Air pressure- the force with which air presses on the earth's surface. Measured in millimeters 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. So, in the region of the equator in the belt low pressure Constantly heated air contains a lot of moisture, it rises, cools and becomes saturated. Therefore, in the equator region there are many clouds and heavy rainfall. There is also a lot of precipitation in other areas of the earth's surface where pressure is low.
In belts high pressure downward air currents predominate. Cold air, sinking, 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 - ratio annual quantity 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.
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- free fall acceleration.
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 total 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.
Graphic representation of magnitude and direction hydrostatic pressure acting on any point on the surface is called the 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.
Strength 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.
Strength 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 origin 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?
The fabric can be pierced with a needle, but not with a pencil (if you apply the same force). The pencil and needle have different shapes and therefore apply different pressure to the fabric. Pressure is omnipresent. It activates the mechanisms (see article ““). It affects. exert pressure on the surfaces they come into contact with. Atmospheric pressure affects the weather. A device for measuring atmospheric pressure -.
What is pressure
When a body is acted upon perpendicular to its surface, the body is under pressure. Pressure depends on how great the force is and the surface area over which the force is applied. For example, if you go out into the snow in ordinary shoes, you may fall through; But this won't happen if we put on skis. The weight of the body is the same, but in the second case the pressure will be distributed over a larger surface. The larger the surface, the less pressure. Reindeer have wide hooves - after all, they walk on snow, and the pressure of the hoof on the snow should be as little as possible. If the knife is sharp, the force is applied to a small surface area. A dull knife distributes force over a larger surface, and therefore cuts worse. Unit of pressure - pascal(Pa) - named after the French scientist Blaise Pascal (1623 - 1662), who made many discoveries in the field of atmospheric pressure.
Pressure of liquids and gases
Liquids and gases take the shape of the container in which they are contained. Unlike solids, liquids and gases exert pressure on all walls of the container. The pressure of liquids and gases is directed in all directions. puts pressure not only on the bottom, but also on the walls of the aquarium. The aquarium itself only presses downwards. presses on the inside of the soccer ball in all directions, and therefore the ball is round.
Hydraulic mechanisms
The action of hydraulic mechanisms is based on fluid pressure. Liquid does not compress, so if force is applied to it, it will be forced to move. And the brakes operate on a hydraulic principle. Reducing the wheel speed is achieved using brake fluid pressure. The driver presses the pedal, the piston pumps brake fluid through the cylinder, then it flows through the tube into the other two cylinders and presses on the pistons. The pistons press the brake pads against the wheel. The resulting slows down the rotation of the wheel.
Pneumatic mechanisms
Pneumatic mechanisms operate due to the pressure of gases - usually air. Unlike liquids, air can be compressed, and then its pressure increases. The action of a jackhammer is based on the fact that a piston compresses the air inside it to a very high pressure. In a jackhammer, compressed air presses on the cutter with such force that even stone can be drilled.
A foam fire extinguisher is a pneumatic device that runs on compressed carbon dioxide. By squeezing the handle, you release the compressed carbon dioxide contained in the canister. The gas presses down with enormous force onto the special solution, displacing it into the tube and hose. A stream of water and foam shoots out of the hose.
Atmospheric pressure
Atmospheric pressure is created by the weight of air above the surface. For every square meter the air presses with a force greater than the weight of an elephant. The pressure is higher near the Earth's surface than high in the sky. At an altitude of 10,000 meters, where jet planes fly, the pressure is low, since there is little air mass pressing from above. 
Normal atmospheric pressure is maintained in the aircraft cabin so that people can breathe freely at high altitudes. But even in a pressurized airplane cabin, people's ears become blocked when the pressure becomes lower than the pressure inside the auricle.
Atmospheric pressure is measured in millimeters of mercury. When pressure changes, so does . Low pressure means worse weather is coming. High pressure brings clear weather. Normal pressure at sea level is 760 mm (101,300 Pa). On hurricane days it can drop to 683 mm (910 Pa).
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 of the assumed constant liquid level in the cup to the level in the tube along the slope of the tube; - 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 liquid in the capillary is higher than in vessel B, and the level of the non-wetting liquid is lower, since the wetting liquid in the capillary forms a concave meniscus, and the non-wetting liquid forms a convex meniscus.
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. It is always directed towards the center of curvature of the surface section, i.e. towards its concavity. In the case of a spherical surface, the excess pressure can be calculated using the formula:
where is the surface tension coefficient, 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. 3a) balances the excess pressure, which in this case is directed upward. Height 0 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 a 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. 3b). 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:
+ p = p out +ρg h = 2σ / r +ρg h
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. 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.
2. Open the aspirator tap so that the pressure change occurs sufficiently slowly. Air bubbles should break off approximately every 10 to 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.
3. Similarly, determine the surface tension coefficient of ethyl alcohol.
4. Find the maximum absolute and relative errors when determining the surface tension of each liquid. Write down the final measurement results for each liquid, taking into account their accuracy using the formula.
Pressure - a quantity equal to the ratio of the force acting perpendicular to the surface is called pressure. A unit of pressure is taken to be the pressure produced by a force of 1 N acting on a surface of 1 m2 perpendicular to this surface.
Therefore, to determine the pressure, the force acting perpendicular to the surface must be divided by the surface area.
It is known that gas molecules move randomly. As they move, they collide with each other, as well as with the walls of the container containing the gas. There are many molecules in a gas, and therefore the number of their impacts is very large. Although the impact force of an individual molecule is small, the effect of all molecules on the walls of the vessel is significant, and it creates gas pressure. So, the pressure of the gas on the walls of the vessel (and on the body placed in the gas) is caused by the impacts of gas molecules.
As the volume of a gas decreases, its pressure increases, and as its volume increases, the pressure decreases, provided that the mass and temperature of the gas remain unchanged.
In any liquid, the molecules are not rigidly bound, and therefore the liquid takes the shape of the container into which it is poured. Like solids, liquid exerts pressure on the bottom of the container. But unlike solids, liquid also exerts pressure on the walls of the container.
To explain this phenomenon, let’s mentally divide the liquid column into three layers (a, b, c). At the same time, you can see that there is pressure inside the liquid itself: the liquid is under the pressure of gravity, and the weight of its upper layers acts on the lower layers of the liquid. The force of gravity acting on layer a presses it towards the second layer b. Layer b transmits the pressure exerted on it in all directions. In addition, gravity also acts on this layer, pressing it towards the third layer c. Consequently, in the third stage the pressure increases, and it will be greatest at the bottom of the vessel.
The pressure inside a liquid depends on its density.
The pressure exerted on a liquid or gas is transmitted without change to every point in the volume of the liquid or gas. This statement is called Pascal's law.
The SI unit of pressure is the pressure produced by a force of 1N on a surface with an area of 1m2 perpendicular to it. This unit is called the pascal (Pa).
The name of the pressure unit is given in honor of the French scientist Blaise Pascal
Blaise Pascal
Blaise Pascal - French mathematician, physicist and philosopher, born June 19, 1623. He was the third child in the family. His mother died when he was only three years old. In 1632, Pascal's family left Clermont and went to Paris. Pascal's father had a good education and decided to pass it on directly to his son. His father decided that Blaise should not study mathematics until he was 15, and all mathematical books were removed from their home. However, Blaise's curiosity pushed him to study geometry at the age of 12. When his father found out, he relented and allowed Blaise to study Euclid.
Blaise Pascal made significant contributions to the development of mathematics, geometry, philosophy and literature.
In physics, Pascal studied barometric pressure and hydrostatics.
Based on Pascal's law, it is easy to explain the following experiment.
We take a ball that has narrow holes in various places. A tube is attached to the ball into which a piston is inserted. If you fill a ball with water and push a piston into the tube, water will flow out of all the holes in the ball. In this experiment, a piston presses on the surface of water in a tube.
Pascal's law
The water particles located under the piston, when compacted, transmit its pressure to other layers that lie deeper. Thus, the pressure of the piston is transmitted to each point of the fluid filling the ball. As a result, some of the water is pushed out of the ball in the form of streams flowing out of all the holes.
If the ball is filled with smoke, then when the piston is pushed into the tube, streams of smoke will begin to come out of all the holes in the ball. This confirms (that gases transmit the pressure exerted on them equally in all directions). So, experience shows that there is pressure inside the liquid and at the same level it is equal in all directions. With depth, pressure increases. Gases are no different from liquids in this respect.
Pascal's law is valid for liquids and gases. However, he does not take into account one important circumstance - the existence of weight.
In earthly conditions this cannot be forgotten. Water also weighs. Therefore, it is clear that two sites located at different depths under water will experience different pressures.
The pressure of water due to its gravity is called hydrostatic.
Under terrestrial conditions, air most often presses on the free surface of a liquid. Air pressure is called atmospheric pressure. Pressure at depth consists of atmospheric and hydrostatic pressure.
If two vessels of different shapes, but with the same levels of water in them, are connected with a tube, then the water will not pass from one vessel to another. Such a transition could occur if the pressures in the vessels differed. But this is not the case, and in communicating vessels, regardless of their shape, the liquid will always be at the same level.
For example, if the water levels in communicating vessels are different, then the water will begin to move and the levels will become equal.
Water pressure is much greater than air pressure. At a depth of 10 m, water presses 1 cm2 with an additional force of 1 kg to atmospheric pressure. At a depth of a kilometer - with a force of 100 kg per 1 cm2.
The ocean in some places is more than 10 km deep. The water pressure forces at such depths are extremely high. Pieces of wood, lowered to a depth of 5 km, are compacted by this enormous pressure so much that after this they sink in a barrel of water, like bricks.
This enormous pressure creates great obstacles for researchers of marine life. Deep-sea descents are carried out in steel balls - the so-called bathyspheres, or bathyscaphes, which have to withstand pressure above 1 ton per 1 cm2.
Submarines descend only to a depth of 100 - 200 m.
The pressure of the liquid at the bottom of the vessel depends on the density and height of the liquid column.
Let's measure the water pressure at the bottom of the glass. Of course, the bottom of the glass is deformed under the influence of pressure forces, and knowing the magnitude of the deformation, we could determine the magnitude of the force that caused it and calculate the pressure; but this deformation is so small that it is almost impossible to measure it directly. Since it is convenient to judge by the deformation of a given body the pressure exerted on it by a liquid only in the case when the deformations are precisely large, then to practically determine the pressure of a liquid, special devices are used - pressure gauges, in which the deformation has a relatively large, easily measurable value. The simplest membrane pressure gauge is designed as follows. A thin elastic membrane plate - hermetically closes an empty box. A pointer is attached to the membrane and rotates around an axis. When the device is immersed in liquid, the membrane bends under the influence of pressure forces, and its deflection is transmitted in an enlarged form to the pointer moving along the scale.
Pressure gauge
Each position of the pointer corresponds to a certain deflection of the membrane, and therefore a certain force of pressure on the membrane. Knowing the area of the membrane, we can move from pressure forces to the pressures themselves. You can directly measure pressure if you calibrate the pressure gauge in advance, that is, determine what pressure a particular position of the pointer on the scale corresponds to. To do this, you need to expose the pressure gauge to pressures, the magnitude of which is known and, noticing the position of the pointer arrow, put the corresponding numbers on the instrument scale.
The shell of air surrounding the Earth is called the atmosphere. The atmosphere, as shown by observations of the flight of artificial Earth satellites, extends to a height of several thousand kilometers. We live at the bottom of a huge ocean of air. The surface of the Earth is the bottom of this ocean.
Due to gravity, the upper layers of air, like ocean water, compress the lower layers. The air layer adjacent directly to the Earth is compressed the most and, according to Pascal's law, transmits the pressure exerted on it in all directions.
As a result of this, the earth's surface and the bodies located on it experience the pressure of the entire thickness of the air, or, as they usually say, experience atmospheric pressure.
Atmospheric pressure is not that low. A force of about 1 kg acts on every square centimeter of body surface.
The reason for atmospheric pressure is obvious. Like water, air has weight, which means it exerts a pressure equal (as for water) to the weight of the column of air above the body. The higher we go up the mountain, the less air there will be above us, which means the lower the atmospheric pressure will become.
For scientific and everyday purposes, you need to be able to measure pressure. There are special devices for this - barometers.
Barometer
Making a barometer is not difficult. Mercury is poured into a tube closed at one end. Holding the open end with your finger, tip the tube over and immerse its open end in a cup of mercury. In this case, the mercury in the tube drops, but does not pour out. The space above the mercury in the tube is undoubtedly airless. The mercury is maintained in the tube by outside air pressure.
No matter what size we take the cup of mercury, no matter the diameter of the tube, the mercury always rises to approximately the same height - 76 cm.
If we take a tube shorter than 76 cm, then it will be completely filled with mercury, and we will not see the void. A column of mercury 76 cm high presses on the stand with the same force as the atmosphere.
