06 Recapitulation

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h pter 6

Recapitulation

6 Summary

Figure 6.1 schematizes the results we have reached thus far in our first five chapters.

Note the dividing line in Fig. 6.1. In Chap. 1 we stated that any problem has two

fundamental aspects to its solution: 1) we must select the correct mathematical tool

to describe the functions and their changes, and b) we must incorporate the appropriate

physical laws that define the problem fully.

Three methods govern the mathematical modeling of any problem: methods of

analysis, description, and approach. When we

n lyze

a problem, we must first select

the appropriate control: we can choose either a system where we identify a fixed

mass), or a control volume where we identify a volume of space). Once we have

selected the control, we then determine the method that describes the fluid s behavior

for that control. we have chosen a system, we select the Lagrangian description,

because we want to study the history of a fixed mass.

we have chosen a control

volume, we select the Eulerian description, because mass can enter, leave, accumulate,

or deplete with respect to the control. Finally, we must decide how to

ppro ch the

description of the control. Two forms of approach are available: the integral form

I.F), or the differential form D.F.). The former is used for large volumes or quantities

of mass, the latter for differential or elemental sizes.

Referring again to Fig. 6.1, we note that whatever choices we have made within

the three methods of modeling, we will eventually have to consider the I.F.) general

property balance Eq. 4.13), or the D.F.) general property balance Eq. 4.16). In

Chap. I we applied the conservation principles of physics to these equations. Applying

the conservation of mass, we derived a mass balance equation that predicts the behavior

of a parcel of mass or mass through a continuous volume of space. Applying the

conservation of linear and angular momentum, we produced a similar result for the

behavior of the fluid flow linear and angular momentum. Applying the conservation

of energy, we found an energy balance for either an elemental volume or a macroscopic

one. The results of applying conservation laws to fluid flow are summarized as follows.

l

The D.F.) equation for the conservation of mass is

ap

V pV

o

6.1)

4



Properties of fluids

onservation law:

onservation of

momentum

Mathematical

modeling

igur

6

chematic introduction to the theory fluid mechanics

Physics

f fluids

avicr Stokes

Eg.6.4

I

om ntum

I

Eg

6.5-68



6.1 Summary /

2. The I.F. equation for the conservation of mass is

i

pdV

+

at

t pV dA = 0

6.2

The one-dimensional incompressible form is

Q = VA

3. The D.F. equation for the conservation of linear momentum is

6.3

av

I I

- +

V·V)V = - -

Vp +

g

+ VV2V + -

vV V·V) 6.4

p 3

For incompressible flow, V V = 0, and the last term vanishes.

4. The I.F. equation for the conservation of linear momentum is

i pV dV + f pV V ·dA = I-F

arJv

A

For incompressible steady one-dimensional flow, the equation reduces to

I- F = pQ.lV

5. The I.F. equation for the conservation of angular momentum is

i

pr X V dA +

f

pr x V V dA)

=

I-r x F

arJv

A

For incompressible steady one-dimensional flow, the equation reduces to

6. The D.F. equation for the conservation of energy is

aw

a

- = - pe

+ V [ e + p/p pV

dt mech at

. .

kV + 2J.l.V·S] - 2J.l. S·V ,V

6.5

6.6

6.7

6.8

6.9



/ Chapter 6 Recapitulation

7. The I.F. equation for the conservation of energy is

. . a

_ V

2

v

2

iQe + iWe =

Jv p

i

+ 2 + gz dV +

J

P i + 2 + gz V .

A

6.10

For incompressible, steady, one-dimensional flow, Eq. 6.10 can be expressed as

p

y p y

CWe mech

+ - + + z . - - + + z = h

f

g I g

e

6.11

The foregoing set of equations represents the governing equations of motion for

fluid flow. We shall next summarize some special forms of the governing equations.

6 Special Forms the Governing Equations

Special forms of some of the governing equations of motion are of value when we

deal with certain types of fluid flow. For example, whenever the flow is

1. Steady:

2. Inviscid:

3. Two-dimensional:

a

t

v

a

az

= 0

= 0

= 0

6.12

6.13

6.14

where

x

and y are the principle direction of flow.

4. One-dimensional:

a

a

= -

az

=

0

6.15

where

x

is the principle direction of flow.

5. Incompressible:

= 0

6.16



6

SpecU

Forms

the Governing Equations / 5

6. Irrotational:

7. Definition of a real average:

x

L

°

dA

6.17)

6.18)

Substitution of these results in any of the governing equations helps reduce the

complexity of the equations. Special forms of the equations can now be constructed.

