advanced materials in fox's fluid mechanics

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CHAPTER 3 FLUID STATICS3-7 Fluids in Rigid-Body Motion

CHAPTER 4 BASIC EQUATIONS IN INTEGRAL FORM FOR A CONTROL VOLUME4-6 Momentum Equation for Control Volume with Arbitrary Acceleration

4-7 Equation for Rotating Control Volume

CHAPTER 6 INCOMPRESSIBLE INVISCID FLOW6-6 Unsteady Bernoulli Equation—Integration of Euler’s Equation along a Streamline

6-7 Irrotational Flow

Bernoulli Equation Applied to Irrotational Flow

Velocity Potential

Stream Function and Velocity Potential for Two-Dimensional, Irrotational,Incompressible Flow: Laplace’s Equation

Elementary Plane Flows

Superposition of Elementary Plane Flows

CHAPTER 9 EXTERNAL INCOMPRESSIBLE VISCOUS FLOW9-3 Laminar Flat-Plate Boundary Layer: Exact Solution

CHAPTER 12 COMPRESSIBLE FLOW12-3 Isothermal Flow

12-6 Supersonic Diffuser

Supersonic Wind Tunnel Operation

Supersonic Flow with Friction in a Constant-Area Channel

Supersonic Flow with Heat Addition in a Constant-Area Channel



3-7 FLUIDS IN RIGID-BODY MOTION S-1

3-7 FLUIDS IN RIGID-BODY MOTION

We are almost ready to begin studying fluids in motion (starting in Chapter 4), butfirst there is one category of fluid motion that can be studied using fluid statics ideas:

rigid-body motion. As the name implies, this is motion in which the entire fluid

moves as if it were a rigid body—individual fluid particles, although they may be inmotion, are not deforming. This means that there are no shear stresses, as in the case

of a static fluid.

What kind of fluid flow has rigid-body motion? You recall from kinematics thatrigid-body motion can be broken down into pure translation and pure rotation. For

translation the simplest motion is constant velocity, which can always be converted to

a fluid statics problem by a shift of coordinates. The other simple translational motion

we can have is constant acceleration, which we will consider here (Example Problem

3.9). In addition, we will consider motion consisting of pure constant rotation (Exam-

ple Problem 3.10). As in the case of the static fluid, we may apply Newton’s second

law of motion to determine the pressure field that results from a specified rigid-bodymotion.

In Section 3-1 we derived an expression for the forces due to pressure and grav-ity acting on a fluid particle of volume d V . We obtained

or

(3.2)

Newton’s second law was written

Substituting from Eq. 3.2, we obtain

(3.16)

If the acceleration is constant, we can combine it with and obtain an effective

“acceleration of gravity,” , so that Eq. 3.16 has the same form as our

basic equation for pressure distribution in a static fluid, Eq. 3.3. This means that wecan use the results of previous sections of this chapter as long as we use in

place of . For example, for a liquid undergoing constant acceleration the pressure

increases with depth in the direction of , and the rate of increase of pressure will

be given by geff ,where geff is the magnitude of . Lines of constant pressure will

be perpendicular to the direction of . The physical significance of each term in

this equation is as follows:

+

=

×

net pressure force

per unit volume

at a point

body force per

unit volume

at a point

mass per

unit

volume

acceleration

of fluid

particle

− ∇ + = p g a r r

r

geff

r

geff

geff

r

g

r

geff

r r r

g g aeff = −

r

gr

a

−∇ + = p g a r r

dF a dm a ar

r r

r

r

= = = dV dF

dV or

dF

dV p g

r

r

= −∇ +

dF p g dV r

r

= − ∇ +( )



S-2 CHAPTER 3 / FLUID STATICS

This vector equation consists of three component equations that must be satisfied

individually. In rectangular coordinates the component equations are

Component equations for other coordinate systems can be written using the appropriate

expression for . In cylindrical coordinates the vector operator, ∇, is given by

(3.18)

where and are unit vectors in the r and directions, respectively. Thus

(3.19)

EXAMPLE 3.9 Liquid in Rigid-Body Motion with Linear Acceleration

As a result of a promotion, you are transferred from your present location. You must

transport a fish tank in the back of your minivan. The tank is 12 in. ϫ 24 in. ϫ 12 in.

