CHEMICAL ENGINEERING

PUMPING OF FLUIDS 119

Water velocityD 0. 0001 / 0. 0000785 D 1 .273 m/s.

Entry pressure drop is:

PfD

u^2

2

(

1

Cc

 1

) 2

(from equation 3.78)

D 1000 ð 1. 2732 / 2 [ 1 / 0. 6 1]^2 D360 N/m^2

ReDud/D 1000 ð 1. 273 ð 0. 01 / 1. 0 ð 10 ^3 D 1. 273 ð 104
Ifeis taken as 0.046 mm from Table 3.1, thene/dD 0 .0046 and from Fig. 3.7:

R/u^2 D 0. 0043.

The pressure drop due to friction is:

PfD 4 R/u^2 l/du^2  (equation 3.18)

D 4 ð 0. 0043  4. 5 / 0. 01  1000 ð 1. 2732 D 12 ,540 N/m^2

Total pressure drop across one tubeD 12 , 540 C 360 D 12 ,900 N/m^2 or 12.9kN/m^2

If tubes are connected in parallel, power requiredD

(

pressure

drop

)

ð

(

volumetric

flowrate

)

D 12 , 900 ð 0. 04 

D516 W

PROBLEM 8.16

75% sulphuric acid, of density 1650 kg/m^3 and viscosity 8.6mNs/m^2 , is to be pumped
for 0.8 km along a 50 mm internal diameter pipe at the rate of 3.0 kg/s, and then raised
vertically 15 m by the pump. If the pump is electrically driven and has an efficiency
of 50%, what power will be required? What type of pump would you use and of what
material would you construct the pump and pipe?

Solution

Cross-sectional area of pipeD/ 4  0. 05 ^2 D 0 .00196 m^2.

Velocity,uD 3. 0 / 1650 ð 0. 00196 D 0 .93 m/s.

ReDud/D 1650 ð 0. 93 ð 0. 05 / 8. 6 ð 10 ^3 D 8900

Ifeis taken as 0.046 mm from Table 3.1,e/dD 0 .00092.

From Fig. 3.7,R/u^2 D 0 .0040.

Head loss due to friction is:

hfDPf/gD 4 R/u^2 l/du^2 /g (equation 3.20)

D 4 ð 0. 004  800 / 0. 05  0. 932 / 9. 81 D 22 .6m

Total headD 22. 6 C 15 D 37 .6m.