Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

101.

Is viscosity effects the pitching moments?(a) True(b) FalseThe question was posed to me during an online exam.I'm obligated to ask this question of Viscous Flow in chapter Incompressible Flow over Airfoils of Aerodynamics

Answer»

The correct choice is (a) True

The BEST I can explain: Viscosity ALSO affects the PITCHING moments of an AIRFOIL. The pitching moment coefficient of an airfoil about chord point, potential flow theory states that this coefficient about this point is ZERO. In reality, a nose down pitching moment develops at a high angle of attack as the lift distributed over the nose part of the airfoil decrease due to viscous effects.

102.

Is lift loss with increase in angle of attack?(a) True(b) FalseThis question was addressed to me in class test.I want to ask this question from Viscous Flow in section Incompressible Flow over Airfoils of Aerodynamics

Answer»

Right choice is (a) True

Explanation: The loss in lift which does the increase in the ANGLE of attack, in most instances stall occurs GRADUALLY, with a slow upward motion of the separation of the boundary layer from the trailing edge to the LEADING edge.

103.

Is initial distribution is given by solid curves?(a) True(b) FalseThis question was addressed to me in unit test.Question is from Modern Low Speed Airfoils topic in section Incompressible Flow over Airfoils of Aerodynamics

Answer»

The CORRECT choice is (a) True

The explanation is: The optimization technique is iterative and requires starting with a pressure distribution that is not the desired, specified one, the initial distribution is given by the solid curves and the AIRFOIL shapes appears distorted because an expanded SCALE is USED for the ORDINATE.

104.

For a cambered airfoil, the center of pressure does not vary with_____(a) Chord length(b) Angle of attack(c) The shape of the airfoil(d) Location of the aerodynamic centerThe question was asked in an interview.I'd like to ask this question from The Cambered Airfoil topic in division Incompressible Flow over Airfoils of Aerodynamics

Answer»

Correct option is (d) LOCATION of the AERODYNAMIC center

To elaborate: This solution is best ANSWERED from the FORMULA for the center of pressure which is xcp=\(\frac {C }{4}\)(1+\(\frac {\pi }{c_l}\)(A1-A2)). Moreover, the aerodynamic center is located at the quarter-chord but the center of pressure keeps changing with the other three parameters.

105.

What is the leading edge radius of LS-0417 airfoil?(a) 0.07c(b) 0.05c(c) 0.08c(d) 0.09cThe question was posed to me during an interview for a job.Asked question is from Modern Low Speed Airfoils topic in chapter Incompressible Flow over Airfoils of Aerodynamics

Answer»

The correct choice is (c) 0.08c

To explain: LS-0417 AIRFOIL has a leading edge RADIUS of 0.08c in comparison to the standard 0.02c in order to FLATTEN the usual peak in pressure COEFFICIENT near the nose, also that the bottom surface near the TRAILING edge is cusped in order to increase the camber.

106.

The fundamental equation for the thin airfoil theory does not encompass which of the following approximations?(a) The angle of attack and slope of the camber line is small(b) Camber line induced velocity distribution is the same for chord line(c) Vortex sheet is placed at the chord line(d) Kutta condition is satisfied at the trailing edgeThis question was posed to me during an internship interview.The above asked question is from The Symmetric Airfoil topic in portion Incompressible Flow over Airfoils of Aerodynamics

Answer»

Right option is (d) Kutta CONDITION is satisfied at the trailing edge

Easy 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.

107.

A vortex sheet placed along the chord line gives the best representation for______(a) Thick Airfoil(b) NACA Airfoils(c) Negatively Cambered Airfoil(d) Thin AirfoilI got this question during a job interview.I need to ask this question from The Symmetric Airfoil in portion Incompressible Flow over Airfoils of Aerodynamics

Answer»

The CORRECT option is (d) Thin Airfoil

Easiest 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.

108.

