Liquids and gases transmit in all directions not only the external pressure exerted on them, but also the pressure that exists inside them due to the weight of their own parts. The upper layers of liquid press on the middle ones, those on the lower ones, and the latter ones on the bottom.
The pressure exerted by a fluid at rest is called hydrostatic.
Let us obtain a formula for calculating the hydrostatic pressure of a liquid at an arbitrary depth h (in the vicinity of point A in Figure 98). The pressure force acting in this place from the overlying narrow vertical pillar liquid can be expressed in two ways:
firstly, as the product of the pressure at the base of this column and its cross-sectional area:
F = pS ;
secondly, as the weight of the same column of liquid, i.e. the product of the mass of the liquid (which can be found by the formula m = ρV, where volume V = Sh) and the acceleration of gravity g:
F = mg = ρShg.
Let us equate both expressions for the pressure force:
pS = ρShg.
Dividing both sides of this equality by area S, we find the fluid pressure at depth h:
p = ρgh. (37.1)
We got hydrostatic pressure formula. Hydrostatic pressure at any depth inside the liquid does not depend on the shape of the vessel in which the liquid is located, and is equal to the product of the density of the liquid, the acceleration of gravity and the depth at which the pressure is considered.
The same amount of water, being in different vessels, can exert different pressure on the bottom. Since this pressure depends on the height of the liquid column, it will be greater in narrow vessels than in wide ones. Thanks to this, even a small amount of water can create very high pressure. In 1648, this was very convincingly demonstrated by B. Pascal. He inserted a narrow tube into a closed barrel filled with water and, going up to the balcony of the second floor of the house, poured a mug of water into this tube. Due to the small thickness of the tube, the water in it rose to high altitude, and the pressure in the barrel increased so much that the fastenings of the barrel could not stand it, and it cracked (Fig. 99).
The results we obtained are valid not only for liquids, but also for gases. Their layers also press on each other, and therefore hydrostatic pressure also exists in them. 
1. What pressure is called hydrostatic? 2. What values does this pressure depend on? 3. Derive the formula for hydrostatic pressure at an arbitrary depth. 4. How can you create a lot of pressure with a small amount of water? Tell us about Pascal's experience.
Experimental task. Take a tall vessel and make three small holes in its wall different heights. Cover the holes with plasticine and fill the vessel with water. Open the holes and watch the streams of water flowing out (Fig. 100). Why does water leak out of the holes? What does it mean that water pressure increases with depth?
Plumbing, it would seem, does not give much reason to delve into the jungle of technologies, mechanisms, or engage in scrupulous calculations for building the most complex schemes. But such a vision is a superficial look at plumbing. The real plumbing industry is in no way inferior in complexity to the processes and, like many other industries, requires a professional approach. In turn, professionalism is a solid store of knowledge on which plumbing is based. Let’s dive (albeit not too deeply) into the plumbing training stream in order to get one step closer to the professional status of a plumber.
The fundamental basis of modern hydraulics was formed when Blaise Pascal discovered that the action of fluid pressure is constant in any direction. The action of liquid pressure is directed at right angles to the surface area.
If a measuring device (pressure gauge) is placed under a layer of liquid at a certain depth and its sensitive element is directed in different directions, the pressure readings will remain unchanged in any position of the pressure gauge.
That is, the fluid pressure does not depend in any way on the change in direction. But the fluid pressure at each level depends on the depth parameter. If the pressure meter is moved closer to the surface of the liquid, the reading will decrease.
Accordingly, when diving, the measured readings will increase. Moreover, under conditions of doubling the depth, the pressure parameter will also double.
Pascal's law clearly demonstrates the effect of water pressure in the most familiar conditions for modern life.
Therefore, whenever the speed of movement of a fluid is set, part of its initial static pressure is used to organize this speed, which subsequently exists as a pressure speed.
Volume and flow rate
The volume of fluid passing through a particular point at a given time is considered as flow volume or flow rate. Flow volume is usually expressed in liters per minute (L/min) and is related to the relative pressure of the fluid. For example, 10 liters per minute at 2.7 atm.
Flow rate (fluid speed) is defined as average speed, at which the fluid moves past given point. Typically expressed in meters per second (m/s) or meters per minute (m/min). The flow rate is important factor when calibrating hydraulic lines.
The volume and speed of fluid flow are traditionally considered “related” indicators. With the same transmission volume, the speed may vary depending on the cross-section of the passage Volume and flow rate are often considered simultaneously. All other things being equal (assuming the input volume remains constant), the flow rate increases as the cross-section or size of the pipe decreases, and the flow rate decreases as the cross-section increases.
Thus, a slowdown in flow speed is observed in wide parts of pipelines, and in narrow places, on the contrary, the speed increases. At the same time, the volume of water passing through each of these control points remains unchanged.
