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Unformatted text preview: ubject: Diffusion of NaCl into stagnant water at 25oC (298 K).
Given: An initial 0.10inch (0.254cm) thickness of NaCl sitting below a 3ft (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 105 cm2/s.
Assumptions: Water acts as a semiinfinite slab. Equilibrium at saltwater 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 solidliquid 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 semiinfinite slab, Eq. (379) 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. (375) to compute concentration profiles, where z = 0 at the saltwater 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 semiinfinite slab is valid. Exercise 3.23
Subject: Diffusion of water into 4inch (10.16cm) 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 106 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 semiinfinite slab solution of Eq. (375).
Consider center concentrations of moisture equal to 1%, 0.1%, and 0.01% of the
equilibrium surface concentration.
From Eq. (375),
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....
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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