transport phenomena from biological processes

7/29/2019 Transport Phenomena from Biological Processes

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MEDE3005 TRANSPORT PHENOMENA IN

BIOLOGICAL SYSTEMS

2008 2009 Academic Year

Part I (11 hours, Dr. K. W. Chow) Governing Equations of Fluid

Dynamics

(A) Elementary concepts of fluid flows

Dependence on space and time

(1) Conditions in a body of fluid can vary from point to point and, at any givenpoint, can vary from one moment of time to the next. A flow is uniform if the

velocity at a given instant is the same in magnitude and direction at every point

in the fluid. If at any instant in time, the velocity changes from point to point,

the flow is nonuniform.

(2) A flow is steady if the fluid properties, e.g. velocity and pressure, at any

given point do not change with time. A flow which is not steady, i.e. with time

dependent properties, is termed unsteady.

(3) Note that all these combinations are possible:(a) Steady uniform flowsflow properties independent on position and time;

(b) Steady nonuniform flows flow properties independent of time but

depending on position;

(c) Unsteady uniform flows flow properties independent of position butdepending on time;

(d) Unsteady nonuniform flows flow properties depending on both position

and time.

Real and ideal fluids(4) When a real fluid flows past a boundary, the fluid immediately in contact

with the boundary will have the same velocity as the boundary (the no slip

boundary condition, or friction will not permit a relative motion along the

boundary). In most problems encountered in practice, this boundary is taken to

be at rest.

(5) A frictionless fluid which permits sliding along the boundary will be termed

anideal fluid

.


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(6) In practice, the no slip condition must be enforced. A thin portion of fluid

around the boundary, termed the boundary layer, will be the region where

viscous effects are the largest. This is also the region where drag force,

momentum transfer and energy loss are important considerations. Tremendous

efforts have been spent over the years to understand this dynamics, and we shallattempt a brief introduction later this course.

Compressible and incompressible flows(7) If the density of a fluid does not change in the flow, the fluid is

incompressible (otherwise, compressible). In practice, a liquid can be taken as

incompressible. The dynamics of a gas will usually require compressible effects

to be taken into account. However, as a working rule, a gas can still be treated

as almost incompressible if the Mach number is small (about 0.1 or 0.2). The

Mach number is the ratio of flow speed to the local sound speed. Given soundspeed in air as roughly 340 m s1

, it is remarkable that a gas is still roughlyincompressible for a speed as fast as say 40 m s

1. Compressibility must be

considered when the Mach number reaches say 0.4 or 0.5.

One, two and three (1D, 2D, 3D) dimensional flows(8) A flow is termed 1D, 2D or 3D if the flow depends on one, two or three

spatial coordinates respectively.

(B) Differential Analysis of Fluid Flow

(1) The main objective of this section is to apply two basic laws of physics,namely, conservation of mass and Newtons law of motion, to the dynamics of a

fluid. Cartesian coordinates will be used first, but once the equations areestablished, one can easily extend the derivation to configurations in cylindrical

or spherical geometry, or for that matter, any other curvilinear coordinates

system.

(2) Review(a) Lagrangian description The motion of a fixed particle is traced. Theusual Newtons law of motion is applied. The subsequent locations of the

particles are to be found as a function of the initial position and time.

(b) Eulerian description The fluid properties and locations are given interms of positions / coordinates fixed in space. This approach is more

convenient but note that different particles will flow through a fixed point at

different values in time.


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In this case the total or material derivative is

z

w

y

v

x

u

tDt

D

where u, v, w are velocities in thex, y, z directions. Roughly speaking, the time

derivative term denotes the local rate of change, while the other terms represent

the contribution from the convection or advection.

