Aerodynamics Questions and Answers – The Velocity Potential Equation

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 + 3z2 – 4y2 + 8x2, the flow field satisfies continuity equation.
a) True
b) False
View Answer

Answer: b
Explanation: The velocity potential is given by ϕ = 2x + 3y – 4y2 + 8x2
The velocity components u and v are calculated as follows:
u = –\(\frac {∂ϕ}{∂x} = -\frac {∂}{∂x}\)(2x + 3y – 4y2 + 8x2) = -2 – 16x
v = –\(\frac {∂ϕ}{∂y} = -\frac {∂}{∂y}\)(2x + 3y – 4y2 + 8x2) = -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

Answer: 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. 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

Answer: a
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 = +2x2 – 4xy2 + \( \frac {8x^2}{y}\), 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

Answer: c
Explanation: The velocity potential is given as ϕ = 2x2 – 4xy2 + \( \frac {8x^2}{y}\)
Velocity component in x – direction is given by u = –\( \frac {∂ϕ}{∂x}\)
u = –\( \frac {∂}{∂x} \big ( \)2x2 – 4xy2 + \( \frac {8x^2}{y} \big ) \) = 4x – 4y2 + \( \frac {16x}{y}\)
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

Answer: 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.
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6. If the Laplace equation is satisfied by the velocity potential, then the fluid flows.
a) True
b) False
View Answer

Answer: a
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
2ϕ = 0

Sanfoundry Global Education & Learning Series – Aerodynamics.


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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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