This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ppm of NaCl. Desalinization to 400
ppm by electrodialysis with a 40% conversion. 4 stages with 3 stacks of 150 cell pairs in each
stage. Fractional desalinization is the same in each stage. Expected current efficiency = 90%.
Applied voltage for first stage = 220 v. Each cell pair has an area = 1160 cm2.
Find: Current density in mA/cm2, current in amperes, and power in kW for the first stage.
Analysis: For the four stages, the overall fractional demineralization is:
f = (3,000  400)/3,000 = 0.8667
For equal fractional demineralization in each stage, (1 f) = (1  fs )4
Therefore, (1  0.8667) = (1  fs )4 . Solving, (1  fs ) = 0.6043
Thus, the salt concentration changes from stage to stage as follows:
Stage
Feed
leaving 1
leaving 2
leaving 3
leaving 4 NaCl concentration, ppm
3,000
1,813
1,096
662
400 For a 50% conversion, the volumetric feed flow rate = Q = 80,000 (0.5)/[24(3,600)] = 0.463 gal/s
or 0.00175 m3/s.
Take the solution density = 1,000,000 g/m3. Then for Stage 1, the NaCl concentrations are:
Feed: 3,000 ppm (assume by weight) or c = 3,000(1,000,000)/1,000,000 = 3,000 g/m3
Leaving Stage 1: 1,813 ppm or c = 1,813(1,000,000)/1,000,000 = 1,813 g/cm3
MW of NaCl = 58.5. Therefore, in mol, ∆c = (3,000  1,813)/58.5 = 20.3 mol/m3 or equiv/m3
Membrane area for Stage 1 = AM = (1,160 cm2/cell pair)(150 cell pairs/stack)(3 stacks/stage) =
522,000 cm2 or 52.2 m2
Use a rearrangement of Eq. (1458) to compute the current density:
i= zFQ (∆c) (1)(96, 520)(0.00175)(20.3)
=
= 73 amp/m 2
AM ξ
(52.2)(0.9) Therefore, the current flow through a cell pair = I = iAM/n = (73)(52.5)/150 = 25.6 amp
From Eq. (1459), for a voltage of 220 volts across all cell pairs,
the power = P = IE = 25.6(220) = 5,630 W = 5.6 kW Exercise 14.16
Subject: Reverse osmosis of sea water.
Given: 30,000,000 gal/day of sea water at 20oC, containing 3.5 wt% dissolved solids. Permeate
is 10,000,000 gal/day of water containing 500 ppm of dissolved solids. Retentate contains 5.25
wt% dissolved solids. Feedside pressure = 2000 psia constant, and permeate pressure = 50 psia
constant. One stage of spiralwound membranes with total membrane area of 2,000,000 ft2.
Assumptions: Crossflow.
Find: Permeance for water, and salt passage for the dissolved solids.
Analysis: First make an approximate calculation assuming perfect mixing on the permeate side
and an arithmetic average of feed and retentate on the feed side. Rearranging Eq. (1469),
nH 2 O
10,000,000
PM H O =
=
(1)
2
AM ∆P − ∆π avg 2,000,000 ∆P − ∆π avg
∆P = 2,000  50 = 1,950 psi
To estimate osmotic pressure, assume dissolved solids are NaCl (M = 58.5).
Salt concentrations are approximately:
Feed: 3.5(1,000)/[58.5(96.5)] = 0.620 mol/L
Retentate: 5.25(1,000)/[58.5(94.75)] = 0.947 mol/L
500
(1,000z 0
Permeate:
(100)
= 0.00855 mol / L
1,000,000
58.5(99.95)
From Eq. (1468), for T = 293 K and mi = 2 mNaCl Feed: π = 1.12(293)(2)(0.620) = 407 psia
Retentate: π = 1.12(293)(2)(0.947) = 622 psia
Permeate: π = 1.12(293)(2)(0.00855) = 5.6 psia
407...
View
Full
Document
This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

Click to edit the document details