Engineering Questions with Answers - Multiple Choice Questions

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# Aerodynamics Questions and Answers – Quantitative Formulation

1 - Question

Which of these is not the assumption for Taylor – Maccoll conical flow?

a) Cone is placed at the zero angle of attack

b) Flow properties along a ray of cone are constant

c) Shock wave is curved

d) Flow is axisymmetric

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Answer: cExplanation: There are certain assumptions made to determine the conical flow. As per Taylor – Maccoll, the flow is assumed to be axisymmetric about z – axis \frac {∂}{∂ϕ} = 0, and the cone is assumed to be at zero angle of attack. If it was kept at any other angle then there will be 3 – dimensional effects that will be hard to account for. Also, the flow properties along the ray of the cone is assumed to be constant \frac {∂}{∂r} = 0 and the shock wave is straight

2 - Question

Which of these is the correct relation for the entropy across a shock for all the streamlines?

a) ∇s = 0

b) ∇ × s = 0

c) (∇s) × s = 0

d) (∇ × s).s = 0

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Answer: aExplanation: The shock wave in the conical flow is assumed to be straight resulting in same increase in the entropy for all streamlines passing through the shock (∇s = 0). This property implies that the conical flow is irrotational as per Crocco’s equation.

3 - Question

Which of these is the continuity equation for an axisymmetric flow?

a) ρVθcotθ + ρ∂(Vθ)∂θ + Vθ∂(ρ)∂θ = 0

b) 2ρVr + ρVθcotθ + ρ∂(Vθ)∂θ + Vθ∂(ρ)∂θ = 0

c) 2ρVr + ρVθcotθ = 0

d) 1r2∂∂r (r2ρVr) + 1rsinθ∂∂θ(ρVθsinθ) + 1rsinθ∂(ρVϕ)∂ϕ = 0

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Answer: bExplanation: The general continuity equation is given by ∂∂t + ∇.(ρV) = 0. Since the flow is assumbed to be steady, ∂∂t = 0. For spherical coordinated of a cone, the del operator is expanded as ∇.(ρV) = 1r2∂∂r(r2ρVr) + 1rsinθ∂∂θ (ρVθsinθ) + 1rsinθ∂(ρVϕ)∂ϕ = 0 Solving the partial derivatives, we get 1r2[r2∂∂r(ρVr) + ρVr∂(r2)∂r]+1rsinθ[ρVθ∂∂θ(sinθ ) + sinθ∂(ρVθ)∂θ]+1rsinθ∂(ρVϕ)∂ϕ = 0 This is equal to 1r2[r2∂∂r(ρVr) + ρVr(2r)]+1rsinθ[ρVθ(cosθ) + sinθ∂(ρVθ)∂θ]+1rsinθ∂(ρVϕ)∂ϕ = 0 Since the flow properties are constant along a ray, ∂∂r(ρVr) = 0 and ∂(ρVϕ)∂ϕ = 0 The equations becomes 1r2[ρVr(2r)] + 1rsinθ[ρVθ(cosθ) + sinθ ∂(ρVθ)∂θ] + 0 Multiplying the final equation with r: 2ρVr + ρVθcotθ + ρ∂(ρVθ)∂θ + Vθ∂(ρ)∂θ = 0

4 - Question

Along the streamline of the conical flow, the total enthalpy stays constant.

a) True

b) False

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Answer: aExplanation: The energy equation is given by ρDh0Dt=∂p∂t + pq˙ + ρ(f.V) Where Dh0Dt is the total derivative of total enthalpy p is the pressure q˙ is per rate of heat added/removed f.V is body forces According to Taylor – Maccoll’s assumptions, the flow is said to be steady (∂∂t = 0), adiabatic (q˙ = 0), inviscid and has no external body forces (f.V = 0). Thus, the equation reduces to ρDh0Dt = 0 Which on integrating gives us h0 = const.

