biotrans 08 ch5_01.pdf

Ch 5 Solute Transport in Biological Systems

5.1 DESCRIPTION OF SOLUTE TRANSPORT IN BIOLOGICAL SYSTEMS

Objective today

Introduction of diffusion in capillary

1st Ficks law and diffusivity

2nd Ficks law

General mass transport equation with flow

Boundary flow

Turbulence

Laminar

5.1 DESCRIPTION OF SOLUTE TRANSPORT IN BIOLOGICAL SYSTEMS

Solute transport process

Solute transport occurs both

through bulk fluid motion, also known as convection,

and by solute diffusion due to the presence of solute concentration gradients.

In biological as well as synthetic membrane systems, the diffusion of a solute will also be affected by the presence of a variety of heterogeneous structures.

e. g., solutes will need to diffuse through

porous structures, such as the capillary wall or

a polymeric membrane, around or through cells within the extravascular space, and

through the interstitial fluid containing a variety of macromolecules.

5.2 CAPILLARY PROPERTIES

CAPILLARY PROPERTIES

This chapter concerns solute transport through the capillary wall as a representative porous semipermeable membrane.

The content is also generally applicable to synthetic membranes that are used in a variety of biomedical device applications.

1st we need to define the physical properties of a typical capillary.

These properties are summarized in Table 5.1.

Each capillary can only supply nutrients and remove waste products from a very small volume of tissue that surrounds each capillary.

Because of

Small surface: capillaries are very small: D: 810 m and

L: lengths < 1 mm.

Short diffusion time: the residence time of blood in a capillary: ~ 12 sec


TABLE 5.1 Capillary Characteristics


Three types of capillaries

Continuous capillaries

found in the muscle, skin, lungs, fat, the nervous system, and in connective tissue.

The capillary lumen lies within a circumferential ring of several endothelial cells, as shown in Fig. 3.2.

Fenestrated capillaries

have the pores that facilitate diffusion across the endothelial lining

are much more permeable to H2O and small solutes in comparison to continuous capillaries.

The endothelium of these capillaries is perforated by numerous small holes called fenestrae.

The fenestrae are sometimes covered by a thin membrane that provides selectivity with regard to the size of solutes that are allowed to pass through.

found in tissues that are involved in the exchange of fluid or solutes such as hormones.

e.g., within the kidney they are found in the glomerulus and allow for a high filtration rate of plasma.

Discontinuous capillaries

have large endothelial cell gaps that readily allow the passage of proteins and even

red blood cells.

http://www.lab.anhb.uwa.edu.au/mb140/corepages/vascular/vascular.htm


5.3 CAPILLARY FLOW RATES

As learnt, some of the blood plasma that enters the capillary will be carried or filtered across the capillary wall by the combined effect of the

hydrodynamic and

oncotic pressure differences that exist between the capillary and the surrounding interstitial fluid.

This perfusion of plasma across the capillary wall is also known as plasmapheresis.

The total flow rate of this fluid across the capillary wall

Can be estimated by Starling equ. (3.4) developed in Chapter 3.

Equ 3.7 that the value of the hydraulic conductance, LP, is given by

S represents the circumferential surface area of a given capillary and

AP is the total cross-sectional area of the pores in the capillary wall

r radius in the capillary wall.

AP/S: the porosity of the capillary wall (or membrane) and is often given the symbol .

(3.7)


Example 5.1

Calculate the convective flow rate of plasma across the capillary wall and compare it with the total flow rate through the capillary.

Solution

Using the capillary properties provided in Table 5.1

plasma viscosity = 1.2 cP

hydraulic conductance, LP, can be shown as in Example 3.2 to be equal to 0.61 cm3 h1m2 mmHg1 (or 1.28 1012 m2 sec kg1).

We can then calculate the filtration rate as follows using Equ. 3.4:

Q

effective pressure drop


Example 5.1

We can compare this value of the plasma filtration flow rate across the capillary wall to the total flow rate of blood entering the capillary, i.e., :

Q capillary >> Q filtration We may assume that the blood flow, Qcapillary, is constant along the length

of the capillary.

5.4 SOLUTE DIFFUSION

In addition to the convective transport, solute can also diffuse down its own concentration gradient.

From

this is also equivalent to stating that a solute diffuses from a region of high i(high) to a region of lower i(low)

5.4.1 FICK'S FIRST LAW AND DIFFUSIVITY

Consider the situation shown in Fig. 5.1.

The surface of a semi-infinite plate of length L contains a solute that maintains a constant

C (y=0) =C0 along the surface of the plate

C (y>0, t=0) = 0

At t = 0, this surface is contacted with a quiescent fluid or a solid material that initially does not contain solute:

V (t=0) = 0

As time progresses, solute diffuses from the surface of the plate into the quiescent fluid or solid material

C (y>0, t>0) > 0

If the fluid has solute at a concentration of C, then the following analysis still applies; however, the concentration that cause the diffusion would need to be defined as C = C0 -C.

Fig. 5.1 shows a thin shell of thickness y.

The rate at which the solute enters and leaves this thin shell by diffusion at y and y + y is proportional to the solute concentration gradient at these locations.

Velocity profile in Couette flow

For the situation shown in Fig. 4.2,

vy = 0

steady state there is no flow in the y direction

vz, = 0

because the plates are very large in the z direction (no wall and corner effect) in comparison to the distance h that

separates them


FIG. 5.1 Solute concentration in the vicinity of a flat plate of constant surface concentration.

Similar to the situation shown in Fig. 4.2 for Couette flow


The diffusion rate of this solute can therefore be described by Fick'sfirst law:

C : concentration of the solute and typical units are moles per liter.

JS: solute diffusion rate (units: moles /s)

S: The surface area normal to the y-direction of solute diffusion

D: solute diffusivity or the diffusion coefficient.

generally depends on the

size of the solute and the

physical properties of the fluid or

material in which the solute is diffusing.

Since the solute is assumed to be diffusing through a homogeneous medium, this solute diffusivity is sometimes referred to as the bulk diffusivity.


Diffusivity

Fig. 5.2 presents diffusivity data for a variety of solutes in dilute aqueous solutions at 37C as a function of solute molecular weight (MW).

The solid line through the data is the result of a linear least squares regression.

The following empirical equation is based on the data in Fig. 5.2,

A useful relationship for estimating the diffusivity of a solute in water at 37C, knowing only the MW of the solute:


FIG. 5.2 Solute diffusivity in water at 37C.

(Based on data from Renkin, E.M. and Curry, F.E., Membrane Transport in Biology, Springer-Verlag, New York, 1979.)


Diffusivity

The diffusivity of solutes in dilute solutions can also be estimated from the Stokes-Einstein equation (Bird et al. 2002):

R: ideal gas constant (8.314 J mol1 K1),

T: temperature in K,

a: solute radius,

NA: Avogadro's number (6.023 1023 mol1),

: solution viscosity.

This equation at best is only accurate to about 20% .

If the diffusivity of a solute is known, Equ. 5.4 may be used to obtain an estimate of the molecular radius (a) of the solute.


Diffusivity

Combining Equ (5.3) and (5.4) give a rough estimation of a relation between a and MW, i.e. a(MW)

If the diffusivity and size of the solute are not known, the solute size can first be estimated from Equ. 5.5.

This equation assumes that the solute of a given MW is

a sphere

with a density ( 1 g cm3) equal to that of the solute in the solid phase


Diffusivity

Example 5.2

Estimate the diffusivity of a spherically shaped protein with a MW of 36,000 in water at 37C.

Solution

From Fig. 5.2, we see that for a solute of this size, D = 8 107 cm2

sec1.

From Equ. 5.3, we find that D = 8.12 107 cm2 sec1.

The Stokes-Einstein equation (Equ. 5.4) can also be used,

but radius a of this molecule is needed

a can be estimated using Equ. 5.5


Diffusivity

Then from Equ. 5.4

Note that

D obtained from the Stokes-Einstein equation is about 40% larger than that estimated from the actual data shown in Fig. 5.2.

Possible reason:

Very large molecules like proteins can be solvated or hydrated,

making the radius of the actual solute-solvent complex larger than the radius estimated from the solute MW alone using Equ. 5.5.


Proteins can be solvated or hydrated

http://bimanbagchi.com/hydration.html

5.4.2 FICK'S SECOND LAW

Unsteady mass transport

Mass conservation equation Equ. 1.8

Accumulation (Rate of increase in mass)

= in out + generation consumption (1.8)

Rate of increase in mass = (rate of mass flowed in

- rate of mass flowed out)

+ (rate of mass diffused in

- rate of mass diffused out)

+ mass production

mass consumption


FIG. 5.1 a Solute concentration in the vicinity of a flat plate of constant surface concentration.

S = x z


For derivation of equ. (5.7)

No flow

z

x

S = x z


Unsteady mass transport without convection

Consider Fig 5.1

There can be two cases in general

Constant c0 Changing c0 Both are und=steady process

Fick's 1st law:

Consider a surface

Fick's 2nd law:

Consider a control volume

Constant c0 Changing c0

c0

c1

c2

cin

coutc

c c

Fig 5.1b


Unsteady mass transport without flow (convection)

unsteady solute balance across the shell of thickness, y, in Fig 5.1,

using Equ. 1.8 as our guide, that

S y: control volume

the accumulation of solute mass within S y.

the net rate at which solute enters or leaves the control volume by diffusion, according to Fick's first law.

as y 0, we obtain the following result, which is known as Fick's 2nd law:

Solution of Equ. 5.7 for the situation shown in Fig. 5.1 requires IC and BC for the solute within the fluid or solid material region:

t

Cy S

Passive scalar transport equation

Scalar fluxes with flow (convection)

Molecular diffusion flux

Convective flux

Mass uScjconv

y+yy+y

viciy

o Di: diffusivityo This figure

shows flux per area

y

cD ii

y

Passive scalar transport equationGeneral mass conservation equation

Rate of increase in mass = (rate of mass flowed in

- rate of mass flowed out)

