(Solved): Solve using mass transport, notes for problem included. Other answer on Chegg is incorrect, please d ...

Consider a polymer sphere of radius \( R \) whose inside is allowed to dissolve steadily in a large bath of solvent. The sphere outer shell has a low solubility in the solvent so its size does not change significantly. However, the inside material that does dissolve does so very quickly so that the bath concentration at the sphere's surface is saturated, Csat. With the large size of bath, the polymer concentration far from the sphere surface is negligible (i.e. zero). A) Find a relationship for the steady-state dissolution flux of the sphere. B) What happens to the dissolution rate if the sphere's diameter is doubled? 6.7.3 Radial Diffusion in Spherical Coordinates In some cases, diffusion occurs through spherical objects; white blood cells, beads containing immobilized proteins, and spherical aggregates of cultured cells are examples. A schematic of diffusion in a spherical volume is shown in Figure 6.14. To perform a material balance, we assume that concentration varies only in the radial \( C=C(r) \) only). For an analysis of radial dendent of the angle of orientation (i.e., with an inner surface area of \( 4 \pi r^{2} \) and an outer surfacider a thin spherical shell volume of this shell is \( 4 \pi r^{2} \Delta r \). A material balance ou the area of \( 4 \pi\left(r+\Delta r^{2}\right) \). The accumulating within this volume is \[ \frac{\partial C_{i}}{\partial t} 4 \pi r^{2} \Delta r=\left(N_{i t} l_{r} r^{2}-N_{i r l} \mid\left(r+\Delta r r(r+\Delta r)^{2}\right) 4 \pi+R_{i} 4 \pi r^{2} \Delta r \quad(6.7 .41)\right. \] Dividing the terms in the material balance by \( 4 \pi r^{2} \Delta r \) yields \[ \frac{\partial C_{i}}{\partial t}=\frac{\left(\left.N_{i r}\right|_{r} r^{2}-N_{i r l}(r+\Delta r)(r+\Delta r)^{2}\right)}{r^{2} \Delta r}+R_{i} . \] Taking the limit as \( \Delta r \rightarrow 0 \) and dividing by \( r^{2} \) results in \[ \frac{\partial C_{i}}{\partial t}=\frac{1}{r^{2}} \frac{\partial}{\partial r}\left(N_{i r} r^{2}\right)+R_{r} \] Sokstiuting Fick's law for diffusion without convection, \( N_{i r}=-D_{i}\left(\partial C_{i} / \partial r\right) \), into Equation \( (6.7 .43) \) gives \[ \frac{\partial C_{i}}{\partial t}=\frac{D_{i j}}{r^{2}} \frac{d}{d r}\left(r^{2} \frac{\partial C_{i}}{\partial r}\right)+R_{i} . \] Example \( 6.8 \) Dissolution and Release of a Drug A porous spherical polymer of radius \( R \) contains a drug to treat brain to Experiments were performed in vitro to characterize the release of the do be the surface of the polymer \( (r=R) \), the drug concentration is \( \Phi C_{0 .} \). Far from \( f \) sero. Determine the surface \( (r \rightarrow \infty) \), the drug concentra state release rate from the polymer Integrating twice results in the following expression: \[ C_{i}=B-\frac{A}{r} \] From the boundary condition at \( r \rightarrow \infty, B=0 \). From the boundary condition at \( r=R, A=-R C_{0} \). The resulting concentration profile and flux are, respestively, \[ C_{i}=\frac{\Phi C_{0} R}{r} \] and \[ N_{i r}=-D_{i} \frac{d C_{i}}{d r}=D_{i j} \frac{\Phi C_{0} R}{r^{2}} \text {. } \] As the distance from the surface increases, the flux decreases, due to the incress in the cross-sectional area through which transport occurs. The steady-stati release rate (in moles per second) at the polymer surface is simply the produtch the flux and the surface area: Release Rate \( =N_{\text {irl }}^{r}=R^{4} 4 R^{2}=4 \pi D_{i j} \Phi C_{0} R \). Note that this analysis is approximate because the drug concentration at the surtiv. loes not remain constant but decreases with time as the drug leaves the polyneh