Aerodynamics - Incompressible Flow over Airfoils Quiz(MCQ)

1 .

What is a chord?

A)

Distance between chord and chamber

B)

Distance between leading edge and chord

C)

Distance between trailing edge and chord

D)

Distance between leading edge and trailing edge

Correct Answer : Distance between leading edge and trailing edge

Explanation : Distance between leading edge and the trailing edge is called chord. Chord refers to the imaginary straight line joining the leading edge and trailing edge of an airfoil. The chord length is the distance between the trailing edge and the point on the leading edge, where the chord intersects the leading edge.

2 .

Purpose of leading edge is to ________

A)

allow the wing to operate at stall condition

B)

allow the wing to operate in level condition

C)

allow the wing to operate at high angle of attack

D)

allow the wing to operate at low angle of attack

Correct Answer : allow the wing to operate at high angle of attack

Explanation : Leading edge is a part of the airfoil. Leading edge allows the wing to operate at a high angle of attack. Slats are placed at the leading edge. These are aerodynamic surfaces on the leading edge of the wing. These are high lift devices used for short takeoff.

3 .

Who designed the nomenclature of airfoil?

A)

NACA series

B)

EULAR series

C)

CLARK series

D)

EPPLER series

Correct Answer : NACA series

Explanation : The nomenclature of airfoil is designed by the NACA series. The shape of the NACA airfoil is described using a series of digits following the word NACA. The NACA identified different airfoils shape with a logical numbering system, such as symmetric airfoil and cambered airfoil.

4 .

Purpose of camber in an airfoil is _______

A)

to decrease maximum lift

B)

to increase maximum lift

C)

to increase maximum drag

D)

to decrease maximum drag

Correct Answer : to increase maximum lift

Explanation : The main purpose of camber is to increase the maximum lift in an airfoil. The maximum lift coefficient can get by increasing the camber in an airfoil. Some recent design use negative camber. That airfoil is called the supercritical airfoil. This type of airfoil is used in the supersonic flight and to produce a higher lift to drag ratio.

5 .

Why NACA 0012 airfoil is said to be symmetric airfoil?

A)

Because of camber

B)

Because of chord line

C)

Because of no camber

D)

Because of no thickness

Correct Answer : Because of no camber

Explanation : The NACA 0012 airfoil indicates the symmetric airfoil, because an airfoil with no camber, that is with the camber line and chord line co-incident in an airfoil is called a symmetric airfoil. For the NACA 0012 airfoil is with a maximum thickness of 12%.

6 .

What is the thickness in NACA 747A315 airfoil?

A)

7%

B)

15%

C)

25%

D)

30%

Correct Answer : 15%

Explanation : In NACA 747A315 airfoil. The 7 denotes the series, 4 provides the location of minimum pressure on the upper surface in a tenth of chord, 7 provides the location of the minimum pressure on the lower surface in tenth of chord, the 4th character i.e. A indicates the thickness distribution and mean line form used. Again 5th digit 3 indicates the design lift coefficient in the tenth and the final two digits indicate the thickness of the chord 15%.

7 .

What is the purpose of trailing edge?

A)

Vortex are created

B)

Airflow separated

C)

Stalling will created

D)

Airflow rejoins

Correct Answer : Airflow rejoins

Explanation : Trailing edge is a part of airfoil. Trailing edge of an aerodynamic surface such as awing is its rear edge. Where the airflow will separate at the leading edge and rejoins at the trailing edge. Ailerons are placed at the trailing edge of the wing and rudder, elevator is placed at the trailing edge of tail.

8 .

What is the shape of the airfoil at the leading edge?

A)

Circular

B)

Curve

C)

Straight

D)

Semi-circular

Correct Answer : Circular

Explanation : The shape of the airfoil at the leading edge is usually circular with a leading edge radius of approximately 0.02c. The shape of all standard NACA airfoils are generated by specifying the shape of the mean chamber line and then wrapping specified symmetric thickness distribution around the mean camber line.

9 .

What is the thickness in NACA 2412 airfoil?

A)

12%

B)

24%

C)

41%

D)

53%

Correct Answer : 12%

Explanation : The NACA 2412 airfoil has a maximum camber of 2% located at 40% from the leading edge with a maximum thickness of 12%. It was the first 4-digit NACA series and it was developed in 1930’s because of the 12% thickness the shape of the airfoil is symmetric.

10 .

What parameter is used to decrease the stall speed?

A)

Chord

B)

Camber

C)

Chord line

D)

Camber line

Correct Answer : Camber

Explanation : Camber is generally used to increase the maximum lift coefficient. Which in turn it decreases the stall speed of the aircraft. Due to decrease in the stall there will be an increase in the lift. In supercritical airfoil. There will be a highly cambered curve after section which used in supersonic flight for their maximum lift.

11 .

How lift and drag ratio can be expressed in a relation?

A)

Dividing the drag coefficient by the lift coefficient

B)

Dividing the lift coefficient by the moment coefficient

C)

Dividing the lift coefficient by the drag coefficient

D)

Dividing the drag coefficient by the moment coefficient

Correct Answer : Dividing the lift coefficient by the drag coefficient

Explanation : The lift-drag ratio is used to express the relation between lift and drag, and is obtained by dividing the lift coefficient by the drag coefficient. We can get the maximum lift coefficient by the drag coefficient for a given airfoil. The characteristics of any airfoil section can conveniently be represented by a graph showing the lift-drag ratio.

12 .

What is the graph that is represented in the airfoil section?

A)

Lift-moment ratio

B)

Lift–angle of attack ratio

C)

Angle of attack-drag ratio

D)

Coefficient of lift-coefficient of drag ratio

Correct Answer : Coefficient of lift-coefficient of drag ratio

Explanation : The characteristics of any particular airfoil section can conveniently be represented by a graph, showing the amount of lift and drag obtained at a various angle of attack, the lift-drag ratio, and the movement of the center of pressure. With these graphs, we can choose suitable airfoil for the aircraft.

13 .

Define stalling angle in the graph.

A)

The curve remains constant at all section

B)

Lift curve will increases and drag curve will remain constant

C)

The curve increases at different section and decreases rapidly at different section

D)

The curve increases maximum at a particular section and decreases rapidly in the same section

Correct Answer : The curve increases maximum at a particular section and decreases rapidly in the same section

Explanation : The lift curve reaches its maximum for any wing section at certain degree of angle of attack and then rapidly decreases at some degree of angle of attack is therefore the stalling angle. Stalling will occurs at particular degree were the maximum lift will be achieved by the aircraft at that point stalling will takes place.

14 .

Define drag curve.

A)

Drag curve remains constant and lift curve will be increasing

B)

Lift curve remains constant and drag curve will be increasing

C)

Drag will increase rapidly at particular degree of angle of attack and overcomes the lift curve at particular degree of angle of attack

D)

Drag curve will decrease at particular degree of angle of attack and lift curve increasing at particular degree of angle of attack

Correct Answer : Drag will increase rapidly at particular degree of angle of attack and overcomes the lift curve at particular degree of angle of attack

Explanation : The drag curve increases very rapidly at a certain degree of angle of attack and completely overcomes the lift curve at some degree of angle of attack, at this point maximum drag will be caused.so, at this point lift will drastically decrease and hence producing the drag over the body.

15 .

How propeller creates a thrust force?

A)

With own power

B)

With supplied power

C)

With transmitted power

D)

With the existing power

Correct Answer : With supplied power

Explanation : A propeller creates a thrust force out of the supplied power. The magnitude of this force is not for a given propeller but depends on the velocity of the incoming air and the rotational velocity of the propeller and helps the propeller to create the forward thrust.

16 .

Is airfoil shape similar to blades of the propeller?

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : True

Explanation : An airfoil is the shape of a wing, blade of a propeller, rotor, and blades of a turbine. A propeller is a type of fan that transmit power by converting rotational motion in its thrust.A pressure difference is produced between the forward and rear surface of the airfoil-shape and fluid is accelerated behind the blade.

17 .

How vortex filament are arrange in the vortex sheet?

A)

Side by side

B)

Up and down

C)

Left and right

D)

Front and back

Correct Answer : Side by side

Explanation : The flow in the planes perpendicular to the vortex filament at zero are identical to the flow induced by the point vortex of strength. An infinite number of straight vortex filaments side by side. Where the strength of each filament is infinite small. These vortex filament are arranged side by side from the vortex sheet.

18 .

At what point circulation between two material points in the sheet remains conserved.

A)

One

B)

Zero

C)

Less than zero

D)

Greater than zero

Correct Answer : Zero

Explanation : Kelvin’s circulation theorem, is the absence of external forces on the sheet, the circulation between any two material points in the sheet remains conserved at zero. The equation of motion of the sheet can be rewritten in terms of circulation and by a change of variable.

19 .