How much water can you leave in the tank and still be reasonably sure that it will not

spill over during the trip?

∇ =∂∂ +

∂∂ +

∂∂ p e

p

r e r

p

k

p

zr ˆ ˆˆ

1

e er

∇ =∂∂

+∂

∂+

∂∂

ˆ ˆ ˆer

er

k z

r

1

∇ p

(3.17)

−∂∂

+ =

−

∂

∂ + =

−∂∂

+ =

p

x g a x

p

yg a y

p

zg a z

x x

y y

z z

direction

direction

direction

EXAMPLE PROBLEM 3.9

GIVEN: Fish tank 12 in. ϫ 24 in. ϫ 12 in. partially filled with water to be transported in an automobile.

FIND: Allowable depth of water for reasonable assurance that it will not spill during the trip.

SOLUTION:The first step in the solution is to formulate the problem by translating the general problem into a more

specific one.

We recognize that there will be motion of the water surface as a result of the car’s traveling over

bumps in the road, going around corners, etc. However, we shall assume that the main effect on the water

surface is due to linear accelerations (and decelerations) of the car; we shall neglect sloshing.

Thus we have reduced the problem to one of determining the effect of a linear acceleration on the

free surface. We have not yet decided on the orientation of the tank relative to the direction of motion.

Choosing the x coordinate in the direction of motion, should we align the tank with the long side parallel,

or perpendicular, to the direction of motion?If there will be no relative motion in the water, we must assume we are dealing with a constant accel-

eration, a x . What is the shape of the free surface under these conditions?

Let us restate the problem to answer the original questions by idealizing the physical situation to ob-

tain an approximate solution.

GIVEN: Tank partially filled with water (to depth d ) subject to constant linear acceleration, a x . Tank height

is 12 in.; length parallel to direction of motion is b.Width perpendicular to direction of motion is c.




FIND: (a) Shape of free surface under constant a x .

(b) Allowable water depth, d , to avoid spilling as a function

of a x and tank orientation.

(c) Optimum tank orientation and recommended water depth.

SOLUTION:Governing equation:

Since p is not a function of

The component equations are:

Recall that a partial

derivative means that

all other independent

variables are held constant

in the differentiation.

The problem now is to find an expression for p ϭ p( x , y). This would enable us to find the equation

of the free surface. But perhaps we do not have to do that.

Since the pressure is p ϭ p( x , y), the difference in pressure between two points ( x , y) and ( x ϩ dx ,

y ϩ dy) is

Since the free surface is a line of constant pressure, p ϭ constant along the free surface, so dp ϭ 0 and

Therefore,

{The free surface is a plane.}

Note that we could have derived this result more directly by converting Eq. 3.16 into an equivalent “accel-

eration of gravity” problem,

where Lines of constant pressure (including the free surface) will then be

perpendicular to the direction of , so that the slope of these lines will be .− = −1

g a

a

g x

x r

gef

r r

g g i a ia jgeff x x = − = − −ˆ ˆ ˆ .

−∇ + = p geff r

0

dy

dx

a

g

x = −

free surface

0 =∂∂

+∂∂

= − − p

x dx

p

ydy a dx g dy x

dp p

x dx p

y dy=∂∂ +

∂∂

∂∂

= −

∂∂

= −

p

x a

p

yg

x

∴ −∂∂

−∂∂

− =ˆ ˆ ˆ ˆi p

x j

p

y j g i a x

z p z g g g g a a x y z y z, / . Also, , , , and .∂ ∂ = = = − = = =0 0 0 0

−∂∂

+∂∂

+∂∂

+ + + = + +ˆ ˆ ˆ (ˆ ˆ ˆ ) (ˆ ˆ ˆ )i

p

x j

p

yk

p

zig jg kg i a ja ka x y z x y z

−∇ + = p g a r r

g

y

b

O x

a x d

←




In the diagram,

d ϭ original depth

eϭ height above original depth

bϭ tank length parallel to direction of motion

Since we want e to be smallest for a given a x , the tank should be aligned so that b is as small as possible.