(3) Vorticity Formally the vorticity vector is the curl of the velocityvector (u), or = curl u. Physically the vorticity represents twice the local

angular velocity of the fluid element, or in other words, it measures roughly

how fast the fluid is rotating and spinning. To keep the analysis to a minimum,we consider again a two-dimensional (2D) setting. First we decompose the fluid

motion into these components: (i) translation, (ii) rotation, (iii) angular

distortion and (iv) volume distortion. We argue that, for the case ofincompressible flow, the volume distortion is zero or small, and thus we focus

of (ii) and (iii).

We now consider a deformation with angles , as shown in the figure, and

regard the whole deformation process as consisting of

(i) a rigid rotation of angle ( )/2 (angle of rotation of the fluidelement) and then

(ii) a bending/squeezing of ( +)/2 where the two sides move in oppositedirections (i.e. the dy side moves in clockwise sense while the dx side

moves in anticlockwise direction (angle of distortion).

Quantitatively we now relate this angle of rotation, ( )/2, to the

spatial gradients of the velocity vector. More precisely, for small angles,

dxdtdxx

v 1

dydtdy

y

u 1

and thus the rate of rotation about thez-axis is

y

u

x

v

dt 2

1

2

1


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which is related to the z-component of curl (u). We define curl (u) to be the

vorticity (vector) of the fluid, and is, by this derivation, twice the local angular

velocity of the fluid.

(C) The Continuity Equation

(1) The continuity equation is simply a statement for the conservation of mass.

For any control surface (or control volume), the net mass outflux of fluid is

equal to the contribution from any source (or sink) located inside and the effect

from a net change in density (if the fluid is compressible).

(2) For simplicity we first derive it in a two dimensional (2D) setting, or one

unit of depth in the spanwise direction (direction perpendicular to the plane ofthis page). In one unit of time, a length of fluid equal to the velocity of the fluidwill flow through a fixed section. The volume flux (volume per unit time) of

fluid is thus velocity times the cross sectional area. The mass flux is the density

times the volume flux.

Mass flux in thex direction u dy

Mass flux out in thex direction on the right hand side: dxdyux

dyu )(

Net mass out in thex-direction

dydxux

)(

Similarly in they-direction the mass out flux is dydxvy

)(

. This mass out flux

must come at the expense of the decrease in mass or density

dydxt

)()(

vy

uxt

or .0)()( yxt vu

For the case of constant density this reduces to


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.0 yx vu

In vector forms the continuity equation can be written as

0)(div ut

whereu = u i + v j

and div is the divergence operator. Again for the constant density case this

reduces to div (u) = 0.

(3) Extension to three dimensions is conceptually straightforward.

(4) A popular approach is to use the summation convention, i.e. repeated indices

denote summation, while a free index, or an index occurring once, implies that

the equation holds for all three directions, e.g. the continuity equation can bewritten as

u j/ x j = 0 .

(P1)

Water flows through a pipe 25 mm in diameter at a velocity of 6 m s1.

Determine whether the flow will be laminar or turbulent, assuming that the

dynamics viscosity of water is 1.3 (103

) kg m1

s1

and its density is

1000 kg m3

. If oil of specific gravity 0.9 and dynamic viscosity 9.6 (102

) kgm

1s1

is pumped through the same pipe, what type of flow will occur?

(Recall from last year (MEDE2005) that the Reynolds number, defined as

Re = (reference velocity)(reference length)/(kinematic viscosity) measures aratio of inertial force to viscous force. The kinematic viscosity is defined as

(dynamic viscosity/density). For pipe flow a transition from laminar to turbulentflows occurs typically around Re around 1000 to 3000, depending on the

experimental configuration (i.e. laminar for Re < Recritical and turbulent for Re >

Recritical with Recritical somewhere between 1000 to 3000 (Note: these are NOT

rigid limits)).

For the flow with water:

Re = (6)(0.025)(1000)/(0.0013) = 115385, therefore turbulent.

For the flow with oil:Re = (6)(0.025)(900)/(0.096) = 1406, probably still laminar)


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(P2) An air duct is of rectangular cross section with dimensions 300 mm by 450

mm. Determine the mean velocity in the duct when the flow rate is 0.42 m3s1

.

