Let us solve the following problem (Banach problem). A person carries two boxes of matches (60 matches each) in his pocket, and whenever a match is needed, he takes the box at random and takes out a match. What is the probability that when the first box is empty, there will still be 20 matches left in the second? Box selection can be thought of as an independent trial in which the first box is selected with probability. Total experiments performed n= 60+40=100, and in these hundred experiments the first box must be chosen 60 times. The probability of this is:

.

From the record it is clear that for large n It is difficult to use Bernoulli's formula due to cumbersome calculations. There are special approximate formulas that allow you to find probabilities
, If n great. One of such formulas is given by the following theorem.

Theorem 2.1. ( Laplace local ). If in the Bernoulli scheme
, then the probability that the event A will come exactly k times, satisfies for large n ratio

Where
.

For convenience, we introduce the function
is the local Laplace function, with the help of which Laplace’s theorem can be written as follows:

There are special function tables
, according to which for any value:
you can find the corresponding function value. These tables were obtained by expanding the function
in a row.

Geometrically, this result means that for large n the distribution polygon fits well into the graph of the function on the right in the formula (Fig. 2.3) and instead of the true probability value
possible for everyone k take the value of a function at a point k.

Rice. 2.3. Local Laplace function

Let's now return to the problem. Using formula (2.1) we find:

,

where is the value
determined from the table.

2.2.2. Laplace's integral theorem

Theorem 2.2(Laplace integral) . The probability that in the circuit n independent tests the event will occur from k 1 to k 2 times, approximately equal

P n (k 1
k 2 )
,

– Laplace integral function, for which tables have been compiled. Function F(x) odd: Ф(-х)=-Ф(х) And F(X 4)=0,5.

Let us consider yet another statement without proof.

Relative frequency deviation from probability p V n independent tests equals

(

.

Comment. The rationale for these facts will be discussed further in Section 7 (Sections 7.2, 7.3). Laplace's theorems are sometimes called the Moivre–Laplace theorems.

Example 2.3.

The probability of an event occurring in each of 900 independent trials is 0.5. 1) find the probability that the event will occur 400 to 500 times, 2) find the probability that the relative frequency of occurrence of the event will deviate from its probability in absolute value by no more than 0.02.

Solution

1) R 900 (400<k<500)=
=

2)

=

2.3. Poisson's formula

If we fix the number of experiments n, and the probability of an event occurring in one experiment r change, then the distribution polygon will have a different appearance depending on the value r(Fig. 2.4). With values p, close to 1/2, the polygon is almost symmetrical and fits well into the symmetric graph of the Laplace function. Therefore, the approximate Laplace formula gives good accuracy.

For small ones r(in practice less ) the approximation is poor due to the asymmetry of the distribution polygon. Therefore, the task arises of finding an approximate formula for calculating the probabilities
in case of large n and small r. The answer to this question is given by Poisson's formula.

So, let's consider an independent test scheme in which n is large (the more the better), and r little (the less the better). Let's denote nr=λ . Then, according to Bernoulli’s formula, we have

.

The last equality is true due to the fact that
(second remarkable limit). When obtaining the formula for the most probable occurrence of an event k 0 the odds ratio was considered. It follows from this that

Thus, when k many smaller ones n we have a recurrence relation

.

For k=0 let’s take into account the result obtained earlier:
, Then

………………

So, if n is large in an independent test design, and r little, then it happens Poisson's formula

R n (To)
, where λ = nr.

Poisson's law is also called the law of rare events.

Example 2.4.

The probability of producing a defective part is 0.02. Parts are packed in boxes of 100 pieces. What is the probability that a) there are no defective parts in the box, b) there are more than two defective parts in the box?

Solution

a) Because n big and r little, we have ; R 100 (0)
;

b)R 100 (k>2)= 1-R 1-

Thus, in an independent trial design to calculate the probability R n (k) Bernoulli's formula should be used if n small, but if n is large, then depending on the size r one of the approximate Laplace formulas or the Poisson formula is used.

Properties of the liquid state. Surface layer. Surface tension. Wetting. Laplace's formula. Capillary phenomena.

Liquids are substances that are in a condensed state, which is intermediate between the solid crystalline state and the gaseous state.

The region of existence of liquids is limited on the high-temperature side by its transition to the gaseous state, and on the low-temperature side by its transition to the solid state.

In liquids, the distance between molecules is much smaller than in gases (the density of liquids is ~ 6000 times greater than the density of saturated vapor far from the critical temperature) (Fig. 1).

Fig.1. Water vapor (1) and water (2). Water molecules are enlarged approximately 5 10 7 times

Consequently, the forces of intermolecular interaction in liquids, unlike gases, are the main factor that determines the properties of liquids. Therefore, liquids, like solids, retain their volume and have a free surface. Like solids, liquids are characterized by very low compressibility and resist stretching.

However, the bonding forces between liquid molecules are not so strong as to prevent the layers of liquid from sliding relative to each other. Therefore, liquids, like gases, have fluidity. In the field of gravity, liquids take the shape of the container into which they are poured.

The properties of substances are determined by the movement and interaction of the particles of which they are composed.