Purpose of vortex method is _________

A)

To reduce the lift of aerodynamic model

B)

To increase the lift of aerodynamic model

C)

To reduce the dimensionality of aerodynamic model

D)

To increase the dimensionality of aerodynamic model

Correct Answer : To reduce the dimensionality of aerodynamic model

Explanation : Vortex method is extensively applied to reduce the dimensionality of these aerodynamic models based on the proper estimation of the strength and distribution of the vortices in the wake this condition can be readily applied to the flat plate or an airfoil with a cusped trailing edge were the flow will leave smoothly.

20 .

Is vortex sheet can go beyond the critical time?

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : False

Explanation : The vortex sheet solution as given by the birkoff-roff equation cannot go beyond the critical time. The spontaneous loss of analyticity in a vortex sheet is a consequence of mathematical modeling since a real fluid with viscosity, however, small will never develop singularity.

21 .

Is Kutta condition is applicable to solid bodies with sharp corners?

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : True

Explanation : The kutta condition is a principle in steady-flow fluid dynamics, especially aerodynamics that is applicable to solid bodies with sharp corners, such as the trailing edge of the airfoil. It is named for German mathematician and aerodynamicist martin Wilhelm kutta.

22 .

Is kutta condition apply to oval shaped body?

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : True

Explanation : When a smooth symmetric body, such as a cylinder with oval cross-section moves with zero angle of attack through a fluid it generates no lift. There are two stagnation points on the body one at the front and the other at the back. Since no lift will be generated by the cylinder at zero angle of attack.

23 .

Vortex flow occurs at _________

A)

Chord

B)

Chamber line

C)

Leading edge

D)

Trailing edge

Correct Answer : Trailing edge

Explanation : Vortex flow occurs at the trailing edge and because the radius of the sharp trailing edge is zero, the speed of the air around the trailing edge should be infinitely fast, real fluid cannot move at infinite speed, they can move extremely fast finite speed.

24 .

How stagnation point on the topside of the airfoil reaches the trailing edge?

A)

Due to drag forces

B)

Due to surface forces

C)

Due to viscous forces

D)

Due to pressure forces

Correct Answer : Due to viscous forces

Explanation : The stagnation point on the topside of the airfoil then moves until it reaches the trailing edge. The starting vortex eventually dissipates due to viscous forces. As the airfoil continues on its way, there is a stagnation point at the trailing edge. The flow over the topside of the airfoil conforms to the upper surface of the airfoil.

25 .

How airfoil increases its speed?

A)

Flow over the leading edge

B)

Flow over the topside

C)

Flow over the bottom side

D)

Flow over the trailing edge

Correct Answer : Flow over the topside

Explanation : As the vorticity increases the bound vortex and also progressively increases and causes the flow over the topside of the airfoil to increase in speed. The starting vortex is soon cast-off the airfoil and is left behind, spinning in the air, where the airfoil left it.

26 .

How starting vortex is formed?

A)

As the airfoil begins to move vortex are formed

B)

As the airflows on the airfoil vortex are formed

C)

As the airflows on the circular body vortex are formed

D)

As the airfoil moves against the relative wind vortex are formed

Correct Answer : As the airfoil begins to move vortex are formed

Explanation : As the airfoil begins to move it carries this vortex, known as the starting vortex along with it, pioneering aerodynamicists were able to photograph starting vortices in liquids to confirm their existence. The vorticity in the starting vortex is matched by the vorticity in the bound vortex in the airfoil.

27 .

Which of the following ensures flow smoothly leaving the trailing edge given the right value of circulation?

A)

Angle of Attack

B)

Kutta Condition

C)

Momentum Theorem

D)

The Shape of the Airfoil

Correct Answer : Kutta Condition

Explanation : According to the Kutta Condition, for a given angle of attack, the value of circulation around the airfoil is such that the flow leaves the trailing edge smoothly. The velocity at the trailing edge is dependent on the shape of the airfoil.

28 .

For an arbitrary inviscid and incompressible flow, with all the body forces zero, what is best described by the given sketch?

A)

Kutta Condition

B)

Generation of Lift

C)

Boundary Layer Formation

D)

Kelvin’s Circulation Theorem

Correct Answer : Kelvin’s Circulation Theorem

Explaination : For the same fluid elements in closed curves C1 and C2, the circulation remains constant with time as the fluid elements move downstream. This is essentially Kelvin’s circulation theorem.

29 .

For which of the following Kelvin's theorem is applicable?

A)

Compressible Flow

B)

Flow with Viscous Stresses

C)

Inviscid, Compressible Barotropic Flow

D)

Flow with Non-Conservative Body Forces

Correct Answer : Inviscid, Compressible Barotropic Flow

Explanation : Kelvin’s Theorem is applicable for the special case of barotropic flow while dealing with inviscid, compressible flows.

30 .

Generation of lift is accompanied by a starting vortex at the trailing edge. If the flow is inviscid, this will not happen. What reason can best describe this?

A)

Kutta Condition is enforced

B)

Kelvin’s Theorem is violated

C)

Starting Vortex dies off instantly

D)

There is no boundary layer formation, hence no vorticity

Correct Answer : There is no boundary layer formation, hence no vorticity

Explaination : For inviscid flows, the boundary layer is not formed. Therefore, in the regions of high velocity, high viscosity is not there and hence no vortex can form. Thus, there is no lift produced. Starting vortex cannot form in inviscid medium and in the viscous medium it dies due to viscosity.

31 .

Which of these is a result of Kelvin's Theorem is essentially?

A)

Lift

B)

Vorticity

C)

Circulation

D)

Frozen Vortex Lines

Correct Answer : Frozen Vortex Lines

Explanation : Kelvin’s theorem can be used to prove Helmholtz theorems, one of which says ‘vortex lines move with the fluid’ which is what is known as “frozen vortex lines”.

32 .

Which type of the following flow is characterized by density being a single-valued function of pressure only?

A)

Inviscid Flow

B)

Barotropic Flow

C)

Viscous Flow

D)

Baroclinic Flow

Correct Answer : Barotropic Flow

Explanation : A barotropic flow is a fluid where density is a function of pressure only, i.e. ρ = ρ(p). Baroclinic flow is the fluid which is not only dependent on the pressure but on other factors also. Viscous and inviscid flows are not necessarily dependent on pressure only.

33 .

For a fluid initially at rest, the formation of starting vortex implies ______

A)

no lift is produced

B)

generation of lift

C)

generation of circulation

D)

generation of lift and circulation

Correct Answer : generation of lift and circulation

Explanation : From Kelvin’s Theorem, circulation remains constant with time. So for initial zero circulation, the formation of starting vortex means there has to be equal and opposite circulation in the form of lift.

34 .

Generation of lift over an airfoil and formation of starting vortex is correctly explained by which of these?

A)

Kutta Condition and Kelvin’s Theorem

B)

Kutta-Joukowski Theorem

C)

Kutta Condition and Helmholtz Theorem

D)

Kutta-Joukowski Theorem and Kelvin’s Theorem

Correct Answer : Kutta Condition and Kelvin’s Theorem

Explanation : Kutta condition enforces smooth flow at the trailing edge. In doing so high velocity gradients formed at the trailing edge generates vorticity and hence circulation is there. From Kelvin’s circulation theorem starting vortex is formed to conserve circulation.

35 .

In reality, the starting vortex dies out. Why?

A)

Due to Viscosity

B)

Lift becomes zero

C)

At later times, Kelvin’s theorem is not applicable

D)

This assumption is wrong. Starting vortex never dies

Correct Answer : Due to Viscosity

Explanation : Starting vortex cannot form in inviscid medium. It can form only in a viscous medium. In a viscous medium, it dies instantly due to viscous effects.

36 .

Is symmetric airfoil with good lift to drag ratio is used for an aircraft wing?

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : True

Explanation : The symmetric airfoil with good lift to drag ratio is used for an aircraft wing, this will be translated into lower fuel consumption, shorter take-off and landing times, and shorter runways. Airfoil With backward facing will get high lift coefficient.

37 .

Is NACA 0012 is symmetric airfoil?

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : True

Explanation : The NACA 0012 is symmetric airfoil because the mean camber line and chord line intersect in the same line.so, there is no camber in the NACA 0012 airfoil. That the reason the first and second digits are will become zero, and the thickness of NACA 0012 airfoil is 12 percent.

38 .

What is the thickness of NACA 0002 airfoil?

A)

0%

B)

1%

C)

2%

D)

3%

Correct Answer : 2%

Explanation : The thickness of the NACA 0002 airfoil is 2 percent. There is no camber in the NACA 0002 airfoil because of the mean camber line and chord line intersect in the same line, so the position maximum camber in the NACA 0002 airfoil is zero.

39 .

How many types the wake structure is classified?

A)

2

B)

3

C)

4

D)

5

Correct Answer : 5

Explanation : The wake structure is classified into five different modes according to their pattern obtained from instantaneous and mean vorticity fields by also taking into account the amplitude spectrum of the lift coefficient.

40 .

How lift is calculated?

