unloaded so that the barge's immersion depth is 1.2 m, if the relative specific gravity of wet sand is 2.0?

Problem 1.24. The volumetric displacement of the submarine is 600 m 3 . In order to submerge the boat, the compartments were filled with sea water in an amount of 80 m 3. The relative specific gravity of sea water is 1.025. Determine: what part of the volume of the boat (in percent) will be immersed in water if all the water is removed from the submarine and it floats up; What is the weight of a submarine without water?

Pressure is a physical quantity that plays a special role in nature and human life. This phenomenon, invisible to the eye, not only affects the state of the environment, but is also very well felt by everyone. Let's figure out what it is, what types it exists and how to find pressure (formula) in different environments.

What is pressure in physics and chemistry?

This term refers to an important thermodynamic quantity, which is expressed in the ratio of the pressure force exerted perpendicularly to the surface area on which it acts. This phenomenon does not depend on the size of the system in which it operates, and therefore refers to intensive quantities.

In a state of equilibrium, the pressure is the same for all points of the system.

In physics and chemistry it is denoted by the letter “P”, which is an abbreviation of the Latin name of the term - pressūra.

When talking about the osmotic pressure of a fluid (the balance between the pressure inside and outside the cell), the letter “P” is used.

Pressure units

According to the standards of the International SI System, the physical phenomenon in question is measured in pascals (Cyrillic - Pa, Latin - Ra).

Based on the pressure formula, it turns out that one Pa is equal to one N (newton - divided by one square meter (unit of area).

However, in practice it is quite difficult to use pascals, since this unit is very small. In this regard, in addition to SI standards, this quantity can be measured differently.

Below are its most famous analogues. Most of them are widely used in the former USSR.

• Bars. One bar is equal to 105 Pa.
• Torrs, or millimeters of mercury. Approximately one torr corresponds to 133.3223684 Pa.
• Millimeters of water column.
• Meters of water column.
• Technical atmospheres.
• Physical atmospheres. One atm is equal to 101,325 Pa and 1.033233 atm.
• Kilogram-force per square centimeter. Ton-force and gram-force are also distinguished. In addition, there is an analogue to pound-force per square inch.

General formula for pressure (7th grade physics)

From the definition of a given physical quantity, one can determine the method for finding it. It looks like in the photo below.

In it, F is force and S is area. In other words, the formula for finding pressure is its force divided by the surface area on which it acts.

It can also be written as follows: P = mg / S or P = pVg / S. Thus, this physical quantity turns out to be related to other thermodynamic variables: volume and mass.

For pressure, the following principle applies: the smaller the space that is affected by the force, the greater the amount of pressing force that falls on it. If the area increases (with the same force), the desired value decreases.

Hydrostatic Pressure Formula

Different states of aggregation of substances provide for the presence of properties that differ from each other. Based on this, the methods for determining P in them will also be different.

For example, the formula for water pressure (hydrostatic) looks like this: P = pgh. It also applies to gases. However, it cannot be used to calculate atmospheric pressure due to the difference in altitude and air density.

In this formula, p is the density, g is the acceleration due to gravity, and h is the height. Based on this, the deeper an object or object is immersed, the higher the pressure exerted on it inside the liquid (gas).

The option under consideration is an adaptation of the classic example P = F / S.

If we remember that the force is equal to the derivative of mass by the speed of free fall (F = mg), and the mass of the liquid is the derivative of volume by density (m = pV), then the formula pressure can be written as P = pVg / S. In this case, volume is area multiplied by height (V = Sh).

If we insert this data, it turns out that the area in the numerator and denominator can be reduced at the output - the above formula: P = pgh.

When considering pressure in liquids, it is worth remembering that, unlike solids, curvature of the surface layer is often possible in them. And this, in turn, contributes to the formation of additional pressure.

For such situations, a slightly different pressure formula is used: P = P 0 + 2QH. In this case, P 0 is the pressure of the non-curved layer, and Q is the tension surface of the liquid. H is the average curvature of the surface, which is determined according to Laplace's Law: H = ½ (1/R 1 + 1/R 2). The components R 1 and R 2 are the radii of the main curvature.

Partial pressure and its formula

Although the P = pgh method is applicable for both liquids and gases, it is better to calculate the pressure in the latter in a slightly different way.

