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Spontaneous drug release rate equation

  1. Mar 8, 2012 #1
    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. jcsd
  3. Mar 8, 2012 #2

    Borek

    User Avatar

    Staff: Mentor

    You are doing strange things, difficult to follow.

    What is

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

    equal to?
     
  4. Mar 8, 2012 #3

    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:

  5. Mar 8, 2012 #4

    Borek

    User Avatar

    Staff: Mentor

    Good. Now multiply it by concentration in ## \frac {mg}{cm^3}##.
     
  6. Mar 8, 2012 #5
    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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