In gases, collisions mainly involve two molecules. Consequently, the theory of gases reduces to the solution of the two-body problem, which can be solved exactly. In solids, molecules undergo vibrational motion at the nodes of the crystal lattice in a periodic field created by other molecules. This problem of particle behavior in a periodic field can also be solved exactly.

In liquids, each molecule is surrounded by several others. A problem of this type (the many-body problem), in general, regardless of the nature of the molecules and the nature of their arrangement, has not yet been precisely solved.

Experiments on diffraction of X-rays, neutrons, and electrons helped determine the structure of liquids. Unlike crystals, in which long-range order is observed (regular arrangement of particles in large volumes), in liquids at distances of the order of 3–4 molecular diameters, the order in the arrangement of molecules is disrupted. Consequently, in liquids there is a so-called short-range order in the arrangement of molecules (Fig. 2):

Fig.2. An example of short-range order of liquid molecules and long-range order of molecules of a crystalline substance: 1 – water; 2 – ice

In liquids, molecules undergo small vibrations within limits limited by intermolecular distances. However, from time to time, as a result of fluctuations, a molecule can receive energy from neighboring molecules that is enough to jump to a new equilibrium position. The molecule will remain in the new equilibrium position for some time until, again, as a result of fluctuations, it receives the energy necessary for the jump. The molecule jumps over a distance comparable to the size of the molecule. Vibrations that give way to jumps represent the thermal movement of liquid molecules.

The average time that a molecule is in a state of equilibrium is called relaxation time. As the temperature increases, the energy of the molecules increases, therefore, the probability of fluctuations increases, while the relaxation time decreases:

(1)

Where τ – relaxation time, B– coefficient having the meaning of the vibration period of the molecule, W – activation energy molecules, i.e. energy required to make a molecular jump.

Internal friction in liquids, as in gases, occurs when layers of liquid move due to the transfer of momentum in the direction normal to the direction of movement of layers of liquid. Momentum transfer from layer to layer also occurs during molecular jumps. However, mainly, momentum is transferred due to the interaction (attraction) of molecules of neighboring layers.

In accordance with the mechanism of thermal movement of liquid molecules, the dependence of the viscosity coefficient on temperature has the form:

(2)

Where A– coefficient depending on the jump distance of the molecule, the frequency of its vibrations and temperature, W – activation energy.

Equation (2) – Frenkel-Andrade formula. The temperature dependence of the viscosity coefficient is mainly determined by the exponential factor.

The reciprocal value of viscosity is called fluidity. As the temperature decreases, the viscosity of some liquids increases so much that they practically stop flowing, forming amorphous bodies (glass, plastics, resins, etc.).

Each liquid molecule interacts with neighboring molecules that are within the range of its molecular forces. The results of this interaction are not the same for molecules inside the liquid and on the surface of the liquid. A molecule located inside a liquid interacts with neighboring molecules surrounding it and the resultant force that acts on it is zero (Fig. 3).

Fig.3. Forces acting on liquid molecules

The molecules of the surface layer are under different conditions. The density of vapor above the liquid is much less than the density of the liquid. Therefore, each molecule of the surface layer is acted upon by a resultant force directed normally into the liquid (Fig. 3). The surface layer exerts pressure on the rest of the liquid like an elastic film. Molecules lying in this layer are also attracted to each other (Fig. 4).

Fig.4. Interaction of surface layer molecules

This interaction creates forces directed tangentially to the surface of the liquid and tending to reduce the surface of the liquid.

If an arbitrary line is drawn on the surface of a liquid, then surface tension forces will act along the normal to the line and tangent to the surface. The magnitude of these forces is proportional to the number of molecules located along this line, therefore proportional to the length of the line:

(3)

Where σ – proportionality coefficient, which is called surface tension coefficient:

(4)

The surface tension coefficient is numerically equal to the surface tension force acting per unit length of the contour delimiting the surface of the liquid.

The surface tension coefficient is measured in N/m. Magnitude σ depends on the type of liquid, temperature, and the presence of impurities. Substances that reduce surface tension are called superficially active(alcohol, soap, washing powder, etc.).

To increase the surface area of ​​a liquid, work must be done against surface tension forces. Let's determine the amount of this work. Let there be a frame with a liquid film (for example, soap) and a movable crossbar (Fig. 5).

Fig.5. The movable side of the wire frame is in equilibrium under the action of external force F ext and the resulting surface tension forces F n

Let's stretch the film with a force F ext by dx. Obviously:

Where F n = σL–surface tension force. Then:

Where dS = Ldx– increment of film surface area. From the last equation:

(5)

According to (5), the surface tension coefficient is numerically equal to the work required to increase the surface area by one unit at a constant temperature. From (5) it is clear that σ can be measured in J/m 2.

If a liquid borders another liquid or a solid, then due to the fact that the densities of the substances in contact are comparable, one cannot ignore the interaction of the molecules of the liquid with the molecules of the substances bordering it.

If, upon contact between a liquid and a solid, the interaction between their molecules is stronger than the interaction between the molecules of the liquid itself, then the liquid tends to increase the surface of contact and spreads over the surface of the solid. In this case, the liquid wets the solid. If the interaction between the molecules of the liquid is stronger than the interaction between the molecules of the liquid and the solid, then the liquid reduces the contact surface. In this case, the liquid does not wet solids. For example: water wets glass, but does not wet paraffin; mercury wets metal surfaces, but does not wet glass.

