CHEMICAL ENGINEERING

122 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS

If the pump efficiency is 50%,

powerDheadðmass flowrateðg/ 0. 5

Dpressure drop (N/m^2 ðvolumetric flowrate (m^3 s/ 0. 5

D 316 , 900 ð 0. 04 / 0. 5 D 25 ,350 W or 25.4W

If, due to impeller erosion, the delivery pressure is halved, the new flowrate may be
found from:
R/u^2 Re^2 DPfd^3 / 4 l^2 (equation 3.23)

The new pressure dropD 316 , 900 / 2 D 158 ,450 N/m^2 and:

R/u^2 Re^2 D 158 , 450 ð 0. 153 ð 705 / 4 ð 2000 ð 0. 52 ð 10 ^6 D 1. 885 ð 108

From Fig. 3.8, whenR/u^2 Re^2 D 1. 9 ð 108 ande/dD 0 .000027,ReD 3. 0 ð 105

and:  3. 0 ð 105 D 705 ð 0. 15 ðu/ 0. 5 ð 10 ^3 and:uD 1 .418 m/s

The volumetric flowrate is now: 1. 418 ð 0. 0177 D 0 .025 m^3 /s

PROBLEM 8.20

Calculate the power required to pump oil of density 850 kg/m^3 and viscosity 3 mN s/m^2
at 4000 cm^3 /s through a 50 mm pipeline 100 m long, the outlet of which is 15 m higher
than the inlet. The efficiency of the pump is 50%. What effect does the nature of the
surface of the pipe have on the resistance?

Solution

Cross-sectional area of pipeD/ 4  0. 052 D 0 .00196 m^2
Velocity of oil in the pipeD 4000 ð 10 ^6 / 0. 00196 D 2 .04 m/s.
ReDud/D 0. 85 ð 1000 ð 2. 04 ð 0. 05 / 3 ð 10 ^3 D 2. 89 ð 104
If the pipe roughnesseis taken to be 0.05 mm,e/dD 0 .001, and from Fig. 3.7,
R/u^2 D 0 .0031.
Head loss due to friction is:
hfD 4 R/u^2 l/du^2 /g (equation 3.20)

D 4 ð 0. 0031  100 / 0. 05  2. 042 / 9. 81 D 10 .5m
The total headD 10. 5 C 15 D 25 .5m
The mass flowrateD 4000 ð 10 ^6 ð 850 D 3 .4 kg/s
Power requiredD 25. 5 ð 3. 4 ð 9. 81 / 0. 5 D1700 W or 1.7kW
The roughness of the pipe affects the ratioe/d. The rougher the pipe surface, the higher
will bee/dand there will be an increase inR/u^2. This will increase the head loss due
to friction and will ultimately increase the power required.