This preview shows page 1. Sign up to view the full content.
Unformatted text preview: , and so the
streamlines are 1r sin ø = const, or y = const—horizontal lines.
F'(z) = A Í1 - 2˙ ,
F'(z) = A Í1 - 2˙
We get stagnation points when the velocity equals zero, so we see that this gives z = ±
Exercise Set 12
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. - + 41 The width of the aperture is set to 2 a . To model this, [Suggestion: Consider what the
inverse sine function does to this region.] 13. Parametrically Representing Streamlines, and Using Technology
Sometimes it is hard to draw the streamlines from the implicit equation that defines them.
An analytic approach (short of seeing directly what the curves are, as we have done up to
now) is to find an equation for dy/dx using implicit differentiation and then drawing the
integral curves using technology. However, a more direct way is the following, which
hinges on inverting the conformal mappings we have been using up to now.
Proposition Streamlines and Equipotentials go to Streamlines and Equipotential
S’pose F:DÆH is a conformal invertible map, with P= ∞ + i§ is a complex
potential on H; P: HÆ C . Then the image, under F of the streamlines and
equipotentials on H are...
View Full Document