Modern Compressible Flow Chapter 4 Notes

Chapter 4 is titled "Oblique Shock and Expansion Waves"
==Introduction==
Anderson provides a general explanation of plane oblique shock compression and plane oblique centered expansion: 
(see figure 4.4)
===Compression Wave===
M1, p1, T1, pt1, Tt1 ahead of the shock turn. θ is the turn angle and β is the shock angle.
Relationships that we might expect are:
*M2 < M1
*p2 > p1
*T2 > T1
*pt2 < pt1
*Tt2 = Tt1
*s2 > s1
Also, there is a limit to θ, for θ above this limit, the shock can no longer be attached to the corner.
Real viscous fluids have boundary layers which modify the shock attachment process, compared to inviscid theory. But for thin boundary layers, inviscid theory holds pretty well.
===Plane Centered Expansion (Prandtl-Meyer)===
M1, p1, T1, pt1, Tt1 ahead of the expansion turn. θ is the turn angle of expansion.
Relationships that we might expect are:
*M2 > M1
*p2 < p1
*T2 < T1
*pt2 = pt1
*Tt2 = Tt1
*s2 = s1
==Source of Oblique Waves==
===Mach Wave===
A Mach wave is an infinitesimal, or very weak, disturbance that is propagating like a sound wave in compressible flow (essentially an acoustic disturbance).
The angle of a Mach wave, μ, is calculated by the Mach wave geometry as: 
Stronger disturbances coalesce into a shock wave which is at angle β, greater that μ.
The Mach wave is a wave of small disturbance, hence, either compression or expansion disturbances of infinitesimal magnitude propagate along the Mach wave.
The book introduces the concept of the "moving beeper" to describe the propagation of a Mach wave. See previous description in Chapter 3 notes.
Anderson presents figure 4.5 that shows a process of disturbances impacting a perforated plate from below, creating periodic disturbances above the plate.
==Oblique Shock Relations==
The key to understanding oblique shock waves is to understand the geometry SEE NOTES TO FINISH THIS PART
Anderson has a conservation of momentum approach for demonstrating that the tangential component of the velocity is preserved across the shock.
:V_{1t} = V_{2t}
or,
:w_1 = w_2
The normal momentum equation is given as:
:(-\rho_1 u_1)u_1 + (\rho_2 u_2)u_2 = -(-p_1+p_2)
or;
:p_1 + \rho_1 u_1^2=p_2 + \rho_2 u_2^2
The energy equation reduces to:
:h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}
However,
:V^2 = u^2 + w^2
Therefore,
:h_1 + \frac{u_1^2}{2} = h_2 + \frac{u_2^2}{2}
The previous equations are identical in form to the equations for normal shock waves.
Therefore, the changes in the flow properties across the oblique shock are controlled by the changes in the normal component only.
All of the equations for normal shock waves apply, where u is now given by u1 = V1sin(β) and M1n = Msin(β). Therefore, for gas with a constant γ, we have:
:\frac{\rho_2}{\rho_1} = \frac{(\gamma +1)M_{n1}^2}{(\gamma -1)M_{n1}^2 +2}
Equations (4.8) - (4.11) give the oblique shock equations for property ratios across the oblique shock. Note that these equations are in terms of the shock angle, β, not the deflection angle, θ.
The downstream Mach number is given by:
:M_2 = \frac{M_{n2}}{\sin (\beta - \theta)}
How do we find θ from knowing M1 and β?
:\tan \beta = \frac{u_1}{w_1}
:\tan (\beta - \theta) = \frac{u_2}{w_2}
Since w1 = w2, we can write
:\frac{\tan(\beta - \theta)}{\tan \beta} = \frac{u_2}{u_1} = \frac{\rho_1}{\rho_2} = \frac{(\gamma -1)M_{n1}^2 +2}{(\gamma +1)M_{n1}^2}
:\frac{\tan(\beta - \theta)}{\tan \beta} = \frac{(\gamma -1)M_1^2 \sin^2\beta +2}{(\gamma +1)M_1^2 \sin^2\beta}
Using trigonometric identities, this equation reduces to: 
This equation is called the θ-β-M Relation, and gives θ as a unique function of M1, γ, and β. β versus θ has been plotted in figure 4.8, p. 137 (and inside the front cover of the text) for different Mach numbers at γ = 1.4.