If the duct tapers to a smaller section with dimensions of 150 mm by 400 mm,fine the new mean velocity, assuming that the density and flow rate remain

constant. (3.11 m s1

; 7 m s1

)

(P3) Determine if the following flows of an incompressible fluid satisfy the

continuity equation.

(a) ,

)(

21 200222

2

22rV

yx

x

yx

u

.)(

2 200222rV

yx

xyv

(Yes)

V0 is a reference velocity and 0r is a reference length. Both are constants.

(b) ,)(

200222rV

yx

xyzu

00222

22

)(

)(rV

yx

zyxv

,

0022rV

yx

yw

. (Yes)

(P4) For the flow of an incompressible fluid the velocity component in the x-direction is

byaxu 2

,

and the velocity component in the z-direction is zero. Find the velocity

component v in the y-direction. To evaluate arbitrary functions that might

appear in the integration, assume that v = 0 aty = 0. )2( axyv


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(P5) An incompressible fluid flows through the circular pipe shown in the figure

at the rate ofQ m3/s.

(a) If it is assumed that the velocities at stations 1, 2, and 3 are uniform, what

are the velocities, given that the diameters of the pipe at the three sections areA,B, C, respectively?

)/(4),/(4),/(4222

CQBQAQ

(b) Compute numerical values for the caseQ = 0.4

m3/s,

A = 0.4 m,B = 0.2 m,

C = 0.65 m.

(P6) The incompressible continuity equation in polar coordinates is (vr, v are

the velocities in the rand directions respectively)

01)(1

rr

rv

r

r .

(a) On the basis of this equation alone, what is the most general type of flow

possible if ?0

Sketch the flow. )/( rCr

(b) Similarly, what is the most general type of flow ifr

= 0? Sketch the flow.

))(( rFv

(P7) A compressible fluid is caused to flow through a tube of constant diameterin such a way that the velocity along the axis is given by

xuuuu

u htan22

1221

,

where1

u and2

u are the velocities whenx is minus or plus infinity, respectively.

The density does not change with time at any point. The density at x = is

1 . Obtain an equation for the distribution of density along the tube.

)/( 11 uu


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(D) Equations of motion of an inviscid fluid

Newtons second law (force = mass(acceleration)) in Eulerian

coordinatesThe total forces acting on a fluid element are those resulting from the pressure

and any body forces arising from stratification, electromagnetic effects and

other relevant physical effects. There is no contribution from viscous stress (i.e.

no friction). For a 2D setting theNewtons law of motion or force = mass times

acceleration is

xf

x

p

y

uv

x

uu

t

u

1

yf

y

p

y

vv

x

vu

t

v

1

wherefx,fyare the body forces per unit mass. These are the Eulersequationof motion.

(P8) The equations of motion in plane, polar coordinates are:

,

1

,

12

fp

rr

vvv

r

v

r

vv

t

v

fr

p

r

vv

r

v

r

vv

t

v

r

r

r

rr

r

r

and can in principle be obtained by considering the forces on a small element

bounded by the lines corresponding to r, r+ dr, and , + d.

(P9) Consider the stream function22

yx for a fluid of constant density

(Recall that ),xy

u

(a) Calculate the total acceleration vector a and show that it is proportional to

the radius vector. (4 xi + 4yj)

(b) Use the result in part (a) to integrate the equation of motion assuming no

friction and no body forces and find the pressure as a function of radius. (p =

2r2 + constant)


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(P10) Assume that a perfect and incompressible fluid is flowing horizontally

over a large surface. The flow is steady. Let thex-direction be the direction of

flow and let the y-direction be parallel to the action of the gravitational forces.Gravitation is the only body force acting. Find the relation between p and y.

),0),(,0( gyppyuux

(P11) Consider a perfect and incompressible fluid flowing in circular path about

a center. The flow is purely two-dimensional and steady. There are no body

forces acting.

