Mar 4, 2018 - 4.4 Examples of simple waves, expansion waves, expansion fans, .... Figure 1.3: (a) Visualization of Mach waves around a bullet in supersonic.
FUNDAMENTALS OF GAS DYNAMICS M.E.H. van Dongen, A. Hirschberg, D.M.J. Smeulders 26 August 2016
Contents 1 Introduction
2
2 Basic Equations 2.1 Conservation Laws . . . . . . . . . . . . . 2.2 Constitutive equations . . . . . . . . . . . 2.3 Conservation laws in differential form . . 2.4 Thermodynamic relations . . . . . . . . . 2.5 Simplification of the laws of conservation; Euler equations . . . . . . . . . . . . . . . 2.6 The Bernoulli equation . . . . . . . . . . . 2.7 Questions . . . . . . . . . . . . . . . . . . 2.8 Answers . . . . . . . . . . . . . . . . . . .
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17 18 21 23
3 Stationary Quasi-1D Gas Dynamics 31 3.1 Laval tube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 3.1.1 Quasi-one dimensional theory . . . . . . . . . . . . . . 31 3.1.2 Pressure distributions inside a nozzle . . . . . . . . . . 34 3.2 Normal shock waves . . . . . . . . . . . . . . . . . . . . . . . 41 3.2.1 Rankine-Hugoniot adiabate . . . . . . . . . . . . . . . 41 3.2.2 Entropy production in shocks . . . . . . . . . . . . . . 43 3.2.3 Mach number dependance . . . . . . . . . . . . . . . . 46 3.2.4 Mach number from pressure measurements . . . . . . 48 3.3 Oblique shocks . . . . . . . . . . . . . . . . . . . . . . . . . . 49 3.3.1 Mach lines in steady flow . . . . . . . . . . . . . . . . 49 3.3.2 Shock relations for plane oblique shocks . . . . . . . . 51 3.3.3 Weak and strong oblique shocks . . . . . . . . . . . . 52 3.4 Pipe flow with friction and heat transfer . . . . . . . . . . . . 54 3.4.1 Integral conservation laws for quasi-one dimensional pipe flow . . . . . . . . . . . . . . . . . . . . . . . . . 54 3.4.2 Choking due to friction and heat transfer . . . . . . . 55 3.4.3 The long pipe-line problem . . . . . . . . . . . . . . . 57 3.4.4 Choking due to heat transfer . . . . . . . . . . . . . . 58 3.5 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
i
3.6
Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
63
4 Instationary 1D compressible flows 69 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 4.2 Non-linear waves . . . . . . . . . . . . . . . . . . . . . . . . . 69 4.3 Simple waves. . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 4.4 Examples of simple waves, expansion waves, expansion fans, compression waves, origin of shock waves. . . . . . . . . . . . 73 4.5 Deformation of an arbitrary disturbance. Conditions of shock wave formation in a medium with arbitrary thermodynamic properties. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 4.6 The shock tube . . . . . . . . . . . . . . . . . . . . . . . . . . 80 4.7 Shock tube applications . . . . . . . . . . . . . . . . . . . . . 84 4.8 Some elementary wave interactions . . . . . . . . . . . . . . . 87 4.9 Traffic waves . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 4.10 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 4.11 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
1
Chapter 1
Introduction In general a change in gas flow velocity involves a change in pressure, which is related to a change in gas density. At low velocities the pressure variations within the flow are small compared to the average absolute pressure. Density variations will be so small that we can assume the gas to be incompressible. As the velocity increases this approximation becomes inaccurate. Density and temperature changes become essential. The study of compressible fluid flows is called Gas dynamics. The most spectacular phenomena in gas dynamics are related to the ratio of the flow velocity V and the speed of sound c. The speed of sound c is the velocity of propagation of a small pressure perturbation relative to the gas. This corresponds to the velocity of the propagation of information in the flow. The ratio M = V /c is called the Mach number. A flow in which M < 1 is called subsonic. A flow with M > 1 is called supersonic. A subsonic flow approaching an obstacle will be warned by acoustic waves and will smoothly flow around the obstacle. A supersonic flow will not receive warnings, because the acoustic waves are washed away by the flow. The flow will collide on the obstacle forming a shock wave. In figure 1.1 we illustrate the difference between a subsonic and a supersonic flow. As we will see later compressibility effects become dominant for M ≥ 1. The phenomenon corresponds to a fundamental change in the type of the differential equations describing the flow. Subsonic flow are described by so called elliptical differential equations. Supersonic flows are described by hyperbolical differential equations. The same mathematics and therefore physics is found in many other phenomena. For example the flow of shallow water and car traffic. In both cases when the flow travel faster than the information one can observe discontinuities. In shallow water, this is called a water jump (see figure 1.2). In car traffic this can be an accident due to fog. In gas flows the existence of such discontinuities was controversial until Ernst Mach obtained a photograph of such shock waves (see figure 1.3).
2
Figure 1.1: Spherical waves generated by point sources in a) a subsonic flow M < 1 and in b) a supersonic flow M > 1. In the subsonic flow the waves travel upstream and can warn the flow for the presence of an obstacle. In the supersonic flow the waves are convected away and only reach a limited conical region of space (Mach cone with opening angle 2 arcsin (c/V ) ) downstream of the origin of the perturbation.
Supersonic flows involve large velocities. The Reynolds number Re = V L/ν (with L a characteristic length scale and ν the kinematic viscosity of the gas) is related to the Mach number because the viscosity of a gas is related to the speed of sound by [Landau (1987)] ¯ ν ∼ cλ
(1.1)
¯ is the mean free path of molecules. This is the average distance where λ which molecules travel between two collisions. The speed of sound is used here as an estimate for the average thermal random velocity of the molecules, which takes care of the viscous transport of momentum within the gas. Hence we see that [Landau (1987)]: L Re ∼ M ¯ (1.2) λ For compressible flows M = O(1) so that we conclude that if we use a con¯ >> 1 the Reynolds number of the flow is very large tinuum hypothesis L/λ Re >> 1. Considering low Reynolds numbers at high Mach numbers involves rarefied gas effects (Knudsen effects) such as the failure of the no-slip boundary condition at the walls. Also the assumption of local thermodynamic equilibrium will fail. The velocity distribution of molecules is not necessarily described by one single temperature T . In these lecture notes we ignore such effects, which are essential in vacuum and space technology. We focus on large Reynolds number flows, for which the fluid can be described as a continuum with locally defined thermodynamic state variables such as the pressure p, temperature T and the density ρ.1 1
We always consider absolute pressures and absolute temperatures.
3
Figure 1.2: Shallow water in a kitchen sink displaying a circular water jump. Within the circle the water flow is supercritical. The flow velocity is larger than the propagation speed of small surface waves. We observe Mach cones on the water surface. The jump is an abrupt transition to subcritical flow. In acoustics and aeroacoustics one considers the behaviour of small pressure perturbations. In gas dynamics perturbations can be arbitrarily large, which involves an essentially non-linear behaviour. The formation of shock waves, which are discontinuities in the flow, is such a non-linear phenomenon.
4
Figure 1.3: (a) Visualization of Mach waves around a bullet in supersonic flight obtained by Ernst Mach. The density changes are made visible by the refraction of light, due to the large density change (schlieren method) [Settles (2001)]. For an observer moving with the bullet the approaching flow is supersonic. (b) Bullet shot by a rifle. Schlieren photograph by Gary Settles. The understanding of gas dynamics has been developed mainly in the nineteenth century by: Poisson (1808) non-linear plane waves, Riemann (1860) method of characteristics, Rankine (1870) and Hugoniot (1887), Mach (1887) shock-waves, Prandtl and Meyer (1908) 2-D steady gas dynamics. The significance of gas dynamics has dramatically increased in the twentieth century with the large scale use of gas turbines for electricity production, combustion car engines and aircraft engines. In the second half of the twentieth century the space exploration and military applications have become important (supersonic aircraft, rockets, scramjets). Also turbocompressors are currently used in modern cars. Supersonic flows are also found in torches used for welding and plasma surface treatment. Also gas dynamics and shock waves are essential in astronomy to understand the evolution of stars and galaxies. As indicated above many non-linear wave phenomena are described by the same mathematical tools as gas dynamic flows, such as: tsunami waves, car traffic waves, compositional waves in chromatography, water breakthrough in secondary oil production, kidney stone destruction by ultrasound waves. The formation of shock waves explains also simple daily phenomena such as the whip crack, the sound of hand clapping, the sound of thunder, the tongue clap and the brassy sound of trombones and trumpets [Hirschberg (1996)]. In these lecture notes we will limit ourselves to a very modest study of a few basic flow phenomena. Chapter 2 describes the basic equations of gas flows starting from the mass, momentum and energy integral balance equations. The differential form of these equations and constitutive equations are then discussed. Thermodynamics is refreshed, which is needed to obtain relations between basic state variables such as pressure, density, temperature, speed of sound, enthalpy
5
and entropy. We then focus on the thermal properties of ideal gases and define perfect gases. Simplified equations for frictionless flows are introduced. In particular we discuss the equation of Bernoulli for steady frictionless flows. Chapter 3 discusses mainly quasi-one dimensional steady flows. We start by discussing the flow through a Laval nozzle (convergent divergent pipe used for rocket engine). We then discuss the normal shocks (perpendicular to the flow), which can appear in the divergent part of the Laval nozzle. After a small excursion to 2-D shock waves (waves around aircraft), we discuss the effect of friction and heat transfer in pipe flows (gas transport pipe lines). Chapter 4 describes the one-dimensional unsteady wave propagation in pipes. This is used to introduce the method of characteristics. We use this to understand the formation of shock discontinuities. Technological applications of this are the water hammer phenomena, the shock tube and rasping sounds of combustion engine mufflers. Chapter 4 also describes the use of the method of characteristics to analyze car traffic waves [Lighthill (1955)], [Whitman (1974)], [Thompson (1972)]. At the end of each chapter we provide questions with detailed answers. Basic questions can be used to study the theory. Additional question marked with a star are useful for students looking for a deeper understanding of the theory. At the end of the lecture notes a collection of old exam questions is provided. In the list of references at the end of these lecture notes we give some references to lecture notes and textbooks. The student can find is these excellent books complementary information or alternative explanations. These lecture notes are a translation of the original lecture notes of M.E.H. van Dongen (published in 2008), with some modifications and the addition of exercises.
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Chapter 2
Basic Equations In this chapter we discuss the fundamental equations of fluid dynamics and thermodynamics. Together they form the underlying basis for an analytical description of compressible fluid motion.
2.1
Conservation Laws
n
F dA
v
B
Figure 2.1: Control volume B, with outer boundary F. We begin with the conservation of mass. Consider an arbitrary fixed volume B with boundary F , as seen in figure ! 2.1. The mass of a fluid (which can be a liquid or gas) in this volume is ρ dτ , with ρ the fluid density and dτ an infinitesimally small volume element within B. Per unit time, a fluid mass ρ(v · n) dA flows through a surface element dA, defined by its unit normal vector n (pointing outward). The fluid velocity is v. Without mass sources and sinks we have that: " " d ρ dτ + ρ(v · n) dA = 0 (2.1) dt B F Using the momentum density ρv, we obtain an expression for momentum balance. The total amount of momentum changes due to volume and surface 7
forces. This leads to two terms on the right hand side of the equation, describing momentum production: " " " " d ρv dτ + ρv(v · n) dA = ρg dτ + t dA, (2.2) dt B F B F
with ρg a body force (e.g., gravity) and t a surface force (e.g., caused by friction or pressure). For the same volume B we can also consider the energy of the fluid per unit mass, which comprises kinetic energy (1/2V 2 ) and internal energy (e) per unit of mass. This energy can flow through F and can be changed by the work of body forces, surface force, by transport and production of heat: d dt
" # B
$ $ " # 1 2 1 2 V + e ρ dτ + V + e ρ(v · n) dA = 2 2 F " " " " −(q · n) dA + ρ(g · v) dτ + (t · v) dA + φE dτ F
B
F
(2.3)
B
with q the heat flow density and V 2 = |v|2 . The rate of energy production per unit volume φE includes any process not accounted for in the definition of e. This could be chemical or nuclear reactions, or electrical heating. Equation (3.39) is a generalization of the first law of thermodynamics to open systems.
2.2
Constitutive equations
The surface force t depends on the choice of the control surface F . One prefers to describe this force in terms of a stress tensor S, independent of F , at a given point in space. The relation between the surface force t, the normal n and the stress tensor S is given by: t = n · S ; S = −pI + S ′ ,
(2.4)
with p the mechanical pressure, I the unit tensor and S ′ the deviatoric stress tensor. For a Newtonian fluid it holds that: S ′ = 2ηD + η ′ (∇ · v)I 1 = 2η(D − (∇ · v)I) + κ(∇ · v)I. (2.5) 3 For such a fluid, the stress tensor S ′ depends linearly on the deformation % & ∂vj ∂vi tensor D. This deformation tensor is defined as [D]ij = 12 ∂x + ∂xi with j i = {1, 2, 3} and vi the components of the velocity v. η is the dynamic viscosity, η ′ is the second viscosity coefficient and κ is the so-called bulk viscosity. They are related by: 8
2 κ = η ′ + η. (2.6) 3 The use of the bulk viscosity instead of the second viscosity coefficient is preferred, because of its direct relation to the compressibility of the fluid, and the fact that it is always positive and vanishes for noble gasses (He, Ne, Ar, Xe, Kr). The heat flow density q satisfies Fourier’s Law: q = −λ∇T,
(2.7)
in which λ is the heat conduction coefficient and the temperature T will be defined in section 2.4 where we discuss the thermal equation of state. This linear relationship between q and ∇T is consistent with Newton’s approximation (2.5) for the stress tensor.
