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# Chapter 12 Compressible Flow 2–Dimensional

## 12.1 Introduction

In Chapter 11 the discussed dealt with one–dimensional and semi one–dimensional flow. In this Chapter the focus is around the two diminsional effect which focus around the oblique shock and Prandtl–Meyer flow (in other word it focus around Theodor Meyer's thesis). This Chapter present a simplified summary of two chapters from the book Fundamentals of Compressible Flow'' by this author.

### 12.1.1 Preface to Oblique Shock

Fig. 12.1 A view of a straight normal shock as a limited case for oblique shock.

In Section , a discussion on a normal shock was presented. A normal shock is a special type of shock wave. In the literature oblique shock, normal shock, and Prandtl–Meyer function are presented However, one can view all these cases as three different regions of a flow over a plate with a deflection section. Clearly, variation of the deflection angle from a zero ($\delta = 0$) to a positive value results in oblique shock (see Figure 12.1). Further changing the deflection angle to a negative value results in expansion waves. The common representation is done by ignoring the boundaries of these models. However, this section attempts to show the boundaries and the limits or connections of these models. A normal shock occurs when there is a disturbance downstream which imposes a boundary condition on the flow in which the fluid/gas can react only by a sharp change in the flow direction. As it may be recalled, normal shock occurs when a wall is straight/flat ($\delta = 0$) as shown in Figure 12.1 due to disturbance. When the deflection angle is increased, the gas flow must match the boundary conditions. This matching can occur only when there is a discontinuity in the flow field. Thus, the direction of the flow is changed by a shock with an angle to the flow. This shock is commonly referred to as the oblique shock.

Fig. 12.2 The regions where oblique shock or Prandtl–Meyer function exist. Notice that both have a maximum point and a no solution'' zone, which is around zero.

Decreasing the deflection angle also requires the boundary conditions to match the geometry. Yet, for a negative deflection angle (in this section's notation), the flow must be continuous. The analysis shows that the flow velocity must increase to achieve this requirement. This velocity increase is referred to as the expansion wave. As it will be shown in the next section, as opposed to oblique shock analysis, the increase in the upstream Mach number determines the downstream Mach number and the negative'' deflection angle. It has to be pointed out that both the oblique shock and the Prandtl–Meyer function have a maximum point for $M_1 \rightarrow \infty$. However, the maximum point for the Prandtl–Meyer function is much larger than the oblique shock by a factor of more than 2. What accounts for the larger maximum point is the effective turning (less entropy production) which will be explained in the next chapter (see Figure 12.2).

### 12.1.1.1 Introduction to Zero Inclination

Fig. 12.3 A typical oblique shock schematic.

What happens when the inclination angle is zero? Which model is correct to use? Can these two conflicting models, the oblique shock and the Prandtl–Meyer function, co-exist? Or perhaps a different model better describes the physics. it was assumed that Mach wave and oblique shock co–occur in the same zone. Previously (see Section ), it was assumed that normal shock occurs at the same time. In this chapter, the stability issue will be examined in greater detail.

## Oblique Shock

The shock occurs in reality in situations where the shock has three–dimensional effects. The three–dimensional effects of the shock make it appear as a curved plane. However, one–dimensional shock can be considered a representation for a chosen arbitrary accuracy with a specific small area. In such a case, the change of the orientation makes the shock considerations two–dimensional. Alternately, using an infinite (or a two–dimensional) object produces a two–dimensional shock. The two–dimensional effects occur when the flow is affected from the side,'' i.e., the change is in the flow direction. An example of such case is creation of shock from the side by deflection shown in Figure . To match the boundary conditions, the flow turns after the shock to be parallel to the inclination angle schematicly shown in Figure 12.3. The deflection angle, $\delta$, is the direction of the flow after the shock (parallel to the wall). The normal shock analysis dictates that after the shock, the flow is always subsonic. The total flow after the oblique shock can also be supersonic, which depends on the boundary layer and the deflection angle. The velocity has two components (with respect to the shock plane/surface). Only the oblique shock's normal component undergoes the shock.'' The tangent component does not change because it does not move'' across the shock line. Hence, the mass balance reads \begin{align} \rho_1\, {U_1}_n = \rho_2\, {U_2}_n \label{2Dgd:eq:Omass} \end{align} The momentum equation reads \begin{align} P_1 + \rho_1 \,{{U_1}_n } ^{2} = P_2 + \rho_2\, {{U_2}_n } ^{2} \label{2Dgd:eq:Omomentum} \end{align} The momentum equation in the tangential direction is reduced to \begin{align} {U_1}_t = {U_2}_t \label{2Dgd:eq:OtangentU} \end{align} The energy balance in coordinates moving with shock reads \begin{align} C_p T_1 + \dfrac{{{U_1}_n } ^{2} }{ 2} = C_p T_2 + \dfrac{{{U_2}_n } ^{2} }{ 2} \label{2Dgd:eq:Oenergy} \end{align} Equations \eqref{2Dgd:eq:Omass}, qref{2Dgd:eq:Omomentum}, and qref{2Dgd:eq:Oenergy} are the same as the equations for normal shock with the exception that the total Yet, the new relationship between the upstream Mach number, the deflection angle, $\delta$, and the Mach angle, $\theta$ has to be solved. From the geometry it can be observed that \begin{align} \tan \theta = \dfrac{{U_1}_n }{ {U_1}_t} \label{2Dgd:eq:Otheta} \end{align} and \begin{align} \tan ( \theta - \delta ) = \dfrac{{U_2}_n }{ {U_2}_t} \label{2Dgd:eq:OthetaAlpha} \end{align} Unlike in the normal shock, here there are three possible pairs of solutions to these equations. The first is referred to as the weak shock; the second is the strong shock; and the third is an impossible solution (thermodynamically). Experiments and experience have shown that the common solution is the weak shock, in which the shock turns to a lesser extent. \begin{align} \dfrac{\tan \theta }{ \tan ( \theta - \delta ) } = {{U_1}_n \over {U_2}_n } \label{2Dgd:eq:OthetaR} \end{align} The above velocity–geometry equations can also be expressed in term of Mach number, as \begin{align} \sin \theta = \dfrac{{M_1}_n }{ {M_1}} \label{2Dgd:eq:OM1n} \end{align} and in the downstream side reads \begin{align} \sin (\theta - \delta ) = \dfrac{{M_2}_n }{ {M_2}} \label{2Dgd:eq:OM2n} \end{align} Equation \eqref{2Dgd:eq:OM1n} alternatively also can be expressed as \begin{align} \cos \theta = \dfrac{{M_1}_t }{ {M_1}} \label{2Dgd:eq:OM1t} \end{align} And equation \eqref{2Dgd:eq:OM2n} alternatively also can be expressed as \begin{align} \cos\, \left(\theta - \delta \right) = \dfrac{{M_2}_t }{ {M_2}} \label{2Dgd:eq:OM2t} \end{align} The total energy across a stationary oblique shock wave is constant, and it follows that the total speed of sound is constant across the (oblique) shock. It should be noted that although, ${U_1}_t = {U_2}_t$ the Mach number is ${M_1}_t \neq {M_2}_t$ because the temperatures on both sides of the shock are different, $T_1 \neq T_2$. As opposed to the normal shock, here angles (the second dimension) have to be determined. The solution from this set of four equations, \eqref{2Dgd:eq:OM1n} through qref{2Dgd:eq:OM2t}, is a function of four unknowns of $M_1$, $M_2$, $\theta$, and $\delta$. Rearranging this set utilizing geometrical identities such as $\sin\alpha = 2\sin\alpha\cos\alpha$ results in

