EAS 4101 Exam 2 Solutions v4

EAS 4101 Exam 2 Solutions v4 - EAS 4101 – Aerodynamics...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EAS 4101 – Aerodynamics – Spring 2010 03/31/10 Exam #2 You are allowed a pen only. Scrap paper will be provided to you if necessary. The following items are NOT allowed: notes, textbook, calculator, crib sheet, cell phones, PDAs, etc. Honor Code: “On my honor, I have neither given nor received unauthorized aid in doing this exam." All cases of academic dishonesty will be immediately reported to the Dean of Students Office. Academic dishonesty includes copying homework from another student or resource, cheating on an exam or quiz, modifying a graded document to gain more points, etc. Print Name:________________________________ Sign Name:________________________________ Date:_____________________________________ EAS 4101 – Aerodynamics – Spring 2010 03/31/10 Exam #2 Name: ________________________ Formula Sheet The following formulas are provided to help you in this exam. You are expected to know what they are and how they are used. 1 2 , 2 2 1 Γ cos 1 cos 1 cos 4 1 Γ / 2 sin 1 , , ,... 2 2 2 Γ / 1 sin cos 2 1 cos sin , , ,… sin sin , , ,… sin cos sin Γ 4 || cos cos sin sin cos 2 8 2 EAS 4101 – Aerodynamics – Spring 2010 03/31/10 Exam #2 Name: ________________________ I. TRUE/FALSE (12 Points total – 1 point each) 1. The lift force is always perpendicular to the chord line. 2. Kelvin's Theorem states that the time rate of change of circulation around a closed curve consisting of the same fluid elements is constant. 3. Wing tip vortices are driven by the pressure difference between the upper and lower surfaces of a wing. 4. The geometric angle of attack on a wing is always less than the effective angle of attack. 5. Aerodynamic twist refers to a finite wing with airfoil sections that vary along the span 6. The induced drag is not a function of wing geometry 7. The zero lift angle of attack for a symmetric airfoil is always zero. 8. Induced drag coefficient is proportional to . 9. Physically, the lift distribution on a finite wing is symmetric about the x-axis and is infinite at the wingtips. 10. The Kutta condition is naturally enforced at the trailing edge due to friction forces. 11. Induced drag acts parallel to the local relative wind of a finite wing. 12. The aerodynamic center of symmetric and cambered thin airfoils are the same. II. F F T F T F T F F T F T SHORT ANSWER (30 points total) 13. (6 pts total) , find the (a) (4 pts) Given the moment at the leading edge of a symmetric airfoil moment coefficient at an arbitrary location / ( below). Write your answer in terms of the moment coefficient at the leading edge and the lift coefficient. M 'x M'LE x L' L' ∑ , , EAS 4101 – Aerodynamics – Spring 2010 03/31/10 Exam #2 Name: ________________________ (b) (2 pts) Based on the definition of the center of pressure, use your answer to (a) to /. solve for its location, At the center of pressure, the moment coefficient goes to zero. Therefore, , 0 , , 14. (7 pts) Identify which type(s) of airfoil(s) or wing(s) correspond to each coefficient of lift curve in the figure below. Circle your answer(s) in the figure legend. a. Thin, symmetric airfoil b. Thin, cambered airfoil with 2.5° c. Finite wing using the airfoil of (a) d. Finite wing with a cambered airfoil 0.4 -0.05 0 Angle of Attack, (rad) 0.05 0.1 0.15 0.3 Coefficient of lift, c l 0.2 0.1 0 (a) (b) (c) (d) (none of above) (a) (b) (c) (d) (none of above) (a) (b) (c) (d) (none of above) (a) (b) (c) (d) (none of above) (a) (b) (c) (d) (none of above) (a) (b) (c) (d) (none of above) (a) (b) (c) (d) (none of above) -0.1 -0.2 -0.3 -0.4 -5 -2.5 0 2.5 5 Angle of Attack, (deg) 7.5 10 EAS 4101 – Aerodynamics – Spring 2010 03/31/10 Exam #2 Name: ________________________ 15. (7 pts) Find the coefficient of lift cl for a cambered airfoil with parabolic mean camber 1 4 line, , where is a constant. Solution: 12 1 4 cos , cos 14 1 cos 2 2 , 16. (8 pts) Provide the symbols for and brief descriptions of the quantities labeled in the following figure, (a) – (h) , demonstrating the effect of downwash over a local airfoil section of a finite wing. (a) α: geometric angle of attack (b) αeff: effective angle of attack (c) αi: induced angle of attack (d) V∞: Free-stream velocity (e) Local