EXAMPLE 6-1 Momentum-Flux Correction Factor for Laminar Pipe Flow Vavg Consider laminar flow through a very long straight section of round pipe. It is shown in Chap. 8 that the velocity profile through a cross-sectional area of the pipe is parabolic (Fig. 6-15), with the axial velocity component given by V = 2V 1 - R2 (1) where R is the radius of the inner wall of the pipe and Vvg is the average velocity. Calculate the momentum-flux correction factor through a cross sec- tion of the pipe for the case in which the pipe flow represents an outlet of the control volume, as sketched in Fig. 6-15. Assumptions 1 The flow is incompressible and steady. 2 The control volume slices through the pipe normal to the pipe axis, as sketched in Fig. 6-15. Analysis We substitute the given velocity profile for V in Eq. 6-24 and inte- grate, noting that dA, = 2mr dr, FIGURE Velocity profile over a cross se of a pipe in which the flow is- developed and lam •R 4 dA 2mr dr R2 (2) TR2 avg Defining a new integration variable y = 1 - r2/R and thus dy = -2r drlR2 (also, y = 1 at r 0, and y = 0 at r = R) and performing the integra- For turbulent flow Bmay h- tion, the momentum-flux correction factor for fully developed laminar flow an insignificant effect at in and outlets, but for laminar flow Bmay be important an (3) should not be neglected. It wise to include Bin all becomes Laminar flow: B = -4 y² dy = -4 Discussion We have calculated B for an outlet, but the same result would have been obtained if we had considered the cross section of the pipe as an inlet to the control volume. momentum control volume problems.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Transcribed Image Text:Hello sir, can you solve in detail? How did the output of 4/3 come out without words, just solve the question? EXAMPLE 6-1 Momentum-Flux Correction Factor for Laminar Pipe Flow CV Vavg Consider laminar flow through a very long straight section of round pipe. It is shown in Chap. 8 that the velocity profile through a cross-sectional area of the pipe is parabolic (Fig. 6-15), with the axial velocity component given by V R V = 2Vavg (1) R where R is the radius of the inner wall of the pipe and Vavg is the average velocity. Calculate the momentum-flux correction factor through a cross sec- tion of the pipe for the case in which the pipe flow represents an outlet of the control volume, as sketched in Fig. 6-15. Assumptions 1 The flow is incompressible and steady. 2 The control volume slices through the pipe normal to the pipe axis, as sketched in Fig. 6-15. Analysis We substitute the given velocity profile for V in Eq. 6-24 and inte- grate, noting that dA. = 2nr dr, FIGURE 6- Velocity profile over a cross secti (2) of a pipe in which the flow is fu developed and lamin 2 CR 4 dA. 1 - 2mr dr TR R2 avg Defining a new integration variable y = 1 - r2/R2 and thus dy = -2r dr/R2 %3D (also, y = 1 at r = 0, and y = 0 at r = R) and performing the integra- For turbulent flow Bmay hav tion, the momentum-flux correction factor for fully developed laminar flow an insignificant effect at inlet and outlets, but for laminar flow B may be important and (3) should not be neglected. It is wise to include Bin all %3D becomes 4. Laminar flow: B = -4 dy = -4 3. Discussion We have calculated B for an outlet, but the same result would momentum control volume have been obtained if we had considered the cross section of the pipe as an inlet to the control volume. problems.

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