A)

Parallel to the direction of motion

B)

opposite to the direction of motion

C)

relative to the direction of motion

D)

Perpendicular to the direction of motion

Correct Answer : Perpendicular to the direction of motion

Explanation : An airfoil shaped body moved through a fluid produces an aerodynamic force. The component of this force perpendicular to the direction of motion is called lift. The lift will oppose the motion and it produces the lift.

41 .

A vortex sheet placed along the chord line gives the best representation for______

A)

Thick Airfoil

B)

Thin Airfoil

C)

NACA Airfoils

D)

Negatively Cambered Airfoil

Correct Answer : Thin Airfoil

Explanation : Thin airfoils can be simulated as a vortex sheet kept along the camber line. Thick airfoils cannot be represented like this. Negatively cambered airfoils may or may not be thin and in that case, this representation might be incorrect. NACA airfoils have all types of airfoils, not thin airfoils only.

42 .

What is the Kutta Condition in terms of strength (γ) of the thin airfoil?

A)

γ(TE) = 0

B)

γ = 0

C)

γ(LE) = 0

D)

\(\frac {D\gamma }{Dt}\)=0

Correct Answer : γ(TE) = 0

Explaination : According to the Kutta condition, the condition for flow leaving smoothly is satisfied always by γ(TE) = 0. The other options may not be true as per the Kutta Condition.

43 .

Given the component of free-stream velocity in the direction perpendicular to the camber line is 15 units and the velocity induced by the vortex sheet is 4 units at a point. The angle of attack for the thin airfoil is α. Then which of the following condition is true in case of thin airfoil theory?

A)

15+4=0

B)

15tanα-4=0

C)

15sinα+4=0

D)

15cosα+4=0

Correct Answer : 15+4=0

Explaination : In thin airfoil theory, the camber line has to be a streamline. Thus, the component of velocity in the normal direction to the camber line is zero at any given point. This means the sum of velocity components by the free stream velocity and the induced velocity is zero.

44 .

The thin airfoil theory assumes the vortex sheet is kept along the chord line and the camber line acts as a________

A)

Vortex Filament

B)

Streamline

C)

Dividing Streamline

D)

Another Vortex Sheet

Correct Answer : Streamline

Explanation : For thin airfoils, camber and chord lines are very close and the vortex sheet can be assumed to be placed at the chord line. And the camber line becomes a streamline.

45 .

Identify the missing angle given by the question mark for the following camber distribution of a thin airfoil.

A)

tan^-1?α

B)

tan^-1\(\frac {dz}{dx}\)

C)

tan^-1?-\(\frac {dz}{dx}\)

D)

tan^-1–\(\frac {dz}{dx}\)+α

Correct Answer : tan^-1?-\(\frac {dz}{dx}\)

Explaination : For a camber distribution where z=z(x), at a given point on the camber line the missing angle is the slope at that point, which is tan^-1?-\(\frac {dz}{dx}\) which is independent of α.

46 .

Identify 'a' as shown in the figure for a thin airfoil.

A)

Camber

B)

Chord Line

C)

Neither Chord Line neither Camber

D)

Camber and Chord Line

Correct Answer : Camber and Chord Line

Explaination : For a symmetric airfoil(\(\frac {dz}{dx}\)=0), camber line is the same as chord line. Therefore, the best option is Camber and Chord Line. Only Camber or Chord Line is an incomplete answer.

47 .

The value of integral ∫\(_0^?\frac {cos?n\theta d\theta }{cos?\theta-cos∅}\)=π\(\frac {sin? n∅}{sin ?∅}\) is valid for the limit 0 to_____

A)

π

B)

–π

C)

2π

D)

nπ

Correct Answer : π

Explaination : This is a standard integral which is used many times in the study of airfoils. This is used to solve the transformed fundamental equation in thin airfoil theory where the chord length is expressed as θ=0 to θ=π.

48 .

The Kutta Condition according to the thin airfoil theory is_____

A)

γ(x)=0

B)

γ(c)=0

C)

γ(ξ )=0

D)

\(\frac {dz}{dx}\)=0

Correct Answer : γ(c)=0

Explaination : The Kutta Condition says the flow leaves the trailing edge smoothly. The trailing edge is at chord length c for a thin airfoil which gives γ(c)=0. Also, γ(x)=0 or γ(ξ)=0 is not true for all x or ξ. \(\frac {dz}{dx}\)=0 is the definition of the symmetric airfoil, not Kutta condition.

49 .

Which of the following is not correct for symmetric airfoil according to the fundamental equation of thin airfoil in transformed coordinates?

A)

γ(θ)=0

B)

0≤θ≤π

C)

\(\frac {dz}{dx}\)=0

D)

ξ=\(\frac {c}{2}\)(1-cosθ)

Correct Answer : γ(θ)=0

Explaination : The transformation for thin airfoil theory uses ξ=\(\frac {c}{2}\)(1-cosθ) to make the coordinate transformation, where 0≤θ≤π. For a symmetrical airfoil (\(\frac {dz}{dx}\)=0), while γ(θ)=0 is not true for all values of θ.

50 .

The fundamental equation for the thin airfoil theory does not encompass which of the following approximations?

A)

Vortex sheet is placed at the chord line

B)

The angle of attack and slope of the camber line is small

C)

Kutta condition is satisfied at the trailing edge

D)

Camber line induced velocity distribution is the same for chord line

Correct Answer : Kutta condition is satisfied at the trailing edge

Explanation : Kutta condition is always satisfied at the trailing edge. This is the boundary condition for the camber line to be a streamline also and is not an approximation. While the other three statements are valid approximations, which have been made under the assumption of a thin airfoil where camber and chord line is very close.

51 .

Which is the component of free-stream velocity normal to the camber line for the given thin airfoil?

A)

i

B)

ii

C)

iii

D)

V∞

Correct Answer : ii

Explaination : V∞ is the free-stream velocity. Here i is not a component of free-stream velocity, iii is the parallel component and ii is the normal component for the camber line.

52 .

Is airfoil with high camber produce the maximum lift?

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : True

Explanation : Camber is usually designed into an airfoil is to increase the maximum lift coefficient. This minimizes the stalling speed of aircraft using airfoil. Aircraft with wings based on cambered airfoils usually have low lower stalling speeds than similar aircraft with wings based on symmetric airfoils.

53 .

How to increase the critical angle of attack?

A)

Reduce the camber

B)

Increase the camber

C)

Reduce the thickness

D)

Increase the thickness

Correct Answer : Reduce the camber

Explanation : An aircraft designer may also reduce the camber of the outboard section of the wings to increase the critical angle of attack at the wing tips. When the wing approaches the stall angle this will ensure that the wing root stalls before the tip, giving the aircraft resistance to spinning and maintaining ailerons effectiveness close to the stall.

54 .

What is reflexed camber airfoil?

A)

Chord is at stationary

B)

Camber and chord are at stationary

C)

Chord line curves backup near the trailing edge

D)

Camber line curves backup near the trailing edge

Correct Answer : Camber line curves backup near the trailing edge

Explanation : An airfoil where the camber line curves back up near the trailing edge is called a reflexed camber airfoil. Such an airfoil is useful in certain situations, such as with tailless aircraft, because the moment about the aerodynamic center of the airfoil is zero.

55 .

Is camber of an airfoil causes an increase in velocity?

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : True

Explanation : The camber of an airfoil causes an increase in velocity and a consequent decrease in pressure of the stream of air moving over it due to the increase in velocity it gets a maximum coefficient of lift. Hence it is used in military aircraft.

56 .

The camber line is not a streamline of flow for a cambered airfoil according to the thin airfoil theory.

A)

Always true

B)

Always false

C)

True only for thin airfoils

D)

Depends on the camber distribution

Correct Answer : Always false

Explanation : The thin airfoil theory solution when subjected to the Kutta condition makes the camber line as a streamline of the flow, irrespective of the airfoil being symmetrical or cambered.

57 .

Which of the following is incorrect for a thin, cambered airfoil?

A)

The angle of attack is small

B)

Vortex sheet is kept at the chord line

C)

The slope of the camber line is zero

D)

The induced velocity distribution for the camber line is the same for the chord line

Correct Answer : The slope of the camber line is zero

Explanation : Slope of the camber line \(\frac {dz}{dx}\) is not zero for a cambered airfoil but is some finite value. All the other statements are valid assumptions of thin airfoil theory.

58 .

The equation \(\frac {1}{2\pi } \int_0^c \frac {\gamma(\xi)d\xi}{x-\xi}\)=V∞α is called the fundamental equation of thin airfoil theory for______

A)

All thin airfoils

B)

Cambered airfoils only

C)

Symmetric airfoils only

D)

Symmetric and positively cambered airfoils

Correct Answer : Symmetric airfoils only

Explaination : The original fundamental equation of thin airfoil theory is \(\frac {1}{2\pi } \int_0^c \frac {\gamma(\xi)d\xi}{x-\xi}\)=V∞(α-\(\frac {dz}{dx}\)). For the symmetric airfoils, \(\frac {dz}{dx}\)=0 and so \(\frac {1}{2\pi } \int_0^c \frac {\gamma(\xi)d\xi}{x-\xi}\)=V∞α is valid. While for the cambered airfoils \(\frac {dz}{dx}\) is non-zero.

