The hydraulic press is based on Pascal's law. Lesson "Pascal's Law
Definition
Hydraulic Press is a machine that operates on the basis of the laws of motion and equilibrium of fluids.
Pascal's law underlies the principle of operation of a hydraulic press. The name of this device comes from Greek word hydraulics - water. A hydraulic press is a hydraulic machine that is used for pressing (squeezing). A hydraulic press is used where a lot of force is needed, such as pressing oil out of seeds. With the help of modern hydraulic presses it is possible to obtain force up to $(10)^8$newtons.
The basis of the hydraulic machine is made up of two cylinders of different radii with pistons (Fig. 1), which are connected by a pipe. The space in the cylinders below the pistons is usually filled with mineral oil.
In order to understand the principle of operation of a hydraulic machine, one should remember what communicating vessels are and what is the meaning of pascal's law.
Communicating vessels
Communicating vessels are interconnected and in which liquid can freely flow from one vessel to another. The shape of the communicating vessels may be different. In communicating vessels, a fluid of the same density is set at the same level if the pressures above the free surfaces of the fluid are the same.
From Fig. 1 we see that structurally a hydraulic machine is two communicating vessels of different radii. The heights of the liquid columns in the cylinders will be the same if there are no forces acting on the pistons.
Pascal's law
Pascal's law tells us that the pressure exerted by external forces on a fluid is transferred to it without change at all its points. The operation of many hydraulic devices is based on Pascal's law: presses, brake systems, hydraulic drives, hydraulic boosters, etc.
The principle of operation of the hydraulic press
One of the simplest and oldest devices based on Pascal's law is the hydraulic press, in which a small force $F_1$ applied to the piston does not large area$S_1$ is converted into a large force $F_2$, which acts on the large area $S_2$.
The pressure that piston number one creates is:
The pressure of the second piston on the liquid is:
If the pistons are in equilibrium, then the pressures $p_1$ and $p_2$ are equal, therefore, we can equate the right parts of expressions (1) and (2):
\[\frac(F_1)(S_1)=\frac(F_2)(S_2)\left(3\right).\]
Let's determine what will be the modulus of the force applied to the first piston:
From formula (4), we see that the value of $F_1$ is greater than the force modulus $F_2$ by $\frac(S_1)(S_2)$ times.
And so, using a hydraulic press, you can balance a much larger force with a small force. The ratio $\frac(F_1)(F_2)$ shows the gain in strength.
The press works like this. The body to be compressed is placed on a platform resting on a large piston. A small piston creates a high pressure on the liquid. A large piston, together with a compressible body, rises, rests against a fixed platform located above them, the body is compressed.
From a small cylinder into a large liquid is pumped by repeated movement of the piston of a small area. Do it in the following way. The small piston rises, the valve opens, and liquid is sucked into the space under the small piston. When the small piston lowers the liquid, exerting pressure on the valve, it closes, and the valve opens, which passes the liquid into the large vessel.
Examples of problems with a solution
Example 1
Exercise. What will be the gain in strength of the hydraulic press if, when acting on a small piston (with an area of $S_1=10\ (cm)^2$) with a force of $F_1=800$ N, a force is obtained, the impact on a large piston ($S_2=1000 \ (cm)^2$) equal to $F_2=72000\ $ H?
What gain in strength would this press have if there were no friction forces?
Solution. The gain in force is the ratio of the modules of the received force to the applied force:
\[\frac(F_2)(F_1)=\frac(72000)(800)=90.\]
Using the formula obtained for the hydraulic press:
\[\frac(F_1)(S_1)=\frac(F_2)(S_2)\left(1.1\right),\]
find the gain in force in the absence of friction forces:
\[\frac(F_2)(F_1)=\frac(S_2)(S_1)=\frac(1000)(10)=100.\]
Answer. The gain in force in the press in the presence of friction forces is equal to $\frac(F_2)(F_1)=90.$ Without friction, it would be equal to $\frac(F_2)(F_1)=100.$
Example 2
Exercise. Using a hydraulic lifting mechanism, a load having a mass $m$ should be lifted. How many times ($k$) must the small piston be lowered in time $t$ if it is lowered by a distance $l$ at one time? The ratio of lift piston areas is: $\frac(S_1)(S_2)=\frac(1)(n)$ ($n>1$). The efficiency of the machine is $\eta$ with the power of its engine $N$.
