Engineering Questions with Answers - Multiple Choice Questions

# MCQs on 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

b
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

b
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

a
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

c
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

b
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