Cell Process
1. Cell and transport phenomena (2)
Liang Yu
Department of Biological Systems Engineering
Washington State University
02. 12. 2013
Main topics
Cell and transport phenomena
High performance computation
Metabolic reactions and C-13 validation
Enzyme and molecular simulation
Physiological Fluid Mechanics
Fluid
A material that deforms continuously when subjected to a force
applied tangentially to a surface
Biological fluid
Quite complex, exhibiting solid and liquid-like behavior and
deforming in a time-dependent fashion
Fluid mechanics
The study of the motion of fluids in response to the application of
stresses
Physiological Fluid Mechanics
Concerned with fluid flow within organisms and the relation between fluid
flow and physiological processes
Conservation relations and
momentum balances
Fluid kinematics: describe fluid motion in
time and space
Eulerian description: observing motion from a
fixed reference frame
Lagrangian description: observing motion
from a moving reference frame
Deference between Eulerian and
Lagrangian description
River bank
If record the motion from the riverbank, Eulerian description (fixed reference frame )
If you are on the boat as it drifts with the current, Lagrangian description (moving)
Eulerian approach
Describes the flow field (velocity, acceleration,
pressure, temperature, etc.) as functions of position
and time
Do not follow particles, i.e. different particles will
flow through the same point at different times
Relatively low computation, most fluid mechanics
textbooks and papers use this system
The time-change of the velocity in such a measurement
v
t
Velocity field: v  v  x, y, z, t 
 x, y ,z 
Called the partial derivative of the temperature with respect to time, or local
accelerate
Fluid element and properties
• The behavior of the fluid is described
in terms of macroscopic properties:
–
–
–
–
–
Velocity u.
Pressure p.
Density r.
Temperature T.
Energy E.
• Properties are averages of a
sufficiently large number of
molecules.
• A fluid element can be thought of as
the smallest volume for which the
continuum assumption is valid.
dz
(x,y,z)
dy
dx
y
z
x
Faces are labeled
North, East, West,
South, Top and Bottom
Lagrangian approach
Follow individual particles, positions of specified particles are
the objectives
Track (or follow) it as it moves, and monitor change in its
properties
The properties may be velocity, temperature, density, mass, or
concentration, etc in the flow field
The recorded properties are associated with the same fluid
particle, but at different locations and at different times
The time-change of the velocity in such a measurement
Dv
Dt
Called material derivative or substantial derivative.
It reflects time change in the velocity (or any other properties) of
the tagged fluid particles as observed by an observer moving
with the fluid.
Lagrangian approach is also called “particle based approach”.
Lagrangian approach
Relatively high computation, difficult to use
for practical flow analysis
Fluids are composed of billions of molecules
Interaction between molecules hard to
describe/model
dx A
 u  xA, yA, zA, t 
dt
dy A
 v  xA, yA, zA, t 
dt
dz A
 w  xA, yA, zA, t 
dt
The above equations determine the path of the particle A if the position is specified
at some initial instant in its path history
Acceleration
Fluid velocity is a function of the three spatial coordinates x, y, z, and if the
flow is unsteady, time t
v  v  x, y , z , t 
The total differential is defined as
v
v
v
v
dv 
dx  dy  dz  dt
x
y
z
t
dv v dx v dy v dz v




dt x dt y dt z dt t
Acceleration is a vector defined as the rate of change of velocity
v
v
v
v
a
vx  v y  vz 
x
y
z
t
Dv v

 v  v
Dt t
Convective acceleration
Local acceleration
Rate of change for a fluid particle
• Terminology: fluid element is a volume stationary in
space, and fluid particle is a volume of fluid moving with
the flow.
• A moving fluid particle experiences two rates of changes:
– Change due to changes in the fluid as a function of time.
– Change due to the fact that it moves to a different location in the
fluid with different conditions.
• The sum of these two rates of changes for a property per
unit mass  is called the total or substantive derivative D
D   dx  dy  dz
/Dt:




Dt
t x dt y dt z dt
• With dx/dt=u, dy/dt=v, dz/dt=w, this results in:
D


 u.grad 
Dt
t
Rate of change for a stationary fluid element
• In most cases we are interested in the changes of a flow
property for a fluid element, or fluid volume, that is
stationary in space.
• However, some equations are easier derived for fluid
particles. For a moving fluid particle, the total derivative
per unit volume of this property  is given by:
D
 

(for moving fluid particle) r
r
 u.grad   (for given location in space)
Dt
 t

• For a fluid element, for an arbitrary conserved property  :
r  div ( r u)  0
t
Continuity equation
( r )
 div ( r u)  0
t
Arbitrary property
Fluid particle and fluid element
• We can derive the relationship between the equations for a
fluid particle (Lagrangian) and a fluid element (Eulerian) as
follows:
 ( r )
D
 

 r

 div( r u)  r   u.grad      div( ru)  r
t
Dt
 t

 t

zero because of continuity
( r )
D
 div( r u)  r
t
Dt
Rate of increase of
 of fluid element
Net rate of flow of
 out of fluid element
=
Rate of increase of
 for a fluid particle
Control volumes
 A convenient region of space
 For examining the flow of
mass, momentum, and energy
 Control volume is arbitrary
 Control volume can be fixed in
size or may change size with
time
 Simple geometry is square (2D)
or cube (3D)
Continuity equation
• Rate of increase of mass in fluid element equals the
net rate of flow of mass into element.
r
• Rate of increase is: t ( rdxdydz)  t dxdydz
• The inflows (positive) and outflows (negative) are
 ( r w) 1 
shown here:

rw 
. d z d xd y


z
2



 ( r v) 1 
r
v

. d y  d xd z

y
2 

( ru) 1 

r
u

. d x  d yd z

x
2 

( ru ) 1 

. dx  dydz
 ru 
x 2 

y

 ( rv) 1 
. dy  dxdz
 rv 

y
2 

z
x
( rw) 1 

. dz  dxdy
 rw 

z
2 

Continuity equation
• Summing all terms in the previous slide and dividing by the
volume dxdydz results in:
r  ( ru)  ( rv)  ( rw)  0
t
x
y
z
• In vector notation:
Change in density
r  div ( r u)  0
t
Net flow of mass across boundaries
Convective term
• For incompressible fluids r / t = 0, and the equation
becomes:
div u = 0.
ui
u  v  w  0
• Alternative ways to write this: x y z
and x  0
i
Navier–Stokes equation (Momentum equation )
Describe the motion of fluid substances
Applying Newton's second law to fluid motion
Assumption
the fluid stress is the sum of a diffusing viscous term
(proportional to the gradient of velocity) and a pressure
term - hence describing viscous flow
Constant density, ρ
Constant viscosity, μ
Continuity (incompressible flow)
A solution of the Navier–Stokes equations is called a
velocity field or flow field, which is a description of
the velocity of the fluid at a given point in space and
time
Momentum equation in three dimensions
• Derive conservation equations for momentum for fluid
particles. Then use the above relationships to transform
those to an Eulerian frame (for fluid elements).
• Start with deriving the momentum equations from
Newton’s second law: rate of change of momentum
equals sum of forces.
Dv
F  ma  r
Dt
• Rate of increase of x-, y-, and z-momentum:
Du
r
Dt
Dv
r
Dt
Dw
r
Dt
• Forces on fluid particles are:
– Surface forces such as pressure and viscous forces.
– Body forces, which act on a volume, such as gravity, centrifugal,
Coriolis, and electromagnetic forces.
Viscous stresses
• Stresses are forces per area.
Unit is N/m2 or Pa.
• Viscous stresses denoted by t.
• Suffix notation tij is used to
indicate direction.
• Nine stress components.
– txx, tyy, tzz are normal stresses.
E.g. tzz is the stress in the zdirection on a z-plane.
– Other stresses are shear
stresses. E.g. tzy is the stress in
the y-direction on a z-plane.
• Forces aligned with the direction
of a coordinate axis are positive.
Opposite direction is negative.
Forces in the x-direction
t yx 1
(t yx 
. dy )dxdz
y 2
(t zx 
t zx 1
. dz )dxdy
t yx 1
z 2
 (t yx 
. dy )dxdz
y 2
p 1
( p  . dx)dydz
x 2
p 1
 ( p  . dx)dydz
x 2
t xx 1
 (t xx 
. dx)dydz
x 2
z
y
x
(t xx 
t xx 1
. dx)dydz
x 2
t zx 1
 (t zx 
. dz )dxdy
z 2
Net force in the x-direction is the sum of all the force components in that direction.
Momentum equation
• Set the rate of change of x-momentum for a fluid particle
Du/Dt equal to:
– the sum of the forces due to surface stresses shown in the previous
slide, plus
– the body forces. These are usually lumped together into a source
term SM:
Du  ( p  t xx ) t yx t zx
r



