Waste water treatment: Desalination

10 236 0
Waste water treatment: Desalination

Đang tải... (xem toàn văn)

Thông tin tài liệu

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 A25o C = Q / Fw = 200,000 / 10.675 = 18,735 ft A10o C = 1.58 × A25o C = 29,602 ft 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 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 A= current 69.8 = = 2640 cm current density 26.4 ×10 −3 Power = RI = (4.5)(69.8) = 21,900 watts

Ngày đăng: 11/06/2016, 14:36

Mục lục

  • PowerPoint Presentation

  • Slide 2

  • Slide 3

  • Slide 4

  • Slide 5

  • Slide 6

  • Slide 7

  • Example

  • Slide 9

  • Slide 10

Tài liệu cùng người dùng

  • Đang cập nhật ...

Tài liệu liên quan