Chapt06

# According to the general rule proposed in eq 689

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Unformatted text preview: effects and take 1000 kg/m3 6 2 and 1.02 10 m /s for water. 1 h=? a = 5 cm b = 3 cm 2 Q, V L = 30 m Water E6.14 Solution Compute the average velocity and hydraulic diameter 0.01 m3/s [(0.05 m)2 (0.03 m)2] Q A V Dh 2(a b) 2(0.05 1.99 m/s 0.03) m 0.04 m Apply the steady-flow energy equation between sections 1 and 2: p1 But p1 p2 pa, V1 12 V1 2 But z1 z2 p2 gz1 0, and V2 12 V2 2 gz2 ghf V in the pipe. Therefore solve for hf f L V2 Dh 2g z1 z2 V2 2g h, the desired reservoir height. Thus, finally, V2 1 2g h f L Dh (1) Since V, L, and Dh are known, our only remaining problem is to compute the annulus friction factor f. For a quick approximation, take Deff Dh 0.04 m. Then ReDh VDh Dh 1.99(0.04) 1.02 10 6 0.046 mm 40 mm 78,000 0.00115 6 Jones and Leung [44] show that data for annular flow also satisfy the effective-laminar-diameter idea. | v v b/a | e-Text Main Menu | Textbook Table of Contents | Study Guide 6.6 Flow in Noncircular Ducts 365 where 0.046 mm has been read from Table 6.1 for commerci...
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