AA210A Fundamentals of Compressible Flow. Chapter 13 - Unsteady Waves in Compressible Flow

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1 AA210A Fundamentals of Compressible Flow Chapter 13 - Unsteady Waves in Compressible Flow

2 The Shock Tube - Wave Diagram

3

4 13.1 Equations for irrotational, homentropic, unsteady flow ρ t + x k ρ U i t ( ρu k ) = 0 + ρ x i P = ρ P 0 ρ 0 γ U k U k 2 + P = 0 x i

5 13.2 The acoustic equations Consider the case where fluctuations in flow variables are very small compared to the mean. There is no mean velocity and the velocity disturbance is small. Linearize the equations

6 Define the condensation The equations of motion in terms of the condensation are.

7 Replace the pressure with the density. where

8 Differentiate both equations. Subtract one from the other The condensation satisfies the linear wave equation.

9 Consider the velocity disturbance. Again, subtract one from the other Use the identity

10 Similarly, all other variables of the flow satisfy the linear wave equation.

11 13.3 Propagation of acoustic waves in one space dimension In one dimension the acoustic equations are The wave equation Can be solved in the general form where F and G are arbitrary functions

12 Right propagating F waves Left propagating G waves

13 The pressure disturbance generated by the wave is In differential form

14 The particle velocity induced by an acoustic disturbance can also be written in a very general form Substitute the expressions for the condensation and the velocity into the acoustic equations.

15 Let The equations for r and U become Add and subtract these relations. The result is From which we can conclude

16 This result gives us the relationship between density and particle velocity in left and right running waves. Now In a right running wave In a left running wave In differential form Recall

17 13.4 Isentropic, finite amplitude waves In a general 1-D isentropic flow Locally, the velocity disturbance can be assumed to be Integrate this result from an initial to a final state beginning at U 1 =0 Thus

18 The local acoustic speed is The wave speed at any point is

19 We can put this result on a somewhat more rigorous foundation for a finite amplitude wave as follows. The equations for 1-D isentropic unsteady flow are: Assume The derivatives of the density are

20 Substitute the derivatives of the density into the 1-D equations. Rearrange The continuity and momentum equations become identical except for the coefficient of the last term. To have a solution the coefficients must be equal.

21 Rearrange Take the square root Substitute Result

22 13.5 Centered expansion wave Consider a piston withdrawn from a compressible fluid at rest. The speed of the fluid in region three is equal to the piston speed. Note that the piston speed is negative. In effect, this gives us the sound speed in region 3.

23 The leading characteristic propagates to the right with the speed of sound in region 4, the undisturbed gas. The tail of the disturbance moves to the right with wave speed

24 The density ratio across the wave is given by the isentropic relation. or and the pressure ratio is

25 13.6 Compression wave Suppose the piston motion is into the gas. x=c 2 t x=c 1 t The wave speed at the surface of the piston is Once the compression waves catch up the isentropic assumption is no longer valid and there is the formation of a shock wave.

26 13.7 The shock tube

27 The conditions at the contact surface are In a frame of reference moving with the shock The shock jump conditions give

28 The shock Mach number can be written in terms of the piston speed. where U p is positive. This can be written in terms of the shock pressure ratio.

29 The velocity behind the expansion is Note that Use the conditions at the contact surface. and

30 The result is the basic shock tube equation

31 The shock Mach number is determined from or

32

33

34

35 Taylor Wave This is an expansion wave that occurs behind a detonation wave. Recall from Chapter 11 the properties of a Chapman-Jouget wave can be determined from the heat addition jump conditions. Recall the ZND model in a frame of reference moving with the detonation.

36 Detonation Wave Mach number Subsonic and supersonic roots. We assume gamma does not change through the wave Pressure, temperature and density directly behind the wave come from Rayleigh theory for thermally choked heat addition.

