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Waste water treatment: Desalination

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



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