Chapt06

# Chapt06

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Unformatted text preview: in 40° 0.18 kg/(m s) (10 m)(0.643) 6.43 m The flow goes in the direction of falling HGL; therefore compute the hydraulic grade-line height at each section HGL1 z1 p1 g 0 HGL2 z2 p2 g 6.43 350,000 900(9.807) 39.65 m 250,000 900(9.807) 34.75 m The HGL is lower at section 2; hence the flow is from 1 to 2 as assumed. Part (b) Ans. (a) The head loss is the change in HGL: hf HGL1 HGL2 39.65 m 34.75 m 4.9 m Ans. (b) Half the length of the pipe is quite a large head loss. Part (c) We can compute Q from the various laminar-flow formulas, notably Eq. (6.47) Q Part (d) gd4hf 128 L (900)(9.807)(0.06)4(4.9) 128(0.18)(10) Q R2 0.0076 (0.03)2 2.7 m/s Ans. (d) 810 Ans. (e) With V known, the Reynolds number is Red 1 Ans. (c) Divide Q by the pipe area to get the average velocity V Part (e) 0.0076 m3/s Vd 2.7(0.06) 0.0002 This is well below the transition value Red 2300, and so we are fairly certain the flow is laminar. Notice that by sticking entirely to consistent SI units (meters, seconds, kilograms, newtons) for all variables we avoid the need for any conversion factors i...
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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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