+ mass production

mass consumption

Rate of mass flowed in along y direction includes

Molecular diffusion flows in

Convective flux (out) in along y direction

Rate of mass flowed out along y direction includes

Molecular diffusion flow out in y direction

Convective flux (out) in y direction

yyyyconv cSVj ,

yyyyyDdy

dCDSj ,

yyyyyyconv cSVj ,

yyyDdy

dCDSj ,

Equ. 1.8a


General mass conservation equation without source and sink

Total mass flow rate in y-direction

yyyyyyyy

yyyyyyyy

yyyconvyyyDyyconvyyD

zVxcdy

dCzxDzVxc

dy

dCzxD

cSVdy

dCDScSV

dy

dCDS

jjjj

)(

)(

,,,,


General mass conservation equation

x and z direction have the similar description

Mass production = 0

Mass consumption = 0

Rate of mass change in the control volume due to 3-D flow and molecular diffusion

zdectioincomponentingcorrespond

diectionxincomponentingCorrespond

cVdy

dCDScV

dy

dCDS

t

Czyx

yyyyyyyy

)(

t

Czyx

t

CV


General mass conservation equation

Divide and assume 0

x and z direction have the similar description

Sum together

2

2

/))((dy

CdDSy

dy

dCDS

dy

dCDS yyy

zyx zyx

y

cVycVcVycVcV

y

yyyyyyyyyy

)(/)/)(

Passive scalar transport equationScalar fluxes and constitutive properties

we get convective-diffusion equation

non-dimensional form

C* = c/C;

t * = t/(l/U)

x* = x/l; y* = y/l; z* = z/l;

Peclet number Pe = Ul/D

Physical meaning: convection flux/diffusion flux

Note there are many Pe as there are many Re

Pe >> 1 negligible molecular diffusion

Pe


To solve c distribution in the convective-diffusion equation

We need to solve u first

Solution of u depends if the flow is laminar or turbulence

Laminar is easier

But turbulence is not easy to solve Statistic method

Turbulence model

We will study velocity field first

Then concentration

cDct

c 2)(

u

4.15 BOUNDARY LAYER THEORY

Turbulence (brief introduction)

It is difficult to give a clear definition of turbulence

But we can understand turbulence by studying its features

Turbulence features

fast diffusion

random motion

high dissipation rate

continuous flow

multiscale eddies

3-D and

high Re

large coherent structures

Turbulence

Fast diffusion

Because vortices and eddies

Much faster than that in laminar

Turbulence

Random motion

Temporal velocity at a point is random

Turbulence

High dissipation rate

Large flow resist or pressure drop

Turbulence

Multiscale eddies

There are many different size of eddies

Can be quantitatively measured by power spectrum density

Turbulence

3-D

In all 3 direction, there are difference


Boundary layer

V changes from 0 at wall to the free stream value over a narrow region near the surface: boundary layer (BL).

A revolutionary concept in 1904 published by Prandtl

Generally, it is due to the effect of

Describing the flow of a fluid near a surface is extremely important in a wide variety of engineering problems.

It is the this BL that affects the rates of

Momentum

mass transfer and

heat transfer between the surface and the fluid.

Analysis of simple flow in BL provide a great deal of insight into how the flow of the fluid affects the transport of mass and energy

leading to the rational development of correlations to describe transport in more complex geometries and flow systems.

We will use these correlations for mass transfer

Turbulence

Study methodology

Statistics view

Random process

Use mature statistics

Coherent structures

Turbulent flows are not purely random

but also have quasi periodic structures, i.e. large vortices

Numerical simulation

Direct numerical simulation

Large eddy simulation

A particular theory in flow dynamics is boundary layer


Boundary layer

V changes from 0 at wall to the free stream value over a narrow region near the surface: boundary layer (BL).

A revolutionary concept in 1904 published by Prandtl

Generally, it is due to the effect of

Describing the flow of a fluid near a surface is extremely important in a wide variety of engineering problems.

It is this BL that affects the rates of

Momentum

mass transfer and

heat transfer between the surface and the fluid.

Analysis of simple flow in BL provide a great deal of insight into how the flow of the fluid affects the transport of mass and energy

leading to the rational development of correlations to describe transport in more complex geometries and flow systems.

We will use these correlations for mass transfer

x

u

4.15.1 FLOW NEAR A WALL THAT IS SET IN MOTION

FIGURE 4.13 Flow of a fluid near a flat plate that is set in motion.


Momentum transport and velocity profile without convection


A semi-infinite quantity of a viscous fluid is contacted from below by a flat, horizontal plate.

For t < 0, the plate and the fluid are not moving V (t



t > 0, momentum is transferred further into the fluid.

This creates a velocity profile in the y direction.

We can arbitrarily define the BL thickness () for this problem as

the distance perpendicular to the plate surface where the fluid has just been set in motion, and

this distance is defined as that point where the local velocity is equal to 1% of V0, i.e. vx = 0.1 V0

Goal: to determine the

vx profile vx(y,t)

(t). (Do vx and (t) depend on x? Why)

The concept of a shell balance can be used to analyze this problem.

