SME
Compressible Flow
Normal and oblique shocks
Use this for “Air at M₁ = 2 meets a normal shock. Find M₂, p₂/p₁, ρ₂/ρ₁, T₂/T₁ and p₀₂/p₀₁.”
Diagram
Supersonic in, subsonic out: M₁ = 2 → M₂ = 0.5774.
The gas slows by 2.667 times, so the arrows shorten; p₀ falls to 0.7209 of its value.
This is an example. Type your own values to start a new problem.
M₂
2
=
1
+
0.2
×
M₁
2
1.4
×
M₁
2
−
0.2
p₂/p₁
=
1
+
2.8
2.4
×
(
M₁
2
−
1)
ρ₂/ρ₁
Density ratio
calculated
T₂/T₁
Temperature ratio
calculated
p₀₂/p₀₁
Stagnation pressure ratio
calculated
Formulas
M₂
=
√(
1 + 0.2 ×
M₁
2
1.4 ×
M₁
2
− 0.2
)
0.5774
=
√(
1 + 0.2 ×
2
2
1.4 ×
2
2
− 0.2
)
p₂/p₁
=
1
+
2.8
2.4
×
(
M₁
2
−
1)
4.5
=
1
+
2.8
2.4
×
(
2
2
−
1)
ρ₂/ρ₁
=
2.4
×
M₁
2
0.4 ×
M₁
2
+ 2
2.667
=
2.4
×
2
2
0.4 ×
2
2
+ 2
T₂/T₁
=
p₂/p₁
ρ₂/ρ₁
1.687
=
4.5
2.667
p₀₂/p₀₁
=
(ρ₂/ρ₁)
3.5
÷
(p₂/p₁)
2.5
0.7209
=
(2.667)
3.5
÷
(4.5)
2.5
Assumptions
•
The shock is thin and adiabatic, so T₀ is the same both sides; entropy rises, so p₀ falls.
•
Calorically perfect air, γ = 1.4: (γ − 1) ÷ 2 = 0.2 and 2γ ÷ (γ + 1) = 2.8 ÷ 2.4.
•
The flow meets the shock square on (normal), supersonic before it and subsonic after.
Step-by-step
WE KNOW
M₁ = 2
FIND
Mach number after (M₂), pressure ratio (p₂/p₁), density ratio (ρ₂/ρ₁), temperature ratio (T₂/T₁), stagnation pressure ratio (p₀₂/p₀₁)
Step 1 · Find Mach number after (M₂)
Use M₂ = √((1 + 0.2 × M₁²) ÷ (1.4 × M₁² − 0.2))
With
γ
=
1.4
the
relation
reads
1 + 0.2
M₁
²
1.4
M₁
² − 0.2
;
take
the
square
root.
M₂
=
√(
1 + 0.2 ×
2
2
1.4 ×
2
2
− 0.2
)
M₂
=
√(
1 + 0.2 × 4
1.4 × 4 − 0.2
)
M₂
=
√(
1 + 0.8
5.6 − 0.2
)
M₂
=
√(
1.8
5.4
)
M₂
=
√
0.3333
M₂
=
0.5774
Step 2 · Find pressure ratio (p₂/p₁)
Use p₂/p₁ = 1 + 2.8 ÷ 2.4 × (M₁² − 1)
The
pressure
jump:
2γ
γ + 1
=
2.8
2.4
for
air.
p₂/p₁
=
1
+
2.8
2.4
×
(
2
2
−
1)
p₂/p₁
=
1
+
2.8
2.4
×
(4
−
1)
p₂/p₁
=
1
+
2.8
2.4
×
3
p₂/p₁ = 1 + 1.1667 × 3
p₂/p₁ = 1 + 3.5001
p₂/p₁ = 4.5
Step 3 · Find density ratio (ρ₂/ρ₁)
Use ρ₂/ρ₁ = 2.4 × M₁² ÷ (0.4 × M₁² + 2)
The
density
jump;
it
never
passes
γ + 1
γ − 1
=
6,
however
strong
the
shock.
ρ₂/ρ₁
=
2.4
×
2
2
0.4 ×
2
2
+ 2
ρ₂/ρ₁
=
2.4
×
2
2
0.4 × 4 + 2
ρ₂/ρ₁
=
2.4
×
2
2
1.6 + 2
ρ₂/ρ₁
=
2.4
×
2
2
3.6
ρ₂/ρ₁
=
2.4
×
4
3.6
ρ₂/ρ₁
=
9.6
3.6
ρ₂/ρ₁ = 2.6667
Step 4 · Find temperature ratio (T₂/T₁)
Use T₂/T₁ = (p₂/p₁) ÷ (ρ₂/ρ₁)
The
ideal
gas
law:
T
∝
p
ρ
.
T₂/T₁
=
4.5
2.6667
T₂/T₁ = 1.687
Step 5 · Find stagnation pressure ratio (p₀₂/p₀₁)
Use p₀₂/p₀₁ = (ρ₂/ρ₁)3.5 ÷ (p₂/p₁)2.5
T₀
is
the
same
both
sides,
so
the
entropy
rise
sets
the
loss
of
p₀:
γ
γ − 1
=
3.5,
1
γ − 1
=
2.5.
p₀₂/p₀₁
=
(2.6667)
3.5
÷
(4.5)
2.5
p₀₂/p₀₁
=
30.97
42.96
p₀₂/p₀₁ = 0.7209
Check
Put the numbers back in. Both sides should match.
0.5774
=
√(
1 + 0.2 ×
2
2
1.4 ×
2
2
− 0.2
)
✓
4.5
=
1
+
2.8
2.4
×
(
2
2
−
1)
✓
2.6667
=
2.4
×
2
2
0.4 ×
2
2
+ 2
✓
1.687
=
4.5
2.6667
✓
0.7209
=
(2.6667)
3.5
÷
(4.5)
2.5
✓
Topic check
See what you know: 5 questions on this topic
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