Lecture notes on Gas Dynamics

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LECTURE NOTES ON GAS DYNAMICS Joseph M. Powers Department of Aerospace and Mechanical Engineering University of Notre Dame Notre Dame, Indiana 46556-5637 USA updated March 16, 2005Contents 1 Introduction 9 1.1 De nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.2 Motivating examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 1.2.1 Re-entry ows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 1.2.1.1 bow shock wave . . . . . . . . . . . . . . . . . . . . . . . . . 10 1.2.1.2 rarefaction (expansion) wave . . . . . . . . . . . . . . . . . 11 1.2.1.3 momentum boundary layer . . . . . . . . . . . . . . . . . . 11 1.2.1.4 thermal boundary layer . . . . . . . . . . . . . . . . . . . . 12 1.2.1.5 vibrational relaxation e ects . . . . . . . . . . . . . . . . . . 12 1.2.1.6 dissociation e ects . . . . . . . . . . . . . . . . . . . . . . . 12 1.2.2 Rocket Nozzle Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 1.2.3 Jet Engine Inlets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2 Governing Equations 15 2.1 Mathematical Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.1.1 Vectors and Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.1.2 Gradient, Divergence, and Material Derivatives . . . . . . . . . . . . 16 2.1.3 Conservative and Non-Conservative Forms . . . . . . . . . . . . . . . 17 2.1.3.1 Conservative Form . . . . . . . . . . . . . . . . . . . . . . . 17 2.1.3.2 Non-Conservative Form . . . . . . . . . . . . . . . . . . . . 17 2.2 Summary of Full Set of Compressible Viscous Equations . . . . . . . . . . . 20 2.3 Conservation Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.3.1 Conservation of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2.3.1.1 Nonconservative form . . . . . . . . . . . . . . . . . . . . . 22 2.3.1.2 Conservative Form . . . . . . . . . . . . . . . . . . . . . . . 22 2.3.1.3 Incompressible Form . . . . . . . . . . . . . . . . . . . . . . 22 2.3.2 Conservation of Linear Momenta . . . . . . . . . . . . . . . . . . . . 22 2.3.2.1 Nonconservative form . . . . . . . . . . . . . . . . . . . . . 22 2.3.2.2 Conservative Form . . . . . . . . . . . . . . . . . . . . . . . 23 2.3.3 Conservation of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 24 2.3.3.1 Nonconservative Form . . . . . . . . . . . . . . . . . . . . . 24 2.3.3.2 Mechanical Energy . . . . . . . . . . . . . . . . . . . . . . . 25 3Chapter 1 Introduction Suggested Reading: Anderson, Chapter 1: pp. 1-31 1.1 De nitions The topic of this course is the aerodynamics of compressible and viscous ow. Where does aerodynamics rest in the taxonomy of mechanics? Aerodynamicsa branch of dynamics that deals with the motion of air and other gaseous uids and with the forces acting on bodies in motion relative to such uids (e.g. airplanes) We can say that aerodynamics is a subset of ()  uid dynamics since air is but one type of uid,  uid mechanics since dynamics is part of mechanics,  mechanics since uid mechanics is one class of mechanics. Mechanicsa branch of physical science that deals with forces and the motion of bodies traditionally broken into:  kinematicsstudy of motion without regard to causality  dynamics (kinetics)study of forces which give rise to motion Examples of other subsets of mechanics: 910 CHAPTER 1. INTRODUCTION  solid mechanics  quantum mechanics  celestial mechanics  relativistic mechanics  quantum-electrodynamics (QED)  magneto-hydrodynamics (MHD) Recall the de nition of a uid: Fluida material which moves when a shear force is applied. Recall that solids can, after a small displacement, relax to an equilibrium con guration when a shear force is applied. Recall also that both liquids and gases are uids The motion of both liquids and gases can be a ected by compressibility and shear forces. While shear forces are important for both types of uids, the in uence of compressibility in gases is generally more signi cant. The thrust of this class will be to understand how to model the e ects of compressibility and shear forces and how this impacts the design of aerospace vehicles. 1.2 Motivating examples The following two examples serve to illustrate why knowledge of compressibility and shear e ects is critical. 