Cell Process 1. Cell and transport phenomena (2) Liang Yu Department of Biological Systems...

Cell Process1. 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

FluidA material that deforms continuously when subjected to a force applied tangentially to a surface

Biological fluidQuite complex, exhibiting solid and liquid-like behavior and deforming in a time-dependent fashion

Fluid mechanicsThe study of the motion of fluids in response to the application of stresses

Physiological Fluid MechanicsConcerned 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

Called the partial derivative of the temperature with respect to time, or local accelerate

, ,x y z

v

t

Velocity field: , , ,v v x y z t

(x,y,z)

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.

xy

z

dy

dx

Faces are labeled North, East, West,

South, Top and Bottom

dz

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

, , ,AA A A

dyv x y z t

dt

, , ,AA A A

dzw x y z t

dt

, , ,AA A A

dxu x y z 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

, , ,v v x y z t

Fluid velocity is a function of the three spatial coordinates x, y, z, and if the flow is unsteady, time t

The total differential is defined as v v v v

dv dx dy dz dtx 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

x y z

v v v va v v v

x y z t

Local acceleration Convective acceleration

Dv vv v

Dt t

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 /Dt:

• With dx/dt=u, dy/dt=v, dz/dt=w, this results in:dt

dz

zdt

dy

ydt

dx

xtDt

D

grad

tDt

D.u

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 f is given by:

• For a fluid element, for an arbitrary conserved property f :

grad

tDt

D.u

0)( u div

t 0)()(

udiv

t

Continuity equation Arbitrary property

(for moving fluid particle) (for given location in space)

Dt

Ddiv

tgrad

tdiv

t

)(.)()(

uuu

zero because of continuity

Dt

Ddiv

t

)(

)(u

Rate of increase of of fluid element

Net rate of flow of out of fluid element

Rate of increase of for a fluid particle=

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:

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.• Rate of increase is:• The inflows (positive) and outflows (negative) are

shown here:

zyxt

zyxt

)(

xy

z

( ) 1.2

ww z x y

z

( ) 1

.2

vv y x z

y

zyxx

uu

2

1.)(

zxyy

vv

2

1.)(

yxzz

ww

2

1.)(

( ) 1.2

uu x y z

x

Continuity equation

• Summing all terms in the previous slide and dividing by the volume xyz results in:

• In vector notation:

• For incompressible fluids r / t = 0, and the equation becomes:

div u = 0.• Alternative ways to write this: and

0)()()(

zw

yv

xu

t

0)( u div

t

Change in density Net flow of mass across boundariesConvective term

0

zw

yv

xu 0

i

ixu

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.

• Rate of increase of x-, y-, and z-momentum:

• 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.

Dt

Dw

Dt

Dv

Dt

Du

Dt

DvmaF

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 z-direction 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

x

z

y

zyxx

pp )

2

1.(

zyx

x

pp )

2

1.(

yxzzzx

zx tt )2

1.(

yxzzzx

zx tt )2

1.(

zxyyyx

yx t

t )21.(

zxyyyx

yx t

t )21.(

zyxxxx

xx tt )2

1.(

zyx

xxx

xx tt )2

1.(

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:

– p is a compressive stress and txx is a tensile stress.

• Similarly for y- and z-momentum:

Mxzxyxxx Szyx

p

Dt

Du

ttt )(

Myzyyyxy Szy

p

xDtDv

ttt

)(

Mzzzyzxz S

zp

yxDtDw

)( ttt

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 flow2. The fluid density is constant, which indicates that the fluid is incompressible3. The flow is steady; that is, pressure, shear stress and velocity do not change with

time4. The fluid is Newtonian5. 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


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.

0yx yxx x x x x xp p y z x zt t

A momentum balance in the x direction yields


Dividing by ∆x∆y∆z and taking the limit as each goes to zero results in the ordinary differential equation

yxddp

dx dy

t

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

1 2p C x C


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

0 0

pp p x x

L

yxd p

dy L

t

Integrate τ to yield

3yx

py C

Lt

Newton’s law of viscosity

xyx

du

dyt


Substitute into Newton’s law of viscosity

3xdu p

y Cdy L

After integrating this equation

2 342x

p C yu y C

L

Apply the boundary condition that ux=0 at y=±h/2

3 0C 24 8

pC h

L

2 2

2

41

8x

ph yu

L h


The velocity is a maximum at y=02

8max

phu

L

2

max 2

41x

yu u

h

The volumetric flow rate is the integral of the velocity over the cross-sectional area

2/2 /2max

max 2/2 0 /2

4 21

3

h w h

xh h

y u whQ u dzdy u w dy

h


Analytic solution

yx

py

Lt

2 2

2

41

8x

ph yu

L h

Numerical solution

yxd p

dy L

t

xyx

du

dyt


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 allclc global deltaP L mu HdeltaP = 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);

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)


% Analysis resultsh = 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 stresstau = 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 distributionplot(h,sol.y(2,:),'b-',h,uAnal,'h-.')xlabel('Plane hight h£¬m')ylabel('Velocity£¬m/s')legend('Numerical results','Analysis results') % Average velocityfprintf('\tAverage velocity: um = %.4f',sol.y(3,end))


function dydh = ODEs(h,y)global deltaP L mu HTau = 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)];


We also can use Matlab PDEtool to provide numerical solution

Cell Process 1. Cell and transport phenomena (2) Liang Yu Department of Biological Systems...

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Transcript of Cell Process 1. Cell and transport phenomena (2) Liang Yu Department of Biological Systems...