Chapt06

# Chapt06

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Unformatted text preview: 2 , deg 30 45 60 0.02 K for gradual contraction 0.04 0.07 References 15, 16, 43, and 46 contain additional data on minor losses. EXAMPLE 6.16 Water, 1.94 slugs/ft3 and 0.000011 ft2/s, is pumped between two reservoirs at 0.2 ft3/s through 400 ft of 2-in-diameter pipe and several minor losses, as shown in Fig. E6.16. The roughness ratio is /d 0.001. Compute the pump horsepower required. 2 Screwed regular 90° elbow z2 = 120 ft Sharp exit 1 z1 = 20 ft Sharp entrance Half-open gate valve 12-in bend radius Pump Open globe valve E6.16 400 ft of pipe, d = 2 ft 12 Solution Write the steady-flow energy equation between sections 1 and 2, the two reservoir surfaces: 2 V1 2g p1 g 2 V2 2g p2 g z1 z2 hf where hp is the head increase across the pump. But since p1 pump head hp z2 z1 hf hm 120 ft 20 ft hm p2 and V1 V2 fL 2g d hp V2 0, solve for the K (1) Now with the flow rate known, calculate V Q A 0.2 ft3/s 1 4 ( 122 ft)2 9.17 ft/s Now list and sum the minor loss coefficients: | v v 374 | e-Text Main Menu |...
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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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