In reality, the starting vortex dies out. Why?(a) Lift becomes zero(b) At later times, Kelvin’s theorem is not applicable(c) Due to Viscosity(d) This assumption is wrong. Starting vortex never diesI got this question in examination.This interesting question is from Kelvin’s Circulation Theorem and the Starting Vortex topic in section Incompressible Flow over Airfoils of Aerodynamics

Answer»

The correct choice is (c) 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.

109.

Is kutta condition apply to oval shaped body?(a) True(b) FalseI had been asked this question in homework.My question is from The Kutta Condition topic in chapter Incompressible Flow over Airfoils of Aerodynamics

Answer»

The correct ANSWER is (a) True

To elaborate: 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.

110.

Is airfoil can be characterized by the relation between the angle of attack, lift coefficient and drag coefficient?(a) True(b) FalseThe question was asked in final exam.My question is taken from Airfoil Characteristics topic in section Incompressible Flow over Airfoils of Aerodynamics

Answer»

Correct option is (a) True

For explanation: Airfoil can be characterized by the relation between an angle of ATTACK, LIFT coefficient and DRAG coefficient. Where a propeller can be described in TERMS of advance ratio, thrust coefficient and power coefficient. The efficiency which corresponds to the lift-drag ratio of a wing can be CALCULATED from these coefficients.

111.

Is lift is created on the airfoil using kutta condition?(a) True(b) FalseI got this question during an internship interview.This question is from The Kutta Condition in portion Incompressible Flow over Airfoils of Aerodynamics

Answer»

Right choice is (a) True

Easy explanation: The kutta condition is significant when using the kutta joukowski theorem to CALCULATE the lift created by an airfoil with a cusped TRAILING edge. The VALUE of CIRCULATION of the flow around the airfoil must be that value, which would cause the kutta condition to exist.

112.

Define drag curve.(a) Drag will increase rapidly at particular degree of angle of attack and overcomes the lift curve at particular degree of angle of attack(b) Drag curve will decrease at particular degree of angle of attack and lift curve increasing at particular degree of angle of attack(c) Drag curve remains constant and lift curve will be increasing(d) Lift curve remains constant and drag curve will be increasingI have been asked this question by my college director while I was bunking the class.The doubt is from Airfoil Characteristics in section Incompressible Flow over Airfoils of Aerodynamics

Answer»

The correct answer is (a) Drag will increase RAPIDLY at PARTICULAR DEGREE of angle of attack and overcomes the lift curve at particular degree of angle of attack

Best 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.

113.

What is the purpose of trailing edge?(a) Airflow rejoins(b) Airflow separated(c) Vortex are created(d) Stalling will createdThe question was asked in homework.Query is from Airfoil Nomenclature topic in chapter Incompressible Flow over Airfoils of Aerodynamics

Answer»

Right choice is (a) Airflow rejoins

To elaborate: 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.

114.

We can get the skin-friction drag coefficient for a laminar flow for a flat plate by using x as chord length in local skin- friction drag coefficient calculation.(a) True(b) FalseI had been asked this question in examination.Enquiry is from Laminar Flow in division Incompressible Flow over Airfoils of Aerodynamics

Answer»

Correct choice is (b) False

Best explanation: The skin-friction drag coefficient is Cf=\(\frac {1.328}{\sqrt{Re_c}}\) and the LOCAL skin-friction drag coefficient is Cf=\(\frac {0.664}{\sqrt{Re_x}}\). Cf is calculated by integrating cf over the WHOLE AIRFOIL chord and as visible by the formula, we cannot GET it directly by PUTTING x = c.

115.

Is thin viscous region forms over the airfoil at low angle of attack?(a) True(b) FalseThis question was posed to me in final exam.I would like to ask this question from Viscous Flow in section Incompressible Flow over Airfoils of Aerodynamics

Answer»

Right option is (a) True

For 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.

116.

