This preview shows page 1. Sign up to view the full content.
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 xaxis 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∞: Freestream 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 2D 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 BiotSavart 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 freetrailing vortices and bound vortex and use the BiotSavart
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 semiinfinite 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 BiotSavart 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 KuttaJoukowski 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 nondimensional 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 ellipticshaped wing. ~ sin , ...
View Full
Document
 Spring '08
 Sheplak

Click to edit the document details