Fluid Dynamics for Engineers

Transcription

1 Fluid Dynamics for Engineers Dominique Thévenin (TEX version) & Gábor Janiga (WWW version) July 10, 2014

2 Copyright c D. Thévenin & G. Janiga. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.3 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the Appendix entitled GNU Free Documentation License.

3 Contents Preface 9 List of symbols 11 1 Introduction Practical importance of fluid dynamics What is a fluid? Continuum assumption Important flow variables and variable-based classification Basic concepts Mathematical operators Gradient Divergence Laplacian Rotor or Curl Time derivatives Characteristic flow structures Control volume Transport theorem Mass conservation Introduction Point of view of physics Point of view of mathematics Integral formulation of mass conservation Local formulation of mass conservation Local formulation of mass conservation in cylindrical coordinates Local mass conservation for an incompressible flow Euler equation: conservation of momentum in a non-viscous flow Introduction Point of view of mathematics Point of view of physics Integral formulation of momentum conservation Local formulation of momentum conservation Local momentum conservation for an incompressible flow Integral formulation of angular momentum conservation

4 2 5 Hydrostatics and Aerostatics Introduction Fundamental equation of hydro- and aerostatics Pressure variation within an incompressible, static fluid Force exerted by an incompressible, static fluid, on a fully immersed body Force exerted on a partially immersed body Stability of a partially immersed body Aerostatics Pressure variation in an isothermal, ideal gas Principle of Archimedes in a gas Bernoulli equations Introduction Bernoulli equation for an irrotational flow Link with hydrostatics Bernoulli equation (for a rotational flow) The Bernoulli triangle Simplification of the Bernoulli equation for a gas flow Dynamic pressure Averaged Bernoulli equation Hydraulic height Generalized Bernoulli equation with losses and energy exchange Computing the exchanged specific work w Computing the friction loss e f Numerical equations used to estimate the friction factor f Computing a localized loss e l Force and torque exerted by a flow Introduction Force exerted by a flow on its surroundings Force exerted by a flow on a pipe wall surrounded by a fluid at constant pressure Torque exerted by a flow on its surroundings Torque exerted by a flow on a pipe wall surrounded by a fluid at constant pressure Movement of a material control volume Introduction Movement of a material control volume Deformation tensor d Rotation tensor Ω Navier-Stokes equation: conservation of momentum in a viscous flow Introduction Point of view of mathematics Point of view of physics Pressure component T p of the stress tensor σ Friction component τ of the stress tensor σ Full stress tensor σ Integral formulation of momentum conservation Local formulation of momentum conservation

5 9.6 Local momentum conservation for an incompressible flow Local formulation of momentum conservation for a non-newtonian fluid Local formulation of momentum conservation for a Newtonian fluid Local formulation of momentum conservation (incompressible flow, Newtonian fluid) Local formulation of momentum conservation in cylindrical coordinates Dimensional analysis and similarity conditions Introduction Non-dimensional conservation equations Non-dimensional mass conservation for an incompressible flow Non-dimensional momentum conservation (incompressible flow, Newtonian fluid) Non-dimensional parameters of Fluid Dynamics Strouhal number St Froude number Fr Euler number Eu Reynolds number Re Further non-dimensional parameters Choosing the reference quantities Summary: non-dimensional conservation equations A faster solution: the Π-theorem Relevant dimensional variables Similarity conditions in Fluid Dynamics One-dimensional isentropic compressible flows Introduction and hypotheses Generic relations, also valid for a real gas Conservation equations What is a compressible gas flow? Influence of a modification of the cross-section A Critical conditions Laval nozzle Specific relations for a compressible flow of a perfect gas What is a perfect gas? Isentropic relations for a perfect gas Speed of sound for a perfect gas Analytical solution for a compressible flow of a perfect gas Critical conditions Solution procedure and remaining difficulties Minimal stagnation pressure for a properly working Laval-nozzle Tables for compressible flows Solution using the critical Mach number M Discharge velocity v d Conclusions Compressible flows with friction and heat exchange Introduction and hypotheses Generic relations, also valid for a real gas Conservation equations Generalized equation of Hugoniot

6 Pressure variation for a constant flow cross-section Influence of heat exchange in a perfect gas Mass conservation Conservation of momentum Energy conservation Solution procedure Flow modifications Thermal choking Influence of friction Mass conservation Conservation of momentum Energy conservation Qualitative analysis Quantitative solution procedure Flow modifications Conclusions Shock waves Introduction Normal shock wave Considered configuration and hypotheses Conservation equations for a real gas Conservation equations for a perfect gas Jump relations involving M 1 and M Relation between the Mach numbers upstream and downstream of the normal shock Jump relations involving only the upstream Mach number M Necessary condition on M 1 for the existence of a shock Shock relation of Prandtl Summary: evolution of all quantities through a normal shock Normal shock tables Solution and graphical representation using the critical Mach number Relation of Rankine-Hugoniot Rayleigh line Propagating shock waves Why shock waves? Oblique shock wave Conservation equations for a real gas Conservation equations for a perfect gas Jump relations involving the upstream Mach number M Summary: evolution of all quantities through an oblique shock Using the shock tables for an oblique shock Determining the shock angle ε Mach angle and Mach wave Polar curve and Busemann diagram Boundary conditions and shock reflections Conclusions

7 14 Introduction to turbulence Turbulence: complexity and importance A first taste of turbulence: the experiment of Reynolds Qualitative properties of turbulent flows A Basic concepts and keywords of fluid dynamics 201 A.1 Archimedes number A.2 Cavitation A.3 Compressible flow A.4 Compressible fluid A.5 Conservative force A.6 Contact force vs. non-contact force A.7 Hydraulic diameter A.8 Incompressible flow A.9 Incompressible fluid A.10 Internal flow A.11 Irrotational flow A.12 Laminar flow A.13 Mach number A.14 Multiphase flow A.15 Newtonian fluid A.16 Non-Newtonian fluid A.17 Non-viscous flow A.18 One-dimensional flow A.19 Open channel flow A.20 Potential flow A.21 Quasi-steady flow A.22 Speed of sound A.23 Standard coordinate system A.24 Steady flow A.25 Stress in a fluid A.26 Turbulent flow A.27 Unsteady flow A.28 vena contracta A.29 Viscosity B Basic thermodynamic concepts needed for fluid dynamics 213 B.1 Adiabatic process B.2 Barotropic state B.3 Enthalpy B.4 Entropy B.5 Gas constant B.6 Heat capacity B.7 Heat capacity ratio B.8 Ideal gas B.9 Isentropic transformation B.10 Isobaric transformation B.11 Isochoric transformation B.12 Isothermal transformation

8 6 B.13 Perfect gas B.14 Polytropic process B.15 Prandtl number B.16 Real gas B.17 Reversible process B.18 Specific quantity B.19 Standard thermodynamic conditions B.20 Thermal conductivity C Basic mathematical concepts needed for fluid dynamics 219 C.1 Angular relations in a right triangle C.2 Conic curves C.3 Divergence theorem C.4 Summation convention of Einstein C.5 Logarithmic differential C.6 Partial derivative C.7 Scalar product C.8 Surfaces and volumes C.8.1 Circle or disk C.8.2 Sphere C.8.3 Cylinder C.8.4 Cone C.9 Vectors C.10 Vector product C.11 Taylor expansion C.12 Tensors D Biography of selected important scientists 225 D.1 Archimedes D.2 Amedeo Avogadro D.3 Daniel Bernoulli D.4 Blasius D.5 Ludwig Boltzmann D.6 Edgar Buckingham D.7 Adolf Busemann D.8 Henry Darcy D.9 Leonard Euler D.10Richard Feynman D.11William Froude D.12Otto von Guericke D.13Galileo Galilei D.14Carl Friedrich Gauß D.15George Green D.16Pierre Henri Hugoniot D.17Martin Knudsen D.18Gustaf de Laval D.19Joseph Louis Lagrange D.20Horace Lamb D.21Pierre-Simon de Laplace

9 D.22Gottfried Wilhelm Leibniz D.23Leonardo da Vinci D.24Ernst Mach D.25Julius Robert von Mayer D.26Claude Louis Marie Henri Navier D.27Isaac Newton D.28Mikhail Vasilievich Ostrogradsky D.29Blaise Pascal D.30Ludwig Prandtl D.31William Rankine D.32Baron Rayleigh D.33Osborne Reynolds D.34George Gabriel Stokes D.35Vincenc Strouhal D.36Aimé Vaschy D.37Theodore von Kármán D.38Julius Weisbach E Application table for subsonic compressible flows 229 F Application table for supersonic flows, shock waves and rarefaction waves 233 G GNU Free Documentation License APPLICABILITY AND DEFINITIONS VERBATIM COPYING COPYING IN QUANTITY MODIFICATIONS COMBINING DOCUMENTS COLLECTIONS OF DOCUMENTS AGGREGATION WITH INDEPENDENT WORKS TRANSLATION TERMINATION FUTURE REVISIONS OF THIS LICENSE RELICENSING

10 8 Preface

11 Preface This Web-book has been written over many years, the first chapter having been released internally in It has been primarily developed as a support of the corresponding lectures given by the main author, Dominique Thévenin, at the University of Magdeburg Otto von Guericke since Most of the chapters dealing with compressible flows have been already published as a paper document during the nineties, as D. Thévenin was still teaching at the Ecole Centrale Paris. Let me thank here Prof. Sébastien Candel: under his kind supervision, I finally learned (I think!) what is Fluid Dynamics...I would very much recommend the reading of his book [Can90] to all those that can understand French. Sethuraman Ramalingam helped writing some of the equations included in this document. Gordon Fru contributed all figures obtained by Direct Numerical Simulations. Further figures have been contributed by Nico Krause. Thomas Hagemeier was an excellent proofreader of the book. Many thanks to all of you! Finally, let me thank also the developers of TEX and L A TEX: without this wonderful tool, I would never have been able to find enough time to write this document. Magdeburg, July 2014 For the authors, Dominique Thévenin 9

12 10 List of symbols

13 List of symbols You will find here a unified and complete description of all notations and symbols used in the present document. Writing conventions Note that, throughout this document, bold symbolds (for example v) correspond to vector variables, while associated standard symbols (for example v) denote a scalar quantity. Tensors (to be exact, second-order tensors) will be written with a so-called Sans Serif police, like for example in τ. Concerning thermodynamic properties, we will stick to the classical convention stating that lowercase symbols correspond to specific quantities (i.e., per unit mass). The summation convention of Einstein will be used every time it is applicable. Thus, repeated identical indices in a term must be interpreted as a summation over all possible values. Symbol Signification constant constant value constant >0 strictly positive constant value := definition approximately equal to proportional to vector product scalar product adding surfaces or volumes nabla (fixed unit: 1/m) [φ] unit of the variable φ ( ) φ for a constant value of φ line (one-dimensional) integral surface (two-dimensional) integral volume (three-dimensional) integral 11

14 12 List of symbols Lowercase latin symbols Symbol Signification Unit c speed of sound m/s c p specific heat capacity at constant pressure J/(kg.K) c v specific heat capacity at constant volume J/(kg.K) d diameter m d h hydraulic diameter m d deformation tensor components in 1/s e specific internal energy J/kg e loss of specific energy J/kg e x,e y,e z unit vectors associated to directions x,y,z - f friction factor - g gravity acceleration vector components in m/s 2 h specific enthapy J/kg k B Boltzmann constant, k B := J/K l length m ṁ mass flow-rate kg/s n unit vector normal to a given fluid-containing - surface, pointing toward outside n polytropic exponent - p pressure Pa p pressure head loss Pa q dynamic pressure Pa q volumetric flow-rate m 3 /s r := R/W specific gas constant J/(kg.K) s unit vector tangential to the local fluid velocity - s specific entropy J/kg t time s t stress components in Pa v velocity vector components in m/s v magnitude of velocity vector, v := v m/s v d discharge velocity m/s v 1,v 2,v 3 or v x,v y,v z components of velocity vector m/s w velocity of control volume components in m/s w specific work J/kg x position vector components in m x or x 1 first spatial coordinate m y or x 2 second spatial coordinate m z or x 3 third spatial coordinate m z elevation head loss m

15 13 Uppercase latin symbols Symbol Signification Unit A surface area m 2 A geometric surface, outer surface - C vc contraction ratio for vena contracta - F force N F impulsion N G unspecified mathematical function - H total height, hydraulic head m I identity matrix - K loss coefficient - M mass kg M shock Mach number - N A Avogadro constant, N A := /mol P momentum (product mass velocity) kg.m/s P w wetted perimeter (fluid/wall contact length) m P power W R universal gas constant R = J/(mol.K) T temperature K T torque N.m T generic stress tensor components in Pa V volume m 3 V three-dimensional body - V c control volume - W molar mass of a gas kg/mol X starting position (Lagrangian system) components in m Lowercase greek symbols Symbol Signification Unit α thermal diffusivity (or thermal diffusion coefficient) ( ) m 2 /s α p isobaric thermal expansion coefficient, α p := 1 ρ 1/K ρ T ( ) p β T isothermal compressibility coefficient, β T := 1 ρ 1/Pa ρ p T γ heat capacity ratio - δdeflection angle ǫ wall roughness height m ε shock angle η efficiency of a conversion process, η 1 - λ thermal conductivity W/(m.K) µ viscosity (or dynamic viscosity) kg/(m.s) or Pa.s ν kinematic viscosity, ν := µ/ρ m 2 /s ρ density kg/m 3 σ stress tensor components in Pa τ friction tensor components in Pa φ or ϕ unspecified variable depends

16 14 List of symbols Uppercase greek symbols Symbol Signification Unit Propagation speed of a normal shock in a quiescent atmosphere m/s Λ mean free path of fluid particles m Π pressure jump through shock - Ω rotation vector components in 1/s Ω rotation tensor components in 1/s Indices Symbol Signification critical condition non-dimensional value reference value 0 isentropic stagnation value a related to air, to the atmosphere b related to a body f related to a fluid g related to gravity gas related to gas i inflow condition liq related to liquid o outflow condition related to pressure p Non-dimensional numbers Symbol Name Describes Ar Archimedes number influence of buoyancy Eu Euler number influence of pressure (variation) Fr Froude number influence of gravity Kn Knudsen number continuum assumption M := v/c Mach number compressibility effects Pr Prandtl number momentum diffusivity vs. thermal diffusivity Re Reynolds number turbulence and viscosity effects Re s Reynolds number based on length-scale s turbulence and viscosity effects St Strouhal number unsteady effects

17 15 Unit conversions Conversion For 1 atm := Pa pressure 1 bar := Pa pressure 1 centipoise := 0.01 Poise dynamic viscosity (or 1 cp) 0 C := K temperature 1 mole := molecules - 1 Poise := 0.1 Pa.s dynamic viscosity

18 16 List of symbols

19 Chapter 1 Introduction This chapter describes a few basic issues associated with fluid dynamics. Note that the concepts listed alphabetically in the appendices might also be useful at this level, in particular for already experienced readers. 1.1 Practical importance of fluid dynamics Fluid dynamics is essential for so many applications that it is impossible to list all of them here. It describes for example atmospheric behavior (weather forecast), is a key element for many environmental and security issues (floods, typhoons, explosions...), is central for all transportation systems (from cars to aircrafts and spaceplanes) as well as energy production (internal combustion engines, turbomachines like gas turbines and wind turbines, cooling of nuclear power plants, pollutant emissions in coal-firing power plants...). For a start, a detailed visit of the reference Web-site called Efluids is probably the best idea. You will find in particular there a large collection of impressive photographs illustrating many aspects of Fluid Dynamics. As a complementary source of information, the application Web-site of the commercial software ANSYS-Fluent is also describing many actual problems of Fluid Dynamics, that might be solved by numerical simulation. 1.2 What is a fluid? Per definition, a fluid is a substance that continuously deforms when a certain stress (i.e., force per unit surface) is applied parallel to its surface (a so-called shear stress or shear force). The fluid does not come back to its original form after disappearance of the stress (at the difference of the elastic deformation of a solid, for example). A pure fluid can be either a liquid or a gas. A real, complex fluid might also involve a mixture of a liquid and gas phase (e.g., in a bubble column), possibly containing also some amount of solid particles (suspensions). The difference between a gas and a liquid is that surface tension will play an important role at the free surface of a liquid, while a gas will always occupy all the available volume, without apparition of a free surface. All gases are fluids. For liquids, the situation is somewhat more complex. Liquids with a simple behavior, which will be defined later as Newtonian, are obviously fluids: in this case, there is a linear relation between applied shear stress and liquid deformation and the corresponding straight line goes through the origin (0, 0). Non-Newtonian liquids might behave in a much more complex way. In particular, such liquids might be able to withstand shear stress without deformation up to a certain level. They therefore constitute 17

20 18 Chapter 1. Introduction Figure 1.1: A few examples of important problems and applications involving Fluid Dynamics. All photos from FreeFoto apart hurricane (from Wikipedia). a link between liquids and solids. Nevertheless, the threshold associated with the onset of deformation is usually quite low, much below the corresponding threshold for a solid (limit of plastic deformation). Therefore, the difference between such a liquid (fluid) and a solid (non-fluid) is still appearent. Note, however, that the separation between a fluid and a solid might still be a subject of controversy for some exotic cases. This is in particular the case for amorphous solids (like glass, which is claimed to be able to flow under certain circumstances), for plasmas (a very special state of matter), or for some polymer products. A funny video illustrating the unusual possibilities associated with Non-Newtonian liquids can be found for instance under Efluids! 1.3 Continuum assumption Any fluid (liquid or gas) is constituted by individual molecules. Therefore, it would be in principle possible to describe the state and movement of this fluid by considering the individual movements of all molecules together with their mutual interactions and by finally summing up all individual contributions. This is indeed realizable in practice, at least for some simple conditions; but this is not what is called Fluid Dynamics, and will therefore not be considered further in the present document. Researchers working at molecular level deal with a (very interesting) part of science called Statistical Mechanics (or Statistical Physics), founded by Boltzmann. Even if this approach is very interesting, and sometimes the only possible way, it is much too difficult and cumbersome for many practical applications: billions of individual molecules must be considered before obtaining the resulting flow conditions at (our) macroscopic scale. Therefore, Fluid Dynamics do not consider individual molecules in a fluid. Instead, the so-called

21 3 Continuum assumption 19 Continuum Assumption is employed. This means that, from the point of view of Fluid Dynamics, there are no molecular bricks and no holes within a fluid: it is a continuum state of matter; all flow variables can be defined at any point within this fluid. How is it possible to move from physical reality (existence of well-seperated molecules at a very small scale) to the Continuum Assumption? Simply by a specific averaging process in space! This means in practice that, from the point of view of Fluid Dynamics, a point is associated with a finite volume, at the difference of the rigorous, mathematical definition of a point (infinitely small, volume is necessarily zero). A point for Fluid Dynamics, which will be called more usually a fluid element, is associated with a volume V c, very very small but nevertheless verifying V c > 0! Indeed, the volume is chosen in such a manner that a huge quantity of individual molecules are always contained within this volume. In this manner, it is possible to smooth out the fast and chaotic variations associated with individual molecules, and to obtain macroscopic fluid properties like density, pressure, temperature or velocity. This is illustrated in figure 1.2, where the correct definition of local density ρ in the framework of Fluid Dynamics is considered for the convective flow above a candle. In a thought experiment, a control volume of varying extent is centered around a fixed point P. The corresponding volume V is measured together with the mass of the fluid contained within V, written M. The ratio M/ V is expressed in kg/m 3 and would be suitable to define the local fluid density. Now, the macroscopic size of the control volume influences of course our measure of density, M/ V. If the control volume is too large, very inhomogeneous flow conditions are found within the control volume. Cold air from the surroundings is found within V together with hot air from the candle plume. As a consequence, the resulting measured density at point P varies with V: this is obviously not acceptable. On the other hand, if V is chosen to be extremely small (near molecular scale), then it will contain only very few molecules. Repeating the experiment several times with the same control volume, one would get perhaps once 6 molecules, once 3, once 9 within V. The corresponding measured density would therefore appear to be different for each measure. This is again not acceptable! Fortunately, there is a (in fact relatively large) region in-between, where a plateau would be found experimentally for our measured density : this is where Fluid Dynamics is applied. This plateau extends down to a lower size limit V c, used from now on to delineate the continuum regime. Let us further illustrate this point by considering air under standard thermodynamic conditions and assuming that the volume V c of a fluid element is typically (1µm) 3, the volume V c being in this case considered as a cube with a side length of 1 µm; clearly, this is extremely small compared to the human scale! But what about molecules? Air being an ideal gas, it is one basic property that 1 mol (containing molecules, the Avogadro constant) will occupy roughly a volume of 22.4 dm 3 (or liter) under such conditions. By a simple proportionality rule, we obtain that the volume V c contains roughly 27 millions of molecules! This is obviously sufficient, by averaging over all the individual properties of the molecules, to obtain a smooth value for all needed fluid properties at macroscopic scale. Obviously, the Continuum Assumption means also that Fluid Dynamics cannot describe accurately effects that take place below the associated scale: microscopic effects must be described appropriately by adding corresponding models to the equation. Furthermore, the appropriate volume V c of a fluid element will depend on the local flow conditions. For example, when considering the upper atmosphere (a very diluted gas, corresponding to an extremely low density), a volume V c of several cubic meters or even more will be required to accumulate a sufficient number of molecules. In order to define in a rigorous manner the boundary defining the validity of Fluid Dynamics concepts, the Knudsen number is introduced. This is one major non-dimensional number associated with Fluid Dynamics, and is defined as: Kn := Λ L (1.1) where Λ is the mean free path of the fluid particles (i.e., the mean travel distance of a molecule between

22 20 Chapter 1. Introduction Control volume: Mass M Volume V Control volume too small: molecular effects! Control volume too large: non-homogeneous conditions! P M V density Continuum approximation: validity of Fluid Dynamics! V c V Figure 1.2: Defining in a thought experiment the fluid density ρ at a point P within a candle plume using a control volume of varying size. two collisions with another molecule) deduced from the kinetic theory and L is a characteristic (macroscopic) length scale of the considered flow. The mean free path can be computed for an ideal gas using following equation: Λ = k BT (1.2) 2πd2 p where all variables are standard and defined in the Nomenclature; in particular, k B is the Boltzmann constant and d is the collision diameter of the considered gas particles. Fluid Dynamics deal with problems corresponding to Kn 1, sometimes up to Kn< 1, while statistical physics must be employed if Kn 1; in the latter case, the typical scale of the problem is comparable with the mean free path, so that individual particle movements at the molecular scale must be taken into account. 1.4 Important flow variables and variable-based classification In order to understand fluid dynamics and classify different applications, it is useful to understand what are the variables really needed to describe the local, instantaneous state of a fluid. In this document, we will employ following variables: the fluid pressure p, the fluid density ρ, the fluid velocity v,

23 4 Important flow variables and variable-based classification 21 Figure 1.3: Space shuttle just before landing (left) or at the beginning of atmospheric re-entry (right). The left picture corresponds to a problem solvable by Fluid Dynamics (mean free path much below typical flow scale). The right picture corresponds to a problem solvable by Statistical Physics (mean free path roughly equal to typical flow scale, the black points representing gas molecules). and a variable describing the internal energy of the fluid, either in the form of the specific enthalpy h or of the temperature T. The fluid pressure p is the normal stress component within a fluid. It is a scalar quantity, since pressure in a fluid is isotropic and thus acting equally in all directions. It is expressed in Pascal (Pa). The fluid density ρ is the ratio between the total mass and the total volume of a fluid element, such as defined in Section 1.3. It is therefore expressed in kg/m 3. The fluid velocity v is the ratio between the total momentum and the total mass of a fluid element, such as defined in Section 1.3. It is therefore expressed in m/s. The fluid specific enthalpy h is related to the fluid specific internal energy e by the fundamental relation h := e+ p (1.3) ρ Temperature T (expressed in Kelvin, K) is a thermodynamic notion, which is directly connected to the specific internal energy e of the considered fluid. It is now possible to classify the different applications we will consider in this document by looking at the important variables for this case. We will begin applications by considering in Chapter 5 Hydrostatics and Aerostatics, i.e. non-flowing flows. In the case of Hydrostatics, only the fluid pressure p will be variable, all other variables being constant. For aerostatics, pressure, density and temperature will vary, while velocity will still be constant and equal to zero. After that, we will consider the Bernoulli equation. In that case, we will consider only incompressible flows, and only pressure p and velocity v will be important. When considering the forces induced by a fluid, or the Navier-Stokes equations, all three variables, p, v and ρ will be considered variable. Finally, for the most complex applications (compressible flows), all variables introduced previously will really vary. The situation is summarized in table 1.1 and in figure 1.4. Considering a flow perpendicular to a given cross-section A associated with an area A, the flow velocity v (of magnitude v) and fluid density ρ introduced previously can be readily combined to compute the mass flow-rate ṁ, expressed in kg/s through: ṁ := ρva (1.4)

24 22 Chapter 1. Introduction Application Important variables Complexity level Hydrostatics p Very low Aerostatics p (and ρ, T through thermodynamic relations) Low Incompressible flow p, v Intermediate Forces exerted by fluids p, v, ρ Intermediate Generic Navier-Stokes p, v, ρ High Compressible flow p,v,ρ,h (or T) High Table 1.1: Important flow variables for different domains of application with a growing level of complexity 0 p a pressure depth problem complexity Figure 1.4: Two flow problems at a very different complexity level, from the hydrostatic pressure distribution in a water volume at rest (photo from FreeFoto) to a starting space rocket of type Ariane 5 (photo from Arianespace). Similarly, they can be employed to compute the volumetric flow-rate q, expressed in m 3 /s through: q := va (1.5) There is obviously a direct link between both flow-rates: ṁ = ρ q (1.6)

25 Chapter 2 Basic concepts This chapter describes some basic concepts of Fluid Dynamics that will be used throughout this document. Note that a much more complete list of useful concepts organized alphabetically is also proposed in Appendix A. 2.1 Mathematical operators In Fluid Dynamics, different mathematical operators will be used very often to compute important flow quantities and to write corresponding conservation equations. All these operators employ partial derivatives. The most important ones are introduced now. They are illustrated by applying them for analyzing a turbulent non-premixed hydrogen flame computed in our research group using Direct Numerical Simulations (figure 2.1). 8 x 10 3 Velocity field (vector plot) Y [m] X [m] x 10 3 Figure 2.1: Instantaneous structure of a mildly turbulent non-premixed flame computed using Direct Numerical Simulations. Left: density; right: velocity vectors Gradient In Fluid Dynamics, the gradient will be introduced to quantify the variation of a function in space. Typically, the gradient operates on a scalar quantity φ and delivers a vector quantity, written grad(φ) 23

26 24 Chapter 2. Basic concepts or more often φ, defined as φ := ( φ x, φ y, φ ) z (2.1) By computing the gradient of a scalar quantity (figure 2.2), one obtains a vector field. By plotting this vector field, one gets directly a very good feeling concerning the spatial evolution of φ: the resulting vectors show the direction of fastest changes of φ; the magnitude of these vectors tells us how fast these changes are. 8 x 10 3 Gradient of density field Y [m] X [m] x 10 3 Figure 2.2: Instantaneous structure of a turbulent non-premixed flame computed using Direct Numerical Simulations. Left: density; right: gradient of density. Later, we will also consider gradients of a vector quantity, resulting in a tensor. Further information can be found for instance under Wikipedia Divergence In Fluid Dynamics, the divergence will be mostly introduced to determine if vectors tend to diverge (pointing in various directions starting from a common origin) or to converge (pointing onto the same point starting from different origins). Mostly, we will compute the divergence of the flow velocity, the vector quantity v, and we will obtain its divergence, a scalar quantity written div(v) or more often v, and defined as v := v x x + v y y + v z z The divergence of the flow velocity (figure 2.3) is particularly interesting, since we will demonstrate later that, for an incompressible flow local mass conservation can be simply written v = 0. Later, we will also consider the divergence of a tensor, resulting in a vector. This is simply the result obtained when considering each line of the tensor (containing three components) as a vector and computing the divergence as usual. Thus, each line leads to a scalar value (divergence of a vector). Combining these 3 scalars, a vector is obtained as a final result. One fundamental relation associated with the divergence reads, when considering the product of a scalar ϕ with a vector φ: (ϕφ) = ϕ φ+φ. ϕ (2.3) Further information can be found for instance under Wikipedia. (2.2)

27 1 25 Mathematical operators Velocity field (vector plot) 3 8 x Y [m] X [m] x 10 Figure 2.3: Instantaneous structure of a turbulent non-premixed flame computed using Direct Numerical Simulations. Left: velocity; right: divergence of velocity Laplacian In Fluid Dynamics, the Laplace operator or Laplacian will be mostly introduced to quantify diffusion processes, in particular diffusive transport of momentum. The Laplacian (figure 2.4) acts mostly on a scalar quantity φ and delivers again a scalar quantity, written φ or more often 2 φ, and defined as 2 φ := 2φ 2φ 2φ x2 y z (2.4) As can be seen, the Laplacian relies on the second partial derivatives in space, at the difference of all other operators, employing only the first partial derivatives. Figure 2.4: Instantaneous structure of a turbulent non-premixed flame computed using Direct Numerical Simulations. Left: density; right: Laplacian of density. Further information can be found for instance under Wikipedia.

28 26 Chapter 2. Basic concepts Rotor or Curl In Fluid Dynamics, the rotor operator (very often called curl) will be mostly introduced to quantify the importance of vortical structures in a flow. For this purpose, we will usually compute the curl of the flow velocity, the vector quantity v, and we will obtain another vector quantity written rot(v) or more often v, and defined as v := ( vz y v y z, v x z v z x, v y x v ) x y (2.5) The curl of the flow velocity (figure 2.5) is particularly interesting, since we will demonstrate later that an irrotational flow, i.e., a flow verifying v = 0 is always particularly simple. 8 x 10 3 Velocity field (vector plot) Y [m] X [m] x 10 3 Figure 2.5: Instantaneous structure of a turbulent non-premixed flame computed using Direct Numerical Simulations. Left: velocity; right: curl of velocity. In order to quantify rotation, we introduce also the rotation vector Ω defined as: Further information can be found for instance under Wikipedia. Ω := 1 v (2.6) Time derivatives One quite unique feature of Fluid Dynamics is that two different and equally useful time derivatives will be introduced. They can be traced back to two very important contributors to this field of science, Euler and Lagrange. These two scientists defended a very different view concerning the most suitable time derivative in a flow: For Euler, a flow is nothing special, so that the time derivative should be defined there as for any other field of physics. Therefore, the observer is sitting at a fixed position x within the fluid, measures there the evolution of some interesting quantity with time, and just computes the time derivative by deriving the resulting curve. This is just the standard partial derivative in time at

29 2 Time derivatives 27 position x! It will therefore be written as usual. The (Eulerian) time derivative of a variable φ is simply written ( ) φ (2.7) t x=constant or simply φ t ForLagrange, thekeypropertyofaflowisthat...itflows! Therefore, Lagrangechoosesanobserver moving with the flow, and therefore behaving himself like a fluid element. While moving with the flow, this observer again measures the evolution of some interesting quantity with time, and now computes the time derivative by deriving the resulting curve. It is probably obvious for you that the resulting time derivative at the same position and at the same time will nevertheless not be the same, since the frame of reference is different! Therefore, this alternative definition of the time derivative will be written differently, as Dφ (2.9) Dt This time derivative is called either Lagrangian time derivative, substantial time derivative or total time derivative. For this approach, the important point is not the current position of the fluid element, point x, but its origin at the beginning of the observation, point X in space. As a consequence, one can also state that, for a Lagrangian observer, the time derivative is computed for a fixed origin of the movement, point X. This is expressed by following equivalence: Dφ Dt = ( ) φ t X=constant (2.8) (2.10) Even if both definitions (and therefore both derivatives) differ, it is nevertheless possible to relate both results by using the flow velocity v. It is first clear that the local, instantaneous flow velocity v at point x is nothing else that the time derivative of its position following the flow, as usual: v = ( ) x t X=constant (2.11) Let us now consider an arbitrary function φ of space and time. This function might represent equally well a scalar quantity, a vector or a tensor, even if it is written as a scalar for the following proof. This arbitrary function can be equally well represented in an Eulerian frame, φ = φ(x,t) and in a Lagrangian frame, φ = φ(x,t). For the same time t and the same instantaneous position, both values are of course identical. For such conditions, where Euler and Lagrange meet at the same point, one can therefore write: φ(x,t) = φ(x,t) (2.12) Let us now compute the Lagrangian derivative of this arbitrary function: Dφ Dt = = ( ) φ t ( X=constant ) φ(x,t) t X (2.13) (2.14)

30 28 Chapter 2. Basic concepts The corresponding derivative is computed while the observer is moving with the flow, thus along a trajectory x(x,t) with X = constant. Hence Dφ Dt = = ( ) φ(x1 (X,t),x 2 (X,t),x 3 (X,t),t) ( ) φ x 1 x t ( ) x1 (X,t) t X + ( ) φ x 2 x X ( ) x2 (X,t) Taking into account Eq.(2.11), one obtains now directly: t X + ( ) φ x 3 x ( ) x3 (X,t) t X + ( ) φ t x Dφ Dt = φ x 1 v 1 + φ x 2 v 2 + φ x 3 v 3 + φ t 1 (2.16) Reordering the right-hand side, the corresponding, very important relation reads finally: or in a longer, but equivalent manner: Dφ Dt = φ +(v )φ (2.17) t (2.15) ( ) t t X Dφ Dt = φ t +v φ x x +v φ y y +v φ z z (2.18) It is very easy to demonstrate mathematically that the last term in Eq.(2.17), i.e., the convective term can be replaced by introducing a gradient and a curl. For example, considering the flow velocity v, it comes: ( ) v 2 (v )v = +[( v) v] (2.19) Characteristic flow structures Three different characteristic lines will be often used to characterize and analyze the flow structures. A pathline (or trajectory) corresponds to the line obtained in the three-dimensional space by following an individual fluid particle during its displacement with time. It is sometimes described also as a long-exposure photograph of one and the same particle. An infinity of different pathlines can be defined, each associated to another fluid particle. Mathematically, if x p is the vector containing the three components of the pathline position, the geometry of the pathline can be obtained by integrating in time the vector relation dx p dt = v(x p,t) (2.20) starting from some chosen position x p0 and eliminating time t. This relation simply states that the movement along the pathline is purely due to the instantaneous local flow velocity v(x p,t). In order to compute a pathline, some finite time duration must be considered: a pathline is not an instantaneous concept; time must elapse!

31 3 Characteristic flow structures 29 Figure 2.6: Streamlines (blue) computed by post-processing two-dimensional velocity fields measured by Particle Image Velocimetry in the rotating impeller of a centrifugal pump. A streamline is a line that is at any point tangent to the local flow velocity v(t). It is quite easy to draw a streamline by hand on top of a plotted field of instantaneous velocity v(t) (figure 2.6). The mathematical definition of a streamline relies on the fact that the vector product between two collinear (i.e. tangential ) vectors is 0. Therefore, if x s defines the geometry of the streamline in space, its computation is based on integrating the differential relation: dx s v(x s,t) = 0 (2.21) starting from some chosen position x s0. This relation simply states that the displacement along the streamline is tangent to the local instantaneous flow velocity v(x s,t). Component-wise, this differential relation can be written as well under the form of three scalar relations: v z (x s,t)dy s v y (x s,t)dz s = 0 (2.22) v x (x s,t)dz s v z (x s,t)dx s = 0 (2.23) v y (x s,t)dx s v x (x s,t)dy s = 0 (2.24) A streamline is first an instantaneous concept. For any fixed time t, we can obtain a full set of streamlines. Now, it is of course possible to compute the resulting streamlines for successive time values and to assemble the resulting pictures to produce a video. Using streamlines, it is easy to define also a streamtube (figure 2.8). For this purpose, we just need to choose a closed one-dimensional curve C in the three-dimensional space. By joigning together all the streamlines going at some point through this curve C, a streamtube is obtained. This notion is particularly interesting, since the boundary of a streamtube cannot be crossed by any fluid particle

32 30 Chapter 2. Basic concepts 8 x 10 3 Velocity field (vector plot) 0.02 Streamlines of velocity field Y [m] 4 Y [m] X [m] x X [m] Figure 2.7: Instantaneous structure of a turbulent non-premixed flame computed using Direct Numerical Simulations. Left: velocity; right: resulting streamlines. (remember that the local direction of the fluid movement, i.e., the flow velocity, is per definition tangential to the local streamline). Therefore, a streamtube is somehow similar to an internal flow within a duct of variable cross-section (that of the streamtube). If the flow can furthermore be considered non-viscous, the flow within the resulting streamtube is almost equivalent to the corresponding internal flow. C streamlines Figure 2.8: Streamtube obtained by joining streamlines through a closed curve. A streakline (also called emission line) associated with some user-chosen point P is the locus of all fluid elements having passed through point P at some previous time instant. The denomination emission line is indeed quite clear: in order to obtain a streakline, a dye tracer will be in practice injectedintotheflowatafixed pointp.takingapictureoftheresulting dyedistributionsomewhat later, the emission line associated with point P can be obtained. As such, the concept of emission line is an instantaneous concept (the picture shows the instantaneous dye distribution) but necessitates a finite time duration in the past. The dye particles visible on the photograph all went through point P, some of them 30 seconds ago, some of them 10 seconds ago, some of them just 1 ms ago; the past history of the flow is made visible on the instantaneous picture.

33 4 Control volume 31 Since the definitions of pathline, streamline and streakline are different, the resulting lines will usually differ, too. Nevertheless, for a steady flow (and only for such a flow) the resulting geometrical lines will look identical when plotted. Further information can be found for instance under Wikipedia. 2.4 Control volume In order to derive conservation equations for the most important flow variables, it is now necessary to introduce control volumes. A control volume V c (figure 2.9) is a three-dimensional volume bounded by a closed surface A c and placed in a fluid. Of course, it exists only as a theoretical object and is not a real body. As such, this control volume does not lead to any modification of the fluid properties; fluid elements can move freely through the boundary A c of a control volume V c. Furthermore, a control volume can move freely with its own velocity within the fluid. To characterize this movement, a velocity w is defined at any point of the control volume. Outer surface A c w n n da Control volume V c dv flow velocity v w n w Figure 2.9: Generic control volume in a fluid. Up to now, the concept we have introduced is a generic control volume V c. Two specific sub-families must now be introduced: 1. a fixed control volume V cf is a control volume that does not move, i.e., with w = 0 (figure 2.10). As a consequence, the geometry of such a control volume cannot change with time; if the control volume is a sphere at the start of time, it will remain a sphere all the time, with fluid entering and leaving freely through the outer surface A cf. 2. a material control volume V cm is a control volume containing always the same fluid elements; if a fluid element is contained within V cm at the beginning of time, it will remain within it all the time; if it is outside of V cm at the beginning, it can never enter it. How is this possible, since we stated previously that, in principle, fluid can freely enter and leave a control volume? Simply by adapting the velocity of the control volume, w, to the local fluid velocity, v. By choosing w = v at any point of the material control volume V cm, we make sure that fluid elements cannot enter

34 32 Chapter 2. Basic concepts or leave any more the control volume. Since the material control volume follows the flow (figure 2.10), it will usually change its geometrical appearance during time. In a turbulent flow, it might indeed evolve to an extremely complex geometry within a short time! Note furthermore that, in order to define such a material control volume in a proper manner, we in fact have to neglect the influence of diffusion compared to the influence of convection (flow velocity). Outer surface A f n n da Control volume V f dv flow velocity v n Outer surface A m n w n da w Control volume V m dv w flow velocity v n Figure 2.10: Fixed (top) and material (bottom) control volume in a fluid.

