Аннотация (русский):
The monograph presents the results of a study of the digital economy as a new paradigm of economic development, a system of economic relations implemented through the use of digital information computer technologies. It is noted that the main problem in the formation of sustainable economic growth and the successful introduction of digital technologies are the challenges of digitalization of the economy. New digital technologies, innovative business models penetrate into all spheres of the economic life of society, influencing the very essence of the economy, forming qualitative structural changes in it. As a result, a digital economy is being formed as a subsystem of the traditional economy, characterized by the active use of digital technologies and the circulation of specific electronic goods. The monograph is intended for researchers, teachers, graduate students, undergraduates, as well as a wide range of readers interested in topical issues of digitalization of the economy.

Ключевые слова:
digital economy , economic development, digital technologies, engineering applications, dissipative functions

1.1 Method of dissipative functions. Any natural process that manifests itself in nature or proceeds in technical devices at a finite rate is non-equilibrium and irreversible. A non-equilibrium change in the state of a macrobody cannot be described using classical thermodynamics. The application of the equation of state is possible only for the equilibrium flow of the process, when the system can be characterized by certain intensive parameters. The imbalance leads to inhomogeneous distribution of temperature, pressure, concentration of ingredients of a multiphase and multicomponent system, as well as the existence of relaxation material and energy flows. Relaxation processes tend to equalize the uneven distribution of parameters, compensating for the external influence that causes the system's inhomogeneity. The non-equilibrium flow of phenomena associated with the transformation of energy and matter leads to states when the system no longer has a single temperature or density, the same on its mass. To describe the state of a thermodynamic system participating in a nonequilibrium process, more detailed characteristics of the state change are required. Therefore, non-equilibrium processes cannot be represented on a thermodynamic diagram in the coordinates PV, NS, TS, as it is traditionally accepted for equilibrium changes in the object under study. Quantitative judgments about a real process are possible only if the system at the beginning and at the end of the process is in certain equilibrium states. It has been established that all real processes are irreversible and can spontaneously proceed in only one direction. This principle of irreversibility applies to all real processes without exception, both natural and technological. The irreversibility of a real process is manifested in the fact that its course is always accompanied by residual effects. The consequences of a real process cannot be completely eliminated, which is confirmed by numerous experiments. Not a single fact has yet been verified that refutes this distinctive feature of the real process, which, apparently, is associated with the peculiarities of energy transformations. The course of the real process consists in a special transformation of energy in the macrobody. The total energy of an isolated thermodynamic system is conserved only quantitatively, while in a qualitative sense, energy is constantly depreciating, i.e. dissipates. Dissipation does not mean the complete dissipation of energy in space, but the loss of its valuable, convertible part, called exergy. All types of energy, including internal, are limited by convertibility and consist of two parts: a convertible, called exergy, and a ballast part, which is associated with entropy: (1.1) The total energy of the system H (enthalpy) is the sum of its two terms: the Gibbs energy G (convertible part) and the ballast part TS associated with entropy. Any real process is accompanied by the loss of some part of the convertible energy, i.e. exergy. Only an ideal process allows you to fully preserve the entire supply of energy in a quantitative and qualitative sense, therefore it is characterized by energy perfection, i.e. no losses during energy conversions. Indicators of an ideal process can serve as initial information for assessing the quality and efficiency of technical devices, which is one of the main tasks of energy technology. The quantitative assessment of the principle of irreversibility is expressed by the second law of thermodynamics, by introducing an increase in entropy as a measure of dissipation. The entropy of an isolated or closed thermodynamic system, when any real process takes place in it, will certainly increase, thus any real process generates an increase in entropy. The change in entropy in a thermodynamic system during a real process in it is the sum of two terms: , (1.2) where – the first term that characterizes the change in entropy due to the equilibrium transfer of entropy along with the fluxes of heat and matter through the boundary of the thermodynamic system. , (1.3) where – the flow of entropy carried with the flow of matter; – the flow of entropy carried with the flow of heat. The second term in equation (1.2) is the rate of entropy production within the process itself due to its irreversibility. This flow of entropy cannot be transferred through the boundaries of the thermodynamic system. The entropy generated within a real process can act as a quantitative measure of its irreversibility. For this, the concept of a local dissipative function is introduced, which is related to the rate of entropy production in a unit volume by the following relation: , (1.4) where – local flow of entropy generated within the process. , (1.5) where – entropy production rate within the process; – local dissipative function, i.e. the value of the dissipative function in a unit volume contracted to a point: . (1.6) According to the second law of thermodynamics of irreversible processes, the rate of entropy production within the process itself is strictly positive: . (1.7) The entropy generated within a real process is a quantitative measure of its irreversibility. To calculate the dissipative function, two calculation methods are used. The first method is based on the integration of the local dissipative function over the entire volume of the thermodynamic system. The second method is to use the integral equations of energy and matter balance for a fixed control volume of a nonequilibrium thermodynamic system. 1.2 Quantification of irreversible processes based on the concept of local dissipation. Engineering applications of irreversible processes are very diverse. The most complex are the transformations of the system, accompanied by changes in the chemical composition. A comprehensive description of the phenomena accompanying irreversible changes in the state of a macroobject begins with the combination of the laws of conservation of mass and energy with the corresponding kinetic laws that describe the mechanisms of these phenomena. Local changes in the rates of heat transfer, a component of matter and momentum, near equilibrium are characterized by linear relationships between the rate and the driving force of the process: q ⃗=-λ∙∇ ⃗T ¬– Fourier's law (1.8) where: q ⃗ – heat flux density, λ – coefficient of thermal conductivity of the medium. J_j^дф=-D_j ∇ ⃗C_j – Fick's law (1.9) where: J_j^дф – diffusion flux density of the j-th component, D_j- diffusion coefficient of the j-th component. σ_ij=-μ((∂ϑ_i)/(∂x_j )+(∂ϑ_j)/(∂x_i )) – Newton's law of outflow of a viscous medium (1.10) where: μ – dynamic viscosity coefficient, σ_ij – stress tensor in a moving medium. J_jr^V=L_rr A_r – the rate of the resulting process, accompanied by a change in the chemical composition equal to the difference between the rates of the forward and reverse reactions (1.11) (1.11) where: L_rr – phenomenological reaction coefficient associated with the forward and backward reaction constants. In particular, we have for the reaction of ideal gases L_rr=(K^+ ∏_i▒〖C_i〗^(V_i ) ) (RT)^(-1) (1.11а) where: K^+ - direct reaction constant, C_i, V_i – concentrations and stoichiometric coefficients of initial reagents. A_r=-∑_(k=1)^n▒〖ν_kr∙〗 μ_k - chemical affinity, reaction driving force (1.11b) where: ν_kr– stoichiometric coefficients of substance k in the considered reaction r, μ_k – chemical potential of all components. In general, the local rate of the i-th process of energy and mass transfer J_i near equilibrium can be represented as a linear relationship: J_i=∑_(k=1)^n▒〖L_ik∙X_k 〗 (1.12) where J_i – the local speed of the i-th process, which generally depends on all driving forces X_k; L_ik – are phenomenological proportionality coefficients, which can be functions of the state of the object, but do not depend on the driving forces X_k. For non-conjugate processes, the speed depends only on its own driving force X_i: J_i=L_ii∙X_i (1.13) where L_ii – phenomenological coefficients of proportionality, which are related to the thermal conductivity coefficients λ, diffusion coefficient Dj of the j-th component, the viscosity of the medium μ, and the rate constants of the forward and backward reactions k+ and k–. In equations (1.8 - 1.11), the values of J_i are identical to the heat flux density q ⃗, diffusion flux J_j^дф, momentum flux σ_ij, chemical reaction rate J_jr^V in a unit volume of the reaction mixture. The magnitude of the driving force X_i is proportional to the temperature gradients ∇ ⃗T, the concentration of the j-th component ∇ ⃗C_j, the strain rate of the medium in two-dimensional space X_ij=-1/2 ((∂ʋ_i)/(∂x_j )+(∂ʋ_j)/(∂x_i )), chemical affinity A_r. In an equilibrium system, J_i=0 and X_i=0, that is, there is a uniform distribution of intense characteristics in the absence of external fields and, as a consequence, the absence of relaxation flows. The irreversibility of real processes in the macrobody means the fundamental impossibility of returning all the bodies involved in the process to their original state, even if the return process is equilibrium. For example, in a compressor, the temperature and pressure of the gas increase due to an external source (electrical network). If the compression is in equilibrium, then the process is certainly thermodynamically reversible, that is, it is possible to return all bodies to their original state: gas and transfer work to an external source in the same amount. If the process is non-equilibrium and is accompanied by friction (σ_ij≠0) and heat transfer (q ⃗≠0), then, basically, it is possible to return the gas to its original state. However, the amount of work returned to an external source will be noticeably less than the original cost. The cause of irreversibility is the dissipation (depreciation) of energy, i.e. transformation of its convertible part into a ballast part due to entropy. The second law of thermodynamics for equilibrium processes introduces entropy as a function of state, which makes it possible to isolate its convertible part from the internal energy U: the Gibbs-Helmholtz functions A=U-TS and Gibbs G=H-TS. In non-equilibrium processes, an internal flow arises (entropy production) , which acts as a quantitative measure of energy dissipation. For processes of flow of a viscous medium, the local dissipative function (the Rayleigh function) Ψ ̇_тр^V is equal to the product of the momentum flux σ_ij in the velocity profile plane of the moving medium and the cause - the strain rate. Taking into account the tensor nature of the momentum, we obtain (1.14): Ψ ̇_тр^V=-(σ∙∇ ⃗ )∙ʋ ⃗=∑_(i=1)^(n=3)▒∑_(j=1)^(m=3)▒σ_ij [-1/2 ((∂ʋ_i)/(∂x_j )+(∂ʋ_j)/(∂x_i ))], (1.14) where: X_ij=-1/2 ((∂ʋ_i)/(∂x_j )+(∂ʋ_j)/(∂x_i )). The dissipation value Ψ ̇ for the thermodynamic system as a whole can be determined by integration, i.e. distribution of the local dissipative function Ψ ̇^V on the entire volume V under consideration (1.15): Ψ ̇=∫_0^V▒〖Ψ ̇^V dV〗=∫_0^V▒〖T∙J_S^V dV〗=T ̅_(т/д)∙(_in^)S ̇ , (1.15) where: T ̅_(т/д) – average thermodynamic temperature, К; (_in^)S ̇ – is the rate of entropy increment in the entire system under consideration due to internal irreversible processes, W/K. Differential equations for the balance of mass, energy, entropy, kinetic ratios of local rates of energy and mass transfer processes and the Gibbs equation for fixed parameters make it possible to obtain a calculated ratio for the dissipation density: Ψ ̇^V=q ⃗∙(-(∇ ⃗T)/T)+∑_(j=1)^m▒〖(J_j^дф ) ⃗∙(-∇ ⃗μ)_(T,p) 〗+∑_(r=1)^f▒〖J_jr^V∙A_r 〗+(-σ∙∇ ⃗ )∙ʋ ⃗ (1.16) So, the driving force for heat transfer is X=-(∇ ⃗T)/T; for mass transfer – X_j=(-∇ ⃗μ)_(T,p), where μ_j – where μ_j is the chemical potential of the j-th component of the system; for chemical transformations X_r=A_r, where A_r – where A_r is the chemical affinity characterizing the degree of completeness of the chemical reaction. To calculate the local and integral values of the dissipative function, information is needed on the distribution of intensive parameters: temperature T (x,y,z,τ), presseure P(x,y,z,τ), concentration C_j(x,y,z,τ), velocity distributions ʋ(x,y,z,τ) and kinetic relations for calculating the thermophysical properties of the system. 1.2.1 Dissipation of kinetic energy in viscous flow. Task 1. To btain the calculated ratio of the local dissipative function for the hydrodynamic steady motion of an incompressible fluid (50% aqueous solution of glycerin). To determine the value of the dissipative function over the entire volume in a smooth round pipe with an inner diameter D and length L under isothermal conditions. The motion mode is stationary, laminar Re=1800. Process parameters: P=20 bar, t=40 ºC. Let's compute at r/R=1. Let us imagine the flow profile of a medium in a round pipe of constant diameter under laminar conditions (Fig. 1). Fig. 1. Medium flow profile in a round pipe of constant diameter under laminar conditions. Under these conditions, the velocity distribution is parabolic: ʋ_X=ʋ_0 [1-(r/R)^2 ] (1.17) where: r - the current value of the flow radius, R - the inner radius of the pipe, ʋ_0=2ʋ ̅ – the speed on the pipe axis equal to the double value of the average speed. Relation (1.17) describes the profile of velocities in the cross section of the pipe and is a parabola equation. However, it should be borne in mind that, in fact, the velocity distribution is a three-dimensional figure and, with laminar motion in a round pipe, is a paraboloid of revolution. Equation (1.17) is written for any longitudinal section of this paraboloid for a plane passing through the pipe axis. The local dissipative function is determined according to expression (1.14). The viscous stress tensor generally has nine components (1.18). σ_ij=■(σ_xx&σ_xy&σ_xz@σ_yx&σ_yy&σ_yz@σ_zx&σ_zy&σ_zz ) (1.18) where: i,j=1 – corresponds to the x-axis; i,j=2 - corresponds to the y axis; i,j=3 - corresponds to the z axis; σ_ij – component of the viscous stress tensor, N/m2. For an incompressible medium, the diagonal terms are equal to zero σ_xx=σ_yy= σ_zz=0, in addition, taking into account the symmetry σ_ij=σ_ji. In the example under consideration, only one velocity component ʋ_x has a non-zero value. Therefore, the local dissipation of kinetic energy is equal to: Ψ ̇_тр^V=2σ_xr [-1/2 ((∂ʋ_x)/∂r)+((∂ʋ_r)/∂x)]=μ((∂ʋ_x)/∂r)^2 (1.19) where: the radial velocity component for a stabilized flow is zero ʋ_r=0; viscous stress in the radial plane xr σ_xr=-μ (∂ʋ_x)/∂r, where μ – the dynamic viscosity of the medium. Ψ ̇_тр^V=4μʋ_0^2 r^2/R^2 =16μʋ ̅^2 r^2/R^4 (1.20) Obviously, in