(a) Find from Eulers equation the differential relation between the pressure,

the tangential velocity , and the radial distance.

rr

p2

(b) Find this same relation by considering the equilibrium of a small elementof fluid in this special case.

(E) Equations of motion for a viscous fluid

Relations between stress and strain and the Navier Stokes equations(1) The consideration of stress is very similar to the one covered in earlier

course work in solid mechanics. Again, on applying force = (mass)

(acceleration) we get (fx,fy,fz are the body forces)

zyxf

Dt

Du zxyxxxx

zyxf

Dt

Dv zyyyxyy

zyxf

Dt

Dw zzyzxzz

Exact relations between the stress and strains are involved, and please consult

the textbooks and references for details. These connections between stress andstrain are generally known as constitutive equations (or relations) in fluid

mechanics. We just state the results:


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z

w

y

v

x

u

y

w

z

v

x

w

z

u

x

v

y

u

z

wp

y

vp

x

up

zyyz

zxxz

yxxy

zz

yy

xx

.

,

,

,23

2

,2

3

2

,23

2

On simplification this gives

.3

11

,3

11

,

3

11

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

zoo

yoo

xoo

fz

vz

w

y

w

x

wv

z

p

Dt

Dw

fy

vz

v

y

v

x

vv

y

p

Dt

Dv

f

x

v

z

u

y

u

x

uv

x

p

Dt

Du

If the volume dilatation term is zero by the continuity equation then we have the

Navier Stokes equations

xof

z

u

y

u

x

uv

x

p

Dt

Du

2

2

2

2

2

2

1

yof

z

v

y

v

x

vv

y

p

Dt

Dv

2

2

2

2

2

2

1

zof

z

w

y

w

x

wv

z

p

Dt

Dw

2

2

2

2

2

2

1


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where

o

v kinematic viscosity,

= dynamic viscosity, = density.

Non-dimensional forms can be obtained by introducing non-dimensionalcoordinates (with *)

,,,***

L

zz

L

yy

L

xx

),//(*

oULtt

oooU

ww

U

vv

U

uu

***,,

)/(2*

oUpp

L and Uo are the reference length and velocity scales. On dropping the * we get

xzzyyxxxzyxtfuuupwuvuuuu )(

Re

1

yzzyyxxyzyxtfvvvpwvvvuvv )(

Re

1

zzzyyxxzzyxt fwwwpwwvwuww )(Re1

where

Re

LU

v

LUo

o

o

is the Reynolds number and measures (roughly) the ratio of inertial force to

viscous force.

Cylindrical coordinates(2) In a cylindrical coordinate system the Navier Stokes equations are:

r-direction

,211

3

11222

2

2

2

2

2

r

rrrr

oo

r

z

rr

r

r

fv

rr

v

z

vv

rr

vr

rrv

rv

r

p

r

v

z

vv

v

r

v

r

vv

t

v


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-direction

,211

3

11222

2

2

2

2

fv

rr

v

z

vv

rr

vrrr

vr

vr

p

r

vv

z

vv

v

r

v

r

vv

t

v

r

oo

r

zr

z-direction

,11

3

112

2

2

2

2

2

z

zzz

oo

zzz

r

z

fz

vv

rr

vr

rrv

zv

z

p

z

vv

v

r

v

r

vv

t

v

Boundary Conditions(3)(a) No penetration condition for both ideal and viscous fluids at a solid wall

which is at rest, the normal component of fluid velocity should be zero (the no

penetration boundary condition), unless there are sources or sinks inside the

wall (ejection or seepage). In the unlikely situation where the wall is moving,

the fluid velocity perpendicular to the wall should match that component of thewall velocity.

(b) Tangential condition for an ideal fluid, wall at rest the tangential

component of the fluid velocity at the wall needs not be zero as there is nofriction. The fluid can slide along the wall.

(c) Tangential condition for a viscous fluid, wall at rest the tangential

component of the fluid velocity at the wall must be zero as there is friction

(the no slip boundary condition).