Anderson notes the following:
For any given M1, there is a maximum deflection angle, θmax. If the physical θ > θmax, then the shock wave detaches and will be curved, as shown in figure 4.9. This note applies to both 2-dimensional wedges and corner flow.
For any given θ < θmax, there are two β values that are predicted by the θ-β-M Relation for a given M1, as sketched in figure 4.10. The larger β value is called the strong shock solution, and the smaller β value is called the weak shock solution. In nature, the weak shock solution usually occurs because the entropy gain is less for the weak solution than for the strong solution. The strong shock can be forced to occur by some means of increasing the back pressure downstream of the shock. In the strong shock solution, M2 < 1, i.e. the flow is subsonic. In the weak shock solution M2 > 1, i.e. the flow is supersonic.
For a fixed θ, as the value of M1 decreases from high to low, β increases.
===The β-θ-M Relation: An Alternative Form for the θ-β-M Relation===
==Supersonic Flow Over Wedges==
==Shock Polar==
(Stuff I missed) 
Note that as M approaches infinity, M* = 2.45, so that in the M* hodograph...
Note that circle with radius M* = 1 defines the "sonic circle." Inside this circle, all velocities are subsonic; outside, velocities are supersonic.
(see Figure 4.16 on p. 151)
For a given deflection angle, θ, the shock polar is cut at two points, one supersonic (weak solution) and one supersonic (strong solution).
The line OC tangent to the shock polar in the hodograph plan defines θmax for a given M1* (or M1).
For θ > θmax there are no oblique shock solutions for M1* (or M1).
For points E and A, there is no deflection. Point E represents the normal shock solution. Point A corresponds to a Mach line.
If a straight line OH is drawn perpendicular to the line through AB, then the angle HOA is the wave angle, β.
In figure 4.17 (p. 152), Anderson shows several Mach-hodographs. He notes that M1* = 2.45 (M1 = ∞) is a circle.
There is an analytical equation for the Mach-hodograph curve as a shock polar:
:\left(\frac{V_y}{a^*}\right)^2 = \frac{\left(M_1^*-\frac{V_x}{a*}\right)^2 \left [ \left(\frac{V_x}{a*}\right)M_1^* -1 \right ]}{\frac{2}{\gamma+1}\left(M_1^*\right)^2 - \left(\frac{V_x}{a*}\right)M_1^* +1}
==Regular Reflection from a Solic Boundary==
Consider a supersonic flow between two parallel walls (see Figure 4.18 on p. 152). At point A, a compression corner initiates a plane oblique shock wave that traverses the flow and impinges on the upper wall at point B. From point B a second oblique shock wave emanates. The flow is divided into 3 distinct regions.
The solution to determine the properties of these regions is to recognize that the flow in region 2 must turn parallel to the compression surface. At point B, the flow in region 2 must be traveling at an angle θ1 with respect to the plane wall at point B. Another shock wave is created at point B when the flow in region 2 is turned back by an angle θ2 (= θ1 in this geometry).
The strength of the reflected shock that emanates from point B depends on M2 and θ2. θ2 is the total turning angle at point B. In the present case, θ2 = θ1. In general the reflected shock angle β2 is not equal to β1, in fact, it is greater than β1; however, for very small deflections (Mach wave type compressions), β2 will essentially be the same as β1.
Please review example problems 4.10 and 4.11.
==Mass Flow per Unit Area==
A very important parameter in high-speed flow is the mass flow per unit area,
:\dot m' \equiv \frac{\dot m}{A} = \frac{\rho u A}{A} = \rho u
Let us look at how ρu changes through the three regions of problem 4.10.
