1. Derivation of Oxygen Diffusion Equations:
• Assume that oxygen is sparingly soluble in an aqueous medium. It dissolves as gas bubbles and the oxygen solute diffuses from the gas phase into the liquid phase. We model this as diffusion across an imaginary film (with a very small thickness), which captures the resistance to diffusive transfer.
Remember that the liquid phase solute concentration,
C (mol/cm3) and the gas phase solute pressure, P, are interrelated.
They can be converted using a simple Henry’s law
constant (which is known, this will be provided).
Thus, C=k . P. Next, we need to consider the transport
equation for oxygen solute (in terms of C, or P).
Gas
z=0
Liquid
z=d
We have examined Fick’s basic laws of diffusion in class. The first law says that J, the rate of movement of
the solute mass per unit area, is given by diffusivity times the concentration gradient of the solute.

2. Derivation of Oxygen Diffusion Equations, Contd.
Next, perform a mass balance on oxygen solute across this thin film with thickness, Dz.
The left terms of this equation is the amount of change in oxygen mass (moles) in the film
over time, while the right side is the reason for that change (I.e., the mass changes because
oxygen diffuses in and out of the film. The rate of oxygen transport was given by the flux,
described earlier. Make sure the units on both terms are consistent.
Divide both sides by (Dz) times (Dt) and apply the limit as these become very small. In that
limit, DC/Dz approaches dC/dz. Also, the flux terms give you a first derivative over space, z.
Thus, you will end up with the following equation:
If you plug in equation (1) into this (in J), you recover Fick’s second law of diffusion. We will use this equation to describe the diffusive flux of oxygen, and in a separate problem, also use an analogous equation to describe the random movement of cells.
Remember, since C and P are interconvertible,

3. Formula Derivation, Contd.
Equation 2 has to be solved for P varying as a functionof distance, z, and time, t. We define z from z=0 at
the gas-liquid interface (top surface of the liquid media pool added to the cells), to z=d, the location of cells
at the bottom of the liquid pool, in contact with the biomaterial.
We can make one more major simplification is the solution of equation (2). We can assume there to be a
steady state, which means that the concentration of dissolved oxygen does not vary with time, only with
position, z. Eliminate the time-variant term in equation (2) (the left side). You now have to solve the
following equation:
There are two boundary conditions required to solve equation (3).
The first is: at z=0, p = ps (the local pressure of oxygen. e.g. for atmospheric pressure, if the gas is air
containing 21% oxygen (by volume or by mol), ps = 0.21 atm pressure.
The second boundary condition is imposed at z=d. We say that the rate of oxygen transport (by diffusion)
at this location must be equal to the rate of its consumption by cells (in order for there to be a steady state- if
there isn’t any steady state, oxygen will accumulate, and its concentration will change over time). The rate of
oxygen uptake by cells has been modeled as a saturation model, which behaves as a linear dependence at
low oxygen concentration, and reaches a plateau at high concentration values. Typical forms of this saturation dependence of uptake on pressure are: Uptake rate = {constant1 . P}/{constant2 + P}, where
constant 1 and constant 2 are parameters that depend on the system, e.g. cell number, metabolism, etc.

4. Thus, we get the following equation (note this is only valid at z=d, since there are no cells present elsewhere (0<z<d) to consume oxygen; consequently, this serves as another “boundary” condition):
Note, that the right hand term has several constant parameters which will need to be known or provided
(P, the oxygen pressure, is the only variable, which you are trying to solve for, to ultimately determine d)
Clarification of terms in the oxygen uptake equation:
Vm is the maximal oxygen uptake rate by cells (in nmol/(s.106 cells)
Km is the oxygen pressure that elicits half maximal oxygen uptake rate (Vm/2) (units: mm Hg.)
r: number of cells seeded per unit area (called seeding density), (units, multiples of 106 cells/cm2)
D, the diffusion coefficient of dissolved oxygen in liquid medium, = 8 x 10-5 cm2/s
Ps= dissolved oxygen pressure at the top surface of the liquid (at z=0) (this should be given)
Pcrit (or Pd) = the dissolved oxygen pressure below which the cells will starve (should be given)
k = conversion factor for converting units from dissolved oxygen concentration to pressure
= 1.19 nmol oxygen per ml/mm Hg.
Make sure the units are consistent on either sides of the boundary condition.
The typical problem will involve finding out the maximum depth (d) of liquid for which a given
system (that is, data on Ps, Pcrit, cell seeding density, diffusivity, Km, K, Vm, will be provided)
can sustain the cells. Alternatively, d may be given to you and you may be asked to find the
maximum number of cells per unit area (r) this system can sustain.

5. Oxygen Equations Contd.
To solve for P(z), you have to first solve the P diffusion equation (2). You may assume a steady state, which
simplifies this equation (3). Integrate (3) twice to solve for P(z). There will be two integration constants,
which will be undetermined. You know one boundary condition for P at z=0. Plug this into the equation to
eliminate one unknown. To find the other, you have to plug the P(z) into the second boundary condition (4),
and solve the rest of the problem.