An important technique for developing mathematical models to describe the transport of such quantities as V, C, and T.

Approach is conceptually easy to use and is based on the application of the generalized balance equation (Equ 1.8) to a given finite volume of interest.


Consider a small volume element of the fluid x y W

W: is the width of the plate in the z direction (normal to the page) and is assumed to be very large.

Recall that

Momentum = (mass) (velocity)

we can write the momentum per unit volume of the fluid as vx.

The rate of accumulation of momentum within this volume element of the fluid is equal to

is constant (incompressible)

This term has units of force and is also equal to the sum of the forces acting on the volume element of the fluid.

The only forces acting on this volume element of fluid are the shear forces acting on the surfaces at y and y + y, i.e.

yx|y and yx|y+ y These terms also, respectively, represent the flux of x momentum in

the y direction at y and y + y, respectively.


FIGURE 4.13 Flow of a fluid near a flat plate that is set in motion.

y W

x


We now use momentum balance equation to establish equation for velocity field

General momentum balance equation in a control volume

Rate of increase momentum = (rate of momentum flowed in rate of momentum flowed out) + external force loaded on the fluid

In this special case

Rate of increase momentum =

Rate of momentum flowed in y direction is 0 ( vy = 0)

Rate of momentum flowed in z direction is 0 ( vz = 0)

Rate of momentum flowed in and out in x direction is the same

Uniform in x direction

The only external force loaded on the fluid = yx|y - yx|y+ y


The momentum shell balance on the volume element can be written as

Eliminating x and W, and then dividing by y and taking the limit as y 0, :

Valid for any fluid

Newtonian or

non-Newtonian.

Can equ. (4.42) be used for turbulence? Why?


For the special case of the Newtonian fluid, we can use Equ. 4.2 for yx and obtain

v is the kinematic viscosity and is defined as /.

or v is seen as a constant here

This is a partial differential equation

One IC and two BCs are required for solution

The initial condition (IC) and boundary conditions (BC) are

Equ. 4.44 can easily be solved using the Laplace transform technique.


Laplace transform of a function

f(t) is defined by

Tables of Laplace transforms can be found in a variety of calculus textbooks and mathematical handbooks.

Table 4.5 summarizes some of the more commonly used Laplace transforms.

Table 4.5 summarizes some of the more commonly used Laplace transforms.


Solution of equ. 4.43

Taking the Laplace transform of Equ. 4.43 results in the following equation:

denotes the Laplace transform of the velocity.

From the IC (Equ 4.44), we have that

vx(t = 0) = 0, and

From the BCs, the transformed BCs become

Equ. 4.46 can then be written as a linear homogeneous 2nd order differential equation

with a2 = s/v.


Equ. 4.48 can be solved with method and has the following general solution:

The constants C1 and C2 can then be found from the transformed BCs given by Equ. 4.47.

Using these BCs, we find that C1 = 0 and C2 = V/s.

solution in the Laplace transform space is

Inverting Equ. 4.50, we find the function vx(t)

From Table 4.5, we use transform pair 25 and then find that the inverse of Equ. 4.50 provides the following solution for vx(t):


Derivation

From ID 1 and assume c2 =0 2


Derivation

From ID 1 and assume c2 =0, c1 = 1/V

At y =

At y =0


Solution of the linear homogeneous 2nd order differential equation


Derivation


The above solution is in terms of a new function called the

error function abbreviated as erf and the

complementary error function, as erfc

The error function is defined by the following equation:

Note: integral of ex2 cannot be integrated analytically.

Since this integral is quite common in the solution of many engineering problems,

this function has been tabulated in mathematical handbooks and mathematical software and

can be treated as a known function


Error function


(y,t)

Defined as that distance y from the surface of the plate where the velocity has decreased to vx/V =1%

From 4.51)

The complementary error function of

provides a value of

vx/V = 0.01.

we can define the boundary layer thickness, (y,t), as

(y,t) = yv/V=0.01 = 1.821*2 (t)1/2

The value of meaning:

the distance to which momentum from the moving plate has penetrated into the fluid at time t.


(y,t)

Rewrite equ 4.53:

t = 2/(16 ) t ~ 2/

Approximate euq.

u/t ~ u/L2

t ~ L2/

Above relation indicate the time is

Proportional to L2

If L


Example 4.5

Calculate the boundary layer thickness 1 sec after the plate has started to move. Assume the fluid is water for which v = 1 106 m2 sec1.

Solution

Using Equ. 4.53, (y,t) can be calculated

4.15.2 LAMINAR FLOW OF A FLUID ALONG A FLAT PLATE

Objective: derive velocity profile in laminar BL

Steady laminar flow of a fluid along a flat plate Fig. 4.14

The plate is assumed to be semi-infinite and

the fluid approaches the plate at a uniform velocity, V.

Since vx (y=0) = 0 a BL is formed, where, vx increases from 0 to V.

Here, we will obtain an approximate solution, which is very close to the exact solution.

Details on exact solutions to the boundary layer equations can be found in Schlichting (1979), an excellent classical book!

FIG. 4.14 Laminar boundary layer flow of a fluid over a flat plate.

V

y W

Is vy = constant within the BL?

Momentum integral equation for boundary layer

Q2

Q1Q

Q2

Q1

http://www.thermopedia.com/content/595/http://www.dept.aoe.vt.edu/~jschetz/fluidnature/unit02/unit2b.html


Mass balance

Consider the shell volume located

From x to x + x and

from y = 0 to y = (x)

Width is W.

1st perform a steady state (d/dt = 0) mass balance on this shell volume, which is given by

The 1st two terms provide the net rate (i.e., In-Out) at which mass is being added to the shell volume.

The 3rd term accounts for the loss of mass from the shell volume at the top of the boundary layer

due to flow in the y direction, i.e. vy 0.

due to BL

due to resistance


Eliminating W and ,

dividing by x,

then taking the limit as x 0,

provides the following equation for vy description within the BL

Relation between vy and vx


Momentum balance

Similarly we have steady x-momentum balance on the shell volume as

1st two terms: the net rate at which x-momentum is being added to the shell volume.

3rd term: rate at which x-momentum at the top of the BL is being lost due to the flow of fluid out of the BL in the y direction.

4th term: loss of momentum as a result of the shear stress generated by the fluid at the surface of the plate.

All items mean per time


After

eliminating W and

dividing by x and

taking the limit as x 0, from Equ. 4.56

Momentum integral equation for boundary layer

http://www.thermopedia.com/content/595/http://www.dept.aoe.vt.edu/~jschetz/fluidnature/unit02/unit2b.html

Wake flow


Now use Equ. 4.55 to eliminate vy in the previous equation and equ. 4.57 becomes

For a Newtonian fluid, we can use Equ. 4.2 once again and obtain Equ. 4.59:

This is well-known von Karman momentum balance equation

It forms the basis for obtaining an approximate solution to the boundary layer flow over a flat plate.


To obtain an approximate solution, we need to

o Approximate the shape of the velocity profile within the boundary layer, i.e., vx(x,y)

o Know BCs.

In equ. 4.59, differential has already vx2

The simplest function that reasonably approximates the shape of the velocity profile is a simple cubic equation:

Need 4 BCs, which are:


1st BC: no slip BC, which requires that the vx(y) of the fluid at the surface of the plate be the same as the V of the plate that,

o In this case, vx(y=0) = 0.

4th BC: expresses that the stress at the surface of the plate only depends on x and not on y, i.e. (x, y=0) = dvx/dy |y=0(x)

o (x, y=0) = (x)

2nd and 3rd BC: state that beyond BL, i.e. y > ,

vx(y) is constant and

vx(y) = V.

The velocity profile of equ. 4.60 has to satisfy the BCs in equ 4.61

When the above BCs are imposed on Equ. 4.60,

the following expression is obtained for vx(y) within the BL


Thickness of BL

Equ. 4.62 indicates that vx depends on unknown (x)

However, we can

o insert Equ. 4.62 into the von Karman momentum balance equation (Equ. 4.59)

o after some simplification we obtain

with BC: x = 0, = 0.

o Integration of Equ. 4.63 results in the following expression for the BL thickness, (x):

V

BC


Equ. 4.64 shows that

(x) grows in proportion to x1/2

(x) is inversely proportional to V1/2

Compare Equ. 4.64 and Equ. 4.53):

They are simila

Here momentum transfer time t = x/V, i.e. convection time from x = 0 to x

Rearrange Equ. 4.64 into the following form:

Rex = Vx/ is defined as the local value (i.e., at location x) of the Reynolds number.

Re 2

Re 1

Re 2 < Re 1

x

(x)


Meaning of Re

Re is a very important dimensionless number in the field of fluid mechanics.

Physically, Re represents the ratio of the inertial forces (V VL2 = V2L2) acting on the fluid to the viscous forces ((V/L) L2 = VL) acting on the fluid,

L : a characteristic dimension

V: velocity

Critical Re for transition from laminar to turbulent flow depends on the geometry of the flow being considered.

e.g, for BL flow over the flat plate, experiments show that the flow is laminar provided Rex


Displacement thickness

We have

displacement

Physical meaning

A concept meaning that the BL represents a deficit in mass flow compared to inviscid flow with slip at the wall.

A distance by which the wall would have to be displaced in the inviscidcase to provide the same total mass flow as the viscous case.

)( *0

1

VdyvQ x

0000

* 1 dyV

vdy

V

vdydy

V

v xxx


Momentum thickness

We have

Momentum thickness = **- * (Note vx/V < (vx/V)2)

Physical meaning

as the loss of momentum flux per unit width divided by V2 due to the BL

In general (~ 13% ) < * (~ 34% ) <

*)*(20

2

VdyvM x

0 2

2

** )1( dyV

vx

00 2

2

*** )1()1()1( dyV

v

V

vdy

V

v

V

v xxxx

M M1

M2 M2

M1

***

Real fluid Ideal fluid


Drag force

Power needs to be considered in design of biomedical devices resist

The approximate solution vx(y) within BL for laminar flow on a flat plate:

combination of Equ. 4.62 and 4.64

Differentiate Equ. 4.66 at y =0, drag force exerted by the fluid on both sides of the plate.

length L and

width W

dzdxFL

yx

w

00


Drag force

Fx of the exact solution as well as the experimental data is about 3% greater than that predicted by the above approximate solution

The constant in Equ. 4.67

being 1.328 for the exact solution.