1.2.1 Re-entry ows A range of phenomena are present in the re-entry of a vehicle into the atmosphere. This is an example of an external ow. See Figure 1.1. 1.2.1.1 bow shock wave  suddenly raises density, temperature and pressure of shocked air; consider normal shock in ideal air 3 3  = 1:16 kg=m  = 6:64 kg=m (over ve times as dense) o s T = 300 K T = 6; 100 K (hot as the sun's surface ) o s1.2. MOTIVATING EXAMPLES 11 far-field acoustic wave rarefaction waves viscous and thermal boundary layers Oblique Shock Wave Ambient Air Normal Shock Wave Figure 1.1: Fluid mechanics phenomena in re-entry P = 1:0 atm P = 116:5 atm (tremendous force change) o s sudden transfer of energy from kinetic (ordered) to thermal (random)  introduces inviscid entropy/vorticity layer into post-shocked ow  normal shock standing o leading edge  conical oblique shock away from leading edge  acoustic wave in far eld 1.2.1.2 rarefaction (expansion) wave  lowers density, temperature, and pressure of air continuously and signi cantly  interactions with bow shock weaken bow shock 1.2.1.3 momentum boundary layer  occurs in thin layer near surface where velocity relaxes from freestream to zero to satisfy the no-slip condition  necessary to predict viscous drag forces on body12 CHAPTER 1. INTRODUCTION 1.2.1.4 thermal boundary layer  as uid decelerates in momentum boundary layer kinetic energy is converted to thermal energy  temperature rises can be signi cant ( 1; 000 K) 1.2.1.5 vibrational relaxation e ects  energy partitioned into vibrational modes in addition to translational  lowers temperature that would otherwise be realized  important for air above 800 K  unimportant for monatomic gases 1.2.1.6 dissociation e ects  e ect which happens when multi-atomic molecules split into constituent atoms  O totally dissociated into O near 4; 000 K 2  N totally dissociated into N near 9; 000 K 2  For T 9; 000 K, ionized plasmas begin to form Vibrational relaxation, dissociation, and ionization can be accounted for to some extent by introducing a temperature-dependent speci c heat c (T) v 1.2.2 Rocket Nozzle Flows The same essential ingredients are present in ows through rocket nozzles. This is an example of an internal ow, see Figure 1.2 viscous and thermal boundary layers burning solid rocket fuel burning solid rocket fuel possible normal shock Figure 1.2: Fluid mechanics phenomena in rocket nozzles Some features:1.2. MOTIVATING EXAMPLES 13  well-modelled as one-dimensional ow  large thrust relies on subsonic to supersonic transition in a converging-diverging nozzle  away from design conditions normal shocks can exist in nozzle  viscous and thermal boundary layers must be accounted for in design 1.2.3 Jet Engine Inlets The same applies for the internal ow inside a jet engine, see Figure 1.3 viscous and thermal boundary layers oblique shock compressor turbine exhaust combustor inlet Figure 1.3: Fluid mechanics phenomena in jet engine inletChapter 2 Governing Equations Suggested Reading: Hughes and Brighton, Chapter 3: pp. 44-64 Liepmann and Roshko, Chapter 7: pp. 178-190, Chapter 13: pp. 305-313, 332-338 Anderson, Chapter 2: pp. 32-44; Chapter 6: pp. 186-205 The equations which govern a wide variety of these ows are the compressible Navier- Stokes equations. In general they are quite complicated and require numerical solution. We will only consider small subsets of these equations in practice, but it is instructive to see them in full glory at the outset. 2.1 Mathematical Preliminaries A few concepts which may be new or need re-emphasis are introduced here. 2.1.1 Vectors and Tensors One way to think of vectors and tensors is as follows:  rst order tensor: vector, associates a scalar with any direction in space, column matrix  second order tensor: tensor-associates a vector with any direction in space, two- dimensional matrix  third order tensor-associates a second order tensor with any direction in space, three- dimensional matrix  fourth order tensor-... 