Is high adverse pressure gradient that occurs at a sharp leading edge?(a) True(b) FalseI got this question during an online exam.My question is from Viscous Flow topic in chapter Incompressible Flow over Airfoils of Aerodynamics

Answer»

The correct answer is (a) True

Easy explanation: In some AIRFOILS with small leading edge RADIUS the stall is rather abrupt. The flow is well behaved at an angle of attack, say 5.5 degrees, but stalls with flow SEPARATION at the leading edge at 5.6 degrees because of the high adverse pressure gradient that occurs at such SHARP leading edge.

117.

The coefficient of lift for a thin, cambered airfoil with A0=0.2 and A1=1.12 is____(a) π1.52(b) π1(c) π0.52(d) π2The question was posed to me in my homework.My question comes from The Cambered Airfoil topic in portion Incompressible Flow over Airfoils of Aerodynamics

Answer»

The correct CHOICE is (a) π1.52

For explanation I would say: The lift COEFFICIENT from the thin AIRFOIL theory for a cambered airfoil is given by

cl=π(2A0+A1). So PUTTING A0 and A1 we get cl = π1.52.

118.

For α=5°, A0=1 and A1=-2 total circulation Γ for a thin cambered airfoil equals______(a) 0(b) 2πcV∞(c) πcV∞(d) 2.5πcV∞The question was posed to me during an interview for a job.Question is from The Cambered Airfoil in portion Incompressible Flow over Airfoils of Aerodynamics

Answer»

Right choice is (a) 0

For explanation I WOULD SAY: From the formula of TOTAL circulation Γ=cV∞(πA0+\(\frac {\pi }{2}\)A1) we GET Γ=0 upon putting the values given in the question.

119.

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) 0.13c, 0.4c(d) 0.04c, 0.03cI had been asked this question in an internship interview.Origin of the question is The Cambered Airfoil topic in chapter Incompressible Flow over Airfoils of Aerodynamics

Answer»

The correct choice is (a) 0.04c, 0.4c

To explain I WOULD say: 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.

120.

Is upper camber refers to the camber of the upper surface?(a) True(b) FalseThe question was asked in examination.Asked question is from The Cambered Airfoil in portion Incompressible Flow over Airfoils of Aerodynamics

Answer»

The CORRECT option is (a) True

Best explanation: The TERM UPPER camber refers to the camber of the upper SURFACE of the airfoil. The term lower surface refers to the lower surface of the airfoil. When the CURVE is away from the chord, the camber is said to be positive.

121.

What is the Kutta Condition in terms of strength (γ) of the thin airfoil?(a) γ = 0(b) γ(LE) = 0(c) \(\frac {D\gamma }{Dt}\)=0(d) γ(TE) = 0I have been asked this question in an international level competition.Query is from The Symmetric Airfoil in division Incompressible Flow over Airfoils of Aerodynamics

Answer»

Right answer is (d) γ(TE) = 0

The best explanation: 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.

122.

Is high airspeed around the trailing edge causes strong viscous forces?(a) False(b) TrueI got this question in examination.The doubt is from The Kutta Condition in portion Incompressible Flow over Airfoils of Aerodynamics

Answer» RIGHT OPTION is (a) False

To explain: The high airspeed AROUND the trailing edge causes strong viscous forces to act on the air adjacent to the trailing edge of the airfoil and the result is that a strong vortex accumulates on the TOPSIDE of the airfoil, NEAR the trailing edge.
123.

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) 2π(c) –π(d) nπThis question was addressed to me in examination.This key question is from The Symmetric Airfoil in portion Incompressible Flow over Airfoils of Aerodynamics

Answer» RIGHT option is (a) π

The best I can explain: 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.
124.

Purpose of vortex method is ____________(a) To reduce the dimensionality of aerodynamic model(b) To increase the dimensionality of aerodynamic model(c) To reduce the lift of aerodynamic model(d) To increase the lift of aerodynamic modelThe question was asked during an interview.My question comes from The Vortex Sheet topic in division Incompressible Flow over Airfoils of Aerodynamics

Answer»

Correct option is (a) To reduce the dimensionality of aerodynamic model

The 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.