Fig.6. Different shapes of a drop on the surface of a solid for the cases of non-wetting (a) and wetting (b) liquids

Consider a drop of liquid on the surface of a solid (Fig. 7):

Fig.7. Schemes for calculating the equilibrium of a drop on the surface of a solid body for the cases of non-wetting (a) and wetting (b) liquids: 1 - gas, 2 - liquid, 3 - solid

The shape of a drop is determined by the interaction of three media: gas - 1, liquid - 2 and solid - 3. All these media have a common boundary - a circle that encloses the drop. Per element length dl of this contour, surface tension forces will act: F 12 = σ 12 dl– between gas and liquid, F 13 = σ 13 dl- between gas and solid, F 23 = σ 23 dl– between liquid and solid. If dl=1m, then F 12 = σ 12 , F 13 = σ 13 , F 23 = σ 23. Let's consider the case when:

This means that<θ = π (Fig. 7, a). The circle that limits the place of contact of the liquid with the solid body will contract to a point and the drop will take an ellipsoidal or spherical shape. This is a case of complete non-wetting. Complete non-wetting is also observed in the case of: σ 23 > σ 12 + σ 13 .

Another edge case will occur if:

This means that<θ = 0 (Fig. 7b), complete wetting is observed. Complete wetting will also be observed in the case when: σ 13 > σ 12 + σ 23. In this case, there will be no equilibrium, at any angle values θ , and the liquid will spread over the surface of the solid up to the monomolecular layer.

If the drop is in equilibrium, then the resultant of all forces acting on the element of the contour length is zero. The equilibrium condition in this case is:

The angle between the tangents to the surface of a solid and to the surface of a liquid, which is measured inside the liquid,called the contact angle.

Its value is determined from (6):

(7)

If σ 13 > σ 23, then cos θ > 0, angle θ sharp - partial wetting occurs if σ 13 < σ 23, then cos θ < 0 – угол θ blunt – partial non-wetting occurs. Thus, the contact angle is a value characterizing the degree of wetting or non-wetting of the liquid

The curvature of the liquid surface results in additional pressure acting on the liquid below this surface. Let us determine the amount of additional pressure under the curved surface of the liquid. Let us select an element of area ∆ on an arbitrary surface of the liquid S(Fig. 8):

Fig.8. To calculate the amount of additional pressure

O′ O– normal to the surface at a point O. Let us determine the surface tension forces acting on the contour elements AB And CD. Surface tension forces F And F′, which act on AB And CD, perpendicular AB And CD and directed tangentially to the surface ∆ S. Let's determine the magnitude of the force F:

Let's break down the power F into two components f 1 and f ′. Strength f 1 parallel O′ O and directed into the liquid. This force increases the pressure on the internal areas of the liquid (the second component stretches the surface and does not affect the amount of pressure).

Let us draw a plane perpendicular to ∆ S through points M, O And N. Then R 1 – radius of curvature of the surface in the direction of this plane. Let us draw a plane perpendicular to ∆ S and the first plane. Then R 2 – radius of curvature of the surface in the direction of this plane. In general R 1 ≠ R 2. Let's define the component f 1. From the picture you can see:

Let's take into account that:

(8)

Strength F′ let us decompose into the same two components and similarly define the component f 2 (not shown in the figure):

(9)

Reasoning similarly, we will determine the components of the forces acting on the elements A.C. And BD, considering that instead R 1 will be R 2:

(10)

Let's find the sum of all four forces acting on the contour ABDC and exerting additional pressure on the internal areas of the liquid:

Let's determine the amount of additional pressure:

Hence:

(11)

Equation (11) is called Laplace's formula. The additional pressure that the curved surface of a liquid exerts on the internal regions of the liquid is called Laplace pressure.

The Laplace pressure is obviously directed towards the center of curvature of the surface. Therefore, in the case of a convex surface, it is directed into the liquid and is added to the normal pressure of the liquid. In the case of a concave surface, the liquid will be under less pressure than the liquid under a flat surface, because Laplace pressure is directed outside the liquid.

If the surface is spherical, then: R 1 = R 2 = R:

If the surface is cylindrical, then: R 1 = R, R 2 = ∞:

If the surface is flat then: R 1 = ∞, R 2 = ∞:

If there are two surfaces, for example, a soap bubble, then the Laplace pressure doubles.

Associated with the phenomena of wetting and non-wetting are the so-called capillary phenomena. If a capillary (a tube of small diameter) is lowered into a liquid, then the surface of the liquid in the capillary takes on a concave shape, close to spherical in the case of wetting and convex in the case of non-wetting. Such surfaces are called menisci.

Capillaries are those tubes in which the radius of the meniscus is approximately equal to the radius of the tube.

Rice. 9. Capillary in wetting (a) and non-wetting (b) liquids

Fig. 10. Rise of liquid in a capillary in case of wetting

In the case of a concave meniscus, the additional pressure is directed towards the center of curvature outside the fluid. Therefore, the pressure under the meniscus is less than the pressure under the flat surface of the liquid in the vessel by the amount of Laplace pressure:

R– meniscus radius, r– radius of the capillary tube.