Angle Relationship

\begin{align} \label {2Dgd:eq:Osol} \tan \delta = 2\, \cot \theta\, \left[\dfrac{{M_1}^{2}\, \sin^2 \theta - 1 }{ {M_1}^{2} \, \left(k + \cos\, 2 \theta \right) +2 }\right] \end{align}

The relationship between the properties can be determined by substituting $M_1 \sin \theta$ for of $M_1$ into the normal shock relationship, which results in

Pressure Ratio

\begin{align} \label{2Dgd:eq:OPbar} \dfrac{P_2 }{ P_1} = \dfrac{2\,k\, {M_1 }^{2} \sin^2 \theta - (k-1) }{ k + 1} \end{align}

The density and normal velocity ratio can be determined by the following equation

Density Ratio

\begin{align} \label{2Dgd:eq:OrhoBar} \dfrac{\rho_2 }{\rho_1} = \dfrac{{U_1}_n }{ {U_2}_n} = \dfrac{ (k+1) {M_1}^{2} \sin^2\theta} {(k-1) {M_1}^2 \sin^2\theta + 2} \end{align}

The temperature ratio is expressed as

Temperature Ratio

\begin{align} \label{2Dgd:eq:OTbar} \dfrac{T_2 }{ T_1} = \dfrac{{2\,k\, {M_1}^2 \sin^2\theta - (k-1) \left[(k-1) {M_1}^2 + 2 \right] } } {{(k+1)^2 \,{M_1}} } \end{align}

Prandtl's relation for oblique shock is \begin{align} U_{n_1}U_{n_2} = c^{2} - \dfrac{k -1 }{ k+1} \, {U_t}^2 \label{2Dgd:eq:Oprandtl} \end{align} The Rankine–Hugoniot relations are the same as the relationship for the normal shock \begin{align} \dfrac{P_2 - P_1 }{ \rho_2 - \rho_1} = k \,\dfrac{ P_2 - P_1 }{ \rho_2 - \rho_1} \label{2Dgd:eq:ORankineHugoniot} \end{align}

### 12.2.1 Solution of Mach Angle

Oblique shock, if orientated to a coordinate perpendicular and parallel shock plane is like a normal shock. Thus, the relationship between the properties can be determined by using the normal components or by utilizing the normal shock table developed earlier. One has to be careful to use the normal components of the Mach numbers. The stagnation temperature contains the total velocity. Again, the normal shock is a one–dimensional problem, thus, only one parameter is required (to solve the problem). Oblique shock is a two–dimensional problem and two properties must be provided so a solution can be found. Probably, the most useful properties are upstream Mach number, $M_1$ and the deflection angle, which create a somewhat complicated mathematical procedure, and this will be discussed later. Other combinations of properties provide a relatively simple mathematical treatment, and the solutions of selected pairs and selected relationships will be presented.

### 12.2.1.1 Upstream Mach Number, $M_1$, and Deflection Angle, $\delta$

Again, this set of parameters is, perhaps, the most common and natural to examine. Thompson (1950) has shown that the relationship of the shock

Governing Angle Equation

\begin{align} \label {2Dgd:eq:Ocubic} x^3 + a_1 \, x^2 + a_2 \,x + a_3=0 \end{align}

where \begin{align} x = \sin^2 \theta \label{2Dgd:eq:Ox} \end{align} and \begin{align} a_1 & = - \dfrac{{M_1}^2 + 2 }{ {M_1}^2} - k\, \sin ^2 \delta \label{2Dgd:eq:Oa1} \\ a_2 & = - \dfrac{ 2{M_1}^2 + 1 }{ {M_1}^4 } + \left[ \dfrac{(k+1)^2 }{ 4}+ \dfrac{k -1 }{ {M_1}^2} \right] \sin ^2 \delta \label{2Dgd:eq:Oa2} \\ a_3 & = - \dfrac{\cos ^2 \delta }{ {M_1}^4} \label{2Dgd:eq:Oa3} \end{align} Equation \eqref{2Dgd:eq:Ocubic} requires that $x$ has to be a real and positive number to obtain a real deflection angle. Clearly, $\sin\theta$ must be positive, and the negative sign refers to the mirror image of the solution. Thus, the negative root of $\sin\theta$ must be disregarded The solution of a cubic equation such as \eqref{2Dgd:eq:Ocubic} provides three roots. These roots can be expressed as

First Root

\begin{align} \label{2Dgd:eq:Ox1} x_1 = - \dfrac{1 }{ 3} a_1 + (S +T ) \end{align}