Hence, we see that this approximate solution to the flat plate boundary layer problem is quite good.

We can also calculate the power needed to overcome the drag force.

power is defined as

P = force velocity

after multiplying Equ. 4.67 by V, the power is given by


Friction factor f of drag force

defined as the ratio of the shear stress at the wall and the kinetic energy per volume of the fluid based on the free stream velocity.

From (4.62)

The local value of f is then given by

Average f: combine equs. (4.69) and (4.64),

integrating the above equation as

ReL = VL/.


Friction factor of drag force

f meaning: the force acting on both sides of the plate (i.e., Fx) is given by the product of the

kinetic energy per volume of the fluid ((1/2)V2)

the area of the flat plate (2 LW)

friction factor (f), and is given by

the same as that given by Equ. 4.67.

As shown by Equ.s 4.70 and 4.71

f is a convenient method for finding the force acting on a surface as a result of fluid motion.

When the flow is across or over an object, f is also known as the drag coefficient CD (f = CD)

since it allows for the calculation of the drag force exerted on the object by the flowing fluid


Unsteady mass transport in quiescent fluid

unsteady solute balance across the shell of thickness, y, in Fig 5.1,

we can then write, using Equ. 1.8 as our guide, that

S y: control volume

the accumulation of solute mass within S y.

the net rate at which solute enters the control volume by diffusion, according to Fick's first law.

as y 0, we obtain the following result, which is known as Fick's 2nd law:

Solution of Equ. 5.7 for the situation shown in Fig. 5.1 requires IC and BC for the solute within the fluid or solid material region:

t

Cy S


FIG. 5.1 Solute concentration in the vicinity of a flat plate of constant surface concentration.

S = x z

5.4.3 SOLUTION FOR THE CONCENTRATION PROFILE FOR DIFFUSION FROM A FLAT PLATE INTO A QUIESCENT FLUID

Derivation


Mass transport equation

Unsteady process:

Need Ficks 2nd law

Similar to momentum transport due to viscous diffusion from a moving flat plate to a quiescent fluid


Similarity between momentum and mass transport

o Momentum transport due to viscous diffusion from a moving flat plate to a quiescent fluid

o Mass transport due to diffusion from a flat plate to a quiescent fluid

for all t > 0

for all y > 0

for all t > 0


we get convective-diffusion equation

non-dimensional form

C* = c/C;

t * = t/(l/U)

x* = x/l; y* = y/l; z* = z/l;

Peclet number Pe = Ul/D

Physical meaning: convection flux/diffusion flux

Note there are many Pe as there are many Re

Pe >> 1 negligible molecular diffusion

Pe


Solution of C profile

Equ. 5.7 and 5.8 are analogous to Equs 4.43 and 4.44 for the flat plate that is set in motion within a semi-infinite fluid that Vx(y, t=0) = 0

A unsteady process

Similarly, here we can replace

vx with C, and

V with C0,

with D in Equ. 4.51.

result for C profile

within the quiescent fluid

or solid material at any location y and time t:

Similar to momentum transport solution

One of the reason why we study momentum transfer first in detail


BL of C

C BL thickness, c, as that distance where C has decreased to 1% of C0,the value at the surface of the plate.

Recall from Chapter 4 that the complementary error function of

vx/V =0.01

provides a value of C/C0 that is equal to 0.01.

Here for C/C0 = 0.01

BL pf vx

Hence, we can define the concentration BL thickness, c(t), as

The value of c can also be interpreted as the distance to which the solute from the plate has penetrated (or mass transferred) into the fluid at time t.


Example 5.3

Calculate the concentration boundary layer thickness 1 sec after the plate has made contact with the fluid. Assume that the fluid is water and that the solute diffusivity is D = 1 105 cm2 sec1.

Solution

Using Equ 5.10, (y,t) can calculated as

5.4.4 DEFINITION OF THE SOLUTE FLUX

The flux of solute diffusing at any location y is defined as

the moles of solute per unit time per unit area normal to the direction of diffusion, i.e.,

js = Js/S.

The flux at the surface of the plate in Fig. 5.1 is

For the situation in Fig. 5.1, differentiating Equ. 5.9 with respect to y

at y = 0

Equ. 5.12 js at the surface of the plate is inversely proportional to the square root of the contact time of the plate with the fluid.

Why?

How many variable in equ. 5.12?

5.4.5 DEFINITION OF THE MASS TRANSFER COEFFICIENT

Solution of mass transfer problems is oftentimes facilitated by defining the mass transfer coefficient, km.

as the proportionality constant that relates the

molar flux of the solute (js) at surface to the overall concentration driving force, i.e.,

js = km(CHighCLow).

js = km(CHighCLow) LDm/LDm = (kmL /Dm)Dm(CHighCLow) /L = ShDm{CHighCLow) /L

km is useful for both

with convection

without convection


km is often given in terms of the dimensionless group, known as the Sherwood number (Sh).

