CHEMICAL ENGINEERING

34 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

∴ uDRe /dD 1. 1 ð 104 ð 0. 01 / 0. 075 D 1. 47 ð 103 kg/m^2 s

∴massflowrateD 1. 47 ð 103 ð 10 ^3 ð 3600 ð 24 ð/ 4  0. 0752 D561 tonne/day

Depletion of storage in 72 hD 561 ð 72 / 24 D1683 tonne

Maximum net gain in capacity in the systemD 561  500 D61 tonne/day and the time
to recharge the tanksD 1683 / 61 D 27 .6days

PROBLEM 3.19

Water is pumped at 1.4 m^3 /s from a tank at a treatment plant to a tank at a local works
through two parallel pipes, 0.3 m and 0.6 m diameter respectively. What is the velocity in
each pipe and, if a single pipe is used, what diameter will be needed if thisflow of water
is to be transported, the pressure drop being the same? Assume turbulentflow with the
friction factor inversely proportional to the one quarter power of the Reynolds number.

Solution

The pressure drop through a pipe is given by equation 3.18:

PD 4

(

R

u^2

)

l

d

u^2

In this case,R/u^2 DKRe^1 /^4 whereKis a constant.

Hence: PD 4 K

(

ud

) 1 / 4

l

d

u^2

DK

u^1.^75 l^0.^75

d^1.^250.^25

DK^0 u^1.^75 /d^1.^25

For pipe 1 in which the velocity isu 1 ,PDK^0 u^11.^75 / 0. 31.^25 and the diameter is
0.3 m. Similarly for pipe 2,PDK^0 u^11.^75 / 0. 61.^25
Henceu 2 /u 1 ^1.^75 D 0. 6 / 0. 3 ^1.^25 D 2 .38 andu 2 /u 1 D 1. 64
The total volumetricflowrateD 1 .4m^3 /sD/ 4 d^21 u 1 Cd^22 u 2 and substituting ford 1
andd 2 andu 2 D 1. 64 u,u 1 D 2 .62 m/sandu 2 D 4 .30 m/s

If a single pipe of diameterd 3 is used for the same duty at the same pressure drop and
the velocity isu 3 , then:

/ 4 d^23 u 3 D 1 .4andd^23 u 3 D 1. 78  1 

and: PDK^1 u^13.^75 /d^13.^25

and: u 3 /u 1 ^1.^75 Dd 3 / 0. 3 ^1.^25

Sinceu 1 D 2 .62 m/s, then:
0. 185 u^13.^75 D 4. 5 d^13.^25  2 

From equations (1) and (2), the required diameter,d 3 D 0 .63 m