Second Root

\begin{align} \label{2Dgd:eq:Ox2} x_2 = - \dfrac{a_1 }{ 3}- \dfrac{(S +T )}{2} + \dfrac{i\, \sqrt{3} \, ( S-T) }{2} \end{align}

and

Third Root

\begin{align} \label{2Dgd:eq:Ox3} x_3 = - \dfrac{a_1}{ 3} - \dfrac{ (S +T )}{2} - \dfrac{i \,\sqrt{3}\, ( S - T ) }{2} \end{align}

Where \begin{align} S = \sqrt[3]{R + \sqrt{D}}, \label{2Dgd:eq:OS} \end{align} \begin{align} T = \sqrt[3]{R - \sqrt{D}} \label{2Dgd:eq:OT} \end{align} and where the definition of the $D$ is \begin{align} D = Q^3 + R^2 \label{2Dgd:eq:OD} \end{align} and where the definitions of $Q$ and $R$ are \begin{align} Q = \dfrac{ 3 a_2 - {a_1 } ^2 }{ 9} \label{2Dgd:eq:OQ} \end{align} and \begin{align} R = \dfrac{ 9 a_1 a_2 - 27 a_3 - 2 {a_1}^3 }{ 54} \label{2Dgd:eq:OR} \end{align} Only three roots can exist for the Mach angle, $\theta$. From a mathematical point of view, if $D>0$, one root is real and two roots are complex. For the case $D=0$, all the roots are real and at least two are identical. In the last case where $D<0$, all the roots are real and unequal. The physical meaning of the above analysis demonstrates that in the range where $D > 0$ no solution can exist because no imaginary solution can exist. $D>0$ occurs when no shock angle can be found, so that the shock normal component is reduced to subsonic and yet parallel to the inclination angle. Furthermore, only in some cases when $D=0$ does the solution have a physical meaning. Hence, the solution in the case of $D=0$ has to be examined in the light of other issues to determine the validity of the solution. When $D< 0$, the three unique roots are reduced to two roots at least for the steady state because thermodynamics dictates that. Physically, it can be shown that the first solution \eqref{2Dgd:eq:Ox1}, referred sometimes as a thermodynamically unstable root, which is also related to a decrease in entropy, is unrealistic.'' Therefore, the first solution does not occur in reality, at least, in steady–state situations. This root has only a mathematical meaning for steady–state analysis.

Fig. 12.4 Flow around spherically blunted $30^\circ$ cone-cylinder with Mach number 2.0. It can be noticed that the

These two roots represent two different situations. First, for the second root, the shock wave keeps the flow almost all the time as a supersonic flow and it is referred to as the weak solution (there is a small section that the Second, the third root always turns the flow into subsonic It should be noted that this case is where entropy increases in the largest amount. In summary, if an imaginary hand moves the shock angle starting from the deflection angle and reaching the first angle that satisfies the boundary condition, this situation is unstable and the shock angle will jump to the second angle (root). If an additional push'' is given, for example, by additional boundary conditions, the shock angle will jump to the third root. These two angles of the strong and weak shock are stable for a two–dimensional wedge (see the appendix of this chapter for a limited discussion on the stability).

### 12.2.2.1 Large deflection angle for given, $M_1$

The first range is when the deflection angle reaches above the maximum point. For a given upstream Mach number, $M_1$, a change in the inclination angle requires a larger energy to change the flow direction. Once, the inclination angle reaches the maximum potential energy,'' a change in the flow direction is no longer possible. As the alternative view, the fluid sees'' the disturbance (in this case, the wedge) in front of it and hence the normal shock occurs. Only when the fluid is away from the object (smaller angle) liquid sees'' the object in a different inclination angle. This different inclination angle is sometimes referred to as an imaginary angle.

#### The Simple Calculation Procedure

For example, in Figure 12.4 and ef{2Dgd:fig:flowView}, the imaginary angle is shown. The flow is far away from the object and does not see' the object. For example, for, $M_1 \longrightarrow \infty$ the maximum deflection angle is calculated when $D = Q^3 + R^2 = 0$. This can be done by evaluating the terms $a_1$, $a_2$, and $a_3$ for $M_1 = \infty$. \begin{align*} a_1 & = -1 - k \sin^2\delta \\ a_2 & = \dfrac{\left( k + 1 \right)^ 2 \sin ^2 \delta }{ 4 } \\ a_3 & = 0 \end{align*} With these values the coefficients $R$ and $Q$ are \begin{align*} R = \dfrac{ - 9 ( 1 + k \sin^2\delta ) \left(\dfrac{\left( k + 1 \right)^ 2 \, \sin ^2 \delta }{ 4 } \right) - (2) (-) ( 1 + k\, \sin^2\delta )^2 }{ 54} \end{align*} and \begin{align*} Q = \dfrac{ ( 1 + k\, \sin^2\delta )^2 }{ 9 } \end{align*}

Fig. 12.5 The view of a large inclination angle from different

Solving equation \eqref{2Dgd:eq:OD} after substituting these values of $Q$ and $R$ provides series of roots from which only one root is possible. This root, in the case $k =1.4$, is just above $\delta_{max} \sim {\pi \over 4}$ (note that the maximum is also a function of the heat ratio, $k$). While the above procedure provides the general solution for the three roots, there is simplified transformation that provides solution for the strong and and weak solution. It must be noted that in doing this transformation, the first solution is lost'' supposedly because it is negative.'' In reality the first solution is not negative but rather some value between zero and the weak angle. Several researchers suggested that instead Thompson's equation should be expressed by equation qref{2Dgd:eq:Ocubic} by $\tan\theta$ and is transformed into \begin{align} \left( 1 + \dfrac{k-1 }{ 2} {M_1}^{2} \right) \tan \delta \tan^3\theta - \left({M_1}^{2} - 1 \right) \tan^2\theta + \left( 1 + {k+1 \over 2} \right) \tan \delta \tan\theta +1 = 0 \label{2Dgd:eq:Oemanuel} \end{align} The solution to this equation \eqref{2Dgd:eq:Oemanuel} for the weak angle is

