Unformatted text preview: mate from Fig. 6.9 (a) the friction velocity u*, (b) the wall
shear stress w, and (c) the average velocity V Q/A. Solution
u ( y) For pipe flow Fig. 6.9 shows that the logarithmic law, Eq. (6.21), is accurate all the way to the
center of the tube. From Fig. E6.3 y R r should go from the wall to the centerline as shown.
At the center u u0, y R, and Eq. (6.21) becomes u0 = 5 m /s
Since we know that u0 5 m/s and R
solution by trial and error or by EES r
r = R = 7 cm u*
where we have taken v v E6.3 | 1
0.41 | e-Text Main Menu | 1.51 10 (1) 0.07 m, u* is the only unknown in Eq. (1). Find the
0.228 m/s 5 5.0 22.8 cm/s 2 m /s for air from Table 1.4. Textbook Table of Contents | Study Guide Ans. (a) 338 Chapter 6 Viscous Flow in Ducts Part (b) Assuming a pressure of 1 atm, we have
( w/ )1/2, we compute
w u*2 1.205 kg/m3. Since by definition u* p/(RT) (1.205 kg/m3)(0.228 m/s)2 0.062 kg/(m s2) 0.062 Pa Ans. (b) This is a very small shear stress, but it will cause a large pressure drop in a long pipe (170 Pa
for every 100 m of pi...
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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.
- Spring '08