Question

What is the center pressure if the lift coefficient and lift curve slope of an aerofoil of percentage camber 0.6 are 1.02 and 2, respectively?

• A. 0.2685
• B. 0.6852
• C. 0.8526
• D. 0.2745

Hint:

In aerodynamic calculations, ensure that all values, especially coefficients and slopes, are derived and applied consistently within their respective theoretical frameworks.

Answer 1

The Correct Option is D

Given:  Lift coefficient (\(C_L\)) = 1.02 \item Lift curve slope (\(a_0\)) = 2 per radian  Percentage camber = 0.6% (0.006 in decimal)  

Step 1: Determine the Zero-Lift Angle of Attack (\(\alpha_0\)) For a cambered aerofoil, the zero-lift angle of attack (\(\alpha_0\)) is given by: \[ \alpha_0 = -\frac{2 \times \text{camber}}{a_0} \] Substitute the given values: \[ \alpha_0 = -\frac{2 \times 0.006}{2} = -0.006 \, \text{radians} \] 
 

Step 2: Calculate the Center of Pressure (CP) The center of pressure for a cambered aerofoil is given by: \[ \text{CP} = \frac{1}{4} - \frac{C_{m_0}}{C_L} \] Where:  \(C_{m_0}\) is the moment coefficient about the aerodynamic center. For a cambered aerofoil, \(C_{m_0}\) is typically \(-0.025\).  \(C_L\) is the lift coefficient.  Substitute the values: \[ \text{CP} = \frac{1}{4} - \frac{-0.025}{1.02} \] \[ \text{CP} = 0.25 + 0.0245 \] \[ \text{CP} = 0.2745 \] 

Final Answer The center of pressure is located at: \[ \boxed{27.45\%} \]