125.

How propeller creates a thrust force?(a) With transmitted power(b) With own power(c) With the existing power(d) With supplied powerThis question was addressed to me in homework.My question is from Airfoil Characteristics in portion Incompressible Flow over Airfoils of Aerodynamics

Answer»

Correct choice is (d) With supplied POWER

The best 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.

126.

What is responsible for creating the aerodynamic drag on the airfoil?(a) Pressure distribution(b) Friction(c) Thrust(d) GravityThis question was posed to me during a job interview.I need to ask this question from Viscous Flow in portion Incompressible Flow over Airfoils of Aerodynamics

Answer»

Correct CHOICE is (B) Friction

To explain I would say: VISCOSITY causes drag on the airfoil. For an airfoil, friction is responsible for causing viscosity and HENCE, causes drag.

127.

The quarter-chord moment coefficient for a cambered airfoil depends on______(a) Chord length(b) Angle of attack(c) The shape of the airfoil(d) Center of pressureI have been asked this question by my college professor while I was bunking the class.My doubt is from The Cambered Airfoil topic in portion Incompressible Flow over Airfoils of Aerodynamics

Answer» RIGHT choice is (c) The shape of the airfoil

Explanation: The coefficient of moment about the quarter-chord is independent of angle of ATTACK and chord length. It depends on the shape of the cambered airfoil (slope of the camber LINE (\(\FRAC {dz}{dx}\))).
128.

The lift curve slope for a thin, cambered airfoil is 2π.(a) True(b) FalseI got this question during an interview.This intriguing question originated from The Cambered Airfoil topic in portion Incompressible Flow over Airfoils of Aerodynamics

Answer»

Right choice is (a) True

To elaborate: The lift curve slope for CAMBERED airfoils is also equal to 2π. This comes from the THIN AIRFOIL THEORY, irrespective of the shape of airfoils.

129.

Modern low-speed airfoils were developed using numerical methods using the computer directly. The given statement is_____(a) Partially true(b) False(c) True(d) IncompleteThe question was posed to me during an interview.The doubt is from Modern Low Speed Airfoils topic in chapter Incompressible Flow over Airfoils of Aerodynamics

Answer»

Right option is (a) PARTIALLY true

The best I can explain: 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.

130.

For γ(θ)=2V∞(A0\(\frac {1+cos\theta }{sin⁡\theta }\) + Σ\(_{n=1}^∞\)sin⁡ nθ An) select the statement which is invalid.(a) The solution is valid only for cambered airfoils(b) The solution is valid for all thin airfoils(c) A0 Is the n=0^th term for the Fourier series(d) Kutta condition is satisfied at the trailing edge i.e. θ=π.This question was posed to me in quiz.My doubt stems from The Cambered Airfoil topic in section Incompressible Flow over Airfoils of Aerodynamics

Answer»

The correct choice is (a) The solution is valid only for cambered AIRFOILS

Easy explanation: The solution is valid for all thin airfoils (for symmetric airfoils \(\frac {dz}{dx}\)=0 which MAKES An=0 and GIVES the required solution). This is the general solution. Kutta CONDITION is also satisfied (γ(π)=0).

131.

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) Cambered airfoils only(b) Symmetric airfoils only(c) All thin airfoils(d) Symmetric and positively cambered airfoilsI got this question by my college director while I was bunking the class.I need to ask this question from The Cambered Airfoil in portion Incompressible Flow over Airfoils of Aerodynamics

Answer»

Correct CHOICE is (b) Symmetric airfoils only

To elaborate: 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.

132.

