Steady waves in compressible flow

Chapter Steady waves in compressible flow. Oblique shock waves Figure. shows an oblique shock wave produced when a supersonic flow is deflected by an angle. Figure.: Flow geometry near a plane oblique shock wave. We can think of the deflection as caused by a planar ramp at this angle although it could be generated by the blockage produced by a solid body placed some distance away in the flow. In general, a 3-D shock wave will be curved, and will separate two regions of non-uniform flow. However, at each point along the shock, the change in flow properties takes place in a very thin region much thinner than the radius of curvature of the shock. If we consider a small neighborhood of the point in question then within the small neighborhood, the shock may be regarded as locally planar to any required level of accuracy and the flows on either -

CHAPTER. STEADY WAVES IN COMPRESSIBLE FLOW - side can be regarded as uniform. With the proper orientation of axes the flow is locally two-dimensional. Therefore it is quite general to consider a straight oblique shock wave in a uniform parallel stream in two-dimensions as shown below. Balancing mass, two components of momentum and energy across the indicated control volume leads to the following oblique shock jump conditions. u = u P + u = P + u u v = u v (.) h + u + v = h + u + v Since u is constant, v = v and the jump conditions become u = u P + u = P + u v = v (.) h + u = h + u. When the ideal gas law P = RT is included, the system of equations (.) closes allowing all the properties of the shock to be determined. Note that, with the exception of the additional equation, v = v, the system is identical to the normal shock jump conditions. The oblique shock acts like a normal shock to the flow perpendicular to it. Therefore almost all of the normal shock relations can be converted to oblique shock relations with the substitution M! M Sin( ) M! M Sin( ). (.3) Recall the Rankine-Hugoniot relation

CHAPTER. STEADY WAVES IN COMPRESSIBLE FLOW -3 = P P + + P P (.4) P P + P P plotted in figure.. Figure.: Plot of the Hugoniot relation (.4) This shows the close relationship between the pressure rise across the wave (oblique or normal) and the associated density rise. The jump conditions for oblique shocks lead to a modified form of the very useful Prandtl relation u u =(a ) v (.5) + where (a ) = RT. From the conservation of total enthalpy, for a calorically perfect gas in steady adiabatic flow C p T t = C p T + U = a + U = + ( ) (a ). (.6) The Prandtl relation is extremely useful for easily deriving all the various normal and oblique shock relations. The oblique shock relations generated using (.3) are

CHAPTER. STEADY WAVES IN COMPRESSIBLE FLOW -4 P = M Sin ( ) ( ) P + = u = ( + ) M Sin ( ) u ( ) M Sin ( )+ T = M Sin ( ) ( ) ( ) M Sin ( )+ T ( + ) M Sin ( ) (.7) M Sin ( ) = ( ) M Sin ( )+ M Sin ( ) ( ). The stagnation pressure ratio across the shock is P t ( + ) M Sin ( ) = P t ( ) M Sin ( )+ +. (.8) M Sin ( ) ( ) Note that (.8) can also be generated by the substitution (.3)... Exceptional relations One all new relation that has no normal shock counterpart is the equation for the absolute velocity change across the shock. U U = 4 M Sin ( ) M Sin ( )+ ( + ) M 4 Sin ( ) (.9) Exceptions to the substitution rule (.3) are the relations involving the static and stagnation pressure, P t /P and P t /P across the wave. The reason for this is as follows. Consider P t P = P t P t P t P = P t + P t M. (.0) Similarly P t P = P t P t P t P = P t + P t M. (.)

CHAPTER. STEADY WAVES IN COMPRESSIBLE FLOW -5 The stagnation to static pressure ratio in each region depends on the full Mach number, not just the Mach number perpendicular to the shock wave... Flow deflection versus shock angle The most basic question connected with oblique shocks is: given the free stream Mach number, M, and flow deflection,, what is the shock angle,? The normal velocity ratio is u = ( ) M Sin + u ( + ) M Sin = u v. u v (.) From the velocity triangles in figure. Tan( )= u v Tan( ) = u v. (.3) Now ( ) M Sin ( )+ Tan( ) =Tan( ) ( + ) M. (.4) Sin ( ) An alternative form of this relation is 0 Tan( ) =Cot( ) @ + + M Sin ( ) A. (.5) M M Sin ( ) The shock-angle-deflection-angle relation (.5) is plotted in figure.3 for several values of the Mach number. Corresponding points in the supersonic flow past a circular cylinder sketched below are indicated on the M =.5 contour. At point a the flow is perpendicular to the shock wave and the properties of the flow are governed by the normal shock relations. In moving from point a to b the shock weakens and the deflection of the flow behind the shock increases until a point of maximum flow deflection is reached at b. The flow solution between a and b is referred to as the strong solution in figure.3. Notice that the Mach number behind the shock is subsonic up to point c where the Mach number just downstream of

