QUEEN'S UNIVERSITY MECH 448 Chapter FiveChapter Five

Mech 448QUEEN'S UNIVERSITYFaculty of Engineering and Applied ScienceDepartment of Mechanical EngineeringMECH 448Chapter FiveCOMPRESSIBLE FLUID FLOWNORMAL SHOCK WAVES2011Mech 448SHOCK WAVES: It has been found experimentally that, undersome circumstances, it is possible for an almost spontaneouschange to occur in a flow, the velocity decreasing and the pressureincreasing through this region of sharp change. The possibilitythat such a change can occur actually follows from the analysisgiven below. It has been found experimentally, and it also followsfrom the analysis given below, that such regions of sharp changecan only occur if the initial flow is supersonic. The extremely thinregion in which the transition from the supersonic velocity,relatively low pressure state to the state that involves a relativelylow velocity and high pressure is termed a shock wave.Mech 448A photograph of a normal shock wave is shown in thefollowing figure:Mech 448The changes that occur through a normal shock wave, i.e.,a shock wave which is straight with the flow at right angles to thewave, is shown in the following figure:Mech 448A shock wave is extremely thin, the shock wave normallyonly being a few mean free paths thick. A shock-wave isanalogous in many ways to a “hydraulic-jump” that occurs infree-surface liquid flows, a hydraulic jump being shownschematically below. A hydraulic jump occurs, for example, inthe flow downstream of a weir.1

Mech 448In the case of a normal shock wave, the velocities bothahead (i.e. upstream) of the shock and after (i.e., downstream) ofthe shock are at right angles to the shock wave. In the case of anoblique shock wave there is a change in flow direction across theshock. This is illustrated in the following figure:Mech 448Because of their own importance and because, as will beshown later, the oblique shock relations can be deduced fromthose for a normal shock wave, the normal shock wave will befirst be considered in the present chapter. Oblique shockwaves will then be discussed in the next chapter. Curved shockwaves are relatively difficult to analyze and they will not bediscussed in detail in the present course.Normal shock waves occur, for example, in the intakesto the engines in some supersonic aircraft, in the exhaustsystem of reciprocating engines, in long distance gas pipe-linesand in mine shafts as a result of the use of explosives.Mech 448A complete shock wave may be effectively normal inpart of the flow, curved in other parts of the flow andeffectively oblique in other parts of the flow as shown in thefollowing figure:Mech 448When a normal shock wave occurs, for example, in asteady flow through duct, it can be stationary with respect tothe coordinate system which is fixed relative to the walls ofthe duct. Such a shock wave is called a stationary shockwave since it is not moving relative to the coordinate systemused. On the other hand, when a sudden disturbance occursin a flow, such as, for example, the sudden closing of a valvein a pipe-line or an explosive release of energy at a point in aduct, a normal shock wave can be generated which ismoving relative to the duct walls. This is illustrated in thefollowing figure.Mech 448Mech 448To illustrate how a shock wave can form, consider the generationof a sound wave as discussed earlier. It was assumed that therewas a long duct containing a gas at rest with a piston at one endof this duct that was initially at rest. Then, at time 0, the pistonwas given a small velocity into the duct giving rise to a weakpressure pulse, i.e., a sound wave, that propagated down the duct(see following figure).2

Mech 448Mech 448If dV is the velocity given to the piston, which is, of course, thesame as the velocity of the gas behind the wave, then the increasein pressure and temperature behind the wave are equal to ρ a dVand [( γ - 1 ) T dV / a] respectively. Since ρ, a, and T are allpositive, this shows that the pressure and temperature bothincrease across the wave. It was also shown that the velocity atwhich the wave moves down the duct is equal to γ RT , which isby definition the speed of sound. Therefore, since the temperatureincreases across the wave, the speed of sound behind the wave willbe a da , where da is positive. Now consider what happens ifsome time after the piston is given velocity dV into the duct, itsvelocity is suddenly again increased to 2 dV. As a result of thesecond increase in piston speed, a second weak pressure wave willbe generated that follows the first wave down the duct as shownin the above figure.Generation of a Normal Shock WaveMech 448Mech 448This second wave will be moving relative to the gas aheadof it at the speed of sound in the gas through which it ispropagating. But the gas ahead of the second wave has velocitydV. Hence, the second wave moves relative to the duct at avelocity of a da dV. But, the first wave is moving at a velocityof a relative to the duct. Therefore, since both da and dV arepositive, the second wave is moving faster than the first waveand, if the duct is long enough, the second wave will overtake thefirst wave. But the second wave cannot pass through the firstwave. Instead, the two waves merge into a single stronger wave.If, therefore, the piston is given a whole series of step increases invelocity, a series of weak pressure waves will be generated whichwill all eventually overtake each other and merge into a singlestrong wave if the duct is long enough, i.e., a moving normalshock wave will be generated.Mech 448STATIONARY NORMAL SHOCK WAVES: Attention will firstbe given to the changes that occur though a stationary normalshock wave. In order to analyze the flow though a stationarynormal shock wave, consider a control volume of the formindicated in the following figure:The analysis of stationary normal shock waves will first beconsidered and then the application of this analysis to movingnormal shock waves will then be discussed.Mech 448This control volume has a cross sectional area of Anormal to the flow direction. The shock wave relations areobtained by applying the laws of conservation of mass,momentum, and energy to this control volume. Conservation ofmass gives:m ρ1V1 A ρ 2V2 Ai.e.:ρ1V1 ρ 2V23