2.3
Conservation laws in differential form
The constitutive equations (2.4) - (2.7) can be substituted into the conservation laws. Next, Gauss’s law can be used to transform the surface integrals into volume integrals. In this way, we obtain a set of integral equations over an arbitrarily chosen control volume. This allows rewriting the conservation laws into their differential form: ∂ρ Dρ + ∇ · ρv = 0, or + ρ∇ · v = 0, (2.8) ∂t Dt D ∂ where it was used that Dt ≡ ∂t +v·∇, defining the convective time derivative. We also have that: Dv ρ = −∇p + ∇(η ′ ∇ · v) + ∇ · (2ηD) + ρg; (2.9) Dt and d ρ1 De ρ + pρ = ∇ · (λ∇T ) + η ′ (∇ · v)2 + 2η(D : D). (2.10) Dt dt We ' recognize (2.9) as Newton’s law applied to a point mass (mDv/Dt = F ). We now have a set of 5 scalar equations with 7 unknowns: ρ, v, p, T, e. From thermodynamics in combination with empirical data, we construct the constitutive equations we need to close the system of equations.
2.4
Thermodynamic relations
For a gas of uniform composition, it is sufficient to use an equation of state of the form: 1 e = e(ρ, s), (2.11) 1 The equation of state (2.11) is obtained from quasi-static measurements. We assume that it remains locally valid in the non-equilibrium flow condition considered. This is
9
with the additional variable s the specific entropy (entropy per unit mass). Indeed, from the fundamental thermodynamical law de = T ds − pd
1 (Gibbs’ relation) ρ
and the differential form of (2.11), we find that: ( ) # $ # $ ∂e ∂e 1 de = ds + d , ∂s ρ ρ ∂ ρ1
(2.12)
(2.13)
s
so that we immediately find the thermodynamic definitions of T and p: # $ ∂e T ≡ , (2.14) ∂s ρ and p≡−
(
∂e ∂ ρ1
)
.
(2.15)
s
The set of equations (2.8) - (2.10), (2.11), complemented with (2.14) and (2.15) now contains 8 equations with 8 variables.2 We will now discuss a number of important terms and definitions we will encounter during the study of gas dynamics.
Enthalpy The enthalpy is defined as:
p h≡e+ . ρ
(2.16)
Substitution into (2.12) gives that: 1 dh = T ds + dp. ρ
(2.17)
Specific heat The specific heats cv and cp at constant volume and constant pressure, respectively, are given by3 : # $ # $ # $ # $ δQ ∂e δQ ∂h cv = = ; cp = = . (2.18) δT ρ ∂T ρ δT p ∂T p the so-called ”local” thermodynamic equilibrium assumption, which is consistent with the choice of Newton’s approximation (2.5) for the viscous stress and Fourier’s law (2.7) for the heat flux, assuming that fluxes are proportional to the gradients in the flow. 2 As explained by [Thompson (1972)] the hydrodynamic definition (2.15) of the pressure is the only useful definition of pressure. 3 We use δQ/δT to stress the fact that Q is not a state variable
10
where δQ is the amount of heat added to the system and δT the corresponding increase in temperature. We have following the first law of thermodynamics δQ = de + pd(1/ρ). The Poisson ratio γ is defined by: γ ≡ cp /cv . The specific heats cv and cp correspond to the amount of heat one should supply to increase the temperature of one kilogram of gas by one degree, at respectively constant volume and constant pressure.
Speed of sound The thermodynamic property c is defined as: # $ ∂p 2 c = , ∂ρ s
(2.19)
which can also be written as: c2 = γ
#
∂p ∂ρ
$
= T
1 ρ
%
−
&
∂h ∂ρ p
%
&
∂h ∂p ρ
(2.20)
Later is will become clear that the property c is the propagation speed of sound waves (speed of sound).
Ideal gas At low densities, gases are found to obey closely the Boyle-Gay Lussac equation of state of the form: p = ρRT (2.21) where R is the specific gas constant: R = R0 /M ; with R0 the universal gas constant (8, 314kJ kmol−1 K−1 ) and M the molar mass (kg kmol−1 ) of the gas. An ideal gas is defined as a gas which obeys the above equation of state. By substituting (2.21) into (2.19) it follows directly that c2 = γRT.
(2.22)
For an ideal gas, it can be shown that the internal energy and enthalpy are functions of temperature only: e = e(T ) and h = h(T ), so that de = cv dT and dh = cp dT with cv (T ) and cp (T ). By definition of h (2.16), the difference between the specific heats is: # $ p dh − de = (cp − cv )dT = d = RdT ρ hence: cp − cv = R 11
(2.23)
Perfect gas A calorically perfect gas is an ideal gas with constant specific heat. As the specific heat is independent of temperature, we find: p 1 c2 = ; ργ−1 γ(γ − 1) p γ c2 h = cp T = = ργ−1 γ−1
e = cv T =
The fundamental law of thermodynamics (2.11) provides an expression for the entropy for an ideal gas: de = cv dT = T ds − pd
1 ρ
ds dT p dρ dT dρ = − = − (γ − 1) cv T cv T ρ ρ T ρ Integrating, assuming cv and γ to be constant, one obtains for a perfect gas: s − s0 T ρ = ln − (γ − 1) ln , cv T0 ρ0
(2.24)
where the index 0 refers to a reference state or initial condition. When we assume an isentropic proces s = s0 , we obtain: #
T T0
$
=
#
ρ ρ0
$γ−1
(2.25)
and using the ideal gas law (2.21), the isentropic gas relation becomes: #
p p0
$
=
#
ρ ρ0
$γ
=
#
T T0
$γ/(γ−1)
=
#
c c0
$(2γ)/(γ−1)
.
(2.26)
Internal energy The specific internal energy of a molecule is determined by the number of degrees of freedom available to store energy: 1 e = cv T = f RT , 2
f = ntr + nrot + 2nvibr + nel
(2.27)
with: ntr the number of translational degrees of freedom, each contributing 1 RT to the internal energy; 2 12
nrot the rotational degrees of freedom, contributing 12 RT each when the temperature is above the characteristic rotation temperature, θr ; nvibr the vibrational degrees of freedom, contributing RT each when the temperature is above the characteristic vibration temperature, θv . More complex molecules have a θv for every vibrational mode; nel takes into account the electron excitation energy, when the temperature reaches θel . Figure 2.2 shows how the specific heat cv depends on temperature for hydrogen.4 Note the temperature ranges in which cv is constant. This is where the gas behaves ’perfectly’. In between are regions where the gas gradually excites new degrees of freedom.
Figure 2.2: cv /R vs. T for hydrogen. ([Owczarek (1964)]). At temperatures of the order of 104 K, next to dissociation, ionization will become significant. 4 Following classical theory, the total number of degrees of freedom of a molecule of N atoms is 3N , which corresponds to the coordinates of the positions of the atoms. We can move a molecule without deformation (coherent motion), by translation, or rotation. Let ntr and nrot be the number of these degrees of freedom. All other motions of a molecule will be associated with deformation. For small deformations this will result into harmonic oscillation around the equilibrium shape. For such a harmonic motion the time average of the kinetic energy is equal to the time average of the potential energy (due to the deformation). Hence for any molecule, nvibr = 3N − ntr − nrot corresponds to the number of vibrational degrees of freedom. A mono-atomic gas (He, Ne, Ar, Xe, Kr) only has f = 3, because N = 1. For a diatomic gas (H2 , N2 , O2 , CO), N = 2 and we have ntr = 3, nrot = 2, so that nvibr = 3N − ntr − nrot = 1 and f = ntr + nrot + 2nvibr = 7. Measurements show that this exact classical result is wrong! This has led to the discovery of Quantum Mechanics, which does predict that below critical temperatures θrot and θvibr the rotational or vibrational degrees of freedom are inactive (frozen).
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With γ = cp /cv , we have γ=
f +2 f
(2.28)
for Poisson’s ratio. At room temperature, linear molecules like O2 , N2 , CO, CO2 have 2 rotational degrees of freedom while the vibrational mode is frozen, leading to γ = 7/5. A number of characteristic temperatures is given in table 2.1 (from Thompson, 1972). Here, θd is the characteristic dissociation temperature at which dissociation starts. As dissociation and ionization start we have strong peaks in the specific heat. Figure 2.3 shows that these peaks depend strongly on the pressure. This is due to the fact that recombination is a three particle process which is more likely to occur at high densities. Therefore, the degree of dissociation or ionization is dependent on the pressure. Under such conditions the internal energy is a function of both temperature and density. The ideal gas law is certainly not valid any more. The specific heat starts to become a function of pressure as well, as can be seen in figure 2.3. Gas H2 N2 O2 CO
θr [K] 87.5 2.89 2.08 2.78
θv [K] 6325 3393 2273 3122
θd [K] 52000 113400 59000 12900
Table 2.1: Characteristic temperatures
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Figure 2.3: cv /R vs. T for hydrogen. ([Owczarek (1964)])
Relaxation time A system does not instantaneously achieve a new equilibrium state. It takes a certain time for the system to adapt to changes in pressure, temperature, etc; especially when these changes occur very fast. The lagging of internal processes such as dissociation, ionization, evaporation, chemical reaction and transfer of energy between molecular modes (translation, rotation, vibration) is called a relaxation process. It is possible to define a relaxation time, which is a measure of the time required for a system τ to reach equilibrium after a disturbance. When the time scale of the process one considers is small, compared to τ or of the order of τ , the equilibrium equation of state is no longer valid. Table 2.2 shows for some common gases the relaxation times for the different molecular processes ([Thompson (1972)]). The difference in relaxation time for translation and rotation of air molecules largely determine the volume viscosity of air. This explains the dissipa15
Gas N2 N2 N2 H2 O O2
mode translation rotation vibration vibration dissociation
T [K] 300 300 3000 486 3100
τ [s] 1.6 · 10−10 1.2 · 10−9 3.0 · 10−5 3.7 · 10−8 2.3 · 10−5
Table 2.2: Characteristic relaxation times for various processes, at p = 1 bar ([Thompson (1972)]) tion of high frequency sound upon propagation through the atmosphere ([Pierce (1989)]).
Relation between conservation of energy and the laws of thermodynamics The law of conservation of energy (3.39) can be seen as the differential form of the fundamental law of thermodynamics: D ρ1 De 1 +p = − ∇ · q + (viscous dissipation) * +, Dt Dt ρ
(2.29)
>0
We follow a material element (fluid particle) with internal energy e, density ρ and pressure p. When we let it undergo a reversible change, 5 there is per definition no viscous dissipation and, for an infinitesimal time step δt, de + pd
1 1 = − (∇ · q)rev δt. ρ ρ
(2.30)
The right hand side can be written as (dQ)rev . From considerations regarding reversible cycle processes, we know that (dQ)rev must equal T ds, thereby defining a new state variable: the specific entropy s. Gibbs’ relation (2.12) is valid also for non-reversible processes. We can use this relation to rewrite equation (2.29): T
Ds 1 = − ∇ · q + (viscous dissipation) Dt ρ
(2.31)
also valid for non-reversible processes. Next, we can make use of the vector identity q 1 1 ∇· = ∇·q+q·∇ , (2.32) T T T 5 Dissipation involves the square of the rate change such as we see in the law of Ohm for a resistor: power = RI 2 , for a resistance R and an electrical current I. Transporting an electrical charge very slowly through a resistor makes the dissipation negligible. Hence a quasi-steady process is reversible.
16
to obtain: ρ
q Ds 1 + ∇ · = − 2 (q · ∇T ) + ρ(viscous dissipation), Dt T * T +, -
(2.33)
>0
or simply:
q ds +∇· ≥0 (2.34) dt T This relation can be seen as the differential formulation of the second law of thermodynamics. By using conservation of mass and Gauss’ law, we obtain the integral version of (2.34): " " " d 1 ρs dτ + ρs(v · n) dA + (q · n) dA ≥ 0. (2.35) dt B T F F ρ
2.5
Simplification of the laws of conservation; Euler equations
Suppose that the magnitude of the velocity V , the temperature T and the pressure p vary over a known distance L. This can be the length of the system or a typical length scale within the system. Then, we can make an approximation for the relative magnitude of friction and heat conduction by defining a Reynolds and P´eclet number, respectively: ρV L η VL Pe = , a
Re =
(2.36) (2.37)
with η the dynamic viscosity and a = ρcλp the thermal diffusivity coefficient. Re compares the stationary inertial forces to the viscous forces; Pe compares convective energy transport to energy transport by conduction. For most situations considered in gas dynamics, Re and Pe will both be much larger than 1, enabling us to neglect the influence of friction and heat conduction in the bulk of the flow6 . However, it is important to note that this approximation becomes invalid near walls or within shock waves. When neglecting the friction and heat conduction terms, equations (2.8 - 2.10) reduce to the 6
As in gas viscous momentum transfer and heat transfer are limited by the same c η molecular collision process we have P r = pλ = 0(1). Hence Re and P e are of the same order of magnitude.
