CHEMICAL ENGINEERING

FLOW IN PIPES AND CHANNELS 29

PuttingnD 0 .5, the volumetricflowrateQis given by:

QD

(

P

4 kl

) 2 (

1

10

)

d^3

(

4

d^2

)

PuttingdD 2 a, then:

QD

(

5

)

a^5

(

P

2 kl

) 2

PROBLEM 3.13

Calculate the pressure drop when 3 kg/s of sulphuric acidflows through 60 m of 25 mm
pipe ( D1840 kg/m^3 , D 0 .025 N s/m^2 ).

Solution

Reynolds numberDud/ D 4 G/ dD 4. 30 /ð 0. 025 ð 0. 025 D 6110

Ifeis taken as 0.05 mm, then:e/dD 0. 05 / 25 D 0 .002.

From Fig. 3.7,R/u^2 D 0 .0046.

Acid velocity in pipeD 3. 0 /[1840ð/ 4  0. 025 ^2 ]D 3 .32 m/s.

From equation 3.18, the pressure drop due to friction is given by:

PD 4 R/u^2 l/du^2

D 4 ð 0. 0046  60 / 0. 025  1840 ð 3. 322 D 895 ,620 N/m^2 or 900 kN/m^2

PROBLEM 3.14

The relation between cost per unit lengthCof a pipeline installation and its diameterd
is given by:CDaCbdwhereaandbare independent of pipe size. Annual charges are
a fractionˇof the capital cost. Obtain an expression for the optimum pipe diameter on
a minimum cost basis for afluid of density and viscosity flowing at a mass rate of
G. Assume that thefluid is in turbulentflow and that the Blasius equation is applicable,
that is the friction factor is proportional to the Reynolds number to the power of minus
one quarter. Indicate clearly how the optimum diameter depends onflowrate andfluid
properties.

Solution

The total annual cost of a pipeline consists of a capital charge plus the running costs. The
chief element of the running cost is the power required to overcome the head loss which
is given by:
hfD 8 R/u^2 l/du^2 / 2 g (equation 3.20)