An open rectangular tank is filled with liquid (Fig. 1) to a depth H. Find the absolute and gauge pressure at the bottom of the tank. The data for the calculation are given in Table 1.

A closed rectangular tank is filled with liquid to a depth H (Fig. 2). The fluid density ρ and the excess pressure on the surface p 0 are set (see Table 2). Determine the piezometric height h p and plot the excess pressure on the wall indicated in Table 2.

Density, kg / m 3

Density, kg / m 3

Density, kg / m 3

Option 1

Vertical distance between horizontal axes tanks filled with water, a = 4 m, while the gauge pressure on the axis of the right. reservoir p 2 = 200 kPa. The difference between the levels of mercury h = 100 cm. The level of mercury in the left knee is located below the axis of the left tank at H = 6 m.

Determine the gauge hydrostatic pressure p 1 on the axis of the left tank, as well as its upper generatrix, if the tank diameter is d = 2 m.

Option 2

The mercury pressure gauge is connected to a tank filled with water.

I) Determine the excess pressure on the water surface in the tank p 0 if h 1 = 15 cm, h 2 \u003d 35 cm. 2) Determine the vacuum above the water surface if the levels of mercury in both knees of the manometer are equal? The density of mercury ρ rt \u003d 13600 kg / m 3.

Option 3

A mercury manometer is attached to a closed tank filled with water to a depth of H = 10 m. The difference between the levels of mercury in the manometer is h = 100 cm, while the free surface of the water in the tank exceeds the level of mercury in the left knee by H = 12 m. Atmospheric pressure p a = 100 kPa.

I. Determine the absolute air pressure p 0 in the space above the free water surface in the tank. 2. Find the absolute hydrostatic pressure at the lowest point of the tank bottom.

Option 4

In a closed tank there is water with a depth of H = 5 m, on the free surface of which the gauge pressure p 0 = 147.15 kPa. To the tank at a depth h = 3 m a piezometer is connected, i.e. a tube that is open at the top and vented to the atmosphere .

1. Determine the piezometric height h p .

2. Find the value of the gauge hydrostatic pressure at the bottom of the vessel.

Option 5

In a differential pressure gauge connected to a closed reservoir, the difference in mercury levels is h = 30 cm. The level of mercury in the left knee of the manometer is in a horizontal plane coinciding with the bottom of the tank.

1) Find the absolute air pressure and vacuum in the space above the free water surface in the tank.

2) Determine the absolute hydrostatic pressure at the bottom of the tank. Water depth in the tank H = 3.5 m.

Option 6

A piezometer is attached to a closed tank with a horizontal bottom. Atmospheric pressure on the water surface in the piezometer p a =100 kPa. Water depth in the tank h = 2 m, water height in the piezometer H = 18 m. Determine the absolute pressure on the water surface in the tank and the absolute and gauge pressure at the bottom.

Option 7

Point A is buried under the water horizon in the vessel by h = 2.5 m, the piezometric height for this point is equal to h Р = 1.4 m.

Determine for point A the magnitude of the absolute pressure, as well as the magnitude of the vacuum on the surface of the water in the vessel, if the atmospheric pressure p a \u003d 100 kPa.

Option 8

Two tubes are connected to the closed vessel, as shown in the drawing. The left tube is lowered into a jar of water, the right tube is filled with mercury.

Determine the absolute air pressure p 0 on the surface of the liquid in the vessel and the height, the mercury column h 2, if the height of the water column h 1 \u003d 3.4 m, and the atmospheric pressure p a \u003d 100 kPa. The density of mercury ρ rt \u003d 13600 kg / m 3.

Option 9

Two closed tanks, the horizontal bottoms of which are located in the same plane, are connected by a differential pressure gauge, the difference in mercury levels in it is h = 100 cm, while the level of mercury in the left elbow coincides with the plane of the bottom of the tank. The left tank contains water with a depth of H 1 = 10 m. The right one contains oil with a depth of H 2 = 8 m. The density of the oil is ρ m = 800 kg/m 3, the density of mercury is ρ \u003d 13600 kg / m 3. On the surface of the water, the gauge pressure p 1 \u003d 196 kN / m 2 . Find the gauge pressure on the surface of the oil p 0 . Determine the gauge pressure at the bottom of each tank.

Option 10

Horizontally arranged round tanks are filled with water. The diameter of each tank is D = 2 m. The difference between the mercury levels in the manometer is h = 80 cm. The gauge hydrostatic pressure p 1 on the axis of the left tank is 98.1 kPa. The axis of the right reservoir is below the axis of the left one by z = 3 m/

Determine the gauge hydrostatic pressure p 2 on the axis of the right tank, as well as on its lower generatrix - at point A.

Option 11

Determine the pressure difference at points located on the axes of cylinders A and B filled with water, if the difference in mercury levels in the differential pressure gauge Δh = 25 cm, difference between the levels of the axes of the cylinders H = 1 m.

Option 12

The tube, closed at the top, is lowered with its open end into a vessel with water. On the free surface of water in the tube, the absolute pressure p 0 =20 kPa. Atmospheric pressure p a \u003d 100 kPa. Determine the height of the rise of water in the tube h.

Option 13

A closed tank with a horizontal bottom contains oil. Oil depth H = 8 m. Find the gauge and absolute pressure at the bottom of the tank if the gauge pressure above the free surface of the oil is p 0 = 40 kPa , Oil density ρ n = 0.8 g/cm 3 . Atmospheric pressure p a = 100 kPa.

Option 14

The absolute pressure on the water surface in the vessel p 0 = 147 kPa.

Determine the absolute pressure and gauge pressure at point A, located from depth h = 4.8 m, also found piezometric; height h p for this point. Atmospheric pressure a = 100 kPa.

Option 15

Determine the excess surface pressure p 0 in a closed vessel with water, if mercury has risen to a height h \u003d 50 cm in the tube of an open manometer. The surface of the water is at a height h 1 \u003d 100 cm from the lower level of mercury. The density of mercury ρ rt \u003d 13600 kg / m 3.

Option 16

Two closed tanks, the axes of which are in the same horizontal plane, are filled with water and connected by a U-shaped tube.

The water levels in the left and right knees are respectively equal, z l = 1.5 m, z p = 0.5 m.

The upper part of the tube is filled with oil, the density of which is ρ m = 800 kg/m 3 . Gauge pressure on the axis of the left tank p l = 78.5 kPa. Determine the gauge pressure on the axis of the right tank and on the line of separation of water and oil in the left tube.