59 .

The extra term appearing in the thin airfoil theory solution for a cambered airfoil is a______

A)

Constant

B)

Full Fourier series

C)

Fourier sine series

D)

Fourier cosine series

Correct Answer : Fourier sine series

Explanation : The cambered airfoil solution of the thin airfoil theory is different from that of symmetric airfoils with the addition of a Fourier sine series term.

60 .

For NACA4313 what is the maximum camber and the position of maximum camber from the leading edge respectively is______

A)

0.04c, 0.4c

B)

0.4c, 0.03c

C)

0.04c, 0.03c

D)

0.13c, 0.4c

Correct Answer : 0.04c, 0.4c

Explanation : For NACA 4 digit airfoils the first digit gives the maximum camber in 100^th parts of the chord length c and the second digit gives the position of maximum camber in 10^th parts of the chord length from the leading edge.

61 .

NACA 0023 is_____

A)

Thin cambered airfoil

B)

Symmetrical airfoil

C)

Positively cambered airfoil

D)

Negatively Cambered Airfoil

Correct Answer : Symmetrical airfoil

Explanation : The first two digits in the NACA nomenclature give the maximum camber and position of maximum camber. For a symmetric airfoil, both of these are zero.

62 .

Select the statement which is not true for the solution of \(\frac {1}{2\pi } \int_0^c \frac {\gamma(\xi)d\xi}{x-\xi}\)=V∞(α-\(\frac {dz}{dx}\)) for a cambered airfoil.

A)

A₀ depends on the angle of attack

B)

A₀ depends on the slope of the camber line

C)

A_n depends on the slope of the camber line

D)

A_n depends on chord length of the airfoil

Correct Answer : A_n depends on chord length of the airfoil

Explaination : The given equation is the fundamental equation of thin airfoil theory. For a cambered airfoil, the solution is in the form γ(θ)=2V_∞(A₀\(\frac {1+cos\theta }{sin?\theta }\) + Σ\(_{n=1}^∞\)sin? nθ A_n) where A_n depends on the slope of camber line and A₀ depend both on the slope of camber line and angle of attack.

63 .

The correct formula for the Fourier sine series appearing in the solution of thin airfoil theory is_____

A)

A_n=\(\frac {2}{\pi }\int_0^{\pi }\frac {dz}{dx}\) cos?n∅ d∅

B)

A_n=\(\frac {1}{\pi }\int_0^{\pi }\frac {dz}{dx}\) cos?n∅ d∅

C)

A_n=\(\frac {2}{\pi }\int_0^{2\pi }\frac {dz}{dx}\) cos?n∅ d∅

D)

A_n=α-\(\frac {1}{\pi }\int_0^{\pi }\frac {dz}{dx}\) cos?n∅ d∅

Correct Answer : A_n=\(\frac {2}{\pi }\int_0^{\pi }\frac {dz}{dx}\) cos?n∅ d∅

Explaination : From the general solution of thin airfoil theory we have A_n=\(\frac {2}{\pi }\int_0^{\pi }\frac {dz}{dx}\) cos?n∅ d∅ and A₀=α-\(\frac {1}{\pi }\int_0^{\pi }\frac {dz}{dx}\) cos?n∅ d∅ where the limits are 0≤∅≤π.

64 .

The lift per unit span for a thin, cambered airfoil with Γ=10\(\frac {m^2}{s}\), ρ∞=1.0255\(\frac {kg}{m^3}\), V∞=10\(\frac {m}{s}\) is____

A)

B)

102.55N

C)

55\(\frac {N}{m}\)

D)

102.55\(\frac {N}{m}\)

Correct Answer : 0

Explaination : The lift per unit span is given by the formula L’=Γρ_∞V_∞ by Kutta-Joukowski Theorem. Putting the respective values given in the question, L’=102.55\(\frac {N}{m}\) (unit is N/m, not N).

65 .

The standard NACA airfoils were based on ____

A)

Theoretical data

B)

Experiment data

C)

Computer modeling results

D)

Numerical techniques and wind- tunnel testing

Correct Answer : Experiment data

Explanation : The earlier standard NACA airfoils were based exclusively on the experimental results from 1930s-40s. Later on, numerical techniques using the computer were used followed by wind-tunnel testing to develop modern airfoils.

66 .

