What do the Navier-Stokes equations mean?

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(ii) Von Neumann boundary conditions: The gradient in normal direction to the ..... The authors would like to thank Professors Urbaan Titulaer, Erich Steinbauer ...
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What do the Navier-Stokes equations mean? Ph ys

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Simon Schneiderbauer1 ‡ and Michael Krieger2 1

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Christian Doppler Laboratory on Particulate Flow Modeling, Johannes Kepler University, Altenbergerstraße 69, 4040 Linz, Austria. 2 Institute for Fluid Mechanics and Heat Transfer, Johannes Kepler University, Altenbergerstraße 69, 4040 Linz, Austria.

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Abstract. The Navier-Stokes equations are non-linear partial differential equations describing the motion of fluids. Due to their complicated mathematical form they are not part of secondary school education. A detailed discussion of the fundamental physics – the conservation of mass and Newton’s second law – may, however, increase the understanding of the behavior of fluids. Based on these principles the Navier-Stokes equations can be derived. This article attempts to make these equations available to a wider readership, especially teachers and undergraduate students. Therefore, in this article a derivation restricted to simple differential calculus is presented. Finally, we try to give answers to the questions “what is a fluid?” and “what do the Navier-Stokes equations mean?”.

‡ corresponding author: [email protected]

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1. Introduction

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The Navier-Stokes equations describe the motion of fluids and are the fundamental equations of fluid dynamics. They are named after Georg Gabriel Stokes (1816–1903) and Louis Marie Henri Navier (1785–1836), who derived these equation independently. The Navier-Stokes equations are based on work of Leonhard Euler (1707–1783). Euler considered the fluid as a continuum allowing him to derive governing equations for the motion of inviscid fluids based on differential calculus. His equations were the first written down non-linear partial differential equations – the Euler equations. Stokes and Navier contributed a viscous diffusion term to account for the viscosity of a fluid. The Navier-Stokes equations are widely used in science and engineering. However, their complicated mathematical form mostly restricts engineers to the numerical solution of these equations [1, 2]. The mathematical proof of the existence of a global solution of the Navier-Stokes equations is still one of the millennium problems [3]. Nevertheless, the Navier-Stokes equations are successfully applied to design airfoils [4], reduce the drag of (racing) cars, optimize particle filters, understand the windthrow of forests [5], analyze ocean currents [6], study environmental particle transport [7] and so forth. Due to the fact that the Navier-Stokes equations are partial differential equations and their solutions are non-trivial, these are commonly not included in secondary school curriculums. A detailed discussion of the fundamental assumptions of the Navier-Stokes equations and of the underlying physics may, however, increase the understanding of the behavior of fluids. Especially, the physics of fluids can be topic of in-depth physics courses. This article aims to introduce the Navier-Stokes equations to secondary school teachers and undergraduate students. We also intend to provide teaching material for extraordinary interested students. Therefore, we restrict our calculations to simple differential calculus, for example, partial derivatives and Taylor series expansions. This is in contrast to standard derivations using integral theorems. This article is organized as follows. In section 2 we begin with a practical definition of fluids. Then, we discuss the underlying assumptions of the Navier-Stokes equations and the basic concepts. In section 5 we derive the two-dimensional Navier-Stokes equations using differential calculus. At the end of the article, some simple examples for the exact solution of the Navier-Stokes equations are discussed. 2. What is a fluid? Each natural or artificial material is characterized by its distinct state of matter. These states are solid, gas and liquid. There is also a fourth state of matter, referred to as plasma, which we do not discuss in more detail. Intuitively, one is tempted to categorize fluids by its state of matter. For example, water is a fluid since it is liquid at room temperature and atmospheric pressure. However, such a classification of fluids seems to be too restrictive since both gaseous and liquid matter is considered as fluid [2, 8].

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In fluid mechanics it is common to assign all materials, which are not clearly solid, to fluids. Thus, each material can be distinctly assigned to either solids or fluids. F

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Figure 1. An external force is applied to a) a solid and b) a fluid, which consists of randomly moving molecules. The amount of deformation of the solid is determined by the balance of the external and the shear forces. The fluid establishes a velocity gradient, which counteracts the external force. The fluid proceeds moving as long as the force is applied. This is indicated by the parallelograms ABC 0 D0 and ABC 00 D00 .

The fluid counterpart of the elasticity of the solids is known as viscosity. The physical difference between elasticity and viscosity is outlined by the following example. Let us consider a block of elastic solid (Figure 1a), which is mounted to the ground at points A and B. The molecules of the solid hold together by exerting attractive forces on each other. When an elastic solid is deformed it tries to deform back to its initial state. Thus, qualitatively we may model these attractive forces by springs (Figure 1a). Furthermore, a force F , which is parallel to AB, is applied at point D. In case of small deformations we obtain for the shear stress τ (= force/area = F/δA) k θ, (1) δA where k denotes the stiffness of the springs. Our illustrating example indicates that in elastic solids the shear stress is proportional to the angle of deformation. In general, the displacement of the atoms in solids by external shear forces is reversible for small deformations (elastic deformation). In contrast to the deformation of solids, in fluids a velocity gradient establishes when an external force is applied (Figure 1b) [9, 10, 2]. The velocity gradient is a consequence of momentum diffusion at molecular scale. According to Newton’s second law molecules at point D are accelerated by the applied force F . In contrast to solids, the molecules move randomly in a fluid. Thus, some of these “fast” molecules near D move down towards A (Figure 1b). Therefore, the momentum of the fluid in the direction of the force F near A increases. Similarly, some “slow” molecules near A move up towards D leading to a decrease of momentum in the region of D. Mathematically, the shear stress on an area δA is written as [11] τ∝

τ =µ

∂u , ∂y

(2)

What do the Navier-Stokes equations mean?

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where u denotes the velocity of the fluid in lateral direction and y the spatial coordinate in vertical direction. The viscosity µ (N s) is a characteristic property of the fluid like the stiffness of a spring. A more detailed derivation of equation (2) is given in section 5.2. When the applied shear force is withdrawn the velocity gradient vanishes until the fluid is at rest. The random nature of the movement of the molecules indicates that the displacement of the molecules under an external force is not reversible, which is in contrast to elastic solids. This example demonstrates that a fluid at rest, in contrast to solids, is not able to support external shear forces. It immediately reacts to the applied shear force by establishing a velocity gradient. Viscous shear forces are completely different from elastic shear forces [10]. Strictly speaking fluids do not transmit shear forces; these forces (≡ time rate of change of momentum) rather appear due to the random motion of the fluid molecules. As a consequence under external forces, i.e. the earth’s gravitational field, fluids need containing walls to keep their geometric shape. Although solids and fluids behave very differently when subjected to shear forces, they behave similarly under the action of pressure, i.e. normal compressive stresses [10]. However, whereas solids are able to support both normal tensile and compressive forces, a fluid usually supports only compression (pressure).