Solution. The schematic diagram of the operation of the hydraulic lift is shown in Fig. 2. It is similar to the operation of a hydraulic press.
As a basis for solving the problem, we use an expression that relates power and work, but at the same time we take into account the efficiency of the lift, then the power is equal to:
The work is done with the aim of lifting the load, which means that we will find it as a change in the potential energy of the load, for zero potential energy we will consider the energy of the load at the place where it began to rise ($E_(p1)$=0), we have:
where $h$ is the height to which the load was lifted. Equating the right parts of formulas (2.1) and (2.2), we find the height to which the load was raised:
\[\eta Nt=mgh\to h=\frac(\eta Nt)(mg)\left(2.3\right).\]
We find the work done by the force $F_0$ when moving the small piston as:
\[A_1=F_0l\ \left(2.4\right),\]
The work of the force that moves the large piston up (compresses the hypothetical body) is equal to:
\[A_2=FL\ .\] \[A_1=A_2\to F_0l=FL\] \[\frac(F_0)(F)=\frac(L)(l)=\frac(S_1)(S_2)\ left(2.5\right),\]
where $L$ is the distance the large piston moves in one stroke. From (2.5) we have:
\[\frac(S_1)(S_2)=\frac(L)(l)\to L=\frac(S_1)(S_2)l\ \left(2.6\right).\]
In order to find the number of strokes of the pistons (the number of times that the small piston goes down or the big one rises), the height of the load should be divided by the distance that the big piston moves in one stroke:
Answer.$k=\frac(\eta Ntn)(mgl)$
OPERATING PRINCIPLE AND CLASSIFICATION
The hydraulic press is a machine-tool of almost static action. The principle of operation of a hydraulic press is based on Pascal's law. IN general view the press consists of two chambers equipped with pistons (plungers) and connected pipelines (Fig. 20.1, a). If to the piston 1 apply force, then pressure is created under it. According to Pascal's law, pressure is transmitted to all points of the liquid volume and, being directed normally to the base of the large piston 2 , creates a force that exerts pressure on the workpiece 3 .
Based on Pascal's law,
The force is so many times greater than the force, how many times the area is greater than the area.
The structural diagram of the hydraulic press is shown in fig. 20.1, b. Working cylinder 4 , in which the working plunger moves 5 , fixed in the upper fixed cross member 6 . The latter with the help of columns 7 connected to fixed crossbar 9 installed on the foundation. Lower 9 and top 6 the crossbars together with the columns form the press frame. working plunger 5 connected to the movable crossbar 8 , which has a direction along the columns, and tells it to move in only one direction - down. Return cylinders are installed to lift the movable cross member. 10 with plungers 11 .
Cylinders are sealed to prevent leakage of pressurized fluid 12 .
The main parameter of a hydraulic press is the nominal force of the press - the product of the nominal pressure of the liquid in the press cylinder and the active area of its working plungers.
Presses, depending on the technological purpose, differ from each other in the design of the main units, their location and number, as well as the value of the main parameters ( Z- open height of the die space; H- full travel of the movable crossbar, - table dimensions).
Rice. 20.1. Hydraulic Press:
A- operating principle; b– constructive scheme; V- scheme of a press with a movable bed
According to the technological purpose, hydraulic presses are divided into presses for metal (Fig. 20.2, A) and for non-metallic materials (Fig. 20.2, b). In turn, presses for metal are divided into five groups: for forging and stamping; for extrusion; for sheet stamping; for straightening and assembly work and for the processing of metal waste. In view of the wide variety of types of presses, we present the values of nominal forces, the most common of them.
From the first group of presses, one can name: forging - free forging with stamping in backing dies,; stamping (see, for example, Fig. 26.3) - hot forging of parts made of magnesium and aluminum alloys,; piercing - deep hot piercing of steel blanks in a closed matrix,; broaching - pulling steel forgings through rings,.
Rice. 20.3. Types of hydraulic press cylinders:
A- plunger type; b- differential plunger type; V- piston type
From the second group of presses, it is possible to note pipe-bar and rod-profile presses - pressing non-ferrous alloys and steel,.
From the third group we will name the presses: single-action sheet-stamping (see, for example, Fig. 26.5),; exhaust - deep drawing of cylindrical parts,; for rubber stamping, ; for beading, flanging, bending and stamping of plate material, ; bending - bending of thick sheet material in a hot state,.