 S Mx
Dt
x
y
z
– p is a compressive stress and txx is a tensile stress.
• Similarly for y- and z-momentum:
Dv t xy ( p  t yy ) t zy
r



 S My
Dt
x
y
z
Dw t xz t yz ( p  t zz )
r



 S Mz
Dt
x
y
z
Application of Momentum balances
Pressure-driven flow through a narrow rectangular
channel
Pressure-driven flow through a Cylindrical tube
 Flow is induced by a pressure gradient, which could be
produced by a pump or gravity.
 These examples also examine the effect of geometry upon
the flow and the shear stress.
 Flow through these kinds of channels are important in
many biomedical devices such as hemodialyzers and
ultrafiltration.
Pressure-driven flow through a
narrow rectangular channel
Assumption:
1. The flow is pressure driven; that is, pressure varies only in the direction of flow
2. The fluid density is constant, which indicates that the fluid is incompressible
3. The flow is steady; that is, pressure, shear stress and velocity do not change with
time
4. The fluid is Newtonian
5. Edge effect are neglected. To meet this assumption, we require a long, wide
channel; that is, h/w<<1 and h/L <<1, where w is the width and L is the length of
the channel
6. The flow is laminar
Pressure-driven flow through a
narrow rectangular channel
Parallel-plate channels are used widely to study
the effect of flow on cell adhesion and cell
function.
Since there is no net momentum flow and the flow
is steady, the sum of all forces must equal zero.
The only forces arising are those due to pressure
and shear stress.
A momentum balance in the x direction yields
p
x

 p x x  yz  t yx  t yx
x
x x
 xz  0
Pressure-driven flow through a
narrow rectangular channel
Dividing by ∆x∆y∆z and taking the limit as each goes to zero
results in the ordinary differential equation
dp dt yx

dx
dy
The pressure changes only in the x direction (i.e., dp/dx=f(x)) and
the shear stress changes in the y direction (i.e., dτ/dy=g(y))
f ( x)  g ( y)
Integrate f(x) to yield
p  C1 x  C2
Pressure-driven flow through a
narrow rectangular channel
The pressure can be specified at two locations, away from both the
entrance and exit. Thus, at x=x0, p=p0, and at x=xL, p=pL. Defining
∆p=p0-pL and L=xL-x0 to remove C1 and C2
p
p  p0 
 x0  x 
L
dt yx
p

dy
L
Integrate τ to yield
t yx
p

y  C3
L
Newton’s law of viscosity
t yx
dux

dy
Pressure-driven flow through a
narrow rectangular channel
Substitute into Newton’s law of viscosity

dux
p

y  C3
dy
L
After integrating this equation
p 2 C3 y
ux  
y 
 C4
2 L

Apply the boundary condition that ux=0 at y=±h/2
C3  0
C4 
p 2
h
8 L
ph 2  4 y 2 
ux 
1 2 

