Compressible flow 297 If a disturbance of large amplitude, e.g. a rapid pressure rise, is set up there are almost immediate physical limitations to its continuous propagation. The accelera- tions of individual particles required for continuous propagation cannot be sustained and a pressure front or discontinuity is built up. This pressure front is known as a shock wave which travels through the gas at a speed, always in excess of the acoustic speed, and together with the pressure jump, the density, temperature and entropy of the gas increases suddenly while the normal velocity drops. Useful and quite adequate expressions for the change of these flow properties across the shock can be obtained by assuming that the shock front is of zero thickness. In fact the shock wave is of finite thickness being a few molecular mean free path lengths in magnitude, the number depending on the initial gas conditions and the intensity of the shock. 6.4.1 One-dimensional properties of normal shock waves Consider the flow model shown in Fig. 6.7a in which a plane shock advances from right to left with velocity u1 into a region of still gas. Behind the shock the velocity is suddenly increased to some value u in the direction of the wave. It is convenient to superimpose on the system a velocity of u1 from left to right to bring the shock stationary relative to the walls of the tube through which gas is flowing undisturbed at u1 (Fig. 6.7b). The shock becomes a stationary discontinuity into which gas flows with uniform conditions, p1, p1, u1, etc., and from which it flows with uniform conditions, p2, p2, u2, etc. It is assumed that the gas is inviscid, and non-heat conducting, so that the flow is adiabatic up to and beyond the discontinuity. The equations of state and conservation for unit area of shock wave are: State (6.35) Mass flow Siationary shock (b) Fig. 6.7 298 Aerodynamics for Engineering Students Momentum, in the absence of external and dissipative forces P1 + Plu: = Pz + P24 Energy 6.4.2 Pressure-density relations across the shock Eqn (6.38) may be rewritten (from e.g. Eqn (6.27)) as which on rearrangement gives From the continuity equation (6.36): and from the momentum equation (6.37): 1 m u2 1 =Tb1 -Pz) Substituting for both of these in the rearranged energy equation (6.39) (6.37) (6.38) (6.39) (6.40) and this, rearranged by isolating the pressure and density ratios respectively, gives the RankineHugoniot relations: 7+1 PZ 1 (6.41) (6.42) Compressible flow 299 Taking y = 1.4 for air, these equations become: (6.41a) and P2 6-+ 1 P2 - P1 P2 P1 6+- PI (6.42~1) Eqns (6.42) and (6.42a) show that, as the value of p2/p1 tends to (y + l)/(*/ - 1) (or 6 for air), p2/p1 tends to infinity, which indicates that the maximum possible density increase through a shock wave is about six times the undisturbed density. 6.4.3 Static pressure jump across a normal shock From the equation of motion (6.37) using Eqn (6.36): P1 P1 PI or E- P1 1 =./M:[1-3 but from continuity u2/u1 = p1/p2, and from the RankineHugoniot relations p2/p1 is a function of (p2/p1). Thus, by substitution: Isolating the ratio p2/p1 and rearranging gives Note that for air P2 - 7M: - 1 -_ P1 6 (6.43) (6.43a) Expressed in terms of the downstream or exit Mach number M2, the pressure ratio can be derived in a similar manner (by the inversion of suffices): (6.44) 300 Aerodynamics for Engineering Students or P1-7@-1 - - for air P2 6 (6.44a) 6.4.4 Density jump across the normal shock Using the previous results, substituting for p2/p1 from Eqn (6.43) in the Rankine- Hugoniot relations Eqn (6.42): -= or rearranged For air -/ = 1.4 and p2 6M: P1 5+M? -=- Reversed to give the ratio in terms of the exit Mach number - P1 - - (Y+ w; p2 2+ (7- 1)M; For air 6.4.5 Temperature rise across the normal shock Directly from the equation of state and Eqns (6.43) and (6.45): For air (6.45) (6.45a) (6.46) (6.46a) (6.47) (6.47a) Compressible flow 301 Since the flow is non-heat conducting the total (or stagnation) temperature remains constant. 