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Euler equations: ∂ρ + ∇ · ρv = 0, (2.38) ∂t Dv ρ = −∇p, (2.39) Dt Ds ρT = 0, (2.40) Dt where we neglected gravity as well. The last equation, conservation of energy, follows from the relation between internal energy and enthalpy, (2.16): ρ
2.6
D ρ1 De Dh Dp Ds + pρ =ρ − = ρT Dt Dt Dt Dt Dt
The Bernoulli equation
The momentum equation (2.39) ∂v + ρ(v · ∇)v = −∇p, ∂t together with the vector identity ρ
1 (v · ∇)v = ∇ V 2 + ∇ × v × v, 2 the energy balance (2.40) ∂s + v · ∇s = 0, ∂t and the Gibbs’ relation for the enthalpy: 1 ∇p = ∇h − T ∇s, ρ can be reduced to the Crocco-Vazsonyi equation: ∂v 1 + ∇( V 2 + h) + ∇ × v × v − T ∇s = 0. (2.41) ∂t 2 from this result we can deduce a number of variants of the Bernoulli equation: a Bernoulli, in case of a stationary, isentropic flow: v · ∇s = 0. Take the inner product of (2.41) with v to obtain an equation that is valid along a streamline: 1 v · ∇( V 2 + h) = 0, or: 2 1 2 V + h = h0 (iso-energetic, valid along a streamline) (2.42) 2 The constant h0 is called the total enthalpy. 18
b Bernoulli, in case of a stationary, homentropic (uniform entropy), rotationfree flow: ∇s = 0 and ∇ × v = 0. Then: 1 2 V + h = h0 2
(homo-energetic, valid in the entire flow field)
(2.43)
c Bernoulli, in case of an instationary, homentropic, rotation-free flow: ∇s = 0 and with ∇ × v = 0. Because ∇ × ⃗v = 0 there exists a scalar potential φ such that v = ∇φ. Substitution into equation (2.41) gives: ∂φ 1 2 + V + h = C(t) ∂t 2
(entire flow field)
(2.44)
d Bernoulli in differential form. From (2.43) it follows that: 1 V dV + dh = V dV + dp = 0. ρ
(2.45)
Stationary Bernoulli equation for a perfect gas For a perfect gas, by making use of h = cp T , 1 2 V + cp T = cp T0 ; 2
(2.46)
and c2 = γRT, R = cp − cv and γ = cp /cv (2.22 - 2.23), we obtain: T0 γ−1 2 =1+ M , T 2
(2.47)
where M is the Mach number: M = Vc . Starting with the same Bernoulli equation, it is also possible by using (2.26) to derive expressions containing p, ρ and c. The quantities T0 , p0 , ρ0 , c0 are the “total” temperature, -pressure, -density and speed of sound. They are the magnitudes of temperature, pressure, etc. that are reached when the fluid is decelerated to V = 0 by means of an hypothetical isentropic, and quasi-steady process.7 They can be different from the real values Ts , ps , ρs , cs in a stagnation point. In general the flow around a stagnation point is not isentropic nor frictionless. In practice a supersonic flow originates from a settling chamber. The state in this reservoir where the velocity is negligible is called the reservoir state pc , ρc and Tc (with the index c stands for settling “chamber”). From (2.47) we derive expressions relating the density, pressure, speed of sound by using the isentropic relations for a perfect gas (2.26). Thanks to this we have now for given total conditions T0 , ρ0 , ... expressed T, ρ, ... as a function of the Mach number M . From (2.46) it follows that for a given T0 , there exists a maximum achievable velocity: . . Vmax = 2cp T0 = 2h0 (2.48) which can be reached in a steady flow. 7
a quasi-steady isentropic process is by definition frictionless
19
The critical state The temperature, pressure and speed of sound at which the Mach number is equal to 1 are referred to as the critical temperature T ∗ , critical pressure p∗ and critical speed of sound c∗ . This is a so-called sonic condition, where the velocity is equal to the speed of sound. Starting with (2.47) and taking M = 1, we obtain for a steady isentropic flow: T∗ 2 = . T0 γ+1
(2.49)
Conditions for p∗ , ρ∗ and c∗ can also be found using (2.49) in combination with (2.26).
The energy ellipse For a perfect gas, the Bernoulli equation describes an ellipse in the (c, V ) plane. This is shown in figure 2.4. For V < V ∗ = c∗ the flow is subsonic. Around V = V ∗ the flow is said to be transonic, and for V > V ∗ supersonic.
c M=1
c0 c*
subsonic
transsonic sup
ers
on
ic
Vmax
V*
V
Figure 2.4: The energy ellipse. We observe from figure 2.4 that for low subsonic flows (M ≤ 0.2) the speed of sound is almost constant hence the flow is almost isothermal. Also the density changes remain small so that the flow is in good approximation “incompressible”. For very large Mach number (M ≥ 5), in √ the so called hypersonic region, the flow velocity is very close to Vmax = 2h0 . Changes in the Mach number M = V /c are dominated by the variation of the speed of sound and associated with a strong density variation. Hypersonic flows are associated with spectacular variations in temperature across the flow field (between the free flow and a stagnation point for example).
20
For some applications an alternative Mach number M ∗ is used instead of M . This Mach number is defined as: M∗ =
V c∗
(2.50)
where c∗ = V ∗ is the critical flow speed, which is reached when the flow is isentropically and steadily accelerated/decelerated to M = 1. As M ∗ is proportional to V it is in the finite range: 0 ≤ M ∗ < Vmax /c∗
(2.51)
while the Mach number varies in the range: 0 ≤ M < ∞.
2.7
(2.52)
Questions
Questions with a star (*) are more advanced and not considered as basic study material. 1. How is the control volume B determined when applying the integral conservation laws (eq. 2.1, 2.2 and 2.3)? 2. How is the unit vector ⃗n determined when applying the integral equations (eq. 2.1, 2.2 and 2.3)? 3. Why is the bulk viscosity more convenient than the second viscosity? 4. What are the main additional limitations to the validity of the differential form of the conservation equations (eq. 2.8, 2.9 and 2.10) compared to the integral formulation (eq. 2.1, 2.2 and 2.3)? 5. What is a state variable? 6. What is the definition of the temperature T ? 7. What is the difference between an adiabatic process and an isentropic process? 8. What is the definition of an ideal gas? * 9. Starting from the Gibbs’ relation (eq. 2.12) show that: # $ # $ ∂(p/ρ2 ) ∂s =− . ∂T ∂ρ ρ T
21
* 10. Show that:
#
∂e ∂ρ
$
T
/ # $ 0 1 ∂p = 2 p−T ρ ∂T ρ
and verify that the internal energy of an ideal gas is independent of its density. * 11. Show that a gas satisfying the equation of state of Gay-Lussac, p = ρRT , is an ideal gas. 12. Show that for an ideal gas: cp − cv = R 13. Starting from the Gibbs’ relation, show that for an ideal gas: c2 =
γp ρ
14. What is a calorically perfect gas? 15. Starting from Gibbs’ relation show that for a perfect gas: # $γ # $ p ρ s − s0 = exp po ρ0 cv 16. Derive equation: p = po
#
ρ ρ0
$γ
Under which conditions is it valid? 17. Why do we assume that the energy in a vibrational degree of freedom at equilibrium is twice that in a translational or rotational degree of freedom? When is this reasonable? 18. How many degrees of freedom has a molecule with N atoms when we assume that electronic states are not excited. How many of these degrees of freedom are vibrational? 19. Why is the formula: cv =
R (ntr + Nrot + 2nvibr ) 2
in general not valid? * 20. Starting from the energy equation (2.10) derive the entropy equation (2.40). 22
21. Using the integral energy equation (2.3) applied to the adiabatic flow through a segment of a pipe one can show that the total enthalpy is conserved: V2 V2 h0 = h1 + 1 = h2 + 2 2 2 What are the differences between this equation and the equation of Bernoulli: V2 h0 = h + ? 2 22. Derive equation: T0 /T = 1 + (γ − 1)M 2 /2 Specify the assumptions you need. 23. Calculate the critical pressure p∗ as a function of the reservoir pressure p0 for a perfect gas. 24. What is the temperature of a gas when it has reached is maximum √ velocity Vmax = 2h0 in a steady flow? What is the physical interpretation at a microscopic level of this result? * 25. Determine the Mach number dependency of the pressure as a function of the velocity in a steady isentropic flow. Find an expression valid for low subsonic flows. 26. Determine the critical velocity V ∗ of a flow as a function of the reservoir conditions (p0 , T0 ). 27. What are the differences between the reservoir pressure, the total pressure and the stagnation pressure in a flow? 28. An aircraft flies at a typical Mach number M = 0.8. Estimate the temperature difference between the stagnation point and the surrounding air. Same question for the Blackbird with M = 3, a scramjet M = 7 and Apollo during re-entry in the atmosphere M = 24. Assume air to be a perfect gas. Will this assumption be realistic. Provide arguments.
2.8
Answers
1. The control volume B is chosen by the user. Usually it is chosen for its convenience. 2. The normal is directed outwards. Choosing the normal to the control surface defines the “interior” of the control surface. 3. The second viscosity is directly related to volume changes.
23
4. The flow is described by continuous fields that allow differentiation. The integral formulations allow discontinuous fields as found in shocks or contact surfaces (which we will discuss later). 5. Thermodynamics is the study of the equilibrium state of systems. A system is in equilibrium when all the macroscopic measurements given time independent results. All measurable quantities that are independent of the way equilibrium was reached are state variables. Such as temperature T , pressure p, density ρ, entropy s, internal energy e, speed of sound c... 6. The zeroth law of the thermodynamics states that T is a state variable. The second law defines further T and the entropy s by stating that T and s obey T ds = dQ for a reversible process. 7. In both cases there is no exchange of heat with the surroundings, but the isentropic process is also reversible (quasi-steady). 8. In an ideal gas the molecules interact by collisions but do not have long range interactions. As a consequence the internal energy depends only on the temperature, not on the fluid density (volume). 9. From (eq. 1.12): T ds = de + pd(1/ρ) ⇒ de = T ds − pd(1/ρ) f = e − T s ⇒ df = −sdT + (p/ρ2 )dρ As the internal energy is a state variable: # $ # $ ∂f ∂f df = dT + dρ ∂T ρ ∂ρ T and:
#
∂f ∂T
$
ρ
= −s ;
#
∂f ∂ρ
=
#
$
= (p/ρ2 ) T
Furthermore: ( implying:
∂ ∂ρ
#
∂f ∂T
#
∂s ∂ρ
$ )
$
ρ
T
T
=−
24
#
∂ ∂T
#
∂f ∂ρ
∂(p/ρ2 ) ∂T
$
$ $ T
ρ
ρ
10. T ds = de + pd(1/ρ) ⇒ de = T ds − pd(1/ρ) = T ds + ds =
#
∂s ∂ρ
$
dρ + T
#
∂s ∂T
$
p dρ ρ2
dT ρ
⇒ 1 # $ 2 # $ ∂s p ∂s de = T + dρ + T dT ∂ρ T ρ2 ∂T ρ From question 1.7 we know that: # $ # $ # $ ∂(p/ρ2 ) 1 ∂p ∂s = 2 =− ∂T ρ ∂T ρ ∂ρ T ρ Hence:
#
∂e ∂ρ
$
11. Using Gay-Lussac:
so that:
#
12.
T
# ∂e ∂ρ
$
T
/ # $ 0 1 ∂p = 2 p−T ρ ∂T ρ ∂p ∂T
$
= ρR = ρ
p T
/ # $ 0 1 ∂p = 2 p−T =0 ρ ∂T ρ
# $ 1 dQ = de + pd ρ # $ # $ δQ ∂e cv ≡ = δT ρ ∂T ρ 1 dQ = dh − dp ρ p h≡e+ ρ # $ # $ # $ # $ δQ ∂h ∂e ∂(p/ρ) cp ≡ = = + δT p ∂T p ∂T ρ ∂T ρ For an ideal gas:
#
∂(p/ρ) ∂T
25
$
=R ρ
13. T ds = de + pd(1/ρ) Isentropic and ideal gas: # $ ∂e p dρ dT R dρ 0= dT − 2 dρ = cv dT − RT ⇒ = ∂T ρ ρ ρ T cv ρ and 1 2 # $ cp − cv dρ cp dρ cp p dp dT dρ ∂p = + = 1+ = ⇒ = p T ρ cv ρ cv ρ ∂ρ s cv ρ 14. A perfect gas is an ideal gas with a constant specific heat. 15. Perfect gas: T ds = de + pd(1/ρ) ⇒
cv R dT − dρ T ρ " " " dT dρ ⇒ ds = cv dT − R T ρ # $ # $ s − s0 T ρ = ln − (γ − 1) ln cv T0 ρ0 # $ # $γ−1 # $ T ρ s − s0 = exp T0 ρ0 cv # $γ # $ p ρ T ρ s − s0 = = exp p0 ρ0 T0 ρ0 cv ds =
16. Starting from: p = po
#
ρ ρ0
$γ
exp
#
#
$γ
We assume an isentropic process : p = p0
ρ ρ0
s − s0 cv
$
This equation is valid for the isentropic flow of a perfect gas. 17. Translation and rotation of a molecule does not involve a deformation. Vibration involves deformation. For small vibrational amplitudes, we can assume a harmonic oscillation around the equilibrium shape of the molecule. In a harmonic oscillation the time average potential energy is exactly equal to the time average kinetic energy. For large amplitudes as we approach dissociation, the vibration becomes un-harmonic and this assumption fails. 26
18. In 3-D space we need 3 coordinates to determine the position of each atom. Hence the total number of degrees of freedom of the molecule is 3N . The number of vibrational degrees of freedom is: nvibr = 3N − ntr − nrot In free space (3-D) a molecule has 3 translational degrees of freedom: ntr = 3 A mono-atomic gas (N = 1) has no rotational nor vibrational degrees of freedom. A linear molecule (such as in diatomic gases) has two rotational degrees of freedom and one vibrational degree of freedom. A non-linear molecule (water, methane, benzene) has 3 rotational degrees of freedom. Hence water with 3 atoms has 3 vibrational modes. Methane with 5 atoms has 9 vibrational modes. 19. Due to quantum mechanical effects degrees of freedom remain “frozen” (inactive) below a critical temperature. The energy in the modes can only increase stepwise. If the energy step is large compared to the average translational kinetic energy, only very few molecules will reach the excited state for that mode. For example at room temperature the vibrations of oxygen and of nitrogen molecules in air are negligible. It is interesting to note that the difference between the measured specific heat (for air cv = 5R/2 and the predicted value following cv = 7R/2) led to the discovery of quantum mechanics (Planck, Einstein). 20.
De D(1/ρ) + pρ = ∇ · (λ∇T ) + η ′ (∇ · ⃗v )2 + 2η(D : D) Dt Dt Reversible process, Gibbs’ relation: ρ
T ds = de − pd(1/ρ) Yields:
Ds = ∇ · (λ∇T ) + η ′ (∇ · ⃗v )2 + 2η(D : D) Dt No friction nor heat transfer: ρT
ρT
Ds =0 Dt
21. The equation of Bernoulli is valid along a stream line in a frictionless isentropic flow. It is valid over the entire flow if the entropy is uniform. In the integral energy equation applied to a pipe of uniform cross section, we assume that the flow is adiabatic. Friction is taken into 27
account so that the flow is not isentropic. Due to the no-slip condition at the wall the friction forces do not carry any work. The flow is steady but not uniform. The velocity term V 2 /2 is the average of this term over the pipe cross section (not a local value). 22. We assume steady frictionless isentropic flow so that the equation of Bernoulli is valid along a streamline: h0 = h1 +
V12 V2 = h2 + 2 2 2
For a perfect gas we have: h = cp T For an ideal gas we have furthermore: R = cp − cv And: c2 = γRT Hence: cp T0 = cp T + T0 V2 γR =1+ =1+ T 2cp T 2cp
#
V c
V2 2
$2
=1+
cp c p − cv 2 M cv 2cp
23. The critical temperature is given by (M = 1): T0 γ−1 γ+1 =1+ = ∗ T 2 2 We have furthermore: #
ρ ρ0
T∗ T0
$
p = p0 p∗ = p0
#
$γ γ γ−1
=
#
p T0 p0 T
$γ
=
#
2 γ+1
$
γ γ−1
24. At M = ∞ the gas has reached the absolute zero (T = 0K), which is impossible following the third law of thermodynamics. This means that all the random kinetic energy (translation, rotation and vibration) of the molecules in the reservoir has been transformed into a macroscopic translation kinetic energy. The density is vanishing small so that we have a kind of “vacuum” in which there is no collision between molecules.
28
25. For a perfect gas:
V2 2 T0 γ−1 2 =1+ M T 2 ( ) γ # $ γ γ−1 p T γ−1 1 = = 2 p0 T0 1 + γ−1 2 M cp T0 = cp T +
Taylor expansion: p0 = p
#
γ−1 2 1+ M 2
$
γ γ−1
γ−1 γ 1 γ ≈1+ M2 + 2 γ−1 2γ−1 # $ γ 2 1 2 = 1 + M 1 + M + ... 2 4
#
$# $ γ γ−1 2 2 −1 M ... γ−1 2
# $ V2 V2 γ−1 2 M = = 1+ M γRT γRT0 2 # # $ $ p0 − p γ V2 γ−1 1 V2 = 1+ + + ... p 2 γRT 2 4 γRT0 # $# $ p0 − p ρ M2 = 1+ + ... 1 2 ρ0 4 2 ρ0 V # $ −1 ρ γ − 1 2 γ−1 M2 = 1+ M =1− + ... ρ0 2 2 # $ p0 − p M2 = 1− + ... 1 2 4 2 ρ0 V 2
26. We look for the critical velocity as a function of the reservoir conditions. . . V ∗ = c∗ = c0 (c∗ /c0 ) = c0 T ∗ /T0 = 2γRT0 /(γ + 1)
27. The reservoir pressure is the pressure in a “reservoir”(settling chamber), which is used in order to generate the flow. We often assume that the flow velocity in the reservoir is negligible and that the flow is steady and isentropic. In that case the reservoir conditions correspond to the “total” pressure and temperature. The “total” pressure and temperature correspond to the thermodynamic state reached by a fluid particle which we bring to “rest” by using a steady, reversible isentropic process. The stagnation conditions are the pressure and temperature, 29
which are reached in a stagnation point. As the flow near a stagnation point is not necessarily isentropic and reversible, the stagnation pressure deviates in general from the total pressure and temperature. In the particular case of a Pitot tube placed in a supersonic flow, the stagnation pressure will be lower than the total pressure upstream of the tube, because of dissipation on a shock wave in front of the tube (see chapter 3). Furthermore the heat conduction along the tube and radiation heat transfer will result into a temperature at the stagnation point which is different from the total temperature upstream of the tube. 28. γ = 1.4 γ−1 2 M = 1 + 0.2M 2 2 T = 300K
(T0 /T ) = 1 +
M = 0.8 ⇒ (T0 /T ) = 1.128 ⇒ T0 − T = 38.4K M = 3 ⇒ (T0 /T ) = 2.8 ⇒ T0 − T = 540K M = 7 ⇒ (T0 /T ) = 10.8 ⇒ T0 − T = 2940K M = 24 ⇒ (T0 /T ) = 116.2 ⇒ T0 − T = 34560K Note that the temperature in the air varies strongly depending on the altitude from 250K (11 km) up to 450K (20 km). Obviously an aircraft flying at M = 3 cannot be build from aluminium. One uses titanium. At M = 7 the theory fails (perfect gas) because air starts dissociating. A classical rocket has to carry for a given mass of hydrogen eight time the mass in oxygen. A scram-jet uses the oxygen from the air for its propulsion. This allows to reach the critical escape velocity (from the earth gravitation, about M = 7) with a much lighter engine than a classical rocket. Until most scram-jet experiments failed due to thermal problems. At M = 24 the theory fails (perfect gas) because the gas is dissociated and ionized. Above a temperature of the order of 12 000 K radiation heat transfer becomes very important. In general there are strong deviations from local thermodynamic equilibrium. The electrons will have another temperature than atoms and ions.
30
Chapter 3
Stationary Quasi-1D Gas Dynamics 3.1 3.1.1
Laval tube Quasi-one dimensional theory
Figure 3.1: A Laval nozzle is a smoothly converging-diverging duct. It is an essential part of rocket engines, of gas turbines and of a supersonic wind tunnel. Consider the situation in figure 3.1 above, a so-called Laval tube (or nozzle): a perfect gas flows steadily from a reservoir with pressure p0 and entropy s0 through a converging channel followed by a diverging channel after passing a throat of section Ak . We assume there is no friction (Re >> 1). The diameter is assumed to change slowly with x, so that it holds that: L dA 1. Now dV dx and dx have identical signs. For a converging duct, the flow decelerates and in a diverging duct it accelerates. The lowest velocity is expected at the “throat”. We will later see that an entirely supersonic quasi-1D flow through a Laval nozzle cannot be realized. A supersonic flow approaching a converging nozzle cannot anticipate on the reduction in duct section. The flow will collide with the wall and will abruptly be slowed down by a so-called shock wave. This will be discussed later in this chapter.
c The flow changes from subsonic to supersonic, or from supersonic to subsonic. This implies that at some position we have M = 1. The only location where M = 1 is possible is inside the “throat”, where dA dx = 0 so that the acceleration dV remains finite. dx In this last case, in order to determine the acceleration dV /dx at the throat, we look at the second derivative of A(x). To find an expression for dV /dx when dA/dx = 0 and M = 1 we use de l’Hˆopital’s rule:
32
4 1 dA 5 33 3 d − dx 1 dV 33 A dx 3 = d 3 V dx 3throat (1 − M 2) 3 dx
=
throat
3
1 d2 A 3 A dx2 3 . 3 3 2M dM dx throat
Rearranging terms and using c = V /M and M = 1 we obtain: 2 dV dM 1 d2 A = c dx dx A dx2
(3.2) 2
dM d A Because dV dx and dx always have the same sign, dx2 will always be positive. Indeed, the Mach number can pass through unity only at an area minimum or throat. Hence we need a Laval nozzle to reach supersonic flows by a steady expansion from a high pressure reservoir.
Next we look at the mass flux through the nozzle. This will provide us with a less mathematical explanation for the fact that M = 1 can only be reached at the throat of the Laval nozzle. Using Bernoulli in terms of temperature and Mach number (2.47) for a perfect gas, together with the isentropic gas relations (2.26) we obtain an expression for the normalized mass flux density ρV /ρ∗ V ∗ : / 0 γ+1 ρV ρ0 c 0 ρ c (γ + 1)/2 2(γ−1) =M ∗ ∗ =M 2 ρ∗ V ∗ ρ c ρ0 c 0 1 + γ−1 2 M
(3.3)
ρV ρ∗ V ∗ 1
0 0
1
2
3
4
M Figure 3.2: Normalized mass flux density as a function of the Mach number. This function is plotted in figure 3.2. Obviously ρV is positive, as we assume the flow velocity to be positive. From the above expression we see that the mass flux vanishes at M = 0 and in the limit M → ∞. The function has one maximum, at M = 1 we have d(ρV )/dM = 0. As in the 1-D steady flow through the nozzle, mass conservation implies that ρV A = ρ∗ V ∗ Ak we see that Ak /A should be a minimum at M = 1. Indeed this confirms that the transition M = 1 can only occur at a throat. As ρ∗ V ∗ is the maximum 33
of the mass flux, the maximum mass flow ρ∗ V ∗ Ak through the nozzle occurring when M = 1 at the throat is reached. We then say that the nozzle is “choked” or “critical”. In a similar way as for the pressure, temperature, velocity... one can in an arbitrary steady duct flow introduce the concept of (local) critical throat section A∗ . This corresponds to the duct section at which the critical state M = 1 would be reached by considering an ideal steady isentropic process (flow) starting for the flow condition considered (ρ, V, A). From A/A∗ if the A/Ak A/Ak
the mass conservation law ρV A = ρ∗ V ∗ Ak we also conclude that = ρV /(ρ∗ V ∗ ) = F (M ) is a function of the Mach number only. Hence flow at the nozzle throat is critical M = 1 (Ak = A∗ ) the geometry determines the Mach number of the flow. We have for a given ratio two solutions for M , one subsonic M < 1 and one supersonic M > 1.
Using the isentropic gas relations in combination with the equation of Bernoulli we can express the ratios p/p0 , T /T0 , ρ/ρ0 ...as functions of the Mach number M (2.47 and 2.26). Hence given a nozzle geometry and the reservoir conditions we can determine the state of the gas along the nozzle. An alternative for the Mach number M is the ratio of the velocity and the critical velocity M ∗ = V /V ∗ = V /c∗ (2.50). The advantage of this normalized velocity is that it has a finite range 0 ≤ M ∗ < Vmax /c∗ = . (γ + 1)/(γ − 1) while for M = 1 we have M ∗ = 1. In terms of the normalized velocity M ∗ we find: ρV = M∗ ρ∗ V ∗
#
γ + 1 γ − 1 ∗2 − M 2 2
$1/(γ−1)
,
(3.4)
where M ∗ ≡ V /c∗ . Because c∗ is a constant along a stream line, the right hand side is now a function of V only. This formula is plotted in figure 3.3. From this graph we see that, as the velocity increases along a stream line, the mass flux increases as long as the flow remains subsonic. For low Mach numbers, this curve rises as a straight line, corresponding to an incompressible flow. In the supersonic range however, the mass flux density diminishes for increasing velocity and vanishes when V → Vmax . The flux has its maximum value ρ∗ V ∗ when M ∗ = 1 (at the critical velocity).
3.1.2
Pressure distributions inside a nozzle
The fully frictionless subsonic steady flow of a fluid from a pipe (without nozzle) into the atmosphere can be described by potential flow theory. Such 34
pre ssi b
le
ρV ρ*V*
inc om
1.0
0.0 0.0
2.0
1.0
4.0
3.0
M*=V/c* Figure 3.3: Normalized [Thompson (1972)].
mass
flux
density
as
function
of
M∗
a flow would follow the sharp angle wall at the exit of the pipe and would expand radially in all directions. Such a flow is never observed in practice. In reality, due to friction losses, the flow will not be able to follow the sharp edges at the outlet. It separates tangentially from the wall, as shown in figure 3.3. A shear layer is formed between the main jet flow and the surrounding stagnant gas. Across this shear layer, pressure and velocity must be continuous.
a)
b)
pe
pe
pa pa Frictionless subsonic outflow
Actual subsonic outflow
Figure 3.4: (a) Potential flow, (b) Separated discharge in a subsonic flow (from [Thompson (1972)]).
35
The discharge of fluid into a reservoir may either be subsonic or supersonic: (a) Subsonic discharge The continuity of pressure across the contact surface (shear layer) requires that the surrounding (atmospheric) pressure pa is equal to the pressure at the exit of the pipe pk . One can demonstrate that for a steady subsonic isentropic flow pk = pa is the only possible solution.1 (b) Supersonic discharge In this situation we also have continuity of pressure across the contact surface. However, because the outflow is supersonic, this pressure does not propagate upstream. In the exit plane of the pipe, the pressure pk can be higher or lower than the atmospheric pressure pa . When it is higher we call the jet under-expanded. When it is lower we call the jet overexpanded. Outside the pipe the flow will immediately expand or compress to make the pressure match with the atmospheric pressure at the contact surface. Expansion or compression flows will be generated from the side edges of the pipe outlet.2 Suppose that a reservoir with pressure p0 and temperature T0 is connected to converging nozzle. In other words, the throat is at the nozzle exit. The flow discharges into the atmosphere, which has a pressure pa . By decreasing the ratio pa /p0 , for instance by increasing the pressure in the reservoir, the velocity of the flow increases until M = 1 is reached at the throat of the nozzle. Figure 3.5 shows the pressure distribution p(x)/p0 of the flow in the nozzle and just outside the nozzle, for different values of pa /p0 . In case a, where 1 Let us assume a subsonic flow. If one assumes that pk > pa the gas will expand as it leaves the nozzle to adapt to the surrounding pressure pa . Hence the streamlines should diverge following the continuity equation applied to a steady flow ⃗v · ∇ρ = −ρ∇ · ⃗v , which demonstrates that ∇ · ⃗v > 0 when ⃗v · ∇ρ < 0. Following our analysis (1/V )(dV /dx) = (1/A)(dA/dx)/(M 2 − 1) (equation 3.1). As sign[(1/A)(dA/dx)] = sign[∇ · ⃗v ] we conclude that the flow velocity should decrease. Following Bernoulli a decrease in velocity implies an increase in pressure. Hence we come to a contradiction. We assumed that the pressure in the jet was decreasing in order to adjust itself to atmospheric pressure and we come to the conclusion that the pressure is increasing. Obviously the same conclusion can be deduced for a steady subsonic flow with pk < pa . Hence pk = pa should prevail. This boundary condition is actually a dynamic boundary condition. When a pressure disturbance arrives at the outlet it is reflected as an inverse pressure perturbation travelling as a sound wave to the reservoir and adjusting the mass flow to match the boundary condition pk = pa . 2 These waves will meet at the center of the flow. In first approximation they add up, resulting into an over expansion or compression on the jet center line. Further downstream these waves reflect on the shear layer, resulting into an inversion of the wave sign. An expansion (compression) wave reflects as a compression (expansion). The process is then repeated, resulting into an oscillating pressure along the jet center-line. A further discussion of this will be provided later because this flow is not one-dimensional and due to non-linearity shock waves can be formed.
36
Figure 3.5: Pressure distributions in a converging nozzle for different ratios pa /p0 of outlet pressure pa to reservoir pressure p0 (from [Thompson (1972)]). pa = p0 , there is no flow. In case b, p0 > pa > p∗ , the flow is everywhere subsonic and the pressure at the throat pk = pa . At the sonic condition, γ 2 case c, the ratio pa /p0 equals the critical pressure ratio p∗ /p0 = ( γ+1 ) γ−1 . Lowering the ratio pa /p0 further will not produce an additional increase in velocity, since the maximum velocity for this configuration is reached when M = 1 at the throat. The flow is choked at the nozzle outlet. In case d, where pa < p∗ , the flow velocity and the ratio p(x)/p0 inside the nozzle is exactly the same as in case c. Outside the nozzle, the flow expands in order to adapt itself to the atmospheric pressure. The maximum mass flow through the nozzle is: ∗
∗
Φmax = ρ V Ak =
#
2 γ+1
$
γ+1 2(γ−1)
ρ0 c0 Ak =
#
2 γ+1
$
γ+1 2(γ−1)
γp0 Ak c0
(3.5)
Next, we consider the case of a settling chamber (reservoir) connected to a 37
Figure 3.6: Pressure distributions in a Laval nozzle for different ratios pa /p0 of outlet pressure pa to reservoir pressure p0 (from [Thompson (1972)]). Laval nozzle, i.e., a converging- diverging nozzle with throat cross section Ak and outlet (exit) cross section Ae . Figure 3.6 shows the geometry and corresponding pressure distribution p(x)/p0 inside the nozzle. Again, lowering pa /p0 by increasing p0 will increase the mass flow through the throat, until the critical condition is reached Φmax (equation 3.5) . Consider the cases a through g, where we vary the ratio of ambient to reservoir pressures pa /p0 . At a, pa = p0 and there is no flow. In case b, the flow is entirely subsonic, and the velocity changes with the cross-sectional area A according to: ρe Ve Ae = ρ(x)V (x)A(x) < Φmax In case c, the sonic condition pk /p0 = p∗ /p0 is reached at the throat and the nozzle is chocked: ρ(x)V (x)A(x) = Φmax . In the diverging section, the flow is subsonic in order to let the exit pressure pe match with the ambient pressure pe = pa . There exists exactly one other pressure (f ), for which the diverging flow leads to pressure matching at the exit, without shock for a supersonic flow. The flow is supersonic downstream of the throat. In this case the jet flow is supersonic and fully expanded. Further lowering of the ambient pressure (g) does not affect the flow inside the nozzle. The flow 38
within the nozzle is fully determined by the reservoir conditions and the fact that the throat is critical. Lowering of the ambient pressure implies, that pe > pa . The jet flow outside the nozzle is under-expanded now.
Figure 3.7: Strongly under-expanded jets at take off of the SR-71 Blackbird (Aerospaceweb.org, Shock Diamonds and Mach Disks)
Figure 3.8: Schlieren photograph of the flow in a Laval nozzle with a slightly under-expanded jet at the outlet (L. Prandtl). We now discuss the flow for the pressure range pa /p0 in between cases c and f. In this case the differential equation we have derived (equation 3.1) has no solution. Nature finds a solution by violating our assumptions of a continuous and isentropic solution. In the diverging section a discontinuous pressure change occurs: a shock wave. We will now simply assume that it is 39
a narrow region, which can be described as a discontinuity with negligible volume 3 .
Figure 3.9: Normal shock in divergent part of Laval nozzle (G. Settles). Please note the forks due to the interaction of the shock with the viscous boundary layers at the wall. As we will demonstrate the flow downstream of the shock is subsonic. Therefore in cases d and e, the exit pressure is equals the ambient pressure pe = pa . Upon lowering of pa , the position of the shock moves downstream from the throat. Close to the throat it is a vanishing small pressure discontinuity. The pressure difference increases first and the decreases again until the shock reaches the nozzle outlet where it vanishes if the jet is fully expanded. Choking does not only occurs in converging and Laval nozzles. Due to friction forces, volume forces, combustion or heat transfer it can also occur 3
In practice, depending on the geometry of the nozzle and the interaction with viscous boundary layers, this can be a complicated structure of interacting oblique shock waves and expansion wave.
40
in a straight pipe. We will discuss this at the end of this chapter.
3.2 3.2.1
Normal shock waves Rankine-Hugoniot adiabate
The existence of “discontinuities” in the state of a gas within a flow has long time been a controversial subject. An interesting discussion of this historical debate is provided by Thompson [Thompson (1972)]. We restrict our discussion now to normal shock, which are in a (y, z) plane normal to the flow direction ⃗v = (u, 0, 0). In the present discussion we simply assume that the shock is infinitely thin4 . In other words its volume is negligible.
Figure 3.10: Control volume used to derive the shock relations from the integral mass, momentum and energy balances. Applying the integral conservation laws over such a shock while we assume a uniform flow upstream and downstream of the shock we find the so called shock relations. Note that because the flow is uniform upstream and downstream of the shock the friction forces and heat flux vanish (the flow outside 4 ¯ For A “strong” shock has a thickness of the order of a few times the mean free path λ. steady shocks the thickness of the shock is not relevant. In that case the steady integral conservation laws hold anyhow!
41
the shock is isentropic). We choose a thin control volume delimited by planes parallel to the shock, which it encloses. The mass conservation law yields: ρu = ρˆu ˆ
(3.6)
where ρ is the density in front of the shock and u the corresponding flow velocity. The density behind the shock is ρˆ and u ˆ is the corresponding velocity. In other words the mass flux φm = ρu is conserved over a normal shock φm = φˆm . The momentum equation yields: p + ρu2 = pˆ + ρˆu ˆ2
(3.7)
where p and pˆ are the pressures respectively in front and behind the shock. The energy equation yields: h+
u2 ˆ u ˆ2 =h+ 2 2
(3.8)
ˆ are the enthalpy respectively upstream and downstream of where h and h ˆ 0. the shock. Obviously the total enthalpy is conserved across a shock h0 = h We now eliminate the velocity in order to obtain a relationship between the thermodynamic states (p, ρ) and (ˆ p, ρˆ) respectively in front and at the back of the shock. Using the mass conservation law we re-write the momentum and energy equations in the form: # $ 1 1 pˆ − p = − φ2m (3.9) ρ ρˆ and:
$ 1 1 − φ2m . (3.10) ρ2 ρˆ2 Taking the ratio of these two equations we eliminate the mass flux φm to obtain the so called Rankine-Hugoniot relation: ˆ − h) = 2(h
#
1 1 ˆ − h). (ˆ p − p)( + ) = 2(h ρ ρˆ
(3.11)
As in a uniform gas the thermodynamic state is fully determined by two state variable such as (p, ρ). The enthalpy is a function of these variables h = h(p, ρ). Given a initial state (p, ρ) in front of the shock, the RankineHugoniot relation provides all the possible states (ˆ p, ρˆ) behind the shock. The actual state (ˆ p, ρ1ˆ ) will be determined by the value of the mass flux φm imposed by the upstream boundary conditions. For a perfect gas we have furthermore: h = cp T = (cp /R)(p/ρ) = (γ/(γ − 1)(p/ρ). After a significant amount of algebra we can rewrite the RankineHugoniot relation for a perfect gas in the form: ρˆ pˆ µ ρ − 1 = p µ − ρρˆ
42
(3.12)
or:
pˆ ρˆ 1 + µ p = ρ µ + ppˆ
(3.13)
γ+1 . γ−1
(3.14)
where we define µ by: µ=
It is interesting to compare the Rankine-Hugoniot adiabate with the isen% &γ pˆ ρˆ tropic perfect gas relation: p = ρ . A first essential difference is that the Rankine-Hugoniot depends on two parameters (p, ρ). Along a Rankine Hugoniot curve in the (p, 1/ρ) plane there is only one point corresponding to the initial state (in front of the shock). All the other points correspond to the state behind a shock (ˆ p, 1/ˆ ρ). Starting a new Rankine-Hugoniot curve from this final state yields a different curve than the one starting at (p, 1/ρ). Along the isentropic equation of state, any point can be used as initial state, because the curve depends only on one parameter, the entropy s. This entropy along a Rankine-Hugoniot adiabate changes following the equation of state: # $ sˆ − s pˆ ρˆ = ln − γ ln . (3.15) cv p ρ From this equation in combination with the Rankine-Hugoniot relation (3.13) we conclude that the entropy increases over a compression shock pˆ > p for a perfect gas. It decreases for an expansion shock, which implies that an expansion shock of a perfect gas does not exist. For real gasses however expansion shocks can exist depending on the sign of (∂ 2 p/∂(1/ρ)2 )s [Landau (1987)], [Thompson (1972)]].
3.2.2
Entropy production in shocks
Defining the relative increase in density ϵ = (ˆ ρ − ρ)/ρ = (ˆ ρ/ρ) − 1 we can compare the isentropic equation of state for a weak compression ϵ ≪ 1: pˆ γ 2 − γ 2 γ 3 − 3γ 2 + 2γ 3 = (1 + ϵ)γ = 1 + γϵ + ϵ + ϵ + ... p 2 6
(3.16)
and the Rankine-Hugoniot equation: pˆ µ(1 + ϵ) − 1 γ 2 − γ 2 γ 3−2 γ 2 + γ 3 = = 1 + γϵ + ϵ + ϵ + ... p µ − (1 + ϵ) 2 4
(3.17)
The two expressions differ only for terms of order ϵ3 or higher. Hence the entropy production of a weak shock scales as ϵ3 : sˆ − s pˆ ρˆ γ3 − γ 3 = ln − ln ( )γ = ϵ + ... cv p ρ 12 43
(3.18)
Rankine−Hugoniot Isentroop Impulswet
10 pˆ p γ = 1.4
5
0 0
γ−1 γ+1
0.5
1
ρ ρˆ
Figure 3.11: Comparison of the Rankine-Hugoniot adiabate with an isentrope starting from the same initial state. Note the vertical asymptote of the Rankine-Hugoniot with limpˆ/p⇒∞ (ρ/ˆ ρ) = (γ − 1)/(γ + 1). As shown in Figure 3.12, the straight line joining the initial state (p, 1/ρ) to the end %state (ˆ ρ) has following the momentum law the slope φ2m = &p, 1/ˆ 1 1 −(ˆ p − p)/ ρˆ − ρ . The slope of the Rankine-Hugoniot adiabate at the
initial point is: (∂p/∂(1/ρ))RH = −ρ2 (∂p/∂ρ)s = −(ρc)2 . For a weak shock we have furthermore (∂ pˆ/∂(1/ˆ ρ))RH = −ˆ ρ2 (∂p/∂ρ)RH ≃ −(ˆ ρcˆ)2 . Because 2 2 for a perfect gas we have (∂ p/∂(1/ρ) )s > 0 we see that at the initial point −(φ2m /(∂ pˆ/∂(1/ˆ ρ)RH ) = (u/c)2 = M 2 > 1 hence the flow is supersonic ahead of the shock M > 1. At the end point of the Rankine-Hugoniot ˆ 2 < 1 so that the flow is subsonic we have (φ2m /(∂p/∂ρ)RH ≃ (ˆ u/ˆ c )2 = M 44
ˆ < 1. This result appears to be much more general as behind the shock M
Figure 3.12: For a weak shock the Rankine-Hugoniot adiabate can be approximated by an isentrope. The Rayleigh line derived from the momentum balance joints the initial point to the state behind the shock. The slope of the Rayleigh line is proportional to the square of the mass flux density φ2m = (ρu)2 = (ˆ ρu ˆ)2 . The tangents at the initial point and the end point have a slope proportional to the square of the speed of sound, (ρc)2 and ˆ =u (ˆ ρcˆ)2 respectively. We clearly see that M = u/c > 1 while M ˆ/ˆ c < 1. demonstrated by Landau and Lifchitz [[Landau (1987)]] by using a stability argument. The flow in front of a normal shock is always supersonic and it is subsonic behind the shock. It is also quite interesting to compare the % &γ 1+µ ppˆ isentrope equation of state ppˆ = ρρˆ with the Rankine-Hugoniot ρρˆ = p ˆ µ+ p
for very strong shocks ϵ >> 1. As γ > 1 we see that following the isentrope relation the density increases without any bound as the pressure ratio is increased to infinity pˆ/p → ∞. In contrast the maximum compression one can achieve with a shock is limpˆ/p→∞ (ˆ ρ/ρ) = µ = (γ + 1)/(γ − 1). Due to the dissipation in the shock an increase in pressure ratio only contributes to the heating of the gas rather than a compression. Therefore strong shocks should generally be avoided.
45
3.2.3
Mach number dependance
The Mach number M in front of the shock is a convenient parameter to relate the state of the gas in front it to the state behind it. The momentum equation (3.9) can be written as: # $ pˆ φ2 ρ −1= m 1− . (3.19) p ρp ρˆ For a perfect gas we have: φ2m /(pρ) = ρu2 /p = γ(u/c)2 = γM 2 .