The fact is that in nature, as a rule, absolutely pure substances are not very often found, because mixtures predominate in it. And this applies not only to liquids, but also to gases. And as you know, each of these components exerts a different pressure, called partial.

It's quite easy to define. It is equal to the sum of the pressure of each component of the mixture under consideration (ideal gas).

It follows from this that the partial pressure formula looks like this: P = P 1 + P 2 + P 3 ... and so on, according to the number of constituent components.

There are often cases when it is necessary to determine air pressure. However, some people mistakenly carry out calculations only with oxygen according to the scheme P = pgh. But air is a mixture of different gases. It contains nitrogen, argon, oxygen and other substances. Based on the current situation, the air pressure formula is the sum of the pressures of all its components. This means that we should take the above-mentioned P = P 1 + P 2 + P 3 ...

The most common instruments for measuring pressure

Despite the fact that it is not difficult to calculate the thermodynamic quantity in question using the above-mentioned formulas, sometimes there is simply no time to carry out the calculation. After all, you must always take into account numerous nuances. Therefore, for convenience, over several centuries a number of devices have been developed that do this instead of people.

In fact, almost all devices of this kind are a type of pressure gauge (helps determine pressure in gases and liquids). However, they differ in design, accuracy and scope of application.

• Atmospheric pressure is measured using a pressure gauge called a barometer. If it is necessary to determine the vacuum (that is, pressure below atmospheric), another type of it is used, a vacuum gauge.
• In order to find out a person's blood pressure, a sphygmomanometer is used. It is better known to most people as a non-invasive blood pressure monitor. There are many varieties of such devices: from mercury mechanical to fully automatic digital. Their accuracy depends on the materials from which they are made and the location of measurement.
• Pressure drops in the environment (in English - pressure drop) are determined using differential pressure meters (not to be confused with dynamometers).

Types of pressure

Considering pressure, the formula for finding it and its variations for different substances, it is worth learning about the varieties of this quantity. There are five of them.

• Absolute.
• Barometric
• Excessive.
• Vacuum metric.
• Differential.

Absolute

This is the name of the total pressure under which a substance or object is located, without taking into account the influence of other gaseous components of the atmosphere.

It is measured in pascals and is the sum of excess and atmospheric pressure. It is also the difference between barometric and vacuum types.

It is calculated using the formula P = P 2 + P 3 or P = P 2 - P 4.

The starting point for absolute pressure under the conditions of planet Earth is the pressure inside the container from which air has been removed (that is, a classic vacuum).

Only this type of pressure is used in most thermodynamic formulas.

Barometric

This term refers to the pressure of the atmosphere (gravity) on all objects and objects found in it, including the surface of the Earth itself. Most people also know it as atmospheric.

It is classified as one and its value varies depending on the place and time of measurement, as well as weather conditions and location above/below sea level.

The magnitude of barometric pressure is equal to the modulus of the atmospheric force over an area of ​​one unit normal to it.

In a stable atmosphere, the magnitude of this physical phenomenon is equal to the weight of a column of air on a base with an area equal to one.

The normal barometric pressure is 101,325 Pa (760 mm Hg at 0 degrees Celsius). Moreover, the higher the object is from the surface of the Earth, the lower the air pressure on it becomes. Every 8 km it decreases by 100 Pa.

Thanks to this property, water in kettles boils much faster in the mountains than on the stove at home. The fact is that pressure affects the boiling point: as it decreases, the latter decreases. And vice versa. The operation of such kitchen appliances as a pressure cooker and autoclave is based on this property. The increase in pressure inside them contributes to the formation of higher temperatures in the vessels than in ordinary pans on the stove.

The barometric altitude formula is used to calculate atmospheric pressure. It looks like in the photo below.

P is the desired value at altitude, P 0 is the air density near the surface, g is the free fall acceleration, h is the height above the Earth, m is the molar mass of the gas, t is the temperature of the system, r is the universal gas constant 8.3144598 J⁄( mol x K), and e is the Eichler number equal to 2.71828.

Often in the above formula for atmospheric pressure, K - Boltzmann's constant is used instead of R. The universal gas constant is often expressed through its product by Avogadro's number. It is more convenient for calculations when the number of particles is given in moles.

When making calculations, you should always take into account the possibility of changes in air temperature due to a change in meteorological situation or when gaining altitude above sea level, as well as geographic latitude.

