Chapt06

Chapt06

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 |...
View Full Document

This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.

Ask a homework question - tutors are online