the center of the pipe (r = 0) Ψ ̇_тр^V= 0, in turn, directly at the pipe wall (r = R) the dissipation of kinetic energy has a maximum value: Ψ ̇_тр^V=16μʋ ̅^2 R^2 (1.21) Integrating Ψ ̇_тр^V on the volume bounded by the inner surface of the pipe with diameter D and length L, between sections 1-1 and 2 – 2, we obtain an expression for calculating the total dissipation Ψ ̇_тр under the given conditions (1.22a), (1.22b): Ψ ̇_тр= ∫_0^V▒〖Ψ ̇_тр^V dV=∬_(0 0)^(L R)▒〖16μʋ ̅^2 r^2/R^4 2πrdrdx〗〗 (1.22a) Ψ ̇_тр=32πμʋ ̅^2 1/R^4 ∫_0^L▒〖dx∫_0^R▒r^3 〗 dr=8πμʋ ̅^2 L (1.22b) The integration used the condition of an incompressible medium, which makes it possible to assume that the distribution of the flow velocity along the entire length of the pipe is unchanged. The numerical values of the parameters and thermophysical characteristics of the outflow of the solution in the pipe section included in the formula (1.22b) are calculated according to the laminar hydrodynamic regime Re=1800 in a pipe with a diameter D=0.06m and a length L=25m: ʋ ̅=μRe/Dρ , at [3] ʋ ̅=(3,5·10^(-3)·1800)/(0,06·1116)=0,094 m/s W It should be noted that an analytical calculation of dissipation during the motion of a medium under conditions of a turbulent regime is impossible in this case, since there is no information on turbulent fluctuations. 1.2.2 Dissipation of internal energy in heat conduction processes. In engineering, the problems of heat transfer at a constant heat flux density of the walls are encountered in many cases: in electric heating, radiation heating, heating in nuclear reactors and in counterflow heat exchangers, when the mass flow heat capacities (the product of the mass flow rate and the heat capacity) of the heat carriers are the same. In our problem, there is also one boundary condition is a constant temperature of the outer surface of the pipe along the entire length of the reactor. Such a boundary condition is also often encountered in practice, for example, in such heat exchangers as evaporators, condensers and in all heat exchangers, when the mass flow heat capacity of one heat carrier is much greater than that of another. The method used is based on the integration of the local dissipative function on the entire volume of the system under consideration. The advantage of this methodological approach lies in its clarity and the ability to reveal the internal logic of the derivation of the main patterns of the process. Let us analyze this case on a concrete example of the dissipation of internal energy in the process of heat conduction. Task 2. To obtain an analytical and numerical solution for the magnitude of local and integral energy dissipation in the wall of a tubular reactor, if the value of the heat flux density on the outer surface of the pipe q^'=-60 kW/m^2 , the temperature of this surface is T_ст=T_нар=873 К, external diameter d_н=0,1 м and internal diameter〖 d〗_вн=0,08 м of the wall, pipe length L=40 m, thermal conductivity value of the wall λ=23,8 W/(m∙К). It should be noted that the relation given in the problem for the dissipative function (1.16) is the simplest linear combination of the terms of each individual gradient (temperature, concentration, velocity), although it is known that various forms of energy transfer are interconnected. In the problem under consideration, there is only a temperature gradient, therefore, according to the Fourier's law, the heat flux density is determined by equation (1.8): q ⃗=-λ∙∇ ⃗T, where the temperature gradient in the one-dimensional (radial) problem is defined in a simplified way as: ∇ ⃗T=(l_r ) ⃗∙∂T/∂r Let us depict the temperature distribution along the pipe radius for the selected boundary conditions (Fig. 2). Fig. 2. Temperature distribution along the radius of the pipe under the condition of constancy q^',T_нар^' и λ The local dissipative function according to equation (1.8) is equal to: Ψ ̇^V=q ⃗∙(-(∇ ⃗T)/T)=λ/T (∇ ⃗T)^2≥0 (1.23) For a one-dimensional problem with a boundary condition of the second kind, the temperature distribution in a cylindrical pipe of radii r has the form (1.24): T(r)=T_нар^'-(q^'∙r_нар^')/λ ln r/(r_нар^' ), (1.24) where the condition q^'=-λ(∂T/∂r)_(r=r_нар^' )=const, is satisfied regardless of the medium flow regime along the entire wall surface of length l. Differentiating and solving equation (1.24) with respect to ∂T/∂r, we get (1.25): ∂T/∂r=-(q^'∙r_нар^')/λ∙1/r (1.25) Substituting the expression ∂T/∂r into equation 1.23, we obtain (1.26): Ψ ̇^V=λ/T [-(q^'∙r_нар^')/λ∙1/r]^2 ((l_r ) ⃗ )^2, (1.26) where the square of the unit radiant vector is 1 ((l_r ) ⃗ )^2=1. The final expression for local dissipation has the following form (1.27): Ψ ̇^V=λ/T [-(q^'∙r_нар^')/λ∙1/r]^2=[-q^'∙r_нар^' ]^2/(T∙λ∙r^2 ) (1.27) We see that the resulting expression for directly determining the numerical value of the local dissipative function on the outer and inner walls of the reactor tube is very convenient. Let's do these calculations: r=r_нар^'=0,05 м T=T_нар^'=873 К Ψ ̇^V=[-q^'∙r_нар^' ]^2/(T_нар^'∙λ∙〖r_нар^'〗^2 )=(q^' )^2/(T_нар^'∙λ)=(-60∙10^3 )^2/(873∙23,8)=173,27 kW/m^3 r=r_вн^'=0,04 м T=T_вн^'' T_вн^''=T_нар^'-(q^'∙r_нар^')/λ ln (r_вн^'')/(r_нар^' )=873-(-60∙10^3∙0,05)/23,8 ln 0,04/0,05=844,88 К Ψ ̇^V=(q^' )^2/(T_нар^'∙λ)∙((r_нар^')/(r_вн^'' ))^2=(-60∙10^3 )^2/(844,88∙23,8)∙(0,05/0,04)^2=279,72 kW/m^3 To determine the integral value of dissipation in the wall of a pipe with a length of l=1 m, under the condition T_нар^'=873 К, q^'=-60 кВт/м^2 and λ=23,8 W/(м∙К) we get: Ψ ̇_l=∫_0^V▒〖Ψ ̇^V dV〗=∫_(r_нар^')^(r_вн^'')▒〖[-q^'∙r_нар^' ]^2/(T∙λ∙r^2 ) 2πr∙l∙dr〗 Ψ ̇_l=[-q^'∙r_нар^' ]^2/λ 2πl∫_(r_нар^')^(r_вн^'')▒dr/(T∙r) To determine the integral∫_(r_нар^')^(r_вн^'')▒dr/(T∙r) it is necessary to make a change of variable from (1.26): dr/r=-λ/(q^'∙r_нар^' ) dT Then: Ψ ̇_l=[-q^'∙r_нар^' ]^2/λ 2πl∫_(Т_нар^')^(Т_вн^'')▒(-λ/(〖T∙q〗^'∙r_нар^' ))dT Dissipation per 1 meter of pipe: Ψ ̇_l=-q^' 2πr_нар^' l∙ln (Т_нар^')/(Т_вн^'' )=-(-60∙10^3∙2∙3,14∙1∙0,05)∙ln 873/844,88= =617,84 W/m The dissipation along the entire length of the reactor tube is: Ψ ̇=L∙Ψ ̇_l=617,84∙40=24713,6 Вт≈25 kW Internal exergy losses (_in^)(D_L ) ̇ , due to the irreversibility of heat transfer, can be determined from the following relation (1.28): (_in^)(D_L ) ̇ =Ψ ̇ T_(o.c.)/T ̅_(т/д) , (1.28 ) where T ̅_(т/д)=(T_cm^'-T_cm^'')/(ln (T_cm^')/(T_cm^'' )) – average thermodynamic temperature of the process. T ̅_(т/д)=(873-844,88)/(ln 873/844,88)=858,86 K (_in^)(D_L ) ̇ =25∙298,15/858,86=8,68 kW Both functional quantities Ψ ̇ and (_in^)(D_L ) ̇ are quantitative characteristics of the efficiency of an irreversible process and have the same dimension. However, an important difference is that it is the exergy losses (_in^)(D_L ) ̇ , due to the entropy produced inside the process, that are final and, thus, can only be compensated by external energy costs. In turn, the difference between Ψ ̇ and (_in^)(D_L ) ̇ amounts to that part of the energy that can still be usefully used as a secondary (internal) energy resource. With the recovery of heat generated in the wall of the cylindrical reactor, the efficiency of the heat conduction process will increase. The value of this internal energy resource is as follows: Ψ ̇-(_in^)(D_L ) ̇ =25-8,68=16,32 kW In case of qualified energy technological utilization of this hidden energy resource generated in the reactor wall, exergy losses will be minimal. The rational choice of exergy optimization of high-temperature processes depends on the characteristics of a particular energy carrier and the technological possibilities of heat recovery. 