Consequently, Laplace pressure will cause the liquid in the capillary to rise to such a height h(Fig.9) until the hydrostatic pressure of the liquid column balances the Laplace pressure:

From the last equation:

(12)

Equation (12) is called Jurin's formula. If the liquid does not wet the capillary walls, the meniscus is convex, cos θ < 0, то жидкость в этом случае опускается ниже уровня жидкости в сосуде на такую же глубину h according to formula (12) (Fig. 9).

Consider a convex surface (Fig. 5.18), the curvature of which at the point ABOUT for each of two mutually perpendicular normal sections is different. Let me be the external normal

to the surface at a point ABOUT; MN And R g R 2- main sections. Let us mentally select a surface element AS U and calculate the surface tension forces acting on the segments AB And CD, AC And B.D. believing that AB = CD And AC~BD. For each unit of contour length ABDC surface tension force A surrounding fluid, tending to stretch the surface element AS n in all directions. All forces acting on the side AB, replace with one resultant force A.F. applied to the middle of the segment AB= A/ in perpendicular parallel p, only in them instead Rx will the radius of curvature be £? 2 perpendicular sections R g R. g. Radius R 2 shown in Fig. 5.18 segment P-fi." Hence the resultant AF-* of all normal forces acting on four sides

surface element A5 P, AF~ = DK. +AF, + afs fAF. = V af, yes (rAS n | - -|- -V

The force AF^ presses the surface element A5 P to the layers located below it. Hence the average pressure p cf, due to the curvature of the surface,

To get the pressure on r a at a point, let us direct AS to zero. Moving to the limit of the ratio of AF^ to area asn, on which this force acts, we get AF^dF.

AS n -*o AS n dS n \ R, R 2

But by definition

p. = about 14-+ 4-\ (5 - 8)

p„ = a I ■

Where Rlt R 2- the main radii of curvature at a given point on the surface.

In differential geometry the expression e = -~ ^--\-

J--) is called the average curvature of the surface at the point R.

It has the same meaning for all pairs of normal sections perpendicular to each other.

Expression (5.8) establishing the dependence of the hydrostatic pressure drop r a at the interface between two phases (liquid - liquid, liquid -■ gas or vapor) from interfacial surface tension A and average!! the curvature of the surface 8 at the point under consideration is called Laplace's formula in honor of the French physicist Laplace.

Magnitude r a is added to the capillary pressure p corresponding to a flat surface. If the surface is concave, then a minus sign is placed in formula (5.8). In the general case of an arbitrary surface, the radii of curvature Rx And R 2 may differ from each other both in magnitude and sign. So, for example, at the surface shown in Fig. 5.19, radii of curvature Rx And R 2 in two mutually perpendicular normal sections are different in magnitude and sign. This case may result in positive or negative values r a depending on absolute value Rx And R2. It is generally accepted that if the center of curvature of a normal section is located below the surface, then the corresponding radius of curvature is positive, if above the surface it is negative. Surfaces whose average curvature

at all points is equal to zero e == ~(~--1" - 0, called minimal surfaces. If at one point of such a surface /? 1 >0, then automatically /? 2<С0.

For a sphere, any normal section is a circle of radius R, therefore in formula (5.8) /? x = R2 = R and additional capillary pressure

R. = ~.(5-9)

For a soap bubble due to the existence of its outer and inner surfaces

P*=-~-(5-Yu)

If for a circular cylinder one of the normal sections is considered to be the section running along the generatrix, then Rx= co. The second section perpendicular to it gives a circle of radius

R (R 2 = R). Therefore, in accordance with formula (5.8), the additional capillary pressure under the cylindrical surface

R. = -)|- (5-I)

From expressions (5.9) - (5.11) it is clear that when the shape of the surface changes, only the coefficient in front of the ratio changes a/R. If the surface of the liquid is flat, then R x ~ R 2 = co and therefore p z = 0. In this case, the total pressure

Р = Pi ± р а = Pi ± 0 = p t .

The additional capillary pressure, determined by Laplace's formula, is always directed towards the center of curvature. Therefore, for a convex surface it is directed inside the liquid, for a concave surface it is directed outward. In the first case, it is added to the capillary pressure p h in the second, it is subtracted from it. Mathematically, this is taken into account by the fact that for a convex surface the radius of curvature is considered positive, for a concave surface it is considered negative.

The qualitative dependence of the additional capillary pressure on the surface curvature can be observed in the following experiment (Fig. 5.20). ends And I'm B glass tee is immersed in a solution of soapy water. As a result, both ends of the tee are covered with soap film. Taking the tee out of the solution, through the process WITH blow two soap bubbles. As a rule, due to various reasons, bubbles have different sizes. If you close hole C, the larger bubble will gradually inflate, and the smaller one will contract. This convinces us that the capillary pressure caused by the curvature of the surface increases with decreasing radius of curvature.

To get an idea of ​​the value of the additional cap:pillar pressure, let’s calculate it for a drop with a diameter of 1 micron (clouds often consist of approximately such drops):

2a 2.72.75-Yu- 3 „ mgt

r --=-==-= 0.1455 MPa.

