Next:
multiphase
Previous:
Oblique
Chapter 12 2Dgd (continue)
12.3 PrandtlMeyer Function
12.3.1 Introduction
MULTILINE START acquiringCaption=0 differenceBraces=1 tmpCaptionString=The definition of the angle for the Prandtl–Meyer function.
Fig. 12.21 The definition of the angle for the Prandtl–Meyer function.
As discussed in Section 12.2 when the deflection turns to the opposite direction of the flow,
the flow accelerates to match the boundary condition.
The transition, as opposed to the oblique shock, is smooth, without any jump in properties.
Here because of the tradition, the deflection angle is denoted
as a positive when it is away from the flow (see Figure 12.21).
In a somewhat a similar concept to oblique shock there exists a
``detachment'' point above which this model breaks and another model has to be implemented.
Yet, when this model breaks down, the flow becomes complicated,
flow separation occurs, and no known simple model can describe the situation.
As opposed to the oblique shock, there is no limitation for the Prandtl–Meyer function to approach zero.
Yet, for very small angles, because of imperfections of the wall
and the boundary layer, it has to be assumed to be insignificant.
Fig. 12.22 The angles of the Mach line triangle.
Supersonic expansion and isentropic compression (PrandtlMeyer function), are an extension of the Mach line concept.
The Mach line shows that a disturbance in a field of supersonic flow moves in an angle of
$\mu$, which is defined as (as shown in Figure 12.22)
\begin{align}
\mu = \sin^{1} \left( \dfrac{ 1 }{ M} \right)
\label{pm:eq:mnu1}
\end{align}
or
\begin{align}
\mu = \tan ^{1} \dfrac{1 }{ \sqrt{M^1 1}}
\label{pm:eq:mu2}
\end{align}
A Mach line results because of a small disturbance in the wall contour.
This Mach line is assumed to be a result of the positive angle.
The reason that a ``negative'' angle is not applicable is that the coalescing of the small Mach wave
which results in a shock wave.
However, no shock is created from many small positive angles.
The Mach line is the chief line in the analysis because of the
wall contour shape information propagates along this line.
Once the contour is changed, the flow direction will change to fit the wall.
This direction change results in a change of the flow properties,
and it is assumed here to be isotropic for a positive angle.
This assumption, as it turns out, is close to reality.
In this chapter, a discussion on the relationship between the flow
properties and the flow direction is presented.
12.3.2 Geometrical Explanation
Fig. 12.23 The schematic of the turning flow.
The change in the flow direction is assume to be result of the change in the tangential component.
Hence, the total Mach number increases.
Therefore, the Mach angle increase and result in a change in the direction of the flow.
The velocity component in the direction of the Mach line is assumed
to be constant to satisfy the assumption that the change is a result of the contour only.
Later, this assumption will be examined.
The typical simplifications for geometrical functions are used:
\begin{align}
\begin{array}{rl}
d\nu & \sim \sin (d\nu) ; \
\cos (d\nu) & \sim 1
\end{array}
\label{pm:eq:angle}
\end{align}
These simplifications are the core reasons why the change occurs only in the perpendicular direction ($d\nu << 1$).