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3. Continuum hypothesis

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Fluids are composed of a huge number of molecules, which are in constant motion and undergoing collisions with each other [10]. Typical orders of magnitudes for air at 1 bar and 0◦ C are dm ≈ 4 · 10−10 m for the diameter of the molecules, ∆m ≈ 3 · 10−9 m for the average distance between two molecules, λmf p ≈ 6 · 10−8 m for the mean free path (the distance between subsequent collisions) and 1025 molecules per m3 . In principle, it is possible to study the behavior of fluids by following the trajectory of each single molecule including its collisions with the surrounding molecules [10, 9], as done by molecular dynamics. However, it is more practical to ask for the macroscopic behavior of the fluid. The idea is to take account of the behavior of the molecules and their properties by considering a “huge” ensemble of molecules [2]. Thus, the discrete molecular structure of fluids is replaced by continuous distributions, called continuum [9]. For example, we can define the macroscopic continuous pressure at a constraining wall as the time rate of change of momentum per unit area of a “huge” ensemble of colliding molecules. Similarly, we can define the temperature of a fluid as the kinetic energy of an ensemble of molecules; the density as the number of molecules per unit volume; the velocity of the fluid as the average velocity of the ensemble. Note, since we consider the properties and behavior of fluids by considering an ensemble of molecules, the molecular nature of fluids is not neglected. It remains to discuss how many molecules we need for such an ensemble and in which cases the continuum hypothesis is valid.

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3.1. Definition of the macroscopic continuous density and pressure

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It is common to define the density of a fluid ρ(x0 , y0 ) at (x0 , y0 ) (see also Figure 2) by considering the aggregated mass of molecules mδV in volume δV fixed in space at (x0 , y0 ) mδV . (3) ρ(x0 , y0 ) = δV Since the molecules are not fixed in a lattice but move freely [9], the fluid density has no precise meaning. The number of molecules in a given volume δV continuously changes. On the one hand, molecules enter the volume and on the other hand, molecules leave the volume. If the volume is very small, that is in the order of ∆3m ≈ 10−26 m3 , estimating the density would, therefore, result in a huge uncertainty (Figure 3a). Increasing the volume reduces the influence of leaving and incoming molecules since their contribution becomes negligible compared to the number of molecules occupying δV . In Figure 2 the density as calculated from the aggregated mass of the molecules is plotted. There is a limiting volume δV ∗ ≈ 10−18 m3 above which the contribution of the leaving and incoming molecules gets negligible. For smaller volumes than δV ∗ considerable uncertainty in the evaluation of the density may be observed. fluid

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Figure 2. Sketch of the definition of the fluid density and fluid pressure.

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According to the definition of ρ, we define the pressure p(x0 , y0 ) by considering the p time rate of change of momentum of the molecules FδA at a plane of area δA at (x0 , y0 ) (Figure 2)

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p FδA p(x0 , y0 ) = . (4) δA Thus, the order of magnitude of the minimum size of δA should be O(δA) ≈ O(δV 2/3 ) = O(10−12 ) to ensure an accurate definition of the pressure. This is also indicated in Figure 3b, where the limiting area δA∗ , above which the pressure can be precisely defined, is approximately 10−12 m2 .

What do the Navier-Stokes equations mean?

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ρ microscopic uncertainty

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density

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Figure 3. Uncertainties in the estimation of a) the fluid density as a function of the volume δV occupied by the considered molecules and b) of the pressure as a function of the area under consideration [9, 2].

3.2. Knudsen number The Knudsen number is a dimensionless number (Note in fluid mechanics there are lots of dimensionless numbers), that is λmf p , (5) L which is the ratio between the mean free path of the molecules λmf p and a characteristic physical length scale L of the problem under consideration. For example, L may be the length of an airfoil. The Knudsen number gives a measure for the validity of the continuum hypothesis. If the Knudsen number is very small (Kn  1) the physical length scale is much larger than the mean free path of the molecules. Thus, the physical obstacle does only “feel” the average molecular behavior and we can apply the continuum hypothesis to the physical description of the flow. In case of our numerical examples (λmf p ≈ 6 · 10−8 m, δA∗ = 10−12 m2 ) the limiting Knudsen which the √ number, below ∗ −2 ∗ continuum hypothesis can be used, yields Kn = λmf p / δA ≈ 6 · 10 . In case of a Knudsen number greater than Kn∗ statistical methods or molecular dynamics must be used. An example is a space shuttle entering the earth’s exosphere, where the mean free path of the molecules is several kilometers (high temperature and low density), and thus, Kn  1. Similarly, the flow around nanofibers in a extra-fine particle filter shows Knudsen numbers of Kn & 1. Hence, the continuum hypothesis cannot be applied to these cases. Kn =

3.3. Finite control volumes – Infinitesimal fluid elements Based on the idea of the continuum hypothesis we are able to derive the governing equations of fluid dynamics. It is, therefore, common to introduce the concept of finite control volumes V , which are very useful in fluid dynamics [1]. According to the discussion about the definition of density (section 3.1), one has to keep in mind that these finite control volume have to be greater than δV ∗ . Then, the fundamental physical principles can be applied to the fluid inside the control volume V . Such a volume can

What the Navier-Stokes equations What do the do Navier-Stokes equations mean? mean?

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Control A Surface AL Control Surface L

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Control A Surface AE Control Surface E

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Control V Volume VE Control Volume E

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Control V Volume VL Control Volume L

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• Eulerian description of the governing equations, which are obtained from the finite fluid element fixed in space (Figure 4b).

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Figure 4.(a) Models of a flow: a) Finite control volume moving with (b)the fluid such that (a) (b)

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the same individual fluid particles are always in this control volume; b) Finite control volume which is fixed in space. The fluid is moving through it.