5.8. Wetting

Surface tension is possessed not only by the free surface of a liquid, but also by the interface between two liquids, a liquid and a solid, and also by the free surface of a solid. In all cases, surface energy is defined as the difference between the energy of molecules at the interface and the energy in the bulk of the corresponding phase. In this case, the value of surface energy at the interface depends on the properties of both phases. So, for example, at the water-air boundary a = 72.75-10 ~ 3 N/m (at 20 °C and normal atmospheric pressure), at the water-ether boundary a= 12-10 3 N/m, and at the water-mercury boundary a = 427-10~ 3 N/m.

Molecules (atoms, ions) located on the surface of a solid body experience attraction from one side. Therefore, solids, like liquids, have surface tension.

Experience shows that a drop of liquid located on the surface of a solid substrate takes on one shape or another depending on the nature of the solid, the liquid and the environment in which they are located. To reduce the potential energy in the gravitational field, a liquid always tends to take a form in which its center of mass occupies the lowest position. This tendency leads to the spreading of liquid over the surface of a solid. On the other hand, surface tension forces tend to give the liquid a shape that corresponds to a minimum of surface energy. The competition between these forces leads to the creation of one form or another.

Spontaneous increase in the area of ​​the solid-liquid or liquid phase boundary A- liquid IN under the influence of molecular cohesive forces is called spreading.

Let us find out the reasons leading to the spreading of a drop over the surface. Per molecule WITH(Fig. 5.21, A), located at the point of contact of a drop of liquid with a solid substrate, with one

sides there are attractive forces of liquid molecules, the resultant of which Fj_ directed along the bisector of the contact angle on the other - molecules of a solid body, the resultant of which F 2 perpendicular to its surface. Resultant R of these two forces is tilted to the left of the vertical, as shown in the figure. In this case, the tendency of the liquid to position its surface perpendicular to R will lead to its spreading (wetting).

The process of liquid spreading stops when the angle Ф (it is called regional) between the tangent to the surface of the liquid at the point WITH and the surface of a solid body reaches a certain limiting value rt k, characteristic of each liquid-solid pair. If the contact angle is acute

(0 ^ ■& ^ -), then the liquid wets the surface of the solid

body and the smaller it is, the better. At $k= 0, complete wetting occurs, in which the liquid spreads over the surface until a monomolecular film is formed. Wetting is usually observed at the interface of three phases, one of which is a solid (phase 3), and the other two - immiscible liquids or liquid and gas (phases / and 2) (see Fig. 5.21, c).

If strength F x more than F. 2, i.e., from the side of the liquid the force of attraction on the selected molecule is greater than from the side of the solid body, then the contact angle $ will be large and the picture looks as shown in Fig. 5.21, b. In this case, the angle Ф is obtuse (i/2< § ^ я) и жидкость частично (при неравенстве) или полностью (при равенстве) не смачивает твердую подложку. По отношению к стеклу такой несмачивающей жидкостью яв­ляется, например, ртуть, гдесозд = - 1. Однако та же самая ртуть хорошо смачивает другую твердую подложку, например цинк.

These considerations can be expressed quantitatively in

based on the following ideas. Let us denote by o"i_ 2, °1-з, 0-2 -3 respectively, surface tension at the boundary of liquid - gas, solid - gas and liquid -■ solid surface. The directions of action of these forces in the section will be depicted by arrows (Fig. 5.22). The following surface tension forces act on a drop of liquid located on a solid substrate: at the boundary / - 3 -ffi-з, tending to stretch the drop, and on the border 2 - 3 -Og-z. tending to pull it towards the center. Surface tension 04-2 at the boundary 1-2 directed tangentially to the surface of the drop at a point WITH. If the contact angle Ф is acute, then the projection of the force cri_ 2 onto the plane of the solid substrate (ov 2 cos Ф) will coincide in the direction with о 2 .-з (Fig. 5.22; A). In this case, the actions of both forces

will add up. If the angle ft is obtuse, as shown in Fig. 5.21, b, then cos ft is negative and the projection cri._ 2 cosft will coincide in direction with O1-.3. When a drop is in equilibrium on a solid substrate, the following equality must be observed:

= 02-3 + SG1-2 soeF. (5.12)

This equation was derived in 1805 Mr. Jung and named after him. Attitude

B =---^- = cos ft

called wetting criterion.

Thus, the contact angle ft depends only on the surface tensions at the boundaries of the corresponding media, determined by their nature, and does not depend on the shape of the vessel and the magnitude of gravity. When equality (5.12) is not complied with, the following cases may occur. If 01-3 greater than the right side of the equation (5.12), then the drop will spread, and the angle ft-■ will decrease. It may happen that cos ft increases so much that the right side of the equation (5,12) becomes equal to o"b_ 3, then equilibrium of the drop will occur in an extended state. If ov_ 3 is so large that even at cos ft = 1 left side of equality (5.12) more right (01 _z > 0 2 -з + o"i_ 2)> then the drop will stretch into a liquid film. If the right side of the equality (5.12) more than o"i 3, then the drop contracts to the center, the angle ft increases, and cos ft decreases accordingly until equilibrium occurs. When cos ft becomes negative, the drop will take the shape shown in Fig. 5.22, b. If it turns out that 0 2 - 3 so great that even at cos ft = -1 (ft = i) right side of equality (5.12) there will be more o"i-z (01 -z <02 h- 01-2)1 then in the absence of gravity the drop will contract into a ball. This case can be observed in small drops of mercury on the surface of glass.