Weak Angle Solution

\begin{align} \label{2Dgd:eq:OweakThompson} \theta_{weak} = \tan^{-1} \left( \dfrac{ {{M_1}^2 -1 + 2 \,f_1(M_1,\delta) \, \cos \left( \dfrac{4\,\pi + \cos^{-1}(f_2(M_1,\delta)) }{ 3} \right)} }{ { 3 \,\left( 1 + \dfrac{k-1 }{ 2} \, {M_1}^{2} \right)\,\tan \delta} } \right) \end{align}

Strong Angle Solution

\begin{align} \label{2Dgd:eq:OstrongThompson} \theta_{strong} = \tan^{-1} \, \dfrac{ {{M_1}^2 -1 + 2 f_1(M_1,\delta) \cos \left(\dfrac{ \cos^{-1}(f_2(M_1,\delta)) }{ 3 } \right)} } { { 3 \,\left( 1 + \dfrac{k-1 }{ 2} \,{M_1}^{2} \right)\,\tan \delta}} \end{align}

where these additional functions are \begin{align} f_1(M_1,\delta) = \sqrt{{\left({M_1}^2 -1 \right)^2 -3 \, \left( 1 + \dfrac{k-1}{ 2} {M_1}^{2}\right) \left( 1+ \dfrac{k+1 }{ 2} {M_1}^{2} \right) \tan^2\delta } } \label{2Dgd:eq:Ofmtheta1} \end{align} and \begin{align} f_2(M_1,\delta) = \dfrac{{ \left({M_1}^2 -1 \right)^3 - 9 \left( 1 + \dfrac{k-1 }{ 2} {M_1}^{2}\right) \left( 1 + \dfrac{k-1 }{ 2} {M_1}^{2} + \dfrac{k+1 }{ 4} {M_1}^{4} \right) \tan^2\delta } } {{ f_1(M_1,\delta)^3} } \label{2Dgd:eq:Ofmtheta1a} \end{align} Figure \eqref{oblique:fig:MyforMxDelta} exhibits typical results for oblique shock for two deflection angle of 5 and 25 degree. Generally, the strong shock is reduced as the increase of the Mach number while the weak shock is increase. The impossible shock for unsteady state is almost linear function of the upstream Mach number and almost not affected by the deflection angle.

Fig. 12.6 The three different Mach numbers after the oblique

#### The Procedure for Calculating The Maximum Deflection Point

The maximum angle is obtained when $D=0$. When the right terms defined in \eqref{2Dgd:eq:Oa1}-qref{2Dgd:eq:Oa2}, qref{2Dgd:eq:OQ}, and qref{2Dgd:eq:OR} are substituted into this equation and utilizing the trigonometrical identity $\sin^2\delta + \cos^2\delta = 1$ and other trigonometrical identities results in Maximum Deflection Mach Number's equation in which is \begin{align} {M_1}^2 \,\left( k + 1 \right)\, ({M_{1n}}^2+1) = 2\,(k\,{M_{1n}}^4 +2\,{M_{1n}}^2 - 1) \label{2Dgd:eq:Omenikoff} \end{align} This equation and its twin equation can be obtained by an alternative procedure proposed by someone who suggested another way to approach this issue. It can be noticed that in equation \eqref{2Dgd:eq:Osol}, the deflection angle is a function of the Mach angle and the upstream Mach number, $M_1$. Thus, one can conclude that the maximum Mach angle is only a function of the upstream Much number, $M_1$. This can be shown mathematically by the argument that differentiating equation \eqref{2Dgd:eq:Osol} and equating the results to zero creates relationship between the Mach number, $M_1$ and the maximum Mach angle, $\theta$. Since in that equation there appears only the heat ratio $k$, and Mach number, $M_1$, $\theta_{max}$ is a function of only these parameters. The differentiation of the equation \eqref{2Dgd:eq:Osol} yields \begin{align} \dfrac{d \tan \delta }{ d\theta} = {k {M_1}^4 \sin^4\theta + \left(2 - \dfrac{(k+1)}{ 2}{M_1}^2 \right) {M_1}^2\sin^2\theta -\left({ 1 + \dfrac{(k+1)}{ 2}{M_1}^2 } \right) \over k\, {M_1}^4 \sin^4\theta - \left[(k-1)+ \dfrac{(k+1)^2 {M_1}^2 }{ 4 } \right] {M_1}^2\sin^2\theta -1 } \label{2Dgd:eq:OsolD} \end{align} Because $\tan$ is a monotonous function, the maximum appears when $\theta$ has its maximum. The numerator of equation \eqref{2Dgd:eq:OsolD} is zero at different values of the denominator. Thus, it is sufficient to equate the numerator to zero to obtain the maximum. The nominator produces a quadratic equation for $\sin^2\theta$ and only the positive value for $\sin^2\theta$ is applied here. Thus, the $\sin^2\theta$ is \begin{align} \sin ^2 \theta_{max} = \dfrac{ -1 + i\dfrac{ k + 1 }{ 4}{M_1}^2+ \sqrt{(k+1) \left[ 1 + \dfrac{k-1 }{ 2} {M_1}^2 + \left( \dfrac{k+1 }{ 2} {M_1} \right)^4 \right]} } { k \, {M_1}^2} \label{2Dgd:eq:OthetaMax} \end{align} Equation \eqref{2Dgd:eq:OthetaMax} should be referred to as the maximum's equation. It should be noted that both the Maximum Mach Deflection equation and the maximum's equation lead to the same conclusion that the maximum $M_{1n}$ is only a function of upstream the Mach number and the heat ratio $k$. It can be noticed that the Maximum Deflection Mach Number's equation is also a quadratic equation for ${M_{1n}}^2$. Once $M_{1n}$ is found, then the Mach angle can be easily calculated by equation \eqref{2Dgd:eq:OM1n}. To compare these two equations the simple case of Maximum for an infinite Mach number is examined. It must be pointed out that similar procedures can also be proposed (even though it does not appear in the literature). Instead, taking the derivative with respect to $\theta$, a derivative can be taken with respect to $M_1$. Thus, \begin{align} \dfrac{d \tan \delta }{ dM_1} = 0 \label{2Dgd:eq:OmaxMa} \end{align} and then solving equation \eqref{2Dgd:eq:OmaxMa} provides a solution for $M_{max}$. A simplified case of the Maximum Deflection Mach Number's equation for large Mach number becomes \begin{align} {M_{1n}} = \sqrt{\dfrac{ k+1 }{ 2\,k } } M_{1} \quad \text{for} \quad M_{1} >> 1 \label{2Dgd:eq:OmenikoffLarge} \end{align} Hence, for large Mach numbers, the Mach angle is $\sin\theta = \sqrt{ k+1\over 2k }$ (for k=1.4), which makes $\theta = 1.18$ or $\theta = 67.79^{\circ}$. With the value of $\theta$ utilizing equation \eqref{2Dgd:eq:Osol}, the maximum deflection angle can be computed. Note that this procedure does not require an approximation of $M_{1n}$ to be made. The general solution of equation \eqref{2Dgd:eq:Omenikoff} is

