CHEMICAL ENGINEERING

FLOW IN PIPES AND CHANNELS 33

The total volumetricflowrateQis obtained by integrating the equation for the velocity
profile to give:

QtotalD

∫r

0

y^2 dux/dydy

From equation (iv): QtotalD

1

k

∫r

0

y^2

(

yRw

r

RY

)

dyD

r^3

k

(

Rw

4



RY

3

)

m^3 /s

Over the central core, the volumetricflowrateQcoreis:

QcoreDr 02 u 0 DrRY/Rw^2 u 0 (from 3 

From equation (vi):

QcoreDrRY/Rw^2 r/ 2 kRwRwRY^2 Dr^3 R^2 Y/ 2 kR^3 wRwRY^2

If half the totalflow is to be within the central core, then:

QcoreDQtotal/ 2

r^3 R^2 Y/ 2 kR^3 wRwRY^2 Dr^3 / 2 kRw/ 4 RY/ 3 

and: R^2 YRwRY^2 DR^3 w

(

Rw

4



RY

3

)

PROBLEM 3.18

Oil of viscosity 10 mN s/m^2 and density 950 kg/m^3 is pumped 8 km from an oil refinery
to a distribution depot through a 75 mm diameter pipeline and is then despatched to
customers at a rate of 500 tonne/day. Allowance must be made for periods of maintenance
which may interrupt the supply from the refinery for up to 72 hours. If the maximum
permissible pressure drop over the pipeline is 3450 kN/m^2 , what is the shortest time in
which the storage tanks can be completely recharged after a 72 hour shutdown? The
roughness of the pipe surface is 0.05 mm.

Solution

From equation 3.23:

R

u^2

Re^2 D

Pfd^3

4 l^2

PfD3450 kN/m^2 D 3. 45 ð 106 N/m^2 , dD 0 .075 m, D950 kg/m^3 , lD8000 m
and D10 m Ns/m^2 0 .01 Ns/m^2.

∴

R

u^2

Re^2 D 3. 45 ð 106 ð 0. 0753 ð 950 / 4 ð 8000 ð 0. 012 D 4. 32 ð 105

e/dD 0. 05 / 75 D 0. 0007

From Fig. 3.8 withR/u^2 Re^2 D 4. 32 ð 105 ,e/dD 0 .0007.

ReD 1. 1 ð 104 Ddu/