The wetting criterion can be expressed in terms of the work of adhesion and cohesion. Adhesion A a is the occurrence of a connection between the surface layers of two dissimilar (solid or liquid) bodies (phases) brought into contact. A special case of adhesion, when the contacting bodies are identical, is called cohesion(denoted A c). Adhesion is characterized by the specific work spent on separating bodies. This work is calculated per unit area of ​​contact between the surfaces and depends on how they are separated: by shear along the interface or by separation in a direction perpendicular to the surface. For two different bodies (phases) A And IN it can be expressed by the equation

A a= hundred +and in-One-in,

Where A A, and in, and A - in- coefficients of surface tension of phases A and B at the boundary with air and between them.

In the case of cohesion, for each of the phases A and B we have:

АШ = 2а A, A<*> = 2a c.

For the drop we are considering

L S| =2a]_ 2 ; A a= ffi^ 3 -f ai_ 2 - sb-z-

Hence the wetting criterion can be expressed by the equality

IN - With

Thus, as the difference increases 2A a-L with wetting improves.

Note that the coefficients cti-z andОо„ 3 are usually identified with the surface tension of a solid at the boundaries with gas and liquid, while in a state of thermodynamic equilibrium the surface of a solid is usually covered with an equilibrium adsorption layer of the substance forming the drop. Therefore, in an exact solution of the problem for equilibrium contact angles, the values ​​of cri_ 3 and (Tg-z., generally speaking, should be attributed not to the solid body itself, but to the adsorption layer covering it, the thermodynamic properties of which are determined by the force field of the solid substrate.

Wetting phenomena are especially pronounced in zero gravity. The study of liquid in a state of space weightlessness was first carried out by the Soviet pilot-cosmonaut P. R. Popovich on the Vostok-4 spacecraft. In the ship's cabin there was a spherical glass flask half filled with water. Since water completely wets clean glass (O = 0), under weightless conditions it spread over the entire surface and closed the air inside the flask. Thus, the interface between glass and air disappeared, which turned out to be energetically beneficial. However, the contact angle i) between the surface of the liquid and the walls of the flask and in a state of weightlessness remained the same as it was on Earth.

The phenomena of wetting and non-wetting are widely used in technology and everyday life. For example, to make a fabric water-repellent, it is treated with a hydrophobizing (impairing water wetting) substance (soap, oleic acid, etc.). These substances form a thin film around the fibers, increasing the surface tension at the water-fabric interface, but only slightly changing it at the fabric-air interface. In this case, the contact angle O increases upon contact with water. In this case, if the pores are small, water does not penetrate into them, but is retained by the convex surface film and collects in drops that easily roll off the material.

The sanding liquid does not flow out through very small openings. For example, if the threads from which the sieve is woven are covered with paraffin, then you can carry water in it, if, of course, the layer of liquid is small. Thanks to this property, waterfowl insects running quickly through the water do not wet their paws. Good wetting is necessary when painting, gluing, soldering, dispersing solids in a liquid medium, etc.

FEDERAL EDUCATION AGENCY

STATE EDUCATIONAL INSTITUTION OF HIGHER PROFESSIONAL EDUCATION

Coursework

Course "Underground hydromechanics"

Topic: “Derivation of the Laplace equation. Plane problems of filtration theory"

Introduction

1. Differential equations of motion of compressible and incompressible fluid in a porous medium. Derivation of Laplace's equation.

2.1 Inflow to a perfect well

2.1.1 Filtration flow from injection well to production well

2.1.2 Inflow to a group of wells with a remote supply circuit

2.1.3 Inflow to a well in a formation with a straight-line feed circuit

2.1.4 Inflow to a well located near an impermeable straight boundary

2.1.5 Inflow to a well in a formation with an arbitrary supply circuit

2.1.6 Inflow to endless chains and ring well batteries

2.1.6.1 Inflow to ring battery wells

2.1.6.2 Inflow to a straight battery of wells

2.1.7 Equivalent filtration resistance method

Literature

Introduction

Underground hydromechanics is the science of the movement of liquids, gases and their mixtures in porous and fractured rocks - the theoretical basis for the development of oil and gas fields, one of the core disciplines in the curriculum of the field and geological faculties of oil universities.

Subsurface hydraulics is based on the idea that oil, gas and water contained in a porous medium form a single hydraulic system.

The theoretical basis of PGD is the theory of filtration - a science that describes a given fluid movement from the standpoint of continuum mechanics, i.e. hypothesis of continuity (continuity) of flow.

A feature of the theory of oil and gas filtration in natural formations is the simultaneous consideration of processes in areas whose characteristic dimensions differ by orders of magnitude: pore size (up to tens of micrometers), well diameter (up to tens of centimeters), formation thickness (up to tens of meters), distances between wells (hundreds of meters), the length of deposits (up to hundreds of kilometers).

In this course work, the basic Laplace equation is derived and plane problems of filtration theory are considered, as well as their solution.

