Analysis of Expansion Waves P M V Subbarao Associate Professor Mechanical Engineering Department I I T Delhi Another Zero Cost Nozzle …..

Theory of Extrapolation of Physics So if  > 0.. Compression around a concave corner So if  = 0.. No Compression

Expansion Wave : Another Shock !!?!! Consider the scenario shown in the adjacent figure. As a supersonic flow turns, the normal component of the velocity increases (w 2 > w 1 ). The tangential component remains constant (v 2 = v 1 ). The corresponding change is the entropy (Δs = s 2 − s 1 ) can be expressed as follows,

As this is an expansion wave :T 2 < T 1 & p 2 < p 1

Expanding Shock is Impossible !!! 1.A Finite Expansion wave shows Δs < 0. 2.Since this is not possible it means that it is impossible to turn a flow through a single shock wave. 3.The argument may be further extended to show that such an expansion process can occur only if we consider a turn through infinite number of expansion waves in the limit. 4.Accordingly an expansion process is an isentropic process.

Pressure and Temperature Change Across Expansion Fan Because each mach wave is infinitesimal, expansion is isentropic - P 0 2 = P T 0 2 = T 0 1 p 2 p 1  P 0 1 p 1  p 2 P 0 2  1   1 2 M   1 2 M 2 2             1 T 2 T 1  T 0 1 T 1  T 2 T 0 2  1   1 2 M   1 2 M 2 2          

Then it follows that  < 0.. We get an expansion wave

Prandtl-Meyer Expansion Waves Flow accelerates around corner. Continuous flow region … sometimes called “expansion fan” consisting of a series of Mach waves. Each Mach wave is infinitesimally weak isentropic flow region. Flow stream lines are curved and smooth through fan.

Analysis of Prandtl-Meyer Expansion Consider flow expansion around an infinitesimal corner Infinitesimal Expansion Fan Flow Geometry V d   Mach Wave  d  d  V V+dV

From Law of Sines

Using the trigonometric identities &

Substitution gives Since d  is considered to be infinitesimal  V cos  cosd  sin  sind   V  dV cos    1  dV V  cos  cos   cosd    sin   sind  

and the equation reduces to Exploiting the form of the power series (expanded about x=0)

Since dV is infinitesimal … truncate after first order term 1 1  dV V  1  dV V  1 1  dV V  1 1  tan   d   Solve for d  in terms of dV/V 1  tan   d    1  dV V  d  1 tan   dV V

Using Mach Wave Relations:

Performing some algebraic and trigonometric voodoo and …. Valid for Real and ideal gas

For a finite deflection the O.D.E is integrated over the complete expansion fan Write in terms of mach by …

Substituting in For a calorically perfect adiabatic gas flow And T 0 is constant

Returning to the integral for  V  M 2  1 d V M 1 M 2   M 2  1 dM M  (  1) 2 MdM  1   1 2 M 2                 M 1 M 2 

Simplification gives

Evaluate integral by performing substitution  dM M  du, M 2  e 2u M 2  1 dM M 1   1 2 M 2         e 2u  1 1   1 2 e 2u        du Let

Standard Integral Table Form From tables (math handbook) x x

Substituting  1  1 tan  1  1  1 M 2  1            tan  1 M 2  1 (M)   1  1 tan  1  1  1 M 2  1            tan  1 M 2  1 Let

More simply (M) “Prandtl-Meyer Function” Implicit function … more Newton!

M 2 versus M 1,   M 1 = 5 M 1 = 3 M 1 = 1

Pressure and Temperature Change Across Expansion Fan Because each mach wave is infinitesimal, expansion is isentropic - P 0 2 = P T 0 2 = T 0 1

Maximum Turning Angle How much a supersonic flow can turn through. A flow has to turn so that it can satisfy the boundary conditions. In an ideal flow, there are two kinds of boundary condition that the flow has to satisfy, Velocity boundary condition, which dictates that the component of the flow velocity normal to the wall be zero. It is also known as no-penetration boundary condition. Pressure boundary condition, which states that there cannot be a discontinuity in the static pressure inside the flow.

If the flow turns enough so that it becomes parallel to the wall, we do not need to worry about this boundary condition. However, as the flow turns, its static pressure decreases. If there is not enough pressure to start with, the flow won't be able to complete the turn and will not be parallel to the wall. This shows up as the maximum angle though which a flow can turn. Lower to Mach number to start with (i.e. small M 1 ), greater the maximum angle though which the flow can turn. The streamline which separates the final flow direction and the wall is known as a slipstream. Across this line there is a jump in the temperature, density and tangential component of the velocity (normal component being zero). Beyond the slipstream the flow is stagnant (which automatically satisfies the velocity boundary condition at the wall). In case of real flow, a shear layer is observed instead of a slipstream, because of the additional no-slip boundary condition.

Maximum Turning Angle p 2  0  M 2  (  )   1  1 tan  1  1  1  2  1            tan  1  2  1   1  1  1        2  max   1  1  1        2   1  1 tan  1  1  1 M 1 2  1            tan  1 M 1 2  1

Plotting as a  max function of Mach number {T 2, p 2 } = 0

Highest Value for Maximum Turning Angles

Anatomy of Prandtl – Meyer Expansion Wave     

Combination of Shock & Expansion Wave An Important Product !!!

Supersonic Flow Over Flat Plates at Angle of Attack

Review: Oblique Shock Wave Angle

Prandtl-Meyer Expansion Waves  <0.. We get an expansion wave (Prandtl-Meyer)