Question

If the velocity potential function ϕ =5 (x² – y²), the velocity components at the points (4, 5) will be
1. u = -35, v = 40
2. u = -40, v = 55
3. u = -40, v = 50
4. u = 40, v = -50

Answer 1

Correct Answer - Option 3 : u = -40, v = 50

Concept:

Velocity potential function (ϕ):

• If ϕ exists → Irrotational flow
• Equation of equipotential lines → \(\frac{dy}{dx}=\frac{-u}{v}\)
• It is a scalar function of space and time such that

\(\frac{{\partial ϕ }}{{\partial x}} = - u, ~\frac{{\partial ϕ }}{{\partial y}} = - v,~\frac{{\partial ϕ }}{{\partial z}} = - w\)       

Here,

u, v, and w - components of velocity vector along x, y, and z directions respectively.

Calculation:

Given,

ϕ = 5(x² - y²)

From (1)

\(u = -\frac{{\partial ϕ }}{{\partial x}} = -5\frac{{\partial {(x^2 - y^2)} }}{{\partial x}} = -10x\)

\(v = -\frac{{\partial ϕ }}{{\partial y}} = -5\frac{{\partial {(x^2 - y^2)} }}{{\partial y}} = 10y\)

Hence,

At (4, 5)

u = -4 × 10 = -40

v = 5 × 10 = 50.