ρ 	7.643e-2	0.1566	0.2836 	lbm/ft3
a 	 	 	ft/s 	
M 	 	 	 	
ρu	 	 	lbm/ft2/s	
(some stuff)
Since mdot is constant, this means that the inviscid cross-sectional area must decrease along the flow path. However, because real flows are viscous, with boundary layers, the actual cross-sectional area along the flow path will be varied to achieve the purposes of the design: 
(some other stuff)
==Comment on Flow through Multiple Shock Systems==
In example 4.12 (p. 155, Anderson), one case uses two oblique shocks (one incident and one reflected) to change the Mach number from an upstream value, M1, of 2.8 to a downstream Mach number, M3, of 1.45. The second case achieves the same Mach number change (2.8 to 1.45) by using a stronger (greater single θ turn) single oblique shock.
The multiple shock system caused much less total pressure loss than the single oblique shock. 
2 shock system:
:\Delta p_t = 27.14 atm - 22.01 atm = 5.13 atm
Single shock system:
:\Delta p_t = 27.14 atm - 17.67 atm = 9.47 atm
This is a general result: a supersonic flow can be decelerated and compressed much more efficiently by a system of multiple oblique shocks than by a single stronger shock. High-speed aircraft inlets and supersonic wind tunnels use this principle to slow down and pressurize supersonic flow.
==Pressure-Deflection Diagrams==
==Intersection of Shocks of Opposite Families==
Consider the intersection of left-running and right-running oblique shocks as shown in Figure 4.23 (p. 160, Anderson).
(referring to the figure)
A is a left-running wave and B is a right-running wave. A results from a compression corner that changes the flow direction by θ2. B turns the flow by θ3. A and B continue as refracted shocks C and D downstream of their intersection E. θ2 > θ3 for our case, and the streamline passing through A (and C) experiences a greater change in entropy than that passing through B (and D). So, the entropy in region 4' is greater than that in region 4; thus the entropy is discontinuous acroos the "slip line."
The following hold across a slip line:
#Static pressure across a slip line is equal (p4 = p4').
#Velocities across a slip line must be in the same direction, but in general have different magnitudes.
#Static temperature and density vary across a slip line.
Conditions 1 and 2, along with θ2 and θ3, the shock interaction can be determined.
==Intersection of Shocks of the Same Family==
Consider two left-running oblique shock waves generated by two compression corners, as described in Figure 4.26 (p. 163, Anderson).
(referring to the figure)
Shock waves AC and BC will intersect. Above the point C (intersection point), a single shock wave CD will propagate. The entropy across the single shock, CD, will be greater than that for the flow passing through AC and BC; thus a slip line must exist downstream of the intersection, C.
If no other shocks existed in this system,
*p5 = p3
*θ5 = θ3
In general, it is not possible for a single shock, CD, to meet these requirements.
(some stuff)
The purpose of this reflected wave is to process the flow so that:
*p5 = p4
*θ5 = θ4
An iterative approach is required to solve this flow field numerically.
==Mach Reflection==
Once again, lets' consider a shock wave reflection from a solid wall, as shown in Figure 4.28 (p. 164, Anderson).
(referring to the figure)
For this flow, the flow in regions 2 and 3 must be turned by θ so that the flow in regions 1 and 3 are in the same direction. The reflected shock must have a strength to meet this condition.
In our previous analysis we assumed that θ is less than θmax for M2, allowing a straight, attached reflected shock wave, but what happens when (θmax for M2) < θ < (θmax for M1)?
For this case:
*The incident shock is an allowable straight shock.
*For the incident shock (θ > θmax), a regular reflection is not possible.
*A normal shock forms at the upper wall to allow streamlines to continue parallel to the upper wall.
(some stuff)
==Detached Shock Wave in front of a Blunt Body==
==Three-Dimensional Shock Waves==
==Prandtl-Meyer Expansion Waves==
When a supersonic flow is turned away from itself, an expansion wave is created. Expansion waves are isentropic, the entropy of the supersonic flow remains constant through the expansion.
A sketch of a centered-expansion or corner-expansion, also called a Prandtl-Meyer expansion, is shown in figure 4.32.
Property changes across the expansion corner:
#M2 > M1
#p2 < p1
#T2 < T1
#pt2 = pt1
#Tt2 = Tt1
#s2 = s1
#The expansion is a continuous process composed of an infinite number of Mach waves in a "fan."
The theoretical solution for the Prandtl-Meyer flow is provided in section 4.8. It is dependent on the geometry of the velocity triangle shown in Figure 4.33 (p. 169, Anderson).
From the law of sines,
:\frac{V+dV}{V} = \frac{\sin \left(\pi/2 + \mu \right)}{\sin \left( \pi/2 - \mu -d\theta \right)}
But a trigonometric identity gives:
:\sin \left( \pi/2 + \mu \right) = \cos \mu
:\sin \left( \pi/2 - \mu -d\theta \right) = \cos \left(\mu + d\theta\right) = \cos \mu \cos d\theta - \sin \mu \sin d\theta
Thus, by substitution,
:1+\frac{dV}{V} = \frac{\cos \mu}{\cos \mu \cos d\theta - \sin \mu \sin d\theta}
For small dθ, sindθ ≈ dθ and cosdθ ≈ 1.
:1+\frac{dV}{V} = \frac{\cos \mu}{\cos \mu - d\theta \sin \mu} = \frac{1}{1 - d\theta \tan \mu}
Since,
:\frac{1}{1-x} = 1+x+x^2+x^3+...
Then,
:\frac{1}{1 - d\theta \tan \mu} = 1+d\theta \tan \mu + H.O.T.
or, solving for dθ and neglecting higher order terms,
:d\theta = \frac{\frac{dV}{V}}{\tan \mu}
But recall that,
:\mu = \arcsin \left(\frac{1}{M}\right)
which can be written as,
:\tan \mu = \frac{1}{\sqrt{M^2-1}}
When this is substituted into a previous equation, we get Equation 4.35 (p. 170, Anderson), 
"Equation 4.35 is the governing differential equation of Prandtl-Meyer flow."
The solution to changes in velocity when passing through a fan of expansion Mach waves is an integral solution.
:\int_{\theta_1}^{\theta_2} d\theta = \int_{M_1}^{M_2} \sqrt{M^2 - 1} \frac{dV}{V}
Note that,
:V = Ma
or,
:\ln V = \ln M + \ln a
:\frac{dV}{V} = \frac{dM}{M} + \frac{da}{a}
The adiabatic constant specific heat energy equation can be written as:
:\left( \frac{a_t}{a} \right) ^2 = \frac{T_t}{T} = \left( 1+\frac{\gamma -1}{2}M^2\right)
or,
:a = a_t\left( 1+\frac{\gamma -1}{2}M^2\right) ^{0.5}
Via differentiation and algebra,
:\frac{da}{a} = -\left({\gamma-1}{2}\right)M\left(1+\frac{\gamma -1}{2}M^2\right)^{-1}dM
Hence, substitute back,
(some stuff)
The constant of integration for Equation 4.44 (p. 171, Anderson) is not important because it always cancels out when we use ν(M) to solve Prandtl-Meyer expansions. For convenience it is defined as ν(M=1) = 0.
The integral solution for the angle-velocity (now Mach number) can now be written,
:\theta_2 = \nu(M_2) - \nu(M_1)
ν(M) has been tabulated for M>1 in Table A.5 (Anderson) and in NACA 1135.
===How to Solve for a Prandtl-Meyer Expansion===
For the known upstream Mach number, M1, look up ν(M1)
Calculate (stuff)
Look up downstream Mach number, M2, knowing ν(M2), in Table A.5
How do we calculate changes in static pressure, static temperature, velocity, and density through the Prandtl-Meyer expansion? For Prandtl-Meyer expansion the total pressure and total temperature stay constant.
:p_2 = p_{t1}\left( 1 +\frac{\gamma - 1}{2} M_2^2\right) ^{\frac{-\gamma}{\gamma -1}}
==Shock-Expansion Theory==
==Historical Note: Prandtl's early Research on Supersonic Flows and the Origin of the Prandtl-Meyer Theory==
==Summary==