1.2.3 Energy dissipation in diffusion processes. Let us analyze the mass transfer process using a specific example of the dissipation of convertible energy in a diffusion membrane process. Task 3. To obtain an analytical and numerical solution for the magnitude of local and integral energy dissipation in the diffusion layer of a membrane separating an ideal mixture of propane (component A) and methane (component B) hydrocarbons at a temperature T_f=279 К and a pressure P_f=4,6 bar in pressure channel of the device. The composition of the gas mixture in the pressure channel (y_A ) ̃^'=0,59 (kmol А)/(kmol of mixture), in the drainage cavity (y_A ) ̃^''=(y_P ) ̃=0,95 (kmol А)/(kmol of mixture). The thickness of the diffusion layer of the polymer membrane δ=0,2 мкм, the working area А=117 m^2; pressure in the drainage cavity P^''=1 bar. The performance of the initial mixture is (N_F ) ̇=0,023 kmol/s. The gas mixture of the initial composition (y_F ) ̃ is introduced into the pressure channel 1 of the membrane apparatus at P=P_F and T_F, passed through the membrane 3 flow (N_p ) ̇ with the concentration (y_A ) ̃^''=(y_P ) ̃ is removed from the drainage cavity 2 ; the waste flow (N_R ) ̇ is discharged from the pressure channel with the concentration (〖 y〗_R ) ̃=(y_A0 ) ̃^'. (a) (b) Fig. 3. a) Distribution and characteristics of flows in the apparatus: 1 - pressure channel; 2 - drainage channel; 3 - silicone membrane. b) Concentration profile of the components in the membrane layer of the apparatus. The process is stationary and isothermal, the gas mixture is ideal. Hydraulic resistance in the pressure and drainage channel is negligible. External diffusion resistance in the pressure and drainage channel is excluded. The structure of the gas flow in the cavity of the pressure channel 1 corresponds to the ideal mixing model (IMS). Based on these assumptions, the following values follow: P^'=P_f=P_R=4,6 bar, where P^' – pressure in cavity 1; P^''=P_p=1 bar, where P^'' pressure in cavity 2; (y_A ) ̃^'=(y_R ) ̃=0,59 (kmol А)/(kmol of mixture); (y_A ) ̃^''=(y_R ) ̃=0,95 (kmol А)/(kmol of mixture), where (y_A ) ̃^',(〖 y〗_A ) ̃^'' – are the compositions of the gas phase near the membrane surfaces from the side of the pressure and drainage channel. The resistance to mass transfer is concentrated exclusively in the diffusion layer of the silicone membrane with a thickness of δ. At the boundary of the gas and the membrane, it is permissible to assume that there is a local equilibrium. The concentrations of components А (〖C_(A,m)〗^') and В (〖C_(B,m)〗^') in the membrane at the boundary with the pressure channel are as follows: 〖С_(A,M)〗^'=σ_(A,M)∙P^'∙(y_A ) ̃^',(kmol А)/m^3 〖С_(B,M)〗^'=σ_(B,M)∙P^'∙(y_В ) ̃^',(kmol В)/m^3 , where σ_(A,M)=8,825∙10^(-5) mol/(m^3∙Pa) and σ_(B,M)=1,683∙10^(-6) mol/(m^3∙Pa) – the values of the solubility coefficients of this membrane (material polydimethylsiloxane[(CH3)2SiO]x) on both components at a temperature value of T=279 К. We obtain the following values of the concentrations of the components at the boundaries with the membrane: 〖С_(A,M)〗^'=8,825∙10^(-8) ∙4,6∙10^5∙0,59=23,951∙10^(-3) (kmol )/m^3 〖С_(A,M)〗^''=8,825∙10^(-8) ∙1∙10^5∙0,95=8,3837∙10^(-3) (kmol )/m^3 〖С_(B,M)〗^'=1,683∙10^(-9) ∙4,6∙10^5∙0,41=0,3174∙10^(-3) (kmol )/m^3 〖С_(B,M)〗^''=1,683∙10^(-9) ∙1∙10^5∙0,05=0,00841∙10^(-3) (kmol )/m^3 The values of diffusion fluxes J_A, J_B in the membrane are determined based on the condition of constant diffusion coefficients of the components in the membrane layer. For a flat and one-dimensional problem, the distribution of the substance concentration in the membrane is linear, and the concentration gradients are constant (1.9). (∂С_(A,m))/∂x=(〖С_(A,m)〗^''-〖С_(A,m)〗^')/δ=((8,3837-23,951)∙10^(-3))/(2∙10^(-7) ) (∂С_(A,m))/∂x=-7,784∙10^4 (kmol )/m^4 (∂С_(B,m))/∂x=(〖С_(B,m)〗^''-〖С_(B,m)〗^')/δ=((0,00841-0,3174)∙10^(-3))/(2∙10^(-7) ) (∂С_(B,m))/∂x=-0,1545∙10^4 (kmol )/m^4 Diffusion flux density values on components A and B: J_A=-D_(A,m) (∂C_(A,m))/∂x=-5,558∙10^(-10)∙(-7,784∙10^4) J_A=4,326∙10^(-5) kmol/(m^2∙s) J_В=-D_(B,m) (∂C_(B,m))/∂x=-1,433∙10^(-9)∙(-0,1545∙10^4) J_B=0,2214∙10^(-5) kmol/(m^2∙s) The total specific flux (total