35 5 Transport theorem Transport theorem Having now defined a control volume, we need a last mathematical tool, in order to be able to quantify the evolution with time of a specific quantity integrated over such a control volume. The corresponding relation is called in the present work transport theorem. You will sometimes find it in the literature as theorem or rule of Leibniz. In order to develop conservation equations, we will integrate some important flow quantity φ over a control volume V c, and we will compute the evolution of this integral with time. In other words, we would like to compute the quantity d φdv (2.25) dt V c Though it is possible of course to develop a rigorous relation, we will just here consider physical interpretations leading to the final solution. What are possible reasons explaining that this integral will change in time? In fact, two very different possibilities must be taken into account: 1. first, the variable φ (perhaps the density, or velocity) may change with time in an unsteady flow (φ = φ(t)). Such a change might of course change the value of the integral, and must be taken into account! This contribution obviously disappears in a steady flow. 2. a second, indirect possibility must also be taken into account. Remember that a control volume V c may freely move within the fluid with a velocity field w; it might therefore freely expand, shrink, incorporate a changing amount of variable φ... This movement of the control volume can therefore be also responsible for a change of the considered integral quantity. This contribution obviously disappears for a fixed control volume. Taking into account both effects leads finally to the transport theorem: d φ(x,t) φ(x,t)dv = dv + φ(x,t)(w n)da (2.26) dt V c V c t A c In this equation, all symbols are standard and can be found in the Nomenclature. In particular, the vector n is the unity vector normal to the outer surface A c of the control volume and pointing toward the outside. The scalar product w.n appearing in the last term is necessary, since a movement of the control volume tangential to its own surface A c locally does not lead to any change of the integrated quantity; only the movement normal to the surface must be taken into account, hence explaining the appearance of the term w.n. The generic form of the transport theorem can be simplified for specific control volumes. For a fixed control volume V cf, since w = 0, one obtains simply: d φ(x,t) φ(x,t)dv = dv (2.27) dt V cf V cf t For a material control volume V cm, since w = v (the local fluid velocity), one obtains: d φ(x,t) φ(x,t)dv = dv + φ(x,t)(v n)da (2.28) dt V cm V cm t A cm

36 34 Chapter 2. Basic concepts

37 Chapter 3 Mass conservation 3.1 Introduction In order to obtain the universal equation describing conservation of mass, we will now employ the concepts introduced in the previous chapter. Mass is indeed a conserved quantity, which means that it does not change for an isolated system, without any exchange with its surroundings. If fluid elements are exchanged with the surroundings, the mass of the considered system can obviously change! We start by choosing an arbitrary material control volume within a fluid. The evolution of the total mass M contained within this control volume V cm vs. time will be quantified. This total mass can be computed by integrating the mass contained by an elementary volume element, dv; the density ρ(x, t) being the ratio between mass and volume, the elementary mass is ρ(x,t)dv, and the total mass is thus: M = ρ(x,t)dv (3.1) V cm Hence, the purpose of this chapter is to compute dm = d ρ(x,t)dv (3.2) dt dt V cm This problem will be solved by considering successively basic results of physics and of mathematics. 3.2 Point of view of physics From a purely physical point of view, the issue considered in Eq.(3.2) is quite simple; the control volume considered here is a material control volume. Per definition(see Section 2.4), this means that fluid elements cannot enter or leave this control volume. Obviously, these fluid elements are the only possibility to transport mass! If it is impossible to exchange any fluid element with the surroundings, then the total mass M contained within the material control volume cannot change. Hence, M = constant and as a consequence dm = 0 (3.3) dt 3.3 Point of view of mathematics By looking at the right-hand side of Eq.(3.2), a mathematician recognizes immediately the possibility of using the transport theorem introduced in the previous chapter for a material control volume (Eq. 2.28): d φ(x,t) φ(x,t)dv = dv + φ(x,t)(v(x,t) n)da (3.4) dt V cm V cm t A cm 35

38 36 Chapter 3. Mass conservation The right-hand side of Eq.(3.2) is indeed identical with the left-hand side of Eq.(3.4) when taking φ = ρ. It comes therefore: dm dt = d ρ(x,t)dv (3.5) dt V cm ρ(x,t) = dv + ρ(x,t)(v(x,t) n)da (3.6) V cm t A cm 3.4 Integral formulation of mass conservation Recognizing that both results found in the two previous sections are of course correct and identical, it is possible to write following equality, taking on the left-hand side the result of mathematics and on the right-hand side the result of physics (simply 0, here): V cm ρ(x,t) t dv + ρ(x,t)(v(x,t) n)da = 0 A cm (3.7) This is indeed the integral formulation of mass conservation, written for an arbitrary material control volume V cm. In order to solve practical problems, it is often useful to write this integral formulation of mass conservation for a fixed control volume V cf. The procedure is similar to that described previously. From the point of view of mathematics, the transport theorem is now given by Eq.(2.27). d ρ(x,t) ρ(x,t)dv = dv (3.8) dt V cf V cf t From the point of view of physics, the change of fluid mass contained within the fixed control volume V cf due to an exchange with the surroundings is simply written as a flux of mass through the volume boundary, A cf : ρ(x,t)(v(x,t) n)da (3.9) A cf where the minus sign is in fact associated with n, since fluid is entering the control volume V cf when (v(x,t) n) > 0 and leaving it when (v(x,t) n) < 0. Finally, the integral formulation of mass conservation written for an arbitrary fixed control volume V cf reads V cf ρ(x,t) t dv + ρ(x,t)(v(x,t) n)da = 0 A cf (3.10) Note that it is a posteriori trivial to evolve from Eq.(3.7) to Eq.(3.10): it is sufficient to assume that the fixed control volume V cf coincides with the material control volume V cm at time t; both formulations are indeed identical. 3.5 Local formulation of mass conservation Equation (3.7) can indeed be useful when considering a macroscopic control volume (though we will mostly employ in practice fixed control volumes instead of material ones), but is awkward when trying to derive local conditions valid for any fluid element. One problem with Eq.(3.7) is that it combines a volume integral (left) with a surface integral (middle), preventing further simplification.

39 6 Local mass conservation for an incompressible flow 37 This can be easily solved by using the divergence theorem, an extremely famous relation called also integral rule or theorem of Gauß, of Ostrogradsky, of Gauß-Ostrogradsky or of Green-Ostrogradsky. With so many possible fathers, you immediately understand the importance of this theorem, allowing a direct relation between a volume integral on an arbitrary volume V c and a surface integral on the associated boundary A c! Using the first formulation of the divergence theorem (Eq. C.5), it is possible to replace the second, surface integral in Eq.(3.7), leading to: V cm ρ(x,t) t Both integration volumes are now identical, allowing to rewrite: V cm dv + (ρ(x,t)v(x,t))dv = 0 V cm (3.11) [ ] ρ(x,t) + (ρ(x,t)v(x,t)) dv = 0 (3.12) t Remember that this relation is valid for an arbitrary material control volume, and thus for an infinite number of different volumes in the fluid! How is it possible to integrate some quantity (that between the [ ] in Eq.3.12) over an infinite number of different volumes, getting always 0 as a result? Only if the integrated quantity is equal to 0 at every point! Hence, the quantity [ ρ(x,t) + (ρ(x,t)v(x,t)) ] must t be identically nil at every point in space. Finally, the local mass conservation equation (also called sometimes continuity equation) can be written: ρ t + (ρv) = 0 (3.13) This is one of the most fundamental relations of Fluid Dynamics and we will use it many times in this document Local formulation of mass conservation in cylindrical coordinates If a cylindrical coordinate system (r,θ,z) with corresponding velocity components v = (v r,v θ,v z ) is used instead of our standard coordinate system, the local formulation of mass conservation reads: ρ t + 1 (ρrv r ) + 1 (ρv θ ) + (ρv z) = 0 (3.14) r r r θ z 3.6 Local mass conservation for an incompressible flow Equation (3.13) can easily be simplified for an incompressible flow. Since, for such a flow, we can safely assume ρ = constant in space as well as in time, the time-derivative of the density is automatically 0: ρ t = 0 (3.15) Furthermore, the density, being constant, can be pulled out of the divergence operator: (ρv) = ρ v (3.16) Finally, the local mass conservation equation (or continuity equation) can be simply written for an incompressible flow: v = 0 (3.17)

40 38 Chapter 3. Mass conservation

41 Chapter 4 Euler equation: conservation of momentum in a non-viscous flow 4.1 Introduction Now that the mass conservation equation (Eq. 3.13) has been established, it is necessary to derive the corresponding conservation equation for momentum (to be exact, for linear momentum). Momentum is indeed, like mass, a conserved quantity, which means that it does not change for an isolated system, without any interaction with its surroundings. If interaction with the surroundings does take place (for example, when external forces act on the fluid elements of the considered system), momentum can indeed change. We start by choosing again an arbitrary material control volume within a fluid. The evolution of the total momentum P contained within this control volume V cm vs. time will be quantified. This total momentum can be computed by integrating the momentum contained within an elementary volume element, dv, i.e., ρ(x,t)v(x,t)dv, and the total momentum is thus: Hence, the purpose of this chapter is to compute P = ρ(x,t)v(x,t)dv V cm (4.1) dp dt = d ρ(x,t)v(x,t)dv (4.2) dt V cm This problem will be solved again by considering successively basic results of physics and of mathematics, in a similar manner to that employed in the previous chapter for mass conservation. This time, we will start with mathematical considerations. 4.2 Point of view of mathematics By looking at the right-hand side of Eq.(4.2), a mathematician recognizes the possibility of using the transport theorem introduced previously for a material control volume (Eq. 2.28): d φ(x,t) φ(x,t)dv = dv + φ(x,t)(v(x,t) n)da (4.3) dt V cm V cm t A cm Nevertheless, it is in the present case even more useful to introduce first a specific variation of the transport theorem, called theorem of Reynolds. Starting from the standard version of transport theorem, 39

42 40 Chapter 4. Euler equation: conservation of momentum in a non-viscous flow Eq.(4.3), one starts by using the first formulation of the divergence theorem(eq. C.5), in order to replace the last, surface integral in Eq.(4.3), leading to: d φ(x,t) φ(x,t)dv = dv + (φ(x,t)v(x,t))dv dt V cm V cm t V cm [ ] φ(x,t) = + (φ(x,t)v(x,t)) dv (4.4) t V cm We now consider a specific case; the variable φ(x,t) is assumed to be written in the form ρ(x,t)ϕ(x,t), with ρ(x,t) the fluid density, as usual: [ ] d (ρ(x,t)ϕ(x,t)) ρ(x,t)ϕ(x,t)dv = + (ρ(x,t)ϕ(x,t)v(x,t)) dv (4.5) dt V cm V cm t The partial derivative and divergence on the right-hand side may now be separated in two contributions: d ρ(x,t)ϕ(x,t)dv = dt V cm I {}} ]{ { [ ρ(x,t) ϕ(x, t) + (ρ(x,t)v(x,t)) V cm t [ ]} ϕ(x,t) +ρ(x, t) +(v(x,t) )ϕ(x,t) dv (4.6) t using the classical vector relation (Eq. 2.3) for the divergence appearing in the last term on the righthand side of Eq.(4.5). Now, it is easy to recognize that the term marked I in Eq.(4.6) is equal to 0, as demonstrated by the local mass conservation equation (Eq. 3.13) derived in the previous chapter. Finally, one obtains the theorem of Reynolds: d ρ(x,t)ϕ(x,t)dv = ρ(x,t) Dϕ(x,t) dv (4.7) dt V cm V cm Dt introducing the total derivative of ϕ (Eq. 2.17) in the last term on the right-hand side of Eq.(4.6). The right-hand side of Eq.(4.2) is indeed identical with the left-hand side of Eq.(4.7) when taking ϕ = v (remember that ϕ represents an arbitrary variable, and might therefore be in principle a scalar as well as a vector or a tensor). It comes therefore: dp dt 4.3 Point of view of physics = d ρ(x,t)v(x,t)dv (4.8) dt V cm = ρ(x,t) Dv(x,t) dv (4.9) V cm Dt From a purely physical point of view, the issue considered in Eq.(4.2) is again quite simple and has been already solved several centuries ago by Newton. As stated in the second law of Newton, the change of momentum for a material system is simply a consequence of external forces, acting in the present case on the fluid elements contained within the material control volume V cm. Therefore, in order to answer the question described by Eq.(4.2), we simply need to know the forces acting on the considered fluid. Further details concerning Newton s laws of motion can be found for example under Wikipedia. What are the usual forces that will be considered throughout this document? To answer this question, we start by differentiating two different families of forces: the contact and the non-contact forces. More details can be found in Appendix A. Let us just recall briefly here that non-contact forces are long-range forces while contact forces take place only when two fluid elements are in direct vicinity to each other.

43 4 Integral formulation of momentum conservation 41 Throughout this document, the only non-contact force that will be taken into account is the gravitational force F g induced by earth gravity, associated with the acceleration vector g, assumed constant ( g = g = 9.81 m/s 2 ). The action of gravity on an elementary fluid volume dv is then simply ρ(x, t)gdv, ρdv being the mass of the fluid element. The resulting gravitational force for the full material control volume V cm reads F g = ρ(x,t)gdv (4.10) V cm Throughout this document, the only real contact force that will be taken into account is the pressure force F p. This is indeed the only important contact force for a non-viscous flow. Note that we will introduce later a viscous force F f as a further contact force, but this is only a helpful (though controversial) simplification; this so-called viscous force is in fact nothing else than the diffusion term for momentum, neglected up to now. But let us first skip this issue, and consider the pressure force as only contact force, assuming a non-viscous flow. The pressure force acts in a very simple way; its direction is normal to the plane on which the force is acting, its magnitude is simply the product of local pressure p and surface area A. The action of pressure on an elementary element da of the surface A cm of our material control volume is then simply n(p(x,t)da), where the magnitude pda and the direction n can be easily recognized. The vector n is as usual the unity vector normal to the surface A cm and pointing toward the outside. Since we consider in this analysis the force exerted by the surroundings on our material control volume, the suitable direction is indeed n; the surrounding fluid presses onto the fluid within our control volume V cm. The resulting, global pressure force working on the full material control volume V cm reads F p = p(x,t)nda (4.11) A cm Finally, the answer given by physics to the question contained in Eq.(4.2) reads, following the second law of Newton: dp dt = F g +F p (4.12) = ρ(x,t)gdv V cm p(x,t)nda A cm (4.13) 4.4 Integral formulation of momentum conservation Recognizing that both results found in the two previous sections are of course correct and identical, it is possible to write following equality, taking on the left-hand side the result of mathematics and on the right-hand side the result of physics: V cm ρ(x,t) Dv(x,t) Dt dv = ρ(x,t)gdv p(x,t)nda (4.14) V cm A cm This is indeed the integral formulation of momentum conservation (in fact, linear momentum), written for an arbitrary material control volume V cm and valid for any non-viscous flow. It is equally possible to use the standard formulation of the transport theorem, Eq.(2.28) instead of the theorem of Reynolds (Eq. 4.7), as done previously. In that case, Eq.(4.14) is now: V cm (ρ(x,t)v(x,t)) t dv + ρ(x,t)v(x,t)(v(x,t) n)da = A cm ρ(x,t)gdv p(x,t)nda (4.15) V cm A cm

44 42 Chapter 4. Euler equation: conservation of momentum in a non-viscous flow It is acceptable to assume that a fixed control volume V cf coincides with the considered material control volume V cm at time t, allowing to rewrite: V cf (ρ(x,t)v(x,t)) t If the flow is steady, one obtains furthermore: dv + ρ(x,t)v(x,t)(v(x,t) n)da = A cf ρ(x,t)gdv p(x,t)nda (4.16) V cf A cf ρ(x,t)v(x,t)(v(x,t).n)da = ρ(x,t)gdv p(x,t)nda (4.17) A cf V cf A cf Observing that all forces acting on the fluid contained within the fixed control volume V cf are now appearing on the right-hand side of this equation, the developed formulation will be used later in Chapter 7 to compute the force induced by a flow. 4.5 Local formulation of momentum conservation Equation (4.14) can indeed be useful when considering a macroscopic control volume (though we will mostly employ in practice fixed control volumes instead of material ones), but is awkward when trying to derive local conditions valid for any fluid element. One problem with Eq.(4.14) is that it combines a volume integral (first one on the right-hand side) with a surface integral (last one on the right-hand side), preventing further simplification. This can be easily solved by using again the divergence theorem, allowing a direct relation between a volume integral on an arbitrary volume V c and a surface integral on the associated boundary A c. Using the second formulation of the divergence theorem (Eq. C.6), it is possible to replace the last, surface integral in Eq.(4.14), leading to: V cm ρ(x,t) Dv(x,t) Dt All integration volumes are now identical, allowing to rewrite: V cm [ dv = ρ(x,t)gdv p(x,t)dv (4.18) V cm V cm ρ(x,t) Dv(x,t) Dt ρ(x,t)g+ p(x,t) ] dv = 0 (4.19) Remember that this relation is valid for an arbitrary material control volume, and thus for an infinite number of different volumes in the fluid! How is it possible to integrate some quantity (that between the [ ] in Eq.4.19) over an infinite number of different volumes, getting always 0 as a result? Only if the integrated quantity is equal to 0 at every point! Hence, the quantity [ ρ(x,t) Dv(x,t) ρ(x,t)g+ p(x,t) Dt must be identically nil at every point in space. Finally, the local conservation equation for linear momentum can be written for a non-viscous flow: ] ρ Dv Dt = ρg p (4.20) This relation, also called Euler equation, is one of the most fundamental relations of Fluid Dynamics and we will use it many times in this document. Later, we will add a diffusion term (also called improperly viscous force ) to this equation, leading to an even more general formulation.

45 6 Local momentum conservation for an incompressible flow Local momentum conservation for an incompressible flow Equation (4.20) can be sligthly modified for an incompressible flow, even if it does not lead to a major simplification. Since, for such a flow, we can safely assume ρ = constant in space as well as in time, it is possible to divide all terms of this equation by the density, leading to: Dv Dt = g 1 p (4.21) ρ Furthermore, the density, being constant, can be pulled into the gradient operator: 1 ρ p = ( ) p ρ (4.22) Finally, the local conservation equation for momentum can be written for an incompressible flow: Dv Dt = g ( ) p ρ (4.23) 4.7 Integral formulation of angular momentum conservation For practical considerations, only linear momentum plays a major role in Fluid Dynamics, while angular momentum is of very minor importance. The only case for which a relation concerning angular momentum is helpful, is when considering the torque exerted by a fluid volume on its surroundings (Chapter 7). Indeed, the integral formulation describing conservation of angular momentum by reference to a pivot P is identical to the integral formulation for linear momentum written for a fixed control volume V cf (Eq. 4.16): V cf (ρ(x,t)v(x,t)) t dv + ρ(x,t)v(x,t)(v(x,t) n)da = A cf ρ(x,t)gdv p(x,t)nda (4.24) V cf A cf Writing r the vector connecting the (fixed) pivot P and the (variable) position of the integration point, this relation is simply modified by introducing a vector product in each integral: [ ] (ρ(x,t)v(x,t)) r dv + r [ρ(x,t)v(x,t)(v(x,t) n)]da = V cf t A cf r [ρ(x,t)g]dv r [p(x,t)n]da (4.25) V cf A cf If the flow is furthermore steady, one obtains: r [ρ(x,t)v(x,t)(v(x,t) n)]da = A cf r [ρ(x,t)g]dv r [p(x,t)n]da (4.26) V cf A cf

46 44 Chapter 4. Euler equation: conservation of momentum in a non-viscous flow

47 Chapter 5 Hydrostatics and Aerostatics 5.1 Introduction In this chapter we consider the simplest of all possible flows, i.e., those, that do not flow at all! Considering back the variable-based classification proposed in the introduction, such flows can be fully described using only one of the main flow variables: the pressure p within the fluid. The velocity is everywhere equal to zero and is therefore not an unknown of the problem. Even if this fact might seem quite disappointing at first, you will probably soon realize how important such configurations are in practice. The results obtained in this chapter will therefore be very useful for many practical cases. To list a few: You will finally understand why it is safe to use the next swimming-pool, since we will be able to compute forces exerted by a static fluid on a body; You will be able to determine if a ship has been wisely laid out, since we will consider in the following part the stability of partially-immersed bodies; You will unterstand how hot-air or helium balloons might be working, since they rely on aerostatic concepts. In the same context, you will get a first idea of pressure and density variations within the earth s atmosphere. Moreover, such non-flowing flows are of course particularly simple, and therefore appropriate to consider first applications at a basic level. To begin with, we will derive the fundamental equation of hydroand aerostatics. 5.2 Fundamental equation of hydro- and aerostatics This fundamental equation is easily obtained starting from the general conservation equations for mass and momentum of a non-viscous flow (Eqs and 4.20) given in the two previous chapters: ρ t + (ρv) = 0 ρ Dv Dt = ρg p Note that it is equally possible to consider as a starting point the Navier-Stokes equation and not the Euler equation to describe conservation of momentum, without changing anything to the results obtained afterwards. This is due to the fact that the so-called viscous forces, describing the diffusion of momentum, are always expressed based on velocity gradients. Since, in the present case, velocity is always zero everywhere, 45

48 46 Chapter 5. Hydrostatics and Aerostatics the same applies to all possible velocity gradients, so that viscous forces are exactly equal to zero. Considering these viscous forces as a diffusion process for momentum, this is also physically logical, since there cannot be a diffusion flux for a quantity that is everywhere the same (here, momentum is everywhere zero). We can now take into account the fundamental properties of hydro- and aerostatics. As the name statics implies, nothing is moving, nothing in changing, in such non-flowing flows. This means in practice that: 1. the flow velocity v is everywhere and at any time equal to 0; 2. the flow is steady, meaning that the partial time derivative of any variable Φ is also 0: Φ t = 0. To be exact, this last result is only true for hydro- and aerostatics when requiring that the flow boundary conditions are also constant in time and that the initial configuration is stable (and not an unstable water/oilemulsion, for example). In that case, a zero flow velocity leads indeed directly to the fact that the flow properties must be constant in time: the flow is steady. Using these results first in the standard mass conservation equation (3.13) ρ t + (ρv) = 0 we obtain due to the second hypothesis that ρ = 0 and due to the first hypothesis that ρv = 0 so that t equally (ρv) = 0. Finally, the equation describing conservation of mass reads in this specific case 0 = 0 a fact which is clearly true but does not give any useful information about that flow! This shows that mass conservation is automatically fulfilled for all applications involving hydro- and aerostatics, we never have to care about that. Looking now at the Euler equation (4.20) ρ Dv Dt = ρg p the left-hand term ρ Dv is again clearly equal to 0 since v = 0, but both terms appearing on the righthand side are not modified by the hypotheses underlying hydro- and aerostatics. This means that the Dt Euler equation reduces in this case to p = ρg (5.1) the fundamental equation of hydro- and aerostatics. This equation states that the pressure gradient is equal to the action of gravity on the fluid, ρg. Note that it is equally possible to take into account within equation (5.1) any further non-contact force, expressed as ρk with k the acceleration vector associated with the force acting on the fluid per unit mass, as long as this force is conservative, meaning that it is possible to find a scalar potential φ(x) verifying k = φ. This is equivalent to saying that the work associated with the action of this force is independant from the way followed during the movement, but only depends on the starting point and endpoint of this displacement. All the non-contact forces considered in this document are conservative, so that equation (5.1) can be easily extended to take these forces into account if needed. It is sufficient to add the action of these supplementary mass forces on the right-hand side of equation (5.1), leading to something like p = ρg+ρk (5.2) It is now possible to consider the first application of equation (5.1): the variation of pressure within a static liquid.

49 3 Pressure variation within an incompressible, static fluid Pressure variation within an incompressible, static fluid We now limit following results to the case of an incompressible fluid, meaning that the density cannot be changed in any way: ρ = constant. This is typically the case for all static liquids, but a case involving a static gas without any change in density (for example an ideal gas with a constant temperature, pressure and composition) would also fulfill this condition. Nevertheless, the results presented in this section are classically associated with hydrostatics. The question we want to solve is the following (figure 5.1): considering an open container filled with an incompressible fluid, what is the pressure distribution within the fluid? z p a z a H g y x Figure 5.1: Pressure distribution within a static, incompressible fluid: configuration. This problem can easily be solved by considering and integrating equation (5.1): p = ρg Using our standard coordinate system and taking gravity as the only mass force acting on the fluid (we have explained in the previous section how to take into account further mass forces if needed), we logically start by splitting the vector equation (5.1) into 3 scalar equations, reminding that the gravity force is pointing towards the direction ( z): p x = 0 p = 0 y p = ρg (5.3) z The first two equations can be directly integrated and show that the pressure does not depend on the x and y spatial directions. Thus, the pressure variable can be given as p(z) instead of p(x): the only pressure variation will take place in the z-direction, which is logical, since the only mass force

50 48 Chapter 5. Hydrostatics and Aerostatics considered in this example also acts only in this direction. The last equation of (5.3) can be easily integrated considering that the density ρ is constant (incompressible fluid), and that the same applies of course to the magnitude of the gravity vector g at ground level: thus, the right-hand side of equation (5.3) is a constant, allowing a direct integration: p(z) = constant ρgz (5.4) If the total height of the fluid column in the container is equal to H, this means that the pressure difference between the surface of the fluid and the bottom of the container is equal to ρgh, with a pressure level growing linearly with the depth within the fluid (figure 5.2), while it is constant in an horizontal plane, since p neither depends on x nor on y. z in gas z a H in liquid g p a p a + ρ gh p Figure 5.2: Pressure distribution within a static, incompressible fluid. In order to determine the constant in equation (5.4), we need to know the value at the top surface of the fluid. In the present problem, we have stated that the container is open, so that the fluid in the container has a free surface in direct contact with the surrounding air, at an air pressure denoted with p a. Since this free surface is horizontal and not moving (statics), this shows that the forces acting on the interface are equal and opposed to each other: the pressure force acting on the interface from the liquid side is thus equal and opposite to the pressure force acting on the interface from the air side. As a consequence, the pressure at the free surface within the fluid must be equal in magnitude to the external pressure in the air, p a, so that we obtain in the liquid: p(z = z a ) = p a (5.5) = constant ρgz a (5.6)

51 4 Force exerted by an incompressible, static fluid, on a fully immersed body 49 If, for some reason, the free surface of the fluid is not planar, a pressure difference will be observed between both sides of the interface, resulting from the existence of a surface tension σ between both fluids(for example water in the container, air above the free surface). Considering equation (5.6), the final solution for the pressure distribution in the fluid is thus p(x) = p(z) = p a +ρg(z a z) (5.7) showing again the linear growth of the pressure with the depth( z) within the container. At the bottom of the container, the pressure level is p a +ρgh, or ρgh above the pressure at the container surface. The fact that the pressure in the container only depends on the local depth of the fluid is completely independent from the form of the container, as observed for instance in a U-tube manometer. There is no need for a direct contact between the fluid particles at a certain depth to ensure pressure homogeneity at this same depth. It is sufficient that the different fluid compartments communicate at a single level or fulfill the same pressure boundary condition at the free surface to ensure this fact. Now, the question remains: what happens within the air above the liquid tank? We will deal with this point in more details later on when considering aerostatics. Let us just briefly mention that, in principle, the same applies to the air as long as its density might be considered to be constant. Since we might safely consider that the air density does not vary noticeably within a few meters under normal conditions, so that ρ a constant for such a thin air sheet, we obtain the same result as equation (5.4): Considering again as boundary condition p = p a for z = z a, we obtain p air (z) = constant ρ a gz (5.8) p air (z) = p a +ρ a g(z a z) (5.9) so that the pressure in the air diminishes linearly with the elevation. The (considerable!) difference with the previous equation (5.4) is of course that, here, the density of the air ρ a is used (typically ρ a 1.2 kg/m 3 under usual conditions) and not the density of an incompressible fluid (i.e., typically a liquid like water with ρ water 1000 kg/m 3 ), so that the pressure variation with the z-coordinate is roughly 1000 times slower in air compared to water (figure 5.2)! 5.4 Force exerted by an incompressible, static fluid, on a fully immersed body As mentioned in the section title we first consider a body which is fully immersed within an incompressible, static fluid, for example water at rest in a container (figure 5.3). In order to determine the motion of this body we simply use Newton s first law of motion relating body acceleration and forces acting on it. In order to be coherent with the rest of this chapter we still consider all fluid elements in the water to be at rest. Clearly, this will not be true any more if the body is moving, but we are only interested here in identifying the stable position of the body, for which static conditions can indeed be found everywhere. As a consequence of this approach, the influence of viscous forces will be again neglected: these forces would of course influence the time-dependent movement of the body, but they play no role when this body comes to rest. In such a case, only two forces need to be considered in order to determine the movement of a body with total mass M b, total volume V b and resulting body density ρ b := M b /V b : 1. the gravity force, pointing towards ( z) with a magnitude M b g, leading to F g = M b g = M b ge z ;

52 50 Chapter 5. Hydrostatics and Aerostatics n Outer surface A da Body V p z g Gravity M b g y x Figure 5.3: Forces acting on a fully immersed body: configuration. 2. the total pressure force F p resulting from the integration of all local pressure forces existing at each contact point between the body and the surrounding liquid. In order to compute this total pressure force caused by the liquid on the body, we just need to recall the properties of the pressure force introduced in the previous chapter (figure 5.3): the magnitude of the local pressure force is equal to the local pressure value in the fluid multiplied by the local body surface area on which the force is acting; the pressure force is always perpendicular to the body surface and points away from the fluid responsible for this force (i.e., in the present case, towards the interior of the body). Writing as usual n the local unit vector, normal to the body surface and pointing towards the outside of the body, the local pressure force exerted by the liquid on the body is therefore once again f p = pnda. Finally, the total pressure force induced by the liquid on the body is thus: F p = A ( pn) da (5.10) It is now possible to switch from a surface to a volume integral in the same manner as in the previous chapter. For this, we employ again the powerful divergence theorem, in its second formulation, leading to: F p = ( p) dv (5.11) V In order to conclude, we just need to employ again the fundamental equation of hydro- and aerostatics (5.1): p = ρg

53 4 Force exerted by an incompressible, static fluid, on a fully immersed body 51 in order to replace the pressure gradient in the fluid by the gravity term, leading finally to: F p = V ( ρg) dv (5.12) It is worth reminding here that the density ρ appearing in this equation is the fluid density, and not the density of the body! In order to use the last relation, we must assume that the fluid conditions are not modified at all due to the presence of the body, which is not always as obvious as it might appear here. First, this is only true for a non-moving body, but we have already accepted this assumption before. Second, if some interactions (i.e., forces) exist between the body material or surface and the surrounding fluid particles, they must be taken into account for a refined analysis. This is not considered further in what follows. The famous result of Archimedes can be easily deduced from equation (5.12), since both the density ρ of the incompressible fluid and the earth gravity acceleration g at ground level are constant: or F p = ρg dv V F p = ρv b g (5.13) with V b the volume of the body. What is now ρv b? This is the density of the fluid multiplied by the body volume, and hence the weight of the fluid that would occupy the place of the body when removing it. Therefore, equation (5.13) just describes the famous principle of Archimedes: The magnitude of the buoyant force (or buoyancy) is equal to the weight of the displaced fluid. Its direction is opposed to the direction of gravity. A few comments are probably useful at this point: First, buoyancy is nothing else than a pressure force! In what follows, we only use the name of buoyancy because it is traditionally employed for the present purpose. But this is indeed a superfluous concept, which is not really needed. Second, buoyancy is somewhat magical. In the present case, gravity leads to a pressure increase for a growing depth within the container. This pressure variation globally induces a resulting force, buoyancy, acting against the gravity, since it is pointing towards the z-direction. To summarize, one can conclude that, in this specific configuration, gravity is indirectly leading to an upwardpointing force. This is clearly a very unusual statement! Third, buoyancy and gravity are directly connected to each other. Without gravity, no pressure variation within the incompressible fluid, and therefore no buoyancy! Fourth, the magnitude of the buoyant force does not depend on the depth of the body within the liquid, but is constant. It will therefore be easy in what follows to identify a stable position. Finally, it is clear that this buoyant force is only a result of the pressure variation induced by the mass forces acting on the fluid (see 5.2). If forces other than gravity are present, the buoyant force will be modified accordingly, and can include a component in the horizontal direction, for example when rotating the container around its axis. This principle could be for example employed in order to separate liquids with different densities using centrifugal forces. Itisnoweasytodetermine thestablepositionofabodyimmersed inafluid. Two forcesareacting on this body in the standard configuration. Gravity, expressed as F g = M b g = M b ge z or F g = ρ b V b ge z introducing the mean body density ρ b defined as ρ b := M b /V b. And buoyancy (or pressure force),

54 52 Chapter 5. Hydrostatics and Aerostatics expressed as F p = ρv b g = ρv b ge z. Both expressions contain identically the acceleration vector of gravity, g, as well as the body volume V b. As a consequence, only the difference between the fluid density ρ and the body density ρ b must be taken into account in order to identify the stable position of the body. Three cases can be found (figure 5.4): 1. ρ = ρ b : in that case, the body stays in its present position, perhaps in the middle of the container, without any movement. This is quite an unstable configuration, since it requires that both density values are exactly equal. But it corresponds for example to the situation observed... when removing the body! In that case the volume V is occupied by water, with exactly the same density as the neigbouring liquid (of course). And, since we consider a static fluid, this volume of water must be at rest, which is indeed obtained for this case. 2. ρ b > ρ: in that case the magnitude of the gravity force is larger than the magnitude of buoancy. The body moves therefore towards the direction z and reaches a stable position when touching the bottom of the container. This is the standard case of a stone thrown into a water pond. 3. ρ b < ρ: then, the magnitude of buoyancy is larger than the magnitude of the gravity force. The body will move upward and reach a stable position at the surface of the liquid (see next section). This is the classical example of an icicle floating on a glass of water (since ice has a lower density, roughly 920 kg/m 3, compared to liquid water, which is indeed a very unusual property). = b > b < b Buoyancy V g Body V Buoyancy V g Body V Buoyancy V g Body V Gravity b V g Gravity b V g Gravity b V g Figure 5.4: Movement of the body due to buoyancy. As a practical example of these equilibrium conditions, let us consider the thermometer of Galileo Galilei, shown in figure 5.5. This is one of the first technical realizations allowing reliable (but not very accurate) temperature measurements. It relies on the principle of Archimedes, combined with the observation that the density of water is indeed not perfectly constant, but decreases slightly for an increasing water temperature (Table 5.1). The thermometer consists of a water column containing a number of solid spheres, all with the same volume V b but with a slightly different mass, so that each sphere is associated to a different density ρ b, varying typically between 1000 and 990 kg/m 3 for measuring usual temperatures. When leaving this thermometer in a room, the water temperature will be identical to the temperature of the surrounding air (unfortunately, this can take some time: the thermometer of Galileo Galilei is not suited to measure rapidly varying temperatures). This specific temperature (say 25 C) corresponds to a given water density (in this case 997 kg/m 3, see Table 5.1). All spheres with a density larger than this value will

55 4 Force exerted by an incompressible, static fluid, on a fully immersed body 53 Figure 5.5: Thermometer of Galileo Galilei. Temperature ( C) Density (kg/m 3 ) Table 5.1: Density of pure water as a function of temperature. See for instance Wikipedia for complementary information. sink to the bottom of the container, all spheres with a density lower than 997 kg/m 3 will float on the water surface. If there is a sphere with a density exactly equal to 997 kg/m 3, it will stay somewhere in the middle. By marking the spheres with a specific label, it is possible in this manner to measure the water (and thus the air) temperature. Nevertheless, in order to obtain an accurate measurement, it would be nessary to control the density of the spheres with a precision of at least 1%, which is indeed a difficult technical task... Finally, let us consider an essential question: is it safe for a human being to use a swimming-pool? Most people have never considered this important point, since they have been accustomed to swimmingpools as a small child, or even as a baby. But, in principle, there is nothing to prove that this is always safe! We know now how to answer this question: it is sufficient to check the respective densities of water and of a human being. As we already know(table 5.1), standard water at room temperature corresponds to a density of roughly 1000 kg/m 3. But what about human beings? With some difficulty, the human density can be found in medical documents, and lies typically around kg/m 3, this value being remarkably constant for most humans (since people with a large volume are typically associated with a high weight, and vice-versa!). Note that this value corresponds typically to a human body at the end of an inhalation (breathing) cycle. Removing all the air from the lungs leads to a human density usually very close to 1000 kg/m 3, almost equal to the water density. Since the (breathing) human density is lower than the density of water, you may now safely use any swimming-pool: following the rule stated above, your

56 54 Chapter 5. Hydrostatics and Aerostatics stable position will be floating on the surface of this pool. Note, for the same reason, that it would be very dangerous to jump into any man-height container filled up with pure alcohol (density around 790 kg/m 3 ): the stable position for this case would be...sitting on the bottom of this container, a clearly very uncomfortable configuration! Getting back to the (usual) case of a swimming-pool filled up with water, it is now time to check one last point: since the stable position corresponds to a case where your body is not completely immersed, will that change anything to the principle of Archimedes, established in the present section for a fully immersed body? 5.5 Force exerted on a partially immersed body We now consider a body which is partially immersed within an incompressible, static fluid, for example water at rest in a container (figure 5.6). The classical example is an iceberg floating on the sea surface. A gas V gas V liq A liq ρ a ρ z g y x Figure 5.6: Forces acting on a partially immersed body: configuration. Since the body is floating on the liquid surface, the upper part of this body is surrounded by air. As always in this chapter, we consider both the liquid and the air to be at rest (statics). In order to determine the resulting forces, we consider exactly the same reasoning as in the preceeding section 5.4. The body is again submitted solely to the action of gravity, like in the previous section, and to the action of pressure (viscous forces disappear when the system is at rest). But the global pressure force acting on the body is now the sum of a pressure force exerted by the liquid along the lower part of the body, plus a pressure force exerted by the gas on the upper part of this body. F p = A ( pn) da = F p,liq +F p,gas = ( pn)da+ A liq Agas ( pn) da (5.14) where we have separated the external surface of the body into the part in contact with the liquid (A liq ) and in contact with the gas (A gas ), with of course A = A liq A gas. Using these two surfaces plus a cut through the body at the level of the liquid surface, it is possible to define two corresponding, closed

57 6 Stability of a partially immersed body 55 volumes V liq and V gas, in contact with each other, and verifying V = V liq V gas. We may now safely use the integration rule of Gauß on each of these closed volumes, leading to: F p = ( p)dv + V liq Vgas ( p) dv (5.15) In order to conclude, we just need to employ again the fundamental equation of hydro- and aerostatics (5.1), once again separately on each volume, leading to: F p = ( ρg)dv + V liq Vgas ( ρ a g)dv Note that, in this equation, the first integral is built with the liquid density ρ, while the second contains the gas density ρ a. We have already assumed the liquid density to be constant. Using the same hypothesis for the gas (ρ a = constant), which means in particular neglecting any temperature effect (see later section 5.7) allows us to easily compute both integrals, leading to the final result: F p = F p,liq +F p,gas, = ρv liq g ρ a V gas g (5.16) This result is in fact not surprising at all. There are now two contributions to the buoyant force, one coming from the liquid, one from the gas. The principle of Archimedes still applies without the slightest modification: the magnitude of the buoyant force is equal to the weight of the displaced fluid (comprising liquid and gas), its direction is opposed to the direction of gravity. Since, in most practical cases (for example for the iceberg mentioned previously), the largest part of the body volume is immersed, and since furthermore the liquid density is typically several hundred times larger than the gas density, the contribution of the second term in (5.16) can usually safely be neglected compared to the first one. This explains why some textbooks do not mention at all the contribution of the gas side, which is indeed an approximation of the exact result. But this approximation would for example not be very accurate for a table tennis ball floating on the sea Stability of a partially immersed body We again consider a body which is partially immersed within an incompressible, static fluid, for example water at rest in a container (figure 5.7). This is the typical configuration for a ship on the sea, neglecting any current and movements (statics). We already know the magnitude and direction of the forces acting on this body (gravity, and pressure force or buoyancy). In order to determine stability conditions, we just need to know at which position these forces will take effect. This is well-known for gravity: the weight of the body acts at the center of gravity. But what about buoyancy? The answer is indeed quite simple when considering the remark already listed in section 5.4 for the configuration ρ = ρ b. This corresponds for example to the situation observed...when removing the body! In that case the volume V is occupied by water, with exactly the same density as the neigbouring liquid (of course). And, since we consider a static fluid, this volume of water must be at rest, which is indeed obtained for this case. Furthermore, this volume V of water must occupy a stable position, in order to fulfill the conditions associated with a static fluid for all times. Therefore, not only gravity and buoyancy must be equal in magnitude and opposed in direction for this case, but they must also take effect at the same point. This means that the buoyant force takes effect at the center of gravity of the displaced fluid. Since we always consider in this chapter that the body does not influence the behaviour of the surrounding liquid, this result will not be modified when introducing the body back into the liquid. Therefore, buoyancy always acts at the center of gravity of the displaced fluid (figure 5.7). Now,

58 56 Chapter 5. Hydrostatics and Aerostatics Buoyancy (-ρ b Vg) Gravity M b g z g Resulting torque: destabilizing y x Buoyancy (-ρ b Vg) Gravity M b g z g y x Figure 5.7: Stability of a partially immersed body, unstable case. Top figure: starting position. Bottom figure: perturbed (unstable) position. everything is known: magnitude, direction and center for both gravity and buoyancy, so that a stability analysis can easily be carried out by considering a small perturbation on top of the initial conditions. If, for example due to a wind gust, the body gets slightly inclined, it appears clearly from figure 5.7 that a torque will result and amplify the initial disturbance. As a consequence, the stable position of this body will be...lying sideways on the liquid surface. On the contrary, for the case considered in figure 5.8, the resulting torque brings the body back toward the unperturbed position, so that this configuration is stable. What is the difference between both? When the center of gravity lies higher than the center of the buoyant force, this leads to an unstable situation. On the contrary, when the center of gravity is

59 6 Stability of a partially immersed body 57 Buoyancy (-ρ b Vg) z g Gravity M b g Resulting torque: stabilizing y x Buoyancy (-ρ b Vg) z g Gravity M b g y x Figure 5.8: Stability of a partially immersed body, stable case. Top figure: starting position (stable). Bottom figure: perturbed position. deeper than the center of buoyancy, the body position is stable. As a consequence, it is advisable to move all heavy weights toward the bottom of a ship in order to obtain a stable configuration. Since this is typically not the case for a sailing-ship (sails and masts must be placed far above the sea surface), stability must usually be improved by adding a heavy keel. This simple analysis relies on the hypothesis that the center of gravity and buoyancy are not noticeably displaced as a result of the initial perturbation. Another, considerably more difficult solution to obtain stability is to implement a dynamic technical process where center of gravity and center of buoyancy are modified by the initial perturbation in such a way that the resulting torque will lead back to the starting position.