What is the total circulation around the symmetric airfoil according to the thin airfoil theory?(a) Γ=πα^2cV∞(b) Γ=π^2αcV∞(c) Γ=2παcV∞(d) Γ=παcV∞I had been asked this question at a job interview.Question is from The Symmetric Airfoil in section Incompressible Flow over Airfoils of Aerodynamics

Answer»

The correct answer is (d) Γ=παcV∞

Explanation: The total circulation around the SYMMETRIC airfoil can be found by integrating the TRANSFORMED SOLUTION γ(θ)=2αV∞\(\FRAC {1+cos\theta }{sin⁡\theta }\) using ξ=\(\frac {c}{2}\)(1-cosθ)er 0≤θ≤π i.e. Γ=\(\int_0^c\)γ(ξ)dξ=παcV∞.

133.

Find the induced velocity in the direction normal to the camber line a point on where the slope is 0.087. Given the angle of attack is 5° and free-stream velocity is 20 units. Use thin airfoil approximation.(a) 0(b) 101.74 units(c) 98.26 units(d) 3.48 unitsThis question was posed to me by my school principal while I was bunking the class.The query is from The Symmetric Airfoil topic in chapter Incompressible Flow over Airfoils of Aerodynamics

Answer»

The correct answer is (a) 0

To explain I would say: For a THIN airfoil, the sum of free stream VELOCITY component (V∞) and the induced velocity normal to the camber LINE is zero. And for small angles, V∞n ≈ V∞(α-\(\frac {dz}{dx}\)) where α is the angle of attack in RADIANS and \(\frac {dz}{dx}\) is the slope at that point. Solving, V∞n ≈ 20(0.087-0.087), this is equal to the induced velocity.

134.

Is inviscid incompressible fluid has constant density?(a) False(b) TrueThis question was addressed to me in an online interview.The query is from Kelvin’s Circulation Theorem and the Starting Vortex topic in section Incompressible Flow over Airfoils of Aerodynamics

Answer»

The correct option is (a) False

For explanation I would SAY: According to the KELVIN’s CIRCULATION theorem an inviscid incompressible fluid of constant density be in motion in the presence of a conservative body force g PER unit mass. Let c (t) denote a closed CIRCUIT that consists of the same fluid particles as time proceeds.

135.

At what point circulation between two material points in the sheet remains conserved.(a) Zero(b) One(c) Greater than zero(d) Less than zeroI have been asked this question in a job interview.I'm obligated to ask this question of The Vortex Sheet topic in section Incompressible Flow over Airfoils of Aerodynamics

Answer» CORRECT option is (a) 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.
136.

Is airfoil shape similar to blades of the propeller?(a) True(b) FalseI had been asked this question during an interview.This is a very interesting question from Airfoil Characteristics topic in chapter Incompressible Flow over Airfoils of Aerodynamics

Answer»

Correct CHOICE is (a) True

For explanation I would say: 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.Apressure DIFFERENCE is produced between the forward and rear surface of the airfoil-shape and fluid is accelerated behind the blade.

137.

The boundary layer thickness for an incompressible, laminar flow at a distance x with Reynolds number Rex is δ. Which is____(a) Directly proportional to √x(b) Inversely proportional to x(c) Directly proportional to Rex(d) Inversely proportional to Rex^2The question was asked in an online quiz.My enquiry is from Laminar Flow topic in section Incompressible Flow over Airfoils of Aerodynamics

Answer»

Right option is (a) Directly PROPORTIONAL to √x

Explanation: The boundary layer thickness for an incompressible, laminar flow over a flat plate at ZERO angle of attack, at a DISTANCE x where Reynolds number is Rex is given by δ=\(\frac {5x}{\sqrt{Re_x}}\). Here Reynolds number is Rex=\(\frac {xVd}{U}\). THUS, the thickness is directly proportional to √x.

138.

