Spontaneous drug release rate equation

[StheevilH]

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!

 

[Borek]

You are doing strange things, difficult to follow.

What is

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

equal to?

 

[StheevilH]

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?

 

[Borek]

I think that should equals to just "cm"

Good. Now multiply it by concentration in $\frac {mg}{cm^3}$.

 

[StheevilH]

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