Chapt06

153600 ft3s 0002 ft2 332 fts pa and v1 z1 z2 2 0 in

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Unformatted text preview: (6.47): hf 1.66 ft 32 (1.0 ft)(3.32 ft/s) (58 lbf/ft3)(0.004 ft)2 32 LV gd2 1.66 114,500 or 114,500 1.45 E-5 slug/(ft s) Ans. Note that L in this formula is the pipe length of 1 ft. Finally, check the Reynolds number: Red (58/32.2 slug/ft3)(3.32 ft/s)(0.004 ft) 1.45 E-5 slug/(ft s) Vd 1650 laminar Since this is less than 2300, we conclude that the flow is indeed laminar. Actually, for this head loss, there is a second (turbulent) solution, as we shall see in Example 6.8. Turbulent-Flow Solution For turbulent pipe flow we need not solve a differential equation but instead proceed with the logarithmic law, as in Example 6.3. Assume that Eq. (6.21) correlates the local mean velocity u(r) all the way across the pipe u(r) u* 1 where we have replaced y by R V Q A 1 R2 0.41 and B r)u* B (6.48) r. Compute the average velocity from this profile R u* 1 ln (R r)u* B 2 r dr 0 1 2 Ru* u* ln 2 Introducing (R ln 2B 3 (6.49) 5.0, we obtain, numerically, V u* 2.44 ln Ru* 1.34 (6.50) | v v This looks only marginally interesting until we re...
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