SME
Aerodynamics · Problem type
Pressure coefficient from the local speed
Use this for “Air at 1.225 kg/m³ and 50 m/s speeds up to 60 m/s over the top of a wing. Find Cₚ and p − p∞.”
Diagram
Cₚ = −0.44: faster than the free stream, the pressure is below p∞ and pulls outward.
This is an example. Type your own values to start a new problem.
Cₚ
=
1
−
V (m/s)
V∞ (m/s)
2
p − p∞ (Pa)
=
Cₚ
×
q∞ (Pa)
ρ∞
Free-stream density
example
Formulas
q∞ = 0.5 × ρ∞ × V∞²
1,531
=
0.5
×
1.225
×
50
2
Cₚ
=
1
−
(
V
V
∞)
2
−0.44
=
1
−
(
60
50
)
2
p − p∞ = Cₚ × q∞
−673.8 = −0.44 × 1,531
Assumptions
•
Incompressible flow (low speed), so Bernoulli’s equation holds along the streamline.
•
The pressure coefficient Cₚ is the pressure change in units of the dynamic pressure q∞.
•
Cₚ < 0 is suction: the surface is pulled outward.
Step-by-step
WE KNOW
ρ∞ = 1.225 kg/m³, V∞ = 50 m/s, V = 60 m/s
FIND
dynamic pressure (q∞), pressure coefficient (Cₚ), pressure change (p − p∞)
Step 1 · Find dynamic pressure (q∞)
Use q∞ = 0.5 × ρ∞ × V∞²
Dynamic pressure: half the density times the speed squared.
q∞
=
0.5
×
1.225
×
50
2
q∞ = 0.5 × 1.225 × 2,500
q∞ = 0.6125 × 2,500
q∞ = 1,531.3 Pa
Step 2 · Find pressure coefficient (Cₚ)
Use Cₚ = 1 − (V ÷ V∞)²
Bernoulli: the faster the local flow, the lower its pressure.
Cₚ
=
1
−
(
60
50
)
2
Cₚ
=
1
−
1.2
2
Cₚ = 1 − 1.44
Cₚ = −0.44
Step 3 · Find pressure change (p − p∞)
Use p − p∞ = Cₚ × q∞
Cₚ is the pressure change in units of q∞.
p − p∞ = −0.44 × 1,531.3
p − p∞ = −673.8 Pa
Check
Put the numbers back in. Both sides should match.
1,531.3
=
0.5
×
1.225
×
50
2
✓
−0.44
=
1
−
(
60
50
)
2
✓
−673.8 = −0.44 × 1,531.3 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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