Option 17

In a closed tank there is water with a depth of H = 2m, on the free surface of which the pressure is equal to p 0 . In a differential pressure gauge connected to the tank, the level difference is h = 46 cm. The level of mercury in the left knee coincides with the bottom of the tank. Determine the absolute pressure p 0 and the absolute hydrostatic pressure at the bottom of the tank if the atmospheric pressure p a = 100 kPa.

Option 18

The spillway of the dam, which retains water in the reservoir, is closed by a segmental gate AE of a circular shape with a radius r = 2 m. Determine the absolute hydrostatic pressure at the bottom of the gate E (R E, abs) and find the height of the dam h, if the excess pressure at the bottom of the reservoir R di = 75 kPa. Atmospheric pressure p a \u003d 101 kPa.

Option 19

Determine the difference between the levels of mercury h in the connecting tube of communicating vessels, if the pressure on the water surface in the left vessel is p 1 = 157 kPa. The rise of the water level above the lower level of mercury H = 5 m. The difference between the levels of water and oil Δh = 0.8 m. p 2 = 117 kPa. Oil density ρ m \u003d 800 kg / m 3. Density of mercury ρrt \u003d 13600 kg / m 3.

Option 20

Two round tanks, located on the same level, are filled with water. Diameter of each tank D = 3 m. The difference between the levels of mercury h = 40 cm. Hydrostatic pressure on the axis of the first tank p 1 = 117 kPa. Determine the hydrostatic pressure on the axis of the second tank p 2 , as well as at the lower point. Density of mercury ρ rt = 13600 kg / m 3.

Option 21

There is water in the tank. The horizontal part of the inner wall of the BC tank is located at a depth h = 5 m. The water depth in the tank is H = 10 m. Atmospheric pressure p a = 100 kPa.

Find the gauge hydrostatic pressure at points B and C, plot this pressure on the ABSD wall and determine the absolute hydrostatic pressure on the tank bottom.

Option 22

The difference in water levels in closed tanks communicating with each other is h = 4 m. In the left tank, the water depth is H = 10 m and the absolute pressure on the free water surface is p 1 = 300 kPa.

Find the absolute air pressure p 2 on the free water surface in the right tank and at the bottom of the tanks.

Option 23

The closed reservoir contains mineral oil having a density ρ = 800 kg/m 3 . Above the free surface of the oil, the excess air pressure p o u = 200 kPa. A manometer is attached to the side wall of the tank, shown in the drawing. Calculate:

1. Excessive pressure on the bottom of the tank and

2. Gauge reading

Option 24

Vacuum gauge B, connected to the tank above the water level, shows the vacuum pressure p vac = 40 kPa. The water depth in the reservoir is H = 4 m. On the right side, a liquid mercury vacuum gauge is attached to the reservoir above the water level.

Calculate:

absolute air pressure in the tank p abs,

the height of the rise of water in a liquid vacuum gauge h,

absolute pressure at the bottom of the tank r dabs,

Atmospheric pressure p a = 98.06 kPa. The density of mercury ρ rt \u003d 13600 kg / m 3.

Option 25

The difference in water levels in the reservoirs is h= 15 m. The water depth in the left reservoir is H = 8 n.

Calculate

gauge air pressure above the water surface in the closed left tank p o,

excess pressure on the bottom of the left tank rdi,

build a diagram of excess pressure on the left vertical wall of a closed tank.

Option 26

There are three different liquids in a closed tank: mineral oil with a density of ρ m = 800 kg/m 3 water and mercury with a density of ρ rt = 13600 kg/m 3 . The level of mercury in the piezometer is 0.15 m higher than in the tank (h 3 = 0.15 m). Atmospheric pressure p a = 101 kPa. Calculate:

1. Absolute air pressure under the tank cover;

2. Vacuum pressure under the tank cover if h 1 = 2 m, h 2 = 3m.

Option 27

In a hermetically sealed tank is mineral oil with a density ρ m = 800 kg/m 3 . Oil depth h 1 \u003d 4 m. A mercury manometer is attached to the tank wall above the oil level, in which the difference in mercury levels h 2 \u003d 20 cm. Atmospheric pressure p a \u003d 101 kPa. The mercury level in the left knee of the manometer and the oil level in the tank are at the same mark.

Determine the absolute air pressure under the tank cover (R oh abs ) and gauge oil pressure at the bottom of the tank (R d, m )

Option 28

Water is contained in a hermetically sealed tank. To the side wall of the tank at depth h = 1.2 m a mechanical pressure gauge is connected, which indicates the hydrostatic pressure p m = 4 atm. Determine the absolute pressure on the free surface of the water in the tank R oh abs and the pressure value shown by the pressure gauge mounted on the tank cap. Atmospheric pressure is 101 kPa.

Option 29

Two water tanks are separated by a vertical wall with a hole at the bottom. The left tank is open. The right tank is closed with a sealed lid. Depth of water in the left tank h 1 = 8 m. Depth of water in the right tank h 2 = 1m.

Atmospheric pressure p a \u003d 101 kPa.

Determine the excess hydrostatic air pressure under the lid of the right tank and the absolute pressure at the bottom of the right tank.

Option 30

Two hermetically sealed water tanks are connected by a mercury manometer. Gauge air pressure above the water surface in the left tank R l, m = 42 kPa. Absolute air pressure above the water surface in the right tank p p, abs =116 kPa. Depth of water above the level of mercury in the left tank h 1 \u003d 4 m. Water depth above the mercury level in the right tank h 3 = 2.5 m. Atmospheric pressure pa =101 kPa. Determine the difference in mercury levels in the manometer h 2 .

When solving problems on the topic of hydrostatic pressure, it is necessary to distinguish and not confuse the concepts of absolute pressure P A, overpressure P, vacuum P VAK, know the relationship between pressure (Pa) and the corresponding piezometric height (h), understand the concept of pressure, know Pascal's law and properties of hydrostatic pressure.

When determining the pressure at a volume point or at a site point, the basic equation of hydrostatics (1.1.13) is used.

When solving problems with a system of vessels, it is necessary to compose an equation of absolute pressures that ensure the immobility of the system, i.e. equality to zero of the algebraic sum of all acting pressures. The equation is drawn up for some surface of equal pressure, chosen as the reference surface.

All units of measurement of quantities should be taken in the SI system: mass - kg; strength - N; pressure - Pa; linear dimensions, areas, volumes - m, m 2, m 3.

EXAMPLES

Example 1.1.1. Determine the change in the density of water when it is heated from t 1 \u003d 7 o C to t 2 \u003d 97 o C, if the thermal expansion coefficient b t \u003d 0.0004 o C -1.

Solution. When heated, the specific volume of water increases from V 1 to V 2.

According to formula (1.1.1), the density of water at initial and final temperatures is:

r 1 \u003d M / V 1, r 2 \u003d M / V 2.

Since the mass of water is constant, the change in density is expressed as:

From formula (1.4) an increase in the volume of water , then

Note: the change in the density of a liquid during compression is determined similarly using the volumetric compression ratio according to the formula (1.1.2). In this case, V 2 \u003d V 1 - DV.

Example 1.1.2. Determine the volume of the expansion tank of the water cooling system with a capacity of 10 liters when heated from temperature t 1 \u003d 15 ° C to t 2 \u003d 95 ° C at a pressure close to atmospheric.

Solution. Without taking into account the safety factor, the volume of the tank is equal to the additional volume of water during thermal expansion. From formula (1.1.4) an increase in the volume of water

.

The density of water is taken according to table 1: r 1 \u003d 998.9 kg / m 3, r 2 \u003d 961.8 kg / m 3. The coefficient of thermal expansion is determined by the formula (1.1.5):

The initial volume V \u003d 10l \u003d 10. 10 -3 m 3 \u003d 0.01 m 3.

Additional water volume:

DV = 10 . 10 -3 (95 -15) 0.46. 10 -3 = 368. 10 -6 m 3 \u003d 0.368 l

Example 1.1.3. In a cooled vessel, a gas having an initial pressure P 1 = 10 5 Pa. and occupying a volume V 1 = 0.001 m 3, is compressed to a pressure P 2 = 0.5. 10 6 Pa. Determine the volume of gas after compression.

Solution. In the case of a cooled vessel, the process is isothermal (t = const), in which the equation of state of the gas (1.1.8) takes the form:

R V = const or R 1 V 1 = R 2 V 2

How do we determine the volume of gas after compression

V 2 \u003d P 1 V 1 / P 2 \u003d 1. 10 5 . 0.001 / 0.5 . 10 6 \u003d 0.0002 m 3 \u003d 0.2 l.

Example 1.1.4. Determine the volume of water that must be additionally supplied to the pipeline with a diameter of d = 500 mm and a length of L = 1 km, filled with water before a hydraulic test at atmospheric pressure and a temperature of t = 20 ° C, to increase the pressure in it by DP = 5. 10 6 Pa. The pipe material is assumed to be absolutely rigid.

Solution. To determine the additional volume of water that must be supplied, we use the ratio (1.1.2):

=

The initial volume of water in the pipeline is equal to the volume of the pipeline:

Assuming, according to reference data, the modulus of volumetric elasticity of water

E \u003d 2. 10 9 Pa, we determine the volumetric compression ratio:

b V \u003d 1 / E \u003d 1 / 2. 109 = 5. 10 -10 , Pa -1

Transforming relation (1.1.2) with respect to DV, we obtain:

b V DP V TP + b V DP DV = DV; b V DP V TP = (1 + b V DP) DV

Expressing DV, we obtain the required additional volume:

Example 1.1.5. Determine the average thickness of deposits d ETL in a pipeline with an internal diameter of d = 0.3 m and a length of L = 2 km, if, when water is released in the amount of DV = 0.05 m 3, the pressure in it drops by DP = 1. 10 6 Pa.

Solution. The interdependence of changes in the volume and pressure of water is characterized by the modulus of volume elasticity.

We accept: E \u003d 2. 10 9 Pa.

From formulas (1.1.2) and (1.1.3) we find the volume of water in the pipeline with deposits:

The same volume is equal to the capacity of the pipeline:

Where we determine the average inner diameter of the pipe with deposits

The average deposit thickness is:

Example 1.1.6. The viscosity of the oil, determined by the Engler viscometer, is 8.5 o E. Calculate the dynamic viscosity of the oil if its density is r = 850 kg/m 3 .

Solution. Using the Ubellode empirical formula (1.1.9), we find the kinematic viscosity of oil:

n \u003d (0.0731 about E - 0.0631 / about E) 10 -4 \u003d

\u003d (0.0731. 8.5 - 0.0631 / 8.5) \u003d 0.614. 10 -4 m 2 / s

Dynamic viscosity is found from relation (1.1.7):

m = n r = 0.614 . 10 -4 . 850 = 0.052 Pa. With.

Example 1.1.7. Determine the height of water rise in a capillary tube with a diameter of d = 0.001 m at a temperature of t = 80 ° C.

Solution. From reference data we find:

density of water at a temperature of 80 ° C r \u003d 971.8 kg / m 3;

surface tension of water at a temperature of 20 ° C s O = 0.0726 N / m;

coefficient b \u003d 0.00015 N / m O С.

According to the formula (1.1.11) we find the surface tension of water at a temperature of 80 ° C:

s \u003d s O - b Dt \u003d 0.0726 - 0.00015. (80 -20) = 0.0636 N/m

According to formula (1.1.12), the change in surface pressure, which determines the height of the capillary rise h CAP, is:

R POV = 2s / r or r g h KAP = 2s / r,

where we find the height of the rise of water in the tube:

h KAP = 2 s / r g r = 2 . 0.0636/971.8. 9.81. 0.0005 =

0.1272 / 4.768 = 0.027 m = 2.7 cm.

Example 1.1.8. Determine the absolute hydrostatic pressure of water at the bottom of an open vessel filled with water. The depth of water in the vessel is h = 200 cm. Atmospheric pressure corresponds to 755 mm Hg. Art. The water temperature is 20 ° C. Express the obtained pressure value with the height of the mercury column (r RT \u003d 13600 kg / m 3) and the water column.

Solution: According to the basic equation of hydrostatics for an open reservoir, the absolute pressure at any point in the volume is determined by the formula (1.1.14):

R A \u003d R a + r g h

According to table 1, we take the density of water at a temperature of 20 ° C:

r \u003d 998.23 kg / m 3.

Converting the units of measurement of atmospheric pressure and the depth of water in the vessel to the SI system, we determine the absolute pressure at the bottom of the vessel:

R A \u003d 755. 133.322 + 998.23 . 9.81. 2=

100658 + 19585 = 120243 Pa = 120.2 KPa

Find the corresponding height of the mercury column:

h A \u003d P / r RT g \u003d 120243 / 13600. 9.81 = 0.902 m.

Find the height of the water column corresponding to the given absolute pressure:

h A \u003d R A / r g \u003d 120243 / 998.23. 9.81 \u003d 12.3 m.

This means that if a closed piezometer (a tube in which an absolute vacuum is created) is attached to the level of the bottom of the vessel, then the water in it will rise to a height of 12.3 m. The pressure of this column of water balances the absolute pressure exerted on the bottom of the vessel by liquid and atmospheric pressure .

Example 1.1.9. In a closed tank with water, the pressure on the free surface Р О =14.7. 10 4 Pa. To what height H will water rise in an open piezometer connected at a depth h = 5 m. Atmospheric pressure corresponds to h a = 10 m of water. Art.

Solution. To solve this problem, it is necessary to compose an equation for the equality of absolute pressures from the side of the reservoir and from the side of the piezometer relative to the chosen plane of equal pressure. We choose a plane of equal pressure 0-0 at the level of the free surface in the tank.

The absolute pressure from the side of the tank at the selected level is equal to the surface pressure:

P A = P O. (1)

The absolute pressure at the same level from the side of the liquid in the piezometer is the sum of the atmospheric pressure P a and the pressure of water height h 1:

R A \u003d R a + r g h 1 (2)

Since the system is in equilibrium (at rest), the absolute pressures from the side of the reservoir and from the side of the piezometer are balanced. Equating the right parts of equalities (1) and (2), we obtain:

R O \u003d R a + r g h 1,

The value of atmospheric pressure in the SI system is:

P a \u003d 9.806. 10,000 mm = 9.806. 10 4 Pa.

We find the height of the excess of the water level in the piezometer above the selected plane of equal pressure:

h 1 \u003d (P O - R a) / r g \u003d (14.7. 10 4 - 9.806. 10 4) / 1000. 9.81 = 5 m.

This excess does not depend on the connection point of the piezometer, since the pressures of liquid columns with a height h below the comparison plane on the left and right are mutually compensated.

The total water height in the piezometer is greater than the height h 1 by the immersion depth of the piezometer attachment point. For this task

H \u003d h 1 + h \u003d 5 + 5 \u003d 10 m.

Note: a similar result can be obtained by choosing the level of connection of the piezometer as the plane of equal pressure.

Example 1.1.10. Construct a diagram of the absolute pressure of a liquid on a broken wall in an open tank.

Solution. The absolute pressure in the case of an open tank is determined by the formula (1.1.14):

R A \u003d R a + r g h, i.e. excess pressure at each point increases by the value of surface pressure (Pascal's law).

Excess pressure is determined:

in t. C: P \u003d r g. 0 = 0

in t. B: P \u003d r g. H 2

in t. A: P \u003d r g (H 2 + H 1)

Let us set aside the value of the overpressure at point B along the normal to the wall NE and connect it to point C. We will get a triangle of the diagram of the overpressure on the wall NE. To plot the absolute pressure at each point, you must add the value of the surface pressure (in this case atmospheric).

Similarly, a diagram is constructed for the segment AB: Let us set aside the values ​​of the overpressure at point B and at point A in the direction of the normal to the line AB, and connect the obtained points. Absolute pressure is obtained by increasing the length of the vector by an amount corresponding to atmospheric pressure.

Example 1.1.11. Determine the absolute pressure of air in a vessel with water, if the indication of a mercury manometer is h = 368 mm, H = 1 m, the density of mercury r RT = 13600 kg / m 3. Atmospheric pressure corresponds to 736 mm Hg.

Solution.

We choose the free surface of mercury as the surface of equal pressure. Atmospheric pressure on the surface of mercury is balanced by the absolute pressure of air in the vessel P A, the pressure of a water column of height H and a mercury column of height h.

Let's compose an equilibrium equation and determine the absolute air pressure from it (translating all units into the SI system):

R a \u003d R A + r B g H + r PT g h, whence

R A \u003d R a - r B g H - r PT g h \u003d

736 . 133.3 - 1000 . 9.81. 1 - 13600 . 9.81. 0.368 = 39202 Pa

Since the absolute pressure of the air in the vessel is less than the atmospheric pressure, there is a vacuum in the vessel equal to the difference between the atmospheric and absolute pressures:

R VAK \u003d R a - R A \u003d 736. 133.3 - 39202 = 58907 Pa = 59 kPa.

Note: The same result can be obtained by choosing the free surface of water in the vessel or the interface between water and mercury as the surface of equal pressure.

Example 1.1.12. Determine the excess pressure P O of air in the pressure tank according to the readings of the mercury battery manometer. The connecting pipes are filled with water. Level marks are given in m. How high must the piezometer be to measure this pressure?

Solution. Excess pressure P O \u003d P A - P a in the tank is balanced by the pressure of the columns of mercury and water in the pressure gauge.

The pressures of mutually balanced heights in the sections of the pressure gauge bend are excluded from consideration. Summing up (taking into account the direction of action of pressure) the readings of the pressure gauge from the open end to the level of the free surface, we compose the equilibrium equation:

P O \u003d r PT g (1.8 - 0.8) - r V g (1.6 - 0.8) + r PT g (1.6 - 0.6) - r V g (2.6 - 0.6) =

R RT g (1.8 - 0.8 +1.6 - 0.6) - r B g (1.6 - 0.8 + 2.6 - 0.6) =

13600 . 9.81. 2 - 1000 . 9.81. 2.8 = 239364 Pa = 0.24 MPa

From formula (1.16) we find the height of the water column corresponding to the excess pressure P O:

h IZB \u003d P O / r B g \u003d 0.24. 10 6 / 1000 . 9.81= 24.5 m

The height of the piezometer is higher by the excess of the free surface of water in the tank above the plane with a zero mark:

H \u003d h IZB + 2.6 \u003d 27.1 m.

Example 1.13. Determine the thickness s of the steel wall of the tank with a diameter D = 4 m for storing oil (r H = 900 kg / m 3) with an oil layer height H = 5 m. The pressure on the oil surface is P O = 24.5. 10 4 Pa. Permissible tensile stress of the wall material s = 140 MPa.

Solution. The calculated wall thickness of a round tank (without a safety factor) is determined from the condition of resistance to the maximum overpressure. Atmospheric pressure in the tank is not taken into account, since it is compensated by atmospheric pressure from the outside of the tank.

The wall experiences the maximum excess pressure P at the bottom:

P \u003d R A - R a \u003d R O + r H g H - R a \u003d

24.5. 10 4 + 900 . 9.81. 5 - 10 . 10 4 \u003d 18.91. 10 4 Pa

The design wall thickness is determined by the formula:

Example 1.1.14. Determine the pressure drop of water in a vertical pipe ring if at point A it heats up to a temperature of t 1 = 95 ° C, and at point B it cools to t 2 = 70 ° C. The distance between the centers of heating and cooling h 1 = 12 m.

Solution. The pressure difference is due to the difference in the hydrostatic pressures of the hot water column in the left pipe and the cooled water in the right pipe.

The pressures of water columns of height h 2 in the left and right pipes are mutually balanced and are not taken into account in the calculation, since the water temperature in them and, accordingly, the density are the same. Similarly, we exclude from the calculation the pressure in the left and right risers with a height h 3.

Then the pressure on the left P 1 \u003d r G g h 1, the pressure on the right P 2 \u003d r O g h 1.

The pressure drop is:

DP \u003d R 2 - R 1 \u003d r O g h 1 - r G g h 1 \u003d g h 1 (r O - r G)

We accept, according to reference data (table 1), the density of water at a temperature of t 1 = 95 ° C and t 2 = 70 ° C: r G = 962 kg / m 3, r O = 978 kg / m 3

Finding the pressure difference

DP \u003d g h 1 (r 2 - r 1) \u003d 9.81. 12 (978 -962) = 1882 Pa.

Example 1.1.15. a) Determine the excess water pressure in the pipe if P MAN = 0.025 MPa, H 1 = 0.5 m, H 2 = 3 m.

b) Determine the pressure gauge readings at the same pressure in the pipe, if the entire pipe is filled with water, H 3 \u003d 5 m.

a) Decision. The excess pressure in the pipe is balanced by the surface pressure Р О = Р MAN at the pressure gauge connection point and by the system of water and air columns in the tube. The pressure of the air columns can be neglected due to its insignificance.

Let us compose an equilibrium equation, taking into account the direction of pressure of water columns in the tube:

P \u003d R MAN + r WOD g H 2 - r WOD g H 1 \u003d

0.025 + 1000 . 9.81. 10 -6 (3 - 0.5) = 0.025 + 0.025 = 0.05 MPa

b) Decision. Equilibrium equation for this case

P \u003d R MAN + r WOD g H 3,

whence R MAN \u003d R - r WOD g H 3 \u003d 0.05 - 1000. 9.81. 10 -6 . 5 \u003d 0.05 - 0.05 \u003d 0 MPa.

37.1. Home experiment.
1. Inflate the rubber balloon.
2. Number the phrases in such order that you get a coherent story about the experiment.

37.2. The vessel under the piston contains gas (Fig. a), the volume of which changes at a constant temperature. Figure b shows a graph of the distance h, at which the piston is located relative to the bottom, on time t. Fill in the gaps in the text using the words: increases; does not change; decreases.

37.3. The figure shows an installation for studying the dependence of gas pressure in a closed vessel on temperature. The numbers indicate: 1 - test tube with air; 2 - spirit lamp; 3 - rubber stopper; 4 - glass tube; 5 - cylinder; 6 - rubber membrane. Put a "+" sign next to true statements and a "" sign next to incorrect ones.

37.4. Consider graphs of pressure p versus time t corresponding to various processes in gases. Fill in the missing words in the sentence.

38.1. Home experiment.
Take a plastic bag and make four holes of the same size in it in different places on the bottom of the bag, using, for example, a thick needle. Pour water into a bag over the bathtub, hold it on top with your hand and squeeze the water out through the holes. Change the position of the hand with the bag, observing what changes occur with the streams of water. Draw the experience and describe your observations.

38.2. Check off the statements that reflect the essence of Pascal's law.

38.3. Add text.

38.4. The figure shows the transfer of pressure by a solid and liquid body enclosed under a disk in a vessel.

a) Check the correct statement.
After installing the weight on the disk, the pressure increases ... .

b) Answer the questions by writing down the necessary formulas and making the appropriate calculations.
With what force will a weight of 200 g placed on it put pressure on a disk with an area of ​​100 cm2?
How will the pressure change and by how much:
at the bottom of the vessel 1
at the bottom of the vessel 2
on the side wall of the vessel 1
on the side wall of the vessel 2

39.1. Mark the correct ending of the sentence.

The lower and side openings of the tube are tightened with identical rubber membranes. Water is poured into the tube and slowly lowered into a wide vessel of water until the water level in the tube matches the water level in the vessel. In this position of the membrane ... .

39.2. The figure shows an experiment with a vessel whose bottom can fall off.

Three observations were made during the experiment.
1. The bottom of an empty bottle is pressed if the tube is immersed in water to a certain depth H.
2. The bottom is still pressed against the tube when water is started to be poured into it.
3. The bottom begins to move away from the tube at the moment when the water level in the tube coincides with the water level in the vessel.
a) In the left column of the table, write down the numbers of observations that allow you to come to the conclusions indicated in the right column.

b) Write down your hypotheses about what might change in the experience described above if:
there will be water in the vessel, and sunflower oil will be poured into the tube; the bottom of the tube will begin to move away when the oil level is higher than the water level in the vessel;
there will be sunflower oil in the vessel, and water will be poured into the tube; the bottom of the tube will begin to move away before the levels of water and oil coincide.

39.3. A closed cylinder with a base area of ​​0.03 m2 and a height of 1.2 m contains air with a density of 1.3 kg/m3. Determine the "weight" air pressure at the bottom of the balloon.

40.1. Write down which of the experiments shown in the figure confirm that the pressure in a liquid increases with depth.

Explain what each experiment demonstrates.

40.2. The cube is placed in a liquid of density p, poured into an open vessel. Match the indicated liquid levels with formulas for calculating the pressure created by a column of liquid at these levels.

40.3. Mark with a "+" the correct statements.

Vessels of various shapes were filled with water. Wherein … .
+ the pressure of water at the bottom of all vessels is the same, since the pressure of the liquid at the bottom is determined only by the height of the liquid column.

40.4. Choose a couple of words missing from the text. “The bottom of vessels 1, 2 and 3 is a rubber film fixed in the instrument stand.”

40.5. What is the pressure of water at the bottom of a rectangular aquarium 2 m long, 1 m wide and 50 cm deep, filled to the top with water.

40.6. Using the drawing, determine:

a) the pressure created by a column of kerosene on the surface of the water:

b) pressure on the bottom of the vessel, created only by a column of water:

c) pressure on the bottom of the vessel created by two liquids:

41.1. Water is poured into one of the tubes of communicating vessels. What happens if the clamp is removed from the plastic tube?

41.2. Water is poured into one of the tubes of communicating vessels, and gasoline is poured into the other. If the clamp is removed from the plastic tube, then:

41.3. Fill in the text with appropriate formulas and draw a conclusion.
The communicating vessels are filled with the same fluid. liquid column pressure

41.4. What is the height of the water column in the U-shaped vessel relative to the level AB if the height of the kerosene column is 50 cm?

41.5. The communicating vessels are filled with engine oil and water. Calculate how many centimeters the water level is below the oil level if the height of the oil column relative to the liquid interface is Nm = 40 cm.

42.1. A 1 liter glass ball was balanced on a balance. The ball is closed with a cork into which a rubber tube is inserted. When air was pumped out of the ball with a pump and the tube was clamped with a clamp, the balance of the scales was disturbed.
a) What mass of weight will have to be placed on the left side of the scale to balance them? Air density 1.3 kg/m3.

b) What is the weight of the air in the flask before evacuation?

42.2. Describe what happens if the end of the rubber tube of the balloon, from which air has been evacuated (see task 42.1), is lowered into a glass of water, and then the clamp is removed. Explain the phenomenon.

42.3. A square with a side of 0.5 m is drawn on the asphalt. Calculate the mass and weight of a 100 m high air column located above the square, assuming that the air density does not change with height and is equal to 1.3 kg/m3.

42.4. As the piston moves upwards inside the glass tube, the water rises behind it. Mark the correct explanation for this phenomenon. Water rises behind the piston ... .

43.1. Circles A, B, C schematically depict air of different densities. Mark on the figure the places where each circle should be placed so that the whole picture is obtained, illustrating the dependence of air density on height above sea level.

43.2. Choose the correct answer.
In order to leave the Earth, any molecule of the Earth's air shell must have a speed greater than ... .

43.3. On the Moon, the mass of which is about 80 times less than the mass of the Earth, there is no air shell (atmosphere). How can this be explained? Write down your hypothesis.

44.1. Choose the correct statement.
In the experiment of Torricelli in a glass tube above the surface of mercury ... .

44.2. In three open vessels there is mercury: in vessel A, the height of the mercury column is 1 m, in vessel B - 1 dm, in vessel C - 1 mm. Calculate the pressure exerted on the bottom of the vessel by a column of mercury in each case.

44.3. Write down the pressure values ​​in the indicated units according to the example given, rounding the result to the nearest whole number.

44.4. Find the pressure at the bottom of a cylinder filled with sunflower oil if the atmospheric pressure is 750 mm Hg. Art.

44.5. What pressure is experienced by a scuba diver at a depth of 12 m under water if the atmospheric pressure is 100 kPa? How many times greater is this pressure than atmospheric pressure?

45.1. The figure shows a diagram of the aneroid barometer. Separate details of the design of the device are indicated by numbers. Fill the table.

45.2. Fill in the gaps in the text.

The figures show an instrument called an aneroid barometer.
This device measures ___ Atmosphere pressure __.
Record the reading of each instrument, taking into account the measurement error.

45.3. Fill in the gaps in the text. "The difference in atmospheric pressure in different layers of the Earth's atmosphere causes the movement of air masses."

45.4. Record the pressure values ​​in the indicated units, rounding the result to the nearest integer.

46.1. Figure a shows a Torricelli pipe at sea level. In figures b and c, mark the level of mercury in the tube placed on the mountain and in the mine, respectively.

46.2. Fill in the gaps in the text using the words given in brackets.
Measurements show that air pressure rapidly (decreases, increases) with increasing altitude. The reason for this is not only (decrease, increase) in air density, but also (decrease, increase) of its temperature when moving away from the Earth's surface at a distance of up to 10 km.

46.3. The height of the Ostankino TV tower reaches 562 m. What is the atmospheric pressure near the top of the TV tower if the atmospheric pressure at its base is 750 mm Hg. Art.? Express the pressure in mm Hg. Art. and in SI units, rounding both values ​​to integers.

46.4. Select from the figure and circle the graph that most correctly reflects the dependence of atmospheric pressure p on height h above sea level.

46.5. For a TV kinescope, the screen dimensions are l \u003d 40 cm and h \u003d 30 cm. With what force does the atmosphere press on the screen from the outside (or what is the pressure force), if atmospheric pressure patm \u003d 100 kPa?

47.1. Build a graph of pressure p, measured under water, from the depth of immersion h, filling in the table first. Consider g = 10 N/kg, patm = 100 kPa.

47.2. The figure shows an open liquid manometer. The division price and the scale of the device are 1 cm.

a) Determine how much the air pressure in the left leg of the pressure gauge differs from atmospheric pressure.

b) Determine the air pressure in the left knee of the manometer, taking into account that the atmospheric pressure is 100 kPa.

47.3. The figure shows a U-shaped tube filled with mercury, the right end of which is closed. What is atmospheric pressure if the difference in liquid levels in the elbows of a U-shaped tube is 765 mm, and the membrane is immersed in water to a depth of 20 cm?

47.4. a) Determine the division value and the reading of the metal pressure gauge (Fig. a).

b) Describe the principle of operation of the device, using the numerical designations of parts (Fig. b).

48.1. a) Cross out unnecessary from the highlighted words to get a description of the operation of the piston pump shown in the figure.

b) Describe what happens when the pump handle moves up.

48.2. With a piston pump, the diagram of which is given in task 48.1, at normal atmospheric pressure, water can be raised to a height of no more than 10 m. Explain why.

48.3. Insert the missing words in the text to get a description of the operation of a piston pump with an air chamber.

49.1. Complete the formulas showing the correct relationship between the areas of the pistons of the hydraulic machine at rest and the masses of the loads.