1. Differential equations of motion of compressible and incompressible fluid in a porous medium. Derivation of Laplace's equation

When deriving the differential equation of motion of a compressible fluid, the initial equations are the following:

law of liquid filtration; As a filtration law, we accept the linear filtration law, expressed by formulas (3.1)

continuity equation (3.2)

equation of state. For a droplet compressible fluid, the equation of state can be represented as (3.3)

Substituting into the continuity equation (3.2) instead of the filtration velocity projections vx, vy and vz their values ​​from the linear law expressed by formula (3.1), we obtain:

state equation (3.3) we have:

Substituting these partial derivative values

Introducing the Laplace operator

Equation (3.7) can be written more briefly as

Considering that

equation (3.7) can be approximately represented as:

Equation (3.7) or an approximate replacement equation (3.10) is the desired differential equation for the unsteady motion of a compressible fluid in a porous medium. The mentioned equations have the form of a “heat equation”, the integration of which under various initial and boundary conditions is considered in every course of mathematical physics.

The solution to various problems about the unsteady motion of a homogeneous compressible fluid in a porous medium, based on the integration of equation (3.7) under various initial and boundary conditions, is given in the books of V. N. Shchelkachev, I. A. Charny and M. Masket. With steady motion of a compressible fluid

Equation (3.11) is called Laplace's equation.

During steady and unsteady filtration of an incompressible fluid, the density of the fluid is constant, therefore, the value on the right side of equation (3.4) is equal to zero. Reducing the left side of this equation by a constant

Thus, steady and unsteady filtration of an incompressible fluid is described by the Laplace equation (3.12).

2. Plane problems of filtration theory

When developing oil and gas fields (OGF), two types of problems arise:

1. The flow rate of the wells is set and it is necessary to determine the bottomhole pressure required for this flow rate and, in addition, the pressure at any point in the formation. In this case, the flow rate is determined by the value of the maximum depression for existing reservoirs, at which their destruction has not yet occurred, or by the strength characteristics of the well equipment, or by its physical meaning. The latter means, for example, the impossibility of establishing zero or negative bottomhole pressure.

2. The bottomhole pressure is set and the flow rate needs to be determined. The last type of condition occurs most often in GPS development practice. The amount of bottomhole pressure is determined by operating conditions. For example, the pressure must be greater than the saturation pressure to prevent oil degassing in the reservoir or condensate precipitation during the development of gas condensate fields, which reduces the productive properties of wells. Finally, if sand can be carried out of the formation to the bottom of the well, then the filtration rate on the well wall must be less than a certain limiting value.

It was noticed that when operating a group of wells under the same conditions, i.e. with the same bottomhole pressure, the production rate of the entire field grows slower than the increase in the number of new wells with the same bottomhole conditions (Fig. 4.1). An increase in flow rate requires a decrease in bottomhole pressure.

To solve the problems, we will solve the problem of plane interference (overlay) of wells. Let us assume that the formation is unlimited, horizontal, has a constant thickness and an impenetrable base and roof. The formation is penetrated by many perfect wells and filled with homogeneous liquid or gas. The fluid movement is steady, obeys Darcy's law and is flat. Plane motion means that the flow occurs in planes parallel to each other and the pattern of motion in all planes is identical. In this regard, the flow in one of these planes is analyzed - in the main plane of the flow.

We will build the solution of problems on the principle of superposition (superposition) of flows. The superposition method based on this principle is as follows.

When several drains (production wells) or sources (injection wells) act together in a formation, the potential function determined by each drain (source) is calculated using the formula for a single drain (source). The potential function due to all sinks (sources) is calculated by algebraically adding these independent values ​​of the potential function. The total filtration rate is defined as the vector sum of the filtration rates caused by the operation of each well (Fig. 4.2b).

Let there be n drains with a positive mass flow rate G and sources with a negative mass flow rate in an unlimited reservoir (Fig. 4.2a). The flow in the vicinity of each well in this case is plane-radial and the potential

Local theorem of Moivre-Laplace. 0 And 1, then the probability P t n that, that event A will occur m times in n independent trials for a sufficiently large number n is approximately equal to

- Gaussian function And

The larger and, the more accurate the approximate formula (2.7), called local Moivre-Laplace formula. Approximate probability values R tpu given by local formula (2.7), in practice they are used as exact ones for prue about two or more dozen, i.e. given that prue > 20.

To simplify the calculations associated with the application of formula (2.7), a table of values ​​of the function /(x) has been compiled (Table I, given in the appendices). When using this table, it is necessary to keep in mind the obvious properties of the function /(x) (2.8).

• 1. Function/(X) is even, i.e. /(-x) = /(x).
• 2. Function/(X) - monotonically decreasing for positive values X, and at x -> co /(x) -» 0.
• (In practice we can assume that already for x > 4 /(x) « 0.)

[> Example 2.5. In some area, out of every 100 families, 80 have refrigerators. Find the probability that out of 400 families, 300 have refrigerators.

Solution. The probability that a family has a refrigerator is p = 80/100 = 0.8. Because n= 100 is large enough (condition prue= = 100 0.8(1-0.8) = 64 > 20 fulfilled), then we apply the local Moivre-Laplace formula.