60 58 Chapter 5. Hydrostatics and Aerostatics 5.7 Aerostatics We now consider a static gas, and not any more a liquid. This might be for example the atmosphere, if we may assume it to be at rest (v = 0), i.e., in particular without any wind. This is indeed a strong assumption but will nevertheless lead to interesting results, in good agreement with experimental observations. The simplest possible analysis for aerostatics has already been briefly considered in section 5.3. If it is possible to assume that the gas density does not change at all (case of a fully incompressible gas), then all the results presented up to now remain exactly valid and involve the (constant) gas density ρ a. This would for example mean for the atmosphere that, assuming a pressure p 0 = 1 atm = Pa (=1 atm) at ground level (z = 0 m), the pressure would decrease linearly with the elevation z in our standard coordinate system, following: p(x) = p(z) = p 0 ρ a gz (5.17) How valid is this hypothesis? Clearly, it cannot be very good. Considering equation (5.17) and using as standard values p 0 = Pa and ρ a = 1.2 kg/m 3, this equation leads to a zero pressure at an elevation of roughly 8.6 km. This would clearly be the upper limit of the atmosphere. Since experimental observations lead to a typical atmosphere thickness of roughly 80 km (note, however, that it is difficult to define clearly where the atmosphere really ends up), the above approximation is far from appropriate and can only be used for a thin gas sheet, for which density variations can indeed be neglected, as done in section 5.3. The problem comes from the fact that the density of a gas can indeed be modified considerably when varying the local thermodynamic conditions, in particular temperature and pressure. Assuming ρ a = constant is therefore not a very good idea. In order to obtain a more accurate result, it is necessary to start again the analysis from the fundamental equation of hydro- and aerostatics (5.1), p = ρg which is still perfectly valid under the present conditions. We then start similarly with the reasoning presented in section 5.3. In the standard coordinate system and since gravity is the only mass force acting on the fluid here, we split this vector equation into 3 scalar equations, reminding that the gravity force is pointing towards the direction ( z): p x = 0 p = 0 y p = ρg (5.18) z The first two equations can be directly integrated and show that the pressure does not depend on the x and y spatial directions. Thus, the pressure variable can be given as p(z) instead of p(x): the only pressure variation will take place in the z-direction, which is logical, since the only mass force considered in this example also acts only in this direction. But, at the difference of section 5.3, it is not possible any more to integrate directly the third scalar equation of this system, since ρ is now varying. Considering a displacement in the z-direction between an elevation z 1 and z 2, it is nevertheless possible to integrate equation (5.18) to obtain: z2 1 p z 1 ρ z dz = z2 gdz (5.19) z 1

61 7 Aerostatics 59 The term on the right-hand-side can readily be integrated, by still considering the norm of the acceleration due to earth gravity g as a constant, leading finally to: z2 1 z 1 p ρ z dz = (z 1 z 2 )g (5.20) This equation can only be solved further if the behaviour of the density is known. This is in particular the case when considering a barotropic fluid, i.e., a fluid for which density is directly known as a function of pressure, ρ = ρ(p). In that case the resulting equation can be readily integrated when knowing exactly the function ρ(p). z2 1 p z 1 ρ(p) z dz = (z 1 z 2 )g (5.21) Pressure variation in an isothermal, ideal gas Asafirstexample, letusassumethatthebehaviouroftheearth satmospherecorrespondstoanideal gas at constant temperature. This hypothesis is indeed quite strong, since everyone accustomed to airplanes knows very well that the atmospheric temperature is not constant and quite low at usual flight heights. Nevertheless, this assumption will already deliver interesting results. An ideal gas is characterized by the equation of state (B.8) p = ρrt = Cρ (5.22) with C = rt = constant, since we assume an isothermal evolution (temperature T = constant). The parameter r is the specific gas constant, with typically r 287 J/(kg.K) for air. The equation of state can thus be inverted to give ρ = 1 C p (5.23) showing that an isothermal ideal gas is indeed a barotropic fluid. Equation (5.21) can now be rewritten as leading to z2 C p z 1 p z dz = (z 1 z 2 )g This can be easily integrated as z2 lnp z 1 z dz = (z 1 z 2 )g C lnp(z 2 ) lnp(z 1 ) = (z 1 z 2 )g C By taking the exponential of this equation, and choosing as reference condition at ground level (z = 0) the atmospheric pressure p a = Pa (1 atm), one obtains finally for the evolution of the pressure ( p(x) = p(z) = p a exp gz ) rt (5.24)

62 60 Chapter 5. Hydrostatics and Aerostatics where the constant C has been replaced by its equivalent, rt. This shows that the pressure decreases exponentially with the elevation z in this specific atmospheric model. Considering equation (5.23), the same is also true for the atmosphere density, following: ρ(x) = ρ(z) = p ( a rt exp gz ) rt (5.25) The corresponding behaviour(exponential decrease) corresponds quite well to experimental observations. An even better result can be obtained when considering an isentropic ideal gas instead of an isothermal evolution Principle of Archimedes in a gas The principle of Archimedes has been previously demonstrated for an incompressible fluid with ρ = constant. What is the corresponding result for a gas? To answer this question, we can start with the result presented in equation (5.12), since it does not require the assumption ρ = constant. The total pressure force exerted by a gas on an immersed body is thus F p = V ( ρg) dv Once again, we use the fact that the acceleration vector associated with earth s gravity, g, is a constant, so that the previous equation can be rewritten: F p = g ρdv (5.26) V Once again, what is V ρdv? This is the integral of the local fluid density (mass per unit volume) multiplied with the local volume dv. Therefore, this is clearly nothing else than the weight of the fluid that would occupy the place of the body V when removing it or, in other words, the weight of the displaced fluid. This proves that the principle of Archimedes is equally valid for a gas and for a liquid, without any change. In analogy to our previous example concerning a swimming-pool: is it safe for a human being to stay outdoors? Buoyancy might lead to an upward-pointing force transforming any human being in a hot-air balloon! To answer this question, it is once again sufficient to check the respective densities of air (roughly1.2kg/m 3 atgroundlevel)andofahumanbeing(somethinglike890kg/m 3 ). Clearly, thestable position of a human being in the atmosphere is...with both feet solidly on the ground! Nevertheless, the buoyant force induced by the surrounding atmosphere at rest leads to an upward pointing force of magnitude ρ a V b g. With ρ a = 1.2 kg/m 3 and for a human being with a typical volume of roughly 0.09 m 3, this means that you feel indeed about 100 g lighter than your real weight, thanks to buoyancy.

63 Chapter 6 Bernoulli equations 6.1 Introduction After having considered in Chapter 5 the simplest of all possible flows, i.e., those that do not flow at all, it is useful now to consider more usual flow conditions, associated with a non-zero flow velocity v! This is the purpose of the present chapter, in which the different forms of the Bernoulli equation will be progressively introduced. Considering back the variable-based classification proposed in the introduction, such flows will now still be associated with a change of pressure p, but also of velocity v. In order to limit the complexity of the resulting cases, the present chapter will assume consistently that the considered flow is always an incompressible flow. In this manner, the density ρ can be safely assumed to be constant throughout, ρ = constant. Density is therefore not an unknown of this problem. Furthermore, we will only consider steady flow conditions, so that all flow variables are constant in time. Finally, the resulting flow can therefore be described by knowing only the fields of pressure p(x) and velocity v(x). In this chapter, we will consider first only non-viscous flows (but we will later introduce a correction for viscous effects by using a side-door!). Finally, any non-contact force associated with an acceleration vector k is assumed to be conservative, meaning that it is possible to find a scalar potential φ(x) verifying k = φ. All the non-contact forces considered in this document are indeed conservative. The starting point to obtain the Bernoulli equation is always the same: it is the Euler equation describing conservation of momentum, equation (4.20), since we assume a non-viscous flow. Mass conservation must be additionally used in practice to solve completely the problem, but is not required to derive the Bernoulli equation. Since different forms of the Bernoulli equation can be written depending on the underlying hypotheses, we will always state explicitly with each result the list of required assumptions. 6.2 Bernoulli equation for an irrotational flow Starting from equation (4.20), ρ Dv Dt = ρg p (6.1) we start be using equation (2.17) to express the material derivative on the left-hand side, leading to: ρ v t +ρ(v )v = ρg p (6.2) 61

64 62 Chapter 6. Bernoulli equations The second term on the left-hand side (convective term) can be replaced by using the mathematical relation, equation (2.19). One obtains: ρ v ( ) v 2 t +ρ +ρ(( v) v) = ρg p (6.3) 2 or, dividing by the density ρ (obviously non-zero): ( ) v v 2 t + +(( v) v) = g 1 p (6.4) 2 ρ Considering the earth gravity field using the standard coordinate system, one obtains for this conservative force g = ge z = φ (6.5) for the following scalar field of potential energy: φ(x) = φ(z) = gz (6.6) Furthermore, since the flow is considered incompressible, the density ρ is constant and it is possible to write ( ) 1 p ρ p = (6.7) ρ Introducing the last two relations in equation (6.4), it comes: ( ) v v 2 t + +(( v) v) = φ 2 Moving all the terms to the left-hand side and grouping all three gradients together: ( ) v p t + ρ + v2 2 +φ +(( v) v) = 0 (6.9) For a steady flow, as assumed here, we obtain then: ( ) p ρ + v2 2 +φ +(( v) v) = 0 (6.10) If we furthermore assume first that the considered flow is irrotational, this equation finally simplifies to: ( ) p ρ + v2 2 +φ = 0 (6.11) What is the meaning of this relation? Since the gradient of a quantity measures its variation in space (it is simply built from the three partial derivatives in space), the fact that the gradient is 0 necessarily means that the corresponding quantity is constant and does not vary spatially. We therefore obtain the first Bernoulli equation, called from now on Bernoulli equation for irrotational flows: ( ) p ρ (6.8) p ρ + v2 +gz = constant in the complete flow (6.12) 2 where the potential φ has been replaced by its value for earth s gravity. This equation is valid for a steady, incompressible, non-viscous, irrotational flow with gravity as only non-contact force (or with the formulation involving the generic value of φ for any conservative force). This is an extremely useful relation to relate the values of pressure p and velocity v through the vertical position of the fluid element, z. Unfortunately, irrotational flows are not found very often in practice... Therefore, this relation usually cannot be used, and an alternative formulation valid for rotational flows would be even more attractive.

65 3 Link with hydrostatics Link with hydrostatics Before considering rotational flows, let us note that the Bernoulli equation indeed contains the results found in Chapter 5 dealing with hydrostatics(i.e., considering in the same manner only incompressible flows). For such non-flowing flows, v = 0 and the Bernoulli equation simplifies to: p +gz = constant in the complete flow (6.13) ρ It is equally possible to multiple this equation with the (constant) flow density ρ, leading to: p+ρgz = constant in the complete flow (6.14) This is indeed the result found in Chapter 5: the pressure decreases linearly with increasing height z (or conversely increases linearly with increasing depth, z). 6.4 Bernoulli equation (for a rotational flow) This is in fact the most useful form. In that case, it is not possible to assume any more v = 0, and we must start back from equation (6.10): ( ) p ρ + v2 2 +φ +(( v) v) = 0 (6.15) We then multiply this vector equation (scalar product) with the unit vector s, tangential to the local fluid velocity v; this is the vector giving the direction of the streamline. One obtains the scalar equation: s ( ) p ρ + v2 2 +φ +s (( v) v) = 0 (6.16) It is now easy to prove that the second term on the left-hand side is equal to zero. For this purpose, two essential properties of vector operations must be combined: 1. Concerning the vector product, the resulting vector is always perpendicular to the plane containing the two starting vectors. Therefore, ( v) v is in particular perpendicular to v. 2. Concerning the scalar product, it is known that the scalar product between two orthogonal vectors is automatically zero. As a consequence, the scalar product s (( v) v) is automatically equal to 0: s is tangential to the flow velocity vector v; as explained previously, ( v) v is simultaneously perpendicular to v; as a consequence, the scalar product between these two perpendicular vectors is zero. It comes thus: s ( ) p ρ + v2 2 +φ = 0 (6.17) What is the meaning of this new relation? As explained in the previous section, the gradient of a quantity measures its variation in space (it is simply built from the three partial derivatives in space). The scalar product appearing in equation (6.17) means that this variation in space is only considered ( projected ) along the direction of the unit vector s, i.e., along the direction of the streamline. Since the gradient is zero along this specific direction, this necessarily means that the corresponding quantity

66 64 Chapter 6. Bernoulli equations is constant and does not vary spatially when following the streamline. We therefore obtain the most important form of the relations derived by Daniel Bernoulli, called from now on just Bernoulli equation: p ρ + v2 +gz = constant along a streamline (6.18) 2 where the potential φ has been replaced by its value for earth s gravity. This equation is valid for a steady, incompressible, non-viscous flow with gravity as only non-contact force (or with the formulation involving the generic value of φ for any conservative force). Note that this relation is still valid for a rotational flow, as mostly found in practice! This is an essential equation to relate along a streamline (figure 6.1) the values of pressure p and velocity v through the vertical position of the fluid element, z. Figure 6.1: Streamlines (blue) computed by post-processing two-dimensional velocity fields measured by Particle Image Velocimetry in the rotating impeller of a centrifugal pump. 6.5 The Bernoulli triangle Let us first note here that all different formulations of the Bernoulli equation are derived by starting from the momentum conservation equation, the Euler equation (4.20). Nevertheless, due to the involved transformations, the final result is reminiscent of a conservation equation for flow energy, as illustrated in figure 6.2. Three different energy forms appear in the relation, all of them being specific (i.e., per unit mass of fluid): 1. the well-known kinetic energy of the fluid, v 2 /2; 2. the equally well-known potential energy of the fluid associated to gravity, gz;

67 6 Simplification of the Bernoulli equation for a gas flow 65 potential energy gz pressure energy p/ρ kinetic energy v 2 /2 Figure 6.2: Possible exchanges between different forms of flow energy according to the Bernoulli equation (6.18). 3. as a first component, a more specific pressure energy of the fluid, p/ρ. All these contributions have of course the right unit for a specific energy, i.e., J/kg. The so-called pressure energy is the only new term; it is of course well-known that it is possible to generate work by using a pressure difference (i.e., letting a pressure force work). But it is the first time that we associate directly with the absolute pressure p in the fluid a contribution similar to a specific energy. Beware! Please note that the possible exchanges between the three different forms of flow energy illustrated in figure 6.2 are still associated with limitations! We will always assume in practice that the lowest vertical position in the considered flow is associated with z = 0 per convention. This means that the upper term in figure 6.2 (potential energy), is always positive: gz 0. It is furthermore obvious that the kinetic energy term on the right is also positive: v 2 /2 0. Finally, the pressure term on the left is also positive since pressure p and density ρ are both positive: p/ρ 0. As a consequence, when equation (6.18) is used to illustrate the possible exchange between potential energy, kinetic energy and pressure energy, it must be kept in mind that, during the exchange process, all three terms must always remain positive! A flow for which one or more of these contributions would be negative is physically impossible to realize. Concerning the pressure term, the real physical condition is even more stringent, p/ρ > 0, due to a phenomenon called cavitation. 6.6 Simplification of the Bernoulli equation for a gas flow To illustrate this point, let us compare the order of magnitude of the first term (pressure energy) and of the last term (potential energy) for a non-flowing flow (static conditions, see equation 6.14). Considering for example a water height H of 10 m, we have demonstrated in Chapter 5 that the resulting hydrostaticoverpressure isρgh 1bar(withρ = 1000kg/m 3 ). Ithasthereforethesameorder of magnitude as the atmospheric pressure and as pressure levels found in most practical applications. If we now consider the same term for an air column, neglecting any change in density, the resulting static overpressure is ρgh bar (with ρ = 1.2 kg/m 3 ), hence roughly thousand times smaller. This simple example illustrates the fact that, when using the Bernoulli equation for a gas flow, the last term in equation (6.18) can usually safely be neglected compared to the other contributions. The only exceptions would be cases where both gas pressure and gas velocity are extremely small, or when considering huge variations of the z-coordinate (a gas column of several kilometers, for example in an

68 66 Chapter 6. Bernoulli equations atmospheric application). For all other standard engineering conditions, the Bernoulli equation for an incompressible gas flow can be simplified as: p ρ + v2 2 = constant along a streamline (6.19) 6.7 Dynamic pressure It is of course possible to multiply the Bernoulli equation (6.18) with the fluid density ρ, leading to: p+ ρv2 2 +ρgz = constant along a streamline (6.20) Obviously, all three terms in this equation have the same unit, the unit of pressure (Pascal, Pa). More specifically, the second term in this equation, ρv 2 /2 will from now on also be called dynamic pressure q: q := ρv2 2 (6.21) If we neglect the contribution of potential energy, the meaning of this dynamic pressure is clear: starting from a flow with conditions p for pressure and v for velocity, if this flow is decelerated until coming to rest (i.e., v = 0, called stagnation conditions) the pressure will simultaneously increase by an amount equal to the initially available dynamic pressure. This is illustrated in Table 6.1. v 0 p p+ ρv2 2 Table 6.1: Change of flow pressure and flow velocity from starting conditions to stagnation conditions 6.8 Averaged Bernoulli equation We have already seen two possible formulations of the Bernoulli equation. The first one (equation 6.12) would be extremely useful, but is only valid for irrotational flows and is therefore of limited application. The second one, equation (6.18) can be applied for a variety of flows, but can only be used when a specific streamline has been determined. Identifying such a streamline is a tedious process and it can sometimes be a complex issue. Would it be possible to compromise between the two formulations, so that we could use an equation similar to equation (6.12) even for a rotational flow? This is indeed possible for internal flows, for which the formulation to come will be particularly useful. For this purpose, we consider an incompressible flow within a pipe of arbitrary geometry (the crosssection area A may vary in an arbitrary manner, the pipe direction can change freely along horizontal and vertical coordinates). We keep the standard assumptions used at the beginning of this chapter: the flow is steady, incompressible, non-viscous, with gravity as only non-contact force (or with the formulation involving the generic value of the potential φ for any conservative force). It is therefore possible to use readily equation (6.18) as long as a suitable streamline has been identified for this purpose. Now, instead of considering the full flow, we limit the analysis to the variation between an inflow cross-section A 1 and

69 8 Averaged Bernoulli equation 67 an outflow cross-section A 2 (figure 6.3). The identified streamline will connect (in an unknown manner) the inflow and the outflow cross-section. Along this streamline, we might write equation (6.18): p ρ + v2 +gz = constant (6.22) 2 or considering only the initial conditions (index 1) and final conditions (index 2) along this streamline: p 1 ρ + v gz 1 = p 2 ρ + v gz 2 (6.23) It is in principle possible to repeat this process for all streamlines in this flow, therefore connecting all fluid elements present along the inflow cross-section A 1 with their corresponding state along the outflow cross-section A 2 (figure 6.3). A 1 A 2 Figure 6.3: Averaging procedure between inflow and outflow cross-sections. If we now average in space along the cross-section all corresponding equations, we obtain the averaged Bernoulli equation: p 1 ρ + v gz 1 = p 2 ρ + v gz 2 (6.24) where the indices 1 and 2 characterize again respectively the inflow and outflow cross-sections, A 1 and A 2, and where the symbols with overline correspond to spatially-averaged values along the corresponding cross-section, for example: p 1 := 1 A 1 A 1 p(x)da (6.25) If the flow is indeed one-dimensional along the inflow and outflow cross-sections, A 1 and A 2, then this averaging procedure is superfluous for flow pressure and flow velocity. In such a case, the pressure and the velocity are constant along a cross-section, so that p 1 = p 1, v 2 1 = v 2 1, p 2 = p 2 and v 2 2 = v 2 2. Nevertheless, even in such a case, the vertical coordinate z must still be averaged (z 1 z 1 ), except in the case where the pipe is running purely along the vertical direction, so that each cross-section corresponds to a plane z = constant. In general, z is simply the z-coordinate of the geometrical center of the corresponding pipe cross-section. For a circular pipe, this will be the value of z associated with the pipe axis. From now on, we will write the averaged Bernoulli equation without the overlines, leading simply to: associated with following conditions: p 1 ρ + v gz 1 = p 2 ρ + v gz 2 (6.26)

70 68 Chapter 6. Bernoulli equations 1. this equation is valid between an inflow cross-section A 1 and an outflow cross-section A 2 ; 2. all terms appearing in this equation have been spatially averaged along each cross-section following equation (6.25); or alternatively the flow is one-dimensional along the inflow and outflow crosssections and the value of z appearing in the equation is the z-coordinate of the geometric center of the corresponding cross-section; 3. as usual, the flow is steady, incompressible, non-viscous, with gravity as only non-contact force (or with the formulation involving the generic value of the potential φ for any conservative force). On the other hand, it is not necessary to have an irrotational flow in order to apply equation (6.26). And it is also not necessary to identify any streamline before using this equation! Since the boundary of a streamtube cannot be crossed by any fluid particle, a streamtube is closely similar to the internal flow within a duct of variable cross-section (that of the streamtube), as considered up to now. The results presented in this section for an internal flow can therefore equally well be applied to any streamtube, as long as the assumptions listed above are valid. Of course, in that case, it is your responsibility to identify correctly the streamtube before applying equation (6.26) between the inflow and outflow cross-sections. Beware! If you have the feeling that you have been cheated during the developments presented in this section, you are quite right! From a purely mathematical point of view, the so-called averaging process introduced to develop equation (6.26) is certainly not a clean proof... It is indeed possible (but relatively difficult) to prove that equation (6.26) is perfectly correct. The corresponding proof will be added in a later version of this document. 6.9 Hydraulic height Equation (6.26) can be equally well rewritten by dividing everything with the gravity acceleration g, leading to: p 1 ρg + v2 1 2g +z 1 = p 2 ρg + v2 2 2g +z 2 (6.27) Obviously, all terms in this equation have now the same unit as z, i.e., meter (m). All can therefore be written as equivalent lengths. The sum of all three terms, being constant as shown by this equation, is called from now on hydraulic head (sometimes also total head) and is written H: H := p ρg + v2 +z = constant (6.28) 2g The three terms building up the (constant) hydraulic head are the pressure head p/(ρg), the dynamic (or velocity) head v 2 /2g and the elevation (or geodetic) head z. All these terms are expressed in meters. It is now possible to use equation (6.28) to analyze graphically the evolution of the flow properties in a practical installation, as shown schematically in figure 6.4. As already explained previously, it must of course be kept in mind that all three separate contributions (elevation head, dynamic head and pressure head) must always remain positive: this is obvious for the dynamic head; it is the result of the chosen convention (z = 0 at the lowest point in the flow) for the elevation head; and it is even more stringent (p/(ρg) > 0) for the pressure head due to cavitation.

71 10 Generalized Bernoulli equation with losses and energy exchange pressure head p 1 /( g) velocity/dynamic head v 12 /(2g) hydraulic or total head H pressure head p 2 /( g) velocity/dynamic head v 22 /(2g) elevation head z 1 elevation head z 2 0 level 1 2 Figure 6.4: Graphical representation of a possible evolution for pressure head, dynamic head and elevation head for a constant total head and a constant pipe diameter (hence constant velocity v) Generalized Bernoulli equation with losses and energy exchange It is now possible, based on the previously introduced averaged Bernoulli equation (6.26), to present the last and most complete form of the relations associated with Daniel Bernoulli. The purpose of this last formulation that we will later call generalized Bernoulli equation, is in particular to take into account the influence of all kinds of losses found in practical flows. Up to now, these losses have been completely neglected and therefore do not appear at all in equation (6.26). In practice such losses might be of considerable importance and have two main origins: 1. they might be a result of friction, whenever velocity gradients are found in a flow. This is usually the case in practice, and such velocity gradients can be particularly large near the walls, in the boundary layer. 2. they might be the result of other flow modifications leading to an irreversible change of thermodynamic state. This is also very often the case, for instance when changing rapidly flow direction or velocity magnitude, or within vortices, as encountered in bends, reducers, junctions, at the entrance of exit or a pipe, within a valve... Globally, all these losses correspond to an entropy increase. In order to quantify them exactly, a

72 70 Chapter 6. Bernoulli equations conservation equation for energy and entropy in the flow would be needed, together with a detailed analysis of the local flow conditions in time and space. Even if this is possible in practice, this is too complex for our present level of knowledge and for most practical applications. Therefore, our purpose is now simply to model such losses in an appropriate manner, allowing a sufficiently accurate (but not exact) determination of the resulting flow properties. As we already know, there are three different possible units that can be used to write the averaged Bernoulli equation: either all quantities are expressed as equivalent to pressure (unit: Pa), equivalent to geodetic height (unit: m) or equivalent to specific energy (unit: J/kg). Equivalently, the head loss appearing in the flow can be expressed as: equivalent energy loss (or dynamic head loss), written e and measured in J/kg. equivalent pressure loss (or pressure head loss), written p and measured in Pa. equivalent geodetic height loss (or elevation head loss), written z and measured in m. Per convention, all these losses will be considered as positive. It is of course very easy to convert one expression to the other, using: z = p ρg = e (6.29) g If we now write with an index f the losses associated to friction and with an index l all other losses (l designates here a localized loss, since such a loss can be associated to a specific location in the flow; note, however, that these losses are most commonly called minor losses in the international literature, a somewhat misleading name, since they might indeed be quite large). Now, the friction loss (considered globally for the complete flow) and a number of localized losses (numbered 1 to n in what follows, and associated to localized flow modifications), can be taken into account in the Bernoulli equation, leading to: [ ] [ ] p2 ρ + v gz p1 2 = ρ + v2 n 1 2 +gz 1 e f e l,j (6.30) or equivalently [ [ ] p2 ρg + v2 2 2g +z 2 p 2 + ρv ρgz 2 ] = = j=1 [ ] p1 ρg + v2 n 1 2g +z 1 z f z l,j, (6.31) j=1 [ ] p 1 + ρv2 n 1 2 +ρgz 1 p f p l,j (6.32) j=1 Since we will always encounter such losses in practical flows, we must find a solution to compensate them. This is simply realized by exchanging work with the fluid. If the user wants to increase the energy level of the fluid, it will communicate work to it, typically using a pump, delivering a positive specific work w > 0 to the fluid. Conversely, if the fluid is used to generate energy, the negative work w < 0 will be lost by the fluid and communicated to the surroundings, usually using a turbine. As you see, the exchanged specific work w (expressed in J/kg) is always considered from the point of view of the fluid; it is positive when increasing fluid energy, negative when fluid energy is reduced. Finally, if a certain number m of such devices exchanging work with the fluid are found between the inflow cross-section A 1 and the outflow cross-section A 2, the generalized Bernoulli equation can be written: [ ] [ ] p2 ρ + v gz p1 2 ρ + v gz 1 = and is associated with following conditions: m n w i e f e l,j (6.33) i=1 j=1

73 10 Generalized Bernoulli equation with losses and energy exchange all terms appearing in this equation have been spatially averaged along inflow and outflow crosssections following equation (6.25); or alternatively the flow is one-dimensional along the inflow and outflow cross-sections and the value of z appearing in the equation is the z-coordinate of the geometric center of the corresponding cross-section; 2. the flow is steady, incompressible, with gravity as only non-contact force (or with the formulation involving the generic value of the potential φ for any conservative force). On the other hand, it is now not necessary to have a non-viscous flow, since the influence of viscous losses appears in the equation ( e f )! This equation can readily be used for a rotational flow, and it is not necessary to identify any streamline before using this relation. In order to solve any practical problem, the only remaining issue is now to determine exactly all terms on the right-hand side, since the left-hand side is well-known, involving only classical contributions Computing the exchanged specific work w The specific work w exchanged between the fluid and the surroundings is mostly well-known; it will be simply regulated by the user (for a pump, delivering a positive specific work w p ) or measured as an output quantity by the user (for a turbine, delivering to the surroundings a positive specific work w t ). If the efficiency η of the apparatus is 100%, the relation between the value set/read by the user and the value relevant for the fluid and appearing in equation (6.33) is simply w = w p for a pump, resp. w = w t for a turbine. In reality, the efficiency η of the apparatus is not perfect, η < 1. For a pump with an efficiency η p, the real relation between the work of the pump and the final amount really available for increasing fluid energy is then: w = η p w p (6.34) Conversely, for a turbine with an efficiency η t, the relation between the amount of energy lost by the fluid and the energy really available for the turbine user reads: w = w t η t (6.35) If needed, the relation between specific work w and associated power P is simply: P = ṁw (6.36) Computing the friction loss e f Remember that we are looking for an acceptably accurate estimation of the losses occuring in practice in such a flow and not for an exact solution. As a consequence, the underlying methodology relies mostly on systematic experimental measurements, supported as far as possible by theoretical considerations leading to major simplifications. The corresponding theory and the resulting complexity reduction will be described later (Chapter 10); we now concentrate only on the information obtained from experimental studies. From a fundamental point of view, the formulation of the generalized Bernoulli equation written as a specific energy equation (6.33) is most appropriate, and will be used in what follows. Nevertheless, you can easily convert the resulting losses as equivalent pressure head loss or elevation head loss using equation (6.29): z = p ρg = e (6.37) g

74 72 Chapter 6. Bernoulli equations As we will discuss later, friction is indeed a result of gradients of flow momentum (or equivalently, for an incompressible flow as considered here, gradients of flow velocity). There is therefore a resulting connection between frictionloss e f andflowvelocity v. As aconsequence, itis very practical toexpress the friction loss as a function of the available specific kinetic energy in the flow, v 2 /2. Finally, the measured friction loss depends of course on the considered flow configuration. Qualitatively, since we have stated previously that velocity gradients will be particularly large near the walls, in the so-called boundary layer, the importance of friction loss will be considerable for flows in contact with a lot of walls (micro-channels, heat exchangers) and quite negligible in the opposite case (smooth straight tube with a very large diameter, without any insert). In practice, cylindrical pipes are mostly encountered. Therefore, we will only give here results for such a configuration: a straight cylindrical pipe of length l, diameter d, with a mean axial flow velocity v. Systematic experiments show that the associated friction loss can be computed using following relation, called Darcy-Weisbach equation: e f = f l v 2 d 2 (6.38) where the dimensional parameter f appearing in this equation is called friction factor, as first introduced by Darcy in the 19th century after carrying out a large amount of pipe flow measurements. Apart from f, all quantities in this equation are already known (note that the norm of the flow velocity v = v will in practice mostly be computed from the volumetric flow-rate, equation (1.5), using v = q/a). The friction factor has been measured experimentally in a highly systematic manner for a variety of configurations. Corresponding contributions, mainly by Darcy, Weisbach, Nikuradse and Moody finally lead to an expression of the friction factor depending from the Reynolds number Re d of the flow and from the relative wall roughness height ǫ/d: f = f ( Re d, ǫ ) d (6.39) The Reynolds number will be discussed in detail in Chapter 10. Let us simply define it here. It is a non-dimensional number computed as: Re d := ρvd µ = vd ν (6.40) where the dynamic viscosity µ or kinematic viscosity ν can be equally well employed, taking into account the relation ν = µ/ρ. The functional dependency described by equation (6.39) is summarized in a well-known graphical representation, called Moody chart (or Moody-Colebrook chart) (figure 6.5). The resulting value of the friction factor is always well below unity (typically, f < 0.1). Now, the final procedure to compute friction loss e f in a straight cylindrical pipe is clear. First, the mean velocity v must be known or computed, together will all geometrical characteristics including the wall roughness height ǫ, a measure of the smoothness of this wall. Then, it is possible to compute the Reynolds number Re d from equation (6.40) and the relative wall roughness height ǫ/d. A large value of ǫ/d means a very rough wall (pipe of bad quality), while a small value of ǫ/d means a higher smoothness and therefore lower friction loss (it is clear from figure 6.5 that f, and therefore e f, increases with ǫ/d). Knowing these two values, figure 6.5 is employed to determine f. The value of the friction factor is then used in equation (6.38) in order to compute the final value of e f. Finally, remember that this procedure is just a sufficiently accurate approximation of the reality. The value of f employed in the computation is probably not so accurate (an error of a few % is considered

75 10 Generalized Bernoulli equation with losses and energy exchange 73 as fully acceptable). And the value of ǫ entering the computation is not that well-known, may vary spatially or in time (corrosion, particle deposition on the walls). The friction loss you determine in this manner is only a relatively accurate estimation. If the considered pipe is not a standard, cylindrical pipe, it is still possible to use the results presented in this section by replacing everywhere and in a systematic manner the diameter d by the hydraulic diameter d h. This leads to an even higher error level, but allows to consider an arbitrary geometry Numerical equations used to estimate the friction factor f Usually, using the Moody-Colebrook chart diagram as described in the previous section in order to determine graphically the solution is fully sufficient. But, for solving practical problems, an equation giving (directly or indirectly) the value of f would be much more practical and/or would speed-up considerably the solution procedure. Many such fitting laws are available. Friction factor f in the laminar regime For laminar flow conditions, which means in practice for a value of the hydraulic Reynolds number Re d below roughly for an axisymmetric pipe (top-left part of the Moody-Colebrook chart), we will show later in Chapter 9 that an exact formula can be obtained for such conditions, reading simply: Approximation of the friction factor f for smooth conditions f = 64 Re d (6.41) Here, different formulas are found, which are simply fitting more or less accurately the experimental measurement represented in the Moody-Colebrook chart diagram. For 4000 Re d 10 5, the law of Blasius is mostly employed and gives directly the value of f: f = Re d 1/4 (6.42) For Re d > 2300, the (implicit) relation of Prandtl-Kármán can be used instead, usually requiring iterations to identify f: ) 1 = 2log ( f Re d 0.8 (6.43) f For highly turbulent conditions (Re d > 10 5 ), the direct relation of Nikuradse can be used instead: f = Re d (6.44) Approximation of the friction factor f for rough conditions Here again, different formulas are found, which are simply fitting more or less accurately the experimental measurement represented in the Moody-Colebrook chart. For Re d > 2300, the law of Colebrook-White is giving the best approximation, but is an implicit relation and will therefore usually require iterations: ( = 2log + ǫ ) f f Red 3.7d (6.45)

76 74 Chapter 6. Bernoulli equations With a slightly lower precision, but allowing a direct estimation of f, the law of Swamee-Jain can be used instead: [ ( 5.74 f = 0.25 log Re ǫ )] 2 (6.46) d 3.7d Computing a localized loss e l Onceagain,wearelookingforanacceptablyaccurateestimationofthislossandnotforanexactsolution. As a consequence, the underlying methodology relies almost exclusively on systematic experimental measurements. Since localized losses are mostly somehow connected to the flow velocity v, and in order to stay as close as possible to the friction loss considered in the previous section, it is very practical to express again the localized loss as a function of the available specific kinetic energy in the flow, v 2 /2. Finally, the measured friction loss depends of course on the considered flow configuration and on the origin of the considered loss. The practical computation involves simply a loss coefficient(sometimes also called resistance coefficient) K, as simple proportionality coefficient between the loss and the available specific kinetic energy: e l = K v2 2 (6.47) The value of K is given by an experimental measurement. Note that this non-dimensional value might well exceed unity. Available technical documents contain thousands of measurement results, corresponding to a huge variety of specific configurations. The only real difficulty for an engineer is to find a reliable source of information corresponding exactly to the considered flow configuration. Now, the final procedure to compute local loss e l is clear. First, the mean velocity v = v must be known or computed. Knowing the flow configuration and the geometry, the value of the loss coefficient K is found in a reference document. Then, equation (6.47) is employed in order to compute the final value of e l. A few selected examples of localized loss coefficients K are listed in Table 6.2. Once again, these values should definitely not be regarded as very accurate, but as mere approximations.

77 10 Generalized Bernoulli equation with losses and energy exchange 75 Configuration K Tee, flanged, line flow Tee, flanged, branched flow Union, threaded 0.08 Elbow, flanged regular Return bend, flanged Globe valve, fully open 6 10 Angle valve, fully open 2 Gate valve, fully open 0.15 Gate valve, 1/4 closed 0.26 Gate valve, 1/2 closed 2.1 Gate valve, 3/4 closed 17 Diaphragm valve, open 2.3 Diaphragm valve, half open 4.3 Diaphragm valve, 1/4 open 21 Water meter 7 Table 6.2: Loss coefficient K for some selected configurations

78 76 Chapter 6. Bernoulli equations Figure 6.5: Approximate Moody chart based on the Swamee-Jain equation as drawn by S. Beck and R. Collins, University of Sheffield, reprinted from Wikipedia.

79 Chapter 7 Force and torque exerted by a flow 7.1 Introduction In the previous chapter we have mainly considered the local evolution of the main flow variables(pressure p and velocity v). Building on top of these results, it is now possible to determine the force and the torque exerted by such a flow on its surroundings, considering a volume of fluid limited by an inflow cross-section A 1 and an outflow cross-section A 2 ; the side surface (i.e., either the wall of the pipe containing this flow or the corresponding outside surface of the associated streamtube) will be written Σ. Considering this configuration (figure 7.1), the surface constituted by assembling A 1,A 2 and Σ is a fixed, closed surface in the fluid (A cf = A 1 A 2 Σ) and can be used to define a fixed control volume V cf, contained within the surface A cf and always filled up with fluid. Outer surface Σ n p 2 ρ2 n 2 v 2 s 2 A 2 Control volume V v 1 A 1 n 1 p 1 ρ 1 s 1 n Figure 7.1: Configuration to compute the force exerted by a flow on the side boundary Σ. In order to compute the force and torque exerted by the corresponding flow, we will only consider steady flow conditions, so that all flow variables are constant in time. As a supplementary assumption, we will consider only one-dimensional flows along the inflow and outflow cross-sections, A 1 and A 2. Furthermore, we will consider first only non-viscous flows (but we will later show that the obtained results are equally valid for viscous flows!). 77

80 78 Chapter 7. Force and torque exerted by a flow Beware! The results obtained in this Chapter do not require at all an incompressible flow. The density ρ of the fluid may change in any manner (or stay constant), this will not affect the results presented later in this Chapter. This is one further attractive feature of the findings presented in what follows: they stay equally valid for a variety of different flow configurations. 7.2 Force exerted by a flow on its surroundings In order to compute first the resulting force, we start from the integral formulation of momentum conservation for a steady flow introduced in Chapter 4 for a fixed control volume V cf, Eq.(4.17). ρ(x,t)v(x,t)(v(x,t) n)da = A cf ρ(x,t)gdv V cf p(x,t)nda A cf (7.1) Remember that A cf = A 1 A 2 Σ, so that each corresponding integral on A cf can be written as the sum of three integrals on A 1, A 2 and Σ respectively. We will also use, as already done several times previously in the Chapters 3 to 6, the unit vectors n (perpendicular to the local external surface A cf and pointing toward the outside, appearing already in Eq.(7.1) and s (colinear, i.e., parallel to the local flow velocity v). From the geometrical construction of the control volume V cf, it can easily be seen that along the inflow cross-section A 1 one obtains n 1 = s 1, while along the outflow cross-section A 2, n 2 = s 2. The last integral involving the pressure force in Eq.(7.1) can now be rewritten as p(x,t)nda = p(x,t)nda+ p(x,t)nda+ p(x,t)nda (7.2) A cf Σ A 1 A 2 The first term on the right-hand side of this equation (integration over the side section Σ) is directly the pressure force exerted by the fluid on its surroundings, denoted from now on F p : this is exactly what we want to determine here! The second term can be easily computed, since the flow has been assumed one-dimensional along the inflow cross-section A 1, leading to: p(x,t)nda = p 1 n 1 da (7.3) A 1 A 1 = p 1 n 1 A 1 da (7.4) = p 1 n 1 A 1 (7.5) = p 1 s 1 A 1 (7.6) where A 1 is the area of A 1. In the same manner, the last integral in Eq.(7.2) can be written: A 2 p(x,t)nda = p 2 n 2 A 2 (7.7) with A 2 the area of A 2. Finally, one gets: = p 2 s 2 A 2 (7.8) A cf p(x,t)nda = F p p 1 s 1 A 1 +p 2 s 2 A 2 (7.9) Following a similar technique, the first integral on the left-hand side in Eq.(7.1) can be split into three parts: A cf ρ(x,t)v(x,t)(v(x,t) n)da =

81 2 Force exerted by a flow on its surroundings 79 Σ ρ(x,t)v(x,t)(v(x,t) n)da + ρ(x,t)v(x,t)(v(x,t) n)da A 1 + ρ(x,t)v(x,t)(v(x,t) n)da (7.10) A 2 Along Σ (the pipe wall or the side section of a streamtube), one observes necessarily a flow velocity v locally tangential to Σ (the flow cannot cross in any manner the surface Σ). As a consequence, the flow velocity v (parallel to Σ) and the normal vector n (normal to Σ) are perpendicular to each other. Due to the fundamental properties of the scalar product, the corresponding contribution is then automatically 0: ρ(x,t)v(x,t)(v(x,t) n) da = 0 Σ }{{} (7.11) =0 Along the inflow cross-section A 1, the one-dimensional assumption leads again to ρ(x,t)v(x,t)(v(x,t) n)da = ρ 1 v 1 (v 1 n 1 )da (7.12) A 1 A 1 = ρ 1 v 1 (v 1 n 1 ) da A 1 (7.13) = ρ 1 v 1 (v 1 n 1 )A 1 (7.14) Since v 1 = v 1 s 1 = v 1 n 1, this can be rewritten: A 1 ρ(x,t)v(x,t)(v(x,t) n)da = ρ 1 v 1 s 1 ( v 1 )A 1 (7.15) = ρ 1 v 2 1 s 1A 1 (7.16) = ṁ 1 v 1 s 1 (7.17) introducing the mass flow-rate in the inflow cross-section A 1, ṁ 1 = ρ 1 v 1 A 1. Similarly, one obtains along the outflow cross-section A 2, A 2 ρ(x,t)v(x,t)(v(x,t) n)da = ṁ 2 v 2 s 2 (7.18) Since the flow is steady, mass conservation implies the conservation of the mass flow-rate in any crosssection, so that ṁ 1 = ṁ 2 = ṁ (7.19) One can therefore write finally: A cf ρ(x,t)v(x,t)(v(x,t) n)da = ṁv 1 s 1 +ṁv 2 s 2 (7.20) The first integral on the right-hand side of Eq.(7.1), describing the influence of gravity, can be very easily computed in a direct manner, remembering that the acceleration vector g can be assumed to be constant for practical applications on the earth s surface: ρ(x,t)gdv = g ρ(x,t)dv (7.21) V cf V cf = Mg (7.22) where M is the total mass of the fluid contained within the fixed control volume V cf, as already shown in Eq.(3.1).

82 80 Chapter 7. Force and torque exerted by a flow Finally, it is possible to rewrite now Eq.(7.1) as: ṁv 1 s 1 +ṁv 2 s 2 = Mg F p +p 1 s 1 A 1 p 2 s 2 A 2 (7.23) This can now be transformed in a suitable manner to determine F p : F p = (ṁv 1 +p 1 A 1 )s 1 (ṁv 2 +p 2 A 2 )s 2 +Mg (7.24) We define a new scalar quantity along each cross-section A in the flow, called impulsion, written F and expressed in N (like a force), by: F := ṁv +pa (7.25) It is then possible to give the expression of the force exerted by the fluid on its surroundings through the separation surface Σ as: F p = F 1 s 1 F 2 s 2 +Mg (7.26) This relation giving the total force exerted by the fluid between the inflow cross-section A 1 and outflow cross-section A 2 is valid for a steady non-viscous flow, assumed moreover one-dimensional along the inflow and outflow cross-sections, A 1 and A 2. It must be noticed here that the force exerted by a volume of fluid (that contained within the fixed control volume V cf ) can be computed by knowing only the main flow parameters in the inflow and outflow cross-sections (change of impulsion F and/or of flow direction s); it is not necessary to know the details of the flow within V cf, only start and end flow conditions are sufficient to compute the resulting force. This is an extremely powerful formulation, and it can be used to obtain a wealth of essential results in practical applications! Note that, for most practical cases, the contribution associated to the weight of the fluid (Mg, last term in Eq.7.26) can usually be safely neglected when considering a gas flow, since gas densities are quite low. On the other hand, it must be taken into account for a liquid, since it might lead to considerable values. At the present level, the results of Eq.(7.26) are equally valid for the flow in a pipe (pipe wall Σ) or for the flow within a streamtube, in which case Σ is just a geometrical surface in the fluid separating in a purely theoretical manner different fluid elements. 7.3 Force exerted by a flow on a pipe wall surrounded by a fluid at constant pressure If we now consider more specifically an internal flow in a pipe, Σ is the pipe wall surrounding the fluid. In most practical cases, outside of the pipe, a constant pressure p = p a is found everywhere (figure 7.2). When trying to compute the resulting force on the pipe wall, it is more useful in practice to take into account as well the influence of the external pressure, p a, and not only the action of the internal fluid pressure, as done in the previous section. The force exerted by the internal flow on the pipe wall, Σ, has been determined in the previous section and is given by Eq.(7.26). The supplementary pressure force exerted by the constant and homogeneous external pressure p a on the same wall is now: F pa = p a nda (7.27) Σ where the minus sign denotes the fact that this pressure force is acting towards the inside of the control volume, i.e., towards n.

83 3 Force exerted by a flow on a pipe wall surrounded by a fluid at constant pressure 81 Outer surface Σ n p 2 ρ2 n 2 p a v 2 s 2 p a Control volume V A 2 v 1 A 1 n 1 p 1 ρ 1 s 1 n p a p a Figure 7.2: Configuration to compute the force exerted by a flow on the pipe wall Σ taking into account the constant and uniform external pressure p a. Let us consider the same integral but integrating now over the complete, closed outer surface of the considered, fixed control volume V cf, i.e., the surface A cf = A 1 A 2 Σ. One gets first: p a nda = p a nda (7.28) A cf A cf Now, using the second formulation of the divergence theorem (Eq. C.6), it is possible to replace this last integral, leading to: nda = ( 1)dV (7.29) A cf V cf Obviously, the gradient of a constant scalar (here 1) is zero. This demonstrates a well-known result: when integrating the normal unit vector n over any closed geometrical surface, one obtains always identically 0. Finally, this demonstrates that: p a nda = 0 (7.30) A cf = p a nda+ p a nda+ p a nda (7.31) A 1 A 2 Σ Or p a nda = p a nda+ p a nda (7.32) Σ A 1 A 2 Finally, the total pressure force exerted simultaneously by the internal flow and by the external pressure p a, written in what follows F p,pa, reads: F p,pa = F 1 s 1 F 2 s 2 +Mg p a nda (7.33) Σ = F 1 s 1 F 2 s 2 +Mg+ p a nda+ p a nda (7.34) A 1 A 2 = F 1 s 1 + p a nda F 2 s 2 + p a nda+mg (7.35) A 1 A 2

84 82 Chapter 7. Force and torque exerted by a flow = F 1 s 1 +p a n 1 da F 2 s 2 +p a n 2 da+mg A 1 A 2 (7.36) = F 1 s 1 +p a n 1 A 1 F 2 s 2 +p a n 2 A 2 +Mg (7.37) = (F 1 s 1 p a A 1 s 1 ) (F 2 s 2 p a A 2 s 2 )+Mg (7.38) Introducing now along each cross-section A in the flow the relative impulsion, written F and expressed in N (like a force), by: F := ṁv+(p p a )A (7.39) it is then possible to give the expression of the force exerted by the internal fluid on the pipe wall Σ taking into account the external pressure p a by: F p,pa = F 1s 1 F 2s 2 +Mg (7.40) This relation is valid for a steady non-viscous flow, assumed moreover one-dimensional along the inflow and outflow cross-sections, A 1 and A 2. Note that, for most practical cases, the contribution associated to the weight of the fluid (Mg, last term in Eq.7.40) can be safely neglected when considering a gas flow, since gas densities are quite low. 7.4 Torque exerted by a flow on its surroundings We will now compute the resulting torque exerted by a flow on its separation surface Σ (figure 7.3) by reference to a pivot P. In order to compute the resulting torque, we start from the integral formulation of angular momentum conservation for a steady flow introduced in Chapter 4 for a fixed control volume V cf, Eq.(4.26): r [ρ(x,t)v(x,t)(v(x,t) n)]da = A cf r [ρ(x,t)g]dv r [p(x,t)n]da (7.41) V cf A cf Remember that A cf = A 1 A 2 Σ, so that each corresponding integral on A cf canbe written as the sum of three integrals on A 1, A 2 and Σ respectively. Here, we employ again the unit vector n (perpendicular to the local external surface A cf and pointing toward the outside, appearing already in Eq.7.41) and s (colinear, i.e., parallel to the local flow velocity v). From the geometrical construction of the control volume V cf, it can easily be seen that along the inflow cross-section A 1 one obtains n 1 = s 1, while along the outflow cross-section A 2, n 2 = s 2. The last integral involving the pressure force in Eq.(7.41) can now be rewritten as r [p(x,t)n]da = A cf r [p(x,t)n]da+ r [p(x,t)n]da+ r [p(x,t)n]da (7.42) Σ A 1 A 2 The first term on the right-hand side of this equation (integration over the side section Σ) is directly the torque exerted by the fluid pressure on its surroundings, denoted from now on T p : this is what we want to determine here! The second term can be easily computed, since the flow has been assumed one-dimensional along the inflow cross-section A 1, leading to: r [p(x,t)n]da = r [p 1 n 1 ]da (7.43) A 1 A 1

85 4 Torque exerted by a flow on its surroundings 83 Outer surface Σ s 2 v 2 G 2 A 1 s 1 v 1 Control volume V G A 2 G 1 r r 1 r 2 P Figure 7.3: Configuration to compute the torque exerted by a flow on the side boundary Σ by reference to the pivot P. [ ] = rda (p 1 n 1 ) (7.44) A [ 1 ] = rda (p 1 s 1 ) (7.45) A 1 The integral appearing in this relation, leading to the mean value of r when covering all the inflow surface A 1, reads simply: A 1 rda = r 1 A 1 (7.46) where A 1 is the area of A 1 and r 1 = PG 1 is the vector connecting the pivot P with the geometrical center G 1 of A 1. Finally, one gets: A 1 r [p(x,t)n]da = r 1 (p 1 A 1 s 1 ) (7.47) In the same manner, the last integral in Eq.(7.42) can be written: r [p(x,t)n]da = r 2 (p 2 A 2 n 2 ) A 2 (7.48) = r 2 (p 2 A 2 s 2 ) (7.49) with A 2 the area of A 2 and r 2 = PG 2 the vector connecting the pivot P with the geometrical center G 2 of A 2.

86 84 Chapter 7. Force and torque exerted by a flow Finally: A cf r [p(x,t)n]da = T p r 1 (p 1 A 1 s 1 )+r 2 (p 2 A 2 s 2 ) (7.50) Following a similar technique, the first integral on the left-hand side in Eq.(7.41) can be split into three parts: r [ρ(x,t)v(x,t)(v(x,t).n)]da = A cf r [ρ(x,t)v(x,t)(v(x,t).n)]da Σ + r [ρ(x,t)v(x,t)(v(x,t).n)]da A 1 + r [ρ(x,t)v(x,t)(v(x,t).n)]da (7.51) A 2 Along Σ (the pipe wall or the side section of a streamtube), one observes necessarily a flow velocity v locally tangential to Σ (the flow cannot cross in any manner the surface Σ). As a consequence, the flow velocity v (parallel to Σ) and the normal vector n (normal to Σ) are perpendicular to each other. Due to the fundamental properties of the scalar product, the corresponding contribution is then automatically 0: r ρ(x,t)v(x,t)(v(x,t) n) da = 0 (7.52) Σ }{{} =0 Along the inflow cross-section A 1, the one-dimensional assumption leads again to r [ρ(x,t)v(x,t)(v(x,t) n)]da A 1 = r [ρ 1 v 1 (v 1 n 1 )]da (7.53) A [ 1 ] = rda (ρ 1 v 1 (v 1 n 1 )) (7.54) A 1 Since v 1 = v 1 s 1 = v 1 n 1, this can be rewritten: r [ρ(x,t)v(x,t)(v(x,t) n)]da A 1 [ ] = rda (ρ 1 v 1 ( v 1 )) (7.55) A [ 1 ] = rda ( ρ 1 v1 2 s ) 1 (7.56) A 1 The integral appearing in this equation can be again computed using Eq.(7.46), leading finally to A 1 r [ρ(x,t)v(x,t)(v(x,t) n)]da = r 1 ( ρ 1 v 2 1 A 1s 1 ) (7.57) = r 1 (ṁ 1 v 1 s 1 ) (7.58) introducing the mass flow-rate in the inflow cross-section A 1, ṁ 1 = ρ 1 v 1 A 1. Similarly, one obtains along the outflow cross-section A 2, A 2 r [ρ(x,t)v(x,t)(v(x,t) n)]da = r 2 (ṁ 2 v 2 s 2 ) (7.59)

87 4 Torque exerted by a flow on its surroundings 85 As already explained previously, since the flow is steady, mass conservation implies the conservation of the mass flow-rate in any cross-section, so that One can therefore write finally: ṁ 1 = ṁ 2 = ṁ (7.60) A cf r [ρ(x,t)v(x,t)(v(x,t) n)]da = r 1 (ṁv 1 s 1 )+r 2 (ṁv 2 s 2 ) (7.61) in which the vectors r 1 = PG 1 and r 2 = PG 2 have been again used. The first integral on the right-hand side of Eq.(7.41), describing the influence of gravity, can be very easily computed in a direct manner, remembering that the acceleration vector g can be assumed to be constant for practical applications on the earth s surface: V cf r [ρ(x,t)g]dv = [ V cf ρ(x,t)rdv ] g (7.62) The integral appearing on the right-hand side (mean value of ρr over the complete fixed control volume V cf ) is precisely the integral used to define the center of mass of the fluid contained within V cf, so that V cf ρ(x,t)rdv = Mr g (7.63) where M is the total mass of the fluid contained within the fixed control volume V cf, as already shown in Eq.(3.1), and r g = PG connects the pivot P with the center of mass of the fluid within V cf (point G). One obtains thus: r [ρ(x,t)g]dv = r g (Mg) (7.64) V cf Finally, it is possible to rewrite now Eq.(7.41) as: r 1 (ṁv 1 s 1 )+r 2 (ṁv 2 s 2 ) = r g (Mg) T p +r 1 (p 1 A 1 s 1 ) r 2 (p 2 A 2 s 2 ) (7.65) This can now be transformed in a suitable manner to determine T p : T p = r 1 [(ṁv 1 +p 1 A 1 )s 1 ] r 2 [(ṁv 2 +p 2 A 2 )s 2 ]+r g (Mg) (7.66) Using again the impulsion F = ṁv +pa it is now possible to give the expression of the torque exerted by the fluid on its surroundings through the separation surface Σ by reference to a pivot P as: T p = r 1 (F 1 s 1 ) r 2 (F 2 s 2 )+r g (Mg) (7.67) This relation giving the total torque exerted by the fluid between the inflow cross-section A 1 and outflow cross-section A 2 is valid for a steady non-viscous flow, assumed moreover one-dimensional along the inflow and outflow cross-sections, A 1 and A 2. It is practically identical to the expression giving the total force, Eq.(7.26), when considering for each contribution the vector product with the associated lever arm r g (resp. r 1,r 2 ) connecting the pivot P with the corresponding center of mass G (resp. geometric center G 1, G 2 ) Again, the torque exerted by a volume of fluid (that contained within the fixed control volume V cf ) can be computed by knowing only the main flow parameters in the inflow and outflow cross-sections; it is not necessary to know the details of the flow within V cf, only start and end flow conditions are sufficient to compute the resulting torque. This is an extremely powerful formulation!

88 86 Chapter 7. Force and torque exerted by a flow Note that, for most practical cases, the contribution associated to the weight of the fluid (Mg, last term in Eq.7.67) can usually be safely neglected when considering a gas flow, since gas densities are quite low. On the other hand, it must be taken into account for a liquid, since it might lead to considerable values. Equation(7.67)isequallyvalidfortheflowinapipe(pipewallΣ)orfortheflowwithinastreamtube, inwhichcaseσisjustageometricalsurfaceinthefluidseparatinginapurelytheoreticalmannerdifferent fluid elements. 7.5 Torque exerted by a flow on a pipe wall surrounded by a fluid at constant pressure If we now consider more specifically an internal flow in a pipe, Σ is the pipe wall surrounding the fluid. In most practical cases, outside of the pipe, a constant pressure p = p a is found everywhere (figure 7.4). When trying to compute the resulting torque on the pipe wall, it is more useful in practice to take into account as well the influence of the external pressure, p a, and not only the action of the internal fluid pressure, as done in the previous section. Outer surface Σ p a s 2 v 2 G 2 A 1 G 1 p a s 1 v 1 Control volume V G p a A 2 r r 1 r 2 P Figure7.4: Configurationto compute thetorque exerted by a flowonthe pipewall Σtaking into account the constant and uniform external pressure p a. The torque exerted by the internal flow on the pipe wall, Σ, has been determined in the previous section and is given by Eq.(7.67). The supplementary pressure torque exerted by the constant and homogeneous external pressure p a

89 5 Torque exerted by a flow on a pipe wall surrounded by a fluid at constant pressure 87 on the same wall is now: T pa = r [p a n]da (7.68) Σ where the minus sign denotes the fact that the associated force is acting towards the inside of the control volume, i.e., towards n. The further process is completely identical to that already used in Section 7.3. Compared to the results introduced in the previous section, the only difference obtained at the end is the fact that the relative pressure (p p a ) appears everywhere in the system of equations instead of the absolute pressure p. As in Section 7.3, this is indirectly due to the fact that the integral of the normal vector n over an arbitrary (but closed) geometrical surface (here A cf = A 1 A 2 Σ) is always zero. Finally, using again the relative impulsion F = ṁv+(p p a )A, it is possible to give the expression of the torque exerted by the internal fluid on the pipe wall Σ taking into account the external pressure p a by: T p,pa = r 1 (F 1s 1 ) r 2 (F 2s 2 )+r g (Mg) (7.69) This relation is valid for a steady non-viscous flow, assumed moreover one-dimensional along the inflow and outflow cross-sections, A 1 and A 2. Again, for most practical cases, the contribution associated to the weight of the fluid (Mg, last term in Eq.7.69) can be safely neglected when considering a gas flow, since gas densities are quite low.

90 88 Chapter 7. Force and torque exerted by a flow

91 Chapter 8 Movement of a material control volume 8.1 Introduction In the present chapter, we will consider the infinitesimal movement of a material control volume V cm, i.e., one control volume containing always the same fluid elements. We will decompose this movement in several elementary steps, useful to understand the global change in position and shape of such an arbitrary material control volume V cm. 8.2 Movement of a material control volume We consider now the infinitesimal movement of a material control volume V cm, as represented in figure 8.1. It is assumed to be of small, rectangular dimensions (dx dy) at the initial time t 0. The analysis is carried out in two dimensions (x and y directions) for the sake of simplicity. A three-dimensional analysis would be slightly more complex but would deliver exactly the same results. y y Time (t 0 +dt) D Time t 0 D Time t 0 D A A A x x Figure 8.1: Configuration employed to investigate the movement of a material control volume. We will determine the new position of this material control volume V cm at a slightly later time, (t 0 + dt). For our present purpose, it is sufficient to trace the movements of point A and of point D, 89

92 90 Chapter 8. Movement of a material control volume constituting the diagonal direction. The velocity of point A, v A is assumed to be known at time t 0, together with its spatial derivatives: v A (t 0 ) = (v Ax,v Ay ) (8.1) At time t 0, the position of point A is defined as (x A,y A ). The movement of any point belonging to the material control volume until the later time (t 0 + dt) is the result of the fluid movement, starting from its initial position and following the local flow velocity. For point A at t 0, one therefore obtains: Point A time t 0 time (t 0 +dt) x coordinate x A x A +v Ax dt y coordinate y A y A +v Ay dt Initially, at time t 0, the point D is located at a distance (dx,dy) from point A. If we call v D its (at present unknown) velocity at time t 0, one obtains in the same manner: (8.2) Point D time t 0 time (t 0 +dt) x coordinate x D x D +v Dx dt y coordinate y D y D +v Dy dt (8.3) Since x D = x A +dx and y D = y A +dy, this can be rewritten: Point D time t 0 time (t 0 +dt) x coordinate x A +dx x A +dx+v Dx dt y coordinate y A +dy y A +dy +v Dy dt (8.4) Now, how is it possible to determine, at least in an approximate manner, the velocity v D at point D? Since the dimensions(dx, dy) have been assumed very small, this is readily possible using a Taylor expansion, since we know the flow velocity and its derivatives at point A! Retaining only a first-order Taylor expansion, one obtains for v D : v Dx v Ax + v Ax x dx+ v Ax dy (8.5) y v Dy v Ay + v Ay x dx+ v Ay dy (8.6) y By replacing this expression in Eq.(8.4), one obtains (assuming from now on an exact relation and not an approximation): Point D time t 0 time (t 0 +dt) ( x coordinate x A +dx x A +dx+ v Ax + v Ax x dx+ v Ax y dy dt ( y coordinate y A +dy y A +dy + v Ay + v Ay x dx+ v ) Ay y dy dt ) (8.7) It is finally possible to rewrite the coordinates giving the new position of point D at time (t 0 +dt) in a somewhat different, but perfectly identical manner: x A +dx + v Ax dt+ v Ax x dxdt+ 1 2 y A +dy + v Ay dt+ v Ay y dydt+ 1 2 ( vax y + v ) Ay dydt 1 ( vay x 2 x v ) Ax dydt (8.8) y ( vax y + v ) Ay dxdt 1 ( vax x 2 y v ) Ay dydt (8.9) x

93 2 Movement of a material control volume 91 Apart from the (arbitrary) initial position of point D (x A +dx,y A +dy), which could be assumed to be (0, 0) without impacting the rest of the analysis, all the other contributions in Eqs.(8.8) and (8.9) can be identified as different, well-defined movements as follows (considering here only the x-direction): v Ax dt+ v Ax }{{} x dxdt + 1 ( vax }{{} 2 y + v ) Ay dydt 1 ( vay x 2 x v ) Ax dydt y I }{{}}{{} II III IV This elementary movements are as follows: (8.10) 1. The first term (I) corresponds to a simple, rigid translation movement, also called convection in Fluid Dynamics. If this is the only contribution appearing here, this would mean that the material control volume is moving without changing its shape nor its orientation following the (in this case) uniform flow velocity (figure 8.2). y y Time (t 0 +dt) D Time t 0 D Time t 0 D A A A x x Figure 8.2: Translation movement. 2. The second term (II) is a dilatation term (figure 8.3). If its value is zero, then there is no dilatation of the control volume. This term appears when the derivative of a velocity component along its own direction is non-zero. 3. The third term (III) is a shear deformation term, (figure 8.4) that will be considered extensively next, Section 8.3. If its value is zero, then there is no deformation of the control volume. This term appears when the derivative of a velocity component along the other directions is non-zero. Typically, this term will lead to a change of shape for the material control volume. 4. Finally, the last term (IV) is a rotation term, that will also be considered extensively next, Section 8.4. If its value is zero, then there is no rotation of the control volume. This term appears again when the derivative of a velocity component along the other directions is non-zero, but involves the differences and not the sum of these derivatives, at the difference of the deformation term. Typically, this term will lead to a rotation of the material control volume (figure 8.5) without changing its shape.

94 92 Chapter 8. Movement of a material control volume y y Time t 0 D Time t 0 D D Time (t 0 +dt) A A x x Figure 8.3: Dilatation movement. y y Time t 0 D Time t 0 D D Time (t 0 +dt) A A x x Figure 8.4: Shear (or angular) deformation movement. These results, obtained for a generic movement of a material control volume in two dimensions, are of general validity, also in a three-dimensional space. We conclude that the movement of a material control volume can always be decomposed as individual contributions corresponding to (translation + dilatation + shear deformation + rotation).

95 3 Deformation tensor d 93 y y D Time t 0 D Time (t 0 +dt) D Time t 0 A A x x Figure 8.5: Rotation movement. 8.3 Deformation tensor d Considering the results obtained in the previous section, it can be seen that two tensors can be usefully defined to describe the last two remaining contributions to the generic movement of a material control volume. The first one is called deformation tensor, is written d, and will obviously be used to describe and quantify the deformation of the material control volume. This deformation is indeed the superposition of the previously described dilatation and shear (or angular) deformation; the dilatation term will be found along the diagonal of the tensor, while the shear deformation corresponds to all off-diagonal terms. It is defined as: d = (d ij ) for (i = 1,2,3;j = 1,2,3) = d 11 d 12 d 13 d 21 d 22 d 23 d 31 d 32 d 33 and each component can be computed from the fluid velocity v following: (8.11) d ij := 1 2 ( vi + v ) j x j x i Obviously, the components of this tensor are in unit 1/s (inverse of a time). Using as velocity components (v 1,v 2,v 3 ) and coordinates (x 1,x 2,x 3 ), one would obtain: d = v 1 1 ( x v2 + v ) 1 2 ( x 1 x 2 1 v3 + v ) x 1 x 3 2 ( v1 + v 2 x 2 x 1 v 2 1 x 2 ) 2 ( v3 x 2 + v 2 x 3 ) ( 1 v1 2 + v ) 3 ( x 3 x 1 v2 + v ) 3 x 3 x 2 v 3 x 3 (8.12) (8.13)

96 94 Chapter 8. Movement of a material control volume It is possible to use different notations for the velocity components. As an often found alternative, writing the velocity components (v x,v y,v z ) and coordinates (x,y,z), one obtains: d = v x ( x 1 vy 2 x + v ) x ( y 1 vz 2 x + v x z ) 1 2 ( 1 vx 2 y + v y x v y ( y vz y + v y z ) ( 1 vx 2 z + v ) z ( x 1 vy 2 z + v ) z y ) v z z (8.14) Considering the definition of the components, Eq.(8.12), it is obvious that the deformation tensor is symmetric, i.e., d ij = d ji (8.15) for any pair (i,j). Therefore, it is sufficient to know the three diagonal terms and three off-diagonal terms (either the upper right triangle or the lower left triangle) to know all the tensor: it involves only 6 unknown components. 8.4 Rotation tensor Ω The last useful tensor defined to describe the two complex contributions to the generic movement of a material control volume is called rotation tensor, written Ω, and will obviously be used to describe and quantify the rotation of the material control volume. It is defined as: Ω = (Ω ij ) for (i = 1,2,3;j = 1,2,3) = Ω 11 Ω 12 Ω 13 Ω 21 Ω 22 Ω 23 Ω 31 Ω 32 Ω 33 and each component can be computed from the fluid velocity v following: (8.16) Ω ij := 1 2 ( vi v ) j x j x i (8.17) Obviously, the components of this tensor are in unit 1/s(inverse of a time). Note that the only difference between the rotation tensor and the deformation tensor introduced in the previous section is the minus sign in Eq.(8.17)! This apparently small difference leads indeed to a completely different result. Using as velocity components (v 1,v 2,v 3 ) and coordinates (x 1,x 2,x 3 ), one would obtain: Ω = 0 ( 1 v2 v ) 1 2 ( x 1 x 2 1 v3 v 1 2 x 1 x 3 ( 1 v1 v ) x 2 x ) ( 1 v3 v 2 2 x 2 x 3 ) 1 2 ( v1 v ) 3 ( x 3 x 1 v2 v ) 3 x 3 x 2 0 (8.18) As an alternative concerning notations, considering the velocity components (v x,v y,v z ) and coordinates

97 4 Rotation tensor Ω 95 (x,y,z), one obtains: Ω = 0 ( 1 vy 2 x v ) x ( y 1 vz 2 x v x z ( 1 vx 2 y v ) y 1 x 2 0 ) ( 1 vz 2 y v y z ) 1 2 ( vx z v ) z ( x vy z v ) z y 0 (8.19) Considering the definition of the components, Eq.(8.17), it is obvious that the rotation tensor is antisymmetric, i.e., Ω ij = Ω ji (8.20) for any pair (i,j). This is the reason why all three diagonal terms must obviously be zero. Furthermore, it is sufficient to know three off-diagonal terms (either the upper right triangle or the lower left triangle) to know all the tensor: it involves only 3 unknown components. Since three components are sufficient to build a vector, the movement of rotation can indeed equivalently be described by a vector, the rotation vector Ω, already introduced previously (Eq. 2.6) as Ω := 1 v (8.21) 2 The components of this vector are indeed the three unknown components of the rotation tensor!

98 96 Chapter 8. Movement of a material control volume

99 Chapter 9 Navier-Stokes equation: conservation of momentum in a viscous flow 9.1 Introduction We have been able to analyze many useful flows up to now without considering in detail the influence of viscosity. The purpose of the present chapter is to fill the gap between viscous and non-viscous flows, introducing momentum diffusion. Following the traditional approach, we will take into account momentum diffusion through an hypothetical friction force, since it might be helpful from the point of view of a simple engineering feeling. We will later show that both derivations (through impulse diffusion or through a friction force) are indeed equivalent. The methods employed in this Chapter are of course identical to that underlying Chapter 4, at least at the beginning. It would be helpful to re-read this Chapter now if you have forgotten its content. We thus start by choosing again an arbitrary material control volume within a fluid. The evolution of the total momentum P contained within this control volume V cm vs. time will be quantified. This total momentum can be computed by integrating the momentum contained within an elementary volume element, dv, i.e., ρ(x,t)v(x,t)dv, and the total momentum is thus: Hence, the purpose of this chapter is to compute P = ρ(x,t)v(x,t)dv V cm (9.1) dp dt = d ρ(x,t)v(x,t)dv (9.2) dt V cm This problem will be solved again by considering successively basic results of physics and of mathematics, in a similar manner to that employed in the previous Chapters 3 and 4. We will start with mathematical considerations. 9.2 Point of view of mathematics Indeed, the results obtained in Section 4.2 of Chapter 4 leading to Eq.(4.7) are still perfectly valid and completely unchanged. The viscosity does not play any role in this purely mathematical analysis, so that the developed relation will be used as well in the present Chapter for the flow velocity v: d ρ(x,t)v(x,t)dv = ρ(x,t) Dv(x,t) dv (9.3) dt V cm V cm Dt 97

100 98 Chapter 9. Navier-Stokes equation: conservation of momentum in a viscous flow 9.3 Point of view of physics We will directly build on top of the results already presented in Section 4.3 of Chapter 4. From a purely physical point of view, the issue considered in Eq.(9.2) has been already solved several centuries ago by Newton. As stated in the second law of Newton, the change of momentum for a material system is simply a consequence of external forces, acting in the present case on the fluid elements contained within the material control volume V cm. Therefore, in order to answer the question described by Eq.(9.2), we simply need to know the forces acting on the considered fluid. What are the usual forces that will be considered throughout this document? To answer this question, we start again, as in Chapter 4, by differentiating two different families of forces: the contact and the non-contact forces. More details can be found in Appendix A. Let us just recall briefly here that non-contact forces are long-range forces while contact forces take place only when two fluid elements are in direct vicinity to each other. Throughout this document, the only non-contact force that will be taken into account is the gravitational force F g induced by earth gravity, associated with the acceleration vector g, assumed constant ( g = g = 9.81 m/s 2 ). The action of gravity on an elementary fluid volume dv is then simply ρ(x, t)gdv, ρdv being the mass of the fluid element. The resulting gravitational force for the full material control volume V cm reads F g = ρ(x,t)gdv V cm (9.4) Throughout this document, the only real contact force that will be taken into account is the pressure force F p. This is indeed the only important contact force for a non-viscous flow. Additionally, we will now introduce also a so-called viscous force F f as a further contact force. Note that, in reality, this viscous force is only a helpful (though controversial) representation of a diffusion term for momentum, neglected up to now. But let us first skip this issue, and consider viscous effects as being represented by a viscous force F f. All contact forces can be indeed advantageously be represented instead by a stress t, i.e., the ratio between the force and the area A of the geometrical surface A on which the force is acting. This stress is therefore a vector quantity with a magnitude expressed in Pa. For the pressure force, we already know that the corresponding stress is a purely normal component (normal stress). Obviously, this will usually not be the case for a friction force and an arbitrary control volume. One expects a friction force to act completely or at least partly in the tangential plane of the associated surface A (shear stress). In order to define the direction of the corresponding stress, it is even more interesting in practice to introduce a generic stress tensor T. The stress associated to the force acting on the geometrical surface A, characterized as always by a unit normal vector n pointing to the outside of the control volume, will then simply be given by: t = Tn (9.5) The tensor-vector product on the right-hand side (equivalent to a matrix-vector product) will indeed deliver a vector, the expected stress. In Fluid Dynamics, the stress tensor is conventionally written σ Pressure component T p of the stress tensor σ To illustrate this point, let us first consider simply the pressure component T p of the stress tensor σ, finally leading to the pressure force. Since we know very well already this pressure force, fully determined in Chapter 4, this should be indeed an easy task! We have established in Chapter 4 that

101 3 Point of view of physics 99 the local pressure force (a contact force) acting from the surroundings on the fluid elements within the considered material control volume reads locally (see Eq.4.11) p(x, t)nda. The associated stress is obtained simply by dividing with the surface (here da) so that the pressure stress is: t p (x,t) = p(x,t)n (9.6) How is it possible to relate Eq.(9.6) with Eq.(9.5)? This is indeed extremely simple, just by taking: with I the identity matrix: This leads then successively to: I := T p := pi (9.7) (9.8) t p = T p n = pin = pn (9.9) which is exactly the solution we already know (Eq. 9.6)! Therefore, the pressure component T p of the stress tensor σ is now known Friction component τ of the stress tensor σ The friction component τ of the stress tensor σ will also be called friction tensor (or shear tensor) τ. From the start, considering the very different behaviors of different fluids (compare water with toothpaste, for instance), one might expect a difficulty with τ: the result might very well depend on the considered fluid! This will be indeed the case and, in what comes, we consider only what is called a Newtonian fluid. Further details concerning Newtonian and non-newtonian fluids can be found in Appendix A.16. As the name Newtonian already suggests, Newton was the first one to consider extensively and with some success this issue. At his time, only limited experimental results pertaining to friction forces were known, almost exclusively using water. Two experimental observations were particularly striking: as usual for all fields of mechanics, friction could only be observed when a velocity difference is present. Considering applications dealing with hydrostatics or aerostatics (fluid velocity v = 0, see Chapter 5), friction forces could not be observed and should therefore not be predicted by the employed model. the same is also true for a fluid involving a uniform velocity field v = constant in space. In that case, there are again no velocity differences within the fluid, and no friction force could be observed. The idea of Newton is therefore, as done in usual mechanics, to associate the friction tensor τ with velocity differences within the fluid. From a mathematical point of view, such velocity differences are simply quantified by the velocity gradients within the fluid, measuring the deformation of this fluid. Indeed, all the different possible velocity gradients within the fluid (involving three velocity components and three spatial directions) appear in the deformation tensor d introduced in section 8.3 (see Eq. 8.12). Finally, Newton assumes the simplest possible relation between the friction tensor τ and the deformation tensor d: a simple proportionality! Since, for a flow with a uniform velocity field (v = 0 or some constant value v 0 ) the deformation tensor d is obviously 0, the friction tensor (and therefore the friction force) will obviously also disappear in such conditions, verifying the already discussed experimental observations.

102 100 Chapter 9. Navier-Stokes equation: conservation of momentum in a viscous flow This proposition of Newton later proved to be perfectly suitable for a large variety of fluids (but not for all)! All fluids verifying Eq.(9.10) are now called Newtonian fluids, and include in particular water, many simple liquids, air, and all ideal gases. Therefore, a wealth of practical applications can be considered for such Newtonian fluids. We will hence only consider Newtonian fluids in the rest of this document. The proportionality rule proposed by Newton must be corrected when it is necessary to take into account compressibility effects, i.e., for high-speed gas flows. This leads to the last, correction term in Eq.(9.10), that finally reads: τ = 2µd 2 µ( v)i (9.10) 3 For an incompressible flow, mass conservation leads directly to the condition v = 0 (see Eq.3.17) and the last term in Eq.(9.10) automatically disappears. The factor µ appearing in Eq.(9.10) is a measurable fluid quantity, the dynamic viscosity (also just called viscosity ). The magnitude of the viscous tensor (and thus of the viscous force F f ) will then increase proportionally to this fluid viscosity, as expected from the start Full stress tensor σ Finally, the answer given by physics to the question contained in Eq.(9.2) reads, following the second law of Newton: dp dt = F g +F p +F f (9.11) = ρ(x,t)gdv + t(x,t)da (9.12) V cm A cm = ρ(x,t)gdv + σnda (9.13) V cm A cm = ρ(x,t)gdv + ( pi+τ) nda (9.14) V cm A cm }{{} σ = ρ(x,t)gdv + ( pi)nda+ τnda (9.15) V cm A cm A cm = ρ(x,t)gdv pnda+ τnda (9.16) V cm A }{{ cm A }} cm {{} F p F f 9.4 Integral formulation of momentum conservation Recognizing that both results found in the two previous sections are of course correct and identical, it is possible to write following equality, taking on the left-hand side the result of mathematics and on the right-hand side the result of physics: V cm ρ(x,t) Dv(x,t) Dt dv = ρ(x,t)gdv p(x,t)nda+ τnda (9.17) V cm A cm A cm This is indeed the integral formulation of momentum conservation (in fact, linear momentum), written for an arbitrary material control volume V cm and valid for any flow, even in the presence of viscous forces.

