Chapt06

For a gradual contraction the loss is very small as

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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...
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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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