Question

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

The question was asked by my school teacher while I was bunking the class.

This key question is from The Velocity Potential Equation in chapter Velocity Potential Equation of Aerodynamics

Answer 1

The correct answer is (c) 36 m/s

The best 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