The ratio of the transport rate of the solute by convection to that by diffusion

Defined as Sh = kmL/Dm = ratio of solute transport rate of the convection/diffusion

Dm: diffusivity

L: a characteristic length.

e.g., for convective mass transfer within a cylindrical tube, L is the tube diameter D.

Another way to understand Sh = kmL/Dm= km(CHighClow)/(Dm(CHighClow)/L)

Another way to understand km kmCHighClow) = js = Dm dC/dy ~ Dm C/ y ~ Dm(CHighClow)/Lc =

(Dm/Lc)(CHighClow)

Km ~ Dm/Lc km /( Dm/Lc) = Sh


General, Equ. 5.13, defines km in terms of the flux of solute at the surface of an object, using Fick's 1st law:

For the problem in Fig. 5.1, CHigh = C0 and CLow = 0.

Comparing Equ. 5.13 with Equ. 5.10, km is given by

From equ. (5.12)

From Equ 5.14, km D/C , For the unsteady diffusion problem shown in Fig. 5.1, km

is not constant, but

decreases as C increases over time.

5.4.5 DEFINITION OF THE MASS TRANSFER COEFFICIENTExample 5.6:

Find the Km and Sh of mass transfer from the surface of a sphere into an infinite surrounding quiescent fluid

Assume that the concentration of solute at the surface of a sphere is given by C0.

Spheres radius: R

C0

r + rr

R

C

r


Example 5.6:

Km and Sh of mass transfer from the surface of a sphere into an infinite surrounding quiescent fluid

Solution

Again thin shell volume in fluid: for sphere generally thickness r extending from r to r + r.

Note

A model including km and Sh is needed

Start from

Need to find partial differential of C

r < or = R, all C is a constant

r > R, C decreases with r

C may not be 0

C0 is related to solubility

C0

r + r

r

R

C

r


Km and Sh of mass transfer from the surface of a sphere into an infinite surrounding quiescent fluid (diffusion only)

Based on

Perform a steady state solute balance (dC/dt =0) on this shell volume using Fick's 1st law

1st term: the rate at which solute enters the control volume by diffusion and

2nd term: the rate at which solute leaves by diffusion.

Dividing by r and taking the limit as r 0, for the solute concentration (C) in the fluid surrounding the sphere:

dC/dt =0



Integrate twice

C1 and C2 can be determined from BCs. Here

C2 = C and

C1 = R (C C0).

profile in the fluid surrounding the sphere is given by

From

the solute flux at the surface of the sphere can be written as



However, from the previous equation for C(r), i.e.

it is easily shown that

Based on definition of km, and substituted it into

we get

km = Dm/R.

Rearrange Sherwood number

Sh = kmdsphere/Dm = 2

based on the sphere diameter (dsphere = 2 R) as the characteristic length


Example 5.7: Application for km and Sh of a drug

A drug has an equilibrium solubility in water of 0.0025 g cm3.

The diffusivity Dm of the drug in water is 0.9 105 cm2 sec1.

One gram of the drug is made into particles that are 0.1 cm in diameter and these particles have a density of 1.27 g cm3.

These particles are then vigorously mixed in a stirred vessel containing 1 L of water.

After mixing for 15 min, the drug concentration in the solution is 0.03 g L1.

Question:

Estimate the value of the mass transfer coefficient km (centimeters per second) for the drug under these conditions.

Also, find the Sherwood number (Sh).


Solution

C of the drug at the surface of the particles is equal to the equilibrium solubility of the drug in water, which is equal to 2.5 g L1,

CHigh is the C on the surface in

C of the drug dissolved in the solution

starts (t= 0) at 0 g L1 and

after t= 15 min, increases to 0.03 g L1.

Let CLow in Equ. 5.13 be given by the average of these values or 0.015 g L1

Clow = (C |t = 0 + C |t = 15 min)/2 = (0+0.03)/2 = 0.015 g/L

The total amount of drug that was dissolved over the 15 min period is equal to 0.03 g L1 1 L = 0.03 g.

The average dissolution rate rdrug is therefore 0.03 g of drug in 15 min or 3.33 105 g sec1.


Solution

Total particle number in the solution Nparticle Each particle surface Sparticle Total particle surface: Nparticle * Sparticle The dissolution rate of the drug rdrug is then given by Equ. 5.13,

where the solute flux, js, is multiplied by the total surface area of all the drug particles to give the drug dissolution rate (rdrug):

The total area available for mass transfer is equal to (Nparticle)*(Sparticle).

The area of a single drug particle is given by 4 R2, where R is the particle radius.

Hence, we find that Sparticle = 0.0314 cm2.


Solution

Nparticle can be found by dividing the mass of drug placed into the solution, by the drug density and then dividing this result by the volume of a given particle Nparticle = (m/)/(v of single particle)

1 g of drug is equivalent to 1504 particles.

