For an irrotational flow having velocity potential ϕ = 2x + 3z^2 – 4y^2 + 8x^2, the flow field satisfies continuity equation.(a) True(b) FalseThis question was posed to me in examination.This intriguing question comes from The Velocity Potential Equation in division Velocity Potential Equation of Aerodynamics

For an irrotational flow having velocity potential ϕ = 2x + 3z^2 –

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) FalseThis question was posed to me in examination.This intriguing question comes from The Velocity Potential Equation in division Velocity Potential Equation of Aerodynamics
Answer» The correct option is (b) False 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.

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