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Analysis of Oblique Shocks P M V Subbarao Associate Professor Mechanical Engineering Department I I T Delhi A Mild, Efficient and Compact Compressor ….

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Presentation on theme: "Analysis of Oblique Shocks P M V Subbarao Associate Professor Mechanical Engineering Department I I T Delhi A Mild, Efficient and Compact Compressor …."— Presentation transcript:

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2 Analysis of Oblique Shocks P M V Subbarao Associate Professor Mechanical Engineering Department I I T Delhi A Mild, Efficient and Compact Compressor ….

3 Non Conical Inlets at Super Sonic Speeds Low Mach Number>x High Mach Number>x

4 High Angle Objects Bluff Bodies at supersonic Speeds Sleek Bodies at supersonic Speeds

5 Mach Waves, Revisited A ‘’point-mass’’ object moving with Supersonic velocity Generates an infinitesimally weak “mach wave”. The direction of flow remains unchanged across Mach wave.

6 Oblique Shock Wave When generating object is larger than a “point”, shockwave is stronger than mach wave …. Oblique shock wave  -- shock angle  -- turning or “wedge angle”

7 Oblique Shock Wave Geometry Shock is A CV & Must satisfy i) continuity ii) momentum iii) energy TangentialNormal Aheadw x, Mt x u x, Mn x Of Shock Behind w y, Mt y u y, Mn y Shock w x & M tx u x & M nx V x & M x V y & M y w y & M ty u y & M ny V y & M y

8 Continuity Equation For Steady Flow  V   ds      C. S.   0  x u x A  y u y A  x u x  y u y wxwx uxux VxVx wywy uyuy VyVy xy

9 Momentum Equation For Steady Flow w/o Body Forces Tangential Component   x u x w x A  y u y w y A  0  x u x  y u y w x  w y Tangential velocity is Constant across oblique Shock wave But from continuity

10 Normal Component Tangential velocity is Constant across oblique Shock wave  x u x 2 A  y u y 2 A  p y  p x  A  p x  x u x 2  p y  y u y 2

11 Energy Equation thus … Write Velocity in terms of components V x 2  u x 2  w x 2  V x 2  u y 2  w y 2  w x  w y h x  u x 2 2  h x  u x 2 2 h x  V x 2 2  h y  V y 2 2

12 Collected Oblique Shock Equations Continuity w x  w y p x  x u x 2  p y  y u y 2  x u x  y u y c p T x  u x 2 2  c p T y  u y 2 2    u x u y w x w y Momentum Energy

13 w x & M tx u x & M nx V x & M x V y & M y w y & M ty u y & M ny V y & M y Defining: M n x =M x sin(  M t x =M x cos(  Then by similarity we can write the solution

14 Similarity Solution Letting Mn x = M x sin(  

15 Properties across an Oblique Shock wave ~ f(M x,  )

16 Total Mach Number Downstream of Oblique Shock Tangential velocity is Constant across oblique Shock wave

17 Tangential velocity is Constant across oblique Shock wave

18 Tangential velocity is Constant across oblique Shock wave Or … More simply.. If we consider geometric arguments

19 Oblique Shock Wave Angle Properties across Oblique Shock wave ~ f(M 1,  )  is the geometric angle that “forces” the flow How do we relate  to  

20 Since (from continuity)

21 Oblique Shock Wave Angle (cont’d) from Momentum

22 Oblique Shock Wave Angle (cont’d) Solving for the ratio u 2 /u 1 Implicit relationship for shock angle in terms of Free stream mach number and “wedge angle”

23 Solve explicitly for tan(  ) tan       tan    1  M 1 sin       2  2  1  M 1 sin       2      1  M 1 sin       2  tan 2   2  1  M 1 sin       2 

24 Oblique Shock Wave Angle (cont’d) Simplify Numerator

25 Simplify Denominator

26 Oblique Shock Wave Angle (cont’d) Collect terms “Wedge Angle” Given explicitly as function of shock angle and freestream Mach number Two Solutions “weak” and “strong” shock wave. In reality weak shock typically occurs; strong only occurs under very Specialized circumstances.e.g near stagnation point for a detached Shock.

27 Oblique Shock Wave Angle M 1 =1.5M 1 =2.0 M 1 =2.5 M 1 =3.0 “strong shock”“weak shock” M 1 =4.0 M 1 =5.0  max curve

28 Limiting Cases of Oblique Shock Wave

29 Maximum Turning Angle  max

30 High Angle Objects  max  max

31 Weak And Strong Oblique Shock

32 Supersoinc Intakes

33 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle As derived “Wedge Angle” Given explicitly as function of shock angle and freestream Mach number For most practical applications, the geometric deflection angle (wedge angle) and Mach number are prescribed.. Need  in terms of  and M 1 Obvious Approach …. Numerical Solution using Newton’s method

34 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (cont’d) Newton method

35 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (cont’d) Newton method (continued) Iterate until convergence

36 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (cont’d) “Flat spot” Causes potential Convergence Problems with Newton Method Increasing Mach

37 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (cont’d) Newton method … Convergence can often be slow (because of low derivative slope) Converged solution

38 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (concluded) Newton method … or can “toggle” to strong shock solution Strong shock solution

39 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (improved solution) Because of the slow convergence of Newton’s method for this implicit function… explicit solution … (if possible).. Or better behaved.. Method very desirable Substitute

40 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (improved solution) (cont’d) But, since

41 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (improved solution) (cont’d) Simplify and collect terms

42 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (improved solution) (cont’d) Again, Since

43 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (improved solution) (cont’d) Regroup and collect terms