Is laminar flow used in the smooth flow of a viscous liquid through a tube?(a) True(b) FalseI had been asked this question in unit test.This key question is from Laminar Flow in portion Incompressible Flow over Airfoils of Aerodynamics

Answer»

The correct CHOICE is (a) True

The EXPLANATION is: The common PALLIATION of laminar flow is in the SMOOTH flow of a viscous liquid through a tube. In that case, the velocity of flow varies from zero at the walls to a maximum along the cross-sectional CENTER of the vessel.

139.

The major breakthrough in high- speed airfoil industry was_____(a) GA (W)-1 airfoil(b) Supercritical airfoil(c) Standard NACA airfoils(d) Symmetrical airfoilsThe question was posed to me in a job interview.This interesting question is from Modern Low Speed Airfoils in portion Incompressible Flow over Airfoils of Aerodynamics

Answer»

Correct CHOICE is (b) Supercritical airfoil

The explanation is: 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.

140.

Given an angle of attack 5°, A1=1.5, A2=2 and the chord length is 5m, the moment coefficient about the quarter-chord is_____(a) \(\frac {\pi }{8}\)(b) \(\frac {\pi }{4}\)(c) \(\frac {-\pi }{8}\)(d) –\(\frac {\pi }{4}\)The question was asked during an internship interview.This interesting question is from The Cambered Airfoil topic in section Incompressible Flow over Airfoils of Aerodynamics

Answer»

Correct option is (a) \(\frac {\pi }{8}\)

Explanation: The coefficient of moment about the quarter-chord is GIVEN by cm,l/4=\(\frac {\pi }{4}\)(A2-A1) which is cm,c/4=cm,LE+\(\frac {c_l}{4}\). In this question, it COMES out as \(\frac {\pi }{8}\).

141.

Which of the following is the correct solution of the transformed fundamental equation of aerodynamics for a symmetrical airfoil?(a) γ(θ)=2αV∞\(\frac {sin⁡\theta }{1+cos\theta }\)(b) γ(θ)=2αV∞\(\frac {1+cos\theta }{sin⁡\theta }\)(c) γ(θ)=2αV∞\(\frac {1-cos⁡\theta }{sin\theta }\)(d) γ(θ)=2αV∞\(\frac {cos\theta }{sin\theta }\)The question was posed to me during an internship interview.This is a very interesting question from The Symmetric Airfoil topic in section Incompressible Flow over Airfoils of Aerodynamics

Answer» RIGHT choice is (b) γ(θ)=2αV∞\(\frac {1+cos\theta }{sin⁡\theta }\)

The explanation is: The solution of the fundamental equation of THIN airfoil THEORY is obtained using the transformation of coordinates. We have α and V∞ and using the standard integrals we can find a solution for γ(x) as γ(θ)=2αV∞\(\frac {1+cos\theta }{sin⁡\theta }\) where 0≤θ≤π for 0≤x≤C.
142.

For an arbitrary inviscid and incompressible flow, with all the body forces zero, what is best described by the given sketch?(a) Boundary Layer Formation(b) Kelvin’s Circulation Theorem(c) Kutta Condition(d) Generation of LiftI have been asked this question during a job interview.Query is from Kelvin’s Circulation Theorem and the Starting Vortex in portion Incompressible Flow over Airfoils of Aerodynamics

Answer»

Right answer is (B) Kelvin’s Circulation Theorem

Easy explanation: 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.

143.

Is vortex sheet is unstable at high Reynolds number?(a) True(b) FalseI got this question in an online quiz.The above asked question is from The Vortex Sheet in portion Incompressible Flow over Airfoils of Aerodynamics

Answer»

The correct CHOICE is (a) True

The BEST I can explain: The discontinuity in the tangential velocity MEANS that the flow has infinite vorticity in a vortex sheet. At high Reynolds numbers vortex sheets tend to be UNSTABLE and they exhibit Kelvin-Helmholtz instability. The Kelvin-Helmholtz instability can occur when there is velocity shear in a single continuous fluid.

144.

