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Unformatted text preview: he flake and solvent be,
Yi = mXi
Combining (1) and (2),
ni = kc A X i − (Yi )b
A differential mass balance on the solute is given by,
ni = − Aa dX i
dt Where, a = the half thickness of the slice
Combining (3) and (4), and separating variables,
= − c dt
Integrating between the limits of (Xi)o at t = 0 and Xi at t = t,
= exp − c
( X i )o − i b
( X i )o − i b
(4) (5) which is (16-20) Exercise 16.9
Subject: Rate of leaching of solute from a cossette when the external resistance to mass transfer
is negligible, such that mass transfer is controlled by the internal resistance of the solid.
Find: Derive Equation (16-24)
Analysis: Treat the cossette as an infinite slab and ignore mass transfer from assumed thin
edges. For diffusion from both sides of the slab when the rate of mass transfer is controlled by
the internal resistance, with a Fourier number for mass transfer, ( N Fo ) M = e , (16-21) applies,
ln Eavgslab = ln 2 − ( N Fo ) M
De = effective diffusivity for the solute in the solid
t = time
a = half thickness of solid
Eavgslab = fractional unaccomplished approach to equilibrium for leaching, which decreases
From (16-22), the time for leaching is given by,
Differentiating (1), using the definition of the Fourier number for mass transfer, gives,
Separating variables and integrating (3),
4a 2 dE
Ein Rearranging (4) gives,
t= 4a 2
Eout which is (16-24) (4) Exercise 16.10
Subject: Determination, from a set of experimental data for leaching of oil from soybeans by
n-hexane, of the effective diffusivity to determine if it is constant with time
Given: A table of data for the average oil content of soybeans with respect to time
Assumptions: Effective diffusivity is constant
Find: Determine whether the effective diffusivity is constant
Analysis: According to (16-21), if ( N Fo ) M = Det
> 0.10, a plot of ln Eavgslab versus t should be a
a2 2 De
, if De is a constant.
A plot of the data, using a spreadsheet is as shown below, where it is seen that a straight line is
not obtained. Therefore, the effective diffusivity is not constant.
straight line with a slope of − Remaining Fraction of Oil 1 0.1
0 2 4 6
Time, minutes 8 10 12 Exercise 16.11
Subject: Estimation of the molecular diffusivity of sucrose in water at infinite dilution at 80oC.
Given: Value of the molecular diffusivity of 0.54 x 10-5 cm2/s at 25oC.
Assumptions: Applicability of the Wilke-Chang equation (3-39)
Find: The diffusivity value at 80oC and compare it to the value given in Example 16.6.
Analysis: From (3-9), DAB is proportional to Therefore, estimated DAB 80oC = 0.54 × 10 −5 T in K
µΗ2O 273 + 80
273 + 25 0.89 cP at 25o C
= 1.63 × 10−5
0.35 cP at 80o C However, in Example 16.6, the diffusivity value is given as 1.1× 10−6 or 0.11× 10−5 cm 2 /s
This is more than an order-of-magnitude less than the estimated value.
A possible explanation for the difference is that in Example 16.6, we don’t have the diffusion of
sucrose in water, but we have the diffusion of sucrose through water and insoluble fiber, where
the fibers may offer an enhanced resistance to molecule movement. Exercise 16.12
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
- Spring '11
- The Land