How Can You Calculate the Drying Time of a Spherical Granule?

[gfd43tg]

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

 

[Chestermiller]

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 $c_{2} = C - c_{1}$
$$N_{1}(C - c_{1}) = -DC \frac {dc_{1}}{dr} $$

You're good up to here. Nice job and nice reasoning.

But, N₁ is a function of r in your problem (so the integrations are incorrect). This is very similar to the problem you were doing the other day for evaporation of a drop. The above equation applies in the region between r = x and r = a. You need to take into account the spherical geometry, just like the problem the other day. 4πr²N₁ is constant in this region. You need to express the C and c in terms of the total pressure P and the partial pressure of water vapor (which is equal to the equilibrium vapor pressure at r = x), respectively. You need to solve for 4πr²N₁, which is the rate at which water is leaving the inner core.

Chet

 

[gfd43tg]

Thanks, solved it now