428 Substitution of 436 into Eq 435b and slight rearrangement of the result

428 substitution of 436 into eq 435b and slight

This preview shows page 137 - 139 out of 164 pages.

the same pressure gradient treatment used in the plane Poiseuille case, Eq. (4.28).) Substitution of (4.36) into Eq. (4.35b), and slight rearrangement of the result, yields ∂r parenleftbigg r ∂u ∂r parenrightbigg = p μL r , (4.37) a second-order differential equation needing two boundary conditions in order to determine inte- gration constants and produce a complete solution. The first of these corresponds to the no-slip condition for a viscous flow, applied at the pipe wall r = R ; that is, u ( R ) = 0 . (4.38) The second condition to be employed in the present case is less intuitive, but it is one that arises often when working with differential equations expressed in polar coordinate systems. Namely, we require that the velocity remain bounded at r = 0, the center of the pipe. This is equivalent to imposing the formal mathematical condition ∂u ∂r vextendsingle vextendsingle vextendsingle vextendsingle r =0 = 0 , (4.39) which is a natural choice that will enforce the expected symmetry of the velocity profile depicted in Fig. 4.13, We now integrate Eq. (4.37) to obtain r ∂u ∂r = p 2 μL r 2 + C 1 , or ∂u ∂r = p 2 μL r + C 1 r . (4.40) Next we observe that at r = 0 this formula for the velocity derivative will become unbounded unless C 1 = 0. Thus, on physical grounds we should set C 1 = 0. But formally, we can apply the second boundary condition given in Eq. (4.39) to show that, indeed, if this condition is satisfied C 1 = 0 will hold, thus eliminating the unbounded term. We also remark that the equation preceding (4.40) leads to this result without any assumptions beyond boundedness of ∂u/partialr —just simple evaluation at r = 0. (Note that if the term C 1 /r were retained, a second integration, as will be performed next, would result in u ln r which is unbounded as r 0.) We can now integrate (4.40) with C 1 = 0 to obtain u ( r ) = p 4 μL r 2 + C 2 , (4.41) and application of the no-slip condition imposed by Eq. (4.38) gives 0 = pR 2 4 μL + C 2 , or C 2 = pR 2 4 μL .
Image of page 137
132 CHAPTER 4. APPLICATIONS OF THE NAVIER–STOKES EQUATIONS Finally, substitution of this back into Eq. (4.41) yields u ( r ) = p 4 μL ( r 2 R 2 ) = pR 2 4 μL bracketleftbigg 1 parenleftBig r R parenrightBig 2 bracketrightbigg , (4.42) the Hagen–Poiseuille velocity profile for fully-developed flow in a pipe of circular cross section. Some Physical Consequences of the Hagen–Poiseuille Formula We observe from the definition given earlier for ∆ p (= p 1 p 2 ), that we expect ∆ p > 0 to hold; and we see that for this case the flow is indeed in the direction indicated in Fig. 4.13. Furthermore, it is clear from Eq. (4.42) that the variation of u with r is quadratic, and this is quite similar to the velocity profile found for plane Poiseuille flow. Two quantities of interest can be derived directly from the velocity profile obtained above. The first of these is the maximum velocity (actually speed in this case) which from the preceding figures we expect will occur on the centerline of the pipe, i.e. , at r = 0. In fact, we know from calculus that the maximum of a (twice-differentiable) function occurs at the location where the first derivative is zero and, simultaneously, the second derivative is negative. From Eq. (4.42) we see that ∂u ∂r = 1 2 pR 2 μL · r R 2 , and setting this to zero implies that r = 0, as expected, for the location of maximum velocity.
Image of page 138
Image of page 139

You've reached the end of your free preview.

Want to read all 164 pages?

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes
A+ icon
Ask Expert Tutors