(3.20)
Combining these equations with the Rankine-Hugoniot relation 3.12 and 3.13 for a perfect gas we find:
and:
pˆ − p 2γ = (M 2 − 1) p γ+1
(3.21)
ρ − ρˆ u ˆ−u 2 1 − M2 = = ρˆ u γ + 1 M2
(3.22)
ˆ =u and with M ˆ/ˆ c: # $2 % & # $2 u ˆ c 2 2 u ˆ T 2 M2 − 1 1−M =1− M =1− M = . 2γ u cˆ u 1 + γ+1 (M 2 − 1) Tˆ (3.23) ˆ < 1. This is From this equation we see that for M > 1 we have indeed M only valid for normal shocks (flow direction normal to the shock front). We will see further that when the shock is oblique to the flow, the Mach number behind the shock can be supersonic. ˆ2
Figure 3.13: In a) we consider a steady flow u in front of a shock in a coordinate system in which a shock has a fixed position us = 0. Changing towards a coordinate system shown in b), in which the fluid ahead of the shock is stagnant, we see a shock travelling at a constant speed us = −u. In the limit case u → c corresponding to M → 1 the shock is an acoustic wave propagating at the speed of sound c. 46
From these equations we can see that in the limit M → 1 the shock strength ϵ = (ˆ ρ − ρ)/ρ → 0 vanishes. In fact this very weak normal shock is an acoustic wave travelling with the velocity u = c with respect to the fluid. This can be best visualized by changing of coordinate system. Until now, the shock had a fixed position. Now we choose a coordinate system in which the fluid in front of the shock is stagnant. This is illustrated in figure 3.13. In figure 3.14 we show the shock relations for the range 1 ≤ M ≤ 5 for a perfect gas with γ = 1.4. We see that limM →∞ ρρˆ = (γ + 1)/(γ − 1) and . ˆ = (γ − 1)/(2γ) limM →∞ M 2
10
γ = 1.4
p ˆ p
1
10
ρ ˆ ρ Tˆ T
ˆ1 M
0
10
ˆ M
−1
10
1
2
3
4
5
M Figure 3.14: Shock relations for a perfect gas γ = 1.4 as functions of the Mach number.
47
3.2.4
Mach number from pressure measurements
We now consider a Pitot tube as show in figures 3.15, 3.16 and 3.17.
Figure 3.15: Schlieren photograph of the flow around the Pitot tube (D. Rouwenhorst and H. Stobbe, UTwente) In front of the shock we assume a uniform flow with Mach number M , stagnation pressure p0 and stagnation temperature T0 . Assuming a perfect gas the stagnation temperature across the shock is conserved Tˆ0 = T0 while due to friction the total pressure decreases pˆ0 < p0 . Using the equation of state for a perfect gas: ( ) pˆ0 sˆ − s Tˆ0 = exp(− ) (3.24) p0 cv T0 we obtain: pˆ0 sˆ − s = exp( )= p0 cv
# $ −1 # $ γ pˆ γ−1 ρ γ−1 p ρ
(3.25)
which can be expressed in terms of the Mach number M in front of the shock: In figure 3.18 we show the Mach number dependency of pˆ0 /p0 . We also show the ratios p/p0 and p/ps . For subsonic flows M < 1 we can use the isentropic gas relation ps /p ≃ γ 2 γ−1 to determine the Mach number. For supersonic flows p0 /p = (1+ γ−1 2 M ) we use the relation ps /p ≃ (p0 /p)(ˆ p0 /p0 ) to determine the Mach number.
48
Figure 3.16: Detail of the shock wave in front of a Pitot tube (D. Rouwenhorst and H. Stobbe, UTwente)
Figure 3.17: A Pitot tube measures the stagnation pressure ps . For supersonic flows M > 1 a normal shock is formed in front of the tube. The stagnation pressure is approximatively equal to the total pressure behind the normal shock ps = pˆ0 .
3.3 3.3.1
Oblique shocks Mach lines in steady flow
Consider a uniform steady flow of velocity 49 ⃗v = (u, 0, 0) and speed of sound c. A small pressure perturbation is generated at time t = 0 at ⃗x0 = (x0 , y0 , z0 ).
γ = 1.4 1
ps p0
0.5
✙ ✟ p p0
0 0
1
p ps
✯ ✟ 2
3
4
M
Figure 3.18: From the measurement of the stagnation point pressure ps ≃ pˆ0 by means of a Pitot tube and the static pressure p by means of a static pressure tube or a pressure hole in a wall, we can calculate the Mach number M. This perturbation forms an expanding spherical wave of radius ct, which is convected by the main flow ⃗v . If the flow is subsonic M = u/c < 1 the spherical wave will, given sufficient time, reach any point in the flow. If the flow is supersonic M > 1 the perturbation will only reach points within a cone around the x-axis with apex at ⃗x0 and an opening angle α such that: sin
α c 1 = = . 2 u M
(3.26)
This is illustrated in figure 3.19.
Figure 3.19: Pressure perturbations expand in a uniform flow as spherical waves. For subsonic flows (a) they will reach any point in the flow. For supersonic flow (b) they only reach points within the Mach cone with apex angle α = 2 arcsin(1/M ).
50
If we consider a continuous source of infinitesimal perturbations at ⃗x0 in a uniform steady supersonic flow, the boundary between the Mach cone and the unperturbed region is a M ach wave. In contrast with a normal shock of vanishing strength a compression Mach wave does not induce a transition from a supersonic to a subsonic flow. The flow behind this very weak oblique shock remains supersonic. We will see later that for finite shock strength ˆ < 1 (subsonic flow an oblique shock can either be a strong shock with M ˆ > 1 (supersonic flow behind the behind the shock) or a weak shock with M shock).
3.3.2
Shock relations for plane oblique shocks
We consider a plane oblique shock with an angle β with respect to the incoming uniform flow ⃗v1 = (u, v, 0). We choose the x-axis normal to the shock. Hence we have: tan β = u/v (see figure 3.20).
Figure 3.20: Definition of approach angle β and flow turning angle θ for a plane oblique shock. The flow ⃗v2 = (ˆ u, vˆ, 0) behind the shock is also uniform, making an angle θ with respect to the approach flow: cos θ = (⃗v1 · ⃗v2 )/(|⃗v1 ||⃗v2 |). Using the integral conservation law for mass over a thin controle volume enclosing the shock, we have: ρu = ρˆu ˆ = φm
(3.27)
From the y- momentum equation we find: φm (v − vˆ) = 0 51
(3.28)
and from the x-momentum: ρu2 + p = ρˆu ˆ2 + pˆ.
(3.29)
u2 + v 2 ˆ u ˆ2 + vˆ2 =h+ . 2 2
(3.30)
The energy equation yields: h+
This implies that v = vˆ and that the shock relations for an oblique shock are identical to those for a normal shock observed in a reference frame moving with the velocity (0, v, 0). Such a Galilean transformation does not affect the flow physics. Hence we can use all the results obtained for a normal shock by simply replacing the Mach number by the normal Mach number Mn = u/c = M sin β. Obviously a shock wave can only be found if Mn > ˆ is related to the normal Mach 1. The Mach number behind the shock M ˆ ˆ ˆ number Mn by the relation: Mn = M tan(β − θ).
3.3.3
Weak and strong oblique shocks
Figure 3.21: Relationship between the approaching angle β and the flow turning angle θ for a planar shock at fixed approach Mach number M (after [Shapiro (1953)]). 52
Figure 3.22: Weak shock attached to a wedge in a supersonic flow (D. Rouwenhorst and H. Stobbe, UTwente) From goniometry and v = vˆ we have (figure 3.20): u ˆ tan(β − θ) = . u tan β
(3.31)
From the normal shock relations for a perfect gas we have: u ˆ 2γ 1 − Mn2 =1+ . u γ + 1 Mn2
(3.32)
Eliminating u ˆ/u we find a relationship between β, θ and Mn = M sin β: tan(β − θ) 2γ 1 − Mn2 =1+ . tan β γ + 1 Mn2
(3.33)
For a fixed Mach number we obtain a curve in the (β, θ) plane. Such curves are shown in figure 3.21. We observe that for a given Mach number there is a maximum flow turning angle θmax . For θ > θmax there is no plane shock solution. A detached shock will be formed ahead of the wedge (see figure 3.24). This is the reason why supersonic aircraft have a sharp nose and wings with sharp leading edges. For any turning angle θ < θmax we have two solutions for β. They both correspond to an attached plane shock. One of the solutions corresponds to ˆ < 1, which we call a strong shock. The a subsonic flow behind the shock M ˆ > 1. Of course the weak shock other solution is a weak shock solution M solution is preferred, as it has less dissipation than the strong shock. 53
Figure 3.23: For large apex angles a detached shock will be formed in front of a wedge placed in a supersonic flow.
3.4 3.4.1
Pipe flow with friction and heat transfer Integral conservation laws for quasi-one dimensional pipe flow
Consider an infinitesimally thin section of a cylindrical pipe between x and x + dx (see figure 3.24).
Figure 3.24: Thin pipe segment for balance equations. We assume a quasi-one dimensional steady flow along the pipe of radius a in the x-direction: ⃗v = (V (x), 0, 0), ρ = ρ(x), p = p(x)... The integral mass
54
conservation implies that the mass flux is conserved along the pipe: d[ρV ] = 0
(3.34)
ρ(x)V (x) = φm
(3.35)
or: where φm is a constant determined by the boundary conditions. The momentum balance implies that: Ap d[ρV 2 + p] = Pp dxτ
(3.36)
where τ is the wall shear stress (viscous force per unit area of the pipe wall on the fluid). The perimeter of the pipe section is Pp = 2πa and the cross sectional area is Ap = πa2 . The integral energy balance applied to the pipe section yields: 1 2 V2 Ap φm d (h + ) = Pp qdx (3.37) 2 where q is the heat flux from the wall to the flow. Using the conservation of mass flux ρV = φm we rewrite the momentum equation as: dp 2τ + V dV = dx ρ ρa
(3.38)
and energy balance equation as: dh + V dV =
2q dx. φm a
(3.39)
Eliminating the flow velocity V by substraction the momentum equation 3.38 from the energy equation 3.39 and using the fundamental Gibbs’ equation yields an equation for the entropy: 1 2 dp 2q 2τ T ds = dh − = − dx. (3.40) ρ φm a ρa Heat flow into the pipe and friction do correspond to entropy production. Note that the sign of τ is negative because the wall friction opposes the flow. Hence friction increases the entropy of the flow.
3.4.2
Choking due to friction and heat transfer
Using the mass conservation dρ/ρ = −dV /V we can also re-write the momentum equation as: dp dρ 2τ −V2 = dx. (3.41) ρ ρ ρa
55
We can furthermore describe the enthalpy as a function of p and ρ, which implies: # $ # $ ∂h ∂h dh = dp + dρ. (3.42) ∂p ρ ∂ρ p Substitution into equation 3.40 we have: # $ # $ ∂h 1 ∂h 2q 2τ [ − ]dp + dρ = [ − ]dx. ∂p ρ ρ ∂ρ p φm a ρa
(3.43)
The two equations 3.41 and 3.43 form a set of linear equations for dp and dρ. The determinant D of this set of equations is given by: ( ) 1 − hp hρ 2ρ D= 1−V (3.44) ρ hρ where we used the short hand notation hp = (∂h/∂p)ρ and hρ = (∂h/∂ρ)p . Using the Gibbs relation dh = T ds + dp/ρ = hp dp + hρ dρ we have that for ds = 0: 1 ( − hp )dp = hρ dρ (3.45) ρ which implies that c2 = (∂p/∂ρ)s = ((1/ρ) − hp )/hρ (2.20) and: D=
hρ (1 − M 2 ). ρ
(3.46)
When D ̸= 0 the system of equations yields a solution so we find expressions for dp/dx and dρ/dx. For M = 1 the determinant vanishes (D = 0), so that we have no solutions for dp/dx and dρ/dx. They become infinitely large. This situation corresponds to choking induced by friction and heat transfer. We consider now a subsonic M < 1 adiabatic q = 0 flow. The pressure decreases due to friction. It seems logical that we need a negative pressure gradient in order to balance the friction forces to maintain a positive velocity (V > 0 implies dp/dx < 0). This corresponds to our intuition based on experience with incompressible flows. An adiabatic decrease of pressure implies a decrease of density dρ/dx < 0. From mass conservation d(ρV )/dx = 0 we see that the flow velocity should increase dV /dx > 0 in order to keep the mass flow constant. Hence due to friction a subsonic flow speed along a pipe increases. This continues until the critical condition M = 1 is reached. Further increase becomes impossible because of the singularity of the pressure gradient dp/dx → −∞. We then have two possibilities: or:
-either we reach the end of the pipe for M ≤ 1
56
-the mass flow decreased for given inlet pressure. As in the case of a Laval nozzle the behaviour for supersonic flows M > 1 is opposite to that found for subsonic flows. This is due to the fact that D > 0. Instead of accelerating the flow will now decelerate, contradicting our intuition. The Mach number decreases approaching M = 1. As choking is approached (the pipe is too long) a shock wave will form and the flow will continue as subsonic flow behind the shock. This subsonic flow accelerates up to M = 1. If the pipe is too long, the shock will move up to the pipe inlet causing the mass flow to change. This complex behaviour has been studied for an adiabatic pipe flow of a perfect gas by [Fanno (1904)]. He obtained an analytical solution [[Shapiro (1953)], [Thompson (1972)], [Owczarek (1964)], [Landau (1987)]].
