Desalination
Fujairah RO Plant
Objective: To reduce the concentration of dissolved
solids.
Methods Distillation (multi-stage)
Reverse Osmosis
Electrodialysis
Distillation
Negative
pressure
Boiler
Condenser
Steam
Cold salt water
Brine
Warm salt water
Heat
Fresh water
In class Exercise
Steam
Brine
Cold salt water
C=?
Q=10MGD
Heat
Fresh water
Q=8 MGD
C=50 mg/l
C=38,000 mg/l
Reverse Osmosis (RO)
Pressure
Membrane
Water movement
Fresh Saline
water water
Osmosis
Basic components of an RO unit
Fresh Saline
water water
Reverse Osmosis
Fujairah SWRO Flow Sheet
Dual-media filters
Cartridge filters
RO units
Spiral-wound Module
Show
video
Hollow-fiber module
RO Design Equation
Fw = K (∆p − ∆ψ )
osmotic pressure difference
pressure difference
mass transfer coefficient per membrane unit area
water flux, gal/(d.ft2)
Example
Determine the membrane area of a reverse osmosis system that is required to
demineralize 200,000 gallon per day with temperature of 10°C. Pertinent data
are as follows: K= 0.035 gal/(day-ft2)(psi) at 25 °C, ∆ψ = 45 psi, ∆p= 350 psi,
lowest operating temperature is 10 °C, and the membrane area correction
factor A 10°C/A 25°C =1.58.
Solution
Fw = K (∆p − ∆ψ ) = 0.035(350 − 45) = 10.675 gal / d − ft 2
A25o C = Q / Fw = 200,000 / 10.675 = 18,735 ft 2
A10o C = 1.58 × A25o C = 29,602 ft 2
Electrodialysis
Fresh water
A
Cathode
⊝
⊝
⊕
C
A
⊝
⊕ ⊕
⊝
C
⊝
⊕ ⊕
⊝
A
⊕
⊝
Saline water
I=current
FQNE r
I=
nE c
F= 96,500 amp-sec per eq
removed
Q=flow rate
N=normality of solution
Er = removal efficiency
Ec = current efficiency
n= number of cells
⊕
Anode
Example
Given that n=200, Q=90,000 gal/d, Co=4000 mg/l, cation or anion content is
0.066 eq/l. Current efficiency= 90%. Salt removal efficiency= 50%. Resistance
4.5 ohms, current density (milli amp/cm2)/normality ratio = 400. What is the
required current, the area of the membrane, and the power needed?
Solution
I=
FQNEr
nEc
96,500 amp ⋅ sec 90,000 gal 0.066 eq
1
1
3.785l
d
×
×
× 0.5 ×
×
×{
×
}
eq
d
l
200 0.9
gal
86,400 sec
= 69.8 amp
=
current density = 0.066 × 400 = 26.4 milli amp / cm 2
A=
current
69.8
2
=
=
2640
cm
current density 26.4 ×10 −3
Power = RI 2 = (4.5)(69.8) 2 = 21,900 watts