mm302 1 lecture note 1

7/24/2019 MM302 1 Lecture Note 1

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MM302E FLUID

MECHANICS II

2015-2016 SPRING


2/18

INTRODUCTION TO DIFFERENTIAL ANALYSIS OF FLUID MOTION

In course Fluid Mechanics I, we developed the basic equations inintegral form for a finite control volume. The integral equations are

particularly useful when we are interested in the gross behavior of a

flow and its effect on various devices. However, the integral approach

does not enable us to obtain detailed point by point data of the flow

field.

To obtain this detailed knowledge, we must apply the equations of fluid

motion in differential form.

In this chapter, we will derive fundamental equations in differential

form and apply this equations to simple flow problems.

EQUATION OF CONSERVATION OF MASS (CONTINUITY EQUATION)

The application of the principle of conservation of mass to fluid flow

yields an equation which is referred as the continuity equation. We shall

derive the differential equation for conservation of mass in rectangular

and in cylindrical coordinates.

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Rectangular Coordinate System

The differential form of the continuity equation may be obtained by applying the

principle of conservation of mass to an infinitesimal control volume. The sizes of

the control volume are dx, dy, and dz. We consider that, at the center, O, of the

control volume, the density is and the velocity is

The word statement of the conservation of mass is

The values of the mass fluxes at each of six faces of the control volume may be

obtained by using a Taylor series expansion of the density and velocity

components about point O. For example, at the right face,

Neglecting higher order terms, we can write

and similarly,

kwjvuV

2

2

2

2 2!2

1

2

dx

x

dx

xdx

x

22

dx

xdx

x

3

0volumecontroltheinside

massofchangeofRatesurfacecontrolethrough th

outfluxmassofrateNet

0

VdtAdV

CVCS

22dx

xuuu dxx

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Corresponding terms at the left face are

To evaluate the first term in this equation, we must evaluate .

Table.Mass flux through the control surface of a rectangular differential control volume

CS AdV

4

222

dx

x

dx

xdx

x

222

dx

x

uu

dx

x

uuu dxx

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The net rate of mass flux out through control surface is

The rate of change of mass inside the control volume is given by

Therefore, the continuity equation in rectangular coordinate is

Since the vector operator, , in rectangular coordinates, is given by

The continuity equation may be simplified for two special cases.

1. For an incompressible flow, the density is constant, the

continuity equation becomes,

2. For a steady flow, the partial derivatives with respect to time are

zero, that is _________. Then, .

dxdydzz

w

y

v

x

uAdV

CS

0 tzw

yv

xu

0

tV

k

z

j

y

x

0 V

5

dxdydzt

VdtCV

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Example:For a 2-D flow in the xy plane, the velocity component in the y

direction is given by

a) Determine a possible velocity component in the x direction for

steady flow of an incompressible fluid. How many possible xcomponents are there?

b) Is the determined velocity component in the x-direction also valid

for unsteady flow of an incompressible fluid?

yxyv 222

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7

Example: A compressible flow field is described by

Determine the rate of change of the density at point x=3 m, y=2 m and

z=2 mfor t=0.

ktejbxyiaxV

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Derivation of Continuity Equation Cylindrical Coordinate System

In cylindrical coordinates, a suitable differential control volume is shown

in the figure. The density at center, O, is and the velocity there is

Figure.Differential control volume in cylindrical coordinates.

To evaluate , we must consider the mass flux through each of

the six faces of the control surface. The properties at each of the six

faces of the control surface are obtained from Taylor series expansion

about point O.

zzrr evevevV

CS

AdV

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Table.Mass flux through the control surface of a cylindrical differential control volume

The net rate of mass flux out through the control surface is given by

The rate of change of mass inside the control volume is given by

In cylindrical coordinates the continuity equation becomes

dzdrdz

vr

v

r

vrvAdV zrr

CS

dzrdrdt

dVt

CV

0

tr

z

vr

v

r

vrv zrr

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Dividing by rgives

or

In cylindrical coordinates the vector operator is given by

Then the continuity equation can be written in vector notation as

The continuity equation may be simplified for two special cases:

1. For an incompressible flow, the density is constant, i.e.,

2. For a steady flow,

01

tz

vv

rr

v

r

v zrr

0)()(1)(1

tzvv

rrvr

rzr

zr ez

er

er

1

0

tV

r

r ee

er

e

and:Note

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Example: Consider one-dimensional radial flow in the r plane,

characterized by vr = f(r) and v = 0. Determine the conditions on f(r)

required for incompressible flow.

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STREAM FUNCTION FOR TWO-DIMENSIONAL INCOMPRESSIBLE

FLOW

For a two-dimensionalflow in the xyplane of the Cartesian coordinate systems,

the continuity equation for an incompressible fluid reduces to

If a continuous function , called stream function, is defined such that

then continuity equation is satisfied exactly, since

Streamlines are tangent to the direction of flow at every point in the flow field.

