Separation Process Principles- 2n - Seader & Henley - Solutions Manual

Separation Process Principles 2n Seader& Henley Solutions Manual

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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. (14-58) 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. (14-59), 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. Feed-side pressure = 2000 psia constant, and permeate pressure = 50 psia constant. One stage of spiral-wound 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. (14-69), 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. (14-68), 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.

Ask a homework question - tutors are online