SME
Compressible Flow · Problem type
Oblique shock: θ–β–M
Use this for “Air at M₁ = 2 meets a wedge and the shock stands at β = 40°. Find θ, p₂/p₁ and M₂.”
Diagram
The shock stands at β = 40°, steeper than the Mach angle 30° and than the wedge’s θ = 10.62°; the streamlines turn by θ as they cross it.
This is an example. Type your own values to start a new problem.
tan
θ (°)
=
2
cot
β (°)
×
M₁
2
sin²
β (°)
−
1
M₁
2
(1.4
+
cos(2
×
β (°)
))
+
2
This is the θ–β–M relation with γ = 1.4. The Formulas section writes 1.4 + cos 2β as 2.4 − 2 sin²β, which is the same number.
B
Branch
calculated
βₘₐₓ
Shock angle at θₘₐₓ
worked out (not typed) · °
Mₙ₁
Normal Mach number before
calculated
Mₙ₂
Normal Mach number after
calculated
M₂
Mach number after
calculated
p₂/p₁
Pressure ratio
calculated
Formulas
Mₙ₁
=
M₁
×
sin(
β
)
1.286 = 2 × sin(40°)
βₘₐₓ
=
sin⁻¹(√((0.6
×
M₁
2
−
1
+
√(2.4
×
(1
+
0.2
×
M₁
2
+
0.15
×
M₁
4
)))
÷
(1.4
×
M₁
2
)))
64.67
=
sin⁻¹(√((0.6
×
2
2
−
1
+
√(2.4
×
(1
+
0.2
×
2
2
+
0.15
×
2
4
)))
÷
(1.4
×
2
2
)))
tan(
θ
)
=
2
tan(
β
)
×
(
M₁
2
×
sin
(
β
)
2
−
1)
÷
(
M₁
2
×
(2.4
−
2
×
sin
(
β
)
2
)
+
2)
tan(10.62°)
=
2
tan(40°)
×
(
2
2
×
sin
(40°)
2
−
1)
÷
(
2
2
×
(2.4
−
2
×
sin
(40°)
2
)
+
2)
B
=
weak
if
β
<
βₘₐₓ,
else
strong
40 < 64.67, so weak
p₂/p₁
=
1
+
2.8
2.4
×
(
Mₙ₁
2
−
1)
1.761
=
1
+
2.8
2.4
×
(
1.286
2
−
1)
Mₙ₂
=
√(
1 + 0.2 ×
Mₙ₁
2
1.4 ×
Mₙ₁
2
− 0.2
)
0.7934
=
√(
1 + 0.2 ×
1.286
2
1.4 ×
1.286
2
− 0.2
)
M₂
=
Mₙ₂
sin(
β
) × cos(
θ
) − cos(
β
) × sin(
θ
)
1.617
=
0.7934
sin(40°) × cos(10.62°) − cos(40°) × sin(10.62°)
Assumptions
•
An attached, straight oblique shock; only the normal part M₁ sin β passes through a normal shock.
•
Calorically perfect air, γ = 1.4; the tangential speed is the same both sides.
•
Each θ below θₘₐₓ has two shock angles: the weak one below βₘₐₓ (usual) and the strong one above.
Step-by-step
WE KNOW
M₁ = 2, β = 40°
FIND
normal Mach number before (Mₙ₁), shock angle at θₘₐₓ (βₘₐₓ), branch (B), pressure ratio (p₂/p₁), normal Mach number after (Mₙ₂), wedge angle (θ), Mach number after (M₂)
Step 1 · Find normal Mach number before (Mₙ₁)
Use Mₙ₁ = M₁ × sin(β)
The part of the flow normal to the shock.
Mₙ₁ = 2 × sin(40°)
Mₙ₁ = 2 × 0.6428
Mₙ₁ = 1.285575
Step 2 · Find shock angle at θₘₐₓ (βₘₐₓ)
Use βₘₐₓ = sin⁻¹(√((0.6 × M₁² − 1 + √(2.4 × (1 + 0.2 × M₁² + 0.15 × M₁⁴))) ÷ (1.4 × M₁²)))
The shock angle that turns the flow the most; with γ = 1.4 the factors are 0.6, 2.4, 0.2 and 0.15.