Certain D.F.) equations of linear momentum are presented in Table 6.1. So important

are these forms in fluid dynamics that we will devote entire chapters to them. For

example, Laplace s equation is the governing equation for potential theory the topic

of Chap. 12). The differential equation for the accelerating flat plate is important as

it introduces us to the concept of the boundary layer the topic of Chap. 14). Poiseuille s

equation is important in describing laminar flow in a pipe the subject of Chap. 10).

A solution of Euler s equation yields one of the most widely used equations in fluid

dynamics, namely, Bernoulli s equation.

All the equations shown in Table 6.1 have exact solutions. Though the list is far

from complete, it does illustrate that, for certain fluid flows, the Navier-Stokes equa

tions call be reduced to a form where we can obtain closed form analytic solutions.

This will be pursued in Chap. 9.

Figure 6.2 shows how we might take one of the two forms of approach the D.F.)

and apply it to a general class of incompressible fluid flows. We start by identifying

the two governing equations: the continuity equation and the Navier-Stokes equation.

The types of flows we can treat depend upon the relative significance of the

various acceleration terms in the Navier-Stokes equation. Referring to Fig. 6.2, we

can identify the following terms:

I.

@

The local acceleration, which is important only if the fluid properties change

at a point in the flow field.

2. ® The convective acceleration, which would be zero if the flow is uniform

because there is no change in the velocities in space.

3.

@

The pressure acceleration, which would vanish because, for a fluid with a

free-surface, the pressure cannot change except hydrostatically).

4.

@

The body force per unit mass. This deceleration would vanish for two

dimensional planar flow.

5.

0

The viscous deceleration, which vanishes for an inviscid fluid.

All incompressible flows using the D.F.) in describing the fluid motion must

satisfy this Navier-Stokes equation. We have already studied some special cases:

• Hydrostatics:

@ = ® = =

0, in Chap. 3.

• Inviscid flows: 0 0, in Chaps. 4 and 5.



ble 6.1

Equations

of

Fluid Flow That Have Exact Solutions

Assumptions

Origin

Equation

Mathematical Expression

Coord. System

Name

References

equation

v =

0

p = constant

a

=

0

a

z

o

4.109)

i)

ii)

iii)

iv)

v

av av

la p

- + v

=

-

at as p as

au au au

- + +

v

at ax ay

aV av av

- +

+

v

at ax ay

aV

r

aV

r

V -

at r ar r

e e r e

at + vra +

p

}

x

ap

y

ar

Intrinsic

Rectangular

Polar

Euler s equation

4.110

p = constant

Vxv o

Potential flow)

p = const.

~

at

g

=

0

V = w = 0

4.24

4.109)

vi)

vii)

viii)

a

2

j a

2

j a

2

j

=0

ax

a

y

az

a

2

j I a j I

a

2

j

j>

=0

ar

r r

2

az

d

u

_

dp

dy

-

l dx

Rectangular

Cylindrical

Rectangular

Laplace s equation

Poiseuille s equation

12.2

9.6



Table 6

Con t.)

Origin

References

Assumptions

Equation Mathematical Expression Coord. System

Name equation

p = canst.

~

4.109) ix)

v

2

_

dp

Axisymmetric

-

at

p dr

g = 0

u, = w - 0

x)

d v + = 0

Polar

Inviscid circular

dr

2

dr r

vortex flow

v = 0

p = canst.

~

4.109) xi)

d

2w

dw dp

Cylindrical

Hagen-Poiseuille

=

Prob.4.69

at

dr? r dr

I l

dz

equation

g = 0

a

- = 0

az

u, = = 0

p = canst.

a

4.109) xii)

v dv, = dp + v a

2v,

+ . .av,

=

at

dr p dr ar

2

r dr

Cylindrical Hamel s equation

= 0

a

2v,

v,

az

+ ?

2

- ?

g = 0

oI l

ap

2 l

Bu,

= w = 0

xiii)

r

2



ble 6.1 Con t.)

Assumptions

p

=

canst.

g = 0

p =

canst.

v = w = 0

p = canst.

g

= 0

Va = w = 0

Axisymmetric

Origin

Equation

4.109

4.109

xiv

xv

Mathematical Expression

au a

u

at ay

Bu

r

1

ap

- + V - =

at r ar p ar

v v

r

+

r

_

r

ar

2

r ar r

2

Coord. System

Rectangular

Polar

Name

Accelerating

flat plate

Bubble dynamics

References

equation

9.24



From Fig.-6.1

D .F . of

Navier-Stokes equation

continuity equation

av

I

V

•

V)V

V

P

- g -

I

V

= 0

V

·v=o

at

p

0

0 0

0

0

=0

L T I

t

0=0

0=0=0

0=0=0

0=0

0=0=0=0

w=o

u

u

v w

=0

Euler s equation

Hydrostatics

« a/

Potential flow

Prandtl boundary

layer equation

0=0

Couette flow

[ 0 0 0 ]

d streamline)

Blasius solution

Poiseuille flow

V

2

c

= 0

Equation of a

Etc.