One kilogram per square centimeter is the value of normal atmospheric pressure.
The figure 76 cm means that such a column of mercury balances the air column of the entire atmosphere located above the same area.
The barometric tube can be given a variety of shapes; only one thing is important: one end of the tube must be closed so that there is no air above the surface of the mercury. Another level of mercury is affected by atmospheric pressure.
A mercury barometer can measure atmospheric pressure with very high accuracy. Of course, it is not necessary to take mercury; any other liquid will do. But mercury is the heaviest liquid, and the height of the mercury column at normal pressure will be the smallest.
Various units are used to measure pressure. Often the height of the mercury column is simply indicated in millimeters. For example, they say that today the pressure is higher than normal, it is equal to 768 mm Hg. Art.
Pressure 760mm Hg. Art. sometimes called the physical atmosphere. A pressure of 1 kg/cm2 is called a technical atmosphere.
A mercury barometer is not a particularly convenient instrument. It is undesirable to leave the surface of the mercury exposed (mercury vapor is poisonous); in addition, the device is not portable.
Metal barometers - aneroids - do not have these disadvantages.
Everyone has seen such a barometer. This is a small round metal box with a scale and arrow. The scale shows pressure values, usually in centimeters of mercury.
The air has been pumped out of the metal box. The box lid is held in place by a strong spring, as otherwise it would be pressed in by atmospheric pressure. When pressure changes, the lid either bends or bulges. An arrow is connected to the lid, and in such a way that when pressed in, the arrow goes to the right.
Such a barometer is calibrated by comparing its readings with a mercury barometer.
If you want to know the pressure, don't forget to tap the barometer with your finger. The dial hand experiences a lot of friction and usually gets stuck at >.
A simple device is based on atmospheric pressure - a siphon.
The driver wants to help his friend who has run out of gas. How to drain gasoline from the tank of your car? Don't tilt it like a teapot.
A rubber tube comes to the rescue. One end of it is lowered into the gas tank, and air is sucked out of the other end with the mouth. Then a quick movement - the open end is clamped with a finger and set at a height below the gas tank. Now you can remove your finger - gasoline will pour out of the hose.
The curved rubber tube is the siphon. The liquid in this case moves for the same reason as in a straight inclined tube. In both cases, the liquid eventually flows downward.
For the siphon to operate, atmospheric pressure is necessary: it > liquid and prevents the liquid column in the tube from bursting. If there were no atmospheric pressure, the column would rupture at the pass point, and the liquid would roll into both vessels.
Pressure siphon
The siphon begins to work when the liquid in the right (so to speak, >) elbow drops below the level of the pumped liquid into which the left end of the tube is lowered. Otherwise, the liquid will flow back.
In practice, to measure atmospheric pressure, a metal barometer is used, called an aneroid (translated from Greek - without liquid. The barometer is called this because it does not contain mercury).
The atmosphere is held in place by gravity acting from the Earth. Under the influence of this force, the upper layers of air press on the lower ones, so the layer of air adjacent to the Earth turns out to be the most compressed and the densest. This pressure, in accordance with Pascal's law, is transmitted in all directions and acts on all bodies located on the Earth and on its surface.
The thickness of the layer of air pressing on the Earth decreases with height, therefore, the pressure also decreases.
The existence of atmospheric pressure is indicated by many phenomena. If a glass tube with a lowered piston is placed in a vessel with water and raised smoothly, then the water follows the piston. The atmosphere presses on the surface of the water in the vessel; according to Pascal's law, this pressure is transferred to the water under the glass tube and drives the water upward, following the piston.
Suction pumps were known to ancient civilization. With their help it was possible to raise water to a considerable height. The water surprisingly obediently followed the piston of such a pump.
Ancient philosophers thought about the reasons for this and came to such a thoughtful conclusion: water follows the piston because nature is afraid of emptiness, which is why there is no free space left between the piston and water.
They say that one master built a suction pump for the gardens of the Duke of Tuscany in Florence, the piston of which was supposed to draw water to a height of more than 10 m. But no matter how hard they tried to suck up the water with this pump, nothing worked. At 10m, the water rose behind the piston, then the piston moved away from the water, and that very void that nature fears was formed.
