Rukhlenko A.P.

HYDRAULICS

Examples of problem solving

Teaching aid

For the preparation of bachelors in the direction

Agroengineering

Tyumen - 2012

Reviewer:

Candidate of Technical Sciences, Associate Professor A. E. Korolev.

G 46 Rukhlenko A.P. Hydraulics. Examples of solving problems of Tyumen State Agricultural Academy. - Tyumen, 2012.

Examples of problem solving in all main sections of the discipline are given. The manual contains 57 tasks with a detailed explanation of the solution to each.

The purpose of this manual is to help students in independent study and assimilation of the methodology for solving problems on all topics of the course.

Published by decision of the methodological commission of the Mechanics and Technology Institute of the TGSHA.

© Tyumen State

Agricultural Academy.

© A.P. Rukhlenko, 2012.

Foreword

An important condition for students to master the theoretical course is the ability to use the knowledge of theoretical foundations in solving specific engineering problems. It is problem solving that develops students' skills for creative engineering thinking, contributes to the development of independence in solving engineering issues related to the study of this discipline.

All tasks in this manual are placed in the order of studying the discipline by subject, according to the work programs for the preparation of bachelors of the direction 110800 - agroengineering.

The manual is intended for full-time and part-time students. Its purpose is to help students master the methodology for solving problems on the topics of the course "Hydraulics". Especially useful, according to the author, the manual will be for students who skip classes, because it will help them in mastering this discipline.

The table below indicates the numbers of problems for each topic and the literature for studying the theoretical material on each topic.

Topics of practical classes

for solving problems

Topic of the lesson	№№ tasks on the topic	Literature, p. No.
Physical properties of liquids	1,2	8..13	8..14	7..12	3..4	3…4
Hydrostatic pressure	3,4,5,6,7,8,	20..25	19..25	17..20	5..7	7..8
The force of hydrostatic pressure on flat and curved surfaces	9,10,11,12,13,14, 15,16,17,19,21	25..31	28..34	21..27	7..9	15..16
Bernoulli equation. Hydraulic resistance	22,23,24,25,26,27 28,29,30,31,32	42..45 55..64	46..52 52..78	44..59	13..16 19..24	30..36
Fluid flow through holes, nozzles, throttles and valves	34,35,36,37,38,39, 40,41	72..79	78..89	63..76	25..29	45..48
Hydraulic calculation of pipelines	42,43,44	64..70	94..104	76..99	31..38	57..63
Vane pumps	45,46,47,48	89..108 131..134	139..158 163..173	146..161	41..59	78..83
Volumetric hydraulic machines	50,51,52,53	141..169	177..204	223..235	59..76	88..91
Volumetric hydraulic drive	54,55,56,57	192..200	204..224	271..279	77..84	95..98

Literature for studying the theoretical part of the discipline

1. Isaev A.P., Sergeev B.I., Didur V.A. Hydraulics and hydromechanization of agricultural processes M: Agroprom Publishing House, 1990 - 400s.

2. N.A. Palishkin Hydraulics and agricultural water supply M: Agroprom publishing house, 1990 - 351s.

3. Sabashvili R.G. Hydraulics, hydraulic machines, agricultural water supply: Proc. allowance for universities M: Kolos 1997-479s.

4. Rukhlenko A.P. Hydraulics and hydraulic machines. Textbook TGSHA-Tyumen 2006 124p.

1. Determine the bulk modulus of elasticity of the liquid,

if, under the action of a load A with a mass of 250 kg, the piston traveled a distance △h=5mm. Piston initial height H=1.5m, piston diameter d=80mm and reservoir D=300mm, reservoir height h=1.3 m. Neglect piston weight. The reservoir is assumed to be absolutely rigid.

Solution: The compressibility of a liquid is characterized by the bulk modulus E, which is included in the generalized Hooke's law: = ,

where \u003d increment (in this case, decrease) of the liquid volume V due to an increase in pressure ∆p . We write the above dependence relative to the desired value:

On the right side of the equation, the unknown quantities must be expressed in terms of the initial data. Pressure increase ∆ due to external load, namely the weight of the load:

The initial volume of liquid is the sum of the volumes of liquid in the cylinder and reservoir:
= · .

Absolute change in liquid volume ∆V:

Substituting the expressions for ∆p, ∆V and V into the right side of the equation, we obtain

E= =

= = .

2. The height of the cylindrical vertical tank h=10m, its diameter D=3m. Determine the mass of fuel oil (ρ m \u003d 920 kg / ), which can be poured into the tank at 15, if its temperature can rise to 40 0 ​​C. Neglect the expansion of the tank walls, the temperature coefficient of volumetric expansion of the liquid β t \u003d 0.0008 1/ 0 C.

Solution: The mass of fuel oil can be expressed as the product of its density and volume, i.e.:

or ,

where h m is the initial level of fuel oil in the tank at t=15 0 C. From the expression for β t we find the absolute change in the volume of fuel oil with increasing temperature, i.e.:

.

On the other hand, the same value can be represented as the difference between the volumes of the reservoir and the initial volume of fuel oil:

Expressing these volumes in terms of geometric parameters, we can write that:

∆V = ·

Equate the right parts of the expressions for :

.

Reducing the left and right sides of the equation by , we obtain

Where = .

Substitute the resulting value into the original equation

Here: △t \u003d t k - t n \u003d 40 - 15 \u003d 25 0 С.

3. Determine the absolute air pressure in the tank, if at atmospheric pressure corresponding to h a \u003d \u003d 760 mm Hg. Art. indication of a mercury vacuum gauge = 0.2 m, height h = 1.5 m. What is the indication of a spring vacuum gauge? Mercury density ρ = 13600kg/.

Solution: To solve this problem, we use the basic equation of hydrostatics, which allows us to determine the pressure at any point in the fluid and the concept of "surface of equal pressure". As is known, for a stationary Newtonian fluid, surfaces of equal pressure represent a set of horizontal planes. In this case, we take two horizontal planes as surfaces of equal pressure - the interface between water and air in the connecting tube and the interface between air and mercury in the right knee of a mercury vacuum gauge. For the first surface, the pressure at points A and B is the same and, according to the basic equation of hydrostatics, is determined as follows:

p A \u003d p B \u003d p 1 + ρ g h,

where p 1 is the absolute air pressure in the tank. From this equation it follows that:

p 1 \u003d p A - ρ · g · h.

If we do not take into account the air density, then we can write that p A \u003d p B \u003d p E, i.e. The pressures at points A, B, and E are the same.

For the second surface, the pressures at points C and D are the same and equal to atmospheric,

p a \u003d p C \u003d p D.

On the other hand, the pressure at t. C can be represented as

whence p e \u003d p a - ρ rt ·g · h rt.

Substituting the expressions for p A into the equation for determining p 1, we get

p 1 \u003d p a - ρ rt g h h rt - ρ g h \u003d ρ rt g (h a - h rt) - ρ g h h.

We find the numerical value p 1 by substituting the numerical values ​​​​of the quantities on the right side of the equation:

p 1 \u003d 13600 9.81 (0.76 - 0.2) - 1000 9.81 1.5 \u003d

74713 - 14715 = 59998Pa = 60kPa.

The vacuum that the vacuum gauge will show:

p wak \u003d p a - p 1 \u003d ρ rt g h h a - p 1 \u003d

13600 9.81 0.76 10 -3 - 60 = 101.4 - 60 = 41.4 kPa.

4. Determine the absolute pressure in the vessel according to the indication of a liquid manometer, if it is known: h 1 \u003d 2m, h 2 \u003d 0.5 m, h 3 \u003d 0.2 m, m \u003d = 880 kg / m 3.

Solution: To solve this problem, it is necessary to write down the basic equation of hydrostatics for two points located on a horizontal plane (surface of equal pressure) passing along the water-mercury interface. Pressure in t. A

r A \u003d r abs + ρ g h 1;

Pressure in t.V

Equating the right parts of these expressions, we determine the absolute pressure

r abs + ρ g h 1 \u003d r a + ρ m g h 3 + ρ rt g h 2,

100000+880 9.81 0.2+13600 9.81 0.5–1000 9.81 2 =

100000+1726.6+66708-19620=148815Pa=148kPa.

5. Closed tank A, filled with kerosene to a depth of H=3m, is equipped with a vacuum gauge and a piezometer. Determine the absolute pressure p 0 above the free surface in the tank and the difference between the levels of mercury in the vacuum gauge h 1 if the height of the rise of kerosene in the piezometer h = 1.5 m.

Solution: Let us write down the basic equation of hydrostatics for t. A, located at the bottom of the tank,

On the other hand, the same pressure at point A can be expressed through the reading of an open piezometer

The resulting expression for p A is inserted into the equation for determining p 0:

then the numerical value of p 0 will be equal to:

The difference between the levels of mercury in the vacuum gauge is determined by writing the basic equation of hydrostatics for two points B and C of the surface of equal pressure, coinciding with the free surface of mercury in the right knee of the vacuum gauge

h 1 = = .

6. Determine the excess water pressure in pipe B, if the pressure gauge reading = 0.025 MPa.

Connecting tube filled with water and

air, as shown in the diagram, with H 1 \u003d 0.5 m, H 2 \u003d 3 m. How will the reading of the pressure gauge 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

Solution: When solving this problem, the basic equation of hydrostatics is used, according to which the pressure in the pipe B is the sum of the pressure on the free surface (in this case, the gauge - p m) and the weight pressure of the water. Air is not taken into account due to its low density compared to water.

So the pressure in pipe B:

Here 1 is taken with a minus sign, because this column of water helps to reduce the pressure in the pipe.

If air is completely removed from the connecting tube, then in this case the basic equation of hydrostatics will be written as follows:

The exact meaning of the answers: and is obtained at g = 10 m/.

7. With the valve of the pipeline K closed, determine the absolute pressure in the tank buried at a depth of H = 5m, if the reading of the vacuum gauge installed at a height of h = 1.7m, . Atmospheric pressure corresponds to the density of gasoline .

Solution: According to the basic equation of hydrostatics, the absolute pressure in the tank will be the sum of the absolute pressure on the free surface and the weight pressure, i.e.

Absolute pressure on the free surface :

or

Taking into account the obtained expression for
We write the original equation as follows:

8. Water and gasoline are poured into a cylindrical tank with a diameter of D \u003d 2m to the level of H \u003d 1.5m. The water level in the piezometer is lower than the gasoline level by h=300mm. Determine the weight in the tank

gasoline, if .

Solution: The weight of the gasoline in the tank can be written as

,

where is the volume of fuel in the tank. We express it in terms of the geometrical parameters of the tank:

.

To determine the unknown value - the level of gasoline in the tank, it is necessary to write down the basic equation of hydrostatics for a surface of equal pressure, which is most appropriate to take the bottom of the tank, since we have information about it in the form of H - the total level of gasoline and water in the tank. Since the tank and the piezometer are open (communicate with the atmosphere), we will take into account only the weight pressure on the bottom.

So, the pressure on the bottom from the side of the tank can be written as

This is the same pressure from the side of the piezometer:

.

Equating the right parts of the obtained expressions, we express the desired value from them:

We reduce the resulting equation by g, removing in both parts of the equation , we write the desired value

From the last equation

We substitute the resulting expressions for and into the original equation and determine the weight of gasoline

9. The hydraulic jack consists of a fixed piston 1 and a cylinder 2 sliding along it, on which a housing 3 is mounted, forming an oil bath of the jack and a manual plunger pump 4 with suction 5 and discharge 6 valves. Determine the pressure of the working fluid in the cylinder and the mass of the lifted load m, if the force on the handle of the pump drive lever is R=150 N, the diameter of the jack piston is D=180 mm, the diameter of the pump plunger is d=18mm, the efficiency of the jack is η = 0.68, the arms of the lever are a =60mm, b=600mm.

Air pressure- the force with which the 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 for the formation of wind is the difference in atmospheric pressure. Wind blows from an area of ​​higher pressure to an area of ​​lower pressure. The greater the difference in atmospheric pressure, the stronger the wind. The distribution of atmospheric pressure on Earth determines the direction of the winds that prevail in the troposphere at different latitudes.

Formed when water vapor condenses in the rising air due to its cooling.
. Water in liquid or solid state that falls on the earth's surface is called precipitation.

There are two types of precipitation:

falling out of the clouds (rain, snow, grains, hail);
formed near the surface of the Earth (, dew, frost).
Precipitation is measured by a layer of water (in mm.), Which is formed if the precipitated water does not drain and does not evaporate. On average, 1130 mm falls on the Earth per year. precipitation.

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

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

In high pressure belts, descending air currents predominate. Cold air, descending, contains little moisture. When lowered, it contracts and heats up, due to which it moves away from the saturation point and becomes drier. Therefore, in areas of high pressure over the tropics and near the poles, there is little precipitation.

By the amount of precipitation it is still impossible to judge the provision of the territory with moisture. It is necessary to take into account the possible evaporation - volatility. It depends on the amount of solar heat: the more it is, the more moisture can evaporate, if any. Evaporation can be large and evaporation small. For example, volatility (how much moisture can evaporate at a given temperature) is 4500 mm/year, and evaporation (how much actually evaporates) is only 100 mm/year. According to the ratio of evapotranspiration and evaporation, the moisture content of the territory is judged. Moisture coefficient is used to determine moisture content. Moisture coefficient - the ratio of annual precipitation to evaporation for the same period of time. It is expressed as a fraction as a percentage. If the coefficient is equal to 1 - sufficient moisture, if less than 1, moisture is insufficient, and if more than 1, then moisture is excessive. According to the degree of moisture, wet (humid) and dry (arid) areas are distinguished.

Pressure is a physical quantity that plays a special role in nature and human life. This phenomenon, imperceptible 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 of it exist and how to find the pressure (formula) in different environments.

What is called pressure in physics and chemistry

This term refers to an important thermodynamic quantity, which is expressed as the ratio of the perpendicularly exerted pressure force 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 in the system.

In physics and chemistry, this is denoted by the letter "P", which is an abbreviation for the Latin name of the term - pressūra.

If we are talking about the osmotic pressure of a liquid (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 under consideration is measured in pascals (in Cyrillic - Pa, in Latin - Ra).

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

However, in practice, it is rather difficult to use pascals, since this unit is very small. In this regard, in addition to the standards of the SI system, this value can be measured in a different way.

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

• bars. One bar is equal to 105 Pa.
• Torres, 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 at.
• Kilogram-force per square centimeter. There are also ton-force and gram-force. In addition, there is an analog pound-force per square inch.

General pressure formula (7th grade physics)

From the definition of a given physical quantity, one can determine the method of finding it. It looks like 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 is related to other thermodynamic variables: volume and mass.

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

Hydrostatic pressure formula

Different aggregate states of substances provide for the presence of their properties that are different 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. At the same time, it cannot be used to calculate atmospheric pressure, due to the difference in altitudes and air densities.

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

The variant under consideration is an adaptation of the classical example P = F / S.

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

If you insert this data, it turns out that the area in the numerator and denominator can be reduced and the output is the above formula: P \u003d pgh.

Considering the pressure in liquids, it is worth remembering that, unlike solids, the 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 \u003d P 0 + 2QH. In this case, P 0 is the pressure of a non-curved layer, and Q is the liquid tension surface. H is the average curvature of the surface, which is determined by Laplace's Law: H \u003d 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 to 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 common, 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 pressure.

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

From this it follows that the partial pressure formula looks like this: P \u003d 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 the air pressure. However, some 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. So, you should take the aforementioned P \u003d 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 under consideration using the above 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, a number of devices have been developed over several centuries to do this instead of people.

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

• Atmospheric pressure is measured using a pressure gauge called a barometer. If it is necessary to determine the vacuum (that is, pressure below atmospheric pressure), another version of it, a vacuum gauge, is used.
• In order to find out the blood pressure in a person, a sphygmomanometer is used. To most, it is better known as a non-invasive tonometer. 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 place of measurement.
• Pressure drops in the environment (in English - pressure drop) are determined using or difnamometers (not to be confused with dynamometers).

Types of pressure

Considering the 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
• Excess.
• Vacuum.
• 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 by the formula P = P 2 + P 3 or P = P 2 - P 4.

For the reference point for absolute pressure under the conditions of the planet Earth, the pressure inside the container from which air is removed (that is, classical vacuum) is taken.

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 under the name atmospheric.

It is reckoned to and its value varies with the place and time of measurement, as well as weather conditions and being above / below sea level.

The value of barometric pressure is equal to the modulus of the force of the atmosphere per unit area along the 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 norm of 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, in the mountains, water in kettles boils much faster than at home on the stove. The fact is that pressure affects the boiling point: with its decrease, the latter decreases. And vice versa. The work of such kitchen appliances as a pressure cooker and an autoclave is built on this property. The increase in pressure inside them contributes to the formation of higher temperatures in the dishes than in ordinary pans on the stove.

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

P is the desired value at height, P 0 is the density of air 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 Eclair number, equal to 2.71828.

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

When making calculations, it is always worth taking into account the possibility of changes in air temperature due to a change in the meteorological situation or when climbing above sea level, as well as geographical latitude.

Gauge and vacuum

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

If it is positive, it is called gauge pressure.

If the result obtained is with a minus sign, it is called a vacuum gauge. It is worth remembering that it cannot be more than barometric.

differential

This value is the pressure difference at different measuring 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 the help of what formulas it is found, we can conclude that this phenomenon is very important, and therefore knowledge about it will never be superfluous.

Do you think a fish, swimming in the ocean, notices that there is water around it? Does the dog feel that he is walking on the bottom of the air ocean? Habit dulls observation. A fish that was born in water and has spent its entire life in it, no doubt does not notice the water and does not feel the pressure caused by its weight. Just like a dog, of course, does not pay attention to the air around him and does not feel its pressure on his body. We wouldn't have noticed it either, unless we heard it from someone or read it in books. Something has to happen for us to pay attention to the air. Either it starts to move quickly and the wind blows in our face, or a clearly visible cloud forms in it. But the most obvious way to verify the presence of air is to see how it presses on the objects in it.

Take a plastic cup or other container and completely submerge it in the bath water. Let's wait until the glass is filled with water and turn it upside down. Slowly begin to pull it out of the water. Look! The water rises with the glass, and its level is much higher than the level of the water in the bath. It would seem that nothing supports water in a glass. But this, of course, is not so, otherwise it would have fallen. What is this force that lifts water? An ocean of air stretches above us for several hundred kilometers. Although the air seems to us completely weightless, it exerts a significant pressure on the surface of the Earth for every square centimeter. Your bath, of course, is no exception, the air presses on the surface of the water in it in the same way as on everything else around.

When we begin to pull out a glass turned upside down, the water in it tends to sink under the influence of the earth's gravity. However, she cannot go down. Why?

To understand this, imagine that the water actually dropped a little, as shown in the picture. What will be in space above dashed line A? Naturally, there is no air here, and therefore, its pressure too. In other words, in a glass at level A, atmospheric pressure does not act on the surface of the water. Now let's look at arrows B and C. They show how atmospheric pressure acts on the surface of the water in the bath. Air presses on water, it is compressed by this air, which means it seeks to fill the resulting empty space. As a result, as soon as the water begins to pour out of the glass, the pressure will drive it back into the space above level A, as shown in the figure by arrows D and E.

There is no atmospheric pressure.

In fact, the water in the glass never sinks enough to be noticeable, atmospheric pressure immediately pushes it back into the glass and holds it there while we pull it out.

But if water is held by atmospheric pressure in a glass 15 cm high, will it also be held in a vessel 30 cm high? And in 60 cm? 3 meters? 5 meters? If you have suitable dishes at home, you will make sure that water is retained in them. However, there is a limit to the height of the water column that can be maintained in this manner. Water has a mass much greater than the mass of air, if we compare their equal volumes. Water is 800 times heavier than air of the same volume. Water, like air, presses on the bodies in it. This means that the pressure of a column of water 10 m high (more precisely, 10 m 33 cm) will just balance the atmospheric pressure, which holds the water in the vessel. Thus, you see that the height of the water column cannot noticeably exceed 10 meters.

Imagine a tall 15-meter "glass" (or rather, a pipe), turned upside down, which we pull out of the water, as shown in the figure. When the closed part of the "glass" reaches a height of about 10 m above the water level, the liquid in the "glass" will stop rising. We continue to raise the "glass", but the water inside it is at the same level. In this case, an empty space is formed in the vessel above the water level.

