Math
Honors Contract: Diffusion through a Membrane
In
this
project I explored diffusion across membranes. First I used a worksheet
from
Georgia Tech. to learn about the process of diffusion and the many
elements
that need to be taken in account when creating a mathematical model.
Then I
expanded beyond the worksheet and instead of using a cylinder like the
worksheet did, I changed the shape to a sphere and applied what I
learned about
diffusion to a new shape.
The
Maple
worksheet I used was found at http://www.math.gatech.edu/~bourbaki/twentyfive.html,
called "Flow
Across Membranes," under Explorations
for Differential
Equations.
In order to use this worksheet – and other
Maple files – with the Citrix online version of Maple at ASU it is
necessary to
first upload the files to one’s personal section of ASU’s Andrew File
System
(AFS), which is disk space that can be accessed from many different
computers.
AFS is also known as the ‘m’ drive which all the computing sites on
campus use
when you save stuff on those PCs. In order to upload something to AFS,
go to
ASU Interactive and on the left-hand sidebar click on the link labeled
AFS
Manager. This will lead to a screen where all the files on AFS can be
transferred among directories, deleted, access, or, more relevantly,
new files
can be uploaded from the computer to the AFS directory using the
‘upload’
button. Once it has been uploaded, simple open the Maple program and
when
attempting to open files, the program looks at AFS.
Part 1:
Diffusion through a Cylindrical Membrane
The
type of
mathematical modeling done in this project serves as an example to
create understanding of biological systems
and how the process of
diffusion works in them such as the diffusion of nutrients out of the
bloodstream or the passage of oxygen across the aveoli in the
lungs.
In
the
worksheet the example used is a thin tube-shaped membrane, such as a
vessel,
which allows the passage of solutes, but not the liquid solvents which
simply
pass across the membrane. Solutes are the substances which are
dissolved in the
liquid, such as oxygen or various nutrients. When trying to create an
equation
that shows how diffusion takes place there are many factors that must
be taken
into consideration, including surface area, concentration both inside
and
outside the tube, and how big the solute is in addition to easily the
membrane
lets things through.
The
equation for the diffusion of a solute through a membrane into a vessel
on the
worksheet is as follows:
C(x) is the
concentration of the solute entering the vessel over a small
section of length (∆x), which is why it is derived in terms of x. The
term 2πa is the circumference of a circle with the radius a, the two
dimensional shape of ∆x. k is the constant related to that particular
diffusing
substance as well as the properties of the membrane under
consideration. S is
the concentration of the solute outside of the cell walls which, for
this
project, remains constant. V represents the volume of the solution as
it passes
through the section of vessel, ∆x.
Finally, C(x) shows how the concentration of solute in the
vessel will
change in respect to the length of the cylindrical cross-section. This
equation
can be used in Maple to create a graph showing the change in
concentration over
varying lengths in regard to all the other variables. As the
length of the cylinder increases, the concentration inside approaches
the
concentration of the solution outside the vessel.
This is the
differential equation to solve using Maple:
>dsolve({diff(C(x),x)=2*Pi*a*k*(S-C(x))/V,C(0)=Co},C(x));
Part of the
worksheet contained exercises for the reader.
1. What is
the total area of the cylinder, with a=1/2 and L = 20?
To solve
for this the term 2aπ∆x can be
used, replacing ∆x with L.
So: 2π(1/2)(20) = 20πm2
≈ 6.83m2.
2. To what extent must V be increased in order
that the level of the concentration at x = 20 is half what it is with V
= 3?
To solve
this problem, the graph found on the worksheet was used.
The
original numbers that were used for the graph are as follows:
v = 3
x = 20
Co =1
S [C(x)]=
5
a = ˝
>plot(subs({S=5,a=1/2,Co=1,V=3,k=1/5},C(x)),x=0..20);
By setting
the value of x to x=20 it is possible to change the final
concentration, aiming for a goal concentration of ˝ C(x) or C(x) = 2.5
g/m3,
by changing the value for V. Because V is in the denominator of the
equation,
increasing the value for V will cause C(x) to decrease.
So the new
values for the graph are as follows:
v = 27.5
x = 20
Co =1
S = 5
a = ˝
C(x) = 2.5
>plot(subs({S=5,a=1/2,Co=1,V=27.5,k=1/5},C(x)),x=0..20);
This means
that when the volume (V) is increased to 27.5, then the
concentration solute that diffuses is 2.5, half of the original
concentration.
3. To what
extent must the radius be increased so that when V is at the
level you computed, then the level of the concentration is the same as
it was
with a = 1/2 and V = 3?