First, we determine by formula (2.9)

Then according to formula (2.7)

(the value /(2.50) was found from Table I of the appendices). The very low probability value /300,400 should not raise doubts, since in addition to the event

“exactly 300 families out of 400 have refrigerators”, another 400 events are possible: “0 out of 400”, “1 out of 400”,..., “400 out of 400” with their own probabilities. All together, these events form a complete group, which means the sum of their probabilities is equal to one. ?

Suppose that in the conditions of example 2.5 it is necessary to find the probability that from 300 to 360 families (inclusive) have refrigerators. In this case, according to the addition theorem, the probability of the desired event

In principle, each term can be calculated using the local Moivre-Laplace formula, but the large number of terms makes the calculation very cumbersome. In such cases, the following theorem is used.

Integral theorem of Moivre - Laplace. If the probability p of the occurrence of event A in each trial is constant and different from 0 And 1, then the probability is, that the number m of occurrence of event A in n independent trials lies in the range from a to b (inclusive), for a sufficiently large number n is approximately equal to

- function(or probability integral) Laplace",

(The proof of the theorem is given in Section 6.5.)

Formula (2.10) is called the integral formula of Moivre-Laplace. The more p, the more accurate this formula is. When the condition is met prue > > 20 integral formula (2.10), like the local one, usually gives an error in calculating probabilities that is satisfactory for practice.

The function Ф(дг) is tabulated (see Table II of the appendices). To use this table you need to know the properties of the function Ф(х).

1. Function f(x) odd, those. Ф(-х) = -Ф(х).

? Shall we make a variable change? = -G. Then (k =

= -(12. The limits of integration for variable 2 will be 0 and X. We get

since the value of the definite integral does not depend on the designation of the integration variable. ?

2. Function Ф(х)monotonically increasing, and at x ->+so f(.g) -> 1 (practically we can assume that already at x > 4 Ф(х)~ 1).

Since the derivative of the integral with respect to the variable upper limit is equal to the integrand at the value of the upper limit, g.s.

, and is always positive, then Ф(х) increases monotonically

on the entire number line.

Let us make a change of variable, then the limits of integration do not change and

(since the integral of an even function

Considering that (Euler integral - Poisson), we get

?

O Example 2.6. Based on the data in Example 2.5, calculate the probability that from 300 to 360 (inclusive) families out of 400 have refrigerators.

Solution. We apply the integral theorem of Moivre - Laplace (pr= 64 > 20). First, we determine using formulas (2.12)

Now, using formula (2.10), taking into account the properties of Ф(.т), we obtain

(according to Table II of the appendices?

Let us consider a corollary of the Moivre-Laplace integral theorem. Consequence. If the probability p of the occurrence of event A in each trial is constant and different from 0 and I, then with a sufficiently large number n of independent trials, the probability is that:

A) the number m of occurrences of event A differs from the product pr by no more than e > 0 (in absolute value), those.

b) the frequency of t/p event A is within the limits from a to p ( I'll turn it on- emphatically, i.e.

V) the frequency of event A differs from its probability p by no more than A > 0 (in absolute value), i.e.

A) Inequality |/?7-7?/?| equivalent to double inequality pr-e Therefore, according to the integral formula (2.10)

• b) Inequality and is equivalent to inequality and when a = pa And b= /?r. Replacing the quantities in formulas (2.10), (2.12) A And b Using the obtained expressions, we obtain the formulas to be proved (2.14) and (2.15).
• c) Inequality mjn-р is equivalent to the inequality t-pr Substituting in formula (2.13) g = Ap, we obtain the formula (2.16) to be proved. ?

[> Example 2.7. Based on the data in Example 2.5, calculate the probability that from 280 to 360 families out of 400 have refrigerators.

Solution. Calculate the probability P 400 (280 t pr = 320. Then according to formula (2.13)

[> Example 2.8. According to statistics, on average 87% of newborns live to be 50 years old.

• 1. Find the probability that out of 1000 newborns the proportion (frequency) of those surviving to 50 years will: a) be in the range from 0.9 to 0.95; b) will differ from the probability of this event by no more than 0.04 (but in absolute value).
• 2. For what number of newborns with a reliability of 0.95 will the proportion of those surviving to 50 years be within the range from 0.86 to 0.88?

Solution. 1, a) Probability r that a newborn will live to be 50 years old is 0.87. Because n= 1000 is large (condition prd=1000 0.87 0.13 = = 113.1 > 20 satisfied), then we use a corollary of the Moivre-Laplace integral theorem. First, we determine by the formulas (2.15)

Now according to the formula (2.14)

1, b) According to formula (2.16)

Since inequality tantamount to inequality

the result obtained means that it is almost certain that from 0.83 to 0.91 out of 1000 newborns will live to be 50 years old. ?

2. By condition or

According to the formula (2.16) at A = 0.01

According to the table II appendices Ф(Г) = 0.95 at Г = 1.96, therefore,

where

those. condition (*) can be guaranteed with a significant increase in the number of newborns under consideration to n = 4345. ?

• The proof of the theorem is given in section 6.5. The probabilistic meaning of the quantities pr, prs( is established in paragraph 4.1 (see note on p. 130).
• The probabilistic meaning of the RF/n value is established in paragraph 4.1.