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4. of Models a flow: Finite volume control moving volume with moving fluid such that Figure Figure 4. Models a flow:of a) Finitea)control thewith fluidthe such that the samecan individual fluid particles always in fluid this volume; control volume; b)control Finite control the individual fluid particles are always in this control b) The Finite Thesame same ideas also be represented byare infinitesimal elements. fluid volumeiswhich is fixed space. fluid is through moving through it. enough volume which fixed in space. fluidThe is calculus moving it. to be large element is infinitesimal in the sense in ofThe differential [1] but has

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δVL

du and a(δVL , t) = , dt δVL

(6)

The where velocity u the acceleration Lagrangian fluid element moving with the The velocity u and theand acceleration of a a Lagrangian fluid element moving with the the position vector x andavelocity uof in a two dimensional cartesian space are areby given by [2, 9, 10, 12] flow areflow given [2, 9, x10, = e12] x x + ey y, du dxu + eydx du = v, and a( andVL , a( , (6) VLe,xt) t)u== t) =VL , t) = , (6) u( VL ,u( dt V dt VL dt V dt VL L

L

where the position x and velocity u in two dimensional cartesian space are where the position vector vector x and velocity u in two dimensional cartesian space are

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δVE = δVL (t0 ).

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u(x, t0 ) = u(δVL , t0 ) if

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with ex and ey denoting the cartesian unit basis vectors. In equation (6) the symbol |δVL indicates that we follow the trajectory of the fluid element δVL of fixed mass, which we highlighted at t = 0. However, in fluid dynamics it is not practical to use the governing equations in Lagrangian form. It is common to obtain these equation based on Eulerian control volumes VE (Eulerian form), which are fixed in space and where the velocity and density fields are functions of x and t. It is straightforward to transform the velocity of a Lagrangian fluid element into Eulerian form at an arbitrary time t0

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Hence, we simply consider a Eulerian control volume VE equal to δVL (t0 ) at t = t0 . For

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Figure 5. Steady flow through a convergent pipe. The time rate of change of the velocity at an arbitrary location x is zero for all t. A fluid element δV moving through the pipe (Figure 5) experiences acceleration due to the area reduction in the middle of the pipe.

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the derivation of the acceleration in Eulerian form the situation is more complicated, since the time rate of change of the velocity at location x at time t0 is in general not equal to the acceleration of the Lagrangian fluid element passing x at t0 . A simple example is shown in Figure 5. The figure illustrates a steady incompressible flow through a convergent pipe. Thus, the time rate of change of the velocity at an arbitrary fixed location x is zero for all t since the flow is stationary. However, the fluid element δV moving through the pipe (Figure 5) has to be accelerated due to the area reduction in the middle of the pipe. The conclusion applies to a stationary waterfall. If we monitor a fixed location at the waterfall we will observe that the velocity of the flow does not change with time. In contrast, the water is accelerated downstream of the observed location by gravity. Thus, a leave floating on the water is accelerated when passing the waterfall. In other words, the leave is equivalent to the fluid element moving with the flow. Since the fluid element is accelerated the time rate of change of the velocity of the fluid element is not zero. Therefore, we have du du 6= 0 and =0 (8) dt dt δVL

δVE

in this case. The symbol |δVE indicates that we evaluate the time rate of change of the fluid velocity for a Eulerian control volume fixed in space. As an example, we derive an expression for the acceleration in x-direction of a Lagrangian fluid element in Eulerian form. Let us consider an infinitesimally small Lagrangian fluid element moving with the flow in two-dimensional cartesian space. The

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velocity components u and v are functions of the space coordinates (x, y) and the time t u = u(x, y, t) and v = v(x, y, t).

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Thus, the total differential of u reads ∂u ∂u ∂u du = dt + dx + dy. (10) ∂t ∂x ∂y Dividing equation (10) by dt yields ∂u ∂u dx ∂u dy du = + + , (11) dt ∂t ∂x dt ∂y dt where (dx/dt, dy/dt) describes the path of the fluid element in space. In case of a Lagrangian finite control element we obtain from equation (6), (dx/dt, dy/dt) = (u, v), which denotes the velocity of the fluid element. Therefore, equation (11) reads du ∂u ∂u ∂u = +u +v . (12) dt δVL ∂t ∂x ∂y (13)

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In fluid mechanics it is common to define Du du . := Dt dt δVL

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The symbol D/Dt is called the material or substantial derivative, which describes the time rate of change of a scalar quantity of the given fluid element as it travels through space. In other words our eyes are locked on the fluid element and we observe that the x-component of the velocity u of the element changes as it moves through a point (x, y). In case of an Eulerian fluid element we take the derivative (11) at a fixed location and, therefore, (dx/dt, dy/dt) = 0. Thus, equation (11) reveals du ∂u = . (14) dt δVE ∂t

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Note that ∂u/∂t, which indicates the time rate of change of u at the fixed point (x, y), is different from the material derivative. In literature the material derivative D/Dt is called the Lagrangian description of the dynamics of a fluid whereas the local partial derivative ∂u/∂t is termed as the Eulerian description of fluid dynamics [1]. From equation (12) we can obtain an expression for the material derivative [9, 10, 1, 2] ∂ ∂ ∂ D = +u +v . (15) Dt ∂t ∂x ∂y ∂ ∂ The expression u ∂x + v ∂y on the right hand side of equation (15) is called the convective derivative, which describes the time rate of change due to the movement of the fluid element from one location to another in the flow field, where generally the considered flow property (i.e. u, ρ, p, etc.) is spatially different. The substantial derivative applies to any flow-field variable [1], for example Dρ ∂ρ ∂ρ ∂ρ = + u +v . Dt ∂t ∂x ∂y |{z} |{z} | {z } material derivative

local derivative

convective derivative

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5. Conservation laws In obtaining the basic equations of fluid dynamics we have to choose the appropriate fundamental physical principles from the laws of physics [1, 9, 10, 2]: (i) Mass is conserved. There are no nuclear reactions involved in fluid dynamics.