From the fifth group, we note baling and briquetting presses for compressing waste such as metal chips and sheet metal scraps,. Hydraulic presses for non-metallic materials include presses for powders, plastics and for pressing chipboard and board.
The technological purpose of the hydraulic press determines the design of the bed (column, two-column, single-column, special), type, design and number of cylinders (plunger, differential-plunger, piston, etc.).
The most widespread is the four-column fixed frame with the movement of moving parts in a vertical plane (see Fig. 20.1, b). Sometimes the frame of the press is made movable (Fig. 20.1, V).
On fig. 20.3 shows the main types of cylinders. Plunger and differential plunger type cylinders are single acting cylinders. The working cylinder of the differential plunger type is used when, for example, a needle must pass through the working plunger (pipe presses). Piston type cylinders are most often used when using oil as the working fluid. In this case, the sealing element of the piston itself will be piston rings. The piston type cylinder is a double acting cylinder.
A press with a lower location of the working cylinder and a fixed bed may not have return cylinders, in which case the moving parts return to their original position under the influence of their weight. The working cylinder is connected to the filling tank.
According to the number of working cylinders, the presses are divided into one-, two-, three- and multi-cylinder.
The action of the press is based on Pascal's law. A hydraulic press consists of two communicating vessels filled with a liquid (usually technical oil) and closed with pistons of various sizes S 1 and S 2 (Fig. 1).
An external force acting on a small piston creates pressure
According to Pascal's law, it is transmitted by the liquid in all directions without change. Therefore, a force acts on the second piston from the side of the liquid
(1)
The hydraulic press gives a gain in force as many times as the area of the larger piston exceeds the area of the small piston.
Force F 1 also changes the potential energy of the fluid in the press. But since the gravity of this fluid is much less than the force F 1. we considered the liquid to be weightless. In this regard, it should be noted that, under real conditions, Eq. (1) is satisfied only approximately.
The press does not give a win in the work. Indeed, when lowering the small piston, the force does work A 1 = F 1 h 1, where h 1 is the stroke of the small piston. Part of the liquid is forced out of the narrow cylinder into the wide one, and the large piston rises h 2 . Work force F 2
(2)
But the liquid is incompressible. Consequently, the volumes of liquids transferred from one cylinder to another are equal, i.e.
Substitute this equation and equation (1) into (2), we get A 1 = A 2 .
The hydraulic press allows the development of colossal forces and is used for pressing products (from metal, plastic, from various powders), for punching holes in metal sheets, for testing materials for strength, for lifting weights, for squeezing oil from seeds at oil mills, for pressing plywood, cardboard, hay. In metallurgical plants, hydraulic presses are used in the manufacture of steel machine shafts, railway wheels, and many other products.
The action of many hydraulic machines, for example, presses (jacks), is based on Pascal's law.
Hydraulic Press(jack) is used to create large forces required to compress the sample material or lifting weights. The press consists of two communicating vessels - cylinders of different cross-sectional area, filled with liquid (oil or water) and closed with pistons from above. Pressure applied to the handle (lever, fig. 2.8, page 70). A force is applied to a piston of small diameter, which, according to Pascal's law, is transferred to a piston of a larger diameter, this piston moves up and performs useful work.
Let's introduce the notation: let F be the force on the press lever, F1- the force acting on the small piston No. 1 with an area S1, F2- the force developed by the large piston No. 2 with an area S2. An analytical representation of the operating principle of a hydraulic press is as follows:
.
Rice. 2.8. Hydraulic Press
If it is necessary to take into account friction in the cuffs of the press, sealing the gaps, the dependence is valid that takes into account the efficiency η of the press:
hydraulic accumulator(Fig. 2.9, p. 71) serves to accumulate the potential energy of the liquid, which is subsequently consumed as needed. Such a battery is used when it is necessary to perform short-term work, for example, during the operation of locks and hydraulic lifts.
The accumulator consists of a twisted cylinder with weights and a fixed piston. The cylinder is filled with a working fluid using a pump, which raises it to the calculated height H.
The reserve of energy for work in the accumulator is equal to:
G- weight of the cylinder with weights; L- lifting height.
To raise the piston, it is necessary to pump fluid into the cylinder with a volume of:
Where S- sectional area of the cylinder.
Lifting force:
Where p is the pressure in the cylinder.
Then the work done to lift the load is:
A=GL=pV.
