Question 4.8: Determination of Rotationality in a Two-Dimensional Flow Con...

We are to determine whether a flow with a given velocity field is rotational or irrotational, and we are to draw some streamlines in the first quadrant.
Analysis Since the flow is two-dimensional, Eq. 4–31 is applicable. Thus,

Vorticity:        \vec{\zeta } = \left(\frac{∂\upsilon }{∂x} – \frac{∂u}{∂y} \right)\vec{k} = (-2y – 0)\vec{k} = -2y\vec{k}         (2)

Since the vorticity is nonzero, this flow is rotational. In Fig. 4–49 we plot several streamlines of the flow in the first quadrant; we see that fluid moves downward and to the right. The translation and deformation of a fluid parcel is also shown: at Δt = 0, the fluid parcel is square, at Δt = 0.25 s, it has moved and deformed, and at Δt = 0.50 s, the parcel has moved farther and is further deformed. In particular, the right-most portion of the fluid parcel moves faster to the right and faster downward compared to the left-most portion, stretching the parcel in the x-direction and squashing it in the vertical direction. It is clear that there is also a net clockwise rotation of the fluid parcel, which agrees with the result of Eq. 2.
Discussion   From Eq. 4–29, individual fluid particles rotate at an angular velocity equal to \vec{\omega } = -y \vec{k} , half of the vorticity vector. Since \vec{\omega } is not constant, this flow is not solid-body rotation. Rather, \vec{\omega } is a linear function of y. Further analysis reveals that this flow field is incompressible; the area (and volume) of the shaded regions representing the fluid parcel in Fig. 4–49 remains constant at all three instants in time.

\vec{\omega }= \frac{1}{2 }\vec{\triangledown } \times \vec{V} = \frac{1}{2}curl(\vec{V} ) = \frac{\vec{\zeta } }{2}              (4.29)