Normal Shock Minikoff Solution

\begin{align} \label {2Dgd:eq:OminikoffSol} M_{1n} = \dfrac{{\sqrt{\sqrt{\left(k+1\right)^2\,{M_1}^4+8\,\left(k^2-1\right)\,{M_1}^2+16 \,\left(k+1\right)}+\left(k+1\right)\,{M_1}^2-4}}}{{2\,\sqrt{k}}} \end{align}

Note that Maximum Deflection Mach Number's equation can be extended to deal with more complicated equations of state (aside from the perfect gas model). This typical example is for those who like mathematics.

# Example 12.1

Derive the perturbation of Maximum Deflection Mach Number's equation for the case of a very small upstream Mach number number of the form $M_1 = 1 + \epsilon$. Hint, Start with equation \eqref{2Dgd:eq:Omenikoff} and neglect all the terms that are relatively small.

# Solution

The solution can be done by substituting ($M_1 = 1+\epsilon$) into equation \eqref{2Dgd:eq:Omenikoff} and it results in

Normal Shock Small Values

\begin{align} \label {2Dgd:eq:OsmallMachMaxDeflction} M_{1n}= {\sqrt{\dfrac{\sqrt{\epsilon (k) } + {\epsilon^2} +2\,\epsilon - 3 + k\, \epsilon^2+2\,k\epsilon+k }{ 4\,k} } } \end{align}

where the epsilon function is \begin{multline} \epsilon(k) = (k^2+2k+1 )\,\epsilon^4+(4\,k^2+8\,k+4)\,\epsilon^3 + \ (14\,k^2+12\,k - 2)\,\epsilon^2+( 20\,k^2+8\,k-12) \,\epsilon + 9\,\left(k+1\right)^2 \label{2Dgd:eq:OepsilonF} \end{multline} Now neglecting all the terms with $\epsilon$ results for the epsilon function in \begin{align} \epsilon(k) \sim 9\,\left(k+1\right)^2 \label{2Dgd:eq:OepsilonFA} \end{align} And the total operation results in \begin{align} M_{1n}= \sqrt{ \dfrac{3\,\left(k+1\right) -3 + k }{ 4\,k} } = 1 \label{2Dgd:eq:OsmallMaxDefA} \end{align} Interesting to point out that as a consequence of this assumption the maximum shock angle, $\theta$ is a normal shock. However, taking the second term results in different value. Taking the second term in the explanation results in \begin{align} M_{1n}= \sqrt{ \dfrac{\sqrt{9\,\left(k+1\right)^2 +( 20\,k^2+8\,k-12) \,\epsilon} -3 + k + 2\,(1 + k) \epsilon }{ 4\,k} } \label{2Dgd:eq:OsmallMaxDefA1} \end{align} Note this equation \eqref{2Dgd:eq:OsmallMaxDefA1} produce an un realistic value and additional terms are required to obtained to produce a realistic value.

### 12.2.2.2 The case of $D\geq 0$ or $0 \geq\delta$

The second range in which $D>0$ is when $\delta < 0$. Thus, first the transition line in which $D=0$ has to be determined. This can be achieved by the standard mathematical procedure of equating $D=0$. The analysis shows regardless of the value of the upstream Mach number $D=0$ when $\delta=0$. This can be partially demonstrated by evaluating the terms $a_1$, $a_2$, and $a_3$ for the specific value of $M_1$ as following \begin{align} \begin{array}{rl} a_1 & = \dfrac{{ M_1}^2 + 2 }{ { M_1}^2 } \ a_2 & = - \dfrac{ 2 { M_1}^2 + 1 }{ { M_1}^4} \ a_3 & = - \dfrac{ 1 }{ { M_1}^4 } \end{array} \label{2Dgd:eq:ODzeroAlphaZero} \end{align} With values presented in equations qref{2Dgd:eq:ODzeroAlphaZero} for $R$ and $Q$ becoming \begin{multline} R = { 9 \left( \dfrac{ { M_1}^2 + 2 }{ { M_1}^2} \right) \left( \dfrac{ 2 { M_1}^2 + 1 }{ { M_1}^4} \right) + 27 \left( \dfrac{ 1 }{ { M_1}^4 } \right) - 2 \left( \dfrac{ { M_1}^2 + 2 }{ { M_1}^2 } \right)^2 \over 54} \\ = { 9 \left( { M_1}^2 + 2 \right) \left( 2 { M_1}^2 + 1 \right) + 27 { M_1}^2 - 2 \, { M_1}^2 \left( { M_1}^2 + 2 \right)^2 \over 54 \,{ M_1}^6} \label{2Dgd:eq:ORdeltaZero} \end{multline} and \begin{align} Q = \dfrac{ 3 \left( \dfrac{ 2 { M_1}^2 + 1 }{ { M_1}^4} \right) - \left( \dfrac{ { M_1}^2 + 2 }{ { M_1}^2 } \right)^3 } { 9 } \label{2Dgd:eq:OQdeltaZero1} \end{align} Substituting the values of $Q$ and $R$ equations \eqref{2Dgd:eq:ORdeltaZero} qref{2Dgd:eq:OQdeltaZero1} into equation \eqref{2Dgd:eq:OD} provides the equation to be solved for $\delta$. \begin{multline} \left[ \dfrac{ 3 \,\left( \dfrac{2 \, { M_1}^2 + 1 }{ { M_1}^4} \right) - \left( \dfrac{ { M_1}^2 + 2 }{ { M_1}^2 }\right)^3 }{ 9 }\right]^3 + \\ \left[ 9\,\left( { M_1}^2 + 2 \right) \left( 2\, { M_1}^2 + 1 \right) + 27\, { M_1}^2 - 2\, { M_1}^2 \left( { M_1}^2 + 2 \right)^2 \over 54 \, { M_1}^6 \right]^2 = 0 \label{2Dgd:eq:ODzero} \end{multline} The author is not aware of any analytical demonstration in the literature which shows that the solution is identical to zero Nevertheless, this identity can be demonstrated by checking several points for example, $M_1= 1., 2.0, \infty$ and addtional discussion and proofs can be found in Fundamentals of Compressible Flow'' by this author.