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(ii) Newton’s second law: F =

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5.1. Conservation of mass – continuity equation

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In general, the continuity equation describes that the time rate of decrease of the fluid density in an arbitrary control volume equals the net flow out of this arbitrary control volume through its surface. Intuitively, the continuity equation in fluid dynamics can be easily derived from the principle of mass conservation. If we consider a Lagrangian control volume moving with the flow, the conservation of mass reads

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d(ρδVL ) dmVL = = 0, (16) dt dt i.e. the mass mVL of a Lagrangian control volume δVL does not change with time. However, δVL is no physical field quantity and, therefore, we cannot use the material derivative to simplify equation (16). It is more practical to consider an arbitrary Eulerian control volume fixed in space (compare with Figure 6). Since the volume and not the

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p(x0 − δ x / 2, y0 , t) δ Ax

p(x0 , y0 , t)

ρ (x0 , y0 , t) u(x0 , y0 , t)

δVE = δ xδ yδ z

p(x0 + δ x / 2, y0 , t)

ρ (x0 + δ x / 2, y0 , t) u(x0 + δ x / 2, y0 , t)

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ρ (x0 − δ x / 2, y0 , t) u(x0 − δ x / 2, y0 , t)

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Figure 6. Fluid enters the Eulerian control volume δVE with the velocity u(x0 − δx/2, y0 , t) from the left side. Fluid leaves δVE with the velocity u(x0 + δx/2, y0 , t) to the right side, which leads to a change of the density of the fluid occupying δVE = δxδyδz. The pressure p(x0 − δx/2, y0 , t) acts on the left face and the pressure p(x0 + δx/2, y0 , t) acts on the right face of the Eulerian control volume δVE , which leads to a non-zero net force on δVE = δxδyδz. In the two-dimensional case δz is equivalent to δz ≡ 1.

mass of an arbitrary Eulerian control volume is fixed we have to account for the in- and outflow of mass per unit time, that is, ∂mδVE = m ˙+ − m ˙− . (17) |{z} |{z} ∂t inflow

outflow

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Note that we have to apply partial (local) derivatives to Eulerian control volumes. The left hand side of equation (17) can be written as

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∂mδVE ∂ρ(x, y, t)δVE ∂ρ(x, y, t) = = δVE . (18) ∂t ∂t ∂t For the last step we made use of the fact that the Eulerian control volume δVE does not change with time. Expressions for the in- and outflow of mass per unit time can be derived from Figure 6. Without loosing generality we assume that the flow is aligned with the positive xaxis. Hence, there is no mass flow through the lower and upper faces. The inflow of mass per unit time into δVE over the leftmost area, therefore, reads     hmi  2 kg kg (19) = density × inflow velocity × area m . m ˙ in s m3 s

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In terms of Figure 6 the inflow and outflow of mass per unit time read m ˙ ± = ρ(x0 ∓ δx/2, y0 , t)u(x0 ∓ δx/2, y0 , t)δAx .

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Thus, substituting equation (20) into equation (17) and using (18) yields         δVE ∂ρ(x0 , y0 , t) δx δx δx δx = ρ x0 − , y0 , t u x0 − , y0 , t −ρ x+ , y, t u x+ , y, t .(21) δAx ∂t 2 2 2 2 We can expand the terms on the right hand side of equation (21) in a Taylor series yielding  δx  ∂ρ δx ∂u δx δx   u x0 ± = ρ(x0 )u(x0 ) ± u(x0 ) ± ρ(x0 ) + O. (22) ρ x0 ± 2 2 ∂x 2 ∂x 2 x0

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It is reasonable to neglect the higher order terms as δx tends to 0. Thus, by substituting the linearized in- and outflow of mass per unit time into equation (21) yields  ∂ρ(x0 , y0 , t) ∂ρ ∂u  δVE = δAx δx u(x0 ) − ρ(x0 ) . (23) ∂t ∂x ∂x x0

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By using δAx = δyδz, δVE = δAx δx equation (23) can be written as ∂ρ(x0 , y0 , t) ∂ρ ∂u = u(x0 ) − ρ(x0 ) . ∂t ∂x ∂x

(24)

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In equation (23) the physical fields ρ and u and their local derivatives are evaluated at (x0 , y0 , t). Since x0 , y0 and t are arbitrary we skip the explicit dependencies. Applying the product rule to the last two terms on the right hand side of equation (24) yields −u

∂ρ ∂u ∂ρu −ρ =− . ∂x ∂x ∂x

Finally, we have ∂ρ ∂ρu =− ∂t ∂x for the continuity equation.

(25)

(26)

What do the Navier-Stokes equations mean?

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In the case of a non-zero gradient of v in y direction, we can apply the same principles we used for the above derivation of the continuity equation, which then reads ∂ρ ∂ρu ∂ρv + + = 0. (27) ∂t ∂x ∂y For incompressible fluids, i.e. ρ = const., the continuity equation can be simplified as follows ∂u ∂v ∂u ∂v + = 0 or =− . (28) ∂x ∂y ∂x ∂y

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5.2. Newton’s second law – momentum equation

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As mentioned at the beginning of section 5, we can derive a momentum equation for the dynamics of the flow by applying Newton’s second law to an arbitrary finite control volume V . There are two sources of forces acting on such a finite control volume V [1, 9, 10, 2]: Body forces, which act directly on the volumetric mass of the fluid element. These forces act throughout the finite control volume; examples are gravitational, electric, and magnetic forces. In most applications, only the gravitational force on the fluid particles is taken into account [1]. At the earth’s surface the gravitational force on a fluid element is simply

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F g = mV g,

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where g = (0, −g) denotes the standard acceleration due to gravity at the earth’s surface. Surface forces, which act across the surface of the fluid element. These result in general from the random motion of the fluid molecules. It is common to distinguish between:

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• Pressure, which is the force acting at right angles on the surface. The pressure is imposed by the outside fluid surrounding the fluid element. On a molecular scale the pressure is the time rate of change of colliding fluid molecules at the surface of the control volume.

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• The shear and normal stress distributions acting on the surface; these are also imposed by the outside fluid. These arise from momentum diffusion driven by the random motion of the molecules. In literature these are also interpreted as “tugging” or “pushing” of the surrounding fluid on the surface by means of friction [1].

If we examine an arbitrary Lagrangian finite control volume moving with the flow, the force balance, therefore, reads dmVL u m = mVL g + |F p + (29) {z F } . | {z } dt | {z } time rate of change of momentum

body forces

surface forces

Note the right hand side of equation (29) does not contain time derivatives. Thus, the applied forces can also be evaluated for the time independent and spatially fixed Eulerian control volume δVE , which requires δVL = δVE at time t.