The change of the velocity in the flow direction, $dx$ is
\begin{align}
dx = (U + dU) \cos\nu U = dU
\label{pm:eq:horizontalLine}
\end{align}
In the same manner, the velocity perpendicular to the flow, $dy$, is
\begin{align}
dy = (U + dU) \sin(d\nu) = U d\nu
\label{pm:eq:verticalLine}
\end{align}
The $\tan \mu$ is the ratio of $dy/dx$ (see Figure \eqref{pm:fig:turnAngle})
\begin{align}
\tan \mu = \dfrac{dx }{ dy} = \dfrac{ dU }{ U\, d\nu }
\label{pm:eq:tanMU}
\end{align}
The ratio $dU/U$ was shown to be
\begin{align}
\dfrac{dU }{ U } = \dfrac{ dM^2 }{ 2\,M^2
\left( 1 + \dfrac{k 1 }{ 2} M^2 \right) }
\label{pm:eq:adiabatic}
\end{align}
Combining equations \eqref{pm:eq:tanMU} and qref{pm:eq:adiabatic} transforms it into
\begin{align}
d\nu =  \dfrac{ \sqrt{M^2  1} dM^2 }
{ 2\,M^2 \left( 1 + \dfrac{k 1 }{ 2} M^2 \right) }
\label{pm:eq:dNu}
\end{align}
After integration of equation \eqref{pm:eq:dNu} becomes
Turnning Angle
\begin{align}
\begin{array}{c}
\nu (M) = \sqrt{\dfrac{k+1 }{ k1 } }
\tan^{1} \sqrt{ \dfrac{k1 }{ k+1} \left( M^2 1\right)} \\
+ \tan^{1} \sqrt{ \left( M^2 1\right)} + \text{constant}
\end{array}
\label{pm:eq:nu}
\end{align}
The constant can be chosen in a such a way that $\nu= 0$ at $M=1$.
Alternative Approach to Governing Equations
Fig. 12.24 The schematic of the coordinate based on the mathematical description.
In the previous section, a simplified version was derived based on geometrical arguments.
In this section, a more rigorous explanation is provided.
It must be recognized that here the cylindrical coordinates are
advantageous because the flow turns around a single point.
For this coordinate system, the mass conservation can be written as
\begin{align}
\dfrac{\partial \left( \rho\, r\, U_r \right) }{ \partial r} +
\dfrac{\partial \left( \rho \, U_\theta \right) }{ \partial \theta} = 0
\label{pm:eq:mass}
\end{align}
The momentum equations are expressed as
\begin{align}
U_r \, \dfrac{\partial U_r }{ \partial r } +
\dfrac{U_\theta }{ r}\, \dfrac{\partial U_r }{ \partial \theta } 
\dfrac{{U_\theta}^2 }{ r} =  \dfrac{ 1 }{ \rho}
\dfrac{\partial P }{ \partial r} =
 \dfrac{ c^2 }{ \rho} \, \dfrac{\partial \rho }{ \partial r}
\label{pm:eq:mom1}
\end{align}
and
\begin{align}
U_r \,\dfrac{\partial U_\theta }{ \partial r } +
\dfrac{U_\theta }{ r} \, \dfrac{\partial U_\theta }{ \partial \theta } 
\dfrac{{U_\theta} U_r }{ r} =  \dfrac{ 1 }{r\, \rho}
\dfrac{\partial P }{ \partial \theta } =
 \dfrac{ c^2 }{ r\, \rho}\, \dfrac{\partial \rho }{ \partial \theta}
\label{pm:eq:mom2}
\end{align}
If the assumption is that the flow isn't a function of the radius,
$r$, then all the derivatives with respect to the radius will vanish.
One has to remember that when $r$ enters to the function, like
the first term in the mass equation, the derivative isn't zero.
