Page 247 - Membranes for Industrial Wastewater Recovery and Re-Use

2 16 Membranes for lndustrial Wastewater Recovery and Re-use

Now hydraulic losses across the retentate are given by
Aplosses = 1.1 5 UL bar for Um m/s and Lm m

where U is the mean retentate velocity

SO
Aplosses = 1.15 x 0.100 x 1 = 0.115 bar

Thus, total pressure is given by TMP+A~I,,,=~. bar.
74
According to Equation (2.23), the hydraulic energy demand per kg product is
given by:
E = (l/pO)AP

Converting to kWh per m3 product:

1000 2.78 x 10-7
E= AP = AP
60x 6Ox100OxpO 0
2.78 x 10-7
- 6.74 io5 = 1.38
-
0.136
For a pumping efficiency of 40%, this figure becomes 3.44 kWh mP3.

4.3.2 Problem in reverse osmosis: array design
Ifelements ofthe specification given in the previous problem are to be used to achieve an
overall conversion of at least 75% within an array, what array design can be used to
deliver 35 llspermeateproduct and what would be the specificenergy demand?

Solution
For a conversion of 75%, the feed and permeate flows are related by:
(& = &/0.75 = 3510.75 11s = 46.7 l/s = 168 m3/h

The overall conversion of 75% suggests a 2:l array, since conversion of 50% at
each stage produces 25% conversion overall. The retentate flow from the module
QR is then half of the feed flow QF.
If the relationship given in Equation (2.3) is extended to a number of elements
in series, then the retentate flow QZR is related to the feed flow QF and the number
of elements per module n by: