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Unformatted text preview: es = 2(15.15)(14.8) = 448 cm2
Rate of drying in constantrate period = 0.151 g H2O/g dry woodh
or 0.15(84) = 12.6 g H2O/hpiece of wood
Therefore, Rc = 12.6/448 = 0.0281 g H2O/hcm2
(d) Estimate the effective diffusivity from a rearrangement of (1862).
From the given data, Xc = 0.968, X* = 0.066, Xavg = 0.121 at t = 22 h
tc = 2 h, therefore, tf = 22 – 2 = 20 h or 72,000 s
DAB = Xc − X *
a2
0.3752
0.968 − 0.066
ln
=
ln
= 1.8 × 10−6 cm2/s
*
X avg − X
3t f
3 ( 72, 000 )
0.121 − 0.066 (c) Now check the Fourier number for mass transfer to see if moisture profile is parabolic at tc.
N FoM −6
DABtc (1.8 × 10 ) 2 ( 3600 )
at tc = 2 =
= 0.092
2
a
( 0.375 ) which is less than 0.5. Therefore, profile is not parabolic. Exercise 18.34
Subject: Fallingrate equations for drying.
Given: Solid undergoing drying in the fallingrate period either by liquid diffusion or capillary
movement.
Find: Derive governing equations. Outline an experimental procedure for determining which
mechanism prevails.
Analysis:
Case of Liquid diffusion controlling in fallingrate period:
From (3) in Example 18.13, which is a simplification of (1849) when N Fo M <0.1,
N Fo M DABt 4
8 Xo − X *
= 2 = 2 ln 2
a
π
π X avg − X * (1) From (1832), the rate of drying in terms of average moisture content is R=−
Rearranging (1),
X avg = X * + ms dX avg
A
dt 8( Xo − X * ) π2
Taking the derivative of (3) with respect to t gives
dX avg
dt = 8( Xo − X * )
π2 − exp − (2) π2 DABt
4a 2 π2 DABt
π2 DABt
π2 DABt
exp −
= ( X avg − X * ) −
4a 2
4a 2
4a 2 (3) (4) Comparing (2) and (4), it is seen that the rate of drying varies inversely as the square of a, which
is the thickness of the solid. Case of capillary flow controlling in fallingrate period:
Capillary flow will be laminar. From (142), the rate of laminar flow is inversely proportional to
the distance in the direction of flow, which is the thickness of the solid.
To determine experimentally which case applies, use solids of different thickness. Exercise 18.35
Subject: Crosscirculation tray drying in the constantrate drying period.
Given: Wet solid in a tray subject to heat transfer by convection to the bottom of the tray, in
addition to the usual heat convection from the hot air to the surface of the wet solid.
Find: Derive the governing equation and show that the temperature of the surface of the wet
solid will be higher than the wetbulb temperature of the hot air used to dry the solid. In
addition, determine the effect of radiation from the bottom of the tray to a tray below.
Assumption: Neglect the conduction resistance of the tray bottom and heat transfer to the sides
of the trays.
Analysis: This situation is discussed in the reference: Shepherd, C.B, C. Hadlock, and R.C.
Brewer, Ind. Eng. Chem., 30, 388397 (1938), which develops the theory and provides
experimental data on the drying of wet sand that supports the theory. Their results clearly show
that when heat is transferred also to the bottom of the tray, the temperatur...
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 Spring '11
 Levicky
 The Land

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