Question 27.3: Recall the drug capsule described in Example 2. The present ...

This problem is readily solved using the charts given in Appendix F.

(a) Spherical capsule: First, calculate the relative time $\left(X_{D}\right)$, relative position $(n)$, and relative resistance $(m)$ based on the spherical coordinate system

$\begin{gathered} X_{D}=\frac{D_{A B} t}{R^{2}}=\frac{\left(3 \times 10^{-7}  \frac{\mathrm{cm}^{2}}{\mathrm{~s}}\right)\left(48 \mathrm{~h} \frac{3600 \mathrm{~s}}{1 \mathrm{~h}}\right)}{(0.326 \mathrm{~cm})^{2}}=0.488 \\ n=\frac{r}{R}=\frac{0 \mathrm{~cm}}{0.326 \mathrm{~cm}}=0(\text { center of sphere }) \\ m=\frac{D_{A B}}{k_{c} R} \approx 0 \end{gathered}$

From Figure F.1 or Figure 18.3, the value for $Y$, which in this case is the unaccomplished concentration change at the center of the spherical bead, is about 0.018. We can now calculate $c_{A}$

$Y=0.018=\frac{c_{A s}-c_{A}}{c_{A s}-c_{A o}}=\frac{0-c_{A}}{0-68.9  \mathrm{mg} / \mathrm{cm}^{3}}$

The residual Dramamine concentration at the center of the bead after $48 \mathrm{~h}\left(c_{A}\right)$ is $1.24  \mathrm{mg} / \mathrm{cm}^{3}$.

(b) For the cube-shaped capsule the distance from the midpoint of the cube to any of the six faces is $0.652 \mathrm{~cm} / 2$. The relative time $X_{D}$ is now defined as

$X_{D}=\frac{D_{A B} t}{x_{1}^{2}}=\frac{\left(3 \times 10^{-7}  \frac{\mathrm{cm}^{2}}{\mathrm{~s}}\right)\left(48 \mathrm{~h} \frac{3600 \mathrm{~s}}{1 \mathrm{~h}}\right)}{(0.326 \mathrm{~cm})^{2}}=0.488$

Values for $n$ and $m$ are unchanged, with $n=0$ and $m=0$. As all of the faces of the cube are of equal dimension, let

$Y=Y_{a} Y_{b} Y_{c}=Y_{a}^{3}$

From the appendix Figure F.4, given $X_{D}=0.488, \mathrm{~m}=0$ and $n=0, Y_{a}$ is 0.4 for a flat plate of semi-thickness $x_{1}=a=0.326 \mathrm{~cm}$. Extending this value to a three-dimensional cube using the above relationship, we have

$Y=Y_{a}^{3}=(0.4)^{3}=0.064$

Finally,

$Y=0.064=\frac{c_{A s}-c_{A}}{c_{A s}-c_{A o}}=\frac{0-c_{A}}{0-68.9  \mathrm{mg} / \mathrm{cm}^{3}}$

with $c_{A}=4.41  \mathrm{mg} / \mathrm{cm}^{3}$ after $48 \mathrm{~h}$.

(c) For a cylindrical capsule with exposed ends, $R=0.652 \mathrm{~cm} / 2$ for the radial coordinate, and $x_{1}=a=0.3 \mathrm{~cm} / 2$ for the axial coordinate. The relative times are

$X_{D}=\frac{D_{A B} t}{R^{2}}=\frac{\left(3.0 \times 10^{-7}  \frac{\mathrm{cm}^{2}}{\mathrm{~s}}\right)\left(48 \mathrm{~h} \frac{3600 \mathrm{~s}}{1 \mathrm{~h}}\right)}{(0.326 \mathrm{~cm})^{2}}=0.488$

for the cylindrical dimension and

$X_{D}=\frac{D_{A B} t}{x_{1}^{2}}=\frac{\left(3.0 \times 10^{-7}  \frac{\mathrm{cm}^{2}}{\mathrm{~s}}\right)\left(48 \mathrm{~h} \frac{3600 \mathrm{~s}}{1 \mathrm{~h}}\right)}{(0.15 \mathrm{~cm})^{2}}=2.30$

for the axial dimension. Values for $n$ and $m$ are unchanged, with $n=0$ and $m=0$. From Figures F.1 and F.2, respectively, $Y_{\text {cylinder }}=0.1$ for the cylindrical dimension and $Y_{a}=$ 0.006 for the axial dimension. Therefore,

$Y=Y_{\text {cylinder }} Y_{a}=(0.1)(0.006)=0.0006$

and finally

$Y=0.006=\frac{c_{A s}-c_{A}}{c_{A s}-c_{A o}}=\frac{0-c_{A}}{0-68.9  \mathrm{mg} / \mathrm{cm}^{3}}$

with $c_{A}=0.413  \mathrm{mg} / \mathrm{cm}^{3}$ after $48 \mathrm{~h}$. As $Y_{a} \ll Y_{\text {cylinder }}$, the flux directed out of the exposed ends of the cylindrical tablet along the axial dimension dominates.

The above calculations assume that convective mass-transfer resistances associated with external fluid flow over the surface of the capsule are negligible. Problems in Chapter 30 will reconsider the drug release for unsteady-state diffusion and convection in series.