Bernoulli s equation

streamline

Boundary layer thickness Kutta-Zhukouski

Drag on flat plate

Etc.

law

From

Fig. 6.1

Figure 6.2 Continuation of schematic of introduction to the theory of fluid mechanics



350

/

Chapter 6 Recapitulation

In the chapter to follow, we shall study

• Poiseuille flows: @ 0, in Chaps. 9 and 10.

• Potential flows: @

0

0, in Chaps. 12 and 13.

• Boundary layer flows:

@

0,

in Chap.

14.

In particular, we shall examine a variety of applications. For example, we shall

study the principle of flight Kutta-Zhukouski law) which arises out of potential flow

theory, or the friction drag on a ship s hull, which arises out of the Blasius solution

of the boundary layer flow theory. We shall learn about pipe flow and free channel

flow. Not all branches of fluid dynamics can be treated in this text, but many shall

be.

6 3 Problem Solving Technique

How do we solve fluid dynamics problems? There is no mutually agreed upon single

way to set up solutions. The choice of problem-solving technique may depend on

nothing more than the familiarity and confidence of the user. For ease and clarity,

especially for beginners, we suggest the following problem-solving methodology:

Step

Identify the characteristics of the fluid and flow field.

1. List the data given and the results required.

2. Check to see if the fluid is

a) incompressible or compressible

b) inviscid or real

c) steady or unsteady

d) one-, two-, or three-dimensional flow

e) rotational or irrotational

Step 2

Select the method of approach.

1. Examine the problem to see if quantities are to be solved at a point in the flow

D .F. ), or at some boundary, or on some volume of a large control

I.F.).

2. Draw and label any necessary diagrams needed for the solution. Examine what

forces or stresses are acting on the control and what transfer of energies through

the control may exist.

3. Determine if the description takes place with respect to a

fix

reference frame

or a moving reference frame. The former utilizes absolute velocities and accel

erations; the latter utilizes relative velocities and accelerations.

a) Select the appropriate coordinate system that best describes the motion.

b) Let one the coordinates be oriented in the direction of the flow.



6 3 Problem Solving Technique

/

35

Step 3

Write the appropriate form of the governing equations of flow see the summary of

equations in Sec. 6.1 .

1. Apply the results of Step 1 to reduce the equations to their simplest

form,

2. Examine all conditions given at a point or boundary.

a Identify all kinematic conditions.

b Identify all stress conditions.

c Identify all inn r conditions.

d Identify all initial conditions.

Step 4

Solve the problem.

1. Think about the problem in terms of its physical significance. Each term of the

reduced governing equation and its boundary conditions have a physical inter

pretation. The dimensions of each term must be identical and correct.

2. Check to see if there are sufficient equations to solve for the unknowns. not,

then additional equations will have to be sought, e.g., the equation of state and

stress-strain relationships.

3. Select a method for solving the reduced governing equations. The following two

sections offer a few aids in selecting an appropriate method.

6 xamples ofProblem Solving Technique

We are now at a position to apply our problem-solving technique to a flow problem.

We have all the necessary tools to commence building a solution: the governing

equations of fluid flow; the various methods of analysis, description, and approach to

treat a problem; ways to fit the equations to the given problem and simplify their

expressions; the mathematical symbols and operators necessary to describe the behavior

of the flow and the methods to solve them; and the step-by-step procedure to bring

all this together. Two problems will be analyzed in detail. The first will involve the

D.F. approach, and the second will involve the LF. approach.

Example 6 The D.F. Approach

Consider a steady two-dimensional incompressible viscous flow with no body

forces, where the pressure gradient ap/ax is zero and the y-component of velocity

v

is equal in magnitude to the kinematic viscosity

v

and is a constant. At the

location y equals zero, the shear stress y is unity. a Find an expression for

the shear stress y and vorticity component about the z-axis that is valid

everywhere in the flow field. b Find the distribution of velocity given

u

equals

zero at y equals zero.



352

hapter

6

ecapitulation

Example 6 1

Can t.)

Solution:

Step 1.

Identify the characteristics of the fluid and flow field:

• The flow is steady:

o

at

• The flow is two-dimensional:

o

• The flow is incompressible:

p const.

• The flow is real:

v¥-O

• There are no body forces:

g 0

• The pressure gradient is zero:

ap

ax

o

• The velocity component v is given:

v

v

dynamic viscosity

• A boundary condition is given:

xy y o

• Find xy x, z,

t

and

x, z,

t

Step 2.

Select the method of approach:

•

D.F.

