This set of Aircraft Design Multiple Choice Questions & Answers (MCQs) focuses on “Airfoil Selection-3”.
1. Pressure on the upper surface is _________
a) lower than lower surface
b) higher than lower surface
c) always 30.25 times lower surface
d) always same
View Answer
Explanation: Pressure on the upper surface of an airfoil is lower than that of the lower surface. This will provide pressure difference which will generate lift force. Pressure will be the same if AOA is 0° in case of symmetric airfoil.
2. What do you mean by double cambered airfoil?
a) Airfoil with flat lower surface
b) Airfoil with flat upper and lower surface
c) Airfoil with curved upper and lower surface
d) Only upper surface is curved
View Answer
Explanation: Traditionally, airfoils with only curved upper surface and flat bottom was termed as camber airfoil. However, if both surfaces are curved then, it is called a double cambered airfoil. In now days not most people use this terms but at some places it is still in use.
3. Laminar bucket is region where ____
a) turbulent is produced
b) flow remains laminar
c) drag increases drastically
d) lift is reduced drastically
View Answer
Explanation: Laminar bucket is phenomena of laminar airfoils. It is the region where flow remains laminar. This laminar flow is very important as it decreases drag and thus improves overall performance. However, for such airfoils Reynolds number is very important consideration. Following diagram shows laminar bucket for a typical laminar airfoil.
4. A cambered airfoil is operating with 200N of lift. The velocity of the flow is 500m/s. Determine lift-induced drag by this camber airfoil.
a) 0N
b) 0
c) 0.5
d) 0.5
View Answer
Explanation: Given, Lift=200N, velocity = 500m/s.
For cambered airfoil the downwash angle = 0°. Hence, there will be no downwash acting on the flow.
Hence, there will be no induced drag. Hence, for any airfoil induced drag = 0N.
5. An airfoil is such that the drag produced by airfoil is 25N. Airfoil has chord of 2m and is experiencing dynamic pressure of 10Pa. Find the value of drag co-efficient for airfoil.
a) 2.51
b) 1.6
c) 2.5
d) 1.25
View Answer
Explanation: Given, Drag = 25N, dynamic pressure q = 10Pa, chord c = 2m.
Now, drag co-efficient Cd = Drag / (q*c) = 25 / (10*2) = 1.25.
6. Let’s consider we need to design supersonic aircraft then, which type of airfoil should be adopted?
a) Bluff shape
b) Symmetric airfoil
c) Supersonic airfoil
d) Thicker airfoil
View Answer
Explanation: Supersonic aircraft needs to fly at much higher speed. Hence, supersonic airfoils are used to provide required velocity. Thicker airfoils cannot generate required amount of flow characteristics such as critical mach number.
7. What will be the location of maximum camber for NACA 13250 with chord length of 2m?
a) 0.32 from trailing edge
b) 0.32m from leading edge
c) 0.32m from trailing edge
d) 0.32 from leading edge
View Answer
Explanation: Given, NACA 13250 airfoil
Based on NACA -5 digit series, location of maximum camber = (second and third digit)/2 in hundredth of chord from leading edge
= (32/2) * 0.01 * chord = (32/2)*0.01*2 = 0.32m from leading edge.
8. The maximum lift co-efficient for NACA 19584 (chord=2.5m) airfoil is____
a) 0.15N
b) 0.315
c) 0.15
d) 0.3N
View Answer
Explanation: Given, NACA 13250 airfoil
Based on NACA -5 digit series, designed lift co-efficient = first digit*0.15 = 1*0.15 = 0.15.
9. An airfoil is operated with flow velocity such that the pressure on upper and lower surface are 30 Pa and 15 Pa respectively. If chord is 1m then, what will be the lift produced by this airfoil?
a) 30 Pa
b) 15KN
c) 25N
d) 15N
View Answer
Explanation: Given, Pressure on upper surface P1 = 30Pa,
Pressure on lower surface P2 = 15Pa, chord = 1m.
Now lift = (P1- P2)*Chord = (30-15)*1 = 15N.
10. A NACA 22104 airfoil is operating at AOA = 3°. The lift-coefficient at this AOA is 0.15. Determine the value of angle of attack at designed lift co-efficient.
a) 4.36°
b) 4.36rad
c) 0.05 rad
d) 0.05°
View Answer
Explanation: Given NACA 22104 airfoil.
AOA = 3°, Cl = 0.15 at 3°
Based on NACA -5 digit series design Cl = first digit*0.15 = 2*0.15 = 0.3
Now, for thin airfoil lift curve slope = 0.11 per degree
0.11 = (design Cl – Cl at 3°) / (AOA –3°)
0.11 = (0.3-0.15) / (AOA -3°)
0.11*(AOA -3°) = 0.15
Hence, at designed Cl, AOA = (0.15/0.11) + 3° = 4.36°.
11. Determine the corrections or otherwise of the following assertion [A] and reason [R]:
Assertion [A]: Most of the wings are using cambered airfoils.
Reason[R]: Cambered airfoils can produce positive lift at 0° AOA.
a) Both [A] and [R] are true and [R] is the correct reason for [A]
b) Both [A] and [R] are true but [R] is not the correct reason for [A]
c) [A] is true but [R] is false
d) [A] is false but [R] is true
View Answer
Explanation: Wings generally uses cambered airfoils. The camber airfoil is used to provide positive lift even at zero angle of attack. Also, it improves flow characteristics.
12. Location of aerodynamic centre for low speed airfoil is ____
a) at quarter chord point
b) at quarter chord from leading edge
c) there is no such place
d) at half of chord
View Answer
Explanation: Aerodynamic center is that point at which pitching moment curve slope becomes zero. Hence, it is the point at where pitching moment is almost independent of AOA. For, low speed airfoils it is located at 25% of chord from leading edge.
13. If I want to locate my minimum pressure at 0.5 of chord c then _____________
a) NACA 6512
b) NACA 652132
c) NACA 65211
d) NACA 0009
View Answer
Explanation: NACA series is used to identify airfoils as per our requirement. NACA -4 and -5 digit series are not used to provide a minimum pressure point. NACA 6 digit series is used for such operations.
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