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Applying the product rule to the left hand side of equation (29) yields du dmVL dmVL u = mVL +u . (30) dt dt dt In order to evaluate the first term on the right hand side we can use the material derivative. The second term is equivalent to the conservation of mass (equation (16)) and, therefore, is zero. Thus, we have dmVL u Du = δVL ρ . (31) dt Dt Due to these manipulations the time derivative of the Lagrangian control volume disappears in equation (31) and we are able to substitute δVL by δVE , which we choose equal to δVL at time t. Thus, in Eulerian form equation (29) reads   ∂u ∂u ∂u +u +v = fxg + fxp + fxm , (32) ρ ∂t ∂x ∂y   ∂v ∂v ∂v ρ +u +v = fyg + fyp + fym , (33) ∂t ∂x ∂y

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where the fij ’s denote the force densities (force per volume ≡ Fij /δVE ) of the different contributions. The first term on the left hand side of equations (32) and (33) is the time rate of change of the momentum. The second and third terms of equations (32) and (33) are the time rate of change of momentum due to the movement of a Lagrangian fluid element from one location to another in the flow field.

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Pressure forces: Before we derive an expression for the force on a fluid element exerted by the fluid pressure, we discuss the pressure in fluids in more detail. To simplify our analysis we focus our attention on semi ideal gases, where the dimensions of the molecules are much smaller than the distances between the molecules. Therefore, these can be considered as point particles. Other properties of semi ideal gases are that interaction forces between the molecules are neglected, the particles move randomly and that particle-particle collisions and particle-wall collisions, respectively, are considered as perfectly elastic.

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Ideal gas

δ Ax

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Figure 7. Semi ideal gas confined by a container: Random motion of molecules.

Per definition (compare with section 3.1) the pressure is the force per unit area applied in the direction perpendicular to a surface. From the discussion in section 3.1

What do the Navier-Stokes equations mean?

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this force can be calculated from the time rate of change of momentum of the molecules at a specific area. Therefore, let us consider an Eulerian control volume in a container filled with a semi ideal gas at rest. Following Figure 7 the number of molecules N hitting the left surface δAx of the Eulerian control volume δVE from the left side within the time interval ∆t can be estimated from [2]

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left Nin = nx δAx um ∆t,

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where nx denotes the number of molecules per unit volume moving in the positive xdirection. Since the molecules move randomly nx can be calculated from 1 nx = n = n−x = ny = n−y = nz = n−z , 6 where n±y and n±z denote the number of molecules moving in ±y and ±z-direction respectively. n is the number of molecules per unit volume and um is the thermal velocity of the molecules in any particular direction [13]. Within a distance of λmf p from δAx these molecules collide with molecules leaving δVE to the left side 1 left Nout = n−x δAx um ∆t = nδAx um ∆t. 6 The time rate of change of the momentum of the incoming colliding particles at surface δAx , i.e. the force perpendicular to δAx , is, therefore, calculated as

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left (2mum )Nin , ∆t where 2mum denotes the change of momentum of a single colliding molecule and m its mass. Note that FδAx takes the same form whether δAx is a wall or located within the fluid. Recognizing that ρ = mn and simplification yields 1 FδAx = ρδAx um 2 . 3 Dividing by δAx finally gives the pressure in a semi ideal gas 1 p = ρum 2 . 3 It is important to note that the above analysis shows that first, the pressure in a fluid is isotropic, i.e. has no direction, since one sixth of the particles within a distance um ∆t from each arbitrary oriented plane δA move from one side to the other and vice versa. Second, the pressure arises from the molecular nature of fluids. The force f p on a fluid element can be calculated similarly to the derivation of the continuity equation. Again, we consider an Eulerian control volume δVE (Figure 6). Noting that the pressure is isotropic the pressure force on the leftmost surface of the control volume δVE is

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FδAx =

fxp, left δVE = δAx p(x0 − δx/2, y0 , t).

(34)

The force on the fluid element due to the pressure on the rightmost surface reads fxp, right δVE = −δAx p(x0 + δx/2, y0 , t).

(35)

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Combining equations (34) and (35) gives   fxp (x0 , y0 , t)δVE = δAx p(x0 − δx/2, y0 , t) − p(x + δx/2, y, t) Taylor series expansion and neglecting higher order terms yields ∂p fxp δVE = −δAx δx . ∂x Since δVE = δAx δx, we finally have ∂p fxp = − , ∂x ∂p fyp = − . ∂y

(36)

(37)

(38) (39)

From equations (38) and (39) we obtain that a non-zero pressure gradient produces a flow in the opposite direction to the gradient. For example, wind is induced by the pressure gradient between high and low pressure areas. Since the pressure gradient is directed from low to high pressure areas (i.e. the slope of the pressure is positive from a low to a high pressure area), the air flows from the high pressure area to the low pressure area. Normal and shear stresses in incompressible Newtonian fluids: In the last section, we have discussed the time rate of change of momentum due to the random molecular motion. However, if the local velocity gradients, these are ∂u ∂u ∂v ∂v , , and , ∂x ∂y ∂x ∂y are non-zero, additional forces on the surface of the control volume arise from the random motion of the molecules, which contribute to f m . The following example illustrates the y

m 21

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Figure 8. Illustration of the molecular momentum diffusion [2]. Railroad workers shovel coal from their own train to the other, which induces a change of momentum of both trains.

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underlying physics, which is referred to as molecular momentum diffusion [2]. Let us consider two parallel trains moving with different velocities v1 and v2 (Figure 8) with v2 > v1 . On train 1 railroad workers shovel coal from their own train to train 2 with a rate m ˙ 12 (in kg s−1 ). On train 2 equally hard working railroad workers shovel coal back to train 1 with a rate m ˙ 21 (kg s−1 ). For simplicity we assume that m ˙ 12 = m ˙ 21 . Thus, the masses of the trains do not change with time. The shoveling of coal causes train 2 to decelerate since the time rate of change of its momentum is negative

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F21 = m ˙ 12 v1 − m ˙ 21 v2 = m ˙ 12 (v1 − v2 ) < 0. | {z }

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0. | {z }

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Similarly, train 1 is accelerated since the time rate of change of its momentum is positive

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In the case of a very small distance 2λmf p between the trains we may write ∂v v1 = v|x=0 − λmf p , ∂x x=0 ∂v v2 = v|x=0 + λmf p . ∂x x=0

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In our example, v|x=0 is the mean velocity of both trains and 2λmf p is the average throwing distance of the railroad workers. The time rate of change of momentum of both trains, therefore, reads ∂v F12 = 2m ˙ 12 λmf p , ∂x F21 = −F12 .

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Applying this “train” model to the random motion of the molecules of a semi ideal gas the shoveling of coal from train 1 to train 2 can be regarded as the number of particles moving in positive x-direction nx . Similarly, n−x corresponds to the shoveling of coal from train 2 to train 1. λmf p is interpreted as the mean free path of the molecules, i.e. the average distance between two subsequent collisions. Thus, adopting the derivation of the pressure in ideal cases we have 1 m ˙ 12 = |{z} nm δAum . 6

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Hence, we obtain the time rate of change of momentum per unit area, that is the shear (1) stress τyx , which arises from a flow in y-direction and from the molecules moving in x-direction, 1 ∂v F12 (1) τyx = = ρum λmf p . (40) δA 3 ∂x (1)

In accordance with τyx a time rate of change of y-momentum per unit area results from a flow in x-direction and from the molecules moving in y-direction [2]   1 (2) τyx = ρum u(yδAy + λmf p ) − u(yδAy − λmf p ) . (41) 6

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Taylor series expansion of the last two terms in equation (41) and neglecting higher order terms gives 1 ∂u (2) = ρum λmf p . (42) τyx 3 ∂y Thus the total time rate of change of momentum reads   1 ∂u ∂v (1) (2) τyx = τyx + τyx = ρum λmf p + . (43) 3 ∂y ∂x

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It is common to define 1 µ = ρum λmf p , (44) 3 where µ denotes the molecular viscosity. From Figure 9 it can be deduced that τyx contributes to fym . Note that the velocity gradient is commonly denoted as shear rate since it has the dimensions of s−1 . τ yy

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Figure 9. Normal and tangential constraints due to flow in i-direction arising from molecular momentum diffusion in j-direction (i, j ∈ {x, y}. These can be taken into account by normal and shear stresses (normal and tangential force per area) acting on the Eulerian control volume δVE = δxδyδz. In the two-dimensional case δz is equivalent to δz ≡ 1.

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An additional contribution to fym is caused by a flow in y-direction and the molecules moving in y-direction. That is [11] ∂v τyy = 2µ , (45) ∂y which can be obtained from equation (43) by replacing x by y and u by v. Note that in case of vanishing velocity gradients the pressure forces remain the only force component resulting from random motion of molecules. By following the derivation of f p in section 5.2a and from Figure 9 we find ∂τyx ∂τyy fym = + . (46) ∂x ∂y Substituting equations (43) and (45) into equation (46) yields   ∂ ∂u ∂v ∂ 2v m + + 2µ 2 . (47) fy = µ ∂x ∂y ∂x ∂y

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Manipulating above equation and applying Schwarz’ theorem yields   ∂ ∂u ∂ 2 v ∂ 2v m + fy = µ +2 2 . ∂y ∂x ∂x2 ∂y

(48)

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By substituting the continuity equation for incompressible fluids (28) we obtain  2  ∂ v ∂ 2v m fy = µ . + ∂x2 ∂y 2

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By similar considerations fxm is calculated as  2  ∂ u ∂ 2u m fx = µ . + ∂x2 ∂y 2

(50)

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Note that the above derivation shows that τxy = τyx holds. Fluids, where τ is proportional to the velocity gradient (i.e. µ = const.), are referred to as Newtonian fluids. For example, air and water can be considered as Newtonian fluids. In contrast, fluids, where the viscosity itself is a function of the velocity gradient, are referred to as non-Newtonian fluids. For instance, ketchup shows non-Newtonian behavior.

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5.3. Summary of the governing equations for incompressible fluids

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Combining the results of sections 5.1 and 5.2 yields the set of incompressible NavierStokes equations, which is a set of partial differential equations. These consist of the continuity equation and the momentum equations. Since we have restricted ourselves to incompressible flows gravity f g can be included in the pressure term yielding (51)

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pˆ = p + ρgy.

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The quantity pˆ is sometimes called “total hydrostatic pressure” [14]. Thus, the NavierStokes equations for an incompressible fluid read ∂u ∂v + = 0, (52) ∂x ∂y   ∂u ∂u ∂u 1 ∂ pˆ µ ∂ 2 u ∂ 2 u +u +v =− + + , (53) ∂t ∂x ∂y ρ ∂x ρ ∂x2 ∂y 2   ∂v ∂v ∂v 1 ∂ pˆ µ ∂ 2 v ∂ 2 v +u +v =− + + . (54) ∂t ∂x ∂y ρ ∂y ρ ∂x2 ∂y 2

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The mathematical properties of solutions of the Navier-Stokes equations is one concern of the millennium problem referred to as ”The Navier-Stokes existence and smoothness problem” [3]. While numerical solutions of the Navier-Stokes equations are widely established in science and applications, the theoretical understanding of its solutions is quite incomplete [15]. Exact solutions are mostly restricted to special cases as discussed in the next section.

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Note on vectorial notation of the Navier-Stokes equations: Equations (52)–(54) can be presented in a more concise vectorial notation, which reads

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∇ · u = 0, ∂u 1 µ 1 + u · ∇u = − ∇ˆ p + ∆u + f b , ∂t ρ ρ ρ

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where u = (u, v), ∇ = (∂/∂x, ∂/∂y), u · ∇u = (u · ∇u, u · ∇v), ∇ˆ p = ∇p − ρg 2 2 2 2 and ∆ = ∂ /∂x + ∂ /∂y . In this notation ’·’ denotes the dot product and f b describes additional body forces, as for example, magnetic or electric forces. This notation allows easily the generalization of equations (52)–(54) to three dimensions by defining u = (u, v, w), ∇ = (∂/∂x, ∂/∂y, ∂/∂z), u · ∇u = (u · ∇u, u · ∇v, u · ∇w), ∆ = ∂ 2 /∂x2 + ∂ 2 /∂y 2 + ∂ 2 /∂z 2 .

