17. The velocity potential function u(r, ?) for steady flow...

17. The velocity potential function u(r, ?) for steady flow of an ideal fluid around a cylinder of radius r = a satisfies the boundary value problem r^2u_rr + ru_r + u_{??} = 0 (r > a); u_r(a, ?) = 0, u(r, ?) = u(r, -?), lim_{r??} [u(r, ?) ? U_0r cos ?] = 0. (a) By separation of variables, derive the solution u(r, ?) = rac{U_0}{r}(r^2 + a^2) cos ?. (b) Hence show that the velocity components of the flow are u_x = rac{?u}{?x} = rac{U_0}{r^2}(r^2 ? a^2 cos 2?) and u_y = rac{?u}{?y} = ?rac{U_0}{r^2}a^2 sin 2?. The streamlines for this fluid flow around the cylinder are shown in Fig. 9.7.9. FIGURE 9.7.9. Streamlines for ideal fluid flow around a cylinder.

Transcript

00:01 Okay, we have an equation of this form.

00:06 We're given these boundary conditions.

00:09 This is laplace's equation in polar coordinates.

00:15 And the boundary condition at infinity tells us that u, because it's a, remember u is a, it's a velocity potential, essentially.

00:32 And what this is telling us is that at large distances that it basically points in only the z direction or the only in the x direction, actually in our case.

00:45 So it's actually a very stringent condition.

00:51 It tells us a lot.

00:54 So there's a lot of ways to do this, but they want us to use separation of variables.

00:58 So we'll do that.

00:59 So we take a function of r times a function of theta.

01:05 We plug them into the equation and we can get it to look like this where lambda is some constant, okay? and so the solutions for r are gonna be of this form.

01:26 So if we assume that it's a power, and it is because it's r squared, r double prime plus r prime.

01:36 So the only way that'll work is if r is a power law where alpha squared equals lambda.

01:52 It tells me why f of theta is some sine plus some cosine of alpha theta, okay? so the next step is this.

02:09 So our radial equation is gonna be some r to the alpha plus something else, r to the minus alpha, because we know alpha squared is lambda.

02:21 So that's basically what it looks like.

02:23 So our u overall is gonna look like this.

02:34 I'm gonna kind of rename my constants.

02:42 So that's basically what our function looks like.

02:44 Now, when r goes to infinity, we'll lose the f term, r to the minus alpha.

02:54 And what we're left with is gonna be, needs to be r times cosine theta.

03:02 So that's actually a very stringent condition.

03:06 In general, we might expect u to be a sum of terms that looks like this.

03:11 But because of the fact that our, what happens at r goes to infinity actually really restricts what our values of alpha can be.

03:23 And it can only be one.

03:32 So alpha is one and capital b is zero, capital c is one.

03:40 And so our u is some e r plus f over r times the cosine of theta, okay? now they do have that requirement that theta goes to minus theta, it keeps u the same, but we don't really need it.

04:07 So all it's really saying is that it's even on theta, but we lose all the odd terms by the simple, by that requirement that is at large r, it has to go like cosine theta.

04:24 So anyway, there's our u.

04:30 Because it goes to u zero, r cos theta at large r, that tells me that e has to be u zero, okay? and then at r equals a, our u r, our r derivative is e minus f over r squared.

04:57 That's true at r equals a.

05:07 So f is equal to u zero times a squared...

Question

Please give Ace some feedback