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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
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* 1 where we have replaced y by R
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
Introducing (R ln 2B 3 (6.49) 5.0, we obtain, numerically,
u* 2.44 ln Ru* 1.34 (6.50) | v v This looks only marginally interesting until we re...
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- Spring '08