Fig. 12.7 The Mach waves that are supposed to be generated at zero inclination.

In the range where $\delta \leq 0$, the question is whether it is possible for an oblique shock to exist? The answer according to this analysis and stability analysis is no. Suppose that there is a Mach wave at the wall at zero inclination (see Figure 12.7). Obviously, another Mach wave occurs after a small distance. But because the velocity after a Mach wave (even for an extremely weak shock wave) is reduced, thus, the Mach angle will be larger ($\mu_2 > \mu_1$). If the situation keeps on occurring over a finite distance, there will be a point where the Mach number will be 1 and a normal shock will occur, according the common explanation. However, the reality is that no continuous Mach wave can occur because of the viscosity (boundary layer). there is the question of boundary layer. It is well known, in the engineering world, that there is no such thing as a perfect wall. The imperfections of the wall can be, for simplicity's sake, assumed to be as a sinusoidal shape. For such a wall the zero inclination changes from small positive value to a negative value. If the Mach number is large enough and the wall is rough enough, there will be points where a weak weak will be created. On the other hand, the boundary layer covers or smooths out the bumps. With these conflicting mechanisms, both will not allow a situation of zero inclination with emission of Mach wave. At the very extreme case, only in several points (depending on the bumps) at the leading edge can a very weak shock occur. Therefore, for the purpose of an introductory class, no Mach wave at zero inclination should be assumed. Furthermore, if it was assumed that no boundary layer exists and the wall is perfect, any deviations from the zero inclination angle creates a jump from a positive angle (Mach wave) to a negative angle (expansion wave). This theoretical jump occurs because in a Mach wave the velocity decreases while in the expansion wave the velocity increases. Furthermore, the increase and the decrease depend on the upstream Mach number but in different directions. This jump has to be in reality either smoothed out or has a physical meaning of jump (for example, detach normal shock). The analysis started by looking at a normal shock which occurs when there is a zero inclination. After analysis of the oblique shock, the same conclusion must be reached, i.e. that the normal shock can occur at zero inclination. The analysis of the oblique shock suggests that the inclination angle is not the source (boundary condition) that creates the shock. There must be another boundary condition(s) that causes the normal shock. In the light of this discussion, at least for a simple engineering analysis, the zone in the proximity of zero inclination (small positive and negative inclination angle) should be viewed as a zone without any change unless the boundary conditions cause a normal shock. Nevertheless, emission of Mach wave can occur in other situations. The approximation of weak weak wave with nonzero strength has engineering applicability in a very limited cases, especially in acoustic engineering, but for most cases it should be ignored.

### 12.2.2.3 Upstream Mach Number, $M_1$, and Shock Angle, $\theta$

The solution for upstream Mach number, $M_1$, and shock angle, $\theta$, are far much simpler and a unique solution exists. The deflection angle can be expressed as a function of these variables as

$\boldsymbol{\delta}$ For $\boldsymbol{\theta}$ and $M_1$

\begin{align} \label {2Dgd:eq:Odelta-theta} \cot \delta = \tan \left(\theta\right) \left[ \dfrac{(k + 1)\, {M_1}^2 }{ 2\, ( {M_1}^2\, \sin^2 \theta - 1)} - 1 \right] \end{align}

or \begin{align} \tan \delta = {2\cot\theta ({M_1}^2 \sin^2 \theta -1 ) \over 2 + {M_1}^2 (k + 1 - 2 \sin^2 \theta )} \label{2Dgd:eq:Odelta-thetaA} \end{align} The pressure ratio can be expressed as

Pressure Ratio

\begin{align} \label {2Dgd:eq:OPR} \dfrac{P_ 2 }{ P_1} = \dfrac{ 2 \,k\, {M_1}^2 \sin ^2 \theta - (k -1) }{ k+1} \end{align}

The density ratio can be expressed as

Density Ratio

\begin{align} \label {2Dgd:eq:ORR} \dfrac{\rho_2 }{ \rho_1 } = \dfrac{ {U_1}_n }{ {U_2}_n} = \dfrac{ (k +1)\, {M_1}^2\, \sin ^2 \theta } { (k -1) \, {M_1}^2\, \sin ^2 \theta + 2} \end{align}

The temperature ratio expressed as

Temperature Ratio

\begin{align} \label {2Dgd:eq:OTR} \dfrac{ T_2 }{ T_1} = \dfrac{ {c_2}^2 }{ {c_1}^2} = \dfrac{ \left( 2\,k\, {M_1}^2 \sin ^2 \theta - ( k-1) \right) \left( (k-1) {M_1}^2 \sin ^2 \theta + 2 \right) } { (k+1)\, {M_1}^2\, \sin ^2 \theta } \end{align}

