Basics of Fluid Mechanics, 2014a

492 CHAPTER 12. COMPRESSIBLE FLOW 2–DIMENSIONAL
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 an imaginary
hand moves the shock angle starting
from the deflection angle and
Normal
Shock
Fig. -12.4. Flow around spherically blunted 30 ◦ conecylinder
with Mach number 2.0. It can be noticed that
the normal shock, the strong shock, and the weak shock
coexist.
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 9 . 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 10 ).
12.2.2 When No Oblique Shock Exist or the case of D>0
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 (12.5), the imaginary angle is shown. The flow
is far away from the object and does not “see’ the object. For example, for, M 1 −→ ∞
the maximum deflection angle is calculated when D = Q 3 + R 2 =0. This can be done
9 See the discussion on the stability. There are those who view this question not as a stability
equation but rather as under what conditions a strong or a weak shock will prevail.
10 This material is extra and not recommended for standard undergraduate students.