relative wind (f) w: downwash (g) L': Local lift (h) Di: Induced drag EAS 4101 – Aerodynamics – Spring 2010 03/31/10 Exam #2 Name: ________________________ Problem 1: (20 points total) Consider ideal flow over a 2-D airfoil to answer the following questions: a) What assumptions must be made to replace this airfoil by a vortex sheet located on the chord line as shown below? (2 pts) b) If is the component of the velocity normal to the camber line induced by the vortex sheet at a point s, and , is the component of the velocity normal to the camber line due to the freestream velocity (as shown below), what boundary condition must be met to have the camber line be a streamline? (2 pts) c) Using the geometry from the figure above, what is , ? Assume small angles. (2 pts) d) What must be true to allow , where is the total induced velocity at a point x due to the vortex sheet normal to the chord line? (2 pts) EAS 4101 – Aerodynamics – Spring 2010 03/31/10 Exam #2 Name: ________________________ e) Using the figure below, determine . Note the direction of the vortices. (4 pts) f) Using the information above, determine the “fundamental equation of thin airfoil theory.” How does it simplify for a symmetric airfoil? (4 pts) g) In order to solve the “fundamental equation of thin airfoil theory,” derived in part (f), the coordinate transformations 1 cos and 1 cos are used and the general solution for the vortex sheet is found to be 1 2 cos sin sin Show that this solution satisfies the Kutta condition at the trailing edge and provide a physical explanation of its meaning. (4 pts) Solutions: a) Thin airfoil assumption implies that the camber line and the chord line are very close together and therefore the vortex sheet on the mean camber line can be approximated by a vortex sheet of the chord line. b) The normal velocity at all points on the camber line must be zero, i.e.: ′ 0 ∞, c) The normal velocity along the camber line is: ∞, Assuming small angles implies sin to: ∞ sin tan tan and the normal velocity component simplifies EAS 4101 – Aerodynamics – Spring 2010 03/31/10 Exam #2 Name: ________________________ ∞, ∞ d) As in part (a), the airfoil must be thin such that the chord line and the chamber line are approximately on top of each other. Therefore, the component of velocity normal to the chord line is approximately the component of velocity normal to the camber line (i.e. ′ ). e) From the Biot-Savart law, the velocity, is: , at a point induced by a vortex at a point 2 Integrating over the chord of the airfoil to find the velocity elemental vortices along the chord, we find: at a point due to all 2 f) The fundamental equation of thin airfoil theory is found by substituting the results from parts (c) and (e) into the equation found in part(b): ′ 0 ∞, ∞, ∞ 2 0. g) The Kutta condition requires that lim 2 lim Use L'Hopital's rule... lim lim 2 1 cos sin 0 0 sin cos lim 2 0. The Kutta Condition is a physical statement that the velocity must leave the airfoil's trailing edge smoothly. EAS 4101 – Aerodynamics – Spring 2010 03/31/10 Exam #2 Name: ________________________ Problem 2: (15 points total) Consider replacing the finite wing by a single horseshoe vortex filament of strength Γ as shown below. a) Determine the induced velocity distribution the bound vortex (or lifting line). Identify the contribution of the free-trailing vortices and bound vortex and use the Biot-Savart Law to justify your answer for the latter. (5 points) b) Physically, why is replacing a finite wing with a single horseshoe vortex an invalid approach? (5 points) Now consider the superposition of an infinite number of horseshoe vortices along the lifting line that yield a net circulation distribution Γ as shown below. EAS 4101 – Aerodynamics – Spring 2010 03/31/10 Exam #2 Name: ________________________ c) Why is this approach an appropriate alternative to a single horseshoe vortex? Please discuss the differences between this approach and the single horseshoe vortex approach. (5 points) Solutions: a) Both semi-infinite vortex filaments in this problem induce a downward velocity along the bound vortex (which induces no downward velocity along itself). The induced velocity along the bound vortex due to the two trailing vortices is: Γ Γ 4 2 4 Γ 4 2 2 The contribution of the bound vortex to