103 5 Local formulation of momentum conservation 101 It is equally possible to use the standard formulation of the transport theorem, Eq.(2.28) instead of the theorem of Reynolds (Eq. 4.7), as done previously. In that case, Eq.(9.17) reads: (ρ(x,t)v(x,t)) dv + ρ(x,t)v(x,t)(v(x,t) n)da = V cm t A cm ρ(x,t)gdv p(x,t)nda+ τnda (9.18) V cm A cm A cm It is now possible to assume that a fixed control volume V cf coincides with the considered material control volume V cm at time t, allowing to rewrite: (ρ(x,t)v(x,t)) dv + ρ(x,t)v(x,t)(v(x,t) n)da = V cf t A cf ρ(x,t)gdv p(x,t)nda+ τnda (9.19) V cf A cf A cm If the flow is steady, one obtains furthermore: ρ(x,t)v(x,t)(v(x,t) n)da = A cf ρ(x,t)gdv V cf p(x,t)nda+ A cf τnda (9.20) A cm Observing that all forces acting on the fluid contained within the fixed control volume V cf are now appearing on the right-hand side of this equation, the developed formulation can be used in Chapter 7 to compute the force induced by a flow. 9.5 Local formulation of momentum conservation Equation (9.17) can indeed be useful when considering a macroscopic control volume (though we will mostly employ in practice fixed control volumes instead of material ones), but is awkward when trying to derive local conditions valid for any fluid element. One problem with Eq.(9.17) is that it combines a volume integral (first one on the right-hand side) with surface integrals (last ones on the right-hand side), preventing further simplifications. This can be easily solved by using again the divergence theorem, allowing a direct relation between a volume integral on an arbitrary volume V c and a surface integral on the associated boundary A c. Using the second formulation of the divergence theorem (Eq. C.6), it is possible to replace the surface integral associated with pressure in Eq.(9.17), leading to: p(x,t)nda = A cf p(x,t)dv V cm (9.21) Using now the first formulation of the divergence theorem (Eq. C.5), it is possible to replace the surface integral associated with viscous effects in Eq.(9.17), leading here to: Finally, one obtains: V cm ρ(x,t) Dv(x,t) Dt τnda = A cm τ(x,t)dv V cm (9.22) dv = ρ(x,t)gdv p(x,t)dv + τ(x,t)dv V cm V cm V cm (9.23)

104 102 Chapter 9. Navier-Stokes equation: conservation of momentum in a viscous flow All integration volumes are now identical, allowing to rewrite: V cm [ ρ(x,t) Dv(x,t) Dt ρ(x,t)g+ p(x,t) τ(x,t) ] dv = 0 (9.24) Remember that this relation is valid for an arbitrary material control volume, and thus for an infinite number of different volumes in the fluid! How is it possible to integrate some quantity (that between the [ ] in Eq.9.24) over an infinite number of different volumes, getting always 0 as a result? Only if the integrated quantity is equal to 0 at every point! Hence, the quantity [ ρ(x,t) Dv(x,t) ρ(x,t)g+ p(x,t) τ(x,t) Dt ] must be identically nil at every point in space. Finally, the local conservation equation for linear momentum can be written for a viscous flow: ρ Dv Dt = ρg p+ τ (9.25) This relation, also called Navier-Stokes equation, is one of the two most fundamental relations of Fluid Dynamics (together with mass conservation, Eq.3.13) and we will use it many times in this document. It is perhaps helpful to write also this equation as three scalar equations. For this purpose, it is important to recognize that the divergence of a tensor (here τ) is simply the divergence obtained when considering each line of the tensor (three components) as a vector, and computing its divergence. This leads to the set of equations: ρ Dv x Dt ρ Dv y Dt ρ Dv z Dt = ρg x p x + τ xx x + τ xy y + τ xz z = ρg y p y + τ yx x + τ yy y + τ yz z = ρg z p z + τ zx x + τ zy y + τ zz z (9.26) (9.27) (9.28) Of course, the symmetry of the tensor τ (i.e., τ ij = τ ji ) can be used to simplify this equation system. 9.6 Local momentum conservation for an incompressible flow Equation (9.25) can be slightly modified for an incompressible flow, even if it does not lead to a major simplification. Since, for such a flow, we can safely assume ρ = constant in space as well as in time, it is possible to divide all terms of this equation by the density, leading to: Dv Dt = g 1 ρ p+ 1 τ (9.29) ρ Furthermore, the density, being constant, can be pulled into the gradient and divergence operators. Finally, the local conservation equation for momentum can be written for an incompressible flow: Dv Dt = g ( ) p + ρ ( ) τ ρ (9.30)

105 7 Local formulation of momentum conservation for a non-newtonian fluid Local formulation of momentum conservation for a non- Newtonian fluid Beforeconsidering theusualcase(anewtonianfluid) inthenext sectionandintherestofthisdocument, let us briefly comment what we should do in the case of a non-newtonian fluid! In that case, the expression of the friction tensor τ must be obtained from a suitable model, usually based on extensive experimental measurements. The corresponding equation for τ will be plugged into Eq.(9.25) and will be employed to obtain the resulting flow conditions. 9.8 Local formulation of momentum conservation for a Newtonian fluid Equation (9.25) can be considerably simplified when considering only Newtonian fluids, as we will do in all this document from now on. In that case, Eq.(9.10) applies and can be used to compute the friction tensor. τ = 2µd 2 µ( v)i (9.31) 3 For this purpose, it will be necessary to compute the divergence, containing the spatial derivatives of τ. It is hence necessary to identify the variables appearing in Eq.(9.31) that must be derived during this process! First, it is clear that the second spatial derivatives of the flow velocity v will appear. But, in principle, the viscosity µ appearing in τ is also a function of space and time, µ = µ(x,t), and should therefore be derived as well! The viscosity of a fluid depends on its composition (this would be important for a mixture involving different components) but also on its temperature T. As you probably know from your own cooking experience, the viscosity of a simple liquid usually decreases when increasing temperature T. On the other hand, it is possible to understand based on the kinetic theory that the viscosity of a gas will normally increase with its temperature. In both cases, viscosity will therefore vary in many practical cases. In order to obtain a simple system of equations, we will get rid of this additional difficulty by assuming from now on that we consider only a pure Newtonian fluid and an isothermal flow. In this manner, the viscosity µ cannot change any more, and it is not necessary to consider its derivatives. Do not forget to modify the equations appropriately if these conditions are not fulfilled any more! For example, if you consider the flow in a flame, the temperature will change considerably and lead to large viscosity gradients, that usually must be taken into account. Now, for a constant viscosity µ, one obtains easily: ρ Dv Dt = ρg p+µ 2 v+ µ ( v) (9.32) Local formulation of momentum conservation(incompressible flow, Newtonian fluid) Equation(9.32) can be furthermore very much simplified when considering an incompressible flow. Since, in this case, mass conservation reduces to Eq.(3.17), v = 0 (9.33)

106 104 Chapter 9. Navier-Stokes equation: conservation of momentum in a viscous flow the last term in Eq.(9.32) obviously disappears. As a consequence, the local formulation of momentum conservation can be simply rewritten for an incompressible flow of a Newtonian fluid (for example water) as: ρ Dv Dt = ρg p+µ 2 v (9.34) This is the simplest form of the Navier-Stokes equation, and it will therefore be used quite often in what follows! It is perhaps helpful to write again this final equation as three scalar equations, developing all the terms involved (including the substantial time derivative on the left-hand side): ( vx ρ t +v v x x x +v v x y y +v z ( vy ρ t +v v y x x +v v y y y +v z ( vz ρ t +v v z x x +v v z y y +v z ) v x z ) v y z v z z ) = ρg x p ( 2 x +µ v x x + 2 v x 2 y + 2 v x 2 z 2 ( 2 v y = ρg y p y +µ = ρg z p z +µ ) ) x + 2 v y 2 y + 2 v y 2 z 2 ( 2 ) v z x + 2 v z 2 y + 2 v z 2 z 2 (9.35) (9.36) (9.37) As an alternative, it is obviously possible to divide each term by the (non-zero) fluid density ρ. In this manner, the dynamic viscosity µ is replaced in the last term by the kinematic viscosity ν := µ/ρ Local formulation of momentum conservation in cylindrical coordinates Using the same assumptions, if a cylindrical coordinate system (r, θ, z) with corresponding velocity components v = (v r,v θ,v z ) is used instead of our standard coordinate system, the local formulation of momentum conservation reads for the velocity components: v r t +v v r r r + v θ v r r θ v2 θ r +v v r z z v θ t +v v θ r r + v θ v θ r θ + v rv θ r +v z v θ z v z t +v v z r r + v θ v z r θ +v v z z z = g r 1 [ ( ) p 1 ρ r +ν (rv r ) v r r r r r 2 θ 2 ] v θ 2 r 2 θ + 2 v r z 2 = g θ 1 [ ( ) p 1 ρr θ +ν (rv θ ) v θ r r r r 2 θ + 2 ] v r 2 r 2 θ + 2 v θ z 2 = g z 1 [ ( p 1 ρ z +ν r v ) z ] v z r r r r 2 θ + 2 v z (9.38) 2 z 2

107 Chapter 10 Dimensional analysis and similarity conditions 10.1 Introduction The general conservation equations introduced previously for mass (Eq. 3.13) and for momentum (Eq. 9.25) are the mathematical form of fundamental physical laws and properties, and are therefore valid for any conditions! The same is not true for the units employed for the quantities appearing in these equations: you might work equally well when expressing a length in thumb, foot or meter units; as long as the employed system is coherent, the resulting equations will deliver meaningful results. This means that, while the equations are intrinsically important, the employed unit conventions are completely arbitrary: you might use any unit system you want! In other words, from the point of view of nature, only non-dimensional quantities are meaningful. Therefore, it would be useful to get conservation equations for mass and momentum without any unit. This is the purpose of the next section Non-dimensional conservation equations A first discussion concerning non-dimensionalization for applications outside Fluid Dynamics can be found for instance by Wikipedia. It is indeed relatively simple to obtain non-dimensional conservation equations for mass and momentum. For this purpose, it is sufficient to introduce a reference quantity (dimensional constant value, written with an index ) for any flow variable appearing in the corresponding conservation equation. The reference quantity will always be measured using standard units from the international system, abbreviated SI (for Système International d Unités in French, since it was first introduced in France during the 18th century). From the corresponding (dimensional) flow variable φ(x, t) and the dimensional reference quantity φ (obviously non-zero), a non-dimensional variable (written with an index, leading to φ (x,t)) can be readily obtained using: φ (x,t) := φ(x,t) φ (10.1) In other words, the non-dimensional flow variable will be obtained as the ratio between the (usual) dimensional flow variable and a constant reference value for this same variable. For example, the non-dimensional velocity will be written: v := v v (10.2) 105

108 106 Chapter 10. Dimensional analysis and similarity conditions In the same manner, the non-dimensional pressure will be: p := p p (10.3) The coordinates of the system in space and time will be transformed in the same manner: x := x l (10.4) t := t t (10.5) by considering a reference length l (the same for all directions, expressed in m) and a reference time t (in s). Obviously, it is possible to invert all these relations in order to express the (usual) dimensional flow variable as a product of the constant reference quantity with the non-dimensional flow variable: For example: or φ(x,t) = φ φ (x,t) (10.6) v(x,t) = v v (x,t) (10.7) p(x,t) = p p (x,t) (10.8) Since the conservation equations involve time-derivatives and space-derivatives, those must also be modified to remove any dimension. For example, the time-derivative must be modified following: φ t = φ (t t ) (10.9) = 1 t φ t (10.10) recognizing that t is a constant value, and may thus be extracted out of the derivative operator. This relation states that the standard time-derivative can be replaced by the non-dimensional time-derivative (deriving by reference to t ) simply by multiplying the latter by the constant 1/t. Exactly the same relation is found for spatial derivatives: φ x = φ (l x ) (10.11) = 1 l φ x (10.12) The standard space-derivative can be replaced by the corresponding non-dimensional space-derivative (deriving by reference to x ) simply by multiplying the latter by the constant 1/l. Since gradient, divergence and curl all involve simply such first-order spatial derivatives, the same relation applies: φ = 1 l φ (10.13) φ = 1 l φ (10.14) φ = 1 l φ (10.15)

109 2 Non-dimensional conservation equations 107 Finally, the Laplace operator, involving second derivatives in space, will require successively twice the above transformation, leading to: 2 φ = 1 l 2 2 φ (10.16) We are now able to start removing all the units from the standard conservation equations established previously. In order to simplify the developments, we will consider (arbitrarily) an incompressible flow in what follows. The same method can be applied as well for flows with a variable density Non-dimensional mass conservation for an incompressible flow The starting point is logically the dimensional (usual) mass conservation for an incompressible flow, Eq.(3.17): v = 0 (10.17) The procedure is extremely simple; we just need to replace each dimensional term appearing in this equation by the corresponding non-dimensional alternative, following all the results listed above. For the very simple equation considered here, this simply leads to: 1 l (v v ) = 0 (10.18) Now, all the reference quantities (index ) are constant (obviously non-zero) values, which can be exchanged with any time- or space-derivatives. It is equally possible to multiply or divide the equation by any of this value. One obtains then: v l v = 0 (10.19) and finally: v = 0 (10.20) where the value 0 appearing on the right-hand side is now a number without any dimension, and not 0 1/s as in the starting Eq.(10.17)! We could write it equally well 0, but this might be confusing for the reader. Finally, the obtained non-dimensional conservation equation for mass, Eq.(10.20) is formally identical to the starting equation (Eq ). It is sufficient to replace all the terms appearing in the original equation with their non-dimensional counterpart. This means that, for the present case (an incompressible flow), there is no special condition to be fulfilled concerning the units. The situation is different when considering momentum conservation, as described in the next section Non-dimensional momentum conservation (incompressible flow, Newtonian fluid) The starting point is again logically the dimensional(usual) momentum conservation for an incompressible flow of a Newtonian fluid with constant viscosity, Eq.(9.34): ρ Dv Dt = ρg p+µ 2 v (10.21) The procedure is the same as employed previously; we just need to replace each dimensional term appearing in this equation by the corresponding non-dimensional alternative, following all the results listed above. Obviously, this equation is much more complex than Eq.(10.17), so that it is this time a somewhat tedious job...

110 108 Chapter 10. Dimensional analysis and similarity conditions As a first step, it is advantageous to split the Lagrange derivative appearing on the left-hand side in its two standard sub-components, the (standard or Euler) time-derivative and the convective term, following Eq.(2.17). One obtains: ρ v +ρ(v )v = ρg p+µ 2 v (10.22) }{{} t I We will just illustrate the transformation procedure for the first term (marked I in Eq.10.22). A similar method is used for each term appearing in this equation. Concerning term I, one obtains successively: ρ v t = (ρ ρ ) 1 (v v ) (10.23) t t ( ) ρ v [ ] v = ρ (10.24) t t Note that we have separated in the results one group between normal parentheses () on the left side and one group between square parentheses [] on the right side. The left group contains only dimensional reference quantities (index ), while the right group contains only non-dimensional variables (index ). Therefore, the term on the right is already fully non-dimensional; the only units appearing there are contained in the left group, between (). The same procedure is now applied to each term appearing in the starting equation, Eq.(10.22). Multiplying the resulting equation by the constant term l /(ρ v 2 ), one gets finally, ( )[ ] l v ρ v t t }{{} I +[ρ (v )v ] = }{{} II ( ) ( ) ( l g p µ [ρ v 2 g ] [ ρ v 2 p ] + ρ v l }{{}}{{} III IV ) [µ 2 v ] } {{ } V (10.25) where the same convention has again been employed. For each of the five terms (I to V) appearing in this equation, the left group, between (), contains only dimensional reference quantities (index ), while the right group, between [], contains only non-dimensional variables (index ). The second term in this equation (term II) contains only non-dimensional quantities. It is therefore fully non-dimensional. Since all the terms appearing in this equation obviously must have the same dimension, the same is therefore true for all the terms. We have therefore been successful; this is indeed the requested non-dimensional equation! But there are still groups of terms, between (), containing dimensional reference quantities. How can this be? Simply, even if each individual reference quantity is indeed dimensional, the groups appearing in Eq.(10.25) are globally non-dimensional. These groups are key concepts in Fluid Dynamics. They are called non-dimensional parameters (or dimensionless parameters) and they will be discussed in detail in the next section Non-dimensional parameters of Fluid Dynamics We will analyze successively all the groups involving dimensional reference quantities in Eq.(10.25) Strouhal number St The dimensionless group appearing in term I is called the Strouhal number and is written St. Note that all non-dimensional parameters will be written in this document using standard conventions, i.e.,

111 3 Non-dimensional parameters of Fluid Dynamics 109 combining an upper-case letter with a following lower-case letter, both being related to the name of a scientist that has delivered a noticeable contribution to the related issue. This is here the case of Vincenc Strouhal, and the Strouhal number reads: St := l v t = l f v (10.26) where either the reference time t or the corresponding reference frequency f := 1/t may be employed. Of course, the Strouhal number is non-dimensional. This can easily be checked: [St] = m m s s = [1] (10.27) The Strouhal number appears as a factor of the only time-derivative involved in Eq.(10.25). As such, it is clear that the Strouhal number quantifies the importance of unsteady effects in the flow, in the form: unsteady acceleration St = (10.28) convective acceleration Following this relation, a large value of St corresponds to a flow with large unsteady effects, a small value of St corresponds to a flow dominated by purely convective acceleration. The Strouhal number will play a key role for all applications involving time-dependent instabilities and structures, characteristic frequencies, vibrations. Further information can be found under Wikipedia Froude number Fr As already mentioned, there is no dimensionless term associated with II. The next non-dimensional parameter is found in term III and is called the Froude number, written Fr: Fr := v2 g l (10.29) As you see, the Froude number is conventionally defined as the inverse of the non-dimensional group appearing in Eq.(10.25). Again, the Froude number is non-dimensional: m 2 [Fr] = s m 2 = [1] (10.30) s 2m The inverse of the Froude number appears as a factor of the gravity term in Eq.(10.25). As such, it is clear that the Froude number quantifies the importance of gravity forces in the flow, in the form: Fr = contribution of inertial force contribution of gravity force (10.31) Following this relation, a small value of Fr corresponds to a flow with large gravity effects, a large value of Fr corresponds to a flow with small or even negligible gravity effects. As a consequence, the Froude number will play a key role in the analysis of all flows controlled by gravity. This is in particular the case for free-surface flows, appearing at the boundary between a liquid and a gas. Typically, the Froude number is essential for understanding waves and properties of ships. Note that you may find in the literature definitions of Fr corresponding to the square root of Eq.(10.29), i.e., Fr= v / g l. This number is obviously also non-dimensional and could equally well be employed in Eq.(10.25). This alternative form is justified when investigating open channel flows, an issue that is not considered in the present document. We will therefore stick to the usual definition, Eq.(10.29). Further information can be found under Wikipedia.

112 110 Chapter 10. Dimensional analysis and similarity conditions Euler number Eu The next non-dimensional parameter is found in term IV. It is called the Euler number, written Eu: Eu := p 1 2 ρ v 2 (10.32) Again, the Euler number is obviously non-dimensional, since it is the ratio between reference pressure and reference dynamic pressure, both expressed in Pa. The Euler number appears as a factor of the pressure term in Eq.(10.25). As such, it is clear that the Euler number quantifies the importance of pressure forces in the flow, in the form: Eu = contribution of pressure contribution of dynamic pressure (convective effect) (10.33) Following this relation, a small value of Eu corresponds to a flow with small pressure effects, a large value of Eu corresponds to a flow with considerable pressure effects. Note that you may often find in the literature definitions of Eu without the factor 1/2, introduced here to show directly the link with the dynamic pressure. TheEuler number canalsobebuiltbyconsidering asreferencequantity thepressure change( p) insteadofthepressureitself. Therefore, youmightforexamplefindinmanytextbookseu:= ( p) /(ρ v 2 ). Finally, let us note that the Euler number typically plays a minor role in the flow analysis, as exemplified later. The only case where the Euler number is really meaningful is when considering liquid flows with cavitation, an issue that will not be discussed further. For all other configurations, the Euler number will typically be replaced by non-dimensional force cofficients (see later), containing also the influence of pressure forces and much more useful for practical considerations. Further information can be found under Wikipedia Reynolds number Re Last but not least, the non-dimensional parameter found in term V is called the Reynolds number, written Re: Re := ρ v l = v l (10.34) µ ν As you see, the Reynolds number can be equally well defined using either the reference (dynamic) viscosity µ or the reference kinematic viscosity ν := µ /ρ. Again, the Reynolds number is non-dimensional. This can easily be shown by considering the last definition above and remembering that a kinematic viscosity is expressed in m 2 /s in the SI system, leading to: m [Re] = s m m = [1] (10.35) s 2 The inverse of the Reynolds number appears as a factor of the viscous term in Eq.(10.25). As such, it is clear that the Reynolds number quantifies the importance of viscous forces in the flow, in the form: Re = contribution of inertial forces contribution of viscous forces (10.36) Following this relation, a small value of Re corresponds to a flow with large viscous effects, a large value of Re corresponds to a flow with small or even negligible viscous effects.

113 3 Non-dimensional parameters of Fluid Dynamics 111 For reasons that we will better understand in the Chapter concerning turbulence, Chapter 14, this is equivalent to state that a small value of Re corresponds to a flow with minor turbulent effects (a laminar flow), while large values of Re correspond to turbulent flows. The Reynolds number is clearly the most important non-dimensional parameter of Fluid Dynamics. Everybody should know its definition, since it will play a key role for a countless number of practical applications. This is the only non-dimensional parameter that we already know: it has been introduced in Chapter 6 to compute the friction factor. Further information can be found under Wikipedia Further non-dimensional parameters The four non-dimensional parameters introduced up to now (St, Fr, Eu, Re) are sufficient to re-write completely our conservation equations, as we will demonstrate in the next section. This does not mean that they are sufficient to describe all applications of Fluid Dynamics! First, we have restricted the present analysis to incompressible flows. Further terms appear when considering compressibility effects. Second, and even more important, further conservation equations would be required to describe the conservation of further quantities, essential to describe the thermodynamical state of the fluid, in particular energy and entropy. Such equations will be partly described in Chapter 11. When writing non-dimensional equations for compressible flows and considering energy and entropy as important flow variables, a wealth of additional non-dimensional parameters would appear! The only one really essential at present is the Mach number Ma, since it will be a key parameter for all compressible flows, as demonstrated in Chapter 11. It is even so important that it is the only nondimensional parameter of Fluid Dynamics written also with a single, capital letter M: where c is the reference speed of sound. M = Ma = M := v c (10.37) flow velocity speed of information transfer (10.38) As such, the Mach number Ma quantifies the importance of compressibility effects, as demonstrated in Chapter 11 (see for instance figure 11.3). A flow with small values of M is nearly incompressible, while a flow with large values of M is associated with considerable compressibility effects, in particular large density changes. In practice, the Mach number will appear in the analysis of high-speed gas flows. Further information concerning the Mach number can be found under Wikipedia. Furthermore, the forces exerted by a flow on an object placed in this flow can be converted into non-dimensional coefficients as well. Conventionally, these forces are separated as drag force F d, in the direction opposed to the relative movement of the object, and lift force F l, perpendicular to the direction of the relative movement. A force, expressed in N, can be easily transformed into a non-dimensional coefficient by normalizing it using the product of a pressure and a surface (Pa.m 2 = N). In this manner, the non-dimensional drag coefficient C d and lift coefficient C l are defined as: C d := F d 1 2 ρv2 A C l := F l 1 2 ρv2 A (10.39) (10.40)

114 112 Chapter 10. Dimensional analysis and similarity conditions with A the projected area of the object in the direction of the incident flow. These non-dimensional force coefficients take into account both pressure and friction forces and will usually be employed instead of the Euler number. Further non-dimensional parameters of Fluid Dynamics can be found in Appendix A Choosing the reference quantities As explained in the previous sections, the values of the non-dimensional parameters might be used to assess a priori the importance of selected physical processes, like gravity or friction. Obviously, the obtained values depend on the retained reference quantities (index ), since they are directly computed from these quantities! As a consequence, these reference quantities are certainly not arbitrary and must be chosen carefully, if meaningful results are to be obtained. Some of them are very easy to choose. If you are working on the earth at ground level, you will certainly select g = 9.81 m/s 2. If your fluid is water under standard conditions, then choose of course ρ = 1000 kg/m 3. If you are considering a low-speed gas flow surrounded by the atmosphere, then p = p a 10 5 Pa is certainly the right choice. Other variables might be more tricky, and will fully depend on the configuration you are investigating. For instance the reference velocity v, the reference length-scale l and reference time t must be built from a first analysis of the problem and of the questions you want to solve! Summary: non-dimensional conservation equations Considering all the results described in the previous sections, it is now clear that we may rewrite the conservation equations for mass (see Eq.10.20) and momentum (see Eq.10.25) in the case of an incompressible flow of a Newtonian fluid with constant viscosity as: St v = 0 (10.41) [ ] v ρ +[ρ (v )v ] = 1 t Fr [ρ g ] 1 2 Eu[ p ]+ 1 [ µ 2 Re v ] (10.42) What is the consequence of this system of equations, useful for a first understanding of the similarity conditions? When keeping the same boundary conditions for a given flow configuration (for instance, the flow around a model inside a water-tunnel), then the flow solution depends only on the values of the non-dimensional parameters St, Fr, Eu, Re! If we consider the same conditions at two different scales (multiplying for example all dimensions of the first model by a factor 2), but keeping the same values for St, Fr, Eu, Re, then the non-dimensional flow (described by the non-dimensional variables v and p ) will be identical in both configurations. This is indeed the first statement of the similarity theory: if the geometry of a model is kept identical while changing its scale, so that the boundary conditions are also preserved, the obtained flow is identical (in the sense of non-dimensional flow variables) if and only if the non-dimensional parameters of Fluid Dynamics are identical for both conditions. The procedure used in this section is logical and interesting, but quite tedious for practical applications. It requires first the identification of the suitable conservation equations, the definition of reference quantities, the transformation into a non-dimensional equation system, and finally the identification of the resulting non-dimensional parameters; this is quite time-consuming. In an effort to speed-up the analysis, two physicist have introduced at the beginning of the twentieth century a short-cut toward the solution of this problem: the Π-theorem.

115 4 A faster solution: the Π-theorem A faster solution: the Π-theorem The Π-theorem, first proposed by Buckingham and later demonstrated by Vaschy (and hence also known as Buckingham theorem or Vaschy-Buckingham theorem) is extremely efficient to identify the relevant non-dimensional parameters controlling a specific flow configuration, and will be used in practice instead of the previously employed analysis. Note that this theorem has not been developed for Fluid Dynamics; its signification is universal and it may be employed for any analysis of a physical problem. Nevertheless, Fluid Dynamics is probably the branch of science with the largest number of results obtained through the Π-theorem, probably because the number of free parameters describing the problem is optimal, neither too small (preventing the need for such an analysis) nor too large (preventing a useful analysis in practically relevant configurations). Now, what states the Π-theorem? If a physical process can be described by considering only n dimensional variables, and if the units of these n dimensional variables can be expressed using m fundamental physical scales, then: 1. there are exactly (n m) independent non-dimensional parameters Π k (with k = 1,...,n m) involved in the description of this process; 2. the solution describing the physical process can be reduced to a relation between these non-dimensional parameters: Π 1 = G(Π 2,Π 3,...,Π n m ) (10.43) If needed, further information concerning the Π-theorem can be found under Wikipedia. Now, what are the fundamental physical scales appearing in the Π-theorem? Those are the measurement scales (or units) that cannot be obtained as a combination from other scales, and are therefore of fundamental importance. For instance, the SI unit of pressure, Pa, can be derived as Pa = N m 2 = kg.m s 2.m 2 = kg m.s 2 (10.44) Therefore, it can be seen that a pressure scale is nothing but the ratio between a mass scale and the product of a length scale with a time scale squared. The pressure unit is therefore not a fundamental unit, but only a derived unit. In practice, what are the fundamental units or physical scales for Fluid Dynamics? These are simply mass (SI: kg), length (SI: m), time (SI: s), and temperature (SI: K). As a whole, there are thus 4 fundamental scales for Fluid Dynamics, so that we will always obtain m 4. If needed, further information concerning the fundamental physical scales can be found under Wikipedia. Hence, the Π-theorem will be applied in practice to obtain the number of controlling non-dimensional parameters, built from the n relevant dimensional variables. Note that the Π-theorem does not give explicitly these non-dimensional parameters; you will have to do that! There are some mathematical methods that would allow a systematic derivation of these parameters. In practice, this is never needed. The best solution is to directly identify from the n relevant dimensional variables the subset of usual non-dimensional parameters associated with Fluid Dynamics problems. So, you will try to see if it is possible to build a Reynolds number, a Froude number, a drag force coefficient... from the n listed values. With some experience, this will be quite easy! Two key rules will help you and prevent most possible errors: each of the n relevant dimensional variables appearing in the list must be used at least once when building the non-dimensional parameters Π k. If one of these variables appear nowhere at all, then there is something wrong.

116 114 Chapter 10. Dimensional analysis and similarity conditions the non-dimensional parameters Π k should be constructed as independent quantities. If you have already identified Π 1 and Π 2, it would be in principle possible to build Π 3 as Π 3 = Π 2 1 /Π 2; this is indeed a non-dimensional parameter built from the n relevant dimensional variables, no? But this is strictly forbidden, since Π 3 does not deliver any new information; it is just a combination of the already avalaible information contained in Π 1 and Π 2. Hence, you must check that the derived non-dimensional parameters are independent from each other. In most practical cases, this independence can be easily proved, since one of the n relevant dimensional variables will often appear in a single non-dimensional parameter Π k. In that case, no combination of the other Π l,(l k) can lead to Π k, ensuring independence. Finally, what is the only difficulty associated with the Π-theorem? It is the first step; finding the right set of dimensional variables controlling the considered physical process. This issue will be discussed in the next section Relevant dimensional variables As a whole, the Π theorem is a higly useful and fairly automatic solution process, that will be used to solve a wealth of different applications. But the success completely depends on the starting conditions... You must find the right set of dimensional variables controlling the considered physical process! Here, some experience and common sense is needed. Fortunately, a check-list can be used to support the identification of these variables. First, what happens if you do it wrong? There are two possible pitfalls that must be avoided. One could be tempted, to avoid forgetting any important controlling variable, to extend the list as much aspossible. Inthis case, you would include inthe list of relevant variables all theflow variables you might think of, to be on the safe side. In this manner, n would increase and take a very large value. But you remember that m (the number of fundamental scales) will be at most 4, and will therefore not increase further when increasing n. As a consequence, you end up in this case with a Π-theorem stating that you need perhaps 15 or 20 non-dimensional parameters to describe the solution of this problem. This is not really wrong, but completely useless in practice. The Π-theorem helps us by reducing the dimension of the solution space. If you keep all the variables in the analysis, then you reduce nothing at all, and you can completely forget the Π-theorem... Now, the other possible error is to choose only a very small number of relevant dimensional parameters, forgetting perhaps 1 or 2 key quantities. Then, you will never get them back! Therefore, the corresponding effects cannot appear in the results obtained by the Π-theorem. In that case, the solution you obtain is indeed wrong. It might describe at best only part of the true physical process; in the worst case, it leads to completely misleading results. Therefore, in order to be successful with the Π-theorem, you must find the right set of relevant dimensional variables! A check-list will help you, since experience shows that all Fluid Dynamics problems lead to lists with similar quantities. Hence, check carefully one after the other all the items of the check-list: 1. The first element of this list is the only one that must appear in the list. All the further ones are optional, but not the quantity you are looking for! If you want at the end some information on the flow velocity, then it must appear in the list. Alternatively, if you are looking for the drag force, then include in the list! 2. In most cases, one or several dimensional variables describing the strength or intensity of the flow will also appear in the list. This will be typically expressed in the form of a mean velocity, volumetric or mass flow-rate.

117 6 Similarity conditions in Fluid Dynamics It is also very common to find in the list one or several fluid properties. For instance, if you consider friction effects, the viscosity will be obviously important. For problems involving heat exchange, thermal conductivity of specific heat capacity might appear. And so on In many cases, one or more geometrical parameters influencing the flow will appear. Be careful here! You should not take in the list all the typical dimensions of the object you are considering! The geometrical similarity must be preserved any way and will be discussed separately in the next section. Here, you only need (possibly) a few, key geometrical parameters, which will change the considered flow features, and nothing else. 5. Finally, some external physical processes might appear in the list. For Fluid Dynamics applications, this is in practice only the magnitude of gravity acceleration g, which should be considered in the list every time gravity is important for the considered flow Similarity conditions in Fluid Dynamics Now, we can finally state the similarity conditions in Fluid Dynamics: Two flows at different scales are similar from the point of view of Fluid Dynamics if and only if: 1. the geometry (and hence also the boundary conditions) are identical at both scales, meaning that all the appearing dimensions must be multiplied by the same scaling factor Σ. This is called geometrical similarity. 2. simultaneously, all but one of the non-dimensional parameters controlling the flow solution (as obtained by the Π-theorem) must be identical for both flows. This is called dynamic similarity Note that the last non-dimensional parameter is then automatically also identical for both flows, due to Eq.(10.43). Therefore, all non-dimensional parameters are indeed identical for both flows. Geometrical similarity and dynamic similarity must both be enforced simultaneously. Then, the nondimensional flow variables v,p,ρ,... are identical in both flows. As already stated previously, only non-dimensional flow variables v,p,ρ... are (quantitatively) identical for both flow conditions. This is nevertheless sufficient to get back to the quantitative value for the application at real scale. Remember Eq.(10.6): For instance: φ(x,t) = φ φ (x,t) (10.45) v(x,t) = v v (x,t) (10.46) Now, assume you have been able to measure the dimensional velocity field v m on a model at reduced scale, where index m denotes model values. How can you get the dimensional velocity field for the real, large scale applications (index a )? Simply start by computing the non-dimensional velocity field for the model: v m (x,t) = v m(x,t) v m (10.47) Since you have of course respected the similarity conditions, you know that v is identical in the model and in the full-scale application. Hence v a (x,t) = v m (x,t) = v m(x,t) v m (10.48)

118 116 Chapter 10. Dimensional analysis and similarity conditions But you want indeed the dimensional solution for the full-scale application! No problem, you just need to multiply with the corresponding reference velocity for the full-scale application: At the end, you obtain: v a (x,t) = v a v a (x,t) (10.49) v a (x,t) = v a v m v m (x,t) (10.50) Therefore, measuring the dimensional velocity field on the model v m is sufficient to get the dimensional velocity field on the full-scale application, v a ; you just need to multiply by the appropriate scale ratio!

119 Chapter 11 One-dimensional isentropic compressible flows 11.1 Introduction and hypotheses This chapter should be considered only as a first introduction to the domain of compressible flows (also sometimes called Gas Dynamics, though in principle liquids with very large pressure changes can also show considerable density variations). Dealing with compressible flows, we consider here only flows for which the variation of density ρ induced by the variation of pressure p in the flow (and thus indirectly by the variation of velocity v) cannot be neglected any more: the density ρ cannot be considered constant. We will define later more precisely the corresponding conditions. Figure 11.1: Two out of many applications involving compressible flows: A380 aircraft (photo from Airbus) and Ariane 5 space rocket (photo from Arianespace). Since adding the density ρ as a variable increases of course the complexity of the flow description 117

120 118 Chapter 11. One-dimensional isentropic compressible flows (see again the variable-based classification presented in the introduction), supplementary simplifying assumptions must be introduced to obtain a system of acceptable complexity. In the present chapter, we will therefore use throughout following hypotheses: 1. the flow is steady; 2. the flow is considered one-dimensional; thus, all flow variables depend only on a single spatial coordinate, x; 3. the flow is considered non-viscous; 4. the flow does not exchange any heat with the surroundings (adiabatic process). Considering these hypothesis, corresponding flows are automatically irrotational, since v = 0 in the present case. Boundary layers do not exist in such flows, since near-wall velocity gradients perpendicular to the wall are assumed impossible. Furthermore, viscous effects have been considered negligible. Since viscous forces are assumed negligible, corresponding flows are thermodynamically reversible. Since they have been furthermore assumed to be adiabatic, they are automatically isentropic as well. During all this chapter, the specific entropy s can therefore be considered a constant. Such isentropic flows are particularly simple. This is the reason why they will be considered first. In what comes, we will consider specifically an internal flow (flow within a duct). Note, however, that most considerations obtained in this manner could also be used to describe the compressible flow enclosed within a streamtube Generic relations, also valid for a real gas We begin by considering generic gas properties. This means that the relations derived in this section are valid even for complex thermodynamic relations, as might be found for a real gas. Note that all statements are valid as long as the flow is one-dimensional, independently from the real geometry perpendicular to the plane considered in the analysis. In particular, the findings are identical for a twodimensional configuration (constant depth l perpendicular to the analysis plane) or for an axisymmetric configuration (diameter d(x)) Conservation equations Taking into account the hypotheses listed above, it is possible to establish generic conservation equations for such simple compressible flows. Mass conservation For a steady flow bounded by a wall (internal flow) or equivalently contained within a streamtube the mass conservation can be represented simply by the conservation of the mass flow-rate through any tube cross-section of area A(x): ṁ = ρ(x)a(x)v(x) = constant (11.1) In what follows, we will usually not write explicitly the dependency on the spatial coordinate x, thus obtaining ṁ = ρav = constant (11.2)

121 2 Generic relations, also valid for a real gas 119 This relation is the integral formulation of mass conservation. As often in fluid dynamics, a differential form would be more useful. In the present case, since all quantities are always positive, a logarithmic differentiation is even more relevant, leading finally to dρ ρ + da A + dv v = 0 (11.3) Energy conservation The conservation of energy can be derived from the First Law of Thermodynamics. For a compressible gas, the most suitable description of the available internal energy is through the specific enthalpy h, since it contains intrinsically the corresponding contribution due to pressure changes. One finally writes that, for such an isolated adiabatic system (the considered fluid), the sum of internal energy and kinetic energy (both per unit mass) is constant: or, suppressing the x-dependency: h(x)+ [v(x)]2 2 h+ v2 = constant (11.4) = constant (11.5) 2 In this formulation, the contribution of potential energy (and hence the action of gravity) have been neglected, as we have always done in the past when considering a gas, for instance for the Bernoulli equation in Chapter 6 (Eq. 6.19). Indeed, this contribution is negligible in all practical applications. The constant value appearing in Eq.(11.4) will be from now on called isentropic stagnation enthalpy, h 0 : h 0 := constant = h+ v2 (11.6) 2 Considering this relation, it is obvious that the local fluid enthalpy equals the stagnation enthalpy if and only if the fluid would be at rest (v = 0), hence the adjective stagnation. If the fluid is flowing (v > 0) the local enthalpy h is necessarily lower than the stagnation enthalpy h 0. Knowing the local enthalpy h and the (constant) stagnation enthalpy h 0 is of course sufficient to get back the fluid velocity by v = 2(h 0 h) (11.7) Note that in the international literature, the isentropic stagnation enthalpy h 0 may also be called total enthalpy and is sometimes written h t (index t for total) or h i (index i for isentropic). In this document, we will always use the index 0 to denote isentropic stagnation quantities. Energy conservation can also immediately be rewritten in a differential form by simply differentiating Eq.(11.4), leading to: dh+vdv = 0 (11.8) Conservation of momentum The conservation of momentum will be obtained indirectly by considering the forces exerted by the fluid on the boundaries of the small control volume dv c represented in figure 11.2, enclosed between the side walls dσ, the inflow section A and outflow section (A + da). Since we consider an infinitesimal volume, the wall dσ can be considered to be straight at this scale, even if the wall boundary is curved at macroscopic scale.

122 120 Chapter 11. One-dimensional isentropic compressible flows n v x x+dx s A A+dA d Figure 11.2: Configuration for establishing conservation of momentum, with flow from left to right. Using the results presented in Chapter 7 and valid here (steady one-dimensional internal flow), one candirectly writeforthepresent casethattheforceexerted by thecontainedfluid ontheside boundaries dσ is: df = d[(ṁv +pa)s] with s the unit vector giving the direction of the local flow velocity. In the present case (one-dimensional flow), the direction of s is automatically the direction of the single spatial direction considered (x), so that, s = e x. As usual for gas flows (see again the discussions in Chapter 7), the contribution of fluid mass can be safely neglected here due to the very low density of usual gases. By projecting this relation onto the x-axis, one thus obtains simply: df x = d(ṁv +pa) (11.9) = ṁdv vdṁ pda Adp (11.10) Due to mass conservation (ṁ = constant, see above), the second term of this relation is automatically zero, and df x = ṁdv pda Adp (11.11) As an alternative, it is possible to compute now directly the forces exerted by the fluid on the side boundaries dσ. We have assumed throughout that the flow is non-viscous, so that the corresponding contribution disappears. Again, the contribution of fluid mass can be safely neglected here due to the very low density of usual gases. Therefore, the only possible force exerted by the fluid on the side

123 2 Generic relations, also valid for a real gas 121 boundaries is the pressure force. By integrating the contribution of this force on all the boundary (total surface dσ), it can be seen that the contribution of this force is again only along the x-direction (contributions along the direction perpendicular to the x-axis compensate when considering opposed sides of the wall), with a magnitude df px = pdσcosα (11.12) where cos α represents the contribution of the projection onto the x-axis. Considering the light green triangle with dotted line in figure 11.2 and applying standard angular relations in a right triangle for it, it is easy to write cos α = da/dσ (11.13) Finally, one obtains thus directly for the contribution of the pressure force df px = pda (11.14) Since both equations (11.11) and (11.14) must deliver the same result, one finds by equating both: pda = ṁdv pda Adp By using ṁ = ρav, this can be rewritten after simplifying: A(dp+ρvdv) = 0 (11.15) and after division by the surface A: dp+ρvdv = 0 (11.16) Summary Considering all these relations plus a suitable thermodynamic relation describing the behavior of the considered gas, one obtains finally: Mass conservation Conservation of momentum Energy conservation ρav = constant (11.17) dp+ρvdv = 0 (11.18) h+ v2 2 = constant (11.19) Entropy conservation Thermodynamic equation of state (different forms are possible) s = constant (11.20) p = G(ρ,s) (11.21) In principle, this set of equations could now be solved using a computer to obtain the corresponding flow solution, knowing the exact thermodynamic behavior p = G(ρ, s).

