Biomedical Engineering Reference
In-Depth Information
Moreover, nondimensional numbers comparing different physical phenomena
occurring in the transport process are also introduced:
Pe =( u 0 ×
a )/ D O 2 ;Re= u 0 ×
a )/ r
Da =( σ cel ×
V max ×
a )/( c 0 ×
D O 2 );
λ M = K M / c 0
(3.11)
As already mentioned, the Peclet number Pe represents the ratio of con-
vective and diffusive effects for problems in mass transport. It is the equivalent
to the Reynolds number Re for the momentum transfer in fluid dynamics phe-
nomena, with v standing for the kinematic viscosity of the fluid in place of
the solute diffusion coecient D O 2 . Furthermore, the Damkolher number Da
represents the ratio of reactive (for instance, O 2 consumption of the cell layer
with surface density σ cel ) and diffusive effects. And finally the dimensionless
Michaelis constant λ M is reduced using the reference concentration.
In this framework, the distribution of the oxygen concentration within the
channel is governed by the following stationary diffusion-convection equation,
written in its dimensionless form:
.
C = 0
U ×
( C )+ Pe
×
(3.12)
Since cells are only attached on the pore wall, their consumption is rep-
resented by a flux (noted N O 2 ) boundary condition following the Michaelis-
Menten kinetics:
N O 2 ·
f ( C )
n =
Da
×
(3.13)
where f ( C )= C /( C + λ M ) (approximately, f ( C )= C for C
λ M and
f ( C )=1 for C
λ M ) and n is the unit vector normal to the element of wall
surface.
Other boundary conditions associated with this transport equations are as
follows:
- At the inlet, the given reference oxygen concentration:
C inlet = C 0 = 1
(3.14)
- A convective flux at the outlet of the pore:
n outlet .
C = 0
(3.15)
- Symmetric boundary conditions everywhere else on connecting pore surfaces:
n sym .
C
U ×
C + Pe
×
(3.16)
The velocity field U appearing in the convective term of equation (3.12)
is obtained by solving the dimensionless Navier-Stokes equations, considering
the steady flow of an incompressible Newtonian fluid:
Re U .
U =
( P )+∆ U
. U = 0
and
(3.17)
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