Unformatted text preview: − 85 1045.8
= 397 min
1206.6 From the above plot, it might take an additional 10 min to go from 10.2 to 10.0 wt %. Therefore,
take tf = 397 + 10 = 407 min. The total drying time = tc + tf = 123 + 407 = 530 min. Exercise 18.32
Subject: Drying time for a piece of hemlock wood
Given: Hemlock wood measuring 15.15 x 14.8 x 0.75 cm is to be dried at the two large faces
from an initial total moisture content of 90% to a final average total moisture content of 10%,
both dry basis. Drying is in the liquiddiffusioncontrolled fallingrate period with a diffusivity of
1.7 x 106 cm2/s.
Assumptions: Equilibrium moisture content with bonedry air is zero.
Find: Drying time if bonedry air is used.
Analysis: Fig. 3.9 applies, but the coordinates are off the chart. Therefore, use (2) of Example
18.13, provided that NFo for mass transfer > 0.1
Eavg = average unaccomplished freemoisture change = N Fo M X avg − X *
Xo − X * = 0.10 − 0
= 0.111
0.90 − 0 −6
DABt 1.7 × 10 ( t in s )
= 2=
= 1.2 × 10 −5 ( t in s )
2
a
( 0.75/2 ) From (2) of Example 18.13, ln X avg − X *
Xo − X * 8
π2
= ln 2 − N Fo M
π
4 Therfore, ln ( 0.111) = −2.197 = −0.210 − 2.467 N FoM Solving, N Fo M = 0.805 > 0.1 Therefore, 0.805 = 1.2 ×10−5 ( t in s ) and solving, t = 67,100 s or 18.6 h Exercise 18.33
Subject: Mode of drying in the fallingrate period for hemlock wood
Given: A piece of hemlock wood measuring 15.15 x 14.8 x 0.75 cm, for which drying takes
place only from the two largest faces. Air at 25oC is passed over the surfaces at 3.7 m/s. Dryingtime data are plotted below starting at an average total moisture content of 127 wt% dry basis
and finishing at an infinite time of 6.6 wt% dry basis.
Assumptions: Air is at 1 atm and a wetbulb temperature of 17oC.
Find: Whether Case 1 or Case 2 applied for the fallingrate period. If Case 1 applies, determine
from the data the effective diffusivity. If Case 2 applies, determine:
(a) Drying rate in g/hcm2 for the constantrate period assuming a dry density of 0.5
3
g/cm and no shrinkage.
(b) Critical moisture content.
(c) Predicted parabolicmoisturecontent profile at the beginning of the fallingrate
period.
(d) Effective diffusivity during the fallingrate period.
Analysis: Assume that the equilibriummoisture content is that measured at infinite time. Thus,
it is 6.6 wt% dry basis. Now find out if Case 1 or Case 2 applies by plotting moisture content
against time with a spreadsheet.
140 % Moisture Content, dry basis 120 100 80 60 40 20 0
0 5 10 15 20 25 Time, hours From this plot, it is seen that the fallingrate period is preceded by a constantrate period of w 2 h
ours. Therefore, Case 2 applies. Exercise 18.33 continued
(b) From the plot, the constantrate period extends for 2 hours to Xc = 96.8 wt% dry basis.
(a) The drying rate in the constantrate period = (127 – 96.8)/[(100)2] = 0.151 lb H2O/lb dry
woodh. Convert this to g/hcm2.
Volume of one piece of wood = (15.15)(14.8)(0.75) = 168 cm3
Mass of dry wood in one piece = 0.5(168) = 84 g
Area of 2 large fac...
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 Spring '11
 Levicky
 The Land

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