Page 248 - Membranes for Industrial Wastewater Recovery and Re-Use

System design aids 2 1 7

Hence:

If a feed flow of 5.6 m3/h and a mean convcrsion of 13.6% per element is
assumed then:

0 Total number ofrows = 168/5.6 = 30, and
0 EIementspermodule=logO.5/log0.864= 4.74- 5.

Hence. an array comprising 30 rows of modules in the first stage and 15 in the
second stage is needed, each module containing 5 elements. This would then
give an overall recovery of:

Although 2:l arrays are most common, it is also possible to employ a 3:2 array,
which in this case would be 30:20:10. A third stage would be required to obtain
the overall conversion of 75% but the elements per module demanded to achieve
3 3% conversion for the first two stages would be less:

Elements per module, stages 1 and 2 = logO.667/logO.864 = 2.77 - 3

Stage 3 would then demand 4 elements per module to achieve the target overall
recovery, since noverall should still be 10 across the whole array.
A comparison of the two schemes (see below) reveals that the 2:l array
employs 15 fewer modules, but that the 3:2 array uses 3 5 fewer elements. '3' ince
the pressure vessels are expensive and extra pipework is demanded by having a
third stage, the 2: 1 array would usually be preferred on the basis of capital costs.
On the other hand, there is a greater decrease in flow velocity across the module
for the 2: 1 array which inevitably leads to greater hydraulic loading at the front
of the module and/or greater concentration polarisation at the back.

~
Scheme No. of modules No. ofelements
2 stages (21 array) 45 225
2 stages (3:2 array) 60 190
__.~.

The specific energy demand relates to the total pressure and conversion, as
before. The total pressure drop across the retentate is given by:
APrpten+ = 1.15 x U,,,,,, x (L x n)