49.2. The area of ​​the small piston of the hydraulic machine is 0.04 m2, the area of ​​the large piston is 0.2 m2. With what force should act on the small piston to evenly lift a load of 100 kg, located on the large piston?

49.3. Fill in the gaps in the text describing the principle of operation of the hydraulic press, the diagram of which is shown in the figure.

49.4. Describe the principle of operation of a jackhammer, the device diagram of which is shown in the figure.

49.5. The figure shows a diagram of the pneumatic brake device of a railway car.

TASKS

To perform settlement and graphic work

Subject "Hydraulics"

Topic: Hydrostatics

Severodvinsk

MAIN THEORETICAL PROVISIONS

Hydraulics, or technical fluid mechanics is the science of the laws of equilibrium and motion of fluids, of the ways in which these laws are applied to solving practical problems;

Liquid called a substance that is in such a state of aggregation, which combines the features of a solid state (very low compressibility) and a gaseous state (fluidity). The laws of equilibrium and motion of dropping 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 over the surface, called superficial. The former include the forces of gravity and inertia, the latter - the forces of pressure and friction.

By pressure is the ratio of the force normal to the surface to the area. With an even distribution

shear stress is the ratio of the friction force tangent to the surface to the area:

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

If R abs< Р а, то имеется vacuum, the value of which:

R wak = R a - R abs

The main physical characteristic of a liquid is densityρ (kg / m 3), determined for a homogeneous liquid by the ratio of its mass m to volume V:

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

Density and specific gravity are related by the ratio:

where g- acceleration of gravity.

For fresh water γ water \u003d 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 is characterized by the modulus of bulk elasticity TO, included in the generalized Hooke's law:

where ΔV- increment (in this case, decrease) of the volume of the liquid V, due to an increase in pressure on Δр. For example, for water K waters ≈2. 10 3 MPa.

Thermal 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. Distinguish dynamic (μ) and kinematic (ν) viscosity. The first enters Newton's law of fluid friction, which expresses the shear stress τ in terms of the transverse velocity gradient dv/dt:

Kinematic viscosity associated with dynamic ratio

The unit of kinematic viscosity is m 2 /s.

Evaporation liquids is characterized by saturated vapor pressure as a function of temperature.

Saturated vapor pressure is 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 off-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 to the normal inside the volume of the liquid;

At any point inside the liquid, it is the same in all directions, that is, 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 only one force of gravity acts on it from among the body forces, is called the basic equation of hydrostatics:

where p0- pressure on any surface of the liquid level, for example, on a free surface; h- the depth of the considered point, counted 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 free surface, measured from the horizontal plane upwards; p/(pg)- piezometric height.

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

In hydrotechnical practice, external pressure is often equal to atmospheric: P 0 \u003d P at

The pressure value P at \u003d 1 kg / cm 2 \u003d 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.

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

In what follows, we keep the notation R for gauge pressure.

Figure 1.1

The sum of the terms gives the value total hydrostatic head

Sum - expresses hydrostatic head H without atmospheric pressure p at /ρg, i.e.

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

A graphical representation of the magnitude and direction of the hydrostatic pressure acting on any point on the surface is called the hydrostatic pressure diagram. To construct a diagram, it is necessary to plot the value of the hydrostatic pressure for the considered point normal to the surface on which it acts. So, for example, the diagram of the gauge 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 display the gauge pressure at the point TO, i.e. pK = ρghK , and in fig. 1.2,b - total hydrostatic pressure

The force of fluid pressure on a flat wall is equal to the product of the hydrostatic pressure ρ with 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

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: u s- coordinate of the center of gravity of the area.

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

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

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

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

Swimming of bodies and their stability. The floating condition of the body is expressed by the equality

G=P (1.6)

where G- body weight;

R- the resulting force of pressure of the liquid on the 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 the body, or displacement.

Strength R directed upwards and passes through the center of gravity of the displacement.

draft body at called the depth of immersion of the lowest point of the wetted surface (Fig. 1.3, a). Under the axis of navigation understand the line passing through the center of gravity FROM and displacement center D, corresponding / to the normal position of the body in a state of equilibrium (Fig. 1.3, a )-

waterline called the line of intersection of the surface of a floating body with the free surface of the liquid (Fig. 1.3, b). float plane ABEF called the plane obtained from the intersection of the body by the free surface of the liquid, or, otherwise, the plane bounded by the waterline.

Figure 1.3

In addition to fulfilling the navigation conditions (1.5), the body (ship, barge, etc.) must satisfy the stability conditions. A floating body will be stable if, when heeling, 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

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

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

If the center of gravity of the 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. The metacentric radius ρ is found by the formula:

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

W- displacement.

If the center of gravity of the body C is located above the center of displacement and the metacenter, then the body is unstable; emerging pair of forces G and R seeks to increase the roll (Fig. 1.4, in).

INSTRUCTIONS FOR SOLVING PROBLEMS

When solving problems in hydrostatics, first of all, it is necessary to master well and not confuse such concepts as pressure R and strength F.

When solving problems to determine 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 it rises, it decreases.

It is necessary to firmly distinguish between absolute pressure, gauge pressure and vacuum, and it is imperative to know the relationship between pressure, specific gravity and the height corresponding to this pressure (piezometric height).

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

Problem solving should be carried out in the international system of units SI.

The solution of the problem must be accompanied by the necessary explanations, drawings (if necessary), a listing of the initial values ​​​​(the “given” column), the 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 a vessel h = 0,60 m.

Solution:

In this case, we have р 0 =р at and therefore we 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 level of the liquid in the closed vessel. The water in the vessel is under absolute pressure p "1 = 1.06 at(drawing for problem 2).

Solution.

Let us compose the equilibrium conditions for a common point BUT(see picture ). Point pressure BUT left:

Pressure right:

Equating the right parts of the equations, and reducing by γg, we get:

This equation can also be obtained by setting the equilibrium condition for points located in any horizontal plane, for example, in the plane OO(see picture). Let us take as the beginning of the piezometer reference scale the plane 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 as the height of a liquid column.

Answer: h = 0.6 meters

Task 3. Determine the height to which the water rises in the vacuum gauge, if the absolute air pressure inside the cylinder p ' in \u003d 0.95 at(Fig. 1-11). Formulate what pressure the vacuum gauge measures.