Hence, the mass equation is reduced to
\begin{align}
\rho U_r +
\dfrac{\partial \left( \rho U_\theta \right) }{ \partial \theta} = 0
\label{pm:eq:massTheta1}
\end{align}
Equation \eqref{pm:eq:massTheta1} can be rearranged as transformed into
\begin{align}
 \dfrac{1 }{ U_\theta} \left( U_r +
\dfrac{\partial U_\theta }{ \partial \theta} \right) =
\dfrac{1 }{ \rho} \dfrac{\partial \rho }{ \partial \theta}
\label{pm:eq:massTheta}
\end{align}
The momentum equations now obtain the form of
\begin{align}
\begin{array}{c}
\dfrac{U_\theta }{ r} \,\dfrac{\partial U_r }{ \partial \theta } 
\dfrac{{U_\theta}^2 }{ r} = 0 \\
U_\theta \,\left( \dfrac{\partial U_r }{ \partial \theta }
 {U_\theta} \right) =0
\end{array}
\label{pm:eq:mom1Theta}
\end{align}
\begin{align}
\begin{array}{c}
\dfrac{U_\theta }{ r} \,\dfrac{\partial U_\theta }{ \partial \theta } 
\dfrac{{U_\theta} \,U_r }{ r} =
 \dfrac{ c^2 }{ r \rho}\, \dfrac{\partial \rho }{ \partial \theta} \\
{U_\theta }\, \left( \dfrac{\partial U_\theta }{ \partial \theta } 
U_r \right) =
 \dfrac{ c^2 }{\rho} \,\dfrac{\partial \rho }{ \partial \theta}
\end{array}
\label{pm:eq:mom2Theta1}
\end{align}
Substituting the term ${1 \over \rho}{ \partial \rho \over \partial \theta}$ from equation
qref{pm:eq:massTheta} into equation \eqref{pm:eq:mom2Theta1} results in
\begin{align}
U_\theta \left( {\partial U_\theta \over \partial \theta } 
{U_r } \right) = {c^2 \over U_\theta} \left( U_r +
{\partial U_\theta \over \partial \theta} \right)
\label{pm:eq:mom2Theta}
\end{align}
or
\begin{align}
{U_\theta}^2
\left( U_r + {\partial U_\theta \over \partial \theta} \right)
= { c^2} \left( U_r +
{\partial U_\theta \over \partial \theta} \right)
\label{pm:eq:massMom}
\end{align}
And an additional rearrangement results in
\begin{align}
\left( c^2  {U_\theta}^2 \right)
\left( U_r + \dfrac{\partial U_\theta }{ \partial \theta} \right) = 0
\label{pm:eq:massMom1}
\end{align}
From equation \eqref{pm:eq:massMom1} it follows that
\begin{align}
U_\theta = c
\label{pm:eq:tangialVelocity}
\end{align}
It is remarkable that the tangential velocity at every turn is at the speed of sound!
It must be pointed out that the total velocity isn't at the speed of sound, but
only the tangential component.
In fact, based on the definition of the Mach angle, the component shown in Figure
qref{pm:fig:turnAngle} under $U_y$ is equal to the speed of sound, $M=1$.
After some additional rearrangement, equation \eqref{pm:eq:mom1Theta} becomes
\begin{align}
\dfrac{U_\theta }{ r}
\left( \dfrac{\partial U_r }{ \partial \theta}  U_\theta \right)
= 0
\label{pm:eq:dUrUtheta}
\end{align}
If $r$ isn't approaching infinity, $\infty$ and since $U_\theta \neq 0$ leads to
\begin{align}
\dfrac{\partial U_r }{ \partial \theta } =
{U_\theta}
\label{pm:eq:Uthetac}
\end{align}
In the literature, these results are associated with the
characteristic line.
This analysis can be also applied to the same equation when they are normalized by Mach number.
However, the non–dimensionalization can be applied at this stage as well.
The energy equation for any point on a stream line is
\begin{align}
h(\theta) + \dfrac{{U_\theta}^2 + {U_r}^2 }{ 2} = h_0
\label{pm:eq:energy}
\end{align}
Enthalpy in perfect gas with a constant specific heat, $k$, is
\begin{align}
h(\theta) = C_p\, T = C_p\,\dfrac{ R }{ R } \,T =
\dfrac{1 }{ (k1)}
\overbrace{ \underbrace{\dfrac{C_p }{ C_v} }_k
\,R\,T}^{c(\theta)^2 }
= \dfrac{ c^2 }{ k1}
\label{pm:eq:bernolliSound}
\end{align}
and substituting this equality, equation \eqref{pm:eq:bernolliSound}, into equation qref{pm:eq:energy}
results in
\begin{align}
\dfrac{ c^2 }{ k1} + \dfrac{{U_\theta}^2 + {U_r}^2 }{ 2} = h_0
\label{pm:eq:energyDE0}
\end{align}
Utilizing equation \eqref{pm:eq:tangialVelocity} for the speed of sound and substituting equation
qref{pm:eq:Uthetac} which is the radial velocity transforms equation \eqref{pm:eq:energyDE0} into
\begin{align}
\dfrac{ {\left(\dfrac{\partial U_r }{ \partial \theta} \right)}^2 }{ k1}
+ \dfrac{\left(\dfrac{\partial U_r }{ \partial \theta} \right)^2
+ {U_r}^2 }{ 2} = h_0
\label{pm:eq:energyDE}
\end{align}
After some rearrangement, equation \eqref{pm:eq:energyDE} becomes
\begin{align}
\dfrac{ k+1 }{ k1} \,
\left( \dfrac{\partial U_r }{ \partial \theta} \right)^2 + {U_r}^2 = 2 h_0
\label{pm:eq:energyDEa}
\end{align}
Note that $U_r$ must be positive.
