Chapt06

# Solution first compute the velocity from the known

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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, hf 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...
View Full Document

## This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.

Ask a homework question - tutors are online