Consider the flow field given by wax²-ay², where a 3 s¹. Show that the flow is irrotational. Determine the velocity poten- tial for this flow. Given: Incompressible flow field with y=ax²-ay², where a=3s-¹ Find: (a) Whether or not the flow is irrotational. (b) The velocity potential for this flow. Solution: If the flow is imotational, V2=0. Checking for the given flow, V² = (ax²-ay²)+(ax²-ay²)=2a-2a=0 so that the flow is irrotational. As an alternative proof, we can compute the fluid particle rotation (in the xy plane, the only com- ponent of rotation is e₂): до ди дуг then So 2002= and и- 1- dx dy dx =(x²-ay²)=-2ay and a D=-- (ax²-ay²)=-2ax дх 200= до ди dx (-2ax)- -(-2ay) = -2a+2a=0. Once again, we conclude that the flow is irrotational. Because it is irrotational, & must exist, and u=-√x аф аф and D= dy 2a Consequently, u=- ap function of y. Then =-2ay and дх дл =2ay. Integrating with respect to x gives =2axy+f(y), where f(y) is an arbitrary аф a D= -2x= - 2axy+f(y)] dy Therefore, -2ax=-2ax- -= 2ax af (y) =0 and f=constant. Thus dydy =2axy+constant We also can show that lines of constant and constant are orthogonal. wax-ay and -2axy For y=constant, dy =0=2axdx-2aydy; hence For constant, d=0=2aydx+2axdy; hence -c 4-0 y The slopes of lines of constant and constant y are negative reciprocals. Therefore lines of constant & are orthogonal to lines of constant w. This problem illustrates the relations among the stream function, velocity potential, and velocity field. The stream function and velocity potential are shown in the Excel workbook. By entering the equations for and 4, other fields can be plotted.