Gauge and vacuum

The difference between atmospheric and measured ambient pressure is called excess pressure. Depending on the result, the name of the quantity changes.

If it is positive, it is called gauge pressure.

If the result obtained has a minus sign, it is called vacuummetric. It is worth remembering that it cannot be greater than barometric.

Differential

This value is the difference in pressure at different measurement points. As a rule, it is used to determine the pressure drop on any equipment. This is especially true in the oil industry.

Having figured out what kind of thermodynamic quantity is called pressure and with what formulas it is found, we can conclude that this phenomenon is very important, and therefore knowledge about it will never be superfluous.

Laboratory work No. 11

BRIEF THEORY. The most important feature of a liquid is the existence free surface. The molecules of the surface layer of the liquid, which has a thickness of about 10 -9 m, are in a different state than the molecules in the thickness of the liquid. The surface layer exerts pressure on the liquid, called molecular, which leads to the appearance of forces called forces surface tension.

Surface tension forces at any point on the surface are directed tangentially to it and normal to any element of a line mentally drawn on the surface of the liquid. Surface tension coefficient-physical quantity showing the force of surface tension acting per unit length of the line dividing the surface of the liquid into parts:

On the other hand, surface tension can be defined as a value numerically equal to the free energy of a unit surface layer of a liquid. Under free energy understand that part of the energy of the system due to which work can be done during an isothermal process.

The surface tension coefficient depends on the nature of the liquid. For each liquid, it is a function of temperature and depends on what medium is located above the free surface of the liquid.

EXPERIMENTAL SETUP. The experimental setup is shown in Fig. 1. It consists of an aspirator A connected to a micromanometer M and a vessel B containing the liquid being tested. Water is poured into the aspirator. Using tap K, aspirator A can be disconnected from vessel B and connected to the same vessel C with another liquid to be tested. Vessels B and C are tightly closed with rubber stoppers each having a hole. A glass tube is inserted into each hole, the end of which is a capillary. The capillary is immersed to a very shallow depth in the liquid (so that it just touches the surface of the liquid). The micromanometer measures the difference in air pressure in the atmosphere and the aspirator, or, what is the same, in the capillary and vessel B or C.

The micromanometer consists of two communicating vessels, one of which is a large-diameter cup, and the other is an inclined glass tube of small diameter (2 - 3 mm) (Fig. 2). If the ratio of the cross-sectional areas of the cup and tube is sufficiently large, the change in level in the cup can be neglected. Then, from the level of liquid in a small-diameter tube, the measured value of the pressure difference can be determined:

Where - density of gauge fluid; - the distance along the tube of the assumed constant liquid level in the cup; - the angle formed by the inclined tube with the horizontal plane.

At the initial moment of time, when the air pressure above the surface of the liquid in the capillary and vessel B is the same and equal to atmospheric pressure, the level of the wetting fluid in the capillary is higher than in vessel B, and the level of the non-wetting fluid is lower, since the wetting fluid in the capillary forms a concave meniscus, and non-wetting - convex.

The molecular pressure under a convex surface of a liquid is greater, and under a concave surface it is less relative to the pressure under a flat surface. The molecular pressure caused by the curvature of the surface is usually called excess capillary pressure (Lapplace pressure). Excess pressure under a convex surface is considered positive, under a concave surface - negative. The force of this pressure is always directed towards the center of curvature of the surface section. In the case of a spherical surface, the excess pressure can be calculated using the formula:

where is the surface tension and is the radius of the spherical surface.

The liquid wetting the capillary rises until the hydrostatic pressure of a liquid column with a height (Fig. 3) balances the excess pressure, which in this case is directed upward. The height is determined from the equilibrium condition:

where is the acceleration of free fall, i.e.