1516 CHAPTER 2. GOVERNING EQUATIONS Here a vector, denoted by boldface, denotes a quantity which can be decomposed as a sum of scalars multiplying orthogonal basis vectors, i.e.: v = ui + vj + wk (2.1) 2.1.2 Gradient, Divergence, and Material Derivatives The \del" operator,r, is as follows: r i + j + k (2.2) x y z Recall the de nition of the material derivative also known as the substantial or total derivative: d  + vr (2.3) dt t where Example 2.1 Does vr =rv =rv? vr = u + v + w (2.4) x y z u v w rv = + + (2.5) x y z 2 3 u v w x x x u v w 4 5 rv = (2.6) y y y u v w z z z So, no. Here the quantityrv is an example of a second order tensor. Also vr v div (2.7) rv div v (2.8) rv grad v (2.9) r grad  (2.10)2.1. MATHEMATICAL PRELIMINARIES 17 2.1.3 Conservative and Non-Conservative Forms T If h is a column vector of N variables, e.g. h = h ;h ;h ;:::h , and f (h ) g (h ) are a i i 1 2 3 N i i i i column vectors of N functions of the variables h , and all variables are functions of x and i t, h = h (x;t), f (h (x;t)), g (h (x;t)) then a system of partial di erential equations is in i i i i i i conservative form i the system can be written as follows: h + (f (h )) = g (h ) (2.11) i i i i i x x A system not in this form is in non-conservative form 2.1.3.1 Conservative Form Advantages  naturally arises from control volume derivation of governing equations  clearly exposes groups of terms which are conserved  easily integrated in certain special cases  most natural form for deriving normal shock jump equations  the method of choice for numerical simulations Disadvantages  lengthy  not commonly used  dicult to see how individual variables change 2.1.3.2 Non-Conservative Form Advantages  compact  commonly used  can see how individual variables change Disadvantages  often dicult to use to get solutions to problems18 CHAPTER 2. GOVERNING EQUATIONS  gives rise to arti cial instabilities if used in numerical simulation Example 2.2 Kinematic wave equation The kinematic wave equation in non-conservative form is u u + u = 0 (2.12) t x This equation has the same mathematical form as inviscid equations of gas dynamics which give rise to discontinuous shock waves. Thus understanding the solution of this simple equation is very useful in understanding equations with more physical signi cance.   2 u u Since u = the kinematic wave equation in conservative form is as follows: x x 2   2 u u + = 0 (2.13) t x 2 2 u Here h = u;f = ;g = 0. i i i 2 Consider the special case of a steady state  0. Then the conservative form of the equation can t be integrated   2 d u = 0 (2.14) dx 2 2 2 u u o = (2.15) 2 2 u =u (2.16) o Now u = u satis es the equation and so does u = u . These are both smooth solutions. In o o addition, combinations also satisfy, e.g. u = u ;x 0;u =u ;x 0. This is a discontinuous solution. o o u Also note the solution is not unique. This is a consequence of the u non-linearity. This is an example x of a type of shock wave. Which solution is achieved generally depends on terms we have neglected, especially unsteady terms. Example 2.3 Burger's equation Burger's equation in non-conservative form is 2 u u u + u =  (2.17) 2 t x x2.1. MATHEMATICAL PRELIMINARIES 19 This equation has the same mathematical form as viscous equations of gas dynamics which give rise to spatially smeared shock waves. Place this in conservative form: u u u + u  = 0 (2.18) t x x x     2 u u u +  = 0 (2.19) t x 2 x x   2 u u u +  = 0 (2.20) t x 2 x Here, this equation is not strictly in conservative form as it still involves derivatives inside the x operator. Consider the special case of a steady state  0. Then the conservative form of the equation can t be integrated   2 d u du  = 0 (2.21) dx 2 dx u Let u u as x1 (consequently 0 as x1) and u(0) = 0 so o x 2 2 u du u o  = (2.22) 2 dx 2  du 1 2 2  = u u (2.23) o dx 2 du dx = (2.24) 2 2 u u 2 o Z Z du dx = (2.25) 2 2 u u 2 o 1 u x 1 tanh = + C (2.26) u u 2 o o   u o u(x) = u tanh x + Cu (2.27) o o 2 u(0) = 0 = u tanh (Cu ) C = 0 (2.28) o o   u o u(x) = u tanh x (2.29) o 2 lim u(x) = u (2.30) o x1 lim u(x) =u (2.31) o x1 Note  same behavior in far eld as kinematic wave equation 2  continuous adjustment from u tou in a zone of thickness o o u o  zone thickness 0 as  0  inviscid shock is limiting case of viscously resolved shock Figure 2.1 gives a plot of the solution to both the kinematic wave equation and Burger's equation.20 CHAPTER 2. GOVERNING EQUATIONS u u uo uo x x -uo -uo Kinematic Wave Equation Solution Burger’s Equation Solution Discontinuous Shock Wave Smeared Shock Wave Shock Thickness 2 ν / uo Figure 2.1: Solutions to the kinematic wave equation and Burger's equation 2.2 Summary of Full Set of Compressible Viscous Equa- tions A complete set of equations is given below. These are the compressible Navier-Stokes equa- tions for an isotropic Newtonian uid with variable properties d + r v = 0 1 (2.32) dt dv  =rP +r + g 3 (2.33) dt de  =r q Pr v +:rv 1 (2.34) dt  T  =  rv +rv +  (r v)I 6 (2.35) q =krT 3 (2.36)  =  (;T) 1 (2.37)  =  (;T) 1 (2.38) k = k (;T) 1 (2.39) P = P (;T) 1 (2.40) e = e (;T) 1 (2.41) The numbers in brackets indicate the number of equations. Here the unknowns are 3  density kg=m (scalar-1 variable)  vvelocity m=s (vector- 3 variables) 2  Ppressure N=m (scalar- 1 variable)  einternal energy J=kg (scalar- 1 variable)2.3. CONSERVATION AXIOMS 21  Ttemperature K (scalar - 1 variable) 2  viscous stress N=m (symmetric tensor - 6 variables) 2  qheat ux vectorW=m (vector - 3 variables) 2   rst coecient of viscosity Ns=m (scalar - 1 variable) 2  second coecient of viscosity Ns=m (scalar - 1 variable) 2  kthermal conductivity W=(m K) (scalar - 1 variable) Here g is the constant gravitational acceleration and I is the identity matrix. Total19 variables Points of the exercise  19 equations; 19 unknowns  conservation axiomspostulates ( rst three equations)  constitutive relationsmaterial dependent (remaining equations)  review of vector notation and operations Exercise: Determine the three Cartesian components of r  for a) a compressible Newtonian uid, and b) an incompressible Newtonian uid, in whichr v = 0. This system of equations must be consistent with the second law of thermodynamics. De ning the entropy s by the Gibbs relation:   1 Tds = de + Pd (2.42)    ds de d 1 T = + P (2.43) dt dt dt  the second law states:   ds q  r (2.44) dt T In practice, this places some simple restrictions on the constitutive relations. It will be sometimes useful to write this in terms of the speci c volume, v  1=. This can be confused with the y component of velocity but should be clear in context. 2.3 Conservation Axioms Conservation principles are axioms of mechanics and represent statements that cannot be proved. In that they provide predictions which are consistent with empirical observations, they are useful.22 CHAPTER 2. GOVERNING EQUATIONS 2.3.1 Conservation of Mass This principle states that in a material volume (a volume which always encompasses the same uid particles), the mass is constant. 2.3.1.1 Nonconservative form d + r v = 0 (2.45) dt This can be expanded using the de nition of the material derivative to form       u v w + u + v + w +  + + = 0 (2.46) t x y x x y z 2.3.1.2 Conservative Form Using the product rule gives  (u) (v) (w) + + + = 0 (2.47) t x y z The equation essentially says that the net accumulation of mass within a control volume is attributable to the net ux of mass in and out of the control volume. In Gibbs notation this is  +r (v) = 0 (2.48) t 2.3.1.3 Incompressible Form I the uid is de ned to be incompressible, d=dt 0, the consequence is r v = 0; or (2.49) u v w + + = 0 (2.50) x y z As this course is mainly concerned with compressible ow, this will not be often used. 2.3.2 Conservation of Linear Momenta P This is really Newton's Second Law of Motion ma = F 2.3.2.1 Nonconservative form dv  =rP +r + g (2.51) dt  : mass/volume2.3. CONSERVATION AXIOMS 23 dv  : acceleration dt  rP;r: surface forces/volume  g: body force/volume Example 2.4 Expand the termr 0 1 T 0 1  +  +  xx yx zx    x y z xx xy xz  B C A  +  +  r =    = (2.52) xy yy zyA yx yy yz x y z x y z    zx zy zz  +  +  xz yz zz x y z This is a vector equation as there are three components of momenta. Let's consider the x momentum equation for example. du P    xx yx zx  = + + + + g (2.53) x dt x x y z Now expand the material derivative: u u u u P    xx yx zx  + u + v + w = + + + + g (2.54) x t x y z x x y z Equivalent equations exist for y and z linear momentum: v v v v P    xy yy zy  + u + v + w = + + + + g (2.55) y t x y z y x y z w w w w P    xz yz zz  + u + v + w = + + + + g (2.56) z t x y z z x y z 2.3.2.2 Conservative Form Multiply the mass conservation principle by u so that it has the same units as the