44 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (improved solution) (cont’d) Finally Regrouping in terms of powers of tan(  )

45 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (improved solution) (cont’d) Letting Result is a cubic equation of the form Polynomial has 3 real roots i) weak shock ii) strong shock iii) meaningless solution (  < 0)

46 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (improved solution) (cont’d) Numerical Solution of Cubic (Newton’s method)

47 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (improved solution) (cont’d) Collecting terms

48 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (improved solution) (cont’d) Solution Algorithm (iterate to convergence) Where again

49 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (improved solution) (cont’d) Properties of Solver algorithm are much improved Improved Algorithm Original Algorithm Original algorithm Improved algorithm

50 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (improved solution) (cont’d) Three Solutions always returned depending on start condition Original Algorithm Improved Algorithm Weak Shock Solution

51 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (improved solution) (cont’d) Three Solutions always returned depending on start condition Improved Algorithm Strong Shock Solution

52 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (improved solution) (cont’d) Three Solutions always returned depending on start condition Improved Algorithm Meaningless Solution

53 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (explicit solution) Improved Algorithm Meaningless Solution Explicit Solution … Using guidance from numerical algorithm, can we find Explicit (non -iterative) solution for shock angle? Cubic equation has three explicit solutions i) weak shock ii) Strong shock iii) non-physical solution

54 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (explicit solution) Improved Algorithm Explicit Solution … Using guidance from numerical algorithm, can we find Explicit (non -iterative) solution for shock angle? Root 1: tan[  ]= Root 2: tan[  ]= Root 3: tan[  ]= Break solutions down into manageable form

55 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (explicit solution) Improved Algorithm Meaningless Solution Explicit Solution (From Anderson, pp. 142,143) …  = 0 ---> Strong Shock  = 1 ---> Weak Shock

56 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (explicit solution) Improved Algorithm Meaningless Solution Explicit Solution Check … let {M=5, ,  =40  = =

57 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (explicit solution) Improved Algorithm Meaningless Solution Explicit Solution Check … let {M=5, ,  =40  = =

58 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (explicit solution) Improved Algorithm Meaningless Solution  = 1 ---> Weak Shock = =  Check! Explicit Solution Check … let {M=5, ,  =40 

59 Solving for Oblique Shock Wave Angle in Terms of Wedge Angle (explicit solution) Meaningless Solution  = 0 ---> Strong Shock = =  Check! … OK.. This works and is Clearly the Best method (concluded) Explicit Solution Check … let {M=5, ,  =40 

60 Oblique Shock Waves: Collected Algorithm Properties across Oblique Shock wave ~ f(M 1,  ) is the geometric angle that “forces” the flow

61 Oblique Shock Waves: Collected Algorithm (cont’d) Can be re-written as third order polynomial in tan(  ) “Very Easy” numerical solution Cubic equation has three solutions i) weak shock ii) Strong shock iii) non-physical solution

62 Oblique Shock Waves: Collected Algorithm (cont’d) “Less Obvious” explicit solution  = 0 ---> Strong Shock  = 1 ---> Weak Shock

63 Oblique Shock Waves: Collected Algorithm (cont’d)... and the rest of the story …

64 Oblique Shock Waves: Collected Algorithm (concluded)... and the rest of the story …

65 Example: M 1 = 3.0, p 1 =1atm, T1=288  K,  =20  =1.4, Compute shock wave angle (weak) Compute P 0 2, T 0 2, p 2, T 2, M 2 … Behind Shockwave

66 Example: (cont’d) M 1 = 3.0, p 1 =1atm,  =1.4, T1=288  K,  =20  Explicit Solver for  = =

67 Example: (cont’d) M 1 = 3.0, p 1 =1atm,  =1.4, T1=288  K,  =20   = 1 (weak shock) = 

68 Example: (cont’d) M 1 = 3.0, p 1 =1atm,  =1.4, T1=288  K,  =20   Normal Component of Free stream mach Number =1.837 Normal Shock Solver p 2 = 3.771(1 atm) = atm Compute Pressure ratio across shock Flow is compressed

69 Example: (cont’d) M 1 = 3.0, p 1 =1atm,  =1.4, T1=288  K,  =20   Temperature ratio Across Shock Normal Shock Solver T 2 = (288  K) =  K

70 Example: (cont’d) M 1 = 3.0, p 1 =1atm,  =1.4, T1=288  K,  =20   Stagnation Pressure ratio across shock Normal Shock Solver =1.837

71 Example: (cont’d) M 1 = 3.0, p 1 =1atm,  =1.4, T1=288  K,  =20   Stagnation Pressure ratio (alternate method) =0.7961

72 Example: (cont’d) M 1 = 3.0, p 1 =1atm,  =1.4, T1=288  K,  =20   Stagnation Pressure =29.24 atm

73 Example: (cont’d) M 1 = 3.0, p 1 =1atm,  =1.4, T1=288  K,  =20   Stagnation Temperature behind shock =806.4 o K

74 What Happens when M 1 = 3.0, p 1 =1atm,  =1.4, T1=288  K,  =  Explicit Solver for  =8.0 =1.0

75 What Happens when (cont’d) M 1 = 3.0, p 1 =1atm,  =1.4, T1=288  K,  =   =19.47  “mach line”

76 What Happens when (cont’d) M 1 = 3.0, p 1 =1atm,  =1.4, T1=288  K,  =   Normal Component of Free stream mach Number = 1.0 (NO COMPRESSION!)

77 Expansion Waves So if  >0.. Compression around corner  =0 … no compression across shock

78 Expansion Waves (concluded) Then it follows that  <0.. We get an expansion wave Next Prandtl-Meyer Expansion waves


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