5 - Question

Conical flow is rotational according to the result obtained from Crocco’s theorem.

a) True

b) False

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Answer: bExplanation: Based on Crocco’s theorem, for a steady floe the relation is given by T∇s = ∇h0 – V × (∇ × V) Therefore the vorticity is related to total enthalpy and gradient as (on rearranging the terms): V × (∇ × V) = ∇h0 – T∇s In a flow having a change in enthalpy or entropy would result in a rotational flow, but since for conical flow, the assumptions result in both change in entropy and enthalpy being zero, we get V × (∇ × V) = 0 Since curl of velocity i.e. vorticity is zero, thus the flow is irrotational.

6 - Question

What is the irrotationally condition for a conical flow?

a) Vθ = ∂(Vr)∂θ

b) Vϕ = ∂(Vr)∂ϕ

c) Vθ = 1r∂(Vθ)∂θ

d) Vθ = ∂(Vr)∂θVr

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Answer: aExplanation: If we apply Crocco’s theorem in spherical coordinates we get, ∇ × V = 1r2sinθ∣∣∣∣∣er∂∂rVrreθ∂∂θrVθ(rsinθ)eϕ∂∂ϕ(rsinθ)Vϕ∣∣∣∣∣ = 0 On expanding this we get, ∇ × V = 1r2sinθ[er (∂∂θ(rsinθ)Vϕ – ∂(rVθ)∂ϕ) – reθ(∂∂r(rsinθ)Vϕ – ∂(Vr)∂ϕ) + (rsinθ)eϕ(∂(rVθ)∂r–∂(Vr)∂θ)] = 0 For this equation to be valid, the terms inside the bracket are zero. Taking the last bracket term, ∂(rVθ)∂r–∂(Vr)∂θ = 0 Using chain rule to expand this, we get r∂(Vθ)∂r + Vθ∂(r)∂r–∂(Vr)∂θ = 0 Based on the conical flow assumptions, ∂∂r = 0 and ∂∂ϕ = 0. Applying this the equation reduced to ∂(Vr)∂θ = 0 Which results in the irrotationally condition for a conical flow as Vθ = ∂(Vr)∂θ.

7 - Question

Conical flow is assumed to be symmetric about which of these axis?

a) X – axis

b) Y – axis

c) Z – axis

d) No symmetry

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Answer: cExplanation: The conical flow is obtained by keeping a wedge in a y – z plane which is rotated about z – axis. This results in an axisymmetric flow in which the flow properties remain constant along the ray in a cone and depend only on radius r and the axis.

8 - Question

What is the flow over right circular cone at zero angle of attack is considered to be?

a) One – dimensional

b) Quasi three – dimensional

c) Three – dimensional

d) Quasi two – dimensional

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Answer: dExplanation: Since the cone is revolved around the z – axis, thus the conical flow is known to be axisymmetric. The spherical coordinate system used to determine the position in this flow is (r, ϕ, θ). But since the flow is axisymmetric, ∂∂ϕ = 0 and thus only (r, θ) coordinate system is used to determine the position in the flow making it quasi two0dimensional flow.

9 - Question

How many unknowns are present in the Taylor Maccoll equation?

a) One

b) Two

c) Three

d) Four

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Answer: aExplanation: Taylor – Maccoll is a one – dimensional equation in which it is dependent on only one unknown. This term is the radial velocity. Thus the radial velocity is a function of angle θ. Vr = f(θ)

10 - Question

The solution proposed by Taylor and Maccoll for supersonic flow over a cone is obtained using which of these techniques?

a) Analytically

b) Graphically

c) Numerically

d) Simulation

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Answer: cExplanation: The supersonic flow over a cone was first obtained by A. Busemann in the year 1929 when the supersonic flow was not studied or achieved practically. Later in the year 1933, Taylor and Maccoll came up with a numerical solution for the supersonic conical flow. The equation obtained is a ordinary differential equation having no closed – form solution thus seeking a numerical solution.