This preview shows page 1. Sign up to view the full content.
Unformatted text preview: n the calculations. 1 ft EXAMPLE 6.5 1 ft d = 0.004 ft A liquid of specific weight g 58 lb/ft3 flows by gravity through a 1-ft tank and a 1-ft capillary tube at a rate of 0.15 ft3/h, as shown in Fig. E6.5. Sections 1 and 2 are at atmospheric pressure. Neglecting entrance effects, compute the viscosity of the liquid.
Q = 0.15 ft3/ h Solution E6.5 | v v Apply the steady-flow energy equation (6.24), including the correction factor : | e-Text Main Menu | Textbook Table of Contents | Study Guide 344 Chapter 6 Viscous Flow in Ducts
1V 1 p1
g 2g 2
2V 2 p2
g z1 2g z2 hf The average exit velocity V2 can be found from the volume flow and the pipe size:
p2 pa, and V1 hf Meanwhile p1 z1 z2 (0.15/3600) ft3/s
(0.002 ft)2 Q
R2 3.32 ft/s 0 in the large tank. Therefore, approximately,
2g 2.0 ft 2.0 (3.32 ft/s)2
2(32.2 ft/s2) 1.66 ft where we have introduced 2 2.0 for laminar pipe flow from Eq. (3.72). Note that hf includes
the entire 2-ft drop through the system and not just the 1-ft pipe length.
With the head loss known, the viscosity follows from our laminar-flow formula...
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.
- Spring '08