Convective transfer of a viscous fluid: еducational manual 9786010429659

Table of contents :
1.2. Fluid Viscosity 7
6.3. Dynamic Boundary Layer Equation 51
6.5. Obtaining an equation for the temperature
boundary layer. 56
7. SELF-SIMILAR REARRANGEMENTS
OF BOUNDARY LAYER EQUATIONS 59
11. MOMENTUM EQUATION FOR THE BOUNDARY LAYER 85
13.1. Dynamic Problem 92
13.2. Application of the Integral Method to the Temperature
Boundary Layer Calculation on a Plate 98
1.2. Liquid Viscosity
In such a jet there is also a steep velocity gradient region - this is the region of the boundary layer that is formed symmetrically on both sides of the jet axis y = 0. The transition of the boundary layer velocity to external flow velocity occurs as...
13.1. Dynamic Problem

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AL-FARABI KAZAKH NATIONAL UNIVERSITY

CONVECTIVE TRANSFER OF A VISCOUS FLUID Educational manual

Almaty «Qazaq university» 2017

UDC 536 (075) LBC 22.317 я 73 C 73 Recommended for publication by the Scientific Council of the Faculty of Physics and Technology and RISO of Al-Farabi Kazakh National University (Protocol №2 dated 03.11.2017) Reviewers: doctor of technical sciences, Professor B.N. Absadykov doctor of technical sciences, Professor V.E. Messerle Authors: A.S. Askarova, S.A. Bolegeneva, V.I. Maximov, S.A. Bolegenova, I.E. Berezovskaya

C 73

Convective transfer of a viscous fluid: еducational manual / A.S. Askarova, S.A. Bolegeneva, V.I. Maximov, [et. al.]. – Almaty: Qazaq university, 2017. – 106 p. ISBN 978-601-04-2965-9 The training manual covers the main issues of heat and mass transfer processes in a viscous fluid and its practical application. First, the manual is intended for the physics students specializing in thermal physics, and may also be of interest to future physicists, chemists, engineers and all those who wish to learn more about the phenomena of heat and mass transfer in a viscous liquid and its applications to a variety of areas of physics and physical and chemical processes technology. Publishing in authorial release.

UDC 536 (075) LBC 22.317 я 73 ISBN 978-601-04-2965-9

© Askarova A.S., Bolegeneva S.A., Maximov V.I., [et. al.], 2017 © Al-Farabi KazNU, 2017

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TABLE OF CONTENTS

INTRODUCTION TO THE VISCOUS FLUID MECHANICS................ 5 1. BASIC DEFINITIONS OF THE VISCOUS FLUID MECHANICS ......... 6 1.1. Fluid .................................................................................................... 6 1.2. Fluid Viscosity .................................................................................... 7 1.3. Fluid Motion ....................................................................................... 10 1.4. Medium Uniformity ............................................................................ 12 1.5. Medium Compressibility..................................................................... 13 2. BASIC EQUATIONS OF VISCOUS FLUID MECHANICS ............... 15 2.1. Substance Conservation Law .............................................................. 15 2.2. Navier-Stokes Equation ...................................................................... 17 2.2.1. Equation of Continuity…. ................................................................ 17 2.2.2. Equation of Motion .......................................................................... 18 2.3. Incompressible Flow with Constant Properties ................................... 23 2.4. Energy Equation. ................................................................................ 25 2.5. Steady Fluid Motion. Divergent Form of Energy Equation ......................................................... 27 3. LAYERED FLOW ................................................................................ 31 3.1. Dynamic Problem ............................................................................... 31 3.2. Heat Problem. ..................................................................................... 32 4. PLANE POISEUELLE FLOW .............................................................. 33 4.1. Dynamic Problem ............................................................................... 33 4.2. Heat Problem ...................................................................................... 36 5. COUETTE FLOW ................................................................................. 41 5.1. Dynamic Problem ............................................................................... 41 5.2. Heat Problem ...................................................................................... 43 6. BOUNDARY LAYER........................................................................... 47 6.1. Dynamic Boundary Layer Properties .................................................. 48 6.2. Dynamic Boundary Layer Thickness .................................................. 50 6.3. Dynamic Boundary Layer Equation .................................................... 51 6.4. Thermal Boundary Layer Properties ................................................... 55 6.5. Obtaining an equation for the temperature boundary layer. .......................................................................................... 56 7. SELF-SIMILAR REARRANGEMENTS OF BOUNDARY LAYER EQUATIONS ................................................. 59 3

7.1. Dynamic Problem ............................................................................... 60 7.2. Heat Problem ...................................................................................... 63 8. PAST-PLATE LONG STREAMLINING WITH A UNIFORM INCOMPRESSIBLE VISCOUS FLUID FLOW. BLASIUS PROBLEM ............................................................................... 64 9. ITERATION METHOD TO SOLVE BLASIUS PROBLEM ............... 73 10. ANALOGY OF THE MOMENTUM AND HEAT TRANSFER PROCESSES ................................................... 82 11. MOMENTUM EQUATION FOR THE BOUNDARY LAYER ......... 85 12. ENERGY EQUATION. ENERGY THEOREM FOR THE BOUNDARY LAYER ............................................................. 88 13. INTEGRAL METHOD ....................................................................... 92 13.1. Dynamic Problem ............................................................................. 92 13.2. Application of the Integral Method to the Temperature Boundary Layer Calculation on a Plate ..................................................... 98 REFERENCES .......................................................................................... 104

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INTRODUCTION TO THE VISCOUS FLUID MECHANICS

Classical fluid dynamics studies the motion of an ideal fluid. Separate layers of such a liquid move without friction. Only normal pressure forces operate between them, while there are no tangential forces (shear stresses). The motion of such a liquid is described by a set of Euler equations. Much of the theoretical research on fluid motion is based on the assumption that the fluid is ideal. This allows us to consider such phenomena as flow past bodies, fluid overflow through channels, as well as applied problems. However, since in the ideal fluid bodies do not experience resistance (the D'Alembert paradox), for these liquids, it is impossible to calculate resistance of a body moving in it, it is impossible to explain the heating of a fluid or gas due to mechanical energy dissipation, etc. To explain these phenomena, it is necessary to abandon the ideal fluid model. In actual real fluids, both normal and tangential forces operate between its separate layers. The same forces operate between the fluid and the surface of the streamlined body. In a real fluid, the presence of tangential forces results in the fact that they are transferred to the wall of the body being streamlined. This leads to the fluid adhesion to this wall. In an ideal fluid, the absence of tangential forces ensures the existence of a difference in the velocities of the fluid and the body. As a result, the fluid slides along the wall, and does not adhere to it. Taking any additional property of a real fluid into account does not come freely; therefore, where the viscosity effect can be neglected, it seems reasonable to use a simpler model of an ideal fluid. In real liquids, internal friction and viscosity can hardly be neglected; most of the interesting things in a fluid behavior have something to do with this property. You can see that the fan blades are collecting a thin layer of dust. Dust is not blown off, because air velocity relative thereto, measured directly on the surface, is zero. The theory must take into account the fact that in all ordinary liquids molecules near the surface have a zero velocity (relative to the surface itself). 5