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Note on initial and boundary conditions: Solutions of the initial value problem (52)– (54) are determined by initial and boundary conditions. The initial conditions for the governing equations for incompressible fluids (52)–(54) are given by

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(56)

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p(x, t0 ) = p0 (x),

(55)

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where u0 and p0 are the initial velocity and pressure fields. The boundary conditions for u and p for a given domain Ω can be expressed in different ways:

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(i) Dirichlet boundary conditions: A fluid property φ, that is u or p, is prescribed at the boundary ∂Ω of Ω as follows (57)

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(ii) Von Neumann boundary conditions: The gradient in normal direction to the boundary of a fluid property is specified as

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∇φ(x, t) · n(x, t) = φvN (x, t),

(58)

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with x ∈ ∂Ω and where φvN (x, t) is a given function. n denotes the outward unit surface normal and ∇ is defined as ∇ = (∂/∂x, ∂/∂y).

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(iii) The Robin boundary conditions represent a linear combination of the above, i.e. aφ(x, t) + b∇φ(x, t) · n = φR (x, t),

(59)

with x ∈ ∂Ω, a 6= 0, b 6= 0 and φR (x, t) given. Physical boundary conditions for pressure and velocity are usually a combination of Dirichlet- and von Neumann-type boundary conditions. At an inlet with specified velocity or a wall (Dirichlet boundary condition) a von Neumann (zero-gradient in normal direction) boundary condition must be supplied for pressure. For a constant pressure outlet, a zero-normal-gradient boundary condition for velocity must be specified. At symmetry planes zero-normal gradient boundary conditions must be used

What do the Navier-Stokes equations mean?

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for all flow quantities. Robin-type boundary conditions are hardly ever used for pressure and velocity but can, for example, occur for temperature boundary conditions when solving additionally the energy equation.

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Note on compressible fluids: In the case of compressible fluids additional terms arise in the momentum equations (53) and (54). These result from the time rate of change of the volume of the control fluid element moving with the flow. When the volume dilates (increases or decreases) the momentum of the control volume changes due to the random motion of the molecules. Additionally, the set of compressible Navier-Stokes equations is not closed. There are 3 partial differential equations but 4 unknowns (u, v, p and ρ). Therefore, we need an additional equation to close the system of equations. It is common to introduce an equation of state, for example the isothermal ideal gas law, which relates the density with the pressure, that is ρ = ρ(p).

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6. Exact solutions of the incompressible Navier-Stokes Equations

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In this section we discuss three simple examples of exact solutions of the NavierStokes equations, which show the influence of the viscosity of the fluid on the velocity distribution. These examples may also be subject of in-depth physic courses at senior high schools.

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6.1. Simple shear flow

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In the first example we consider the flow between two infinite parallel confining walls parallel to the x-axis. The flow is driven by the upper moving wall and a constant pressure is assumed. In literature such a flow is referred to a simple shear flow. Furthermore, in steady state the flow does not vary in x-direction. Thus, the velocity and pressure gradients read ∂v ∂ pˆ ∂u = 0, = 0, = 0. (60) ∂x ∂x ∂x Since ∂u/∂x = 0, it follows from the continuity equation (equation (52)) that

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∂v = 0. (61) ∂y Integration yields that v is constant. It is clear that v has to be zero because v 6= 0 would imply a flow through the confining walls, which is not possible. Thus, for the steady state, when the velocities do not vary with time, we have ∂u ∂u ∂ pˆ = 0, = 0, = 0, ∂t ∂x ∂x ∂v ∂v ∂v = 0, = 0, = 0. (62) ∂t ∂x ∂y

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Substituting equation (62) into equations (53) and (54) yields

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µ ∂ 2u = 0, (63) ρ ∂y 2 ∂ pˆ = 0. (64) ∂y Integration of equation (64) and using the definition of pˆ yields p(y) = −ρgy. The pressure varies linearly with y in order to counteract the gravitational force but has no influence on the velocity field. Hence, equation (54) is identically zero since v(x, y) = 0 and u is solely a function of y. Integration of the u-momentum equation gives (65)

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u(y) = ay + b,

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where a and b are constants of integration. These can be calculated by substituting the boundary conditions u(0) = 0 and u(h) = uh

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(66) (67)

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These boundary conditions assume that the relative velocity between the wall and the adjacent fluid particles is zero (no-slip condition). Solving the linear system of equations yields a = uh y/h and b = 0. Thus, we finally have a linear velocity profile uh (68) u(y) = y with y ∈ [0, h]. h The solution is shown in Figure 10a. It is noteworthy that the steady solution (t → ∞) for u does not depend on the viscosity, whereas the shear stresses acting on the walls are ∂u ∂u µuh τxy = µ . (69) =µ = ∂y y=0 ∂y y=h h

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Thus, increasing the viscosity of the fluid increases the force required to move the upper constraining wall with uh , which is µuh A F = , h where A is the area of the wall. However, for inviscid fluids, i.e. µ = 0 and therefore no momentum diffusion, we would obtain u(y) = 0 since there is no physical process, which transfers the momentum of the constraining walls to the fluid. 6.2. Two-dimensional shear flow with pressure gradient

Now the previous example will be modified by dropping the assumption ∂p/∂x = 0, so that the fluid will not only be driven by the wall movement, but also due to the pressure gradient. We consider a two-dimensional channel of length l and height h, where we apply at the inlet a pressure pin and at the outlet a pressure pout . Thus, the pressure gradients read ∂ pˆ pout − pin ∂ pˆ = , = 0. (70) ∂x l ∂y

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Figure 10. a) Simple steady shear flow (Couette-flow) between two constraining walls at a distance h. The upper wall moves with a velocity uh in positive x-direction; b) Steady flow through a two-dimensional channel of length l and height h (Poiseuilleflow). At the inlet a pressure pin and at the outlet a pressure pout are applied.

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By using the definition of pˆ (gravity acts in negative y-direction) equation (63) now becomes µ ∂ 2u 1 ∆p = (71) ρ l ρ ∂y 2 by introducing ∆p = pout − pin . Integration gives ∆p 2 u(y) = y + by + c, (72) 2µl

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where b and c are constants of integration. Applying the boundary conditions u(0) = 0 and u(h) = uh yields

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u(0) = c = 0, (73) ∆p 2 u(h) = uh = h + bh. (74) 2µl The constant b can easily be calculated from the second equation and we finally obtain uh ∆p u(y) = − y(h − y) + y. (75) 2µl h The result is a superposition of a quadratic velocity profile due to the pressure gradient and the linear profile due to the wall movement already known from equation (68). The two special cases, when one of the two velocity components vanishes, are both named after french physicists. The linear profile when ∆p = 0 is called Couette-flow. For uh = 0 and ∆p 6= 0 the velocity profile is of a parabolic shape and is called Poiseuille-flow (compare with Figure 10b). Its maximum velocity in the center of the channel is umax = u(h/2) = −(∆p)h2 /8µl. Increasing the viscosity (for example using honey instead of water) leads to a decrease of the maximum velocity and to a decrease

What do the Navier-Stokes equations mean?

