Compressible Flow Free Textbook

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The case of D >= 0 or 0 >= delta

The second range in which

is when $\delta < 0$ . Thus, first the transition line in which

has to be determined. This can be achieved by the standard mathematical procedure of equating

. The analysis shows regardless of the value of the upstream Mach number

when $\delta = 0$ . This can be partially demonstrated by evaluating the terms

, and

for the specific value of

as following

$\displaystyle a_1 = {{ M_1}^2 + 2 \over { M_1}^2 }$
$\displaystyle a_2 = - { 2 { M_1}^2 + 1 \over { M_1}^4}$
$\displaystyle a_3 = - { 1 \over { M_1}^4 }$	(13.47)

With values presented in equations (13.47) for

and

becoming

$\displaystyle R = { 9 \left( { M_1}^2 + 2 \over { M_1}^2 \right) \left( { 2 { M... ...ver { M_1}^4 \right) - 2 \left( { M_1}^2 + 2 \over { M_1}^2 \right)^2 \over 54}$
$\displaystyle = { 9 \left( { M_1}^2 + 2 \right) \left( 2 { M_1}^2 + 1 \right) + 27 { M_1}^2 - 2 { M_1}^2 \left( { M_1}^2 + 2 \right)^2 \over 54 { M_1}^6}$	(13.48)

and

$\displaystyle Q = { 3 \left( { 2 { M_1}^2 + 1 \over { M_1}^4} \right) - \left( { M_1}^2 + 2 \over { M_1}^2 \right)^3 \over 9 }$

(13.49)

Substituting the values of

and

equations (13.48) (13.49) into equation (13.28) provides the equation to be solved for $\delta$ .

$\displaystyle \left[ 3 \left( { 2 { M_1}^2 + 1 \over { M_1}^4} \right) - \left( { M_1}^2 + 2 \over { M_1}^2 \right)^3 \over 9 \right]^3 +$
$\displaystyle \hspace{16mm}\left[ 9 \left( { M_1}^2 + 2 \right) \left( 2 { M_1}... ...1}^2 - 2 { M_1}^2 \left( { M_1}^2 + 2 \right)^2 \over 54 { M_1}^6 \right]^2 = 0$	(13.50)

The author is not aware of any analytical demonstration in the literature which shows that the solution is identical to zero for $\delta = 0$ ^13.17. Nevertheless, this identity can be demonstrated by checking several points for example, $M_1= 1., 2.0, \infty$ . Table (13.6) is provided for the following demonstration. Substitution of all the above values into (13.28) results in

Utilizing the symmetry and antisymmetry of the qualities of the $\cos$ and $\sin$ for $\delta < 0$ demonstrates that regardless of Mach number. Hence, the physical interpretation of this fact is that either no shock exists and the flow is without any discontinuity or that a normal shock exists^13.18. Note that, in the previous case, with a positive large deflection angle, there was a transition from one kind of discontinuity to another.

M₁	a₁	a₂	a₃
1.0	-3	-1	-2/3
2.0	3	0	9/16
∞	-1	0	-1/16

The various coefficients of three different Mach numbers to demonstrate that is zero

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. And according to this analysis, no Mach wave can be generated from the wall with zero deflection. In other words, the wall does not emit any signal to the flow (assuming zero viscosity), which contradicts the common approach. Nevertheless, in the literature, there are several papers suggesting zero strength Mach wave; others suggest a singular point^13.19. The question of singular point or zero Mach wave strength are only of mathematical interest.

**Figure 13.7:**

The ``imaginary'' Mach waves at zero inclination.

Suppose that there is a Mach wave at the wall at zero inclination (see Figure (13.7)). 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).

In reality, there are imperfections in the wall and in the flow and 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^13.20 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.

**Figure 13.8:** The D, shock angle, and for

Next: Upstream Mach Number, M1, Up: Upstream Mach number, M1, Previous: The Procedure for Calculating Index

Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21

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