37 Suppose a detonation is initiated in a tube filled with a fuel/oxidizer mixture. t < 0 Δh / C p T 1 (1) P 1,T 1,ρ 1 Ignition t = 0 (1) (2) t > 0 (3) c detonation U 3 = 0 U 1 = 0 (1) P 2 = P Chapman Jouget U p c 3 = a 3 t P P 3 t (3) Taylor wave (2) P 1 x (1) C-J detonation x

38 Properties of the Taylor wave connecting regions 2 and 3 (2) U 3 = 0 U 1 = 0 (3) c detonation (1) U ' 2 = c detonation +U p = a 2 U p = c detonation a 2 = M 1 a 1 a 2 U ' 1 = c detonation Velocities in a frame of reference moving with the detonation U ' 1 = ( 1+ γ )M U ' 2 1+ γ M 1 ( ) = ρ 2 2 ( ) ρ 1 T 2 ( ) 2 = 1+ γ M 2 1 T 1 1+ γ ( ) 2 M ( ) a 2 = 1+ γ M 1 P 2 = 1+ γ M a 1 ( 1+ γ )M 1 P 1 1+ γ a 3 = a 2 γ 1 2 U p ρ 3 = a γ 1 3 γ 1 ρ 2 = 1 2 U p a 2 P 3 = 1 γ 1 P 2 2 U p 2 a 2 2γ γ 1 a 2 ( ) 2 γ 1 Pressure and temperature in region 2 just behind the detonation wave comes from steady flow Rayleigh line theory. Pressure and temperature in region 3 comes from unsteady expansion wave and isentropic flow theory.

39 Example - mixture of 5% hydrogen and 5%fluorine in 90% nitrogen. Behind the Taylor wave expansion. U p = 700 m / sec a 2 = 1050 m / sec γ = 1.4 a 3 = a 2 γ 1 2 U p = = 910 m / sec ρ 3 = a γ 1 3 γ 1 ρ 2 = 1 2 U p γ = a 2 2 P 3 = 1 γ 1 P 2 2 U p γ = a 2 2γ a = = T 3 T 2 = P 3 P 2 ρ 2 ρ 1 = 0.751

40 Properties of the reflected shock Incident shock c s > 0 P 4 = f γ 1,γ 4, a 1, P 2 P 1 a 4 P 1 +x U 2 = U p ( 2) c s ( 1) U = 0 M s = c s a 1 = M 1 M 1 = M s = γ γ 1 1/2 P 2 + γ 1 1 P 1 γ /2 M 2 U p = a γ M 1 a 2 a 1 = 1/2 γ 1 M 2 1 γ γ γ M 1 M 1 2 1/2 P 4 P 1 P 2 P 1 M 1 U p, a 2 a 1 c rs < 0 Reflected shock +x U 2 = U p ( 2) c rs ( 5) U = 0 M rs = c rs +U p a 2 = M 2 M 2 U p = a γ M 2 Take the positive root M 2 2 U p γ 1 +1 a 2 2 M 2 1 = 0 U p, a 2 a 1 M 2 a 5 a 2, P 5 P 2 a 5 a 2 = 1/2 γ 1 M 2 2 γ γ γ M 2 2γ 1 P 5 γ = 1 1 M P 2 γ 1 +1 γ 1 1 M 2 2 1/2

41 Relected shock with a high Mach number incident shock Take the limit Take the positive root M 2 2 U p γ 1 +1 a 2 2 M 2 1 = 0 M γ 1 +1 γ M 1 2 M 2 2 γ 1 M 2 1 γ 1/ γ γ M 1 lim M 1 M 2 2 γ 1 γ ( ) 1/2 γ M 1 2 1/2 M 1 = 0 1/2 2 M 2 1 = 0 U p, a 2 a 1 M 2 a 5 a 2, P 5 P 2 M 2 2 M 2 = γ 1 γ ( ) 1/2 γ γ 1 a 5 a 2 = 2 γ ( ) 1/2 γ 1 1 2γ M 2 = 1 γ 1 1 1/2 2 M 1 = 0 1/2 2 1/2 γ 1 M 2 2 γ γ γ M 2 2γ 1 P 5 γ = 1 1 M P 2 γ 1 +1 γ 1 1 ( ) 2 + 8( γ 1 )( γ 1 1) ( )( γ 1 1) + 1 γ /2 2 2 γ 1 M 2 2 1/2 1/2

42 Problem 13.5

43 From a previous final exam

44

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47 13.8 Problems

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