Spontaneous drug release rate equation


by StheevilH
Tags: drug, equation, rate, release, spontaneous
StheevilH
StheevilH is offline
#1
Mar8-12, 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!!!!
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Borek
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#2
Mar8-12, 08:44 AM
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You are doing strange things, difficult to follow.

What is

$$ \sqrt {\frac {cm^2} {min} \times {min} }$$

equal to?
StheevilH
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#3
Mar8-12, 03:34 PM
P: 13
Quote Quote by Borek View Post
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
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Borek
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#4
Mar8-12, 04:08 PM
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Spontaneous drug release rate equation


Quote Quote by StheevilH View Post
I think that should equals to just "cm"
Good. Now multiply it by concentration in ## \frac {mg}{cm^3}##.
StheevilH
StheevilH is offline
#5
Mar8-12, 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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