6.4.6 Entropy change across the normal shock Recalling the basic equation (1.32) = (~~~~-'= e) (E)' from the equation of state which on substituting for the ratios from the sections above may be written as a sum of the natural logarithms: These are rearranged in terms of the new variable (M: - 1) On ex anding these logarithms and collecting like terms, the first and second powers of (M, !i? - 1) vanish, leaving a converging series commencing with the term (6.48) Inspection of this equation shows that: (a) for the second law of thermodynamics to apply, i.e. AS to be positive, M1 must be greater than unity and an expansion shock is not possible; (b) for values of M1 close to (but greater than) unity the values of the change in entropy are small and rise only slowly for increasing MI. Reference to the appropriate curve in Fig. 6.9 below shows that for quite moderate supersonic Mach numbers, i.e. up to about M1 = 2, a reasonable approximation to the flow conditions may be made by assuming an isentropic state. 6.4.7 Mach number change across the normal shock Multiplying the above pressure (or density) ratio equations together gives the Mach number relationship directly: =1 p2 xp' = 2YM; - (7- 1) 2YM; - (7 - 1) P1 P2 Y+l Y+l Rearrangement gives for the exit Mach number: (6.49) 302 Aemdynamics for Engineering Students For air M:+5 M2 - 2-7Mf-1 (6.49a) Inspection of these last equations shows that M2 has upper and lower limiting values: For M1 403 M2 + E = (l/& = 0.378 for air) ForM1+ 1 M2+ 1 Thus the exit Mach number from a normal shock wave is always subsonic and for air has values between 1 and 0.378. 6.4.8 Velocity change across the normal shock The velocity ratio is the inverse of the density ratio, since by continuity u2/u1 = p1/p2. Therefore, directly from Eqns (6.45) and (6.45a): or for air (6.50) (6.50a) Of added interest is the following development. From the energy equations, with cpT replaced by [r/(-y - l)lp/p, pl/p~ and p2/p2 are isolated: !!! - - (cPTo - $) ahead of the shock P1 Y and E = P2 7 (cPTo - 2) downstream of the shock The momentum equation (6.37) is rearranged with plul = p2u2 from the equation of continuity (6.36) to P2 P1 u1 2= p2u2 PlUl and substituting from the preceding line Compressible flow 303 PlMl UIT( Disregarding the uniform flow solution of u1 = 242 the conservation of mass, motion and energy apply for this flow when P24 U2G (6.51) i.e. the product of normal velocities through a shock wave is a constant that depends on the stagnation conditions of the flow and is independent of the strength of the shock. Further it will be recalled from Eqn (6.26) that where a* is the critical speed of sound and an alternative parameter for expressing the gas conditions. Thus, in general across the shock wave: 241242 = a*2 (6.52) This equation indicates that u1 > a* > 242 or vice versa and appeal has to be made to the second law of thermodynamics to see that the second alternative is inadmissible. 6.4.9 Total pressure change across the normal shock From the above sections it can be seen that a finite entropy increase occurs in the flow across a shock wave, implying that a degradation of energy takes place. Since, in the flow as a whole, no heat is acquired or lost the total temperature (total enthalpy) is constant and the dissipation manifests itself as a loss in total pressure. Total pressure is defined as the pressure obtained by bringing gas to rest isentropically. Now the model flow of a uniform stream of gas of unit area flowing through a shock is extended