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218 Membranesfor industrial Wastewater Recover!! and Re-use

where n is the number of elements across the module. Estimation of the mean
retentate flow velocity presents a problem. If the velocity at the outlet must
be kept above 0.927 m/s to suppress CP then this implies that the module
inlet velocity must be double this value, assuming 50% conversion, giving a
mean velocity of (3/2)x0.0927=0.139 m/s. However, maintaining this velocity
would necessitate an increase in the retentate flow through the module,
which would then reduce the conversion unless this flow increase were to be
produced by retentate recirculation. For the purposes of this calculation a mean
retentate velocity of 0.139 m/s is assumed, notwithstanding the above
implications.
APretent = 1.15 x 0.139 x (10 x 1) = 1.60 bar

The maximum osmotic pressure difference is given as before by:
An* 1.14*70000
-- = 344 000 Pa = 3.44 bar
1 - 0 - (1 - 0.768)

Thus, the total pressure, including the pressure drop due to the membrane
itself, is:

P = 5.70 + 1.60 + 3.44 = 10.7 bar

and the specific energy consumption in kWh m-3 product is

2.78 x lopi 2.78 x
E= AP = 10.7 x 10’ = 0.391
0 0.768
For a pumping efficiency of 40%, this figure becomes 0.978 kWhm-3,
considerably lower than the single element case because of the increased
conversion and minimal extra energy expenditure.

4.3.3 Problem in reverse osmosis: CAD array design

Produce a complete array design based on the previous design specification for a
feedwater of the following composition: 200 rng I-l hardness “as CaC03”, 150 rng 1-l
alkalinity “as CaC03”, 75 mg r1 sulphate, -7 200 mg 1-’ total dissolved solids, p1-I
7.5, temperature 15”C, withanoverall rejection ofat least 98% andaproduct waterpH
of -7.0.

Solution
As already stated (Section 4.1), all of the leading RO manufacturers produce
CAD packages which allow the designer to produce a complete array design for
any specified combination of feed and product water quality. The key design
parameters for any RO CAD package are as follows: