Chapter For the solution of the Euler equations to represent adequately a given large-reynolds-number flow, we need to consider in general the existence of discontinuity surfaces, across which the fluid properties or their derivatives exhibit finite jumps. Mach cones, for instance, represent an example of weak discontinuities with jumps in derivatives. Boundary layers and also vortex sheets and mixing layers) are examples of tangential discontinuities. We shall focus below on the analysis of normal discontinuities, showing that they are always compression waves, also called shock waves, with specific results given below for the jump conditions across normal and oblique shocks. Conversely, expansion waves cannot be discontinuities; rather, they are continuous isentropic waves, to be analyzed below in the specific case of steady flow around a sharp corner the so-called Prandtl-Meyer expansion flow). Jump conditions across discontinuity surfaces To determine the conditions that relate the different flow properties that exist on the sides of a given discontinuity surface we shall apply the conservation equations to the infinitesimally small control volume shown in figure.. As can be seen, the control volume includes faces of area dσ, lying on the two sides of the surface, separated by a small distance ε. In a reference frame moving with the discontinuity, mass conservation can be expressed in the form d ρdv + ρ v ndσ 0..) dt V c Σ c The first integral is proportional to the volume of the control volume, of order εdσ, and therefore becomes negligibly small when a infinitesimally thin control volume satisfying ε dσ) / is considered. In that limit, only fluxes across the two faces of area dσ need to be computed when evaluating the second integral, which therefore provides the simplified continuity equation ρ v n ρ v n m,.) where m denotes the local value of the mass flux across the discontinuity and the subscript n is used to denote normal velocity components, with the subscript t employed below for the tangential components. Volume integrals can also be neglected when evaluating the momentum
and energy conservation equations, which can be seen to provide ρ v n +p ρ v n +p.3) ρ v n v t ρ v n v t.4) and m e + p ) + v m e + p ) + v..5) ρ ρ v t n v n dσ v ε n v n v v t Figure.: Jumps across a discontinuity surface. Equations.).5) apply to all discontinuities. Tangential discontinuities, such as boundary layers and vortex sheets, are defined by the condition that the mass flux is identically zero, that is, m 0. From.) it then follows that v n v n 0 and, from.3), it is seen that p p, indicating that the jump in pressure across tangential discontinuities is zero, a condition used before when analyzing jets and boundary layers. With m 0, the conservation of tangential momentum and energy, expressed in.4) and.5), are identically satisfied, regardless of the values of v t v t, ρ ρ, and T T. Normal discontinuities When the mass flux across the discontinuity is nonzero, then it follows from.4) that whereas.5) reduces to v t v t,.6) h + v h + v,.7) indicating that, in a reference frame moving with the discontinuity, the stagnation enthalpy and, therefore, the stagnation temperature) is conserved. 0
For given values of v n, ρ, and p, equations.),.3), and γ p + v n γ ρ γ p + v n γ ρ determine v n, ρ, and p, with.8) corresponding to the energy conservation balance.7) written with use made of.6) and of the thermodynamic identity h [γ/γ )]p/ρ. Straightforward manipulation of.) and.3) leads to the equation of the so-called Rayleigh line p m ρ )..9) p ρ p ρ On the other hand, writing.8) in the form γ p p ) γ ρ ρ vn v n) ) ) m ρ ρ + ρ ρ and using.9) to eliminate m yields p p γ+ γ ρ ρ γ+ ρ γ.8).0) ρ..) This last equation, relating the jumps in pressure and density across the discontinuity, is known as the Hugoniot curve. Observation of the numerator and denominator in.) indicates that the solution for the density ratio across the discontinuity must lie in the range γ γ + < ρ < γ + ρ γ.) while the range of possible pressure ratios extends in principle