The Mach number after the shock is

Exit Mach Number

\begin{align} \label{2Dgd:eq:OM2_0} {M_2}^2 \sin (\theta -\delta) = { (k -1) {M_1}^2 \sin ^2 \theta +2 \over 2 \,k\, {M_1}^2 \sin ^2 \theta - (k-1) } \end{align}

or explicitly \begin{align} {M_2}^2 = {(k+1)^2 {M_1}^4 \sin ^2 \theta - 4\,({M_1}^2 \sin ^2 \theta -1) (k {M_1}^2 \sin ^2 \theta +1) \over \left( 2\,k\, {M_1}^2 \sin ^2 \theta - (k-1) \right) \left( (k-1)\, {M_1}^2 \sin ^2 \theta +2 \right) } \label{2Dgd:eq:OM2} \end{align} The ratio of the total pressure can be expressed as

Stagnation Pressure Ratio

\begin{align} \label {2Dgd:eq:OP0R} {P_{0_2} \over P_{0_1}} = \left[ (k+1) {M_1}^2 \sin ^2 \theta \over (k-1) {M_1}^2 \sin ^2 \theta +2 \right]^{k \over k -1} \left[ k+1 \over 2 k {M_1}^2 \sin ^2 \theta - (k-1) \right] ^{1 \over k-1} \end{align}

Even though the solution for these variables, $M_1$ and $\theta$, is unique, the possible range deflection angle, $\delta$, is limited. Examining equation \eqref{2Dgd:eq:Odelta-theta} shows that the shock angle, $\theta\;$, has to be in the range of $\sin^{-1} (1/M_1) \geq \theta \geq (\pi/2)$ (see Figure 12.8). The range of given $\theta$, upstream Mach number $M_1$, is limited between $\infty$ and $\sqrt{1 / \sin^{2}\theta}$.

Fig. 12.8 The possible range of solutions for different parameters for given upstream Mach numbers.

### 12.2.2.4 Given Two Angles, $\delta$ and $\theta$

It is sometimes useful to obtain a relationship where the two angles are known. The first upstream Mach number, $M_1$ is

Mach Number Angles Relationship

\begin{align} \label{2Dgd:eq:OM1} {M_1}^2 = \dfrac{ 2 \,( \cot \theta + \tan \delta ) } { \sin 2 \theta - (\tan \delta)\, ( k + \cos 2 \theta) } \end{align}

The reduced pressure difference is \begin{align} \dfrac{2\,(P_2 - P_1) }{ \rho\, U^2} = \dfrac{2 \,\sin\theta \,\sin \delta }{ \cos(\theta - \delta)} \label{2Dgd:eq:OreducedPressure} \end{align} The reduced density is \begin{align} \dfrac{\rho_ 2 -\rho_1 }{ \rho_2} = \dfrac{\sin \delta }{ \sin \theta\, \cos (\theta -\delta)} \label{2Dgd:eq:OreducedDensity} \end{align} For a large upstream Mach number $M_1$ and a small shock angle (yet not approaching zero), $\theta$, the deflection angle, $\delta$ must also be small as well. Equation \eqref{2Dgd:eq:Odelta-theta} can be simplified into \begin{align} \theta \cong {k +1 \over 2} \delta \label{2Dgd:eq:OlargeM1theta} \end{align} The results are consistent with the initial assumption which shows that it was an appropriate assumption.

Fig. 12.9 Color-schlieren image of a two dimensional flow over a wedge. The total deflection angel (two sides) is $20^\circ$ and upper and lower Mach angel are $\sim 28^\circ$ and $\sim 30^\circ$, respectively. The image show the end–effects as it has thick (not sharp transition) compare to shock over a cone. The image was taken by Dr.~Gary Settles at Gas Dynamics laboratory, Penn State University.

# Example 12.2

In Figure 12.9 exhibits wedge in a supersonic flow with unknown Mach number. Examination of the Figure reveals that it is in angle of attack. 1) Calculate the Mach number assuming that the lower and the upper Mach angles are identical and equal to $\sim 30^\circ$ each (no angle of attack). 2) Calculate the Mach number and angle of attack assuming that the pressure after the shock for the two oblique shocks is equal. 3) What kind are the shocks exhibits in the image? (strong, weak, unsteady) 4) (Open question) Is there possibility to estimate the air stagnation temperature from the information provided in the image. You can assume that specific heats, $k$ is a monotonic increasing function of the temperature.

# Solution

#### Part (1)

The Mach angle and deflection angle can be obtained from the Figure 12.9. With this data and either using equation \eqref{2Dgd:eq:OM1} or potto-GDC results in

 Oblique Shock Input: $\theta_w$ and $\delta$ k = 1.4 $M_1$ $M_x$ ${{M_y}_s}$ ${{M_y}_w}$ $\theta_{s}$ $\theta_{w}$ $\delta$ $\dfrac{{P_0}_y}{{P_0}_x}$ 2.6810 2.3218 0 2.24 0 30 10 0.97172

The actual Mach number after the shock is then \begin{align*} M_2 = \dfrac{{M_2}_n}{\sin\left(\theta-\delta\right)} = \dfrac{0.76617}{\sin(30-10)} = 0.839 \end{align*} The flow after the shock is subsonic flow.

#### Part (2)

For the lower part shock angle of $\sim 28^\circ$ the results are

 Oblique Shock Input: $\theta_w$ and $\delta$ k = 1.4 $M_1$ $M_x$ ${{M_y}_s}$ ${{M_y}_w}$ $\theta_{s}$ $\theta_{w}$ $\delta$ $\dfrac{{P_0}_y}{{P_0}_x}$ 2.9168 2.5754 0 2.437 0 28 10 0.96549

From the last table, it is clear that Mach number is between the two values of 2.9168 and 2.6810 and the pressure ratio is between 0.96549 and 0.97172. One of procedure to calculate the attack angle is such that pressure has to match by guessing'' the Mach number between the extreme values.

#### Part (3)

The shock must be weak shock because the shock angle is less than $60^\circ$.