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of the mass flow through the channel, that is   h Z h ∆p y 2 h y 3 ρu(y)dy = −δz − q0 = δz 2µl 2 3 0 0 δz ∆p 3 =− h. (76) 12 µl δz denotes the “width” of the channel in the third spatial direction to receive the right dimensions of the mass flow (kg s−1 ). This behavior sounds plausible since viscosity is commonly denoted as internal friction. Finally, note that if pout < pin the flow is aligned in positive x-direction and if pin < pout the flow is aligned in negative x-direction (compare with section 5.2). Before visualizing the general superposed solution, it is common to nondimensionalize the solution by dividing equation (75) by uh y y y u(y) = −P 1− + , (77) uh h h h p h2 with P = − , (78) l 2µuh where P is a dimensionless pressure gradient. The velocity profiles for various values of P are shown in Figure 11a. For P = ±1 the profiles are vertical in the vicinity of the upper and lower wall, respectively. For P > 1 the velocity in the channel exceeds the wall velocity. For P < −1 the strong adverse pressure gradient leads to local backflow near the lower wall. The viscous force between the fluid layers is not strong enough to overcome the adverse pressure gradient. In pratice, simple Couette flows are very important in rheology, viscosimetry and rheometry, the studies of the viscous shear behaviour of liquids and their experimental investigation, respectively. The main advantage of Couette viscometers is linear relation between the wall force and the fluid viscosity (see equation (69)), which allows for a simple determination of the viscosity by force or moment measurements. Poiseuille-Couette flows are an important aspect in lubrication theory, or, more generally, for flows through small gaps and channels with moving boundaries, encountered e.g. in the coating process of surfaces, non-hermetic sealing, etc.

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6.3. Poiseuille flow with wall suction/injection

A simple generalization of the Poiseuille flow is possible, if we consider the flow through a channel made of porous walls. Fluid can enter or exit the channel via these porous walls. We consider the same setting as before, but with the upper wall at rest (uh = 0). Fluid is now injected through the lower wall (y = 0) with a constant velocity vw and also exits at the upper wall (y = h) with the same constant velocity. If the channel is assumed to be infinitely long, we can still chose the velocity independent of x. The continuity equation then reads ∂u ∂v ∂v + = = 0 → v(y) = vw = const. (79) ∂x ∂y ∂y

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What do the Navier-Stokes equations mean?

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Figure 11. a) Different solutions for a Poiseuille-Couette-flow between two constraining walls at a distance h. P is the dimensionless pressure gradient according to equation (78); b) Velocity profiles of a steady flow through a two-dimensional channel height h with uniform crossflow at different crossflow Reynolds numbers Rev for a given pressure gradient in x-direction.

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Hence, there is a constant vertical fluid motion in the channel. The influence of this uniform crossflow on the flow in x-direction will be calculated in the following. In contrast to the previous examples the term v ∂u in equation (53) is no longer zero. ∂y Hence, the x-momentum equation becomes

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∂u 1 ∆p µ ∂ 2 u =− + . (80) ∂y ρ l ρ ∂y 2 Thus, we have to deal with an inhomogeneous ordinary differential equation of second order with constant coefficients. First we solve the homogeneous part of the differential equation

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∂ 2u ∂u −a = 0, 2 ∂y ∂y

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with a = vw ρ/µ. The characteristic polynomial of this equation is λ2 − aλ = 0

with the solutions λ1 = 0 and λ2 = a. The solution of the homogeneous equation, therefore, is uhom (y) = c1 eλ1 y + c2 eλ2 y = c1 + c2 eay . Since the inhomogeneous pressure term is constant and λ1 = 0, we assume a linear particular solution ypar = c3 y. Substitution into the differential equation yields c3 = −

1 ∆p y ∆p → upar (y) = − . ρ vw l ρ vw l

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Superposition of the homogeneous and the particular solutions leads to vw y y ∆p . u(y) = c1 + c2 e ρµ − ρ vw l The constants c1 and c2 can be determined from the no-slip condition at the walls u(0) = u(h) = 0. The solution finally becomes vw y

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y ∆p h ∆p 1 − e ρµ . (81) u(y) = vw h − ρ vw l 1 − e ρµ ρ vw l If we introduce dimensionless quantities this solution can be written in a more simple and general form. By relating the x-velocity to the maximum velocity of the Poiseuille-flow, that is u0,max = −(∆p)h2 /8µl and by introducing a dimensionless crossflow Reynolds number ρ vw h , (82) Rev = µ the solution reads   u(y) 8 y 1 − eRev y/h = − . (83) u0,max Rev h 1 − eRev Calculating the mass flow and relating it to the result for the Poiseuille-flow q0 from equation (76) yields   12 1 1 q 1 = − − . (84) q0 Rev 2 Rev 1 − eRev Several solutions for the velocity profiles at different cross-flow Reynolds numbers are shown in Figure 11b. The velocity profile is shifted upwards, away from the wall with fluid injection. The maximum velocity and the mass flow decrease with higher injection/suction rates. At higher injection/suction rates, i.e. higher crossflow Reynolds numbers, the profile becomes nearly linear throughout most of the channel height. Applications of channel flows with superposed crossflow are common in the chemical industry, for filter applications, etc.

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6.4. Further analytic solutions

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A wide range of other problems exist, where it has been possible to solve the NavierStokes-equations analytically. Covering them here exceeds the scope of this paper, especially since these problems are well covered in existing literature [2, 16, 17, 14, 18, 19]. Some examples are the stagnation point flow in the vicinity of a wall positioned normal to the oncoming flow (Hiemenz-flow), the flow through converging and diverging channels (Jeffrey-Hamel-flow), the diffusion of a vortex in a viscous fluid over time (Lamb-Oseen vortex), etc. 7. Summary In this article we have attempted to make the Navier-Stokes available to a wider readership, especially teachers and undergraduate students, by outlining the underlying physical principles and assumptions.

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Acknowledgments The authors would like to thank Professors Urbaan Titulaer, Erich Steinbauer and Stefan Pirker for many useful discussions and helpful comments.

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References

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