density of the substance) that has penetrated through the separating membrane is: J=J_A+J_B=4,326∙10^(-5)+0,2214∙10^(-5) J=4,5474∙10^(-5) kmol/(m^2∙s) The total flow, taking into account the active surface of the membrane, is: (N_p ) ̇=∫_0^A▒JdA=J∙A=4,5474∙10^(-5)∙117=0,532∙10^(-2) kmol/s Waste flow, according to the material balance equation: (N_R ) ̇=(N_F ) ̇-(N_p ) ̇=0,023-0,00532=0,01768 kmol/s Let's check the composition of the flow that penetrated through the diffusion layer of the membrane: (y_A ) ̃^''=J_A/J=0,95 (kmol А)/(kmol of mixture) Initial composition: (y_F ) ̃=(N_R ) ̇/(N_F ) ̇ (y_R ) ̃+(N_P ) ̇/(N_F ) ̇ (y_P ) ̃=0,01768/0,023∙0,59+0,00523/0,023∙0,95=0,673 (kmol А)/(kmol of mixture) The local dissipative function in the isothermal membrane diffusion process is determined according to relation (1.16): Ψ ̇^V=(J_A ) ⃗∙(-∆(μ_(A,M) ) ⃗ )+(J_B ) ⃗∙(-∆(μ_(B,M) ) ⃗ ) For ideal solutions of the component in the membrane (the value of the activity coefficient γ is equal to unity), under the conditions of the flat geometry of the membrane, we have: (∂μ_A)/∂x=(∂μ_A)/(∂c_(A,M) )∙(∂c_(A,M))/∂x, , where, taking into account the assumptions: (∂μ_A)/(∂c_(A,M) )=(∂ln⁡(γ∙c_(A,M)))/(∂c_(A,M) )=RT/c_(A,M) Let us present the calculated relation for local dissipation and calculate its value for the given boundary conditions (x=0,x=δ): Ψ ̇^V=RT[D_(A,M) ((∂c_(A,M))/∂x)^2 1/c_(A,M) +D_(B,M) ((∂c_(B,M))/∂x)^2 1/c_(B,M) ] Ψ ̇_(x=0)^V=8,314∙279∙[5,558∙10^(-10)∙(-7,784∙10^4 )^2∙1/(23,951∙10^(-3) )+ +1,433∙10^(-9)∙(-0,1545∙10^4 )^2∙1/(0,3174∙10^(-3) )]=3,51∙10^5 kW/m^3 Ψ ̇_(x=δ)^V=8,314∙279∙[5,558∙10^(-10)∙(-7,784∙10^4 )^2∙1/(8,3837∙10^(-3) )+ +1,433∙10^(-9)∙(-0,1545∙10^4 )^2∙1/(0,00841∙10^(-3) )]=18,7525∙10^5 kW/m^3 The local dissipation of the convertible Gibbs energy is proportional to the square of the driving force and inversely proportional to the local concentration of the components; therefore, the largest value of Ψ ̇^V corresponds to the membrane region adjacent to the drainage cavity. The integral value of the dissipative function, referred to the unit area of the membrane, is determined by integrating its local value along the coordinate x : Ψ ̇_(δ=1м^2 )=∫_0^δ▒〖Ψ ̇^V dx〗 For planar geometry, the linear distribution of the concentration of components in the membrane obeys the expression: c_(j,M)=c_(J,M) (x=0)+bx_J, где b=(∂c_j)/∂x=const. Taking into account this assumption, we have: Ψ ̇_(δ=1м^2 )=∫_0^δ▒〖Ψ ̇^V dx〗=RT(∑_(j=1)^2▒{∫_(c_(j,M)^')^(c_(j,M)^'')▒〖D_(j,M)∙((∂c_(j,M))/∂x) (dc_(j,M))/c_(j,M) 〗} )= =RT(∑_(j=1)^2▒{D_(j,M)∙((∂c_(j,M))/∂x)ln (c_(j,M)^'')/(c_(j,M)^' )} ) Ψ ̇_(δ=1м^2 )=RT(∑_(j=1)^2▒{(-J_j )ln (c_(j,M)^'')/(c_(j,M)^' )} ) In the conditions of our task, we assume that j=1 (component A), j=2 (component B): Ψ ̇_(δ=1м^2 )=RT[J_A (-ln (c_(A,M)^'')/(c_(A,M)^' ))+J_B (-ln (c_(B,M)^'')/(c_(B,M)^' ))] Ψ ̇_(δ=1м^2 )=8,314∙279[4,325∙10^(-5) (-ln (8,3837∙10^(-3))/(23,951∙10^(-3) ))+┤ ├ +0,2214∙10^(-5) (-ln (0,00841∙10^(-3))/(0,3174∙10^(-3) ))]= =0,1053+0,00186=0,1239 кВт/м^2 >0 The integral value of the dissipation of convertible energy when passing through a membrane with a contact surface area A=117 м^2 is determined by integrating Ψ ̇_(δ=1м^2 ) over the entire working surface: Ψ ̇^V=∫_0^A▒〖Ψ ̇_(δ=1м^2 ) dA〗=0,1239∙117=14,5 kW The task is devoted to diffusion in a binary gaseous system obeying Fick's law: the intensity of mass transfer of an individual component is proportional to its concentration gradient and does not depend on the potentials of other intensive properties of the system. Often such restrictions in applied matters are justified, despite the fact that engineering applications of diffuse phenomena, in particular in gaseous systems, are very diverse, and therefore require consideration of the inseparable connection between the mechanisms of heat and mass transfer convection in each individual case.

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