Thus, if dris an element of length along a streamline, the equation of streamline is

given by

Substituting for the velocity components of u and v, in terms of the stream

function

(A)

At a certain instant of time, t0, the stream function may be expressed as

. At this instant, the streamfunction

(B)

Comparing equations (A) and (B), we see that along instantaneous streamline

= constant. In the flow field, 2-1, depends only on the end points of

integration, since the differential equation of is exact.

0

yv

xu

),,( tyx

xv

yu

and

022

xyyxyv

xu

kvdxudyjdydxjvurdV

)()()(0

0

vdxudy

0

dy

ydx

x

),,( 0tyx

dyy

dxx

d

0d

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13

For a unit depth, the flow rate across AB is

Along AB,x = constantand . Therefore,

For a unit depth, the flow rate across BC is

Along BC, y = constantand . Therefore,

Thus, the volumetric flow rate per unit depth between any twostreamlines, can be expressed as the difference between

constant values of defining the two streamlines.

Now, consider the two-dimensionalflow of an incompressible fluid

between two instantaneous

streamlines, as shown in the Figure.

The volumetric flow rate across

areas AB, BC, DE, and DF must be

equal, since there can be no flow

across a streamline.

2

1

2

1

y

y

y

ydy

yudyQ

dyy

d

12

2

1

2

1

ddyy

Qy

y

2

1

2

1

x

x

x

xdx

xvdxQ

dxx

d

2

1

1

212

x

xddx

xQ

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In r plane of the cylindrical coordinate system, the incompressible

continuity equation reduces to

The streamfunction (r, ,t) then is defined such that

Example: Consider the stream function given by = xy. Find the

corresponding velocity components and show that they satisfy the

differential continuity equation. Then sketch a few streamlines andsuggest any practical applications of the resulting flow field.

0

v

r

rvr

r

vr1

rv

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MOTION OF A FLUID ELEMENT (KINEMATICS)

Before formulating the effects of forces on fluid motion (dynamics), let

us consider first the motion (kinematics) of a fluid in a flow field. When a

fluid element moves in a flow field, it may under go translation, linear

deformation, rotation, and angular deformation as a consequence of

spatial variations in the velocity.

Figure.Pictorial representation of the components of fluid motion.

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Translation

Acceleration of a Fluid Particle in a Velocity Field

Figure.Motion of a particle in a flow field.

Consider a particle moving in a velocity field. At time t, the particle is at

a positionx, y, zand has a velocity .

At time t+dt, the particle has moved to a new position, with

coordinatesx+dx, y+dy, z+dz,and has a velocity given by

The change in the velocity of the particle moving from location to

is given by

Dividing both sides by dt, the total acceleration of the particle is

obtained as

Since

then

),,,( tzyxVVtp

),,,( dttdzzdyydxxVVdttp

rdr

dt

t

Vdz

z

Vdy

y

Vdx

x

VVd pppp

t

V

dt

dz

z

V

dt

dy

y

V

dt

dx

x

V

dt

Vda

pppp

p

w

dt

dzandv

dt

dyu

dt

dx ppp ,

t

V

z

Vw

y

Vv

x

Vu

dt

Vda

p

p

r

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Acceleration of a fluid particle in a velocity field requires a special

derivative, it is denoted the symbol .

Thus,

This derivation is called the substantial, the material or particle

derivative.

The significance of the terms,

The convective acceleration may be written as a single vector expression

using the vector gradient operator,.

Thus,

It is possible to express the above equation in terms of three scalar

equations as

The components of acceleration in cylindrical coordinates may be

obtained by utilizing the appropriate expression for the vector operator

. Thus

Dt

VD

t

V

z

Vw

y

Vv

x

Vua

Dt

VDp

onacceleratilocal

onacceleraticonvective

particleaofonaccelarati

totalt

V

z

Vw

y

Vv

x

Vu

Dt

VDap

VV

z

Vw

y

Vv

x

Vu

)(

t

VVVa

Dt

VDp

)(

t

w

z

ww

y

wv

x

wu

Dt

Dwa

t

v

z

vw

y

vv

x

vu

Dt

Dva

t

u

z

uw

y

uv

x

uu

Dt

Dua

p

p

p

zx

y

x

t

V

z

VV

V

r

V

r

VVa

t

V

z

VV

r

VVV

r

V

r

VVa

t

V

z

V

Vr

VV

r

V

r

V

Va

zzz

zzrz

zr

r

rr

z

rr

rr

p

p

p

2

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Example:The velocity field for a fluid flow is given by

Determine

a) the acceleration vector,b) the acceleration of the fluid particle at point P(1,2,3) at time

t = 1 sec.

kztjxyxtzyxV

32),,,( 2

mm302 1 lecture note 1

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