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Unformatted text preview: f 0.0227]. Then the head loss is
hf f L V2
d 2g (0.0225) 500 m (6.4 m/s)2
0.2 m 2(9.81 m/s2) 117 m Ans. (a) From Eq. (6.25) for the inclined pipe,
or g[hf p p
g z1 (500 m) sin 10°] p
g z2 g(117 m (900 kg/m3)(9.81 m/s2)(30 m) L sin 10°
87 m) 265,000 kg/(m s2) 265,000 Pa Ans. (b) EXAMPLE 6.8
Repeat Example 6.5 to see whether there is any possible turbulent-flow solution for a smoothwalled pipe. Solution
In Example 6.5 we estimated a head loss hf
this condition the friction factor is
f hf d 2g
L V2 (1.66 ft) 1.66 ft, assuming laminar exit flow (
(0.004 ft)(2)(32.2 ft/s2)
(1.0 ft)(3.32 ft/s)2 2.0). For 0.0388 For laminar flow, Red 64/f 64/0.0388 1650, as we showed in Example 6.5. However, from
the Moody chart (Fig. 6.13), we see that f 0.0388 also corresponds to a turbulent smooth-wall
condition, at Red 4500. If the flow actually were turbulent, we should change our kineticenergy factor to
1.06 [Eq. (3.73)], whence the corrected hf 1.82 ft and f 0.0425. With
f known, we can estimate the Reynolds number fro...
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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.
- Spring '08