The solution of the differential equation \eqref{pm:eq:energyDEa} incorporating the constant becomes
\begin{align}
U_r = \sqrt{2h_0} \sin \left( \theta \sqrt{ \dfrac{ k1 }{ k+1 } } \right)
\label{pm:eq:energySolution}
\end{align}
which satisfies equation \eqref{pm:eq:energyDEa} because $\sin^2\theta + \cos^2\theta = 1$.
The arbitrary constant in equation \eqref{pm:eq:energySolution} is chosen such that $U_r (\theta=0) =0$.
The tangential velocity obtains the form
\begin{align}
U_\theta = c = {\partial U_r \over \partial \theta} =
\sqrt{k1 \over k+1 } \sqrt{2\;h_0} \;\;\cos
\left( \theta \sqrt{k1 \over k+1} \right)
\label{pm:eq:veloctyRadious}
\end{align}
The Mach number in the turning area is
\begin{align}
M^2 = {{U_\theta}^2 + {U_r}^2 \over c^2} =
{{U_\theta}^2 + {U_r}^2 \over {U_\theta}^2 } =
1 + \left( {U_r} \over U_\theta \right) ^2
\label{pm:eq:Mach}
\end{align}
Now utilizing the expression that was obtained for $U_r$ and $U_\theta$ equations
qref{pm:eq:veloctyRadious} and \eqref{pm:eq:energySolution} results for the Mach number is
\begin{align}
M^2 = 1 + {k+1 \over k1 } \tan^2
\left( \theta \sqrt{k1 \over k+1} \right)
\label{m:eq:Mtheta}
\end{align}
or the reverse function for $\theta$ is
Reversed Angle
\begin{align}
\label{pm:eq:reverseTheta}
\theta = \sqrt{\dfrac{ k+1 }{ k1 } }\, \tan^{1}
\left( \sqrt{\dfrac{k1 }{ k+1} } \, \left( M^2 1 \right) \right)
\end{align}
What happens when the upstream Mach number is not 1?
That is when the initial condition for the turning angle doesn't start
with $M=1$ but is already at a different angle.
The upstream Mach number is denoted in this segment as
$M_{starting}$.
For this upstream Mach number (see Figure \eqref{pm:fig:MachLineAngle})
\begin{align}
\tan \nu = \sqrt{{M_{starting}}^2  1}
\label{pm:eq:nuInfty}
\end{align}
The deflection angle $\nu$, has to match to the definition of
the angle that is chosen here ($\theta =0$ when $M=1$), so
\begin{align}
\nu (M) = \theta(M)  \theta(M_{starting})
\label{pm:eq:nuTheta1}
\end{align}
Deflection Angle
\begin{align}
\label{pm:eq:nuTheta}
\nu (M) = \sqrt{k+1\over k1}
\tan^{1} \left( \sqrt{k1\over k+1} \sqrt{ M^2 1}\right)
 \tan^{1} \sqrt{ M^2 1}
\end{align}
These relationships are plotted in Figure 12.26.
Comparison And Limitations between the Two Approaches
The two models produce exactly the same results, but the assumptions
for the construction of these models are different.
In the geometrical model, the assumption is that the velocity change in the radial direction is zero.
In the rigorous model, it was assumed that radial velocity is only a function of $\theta$.
The statement for the construction of the geometrical model can be improved by assuming that
the frame of reference is moving radially in a constant velocity.
Regardless of the assumptions that were used in the construction of these models, the fact remains that
there is a radial velocity at $U_r(r=0)= constant$.