We should align the tank with the long side perpendicular to the direction of motion. That is, we should

choose b ϭ 12 in. b

With b ϭ 12 in.,

The maximum allowable value of e ϭ 12 Ϫ d in. Thus

If the maximum a x is assumed to be , then allowable d equals 8 in.

To allow a margin of safety, perhaps we should select d ϭ 6 in. d

Recall that a steady acceleration was assumed in this problem. The car would have to be driven very care-

fully and smoothly.

23g

12 6 12 6− = = −d a

gd

a

g

x x and max

ea

g

x = 6 in .

Only valid when the free surface intersects

the front wall at or above the floor

eb b dy

dx

b a

g

x = = −

=

2 2 2tan

freesurface

b

a x d

θ 12 in.

e

←

←

This Example Problem shows that: Not all engineering problems are clearly defined, nor do

they have unique answers.

For constant linear acceleration, we effectively have a hy-

drostatics problem, with “gravity” redefined as the vector

result of the acceleration and the actual gravity.

g

R

ω

EXAMPLE 3.10 Liquid in Rigid-Body Motion with Constant Angular Speed

A cylindrical container, partially filled with liquid, is rotated at aconstant angular speed, , about its axis as shown in the diagram.After a short time there is no relative motion; the liquid rotates

with the cylinder as if the system were a rigid body. Determine

the shape of the free surface.




g

R

r

z

ω

h1

h0

EXAMPLE PROBLEM 3.10

GIVEN: A cylinder of liquid in rigid-body rotation with angular

speed about its axis.

FIND: Shape of the free surface.

SOLUTION:Governing equation:

It is convenient to use a cylindrical coordinate system, r , , z. Since gr ϭ g ϭ 0 and g z ϭ Ϫg, then

Also, a ϭ a z ϭ 0 and ar ϭ Ϫ 2r .

The component equations are:

From the component equations we see that the pressure is not a function of ; it is a function of r and

z only.

Since p ϭ p(r , z), the differential change, dp, in pressure between two points with coordinates

(r , , z) and (r ϩ dr , , z ϩ dz) is given by

Then

To obtain the pressure difference between a reference point (r 1, z1), where the pressure is p1, and the

arbitrary point (r , z), where the pressure is p, we must integrate

Taking the reference point on the cylinder axis at the free surface gives

Then

p pr

g z h− = − −atm ( )

2 2

12

p p r z h1 1 1 10= = =atm

p p r r g z z− = − − −1

22

12

12

( ) ( )

dp r dr g dzr

r

p

p

z

z

= −∫ ∫ ∫ 2

11 1

dp rdr gdz= − 2

dp p

r

dr p

z

dz

z r

=∂

∂

+∂

∂

∂∂

=∂∂

=∂∂

= − p

r r

p p

zg 2 0

∴ −∂∂ +

∂∂ +

∂∂

= − +ˆ ˆ ˆ ˆ ˆe

p

r e r

p

k

p

z e r k gr r r

1 2

−∂∂

+∂∂

+∂∂

− = + +ˆ ˆ ˆ ˆ ( ˆ ˆ ˆ )e

p

r e

r

pk

p

zk g e a e a kar r r z

1

−∇ + = p g a r r




Since the free surface is a surface of constant pressure ( p ϭ patm), the equation of the free surface is

given by

or

The equation of the free surface is a parabaloid of revolution with vertex on the axis at z ϭ h1.

