CHEMICAL ENGINEERING

FLOW IN PIPES AND CHANNELS 31

Massflow rate per tubeD 0. 84 ð 0. 00049 D 0 .000412 m^3 /sD 0 .412 kg/s

Hence the number of tubes requiredD 47. 8 / 0. 412 D116 tubes

PROBLEM 3.16

Sulphuric acid is pumped at 3 kg/s through a 60 m length of smooth 25 mm pipe. Calcu-
late the drop in pressure. If the pressure drop falls by one half, what will be the new
flowrate? Density of acidD1840 kg/m^3. Viscosity of acidD25 mN s/m^2.

Solution

Cross-sectional area of pipeD/ 4  0. 0252 D 0 .00049 m^2.

Volumetricflowrate of acidD 3. 0 / 1840 D 0 .00163 m^3 /s.

Velocity of acid in the pipeD 0. 00163 / 0. 00049 D 3 .32 m/s.

Reynolds number,ud/ D 1840 ð 3. 32 ð 0. 025 / 25 ð 10 ^3 D 6120

From Fig. 3.7 for a smooth pipe andReD6120,R/u^2 D 0 .0043.

The pressure drop is calculated from equation 3.18:

PfD 4 R/u^2 l/du^2 

D 4 ð 0. 0043  60 / 0. 025  1840 ð 3. 322 D 837 ,200 N/m^2 or 840 kN/m^2

If the pressure drop falls to 418,600 N/m^2 , equation 3.23 and Fig. 3.8 may be used to
calculate the newflow.

From equation 3.23 :R/u^2 Re^2 DPfd^3 / 4 l^2

D 418 , 600 ð 0. 0253 ð 1840 / 4 ð 60 ð 252 ð 10 ^6 

D 8. 02 ð 104

From Fig. 3.8:ReD3800 and the new velocity is:

u^0 D 3800 ð 25 ð 10 ^3 / 1840 ð 0. 025 D 2 .06 m/s

and the massflowrateD 2. 06 ð 0. 00049 ð 1840 D 1 .86 kg/s

PROBLEM 3.17

A Bingham plastic material isflowing under streamline conditions in a pipe of circular
cross-section. What are the conditions for one half of the totalflow to be within the central
core across which the velocity profile isflat? The shear stress acting within thefluid,Ry,
varies with velocity gradient dux/dyaccording to the relation:RyRcDkdux/dy
whereRcandkare constants for the material.