upstream, by assuming the gas to have acquired the conditions of suffix 1 by expansion from a reservoir of pressure pol and temperature TO, and downstream, by bringing the gas to rest isentropically to a total pressurep02 (Fig. 6.8) Isentropic flow from the upstream reservoir to just ahead of the shock gives, from Eqn (6.18a): (6.53) Fig. 6.8 304 Aerodynamics for Engineering Students and from just behind the shock to the downstream reservoir: (6.54) Eqn (6.43) recalled is and Eqn (6.49) is These four expressions, by division and substitution, give successively Rewriting in terms of (M? - 1): x [l + (M: - 1)]-7/(74 Expanding each bracket and multiplying through gives the series For values of Mach number close to unity (but greater than unity) the sum of the terms involving M; is small and very close to the value of the first term shown, so that the proportional change in total pressure through the shock wave is APO -pol -pOz e- 2y (M: - 113 (Y+ 3 Po 1 Po 1 (6.55) It can be deduced from the curve (Fig. 6.9) that this quantity increases only slowly from zero near M1 = 1, so that the same argument for ignoring the entropy increase (Section 6.4.6) applies here. Since from entropy considerations MI > 1, Eqn (6.55) shows that the total pressure always drops through a shock wave. The two phenomena, Compressible flow 305 Fig. 6.9 i.e. total pressure drop and entropy increase, are in fact related, as may be seen in the following. Recaiiing Eqn (1.32) for entropy: eAS/c, =E (cy= E (cy P1 P2 POI Po2 since 5 =Po' etc, P: 4, But across the shock To is constant and, therefore, from the equation of state pol /pol = p02/po;? and entropy becomes and substituting for AS from Eqn (6.48): (6.56) 306 Aerodynamics for Engineering Students Now for values of MI near unity /3 << 1 and = 1 - e-P APO -Pol - Po2 Po1 Pol ~ APO - 2n' (M: - 'I3 (as before, Eqn (6.55)) Po1 (y+ 1)2 3 6.4.10 Pit& tube equation The pressure registered by a small open-ended tube facing a supersonic stream is effectively the 'exit' (from the shock) total pressure p02, since the bow shock wave may be considered normal to the axial streamline, terminating in the stagnation region of the tube. That is, the axial flow into the tube is assumed to be brought to rest at pressure p02 from the subsonic flow p2 behind the wave, after it has been compressed from the supersonic region p1 ahead of the wave, Fig. 6.10. In some applications this pressure is referred to as the static pressure of the free or undis- turbed supersonic streampl and evaluated in terms of the free stream Mach number, hence providing a method of determining the undisturbed Mach number, as follows. From the normal shock static-pressure ratio equation (6.43) P2 -= 2-M: - (7 - 1) P1 -!+I From isentropic flow relations, - MlPl PI Y- Shock.assumed normal ond plane lmlly to the oxiol streamline Fig. 6.10 [...]... and with q2 = (v2 a2) (Eqn (6. 17)): + + (6. 65) which gives the differential equation (6. 66) Equation (6. 66) may now be integrated Thus or (6. 67) From Eqn (6. 65) 314 Aerodynamics for Engineering Students which allows the flow deflection in Eqn (6. 67) to be expressed as a function of Mach angle, i.e vp -a =p + $ tan-' 2 $scot Y+l -K 2 (6. 68) or vp - = f(p) (6. 68a) In his original paper Meyer" used the... p=arctan- and d p = 1 V + 1 du v - = -& (u/v)2 v q2 but the change in deflection angle is the incremental change in Mach angle Thus V dvp = dp = -du (6. 60) q2 Combining Eqns (6. 59) and (6. 60) yields 24 dq _ - q- dvp v U since - = arc tan p = V 1 m (6. 61) where q is the flow velocity inclined at vp to some datum direction It follows from Eqn (6 lo), with q substituted for p, that (6. 62) 312 Aerodynamics for. .. complementary angle to the Mach wave (+) = [(7c/2) - p] and expressed the function as the angle q5 to give Eqn (6. 