Basics of Fluid Mechanics, 2014a

12.2. OBLIQUE SHOCK 487
12.2 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
12.3.
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, δ,
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
The momentum equation reads
ρ 1 U 1n = ρ 2 U 2n (12.1)
P 1 + ρ 1 U 1n 2 = P 2 + ρ 2 U 2n
2
(12.2)
The momentum equation in the tangential direction is reduced to
U 1t = U 2t (12.3)
The energy balance in coordinates moving with shock reads
C p T 1 + U 1n 2
= C p T 2 + U 2n 2
(12.4)
2
2
Equations (12.1), (12.2), and (12.4) are the same as the equations for normal shock
with the exception that the total velocity is replaced by the perpendicular components.
Yet, the new relationship between the upstream Mach number, the deflection angle, δ,
and the Mach angle, θ has to be solved. From the geometry it can be observed that
and
tan θ = U 1n
U 1t
(12.5)
tan(θ − δ) = U 2n
U 2t
(12.6)