Chapt06

# It was proposed by julius weisbach a german professor

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Unformatted text preview: r laminar or turbulent flow, the continuity equation in cylindrical coordinates is given by (App. D) 1 r r 1 r (r r) u x () 0 (6.31) We assume that there is no swirl or circumferential variation, developed flow: u u(r) only. Then Eq. (6.31) reduces to 1 r or r r r (r r) / 0, and fully 0 const (6.32) But at the wall, r R, r 0 (no slip); therefore (6.32) implies that υr 0 everywhere. Thus in fully developed flow there is only one velocity component, u u(r). The momentum differential equation in cylindrical coordinates now reduces to u u x dp dx 1 r gx r (r ) (6.33) where can represent either laminar or turbulent shear. But the left-hand side vanishes because u u(r) only. Rearrange, noting from Fig. 6.10 that gx g sin : 1 r r (r ) d (p dx gx sin ) d (p dx gz) (6.34) Since the left-hand side varies only with r and the right-hand side varies only with x, it follows that both sides must be equal to the same constant.2 Therefore we can integrate Eq. (6.34) to find the shear distribution across the pipe, utilizing the fact that 0 at r 0 1d r (...
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## This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.

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