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Engineering LibreTexts 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
    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 }   
    \label{2Dgd:eq:ODzeroAlphaZero}  \tag{47}
With values presented in equations (47) for \(R\) and \(Q\) becoming
    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}  \tag{48}
    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}  \tag{49}
Substituting the values of \(Q\) and \(R\) equations (48) (49) into equation (28) provides the equation to be solved for \(\delta\).
    \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
             54 \, { M_1}^6 \right]^2
                  = 0   
    \label{2Dgd:eq:ODzero}  \tag{50}
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.