124 122 Chapter 11. One-dimensional isentropic compressible flows What is a compressible gas flow? To finally characterize quantitatively the difference between a compressible and an incompressible flow, we start back from the conservation of momentum (Eq ). By dividing with the density ρ(obviously always verifying ρ > 0), one obtains trivially: dp ρ +vdv = 0 (11.22) The first term can then be rewritten as: dp ρ = dp dρ dρ ρ ( ) dp = dρ s dρ ρ (11.23) (11.24) where the ratio between pressure variation and density variation corresponds to a thermodynamic change of state for a constant specific entropy s. Why is it allowed to write such a relation? Simply because the present flow is completely isentropic! As a consequence, any change of pressure and density from an initial state (p 1,ρ 1 ) to a final state (p 2,ρ 2 ) within the flow occurs necessarily at constant specific entropy s. Now, Eq.(11.24) is very interesting because the definition of the speed of sound c appears directly in the relation, since: ( ) dp c 2 := (11.25) dρ s Starting back from Eq.(11.24), one obtains now: dp ρ = c2dρ ρ (11.26) Substituting now this relation in Eq.(11.22) leads to: c 2dρ ρ +vdv = 0 (11.27) or equivalently (velocity v being non-zero, excluding aerostatics, since we only consider compressible flows, i.e., flows at large Mach numbers, as demonstrated in a few seconds): c 2dρ ρ +v2dv v = 0 (11.28) Moving the second term on the right hand side and introducing the definition of the Mach number, ratio between flow velocity v and speed of sound c, M := v c (11.29) one obtains finally the important relation dρ ρ = M2dv v (11.30)

125 2 Generic relations, also valid for a real gas 123 M (dρ/ρ)/(dv/v) (in %) Table 11.1: Ratio between the relative variation of density and the relative variation of velocity magnitude as a function of the Mach number M As a function of the Mach number, the ratio between the relative variation of density and the relative variation of velocity magnitude can then be directly computed, leading e.g. to Table Now, the threshold between a compressible flow and an incompressible flow can be defined more clearly. If, as usual for engineering purposes, 10% is considered as some kind of magical limit under which an effect can be neglected, then a Mach number of 0.3 (strictly speaking, of ) corresponds to the limit between incompressible (M 0.3) and compressible (M > 0.3) flows. If, for some reason, the accuracy of the computation must be very high, so that only a maximal error of 1% can be tolerated, then incompressible flows take place only up to M = 0.1. This is illustrated in figure subsonic sonic supersonic incompressible flow compressible flow Value depends on accuracy requirements, typically between M=0.1 and Mach number M Figure 11.3: Compressible vs. incompressible flow conditions Influence of a modification of the cross-section A For the flow considered in the present chapter, the local cross-section A(x) of the tube or pipe (resp. of the streamtube) can change freely between 0 and infinity. The purpose of the present section is to investigate how velocity and pressure would be modified when increasing or reducing A(x). For this purpose, we start back from Eq.(11.30): dρ ρ = M2dv v

126 124 Chapter 11. One-dimensional isentropic compressible flows Using this expression to replace the relative density variation in the equation describing conservation of mass, Eq.(11.3): dρ ρ + da A + dv v = 0 one finally obtains a relation linking the relative change of flow velocity with the relative change of the cross-section: ( 1 M 2 ) dv v + da A = 0 (11.31) This fundamental relation is often called Hugoniot equation and involves the value of the Mach number M compared to unity. Considering this relation, one can therefore distinguish following possibilities: Forasubsonicflow(i.e., M < 1), oneobtainsobviouslythat(1 M 2 ) > 0. Undersuchconditions, a reduction oftheflowcross-section A(i.e., da < 0)obviously leadstoflowacceleration (i.e., dv > 0) since both A and v are always positive in Eq.(11.31). An increase of flow cross-section, da > 0 leads to flow deceleration (dv < 0). Conversely, for a supersonic flow (i.e., M > 1), one obtains (1 M 2 ) < 0. Then, a reduction of the flow cross-section A (i.e., da < 0) now leads to flow deceleration (i.e., dv < 0), while an increase of the flow cross-section, da > 0, results in flow acceleration, dv > 0. This situation is summarized in figure This figure contains also the associated evolution of pressure, which can be obtained by starting again from the conservation of momentum, Eq.(11.16) dp+ρvdv = 0 Dividing both terms by pressure p (obviously, p > 0) and artificially multiplying the second term by v/v (obviously, v > 0 for compressible flows), as already done in Eq.(11.28), one obtains: dp p + ρv2 dv p v = 0 (11.32) It is easy to combine this relation with the standard equation of Hugoniot, Eq.(11.31), in order to eliminate the relative velocity variation dv/v. This leads finally to: ( M 2 1 ) dp p + ρv2 da p A = 0 (11.33) Since p,v 2 and A are obviously always positive, this relation is a direct link between the sign of da and the sign of dp, depending on the sign of (M 2 1). Corresponding evolutions are shown in figure Note that the obtained link between pressure and velocity (a pressure decrease for flow acceleration, respectively a pressure increase for flow deceleration) is absolutely standard: we have found exactly the same for incompressible flows when analyzing the Bernoulli equation in Chapter 6! Critical conditions Per definition, critical conditions are the conditions found locally when the flow reaches the sonic state, i.e., M(x) = 1. Critical flow parameters are written with the index, for example (v,c,p,t,...). For the compressible flows considered in this chapter, there will be at most one location where M = 1. Therefore, the critical pressure p or critical temperature T is indeed a constant, scalar value (perhaps

127 2 Generic relations, also valid for a real gas 125 acceleration v deceleration v M < 1 M > 1 p deceleration v v p acceleration p p Figure 11.4: Influence of a modification of the cross-section A. p = 1.5 bar and T = 305 K, for instance). The flow cross-section for these conditions is similarly written A. There is one exception to this definition: the critical Mach number M is not a constant, scalar value but a flow variable, depending on x. Indeed, the critical Mach number is defined as: M (x) := v(x) c (11.34) This new flow variable can be used in practice to parametrize the local flow conditions, as explained later. Beware! Note that critical conditions can be usefully employed to solve compressible flow problems, even if the condition M = 1 is not found anywhere in this flow! In that case, we can call these conditions theoretical critical conditions, to mark the fact that they are not found in reality. Nevertheless, it is always possible to imagine a related isentropic flow, in which the sonic state would be achieved at some point: the flow conditions that would be found at that point are then the theoretical critical conditions, and can be employed as well to understand the real flow. Note that the critical speed of sound c appearing in Eq.(11.34) can directly be deduced from the stagnation temperature T 0 defined later in Eq.(11.55), since: c := γrt

128 126 Chapter 11. One-dimensional isentropic compressible flows T 0 T = 1+ γ 1 M 2 2 (11.35) The second relation delivers for the critical temperature T = 2T 0 γ +1 (11.36) Reporting now in the first relation: c = 2 γ γ +1 rt 0 (11.37) As a consequence, the critical speed of sound is constant as long as the considered flow is adiabatic, since under such conditions the stagnation temperature T 0 is constant as well Laval nozzle At the end of the 19th century, the swedish engineer Gustaf de Laval considered the challenge of accelerating an initial high-pressure, low-speed gas flow to the highest possible level, and finally invented the Laval nozzle. His purpose was also to convert the flow energy available at the beginning in the form of a high pressure into kinetic energy (i.e.,, a high velocity), a process that we have already considered extensively in Chapter 6 for incompressible flows. But now, the density ρ is varying and we must take this effect into account. In order to solve this problem, de Laval was already aware of the relationships derived from the equation of Hugoniot (Eq ), discussed in Section and summarized in figure Since the initial conditions correspond to a low-speed flow (therefore associated with M < 1), de Laval decided to develop a system leading to a reduction of the flow cross-section A(x): a converging nozzle (see figure 11.5). Considering the previous results, this indeed leads to a flow acceleration, the flow remaining subsonic throughout. But this was not sufficient to really satisfy de Laval. Therefore, considering again the results of Section summarized in figure 11.4, he had a brilliant idea: if it would be possible to jump in the table of figure 11.4 from the upper left quadrant (acceleration in a subsonic flow) to the lower right quadrant (acceleration in a supersonic flow), the level of acceleration that could be obtained would be tremendously increased! But, of course, this is only possible if the flow can evolve somehow from subsonic to supersonic conditions. Considering again the equation of Hugoniot (Eq ), it is easy to see that the sonic state can only be found at a place associated with a locally constant flow cross-section A. In Eq.(11.31), M = 1 implies directly da = 0! This is perfect for the realization of the Laval-nozzle(figure 11.6): the transition between subsonic flow and supersonic flow (and hence the sonic state) can only take place where it is needed, at the throat (smallest cross-section) of the nozzle. Beware! Nevertheless, it would be a mistake to believe that the reciprocal relation is true. If we build a nozzle with a throat and let a gas flow through it, this will not lead always to sonic conditions at the smallest cross-section! This is due to the fact that, considering once again the equation of Hugoniot (Eq ), a locally constant cross-section da = 0 implies either M = 1 (in very specific cases, like a properly working Laval-nozzle, figure 11.6), or much more often simply dv = 0, i.e., the fact that the flow velocity locally does not change...a throat in a gas flow is certainly no guarantee that the flow will be sonic there, or our own throat would continuously produce supersonic flows while talking to our neighbors! The geometry of a Laval-nozzle is not sufficient to generate a supersonic flow: for most conditions, the convergent part of the Laval-nozzle will lead to flow acceleration in the subsonic

129 2 Generic relations, also valid for a real gas 127 v p>p a p a p Figure 11.5: First step toward a Laval nozzle: the minimum nozzle. range, the Mach number at the throat will be in the far subsonic range (M(x) 1), so that the final, diverging part of the Laval-nozzle will lead to a strong flow deceleration (again in the subsonic range), in agreement with the equation of Hugoniot (Eq ), summarized in figure 11.4: as a whole, the flow will probably be slower when leaving the nozzle than when entering it; this is clearly not the purpose of a properly-working Laval-nozzle! The missing condition will be explained in Section Finally, in a properly working Laval-nozzle (figure 11.6), the flow is initially subsonic, is accelerated (while staying subsonic) until reaching the throat; at this level, it reaches the sonic condition and goes on accelerating as a supersonic flow until leaving the nozzle. Therefore, this Laval-nozzle is indeed a perfect way of accelerating a flow! What about pressure? As explained in Section , a flow acceleration is always coupled to a decrease of pressure. Therefore, the pressure is steadily decreasing throughout the Laval-nozzle, from an initially high value to a low value. At present and considering the hypotheses retained in this chapter, we have no solution to modify abruptly the pressure. Therefore, the value of p at the end of the nozzle must be equal to the pressure in the surroundings, which we will write p a (atmospheric pressure). Later, we will call such specific conditions an adapted or ambient Laval-nozzle. Note that another, major advantage of the Laval-nozzle (figure 11.6) is that it does not require any moving part. It is therefore relatively easy to conceive and can be very robust. This is essential, since such a nozzle will usually encounter very high levels of pressure and possibly of temperature (for instance in steam turbines or rocket engines, figure 11.7). Further information can be found for instance under Wikipedia.

130 128 Chapter 11. One-dimensional isentropic compressible flows p a v M=1 p>p min subsonic supersonic p a A * p Figure 11.6: Adapted Laval nozzle working properly Specific relations for a compressible flow of a perfect gas What is a perfect gas? It is difficult to go further without specifying the actual gas properties. Therefore, we will go on now by considering only a perfect gas. A perfect gas is not really a thermodynamic concept, but is used mainly for Fluid Dynamics applications. In order to be called a perfect gas, a gas must fulfill simultaneously two conditions: 1. A perfect gas must behave like an ideal gas. We can therefore employ the equation of state p = ρrt (11.38) connecting pressure p, density ρ and temperature T through the specific gas constant r. 2. A perfect gas must furthermore be associated with a constant value of the specific heat, (for an isobaric evolution), or equivalently c p = constant (11.39) c v = constant (11.40) (for an isochoric evolution). Since the thermodynamic relation dh = c p dt (11.41)

131 3 Specific relations for a compressible flow of a perfect gas 129 Figure 11.7: Vulcain 2 rocket engine, main engine of the Ariane 5 space rocket (photo from EADS Astrium, Space Propulsion). The diverging part of the Laval-nozzle is clearly visible, the (smaller) converging part is hidden behind the gas supply systems. is an exact connection between variation of specific enthalpy and temperature, it is easily possible to integrate this relation for a constant value of c p, leading to h = c p T (11.42) by choosing in an appropriate manner the value of the integration constant Isentropic relations for a perfect gas Using the basic laws of thermodynamics, it is possible to demonstrate that the specific entropy of a perfect gas is given by: ( ) p s = c v ln + constant (11.43) ρ γ

132 130 Chapter 11. One-dimensional isentropic compressible flows or equivalently by taking the exponential of this equation and switching left and right side ) p s ρ = constant γ >0 exp( c v (11.44) Both relations involve the (constant) heat capacity ratio γ. It is now very easy to determine the socalled isentropic relations verified by a perfect gas. First comes, by setting s = constant in Eq.(11.44): p = constant (11.45) ργ Using the ideal gaslaw(eq ) to eliminate either pressure p ordensity ρfromthe aboverelation, one obtains the equivalent formulations: Speed of sound for a perfect gas p γ 1 = constant T γ (11.46) T = constant ρ γ 1 (11.47) Reminding the definition of the speed of sound c: c 2 := ( ) p ρ s (11.48) this quantity can now easily be calculated for a perfect gas. Since, for s = constant, Eq.(11.45) leads to p = constant ρ γ (11.49) one obtains directly Hence, p ρ = constant γργ 1 = γ p ρ c = (11.50) γp ρ = γrt (11.51) where the second relation has been obtained by replacing p/ρ by rt thanks to the ideal gas law, Eq.(11.38). For air at ambient conditions, one obtains for example for the speed of sound c = γrt = = m/s = 1246 km/h (11.52) Analytical solution for a compressible flow of a perfect gas In order to determine now analytically the full solution for the flow of a perfect gas (still verifying of course the assumptions listed in section 11.1: the flow is steady, one-dimensional, non-viscous, adiabatic), we will now express all the flow variables as a function of the local Mach number M(x) and of the stagnation quantities. In a first step, the energy conservation equation(eqs and 11.6) is considered: h 0 = h+ v2 2 = constant (11.53)

133 3 Specific relations for a compressible flow of a perfect gas 131 Using Eq.(11.42), one obtains: c p T 0 = c p T + v2 2 where the stagnation temperature T 0 is defined as (11.54) T 0 := h 0 c p (11.55) and is therefore a constant value within our isentropic flow. As for the stagnation enthalpy, the local value of the temperature, T(x) is necessarily lower than T 0, both values being equal in theory for v = 0 (stagnation conditions). By dividing Eq.(11.54) by c p T, one gets: It is then possible to use Eq.(B.6) to eliminate c p from this equation, leading to: T 0 T T 0 T = 1+ v2 2c p T c p = γr γ 1 = 1+ γ 1 2 v 2 γrt Considering now the expression of the speed of sound, Eq.(11.51): the denominator on the right-hand side can be replaced by c 2, leading to In this manner, the Mach number M := v c appears naturally in this equation, which takes the final form: T 0 T T 0 T (11.56) (11.57) (11.58) c 2 = γrt (11.59) γ 1v 2 = 1+ (11.60) 2 c 2 (11.61) γ 1 = 1+ M 2 (11.62) 2 Considering that γ and T 0 are constant values, this relation is a direct connection between the local Mach number (M, or indeed M(x)) and the local temperature (T, or indeed T(x)). Now, remembering that our compressible flow is fully isentropic, we can readily use the isentropic relations (Eqs to 11.47) introduced in Section for a change of thermodynamic state between the (real or theoretical) stagnation state (zero velocity, stagnation enthalpy h 0, stagnation temperature T 0, stagnation pressure p 0, stagnation density ρ 0, stagnation speed of sound c 0...) and the local flow conditions (velocity v(x), enthalpy h(x), temperature T(x), pressure p(x), density ρ(x), speed of sound c(x)..., for which the x-dependency will not be written explicitly to gain space), leading to: p = p 0 (11.63) ρ γ ρ γ 0 p γ 1 T γ = pγ 1 0 T γ 0 T ρ γ 1 = T 0 ρ γ 1 0 (11.64) (11.65)

134 132 Chapter 11. One-dimensional isentropic compressible flows As for the stagnation temperature, the local value of the pressure, p(x) is necessarily lower than p 0, both values being equal in theory for v = 0 (stagnation conditions). Combining now Eq.(11.64) with Eq.(11.62), one obtains directly: and therefore p 0 p = ( ( ) p γ 0 p = T0 γ 1 T 1+ γ 1 2 M 2 ) γ γ 1 (11.66) (11.67) This relation is again a direct connection between the local Mach number (M, or indeed M(x)) and the local pressure (p, or indeed p(x)). In what comes, the inverse of this relation will be defined as the function π, depending only on the Mach number, as π(m) := p/p 0 (11.68) with π(m) 1. In the same manner, combining Eq.(11.65) with Eq.(11.62), one obtains: ( ) ρ 1 0 ρ = T0 γ 1 T (11.69) where, as for all other stagnation values, the local value of the density, ρ(x) is necessarily lower than ρ 0, both values being equal in theory for v = 0 (stagnation conditions). As a consequence, ρ 0 ρ = ( 1+ γ 1 2 M 2 ) 1 γ 1 (11.70) This relation is again a direct connection between the local Mach number (M, or indeed M(x)) and the local density (ρ, or indeed ρ(x)). Note that it is equally possible to obtain this last equation (Eq ) by combining the two previous relations (Eqs and 11.67) with the ideal gas relation (Eq ). To summarize, we have now relations connecting directly the local value of the Mach number, M(x) with the local temperature, pressure and density, provided the (constant) stagnation values for temperature (T 0 ), pressure (p 0 ) and density (ρ 0 ) are known. Unfortunately, the local Mach number M(x) is not known yet. In order to solve this last issue, it is convenient to consider again mass conservation in our flow. Starting from the definition of the mass flow-rate: ṁ = ρav (11.71) it is possible to substitute the density by ρ = p/(rt) (using the ideal gas law, Eq.11.38), as well as to substitute the flow velocity by v = Mc = M(γrT) 1/2 (using the definition of the Mach number, Eq.11.61). This leads now to: ṁ = p rt AM(γrT)1/2 (11.72) ( ) γ 1/2 = AMp (11.73) rt Now, we introduce artificially the ratios p 0 /p and T 0 /T in this equation, by ( ) ( ) p γ 1/2 ( ) 1/2 T0 ṁ = AMp 0 (11.74) p 0 rt 0 T

135 3 Specific relations for a compressible flow of a perfect gas 133 Theratiosp/p 0 andt 0 /T cannowbereplacedbytheinverseofeq.(11.67)andbyeq.(11.62)respectively. One gets: ( ) γ 1/2 ( ṁ = AMp 0 1+ γ 1 ) (γ+1) M 2 2(γ 1) (11.75) rt 0 2 This relation can now identically be written for the critical conditions, associated with the critical value of the cross-section, A. Let us stress once again here that it is not absolutely necessary to have really critical conditions at some position in the flow in order to use this relation; if not, the corresponding conditions correspond to theoretical critical conditions, as already explained in Section , but can nevertheless be used to ease problem resolution. One obtains, taking into account that M = 1 there, ṁ = A p 0 ( γ rt 0 ) 1/2 ( γ +1 2 ) (γ+1) 2(γ 1) (11.76) Since mass conservation obviously leads to the relation ṁ = ṁ, dividing the last two relations side by side (on the left-hand side, leading just to 1) and re-arranging, one obtains easily the final equation: A = 1 [ 2 A M γ +1 ( 1+ γ 1 2 M 2 ) ] (γ+1) 2(γ 1) (11.77) Inwhatcomes, thisrelationwillalsobeusedtodefinethefunctionσ,dependingonlyonthemach number, by Σ(M) := A A (11.78) Considering again Eq.(11.76) and the fact that ṁ = ṁ, it is also possible to write ṁ = C p 0A (rt 0 ) 1/2 (11.79) where C is simply a constant value, function of the heat capacity ratio γ, with ( ) γ+1 2 C := γ 1/2 2(γ 1) γ +1 (11.80) For usual conditions (i.e., γ = 1.4), one obtains C = The relation (11.79) is by far the most accurate relation to determine the mass flow-rate in such a compressible flow, since it involves only fluid properties, geometrical data and stagnation quantities, which are mostly given as exact information at the start of the problem. The relation (11.79) explains also why Laval-nozzles (figure 11.6) are also used as mass flow-rate controllers or as reference calibration for such systems. If the nozzle is working properly and if the stagnation conditions and throat section are kept constant, the mass flow-rate through the Laval-nozzle is constant as well, independently from the ambient pressure at the nozzle outlet! Hence, properly working Laval-nozzles are typically employed to calibrate mass flow-rate measurement and control devices for gas flow applications Critical conditions Using the previously introduced relations, Eqs.(11.62), (11.67) and (11.70), it is directly possible now to obtain the critical conditions, i.e., those associated with M = 1: T = 2 γ +1 T 0 (11.81)

136 134 Chapter 11. One-dimensional isentropic compressible flows p = ρ = ( ) γ/(γ 1) 2 p 0 γ +1 (11.82) ( ) 1/(γ 1) 2 ρ 0 γ +1 (11.83) For usual conditions (i.e., γ = 1.4), the critical temperature is thus roughly 17% lower than the stagnation temperature (T = 0.83 T 0 ), while the critical pressure is about half the stagnation pressure (p = 0.53 p 0 ) Solution procedure and remaining difficulties Using the previous relations, it is now possible to fully determine the resulting compressible flow, with all flow quantities. The procedure is as follows: 1. first, the geometry of the system (i.e., the distribution of the available flow cross-section A(x) and the corresponding throat A for the case of the standard Laval-nozzle, figure 11.6) must of course be known! 2. Using Eq.(11.77), it is now possible to determine in the full system the distribution of the Mach number, M(x). Beware! Note that Eq.(11.77), a relatively complex non-linear equation, leads in fact to two different solutionsforagivenvalueofa/a : oneisalwaysinthesubsonicrange(seelaterappendix E), one in the supersonic range (see later Appendix F). This is indeed necessary. If you look again at figure 11.6, describing the correctly working, adapted Laval-nozzle, you see that you will have on the left side of the throat a subsonic condition and on the right side of the throat a supersonic condition, both for the same value of A/A. Therefore, in order to connect the value of A/A with the local Mach number M(x), some user input is necessary: you must decide if a subsonic or a supersonic value is expected! 3. Knowing now the complete distribution of M(x), you can get the temperature field T(x) from Eq.(11.62) knowing the stagnation temperature, T 0. Beware! This is of course only possible if you know the stagnation conditions. They are mostly given values. If not, you must be able to compute these stagnation quantities from a flow section for which both Mach number and temperature, pressure or density are known In the same manner, you can get the full pressure field p(x) from Eq.(11.67) knowing the stagnation pressure, p Knowing temperature and pressure, the ideal gas law (Eq ) can be employed to compute the density field, ρ(x). 6. Since the temperature is known everywhere, it is equally possible to compute the local speed of sound using: c(x) = γrt(x) (11.84) 7. Finally, the velocity field can be determined from the definition of the Mach number: v(x) = M(x)c(x) (11.85) At that point, all flow variables have been successfully determined: congratulations!

137 3 Specific relations for a compressible flow of a perfect gas Minimal stagnation pressure for a properly working Laval-nozzle Using the previous relations, it is now possible to fully determine the conditions leading to a properlyworking Laval-nozzle. This is indeed a condition on the starting (stagnation) pressure, p 0, compared to the pressure obtained when leaving the nozzle, the atmospheric pressure p a (at least for the present case, that of an adapted or ambient nozzle). For the shortest possible Laval-nozzle shown in figure 11.5, the outlet pressure is simultaneously the critical pressure, p (Eq ). The corresponding condition reads therefore p = p a = By inverting this relation, one obtains directly: ( ) γ/(γ 1) 2 p 0 (11.86) γ +1 ( ) γ +1 γ/(γ 1) p 0 = p a (11.87) 2 For usual condition (i.e., γ = 1.4), one obtains the minimum value acceptable for the stagnation pressure p 0 = p a (11.88) For a real Laval-nozzle with an outflow cross-section A o > A, this condition is of course increasingly difficult to determine and to fulfill. Equation (11.77) must first be inverted to obtain the corresponding outflow Mach number, M o. This value of M o is then introduced into Eq.(11.67), leading to the minimum stagnation pressure p 0 associated with the corresponding geometry. Clearly, p p a is only the minimum threshold, and the required value for p 0 will increase rapidly when increasing the outflow cross-section, A o > A! Tables for compressible flows The main formulas introduced previously have been solved for the usual condition (i.e., γ = 1.4); corresponding values are listed in Appendix E for subsonic conditions (M(x) < 1), respectively in Appendix F for supersonic conditions (M(x) > 1). In order to solve an isentropic compressible flow problem, following possibilities can now be used: Solve directly the equations introduced previously (Eqs , 11.62, ). This is of course by far the most accurate solution, and the only possible solution for a gas that does not verify γ = 1.4. Use the Tables presented in Appendix E and F to solve the problem (requires γ = 1.4). Use the corresponding graphical representation as a function of the critical Mach number (see sections and ; this requires again γ = 1.4). Use Internet-based Java scripts, as available for example under Efluids Solution using the critical Mach number M The critical Mach number M can easily be related to the usual Mach number M: M 2 = v2 c 2 = v2 c 2 c 2 c 2 = M2 c2 c 2 = M2 T T = M 2 T T 0 T 0 T (11.89)

138 136 Chapter 11. One-dimensional isentropic compressible flows Now, the two temperature ratios appearing last on the right-hand side can easily be replaced by using Eq.(11.62), once involving the local Mach number M, once involving the Mach number at critical conditions (i.e., simply 1!), leading finally to: This relation can as well be inverted: M 2 = M 2 = (γ +1)M2 2+(γ 1)M 2 (11.90) 2M 2 (γ +1) (γ 1)M 2 (11.91) Introducing this last relation to replace the local Mach number M in all previously developed equations, they can be rewritten as a function of the critical Mach number M instead. For example, this leads for Eq.(11.62) to the alternative, fully equivalent formulation: ( T 0 T = 1 γ 1 ) 1 γ +1 M 2 (11.92) The corresponding results can be represented compactly in a graphical manner (figure 11.8), since the value of M 2 remains bounded when M takes very large values. Therefore, this might be a practical alternative to solve corresponding problems. Nevertheless, it must be recognized that a graphical solution leads to a relatively large level of uncertainty. When a better precision is needed, the Tables presented in the Appendix (Appendix E and F) might be used. Even better is of course a direct numerical solution of the corresponding equations! This is the only possible solution if the heat capacity ratio γ is not equal to Discharge velocity v d By definition, this is the velocity obtained by accelerating a flow starting from stagnation conditions up to reaching a final (obviously lower) pressure p, all the flow being isentropic. In order to get the corresponding value, it is convenient to start back from energy conservation for a perfect gas, Eq.(11.54): One obtains: v d = c p T 0 = v d 2 2 +c pt (11.93) (2c p T 0 [ 1 T T 0 ]) 1/2 (11.94) Introducing now the link between p/p 0 and T/T 0 documented by the equations (11.67) and (11.62), which are indeed identical apart from the exponents on the right-hand side, it is possible to write directly: ( ) (γ 1)/γ 1/2 p v d = 2c p T 0 1 (11.95) p Conclusions Now, one-dimensional compressible flows can be solved very accurately as long as the associated transformation is isentropic. More complex cases involving friction and heat exchange and leading therefore to non-isentropic flows will be considered in the next chapter. All flows in the supersonic regime, will be often strongly modified due to the occurrence of shock waves: this is the subject of Chapter 13.

139 4 Conclusions Ma/10 p/p0 ρ/ρ0 T/T0 p1/p2 A*/A p02/p Ma* 2.45 Figure 11.8: Graphical solution involving the critical Mach number M, assuming γ = 1.4.

140 138 Chapter 11. One-dimensional isentropic compressible flows

141 Chapter 12 Compressible flows with friction and heat exchange 12.1 Introduction and hypotheses After having computed very simple (isentropic) compressible flows in the previous Chapter 11, more realistic (but also more complex) conditions are considered now, involving heat exchange and friction. They will lead to a flow with variable specific entropy s. Typical applications (figure 12.1) will involve for instance rocket engines and ramjets when looking at compressible flows with heat exchange. The influence of friction in compressible flows is essential for large-scale installations, for instance pressurized gas lines on industrial sites. Figure 12.1: Example of applications for non-isentropic compressible flows involving heat exchange and friction: ramjets used for propulsion of the Lockheed SR-71 Blackbird (left, photo from Wikipedia) and pressurized gas pipelines (right, photo from Biofuels Energy). Since adding the specific entropy s as an additional variable increases of course again the complexity of the flow description, supplementary simplifying assumptions must be introduced to obtain a system of acceptable complexity. In the present chapter, we will therefore use throughout following hypotheses, partly similar to those considered in the previous Chapter 11: 1. The flow is steady; 2. The flow is considered one-dimensional; thus, all flow variables depend only on a single spatial coordinate, x. 139

142 140 Chapter 12. Compressible flows with friction and heat exchange On the other hand, friction is now taken into account and heat exchange with the surroundings is considered possible. Hence, the flow is neither thermodynamically reversible any more, nor adiabatic. In what comes, we will consider specifically an internal flow (flow within a duct). Note, however, that most considerations obtained in this manner could also be used to describe the compressible flow enclosed within a streamtube. The typical configuration corresponding to the present chapter for a duct element is sketched in figure Generic relations, also valid for a real gas In this first section we follow very closely the developments presented in the previous Chapter 11 and introduce only modifications induced by friction and by heat exchange. We begin by considering generic gas properties. This means that the relations derived in this section are valid even for complex thermodynamic relations, as might be found for a real gas. Note that all statements are again valid as long as the flow is one-dimensional, independently from the real geometry perpendicular to the plane considered in the analysis. In particular, the findings are identical for a twodimensional configuration (constant depth l perpendicular to the analysis plane) or for an axisymmetric configuration (diameter d(x)) Conservation equations Taking into account the hypotheses listed above, it is possible to establish generic conservation equations for such more complex compressible flows. Mass conservation In fact, mass conservation is not impacted by friction or heat exchange. For a steady flow bounded by a wall (internal flow) or equivalently contained within a streamtube the mass conservation is thus unchanged and can be represented simply by the conservation of the mass flow-rate through any tube cross-section of area A(x): ṁ = ρ(x)a(x)v(x) = constant (12.1) Again, we will usually not write explicitly the dependency on the spatial coordinate x, thus obtaining ṁ = ρav = constant (12.2) The logarithmic differential form of this relation is even more useful: dρ ρ + da A + dv v = 0 (12.3) Energy conservation As always, the conservation of energy is derived from the First Law of Thermodynamics. For a compressible gas, the most suitable description of the available internal energy is again through the specific enthalpy h, since it contains intrinsically the corresponding contribution due to pressure changes. In the present case, the system is not adiabatic any more, the specific heat quantity dq (in J/kg) being exchanged with the surroundings. A positive value of dq means that heat is transferred to the fluid, dq < 0 means that heat is extracted out of the fluid. Hence, the differential form of the energy conservation is slightly different from Eq.(11.8): dh+vdv = dq (12.4)

143 2 Generic relations, also valid for a real gas 141 Conservation of momentum Once again, the conservation of momentum will be obtained indirectly by considering the forces exerted by the fluid on the boundaries of the small control volume dv c represented in figure 12.2 by dashed blue lines, enclosed between the side walls dσ, the inflow section A and outflow section (A+dA). The friction force df f is now included in the analysis. It is represented in an idealized manner in figure 12.2 by a near-wall contribution. n dq v x x+dx s A A+dA Friction force df f d Figure 12.2: Configuration for establishing conservation of momentum, with flow from left to right. The contribution of the friction force appears through a near-wall contribution and the heat exchange with the surroundings is represented with a dashed line. Using the results presented in Chapter 7 and valid here (steady one-dimensional internal flow), one candirectly writeforthepresent casethattheforceexerted by thecontainedfluid ontheside boundaries dσ is: df = d[(ṁv +pa)s] with s the unit vector giving the direction of the local flow velocity. This relation is not modified in any manner by the additional friction force, as discussed in Chapter 7. In the present case (one-dimensional flow), the direction of s is automatically the direction of the single spatial direction considered (x), so that, s = e x. As usual for gas flows (see again the discussions in Chapter 7), the contribution of fluid mass can be safely neglected here due to the very low density of usual gases.

144 142 Chapter 12. Compressible flows with friction and heat exchange By projecting this relation onto the x-axis, one thus obtains simply: df x = d(ṁv +pa) (12.5) = ṁdv vdṁ pda Adp (12.6) Due to mass conservation (ṁ = constant, see above), the second term of this relation is automatically zero, and df x = ṁdv pda Adp (12.7) As an alternative, it is possible to compute now directly the forces exerted by the fluid on the side boundaries dσ. Again, the contribution of fluid mass can be safely neglected here due to the very low density of usual gases. One force exerted by the fluid on the side boundaries is the pressure force. By integrating the contribution of this force on all the boundary dσ, it can be seen that the contribution of this force is again only along the x-direction (contributions along the direction perpendicular to the x- axis compensate when considering opposed sides of dσ). The finally resulting magnitude of the pressure force reads: df px = pdσcosα (12.8) where cos α represents the contribution of the projection onto the x-axis. Considering the light green triangle with dotted line in figure 12.2 and applying standard angular relations in a right triangle for it, it is easy to write cos α = da/dσ (12.9) Finally, one obtains thus directly for the contribution of the pressure force df px = pda (12.10) Inadditiontothepressureforce,itisnownecessarytoconsideraswellthefrictionforcedF f. Onceagain, due to the symmetry of the considered geometry, the resulting friction force has only one component along the main flow direction, x, with a positive magnitude written df fx. Both approaches have been used to compute the force acting on the boundary. They must therefore deliver the same result. One thus finds by equating the combination of pressure force and friction along the x-direction (left-hand side) with the result of Eq.(12.7) on the right-hand side: pda+df fx = ṁdv pda Adp By using ṁ = ρav, this can be rewritten after simplification: A(dp+ρvdv)+dF fx = 0 (12.11) and after division by the surface A: dp+ρvdv+df fx = 0 (12.12) with df fx := df fx A the ratio between friction force and flow cross-section, expressed in Pa. (12.13)

145 2 Generic relations, also valid for a real gas 143 Summary Considering all these relations plus a suitable thermodynamic relation describing the behavior of the considered gas, one obtains finally in differential form: Mass conservation dρ ρ + da A + dv v = 0 (12.14) Conservation of momentum dp+ρvdv+df fx = 0 (12.15) Energy conservation Thermodynamic equation of state (different forms are possible) dh+vdv = dq (12.16) p = G(ρ,s) (12.17) In principle, this set of equations could now be solved using a computer to obtain the corresponding flow solution, knowing the exact thermodynamic behavior p = G(ρ, s) Generalized equation of Hugoniot In this section, a generic relation will be derived as an extension of that proposed in section With this generalized Hugoniot equation, it will be immediately possible to identify in a qualitative manner the influence of a variation of the cross-section A, but also of friction and heat exchange. For this purpose, we start again from a general thermodynamic equation of state p = G(ρ,s). By differentiating this equation, it comes: dp = ( ) dp dρ+ dρ s ( ) dp ds (12.18) ds ρ The first parenthesis is already known and is equal to c 2, see Eq.(11.25). We will( accept ) (it is possible dp to demonstrate this, but it would bring us too far) that the second parenthesis, is also always ds ρ positive. Therefore, it is possible to introduce a new parameter k and to define it similarly to Eq.(11.25) through its square value: ( ) dp k 2 := (12.19) ds ρ It is easy to prove the validity of this hypothesis for a perfect gas. In that case, the specific entropy is given by Eq.(B.10): ( ) p s = c v ln + constant (12.20) By differentiating this relation, one obtains: ρ γ ds = dp c v p γdρ ρ (12.21)

146 144 Chapter 12. Compressible flows with friction and heat exchange Hence, and ( ) dp = p > 0 (12.22) ds c ρ v k 2 = p c v (12.23) in the case of a perfect gas. Following this demonstration and to summarize, we therefore accept from now on that it is always possible to write in a general manner, for any gas: By combining this relation with Eq.(12.12) and eliminating dp: dp = c 2 dρ+k 2 ds (12.24) c 2 dρ+k 2 ds+ρvdv +df fx = 0 (12.25) Now, dρ can be replaced by using the mass conservation equation (12.3), leading to: c 2 ρ dv v c2 ρ da A +k2 ds+ρvdv +df fx = 0 (12.26) Let us remember the generic thermodynamic relation between specific enthalpy and specific entropy, Eq.(B.4): dh = Tds+ dp ρ Replacing in the energy conservation equation (12.4), one obtains Tds+ dp ρ (12.27) +vdv = dq (12.28) By comparing this equation with that expressing conservation of momentum, Eq.(12.12), divided by ρ: dp ρ +vdv = df fx ρ (12.29) one recognizes immediately that both equations contain a similar group on the left-hand side, so that it is now possible to write: ds = 1 ( ) dffx T ρ +dq (12.30) This equation describes the two possibilities existing in such flows to modify specific entropy s, either by 1) friction (df) or 2) heat exchange (dq). In the absence of friction and of heat exchange, one would obtain an isentropic compressible flow, as discussed extensively in the previous Chapter 11. Now, reporting Eq.(12.30) into Eq.(12.26) to eliminate ds, one obtains: c 2 ρ dv v c2 ρ da A + k2 T ( dffx ρ +dq ) +ρvdv +df fx = 0 (12.31) After dividing all the equation by ρc 2 and rearranging, the generalized Hugoniot equation is obtained as an extension of Eq.(11.31): ( ) 1 M 2 dv v + da A = 1 [( ) ] 1+ k2 df ρc 2 fx + k2 ρt T dq (12.32)