For a real flow over an airfoil, with Reynolds number in the laminar region as it starts, select the incorrect choice.(a) Flow at trailing edge is laminar(b) Flow is both turbulent and laminar at different places(c) There is a transition region between leading and trailing edge(d) Flow at trailing edge is turbulentThe question was asked in exam.This key question is from Turbulent Flow topic in section Incompressible Flow over Airfoils of Aerodynamics

Answer»

The correct option is (a) Flow at trailing edge is LAMINAR

For explanation I would say: In ACTUAL cases, where we don’t have an IDEAL flow, no flow is fully laminar or TURBULENT. It is laminar as it starts and becomes turbulent near the trailing edge. It has a transition REGION where the transition from laminar to turbulent flow takes place.

145.

When the free-stream velocity is made one-third, the transition point moves downstream.(a) The given statement is true or false?(b) True(c) FalseThis question was posed to me by my college director while I was bunking the class.Question is taken from Turbulent Flow in chapter Incompressible Flow over Airfoils of Aerodynamics

Answer»

Right answer is (a) The given statement is true or FALSE?

For explanation I would say: The transition point is given as xcr=\(\FRAC {Re_{x_{cr}} μ_∞}{\rho_∞V_∞ }\). Thus, from the formula, we can see that as the VELOCITY is DECREASED xcr increases i.e. the transition point moves AWAY from the leading edge (or moves downstream).

146.

Is outer flow over an airfoil has less camber than the original airfoil?(a) False(b) TrueThe question was asked in an international level competition.This question is from Viscous Flow in division Incompressible Flow over Airfoils of Aerodynamics

Answer» CORRECT option is (b) True

Explanation: The outer flow SEES an equivalent airfoil that has less camber than the original airfoil, due to the disproportionate growth of the boundary LAYER on the two side and that has an open trailing EDGE. Such an airfoil produce less lift than the original airfoil.
147.

For finding the skin- friction drag we need to only measure shear-stress at the top or bottom surface.(a) Always true(b) Always false(c) True for flat plate, which is a symmetric airfoil(d) Depends on the Reynolds numberThis question was addressed to me in an internship interview.Origin of the question is Laminar Flow topic in chapter Incompressible Flow over Airfoils of Aerodynamics

Answer»

The correct answer is (c) True for flat plate, which is a SYMMETRIC AIRFOIL

Explanation: The shear stress distribution on the airfoil is the same at the top and BOTTOM SURFACE for a flat plate. And we can just integrate the local shear stress on one side and double the RESULT to get the total skin-friction drag.

148.

Is 17% is the thickness of NASA LS-0417 airfoil?(a) True(b) FalseThis question was posed to me in an online quiz.My question is based upon Modern Low Speed Airfoils in division Incompressible Flow over Airfoils of Aerodynamics

Answer»

The correct OPTION is (a) True

Explanation: The NASA LS- 0417 airfoil has a maximum thickness of 17% and a DESIGN LIFT coefficient of 0.4. USING the same camber line, NASA has extended this airfoil into a family of a low-speed airfoil of different thickness for example LS-0409 and the LS-0413.

149.

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) 0(b) 102.55\(\frac {N}{m}\)(c) 102.55N(d) 55\(\frac {N}{m}\)This question was posed to me during an online exam.My query is from The Cambered Airfoil topic in section Incompressible Flow over Airfoils of Aerodynamics

Answer»

The correct OPTION is (a) 0

The best I can explain: 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).

150.

Is camber is the asymmetry between the two acting surfaces of an airfoil?(a) True(b) FalseI got this question during an interview for a job.My query is from The Cambered Airfoil in division Incompressible Flow over Airfoils of Aerodynamics

Answer» CORRECT option is (a) True

Easiest explanation: Camber is the asymmetry between the two acting SURFACES of an airfoil, with the top SURFACE of a wing or corresponding the front surface of a propeller BLADE, COMMONLY being more convex, positive camber.