Engineering Questions with Answers - Multiple Choice Questions
Aerodynamics – The Velocity Potential Equation
1 - Question
For an irrotational flow having velocity potential ϕ = 2x + 3z2 – 4y2 + 8x2, the flow field satisfies continuity equation.
a) True
b) False
View Answer
Explanation: The velocity potential is given by ϕ = 2x + 3y – 4y2 + 8x2 The velocity components u and v are calculated as follows: u = –∂ϕ∂x=−∂∂x(2x + 3y – 4y2 + 8x2) = -2 – 16x v = –∂ϕ∂y=−∂∂y(2x + 3y – 4y2 + 8x2) = -3 + 8y The continuity equation is given by: ∂u∂x+∂v∂y = 0 Substituting the values of u and v, ∂∂x(-2 – 16x) + ∂∂y(-3 + 8y) = -16 + 8 = -8 Since this is not equal to zero, hence continuity equation is not satisfied.
2 - Question
For an irrotational flow, what is the relation for the velocity potential?
a) V = ∇ × ϕ
b) V = ∇ϕ
c) V = -∇ × V
d) V = (∇ × V)ϕ
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Explanation: The irrotational flow is given by the curl of velocity vector. If we take a gradient of the scalar function, we get zero as a result. ∇ × (∇ϕ) = 0 Thus, the velocity potential is describing as the scalar function ∇ϕ.
3 - Question
Which of these equations is satisfied by the velocity potential equation?
a) Laplace equation
b) Fano’s equation
c) Bernoulli’s equation
d) Rayleigh equation
View Answer
Explanation: The velocity component is the negative derivative of the velocity potential in that direction. According to this, u = –∂ϕ∂x, v = –∂ϕ∂y, w = –∂ϕ∂z The continuity equation for three – dimensional flow is given by: ∂u∂x+∂v∂y+∂w∂z = 0 Substituting the velocity components in the continuity equation, we get ∂∂x(–∂ϕ∂x)+∂∂y(–∂ϕ∂x)+∂∂z(–∂ϕ∂x) = 0 ∂2ϕ∂x+∂2ϕ∂y+∂2ϕ∂z = 0 The above final equation is known as Laplace equation, thus velocity potential satisfies the Laplace equation.
4 - Question
If the velocity potential is given by ϕ = +2x2 – 4xy2 + 8x2y, then what is the value of velocity component in x – direction at point (2,1)?
a) 30 m/s
b) 15 m/s
c) 36 m/s
d) 24 m/s
View Answer
Explanation: The velocity potential is given as ϕ = 2x2 – 4xy2 + 8x2y Velocity component in x – direction is given by u = –∂ϕ∂x u = –∂∂x(2x2 – 4xy2 + 8x2y) = 4x – 4y2 + 16xy For calculating the velocity component at point (2,1), we substitute these points in the above equation u = 4(2) – 4(1)2 + 16(2)(1) = 8 – 4 + 32 = 36 m/s
5 - Question
What is the nature of the flow having a velocity potential?
a) Rotational
b) Irrotational
c) Inviscid
d) Viscous
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Explanation: For irrotational flow, curl of velocity vector yields zero. In case the curl of any vector is zero i.e.∇ × V = 0, where V is a vector, it is also expressed in the form of ∇ζ where ζ is a scalar function. In case of irrotational flow, velocity potential ϕ is the scalar function. Hence if the flow has a velocity potential, it automatically implies that it is irrotational.
6 - Question
If the Laplace equation is satisfied by the velocity potential, then the fluid flows.
a) True
b) False
View Answer
Explanation: For the fluid to flow, it is essential for the velocity potential ϕ to satisfy the Laplace equation. If the condition is not met, the fluid does not flow and has zero velocity. Thus, the condition for the fluid to flow is: ∂2ϕ∂x+∂2ϕ∂y+∂2ϕ∂z = 0 Or ∇2ϕ = 0