1. BASIC DEFINITIONS OF THE VISCOUS FLUID MECHANICS

1.1. Fluid – it’s a body, which has the property of fluidity, – light motion, capable of changing its shape under the influence of external forces and temperature changes. – state of aggregation of matter – intermediate between the solid and gaseous states. The fluid, while retaining certain features of both a solid body and a gas, has, however, a number of its own inherent features, of which fluidity is the most typical. – Like a solid body, the liquid retains its volume, has a free surface, has a certain tensile strength under uniform extension, etc. On the other hand, a sufficient amount of fluid takes the form of the vessel containing it. The possibility in principle of a continuous fluid transition into a gas also indicates the proximity of the liquid and gaseous states. – Ideal – non-viscous (perfect) fluids have the following properties: absolute mobility, i.e., absence of friction forces and tangential stresses; absolute invariance in the volume under the influence of external forces. – Real – viscous (real) fluids have: compressibility, resistance, tensile and shearing forces, viscosity, i.e., the onset internal friction between particles in motion. There is no ideal fluid in nature; it is a model of a real liquid. A fluid in a state of equilibrium or motion is subjected to mass and surface forces. – Mass or volume forces are proportional to the mass, and for a homogeneous fluid – to the volume, too. These forces include: own fluid weight, inertia forces and centrifugal forces. – Surface forces are proportional to the area of the surface they effect. These forces are: a) normal forces to the fluid surface (compressive, pressure, and tensile forces); b) tangential (frictional forces that only originate when the fluid moves). 6

1.2. Liquid Viscosity Viscosity is the property of a fluid to resist the movement of its particles that characterizes the degree of its fluidity and mobility. Viscosity is explained by occurrence of internal friction motion between particles. The internal friction forces are directed along the surface of the contacting layers and depend on their relative velocities. As can be seen from the definition, viscosity is a property of a moving fluid and is not detected at rest. The presence of internal friction forces of a moving fluid was first established by Newton; Later, the Russian scientist V.N. Petrov in 1888 gave a mathematical expression for the friction force. The viscous friction forces operating between moving fluid particles can radically affect both distribution of velocities in the fluid flow and streamlining of bodies placed in a moving stream. Newton established experimentally that when two parallel planes slide relative to each other, with the space between them filled with fluid, the viscous friction forces prevent this sliding (Fig. 1). It is important to note that the fluid particles adjacent to the upper plate move with it at a U0 velocity (they are entrained by the plate). On the contrary, fluid particles near the bottom (fixed) plate are at rest (they adhere to the plate).

Figure 1.

If we mentally divide fluid into parallel plane layers moving uniformly, then it is not difficult to understand that each overlying layer entrains the lower adjacent layer. In turn, this lower layer inhibits movement of the upper layer. Due to the chaotic motion between adjacent layers, a continuous exchange of particles takes place. Moreover, molecules from layers moving at higher velocities transmit momentum to slower layers. 7

Each layer is affected from above and from below by two equal, but opposite forces. Velocity of the layers increases linearly with their height (Fig. 2), and friction force is transmitted from one layer to the next. As a result, the force applied to the upper plate is transferred to the lower plate. Let the lower plate be stationary, and the upper one move with U0 velocity. The distance between plates is h. Let the pressure in the entire space occupied by fluid be constant. Experience shows that the fluid adheres to both plates. Consequently, the velocity of the fluid on the lower plate is zero, while on the upper plate it is U0. In the space between the plates, there is a linear velocity distribution (Fig. 3), i.e. the flow velocity is proportional to the distance y from the lower plate:

y u( y ) = U 0 . h

Figure 2.

Figure 3. 8

(1)

In order to explain this fluid behavior, it must be assumed that a tangential force must be applied to the fluid from the upper plate in the direction of motion, balancing the friction force of the fluid. Experience shows that this force is proportional to the upper plate motion velocity – U0 and is inversely proportional to the distance h between the plates: U τ~ 0 . h

(2)

du ( y ) U 0 = and taking (1) and (3) into dy h account, we can write the following: It follows from (1):

τ =µ

du , dy

(3)

(3) is Newton's friction law. The µ factor depends on the nature of a fluid. It is a physical characteristic and is called viscosity. A distinction I made between dynamic and kinematic viscosity. – Dynamic viscosity – μ is determined from the formula (3):

µ =τ

dy . du

In the SI system, viscosity units are 1 N·c/m2 or 1 Pa·s. – Kinematic viscosity – v. Hydraulic calculations apply viscosity related to density:

v=

µ , ρ

where ρ is density. Dimension v in the SI system: [ν] = 1m2/s. Fluid viscosity decreases with increasing temperature, while gases viscosity increases. Fluid viscosity depends on pressure. At a 9

pressure of up to 2·107 Pa, change in the viscosity of water is insignificant and is often not taken into account in calculations. 1.3. Fluid Motion In real fluids, it adheres to a hard surface, i.e. there is a slowdown of a thin layer of fluid near the surface. In this layer, the velocity of the fluid leaps from zero on the fixed surface to U ∞ – the oncoming flow velocity. This thin wall region is called the boundary layer. The velocity gradient here is high, so it is necessary to consider friction forces. The region outside the boundary layer is the external flow region. Here the friction forces are small and can be ignored. This flow region can be regarded as an ideal fluid flow region and can be used to describe its motion by Euler equations for an ideal fluid. Experience reveals two completely different patterns of fluid motion. At low speeds, a calm, layered flow is observed, which is called a laminar flow. At high velocities, the flow becomes chaotic, and particles and individual regions of the fluid move in a disorderly fashion, spinning into vortices; such a flow is called turbulent. Transition from laminar flow to turbulent flow and back takes place at a certain fluid velocity and also depends on the viscosity and density of the fluid and the characteristic dimension of the body streamlined by the fluid. It is still not clear whether vortices originate from the very beginning and simply have very small dimensions that are invisible to us, or vortices arise from a certain velocity of fluid motion. Laminar flow (Latin Lamina – strip) is an ordered flow of viscous fluid or gas, characterized by an absence of mixing between adjacent layers. The existence of a laminar flow is only possible up to a certain, critical value of the Reynolds number, after which the flow passes into turbulent flow. Turbulent flow (Latin Turbulentus – disorderly) is the flow of a fluid or gas, in which particles perform disordered, chaotic movements along complex trajectories, and the velocity, temperature, pressure, and density of the medium undergo chaotic changes. The turbulent flow differs from the laminar flow by intensive mixing, heat transfer, 10

large values of the friction coefficient, etc. Turbulent motion is accompanied by mixing of layers and generation of vortices. A dimensionless combination of values characterizing the motion of a body in a viscous medium or a flow of viscous fluids (gases) is called the Reynolds number:

Re =

ρυ d , µ

where υ is the characteristic velocity of the fluid or gas flow (or body velocity relative to the medium), d is the characteristic linear dimension (of a body or a flow), ρ is the medium density, and μ is the dynamic viscosity of fluid or gas. At low Reynolds numbers, the flow of a fluid or a gas is laminar, at large it is turbulent. The boundary is the critical Reynolds number, but this boundary is not strict. As soon as flow velocity increases so much that the Reynolds number exceeds the critical value, the laminar flow becomes turbulent. For example, for the flow of a viscous incompressible fluid along a circular cylindrical tube (in this case, d is the diameter of the tube, and υ is the average flow velocity over cross section of the tube) Reкр = 2300. – Laminar Flow: Re > Reкр. Physical meaning of the Re number is:

Re =

inertia force . friction force

Let us consider a simple problem: a fluid flow along a flat solid surface. While far from the wall the Re number is high (d is big, or υ is big), the fluid motion can be described using the ideal fluid model. On the other hand, it is clear that the wall slows down the fluid flow and, if we accept the boundary condition of adhesion, then the flow velocity on the surface must be zero. The Re number near the wall is thus small, and the ideal fluid model ceases to work. In this case, a viscous fluid model must be used. 11

Figure 4.

