This set of Aircraft Design Questions and Answers for Aptitude test focuses on “Configuration Layout – Wetted Area and Volume Determination”.

1. Total exposed area that would get wet if immersed in fluid is called ____________

a) wetted area

b) volume

c) length

d) planform of wing

View Answer

Explanation: Wetted area is the total exposed area which would get wet if we immersed body in fluid. Volume is product of area and length or depth. Length is measurement of how long an object is. Wing planform is shape of wing when viewed from top.

2. Wetted area has effect on drag estimation.

a) True

b) False

View Answer

Explanation: Wetted area will affect the amount of the fluid friction when it touches the skin of the body. Hence, it affects skin friction of the body. As skin friction is affected, the drag estimation is also been affected.

3. Wing has planform area of 2m^{2}. If wing is paper thin then, find wetted area of the wing.

a) 4m^{2}

b) 10m^{2}

c) 5m^{2}

d) 20m^{2}

View Answer

Explanation: Given, planform area = 2m

^{2}.

Since, wing is paper thin wetted are = 2*planform area = 2*2 = 4m

^{2}.

4. Find the wetted area of given body.

a) 65m^{2}

b) 75

c) 85m^{2}

d) 200

View Answer

Explanation: For given diagram,

Wetted area Aw = area under the curve = area of triangle + area of rectangle

Aw = (0.5*6*5) + (10*5) = 15+50 = 65m

^{2}.

5. If thickness ratio t/c is less or equal to the 0.05 then, the wetted area is ____

a) S_{wet} = 2.003*S_{exposed}

b) S_{wet} = 2.003*4*S_{exposed}

c) S_{wet} = 2/S_{exposed}

d) S_{wet} = 2*S_{exposed}

View Answer

Explanation: Thin body has thickness tends to negligible. For a paper thin body wetted area is twice of the exposed or true planform area. However, for finite thickness of t <= 0.05*c wetted area is approximated as S

_{wet}= 2.003*S

_{exposed}.

6. A wing has thickness of 15% of chord then, find the approximated wetted area. Given exposed area A is 10m^{2}.

a) 20.55m^{2}

b) 26.55

c) 36

d) 56

View Answer

Explanation: Given thickness t = 15% of chord = 0.15*c, A = 10m

^{2}.

For given thickness, wetted area Aw is given by,

Aw = A*(1.977+0.52*(t/c)) = 10*(1.977+0.52*0.15) = 20.55m

^{2}

7. For wetted area Aw, which of the following is correct?

a) Aw = 3.4*(S_{top} + A_{side}) / 2

b) Aw = 3.4*(S_{top} + A_{side}) / 5

c) Aw = 3.4*(S_{top} + A_{side}) / 4

d) Aw = 3.4*(S_{top} + A_{side}) / 3

View Answer

Explanation: Typically, wetted area can be estimated through top view and side view of the fuselage. Typically we consider average of the area viewed from top and side to find approximated wetted area. Wetted area is given as, Aw = 3.4*(S

_{top}+ A

_{side}) / 2. S

_{top}= area viewed from top and A

_{side}= area viewed from side.

8. A fuselage is designed with circular C/S only. Average projected area is 28.2m^{2} then, find the wetted area of this fuselage.

a) 88.6m^{2}

b) 96 m

c) 540cm

d) 253m

View Answer

Explanation: Given, average projected area A = 9 m

^{2}.

Now, wetted area Aw = π*A = π*28.2 = 88.59 = 88.6m

^{2}.

9. Wetted area of a rectangular body is 64.64m^{2}. Find the average projected area.

a) 16.16m^{2}

b) 15.85 m^{2}

c) 25.368 km

d) 45.45 cm

View Answer

Explanation: Given, Aw = 64.64m

^{2}.

Now, average projected area A = Aw/4 = 64.64/4 = 16.16m

^{2}.

10. Fuselage has wetted area of 20m^{2}. Given Aright = 5m^{2}. Find the value of area viewed from top.

a) 6.8m^{2}

b) 9.8m^{2}

c) 6.8m

d) 68m

View Answer

Explanation: Given wetted area Aw = 20m

^{2}, Aright = 5m

^{2}.

Now, area vied from top A is given by,

A = 0.588*Aw – Aright = 0.588*20 – 5 = 11.764-5 = 6.764 = 6.8m

^{2}.

11. Find the approximated volume for body as shown in below.

a) 77m^{3}

b) 77m

c) 7.7m

d) 77 cm

View Answer

Explanation: Volume = area under the curve

= area of triangle + rectangle area + 8 + 4 + 6 = 0.5*6*3 + 5*10 + 8 + 4 + 6 = 9+50+8+4+6 = 77m

^{3}.

12. A fuselage is configure to have length L. What will be the approximated internal volume (V)?

a) V = 3.4*(S_{top}*A_{side}/4L)

b) V = 3.4/4L

c) V = 3.4*(S_{top}*A_{side}/4)

d) V = 0.4*(S_{top}*A_{side}/L)

View Answer

Explanation: Internal volume of fuselage will be total available volume for crew, cargo, passenger etc. Typically, from side and top view we can estimate internal volume. If area of side is Aside and area of top is Stop then, internal volume is given as, V = 3.4*(S

_{top}*A

_{side}/4L), where L = length of the fuselage.

13. If a fuselage has length of 12m and area when viewed from top and side are 4m^{2} and 8m^{2} respectively. Find the approximate value of volume.

a) 2.26m^{3}

b) 2.65m^{2}

c) 2.285

d) 2.65

View Answer

Explanation: given, S

_{top}= 4m

^{2}, A

_{side}= 8m

^{2}, L = 12m

Now, internal volume V is given by,

V = 3.4*(S

_{top}*A

_{side}/4L) = 3.4*(4*8/4*12) = 3.4*32/48 = 2.26m

^{3}.

14. My fuselage has length of 10.12m and volume of 8m^{3}. Now if I want to design same fuselage but with volume of 16m^{3} then, what should be the length of fuselage? Given, all others parameters are held constant.

a) 5.06m

b) 1012cm

c) 1000mm

d) 6.025m

View Answer

Explanation: Given, length of old fuselage L1 = 10.12m, volume of old fuselage V1 = 8m

^{3}.

Now, we are asked to design same fuselage which has volume of 16m

^{3}.

Hence, new volume V2 = 16m

^{3}.

Internal volume V is inversely proportional to length L of the fuselage.

Here, we have been asked held all other parameters constant.

Hence, L2 = (V1/V2)*L1 = (8/16)*10.12 =5.06m

^{3}.

15. If for a given fuselage S_{top}=10m^{2}, A_{side}=14m^{2}, L = 10m then, find the ratio of wetted area to the volume of fuselage.

a) 3.42 per m

b) 6.68 per m^{3}

c) 8.36 m

d) 4.001

View Answer

Explanation: Given, S

_{top}=10m

^{2}, A

_{side}=14m

^{2}, L = 10m.

Now ratio of wetted area Aw and volume V is given by,

Aw/V = 2*L*(S

_{top}+ A

_{side}) / (S

_{top}*A

_{side})

Aw/V = 2*10*(10+14) / (10*14) = 20*24/140 = 3.42 per m.

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