This set of Aerodynamics Multiple Choice Questions & Answers (MCQs) focuses on “The Velocity Potential Equation”.

1. For an irrotational flow having velocity potential ϕ = 2x + 3z^{2} – 4y^{2} + 8x^{2}, the flow field satisfies continuity equation.

a) True

b) False

View Answer

Explanation: The velocity potential is given by ϕ = 2x + 3y – 4y

^{2}+ 8x

^{2}

The velocity components u and v are calculated as follows:

u = –\(\frac {∂ϕ}{∂x} = -\frac {∂}{∂x}\)(2x + 3y – 4y

^{2}+ 8x

^{2}) = -2 – 16x

v = –\(\frac {∂ϕ}{∂y} = -\frac {∂}{∂y}\)(2x + 3y – 4y

^{2}+ 8x

^{2}) = -3 + 8y

The continuity equation is given by:

\(\frac {∂u}{∂x} + \frac {∂v}{∂y}\) = 0

Substituting the values of u and v,

\(\frac {∂}{∂x}\)(-2 – 16x) + \(\frac {∂}{∂y}\)(-3 + 8y) = -16 + 8 = -8

Since this is not equal to zero, hence continuity equation is not satisfied.

2. 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**

View Answer

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. 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 = –\(\frac {∂ϕ}{∂x}\), v = –\(\frac {∂ϕ}{∂y}\), w = –\(\frac {∂ϕ}{∂z}\)

The continuity equation for three – dimensional flow is given by:

\(\frac {∂u}{∂x} + \frac {∂v}{∂y} + \frac {∂w}{∂z}\) = 0

Substituting the velocity components in the continuity equation, we get

\(\frac {∂}{∂x} \big ( – \frac {∂ϕ}{∂x} \big ) + \frac {∂}{∂y} \big ( – \frac {∂ϕ}{∂x} \big ) + \frac {∂}{∂z} \big ( – \frac {∂ϕ}{∂x} \big )\) = 0

\(\frac {∂^2 ϕ}{∂x} + \frac {∂^2 ϕ}{∂y} + \frac {∂^2 ϕ}{∂z}\) = 0

The above final equation is known as Laplace equation, thus velocity potential satisfies the Laplace equation.

4. If the velocity potential is given by ϕ = +2x^{2} – 4xy^{2} + \( \frac {8x^2}{y}\), 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 = 2x

^{2}– 4xy

^{2}+ \( \frac {8x^2}{y}\)

Velocity component in x – direction is given by u = –\( \frac {∂ϕ}{∂x}\)

u = –\( \frac {∂}{∂x} \big ( \)2x

^{2}– 4xy

^{2}+ \( \frac {8x^2}{y} \big ) \) = 4x – 4y

^{2}+ \( \frac {16x}{y}\)

For calculating the velocity component at point (2,1), we substitute these points in the above equation

u = 4(2) – 4(1)

^{2}+ \( \frac {16(2)}{(1)}\) = 8 – 4 + 32 = 36 m/s

5. What is the nature of the flow having a velocity potential?

a) Rotational

b) Irrotational

c) Inviscid

d) Viscous

View Answer

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. 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:

\(\frac {∂^2 ϕ}{∂x} + \frac {∂^2 ϕ}{∂y} + \frac {∂^2 ϕ}{∂z}\) = 0

Or

∇

^{2}ϕ = 0

**Sanfoundry Global Education & Learning Series – Aerodynamics.**

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