68a) in the form vp-a=q 5-+ (6. 68b) The local velocity may also be expressed in terms of the Mach angle p by rearranging the energy equation as follows: q2 a2 -+ - =- 2 7-1 c2 2 but a2 = q2sin2p Therefore or (6. 69) Equations (6. 68) and (6. 69) give expressions for the flow velocity and direction at any... =-= u1 a1 VI a1 u1 or (6. 73) Similarly (6. 74) The results of Section 6. 4.2 may now be used directly, but with M I replaced by p, and M by A42 sin (p - 6) The following ratios pertain: 2 Static pressure jump from Eqn (6. 43): M sin 1 (6. 75) or as inverted from Eqn (6. 44): (6. 76) Density jump from Eqn (6. 45): (6. 77) or from Eqn (6. 46) (6. 78) Compressible flow 321 Static temperature change from Eqn (6. 47):... P1 -= - by continuity Thus (6. 81) Equations (6. 77) and (6. 81) give the different expressions for pz/p1, therefore the right-hand sides may be set equal, to give: (6. 82) Algebraic rearrangement gives (6. 83) -1 Mi 2 -( Plotting values of p against S for various Mach numbers gives the carpet of graphs shown in Fig 6. 24 It can be seen that all the curves are confined within the M = 00 curve, and that 1 for. .. maximum value 6~ may result in a smaller (weak) or larger (strong) wave angle p To solve Eqn (6. 83) algebraically, i.e to find P for a given M1 and 6, is very difficult However, Collar* has shown that the equation may be expressed as the cubic - cx2* A.R Collar, J R Ae S, Nov 1959 + ( B- AC) = 0 (6. 84) 322 Aerodynamics for Engineering Students Fig 6. 24 where x = Cot/?, A = M 2 - 1, B =- ;tanS T+lM 2... direction is given by arctan(qt/q,) For the wave angle p (recall the additional notation of Fig 6. 25 323 324 Aerodynamics for Engineering Students Fig 6. 23), linear conservation of momentum along the wave front, Eqn (6. 72), gives vi = v2, or, in terms of geometry: VI cos p = V2 cos@ - 6 ) (6. 89) Expanding the right-hand side and dividing through: V I = V2[cosS+ tanpsin6I (6. 90) or, in terms of the polar... ( V I-qn) (6. 93) Again from continuity (expressed in polar components): pi Vi sin P = p2 V2 sin(P - S) = mqn(sin, - cos ,8tan 6 B ) or (6. 94) Divide Eqn (6. 93) by Eqn (6. 94) to isolate pressure and density: Again recalling Eqn (6. 91) to eliminate the wave angle and rearranging: (6. 95) Compressible flow Finally from the energy equation expressed in polar velocity components: up to the wave (6. 96) and... Fig 6. 16 Prandtl-Meyer expansion with finite deflection angle wave angle p = 4 2 In the general case, the datum (sonic) flow may be inclined by some angle a to the coordinate in use Substitute dvp for (l/q)dq/tanp from Eqn (6. 61) and, since qsinp = a, Eqn (6. 64) becomes dvp - dp = dv/a But from the energy equation, with c = ultimate velocity, a2/( 7- 1) (q2/2)= (c2/2) and with q2 = (v2 a2) (Eqn (6. 17)):... the expansion may be obtained, with reference to Fig 6. 16 Algebraic expressions for the wavelets in terms of the flow velocity be obtained by further manipulation of Eqn (6. 61) which, for convenience, is recalled in the form: Introduce the velocity component v = q cos p along or tangential to the wave front (Fig 6. 13) Then dv=dqcosp-qsinpdp=qsinp (6. 64) It is necessary to define the lower limiting or . (Eqn (6. 17)): which gives the differential equation Equation (6. 66) may now be integrated. Thus or From Eqn (6. 65) (6. 65) (6. 66) (6. 67) 31 4 Aerodynamics for Engineering Students. 2YM; - ( 7- 1) 2YM; - (7 - 1) P1 P2 Y+l Y+l Rearrangement gives for the exit Mach number: (6. 49) 302 Aemdynamics for Engineering Students For air M:+5 M2 - 2-7 Mf-1 (6. 49a). suffices): (6. 44) 300 Aerodynamics for Engineering Students or P 1-7 @-1 - - for air P2 6 (6. 44a) 6. 4.4 Density jump across the normal shock Using the previous results, substituting for p2/p1