for 0 < p /p <, giving the solution shown in Fig.. for γ.4. For a given mass flux m, the values of p /p and ρ /ρ are determined by the point where the Rayleigh line.9) and the Hugoniot curve.) cross in Fig.. of course, a second solution p /p ρ /ρ always exists, but that is just the trivial solution corresponding to a continuous flow with no jumps). To investigate the different possible solutions, we begin by noting that the slope of the Rayleigh line can be written in terms of the Mach number associated with the normal component of the incident velocity M n v n /a according to m ρ p v n p /ρ γm n..3) This value is to be compared with the slope of the Hugoniot curve at ρ /ρ, which can be computed by straightforward differentiation of.) to give ) dp /p ) γ..4) dρ /ρ ) ρ /ρ For M n the Rayleigh line is tangent to the Hugoniot curve at p /p ρ /ρ, which is therefore the only solution of.9) and.), whereas a nontrivial crossing point appears for M n according to
0 9 8 Hugoniot curve 7 6 p/p 5 4 3 Rayleigh line M n > ) Rayleigh line M n < ) p /p ρ /ρ ) γ 0 0 3 4 5 6 γ )/γ+) ρ /ρ γ+)/γ ) Figure.: Hugoniot curve and Rayleigh line. If the normal component of the incident velocity is supersonic M n > ), then the Rayleigh line crosses the Hugoniot curve along the upper stretch, so that the solution corresponds to a compression with > p > and γ p γ + < ρ < ρ If the normal component of the incident velocity is subsonic M n < ), then the Rayleigh line crosses the Hugoniot curve along the lower stretch, so that the solution corresponds to an expansion with > p > 0 and < ρ < γ + p ρ γ Although both compressions and expansions are apparently valid solutions of.9) and.), only compression waves, termed shock waves, may exist in reality. To prove this point, one may use thermodynamic arguments based on the second principle, which indicates that, in an adiabatic process, the entropy change s s can never be negative, i.e., the entropy can never
be destroyed. For a perfect gas, s s c v ) p /p ln ρ /ρ ) γ,.5) and one may use.) to compute the entropy variation. As seen in Fig.., for ρ /ρ < the Hugoniot curve.) lies above the isentrope p /p ρ /ρ ) γ. Compression waves, therefore, are associated with positive entropy increments s s c v ln γ+ γ ρ ρ )/ γ+ ρ γ ρ /ρ ) γ ) ρ > 0,.6) as can be calculated from.5). Conversely, in expansion waves with ρ /ρ < the Hugoniot curve.) lies below the isentrope p /p ρ /ρ ) γ, so that the entropy change computed from.5) would be negative s s < 0), thereby violating the second principle of thermodynamics. Hence, normal discontinuity surfaces are always shock waves, with jumps of thermodynamic properties in the ranges > p p >, γ γ + < ρ ρ <, and > T T >,.7) the latter to be computed from the equation of state T /T p /p )/ρ /ρ ). The normal component of the incident velocity v n must be supersonic, so that the inequalities M > M n >.8) are always satisfied. Use of.3) and.6) leads to the additional conditions It can also be seen that γ γ + < v n v n < and v v <..9) M n <..0) It should be noted that, despite this last result, in oblique shock waves the downstream velocity can be however supersonic, i.e., M >. Jump conditions across shock waves To resolve for the jumps across a generic shock wave, it is convenient to consider first the change in the normal component of the velocity along with the jumps in thermodynamic properties, and address subsequently the deflection of the flow δ as a function of the incident angle β see Fig..3). The development begins by using.),.3), and.8), which are rewritten below for convenience ρ v n ρ v n,.) ρ v n +p ρ v n +p,.) p γ + v n γ ρ p γ + v n γ ρ..3) 3