1.4. Medium Continuity In our course, we are going to study the motion of fluids and gases in the approximation, when they can be considered as continuous media, i.e. a medium, continuously filling the flow space. To solve mathematical problems associated with the calculation of the motion of various objects (airplanes, rockets, ships, etc.) in air or water, with the study of wave processes in liquids and gases, with their flows through pipes and channels, etc., we need a mathematical tool that describes these phenomena. This tool is the equations of the viscous liquids mechanics, which are based on the hypothesis of medium continuity, i.e. the hypothesis that fluid or gas particles continuously fill the part of the physical space they occupy. The natural question arises: under what assumptions is this hypothesis valid? While for fluids (water, liquid metals, etc.) this hypothesis is more or less obvious, for sufficiently low density gases (for example, those occupying outer space, including atmospheres of stars, planets and the Sun), which consist of individual atoms or molecules, as well as for other physical objects, it requires its support. For example, when calculating inhibition of artificial earth satellites, it is not possible to use the mathematical tool of the viscous fluids mechanics, while it is this tool that is used in calculating the inhibition of cosmic objects entering dense atmospheres of the Earth and planets (for example, meteorites or Earth-returned space-crafts, etc.). The hypothesis of medium continuity is valid, in particular, in cases where the characteristic dimension of a streamlined body L (for example, the radius of a 12

spherical satellite) is much larger than the mean free path of atoms or gas molecules l, i.e. the length between successive collisions thereof. – A continuous variable medium is a concept that applies when the molecular structure of a medium can be neglected when studying the motion of a variable medium (in our case a liquid). The criterion here is the so-called Knudsen number K, which is defined as the ratio of the mean molecular free path λ in the substance under consideration to the spatial scale of the studied phenomenon L:

K=

λ

L

, where ( λ ∼

1

ρ

).

As a characteristic dimension, we can consider the length, thickness, and diameter of a streamlined body, in some problems it can be the thickness of the boundary layer. The available theoretical and experimental data indicate that for very small values of the Knudsen number (0,015 – Hyper-sonic. We shall consider laminar flows, although the theory of turbulent flows has the main significance for numerous engineering applications. The study of laminar flow is of considerable interest, both independent and as original models of turbulent flows. Analysis of laminar flows, performed on the basis of physically rigorous boundary layer equations, makes it possible in a "pure form" to reveal the qualitative side of the influence of various effects on the turbulent flow occurring under similar conditions. In many cases, the equations of asymptotic turbulent boundary layer are reduced by corresponding transformation of variables to similar equations for laminar flows. This allows us to use solutions known from the theory of a laminated boundary layer for turbulent flows.

14

2. BASIC EQUATIONS OF VISCOUS FLUID MECHANICS

The equations of viscous fluid mechanics are based on several fundamental conservation laws. These laws are reliable, and the variables in these equations are mass, momentum and energy (m, mυ, E). These laws form the basis for the equations of viscous fluid dynamics – the Navier-Stokes equations. The Navier-Stokes equations form the basis of all hydrodynamics. For compressible flows, it is necessary to add the equation of state, and for nonisothermal flows, the viscosity dependence on the temperature: μ = μ(Т). 2.1. Substance Conservation Law (N) Let N be a substance, ρN is the density of the substance. We select in a continuous medium a fixed volume V bounded by the surface S (Fig. 5).

Figure 5.

The amount of substance N in the volume V: ∫ ρ N dV . For a V

unit of time, the amount of substance will change by the value:

∂ ∫ ρ N dV . ∂t V 15

The change of substance in the allocated volume can occur due to its inflow or outflow, i.e. due to the substance flow through the surface S. To calculate this flow of substance through the surface S,    we divide it into dS surface elements. Then, dS = dS ⋅ n , where n is → the unit vector directed along the normal line to the dS area. Let Р N be the substance flow density per unit time through the surface S. The amount of substance flowing out of the V volume through the S →  surface equals − ∫ P dS . N S Substance may not only flow out of the volume, but also appear inside it (for example, energy and mass due to chemical reactions or phase transitions). Let qN be the density of the emerging substance in a unit of time, then the law of conservation of a certain substance in integral form can be written as follows:

 ∂ ∫ ρ N dV = − ∫ PN dS + ∫ q N dV . V S ∂t V

(4)

Change, outflow, occurrence in s volume with time through the surface. →





Using the Ostrogradskii-Gauss theorem ∫ P dS = ∫ divP dV , we N N S

V

write equation (3) in a differential form, and then (3) takes the following form:

 ∂ ∫ ρ N dV = − ∫ divPN dV + ∫ q N dV , ∂t V V V or ∫(

V

∂ρ N ∂t

 + divPN − q N )dV = 0 . 16

Since the volume is chosen arbitrarily, this equality is satisfied under the condition that the integrand element is zero:

∂ρ N ∂t

 + divPN − q N = 0 .

(5)

(5) – The substance conservation law in a differential form. 2.2. The Navier-Stokes Equations The Navier-Stokes equations are a set of equations that includes the continuity equation (the law of conservation of mass) and the motion equation (the law of conservation of momentum) for the velocity components u, υ, w. 2.2.1. Equation of Continuity Let the substance N be a mass, a scalar quantity, then ρ N = ρ is density, scalar quantity. – the amount of substance occurring in a unit of volume per →   unit time. PN substance flow density vector: PN = ρυ . Then (5) will be written as follows:

∂ρ ∂t

 + div( ρυ ) = q N .

(6)

(6) is the law of conservation of matter or the equation of continuity. For a homogeneous medium, when there are no sources: qN = 0. Then from (6) we have the following:

∂ρ ∂t

 + div ( ρυ ) = 0

is the equation of continuity in the absence of matter sources. 17

Let us denote: 3 ∂υ ∂υ y ∂υ z  ∂υ k = ∂υ k + = ∑ divυ = x + . ∂z ∂xk ∂x ∂y k =1 ∂xk

Then we will have the following:

∂ρ ∂ρυ k + =0. ∂t ∂xk

For an incompressible fluid ρ = const and the following can be written:  ∂υ k = 0 is the continuity equation for an divυ = 0 or ∂xk incompressible fluid. In the Cartesian coordinate system, this equation has the form:

∂υ x ∂υ y ∂υ z ∂u ∂υ ∂w + + =0, + + =0 . ∂x ∂y ∂z ∂x ∂y ∂z For a stationary flow we have the following:

∂ρ = 0, ∂t then the continuity equation (6) takes the following form:  div( ρυ ) = q N . 2.2.2. Motion Equations Let us write down the law of conservation of substance (5) in a general form:

 ∂ρ N + divPN − q N = 0 . ∂t 18

→

→

Let substance N be a momentum, i.e. N = m υ .  Then ρ N = ρυ is the momentum density, q N – the amount of momentum appearing per unit volume per unit time. Here q N is the vector quantity, and

 PN

is the second rank tensor. A 2nd rank tensor

is a plurality of 9 quantities, composed as all possible products of   components of 2 real vectors a and b :

 a1b1   a2 b1 a b  3 1

a1b2 a2 b2 a3b2

a1b3   a2 b3  = aibk = Tik , a3b3 

i=1,2,3 is the line number, and k = 1,2,3 is the column number. An example of a second-rank tensor is the Kronecker symbol (tensor unit) δ ik :

1 0 0   δ ik =  0 1 0  . 0 0 1  

0 i ≠ k δ ik =  , 1 i = k

In contrast to the discrete points dynamics, in the dynamics of continuous media, we are dealing not with the forces themselves, but  with the density of their distribution in a space, i.e. ∑ Fρi is the sum of all the forces per unit volume. The fluid particles are affected by volume and surface forces. →

a) q N – volume forces; q N = Fi



→

b) divPN – surface forces. PN = Pik is the momentum flow density per unit time, the second-rank tensor, which determines the surface forces (pressure p and friction force τ). The flow is created due to the convective transfer of substance N through the surface S and due to the effect of surface forces: 19

Pik = ρυiυ k − σ ik .