momentum equation and add to the x momentum equation:  (u) (v) (w) u + u + u + u = 0 (2.57) t x y z u u u u P    xx yx zx + + u + v + w = + + + + g (2.58) x t x y z x x y z24 CHAPTER 2. GOVERNING EQUATIONS Using the product rule, this yields: (u) (uu) (vu) (wu) P    xx yx zx + + + = + + + + g (2.59) x t x y z x x y z The extension to y and z momenta is straightforward: (v) (uv) (vv) (wv) P    xy yy zy + + + = + + + + g (2.60) y t x y z y x y z (w) (uw) (vw) (ww) P    xz yz zz + + + = + + + + g (2.61) z t x y z z x y z In vector form this is written as follows: (v) +r (vv) =rP +r + g (2.62) t As with the mass equation, the time derivative can be interpreted as the accumulation of linear momenta within a control volume and the divergence term can be interpreted as the ux of linear momenta into the control volume. The accumulation and ux terms are balanced by forces, both surface and body. 2.3.3 Conservation of Energy This principle really is the rst law of thermodynamics, which states the the change in internal energy of a body is equal to the heat added to the body minus the work done by the body; E E = Q W (2.63) 2 1 12 12 The E here includes both internal energy and kinetic energy and is written for an extensive system:   1 E = V e + v v (2.64) 2 2.3.3.1 Nonconservative Form The equation we started with (which is in non-conservative form) de  =r q Pr v +:rv (2.65) dt is simply a careful expression of the simple idea de = dq dw with attention paid to sign conventions, etc. de   : change in internal energy /volume dt  r q: net heat transfer into uid/volume  Pr v: net work done by uid due to pressure force/volume (force deformation)   :rv: net work done by uid due to viscous force/volume (force deformation)2.3. CONSERVATION AXIOMS 25 2.3.3.2 Mechanical Energy Taking the dot product of the velocity v with the linear momentum principle yields the mechanical energy equation (here expressed in conservative form):     1 1  (v v) +r v (v v) =vrP + v (r) + v g (2.66) t 2 2 This can be interpreted as saying the kinetic energy (or mechanical energy) changes due to  motion in the direction of a force imbalance vrP v (r)  motion in the direction of a body force 2 Exercise: Add the product of the mass equation and u =2 to the product of u and the one dimensional linear momentum equation:     u u P  xx u  + u = u + + g (2.67) x t x x x to form the conservative form of the one-dimensional mechanical energy equation:     1 1 P  xx 2 3 u + u =u + u + ug (2.68) x t 2 x 2 x x 2.3.3.3 Conservative Form When we multiply the mass equation by e, we get  (u) (v) (w) e + e + e + e = 0 (2.69) t x y z Adding this to the nonconservative energy equation gives (e) +r (ve) =r q Pr v +:rv (2.70) t Adding to this the mechanical energy equation gives the conservative form of the energy equation:       1 1  e + v v +r v e + v v =r qr (Pv) +r ( v) (2.71) t 2 2 which is often written as       1 1 P  e + v v +r v e + v v + =r q +r ( v) (2.72) t 2 2 26 CHAPTER 2. GOVERNING EQUATIONS 2.3.3.4 Energy Equation in terms of Entropy Recall the Gibbs relation which de nes entropy s:   ds de d 1 de P d T = + P = (2.73) 2 dt dt dt  dt  dt so de ds P d  = T + (2.74) dt dt  dt also from the conservation of mass 1 d r v = (2.75)  dt Substitute into nonconservative energy equation: ds P d P d T + =r q + + :rv (2.76) dt  dt  dt Solve for entropy change: ds 1 1  = r q +  :rv (2.77) dt T T Two e ects change entropy:  heat transfer  viscous work Note the work of the pressure force does not change entropy; it is reversible work. If there are no viscous and heat transfer e ects, there is no mechanism for entropy change; ds=dt = 0; the ow is isentropic. 2.3.4 Entropy Inequality The rst law can be used to reduce the second law to a very simple form. Starting with   q 1 q r = r q rT (2.78) 2 T T T so   1 q q r q =r rT (2.79) 2 T T T Substitute into the rst law:   ds q q 1  =r rT +  :rv (2.80) 2 dt T T T2.3. CONSERVATION AXIOMS 27 Recall the second law of thermodynamics:   ds q  r (2.81) dt T Substituting the rst law into the second law thus yields: q 1 rT +  :rv 0 (2.82) 2 T T Our constitutive theory for q and  must be constructed to be constructed so as not to violate the second law.

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