3.4.3
The long pipe-line problem
We consider a gas transport pipe line buried under the ground. This can be a pipe of 1 m diameter and a few hundred kilometers length transporting natural gas from a gas field to costumers. The gas is in good thermal contact with the ground. We therefore assume an isothermal flow. The mass conservation law is: dρV = 0. (3.47) dx The momentum equation is: ρV
dV dp 2τ =− + . dx dx a
(3.48)
The energy equation reduces to: dT = 0. dx
(3.49)
We complement this set of equation with the ideal gas law p = ρRT , written in differential form: 1 dp 1 dρ 1 dT = + . (3.50) p dx ρ dx T dx At high Reynolds numbers for subsonic flows it is reasonable to approximate the wall shear stress by means: τ =−
cf ρV 2 2
(3.51)
where the friction factor cf is constant. After some algebra we find: (1 −
cf ρV 2 ρV 2 1 dV ) = . p V dx a p 57
(3.52)
We recognize the isothermal Mach number MT = V /cT with the isothermal . speed of sound cT = p/ρ. Hence we have: cf MT2 1 dV = . V dx a(1 − MT2 )
(3.53)
As for friction in an adiabatic flow (Fanno flow) we observe choking when the flow is accelerated from a subsonic velocity MT < 0 up . to MT = 1. The fact that this involves the isothermal speed of sound c = p/ρ rather T . than the adiabatic speed of sound c = γp/ρ is logical because the flow is isothermal. Hence acoustic waves transmitting information to the reservoir to regulate the mass flow travel at the speed cT rather than c.5 By integrating between position x1 with MT = MT,1 and x2 with MT = MT,2 we find: / # $0 MT,1 x2 − x1 1 1 1 = (3.54) 2 − 2M 2 + ln M a cf 2MT,1 T,2 T,2 As one approaches MT = 1 one expects that the isothermal approximation will fail because the gas pressure varies very fast, so that heat transfer from the wall is not fast enough to maintain a constant temperature.
3.4.4
Choking due to heat transfer
We now consider the effect of a constant heat transfer q into a frictionless steady pipe flow of an ideal gas. The conservation equations become:
ρV
dρV = 0, dx
(3.55)
dV dp =− dx dx
(3.56)
and
dT 1 dp 2q − = . dx ρ dx aφm Assuming an ideal gas we find after some algebra: cp
1 dV 2q = . V dx acp T φm (1 − M 2 )
(3.57)
(3.58)
We see that a subsonic flow is accelerated by heat addition until M = 1 is reached. Again this involves choking. The heat addition can also be due to a combustion process. Hence a subsonic combustion in a straight pipe can accelerate the flow up to M = 1. To reach a higher Mach number we need a Laval nozzle. We use such a laval nozzle downstream of a rocket engine as shown in Figure 3.25. 5
The isothermal speed of sound can also prevail in a flow with very efficient heat transfer. This can be for example radiative heat transfer in diluted gasses at very high temperature (astronomy).
58
Figure 3.25: Laval nozzle downstream of the combustion chamber of the solid propellant motor of the Ariane V rocket.
3.5
Questions
Questions with a star (*) are more advanced and not considered as basic study material. 1. Following the quasi-one dimensional theory for steady flows the acceleration dV /dx along a pipe is related to the change in cross-section area dA/dx by: 1 dV 1 dA = . 2 V dx (1 − M )A dx
where M = V /c is the Mach number. Is this theory also valid for a liquid flow? 59
2. Following equation 1 there is a singularity of the factor (M 2 − 1)−1 at M = 1. What does this implies for the actual flow in a Laval nozzle? 3. Which parameters determine the maximum mass flow through a chocked Laval nozzle Φmax = ρ ∗ V ∗ Ak ? 4. Make a sketch of the mass flow Φ = ρV A through a Laval nozzle for given reservoir conditions as a function of pa /p0 in the range pa ≤ p0 . 5. Make a sketch of the pressure profile p(x) along a Laval nozzle for various values of the parameter pa /p0 . 6. Is it true that the mass flow Φ = ρV A through a chocked Laval nozzle cannot be changed? 7. How much does the mass flow Φ of a perfect gas through a choked Laval nozzle increase if we double the reservoir pressure p0 at fixed T0 ? 8. How much does the mass flow Φ of a perfect gas through a choked Laval nozzle increase if we double the reservoir temperature T0 at fixed p0 ? 9. Given that at launch-off the pressure in the combustion chamber of the solid propellant motor of the Ariane V rocket rises up to p0 = 60 bar at a temperature T0 = 3000 K and that we assume a perfect gas flow with γ = 1.3: -Estimate the Mach number M of the nozzle needed to obtain a fully expanded free jet at the outlet. -Estimate the ratio Ak /Ae of the throat to exit cross sections of this nozzle. -Estimate the free jet gas temperature Te . 10. A blow down supersonic wind tunnel consists of an air reservoir with T0 = 298 K and p0 = 60 bar. Such tunnels are commonly used up to M = 4. Why are they not used at M = 24? * 11. As a result of the curvature of the streamlines, the pressure at the wall in the throat of a Laval nozzle will be different from that on the center-line. Do you expect the pressure on the center-line to be higher or lower than that at the wall? For a radius of curvature rk of the wall and a throat cross section radius ak estimate the order of magnitude of this pressure difference. * 12. Due to the non-uniformity of the flow at the throat discussed in the previous exercise, the critical line M = 1 will be a curve. Make a sketch of this curve in a Laval nozzle.
60
13. What are the conditions of validity of the integral balances for mass, momentum and energy taken over a normal shock wave? ρu = ρˆu ˆ
p + ρu2 = pˆ + ρˆu ˆ2
h+
u2 ˆ u ˆ2 =h+ 2 2
Did we neglect friction and heat transfer to derive these equations? 14. Why are the shock relations for mass, momentum and energy discussed in the previous exercise also valid for a moving shock (speed us )? Are they also valid for a time dependent shock velocity us (t)? 15. What are the fundamental differences between an isentropic compression of a perfect gas described by (ˆ p/p = (ˆ ρ/ρ)γ and the RankineHugoniot adiabate: pˆ ρˆ 1 + µ p = ρ µ + pˆ p
with µ = (γ + 1)/(γ − 1)? 16. What is the maximum of the compression ρˆ/ρ that can be reached by a single shock compression? What is the maximum of compression that can be reached by two successive shock compressions? 17. Make a sketch comparing a single strong shock compression from (p, 1/ρ) to (ˆ p, 1/ˆ ρ) in a (p, 1/ρ) diagram with the same increase in pressure obtained by two successive shocks. 18. Determine the ratio Tˆ0 /T0 of the total temperatures across a normal shock in a perfect gas. 19. Is an expansion shock possible? 20. Given the initial point (p, ρ1 ) and the mass flow φm through a shock, draw the line along which one can find all the possible end states (ˆ p, ρ1ˆ ) behind a normal shock as imposed by the momentum balance equation (so called Rayleigh line). What is the physical meaning of the slope dˆ p/d(1/ˆ ρ) of the Rayleigh line?
61
21. The increase in entropy across a shock can be calculated for a perfect gas by combining the equation of state: sˆ − s pˆ ρˆ = ln( ) − γ ln( ) cv p ρ with the Rankine-Hugoniot. Compare, for ρˆ/ρ = 2 and γ = 1.4, the value of (ˆ s − s)/cv calculated in this way to the result obtained by means of the weak shock approximation: sˆ − s γ3 − γ 3 = ϵ cv 12 What happens for ρˆ = 1.2. 22. Which additional information do we need in order to determine the Mach number M from the Pitot tube pressure measurement ps ? Can we determine the Mach number from the ratio ps /p0 of Pitot tube pressure and reservoir pressure? * 23. We place a Pitot tube in the diverging section of a Laval nozzle. Make a sketch of the variation of the Pitot pressure ps as a function of the variation in atmospheric pressure pa for fixed reservoir pressure p0 . 24. Derive the expression for the angle α of Mach waves with the flow velocity direction as a function of the Mach number M . What occurs when M < 1? 25. What is the difference between a weak and a strong planar oblique shock for given flow turning angle θ? 26. Estimate the Mach numbers of these two flows, shown in figure 3.26: (a) Flow in supersonic wind tunnel HST of NLR (A. Elsenaar [Elsenaar (2012)], 50 jaar HST) (b) Flow in supersonic wind tunnel (Spaceex, Space Launcher, DNW) 27. What is the difference between MT and M ? When is MT more useful than M ? 28. Consider an underground natural gas transport pipe line of pipe diameter 2a = 1 m and friction coefficient cf = 2 × 10−3 . Assume for the gas properties γ = 1.3 and Mm = 16 kg/kmol. The temperature of the soil is T = 278K. The initial velocity of the flow is V1 = 10 m/s. The initial pressure is p1 = 60 bar. Estimate the distance x∗ = x2 − x1 at which choking will occur. 29. Does chocking occur earlier for an isothermal pipe flow or for an adiabatic pipe flow of a perfect gas (Fanno process)? 62
a
b
Figure 3.26: (a) Supersonic HST wind tunnel of NLR (Amsterdam); (b) Supersonic flow around a space launcher, DNW.
3.6
Answers
1. The theory is valid for a steady frictionless quasi-one dimensional fluid flow. We assumed local thermodynamic equilibrium. No other assumptions where made concerning the type of fluid. It can be a gas or a liquid. Hence the theory is valid for a liquid. 2. The critical state corresponding to M = 1 can only be reached at a stationary point A1 dA dx = 0. Further analysis demonstrates that this should be a minimum of A (throat). When M = 1 is reached at the throat, the nozzle is chocked. The flow is then independent of the downstream conditions, because acoustic waves generated downstream cannot pass the throat. 3. The critical density ρ∗ and critical velocity V ∗ = c∗ are determined by the reservoir conditions. 4. See figure 3.27. 5. See figure 3.6. 6. Φ = Φmax = ρ∗ V ∗ Ak is independent of downstream boundary conditions. It is determined by the reservoir conditions. By changing the reservoir conditions we can change Φ in a chocked Laval nozzle. 7. For a perfect gas ρ∗√∼ ρ0 . At constance temperature T0 we have ρ0 ∼ p0 while c∗ ∼ T0 remains constant. Doubling p0 at fixed T0 doubles the mass flow Φ of a chocked Laval nozzle. 8. For a perfect gas have ρ∗√∼ ρ0 ∼ (1/T0 ) √ at constant pressure p0∗ we ∗ ∗ while c ∼ c0 = γRT0 . Hence √ Φmax = ρ c Ak ∼ (1/ T0 ). The mass flow decreases by a factor 2 when we increase the reservoir temperature by a factor 2, while keeping the reservoir pressure constant.
63
Figure 3.27: Comparison of a strong single shock compression from (p, 1/ρ) to (ˆ p/ˆ ρ) with a double shock compression with an initial shock from (p, 1/ρ) to (ˆ p1 , 1/ˆ ρ1 ) followed by a second shock from (ˆ p1 , 1/ˆ ρ1 ) to (ˆ p2 , 1/ˆ ρ2 ) in the (p, 1/ρ) diagram. 9.
T0 γ−1 2 =1+ M T 2 p0 ρ0 p0 T = ( ) γ = ( )γ ( )γ p ρ p T0 T0 γ−1 2 p0 γ−1 =1=+ M =( ) γ T 2 p
Figure 3.28: Sketch of stagnation pressure measured as a function of pa in the diverging part of a Laval nozzle at fixed reservoir pressure p0 .
64
M=
6
2 γ−1
#
p0 γ−1 ( ) γ −1 p
$
= 3.24
T = 1166 K Ak ρ V ρ0 ρ c 0 c = ∗ ∗ = ∗ M Ae ρ c ρ ρ0 c ∗ c 0 / 0 γ+1 Ak (γ + 1)/2 2(γ−1) =M = 0.05 2 Ae 1 + γ−1 2 M 10. At M = 24 we have for γ = 1.4: T0 γ−1 2 =1+ M = 116 T 2 so that T = 2.6 K. At such a low temperature the air liquifies. Furthermore at M = 24 with T = 300 K we have T0 = 35 × 103 K so that air is dissociated and ionized. 11. The curvature of streamlines following the wall involves a pressure gradient which should compensate the centrifugal acceleration. Hence the pressure on the center line is larger than at the wall. The pressure difference is of the order pcenter −pwall ≃ [ρ∗ (c∗ )2 /rk ]ak = (ak /rk )γp0 (p∗ /p0 ) = γ 2 (ak /rk )γp0 ( γ+1 ) γ−1 . 12. The critical condition is reached near the wall earlier than on the center line. Hence the critical line is convex seen from downstream. 13. We assume a stationary shock us = 0 of negligible thickness bounded by uniform flows (up- and downstream). Hence we neglect heat transfer and friction at the control surface enclosing the shock. Friction and heat transfer do occur within the shock, which explains the increase of entropy over the shock. 14. A thin shock has !a negligibly small volume. Hence the time derivative ! ! 2 of the integrals ρdτ , ρ⃗v dτ and ρ(h + u2 )dτ vanish and there is within a coordinate system moving with the shock no difference between a shock moving at constant speed us = const or a shock moving with time dependent speed us = us (t). 15. An isentrope curve depends on a single parameter (the entropy). Each point can be an initial point for a process ending along the same curve (compression or expansion). A Rankine-Hugoniot adiabate has one 65
(and only one) initial point and depends on two parameters (the coordinates of this initial point). Furthermore some processes are forbidden because they would reduce the entropy. This is the case for expansion shocks for ideal gasses. The Rankine-Hugoniot describes irreversible processes (with increase in entropy). 16. For γ = 1.4 the maximum compression with a single shock compression is ρˆmax /ρ = (γ + 1)/(γ − 1) = 6. For two successive shocks we have the total temperature is conserved ρˆmax /ρ = [(γ + 1)/(γ − 1)]2 = 36. 17. See Figure 3.27. ˆ 0 . For a 18. From the energy balance across a shock we have: h0 = h ˆ perfect gas h = cp T . Hence we have T0 = T0 over a shock in a perfect gas. 19. An expansion shock is possible if (∂ 2 p/∂(1/ρ)2 )s < 0. This can occur near the critical region of a fluid. dp 20. The slope is d(1/ρ) = −φ2m . As this is a constant imposed by the upstream boundary conditions, the Rayleigh line is a straight line in the (p, ρ1 ) plane.