βₘₐₓ
=
sin⁻¹(√((0.6
×
2
2
−
1
+
√(2.4
×
(1
+
0.2
×
2
2
+
0.15
×
2
4
)))
÷
(1.4
×
2
2
)))
βₘₐₓ = 64.67°
Step 3 · Find branch (B)
Use B = weak if β < βₘₐₓ, else strong
Below
βₘₐₓ
the
shock
is
weak
(
M₂
mostly
above
1);
above
it,
strong
(
M₂
below
1).
40 < 64.67: weak
B
=
weak
if
40
<
64.67,
else
strong
B
=
weak
Step 4 · Find pressure ratio (p₂/p₁)
Use p₂/p₁ = 1 + 2.8 ÷ 2.4 × (Mₙ₁² − 1)
The normal-shock pressure jump, on the normal Mach number.
p₂/p₁
=
1
+
2.8
2.4
×
(
1.285575
2
−
1)
p₂/p₁
=
1
+
2.8
2.4
×
(1.6527
−
1)
p₂/p₁
=
1
+
2.8
2.4
×
0.6527
p₂/p₁ = 1 + 1.1667 × 0.6527
p₂/p₁ = 1 + 0.76151
p₂/p₁ = 1.761
Step 5 · Find normal Mach number after (Mₙ₂)
Use Mₙ₂ = √((1 + 0.2 × Mₙ₁²) ÷ (1.4 × Mₙ₁² − 0.2))
The normal-shock relation, on the normal Mach number.
Mₙ₂
=
√(
1 + 0.2 ×
1.285575
2
1.4 ×
1.285575
2
− 0.2
)
Mₙ₂
=
√(
1 + 0.2 × 1.653
1.4 × 1.653 − 0.2
)
Mₙ₂
=
√(
1 + 0.3306
2.3142 − 0.2
)
Mₙ₂
=
√(
1.3306
2.1142
)
Mₙ₂
=
√
0.6294
Mₙ₂ = 0.7934
Step 6 · Find wedge angle (θ)
Use tan(θ) = 2 ÷ tan(β) × (M₁² × sin(β)² − 1) ÷ (M₁² × (2.4 − 2 × sin(β)²) + 2)
The
θ
–
β
–M
relation:
cot
β
is
1
tan
β
and
γ
+
cos
2
β
is
2.4
−
2
sin²
β
;
take
tan⁻¹
of
the
right
side.
40 < 64.67, on the weak branch: weak
θ
=
tan⁻¹(
2
tan(
β
)
×
(
M₁
2
×
sin
(
β
)
2
−
1)
÷
(
M₁
2
×
(2.4
−
2
×
sin
(
β
)
2
)
+
2))
θ
=
tan⁻¹(
2
tan(40°)
×
(
2
2
×
sin
(40°)
2
−
1)
÷
(
2
2
×
(2.4
−
2
×
sin
(40°)
2
)
+
2))
θ
=
10.62°
Step 7 · Find Mach number after (M₂)
Use M₂ = Mₙ₂ ÷ (sin(β) × cos(θ) − cos(β) × sin(θ))
Behind
the
shock
the
flow
runs
at
β
−
θ
to
it;
Mₙ₂
is
its
normal
part.
sin(
β
−
θ
)
=
sin
β
cos
θ
−
cos
β
sin
θ
.
M₂
=
0.7934
sin(40°) × cos(10.62°) − cos(40°) × sin(10.62°)
M₂
=
0.7934
0.6428 × 0.9829 − 0.766 × 0.1843
M₂
=
0.7934
0.6318 − 0.1412
M₂
=
0.7934
0.4906
M₂
=
1.617
Check
Put the numbers back in. Both sides should match.
1.285575 = 2 × sin(40°) ✓
64.67
=
sin⁻¹(√((0.6
×
2
2
−
1
+
√(2.4
×
(1
+
0.2
×
2
2
+
0.15
×
2
4
)))
÷
(1.4
×
2
2
)))
✓
tan(10.62°)
=
2
tan(40°)
×
(
2
2
×
sin
(40°)
2
−
1)
÷
(
2
2
×
(2.4
−
2
×
sin
(40°)
2
)
+
2)
✓
40 < 64.67, so weak ✓
1.761
=
1
+
2.8
2.4
×
(
1.285575
2
−
1)
✓
0.7934
=
√(
1 + 0.2 ×
1.285575
2
1.4 ×
1.285575
2
− 0.2
)
✓
1.617
=
0.7934
sin(40°) × cos(10.62°) − cos(40°) × sin(10.62°)
✓
Topic check
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