Limiting cases(4) The equations of motion can be simplified considerably in certain regimes:

(a) Reynolds number much, much larger than one (Recall that the Reynoldsnumber is the ratio of inertial force to viscous force) in this case the viscous

effect is only important in the boundary layer. The flow then consists mainly ofan inviscid core outside of the bluff body, and a thin boundary layer around the

body. The difficulties are then (a) the analysis of the boundary layer and (b) the


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matching of the two regions. Despite the efforts in the past 100 years, some

scientific and technical issues are still unresolved.

(b) Reynolds number much, much less than one Viscous forces dominate.

The regime is usually called Stokes flow.

(P12) Flow down an inclined plane, gravity being the body force and friction

opposes the motion Liquid of density and viscosity flows down a wide

plate inclined at the angle to the horizontal under the influence of gravity. The

depth of the liquid normal to the plate is h. The flow is steady and everywhere

parallel to the plate. The viscosity of the air in contact with the upper surface of

the liquid may be neglected. Determine the velocity profile parallel to the plate,

the shear stress at the wall, and the average velocity. (Hint: use a coordinatesystem parallel (x) and perpendicular (y) to the inclined plane. Boundaryconditions: no slip at the wall, and no shear stress at the free surface)

)3

sin

sin

)2(2

sin(

2

average

2

ghu

gh

yhyg

u

w

(P13) Plane Poiseuille flow in a rectangular channel. (No body force,

pressure gradient drives the flow) We consider the steady flow of a viscous,

incompressible fluid in an infinitely long, two dimensional stationary channel of

breadth h with no body forces present. The flow is everywhere parallel to thex-axis and they-axis is placed at the bottom of the channel. The velocity profile

and shear stress distribution are to be determined.

Solution: With v = w = 0 the equations of motion become

(a) ,1

02

2

y

uv

x

po

(b) ,11

0z

p

y

p

in thex-,y-, andz-directions, respectively, and the continuity equation is

(c) .0

x

u


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In addition, we have the boundary conditions u = 0 aty = 0, h. From Eq. (c) it

follows that u can be a function ofy only. Furthermore, Eq. (b) shows that p

cannot depend ony orz, but only onx. Finally, from (a) it is seen that

(d)

ov

dyud

dxdp .

2

2

The right-hand side of this equation can only depend ony and the left side only

on x. These conflicting requirements can only be satisfied if both sides are

equal to the same constant. Thus the pressure gradient must be a constant for

this flow. Equation (d) is now easily integrated, and we obtain

,

2

1 2BAyy

dx

dpu

where A and B are constants. When these are evaluated with the use of the

boundary conditions, we obtain

.)(2

1 2yhy

dx

dpu

Thus the velocity profile is parabolic with the maximum velocity at the center of

the channel. The shear stress is

),2(2

1yh

dx

dp

dy

duyx

so that on the upper surface the stress dxdph /)2/(0

acts on the fluid.

Similarly, on the lower surface the stress on the fluid is dxdph /)2/(0

(recall

the convention of positive shear stresses). Of course, these stresses could have

been determined directly from static consideration since there is no acceleration

of the fluid.

The existence of the motion in this case depends on the pressure gradient dp/dx.

The volume flow rate per unit depth of channel is

h

dx

dphdyuq

0

3

,12

and we see that a discharge in the positive x-direction requires a negative

pressure gradient or a pressure that decreases in the direction of flow. The


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pressure drop in a channel of lengthL can be expressed in terms of the average

velocity q/h =Vas follows:

.