If you turn the tap of aspirator A and slowly release water from it, then the air pressure in the aspirator, in the vessel B connected to it and in the inclined elbow of the micromanometer will begin to decrease. In the capillary above the surface of the liquid the pressure is equal to atmospheric pressure. As a result of the increasing pressure difference, the liquid meniscus in the capillary will lower, maintaining its curvature, until it drops to the lower end of the capillary (Fig. 3c). At this moment, the air pressure in the capillary will be equal to:

where is the air pressure in vessel B, is the depth of immersion of the capillary in the liquid, - Laplace pressure. The difference in air pressure in the capillary and vessel B is equal to:

From this moment on, the curvature of the meniscus begins to change. The air pressure in the aspirator and vessel B continues to decrease. As the pressure difference increases, the radius of curvature of the meniscus decreases and the curvature increases. There comes a moment when the radius of curvature becomes equal to the internal radius of the capillary (Fig. 3c), and the pressure difference becomes maximum. Then the radius of curvature of the meniscus increases again, and the equilibrium will be unstable. An air bubble is formed, which breaks away from the capillary and rises to the surface. The liquid closes the hole. Then everything is repeated. In Fig. Figure 4 shows how the radius of curvature of the liquid meniscus changes, starting from the moment it reaches the lower end of the capillary.

From the above it follows that:

, (1)

where is the internal radius of the capillary. This difference can be determined using a micromanometer, since

Where - the density of the manometric liquid, - the maximum displacement of the liquid level in the inclined tube of the micromanometer, - the angle between the inclined elbow of the micromanometer and the horizontal (see Fig. 2).

From formulas (1) and (2) we obtain:

. (3)

Since the depth of immersion of the capillary in the liquid is negligible, it can be neglected, then:

or , (4)

where is the internal diameter of the capillary.

In the case when the liquid does not wet the walls of the capillary, the outer diameter of the capillary is taken as in formula (4). Water is used as a manometric fluid in a micromanometer ( = 1×10 3 kg/m 3).

MEASUREMENTS. 1. Close the capillary tightly with a rubber stopper, having first measured its internal diameter using a microscope. Insert the capillary into the hole of the plug. Bring the end of the tube into contact with the liquid.

2. Fill the aspirator with water up to the mark and close it. To achieve equal pressure in both elbows of the micromanometer, for a short time remove valve K. Place it in a position in which it connects the vessel to the aspirator.

3. Open the aspirator tap so that the pressure change occurs sufficiently slowly. Air bubbles should break off approximately every 10-15 seconds. Once the specified bubble formation frequency has been established, measurements can be made.

EXERCISE.

1. Using a thermometer, determine and record the room temperature T.

2. Determine the maximum displacement of the liquid level in the inclined elbow of the micromanometer nine times. To calculate the surface tension coefficient, take the average value N avg.

Problem 1

A tourist rode a bicycle 40 km in one day. Moreover, from 9.00 to 11.20 he drove at a speed that gradually increased over time from 10 km/h to 14 km/h. Then the tourist sunbathed on the beach. He spent the rest of the journey from 18.30 to 20.00. Determine the average speed of the tourist during the evening portion of the trip.

Possible Solution

From 9.00 to 11.20 the tourist drove at an average speed of (10 + 14)/2 = 12 km/h (since the speed increased uniformly over time). This means that during this time the tourist traveled a distance

During the time from 18.30 to 20.00 the cyclist traveled 40 – 28 = 12 km. Therefore, the average speed of a tourist during the evening portion of the trip is equal to:

Evaluation criteria

• Average speed of a tourist on the morning section of the trip (12 km/h): 4 points
• Distance traveled by the tourist from 9.00 to 11.20 (28 km): 2 points
• Distance traveled by the tourist from 18.30 to 20.00 (12 km): 2 points
• Average speed of a tourist during the evening section of the trip (8 km/h): 2 points

Maximum per task- 10 points.

Problem 2

A system consisting of two homogeneous rods of different densities is in equilibrium. Top rod weight m 1 = 1.4 kg. Friction is negligible.

Determine at what mass m 2 lower rods such an equilibrium is possible.

Possible Solution

Since the lower rod is suspended by the ends, is in equilibrium and its center of gravity is located in the middle, the reaction forces of the threads acting on it are the same and equal in magnitude m 2 g/2. Let us write the equation of moments for the upper rod relative to the attachment point of the left (upper) thread:

Evaluation criteria

The reaction forces of the threads acting on the lower rod are equal to: 3 points

The values ​​of the moduli of these reaction forces ( m 2 g/2): 2 points

Moment equation: 4 points

m 2 = 1.2 kg: 1 point

Maximum per task- 10 points.