(7)

Here σ ik is the momentum flow density due to surface forces, while ρυiυ k is the momentum flow density due to the convective fluid transfer through the surface S. σ ik is the stress tensor in a viscous medium, which takes into account the forces of pressure and friction. Since pressure force is directed to the surface, and friction reduces (inhibits) the flow, σ ik is taken with a minus sign. We write this tensor as follows:

 σ xx  σ ik =  σ yx   σ zx

σ xy σ xz   σ yy σ yz  .  σ zy σ zz 

This tensor includes surface forces that can be directed both perpendicular to the surface and tangentially thereto.

z 0

х

y Figure 6.

σ xx is a component perpendicular to the area element, which is perpendicular to the x-axis (Fig. 6). It is directed along the x axis and is a normal force, i.e. pressure force. Similarly with σ yy and σ zz . 20

σ xy is the component directed along the y axis to the area perpendicular to the x axis, it is tangent to the surface and is the frictional force. Similarly for σ xz , σ yz , σ yх , σ zх , σ zу . If there is no friction, the tangential forces equal zero. Consequently, in a liquid without friction . Only normal forces σ xy = σ xz = σ yz = σ yх = σ zх = σ zу = 0 remain, acting perpendicularly to the selected area elements. The negative value of any of the normal voltages is called the fluid pressure: σ xx = σ yy = σ zz = − p . The minus sign takes into account the fact that the pressure is always directed inside the selected volume of fluid, and the normal line to the surface – outwardly. The fluid pressure is equal to the arithmetic mean of the normal stresses, taken with the minus sign:

1 p = − ( σ xx + σ yy + σ zz ) . 3 Now we find the tangential stresses. To do this, we isolate the pressure from the stress tensor in a viscous medium, and denote the remainder by σ ik′ – the viscous stress tensor, i.e., this is the impulse flow density caused by frictional forces: σ ik = − pδ ik + σ ik′ , δik – is the tensor unit The

σ ik′

viscous stress tensor is due to frictional forces and its

effect is only manifested when there is a relative movement of the fluid layers, i.e. the fluid layers must have different velocities. To create a momentum flow, there must be a velocity gradient. It follows from the Newton's friction law (3) that the viscosity forces depend not on the velocity υ, but rather on the velocity gradient grad υ. The most general form of the viscous stress tensor is presented below: 21

′ = a( σ ik

∂υ j ∂υi ∂υ k + ) + bδ ik . ∂xk ∂xi ∂x j



Here a, b are coefficients that do not depend on υ and  gradυ . Otherwise, this equation would be nonlinear. a and b may depend on the properties of the medium, on its thermodynamic characteristics (p, T). Landau called a and b the viscosity coefficients. For them the following is accepted: a = µ , 2 b=− µ. 3 Then the viscous stress tensor is:

 ∂υi ∂υ k 2 ∂υ j  . + − δ ik  ∂xk  x 3 x ∂ ∂ i j  

′ = µ σ ik

Thus, inserting this equation in (7), we obtain the following:

 ∂υ ∂υ j  ∂υ 2  + pδ ik . Pik = ρυiυ k − µ  i + k − δ ik  ∂xk  ∂ ∂ x 3 x i j   Then the substance conservation law (5) takes the following form:

 ∂υi ∂υk ∂ ∂  ( ρυi ) + +  ρυkυi + pδ ik − µ  ∂t ∂xk   ∂xk ∂xi 2 ∂υ j   Fρi . − δ ik  = 3 ∂x j   Taking into account the continuity equation, we obtain the following: 22

ρ

∂υi ∂t

∂ − ∂xk

+ ρυk

∂υi ∂p + − ∂xk ∂xi

  ∂υ ∂υ 2 ∂υ j   i + k − δ ik Fρi .  =  µ  ∂x ∂xi 3 ∂x j     k

(8)

Equations (6) and (8) together represent a Navier-Stokes motion equation system, which describes any motion of a viscous fluid. 2.3. The flow of an incompressible fluid with constant properties For a fluid with constant physical characteristics:

µ = const ; ρ = const . For an incompressible fluid the equation of continuity looks as follows:  ∂υ k =0. divυ = 0 or ∂xk Then equation (8) can be written as follows:

     ∂υ ρ + ρ (υ ∇ )υ = −∇p + µ∆υ + F . ∂t

(9)

This is the law of conservation of momentum (or the equation of motion) in a vector form for an incompressible fluid with constant properties. For example, for a two-dimensional flow in projections onto the reference frame, equation (9) splits into 2 scalar equations.

∂u ∂u ∂u + ρu + ρu = ∂t ∂x ∂y i=1; k=1,2.  ∂ 2u ∂ 2u  ∂p = − + µ  2 + 2  + Fx ∂x ∂y   ∂x

ρ

23

ρ

∂υ ∂υ ∂υ + ρυ + ρυ = ∂t ∂x ∂y

 ∂ 2υ ∂ 2υ  ∂p = − + µ  2 + 2  + Fy ∂y ∂y   ∂x

i=1; k=1,2.

(10)

The equation of continuity in coordinate form:

∂u ∂υ + =0. ∂x ∂y This system of equations contains 3 equations and 3 unknown quantities: u, υ, p. If fluid is compressible, with a variable viscosity, then μ ≠ const, ρ ≠ const, in this case the variables will be 5: u, υ, p, μ, ρ. To determine μ and ρ, it is necessary to add the equation of state: p = p( ρ ,T ) and the viscosity dependence of the temperature: µ = µ( T ) . (8) is a nonlinear system of partial differential second order equations. To solve it, you need to know the initial and boundary conditions, which are set depending on the type of a particular problem. Usability conditions for the Navier-Stokes equations. The solution of the Navier-Stokes equations even for an incompressible fluid is a very complicated problem. Up to now it has been possible to solve these equations only for some simplest cases of motion. For example, for the viscous fluid flow in a straight pipe, it is the Poiseuille problem; for the flow between two parallel planar plates, one of which is fixed and the other one is moving at a constant speed – the Couette problem, and for the flow near the critical point – the Hiemenz-Howarth problem, etc. The problems of viscous fluid hydrodynamics are usually solved approximately by dropping some terms in the Navier-Stokes equations, which under certain specific conditions can be small in comparison with other terms. 24

2.4. Energy Equation If the problem is not isothermal, then the equation of conservation of energy must be solved. Let the substance N in the equation (5) be the energy. Then ρN is the energy density, i.e. Energy per unit volume of liquid. a) As the medium moves, it has kinetic energy ρ

υ2 2

.

b) Since the medium is continuous, it has an internal energy

ρε , then:

ρN = ρ

υ2 2

+ ρε = ρU ,

υ2

+ ε is the specific energy, i.e. the energy per unit 2 mass of the medium. q N – is the rate of onset (occurrence) of energy (the amount of energy arising per unit time in a unit of volume). For example, the heat that occurs during chemical reactions, due to phase transitions,

where U =

→ transformations or due to combustion. P is the energy flow N density. Let us consider the causes of energy flow density. – Due to fluid motion - convective energy transfer: ρUυ k . – Work of pressure forces (compression, underpressure) per unit volume: pυ k .

