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

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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