### 12.2.2.6 Close and Far Views of the Oblique Shock

Fig. 12.10 A local and a far view of the oblique shock.

In many cases, the close proximity view provides a continuous turning of the deflection angle, $\delta$. Yet, the far view shows a sharp transition. The traditional approach to reconcile these two views is by suggesting that the far view shock is a collection of many small weak shocks (see Figure 12.10). At the local view close to the wall, the oblique shock is a weak weak oblique'' shock. From the far view, the oblique shock is an accumulation of many small (or again weak) weak shocks.'' However, these small shocks'' are built or accumulate into a large and abrupt change (shock). In this theory, the boundary layer (B.L.) does not enter into the calculation. In reality, the boundary layer increases the zone where a continuous flow exists. The boundary layer reduces the upstream flow velocity and therefore the shock does not exist at close proximity to the wall. In larger distance from the wall, the shock becomes possible.

### 12.2.2.7 Maximum Value of Oblique shock

The maximum values are summarized in the following Table .

 $M_x$ 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 5 6 7 8 9 10 $M_y$ 0.97131 0.95049 0.93629 0.92683 0.92165 0.91941 0.91871 0.91997 0.92224 0.92478 0.93083 0.93747 0.94387 0.94925 0.95435 0.95897 0.96335 0.9663 0.96942 0.97214 0.98183 0.98714 0.99047 0.99337 0.9944 0.99559 $\delta_{max}$ 1.5152 3.9442 6.6621 9.4272 12.1127 14.6515 17.0119 19.1833 21.1675 22.9735 26.1028 28.6814 30.8137 32.5875 34.0734 35.3275 36.3934 37.3059 38.0922 38.7739 41.1177 42.4398 43.2546 43.7908 44.1619 44.429 $\theta_{max}$ 76.2762 71.9555 69.3645 67.7023 66.5676 65.7972 65.3066 64.9668 64.7532 64.6465 64.6074 64.6934 64.8443 65.0399 65.2309 65.4144 65.5787 65.7593 65.9087 66.0464 66.5671 66.902 67.1196 67.2503 67.3673 67.4419

It must be noted that the calculations are for the perfect gas model. In some cases, this assumption might not be sufficient and different analysis is needed. Henderson and Menikoff suggested a procedure to calculate the maximum deflection angle for arbitrary equation of state. When the mathematical quantity $D$ becomes positive, for large deflection angle, there isn't a physical solution to an oblique shock. Since the flow sees'' the obstacle, the only possible reaction is by a normal shock which occurs at some distance from the body. This shock is referred to as the detach shock. The detached shock's distance from the body is a complex analysis and should be left to graduate class and researchers in this area.

# Example 12.3

Air flows at Mach number ($M_1$) or $M_x= 4$ is approaching a wedge. What is the maximum wedge angle at which the oblique shock can occur? If the wedge angle is $20^\circ$, calculate the weak, the strong Mach numbers, and the respective shock angles.

# Solution

The maximum wedge angle for ($M_x= 4$) $D$ has to be equal to zero. The wedge angle that satisfies this requirement is by equation qref{2Dgd:eq:OD} (a side to the case proximity of $\delta=0$). The maximum values are:

 Oblique Shock Input: $M_x$ k = 1.4 $M_x$ $M_y$ $\delta_{max}$ $\theta{max}$ 4.000 0.97234 38.7738 66.0407

To obtain the results of the weak and the strong solutions either utilize the equation \eqref{2Dgd:eq:OD} or the GDC which yields the following results

 Oblique Shock Input: $M_x$ k = 1.4 $M_x$ ${{M_y}_s}$ ${{M_y}_w}$ $\theta_{s}$ $\theta_{w}$ $\delta$ 4.000 0.48523 2.5686 1.4635 0.56660 0.34907

Fig. 12.11 Oblique shock occurs around a cone. This photo is courtesy of Dr.~Grigory Toker, a Research Professor at Cuernavaco University of Mexico. According to his measurement, the cone half angle is $15^\circ$ and the Mach number is 2.2.

# Example 12.4

A cone shown in Figure 12.11 is exposed to supersonic flow and create an oblique shock. Is the shock shown in the photo weak or strong shock? Explain. Using the geometry provided in the photo, predict at which Mach number was the photo taken based on the assumption that the cone is a wedge.

# Solution

The measurements show that cone angle is $14.43^\circ$ and the shock angle is $30.099^\circ$. With given two angles the solution can be obtained by utilizing equation \eqref{2Dgd:eq:OM1} or the Potto-GDC.

 Oblique Shock Input: $M_1$ k = 1.4 $M_1$ ${{M_y}_s}$ ${{M_y}_w}$ $\theta_{s}$ $\theta_{w}$ $\delta$ $\dfrac{{P_0}_y}{{P_0}_x}$ 3.2318 0.56543 2.4522 71.0143 30.0990 14.4300 0.88737

Because the flow is around the cone it must be a weak shock. Even if the cone was a wedge, the shock would be weak because the maximum (transition to a strong shock) occurs at about $60^{\circ}$. Note that the Mach number is larger than the one predicted by the wedge.

Fig. 12.12 Maximum values of the properties in an oblique shock.

### 12.2.3 Application of Oblique Shock

Fig. 12.13 Two variations of inlet suction for supersonic flow.

One of the practical applications of the oblique shock is the design of an inlet suction for a supersonic flow. It is suggested that a series of weak shocks should replace one normal shock to increase the efficiency (see Figure \eqref{oblique:fig:inletSuction}). Clearly, with a proper design, the flow can be brought to a subsonic flow just below $M=1$. In such a case, there is less entropy production (less pressure loss). To illustrate the design significance of the oblique shock, the following example is provided.

# Example 12.5

Fig. 12.14 Schematic for Example \eqref{oblique:ex:inletEx}.

The Section described in Figure and efoblique:fig:inletEx} air is flowing into a suction section at $M=2.0$, $P=1.0[bar]$, and $T=17^{\circ}C$. Compare the different conditions in the two different configurations. Assume that only a weak shock occurs. }

# Solution

The first configuration is of a normal shock

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