At this point ($r=0$) these models fail to satisfy the boundary conditions and something else happens there.
On top of the complication of the turning point, the question of boundary layer arises.
For example, how did the gas accelerate to above the speed of sound when there is no nozzle (where is the nozzle?)?
These questions are of interest in engineering but are beyond the scope of this book (at least at this stage).
Normally, the author recommends that this function be used everywhere
beyond 24 the thickness of the boundary layer based on the upstream length.
In fact, analysis of design commonly used in the industry and even
questions posted to students show that many assume that the turning point can be sharp.
At a small Mach number, $(1+\epsilon)$ the radial velocity is small $\epsilon$.
However, an increase in the Mach number can result in a very significant radial velocity.
The radial velocity is ``fed'' through the reduction of the density.
Aside from its close proximity to turning point,
mass balance is maintained by the reduction of the density.
Thus, some researchers recommend that, in many instances, the sharp
point should be replaced by a smoother transition.
12.4 The Maximum Turning Angle
Fig. 12.25 Expansion of PrandtlMeyer function when it exceeds the maximum angle.
The maximum turning angle is obtained when the starting Mach number is 1 and the end Mach number approaches infinity.
In this case, Prandtl–Meyer function becomes
Maximum Turning Angle
\begin{align}
\label{pm:eq:MaxTurning}
\nu_{\infty} = \dfrac{\pi }{ 2} \left[
\sqrt{\dfrac{k+1 }{ k 1 }}  1 \right]
\end{align}
The maximum of the deflection point and the maximum turning point are only a function of the specific heat ratios.
However, the maximum turning angle is much larger than the maximum deflection point because the process is isentropic.
What happens when the deflection angel exceeds the maximum angle? The flow in this case behaves as if there is almost
a maximum angle and in that region beyond the flow will became vortex street see Figure 12.25
12.5 The Working Equations for the PrandtlMeyer Function
The change in the deflection angle is calculated by
\begin{align}
\nu_2  \nu_1 = \nu(M_2)  \nu(M_1)
\label{pm:eq:omega}
\end{align}
Fig. 12.26 The angle as a function of the Mach number and spesfic heat.
12.6 d'Alembert's Paradox
Fig. 12.27 A simplified diamond shape to illustrate the supersonic d'Alembert's Paradox.
In ideal inviscid incompressible flows, the movement of body does not encounter any resistance.
This result is known as d'Alembert's Paradox, and this paradox is examined here.
Supposed that a two–dimensional diamond–shape body is stationed in a supersonic flow as shown in Figure
12.27.
Again, it is assumed that the fluid is inviscid.
The net force in flow direction, the drag, is
\begin{align}
D = 2 \left( \dfrac{w }{ 2} \, (P_2  P_4)\right) = w \, (P_2  P_4)
\label{pm:eq:dragG}
\end{align}
It can be observed that only the area that ``seems'' to be by the flow was used in expressing equation \eqref{pm:eq:dragG}.
The relation between $P_2$ and $P_4$ is such that the flow depends on the upstream Mach number, $M_1$, and the specific heat, $k$.
Regardless in the equation of the state of the gas, the pressure at zone 2, $P_2$, is larger than the pressure at zone 4, $P_4$.
Thus, there is always drag when the flow is supersonic which depends on the upstream Mach number, $M_1$, specific heat,
$k$, and the ``visible'' area of the object.
This drag is known in the literature as (shock) wave drag.
12.7 Flat Body with an Angle of Attack
Fig. 12.28 The definition of attack angle for the Prandtl–Meyer function.
Previously, the thickness of a body was shown to have a drag.
Now, a body with zero thickness but with an angle of attack will be examined.
As opposed to the thickness of the body, in addition to the drag, the body also obtains lift.
Again, the slip condition is such that the pressure in region $\bbb{5}$ and $\bbb{7}$
are the same, and additionally the direction of the velocity must be the same.
As before, the magnitude of the velocity will be different between the two regions.
12.8 Examples For Prandtl–Meyer Function
Fig. 12.29 Schematic for Example .
A flow of air with a temperature of $20^\circ C$ and a speed of
$U = 450 m/sec$ flows (see Figure 12.29).
Calculate the pressure reduction ratio, and the Mach number after the bending point.
If the air flows in an imaginary two–dimensional tunnel with width
of 0.1$[m]$ what will the width of this imaginary tunnel after the bend?
Calculate the ``fan'' angle.
Assume the specific heat ratio is $k=1.4$.
Solution
First, the initial Mach number has to be calculated (the initial
speed of sound).
\begin{align*}
c = \sqrt{k\,R\,T} = \sqrt{1.4*287*293} = 343.1 m/sec
\end{align*}
The Mach number is then
\begin{align*}
M = \dfrac{450 }{ 343.1} = 1.31
\end{align*}
this Mach number is associated with
Prandtl — Meyer 
Input: $M$ 
k = 1.4 
$M$ 
$\nu$ 
$\dfrac{P}{P_0}$ 
$\dfrac{T}{T_0}$ 
$\dfrac{\rho}{\rho_0}$ 
$\mu$ 
1.3100 
6.4449 
0.35603 
0.74448 
0.47822 
52.6434 
The ``new'' angle should be
\begin{align*}
\nu_2 = 6.4449 + 20 = 26.4449^\circ
\end{align*}
and results in
Prandtl — Meyer 
Input: $M$ 
k = 1.4 
$M$ 
$\nu$ 
$\dfrac{P}{P_0}$ 
$\dfrac{T}{T_0}$ 
$\dfrac{\rho}{\rho_0}$ 
$\mu$ 
2.0024 
26.4449 
0.12734 
0.55497 
0.22944 
63.4620 
Note that $ {P_0}_1 = {P_0}_2$
\begin{align*}
\dfrac{P_2 }{ P_1} = \dfrac{ {P_0}_1 }{ P_1} \, \dfrac{ P_2 }{ {P_0}_2 }
= \dfrac{0.12734 }{ 0.35603} = 0.35766
\end{align*}
The ``new'' width can be calculated from the mass conservation equation.
\begin{align*}
\rho_1 x_1 M_1 c_1 =
\rho_2 x_2 M_2 c_2 \Longrightarrow x_2 = x_1 \,\dfrac{\rho_1 }{ \rho_2 }
\dfrac{M_1 }{ M_2} \sqrt{\dfrac{T_1 }{ T_2} }
\end{align*}
\begin{align*}
x_2 =
0.1 \times \dfrac{0.47822 }{ 0.22944}
\times \dfrac{1.31 }{ 2.0024} \sqrt{ \dfrac{0.74448 }{ 0.55497 } }
= 0.1579 [m]
\end{align*}
Note that the compression ``fan'' stream lines are note and their function
can be obtain either by numerical method of going over small angle
increments.
The other alternative is using the exact solution.
The expansion ``fan'' angle changes in the Mach angle between the
two sides of the bend
\begin{align*}
\text{fan angle} = 63.4 + 20.0  52.6 = 30.8^\circ
\end{align*}
Reverse the example, and this time the pressure on both sides are
given and the angle has to be obtained.
'>
Fig. 12.30 Schematic for Example .
Gas with $k=1.67$ flows over bend (see Figure 12.17).
The gas flow with Mach 1.4 and Pressure 1.2[$Bar$].
It is given that the pressure after the turning is 1[$Bar$].
Compute the Mach number after the bend, and the bend angle.
Solution
The Mach number is determined by satisfying the condition that the
pressure downstream and the Mach are given.
The relative pressure downstream can be calculated by
the relationship
\begin{align*}
\dfrac{P_2 }{ {P_0}_2} = \dfrac{P_2 }{ P_1} \,\dfrac{P_1 }{ {P_0}_1}
= \dfrac{ 1 }{ 1.2}\times{ 0.31424 } = 0.2619
\end{align*}
Prandtl — Meyer 
Input: $M$ 
k = 1.67 
$M$ 
$\nu$ 
$\dfrac{P}{P_0}$ 
$\dfrac{T}{T_0}$ 
$\dfrac{\rho}{\rho_0}$ 
$\mu$ 
1.4000 
7.7720 
0.28418 
0.60365 
0.47077 
54.4623 
With this pressure ratio $\bar{P} = 0.2619$ require either locking in
the table or using the enclosed program.
Prandtl — Meyer 
Input: $M$ 
k = 1.67 
$M$ 
$\nu$ 
$\dfrac{P}{P_0}$ 
$\dfrac{T}{T_0}$ 
$\dfrac{\rho}{\rho_0}$ 
$\mu$ 
1.4576 
9.1719 
0.26190 
0.58419 
0.44831 
55.5479 
For the rest of the calculation the initial condition is used.
The Mach number after the bend is $M= 1.4576$.
It should be noted that specific heat isn't $k=1.4$ but $k=1.67$. The bend angle is
\begin{align*}
\Delta\nu = 9.1719  7.7720 \sim 1.4^\circ
\end{align*}
\begin{align*}
\Delta\mu = 55.5479  54.4623 = 1.0^\circ
\end{align*}
12.9 Combination of the Oblique Shock and Isentropic Expansion
Consider two–dimensional flat thin plate at an angle of attack of
$4^\circ$ and a Mach number of 3.3.
Assume that the specific heat ratio at stage is $k=1.3$, calculate
the drag coefficient and lift coefficient.
Solution
For $M=3.3$, the following table can be obtained:
Prandtl — Meyer 
Input: $M$ 
k = 1.4 
$M$ 
$\nu$ 
$\dfrac{P}{P_0}$ 
$\dfrac{T}{T_0}$ 
$\dfrac{\rho}{\rho_0}$ 
$\mu$ 
3.300 
62.3113 
0.01506 
0.37972 
0.03965 
73.1416 
With the angle of attack the region 3 will be at $\nu \sim 62.31+4$
for which the following table can be obtained (PottoGDC)
Prandtl — Meyer 
Input: $\nu$ 
k = 1.4 
$M$ 
$\nu$ 
$\dfrac{P}{P_0}$ 
$\dfrac{T}{T_0}$ 
$\dfrac{\rho}{\rho_0}$ 
$\mu$ 
3.4996 
66.3100 
0.01090 
0.35248 
0.03093 
74.0528 
On the other side, the oblique shock (assuming weak shock) results in
Oblique Shock 
Input: $M_x$ 
k = 1.4 
$M_x$ 
${{M_y}_s}$ 
${{M_y}_w}$ 
$\theta_{s}$ 
$\theta_{w}$ 
$\delta$ 
$\dfrac{{P_0}_y}{{P_0}_x}$ 
3.300 
0.43534 
3.1115 
88.9313 
20.3467 
4.0000 
0.99676 
and the additional information, by clicking on the minimal button, provides
Oblique Shock 
Input: $M_x$ 
k = 1.4 
$M_x$ 
${{M_y}_w}$ 
$\theta_{w}$ 
$\delta$ 
$\dfrac{{P}_y}{{P}_x}$ 
$\dfrac{{T}_y}{{T}_x}$ 
$\dfrac{{P_0}_y}{{P_0}_x}$ 
3.300 
3.1115 
20.3467 
4.0000 
1.1157 
1.1066 
0.99676 
The pressure ratio at point 3 is
\begin{align*}
\dfrac{P_3 }{ P_1} = \dfrac{P_3 }{ P_{03} } \, \dfrac{P_{03} }{ P_{01}} \, \dfrac{P_{01} }{ P_{1}}
= 0.0109 \times 1 \times \dfrac{1 }{ 0.01506} \sim 0.7238
\end{align*}
The pressure ratio at point 4 is
\begin{align*}
{P_3 \over P_1} = 1.1157
\end{align*}
\begin{align*}
d_L = {2 \over k P_1 {M_1}^2 }(P_4  P_3) \cos \alpha =
{2 \over k {M_1}^2 } \left( {P_4 \over P_1}  {P_3 \over P_1} \right) \,\cos \alpha
\end{align*}
\begin{align*}
d_L = { 2 \over 1.3 3.3^2 } \left( 1.1157  0.7238 \right) \cos 4^\circ \sim .054
\end{align*}
\begin{multline*}
d_d = \dfrac{2 }{ k\, {M_1}^2 } \left( \dfrac{P_4 }{ P_1}  \dfrac{P_3 }{ P_1} \right) \sin \alpha =
\dfrac{ 2 }{ 1.3\, 3.3^2 } \left( 1.1157  0.7238 \right) \sin 4^\circ
\sim .0039
\end{multline*}
This shows that on the expense of a small drag, a large lift can be obtained.
Discussion on the optimum design is left for the next versions.
Fig. 12.31 Schematic of the nozzle and Prandtl–Meyer expansion.
To understand the flow after a nozzle consider a flow
in a nozzle shown in Figure 12.31.
The flow is choked and additionally the flow pressure
reaches the nozzle exit above the surrounding pressure.
Assume that there is an isentropic expansion (Prandtl–Meyer
expansion) after the nozzle with
slip lines in which there is a theoretical angle of expansion
to match the surroundings pressure with the exit.
The ratio of exit area to throat area ratio is 1.4.
The stagnation pressure is 1000 [kPa].
The surroundings pressure is 100[kPa].
Assume that the specific heat, $k=1.3$.
Estimate the Mach number after the expansion.
Solution
The Mach number a the nozzle exit can be calculated using
PottoGDC which provides
Insentropic Flow 
Input: $\dfrac{A}{A^{\star}}$ 
k = 1.3 
$M$ 
$\dfrac{T}{T_0}$ 
$\dfrac{\rho}{\rho_0}$ 
$\dfrac{A}{A^{\star}}$ 
$\dfrac{P}{P_0}$ 
$\dfrac{{A\,P}_y}{{A^{\star}\,P_0}_x}$ 
$\dfrac{F}{F_0}$ 
1.7285 
0.69052 
0.29102 
1.4000 
0.20096 
0.28134 
0.59745 
Thus, the exit Mach number is 1.7285 and the pressure at the exit is
\begin{align*}
P_{exit} = P_0 \dfrac{P_{\text{exit}}}{P_0} = 1000 \times 0.20096 = 200.96 [kPa]
\end{align*}
This pressure is higher than the surroundings pressure and an expansion must occur.
This pressure ratio is associated with a expansion angle that PottoGDC provide as
Oblique Shock 
Input: $M_x$ 
k = 1.3 
$M_x$ 
${{M_y}_w}$ 
$\theta_{w}$ 
$\delta$ 
$\dfrac{{P}_y}{{P}_x}$ 
$\dfrac{{T}_y}{{T}_x}$ 
$\dfrac{{P_0}_y}{{P_0}_x}$ 
3.300 
3.1115 
20.3467 
4.0000 
1.1157 
1.1066 
0.99676 
The final pressure ratio ultimately has to be
\begin{align*}
\dfrac{ P_{\text{surroundings}}} {P_{0}} =
\dfrac{100}{1000} = .1
\end{align*}
Hence the information for this pressure ratio can be provided by
PottoGDC as
Oblique Shock 
Input: $M_x$ 
k = 1.3 
$M_x$ 
${{M_y}_w}$ 
$\theta_{w}$ 
$\delta$ 
$\dfrac{{P}_y}{{P}_x}$ 
$\dfrac{{T}_y}{{T}_x}$ 
$\dfrac{{P_0}_y}{{P_0}_x}$ 
3.300 
3.1115 
20.3467 
4.0000 
1.1157 
1.1066 
0.99676 
The change of the angle is
\begin{align*}
\Delta \text{angle} = 30.6147  20.0641 = 10.5506
\end{align*}
Thus the angle, $\beta$ is
\begin{align*}
\beta = 90  10.5506 \sim 79.45
\end{align*}
The pressure at this point is as the surroundings.
However, the stagnation pressure is the same as originally was enter the nozzle!
This stagnation pressure has to go through serious of oblique shocks and Prandtl–Meyer expansion
to match the surroundings stagnation pressure.