147 2 Generic relations, also valid for a real gas 145 Intheabsenceoffriction(df fx = 0)andofheatrelease(dq = 0), thestandardformulation(eq ) is obviously retrieved. Looking at this relation, it is important to note that the coefficients of da, df fx and dq are all strictly positive. However, da appears on the left-hand side, while df fx and dq appear on the right-hand side. In the previous Chapter 11, the influence of a change in flow cross-section A on the flow velocity v has been discussed extensively for isentropic conditions. It has been found that: For a subsonic flow (i.e., M < 1), one obtains obviously that (1 M 2 ) > 0. Under such conditions, a reduction of the flow cross-section A (i.e., da < 0) obviously leads to flow acceleration (i.e., dv > 0). An increase of flow cross-section, da > 0 leads to flow deceleration (dv < 0). Conversely, for a supersonic flow (i.e., M > 1), one obtains (1 M 2 ) < 0. Then, a reduction of the flow cross-section A (i.e., da < 0) now leads to flow deceleration (i.e., dv < 0), while an increase of the flow cross-section, da > 0, results in flow acceleration, dv > 0. This situation has been previously summarized in figure Now, considering the coefficients of da, df fx and dq in Eq.(12.32), a qualitative analogy can be drawn when trying to find the impact of friction (df fx > 0) or heat addition (dq > 0) onto the velocity magnitude v in a compressible flow: it will be the same as the impact of a reduction of the cross-section (da < 0)! Hence, we can state here in a qualitative manner: Friction leads to modifications of the flow velocity similar to a reduction of flow cross-section; Heat addition leads to modifications of the flow velocity similar to a reduction of flow cross-section; Heat extraction leads to modifications of the flow velocity similar to an increase of flow crosssection. These findings are summarized in Figure This figure contains also the associated evolution of pressure, as obtained from the next section. The obtained link between pressure and velocity (a pressure decrease coupled to flow acceleration, respectively a pressure increase for flow deceleration) is absolutely standard. It is too complex trying to consider further analytically the combined influence of a change in flow cross-section, of friction, and of heat exchange with the surroundings. It is only possible to consider one process at a time, neglecting the two further ones. The influence of a modification of the flow cross-section A(x) without friction and heat exchange has been considered in section Now, the influence of heat exchange and of friction will be considered separately, keeping a constant flow cross-section A Pressure variation for a constant flow cross-section Mass conservation (ṁ = constant) for a constant flow cross-section A = constant simply corresponds to The logarithmic differential form of this relation reads: ρv = constant (12.33) dρ ρ + dv v = 0 (12.34) Starting from this equation, it is possible to use Eq.(12.32) to replace the relative velocity variation dv/v, leading to: [( ) ] dρ ρ = 1 1+ k2 df ρc 2 (M 2 fx + k2 1) ρt T dq (12.35)

148 146 Chapter 12. Compressible flows with friction and heat exchange acceleration v deceleration v M < 1 M > 1 p decreases deceleration v or friction, heat addition or heat extraction p increases acceleration v p increases or friction, heat addition or heat extraction p decreases Figure 12.3: Influence of a modification of the cross-section A, of friction and of heat exchange. Considering again Eq.(12.24): combined with Eq.(12.30): it is possible to write: Using Eq.(12.35) to eliminate dρ/ρ leads to: dp = = = [( 1 M 2 1 [( df fx M 2 1 df fx M 2 1 ( dp = c 2 dρ+k 2 ds (12.36) ds = 1 T ( ) dffx ρ +dq (12.37) dp = ρc 2dρ ρ + k2 df fx T ρ + k2 dq (12.38) T ) 1+ k2 1+ k2 ρt 1+ k2 M 2 ρt ] df fx + k2 T dq + k2 ] ρt ρt df fx + k2 T dq ) + ( M 2 1 ) k 2 + dq [ k 2 ρt M 2 1 ) + dq ( k 2 M 2 ) M 2 1 T T + ( M 2 1 ) k 2 T ] (12.39) Both coefficients between parenthesis are strictly positive. Considering separately friction (df fx > 0) or heat addition(dq > 0), itfollows thatthose leadtoapressure decrease inasubsonic flow (since M 2 1 < 0), respectively to a pressure increase in a supersonic flow (since M 2 1 > 0). The opposite applies

149 2 Generic relations, also valid for a real gas 147 when considering heat removal (dq < 0). Therefore, the obtained link between pressure and velocity corresponds to that already presented in figure 12.3 (a pressure decrease coupled to flow acceleration, respectively a pressure increase for flow deceleration): this is absolutely standard! 12.3 Influence of heat exchange in a perfect gas In the absence of friction df = 0 and for a constant flow cross-section A(x) = constant, it is possible to compute quantitatively the influence of heat exchange with the surroundings for a perfect gas. The typical configuration corresponding to the present section is shown in figure A typical application would be a combustion chamber placed between inflow section A 1 and outflow section A 2 (for dq > 0) or a heat exchanger acting through the duct walls between these same sections (for dq < 0 or dq > 0, depending on the temperature difference between central flow and walls). q v 1 v 2 x A 1 A 2 Figure 12.4: Typical configuration considered for heat exchange with the surroundings. It is important to keep in mind that the flow cross-section A is now considered to be constant: A 1 = A 2 = A = constant (12.40) Under such conditions, and in the absence of friction, the conservation equations listed previously can be simplified considerably Mass conservation Again, mass conservation is represented simply by the conservation of the mass flow-rate through any tube cross-section of area A(x): ṁ = ρ(x)a(x)v(x) = constant (12.41) But, since A = constant, this can now be written: ρv = constant (12.42)

150 148 Chapter 12. Compressible flows with friction and heat exchange Comparing the flow conditions when entering and leaving the domain of interest, it comes simply: Conservation of momentum In the absence of friction, it comes first: ρ 1 v 1 = ρ 2 v 2 (12.43) dp+ρvdv = 0 (12.44) But, again due to the fact that A = constant, it is now possible to obtain an integrated form of this differential relation: p+ρv 2 = constant (12.45) This can be easily understood by differentiating again this relation, leading to: Rewriting successively: dp+v 2 dρ+2ρvdv = 0 (12.46) dp+ρvdv+v 2 dρ+ρvdv = 0 dp+ρvdv+v(vdρ+ρdv) = 0 dp+ρvdv +d(ρv) = 0 (12.47) The original result Eq.(12.44) is finally retrieved by using the mass conservation for a constant crosssection, Eq.(12.42), so that d(ρv) = 0. Finally, the integral formulation comparing flow momentum when entering and leaving the domain of interest reads: Energy conservation p 1 +ρ 1 v 2 1 = p 2 +ρ 2 v 2 2 (12.48) To close the system, it is enough to write now the conservation equation for energy. Taking into account the specific heat q (in J/kg) exchanged with the surroundings between inflow (A 1 ) and outflow (A 2 ), it reads: h 1 + v q = h 2 + v 2 2 (12.49) 2 From the exchanged specific heat q, the associated heating power (in W) can be readily computed as: Now, considering a perfect gas, Equation (B.16) reads: so that we can rewrite: or alternatively P = ṁq (12.50) h = c p T (12.51) c p T 1 + v q = c pt 2 + v (12.52) c p T 01 +q = c p T 02 (12.53) by introducing now the stagnation temperature T 0. Remember that the stagnation temperature is a conserved quantity in an isentropic flow; however, the present flow is certainly not isentropic, since heat exchange is taking place with the surroundings. Therefore, the stagnation temperature T 01 before heat exchange is not equal to T 02 (after completion of heat exchange). Indeed, Eq.(12.53) shows that T 02 > T 01 for q > 0, while T 02 < T 01 for q < 0.

151 2 Generic relations, also valid for a real gas Solution procedure Now, how is it possible to compute the outflow conditions (index 2) knowing the inflow conditions (index 1) and the heat q exchanged between A 1 and A 2? The answer is quite simple when starting from the alternative formulation for the mass flow-rate introduced in Chapter 11, Eq.(11.79): ṁ = C p 0A (rt 0 ) 1/2 (12.54) In the present case, this relation can be written before and after the region associated with heat exchange: ṁ 1 = C p 01A 1 (rt 01 ) 1/2 ṁ 2 = C p 02A 2 (rt 02 ) 1/2 Pleasekeepinmindthatthequantitiesappearingontheright-handside(p 0,T 0,A ),whilebeingconstant in an isentropic flow, are now expected to change due to heat exchange and to the corresponding change in specific entropy s! This is true for the stagnation pressure p 0 and stagnation temperature T 0, but also for the critical cross-section A. In the present configuration, the value of A 1 or A 2 is probably only a theoretical critical cross-section, with no real geometrical signification. Anyway: p 01 p 02 T 01 T 02 A 1 A 2 To solve the problem, the energy conservation equation (12.53) is first inverted, knowing the initial conditions (and hence T 01 ) and the exchanged heat q: T 02 = T 01 + q c p (12.55) Using now the fact that the mass flow-rate is not modified by the heat release for such a steady internal flow (ṁ 1 = ṁ 2 ), Equation (12.54) can be used before and after heat exchange to obtain: Simplifying C and r leads to: C p 01A 1 (rt 01 ) 1/2 = C p 02A 2 (rt 02 ) 1/2 (12.56) p 01 A 1 (T 01 ) 1/2 = p 02A 2 (T 02 ) 1/2 (12.57) This can be rewritten as: ( ) 1/2 T02 = p 02 A 2 (12.58) T 01 p 01 A 1 Now, it is enough to modify artificially the right-hand side as: ( ) 1/2 T02 = T 01 ( p1 p 01 p 2 p 1 p 02 p 2 ) ( ) A1 A 2 A 2 A 1 A 1 A 2 (12.59) Since the cross-section is constant, the term A 2 /A 1 on the right-hand side is unity and disappears. The ratios p/p 0 before (index 1) and after heat exchange (index 2) can be expressed by the function π(m) defined in Eq.(11.68) for isentropic compressible flows in Chapter 11: π(m) = p/p 0 (12.60)

152 150 Chapter 12. Compressible flows with friction and heat exchange This relation is still valid in the present case, since it is only a relation between local flow conditions, and is therefore not impacted by global entropy variations between inflow section A 1 and outflow section A 2. For the same reason, the ratios A/A appearing in Eq.(12.59) can be expressed by the function Σ(M) defined in Eq.(11.78) for isentropic compressible flows in Chapter 11: Therefore, Eq.(12.59) can be rewritten as: Σ(M) = A A (12.61) ( T02 T 01 ) 1/2 = p 2 p 1 π 1 π 2 Σ 1 Σ 2 (12.62) The pressure ratio, first term on the right-hand side, is the only one that still should be replaced in order to obtain the final solution. For this purpose, the suitable starting point is the conservation equation for momentum, Eq.(12.64). It can be rewritten successively: ( ) p+ρv 2 = p 1+ ρv2 p ( ) = p 1+ γρv2 γp ( ) = p 1+γ v2 γp/ρ ( ) = p 1+γ v2 c 2 = p ( 1+γM 2) (12.63) Hence, conservation of momentum can be rewritten for a configuration without friction and for a constant cross-section A as: p ( 1+γM 2) = constant (12.64) Applying this relation to both sides of Eq.(12.48), one obtains: p 1 ( 1+γM 2 1 ) = p2 ( 1+γM 2 2 ) (12.65) Finally, the pressure ratio between inflow and outflow can be rewritten as: p 2 p 1 = 1+γM2 1 1+γM 2 2 (12.66) Replacing in Eq.(12.62), one obtains the final solution: ( ) T 02 1+γM 2 2 = 1 π 1 Σ 1 (12.67) T 01 1+γM2 2 π 2 Σ 2 All the terms appearing on the right-hand side of Eq.(12.67) can be directly computed or read from the Tables in Appendix E or Appendix F (provided that γ = 1.4), as long as the local Mach number is known. Defining as new function: Φ(M) := ( 1+γM 2) π(m)σ(m) (12.68)

153 2 Generic relations, also valid for a real gas 151 Equation (12.67) can be equivalently recast as: Now, the solution procedure is clear: T 02 T 01 = ( ) 2 Φ(M1 ) (12.69) Φ(M 2 ) Knowing the starting conditions before heat exchange (index 1), compute Φ(M 1 ) and T 01 ; Using the energy conservation equation (12.55) and the heat exchanged q, deduce T 02 ; Knowing now the ratio T 02 /T 01 and Φ(M 1 ), deduce Φ(M 2 ) from Eq.(12.69); Invert to obtain M 2 from Φ(M 2 ); Knowing now the Mach number M 2 after heat exchange, all other flow quantities can be directly deduced! Forinstance, thepressure p 2 is obtainedfromthemachnumbers andp 1 using Eq.(12.66). The temperature T 2 is obtained from T 02 and M 2 thanks to Eq.(11.62). And so on Function Phi Mach number Figure 12.5: Evolution of the function Φ with the Mach number M Flow modifications As discussed in section , heat addition leads to the same flow modifications as a reduction in cross-section A, while heat removal leads to the same flow modifications as an increase in cross-section

154 152 Chapter 12. Compressible flows with friction and heat exchange Quantity For M 1 < 1 For M 1 > 1 T 0 ր ր M ր 1 ց 1 v ր ց p ց ր p 0 ց ց T? ր Table 12.1: Evolution of the main flow quantities for heat addition (q > 0) without friction and for a constant cross-section A, including limit values for the Mach number A. Using this analogy and possibly computing all resulting quantities following the explanations of the previous section, it is possible to summarize the corresponding impact for heat addition (q > 0), this being probably the most usual application (Table 12.1). The question mark appearing for the temperature T when starting from subsonic conditions might be surprising. Indeed, it is possible to cool down a compressible flow by heat addition! To understand this point, let us start again from the mass flow-rate: ṁ = ρav (12.70) The density might be replaced by p/(rt) using the ideal gas law, Eq.(B.8). From the definition of the Mach number, Eq.(11.29), it is easy to replace v by (Mc) = (M γrt). Finally, it comes in this manner: ṁ = ( γ r A ) pm T (12.71) Since all terms in the first parenthesis (γ,r,a) are here constant, and since the mass flow-rate is also constant, it ensues that the last ratio on the right hand-side must be constant as well: Using logarithmic differentiation, one obtains: pm T = constant (12.72) dp p + dm M 1 dt 2 T = 0 (12.73) Or alternatively: dt T = 2dp p +2dM M Using again logarithmic differentiation starting from Eq.(12.64), one obtains now: (12.74) Combining the last two relations to eliminate dp/p leads to: dp p + 2γM dm 1+γM 2 = 0 (12.75) dt T = 2 dm 4γM dm M 1+γM 2 2dM ( = 1+γM 2 2γM 2) M (1+γM 2 )

155 2 Generic relations, also valid for a real gas 153 Finally: dt T = 2dM 1 γm 2 (12.76) M 1+γM 2 This explains the question mark in Table Indeed, heat addition (q > 0) in a subsonic flow (M 1 < 1, so that dm > 0) leads to a temperature increase (dt > 0) as long as M 1 < 1/ γ, which is probably intuitive. However, the same heat addition just before reaching sonic conditions (for 1/ γ M 1 1) now leads to cooling: T 2 < T 1! Physically, this is due to the fact that both contributions in the energy conservation equation (12.52), interact with each other. While it is directly clear that T 0, a measure for the total energy, will increase in case of heat addition (Eq ), this is not necessarily so for T, which is only a measure of the internal energy of the fluid Thermal choking Let us close this section by a last but important qualitative remark. Looking again at Eq.(12.69), it is clear that heat addition(q > 0) will always leadtoadecrease ofthefunction Φ(M) i.e., Φ(M 2 ) < Φ(M 1 ), since T 02 > T 01. Looking now at figure 12.5, this condition implies - as expected from the qualitative process described in section , also documented in Table that the Mach number will increase (M 2 > M 1 ) due to heat addition in a subsonic flow, while it will decrease (M 2 < M 1 ) for heat addition in a supersonic flow. Now, Figure 12.5 clearly shows that the function Φ(M) reaches a minimum for M = 1. At that point, it is impossible to find any flow evolution allowing a further decrease of Φ. This is the explanation for the problem called thermal choking. This process corresponds to excessive heat addition, usually in a subsonic flow. Under such conditions, the Mach number will progressively increase. If more heat is added, it will increase further, and so on. But, when reaching the limit value M = 1, it is absolutely impossible to add more heat into this flow. When nevertheless attempting to do so, the flow will abruptly react by... stopping completely! Remember that heat addition is qualitatively similar to a reduction of the flow cross-section, A. Thermal choking corresponds in this analogy to a duct being completely closed. Clearly, thermal choking will have extremely severe consequences (flame extinction due to interruption of the air flow, very large forces acting suddenly on the system...) and must be avoided by all means. In modern systems, the electronic regulation will prevent the user from adding additional heat when coming too close to the sonic conditions Influence of friction After this extensive analysis of heat exchange, let us now consider the alternative process, associated to friction. In the absence of heat exchange (dq = 0) and for a constant flow cross-section (A(x) = A = constant), it is indeed possible to compute quantitatively the influence of friction. We will nevertheless start with a qualitative analysis. The typical configuration corresponding to the present section is shown in figure Typical applications involve pressurized gas lines with a noticeable length, l 1 m. For a qualitative analysis, the generalized Hugoniot equation (12.32) can now be simplified by taking into account the hypotheses listed above as: ( 1 M 2 ) dv v = 1 ρc 2 ( 1+ k2 ρt ) df fx (12.77) This equation reveals, that in the absence of friction (df fx = 0), the flow velocity will remain constant (dv = 0), as expected for such conditions (no change of flow cross-section, no heat exchange, no friction). Introducing now the impact of friction in the analysis (hence df fx > 0), a subsonic condition (M 1 < 1 at the beginning of the analysis, and hence (1 M 2 1) > 0) leads to dv > 0. In other words, taking friction

156 154 Chapter 12. Compressible flows with friction and heat exchange v 1 v 2 x Friction force df f A 1 A 2 Figure 12.6: Typical configuration considered when taking into account friction. into account in a subsonic flow leads to flow acceleration, which appears to be clearly counter-intuitive: systems with friction are usually slower, not faster, than comparable friction-free conditions! Let us explain this apparent contradiction. For this purpose, the conservation equations will be first specified for the conditions considered presently Mass conservation As discussed previously when considering heat exchange, mass conservation (ṁ = constant) for a constant flow cross-section A = constant simply corresponds to ρv = constant (12.78) The logarithmic differential form of this relation will be even more useful: dρ ρ + dv v = 0 (12.79) Comparing the flow conditions when entering and leaving the domain of interest, it comes simply: ρ 1 v 1 = ρ 2 v 2 (12.80) Conservation of momentum The general equation (12.12) derived at the beginning of this chapter stays unchanged. dp+ρvdv+df fx = 0 (12.81)

157 2 Generic relations, also valid for a real gas Energy conservation Here, the solution for q = 0 simply reads: h+ v2 2 = constant (12.82) In differential form, one obtains: dh+vdv = 0 (12.83) Comparing again the flow conditions when entering and leaving the domain of interest, it comes simply: h 1 + v = h 2 + v (12.84) Qualitative analysis Starting from Eq.(12.39), it is now possible to suppress the contribution of heat, since dq = 0. It comes: dp = df ( fx 1+ k2 M 2 ) M 2 1 ρt (12.85) From this relation, it is clear that the observed acceleration (dv > 0) for a subsonic flow (M < 1) involving friction (df fx > 0) is indeed simply a consequence of pressure loss, dp < 0, and therefore nothing surprising. In the same manner, a supersonic flow (M > 1) with friction (df fx > 0) will lead to a deceleration (dv < 0) coupled to a pressure increase (dp > 0). Such combinations are absolutely standard: we have found exactly the same for incompressible flows when analyzing the Bernoulli equation in Chapter 6! Quantitative solution procedure Pursuing now the aim of a quantitative analysis, the conversation equation for momentum (Eq ) divided by the density is a perfect starting point: dp ρ +vdv+ df fx ρ = 0 (12.86) The last term on the left-hand side describes the friction loss. A suitable description for this friction loss has been already presented in Chapter 6 (see Eq.6.38 when deriving the generalized Bernoulli equation: the law of Darcy-Weisbach): df fx ρ = de f = f dx d v 2 2 (12.87) In this equation, the only difference with the original Eq.(6.38) is the differential flow length dx appearing instead of the integral flow length l, since the present equation is in differential form. Using now this model, the conservation equation for momentum reads: with f the friction factor. dp ρ +vdv +fdx d v 2 2 = 0 (12.88)

158 156 Chapter 12. Compressible flows with friction and heat exchange A logarithmic differentiation of the ideal gas law, Eq.(B.8), leads to: since r = constant. Hence: dp p = dρ ρ + dt T (12.89) dp = p dρ ρ + p dt (12.90) T Since, from Eq.(12.79): it comes: Hence: dρ ρ = dv v (12.91) dp = p dv v + p dt (12.92) T dp ρ = p dv ρ v + p dt (12.93) ρt Using again twice the ideal gas law, Eq.(B.8), one obtains finally for the right-hand side: dp ρ = rtdv +rdt (12.94) v This relation is now used to replace the first term in Eq.(12.88), leading to: rt dv v +rdt +vdv+fdx d v 2 2 = 0 (12.95) or equivalently: ( v 2 rt ) dv v +rdt +fdx v 2 d 2 = 0 (12.96) Now, the energy equation (12.83) can be transformed for a perfect gas into: vdv = dh = c p dt (12.97) It is then possible to write successively, using Eq.(B.6): dt = v c p dv rdt = r c p vdv = r γr vdv γ 1 = γ 1 γ vdv = γ 1 γ v2dv v This relation is now used to replace the second term in Eq.(12.96), leading to: (12.98) ( v 2 rt ) dv v γ 1 γ v2dv v +fdx d v 2 2 = 0 (12.99)

159 2 Generic relations, also valid for a real gas 157 By combining the first two terms, one obtains: ( v 2 γ rt ) dv v +fdx d Now, let us introduce again the critical Mach number: Inverting this relation: v 2 2 = 0 (12.100) M = v c (12.101) v = c M (12.102) Since the present flow is adiabatic, the stagnation temperature T 0 and as a consequence the critical speed of sound c are constant (see again Eq.11.37). A logarithmic differentiation of Eq.(12.102) leads therefore to: dv v = dm (12.103) M It is now useful to introduce simultaneously Eqs.(12.102) and (12.103) into Eq.(12.100) in order to replace everywhere the flow velocity v (respectively the ratio dv/v): ( c 2 M 2 γ rt ) dm M +f dx d c 2 M2 2 = 0 (12.104) Multiplying now this relation by γ/c 2, it comes: ( M 2 γrt c 2 ) dm M +f dx d γm 2 2 = 0 (12.105) Considering Eq.(11.37): c = the second term in the parenthesis of Eq.(12.105) reads: 2 γ γ +1 rt 0 (12.106) γrt c 2 = γrt γ +1 2γrT 0 = γ +1 T (12.107) 2 T 0 It is then possible to rewrite Eq.(12.105): ( M 2 γ +1 2 T T 0 ) dm M +f dx d γm 2 2 = 0 (12.108) Reminding now Eq.(11.92): Equation (12.108) can be modified into: ( M 2 γ +1 2 T T 0 = 1 γ 1 γ +1 M 2 (12.109) + γ +1 ) γ 1 2 γ +1 M 2 dm +f dx M d γm 2 2 = 0 (12.110)

160 158 Chapter 12. Compressible flows with friction and heat exchange which simplifies successively into: ( γ +1 2 M 2 γ +1 2 γ +1 2 ) dm +f dx M d ( 2 M 1 ) dm +f dx M d γm 2 2 γm 2 2 = 0 = 0 (12.111) Dividing finally by γ +1 M 2 2 : ( 2 M 1 ) dm γ dx +f M 3 γ +1 d = 0 (12.112) In preparation for integration between inflow cross-section (index 1) and outflow cross-section (index 2), this relation is expanded into: dm dm [ ] γ f + dx = 0 (12.113) M M 3 γ +1d Noticing that the last term between square brackets is constant, integration in space leads to: 2 1 dm M 2 1 dm M 3 + [ γ f γ +1d ] 2 dx = constant (12.114) 1 The integration of all three terms on the left-hand side is straightforward and leads to: [lnm ] [ M 2 ] γ γ +1 f l = constant (12.115) d introducing in the last term l := x 2 x 1, the total length of the system between inflow cross-section (index 1) and outflow cross-section (index 2). One therefore obtains: lnm 2 lnm ( 1 1 ) + γ 2 M 2 2 M 1 2 γ +1 f l = constant (12.116) d Whenconsidering aninfinitelyshorttube(sothatl 0), onewouldobviouslyobtainaswell M 2 M 1 and the left-hand side of this equation would naturally vanish. Therefore, the only possible value for the constant appearing on the right-hand side is 0. The final solution allowing to compute the influence of friction in a compressible flow without heat exchange and for a constant cross-section therefore reads: ln ( ) M 2 M Flow modifications ( 1 M ) + γ M 1 2 γ +1 f l d = 0 (12.117) As discussed in section , friction leads qualitatively to the same flow modifications as a reduction in cross-section A. However, the flow remains adiabatic. Using this analogy and possibly computing all resulting quantities following the explanations of the previous section, it is possible to summarize the corresponding impact of friction (df fx > 0), this being the most usual application (Table 12.2). Considering that there is again an extremal value for sonic conditions, flow choking can be obtained as a consequence of friction, similarly to the thermal choking discussed previously Conclusions Now, one-dimensional compressible flows can be solved accurately even with friction or heat exchange. Such flows, when in the supersonic regime, will be often strongly modified due to the occurrence of shock waves: this will be the subject of the next chapter.

161 2 Generic relations, also valid for a real gas 159 Quantity For M 1 < 1 For M 1 > 1 T 0 = = M ր 1 ց 1 v ր ց p ց ր p 0 ց ց T ց ր Table 12.2: Evolution of the main flow quantities for friction without heat exchange and for a constant cross-section A, including limit values for the Mach number

162 160 Chapter 12. Compressible flows with friction and heat exchange

163 Chapter 13 Shock waves 13.1 Introduction In many practical supersonic flows, discontinuities are observed. This means that the parameters describing the flow will vary strongly and over very short distances, typically of the order of magnitude of the mean free path. Under such conditions, the hypotheses underlying the continuum assumption discussed in Section 1.3 are not valid any more. It is thus impossible to use the conservation equations developed previously to investigate the inner structure of such discontinuities. However, it is still possible to relate the flow conditions encountered in front of and behind such discontinuities, as will be done in what follows. This is in practice sufficient to solve most relevant problems. If the solution within the discontinuity is of interest, a specific description based on Statistical Physics must be used for this purpose. Figure 13.1: Example of shock waves in practical flows: attached to a plane breaking the sound barrier (left, photo from RC Clubs); behind an engine of the Space Shuttle at start (center, photo from NASA); when firing a bullet (right, photo from Subtopia). The only discontinuities we will consider in this document are shock waves (Figure 13.1). Two different kinds of shock waves will be solved for. In a first step, the normal shock wave, perpendicular to the main flow direction, will be described in detail. The obtained results will then be extended towards the oblique shock wave, inclined towards the main flow direction. Note that only straight shock waves will be accepted. In practice, curved shock waves can also be found, but this is a less relevant configuration that can hardly be solved by analytical means. 161

164 162 Chapter 13. Shock waves 13.2 Normal shock wave Considered configuration and hypotheses The generic configuration considered for the analysis of a normal shock wave is depicted in figure v 1 v 2 p 1 1 p 2 2 T 1 T 2 Figure 13.2: Configuration employed for the analysis of a fixed, normal shock wave (shown in red), together with the control volume (in green) used to derive the conservation equations. The hypotheses underlying the analysis are as follows, and correspond indeed quite well to the physical reality: The shock wave is a discontinuity without inner structure, separating the upstream flow (with conditions marked by an index 1) from the downstream flow (with conditions marked by an index 2); The shock wave is a straight structure, normal to the flow direction; hence, all streamlines are parallel to each other and perpendicular (or normal) to the shock wave; The configuration is steady Conservation equations for a real gas We begin by considering generic gas properties. This means that the relations derived in this section are valid even for complex thermodynamic relations, as might be found for a real gas. Taking into account the hypotheses listed above, it is possible to establish generic conservation equations.

165 2 Normal shock wave 163 Mass conservation For the steady flow bounded by the streamtube shown in green in figure 13.2 the mass conservation is represented simply by the conservation of the mass flow-rate through any tube cross-section of area A(x): ṁ = ρ(x)a(x)v(x) = constant (13.1) Here, all streamlines are parallel to each other and horizontal, so that the cross-section of the streamtube is constant, A(x) = A 1 = A 2 = constant. Hence, the mass conservation relating upstream and downstream conditions simply reads: Conservation of momentum ρ 1 v 1 = ρ 2 v 2 (13.2) The developments in this section follow closely those presented when considering heat exchange in a compressible flow, see Section Friction plays obviously no role for the configuration considered in figure 13.2, since there is no velocity gradient upstream and downstream of the discontinuity (homogeneous flow conditions). In the absence of friction, conservation of momentum in a compressible flow reads as usual in differential form (see also Eqs. (11.16) and (12.44)): dp+ρvdv = 0 (13.3) But, again due to the fact that A = constant, it is now possible to obtain an integrated form of this differential relation, as demonstrated in Section : p+ρv 2 = constant (13.4) Finally, conservation of momentum through the normal shock wave simply reads: Energy conservation p 1 +ρ 1 v 2 1 = p 2 +ρ 2 v 2 2 (13.5) The conservation of energy is again simply derived from the First Law of Thermodynamics. For such an isolated adiabatic system, the total energy is constant, which can be written when neglecting as usual the contribution of potential energy for this gaseous flow: h 1 + v2 1 2 = h 2 + v2 2 2 (13.6) Conservation equations for a perfect gas The previous relations are only slightly modified when considering the special case of a perfect gas, as done in the rest of this section. Using Eq.(11.42), one obtains for: Mass conservation (unchanged): ρ 1 v 1 = ρ 2 v 2 (13.7) Conservation of momentum (unchanged): p 1 +ρ 1 v 2 1 = p 2 +ρ 2 v 2 2 (13.8)

166 164 Chapter 13. Shock waves Energy conservation: c p T 1 + v2 1 2 = c pt 2 + v2 2 2 (13.9) which might also be rewritten using the stagnation temperature T 0 that takes into account both the internal energy and the kinetic energy of the fluid: c p T 01 = c p T 02 (13.10) Sincec p = constant,itisthusagainclearthat,inthisadiabaticflow, thestagnation temperature T 0 is the same upstream and downstream of the shock wave. This simple observation will be very useful to solve the problem: T 01 = T 02 (13.11) Jump relations involving M 1 and M 2 Using Eq.(11.62) to relate temperature T to stagnation temperature T 0, once upstream of the shock (link between T 1 and T 01 ) and once downstream of the shock (link between T 2 and T 02 ), it comes: T 01 T 1 T 02 T 2 = 1+ γ 1 M1 2 2 (13.12) = 1+ γ 1 M2 2 2 (13.13) Dividing now the first equation by the second one and keeping in mind that T 01 = T 02, one obtains simply the relation quantifying the temperature jump through the normal shock: γ 1 T = M2 1 T 1 1+ γ 1 2 M2 2 (13.14) Reminding now Eq.(12.64), an alternative form of the conservation of momentum (Eq. 13.5) in the absence of friction and for a constant cross-section A, both hypotheses being valid as well in the present case, it is possible to relate upstream and downstream conditions by: p 1 ( 1+γM 2 1 ) = p2 ( 1+γM 2 2 ) Hence, the relation quantifying the pressure jump through the normal shock is simply: (13.15) p 2 p 1 = 1+γM2 1 1+γM 2 2 (13.16) Using now the ideal gas relation (Eq ), one obtains directly for the density jump through the normal shock: p ρ 2 2 rt = 2 p ρ 1 = p 2T 1 (13.17) 1 rt 1 p 1 T 2 which is sufficient knowing Eqs.(13.14) and (13.16).

167 2 Normal shock wave Relation between the Mach numbers upstream and downstream of the normal shock The relations written in the previous section still involve both Mach numbers M 1 (before the normal shock) and M 2 (after the normal shock). However, in most cases, only the conditions upstream of the shock are known. To close the problem, it is therefore necessary to solve for M 2 knowing the conditions before the normal shock (index 1). For this purpose, we start again from the mass conservation, Eq.(13.2): ρ 1 v 1 = ρ 2 v 2 (13.18) We then rewrite, using the obvious relation v = Mc: ρ 2 ρ 1 = v 1 v 2 = M 1c 1 M 2 c 2 (13.19) On the left-hand side, the density jump is replaced using Eq.(13.17). On the right-hand side, the speed of sound c is replaced using Eq.(11.51), obtaining finally: p 2 T 1 = M 1 γrt1 (13.20) p 1 T 2 M 2 γrt2 Simplifying γ and r and lumping together the temperature jumps on the left and right side of the equation, it comes: p 2 = M 1 T2 (13.21) p 1 M 2 T 1 Replacing the pressure jump and temperature jump appearing in this equation using Eqs.(13.16) and (13.14), one obtains finally: 1+γM1 2 = M 1 γ M2 1 (13.22) 1+γM2 2 M 2 1+ γ 1 2 M2 2 Taking the whole equation to the square and rewriting it using M 1 as a known value and M 2 as the unknown, onerecognizesfinallyasimple quadratic polynomial equation involving M2 2 asunknown value, in the form: am2 4 +bm2 2 +c = 0 (13.23) It is easy to show that this equation delivers two possible solutions for M2 2 and, reminding that both M 1 and M 2 must be positive values, the two solutions correspond simply to: 1. The case with no shock at all, M 2 = M 1! Reporting in all jump relations derived in the previous section, this leads of course to T 2 = T 1, p 2 = p 1, ρ 2 = ρ 1...This should not be a surprise: the shock wave represented in figure 13.2 cannot be forced to be there. In other words, the case without a shock wave is also a physically possible solution. In that case, the flow remains obviously unchanged. We will not consider further this case in the rest of this chapter. 2. The really interesting case, i.e., that with a normal shock wave, corresponds to the second solution of the quadratic equation, which reads: M 2 = 2+(γ 1)M2 1 2γM γ (13.24)

168 166 Chapter 13. Shock waves Figure 13.3 shows the relation between M 1 and M 2 for a supersonic upstream condition (M 1 1), since this is the only relevant case, as will be shown later. In this figure, it is clearly observed that, for M 1 1, one obtains M 2 1, so that the normal shock wave separates a supersonic region (upstream) from a subsonic region (downstream of the normal shock). When increasing M 1, the value of M 2, the Mach number downstream of the normal shock, decreases monotonically. Additionally, as directly visible from Eq.(13.24), there is a minimum value for M 2 at very high incoming Mach number M 1 : M 2 lim = lim M 1 M 2 = γ 1 2γ (13.25) For γ = 1.4, this minimum value corresponds to M 2 lim = Mach number M 2 behind shock Mach number in front of shock wave M 1 Figure 13.3: Relation between the Mach numbers upstream and downstream of the normal shock Jump relations involving only the upstream Mach number M 1 Using now Eq.(13.24) in order to eliminate the value of M 2 in all the jump relations presented in Section and simplifying the obtained equation, it is slightly tedious but very straightforward to obtain all the jump relations as a direct function of the Mach number M 1 upstream of the normal shock only. It

169 2 Normal shock wave 167 comes successively, first combining Eq.(13.24) and Eq.(13.16): p 2 = 2γM2 1 γ +1 p 1 γ +1 (13.26) The corresponding evolution of the pressure jump is shown in figure 13.4 as a function of the upstream Mach number M 1 for M 1 1. Very large values are found for p 2 /p 1 when increasing M 1, showing that a normal shock wave typically leads to a very strong increase in pressure. The value of p 2 tends toward infinity when M 1. This, together with the strong deceleration discussed previously in connection with figure 13.3 and quantified by Eq.(13.24), is typically the most noticeable effect of a normal shock wave. Obviously, it can also be a very dangerous effect if this shock wave is associated to an explosion, as discussed later! Pressure jump p 2 /p Mach number in front of shock wave M 1 Figure 13.4: Pressure jump through the normal shock wave as a function of the upstream Mach number M 1. The dashed blue line denotes the level p 2 /p 1 = 1. Combining now Eq.(13.24) and Eq.(13.14), it comes: ( T 2 2γ = T 1 γ +1 M2 1 γ 1 )( ) γ 1 γ +1 γ (γ +1)M1 2 (13.27) The corresponding evolution of the temperature jump is shown in figure 13.5 as a function of the upstream Mach number M 1 for M 1 1. Relatively large values are found for T 2 /T 1 when increasing

170 168 Chapter 13. Shock waves M 1, showing that a normal shock wave typically leads to a large increase in temperature. The value of T 2 tends toward infinity when M 1. 6 Temperature jump T 2 /T Mach number in front of shock wave M 1 Figure 13.5: Temperature jump through the normal shock wave as a function of the upstream Mach number M 1. Following Eq.(13.17), it is now sufficient to divide Eq.(13.26) by Eq.(13.27) to obtain the density jump, leading to: ρ 2 = (γ +1)M2 1 (13.28) ρ 1 2+(γ 1)M1 2 The corresponding evolution of the density jump is shown in figure 13.6 as a function of the upstream Mach number M 1 for M 1 1. Relatively large values are found for ρ 2 /ρ 1 when increasing M 1, showing that a normal shock wave typically leads to a large increase in density. Using mass conservation, Eq.(13.2), it follows immediately that: so that the velocity jump is given by: v 1 v 2 = ρ 2 ρ 1 (13.29) v 2 v 1 = 2+(γ 1)M2 1 (γ +1)M 2 1 (13.30)

171 2 Normal shock wave Density jump ρ 2 /ρ Mach number in front of shock wave M 1 Figure 13.6: Density jump through the normal shock wave as a function of the upstream Mach number M 1. The obtained curve illustrating the deceleration observed in a normal shock wave (v 2 v 1 ) behaves in opposite manner to that shown in figure 13.6, and is shown in figure Looking at the last two equations, it is also obvious that density and velocity jump converge to a limit value when increasing M 1 : ρ 2 v 1 lim = lim = γ +1 (13.31) M 1 ρ 1 M 1 v 2 γ 1 For γ = 1.4, one obtains therefore lim M1 ρ 2 ρ 1 = 6 and lim M1 v 2 v 1 = 1/6 = The change in stagnation pressure through the normal shock wave can be directly computed by writing: p 02 = p 02p 2 p 1 (13.32) p 01 p 2 p 1 p 01 On the right-hand side of this equation, the first and third ratios correspond to the local relation between pressure and stagnation pressure, discussed extensively in Chapter 11 and defined by Eq.(11.67) as the inverse of function π, see Eq.(11.68). The second ratio on the right-hand side of the previous equation is the pressure jump through the normal shock wave, already given by Eq.(13.26). Hence, it comes: [ ( p 02 = 1+ γ 1 ) γ [ γ 1] 2γM M2 2 2 ( 1 γ γ 1 ) γ ] M 2 γ 1 1 (13.33) p 01 2 γ +1 2

172 170 Chapter 13. Shock waves 1 Velocity jump v 2 /v Mach number in front of shock wave M 1 Figure 13.7: Velocity jump through the normal shock wave as a function of the upstream Mach number M 1. This can be first rewritten as: p 02 p 01 = ( 1+ γ 1 2 M γ 1 2 M2 2 ) γ ( ) γ 1 γ +1 2γM1 2 γ +1 1 γ 1 (13.34) Recognizing now in the first parenthesis the temperature jump T 2 /T 1, see Eq.(13.14), it is possible to reformulate this parenthesis using Eq.(13.27). Simplifying all similar terms with each other, one obtains finally: [( p 02 2γ = p 01 γ +1 M2 1 γ 1 )( ) γ ] 1 γ 1 γ +1 γ +1 + γ 1 2 (13.35) (γ +1)M1 2 The corresponding evolution of the jump in stagnation pressure through the normal shock wave is shown in figure 13.8 as a function of the upstream Mach number M 1 for M 1 1. The stagnation pressure is lower behind the shock, showing that the normal shock is a dissipative structure. Now, onelast question remains: why have we decided toplot figures 13.4to 13.8only forasupersonic upstream condition, M 1 1? This will be justified in the next section.