′ υ k ; σ ik ′ is the viscous – Work of friction (viscosity) forces: σ ik stresses tensor. ∂T – the – Heat transfer due to thermal conductivity: − λ ∂x j Fourier’s law. This is the heat flow per unit surface per time unit, λ is thermal conductivity coefficient; U0 . h

– If the medium is conductive, then it is necessary to take the flow of electromagnetic energy into account (Umov25

Poynting vector). We are considering a non-conducting medium. Then:

′ υk − λ PN = ρUυ k + ρυ k − σ ik

∂T . ∂x j

We substitute the designation of quantities in the law of conservation of substance (5):

∂ ( ρU ) = ∂t ∂  ∂T  = −  ρUυk + pυk − σ ik′ υk − λ  + qN . ∂xk  ∂x j 

(11)

The heat supplied by friction and heat conductivity goes towards the change in internal energy and the change in volume. We transform this equation. To do this, we introduce the enthalpy i and the total enthalpy i0: i = CpT = (CV+R)T, i0 = i +

i =ε +

p

ρ

υ2 2

υ p + i − = i0 ρ 2 p U = i0 − . ρ

, U=

2

−

p

ρ

;

The total enthalpy i0, or stagnation enthalpy arises in problems with high velocities. When the fluid moves, i and υ change. If the velocity in some place is zero, then i = i0. We substitute the expressions for υ and σ’ik into equation (11) and taking into account the continuity equation (6), we introduce the v Prandtl number Pr = , where А is temperature conductivity a coefficient: 26

a=

λ λ = 1 . or ρс p µс p Pr

For gases Pr ≈ 1; for air Pr = 0,72. Then we obtain the following equaiton:

  2 i  diî ∂p + divµ ∇υ + ρ = − dt ∂t Pr  

   2   − υ ⋅rotυ  − δ ik υdivυ  + qN 3 

(12)

.

(12) is the energy equation. It is linear relatively to

i = с pT . 2.5. Stationary fluid motion. Divergent form of energy equation We write the complete set of equations:

∂υi ∂υi + ρυ = − ∂p − k ∂x ∂t ∂xi k   ∂υ ∂υ j  ∂υ   − ∂  µ  i + k  − 2 µδ +F ; ∂x   ∂x ∂xi  3 ik ∂x j  i k  k   ∂ρ  + div(ρυ ) = 0

ρ

∂t

dio ∂p ∂    2 i   ρ = µ ∇ υ +  − + dt ∂t ∂x    Pr   k

  2   − υ ⋅rotυ  − δ ik υdivυ  + qqNN. . 3  27

(13)

The equation of continuity for incompressible fluid looks as  follows: divυ = 0 . The motion equation in the stationary case is written as follows:

ρυk

∂υi ∂ 2υ ∂p = − + µ 2i + Fi ∂xk ∂xi ∂xk

or in the vector form:

( )

 ρ υ∇ υ = −∇p + µ∆υ + F . Then the energy equation is rewritten as follows:

[

( )

]

  i   N. . div  ρi0 υ − µ∇υ 2 +  + µ υ ⋅ rotυ  = qqN Pr     Since total inhibition enthalpy is i0 = i + υ 2 2 we have the following: 2    υ  (14) div  ρυ  i +  − 2        i   − µ∇υ 2 + + µ υ ⋅rotυ   = qqNN. Pr    (14) is the divergent form of energy equation for stationary flow of an incompressible fluid. We have 5 equations and 8 unknown quantities: u ,υ , w, p , ρ , µ ,T , λ . We need to add 3 more equations: 1) Thermodynamic equation of the state of the medium under consideration. For a gas (an ideal fluid), the Boyle Marriott equation can be used: 28

pV = m RT ; М

2) Dependence µ = µ (T ) For gases: µ = сT n , 0 ≤ n ≤ 1 ; For fluids: −b( T −T0 ) µ µ 1 =e , . = µ0 1 + R( T − T0 ) µ0

3) Relationship between μ and λ.

µс p µ ρ Pr = v = = = 0 ,72 , a λ ρ λ ρс p

a=

λ ρC p

,

for gases с p = const over a wide range of temperatures. If ρ = const ; µ = const ; λ = const , the number of unknown quantities is 5 and the number of equations is 5. To determine the integration constants, it is necessary to indicate the boundary and initial conditions. For example, conditions for velocity and temperature when surface is streamlined by a fluid, may look like this (Fig. 7): u = 0 , T = Tw with y = 0 ; u = u∞ , T = T∞ with y → ∞ .

Figure 7. 29

These are boundary conditions of first kind. For a temperature on the wall, boundary conditions of second kind can be set when the wall temperature is found from the condition of the thermal balance on the surface:

q w = −λ qw must be known. If the wall is adiabatic, q w = 0.

30

∂T w, ∂y

3. LAYERED FLOWS

It is difficult to obtain an exact solution of the Navier-Stokes equations because of nonlinearity of these equations (they contain 2nd order terms with respect to velocity). In cases where the quadratic terms disappear, the problem can be solved analytically. Examples of such a case are layered flows. A laminated flow is a motion where the fluid moves in layers parallel to each other. In this case, the velocity only has one component, while the other two are equal to zero:

υ ( u ,0 ,0 ) ; υ x = u , υ y = υ z = 0 ; u = u( x , y , z , t ) . Let us consider the stationary flow of an incompressible viscous fluid with constant properties.  Incompressible fluid: ρ = const , hence divν = 0 . The fluid has permanent properties: µ = const . This can be taken if gradT is not high. There are no energy sources: qN = 0. There are no volume forces: Fρi = 0. 3.1. Dynamic Problem We write down the Navier-Stokes equations for the case under consideration:

ρ

∂u ∂u ∂u ∂u + ρu + ρυ + ρw = ∂t ∂x ∂y ∂z

 ∂ 2u ∂ 2u ∂ 2u  ∂p = − +µ 2 + 2 + 2   ∂x ∂x ∂y ∂z  

∂p = 0 , ∂p = 0 , therefore, р = р(х). ∂y ∂z 31

∂u ∂υ ∂w + + = 0 , ∂u = 0 . ∂x ∂y ∂z ∂x Motion Equation takes on the following form:

 ∂ 2u ∂ 2u ∂ 2u  ∂p = µ  2 + 2 + 2  or  ∂x ∂x ∂y ∂z  

∂p = ∇p , we have: ∂x

∇p = µ∆u .

(15)

(15) is the equation of motion for a layered flow. 3.2. Heat Problem We will write the energy equation in a divergent form:

   υ 2  div  ρυ  i + −  2       i   − µ∇  υ 2 + 0.  + µ υ ⋅ rotυ   = Pr    Taking into account the continuity equation and the fact that the flow is plane-parallel, we have the following: 2

1 d 2i d 2u  du  +u +   = 0 . Pr dy 2 dy 2  dy  (14) is the energy equation for a layered flow.

32

(16)

4. PLANE POISEUELLE FLOW

Let us consider the layered flow of a viscous incompressible fluid in a flat channel formed by two infinite parallel plates (Fig. 8). The fluid comes into motion under the pressure drop. ∂p Let the distance between the walls be 2b. ≠ 0 , p1 ≠ p2 . ∂x We set this quantity, since the fluid is not compressible, then we can separate dynamic and thermal problems. y

Figure 8.

4.1. Dynamic Problem We will write the motion equation:

dp d 2u dp dp = =0. =µ 2 ; dy dz dx dy To solve the equation of motion, it is necessary to set the boundary conditions that allow to determine the integration constants. As boundary conditions, we take adhesion conditions of fluid to the channel walls: 33

u = 0 with y = +b u = 0 with y = −b . Since both parts of the equation of motion depend on different variables, they are both constant, i.e. dp = const = А ; p = Ax + C dx – the pressure varies linearly along the x axis. We integrate the equation of motion:

d 2u 1 dp . = dy 2 µ dx

(17)

From the boundary conditions we define the integration constants and obtain the following equation for the velocity:

y 2  . dp 2  b 1 − u=− 1 2 µ dx  b 2 

(18)





(18) is the velocity profile for the layered Poiseuille flow. We have a parabolic velocity distribution. If y=0, u( 0 ) = u max , for

the

maximum 1 dp 2 u max = − b .