v n v β v t v t v n v δ Figure.3: Velocity jumps across a normal discontinuity. For a gas of known specific-heat ratio γ, these three equations determine the downstream values v n, p, and ρ in terms of v n, p, and ρ. The dependence can be simplified by using the Π theorem of dimensional analysis to give v n f v n,p,ρ,γ) p f v n,p,ρ,γ) ρ f 3 v n,p,ρ,γ) v n /v n ϕ M n,γ) p /p ϕ M n,γ) ρ /ρ ϕ 3 M n,γ).4) The functions ϕ, ϕ, and ϕ 3 can be determined by manipulation of.).3) to give we leave the needed development as an exercise for the reader) v n v n ρ ρ ) +γ )M n γ +)M n and p γm n + γ,.5) p γ + which can be combined with T /T p /p )/ρ /ρ ) and M n M n ρ /ρ ) p /p ) to give T γm n + γ)[+γ )M n ] T γ +) Mn and Mn +γ )M n γmn..6) + γ The above formulae take simplified forms in limiting cases of interest, including very strong shocks with M n, for which v n v n ρ ρ ) γ γ +, p p γm n γ +, T T γγ )M n γ +) and M n γ γ,.7) 4
and weak shocks with ε M n, for which v n v n ρ ρ ) ε γ +, p + γε p γ +, T γ )ε + T γ +.8) and M n ε. The associated entropy increase can be computed from.5) to give s s )/c v ε 3, indicating that these weak shocks are nearly isentropic. For normal shock waves, such that the incident flow forms a angle β π/ with the shock, no flow deflection occurs, so that M M n and M M n, and the above equations.5) and.6) are all we need to determine the jumps across the shock. As an illustrative example, we can use them to determine the flow induced by a piston moving with constant velocity in a long cylindrical tube containing a gas. The piston, initially at rest, is suddenly accelerated to reach a constant velocity V P, generating a shock wave that moves with velocity V SW along the tube. The gas ahead of the shock is at rest, with initial thermodynamic properties characterized for instance by the sound velocity a and pressure p. The shock wave sets the fluid in motion, so that the gas between the shock and the piston moves with uniform constant velocity, equal to that of the piston. FIXED REFERENCE FRAME MOVING REFERENCE FRAME V P p V SW p v V V SW P v V SW Figure.4: Piston-supported shock wave. The velocities ahead and behind the shock v v n and v v n and the corresponding Mach numbers M n and M n appearing in.5) and.6) are measured relative to the shock, so in analyzing the problem we need to use a reference frame moving to the right with velocity V SW. In that reference frame, the incident and downstream velocities are v V SW and v V SW V P, whereas the incident Mach number is M n M V SW /a. In terms of these quantities, the first equation in.5) yields which can be rewritten in the form ) VSW γ + V P a a v v V P V SW +γ )V SW/a ) γ +)V SW /a ),.9) VSW a ) 0..30) Solving now for V SW /a gives [ M n V SW γ + ) V P γ + ) ] / VP + +,.3) a 4 a 4 which can be used in.5) and.6) to determine the pressure, density and temperature in front of the piston as a function of V P /a. a 5
The above piston problem can be used as a simplified representation of the flow induced by a high-speed train entering a tunnel, the main simplification being that our piston occupies the whole transverse section of the tube, whereas in reality the train occupies only a fraction of the tunnel section. The results indicate that, as the train enters the tunnel, a sudden overpressure develops immediately ahead. The shock is weak for V P a, when M n γ + V P and p γ V P, 4 a p a which explain why, in conventional trains, the problems encountered at the entrance of a tunnel are minor. Relative pressure differences of order unity appear however as V P approaches values of the order of the sound velocity a. That is the case of high-speed trains, for which special measures need to be taken to remedy the overpressure problem at the tunnel entrance. Oblique shocks For known values of M and β one may compute M n M sinβ,.3) which in turn determines from.5) and.6) the jumps of thermodynamic properties and the values of Mn +γ )M sin β γm.33) sin β + γ and v n +γ )M sin β v n γ +)M sin β..34) To determine the flow deflection, we use.6) written in the form v t v cosβ) v cosβ δ)..35) Using the equations v n v sinβ) and v n v sinβ δ) to give tanβ δ) tanβ) v n v n.36) and solving for δ with use made of.34) yields Once δ is known one may easily compute with M n determined from.33). M tanδ) cotβ) sin β) +M.37) [γ +cosβ)]. M M n sinβ δ),.38) 6
90 80 70 60 50 40 30 0 0 0...3.4.5.6.7.8.9.0..4.6.8 3.0 3. 3.4 3.63.8 4.0 4.5 5 6 8 0 0 M < M >! γ.40 M β δ 0 5 0 5 0 5 30 35 40 45 50 Figure.5: The variation of β with δ for constant values of M. 7
