An NACA 4415 airfoil is mounted in a high-speed subsonic wind tunnel. The lift coefficient is measured as 0.85. If the test-section Mach number is 0.7, at what angle of attack is the airfoil?
The correct answer and explanation is :
To determine the angle of attack (α) of an NACA 4415 airfoil given the following conditions:
- Lift coefficient, ( C_L = 0.85 )
- Mach number, ( M = 0.7 ) (subsonic)
- Airfoil: NACA 4415
Step 1: Understand the airfoil characteristics
The NACA 4415 airfoil has:
- Maximum camber = 4% of chord (first digit)
- Camber location = 40% of chord (second digit)
- Thickness = 15% of chord (last two digits)
For subsonic conditions, the lift coefficient can be approximated by:
[
C_L = C_{L_\alpha} (\alpha – \alpha_0)
]
Where:
- ( C_{L_\alpha} ) = Lift curve slope (per radian)
- ( \alpha ) = Angle of attack (radians)
- ( \alpha_0 ) = Zero-lift angle of attack (radians)
Step 2: Estimate lift curve slope
For thin airfoils in compressible subsonic flow:
[
C_{L_\alpha} = \frac{2\pi}{\sqrt{1 – M^2}} \quad \text{(per rad)}
]
[
C_{L_\alpha} = \frac{2\pi}{\sqrt{1 – 0.7^2}} = \frac{2\pi}{\sqrt{1 – 0.49}} = \frac{2\pi}{\sqrt{0.51}} \approx \frac{6.283}{0.714} \approx 8.8 \, \text{per rad}
]
Step 3: Use known zero-lift angle for NACA 4415
For NACA 4415, zero-lift angle ( \alpha_0 \approx -2^\circ ) = (-0.035) radians.
Step 4: Solve for angle of attack
[
0.85 = 8.8(\alpha – (-0.035))
]
[
0.85 = 8.8(\alpha + 0.035)
]
[
\alpha + 0.035 = \frac{0.85}{8.8} \approx 0.0966
]
[
\alpha = 0.0966 – 0.035 = 0.0616 \, \text{rad}
]
Convert to degrees:
[
\alpha = 0.0616 \times \frac{180}{\pi} \approx 3.53^\circ
]
✅ Final Answer:
Angle of attack ( \alpha \approx 3.5^\circ )
Explanation (Approx. 300 Words):
The angle of attack at which an airfoil generates a given lift coefficient depends on its aerodynamic shape, flow conditions, and compressibility effects. For a NACA 4415 airfoil, which has moderate camber and thickness, the lift characteristics are predictable using classical thin airfoil theory modified for compressibility.
In subsonic compressible flows (Mach < 1), the lift curve slope increases compared to incompressible flow. This effect is corrected by the Prandtl-Glauert factor, where the lift slope increases by ( 1/\sqrt{1 – M^2} ). At a Mach number of 0.7, this results in a higher lift curve slope, roughly 8.8 per radian instead of the incompressible ( 2\pi \approx 6.28 ).
Given a lift coefficient of 0.85, and knowing that the NACA 4415 has a zero-lift angle around −2 degrees, we solve the linear lift equation:
[
C_L = C_{L_\alpha} (\alpha – \alpha_0)
]
After plugging in the known values and solving, we find the angle of attack is approximately 3.5 degrees. This indicates the airfoil must be pitched slightly above the chord line to counteract its negative zero-lift angle and produce the required lift.
This method assumes linear behavior and is valid up to moderate angles of attack where flow remains mostly attached. It’s a good approximation in wind tunnel tests and early design phases of aircraft performance analysis.