21. For ρˆ/ρ = 2: ρˆ pˆ µ ρ − 1 = = 2.75 p µ − ρρˆ
#
#
sˆ − s cv
sˆ − s cv
$
$
= 0.041 exact
= 0.112 weak−shock
While the shock is almost isentropic, the prediction of the entropy is not accurate. For ρˆ/ρ = 1.2:
#
#
ρˆ pˆ µ ρ − 1 = = 1.29 p µ − ρρˆ
sˆ − s cv
sˆ − s cv
$
$
exact
= 6.8 × 10−4 = 9.010−4
weak−shock
The shock is almost perfectly isentropic and the predicted entropy increase is reasonably accurate. 66
22. From a measurement of the stagnation pressure ps (Pitot tube) and the static pressure p (wall pressure) we can determine the Mach number because ps /p is a function of M . For M >> 1 we can also determine the Mach number from ps /p0 . For M ≤ 1 we have ps /p0 = 1 so we cannot determine the Mach number from a measurement of pS /p0 . 23. See figure 3.28. 24. α = 2 arcsin (c/V ) = 2 arcsin (1/M ) For M < 1 this equation cannot be satisfied because sin(2α) ≤ 1. There are no Mach waves in subsonic flows. ˆ > 1. 25. For weak oblique shocks M ˆ < 1. For strong oblique shocks M 26. (a) M ≃ 1.8 (b) M ≃ 1.3.
. 27. M = V /c with . c = (∂p/∂ρ)s (isentropic wave speed). MT = V /cT with cT = (∂p/∂ρ)√ T (isothermal wave √ speed). For an ideal gas c = γRT and 7 cT = RT . As 1 < γ ≤
5 3
we have 1
0 and t > x/c0 we find: u′ = −u0 and p′ = −ρ0 c0 u0 . For a water flow with u0 = 1 m/s we find p′ = 15 bar at the left of the valve and p′ = −15 bar at the right. As p′ /(ρ0 c20 ) = O(10−3 ) −x/c0 with: p′ = p1 − p0 and: u′ =
p0 − p1 . ρ0 c 0
At a point (x, t) within the simple wave region we have a perturbation ! " % & p0 −p1 t− cx x ′ ′ ′ ′ 0 (p , u ) given by p (x, t) = p1 t − c0 − p0 and u (x, t) = . ρ0 c0 For large pressure differences p0 − p1 we consider a perfect gas. In that
Figure 4.31: Porous wall imposing a pressure p(0, t) at x = 0. case (see figure 4.31): u+
2c 2 c0 = γ−1 γ−1
And at x = 0 the boundary condition p(0, t) = p1 . Using the isentropic law for a perfect gas we find with c2 = γRT and p = ρRT : # $γ # $γ % & p ρ p c0 2 γ = = . p0 ρ0 p0 c Hence we have: c = c0
#
p p0
$ γ−1 2 γ
.
Consequently at x = 0 for t > 0 we have: c = c0
#
109
p1 p0
$ γ−1 2 γ
.
And:
/ # $ γ−1 0 2 c0 p1 2 γ u= 1− . γ−1 p0
A general solution of the simple wave region can be obtained by following C − characteristics with slope dx dt = u(0, tw ) − c(0, tw ) along which u = u(0, tw ) and c = c(0, tw ) where tw is the time at which the C − leaves the wall at x = 0. If p1 is contant the state found at the porous wall can be extended to a uniform region t > (u − c)t left of the wall. The original uniform region is delimited by t < −x/c0 . Between these two uniform regions, we a a simple wave region corresponding to an expansion fan. In the expansion fan the C − characteristics are straight lines emanating from the origin, so that: x = u − c. t Combining this equation for given (x, t) with the message carried by the C + emanating from the original uniform region: u+ yields: c= And:
2c 2 c0 = γ−1 γ−1
2 c0 γ − 1 %x& − γ+1 γ+1 t
u=c+
x 2 x = (c0 + ). t γ+1 t
19. As there are no independent time nor length scales in this problem we expect a self similar solution, which can consist of two pressure waves (expansion fan or shock wave) and a contact discontinuity. We do not have a complex wave region. As we have an expansion to vacuum, the compression wave (shock) and the contact discontinuity vanish. We have only an expansion fan, which is a simple wave. As we have a uniform region for x > 0 the characteristics travelling from the uniform region into the simple wave region are C − and they carry the message (see figure 4.32): 2c 2 c0 u− =− . γ−1 γ−1 With γ = 5/3 we have 2/(γ − 1) = 3:
u − 3 c = −3 c0 . The other family characteristics in the simple wave region are the C + emerging from the origin (x, t) = (0, 0). As these characteristics are 110
straight lines with the slope dx/dt = u + c we have: x = u + c. t From these equations we find for the expansion fan: & 3 %x u= − c0 4 t and:
& 1 %x + 3 c0 . 4 t The fan is delimited on the right by the line x = c0 t on which u = 0 (limit of uniform stagnant region). On the left it is limited by the line x = −3 c0 t corresponding to the maximum expansion (c = 0 and umax = −3 c0 ). The pressure follows from: # $5 c p = p0 . c0 c=
Figure 4.32: Expansion to vacuum. 20. We assume that the wave in the trombone is a simple wave propagating in a uniform region (p0 , c0 , u0 = 0). This assumption is reasonable 0 because p−p p0 < 1. We assume air to be a perfect gas with γ = 7/5. − The C characteristics emerging from the uniform region carry the message: 2c 2 c0 u− =− . γ−1 γ−1 Assuming an isentropic flow, we have: #
c c0
$
=
111
#
p p0
$ γ−1 2γ
.
Figure 4.33: Pressure measured at the end of the slide of a trombone. We clearly observe a shock wave at fortissimo level. This implies that: u + c = c0
/
γ+1 γ−1
#
p p0
$ γ−1 2γ
0 2 − . γ−1
Hence we have for a small perturbation ∆p: ∆(u + c) = c
γ + 1 ∆p . 2γ p
Furthermore we have: u+c ts = lim∆t→0 ∆(u+c) . c ∂p = u+c γ+1 p ∂t ∆t 2γ
As for small perturbations as considered here u 0 we have an expansion of the gas in the pipe. As there are no independent time or length scales we expect an expansion fan as simple wave in the region −c0 t ≤ x ≤ 0. At x = 0 we have the message carried by the C + emerging from the uniform region: u + 5c = 5c0 . Using u(0, t)/c = M0 we find at x = 0: u=
5 c0 M0 5 + M0
and M0 = u/c yields: c = c0 −
u 5 c0 = . 5 M0 + 5
The C − characteristics emanating from the x = 0 bring the message: u − 7c = c(0, t)[M0 − 5] = 5 c0
M0 − 5 M0 + 5
Along the C + characteristics we have: u + 5c = 5c0 . As a consequence we find a uniform region extending between the lines 0 0 x = 0 and x = M5c0 +5 (M0 −1)t. In the region −c0 t < x < M5c0 +5 (M0 −1)t − we have an expansion fan with straight C characteristics emanating from the origin (x, t) = (0, 0) so that: x = u − c. t Combined with the message from the C + emanating from the uniform region: u + 5c = 5c0 we find:
1 x c = (5c0 − ) 6 t
and: p = p0
#
c c0
$7
.
23. For x = ϵ > 0 we have a uniform region at the wall and between this uniform region and the original uniform region x > c0 t we have an expansion fan. The C − bring the message to the wall: u − 5c = u0 − 5c0 . 113
At the wall we have u = 0 so that c = c0 − u0 /5. The limiting C + characteristics of the new uniform region has the slope dx/dt = u+c = c0 − u0 /5 and passes through the origin: x = (c0 − u0 /5)t. Between this characteristics and the limit of the old uniform region x = c0 t we have an expansion fan with straight characteristics: x = u + c. t In this fan region we have: & 1% x 5c0 + − u0 6 t
c= and:
p = p0
#
c c0
$7
.
On the other side of the valve x = −ϵ < 0 we have a sudden compression involving a shock wave. We cannot use the method of characteristics. As we expect a self similar solution, the shock should have a time independent velocity us . We expect furthermore a uniform region between the shock and the valve −us t < x < 0. Using the continuity of mass flux over a shock ρw = ρˆw ˆ we find a shock relation for the velocities w and w ˆ in the reference frame moving with the shock. The valve the flow velocity vanishes u = 0 for t > 0 . We find using the shock relations in a framework moving with the speed us of the shock (w = u0 − us and w ˆ = −us ): −us − (u0 − us ) −u0 2 c20 − (u0 − us )2 = = u0 − us u0 − u s γ + 1 (u0 − us )2 After some algebra we find: M=
w u0 − us γ+1 = = [M0 + c0 c0 4
6
M02 +
#
4 γ+1
$2
].
The pressure behind the shock is uniform and given by: p − p0 2γ = (M 2 − 1). p0 γ+1 The shock velocity us is given by us = u0 − M c0 . Notice that for a weak shock we find us = −c0 . The shock degenerates into an acoustic wave. 24. We have four lines delimiting the various wave regions: Two C + characteristics: -Line I defined by x = L + (u0 + c0 )t. 114
-Line II defined by x = (u0 + c0 )t. Two C − characteristics: -Line III defined by x = (u0 − c0 )t. -Line IV defined by x = L + (u0 − c0 )t. Right from line I (x > L + (u0 + c0 )t) we have the original uniform region. Left from line III (x < (u0 − c0 )t) we have the original uniform region. Below lines II and IV (t < L/(2c0 )) we have a uniform perturbation p′ = Deltap and u′ = 0. Above lines II and IV (t > L/(2c0 )) we have a uniform region p′ = 0 and u′ = 0. The two regions left are simple waves. In the simple wave above line IV and between lines I and II we have a perturbation p′ = ∆p/2 and u′ = ∆p/(2ρc0 ). In the simple wave above line II and between lines III and IV we have a perturbation p′ = ∆p/2 and u′ = −∆p/(2ρ0 c0 ). 25. For γ = 3 we have 2/(γ − 1) = 1 and 2γ/(γ − 1) = 3.. Hence we have: p = p0
#
c c0
$3
.
At le right side of the piston the C − from the uniform region x > c0 t bring the message: u − c = −c0 . At the left side of the piston the C + from the uniform region x < −c0 t bring the message: u + c = c0 . The equation of motion of the piston is: 1 2 dup up 3 1 up 3 − + m = A[p − p ] = Ap0 (1 − ) − (1 + ) . dt c0 2 c0 For short times we find by Taylor expansion: m
dup up Ap0 = [1 − 9 ] dt 2 c0
The initial acceleration a0 = du/dt = Ap0 /(2m). For short times up ≃ a0 t. Hence on the compression side of the piston x > 0 we have c ≃ c0 + u and for the slope of the C + characteristics dx/dt = u + c ≃ c0 + 2u = c0 + A mp t . The shock formation distance is xs = c0 ts = c0 (u + c)/(∂(u + c)/∂t) ≃ m c20 /(A p0 ). 26. We assume a quasi-one dimensional flow ⃗v = (u, 0, 0). The equations motion for a frictionless incompressible fluid under influence of gravity are: 1 2 ∂u ∂u ∂p ρ +u =− ∂t ∂x ∂x 115
and:
∂p − ρg. ∂z Integration of the z-component yields: 0=−
p − patm = ρg(h − z) where z = 0 is the bottom of the flow and h the height of the free surface. Applying a integral mass conservation on a slab between x and x + ∆x we have: ∂h∆x = u(x)h(x) − u(x + ∆x)h(x + ∆x). ∂t Dividing by ∆x and taking the limit for ∆x → 0 we find: ∂h ∂u h =− . ∂t ∂x Applying the momentum equation for z = 0 we find: ∂u ∂u ∂h +u = −g ∂t ∂x ∂x √ Using the notation c = gh we can write these equations as: A ∂u ∂u g ∂h +u + c =0 ∂t ∂x h ∂x and:
A 1 2 g ∂h ∂h ∂u +u +c = 0. h ∂t ∂x ∂x
Addition and substraction yields: # $1 2 " A ∂ ∂ g + (u ± c) u± dh . ∂t ∂x h Hence:
" A
g dh = 2 c. h √ 27. We have an expansion fan for x < t g h0 . The maximum velocity √ − of the water is given by the √ C message u − 2c = −2c0 = −2 gh0 , which√implies umax = −2 √ gh0 . Hence the expansion fan is delimited by t gh0 > x > −2 t gh0 . Within the expansion fan we have: Γ=
x =u+c t
and: u − 2c = −2c0 . 116
Hence: c= or:
.
@ 1 ?x @ . 1 ?x + 2c0 = + 2 g h0 . 3 t 3 t @2 . 1 ?x h= + 2 g h0 . 9 t
g h=
and:
u= Appendix A
@ x 2 ?x . −c= − g h0 . t 3 t
Shock relations for a perfect gas. A1 :
pˆ p
A2 :
ρ−ˆ ρ ρ
A3 :
Tˆ−T T
A4 :
ˆ2 = 1−M
M 2 −1 2γ 1+ γ+1 (M 2 −1)
A5 :
ˆ1 ≡ M
=
=1+
2γ 2 γ+1 (M
u ˆ−u u
= =
=
− 1)
2 1−M 2 γ+1 M 2
2(γ−1)(γM 2 +1)(M 2 −1) (γ+1)2 M 2
u−ˆ u cˆ
2 M 2 −1 γ+1 M
ˆ = ; M 7
u ˆ cˆ
T Tˆ
These relations can be deduced from the elementary conservation laws: A6 :
ρu = ρˆu ˆ,
A7 :
p + ρu2 = pˆ + ρˆu ˆ2 ,
A8 :
1 2 2u
ˆ + h = 12 u ˆ2 + h,
combined with the following expression for the enthalpy of a calorically perfect gas: γ p γ γ pˆ γ ˆ ˆ h = γ−1 ρ = γ−1 RT , h = γ−1 ρˆ = γ−1 RT . The solution of the problem is very much simplified by the condition that the medium in the tube is initially at rest. The disturbance generated at x = 0 propagates in the +x-direction. There are no reflections or other disturbances which travel in the −x-direction. This is an example of a so called simple wave, a concept that will appear to be very important in the analytical description of non-linear waves. In this case we have a so-called right-running simple wave. Such a simple wave has two important properties:
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