∂u = 0 is the symmetry condition, then ∂y

velocity

we

have

the

following:

2 µ dx

Therefore:

 u = u max  1 −  

y 2  . b 2 

y2 = 1 − 2 is a dimensionless velocity profile. umax b The dimensionless velocity profile for the layered motion of a fluid in a plane channel is independent of the viscosity value and the longitudinal pressure gradient and is a parabola. Here

u

34

Liquid flow rate Q is the amount of fluid flowing every second through the cross section of the channel. Let's find Q .

+b +b  y 2  4 Q = ∫ udy = u max ∫  1 − dy = 3 bu max 2   −b −b  b  or

Q=−

2 b 3 dp . 3 µ dx

The solution obtained is valid for any Re grad u, and on the axis in the center of the channel, on the contrary. 38

Figure 10.

Let us find the heat flow from the upper wall to the fluid:

q = −λ

dT 4 u2 ; q = µ max . dy y =0 3 b

(22)

b) Let us consider other boundary conditions: y = −b : i = i1 ;

y = b : i = i2 ;

i2>i1.

Let us find constants С3 and С4:

C3 = −

i +i 1 i1 − i2 2 ; C4 = 1 2 + Pr u max 2 3 2b

Then:  4 Pr 2  y  i = − um 1− − 3  b4    i1 − i2 i1 + i2 y+ − . 2b 2 39

(23)

с) Let the lower wall be thermally non-conducting: y = b i = i1 , dT dT di y = −b , q = −λ =0; = =0, dy dy dy

Then C3 = −

5 u2 4 2 and solution Pr max ; C4 = i1 + Pr u max 3 3 b

shall take on the following form:

i = i1 −

with y = 0 : i = i0 = i1 + with y = −b :

Pr 2 u max 3

y  + 4 − 5  . b 

2 5 Pr u max 3

i = iw = i1 +

Then: i = i0 = i1 +

 y4   b4 

8 2 Pr umax . 3

5 2 Pr u max . 3

Figure 11.

40

5. COUETTE FLOW

Let us consider the flow between two parallel plane plates, one of which is at rest and the other moves at a constant velocity U. This flow is called the Couette flow (Fig. 12).

Figure 12.

Let the distance between the walls be h. In this case, the flow is

layered, therefore: υ (u ,0 ,0 ) . 5.1. Dynamic Problem

We write the equation of motion for a layered flow:

dp d 2u =µ 2 , dx dy with the following boundary conditions: with y =0, u =0; Let

u=

us

integrate

the

2

при y =h, u =U. equation

of

(24) motion:

1 dp y + C1 y + C2 and using boundary conditions define the µ dx 2

integration constants C1 and C2: 41

С2 =0, С1 =

U 1 dp − h . h 2 µ dx

Inserting the obtained values of the constant quantities in equation (24), we obtain:

u=

U h 2 dp y  y y− 1 −  . h 2 µ dx h  h

(25)

(25) is the velocity profile in the Couette flow. The velocity profiles for different pressure drop values look different (Fig. 13). y dp А) with =0 ⇒ u= U . h dx The profile is determined by the velocity of the upper plate (straight line 1). The shape of other profiles is determined by the dp pressure gradient . dx

Figure 13.

b) For dp < 0 , i.e. р2 < р1 – are the freehand curves 2, 3, 4. Here dx the velocity is everywhere positive across the entire width of the channel. 42

c) dp > 0 , i.e. р2 > р1 – are the freehand curves 5, 6. dx With a positive pressure gradient, in some area of the channel, a return flow of fluid is possible. This can be explained by the fact that some particles of the fluid that are near the bottom stationary plate are not able to overcome the pressure drop operating in the direction opposite to the motion of the upper plate. The results of the Couette flow calculation are used in the hydrodynamic theory of lubrication. Let us find the flow rate through the channel cross-section for dp =0: the case dx h

hU

0

0h

Q = ∫ udy = ∫

ydy =

Uh . 2

Viscosity does not affect the flow rate. Consequently, for air or glycerin, the volume that is entrained by the current will be the same. dp =0: Let us find friction on the lower plate for the case dx

τw = µ

du U y =0 = µ . dy h

5.2. Heat Problem Let us find the temperature distribution for the Couette flow.

dp = 0 ; then from (25) the velocity profile has the dx following form: u = U y . h We will take

Let us write the energy equation for the layered flow: 2

1 d 2i d 2 u  du  + u 2 +   = 0 . Pr dy 2 dy  dy  43

(26)

Let the boundary conditions be given as follows: y = 0 : T = T1 or i = i1 ; y = h : T = T2 or i = i2 .

dp = 0 , we have the following from the equation of dx d 2u u motion: dy 2 = 0 . Since

Energy equation takes on the following form: 2

1 d 2 i  du  +  =0. Pr dy 2  dy 

(27)

The first term of this equation is the heat flow due to thermal conductivity, the second term is due to friction. Thus, the temperature field is only due to heat conductivity in the transverse direction and heat generated by friction. Inserting in the energy equation (27) the expression for velocity (1), we obtain the following: 2

d 2i U  = − Pr   . 2 dy h After integration, we have the following:

i = − Pr

U2 2 y + C1 y + C2 . 2h 2

(28)

For an ideal gas, i and T only differ in ср multiplier. From the boundary conditions we have the following:

44

C1 =

 1  Pr U 2  + i2 − i1  ; С 2 = i1 . h 2 

The solution (26) shall look as follows:

i = i1 +

2 y (i2 − i1 ) + Pr U y  1 − y  2 h h h

y = η ; 0< η < 1, we write the h expression for the enthalpy in the non-dimensional form:

Introducing the designation:

i − i1 Pr U 2 =η + η (1 − η ) ; i2 − i1 2(i2 − i1 ) Let us designate

(29)

U2 = Ec – the Eckert's number. i2 − i1

The physical meaning of the Eckert's number:

Eс =

fluid mass unit kinetic energy . heat content variation

Thus, the temperature field depends on the product of Pr⋅ Ec :

Pr Ec i − i1 =η + η (1 − η ) . i2 − i1 2 1) If we neglect the heat of friction, then linear relationship. 45

i − i1 = η . This is a i2 − i1

2) If we take the friction heat into account, linear temperature distribution is superimposed by a parabolic distribution, depending on the heat produced by friction. Let, i2 > i1 , ( T2 > T1 ), that is, the upper plate is more heated. Let us calculate the heat flux on the wall:

∂T λ ∂i ∂η −λ = − q= = y h ∂y c p ∂η ∂y

η =1

=

λ PrU 2  . = −  T2 − T1 −  h  2c p  А) if T2 − T1 >

Pr U 2 , then q < 0, i.e. with T2 > T1 the heat 2c p

flux is directed from the hot top plate to the fluid, i.e. the upper plate is cooled.

Pr U 2 , then q > 0 . The heat flux is directed В) if T2 − T1 < 2C p from the liquid to the upper plate. Where in T2 > T1 . The hot plate is heated. Thus, we have obtained that the hot top wall will be heated by cold fluid due to friction. This phenomenon is of great importance in solving the problem of cooling the streamlined wall at high velocities (protection of rocket or aircraft surfaces in motion).

