# Spontaneous drug release rate equation

by StheevilH
Tags: drug, equation, rate, release, spontaneous
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 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!!!!
 Admin P: 23,600 You are doing strange things, difficult to follow. What is $$\sqrt {\frac {cm^2} {min} \times {min} }$$ equal to?
P: 13
 Quote by Borek You are doing strange things, difficult to follow. What is $$\sqrt {\frac {cm^2} {min} \times {min} }$$ equal to?

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?
Attached Files
 exp 1 phrm3021.pdf (883.2 KB, 0 views)

Admin
P: 23,600
Spontaneous drug release rate equation

 Quote by StheevilH I think that should equals to just "cm"
Good. Now multiply it by concentration in ## \frac {mg}{cm^3}##.
 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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