302 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS
Since near a surface dux/dyis positive and assumingRyis approximately constant at a
value at the pipe wall, that isRyDRoDR, then:
RD^2 Edux/dy^2
or:
√
R/DEdux/dy (equation 12.26)
Here,
p
R/, the shearing stress or friction velocity is usually denoted byuŁ.
Since from equation 12.35,ED 0. 4 y, then:
uŁD 0. 4 ydux/dy
Rearranging: dux/uŁDdy/ 0. 4 y
and integrating: ux/uŁD 2 .5lnyCconst. (i)
AtyDr:
umax/uŁD 2 .5lnrCconst.
or: const.Dumax/uŁ 2 .5lnr
Substituting in equation (i):
umax/uŁ 2 .5lnrDux/uŁ 2 .5lny
and: umaxux/uŁD 2 .5lnr/y (ii)
The mean velocity: uD
∫r
0
2 rydyux/r^2
and dividing byr: uD 2
∫ 1
0
1 y/rdy/rux
Substituting foruxfrom equation (ii):
uD 2
∫ 1
0
1 y/rdy/rumaxC 2. 5 uŁlny/r
D 2
(
[
[umaxC 2 .5lny/r][y/r 0. 5 y/r^2 ]
] 1
0
uŁ
∫ 1
0
2. 5 r/y[y/r 0. 5 y/r^2 ]dy/r
)
D 2
(
umax 0. 5 2. 5 uŁ
[
y/r 0. 25 y/r^2
] 1
0
)
Dumax 3. 75 uŁ
Whenu/umaxD 0. 8 ,uDu/ 0. 8 3. 75 u
√
R/u^2
∴
√
R/u^2 D 0 .0667 andR/u^2 D 0. 00444
PROBLEM 12.8
The velocity profile in the neighbourhood of a surface for a Newtonian fluid may be
expressed in terms of a dimensionless velocityuCand a dimensionless distanceyCfrom