CHAPTER. STEADY WAVES IN COMPRESSIBLE FLOW -6 Figure.3: Flow deflection versus shock angle for oblique shocks. the shock is one. Between c and d the flow corresponds to the weak solutions indicated in figure.3. If one continued along the shock to very large distances from the sphere the shock will have a more and more oblique angle eventually reaching the Mach angle! µ = Sin (/M ) corresponding to an infinitesimally small disturbance. Figure.4: Supersonic flow past a cylinder with shock structure shown. Note that as the free-stream Mach number becomes large, the shock angle becomes independent of the Mach number.

CHAPTER. STEADY WAVES IN COMPRESSIBLE FLOW -7 lim M! Cos( ) Sin( ) Tan( ) = (.6) Sin +. Weak oblique waves In this section we will develop the di erential equations that govern weak waves generated by a small disturbance. The theory will be based on infinitesimal changes in the flow and for this reason it is convenient to drop the subscript 0 0 on the flow variables upstream of the wave. The sketch below depicts the case where the flow deflection is very small d <<. Note that M is not close to one. Figure.5: Small deflection in supersonic flow. In terms of figure.3 we are looking at the behavior of weak solutions close to the horizontal axis of the figure. For a weak disturbance, the shock angle is very close to the Mach angle Sin(µ) =/M. Let and make the approximation Sin( )= M + " (.7) Using (.8) we can also develop the approximation M Sin ( ) = +M". (.8) Cot( ) = M / M 3 M ". (.9) Using (.8) and (.9) the (, ) relation (.5) can be expanded to yield

CHAPTER. STEADY WAVES IN COMPRESSIBLE FLOW -8 Tan(d ) = d = 4 + M / ". (.0) M The velocity change across the shock (??) is expanded as U U + = 4 U M M + " M M + " + ( + ) M 4 M + ". (.) Retaining only terms of order " the fractional velocity change due to the small deflection is du U + = 8 ". (.) ( + ) M Equation (.) is approximated as Write (.3) in terms of the deflection angle du U = 4 ". (.3) ( + ) M or du U = 4 ( + ) M " = 4 ( + ) M + 4 M d (.4) (M / ) du U = d (.5) (M / ) where d is measured in radians. Other small deflection relations are dp P = M d (M / ) d = M d (M / ) (.6) dt T = ( ) M (M ) / d

CHAPTER. STEADY WAVES IN COMPRESSIBLE FLOW -9 and dp t P t = 3( + ) M Sin 3 = 6 M 3 3( + ) "3 = ds R (.7) or using (.0) dp t P t = ( + ) M 6 (M ) 3/ (d )3 = ds R. (.8) Note that the entropy change across a weak oblique shock wave is extremely small; the wave is nearly isentropic. The Mach number is determined from dm M = du U dt T = (M ) / d ( ) M d. (.9) (M / ) Adding terms dm + M = M d. (.30) (M / ) Eliminate d between (.5) and (.30) to get an integrable equation relating velocity and Mach number changes. du U = + M The weak oblique shock relations (.6) are, in terms of the velocity. dm M (.3) dp P = M du U dt T = M du U (.3) d = M du U

CHAPTER. STEADY WAVES IN COMPRESSIBLE FLOW -0 These last relations are precisely the same ones we developed for one dimensional flow with area change in the absence of wall friction and heat transfer in chapter 9. From that development we had M du U = da A M dm + M M = da A. (.33) If we eliminate da/a between these two relations, the result is du U = + M which we just derived in the context of weak oblique shocks. dm M (.34).3 The Prandtl-Meyer expansion The upshot of all this is that du U = d (.35) (M / ) is actually a general relationship valid for steady, isentropic flow. In particular it can be applied to negative values of d. Consider flow over a corner. Figure.6: Supersonic flow over a corner.

