Question

. Saturated drug is encapsulated in a membrane-controlled reservoir device. The shape of the device is cylindrical. The total amount of released drug at time t can be calculated according to the following equation. M_t=(2πHDKC_s)/(ln⁡(R_0/R_i)) t The saturated drug concentration in the polymer matrix is 150 mg/cm3. The diffusion coefficient of the drug D is 2.2×10-10 cm2/s, and the partition coefficient K is 10. The length (H) and diameter (2R0) of the device are 1 cm and 500 µm respectively. The saturated drugs are in a cylindrical space with a diameter (2Ri) of 200 µm. 1) Calculate the amount of released drug in 48 hours. (25 points) 2) In order to apply this delivery system to children, the release rate should be reduced by 60%. How to design the device if the volume of the space for concentrated drugs and the volume of the device are both unchanged? (25 points)

Answer 1

Answer:

1) 0.391 mg 2) By resizing the semi-permeable membrane and reducing the saturated drug concentration from 100% to 60%  (i.e. from $150 mg/cm^{3}$ to $90 mg/cm^{3}$).

Explanation:

1) The total amount of released drug at time 't' is:

$M_{t} = [\frac{2*3.142*H*D*K*C_{s} }{ln\frac{R_{0} }{R_{i} } }]*t$

Where:

H is the length of the device = 1 cm,

D is the diffusion coefficient of the drug $= 2.2*10^{-10} cm^{2}/s$

K is the partition coefficient = 10

$C_{s}$ is the saturated drug concentration in the polymer matrix $= 150 mg/cm^{3}$

$R_{0}$ is the radius of the device = 250 μm

$R_{i}$ is the radius of the cylindrical space = 100 μm

t is the released time = 48*3600 s = 172800 s

Therefore:

$M_{t} = [\frac{2*3.142*1*2.2*10^{-10}*10*150 }{ln\frac{250}{100} }]*172800 = \frac{2.074*10^{-6}*172800 }{0.9163} = 0.391 mg$

2) For a membrane-controlled device, the released rate can be controlled (i.e. decrease or increase) by resizing the semi-permeable membrane. In addition, the saturated drug concentration should be reduced from 100% to 60% (i.e. from $150 mg/cm^{3}$ to $90 mg/cm^{3}$)