34 since the left hand side varies only with r and

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Unformatted text preview: p 2 dx gz) (const)(r) (6.35) 2 | v v Ask your instructor to explain this to you if necessary. | e-Text Main Menu | Textbook Table of Contents | Study Guide 6.4 Flow in a Circular Pipe 341 Thus the shear varies linearly from the centerline to the wall, for either laminar or turbulent flow. This is also shown in Fig. 6.10. At r R, we have the wall shear 1 R 2 w p gz L (6.36) which is identical with our momentum relation (6.27). We can now complete our study of pipe flow by applying either laminar or turbulent assumptions to fill out Eq. (6.35). Laminar-Flow Solution Note in Eq. (6.35) that the HGL slope d( p gz)/dx is negative because both pressure and height drop with x. For laminar flow, du/dr, which we substitute in Eq. (6.35) du dr 1 rK 2 K d (p dx gz) (6.37) Integrate once 1 2K r 4 u C1 (6.38) The constant C1 is evaluated from the no-slip condition at the wall: u 1 2K R 4 0 0 at r C1 R (6.39) 12 or C1 4 R K/ . Introduce into Eq. (6.38) to obtain the exact solution for laminar fully developed pipe flow 1 4 u d (p dx gz)...
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