Mech 448Mech 448Next consider conservation of momentum. Since the onlyforces acting on the control volume in the flow direction are thepressure forces, conservation of momentum applied to thecontrol volume gives:p1 p2 ρ1V1 (V2 V1 )orThese two equations can be rearranged to give:V1 V2 V12 p1 p2ρ1and:V22 V2 V1 p1 p2 ρ 2V2 (V2 V1 )p1 p2ρ2Adding these two equations together then gives:m 1 1 V22 V12 ( p1 p2 ) ρ1 ρ 2 Mech 448Lastly, consider the application of conservation of energyto the flow across the shock wave. Because one-dimensional flowis being considered there are no changes in the flow propertiesin any direction that is normal to that of the flow and, becausethe upstream and downstream faces of the control volume lieupstream and downstream of the shock wave, there are notemperature gradients normal to any face of the control volume.The flow through the control volume is, therefore, adiabatic andthe energy equation, therefore, gives:V12V2 c pT1 2 c pT2 c pT0 constant22Mech 448Applying the equation of state across the wave gives:p1ρ1T1 p2ρ 2T2The above four equations obtained by applyingconservation of mass, conservation of momentum, conservationof energy and the equation of state can be combined to give thefollowing:The stagnation temperature therefore does not change acrossthe shock.Mech 448Mech 448 γ 1 p2 1 ρ 2 γ 1 p1 ρ1 γ 1 p2 γ 1 p1 γ 1 p2 1 V1 γ 1 p1 V2 γ 1 p2 γ 1 p1 γ 1 T2 γ 1 T1 γ 1 γ 1 The above three equations give the density, velocity andtemperature ratios, ρ 2 / ρ1 , V2 / V1 , and T2 / T1 , across a normalshock wave in terms of the pressure ratio, p2 / p1 , across theshock wave. The pressure ratio, p2 / p1 , is often termed thestrength of the shock wave. This set of equations is often termedthe Rankine-Hugoniot normal shock wave relations.p2 p1 p1 p2 4

Mech 448Mech 448While the application of conservation of mass, momentumand energy principles shows that a shock wave can exist, it doesnot indicate whether the shock can be either compressive (i.e., p2 /p1 1 ) or expansive (i.e., p2 / p1 1 ). To examine this, the secondlaw of thermodynamics must be used. Now the entropy changeacross the shock wave is given by: Ts2 s1 c p ln 2 T1 p2 R ln p1 Using the relations for T2 / T1 and p2 / p1 given above then gives:Mech 448 γ p2 γ 1 1 (γ1)(γ1) p γ 1 s2 s1p1 ln 2 R p1 (γ 1) (γ 1) p2 p1 The variations of ( s2 - s1 ) / R for various values of γ (γ is alwaysgreater than 1) as given by the above equation are shown in thefollowing figure:Mech 448Now, the second law of thermodynamics requires that for anadiabatic process the entropy must remain unchanged or mustincrease, i.e., it requires that:s2 s1 0RUsing the above equation for the entropy change, or the figuregiven above, it is found that this will only occur if:p2 0p1Mech 448It therefore follows that the shock wave must always becompressive, i.e., that p2 / p1 must be greater than 1, i.e., thepressure must always increase across the shock wave. Using theequations for the changes across a normal shock then shows thatthat the density always increases, the velocity always decreases andthe temperature always increases across a shock wave.Mech 448The entropy increase across the shock is, basically, theresult of the fact that, because the shock wave is very thin, thegradients of velocity and temperature in the shock are very high.As a result, the effects of viscosity and heat conduction areimportant within the shock leading to the entropy increaseacross the shock wave.Because the flow across a shock is adiabatic, thestagnation temperature does not change across a shock wave.However, because of the entropy increase across a shock, thestagnation pressure always decreases across a shock wave.5

Mech 448Mech 448NORMAL SHOCK WAVE RELATIONS IN TERMS OFMACH NUMBER: While the relations derived in the previoussection for the changes across a normal shock in terms of thepressure ratio across the shock, i.e., in terms of the shockstrength, are the most useful form of the normal shock waverelations for some purposes, it is often more convenient to havethese relations in terms of the upstream Mach number M1 . Toobtain these forms of the normal shock wave relations, it isconvenient to start again with a control volume across the shockwave such as that shown in the following figure and to againapply conservation of mass, momentum and energy to thiscontrol volume but in this case to rearrange the resultingrelations in terms of Mach number.Mech 448Mech 448In writing the conservation laws, no generality is lost bytaking the area of the control volume parallel to the wave asunity. Conservation of mass then gives:ρ1V1 ρ 2V2Dividing this equation by a1 then gives:ρ1V1V a ρ2 2 2a1a2 a1which can be rewritten in terms of Mach numbers as:ρ 2 M 1 a1 ρ1 M 2 a2Mech 448Mech 448Next consider conservation of momentum applied to thecontrol volume shown earlier. This gives:p1 p2 ρ 2V22 ρ1V12Hence, since:a2 γpρi.e.,p a2 ρthe above equation gives:ρ 2 1 γ M 12 a2 ρ1 1 γ M 22 a1 γLastly, consider the application of the conservation ofenergy principle to the control volume used above. This gives: 2 2 2 22V12 a1 V2 a2 γ 1 γ 1 2Dividing this equation by 2a1 / ( γ 1) gives on rearrangement: a2 2 (γ 1) M 12 2 a1 2 (γ 1) M 2 6