8 L 
h 
Pressure-driven flow through a
narrow rectangular channel
The velocity is a maximum at y=0
umax
ph 2

8 L
 4 y2 
ux  umax 1  2 
h 

The volumetric flow rate is the integral of the velocity over the
cross-sectional area
Q
h /2

w
 h /2 0
 4 y2 
2umax wh
ux dzdy  umax w 1  2  dy 
 h /2
h 
3

h /2
Pressure-driven flow through a
narrow rectangular channel
Analytic solution
t yx  
p
y
L
ph 2  4 y 2 
ux  
1 2 

8 L 
h 
Numerical solution
dt yx
p

dy
L
t yx  
dux
dy
Pressure-driven flow through a
narrow rectangular channel
Use Matlab to provide analytic solution and numerical solution
∆ p= 1(Pa)
H=0.1(m)
L=xL-x0=1.5(m)
µ= 1.0 x 10-3 (Pa s, N s/m2)
function NewtonianFluidFlowRectangular
% Pressure-driven flow through a cylindrical tube
% Solve momentum equation to obtain shear stress and velocity
distribution
% Laminar Flow in a Horizontal rectangular channel (Newtonian
Fluid)
%
clear all
clc
global deltaP L mu H
deltaP = 1;
L = 1.5;
mu = 1.0e-3;
H = 0.1;
a = 0;
b = H;
% Sovle the problem of ODE-BVP
% initialize of solution with a guess of y1(r)=0,y2(r)=0,y3(r)=0
solinit = bvpinit(linspace(a,b,100),[0 0 0]);
sol = bvp4c(@ODEs,@BCfun,solinit);
Pressure-driven flow through a
narrow rectangular channel
% Analysis results
h = sol.x/2;
TauAnal = deltaP*h/L;
uAnal = (deltaP*H^2/(8*mu*L))*(1-(2*h/H).^2);
umAnal = deltaP*H^2/(8*mu*L);
% Plot
% Shear stress
tau = sol.y(1,:);
%tau = [0 tau];
plot(h,tau,'b-',h,TauAnal,'h-.')
xlabel('Plane hight h£¬m')
ylabel('Shear stress£¬kg/(m s^2)')
legend('Numerical results','Analysis results')
figure
% Velocity distribution
plot(h,sol.y(2,:),'b-',h,uAnal,'h-.')
xlabel('Plane hight h£¬m')
ylabel('Velocity£¬m/s')
legend('Numerical results','Analysis results')
% Average velocity
fprintf('\tAverage velocity: um = %.4f',sol.y(3,end))
Pressure-driven flow through a
narrow rectangular channel
function dydh = ODEs(h,y)
global deltaP L mu H
Tau = y(1);
u = y(2);
um = y(3);
dTaudh = deltaP/L/2;
dudh = -Tau/mu/2;
dumdh = u*2*h/H^2;
dydh = [dTaudh; dudh; dumdh];
% -----------------------------------------------------------------function bc = BCfun(ya,yb)
bc = [ya(1); yb(2); ya(3)];
Pressure-driven flow through a
narrow rectangular channel
Pressure-driven flow through a
narrow rectangular channel
We also can use Matlab PDEtool to provide numerical solution
Pressure-driven flow through a
narrow rectangular channel
Pressure-driven flow through a
narrow rectangular channel
Pressure-driven flow through a
narrow rectangular channel
Pressure-driven flow through a
narrow rectangular channel
Pressure-driven flow through a
narrow rectangular channel
Pressure-driven flow through a
narrow rectangular channel
Pressure-driven flow through a
narrow rectangular channel
Pressure-driven flow through a
narrow rectangular channel
Pressure-driven flow through a
narrow rectangular channel
Pressure-driven flow through a
narrow rectangular channel
Pressure-driven flow through a
narrow rectangular channel
Pressure-driven flow through a
narrow rectangular channel
Pressure-driven flow through a
narrow rectangular channel
Pressure-driven flow through a
narrow rectangular channel
Pressure-driven flow through a
narrow rectangular channel
Pressure-driven flow through a
narrow rectangular channel