Unsteady state mass transfer in a spherical bead of drug A small spherical bead

Question

Unsteady state mass transfer in a spherical bead of drug A small spherical bead is used as a controlled drug release capsule in the gastrointestinal system (your stomach). In this particular case, the 0.1 cm diameter head has a uniform initial concentration of 0.2 mmol/L of the drug grisefulvin. The diffusivity of the drug within the bead material is 1.5 times 10^-7 cm^2/s. Upon release from the bead, the drug is immediately consumed in the stomach so that the surface concentration is essentially zero. Determine the time (in hours) it will take for the concentration of the drug in the center of the bead to reach 10% of its initial value. Ignore changes in the capsule dimensions with time, ignore surface resistance (kc=infinity) and use K=1.0. Assume that the stomach has no drug in it at time zero.

Explanation / Answer

D=diffusivity=1.5*10^-7 cm2/s

co=initial concentration=0.2mmol/L

cs=surface concentration=0

t=time taken by the drug at the centre to reach 10%

cx=10%*co=0.1Co

x=distance travelled =radius=0.1/2=0.05 cm

using fick's second law,

cx-co/cs-co =1-erf(x/2(Dt)^1/2)

0.1co-co/0-co=1-erf(0.05cm/2(1.5*10^-7 cm2/s*t)^1/2)

0.9-1=-erf(0.05cm/2(1.5*10^-7 cm2/s*t)^1/2)

0.1=erf(0.05cm/2(1.5*10^-7 cm2/s*t)^1/2)

from tables erf z=0.1125,z=0.1

(0.05cm/2(1.5*10^-7 cm2/s*t)^1/2)=0.1

0.025cm*(1/(1.5*10^-7 cm2/s*t)^1/2))=0.1

(1/(1.5*10^-7 cm2/s*t)^1/2))=4

4((1.5*10^-7 cm2/s*t)^1/2))=1

squaring both sides,

(1.5*10^-7 cm2/s*t)=1/16

t=0.0625/1.5*10^-7 =0.0416*10^7 s=416666.67s=416666.67s/3600s/h=115.7 hrs

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