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# Chapter 12 2Dgd (continue)

## 12.3 Prandtl-Meyer 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 (Prandtl-Meyer 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:d-Nu} \end{align} After integration of equation \eqref{pm:eq:d-Nu} becomes

Turnning Angle

\begin{align} \begin{array}{c} \nu (M) = -\sqrt{\dfrac{k+1 }{ k-1 } } \tan^{-1} \sqrt{ \dfrac{k-1 }{ 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 }{ (k-1)} \overbrace{ \underbrace{\dfrac{C_p }{ C_v} }_k \,R\,T}^{c(\theta)^2 } = \dfrac{ c^2 }{ k-1} \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 }{ k-1} + \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 }{ k-1} + \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 }{ k-1} \, \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{ k-1 }{ 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{k-1 \over k+1 } \sqrt{2\;h_0} \;\;\cos \left( \theta \sqrt{k-1 \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 k-1 } \tan^2 \left( \theta \sqrt{k-1 \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 }{ k-1 } }\, \tan^{-1} \left( \sqrt{\dfrac{k-1 }{ 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 k-1} \tan^{-1} \left( \sqrt{k-1\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 2-4 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 Prandtl-Meyer 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 Prandtl-Meyer 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.

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.

# Example 12.16

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.

# Example 12.17

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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*}

# Example 12.18

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 (Potto-GDC)

 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.

# Example 12.19

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 Potto-GDC 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 Potto-GDC 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 Potto-GDC 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.