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gfd43tg
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Homework Statement
Fig. below shows the cross-section of a porous spherical granule of radius a. The pores are initially saturated with water. The granule dries in air at pressure P and temperature T. The drying rate is controlled by diffusion of water vapor through the dry region B; the shrinking core A has the original moisture content ##\rho## kmol per unit granule volume.
Using the equation within the region B, ##N_{1}c_{2} - N_{2}c_{1} = -DC \frac {dc_{1}}{dr}## show that the drying
time is
$$ \frac {\frac {\rho a^{2} R T}{6DP}}{ ln \Big(\frac {P}{P-P^{*}} \Big)} $$
where p* is the vapor pressure.
For a granule 10 mm in diameter, the drying time at 25 °C is 20 hours, the initial
water content being 50 mg; P = 1 bar, p* = 0.032 bar. Estimate D.
Homework Equations
The Attempt at a Solution
I am having a lot of problems solving these mass transfer problems, I am lacking intuition for the problem
So I start with the equation
$$N_{1}c_{2} - N_{2}c_{1} = -DC \frac {dc_{1}}{dr}$$
I am thinking 1 refers to water and 2 refers to air. I think the flux of air into the water, ##N_{2}##, is negligible since air is not very soluble in water, thus I am saying that term is zero, leaving me with
$$N_{1}c_{2} = -DC \frac {dc_{1}}{dr} $$
substituting ##c_{2} = C - c_{1}##
$$N_{1}(C - c_{1}) = -DC \frac {dc_{1}}{dr} $$
rearranging and integrating,
$$ \int_{\rho}^{0} \frac {dc_{1}}{C-c_{1}} = - \frac {N_{1}}{DC} \int_0^a dr $$
$$ln \Big(\frac {C}{C - \rho} \Big) = - \frac {N_{1}}{DC}a$$
From here I don't know where I will get those pressure terms
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