Upstream Mach number, M1, and deflection angle, δ

Again, this set of parameters is, perhaps, the most common and natural to examine. Thompson (1950) has shown that the relationship of the shock angle is obtained from the following cubic equation:

$$x^3 + a_1 x^2 + a_2 x + a_3=0$$ (13.18)

where

$$x = \sin^2 \theta$$ (13.19)

and

$$a_1 = - {{M_1}^2 + 2 \over {M_1}^2} - k \sin ^2 \delta$$ (13.20)

$$a_2 = - { 2{M_1}^2 + 1 \over {M_1}^4 } + \left[ {(k+1)^2 \over 4}+ {k -1 \over {M_1}^2} \right] \sin ^2 \delta$$ (13.21)

$$a_3 = - {\cos ^2 \delta \over {M_1}^4}$$ (13.22)

Equation (13.18) requires that $x$ has to be a real and positive number to obtain a real deflection angle^13.8. 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 (13.18) provides three roots^13.9. These roots can be expressed as

$$x_1 = - {1 \over 3} a_1 + (S +T )$$ (13.23)

$$x_2 = - {1 \over 3} a_1 - \half (S +T ) + \half i \sqrt{3} ( S-T)$$ (13.24)

and

$$x_3 = - {1 \over 3} a_1 - \half (S +T ) - \half i \sqrt{3} ( S-T )$$ (13.25)

Where

$$S = \sqrt[3]{R + \sqrt{D}},$$ (13.26)

$$T = \sqrt[3]{R - \sqrt{D}}$$ (13.27)

and where the definition of the $D$ is

$$D = Q^3 + R^2$$ (13.28)

and where the definitions of $Q$ and $R$ are

$$Q = { 3 a_2 - {a_1 } ^2 \over 9}$$ (13.29)

and

$$R = { 9 a_1 a_2 - 27 a_3 - 2 {a_1}^3 \over 54}$$ (13.30)

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^13.10. $D > 0$ 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^13.11 that. Physically, it can be shown that the first solution(13.23), 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^13.12.

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 flow is subsonic). Second, the third root always turns the flow into subsonic and it is referred to as the strong solution. It should be noted that this case is where entropy increases in the largest amount.

In summary, if a 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^13.13. 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^13.14).

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. In 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.