2

24122

2

V

h

L

Vh

vV

h

Lp

(P14) Poiseuille flow in a CIRCULAR channel (again no body force, pressure

gradient drives the flow, but the geometry is now cylindrical). Consider circularpipe flow in a cylinder of radius a. Derive the corresponding formulae for

velocity profile and volume flow rate

)(

4

1 22ra

dx

dpu

L

pa

dx

dpaQ

8

8

4

4

L = length of the pipe, p = (numerical value for the) drop in pressure

(P15) (A 2layer fluid, no body force, no pressure gradient, motion driven by a

moving plate, velocity profile piecewise linear) Consider the steady laminar

incompressible flow between two parallel plates. The upper plate moves at

velocity Uo to the right and the lower plate is stationary. The pressure gradient

is zero. The lower half of the region between the plates (i.e., 0 yh/2) is

filled with a fluid with density1

and viscosity1

, and the upper half (h/2 y

h) is filled with a fluid of density2

and viscosity2

.

(a) State the condition that the shear stress must satisfy for 0


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(d) Calculate the shear stress at the lower wall.

)(

2

211

21

h

Uo

( = constant; u = 0 aty = 0, u = Uo aty = h, u is continuous aty = h/2;

hyh

h

yhUU

hyy

h

Uu

oo

o

2)(

)(2

20

)(

2

21

1

21

2

(P16) (Co- or counter- rotation cylinders) A cylinder of radius a rotates with

angular velocity concentrically inside a larger stationary one of radius b.

Fluid of viscosity and constant density fills the gap between the cylinders.

Assume that the flow is steady and the motion is purely circular so that no

quantities change with angular position. Obtain the velocity distribution

between the cylinders and the shear stress on the two cylinders in two ways: (a)

by setting up the equations of equilibrium for an annulus consisting of the inner

cylinder and an outer one of arbitrary radius ror (b) by solution of the Navier-

Stokes equations in cylindrical coordinates.

(Perhaps it is easier to proceed directly with (b). The equation in v is

equidimensional and can be solved by v= rn. Solve for n)

)(

(

2

22

2

r

br

ba

a

on inner cylinder22

2b2

ba

on outer cylinder )a

222

2

b

a

(P17) (Stream function of planar flows) A viscous flow of viscosity is given

by the stream function

.Axy


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(a) Sketch the streamlines accurately near the pointx =y = 0. TakeA to be

positive.

(b) Assuming that there is no body force f, integrate the Navier-Stokes

equations for this special case and obtain a relationship between the pressure,A,x, andy.

(c) If the average pressure at the point x = y = 0 is0

p , what is the normal

stressxx

at the same point ?

(d) Assume that the line x = 0 is a solid surface. Can the stream function

above represent correctly the flow near x =y = 0 ? Explain very briefly. (Are

the no-slip and no penetration conditions satisfied?)

((b)2

)(222

yxAp

(c) )2 Apoxx

(P18) Consider again the planar flow of an incompressible fluid given by

.Axyt

Assume that the flow is inviscid and that there are no body forces. Find thepressure,p, within the flow as a function ofA, (fluid density), x, y, and tby

integrating the equations of motion. (The result contains an arbitrary constantwhich could be evaluated by assuming that the pressure at any one point in the

flowfor example, at the originis known.)

(P19) (Flow outside of a rotating cylinder) An incompressible liquid is

contained in the annulus between an inner rotating cylinder of radiusirand an

outer stationary one. The outer radius or is much larger than ir ( or mayessentially be taken to be infinite). The inner cylinder rotates at a constant

angular speed , so that the surface velocity isiir

V . The velocity at r= ro

may be taken to be zero. The flow is steady, the componentsz

andr

are

assumed to be zero, and all derivatives with respect to are also zero. The flow

is laminar and the viscosity is constant.

(a) Determine the velocity

as a function of the radius r.


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(b) What is the torque (surface force times radius) required to turn thecylinder?

).2,(

2

r

rv i

(P20) (Vorticity, or the dynamics of rotation in fluids) Consider the two

equations of motion for the planar flow of an incompressible, frictionless fluid

under the action of conservative body forces (= body forces which are results of

a gradient vector of a scalar potential). By eliminating the pressure from these

two equations and using the equation of the continuity )0//( yvxu , show

that

,0 yv

xu

t

where is the magnitude of the vorticity ).//( yuxv What can you

conclude about the vorticity of a particular element of fluid in such a flow ?

