{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solution4

# solution4 - 650:460 Aerodynamics Homework Assignment No 4...

This preview shows pages 1–7. Sign up to view the full content.

650:460 Aerodynamics Homework Assignment No. 4 Prof. Doyle Knight Tel: 732 445 4464, Email: [email protected] Office hours: Tues and Thur, 4:30 pm - 6:00 pm and by appointment Fall 2009 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Problem 6.1 Show that the vorticity distribution γ ( θ ) = 2 α U 1 + cos θ sin θ satisfies the condition that the flow is parallel to the surface. Show that the Kutta condition is satisfied. Sketch the 2 γ/ U distribution as a function of x / c for a lift coefficient of 0.5. What is the physical significance of 2 γ/ U ? What angle of attack is required for a symmetric airfoil to develop a lift coefficient of 0.5 ? Solution Substituting the vorticity distribution into Eq (6.8), 1 2 π integraldisplay π o parenleftbigg 2 α U 1 + cos θ sin θ parenrightbigg sin θ cos θ cos θ o d θ = U α 2
Problem 6.1 Thus 1 π integraldisplay π o 1 + cos θ cos θ cos θ o d θ = 1 Using Eq (6.10) integraldisplay π o cos n θ d θ cos θ cos θ o = π sin n θ o sin θ o the integral on the left equals 1. Thus, the equation is satisfied. 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Problem 6.1 The Kutta condition requires lim θ π γ ( θ ) = 0 Using L’Hospital’s rule lim θ π γ ( θ ) = 2 α U lim θ π sin θ cos θ = 0 Thus, the Kutta condition is satisfied. The lift per unit depth is l = integraldisplay c o ρ U γ dx and thus ρ U γ represents the lift per infinitesimal chordwise distance per unit depth. 4
Problem 6.1 Since the lift is due to the pressure difference ρ U γ = p l p u where l and u denote lower and upper surfaces. Equivalently ρ U γ = ( p l p ) ( p u p ) and dividing by the dynamic pressure ρ U γ 1 2 ρ U 2 = c p l c p u and thus 2 γ U = c p l c p u 5

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Problem 6.1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern