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
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
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
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
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