CHEMICAL ENGINEERING

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