{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapt06

# For a gradual contraction the loss is very small as

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: Textbook Table of Contents | Study Guide 6.8 Multiple-Pipe Systems Loss 375 K Sharp entrance (Fig. 6.21) Open globe valve (2 in, Table 6.5) 12-in bend (Fig. 6.20) Regular 90° elbow (Table 6.5) Half-closed gate valve (from Fig. 6.18b) Sharp exit (Fig. 6.21) 0.5 6.9 0.15 0.95 2.7 1.0 12.2 K Calculate the Reynolds number and pipe-friction factor Red For /d Vd 9.17( 122 ) 0.000011 0.001, from the Moody chart read f 139,000 0.0216. Substitute into Eq. (1) 100 ft (9.17 ft/s)2 2(32.2 ft/s2) 0.0216(400) 100 ft hp 84 ft pump head 184 ft 2 12 12.2 The pump must provide a power to the water of P gQhp [1.94(32.2) lbf/ft3](0.2 ft3/s)(184 ft) The conversion factor is 1 hp 2300 ft lbf/s 550 ft lbf/s. Therefore P 2300 550 4.2 hp Ans. Allowing for an efficiency of 70 to 80 percent, a pump is needed with an input of about 6 hp. 6.8 Multiple-Pipe Systems8 If you can solve the equations for one-pipe systems, you can solve them all; but when systems contain two or more pipes, certain basic rules make the calculations very smooth. Any resemblance...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online