M.0 M.5 M.0 M.5 M 3.0 M 3.5 M 4.0 M 4.5 30 8 6 4 0 8 6 4 9 0 8 7 6 5 4 3 γ.40 M,p M,p δ 40 35 30 5 0 Normal shock 5 0 5.5.5 3 3.5 4 4.5 5 M p/p Figure.6: The variation of p /p with M for constant values of M and δ. 8
The results are summarized in Figs..5 and.6, where the angles are given in degrees. In particular, Fig..5 indicates that for given values of M and δ two different solutions are found if δ < δ max and no solution exists for δ > δ max, where the value of δ max increases for increasing M. For δ < δ max, the solution with larger β corresponds to a stronger shock wave i.e., larger value of M n M sinβ). The corresponding value of M n is sufficiently small that the flow downstream is subsonic, that is, M <. On the other hand, the solution lying along the lower part of the curve of constant M in Fig..5 corresponds to a lower value of β, and therefore a weaker solution including a values of M that are generally supersonic except for a small region near δ δ max, as indicated in the figure). Because the associated downstream flow is supersonic, the weak-shock solution is more stable, in that perturbations can never reach it from downstream. Because of their stable character, when both strong or weak solutions may exist, weak solutions tend to prevail in realistic configurations. p β θ M M p δ Figure.7: Supersonic flow over a wedge. For each value of M, there exist two solutions with zero deflection angle δ 0. One is the normal shock wave, corresponding to β π/ and M M n, which is the strongest possible wave for that value of M. The other, corresponding to β µ with µ sin M ),.39) is the weakest shock wave for that value of M ; the incident angle is such that M n M sinβ. These limiting solutions are termed Mach waves. According to what we mentioned below.8), these weak shocks are effectively isentropic compressions. A Mach wave is seen to bound, for instance, the Mach cone sketched in Fig.. Figure.5 can be used, for instance, to solve the supersonic flow over a wedge, characterized by the appearance of an oblique shock, that forms an angle θ with the wall, deflecting the incoming 9
Figure.8: Supersonic flow over a blunt body. stream to make it parallel to the wedge, as sketched in Fig..7. If the incident Mach number is M and the wedge semi angle is δ 5 o, then two solutions are possible, as can be seen in Fig..5. Following the arguments given above, we select the weaker shock, corresponding to β 45 o, thereby giving θ β δ 30 o and M n M sinβ. Using now.5) and.6) yields p /p.7 and M n 0.734. This last value can be substituted into.38) to give M.47. The problem could also be solved by using Fig..6 instead, an exercise left for the reader to attempt. Also of interest is that, for the flow over a wedge, no solution involving an oblique shock exists for δ > δ max 3 o. In that case, the shock detaches to form a bow shock wave, which is locally planar at the center line and becomes increasing inclined farther away. These bow shock waves are found in general in front of blunt bodies, as seen in the experimental visualization shown below in Fig..8. Regarding the flow over a wedge or rather the equivalent case of the flow around a concave corner), it is of interest to consider what happens when, instead of a sharp corner, one encounters a smoothly curved wall, as shown in Fig..9. The perturbations associated with the continuous increase of slope are necessarily weak near the wall, and therefore give rise to a system of Mach waves that propagate into the stream with a local angle of incidence β µ such that, at each position, M n M sinβ. For the concave wall, the value of M decreases and the pressure increases as the flow crosses the system of Mach waves, giving an increasing value of µ and causing the Mach waves to approach one another. The compression reinforces as a result of the interaction between the different Mach waves, so that a shock wave of finite strength eventually emerges. This complex interaction process is at the origin of the oblique shock that is seen to emerge from the corner when the observation distance is sufficiently large for the curved wall to appear as a sharp corner see Fig..9). Prandtl-Meyer expansions Let us now consider the case of supersonic flow over a convex wall, represented in Fig..0. A system of Mach waves originates from the wall as it curves downwards. In this case, however, the value of M increases and the pressure decreases as the flow crosses the system of Mach waves, giving an decreasing value of µ and causing the Mach waves to open up. As a result, the 0
