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Spontaneous drug release rate equation 
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#1
Mar812, 06:13 AM

P: 13

Okay I found some references and the equation I am after is
Q = 2 * C * (D * t / ∏)^(1/2) where Q is the weight of drug released per unit area (hence unit is mg/cm^2) C is the initial drug concentration D is the diffusion coefficient (unit of cm^2 min^1) and t is release time in min. What I am not sure about is the units. Logically the concentration of drug would be in g/L. but when I merge all the units together, I get something else. This is my working. The units on the right hand side should equals to Q. = 2 (constant) * C (g/L) * [D (cm^2 * min^1) * t (min) / ∏ (constant)]^(1/2) only considering units (ie discard constants) = [g * cm^[2*(1/2)] * min^(1/2)] / [L * min^(1 * 1/2)] = g * cm * min^(1/2) / L * min^(1/2) = g * cm * min^(1/4) / L which does not equals to unit of Q (mg/cm^2). The units given are all correct but is there something I am missing here? Breaking rules of powers perhaps? Thank you!!!! 


#2
Mar812, 08:44 AM

Admin
P: 23,731

You are doing strange things, difficult to follow.
What is $$ \sqrt {\frac {cm^2} {min} \times {min} }$$ equal to? 


#3
Mar812, 03:34 PM

P: 13

I think that should equals to just "cm" If you look at the attachment, equation 24, that is what I am after and the units were taken from other sources. But to me, equation makes sense but units don't from what I did. Is there a problem with my algebra skill? 


#4
Mar812, 04:08 PM

Admin
P: 23,731

Spontaneous drug release rate equation



#5
Mar812, 08:20 PM

P: 13

then it would equals to mg / cm^2.... which is the unit of Q.
where did you get mg/cm^3? oh wait... damn it.... you changed L into cm^3... thank you so much 


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