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

C 90 of the salt has dissolved analysis basis area for

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: ubject: Diffusion of NaCl into stagnant water at 25oC (298 K). Given: An initial 0.10-inch (0.254-cm) thickness of NaCl sitting below a 3-ft (91.44 cm) depth of water. Solubility of NaCl in water of 36 g NaCl/100 g water. Diffusivity of NaCl in water = 1.2 x 10-5 cm2/s. Assumptions: Water acts as a semi-infinite slab. Equilibrium at salt-water interface. Find:. Time and concentration profile of salt in the water, when: (a) 10% of the salt has dissolved. (b) 50% of the salt has dissolved. (c) 90% of the salt has dissolved. Analysis: Basis: area for mass transfer = 1 cm2 = A Volume of salt = 1(0.254) = 0.254 cm3 From Perry's Handbook, density of NaCl = 2.165 g/cm3 Mass of salt = 2.165 (0.254) = 0.550 g Molecular weight of NaCl = 58.44 g/mol Moles of salt = 0.550/58.44 = 0.00941 mol/cm3 At the solid-liquid interface, NaCl concentration = 36/(100 + 36) x 100% = 26.5 wt% From Perry's Handbook, density of this solution at 25oC = 1.2 g/cm3 or cs = 0.265 (1.2)/58.44 = 0.00544 mol NaCl/cm3 Initial concentration of salt in the bulk of the water = co = 0 For diffusion into a semi-infinite slab, Eq. (3-79) applies, for the number of moles of NaCl, N, transferred in time t. Solving that equation for time t gives, t= N 2π 4 A ( cs − co ) DNaCl 2 2 = N 2 (3.14) 4(1) ( 0.00544 − 0 ) (1.2 ×10 2 2 −5 ) = 2.21× 109 N 2 s Using Eq. (1), the following times are computed: Part (a) 10% transferred (b) 50% transferred (c) 90% transferred Moles transferred 0.000941 0.004705 0.008469 Time, s 1,960 48,900 159,000 Time, h 0.544 13.6 44.0 Use Eq. (3-75) to compute concentration profiles, where z = 0 at the salt-water interface, c = cs erfc z z = 0.00544 erfc = 0.00544 erfc( x ) 2 DNaCl t 2 1.2 × 10 −5 t (2) (1) Exercise 3.22 (continued) Analysis: (continued) Solving Eq. (2) for values of x, which fixes the value of c, the following results are obtained: z, cm for value of c______ x 0.5 1.0 1.5 2.0 erfc (x) 0.4795 0.1573 0.0339 0.0047 c, mol/cm3 0.00261 0.000856 0.000184 0.0000256 10% transf. 0.157 0.313 0.470 0.627 50% transf. 90% transf. 0.782 1.408 1.565 2.816 1.173 4.224 3.130 5.632 Since the water layer is 91.44 cm thick, we see that the penetration of the salt is far less than this, such that the assumption of a semi-infinite slab is valid. Exercise 3.23 Subject: Diffusion of water into 4-inch (10.16-cm) thick wood, with sealed edges, from air of 40% relative humidity. Given: Equilibrium moisture content for air conditions = 10 lb water/100 lb of dry wood. Diffusivity of water in wood = 8.3 x 10-6 cm2/s. Initially, wood is dry. Find:. Time for water to penetrate to the center of the wood Analysis: To determine the time, apply the semi-infinite slab solution of Eq. (3-75). Consider center concentrations of moisture equal to 1%, 0.1%, and 0.01% of the equilibrium surface concentration. From Eq. (3-75), θ= ccenter − co ccenter z 5.08 = = erfc = erfc = erfc x cs − co cs 2 Dwater t 2 8.3 × 10−6 t (1) Using Eq. (1), the following results are obtained, θ 0.01 0.001 0.0001 x 1....
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