Question 27.4: An oak pole having an initial uniform moisture content of 45...
An oak pole having an initial uniform moisture content of 45 \mathrm{wt} \% is placed in a drying kiln where its surface moisture is maintained at 15 \mathrm{wt} \%. Under the specified drying conditions, the drying was controlled by the internal diffusion of liquid water to the surface. If the maximum moisture content of the pole is set at 25 \mathrm{wt} \%, how long must the 10 -cm-diameter by 45 -cm-length pole be dried when the ends of the pole are sealed with a vapor barrier? In comparison, how long must the same cylinder be dried if the cylinder surface is sealed with a vapor barrier and the ends are exposed?
The diffusivity of moisture through the oak is 1.04 \times 10^{-5} \mathrm{~cm}^{2} / \mathrm{s} under the specified drying conditions.
This problem is readly solved using the charts given in Appendix F. The initial uniform moisture content is 45 \mathrm{wt} \%, the surface moisture content is 15 \mathrm{wt} \%, and the maximum moisture content is 25 \mathrm{wt} \%.
Y=\frac{w_{A}^{\prime}-w_{A s}^{\prime}}{w_{A o}-w_{A s}}=\frac{\frac{0.25}{1.0-.25}-\frac{0.15}{1.0-0.15}}{\frac{0.45}{1.0-0.45}-\frac{0.15}{1.0-0.15}}=0.244
(a) When the ends of the cylinder are sealed, the moisture (species A) will diffuse to the cylindrical surface, transferring in the r direction only. As the drying was controlled by the internal diffusion of the water, m=0. The 25 \mathrm{wt} \% moisture content will be at the center of the cylinder.
n=\frac{r}{R}=\frac{0 \mathrm{~cm}}{5 \mathrm{~cm}}=0
For n=0, \mathrm{~m}=0, and Y=0.244, X=0.32, or
X=\frac{D_{A B} t}{R^{2}}=0.32
and
t=\frac{0.32 R^{2}}{D_{A B}}=\frac{0.32(5.0 \mathrm{~cm})^{2}}{1 \times 10^{-5} \mathrm{~cm}^{2} / \mathrm{s}}=7.81 \times 10^{4} \mathrm{~s}(21.7 \mathrm{~h})
(b) When the cylinder surface is sealed and the ends are exposed, the transfer will be from the two flat ends. With the specified moisture contents, Y=0.244, m will still be 0 , and n=0 ; X=0.68, or
X=\frac{D_{A B} t}{R^{2}}=0.68
and
t=\frac{0.68 x_{1}^{2}}{D_{A B}}=\frac{0.68(22.5 \mathrm{~cm})^{2}}{1 \times 10^{-5} \mathrm{~cm}^{2} / \mathrm{s}}=3.44 \times 10^{7} \mathrm{~s}(398 \text { days })