CHAPTER. STEADY WAVES IN COMPRESSIBLE FLOW - Express the angle in terms of the Mach number. d = M / dm + M M (.36) Now integrate the angle between the initial and final Mach numbers. Z 0 d 0 = Z M / M M + M dm M (.37) Let! be the angle change beginning at the reference mach number M =. The integral (.37) is! (M) = + / Tan! / M / + Tan M /. (.38) This expression provides a unique relationship between the local Mach number and the angle required to accelerate the flow to that Mach number beginning at Mach one. The straight lines in figure.5 are called characteristics and represent particular values of the flow deflection. According to (.38) the Mach number is the same at every point on a given characteristic. This flow is called a Prandtl-Meyer expansion and (.38) is called the Prandtl-Meyer function, plotted below for several values of. Figure.7: Prandtl-Meyer function for several values of.

CHAPTER. STEADY WAVES IN COMPRESSIBLE FLOW - Note that for a given there is a limiting angle at M!.! max = + /! (.39) For =.4,! max =.45. The expansion angle can be greater than90. If the deflection is larger than this angle there will be a vacuum between! max and the wall..3. Example - supersonic flow over a bump Air flows past a symmetric -D bump at a Mach number of 3. The aspect ratio of the bump is a/b = p 3. Figure.8: Supersonic flow over a bump. Determine the drag coe cient of the bump assuming zero wall friction. C d = Drag force per unit span U b (.40) Solution The ramp angle is 30 producing a 5 oblique shock with pressure ratio P P =6.356. (.4) The expansion angle is 60 producing a Mach number M =.406! =9.6. (.4) The stagnation pressure is constant through the expansion wave and so the pressure ratio over the downstream face is

CHAPTER. STEADY WAVES IN COMPRESSIBLE FLOW -3 M 3 =4.68! = 69.6 (.43) and P 3 P = 0 @ + + M M 3 A! 3.5 = +0.(.406) +0.(4.68) =.395 3.5 =0.049 (.44) 4.643 and P 3 P = P 3 P P P =0.049 6.356 = 0.0945. (.45) The drag coe cient becomes C d = P b Sin(30) P 3 bsin(30) M P b = 6.356 0.0945.4 (9) =0.994. (.46).4 Problems Problem - Use the oblique shock jump conditions (.) to derive the oblique shock Prandtl relation (.5). Problem - Consider the supersonic flow past a bump discussed in the example above. Carefully sketch the flow putting in the shock waves as well as the leading and trailing characteristics of the expansion. Problem 3 - Consider a streamline in compressible flow past a -D ramp with a very small ramp angle. Figure.9: Supersonic flow past a -D ramp.

CHAPTER. STEADY WAVES IN COMPRESSIBLE FLOW -4 Determine the ratio of the heights of the streamline above the wall before and after the oblique shock in terms of M and d find the unknown coe cient in (.47). Pay careful attention to signs. H H =+(???)d (.47) Problem 4 - Consider a body in subsonic flow. As the free-stream Mach number is increased there is a critical value, M c, such that there is a point somewhere along the body where the flow speed outside the boundary layer reaches the speed of sound. Figure.0 illustrates this phenomena for flow over a projectile. Figure.0: Projectile in high subsonic flow. In this figure.0 the critical Mach number is somewhere between 0.840 and 0.885 as evidenced by the weak shocks that appear toward the back of the projectile in the middle picture. The local pressure in the neighborhood of the body is expressed in terms of the pressure coe cient. C P = P P U (.48) Show that the value of the pressure coe cient at the point where sonic speed occurs is

CHAPTER. STEADY WAVES IN COMPRESSIBLE FLOW -5 C Pc = + Mc + M c. (.49) State any assumptions needed to solve the problem. Problem 5 - Consider frictionless (no wall friction) supersonic flow over a flat plate of chord C at a small angle of attack as shown in figure.. Figure.: Supersonic flow past a flat plate at a small angle of attack. The circulation about the plate is defined as I = Uds. (.50) where the integration is along any contour surrounding the plate. ) Show that, to a good approximation, the circulation is given by = U C M / (.5) where the integration is clockwise around the plate. ) Show that, to the same approximation, Liftperunitspan = U. Problem 6 - Consider frictionless (no wall friction) flow of air at M = over a flat plate of chord C at 5 angle of attack as shown in figure.. Figure.: Supersonic flow over a flat plate at 5 angle of attack.