Problem 3

In a cylindrical vessel with water there is a body partially immersed in water, tied by a stretched thread to the bottom of the vessel. In this case, the body is immersed in water by two-thirds of its volume. If you cut the thread, the body will float up and float half submerged in water. How much will the water level in the vessel change? Body mass m= 30 g, density of water ρ = 1.0 g/cm 3, area of ​​the bottom of the vessel S= 10 cm 2.

Possible solution 1

The force of pressure of the glass on the table (after cutting the thread) will not change, therefore,

T= ρ g∆h · S, where ̶T is the reaction force on the part of the thread, ∆h is the change in water level. Let us write the equilibrium equation of the body in the first case:

Mg = ρg·(1/2)·V

From the last two equations we find that ͶT = 1/3 mg

Finally we get:

Evaluation criteria

• The force of pressure of the glass on the table will not change: 2 points
• The equilibrium equation of the body in the first case: 2 points
• The equilibrium equation of the body in the second case: 2 points
• T = 1/3 mg:1 point
• ∆h = T/( ρ g· S): 2 points
• ∆h = 0.01m: 1 point

Possible solution 2

The equilibrium equation of the body in the second case:

mg = ρg ½ V⟹V = 2m/ ρ, where ͸V – body volume.

The change in volume of the immersed part of the body is equal to:

Finally we get:

Evaluation criteria

• mg = ρg ½ V: 4 points
• ∆V = 1/6 V:2 points
• ∆h = ∆V/S: 3 points
• ∆h = 0.01 m: 1 point

Maximum per task- 10 points.

Problem 4

Determine the air pressure above the surface of the liquid at the point A inside the closed section of a curved tube, if ρ = 800 kg/m 3, h= 20 cm, p 0 = 101 kPa, g= 10 m/s 2. Liquid densities ρ and 2 ρ do not mix with each other.

Air pressure- the force with which air presses on the earth's surface. It is measured in millimeters of mercury, millibars. On average, it is 1.033 g per 1 cm2.

The reason that causes wind formation is the difference in atmospheric pressure. The wind blows from an area of ​​higher atmospheric pressure to an area of ​​lower. The greater the difference in atmospheric pressure, the stronger the wind. The distribution of atmospheric pressure on Earth determines the direction of the winds prevailing in the troposphere at different latitudes.

They are formed when water vapor condenses in rising air due to its cooling.
. Liquid or solid water that falls on the earth's surface is called precipitation.

Based on their origin, there are two types of sediment:

falling from clouds (rain, snow, graupel, hail);
formed at the surface of the Earth (dew, frost).
Precipitation is measured by the layer of water (in mm) that forms if the fallen water does not drain and evaporate. On average, 1130 mm falls on the Earth per year. precipitation.

Precipitation distribution. Atmospheric precipitation is distributed very unevenly over the earth's surface. Some areas suffer from excess moisture, others from its lack. Territories located along the northern and southern tropics, where air quality is high and the need for precipitation is especially great, receive especially little precipitation.

The main reason for this unevenness is the placement of atmospheric pressure belts. Thus, in the region of the equator in a low-pressure zone, constantly heated air contains a lot of moisture, it rises, cools and becomes saturated. Therefore, in the equator region many clouds form and there is heavy rain. There is also a lot of precipitation in other areas of the earth's surface where pressure is low.

In high pressure zones, downward air currents predominate. Cold air, as it descends, contains little moisture. When lowered, it contracts and heats up, due to which it moves away from the saturation point and becomes drier. Therefore, areas of high pressure over the tropics and near the poles receive little precipitation.

The amount of precipitation cannot yet be used to judge the moisture supply of an area. Possible evaporation - volatility - must be taken into account. It depends on the amount of solar heat: the more heat there is, the more moisture, if any, can evaporate. Volatility can be high, but evaporation can be small. For example, evaporation (how much moisture can evaporate at a given temperature) is 4500 mm/year, and evaporation (how much moisture actually evaporates) is only 100 mm/year. The moisture content of the area is judged by the ratio of evaporation and evaporation. To determine moisture, the moisture coefficient is used. Humidity coefficient is the ratio of annual precipitation to evaporation over the same period of time. It is expressed as a fraction as a percentage. If the coefficient is 1, the moisture is sufficient; if it is less than 1, the moisture is insufficient; and if it is greater than 1, the moisture is excessive. Based on the degree of moisture, wet (humid) and dry (arid) areas are distinguished.