Math144Notes

# In other words zt k it t0 therefore the corresponding

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: , and so the streamlines are 1r sin ø = const, or y = const—horizontal lines. Velocity Field: È 1˘ F'(z) = A Í1 - 2˙ , Î z˚ whence È 1˘ F'(z) = A Í1 - 2˙ Î z– ˚ 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. - + 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 -1 potential on H; P: HÆ C . Then the image, under F of the streamlines and I equipotentials on H are...
View Full Document

## This document was uploaded on 03/20/2014 for the course MATH 144 at Hofstra University.

Ask a homework question - tutors are online