Solution:

Let us compose the equilibrium condition relative to the horizontal plane O-O:

hydrostatic pressure acting from the inside:

Hydrostatic pressure in plane O-O, operating from the outside

Since the system is in equilibrium,

Task 4. Determine the gauge pressure at a point BUT pipeline, if the height of the mercury column according to the piezometer h 2 \u003d 25 cm. The center of the pipeline is located h 1 \u003d 40 cm below the dividing line between water and mercury (figure for the task).

Solution: Find the pressure at point B: p "B \u003d p" A-γ h1, since the dot AT located above the point BUT by the amount h1. At point C, the pressure will be the same as at point AT, since the pressure of the water column h mutually balanced, i.e.

hence the gauge pressure:

Substituting Numeric Values , we get:

p "A -p atm=37278 Pa

Answer: r "A -r atm=37278 Pa

TASKS

Task 1.1. A canister filled with gasoline and containing no air heated up in the sun to a temperature of 50 ° C. By how much would the pressure of gasoline inside the can rise if it were absolutely rigid? The initial temperature of gasoline is 20 0 С. The modulus of volume elasticity of gasoline is assumed to be K=1300 MPa, the thermal expansion coefficient β = 8. 10 -4 1/deg.

Task 1.2. Determine the overpressure at the bottom of the ocean, the depth of which is h=10 km, assuming the density of sea water ρ=1030 kg/m 3 and assuming it to be incompressible. Determine the density of water at the same depth, taking into account compressibility and assuming the bulk modulus K = 2. 10 3 MPa.

Task 1.3. Find the law of pressure change R atmospheric air height z , considering the dependence of its density on pressure isothermal. In reality, 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.

Task 1.4. Determine the excess water pressure in the pipe AT, if the pressure gauge reading 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 \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 H 3 \u003d 5 m.

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

Problem 1.6. Water and gasoline are poured into a cylindrical tank with a diameter D = 2 m to the level H = 1.5 m. The water level in the piezometer is lower than the gasoline level by h = 300 mm. Determine the gasoline springing in the tank if ρ b \u003d 700 kg / m 3.

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

Problem 1.8. Determine the overpressure p 0 of air in the pressure tank according to the pressure gauge, made up of two U-shaped tubes with mercury. The connecting pipes are filled with water. Level marks are given in meters. What height H 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 pressure force of the liquid (water) on the 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 \u003d 1.5 m;

2) indication of a mercury vacuum gauge h= 73.5 mm at a= 1m; ρrt \u003d 13600 kg / m 3; H 0 \u003d 1.5 m.

Problem 1.10. Determine the volumetric modulus of elasticity of the liquid, if under the action of a load BUT with a mass of 250 kg, the piston traveled a distance Δh = 5 mm. The 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. Ignore the weight of the piston. The reservoir is assumed to be absolutely rigid.

Problem 1.11. A manual piston pump is used to pressurize the underground pipeline with water (tightness test). 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 to be 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/c= 5?

Task 1.12. Determine the absolute air pressure in the tank p 1, if at atmospheric pressure corresponding to h a \u003d 760 mm Hg. Art., indication of a mercury vacuum gauge h RT = = 0.2 m, height h = 1.5 m. What is the indication of a spring vacuum gauge? Mercury density ρ=13600 kg/m 3 .

Task 1.13. When the pipeline valve 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 p vac = 0.02 MPa. Atmospheric pressure corresponds to p a = 740 mm Hg. Art. The density of gasoline ρ b \u003d 700 kg / m 3.

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

Problem 1.15. Define a vacuum r wack and absolute pressure inside the cylinder p" in(Fig. 1-11) if the 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 - 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 h1 = 70 cm, h 2 = = 50 cm.

Problem 1.17. What will be the height of the mercury column h 2 (Fig. to problem 1.16), if the gauge pressure of oil in the cylinder And p a \u003d 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 h2, (figure) if the location of the center of the pipeline BUT will increase compared to that shown 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 37 278 Pa .

Problem 1.19. Determine how high z the level of mercury in the piezometer will be established if, at gauge pressure in the pipe R A \u003d 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 sediment 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.

Task 1.22 A wooden bar measuring 5.0 x 0.30 m and 0.30 m high was 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 bar so that the upper surface of the bar is flush with the free surface of the water, assuming that each person has an average mass of 67.5 kg.

Task 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 must be unloaded so that the barge immersion depth is 1.2 m if the relative specific gravity of wet sand is 2.0?

Problem 1.24. The displacement of the submarine is 600 m 3 . In order to submerge the boat, the compartments were filled with sea water in the 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 water is removed from the submarine and it floats; What is the weight of a submarine without water?

Solved problems from the textbook PHYSICS. Methodical instructions and control tasks. Edited by A. G. Chertov

Below are the conditions of the problems and scanned sheets with solutions. The page may take some time to load.

209. Determine the relative molecular weight Mr 1) of water; 2) carbon dioxide; 3) table salt.

219. In a vessel with a volume of V = 40 liters, there is oxygen at a temperature of T = 300 K. When part of the oxygen was used up, the pressure in the cylinder decreased by Δp = 100 kPa. Determine the mass Δm of the consumed oxygen. The process is considered isothermal.

229. The smallest dust particles are suspended in nitrogen, which move as if they were very large molecules. The mass of each dust particle is 6×10-10g. The gas is at a temperature T=400 K. Determine the mean square velocities, as well as the mean kinetic energies of the translational motion of a nitrogen molecule and a grain of dust.

239. A triatomic gas under pressure P = 240kPa and temperature T = 20°C occupies a volume V=10l. Determine the heat capacity Cp of this gas at constant pressure.

249. The mean free path of a hydrogen molecule under certain conditions is 2 mm. Find the density ρ of hydrogen under these conditions.

259. What proportion ω1 of the amount of heat Q supplied to an ideal diatomic gas in an isobaric process is spent on increasing ΔU of the internal energy of the gas and what proportion ω2 is spent on the work A of expansion? Consider three cases if the gas is: 1) monoatomic; 2) diatomic; 3) triatomic.

269. A gas making a Carnot cycle receives heat Q1 = 84 kJ. Determine the work A of the gas if the temperature T1 of the heat sink is three times higher than the temperature T2 of the heat sink.

279. An air bubble with a diameter of d \u003d 2.2 microns is located in water at its very surface. Determine the density ρ of the air in the bubble if the air above the water surface is under normal conditions.