the downwash is zero. It has been frequently said that "the bound vortex induces no velocity along itself." The reason for this is that both and are parallel along the bound vortex and thus the cross product in the numerator of the Biot-Savart Law is zero. b) This approach is invalid because as 2, the induced velocity approaches infinity. This is not physically possible. c) A vortex sheet is an appropriate alternative because it allows the lift to go to zero at the wingtips, which is a physically valid condition. This is not the case when using a single horseshoe vortex which results in the physically invalid condition of an infinite induced velocity at the wingtips. EAS 4101 – Aerodynamics – Spring 2010 03/31/10 Exam #2 Name: ________________________ Problem 3: (25 points total) Consider an elliptical circulation distribution on a finite wing of wingspan . a) The circulation distribution is given by Γy Γ 1 2 where Γ is the circulation at the center of the wing as shown in the figure below. Calculate the lift per unit span . Comment on whether or not the lift at the wingtips (i.e. at ) is physically valid. (4 pts) b) Calculate the total lift coefficient, , making use of the coordinate transformation cos θ. Leave your answer in terms of the constants, Γ and . (6 pts) c) The downwash is found to be constant for an elliptic wing and has a value: Γ 2 Using this result, calculate the induced angle of attack, , assuming small angles.(4 pts) d) Determine the induced drag coefficient for the finite elliptic wing. Express your answer in terms of and . (6 pts) The generalized Fourier series for the circulation distribution over a finite wing is Γ 2 ∑ , , ,... sin . EAS 4101 – Aerodynamics – Spring 2010 03/31/10 Exam #2 Name: ________________________ Determine the value of the (2 pts) coefficients if the circulation is written as a Fourier series. e) Explain why a finite wing with no aerodynamic or geometric twist must be of an elliptic shape in order to generate the elliptic circulation distribution given in (a). Asked another way, why must the chord be given as c y 1 c , where is the chord at the center of the elliptic wing? (Hint: Refer to the Monoplane Equation.) (3 pts) Solutions: a) The lift per unit span is related to the circulation by the Kutta-Joukowski theorem: Γ Γ 2 1 The lift per unit span at the wingtips (i.e. 2) is zero, which is required in order for the pressure on the top and bottom of the wing to be matched at the wingtips. b) First, we must calculate the total lift on the wing by integrating the lift per unit span over the entire span of the wing: ⁄ Γ 1 2 ⁄ In order to perform the integral, we will use the given coordinate transformation. The resulting integral is: ⁄ Γ 1 ⁄ 2 2 cos θ sin 2 Γ 4 The lift coefficient is defined as: Γ 2 sin Γ 1 2 4 Γ 2 1 2 Γ 2 c) The induced angle of attack is angle between the freestream and downwash velocity vectors: EAS 4101 – Aerodynamics – Spring 2010 03/31/10 Exam #2 Name: ________________________ tan For small angles of , the induced angle of attack can be approximated by: From part (b), the circulation at the center of the wing, Γ , is: Γ 2 Γ 2 Therefore, the downwash and induced angle of attack are expressed in terms of the lift coefficient: Γ 2 The aspect ratio is defined as: AR Therefore, the induced angle of attack can be written as: AR d) The induced drag coefficient is found from From answer (b), Γ 4 and now writing in terms of the non-dimensional coefficients, Γ Γ 1 1 2 4 2 2 AR e) The transformed elliptic distribution is simply Γ Γ sin θ, so it is clear by inspection 2bV A and all other coefficients are zero. of the Fourier series form that Γ Γ A 2bV f) With no geometric or aerodynamic twist, both and must be constants (i.e. not functions of ). Using this information and the values of the coefficients we found in the previous problem, the monoplane equation becomes EAS 4101 – Aerodynamics – Spring 2010 03/31/10 Exam #2 Name: ________________________ 2 sin 1 The only way to satisfy this equation such that no twist is required is for which is the case for an elliptic-shaped wing. ~ sin , ...
View Full Document

This note was uploaded on 09/05/2011 for the course EAS 4101 taught by Professor Sheplak during the Spring '08 term at University of Florida.

Ask a homework question - tutors are online