Furthermore since p 0 it is clear that the second derivative ofu is uniformly

Furthermore since p 0 it is clear that the second

This preview shows page 138 - 140 out of 164 pages.

= 0, as expected, for the location of maximum velocity. Furthermore, since ∆ p > 0 it is clear that the second derivative of u is uniformly negative implying that r = 0 is the location of the maximum of u . Then evaluation of Eq. (4.42) at r = 0 yields the maximum velocity U max = pR 2 4 μL . (4.43) A second quantity of even more importance for later use is the average flow velocity, U avg . We obtain this by integrating the Hagen–Poiseuille profile over the cross section of the pipe and dividing by the area of the cross section. Thus, we have U avg = 1 πR 2 integraldisplay 2 π 0 integraldisplay R 0 pR 2 4 μL bracketleftbigg 1 parenleftBig r R parenrightBig 2 bracketrightbigg rdrdθ = p 2 μL integraldisplay R 0 r r 3 R 2 dr = p 2 μL bracketleftbigg r 2 2 r 4 4 R 2 bracketrightbigg vextendsingle vextendsingle vextendsingle vextendsingle R 0 . Thus, it follows that U avg = pR 2 8 μL , (4.44) showing that in the case of a pipe of circular cross section the average flow speed is exactly one half the maximum speed for fully-developed flow. It should be emphasized, however, that the factor 1 / 2 arises from the circular geometry and does not hold in general. Finally, we observe that we can use Eq. (4.40) written as (since C 1 = 0) μ ∂u ∂r = p 2 L r
Image of page 138
4.5. PIPE FLOW 133 to calculate the shear stress at any point in the flow. In particular, we see that τ = 0 on the pipe centerline, and at the pipe wall we have τ w = p 2 L R . 4.5.3 Practical Pipe Flow Analysis In the practical analysis of piping systems the quantity of most importance is the pressure loss due to viscous effects along the length of the system, as well as additional pressure losses arising from other physics (that in some cases involve viscosity indirectly). The viscous losses are already embodied in the physics of the Hagen–Poiseuille solution since Eq. (4.37) is actually a force balance needed to maintain fully-developed, steady flow. In particular, the viscous forces represented by the left-hand side of this equation (after multiplication by μ/r ) must be balanced by the pressure forces given on the right-hand side. Thus, a pressure change ∆ p will occur over a distance L due to viscous action throughout the flow, but especially near the pipe walls. This change in pressure is a loss (recall the definition of ∆ p ), and it must be balanced ultimately by a pump. Hence, analyses of the sort we will undertake in this section will provide information on how large a pump should be in order to move a specified fluid (with given density and viscosity) through a piping system. We will begin the section with a rearrangement of the Hagen–Poiseuille solution to express ∆ p in terms of Reynolds number and then obtain formulas for the friction factor in circular cross section pipes, including effects of turbulence. We than introduce a modification of Bernoulli’s equation that permits account of viscosity to be included in an empirical way and quantify this with a physical parameter known as the head loss, which is then related to the friction factor. Following this we will treat so-called minor losses, consideration of which allows analyses of general geometric shapes in piping systems.
Image of page 139
Image of page 140

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