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2.75 t, s
104,400 Time, h
29 If the penetration depth is defined as x = 2, then time = 54 h. Regardless, it takes a long time Exercise 3.24
Subject: Determination of moisture diffusivity in clay brick from measurements of drying rate.
Given: Brick dimensions of 2 in (5.08 cm) by 4 in (10.16 cm) by 6 in (15.24 cm). Uniform
initial moisture content of 12 wt%. Equilibrium surface moisture content maintained at 2 wt%.
After 5 h, average moisture content is 8 wt%.
Assumptions: Diffusivity is uniform in all directions.
Find: (a) Diffusivity of moisture in clay.
(b) Additional time for average moisture content to reach 4 wt%.
Analysis: (a) By definition in Eq. (3-80), the unaccomplished fractional moisture concentration
change after 5 hours = Eavg = (cs - cavg)/( cs - co) = (2-8)/(2-12) = 0.60.
Apply Newman's method for a rectangular slab, given by Eq. (3-86). Thus,
Eavg = Ex Ey Ez
where the three E values are average values.
These values are plotted in Fig. 3.9 as a function of the Fourier number for mass transfer,
using half dimensions, a, b, and c in the x, y, and z directions. Thus, using units of cm and
seconds, the three Fourier numbers are:
Dt/a2 = D (5 x 3600)/(2.54)2 = 349 D to get Ex
Dt/b2 = D (5 x 3600)/(5.08)2 = 698 D to get Ey
Dt/c2 = D (5 x 3600)/(7.62)2 = 1047 D to get Ez
An iterative procedure is used to determine the diffusivity, D.
1. Assume a value of D in cm2/s.
2. Compute the three Fourier numbers.
3. From Fig. 3.9, read off values of Ex , Ey , and Ez for the corresponding Fourier
4. Calculate Eavg from Eq. (1). If it does not equal 0.60, repeat the four steps.
This procedure leads to the following results, where the first assumed value of D is obtained by
reading the Fourier number for Eavg = 0.61/3 = 0.84 for Fig. 3.9 and using it with the Fourier
number for the y dimension to compute D.
Assumed D, cm2/s
3.2 x 10-5
3.4 x 10-5 Dt/a2
0.604 Therefore, D = 3.4 x 10-5 cm2/s.
(b) At 4 wt% average moisture, Eavg = (2-4)/(2-12) = 0.2
The three Fourier numbers, with time in hours, are 2.37 x 10-3 t, 4.74 x 10-3 t, and 7.11x 10-3 t.
By iteration, to satisfy Eq. (1), using Fig. 3.9, time = 32 h or additional time = 27 h. Exercise 3.25
Subject: Diffusion of moisture from a 2-inch (5.08 cm) diameter ( a = 2.54 cm radius) spherical
ball of clay into air.
Given: Initial moisture content of clay = 10 wt%. Equilibrium moisture content of clay-air
interface = 3 wt%. Diffusivity of water in clay = D = 5 x 10-6 cm2/s.
Assumptions: Diffusivity is uniform throughout clay.
Find: Time for average moisture content to drop to 5 wt%.
Analysis: By definition in Eq. (3-80), the unaccomplished fractional moisture concentration
change after time, t is:
Eavg = (cs - cavg)/( cs - co) = (3-5)/(3-10) = 0.286 The lower curve in Fig. 3.9 relates that change to the Fourier number. From the figure,
N Fo M 5 × 10−6
= 0.08 = 2 =
t = 7.75 × 10 −7...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
- Spring '11
- The Land