Rice. 2.9. hydraulic accumulator
efficiency battery:
Multiplier serves to increase the pressure in the oil lines of lubricators, etc.
The simplest multiplier in design consists of a cylinder, a piston with a rod, and stuffing box seals for the piston and rod (Fig. 2.10).
Rice. 2.10. Multiplier
In container A behind the piston fluid is supplied under some pressure p1 which pushes the piston with a force:
D is the diameter of the inner surface of the cylinder.
The movement of the piston and rod is resisted by forces
Where f 1 , f 2- coefficients of friction of sealing rings; n 1 , n 2 b 1 , b 2- the number of sealing rings; d– diameter.
The resultant force acting on the piston creates pressure on the liquid in cavity B - behind the piston. The fluid pressure in this cavity will be greater, since the pressure area behind the piston is smaller than in front of the piston.
The action of a force on a rigid body depends not only on the modulus of this force, but also on the surface area of the body on which it acts. Interaction of liquids and gases with solid bodies, as well as the interaction between adjacent layers of liquid or gas also occurs not at separate points, but on a certain surface of their contact. Therefore, to characterize such interactions, the concept of pressure is introduced.
pressure p name the quantity equal to the ratio modulus of the pressure force F, acting perpendicular to the surface, to the area 5 of this surface:
p=F/S. (5.1)
At uniform distribution pressure forces, the pressure on all parts of the surface is the same and numerically equal to the pressure force acting on the surface of a unit area.
The pressure unit is set from formula (5.1). In SI, the unit of pressure is the pressure caused by a force of 1 N, uniformly distributed over a surface perpendicular to it with an area of 1 m 2. This unit of pressure is called pascal (Pa): 1 Pa=1 N/m2.
The following non-systemic units of pressure are often used:
- technical atmosphere (at): 1 at = 9.8 10 4 Pa;
- physical atmosphere (atm) equal to the pressure produced by a column of mercury 760 mm high. As shown in § 24, 1 atm \u003d 1.033 atm \u003d 1.013 10 5 Pa;
- millimeter of mercury (mmHg): 1 mmHg Art. » 133.3 Pa;
- bar (millibar is used in meteorology); 1 bar=10 5 Pa, 1 mbar=10 2 Pa.
Pascal's law for liquids and gases
Solids transfer the pressure produced on them from the outside in the direction of the force that causes this pressure. Liquids and gases transmit external pressure quite differently.
Consider the following experiment (Fig. 48). A vessel with a stopper contains water. Three tubes of the same diameter are inserted into the cork, the lower holes of which are in the water at the same depth, but directed in different directions (down, sideways and up), as well as a tube that does not reach the water, to which a rubber balloon is connected from the spray gun. By pumping air into the vessel with it, we increase the pressure exerted by air on the surface of the water in the vessel. We note that in this case, in all three tubes, the water rises to the same height. Hence, a stationary liquid in a closed vessel transmits the external pressure produced on it in all directions equally(i.e. no change).
Observations show that external pressure and gases in a closed vessel also transmit. The described pattern was first discovered by the French scientist Pascal and was called pascal's law.
hydrostatic pressure
Every molecule of liquid in the Earth's gravitational field is affected by the force of gravity. Under the action of these forces, each layer of liquid presses on the layers located below it. According to Pascal's law, this pressure is transferred by the liquid in all directions equally. Hence, Liquids have pressure due to gravity.
Observations show that a liquid in a vessel at rest puts pressure on the bottom and walls of the vessel and on any body immersed in this liquid. The pressure exerted by a fluid at rest on any surface in contact with it is called hydrostatic.
Hydrostatic pressure formula
Hydrostatic pressure can be determined using an instrument called Pascal's hydrostatic balance (Fig. 49). In the stand P, through which the annular pipe K passes, it is possible to hermetically fix vessels C of any shape that do not have a bottom. The movable bottom of these vessels is a flat round platform D, suspended on a balance beam of equal arms, located near the lower opening of the nozzle K. This platform is pressed against the end of the nozzle by a force caused by the fact that a weight G is placed on the balance pan, suspended on their other balance beam. P attached ruler L, which determines the height h of the liquid in the vessel, mounted on a stand.
This is how experience is made. A vessel in the form of a straight circular cylinder is fixed on a stand. Water is poured into it until the weight of this water becomes equal to the weight of the weight placed on the right scale pan, i.e. R f = R g. (Maintenance of this amount of water is automatically provided by the device itself, since if the weight of water in the vessel exceeds the weight of the weight, the bottom will open slightly and excess water will flow out.)