Modern low-speed airfoils were developed using numerical methods using the computer directly. The given statement is_____

A)

True

B)

False

C)

Incomplete

D)

Partially true

Correct Answer : Partially true

Explanation : The modern low-speed airfoils were developed using numerical techniques on the computer which was followed by wind- tunnel testing to confirm the computer results. This gave the definite airfoil properties for the new airfoils.

67 .

The main motive behind redesigning standard airfoils to get the new airfoils was higher maximum lift coefficient and______

A)

Lesser drag

B)

Better shapes

C)

Strength of material

D)

Handling flow separation effects

Correct Answer : Handling flow separation effects

Explanation : The use of computers led to the design of better airfoils since it made possible to get the definitive properties of the airfoils. This had many advantages like a higher coefficient of lift and shape to tackle the flow separation effects at high angles of attack.

68 .

Which of the following is a new NACA airfoil?

A)

NACA 0021

B)

NACA 2212

C)

NACA LS (1)-0407

D)

Both NACA LS (1)-0407 and NACA 2412

Correct Answer : NACA LS (1)-0407

Explanation : The new airfoils are the low-speed airfoils (designated by LS). So NACA LS (1)-04XX are the new airfoils while the NACA XXXX are the standard airfoils.

69 .

The NACA LS (1) – 04XX airfoils when compared to NACA airfoils with same thickness had higher L/D ratios. For a lift coefficient of 1.0, what was this increase approximately in percentage?

A)

Less than 10%

B)

20%

C)

50%

D)

None of the above

Correct Answer : 50%

Explaination : The lift coefficient of 1 is vital for the aviation sector. The new low-speed airfoils developed had higher L/D ratios. For a lift coefficient equal to 1, the increase was about 50%.

70 .

The major breakthrough in high- speed airfoil industry was_____

A)

Supercritical airfoil

B)

GA (W)-1 airfoil

C)

Symmetrical airfoils

D)

Standard NACA airfoils

Correct Answer : Supercritical airfoil

Explanation : The GA (W)-1 (also known as Whitcomb airfoil) airfoil was the first low-speed airfoil obtained under the new airfoils. It led to the development of supercritical airfoils which had an almost similar shape. The supercritical airfoils had lesser drag at high subsonic speeds, which was a major performance improvement.

71 .

Which is not a feature of the first new NACA airfoil (GA(W)-1 or the Whitcomb airfoil) developed?

A)

Cusped trailing edge

B)

Higher symmetry

C)

Large leading-edge radius

D)

Higher maximum lift coefficient

Correct Answer : Higher symmetry

Explanation : The larger leading edge radius gave a reduced peak in pressure coefficient at the leading edge. The trailing edge was cusped which increased camber, thus decreasing symmetry. These features reduced flow separation and gave a higher value of maximum lift coefficient.

72 .

Enlist the computer methods used which benefitted the airfoil performance along with wind- tunnel testing.

A)

Panel method

B)

Hit – and – trial methods

C)

Advanced viscous flow solutions

D)

Panel method and advanced viscous flow solutions

Correct Answer : Panel method and advanced viscous flow solutions

Explanation : The numerical methods used were like source and vortex panel methods and numerical predictions of the viscous flow behavior, to analyze skin friction and flow separation effects. This was followed by experimental testing for verification of computer results.

73 .

Is thin viscous region forms over the airfoil at low angle of attack?

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : True

Explanation : At low angle of attack, a thin viscous region forms over the airfoil, and grows from the leading edge to the trailing. On the upper surface, where adverse pressure gradients exists the boundary layer grows more rapidly.

74 .

Is lift is generated due to large pressure gradients on the upper side of the airfoil?

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : True

Explanation : At higher angles, very large adverse pressure gradients that develop on the upper side as the airfoil attempts to generate more lift causes the boundary layer to separate, leading to a major disruption of the flow over the airfoil and the wing stalls.

75 .

Example of turbulent flow?

A)

Laminar flow

B)

Flow on a symmetric airfoil

C)

Smoking rises from cigarette

D)

Turbulent flow on the airfoil

Correct Answer : Smoking rises from cigarette

Explanation : Smoke rising from a cigarette is mostly turbulent flow. However, for the first few centimeters, the flow is laminar. The smoke plume becomes turbulent as its Reynolds number increases, due to its flow velocity and the characteristic length.

76 .

Is laminar flow is used to increase the drag and reduce the lift?

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : True

Explanation : The effect can also be exploited by devices such as aerodynamic spoilers on aircraft, which deliberately spoil the laminar flow to increase the drag and reduce the lift on the aircraft, which also used as a break to stop the aircraft