What will happen to the water in the vessel if the atmospheric pressure decreases for any reason? The new atmospheric pressure will be able to hold an already smaller column of water, the water level in the "glass" will drop. What if the outside air pressure increases? It will be able to hold the height of the pillar greater than 10 m, and the water in the vessel will begin to rise.

In essence, we have analyzed the principle of operation of the device - a barometer, with which atmospheric pressure is measured. In our case, atmospheric pressure is balanced by a column of water of a certain height. Air pressure can be measured by the height of the water column it can hold.

This type of water barometer was invented by Otto von Guericke several centuries ago. As a "glass" he used a glass pipe, closed at the upper end, which he filled with water and installed near his house. The pipe was lowered into a tank of water. Guericke set up the barometer so that the level of the upper part of the pipe was visible from everywhere to the inhabitants of the town, and they could observe how the float on the surface of the water in the pipe, marking its level, rose and fell according to changes in atmospheric pressure. If the float in the barometer dropped sharply, the townspeople already knew that the air pressure was dropping, and, most likely, bad weather was coming, and when the float rose in the tube, it meant that good weather would soon come to the town.

Why does a change in barometric pressure mean a likely change in the weather? It turns out that warm, moist air, which usually brings cloudy weather, is lighter than cold and dry air - a harbinger of clear and good weather, which means that when the weather worsens, the pressure should fall, and when it improves, it should rise. The barometer is a widely used instrument. True, a pipe 10 meters high, and even filled with water, is obviously very inconvenient for use.

You can significantly shorten the pipe if you use mercury instead of water - a liquid metal that is 13.6 times heavier than water. In a mercury barometer, the pressure that equalizes atmospheric pressure is created by a column of liquid with a height of only 1033/13.6 = 76 (cm). This, of course, is much more convenient than more than 10 meters, so it is better to use mercury instead of water in barometers. Such a device in its design is no different from a water one, only it is much smaller, and it is not necessary to hold the pipe with your hand - it is fixed in the required position, in some more convenient way.

The fabric can be pierced with a needle, but not with a pencil (if you apply the same force). The pencil and the needle have different shapes and therefore exert unequal pressure on the tissue. The pressure is omnipresent. It activates the mechanisms (see the 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 acting perpendicular to its surface, the body is under pressure. The pressure depends on how big the force is and on the area of ​​the surface that the force is acting on. For example, if you go out into the snow in ordinary shoes, you can fail; this will not happen if we put on skis. The weight of the body is the same, but in the second case, the pressure is distributed over a larger surface. The larger the surface, the lower the pressure. The reindeer has wide hooves - after all, he walks on the snow, and the pressure of the hoof on the snow should be as low as possible. If the knife is sharp, force is applied to the surface of a small 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 vessel in which they are contained. Unlike solids, liquids and gases press on all walls of the vessel. The pressure of liquids and gases is directed in all directions. presses not only on the bottom, but also on the walls of the aquarium. The aquarium itself only pushes down. presses from the inside on the soccer ball in all directions, and therefore the ball is round.

Hydraulic mechanisms

The action of hydraulic mechanisms is based on fluid pressure. The liquid does not compress, so if you apply force to it, it will be forced to move. And the brakes work on the hydraulic principle. Reducing the speed of the track is achieved with the help of brake fluid pressure. The driver presses the pedal, the piston pumps the brake fluid through the cylinder, then it enters the other two cylinders through the tube and presses on the pistons. The pistons press the brake pads against the wheel disk. The resulting slows 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 the 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 powered by compressed carbon dioxide. By squeezing the handle, you release the compressed carbon dioxide in the canister. The gas with great force presses down on a special solution, displacing it into the tube and hose. A stream of water and foam escapes from the hose.

Atmosphere pressure

Atmospheric pressure is created by the weight of the air above the surface. For every square meter, the air presses with a force greater than the weight of an elephant. Near the surface of the Earth, the pressure is higher than high in the sky. At an altitude of 10,000 meters, where jet planes fly, the pressure is small, since an insignificant air mass presses from above. Normal atmospheric pressure is maintained in the cabin so that people can breathe freely at high altitude. But even in a pressurized cabin, people get stuffy ears when the pressure is lower than the pressure inside the auricle.

Atmospheric pressure is measured in millimeters of mercury. When the pressure changes, so does . Low pressure means that worsening weather is ahead. 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).