LECTURE 1_Fluid Dynamics

7/29/2019 LECTURE 1_Fluid Dynamics

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LECTURE 1

INTRODUCTION TO DIFFERENTIAL ANALYSIS OF

FLUID MOTION

5.1. Review of the Field Concept.

The property fields are defined by continuous functions of the space coordinates and time.

a scalar field [ = (x, y, z, t].

a vector field ( )

=

tzyxVV ,,, or a tensor field (the stress field).

5.2. The Continuity Equation.

We shall drive the differential equation for the conservation of mass in both rectangular

and cylindrical coordinates. In both cases the derivation is carried out by applying

conservation of mass to a differential control volume.

5.2.1 Rectangular Coordinate System

In rectangular coordinates, the control volume chosen is an infinitesimal cube with sides

of length dx, dy, dz as shown in Figure 1. The density at the center, O, of the control

volume is and the velocity there is wkvjuiV

++= .

To determine the values of the properties at each of the six faces of the control

surface, we must use a Taylor series expansion of the properties about the point O. For

example, at the right face,

) ..........2!2

1

2

2

2

2

2/ +

+

+=+dx

x

dx

xdxx

Neglecting higher order terms, we can write

)2

2/

dx

xdxx

+=+

Similarly, )2

2/

dx

x

uuu dxx

+=+

The corresponding terms at the left face are

)22

2/

x

x

dx

xdxx

=

+=

)

22

2/

dx

x

uu

dx

x

uuu dxx

=

+=


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Fig. 5.1 Differential control volume in rectangular coordinates

A word statement of the conservation of mass is

0volumecontroltheinside

massofchangeofrate

surfacecontrolethrough th

efluxmassofratenet

=

+

To evaluate the first term in this equation, we must consider the mass flux through each

of six surfaces of the control surface, that is, we must evaluate

CSAdV . The details of

this evaluation are shown in the table 5.1.

We see that the net rate of mass efflux through the control surface is given by

dxdydzz

w

y

v

x

u

+

+

The mass inside the control volume at any instant is the product of the mass per unit

volume, , and the volume, dx dy dz. Thus the rate of change of mass

x

z

y

dxdz

dy

Control Volume

O

uv

w


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Table 5.1 Mass Flux through the Control Surface of a Rectangular Differential Control Volume

Inside the control volume is given by

dx dy dzt

In rectangular coordinates the deferential equation for the conservation of mass is then

0=

+

+

+

tz

w

y

v

x

u (5.1a)

Equation 5.1a may be written more compactly in vector notation, since

=

+

+

Vz

w

y

v

x

u

where in rectangular coordinates is given by

z

k

y

j

x

i

+

+

=

Then the conservation of mass may be written as


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0=

+

tV

(5.1b)

Two flow cases for which the differential continuity equation may be simplified

are worthy of note. For incompressible flow, = constant; that is, the density is neither a

function of space coordinates, nor time. For incompressible flow, the continuity equation

simplifies to

0=

+

+

z

w

y

v

x

u

or 0=

V

Thus the velocity field, ( ),tz,y,x,V

for incompressible flow must satisfy 0=

V .

For steady flow, all fluid properties are, by definition, independent of time. Thus

at most ( )zy,x, , and for steady flow, the continuity equation can be written as

0=

+

+

z

w

y

v

x

u

or 0=

V

Fig. 5.2 Differential control volume in cylindrical coordinates

5.2.2 Cylindrical Coordinate System

In cylindrical coordinates, a suitable differential control volume is shown in Fig. 5.2. The

density at the center, 0, of control volume is and the velocity there is

zzrr

ViViViV ++=

, where zr iii ,, are unit vectors in the r, , and z directions,

continuity

Mass conservation


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respectively, and Vr, V, and Vz are the velocity components in the r, , and z directions,

respectively.

The conservation of mass states that:

0volumecontroltheinsidemassofchangeofrate

surfacecontrolethrough theffluxmassofratenet =

+

To evaluate the first term in this equation, we must consider the mass flux through each of six

faces of control surface; that is, we must evaluate

CSAdV . The properties at each of the

six faces of control surface are obtained from a Taylor series expansion of properties

about the point O. The details of the mass flux evaluation are shown in table 5.2.

Table 5.2 Mass Flux through the Control Surface of a Cylindrical Differential Control Volume


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We see that the net rate of mass efflux through the control surface is given by

dzddrz

Vr

V

r

VrV

zr

r

+

+

+

The mass inside the control volume at any instant is the product of mass per unit volume,

, and the volume, r d dr dz. Thus the rate of change of mass inside the control volume

is given by

dzdrdrt

In cylindrical coordinates the differential equation for the conservation of mass is then

0=

+

+

+

+ trzV

r

V

r

V

rVzr

r

Dividing by rgives

01

=

+

+

+

+

tz

VV

rr

V

r

V zrr

or

011

=

+

+

+

tz

VV

rr

Vr

r

zr

(5.2)

In cylindrical coordinates the vector operator, , is given by

zi

ri

ri zr

+

+

=

1

Thus in vector notation the conservation mass may be written

0=

+

tV

For incompressible flow, = constant, and Eq. 5.2 reduces to

0

11

=

+

+

z

VV

rr

rV

r

zr

For steady flow, Eq 5.2 reduces to

011

=

+

+

z

VV

rr

Vr

r

zr

Example 5.1


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For a two-dimensional flow, the x component of velocity is given by u = ax2 bx +by.

Determine a possible y component for steady, incompressible flow. How many possible y

components are there?

Solution:

Basic equation: 0=

+

tV

(mass conservation)

For incompressible flow, = constant, and we can write 0=

V . In rectangular

coordinates

0=

+

+

z

w

y

v

x

u

Assume two-dimensional flow in the xy plane. Then partial derivatives with respect to z

are zero, and

0=

+

y

v

x

u

Then baxx

u

y

v+=

=

2 where: u = ax2 bx +by.

{ }componentxholdingvofchangeofratefor theexpressionangivesThis

This equation can be integrated to obtain an expression for v. On integrating , we obtain

for steady flow.

( )xfbyaxyv ++= 2

ytorespectwithvofderivativepartialthehadwe

becauseappearsxfxoffunctionthe

),(,

Since any function f(x) is allowable, any number of expressions for v could satisfy the

deferential continuity equation under the given conditions. The simplest expression for v

would be obtained by setting f(x) = 0, that is,

byaxyv += 2

Example 5.2

A compressible flow field is described by

[ ] ktebxyjaxiV

=

Where x and y are coordinates in meters, t is time in seconds, and a, b, and k are constants

with appropriate units so that and

Vare in kg/m3 and m/sec, respectively. Calculate the

rate of change of density per unit time at the point x = 3 m, y = 2 m, z = 2 m, for t = 0.


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Solution:

Basic equation: 0=

+

tV

[ ] ktebxyjaxiz

wk

y

vj

x

uiV

t

+

+

==

[ ] ( ) ktkt eabxebxat

==

For t = 0, at the point (3,2,2)

( ) ( )sec

3sec

sec

3

3

0

34 m

kgabe

m

kga

m

kgbm

t

k =

=

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LECTURE 1_Fluid Dynamics