With these parameters found, the previous equation can be solved for

km:

Sh

In this case, Sh > 2 (see Example 5.6, where Sh = 2),

indicating that solute transport from the drug particles in this example is a result of

convection and

diffusion

5.4.6 MASS TRANSFER IN LAMINAR BOUNDARY LAYER FLOW OVER A FLAT PLATE

Equation of concentration boundary layer (CBL)

Fig. 5.3: shows the laminar flow of a fluid across a semi-infinite flat plate of length L.

The surface of the plate maintains a constant concentration of a solute (C0) and the concentration of solute in the fluid is given by C.

Unlike the situation shown in Fig. 5.1, here the solute diffuses from the surface of the plate and is then swept away by the flowing fluid.

solute is transported away from the flat plate by a combination of

Diffusion

Convection

Assume that the bulk of the fluid is free of solute except in the region adjacent to the flat plate, which is defined as the CBL

If the fluid approaching the plate has a solute concentration of C, then the following analysis still applies;

however, the C will need to be defined as C = CC.


FIGURE 5.3 Laminar boundary layer flow in the vicinity of a flat plate of constant surface concentration.

Velocity

Concentration

Questions:

Is this a steady process?

What parameter will determine c

Which BL is thicker for a given x?


In Section 4.15.2: u(y) is within the BL along the length of the plate.

Here, extend u(y) to determine C(y) of the solute in CBL that is also formed along the surface of the flat plate.

Consider the shell volume shown in Fig. 5.3:

width W

located from x to x + x and

from y = 0 to y = c(x),

where c is C BL thickness.

Steady state solute (mass) balance on this shell volume, which is given by

1st and 2nd terms: net rate at which solute is being added to the shell volume by flow (convection) of the fluid in the x-direction.

3rd term: loss of solute from the top of the shell volume as a result of flow in the y-direction.

4th term: the rate at which solute is diffusing away from the surface of the flat plate.


After

eliminating the plate width, W, and

dividing by x,

taking the limit as x 0,

Using Fick's first law, we can insert Equ. 5.11 on the left-hand side of Equ. 5.16 for and obtain

Solution of Equ. 5.17 requires that we know how vx and C depend on x and yin their respective BLs.

From Equ. 4.62 vx(x, y)


In a similar fashion propose that C(y) profile in the CBL is approximately described by

BCs of equ. (5.18):

1ST BC: C is constant along the surface of the flat plate.

2nd and 3rd BCs: beyond CBL, the solute is not present.

we then can eliminate the last term in Equ. 5.17.

Hence, Equ. 5.17 becomes

4th BC: the flux of solute along the surface of the plate is only a function of x


Imposing these BCs on Equ. 5.18, C(y) within the CBL in terms of the concentration boundary layer thickness, c(x)

c(x) at this time is still unknown

We now can substitute Equs 4.62 and 5.21 into Equ. 5.20.

Define ratio of the BL thicknesses = (C(x)/(x)),

Assume is a constant.

The algebra is a bit overwhelming, but one can obtain the following differential equation for the thickness CBL:

BC: at x = 0, C = 0

Solution of Equ. 5.22 is

3

2

1

2

31

0

xC

y

xC

y

C

C(x,y) (5.21)


Recall from Equ 4.64 that (x) is given by

Dividing Equ 5.23 by Equ. 4.64 and simplifying results in the following equation for = (C(x)/(x)):

Sc = /Dm = /Dm is a dimensionless number known as the Schmidt number.

In general,

can be calculated for a given Sc through Equ. 5.24

(x) can be calculated from equ. (4.64)

C(x) can be calculated by (x)


Effect of Schmidt number Sc ratio of momentum diffusion ( = /) to mass diffusion (Dm)

Sc= /Dm For solutes diffusing through liquids, generally SC >> 1

from Equ. 5.24, we have that

< 1 or

> C.

for liquids, the CBL lies within the velocity or momentum (x).

Diffusing through gases, SC ~ 1

1 or

C.

Diffusing through materials like liquid metals, SC .

C

x

x

x


For most mass transfer problems of interest to biomedical engineers, Sc >> 1

From Equ. 5.24


Mass transfer coefficient km Here, CHigh = C0 and CLow = 0.

Using Equs 5.11 and 5.13,

km becomes

Substituting Equ. 5.21 into Equ. 5.27

and replacing C(x) with Equ. 5.26,

we obtain the following expression for the local mass transfer coefficient:

Shx : local Sherwood number at location x, and ratio of the transport rate of solute by convection to that by diffusion.

3

2

1

2

31

0

xC

y

xC

y

C

C(x,y)

(5.21)


As the fluid progresses down the length of the plate,

C(x) increases and

km (a local value) decreases in inverse proportion to x1/2.

For a plate of length L, the average mass transfer coefficient is given by

Substituting Equ. 5.28 into Equ. 5.29

length averaged mass transfer coefficient

Rex < 300,000 (only for laminar)

For flow over one side of a flat plate of length L and width W, the amount of solute transported can then be written as follows

CHigh : concentration of the solute at the surface of the plate and

CLow : concentration of the solute in the fluid outside the boundary layer

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