(Material derivative of the vorticity = 0, or vorticity following a particle is

constant)


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Supplementary Problems

(P21) (Acceleration) The velocity in a certain flow field is given by the

equation

kji 32

yxzyzV

Determine the expressions for the three rectangular components of acceleration.

(From expressions for velocity, ,,32

xzyzu and yw .

Since

zuw

yuv

xuu

tua

x

then

zyxz

yzyzxzyzax

23

22

63

)6()()3()()0()3(0

Similarly,

xzaxyyzazy

3

3 ).

(P22) (Vorticity) Determine an expression for the vorticity of the flow field

described by

ji 22

xyyxV

Is the flow irrotational?

The vorticity is twice the rotation vector:

k)()v(22

yx

As this is non zero, the flow is rotational.

(P23) (Conservation of mass) For a certain incompressible, two-dimensional

flow field the velocity component in they-direction is given by the equation

xyxv 2

2


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Determine the velocity component in the x-direction so that the continuity

equation is satisfied

)(2

yfxu

(wheref(y) is an undetermined function ofy.)

(P24) For a two dimensional flow in the x, y plane, the y component of the

velocity is given by v =y22x + 2y . Determine a possible x component for

this steady, incompressible flow. ( u =2x y2x +f(y) )

(P25) Thex component of velocity in a steady, incompressible flow field in thex,y plane is u = A/x , whereA is a constant. Find the simplest y component of

velocity in this flow field. (Ay/x2)

(P26) Are the following velocities fields consistent with the continuity

equations?

(a) u = U0 (x3 +xy2), v = U0 (y

3 +yx2), (U0 constant) (Yes)

(b) u = 10 x t, v =10 y t. (Yes)

(P27) (Mass flow) Air entering a compressor has a density of 1.2 kg m3

and

velocity 5 m s1

. If the cross sectional area is 20 cm2, find the mass flow rate.

(1.2 102

kg s1

)

(P28) (Acceleration) The velocity in a certain two-dimensional flow field is

given by the equation

ji 22 ytxtV

Determine expressions for the local and convective components of accelerationin thex andy directions.


21/22

21

Then

24

)(

2)(

xt

yuv

xuuconvectivea

xt

ulocala

x

x

and

24

)(

2)(

yt

y

vv

x

vuconvectivea

yt

vlocala

y

y

(P29) (Kinematic description of fluid motion) The three components of velocity

in a flow field are given by

42/32

2

222

zxzw

zyzxyv

zyxu

(a) Determine the volumetric dilatation rate, and interpret the results. (b)

Determine an expression for the rotation vector. Is this an irrotational flow

field?

(b) Volumetric dilatation rate = 0

z

w

y

v

x

u

This result indicates that there is no change in the volume of a fluid element as

it moves from one location to another.

02

2

5

2)v(

2

1

kji

yzz

y

No, this is not an irrotational field.

(P30) (Motion driven by moving boundaries in the absence of body forces and

pressure gradient, Couette flows) An incompressible viscous fluid is placedbetween two large parallel plates. The bottom plate is fixed and the upped plate


22/22

moves with a constant velocity, U. For these conditions the velocity distribution

between the plates is linear, and can be expressed as

b

yUu

Determine: (a) the volumetric dilatation rate, (b) the rotation vector, (c) the

vorticity, and (d) the rate of angular deformation.

(a) Volumetric dilatation rate 0

z

w

y

v

x

u

(b) kb

U

2

(c) kbU

(d)y

u

x

v

Thus,

b

U

(P31) (Irrotational flows) The velocity components of an incompressible, two-dimensional velocity field are given by the equations

22

2

yxv

xyu

Show that the flow is irrotational and satisfies conservation of mass.

0( z

and the flow is irrotational).

transport phenomena from biological processes

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