173 2 Normal shock wave 171 Jump of total pressure p 02 /p Mach number in front of shock wave M 1 Figure 13.8: Jump in stagnation pressure through the normal shock wave as a function of the upstream Mach number M Necessary condition on M 1 for the existence of a shock As a last relation between upstream and downstream conditions, it is very interesting to look at the specific entropy of a perfect gas, given by Eq.(13.36): ( ) p s = c v ln + constant (13.36) ρ γ One therefore obtains immediately: s 2 s 1 c v = ln ( [ ] γ ) p2 ρ1 p 1 ρ 2 (13.37) Replacing the pressure jump using Eq.(13.26) and the density jump using Eq.(13.28), one obtains: {( )( ) s 2 s 1 2γM 2 = ln 1 γ +1 (γ +1)M 2 γ } 1 (13.38) c v γ +1 2+(γ 1)M1 2 The corresponding evolution of the change in specific entropy through the normal shock wave is shown in figure 13.9 as a function of the upstream Mach number M 1. For the first time, the full range

174 172 Chapter 13. Shock waves of values for M 1 is considered, also for subsonic upstream conditions! As can be seen, cases with M 1 > 1 correspond to an increase in entropy, s 2 > s 1, which is fully OK. However, cases with a subsonic upstream condition, M 1 < 1 would lead to a decrease in entropy, which would directly violate the Second Law of Thermodynamics for the isolated system we are considering. Therefore, it is absolutely impossible to observe a steady normal shock wave in a subsonic flow. This explains why we have only considered the range M 1 for all figures 13.4 to 13.8: this is the only valid range for a fixed, steady normal shock! 2 Entropy difference (s 2 s 1 )/c v Mach number in front of shock wave M 1 Figure 13.9: Change in specific entropy through a normal shock wave as a function of the upstream Mach number M 1. Beware! This does not mean that a shock cannot exist in a subsonic flow, or even under quiescent conditions! But, in that case, the shock cannot remain steady and must propagate, so that the upstream velocity relative to the propagating shock is indeed in the supersonic range. This issue will be discussed later in Section Shock relation of Prandtl We start by recalling that the stagnation temperature is the same upstream and downstream of the shock wave, T 01 = T 02. Now, considering Eq.(11.37) demonstrated in Section , it is clear that the critical speed of sound is also unchanged when crossing the normal shock wave, c 2 = c 1. As a

175 2 Normal shock wave 173 Quantity Evolution T 0 T 01 = T 02 M (from M 1 1) ց (M 2 1) M (M 1 1) ց [M 2 = (1/M 1 )] 1 v v 1 ց v 2 p p 1 ր p 2 p 0 p 01 ց p 02 T T 1 ր T 2 s s 1 ր s 2 Table 13.1: Evolution of the main flow quantities through a normal shock wave consequence, it is possible to write using the critical Mach number defined in Eq.(11.34): v 2 v 1 = v 2/c 2 v 1 /c 1 = M 2 M 1 (13.39) At the same time, comparing Eq.(13.30) with Eq.(11.90) demonstrated at the end of Chapter 11, it is immediately clear that v 2 = 1 (13.40) v 1 M 1 2 Hence, it comes by equating the last two relations: 1 M 2 1 = M 2 M 1 (13.41) leading directly to the useful shock relation of Prandtl for normal shock waves: M 1 M 2 = 1 (13.42) Summary: evolution of all quantities through a normal shock All observations described in the previous sections are summarized in Table Normal shock tables In order to facilitate later computations, the most important relations have been computed and the results are tabulated in Appendix F (Appendix E is obviously not relevant, since a normal shock wave can only exist for the upstream condition M ). Remember that all values listed in Appendix F are only valid for γ = 1.4. If this condition does not hold, it is necessary to use the equations listed previously in this chapter. Now, for γ = 1.4, and considering in the first column of this Table M = M 1, the Mach number upstream of the steady normal shock wave: theeighthcolumngivesthemach number downstreamoftheshockwave, M 2, followingeq.(13.24); the ninth column gives the pressure jump, Eq.(13.26); the tenth column gives the density jump, Eq.(13.28); the eleventh column gives the temperature jump, Eq.(13.27); the twelfth column gives the jump in stagnation pressure, Eq.(13.35).

176 174 Chapter 13. Shock waves Solution and graphical representation using the critical Mach number Since there is a direct connection between the Mach number and the critical Mach number (Eqs and 11.91), as shown in Section (see Chapter 11 for further details), all the relations established previously as function of the Mach number can be simply rewritten as function of the critical Mach number. The shock relation of Prandtl (Eq ) can also be used for this purpose. Finally, one obtains an alternative set of solution equations describing pressure jump, temperature jump, density jump..., as function of the critical Mach number M 1 upstream of the normal shock. Since a graphical representation cannot be overloaded, only the pressure jump and the jump of stagnation pressure have been represented as function of critical Mach number M (meaning here M 1 ) in figure Relation of Rankine-Hugoniot It is sometimes useful to analyze a normal shock wave without direct reference to the Mach numbers. For this purpose, the Rankine-Hugoniot relation makes a direct connection between the pressure jump p 2 /p 1 and the inverse of the density jump ρ 1 /ρ 2 (which happens to be equal to the velocity jump v 2 /v 1 due to mass conservation, as discussed previously). In order to derive this relation, we start back from Eq.(13.26): p 2 = 2γM2 1 γ +1 (13.43) p 1 γ +1 We first invert this relation to obtain the upstream Mach number M 1 as a function of the pressures p 1 and p 2, leading to: This relation is used to replace M 2 1 M 2 1 = (γ +1)p 2 +(γ 1)p 1 2γp 1 (13.44) in the density jump, Eq.(13.28): ρ 2 ρ 1 = (γ +1)M2 1 2+(γ 1)M 2 1 (13.45) After simplification, one obtains a relation where M 1 does not appear any more: ρ 2 ρ 1 = (γ +1)p2 p 1 +γ 1 (γ 1) p 2 p 1 +γ +1 (13.46) We finally invert this relation to obtain the pressure jump p 2 /p 1 as function of ρ 1 /ρ 2. This delivers the Rankine-Hugoniot relation, valid for any normal shock wave: p 2 p 1 = γ +1 γ 1 ρ 1 ρ 2 γ +1ρ 1 1 γ 1ρ 2 (13.47) The corresponding relation is displayed graphically as a red curve in figure It is an hyperbolic curve. Any normal shock (for γ = 1.4) will correspond to a single point along this red curve. The evolution corresponding to an isentropic process is shown as a black dashed line in the same figure. Obviously, while the Rankine-Hugoniot relation only exists for p 2 p 1 and ρ 2 ρ 1 (as found in a normal shock), the isentropic evolution covers the whole set of density ratio ρ 1 /ρ 2, since it can describe a compression as well as an expansion process. Confirming the results presented in figure 13.9, it appears

177 2 Normal shock wave 175 pressure ratio p 2 /p Rankine Hugoniot relation Isentropic relation Rayleigh line (M 1 =4) inverse density ratio ρ 1 /ρ 2 Figure 13.10: Relation of Rankine-Hugoniot for γ = 1.4, valid for any normal shock (red line), compared with an Isentropic evolution(dashed black line). Additionally, the Rayleigh line for a specific shock(here, M 1 = 4) is shown as a straight blue line. again that a weak shock (i.e., a shock for M 1 1, or in other words p 2 p 1 is nearly isentropic: the red and black curves in cannot be distinguished in this region, left of the point (1,1). However, this is not true any more for a strong shock (M 1 1, or p 2 p 1 )! In this part of the diagram, the two curves lie far from each other, and the pressure increase is much more considerable in the normal shock for the same density variation Rayleigh line As explained in the previous section, the Rankine-Hugoniot relation is valid for all normal shock waves. An additional condition is needed to obtain the conditions found in a specific shock, for a given value of M 1. For this purpose, the Rayleigh line will be plotted in the same graph. In figure 13.10, this has been done exemplarily for the case M 1 = 4. The Rayleigh line can be simply obtained based on momentum conservation, Eq.(13.5): p 1 +ρ 1 v 2 1 = p 2 +ρ 2 v 2 2 (13.48)

178 176 Chapter 13. Shock waves This can be rewritten: p 2 p 1 = ρ 1 v1 2 ρ 2 v2 2 (13.49) ( = ρ 1 v1 2 1 ρ ) 2v 2 v 2 (13.50) ρ 1 v 1 v 1 Considering mass conservation, Eq.(13.2), the ratio ρ 2v 2 ρ 1 v 1 within the parenthesis is equal to unity and therefore disappears. For the same reason, the last ratio in the parenthesis, v 2 /v 1, can be immediately replaced by ρ 1 /ρ 2, finally leading to: p 2 p 1 = ρ 1 v 2 1 ( 1 ρ ) 1 ρ 2 (13.51) Dividing the whole relation by p 1, one obtains: p 2 1 = ρ 1v 2 ( 1 1 ρ ) 1 p 1 p 1 ρ 2 = γ ρ 1 v 2 ( 1 1 ρ ) 1 γ p 1 ρ 2 (13.52) (13.53) Now, ontheright-handside, thecombinationγp 1 /ρ 1 appears, whichisequaltoc 2 1 considering Eq.(11.51): ( p 2 1 = γv2 1 1 ρ ) 1 p 1 c 2 1 ρ 2 (13.54) Replacing on the right-hand side the ratio v 1 over c 1 by the Mach number M 1 considering Eq.(11.29), the equation of the Rayleigh line is finally obtained: ( p 2 1 = γm1 2 1 ρ ) 1 p 1 ρ 2 (13.55) The point (1,1) obviously always belongs to the Rayleigh line. Since the equation of this straight line still contains the Mach number M 1, it is only valid for a specific normal shock, that occurring for an upstream Mach number equal to M 1. Plotting the Rayleigh line in the same graph as the Rankine- Hugoniot relation, as done exemplarily in figure for M 1 = 4 (straight blue line), the intersection between the blue line (Rayleigh) and the red curve (Rankine-Hugoniot) delivers the pressure and density jumps really found for this specific normal shock. Looking at figure 13.10, one reads p 2 /p 1 18 and ρ 1 /ρ 2 0.2, which corresponds well to the true values that can be read in Appendix F for the same conditions: p 2 /p 1 = 18.5 and ρ 1 /ρ 2 = The graphical representation shown in figure can therefore be used to obtain graphically pressure and density jump through a normal shock Propagating shock waves It must be kept in mind that all the results obtained up to now are only valid for a fixed normal shock wave, without any movement of its own. While this situation can indeed be found, for instance in a supersonic wind tunnel, it does not correspond to the case of an explosion, for instance, in which a shock wave will be propagating and expanding into a quiescent atmosphere, as seen in figure This is the typical application considered in the present section. As you will soon understand, it is fortunately possible to use all the relations derived up to now to understand and quantify a propagating shock wave. For this purpose, it is sufficient to introduce a

179 2 Normal shock wave 177 Figure 13.11: Shock wave during the first test of a nuclear bomb in the USA in July 1945 (photo from The Atlantic). change in reference frame: when the observer is moving together with the shock wave, then the situation is exactly similar to that we have considered up to now (figure 13.2). Figure explains the different steps needed to convert the analysis from a real explosion to the system finally considered in what follows. The conditions found in the quiescent atmosphere, and thus upstream of the (propagating) normal shock, areasusual denotedwiththeindex 1: p 1,T 1,ρ 1...Similarly, theconditions observed afterpassing through the normal shock are indexed 2: p 2,T 2,ρ 2... The propagation speed of the normal shock in the quiescent atmosphere is denoted. Using and the speed of sound upstream of the shock c 1, it is possible to build the Mach number of the shock wave M 1 by: M 1 := c 1 (13.56) Lookingcloselyatfigure13.12, M 1 istheexactequivalent ofm 1 forthestandardconfigurationdescribed in figure All the other equations and properties described previously apply as well. For instance, the conservation equations now read: For mass: ρ 1 = ρ 2 ( v 2 ) (13.57) For momentum: p 1 +ρ 1 2 = p 2 +ρ 2 ( v 2 ) 2 (13.58)

180 178 Chapter 13. Shock waves p 1 1 T 1 v 2 p 2 2 p 1 1 T 1 T 2 a) p 1 1 T 1 v 2 p 2 2 T 2 b) -v 2 p 1 1 p 2 2 T 1 T 2 c) Figure 13.12: Process allowing to analyze the propagating shock wave produced by an explosion: a) Explosion process; b) Converting to a propagating, locally straight normal shock wave, as a local approximation of the expanding shock by zooming using the magnifying class of subfigure (a); c) Change of reference frame, using the observer traveling with the shock wave depicted in subfigure (b), allowing to obtain a fixed normal shock wave and get back to the configuration considered in figure 13.2.

181 3 Why shock waves? 179 For energy: c p T This last equation simplifies also to: = c p T 2 + ( v 2) 2 2 (13.59) = c p T v2 2 2 v 2 (13.60) c p T 1 = c p T 2 + v2 2 2 v 2 (13.61) All the other equations derived for a fixed normal shock can readily be applied to the current problem of a moving shock by considering the change in reference frame described in figure In particular, the pressure jump through the shock, p 2 /p 1, also written Π, can be computed as: p 2 = Π p 1 (13.62) = 2γM2 1 γ +1 γ +1 (13.63) which is identical to Eq.(13.26) when replacing M 1 by M 1. In the same manner, it comes for the density jump: ρ 2 = (γ +1)M2 1 (13.64) ρ 1 2+(γ 1)M 2 1 When analyzing an explosion, it is also useful to invert those relations in order to derive all important quantities as function of the pressure jump Π. When doing so, one obtains in particular for the shock Mach number: (γ +1)Π+γ 1 M 1 = (13.65) 2γ The density ratio is given by Eq.(13.46), already obtained when demonstrating the Rankine-Hugoniot relation: ρ 2 (γ +1)Π+γ 1 = (13.66) ρ 1 (γ 1)Π+γ +1 The solution process is then typically as follows: knowing explosive type and quantity, it is possible to assess the value of p 2. Together with the value of p 1 = p a, this is sufficient to determine the other properties of the normal shock, in particular its propagation speed. Sometimes, the analysis goes the other way round. Knowing, for instance from high-speed images, the propagation speed of the shock, the value of p 2 can be determined from Eq.(13.63) Why shock waves? Now that we have some knowledge on normal shock waves, and before considering in the next section the (similar) oblique shock waves, it is time to sit back andthink about the purpose of such structures. As explained previously, the most visible effect of a shock wave is, beyond deceleration, the resulting increase in pressure. This is also the first reason why a shock wave will appear: separate two regions at different pressures. Indeed, there are two major compatibility conditions for a fluid considered as a continuum (Section 1.3), illustrated in figure 13.13:

182 180 Chapter 13. Shock waves at micro-scale, the pressure is homogeneous and isotropic, so that it is impossible to get pressure differences at distances below that characterizing a fluid element; again at micro-scale, there is a single flow velocity for a fluid element, so that it is impossible to observe velocity differences at such scales. Finally, this explains why shock waves are needed and will indeed appear: this is the only solution we know allowing to separate in a physically correct manner regions with an abrupt change in pressure and/or velocity! p 1 p 2 >> p 1 p 1 p 2 >> p 1 v 1 v 2 v 1 v 2 v 1 v 2 v 1 v 2 impossible possible Figure 13.13: Compatibility conditions at micro-scale for a fluid considered as a continuum. The shock wave is the only solution to separate in a physically correct manner regions with an abrupt change in pressure and/or velocity. The corresponding shock wave (normal or oblique) is sketched as a thick red line Oblique shock wave The hypotheses underlying the analysis follow those employed for the normal shock wave (section 13.2), and correspond indeed quite well to the physical reality: The shock wave is a discontinuity without inner structure, separating the upstream flow (with conditions marked by an index 1) from the downstream flow (with conditions marked by an index 2);

183 4 Oblique shock wave 181 The shock wave is a straight structure, with an inclination angle ǫ compared to the upstream flow direction (given by v 1 ); All streamlines are parallel to each other before the shock wave, and after the shock wave; The configuration is steady. The corresponding configuration is shown in figure Flow deflection might be due to the geometry of the system, for instance if the lower wall of a straight duct is indeed turned by an angle δ. In that case, an oblique shock wave will appear in a supersonic flow, as discussed below, leading additionally to changes in pressure, density, temperature... Looking back at figure 13.13, the shock wave is needed here to fulfill compatibility conditions in terms of velocity. y v 1 n n v 2 p 2 p 1 n n 2 1 T 1 T 2 n v n1 v t1 v n2 v t2 n 90 x Figure 13.14: Configuration employed for the analysis of a fixed, oblique shock wave (shown in red), together with the control volume (in green) used to derive the conservation equations. The angle δ is the deflection angle, the angle ε the shock angle. Here, the shock wave is needed to ensure compatibility in terms of velocity. Alternatively, an oblique shock wave might appear in order to fulfill compatibility conditions in terms of pressure, forinstance atthe end of anoverexpanded Laval nozzle (figure 13.15). Inthat case, a change in velocity direction and magnitude will be the result. Both configurations are of course solved in the same manner. For the developments presented in this section, the configuration shown in figure has been chosen.

184 182 Chapter 13. Shock waves p a jet boundary v 1 p 1 p 2 =p a 1 T 1 v 2 2 T 2 p a Figure 13.15: Oblique shock wave (shown in red) appearing at the end of an overexpanded Laval nozzle (i.e.,, p 1 < p a at the end of the nozzle). Here, the shock wave is needed to ensure compatibility in terms of pressure Conservation equations for a real gas Following again the same procedure as for the normal shock wave, we begin by considering generic gas properties. This means that the relations derived in this section are valid even for complex thermodynamic relations, as might be found for a real gas. Taking into account the hypotheses listed above, it is possible to establish generic conservation equations. The considered flow configuration is now two-dimensional. It is important to have a close look at figure 13.14, since it will be very helpful to derive suitable relations. First, please note how the control volume (green dashed line in figure 13.14) has been chosen. The horizontal limits of this control volume follow as usual the streamlines upstream and downstream of the oblique shock. Hence, there will be no exchange of fluid through these horizontal boundaries. Additionally, the top and bottom horizontal boundaries of the control volume are identical in length, and exactly the same fluid properties will be found when following those boundaries. However, there is one important difference: the normal vector n, pointing always towards the outside of the control volume, is opposed along both lines (figure 13.14). The left and right lines bounding the control volume have been chosen parallel to the oblique shock wave itself. They therefore do not follow the horizontal or vertical direction. Please keep in mind this unusual convention.

185 4 Oblique shock wave 183 In the same manner, the coordinate system is different from our usual choice: the y-direction follows again the oblique shock wave; the x-direction is of course perpendicular to the y-axis and roughly in flow direction, but it is not horizontal. In order to account for this unusual axis system, the flow velocity v has been decomposed into its normal component v n, normal to the oblique shock wave, and its tangential component v t, parallel to the oblique shock wave. The corresponding vector relation simply reads: v = v n +v t (13.67) Now, using the Pythagorean theorem in the right triangle formed by v,v n,v t, one obtains: v 2 1 = v n1 2 +v t1 2 v 2 2 = v n2 2 +v t2 2 (13.68) (13.69) Additionally, looking back at figure 13.14, it is easy to recognize that the angle between v and v n is a complementary angle to the shock angle ε. Therefore, using the standard angular relations in a right triangle, it is possible to write successively: v n1 = v 1 cos(90 ε) = v 1 sinε (13.70) v t1 = v 1 cosε (13.71) v n2 = v 2 sin(ε δ) (13.72) v t2 = v 2 cos(ε δ) (13.73) Mass conservation For the steady flow bounded by the streamtube shown in green in figure 13.2 the mass conservation is again represented simply by the conservation of the mass flow-rate (ṁ = constant) through any tube cross-section. Here, all streamlines are parallel to each other, so that the cross-section of the streamtube is constant in spite of the flow deflection (angle δ between inflow and outflow directions), A(x) = A 1 = A 2 = constant. There is no fluid leaving or entering the control volume through the horizontal boundaries. On the left and right side, the boundaries being parallel to the oblique shock wave, only the normal velocity component v n must be taken into consideration. The tangential velocity component v t, being parallel to the boundaries of the control volume, does not lead to any exchange of fluid with the surroundings. Hence, the mass conservation relating upstream and downstream conditions simply reads: ρ 1 v n1 = ρ 2 v n2 (13.74) Conservation of momentum Friction plays obviously no role for the configuration considered in figure 13.14, since there is no velocity gradient upstream and downstream of the discontinuity (homogeneous flow conditions). Therefore, the suitable expression for conservation of momentum in this two-dimensional, steady compressible flow is that introduced in Chapter 4 (Eq. 4.17): ρ(x,t)v(x,t)(v(x,t) n)da = A cf ρ(x,t)gdv V cf p(x,t)nda A cf (13.75)

186 184 Chapter 13. Shock waves Asusual foragasflow, thecontribution of gravity will becompletely neglected dueto thelow density of gases, even at relatively high pressure. This vector relation will be written instead using two scalar relations along the direction x (normal to the shock wave) and y (tangential to the shock wave). It is clear that all contributions along the horizontal boundaries of the control volume are zero(for the velocity term, due to the fact that (v n) = 0) or compensate each other exactly (for the pressure term, due to the opposite directions of n along the top and bottom boundaries). Therefore, the only remaining contributions in Eq.(13.75) are found along the left and right boundaries. Along the x-direction, those terms involve only the normal velocity component, both in v and in the scalar product (v n). One thus obtains for momentum conservation along x: p 1 +ρ 1 v 2 n1 = p 2 +ρ 2 v 2 n2 (13.76) Considering now the y-direction for the terms along the left and right boundaries, v now involves only v t while the scalar product (v n) still delivers v n, so that the momentum conservation along y finally reads: ρ 1 v n1 v t1 = ρ 2 v n2 v t2 (13.77) Dividing now Eq.(13.77) by Eq.(13.74), one obtains immediately the important relation: v t1 = v t2 (13.78) The tangential velocity component v t remains unchanged when crossing the oblique shock wave! Energy conservation The conservation of energy is again simply derived from the First Law of Thermodynamics. For such an isolated adiabatic system, the total energy is constant, which can be written when neglecting as usual the contribution of potential energy for this gaseous flow: h 1 + v2 1 2 = h 2 + v2 2 2 (13.79) It is identical to that obtained for the normal shock wave Conservation equations for a perfect gas The previous relations are only slightly modified when considering the special case of a perfect gas, as done in the rest of this section. Using Eq.(11.42), one obtains for: Mass conservation (unchanged): ρ 1 v n1 = ρ 2 v n2 (13.80) Conservation of momentum in x-direction (unchanged): p 1 +ρ 1 v 2 n1 = p 2 +ρ 2 v 2 n2 (13.81) Conservation of momentum in y-direction (unchanged): v t1 = v t2 (13.82)

187 4 Oblique shock wave 185 Energy conservation: c p T 1 + v2 1 2 = c pt 2 + v2 2 2 (13.83) Ononehand, thisrelationcanbetransformedinamannersimilartothatdoneforthenormalshock wave, replacing temperature and kinetic energy together using the stagnation temperature T 0, leading to: c p T 01 = c p T 02 (13.84) Sincec p = constant,itisthusagainclearthat,inthisadiabaticflow, thestagnation temperature T 0 is the same upstream and downstream of the oblique shock wave: T 01 = T 02 (13.85) On the other hand, it is also possible to rewrite Eq.(13.83) using again the Pythagorean theorem v 2 = v 2 n +v 2 t and simplifying the tangential component v t since vt1 2 = v2 t2, leading finally to: c p T 1 + v2 n1 2 = c pt 2 + v2 n2 2 (13.86) Summarizing, the final conservation equations for mass, momentum (in x-direction) and energy read as a whole: ρ 1 v n1 = ρ 2 v n2 (13.87) p 1 +ρ 1 v 2 n1 = p 2 +ρ 2 v 2 n2 (13.88) c p T 1 + v2 n1 2 = c p T 2 + v2 n2 2 (13.89) Comparing these equations to those obtained for the normal shock (i.e., comparing Eq.(13.87) to Eq.(13.7), comparing Eq.(13.88) to Eq.(13.8) and comparing Eq.(13.89) to Eq.(13.9)), it is easy to see that both equation sets are identical, provided that the velocities v are replaced everywhere by the respective normal components of the same velocity, v n. This therefore delivers the practical solution for our problem. When considering an oblique shock, a decomposition of the flow field in directions normal (index n) and tangential (index t) to the shock will be first realized. Then: In the tangential direction, the velocity remains unchanged by the shock: v t1 = v t2. In the normal direction, i.e., when considering only the normal components of the velocity, the oblique shock behaves exactly like the normal shock considered in Section There is no need to develop a special solution for this problem. Beware! Unfortunately, this statement does not necessarily apply for relations involving several intermediate steps and dynamic flow properties, switching back and forth between velocity magnitude and normal velocity, or between Mach number and normal Mach number... For instance, the shock relation of Prandtl (see Eq.13.42) cannot directly be transformed into a form valid for oblique shocks; in order to derive a suitable equation, the tangential velocity component must be taken into account as well, limiting its applicability.

188 186 Chapter 13. Shock waves Jump relations involving the upstream Mach number M 1 As explained above, all the relations derived for the normal shock wave can immediately be used for the oblique shock wave as long as the normal velocity components v n are used instead of the velocity magnitude v. Let us exemplify the process for the pressure jump. We know for the normal shock (Eq ): p 2 = 2γM2 1 γ +1 p 1 γ +1 (13.90) Now, the flow velocity v does not appear in this equation, so it would be possible to believe that there is nothing to do. This is indeed misleading. The right answer is that the flow velocity v does not appear directly. But, considering the definition of the Mach number M = v/c, it is clear that the flow velocity is contained as well within M, every time the Mach number appears. Therefore, it is necessary to reformulate slightly the rule mentioned previously: All the direct relations derived for the normal shock wave can immediately be used for the oblique shock wave as long as the normal velocity components v n are used instead of the velocity magnitude v. This rule applies as well to the Mach number, which must be replaced by the normal Mach number M n defined as: M n := v n c (13.91) As a first step, the relation between the normal Mach numbers upstream and downstream of the shock wave is simply: M n2 = 2+(γ 1)M2 n1 (13.92) 2γMn γ All thejumprelationscanbereadilyobtained fortheoblique shock using thissame rule. Oneobtains successively: For the pressure jump: p 2 = 2γM2 n1 γ +1 p 1 γ +1 (13.93) For the temperature jump: ( T 2 2γ = T 1 γ +1 M2 n1 γ 1 )( ) γ 1 γ +1 γ (γ +1)Mn1 2 (13.94) For the density jump: For the velocity jump: ρ 2 ρ 1 = (γ +1)M2 n1 2+(γ 1)M 2 n1 v 2 v 1 = 2+(γ 1)M2 n1 (γ +1)M 2 n1 (13.95) (13.96) It is not necessary to plot again the corresponding evolutions, since they would be identical to figures 13.4 to 13.7 when replacing M 1 by M n1 along the x-axis. The qualitative statements remain the same: the jump in pressure is the most noticeable effect of an oblique shock; pressure and temperature jump

189 4 Oblique shock wave 187 tend toward infinity when increasing the upstream normal Mach number, while density and velocity jump tend toward a finite limit value. The analogy can be pursued in a similar manner concerning the stagnation pressure and specific entropy. Concerning in particular this last quantity, it has been demonstrated in Section that a normal shock can only exist for an upstream Mach number M 1 1. Using the rule derived previously, the condition on specific entropy connected to the Second Law of Thermodynamics for the oblique shock now reads M n1 1. In order to really understand the meaning of this relation, it is necessary to look back at figure Due to geometrical considerations, we have already written Eq.(13.70): Since M n := v n /c, the normal Mach number is thus simply: v n1 = v 1 sinε (13.97) M n1 = M 1 sinε (13.98) Similarly, behind the oblique shock wave, Eq.(13.72) relates normal velocity and velocity magnitude by v n2 = v 2 sin(ε δ), so that the normal Mach number behind the oblique shock is: M n2 = M 2 sin(ε δ) (13.99) The conditions found in Section for the normal shock (M 1 1 before the normal shock, and therefore M 2 1 after the normal shock) are therefore written for the oblique shock: M n1 1 and M n2 1. Taking into account the angular relations written above, one obtains finally upstream of the oblique shock M 1 sinε 1 (13.100) Since (sinε) is obviously below 1, this relation therefore necessitates as well M 1 1. As a consequence, the statement listed in Section for the normal shock remains valid: a steady oblique shock can only exist in a supersonic flow. The situation is different after the oblique shock. The corresponding condition reads: M 2 sin(ε δ) 1 (13.101) But, since again sin(ε δ) is obviously below 1, this condition does not necessitate M 2 1. As a consequence and as will be shown later, the Mach number M 2 behind the oblique shock can be either supersonic or subsonic. In practice, the supersonic regime is often found, though the subsonic regime exists as well. Finally, there is no clear rule concerning the value of M 2 behind the oblique shock, apart from the fact that M 2 must be always smaller than M 1 (deceleration induced by the shock wave) Summary: evolution of all quantities through an oblique shock All observations described in the previous sections are summarized in Table 13.2, which is nearly identical to Table 13.1, apart for the fact that the Mach number M 2 behind the oblique shock can be either supersonic or subsonic Using the shock tables for an oblique shock As discussed previously, all the relations derived for the normal shock wave can immediately be used for the oblique shock wave as long as the normal velocity components v n are used instead of the velocity magnitude v. This rule applies as well to the Mach number, which must be replaced by the normal Mach number M n.

190 188 Chapter 13. Shock waves Quantity Evolution T 0 T 01 = T 02 M (from M 1 1) ց (M 2 : subsonic or supersonic) v v 1 ց v 2 p p 1 ր p 2 p 0 p 01 ց p 02 T T 1 ր T 2 s s 1 ր s 2 Table 13.2: Evolution of the main flow quantities through an oblique shock wave Therefore, the tabulated values listed in Appendix F for the normal shock (in the case γ = 1.4) can be used for the oblique shock, provided the normal Mach number M n1 = M 1 sinε (and not the Mach number!) is used to enter the Table in the first column (entitled M). Then: the eighth column gives the normal Mach number M n2 = M 2 sin(ε δ) downstream of the shock wave (and not M 2!); the ninth column gives the pressure jump through the oblique shock; the tenth column gives the density jump; the eleventh column gives the temperature jump; the twelfth column gives the jump in stagnation pressure Determining the shock angle ε Now, it is clear that the normal velocity v n and even more the normal Mach number M n must be known to solve the problem. Since M n1 = M 1 sinε, the only real unknown is the shock angle ε. A relation is needed to compute this angle. For this purpose, it is appropriate to start again from the mass conservation, Eq.(13.87), rewritten as: ρ 2 = v n1 (13.102) ρ 1 v n2 Now, combining Eq.(13.70) with Eq.(13.71), it is easy to see that: Using now Eq.(13.72) with Eq.(13.73), one obtains in the same way: Considering those last two relations, Eq.(13.102) can now be rewritten: v n1 = v t1 tanε (13.103) v n2 = v t2 tan(ε δ) (13.104) ρ 2 ρ 1 = v t1 tanε v t2 tan(ε δ) (13.105) Keeping in mind Eq.(13.78), v t1 = v t2, this equation simplifies to: ρ 2 ρ 1 = tanε tan(ε δ) (13.106)

191 4 Oblique shock wave 189 Using Eq.(13.95) to replace the density jump and exchanging left-hand and right-hand side, it comes: tanε tan(ε δ) = (γ +1)M2 1 sin2 ε 2+(γ 1)M 2 1 sin 2 ε (13.107) Using now the standard trigonometric relation: tan(φ ϕ) = tanφ tanϕ 1+tanφtanϕ (13.108) it is slightly tedious but straightforward to reformulate Eq.(13.107) into: 1 tanδ = (γ +1)M ( M1 2 sin 2 ε 1 ) 1 tan ε (13.109) Hence, providing a value for the deflection angle δ and the upstream Mach number M 1, Eq.(13.109) can be used to compute the corresponding value of the shock angle ε. This has been done in figure 13.16, where even more information has been shown: Only the lower part of the diagram, plotted in solid red and blue lines, should be considered. It corresponds to so-called weak solutions, and those are the ones we are presently computing. The upper part, associated to strong oblique shock waves, is far less relevant for practical purposes and will not be further discussed. Following now one line M = constant, a turning point is observed for some value of the deflection angle, δ = δ max (M). The corresponding value is marked in figure with. The existence of δ max (M) shows that there is a maximum possible deflection in any supersonic flow leading to an oblique shock wave as shown in figure When trying to deflect such a supersonic flow by an angle δ > δ max (M), a different physical solution will appear. Since this is not often found in practice, such a configuration will not be further considered in what follows. Finally, following again one line M = constant, a + symbol is plotted slightly below the symbol. This is the limit beyond which a subsonic flow (M 2 < 1) will be found after the oblique shock. For values of ǫ below the + symbol, the flow beyond the oblique shock wave is supersonic (M 2 > 1). In-between the + and the symbol, subsonic conditions are found downstream of the oblique shock (M 2 < 1).

192 190 Chapter 13. Shock waves Deflection angle δ (in ) Shock angle ε (in ) Figure 13.16: Graphical relation between the deflection angle δ (x-axis) and the shock angle ε (y-axis) as a function of the upstream Mach number M 1 (lines), corresponding to the solution of Eq.(13.109). Only the lower part of the diagram, shown alternatively with solid red and blue lines, should be considered.

193 4 Oblique shock wave Mach angle and Mach wave It has been discussed previously that thesecond Law of Thermodynamics leads to the condition M 1 1 for a normal shock (see Section ). The similar condition for an oblique shock is now M n1 1, or M 1 sinε 1. Therefore, it is possible to define for any supersonic flow (M 1) the so-called Mach angle µ by the following relation: sinµ := 1 M (13.110) The Mach angle µ is listed in the seventh column of Appendix F for γ = 1.4. Now, an oblique shock wave associated to a shock angle ε = µ automatically leads to the condition M sinε = M sinµ = M n = 1: this is the weakest possible oblique shock, leading to no change at all in the flow, since under such conditions p 2 = p 1, T 2 = T 1, ρ 2 = ρ 1... An oblique shock wave with the shock angle µ is called a Mach wave. It is an invisible structure present in any supersonic flow (remember that the Mach wave does not lead to any change in the flow). An oblique shock wave becoming weaker and weaker (i.e., when M n1 1), for instance because the deflection angle is progressively reduced (δ 0) tends toward such a Mach wave. Therefore, measuring (perhaps from a simple picture) the shock angle of a very weak oblique shock wave, produced for instance by a minute deflection, it is possible to determine indirectly the Mach number of the flow by M = lim δ 0 ( ) 1 arcsin ε 13.5 Polar curve and Busemann diagram (13.111) As a final step, it is possible to combine all possible shock waves existing in a supersonic flow by plotting the polar curve. The objective of this graphical representation is indeed to show at a glance all steady shock structures that might be found in a supersonic flow with fixed conditions (index 1) upstream of the shock. In practice, only the upstream Mach number M 1 and the upstream temperature T 1 are fixed. As usual, the representation shown in figure considers the case γ = 1.4, but a similar picture can obviously be obtained for any value of γ. The underlying process is as follow: Fixing M 1, T 1 and the gas (air in figure 13.17), the corresponding upstream velocity is fixed as well by Eqs.(11.29) and (11.51): v 1 = M 1 γrt 1 (13.112) The resulting upstream velocity vector v 1, which is constant in the present case, is now used to define the x-axis of the graphical representation. The retained origin for this vector, written Ω, is the origin of the coordinate system and will be chosen as well as origin to plot all other vectors in this figure. Note also that there is no need to extend the x-axis beyond v 1, since it corresponds to the maximum downstream velocity for any possible shock in this flow. Now, the only remaining task is to plot in this same coordinate system the downstream conditions for all shocks possibly occurring in this flow for the same gas, Mach number M 1 and upstream temperature T 1. Let us begin with the case where the shock does not lead to any deflection, δ = 0. This means that the velocity vector v 2 downstream of the shock is simply parallel to v 1 and will therefore itself be part of the x-axis. There are two possible shock solutions without deflection: The Mach wave, or invisible shock, leading to no modification at all of the flow. In that case, one obtains directly v 2 = v 1.

194 192 Chapter 13. Shock waves The normal shock discussed in Section In that case, the downstream velocity is given by Eq.(13.30), which has been used to compute the velocity magnitude v 2 in figure 13.17: v 2 v 1 = 2+(γ 1)M2 1 (γ +1)M 2 1 (13.113) Finally, let us consider all further oblique shocks with a real deflection, δ > 0. In that case, the direction of the vector v 2 is already known (angle δ between the x-axis, i.e., the vector v 1 and the downstream velocity vector v 2 ). Only the magnitude is now needed. The computation of v 2 involves several successive steps: Knowing M 1 and δ, determine the shock angle ε, either graphically from figure or numerically from Eq.(13.109); Then, compute the normal Mach number M n1 = M 1 sinε; The normal Mach number M n2 downstream of the shock is given by Eq.(13.92): M n2 = 2+(γ 1)M2 n1 2γMn γ (13.114) The downstream Mach number is now M 2 = M n2 /sin(ε δ); In parallel, the temperature downstream of the shock is given by Eq.(13.94): T 2 T 1 = ( 2γ γ +1 M2 n1 γ 1 γ +1 )( ) γ 1 γ (γ +1)Mn1 2 (13.115) Knowing nowm 2 andt 2, thedownstreamvelocity magnitudecanbeagaindirectlycomputed using Eqs.(11.29) and (11.51): v 2 = M 2 γrt 2 (13.116) Knowing angle and magnitude, the downstream velocity vector v 2 can be finally plotted in figure Repeating this process for any possible deflection angle δ δ max (see also figure 13.18), the polar curve is obtained, as presented in figure Keeping in mind the condition v t1 = v t2 (Eq ), the polar curve can also be used a posteriori to determine graphically the shock angle ε (see now figure 13.18). For this purpose, it is sufficient to first draw the straight line connecting the tip of the upstream velocity vector v 1 with the tip of the downstream velocity vector v 2. Drawing now the straight line starting from Ω and perpendicular to the first line, it is clear that this second line gives the direction of the oblique shock, since the projections of v 1 and v 2 onto this line, which represent the tangential part of those vectors, are equal, hence fulfilling the condition v t1 = v t2, as written in figure Along the polar curve, the oblique shock waves leading to a supersonic downstream flow are shown in red, while those leading to subsonic downstream conditions are in blue in figure It is again visible that most cases lead to supersonic flows behind the oblique shock. The maximum possible deflection angle δ max is also directly visible from figure (last solution leading to an oblique shock, i.e., last point along the polar curve that associated to the lowest velocity v 2 ). The Busemann diagram is that obtained when representing on the same figure the polar curves for a variety of upstream flow conditions. It can be used to determine all relevant properties of any shock wave

195 6 Boundary conditions and shock reflections 193 Ω δ Figure 13.17: Graphical representation of the polar curve showing all possible shock structures found in a fixed supersonic flow. For the present case, air is considered (γ = 1.4 and r = 287 J/(kg.K)). A Mach number M 1 = 2 has been arbitrarily selected here, together with an upstream temperature T 1 = 300 K. in a flow. Its advantage is that it is a very compact representation: on a single figure, everything can be determined. One drawback is the comparatively low accuracy of the obtained information, stemming from the difficulty of reading accurately the value of small angles. Therefore, as far as possible, it is recommended to use instead the shock tables or even better a numerical solution of the equations listed in this chapter Boundary conditions and shock reflections Let us first have a look back at figure 13.13, on which the relevant compatibility conditions leading to shock waves have been explained. Compatibility conditions must be obtained for velocity direction, pressure, and (at least to some extent) velocity magnitude. Very often, these very same boundary conditions will lead to reflected oblique shocks in practical cases involving walls. To understand this point, let us redraw now figure with an upper wall, leading to figure Obviously, the deflection obtained after the first oblique shock is not compatible with the upper flow, since the flow would point into the wall, which is unphysical. Therefore, a reflected shock will appear at the contact point between the first oblique shock and the upper wall. The physical purpose of this

196 194 Chapter 13. Shock waves Ω ε δ max δ Figure 13.18: Graphical representation of the polar curve showing all possible shock structures found in a fixed supersonic flow. For the present case, air is considered (γ = 1.4 and r = 287 J/(kg.K)). A Mach number M 1 = 2 has been arbitrarily selected here, together with an upstream temperature T 1 = 300 K. The maximum deflection angle δ max and the graphical procedure used to determine the shock angle ǫ are shown as well in this figure. second shock is to lead back to a horizontal flow, compatible with the (horizontal) direction of the upper wall. But, at the same time, this second shock will lead to a further increase in pressure, temperature, density, as well as to another flow deceleration. Now, this second shock, when meeting the lower wall, will again be reflected, so that the flow after the third shock will again be parallel to the lower wall. Those reflections will repeat until perfect compatibility conditions are reached, or very often, will stop by reaching a last, normal shock wave when entering a region with parallel walls. The real shock system depends of course on the real geometrical configuration Conclusions Using the relations derived in this chapter, it is now possible to compute all relevant properties for the normal and oblique shock waves that might be found in any supersonic flow. In fact, there are further possible structures that have not been discussed here; in particular, rarefaction waves will be considered