SME
Compressible Flow · Problem type
Pitot tube in supersonic flow
Use this for “A pitot probe flies at M₁ = 2 where p₁ = 20 kPa. What does it read? What M₁ gives a reading of 112.8 kPa?”
Diagram
The probe’s bow shock is normal on the axis: the flow reaches the tip subsonic at M = 0.5774, then stops isentropically, so the probe reads p₀₂, not p₀₁.
This is an example. Type your own values to start a new problem.
p₀₂/p₁
=
5.76
×
M₁
2
5.6
×
M₁
2
−
0.8
3.5
×
2.8
×
M₁
2
−
0.4
2.4
p₀₂ (kPa)
=
p₀₂/p₁
×
p₁ (kPa)
Formulas
p₀₂
p₁
=
(5.76 ×
M₁
2
5.6 ×
M₁
2
− 0.8
)
3.5
×
2.8 ×
M₁
2
− 0.4
2.4
5.64
=
(5.76 ×
2
2
5.6 ×
2
2
− 0.8
)
3.5
×
2.8 ×
2
2
− 0.4
2.4
p₀₂
=
(
p₀₂
p₁
)
×
p₁
112.8 = 5.64 × 20
Assumptions
•
A bow shock stands in front of the probe; on its axis it is a normal shock.
•
Behind the shock the air stops isentropically at the tip, which reads p₀₂.
•
Calorically perfect air, γ = 1.4, which gives the factors 5.76, 5.6, 0.8, 2.8 and 2.4.
Step-by-step
WE KNOW
M₁ = 2, p₁ = 20 kPa
FIND
reading over static pressure (p₀₂/p₁), probe reading (p₀₂)
Step 1 · Find reading over static pressure (p₀₂/p₁)
Use p₀₂/p₁ = (5.76 × M₁² ÷ (5.6 × M₁² − 0.8))3.5 × (2.8 × M₁² − 0.4) ÷ 2.4
The Rayleigh pitot formula: the normal shock’s pressure rise, then the isentropic stop behind it.
p₀₂
p₁
=
(5.76 ×
2
2
5.6 ×
2
2
− 0.8
)
3.5
×
2.8 ×
2
2
− 0.4
2.4
p₀₂
p₁
=
(5.76 ×
2
2
5.6 × 4 − 0.8
)
3.5
×
2.8 ×
2
2
− 0.4
2.4
p₀₂
p₁
=
(5.76 ×
2
2
22.4 − 0.8
)
3.5
×
2.8 ×
2
2
− 0.4
2.4
p₀₂
p₁
=
(5.76 ×
2
2
21.6
)
3.5
×
2.8 ×
2
2
− 0.4
2.4
p₀₂
p₁
=
(5.76 ×
4
21.6
)
3.5
×
2.8 × 4 − 0.4
2.4
p₀₂
p₁
=
(
23.04
21.6
)
3.5
×
11.2 − 0.4
2.4
p₀₂
p₁
=
1.0667
3.5
×
10.8
2.4
p₀₂
p₁
=
1.2536
×
10.8
2.4
p₀₂
p₁
=
13.539
2.4
p₀₂
p₁
=
5.64
Step 2 · Find probe reading (p₀₂)
Use p₀₂ = (p₀₂/p₁) × p₁
Multiply the ratio by the static pressure.
p₀₂
=
5.64
×
20
p₀₂
=
112.8
kPa
Check
Put the numbers back in. Both sides should match.
5.64
=
(5.76 ×
2
2
5.6 ×
2
2
− 0.8
)
3.5
×
2.8 ×
2
2
− 0.4
2.4
✓
112.8 = 5.64 × 20 ✓
Topic check
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