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Sgstcrn design aids 2 13

4.3 Design examples

4.3.7 Problem in reverse osmosis: film theory and energy demand
A maximum concentration polarisation parameter value of 1.14 is recommended for
operation of a membrane element ofthefollowing specifications:

e Length (L): 1 m
e Channel thickness (h): 0.8 mm
e Membrane area (A): 36 nt'
Spacer mesh width (m): 0.6 mm

If can additionally be assumed that the ion diflusion coefficient is 8 x m2 s-',
and thefluid viscosity and density values are 1.1 5 x I 0-3 kg m-I s-' and 1000 kg mP3
respectivelg.
If the element operates at a meanflux of 21 LMH, what is the minimum feedflow rate
and what conversion does this yield?
l$

(a) the hydraulic losses amount to 1.15 bar per m/s cross-flow velocity per m path
If ng t 17,
(b) the membrane resistance is 8.5 x 1013per nt,
(c) the feedwatercontains 850 mg l-lsodium chloride, and
(d) the membrane has a rejection of 98%, the water temperature is 15°C and y = 0.9

what feed pressure is required, and what energy demand does this equate to for a
pumping eflciency of 40%?

Solution
Film theory states that the flux J and concentration polarisation parameter 6.e.
c*/c) are related by (Equation (2.14)):
= kin9

The mass transfer coefficient is thus:
21/(1000 x 60 x 60) - 5.83 x lo-'
k=-= - = 4.45 x 10-j m/s
In d, In 1.14 0.131

The Sherwood number, according to Equation (2.1 5), is then:

kd
Sh=-= 4.45 x lo-' x 1.6 x lop3 = 89.0
D 8 x 10-10

since, for a parallel flow channel, d is equal to twice the channel thickness h.