46

6. BOUNDARY LAYER

In practice, we have to encounter the motion of fluids (gas, water) with low viscosity. Since the Reynolds number is inversely proportional to viscosity, for such flows it is high. The flows with high Re numbers were studied by L. Prandtl. He showed that in this case the Navier-Stokes equations can be simplified to obtain their approximate solution. Let there be a body streamlined by a fluid. Fluid approach velocity to a body is U∞ . Far from the body, inside the fluid, due to the absence of a transverse velocity gradient, the effect of viscosity is low and only inertia forces operate here. The flow velocity here is U∞, this value remains almost to the very surface of the body. Only in a very thin layer near the body does the velocity change, since the fluid does not move over the surface of the body, but adheres to it. In a very thin layer, called the boundary layer, the fluid velocity changes from 0 on the surface to U∞ in the external flow. As a result, the entire flow can be divided into two areas: 1. The boundary layer is a thin layer near a body. Here the ∂u velocity gradient is high in the line of perpendicular to the wall. ∂y Therefore, despite the fact that the viscosity is small, the tangential frictional stress τ = µ

du is caused by the velocity gradient can dy

assume large values. 2. The region outside the boundary layer is the region where the ∂u velocity gradient is small, the friction and viscosity effect is ∂y small. This potential flow for can be regarded as an ideal fluid flow with a known velocity of U∞. If the fluid temperature in the flow is constant, for example, Т∞, and the surface temperature is, for example, Т1 , then a temperature boundary layer appears in the flow, similar to the velocity layer. The boundary layer, determined from the distribution of velocity, is 47

called dynamic, and the layer determined from the distribution of temperature, is thermal. In the general case, dynamic and thermal boundary layers do not coincide.

Figure 14.

6.1. Dynamic boundary layer properties Let us estimate the thickness of dynamic boundary layer δd. Outside the boundary layer, due to low viscosity, frictional forces can be neglected in comparison with the inertia forces. Inside the boundary layer, friction forces and inertia forces operate. Here these forces are the magnitudes of the same order. The friction force per unit of volume:

τ V

=

µ du V dy

=µ

du 1 . dy δ d

The inertia force inside the boundary layer, per unit volume:

F du du =ρ ≈ ρu . V dt dx Let us write the force of friction and the force of inertia through characteristic dimensions. Equating these two forces, we have the following:

vL U U2 µ ∞ ~ρ ∞ or δ d ~ . U∞ L δ d2 48

We introduce the non-dimensional thickness of the boundary layer:

δd = δd L .

Then we have:

δd L

~

Since number Re =

δd ν 1 νL ~ or . L U∞L L U∞

U∞L

ν

, we have the following: δ d ~

1 Re

.

To put an equal sign, it is necessary to enter a proportionality factor k and write: δ d = k

1 Re

.

It is different for different

problems. Thus, for the problem of a flow past a flat plate by a homogeneous fluid flow k = 5 and then δ d = 5

1 Re

:

Analyzing the obtained relationships, we can say that: 1) If Re increases, then δd decreases. 2) With the distance from the leading edge plate (x increases) along the plate δd, δ d ~ x increases. 3) The greater the viscosity v is, the greater is the thickness of the boundary layer. 4) In the boundary layer, the velocity varies: 0δ(x). And we shall take into account that at the upper boundary of the boundary layer: 1 dp = U dU , ρ dx dx y ∂u

and from the continuity equation: υ = − ∫

0 ∂x

dy .

We shall substitute these expressions in (66) and integrate: h

y dU  ∂u ∂u ∂u dy − uU −u ρ ∫ u dy = ∫ dx  ∂y ∂y 0 ∂x 0  h

2

= µ∫u 0

∂ 2u ∂y 2

dy .

After rearrangement we have the following:

ρ h

(U 2 ∫ 0

2

)

∂u − u 2  dy +  ∂x

ρh d

2

h  ∂u   + ∫u  µ ∫   dy U − u  dy = . 2 0  dx  0  ∂y 

(

2

2

)

88

Put in the substitution: h

h

)

(

(

)

d du u U 2 − u 2 dy = ∫ U 2 − u 2 dy + ∫ dx 0 dx 0 h

(

)

d U 2 − u 2 dy . 0 dx

+∫ u

Then equation (66) will be:

ρ d

∞

∫u

2 dx 0

(U

2

−u

2

∞ ∂u 

) dy = µ 0∫  ∂y  



2

dy .

(67)

We have replaced the upper limit of integration y=h with y=∞, since outside the boundary layer the integration elements in the left and right members are zero. 2

 ∂u  Here µ  ∂y  is the dissipation of energy, i.e. this is the amount   of energy per unit volume, which per unit time is converted into heat due to friction; ρu U 2 − u 2 – mechanical energy is the sum of the pressure 2 energy and kinetic energy. This energy is lost in the boundary layer due to the decrease in the flow velocity therein, in comparison with the potential flow velocity.

(

)

ρ ∫ u (U 2 − u 2 )dy – energy loss flow; ∞ 0

ρ d

∞

(

)

2 2 ∫ u U − u dy is the change in the energy flux per unit 2 dx 0 length in the x direction. Let us put in the energy thickness concept δ***, which is determined by the following equation:

89

∞

(

)

U 3δ *** = ∫ u U 2 − u 2 dy . 0

(68)

We denote dimensionless energy dissipation by D*:

∂ u  y  D = ∫   d . y U  δ **  0 ∂ δ ** *

2

∞

Let us divide y by δ** and u by U, then equation (67) shall be rewritten as:

ρ δ ** d ∞ 2 2 ** ∫ u (U − u )dy = D . 2 U 2 µ dx 0 Or taking (68) into account, we obtain the following:

(

)

1 δ ** d U 3δ *** = D** . 2 2 νU dx

Figure 28.

90

For the linear velocity distribution, we have: ∞ u2 δ *U = ∫ (U − u )dy = U ∞ − 0 2 0

∞

* , δ = 0

1 δ; 2

1 1 1 δ ** = δ ; δ *** = δ ; D* = δ . 4 6 6

91

13. INTEGRAL METHOD

Let us consider another approximate solution for motion and energy equations – the integral method. To do this, we use the momentum and energy theorem. Let us apply the integral method to the problem of flow around a semi-infinite plate by an incompressible fluid flow (the Blasius problem). The results obtained by this method are comparable with the exact solution and the experimental data. This method, first proposed in 1921 by Karman and Pohlhausen, was further developed. Many researchers later engaged in clarifying and simplifying this method. The principle of the integral method is that an appropriate expression is chosen for the velocity distribution u(y) in the boundary layer. This velocity distribution must satisfy the boundary conditions and contain one free parameter, for example, the boundary layer thickness, or the momentum thickness. This free parameter is then determined from the momentum theorem, which will allow us to find the displacement thickness and friction stress. 13.1. Dynamic Problem We write the equations of motion for the boundary layer on the plate.

u

1 dp ∂ 2u ∂u ∂u +υ =− +v 2 ρ dx ∂x ∂y ∂y

(69)

∂u ∂υ + =0. ∂x ∂y

Boundary conditions take the following form: y =0, u =0, υ =0; y =∞, u =U∞ 92

(70)

If we write equation (69) on the upper boundary of the boundary layer: U∞

dU ∞ dx

=−

1 dp dp = 0 , then =0 . ρ dx dx

Therefore, we have the following:

u

∂u ∂u ∂ 2u +υ =v 2 ∂x ∂y ∂y ∂u ∂υ + =0. ∂x ∂y

Let the velocity u be represented as a polynomial of degree 3: и = А0+А1у+А2у2+А3у3

(71)

To determine the invariables we use the boundary conditions. 1) y=0, u=0, υ=0 from (69) we have:

∂ 2u =0; ∂y 2 y =0 ∂ 2u = 2a 2 + 6 a 3 y = 0 . ∂y 2 y =0 Then: А0=0, А2=0. 2) Since the boundary layer transition to the potential flow occurs asymptotically, we replace the condition y=∞ with у=δ(х). This means that the closure of the boundary layer with the potential flow occurs at a finite distance of у=δ(х) from the wall: u (∞ ) = u (δ ) = U ∞ = a1δ + a3δ . 93

(72)

3) Since friction at the outer boundary of the boundary layer is zero, ∂u = 0 or А1+3А3δ 2=0 ∂y y =δ

(73)

We combine equations (72) and (73):

U ∞ = a3δ 3 + 3a3δ 3 . Hence, a3 = −

U

∞ .

2δ 3

From (73) we have: a1

=3

U

∞ .

2 δ

Now we know all the values of the invariables: А0, А1, А2, А3. We substitute them in (71) and obtain:

u =U

We put in:

y

δ

y ∞ 2δ

 y2   3 − 2  . δ  

= η , then we will have:

(

)

u 1 = η 3 −η 2 . U 2

(74)

∞

We found a velocity distribution that satisfies the boundary conditions. This distribution depends on δ(х). δ(х) is found from the integral condition – the momentum theorem.

)

(

τ w d ** 2 = δ U ∞ + δ *U ∞ dU ∞ . ρ dx dx 94

(75)

Let us rewrite this expression: τ ρ

w δ ** =

d 2 U ∞ + δ *U ∞ dx

dU∞ + dx

** dδ ** U ∞2 +U ∞2 dδ = dx dx

.

Thus, the momentum equation for the Blasius problem:

dδ * * dx

=

τw

.

2 ρU ∞

δ u  u   1 − dy , then in the integral form the Since δ ** = ∫ U ∞  0U∞  momentum equation takes the following form:

d δ u ∫ dx 0 U ∞

then

 u  1 −  U∞

 τ ∂u dy = w ; but τ w = µ y =0 , 2 ∂y ρU ∞ 

d δ u ∫ dx 0 U ∞

 u  1 −  U∞

 ν ∂u dy = . 2 ∂y y = 0 U∞ 

Let us substitute in (76) equation (74):

(

d 11 δ ∫ η 3 −η 2 dx 0 2

) 1 − 32 η + η2 dη = 32ν δUU∞2 . 3





∞

Let us put in the designation:

(

)

3

1 3 η  δ ∫ η 3 −η 2 1 − η + dη = I = const . 1

02



2 

2

95

(76)

Then we will have:

δ

3ν dδ 3 1 ; δ2 = x+C. = ν dx 2 U ∞ I U∞I

From the boundary conditions of the problem: with x =0, δ =0 we have C=0, δ 2 =

3νx I. U∞

Let us determine the integral I:

I=

δ2 =

(

11 3 ∫ 3η − η 20

39 ) 1 − 32 η + η2 dη = 280 . 3





280 ⋅ 3νx νx δ 4 ,64 ; δ = 4 ,64 , = . U∞ x 39U ∞ Re x

From exact solution based on Blasius

δ x

=

5 Re x

. Integral

method error ∼ 7%. Let us construct a dimensionless velocity profile: u : F' ( ϕ ) = U∞

Figure 29. 96

Figure 30.

Let us find friction stress:

τw = µ

U∞ . ∂u = 0 ,323µU ∞ y 0 = νx ∂y

Based on Blasius: τ w = 0 ,332 µU ∞ U ∞

νx

The difference is ∼2,7%. Let us find local resistance factor:

Сf =

τw ρ

U ∞2 2

=

0 ,646 µU ∞

ρU ∞2

U∞ νx = 0 ,646 . Re x

Based on Blasius С = 0 ,664 . f Re x Thus, comparing the results of the approximative method with the exact solution of the Blasius problem, we see that they qualitatively coincide and correctly determine the boundary layer parameters. The only difference is in the coefficients. Obviously, the accuracy of the integral method can be increased if we represent the velocity distribution in the form of a higher degree polynomials. Of course, additional coefficients appear in this case, as a result of which the chosen velocity distribution must 97

satisfy a greater number of boundary conditions on the wall and on the outer border of the boundary layer. Paulhausen himself used a fourth degree polynomial. G. Schlichting and A. Ulrich (1942) used a sixth degree polynomial. The results obtained using these two methods for the boundary layer parameters and for the separation point position differ little. However, the use of a sixth degree polynomial provides the following advantage: higher velocity derivatives of the boundary layer, taken from the distance from the wall, can be determined much more accurately than by means of a fourth degree polynomial. And this is important in the study of velocity profile stability in the boundary layer. It should be noted that in this method it is possible to apply not only polynomials. Thus, D. Hartree approximated the velocity profiles by means of exponential expressions with fractional exponents, A. Beth (1950) – by means of a simple series. All these methods are based on the replacement of boundary layer differential equations by the momentum equation, i.e. an integral relation that satisfies the equation of motion only on the average over the thickness of the boundary layer. Advantages of the Method: – good in studying new problems and qualitatively correctly reflects the patterns of the flow; – the method is suitable for solving complex turbulent flows problems, with large pressure gradients, when moving along a porous surface; – simplicity and rapidity of the method; Disadvantages of the method: the accuracy with which the results are obtained cannot be determined in advance. 13.2. Application of the integral method to the calculation of the temperature boundary layer on the plate Just as in the study of the dynamic boundary layer, in the case of a thermal boundary layer, one can use the concept of a finite thickness layer. This allows us to apply approximate methods, similar to the Pohlhausen method, for calculating the temperature boundary layers. 98

In 1936 G.N. Kruzhilin considered the problem of longitudinal flow past a plate with a liquid having a Т∞ temperature different from the wall temperature of Тw. He used two integral conditions: the momentum equation and the heat balance equation, which is obtained from continuity equation and energy equation. The heat balance equation for a non-isothermal boundary layer has the meaning of a heat flux balance. Let us write the energy equation and the continuity equation:

u

∂ (T − T∞ ) ∂ (T − T∞ ) ∂ 2 (T − T∞ ) =a +υ , ∂x ∂y ∂y 2

∂u ∂υ + =0. ∂x ∂y

(77)

(78)

Let us rewrite the equation (77), using (78): 2 ∂ [u (T − T∞ )] + ∂ [u (T − T∞ )] = a ∂ T2 . ∂x ∂y ∂y

Integrating both members of this equation across the temperature layer from у = 0 to у = δТ, where δТ is the temperature boundary layer thickness and, taking into account that by the ∂T definition of the temperature boundary layer: = 0 with у = ∞ or ∂y у = δТ, we shall obtain:

∂ δT ∂T ∫ u (T − T∞ )dy = − a y =0 . ∂x 0 ∂y Since у = 0, υ = 0; у=δ, Т = Т∞ = 0. T − Tw ∆T 1 Let Т(η) = = = ηТ 3 − ηТ2 . T∞ − Tw ∆T∞ 2

(

99

)

(79)

Let us rewrite the equation (79): T − Tw  ∂ δT u  3 a  1 − dy = . ∫ ∂x 0 U ∞  T∞ − Tw  2 δT U ∞

(80)

From dynamic and heat problem resolution we have the following:

u 1 ∆T 1 = η T 3 − η T2 , = η 3 −η 2 ; U∞ 2 ∆T∞ 2 y y where = η , = ηT . δ δT

(

)

(

)

Then from (79) we obtain:

d 1 3 a  1  δ T ∫ η ( 3 − η 2 ) 1 − ηT ( 3 − ηT2 )  dηT = . dx 0 2 2 U ∞δ T  2  1

Let us set: 1

∫η T ξ

0

(3 −η ξ )1 − 21 η (3 −η ) dη 2 T

2



T

2 T



T

= J (ξ )

Then we will have:

d 3 a , [δ T J (ξ )] = dx 2 U ∞δ T d 3 a . δ T [δ T J (ξ ) ] = dx 2 U∞ Let us compute:

3 3 3 3  1  J (ξ ) = ξ − ξ = ξ 1 − ξ 2  . 20 280 20  14  100

(81)

Taking into account that

δ T = ξδ , i.e. ξ