Oblique Shock Waves
Here’s a quick refresher on oblique shock waves. We start with the oblique shock
as shown below:
w2 , M t 2
(1)
(2)
( )1 : upstream flow
u
,
M
2
n
2
y
condition
u1 , M n1
w1 , M t1
( )2 : downstream flow
v2 , M 2
θ
condition
x
β : angle of shock wave
v1 , M 1
w.r.t. upstream flow
θ : deflection angle of flow
β
Also, the specific flow quantities above are:
v
a
u
M n : Normal Mach # =
a
v : flowspeed
M : Mach number =
u : normal velocity to shock
w
a
The next step is to apply the 2-D Euler equations to derive jump conditions.
M t : Tangential Mach # =
w : tangential velocity to shock
Let’s consider the following (well-chosen) control volume across the shock:
v
v
na = −ng
c
b
v
ts
a
v1
v
ns
e
f
Shock wave
Where: a & d are parallel to shock
b, f , c, e are parallel to local flow
Apply conservation of mass:
v v
ρV • nds = 0
∫
v v s
But V • n = 0 on b, f , c & e, thus:
v v
v v
ρ
•
+
ρ
V
n
ds
V
∫a
∫d • nds = 0
v v
v v
− ∫ ρ1V1 • n s ds + ∫ ρ 2V2 • n s ds = 0
a
d
u1
u2
d
v2
v
v
nd = + ns
Control volume
s
Oblique Shock Waves
− ρ1u1 ∫ ds + ρ 2 u 2 d ∫ ds = 0
a
d
A1
A2
But, A1 = A2 since all lines of control volume edges b, c, e, f are parallel.
ρ1u1 = ρ 2 u 2
⇒
The next equation we’ll look at is tangential momentum.
v v
v v
ρ
w
V
•
n
ds
=
−
p
n
∫
∫ • t s ds
s
s
v v
v v
− ρ1 w1u1 A1 + ρ 2 w2 u 2 A2 = − ∫ pn • t s ds − ∫ pn • t s ds
b
f
v v
v v
− ∫ pn • t s ds − ∫ pn • t s ds
c
e
Plugging into the pressure terms:
v v
v v
− ρ1 w1u1 A1 + ρ 2 w2 u 2 A2 = − p1 nb • t s ∫ ds − p1 n f • t s ∫ ds
b
f
v
v
v
v
v v
v v
But nb = −n f & nc = −ne
− p2 nc • ts ∫ ds − p2 ne • ts ∫ ds
c
e
⇒ ρ1u1 w1 A1 = ρ 2 u 2 w2 A2
Using ρuA = const .
⇒ w1 = w2
Tangential velocity is unchanged across a shock wave!
The last two equations (see Anderson for more) give:
Normal momentum : p1 + ρ1u12 = p 2 + ρ 2 u 22
1
1
Energy: h1 + u12 = h2 + u22
2
2
These equations can be solved and results are displayed in graph and tables in
Anderson.
ramp
β
Find β & M 2 ?
Example:
M2
θ = 25°
M1 = 5
16.100 2002
2