When Galileo was asked to explain the reason for the failure, he replied that nature really does not like emptiness, but up to a certain limit. Galileo's student Torricelli apparently used this incident as a reason to perform his famous mercury tube experiment in 1643. We have just described this experiment - the production of a mercury barometer is Torricelli's experience.
Taking a tube more than 76mm high, Torricelli created a void above the mercury (often called after the Torricelli void) and thus proved the existence of atmospheric pressure.
With this experience, Torricelli resolved the bewilderment of the master of the Tuscan Duke. Indeed, it is clear for how many meters the water will obediently follow the piston of the suction pump. This movement will continue until a column of water with an area of 1 cm2 becomes equal in weight to 1 kg. Such a column of water will have a height of 10 m. This is why nature is afraid of emptiness. , but more than 10m.
In 1654, 11 years after Torricelli’s discovery, the effect of atmospheric pressure was clearly demonstrated by the Magdeburg burgomaster Otto von Guericke. What brought the author fame was not so much the physical essence of the experience as the theatricality of its production.
The two copper hemispheres were connected by a ring gasket. Through a tap attached to one of the hemispheres, the air was pumped out of the assembled ball, after which it was impossible to separate the hemispheres. A detailed description of Guericke's experience has been preserved. The atmospheric pressure on the hemispheres can now be calculated: with a ball diameter of 37 cm, the force was approximately one ton. To separate the hemispheres, Guericke ordered two eight horses to be harnessed. The harness came with ropes threaded through a ring and attached to the hemispheres. The horses were unable to separate the hemispheres.
The power of eight horses (precisely eight, not sixteen, since the second eight, harnessed for greater effect, could be replaced by a hook driven into the wall, maintaining the same force acting on the hemispheres) was not enough to tear apart the Magdeburg hemispheres.
If there is an empty cavity between two contacting bodies, then these bodies will not disintegrate due to atmospheric pressure.
At sea level, the value of atmospheric pressure is usually equal to the pressure of a column of mercury 760 mm high.
By measuring atmospheric pressure with a barometer, you can find that it decreases with increasing height above the Earth's surface (by about 1 mm Hg when increasing in height by 12 m). Also changes in atmospheric pressure are associated with changes in weather. For example, an increase in atmospheric pressure is associated with the onset of clear weather.
The value of atmospheric pressure is very important for predicting the weather for the coming days, since changes in atmospheric pressure are associated with changes in weather. A barometer is a necessary instrument for meteorological observations.
Pressure fluctuations due to weather are very irregular. It was once thought that pressure alone determined the weather. That’s why barometers are still labeled: clear, dry, rain, storm. There is even an inscription: >.
Pressure changes do play a big role in weather changes. But this role is not decisive.
The direction and strength of the wind are related to the distribution of atmospheric pressure.
The pressure in different places on the earth's surface is not the same, and higher pressure brings air to places with lower pressure. It would seem that the wind should blow in a direction perpendicular to the isobars, that is, where the pressure drops most quickly. However, wind maps show otherwise. The Coriolis force intervenes in matters of air pressure and makes its own correction, a very significant one.
As we know, any body moving in the northern hemisphere is acted upon by a Coriolis force directed to the right in motion. This also applies to air particles. Squeezed from places of greater pressure to places of less pressure, the particle should move across the isobars, but the Coriolis force deflects it to the right, and the direction of the wind forms an angle of approximately 45 degrees with the direction of the isobars.
Amazingly large effect for such a small force. This is explained by the fact that interference with the Coriolis force - friction of air layers - is also very insignificant.
Even more interesting is the influence of the Coriolis force on the direction of winds in > and > pressure. Due to the action of the Coriolis force, the air, moving away from > pressure, does not flow in all directions along radii, but moves along curved lines - spirals. These spiral air flows twist in the same direction and create a circular vortex in the pressure area, moving the air masses clockwise.
The same thing happens in the area of low pressure. In the absence of the Coriolis force, the air would flow towards this area evenly along all radii. However, along the way, the air masses deviate to the right.
Winds in areas of low pressure are called cyclones, winds in areas of high pressure are called anticyclones.
Do not think that every cyclone means a hurricane or storm. The passage of cyclones or anticyclones through the city where we live is a common occurrence, associated, however, mostly with variable weather. In many cases, the approach of a cyclone means the onset of bad weather, and the approach of an anticyclone means the onset of good weather.
However, we will not take the path of weather forecasters.