Bernoulli's principle
The well-known Bernoulli principle is built on the logic that a rise (fall) in the pressure of a fluid fluid is always accompanied by a decrease (increase) in speed. Conversely, an increase (decrease) in fluid velocity leads to a decrease (increase) in pressure.
This principle underlies a number of common plumbing phenomena. As a trivial example, Bernoulli's principle is responsible for causing the shower curtain to "retract inward" when the user turns on the water.
The pressure difference between the outside and inside causes a force on the shower curtain. With this forceful effort, the curtain is pulled inward.
To others a clear example is a bottle of perfume with a spray when an area of low pressure is created due to high air speed. And the air carries the liquid with it.
Bernoulli's principle for an aircraft wing: 1 - low pressure; 2 — high pressure; 3 — fast flow; 4 — slow flow; 5 - wing Bernoulli's principle also shows why windows in a house tend to break spontaneously during hurricanes. In such cases, the extremely high speed of air outside the window leads to the fact that the pressure outside becomes much less than the pressure inside, where the air remains practically motionless.
A significant difference in force simply pushes the windows outward, causing the glass to break. So when a major hurricane approaches, you essentially want to open the windows as wide as possible to equalize the pressure inside and outside the building.
And a couple more examples when Bernoulli’s principle operates: the rise of an airplane with subsequent flight due to the wings and the movement of “curve balls” in baseball.
In both cases, a difference in the speed of air passing past the object from above and below is created. For airplane wings, the difference in speed is created by the movement of the flaps; in baseball, it is the presence of a wavy edge.
Home Plumber Practice
Let's take cylindrical vessel with a horizontal bottom and vertical walls, filled with liquid to a height (Fig. 248).
Rice. 248. In a vessel with vertical walls, the pressure force on the bottom is equal to the weight of the entire poured liquid
Rice. 249. In all the vessels depicted, the pressure on the bottom is the same. In the first two vessels it is more than the weight of the liquid poured, in the other two it is less
The hydrostatic pressure at each point on the bottom of the vessel will be the same:
If the bottom of the vessel has an area, then the force of pressure of the liquid on the bottom of the vessel, i.e., is equal to the weight of the liquid poured into the vessel.
Let us now consider vessels that differ in shape, but with the same bottom area (Fig. 249). If the liquid in each of them is poured to the same height, then the pressure is on the bottom. it is the same in all vessels. Therefore, the pressure force on the bottom is equal to
is also the same in all vessels. It is equal to the weight of a column of liquid with a base equal to the area of the bottom of the vessel and a height equal to the height of the liquid poured. In Fig. 249 this pillar is shown next to each vessel with dashed lines. Please note that the force of pressure on the bottom does not depend on the shape of the vessel and can be either greater or less than the weight of the liquid poured.
Rice. 250. Pascal's device with a set of vessels. The cross sections are the same for all vessels
Rice. 251. Experiment with Pascal's barrel
This conclusion can be verified experimentally using the device proposed by Pascal (Fig. 250). You can attach vessels to the stand various shapes, having no bottom. Instead of a bottom, a plate suspended from the balance beam is tightly pressed against the vessel from below. If there is liquid in the vessel, a pressure force acts on the plate, which tears the plate off when the pressure force begins to exceed the weight of the weight standing on the other pan of the scale.
In a vessel with vertical walls (cylindrical vessel), the bottom opens when the weight of the poured liquid reaches the weight of the weight. In vessels of other shapes, the bottom opens at the same height of the liquid column, although the weight of the poured water may be greater (a vessel expanding upward) or less (a vessel narrowing) than the weight of the weight.
This experience leads to the idea that with the proper shape of the vessel, it is possible to obtain enormous pressure forces on the bottom using a small amount of water. Pascal attached a long thin vertical tube to a tightly caulked barrel filled with water (Fig. 251). When the tube is filled with water, the force of hydrostatic pressure on the bottom becomes equal to the weight of a column of water, the base area of which is equal to the area of the bottom of the barrel, and the height is equal to the height of the tube. Accordingly, the pressure forces on the walls and upper bottom of the barrel increase. When Pascal filled the tube to a height of several meters, which required only a few cups of water, the resulting pressure forces ruptured the barrel.
How can we explain that the force of pressure on the bottom of a vessel can be, depending on the shape of the vessel, greater or less than the weight of the liquid contained in the vessel? After all, the force acting on the liquid from the vessel must balance the weight of the liquid. The fact is that the liquid in the vessel is affected not only by the bottom, but also by the walls of the vessel. In a container expanding upward, the forces with which the walls act on the liquid have components directed upward: thus, part of the weight of the liquid is balanced by the pressure forces of the walls and only part must be balanced by the pressure forces from the bottom. On the contrary, in a vessel that tapers upward, the bottom acts upward on the liquid, and the walls act downward; therefore, the force of pressure on the bottom is greater than the weight of the liquid. The sum of the forces acting on the liquid from the bottom of the vessel and its walls is always equal to the weight of the liquid. Rice. 252 clearly shows the distribution of forces acting from the walls on liquid in vessels of various shapes.
