Environmental Engineering Reference
In-Depth Information
Example 9.2 [9]
Problem:
The petroleum industry and regulatory agencies are searching for alternative methods
for treating drainwater from oil rigs. Membrane separation has proven to be an
effective and economic alternative to conventional methods for oil-water separation.
Drainwater from oil rigs may contain up to 5% oil in addition to different chemicals
in varying concentrations. Drainwater typically is collected in a burner tank and burnt
by flare to the atmosphere. This emission to air causes an unwanted environmental
impact.
The membrane separation plant is tubular ultrafiltration (UF) and the pilot-plant
operation was on a batch basis with a volume reduction factor approaching 40. The
UF membrane had a maximum permeate flux of around 300 L
m 2
/
·
hr at maximum
cm 2
6kg
s fluid velocity with a clean membrane. The flux
typically dropped and approached 80 L
/
inlet pressure and 3.8 m
/
m 2
hr at the end of a day's operation. The
retentate from UF separation was returned to the feed tank whereas the permeate was
routed to the sewer. Design of a full-scale plant was performed using a flux value of
40 L
/
·
/
m 2
·
×
.
Cleaning of the membrane was performed with 0.3 wt% nitric acid at 45 C and the
alkaline detergent Ultrasil 11 at 0.25 wt% and 55 C. The membrane flux did not always
fully recover after each chemical cleaning. A 200 mg/L Cl 2 solution administered as
sodium hypochlorite (NaOCl) would, however, restore the original flux.
For the full-scale membrane plant, calculate:
(a) permeate and retentate flowrates;
(b) the membrane area required.
Solution:
Given: permeate flux
hr and volume reduction of 20
m 2
hr;
volume reduction of concentrate
=
40 L
/
·
20 times.
Mass balance on membrane unit (refer to Figure 9.9):
=
38 m 3
10 3 L
m 3
10 4 L
( a ) Permeate flowrate
=
x
=
/
hr permeate
×
/
=
3
.
8
×
/
hr;
2m 3
10 3 L
m 3
10 3 L
Retentate flowrate
=
/
hr
×
/
=
2
×
/
hr
.
permeate flowrate
permeate flux
(b) Membrane area
=
10 4
3
.
8
×
L
/
hr
950 m 2
=
hr =
.
m 2
4
.
0
×
10 1
L
/
·
[Note: assumes continuous operation.]
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