CHAPTER. STEADY WAVES IN COMPRESSIBLE FLOW -6 Evaluate the drag coe wave approximation. cient of the plate. Compare with the value obtained using a weak Problem 7 - Consider frictionless (no wall friction) supersonic flow of Air over a flat plate of chord C at an angle of attack of 5 degrees as shown in figure.3. Figure.3: Supersonic flow over a flat plate at 5 angle of attack. Determine the lift coe cient L C L = U C (.5) where L is the lift force per unit span. Problem 8 - Figure.4 shows a symmetrical, diamond shaped airfoil at a 5 attack in a supersonic flow of air. angle of Figure.4: Supersonic flow past a diamond shaped airfoil. Determine the lift and drag coe cients of the airfoil. C L = C D = Lift per unit span U C Drag per unit span U C (.53)

CHAPTER. STEADY WAVES IN COMPRESSIBLE FLOW -7 What happens to the flow over the airfoil if the free-stream Mach number is decreased to.5? Compare your result with the lift and drag of a thin flat plate at 5 angle of attack and free-stream Mach number of 3. Problem 9 - The figure below shows supersonic flow of Air over a 30 a 0 wedge. The free stream Mach number is 3. wedge followed by Figure.5: Supersonic over a wedge with a shoulder. ) Determine M, M 3 and the included angle of the expansion fan,. ) Suppose the flow was turned through a single 0 wedge instead of the combination shown above. Would the stagnation pressure after the turn be higher or lower than in the case shown? Why? Problem 0 - Figure.6 shows a smooth compression of a supersonic flow of air by a concave surface. The free-stream Mach number is.96. The weak oblique shock at the nose produces a Mach number of.93 at station. From station to station the flow is turned 0 degrees. Figure.6: Supersonic flow compressed by a concave surface. ) Determine the Mach number at station. ) Determine the pressure ratio P /P. 3) State any assumptions used.

CHAPTER. STEADY WAVES IN COMPRESSIBLE FLOW -8 Problem - Figure.7 shows supersonic flow of air in a channel or duct at a Mach number of three. The flow produces an oblique shock o a ramp at an angle of 6 degrees. The shock reflects o the upper surface of the wind tunnel as shown below. Beyond the ramp the channel height is the same at the height ahead of the ramp. Figure.7: Mach 3 flow in a duct with a ramp. ) Determine the Mach number in region. ) Determine the Mach number in region 3. 3) Describe qualitatively how P t and T t vary between regions, and 3. 4) Suppose the channel height is 0 cm. Precisely locate the shock reflections on the upper and lower walls. 5) Suppose the walls were lengthened. At roughly what point would the Mach number tend to one? Problem - Figure.8 shows supersonic flow of air turned through an angle of 30. The free stream Mach number is 3. Figure.8: Supersonic flow turned 30. In case (a) the turning is accomplished by a single 30 wedge whereas in case (b) the turning is accomplished by two 5 degree wedges in tandem. Determine the stagnation pressure change in each case,(p t /P t ) (a) and (P t3 /P t ) (b) and comment on the relative merit of one design over the other. Problem 3 - Figure.9 shows the flow of helium from a supersonic over-expanded round jet. If we restrict our attention to a small region near the intersection of the first two oblique shocks and the so-called Mach disc as shown in the blow-up, then we can use

CHAPTER. STEADY WAVES IN COMPRESSIBLE FLOW -9 oblique shock theory to determine the flow properties near the shock intersection (despite the generally non-uniform 3-D nature of the rest of the flow). The shock angles with respect to the horizontal measured from the image are as shown. Figure.9: Supersonic flow from an over expanded round jet. ) Determine the jet exit Mach number. Hint, you will need to select a Mach number that balances the pressures in regions and 4 with a dividing streamline that is very nearly horizontal as shown in the picture. ) Determine the Mach number in region. 3) Determine the flow angles and Mach numbers in regions 3 and 4. 4) Determine P /P and P 4 /P. How well do the static pressures match across the dividing streamline (dashed line) between regions and 4? Problem 4 - Figure.0 shows the reflection of an expansion wave from the upper wall of a -D, adiabatic, inviscid channel flow. The gas is helium at an incoming Mach number, M =.5 and the deflection angle is 0. The flow is turned to horizontal by the lower wall which is designed to follow a streamline producing no reflected wave. Determine M, M 3 and H/h. Figure.0: Supersonic flow in an expansion.