459
Chapter 6 x Viscous Flow in Ducts
The match-point at the center gives us a log-law estimate of the shear stress:
1 § hu* ·
V
| ln ¨
B, N | 0.41, B | 5.0, u*
2u* N © 2Q ¸¹
(W w U )12
Ans.
This is one form of “dimensionless shear stress.” The more normal form is friction
coefficient versus Reynolds number. Calculations from the log-law fit a Power-law
curve-fit expression in the range 2000 Reh 1E5:
Cf
Ww
0.018
|
2
(1/2)UV
(UVh/Q )1/4
0.018
Re1h4
Ans.
6.38 Suppose in Fig. P6.37 that h 3 cm, the fluid is water at 20°C (U 998 kg/m3, P
0.001 kg/m˜s), and the flow is turbulent, so that the logarithmic law is valid. If the shear
stress in the fluid is 15 Pa, estimate V in m/s.
Solution: Just as in Prob. 6.37, apply the log-law at the center between the wall, that is,
y h/2, u V/2. With Ww known, we can evaluate u* immediately:
u*
or:
V /2
0.123
Ww
U
15
998
0.123
m V /2 1 § u * h/2 ·
| ln ¨
,
¸ B,
s
u* N © Q ¹
ª 0.123(0.03/2) º
1
ln «
5.0
0.41 ¬ 0.001/998 »¼
23.3, Solve for V | 5.72
m
s
Ans.
6.39 By analogy with laminar shear, W P du/dy. T. V. Boussinesq in 1877 postulated
that turbulent shear could also be related to the mean-velocity gradient Wturb H du/dy,
where H is called the eddy viscosity and is much larger than P. If the logarithmic-overlap
law, Eq. (6.28), is valid with W | Ww, show that H | NUu*y.
Solution: Differentiate the log-law, Eq. (6.28), to find dudy, then introduce the eddy
viscosity into the turbulent stress relation:
If
Then, if
u
u*
du
§ yu ·
B, then
ln ¨
¸
N © Q ¹
dy
1
W | W w { Uu *2
H
du
dy
H
u*
Ny
u*
, solve for H
Ny
NU u * y Ans.