Rice. 252. Forces acting on liquid from the walls of vessels of various shapes
Rice. 253. When water is poured into the funnel, the cylinder rises up.
In a vessel that tapers upward, a force directed upward acts on the walls from the liquid side. If the walls of such a vessel are made movable, the liquid will lift them. Such an experiment can be carried out using the following device: the piston is fixedly fixed, and a cylinder is put on it, turning into a vertical tube (Fig. 253). When the space above the piston is filled with water, pressure forces on the areas and walls of the cylinder lift the cylinder upward.
Let's consider how you can calculate the pressure of a liquid on the bottom and walls of a vessel. Let's first solve the problem with numerical data. A rectangular tank is filled with water (Fig. 96). The bottom area of the tank is 16 m2, its height is 5 m. Let us determine the water pressure at the bottom of the tank.
The force with which the water presses on the bottom of the vessel is equal to the weight of a column of water with a height of 5 m and a base area of 16 m2, in other words, this force is equal to the weight of all the water in the tank.
To find the weight of water, you need to know its mass. The mass of water can be calculated from its volume and density. Let's find the volume of water in the tank by multiplying the area of the bottom of the tank by its height: V= 16 m2*5 m=80 m3. Now let’s determine the mass of water; to do this, multiply its density p = 1000 kg/m3 by volume: m = 1000 kg/m3 * 80 m3 = 80,000 kg. We know that to determine the weight of a body, its mass must be multiplied by 9.8 N/kg, since a body weighing 1 kg weighs 9.8 N.
Therefore, the weight of water in the tank is P = 9.8 N/kg * 80,000 kg ≈ 800,000 N. With such force the water presses on the bottom of the tank.
Dividing the weight of water by the area of the bottom of the tank, we find the pressure p :
p = 800000 N/16 m2 = 50,000 Pa = 50 kPa.
The pressure of the liquid at the bottom of the vessel can be calculated using the formula, which is much simpler. To derive this formula, let’s return to the problem, but only solve it in general form.
Let us denote the height of the liquid column in the vessel by the letter h, and the area of the bottom of the vessel S.
Liquid column volume V=Sh.
Liquid mass T= pV, or m = pSh.
The weight of this liquid P=gm, or P=gpSh.
Since the weight of a column of liquid is equal to the force with which the liquid presses on the bottom of the vessel, then by dividing the weight P To the square S, we get pressure R:
p = P/S, or p = gpSh/S
p =gph.
We have obtained a formula for calculating the pressure of the liquid at the bottom of the vessel. From this formula it is clear that the pressure of the liquid at the bottom of the vessel is directly proportional to the density and height of the liquid column.
Using this formula, you can calculate the pressure on the walls of the vessel, as well as the pressure inside the liquid, including the pressure from bottom to top, since the pressure at the same depth is the same in all directions.
When calculating pressure using the formula:
p =gph
density p must be expressed in kilograms per cubic meter(kg/m3), and the height of the liquid column h- in meters (m), g= 9.8 N/kg, then the pressure will be expressed in pascals (Pa).
Example. Determine the oil pressure at the bottom of the tank if the height of the oil column is 10 m and its density is 800 kg/m3.
Questions. 1. On what values does the pressure of the liquid at the bottom of the vessel depend? 2. How does the pressure of the liquid at the bottom of the vessel depend on the height of the liquid column? 3 . How does the pressure of a liquid at the bottom of a vessel depend on the density of the liquid? 4. What quantities do you need to know to calculate the pressure of a liquid on the walls of a vessel? 5. What formula is used to calculate the pressure of a liquid on the bottom and walls of a vessel?
Exercises. 1. Determine the pressure at a depth of 0.6 m in water, kerosene, and mercury. 2. Calculate the water pressure at the bottom of one of the deepest sea trenches, the depth of which is 10,900 m, Density sea water 1030 kg/m3. 3. Figure 97 shows a football camera connected to a vertical glass tube. There is water in the chamber and tube. A board is placed on the camera, and a weight of 5 kg is placed on it. The height of the water column in the tube is 1 m. Determine the area of contact of the plate with the chamber.
Tasks. 1. Take a tall vessel. Make three small holes in the side surface of it in a straight line, at different heights from the bottom. Seal the holes with matches and fill the vessel to the top with water. Open the holes and watch the streams of water flowing out (Fig. 98). Answer the questions: why does water flow out of the holes? What does it mean that pressure increases with depth? 2. Read the paragraphs “Hydrostatic paradox” at the end of the textbook. Pascal's experiment", "Pressure at the bottom of seas and oceans. Exploration of the depths of the sea."