• No external forces, no energy transfers

• Fixed reference frame Cartesian



6.4 Examples ofProblem-Solving Technique / 353

Example 6 Con t.)

y

........ u x y z t

pxy= 1.0, u = 0

Figure E6.1

Step 3.

Write the appropriate form of the governing equations of flow:

• Continuity equation:

J 0

au

+

?1. =

0

ax

/y

I

• Navier-Stokes equations:

au au

~ j

a

u

a

u

+

v

v

ax ay p x

ax

2

ay

o

0 0 0

+

vj

=

1 ap

J

+

V

2

+

2

pay

since we were given

ap ax

= 0 and v = v.

Equation i becomes

au

= 0

ax

i

ii

iii

iv

which means that since the velocity component u is not a function of x Z and

t it is at most a function only of y.

In a similar fashion, from Eq. iii ,

ap

=

0

Thus, the pressure everywhere in the flow field is constant.

v



354 hapter 6 ecapitulation

Example 6 1

Con t.)

The Navier-Stokes equation, Eq. ii , reveals that

vi

Step

Solve the problem.

We can solve this problem analytically. Integrating Eq. vi once yields

vii

where

is an arbitrary constant of integration. To evaluate it, we use the

boundary condition that

du I

xy = l dy

y

=

From Eqs. vii and viii , we obtain

I

l

a The shear stress distribution is from Eqs. vii and ix ,

The vorticity component is now easily obtained as

viii

ix

x

or

=

=

du

dy

xi

xii

The quantities

xy

and are shown plotted in Fig. 6.3 for values of negative

y

b If we impose the condition that the velocity

u

is zero at

y

equals zero,

we obtain



6.4 Examples of Problem-Solving Technique

/ 355

Example 6 1

Con t.]

Velocity profile

6.0

.0

.0

.0

{

Shear stress profile

Vorticity profile

2.0

.0

.

/

/

I

I

I

I

I

I

I

O L . _ _ L

L J

L _ J

o

1.0

0.8

pxy

0.6

1L z

0.4

-ILU

0.2

y

Figure 6.3 Example of the D.F. problem.

xiii

1

u

=

e

Y

- 1

fl.

as the velocity profile. The distribution is shown plotted in Fig. 6.3 above.

This completes the solution.

Example

6.2.

The

LF.

Approach

Water enters a 60° horizontal reducing elbow with a velocity of 15 ftls at a

pressure of 12 psi. Calculate the x and y-components of the reaction force acting

on the elbow. The entering diameter is 12 in., and the exit diameter is 8 in.

Neglect viscous effects.

Solution:

Step 1.

Identify the characteristics of the fluid and flow field:

• The flow is steady:

o



6

I Chapter

6

Recapitulation

Example 6 Con r.)

• The flow is one-dimensional:

• The flow is incompressible:

p const.

• The flow is ideal:

v = 0

• There are no body forces:

W

• Given data:

15 ft/s

Pi

12 psi

D 12 in.

De 8 in.

e

°

Step :

Select the method of approach:

•

I.F.

• External forces are pressure forces; no energy transfers

• Fixed reference frame Cartesian

t

12in

p = 12 psi

;

f s iii

8 n

Figure E6



6.4 Examples

of

Problem-Solving Technique I 7

x mple 6.2

Con t.)

Step 3.

Write the appropriate form of the governing equations of flow:

• Continuity equation:

-

-

Q = VA ; = VA e

• Linear momentum:

• Energy:

p

V2

p

V

2

z

=

z

Y 2g i Y 2g

e

Step 4.

Solve the problem:

From continuity equation i ,

11 11

15 X - X 12

2

= V

X - X 8

2

4

2

4

or

- 9

V

2

= -

X 15

=

33.75 fps

4

From Bernoulli equation iv ,

PI

P2

V

= -

Y 2g Y 2g

P2

PI V

=

Y Y 2g 2g

or

P2 12 X 144

_ _ 52

_ 33.752

Y 62.4 64.4

P2

=

5.85 psi

From the momentum equation ii ,

z F

x

=

R P l P2A2 cos 60°

- -

=

pQ V2 cos e - VI )

pQV

2

=

1.94

X

11.8

X

33.75

=

773

i

ii

iii

iv

v

vi



8 /

Chapter 6 Recapitulation

Example 6.2

Con t.)

pQV

= 1.94 x 11.8 x 15 = 343

P A

=

5 . 8 5 ~ 8 2

=

294

p,A,

=

12

~

12

2

=

1357

Therefore

R =

-773 .5

343 - 294 .5 1357

or

R = 11671bf

From the momentum Eq. iii ,

I F

= R; P A sin 60° = p V

sin 60° - 0

therefore,

R; = 294 .866 773 .866 = 254.5

670

or

s = 924.5 Ibf

This completes the solution.

vii

viii

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