This
is done in much the same way as the second problem was, changing the
value of a this time, until the
concentration [C(x)]
at x=20 is once again 5 g/m3. Because a
is
in the nominator of the equation, increasing the value of a will also
increase
the concentration. This also makes sense because increasing the radius
creates
more surface area for the solute to diffuse through.
The graph
of the concentration curve after changing the radius:
>plot(subs({S=5,a=3.5,Co=1,V=27.5,k=1/5},C(x)),x=0..20);
The
new
values for the graph are:
v = 27.5
x = 20
Co =1
S [C(x)]=
5
a =3.5
Part 2:
Diffusion through a Spherical
Membrane
For the
second part of this project I chose a new
shape, one also found in the body, a sphere, the approximate shape of
the
alveoli found in the lungs. The equation for a surface area in a sphere
is
4πr2 rather than 2πrL, which is the surface area of a
cylinder, as in the last part. This needs to be taken into account when
creating a new model for diffusion. One challenge of this is the fact
that you
can no longer integrate along one axis and have a ∆x that is the same
shape all the way. For instance if it was integrated along the x-axis
we would
have infinitesimal slices the shape of circle that decrease in size the
as that
circle is located farther away from the circle’s center, until it is
simply a
point. To adjust for this, instead of taking small sections of length
(∆x),
∆θ is used instead with the limits of 0≤θ≤2π. This
gives a small section of surface area based on angular displacement, as
shown
below:
Therefore
the surface area of the slice ∆θ
is:
The 2π
cancels out and the final equation is
A(θ+∆θ) – A(x) = k 2a2∆θ(S-C(θ)).
All the
variables retain the same meaning as they did
with the cylindrical model.
The
difference S-C(θ) is the difference in concentration
of solute across the membrane, while at the same time a volume V of the
solution passes through the small section causing the concentration to
increase, just like with the cylinder, by:
C(θ+∆θ) –
C(θ) = A(θ+∆θ) – A(θ)/V = kπa2∆θ(S-C(θ))/V.
Taking the
limit of
this:
This is the
differential equation to
solve. To solve this without first having had differential equations,
Maple can
be used:
>
dsolve({diff(C(theta),theta)=2*a^2*k*(S-C(theta))/V,C(0)=Co},C(theta));
The differential
equation is:
Maple uses
the data from the last command that was entered, so it may be
necessary to clear all the previous data by insert the Maple command
“restart”
before using the new statements.
We can use
this to answer the same questions from the first part of this
project. Simply replacing the original statement with the one above, it
will
first solve the differential equation, and then changing the ‘x’s to
‘theta’ in
the plot commands and it will graph the change in concentration over
increasing
values of θ:
>C:=unapply(rhs(%),theta);
>plot(subs({S=5,a=2,Co=1,V=3,k=1/5},C(theta)),theta=0..2*Pi);
This will
give the following plot (I changed the value of a
to
a=2, because it showed the exponential graph much better):
1. What is the total area of the cylinder,
with a = 1/2 and with the angular displacement in radians, D = π /2?
To solve
for area the term 2a2∆θ
can be used, replacing ∆θ with D.
2(1/2)(π/2) = ˝ πm2
≈ 1.5708m2
2. To what extent must V be increased in order
that the level of the concentration at θ = π/2 is half what it is
with V = 3?
The plot
that was made can be used to answer this question.
By setting
the value of θ to θ = 2π it is possible to change
the final concentration, aiming for a goal concentration of ˝ C(θ) or C(θ) = 2.5 g/m3, by changing
the value for V. Because V is in the denominator of the equation,
increasing
the value for V will cause C(θ) to decrease.
>plot(subs({S=5,a=2,Co=1,V=22,k=1/5},C(theta)),theta=0..2*Pi);
The
new
values for the graph are:
v = 22
θ = 2π
Co =1
S [C(θ)]=
2.5
a =2
3. To what
extent must the radius be increased so that when V is at the
level you computed, then the level of the concentration is the same as
it was
with a = 1/2 and V = 3?
This
is done in much the same way as the second problem was, changing the
value of a this time, until the concentration [C(θ)]
at θ =2π is once again 5 g/m3.
Because a is in the nominator of the
equation,
increasing the value of a will also increase the concentration. This
also makes
sense because increasing the radius creates more surface area for the
solute to
diffuse through.
The graph
of the concentration curve after changing
the radius:
>plot(subs({S=5,a=5.5,Co=1,V=22,k=1/5},C(theta)),theta=0..2*Pi);
The
new
values for the graph are:
V = 22
θ = 2π
Co =1
S[C(θ)]=
2.5
a =5.5
This idea
of changing shapes and diffusion through membranes can be
expanded even further than is done here. The worksheet shows how to use
mathematically model both the absorption and release of solute through
the
membrane, and an analytic periodic solution. This could be extrapolated
to
other shapes, such as a sphere, as well.