Question 4.9: If a velocity potential exists for the velocity field of Exa...

Since w = 0, the curl of V has only one z component, and we must show that it is zero:

(\nabla \times V)_z=2\omega_z=\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=\frac{\partial}{\partial x}(-2axy)-\frac{\partial}{\partial y}(ax^2-ay^2)=-2ay+2ay=0                                   checks

The flow is indeed irrotational. A velocity potential exists.

To find Φ(x, y), set

u=\frac{\partial \phi}{\partial x}=ax^2-ay^2                                 (1)

v=\frac{\partial \phi}{\partial y}=-2axy                                     (2)

Integrate (1)

\phi=\frac{ax^3}{3}-axy^2+f(y)                (3)

Differentiate (3) and compare with (2)

\frac{\partial \phi}{\partial y}=-2axy+f^{\prime}(y)=-2axy                  (4)

Therefore f´ = 0, or f = constant. The velocity potential is

Letting C = 0, we can plot the Φ lines in the same fashion as in Example 4.7. The result is shown in Fig. E4.9 (no arrows on Φ). For this particular problem, the Φ lines form the same pattern as the ψ lines of Example 4.7 (which are shown here as dashed lines) but are displaced 30°. The Φ and ψ lines are everywhere perpendicular except at the origin, a stagnation point, where they are 30° apart. We expected trouble at the stagnation point, and there is no general rule for determining the behavior of the lines at that point.