Advanced Engineering Math II phần 5 ppsx

41

This region D maps into H via w = z + 1/z, and, on H, § can again be taken to be

§ = Ay

Giving us a complex potential

F(z) = AÎ

Í

È

˚

˙

˘

z"+"

1

z

To see the real and imaginary parts, use polar form: z = re

iø. This gives

F(z) = AÎ

Í

È

˚

˙

˘

re

iø"+"

1

r

e

-iø

= AÎ

Í

È

˚

˙

˘

r"+"

1

r

cos ø + i AÎ

Í

È

˚

˙

˘

r"-"

1

r

!sin ø

So we can now get the potentials and streamlines in polar form:

Equipotentials:

Î

Í

È

˚

˙

˘

r"+"

1

r

cos ø = const

Kind of complicated to draw these piggies - wait until we do things parametrically in the

next section.

Streamlines:

Î

Í

È

˚

˙

˘

r"-"

1

r

!sin ø = const

Again, these are not standard curves. However, at large distances, 1/r ‡ 0, and so the

streamlines are 1r sin ø = const, or y = const—horizontal lines.

Velocity Field:

F'(z) = A Î

Í

È

˚

˙

˘

1"-"

1

z

2 ,

whence

F'(z) = A Î

Í

È

˚

˙

˘

1"-"

1

z–2

We get stagnation points when the velocity equals zero, so we see that this gives z = ±

1.

Exercise Set 12

Hand In:

1. Compute all the details for the flow around a corner of 60º.

2. Flow through an Aperture: Use a conformal map to model the following flow.

- +