In a cylindrical vessel, the weight of the liquid P W = r f ghS, where f = r w is the density of the liquid, g is the acceleration of free fall, h is the height of the liquid column, S is the area of the base of the cylinder, so the liquid exerts pressure on the bottom of the vessel
p \u003d P w / S \u003d r w gh. (5.2)Formula (5.2) determines the value of hydrostatic pressure.
Theoretical derivation of the hydrostatic pressure formula
We single out a fixed element of its volume inside a fluid at restDV in the form of a straight circular cylinder of height h with bases having a small areaDS parallel to the free surface of the liquid (Fig. 50). The upper base of the cylinder is located at a depth h 1 from the liquid surface, and the lower base is at a depth h 2 >h 1 .
Three forces act vertically on the selected element of the liquid volume: pressure forces F 1 \u003d p 1 DS and F 2 = p 2 DS (where p 1 and p 2 are the values of hydrostatic pressure at depths h 1 and h 2) and gravity F t \u003d rg DV = rgh DS.
The fluid volume element we have identified is at rest, which means that F 1 + F 2 + F t \u003d 0, and therefore the algebraic sum of the projections of these forces onto the vertical axis is equal to zero, i.e. p 2 DS-p 1 DS-rgh DS=0, whence we get
p 2 -p 1 = rgh. (5.3)Let now the upper face of the selected cylindrical volume of the liquid coincide with the surface of the liquid, i.e. h1=0. Then h 2 \u003d h and p 2 \u003d p, where h is the depth of immersion, and p is hydrostatic pressure at this depth. Assuming that the pressure on the surface of the liquid is p 1 = 0 (i.e. without taking into account the external pressure on the surface of the liquid), from (5.3) we obtain the formula for the hydrostatic pressure p =rgh, which coincides with formula (5.2).
Communicating vessels
Communicating vessels are vessels that have a channel between them filled with liquid. Observations show that in communicating vessels of any shape, a homogeneous liquid is always set at the same level.
Dissimilar liquids behave differently even in communicating vessels of the same shape and size. Let's take two cylindrical communicating vessels of the same diameter (Fig. 51), pour a layer of mercury (shaded) on their bottom, and pour liquid with different densities into the cylinders on top of it, for example, r 2
Mentally select inside the tube connecting the communicating vessels and filled with mercury, an area of area S, perpendicular to the horizontal surface. Since the liquids are at rest, the pressure on this area from the left and right is the same, i.e. p1=p2. According to formula (5.2), hydrostatic pressure p 1 = r 1 gh 1 and p 2 = r2gh2. Equating these expressions, we obtain r 1 h 1 2 h 2 , whence= r
h 1 / h 2 \u003d r 2 / r 1. (5.4)Consequently, heterogeneous liquids at rest are installed in communicating vessels in such a way that the heights of their columns are inversely proportional to the densities of these liquids.
If r 1 =r 2 , then formula (5.4) implies that h 1 =h 2 , i.e. homogeneous liquids are installed in communicating vessels at the same level.
The principle of operation of the hydraulic press
The hydraulic press consists of two communicating vessels of cylindrical shape and different diameters, in which there are pistons, the areas of which S 1 and S 2 are different (S 2 >> S 1). The cylinders are filled with liquid oil (usually transformer oil). Schematically, the device of the hydraulic press is shown in fig. 52 (this figure does not show the oil reservoir and valve system).
Without load, the pistons are at the same level. The piston S 1 is acted upon by the force F 1 , and between the piston S 2 and the upper support a body is placed to be pressed.
The force F 1 acting on the piston S 1 creates additional pressure in the liquid p=F 1 /S 1 . According to Pascal's law, this pressure is transmitted by the fluid in all directions without change. Therefore, the pressure force acts on the piston S 2 F 2 \u003d pS 2 \u003d F 1 S 2 /S 1.From this equality it follows that
F 2 /F 1 \u003d S 2 / S 1. (5.5)Therefore, the forces acting on the pistons of a hydraulic press are proportional to the areas of these pistons. Therefore, with the help of a hydraulic press, it is possible to obtain a gain in strength the greater, the more S 2 is greater than S 1 .
The hydraulic press is widely used in engineering.
