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 [tex]150 mg/cm^{3}[/tex] to [tex]90 mg/cm^{3}[/tex]).
Explanation:
1) The total amount of released drug at time 't' is:
[tex]M_{t} = [\frac{2*3.142*H*D*K*C_{s} }{ln\frac{R_{0} }{R_{i} } }]*t[/tex]
Where:
H is the length of the device = 1 cm,
D is the diffusion coefficient of the drug [tex]= 2.2*10^{-10} cm^{2}/s[/tex]
K is the partition coefficient = 10
[tex]C_{s}[/tex] is the saturated drug concentration in the polymer matrix [tex]= 150 mg/cm^{3}[/tex]
[tex]R_{0}[/tex] is the radius of the device = 250 μm
[tex]R_{i}[/tex] is the radius of the cylindrical space = 100 μm
t is the released time = 48*3600 s = 172800 s
Therefore:
[tex]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[/tex]
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 [tex]150 mg/cm^{3}[/tex] to [tex]90 mg/cm^{3}[/tex])