p p >p M <M M M n M MACH WAVES Figure.9: Supersonic flow over a concave wall. expansion wave generated as the supersonic stream turns around the convex corner is formed by a fan of Mach waves that cause the stream lines to evolve smoothly. Since the resulting flow is steady and isentropic, all stagnation magnitudes are preserved across the expansion wave. Besides the condition of conservation of all stagnation properties, T 0 T ) γ )/γ p0 p ) γ ρ0 ρ a0 a ) γ + M,.40) the solution for the convex-corner expansion, commonly known as Prandtl-Meyer expansion, requires knowledge of the relationship between the flow deflection and the flow acceleration, which comes from consideration of the infinitesimal evolution due to a given Mach wave. As shown schematically in Fig.., the condition that the tangential component of the velocity must be conserved as the flow is deflected can be used to relate the velocity increment and the deflection according to dθ dv/tanµ..4) v From.39) it follows that tanµ To obtain dv/v we begin by differentiating v Ma to give M..4) dv v dm M + da a..43)
p M µ M n M µ M n M >M p <p θ MACH WAVES M >M Figure.0: Supersonic flow over a convex wall. On the other hand, differentiating the last equation in.40) for a constant value of a 0 yields which can be combined with.43) to provide Substituting.4) and.45) into.4) finally gives da a γ )MdM +γ )M /,.44) dv v dm +γ )M / M..45) dθ M dm +γ )M / M.46) as a relationship between the deflection and the Mach number increment. This last equation can be integrated to relate the total deflection θ with the change in Mach number as the supersonic stream in Fig..0 turns around the corner according to M M θ dm +γ )M / M,.47) which can be alternatively written as M in terms of the so-called Prandtl-Meyer function θ νm ) νm ).48) νm) M M dm +γ )M / M.49)
v µ dθ dv{ tan µ v µ µ { dv Figure.: Differential evolution across a Mach wave. given in explicit form by νm) [ ] γ + γ γ tan γ + M ) tan M ).50) with tan representing the arctangent the inverse of the tangent function). For the problem represented in Fig..0, if the corner angle θ and the Mach number of the incoming stream M are known, then one may use.48) to determine the value of M and then use.40) written in the form T T p p ) γ )/γ ρ ρ ) γ a a ) +γ )M / +γ )M /.5) to obtain the ratios of thermodynamic properties across the expansion. For instance, if θ 5 o and M, then νm ) 6.38 o. Using this value in.48) gives νm ) 4.38 o, which can be employed to obtain from.50) the value M.6. M > p 0 M Figure.: Supersonic stream discharging to the vacuum. In other occasions, what determines the flow deflection is the pressure that exists downstream. An example of interest occurs at the exit of nozzles of thrust rockets used for satellite attitude 3
control. Near the nozzle edge the flow is locally planar and corresponds in the first approximation to a Prandtl-Meyer expansion, as indicated in Fig... The surrounding pressure is very small, so that the expanding gas reaches very high Mach numbers M. If we assume for simplicity complete vacuum by using p 0, then for any finite values of M and p one obtains M, as can be seen from.5). The associated flow deflection must be calculated from.48) with νm ) ν ) 30.45 o. If the value of M is sufficiently small, then the deflection may exceed 90 o, giving rise to a reverse jet flow that may impinge on the satellite wall, possibly damaging it for instance, for M one obtains νm ) 6.38 o and θ 04.07 o ). This could be avoided by ensuring that the value of M remains above M.56, so that θ < 90 o. 4