We can solve for the height h1 under conditions of rotation in terms of the original surface height, h0,

in the absence of rotation. To do this, we use the fact that the volume of liquid must remain constant. With

no rotation

With rotation

Then

Finally,

Note that the expression for z is valid only for h1 Ͼ 0. Hence the maximum value of is given by

max ϭ 2 gh R0 .

z h R

g

r

g

h R

g

r

R

= − + = − −

←

0

2 2

0

2 2

4 2 2

1

2

( ) ( ) ( )

z r ( )

R h h R R

gh h

R

g

20 1

22 4

1 0

2

4 4= +

= −and

( )

V hr r

gh R

R

g

R

= +

= +

2

2 8 41

2 2 4

0

12

2 4

V r dz dr zr dr hr

gr dr

R R z R

= = = +

∫ ∫ ∫ ∫ 2 2 2

21

2 2

0000

V R h= 2

0

z hr

g= +1

2

2

( )

02

2 2

1= − −

r

g z h( )

This Example Problem shows: The effect of centripetal acceleration on the shape of con-

stant pressure lines (isobars).

Because the hydrostatic pressure variation and variation

due to rotation each depend on fluid density, the final free

surface shape is independent of fluid density.



4-6 MOMENTUM EQUATION FOR CONTROL VOLUME WITH ARBITRARY ACCELERATION S-7

4-6 MOMENTUM EQUATION FOR CONTROL VOLUMEWITH ARBITRARY ACCELERATION

In Section 4-5 we formulated the momentum equation for a control volume with

rectilinear acceleration. The purpose of this section is to extend the formulation for

completeness to include rotation and angular acceleration of the control volume, inaddition to translation and rectilinear acceleration.First, we develop an expression for Newton’s second law in an arbitrary, nonin-

ertial coordinate system. Then we use Eq. 4.25 to complete the formulation for a

control volume. Newton’s second law for a system moving relative to an inertial co-

ordinate system is given by

(4.27)

where, as in the previous section, XYZ denotes the inertial (e.g., stationary) reference

frame. Since

and M (system) is constant,

or

(4.35)

The basic problem is to relate to the acceleration , measured relative to a

noninertial coordinate system. For this purpose, consider the noninertial referenceframe, xyz, shown in Fig. 4.5.

The noninertial frame, xyz, itself is located by position vector relative to the

fixed frame XYZ . The noninertial frame rotates with angular velocity . A particle

is instantaneously located relative to the moving frame by position vector

r

r

R

r

a xyz

r

a XYZ

r

r

F a dm XYZ M

= ∫ (system)

r r

r

F d

dt V dm

dV

dt dm XYZ

XYZ

M M = = ∫ ∫ (system)(system)

r r

P V dm XYZ XYZ M ) = ∫ system system( )

r

r

F d P

dt

XYZ =

system

X

r ω

R

y

z

x

Y

X

Z

Particle

Fig. 4.5 Location of a particle in inertial

(XYZ ) and noninertial (xyz ) referenceframes.



S-8 CHAPTER 4 / BASIC EQUATIONS IN INTEGRAL FORM FOR A CONTROL VOLUME

Relative to inertial reference frame XYZ , the position of the particle

is denoted by position vector . From the geometry of the figure, .The velocity of the particle relative to an observer in the XYZ system is

(4.36)

where, as in the previous section, is the instantaneous velocity of the control vol-ume frame itself relative to the inertial XYZ reference frame.

We must be careful in evaluating d /dt because both the magnitude, | |, and the

orientation of the unit vectors, , , and , are functions of time. Thus

(4.37a)

The terms dx / dt , dy / dt , and dz / dt are the velocity components of the particle relativeto xyz. Thus

(4.37b)

You may recall from dynamics (and as we will see in Example Problem 4.13), for a

rotating coordinate system

(4.37c)

Combining Eqs. 4.37a, 4.37b, and 4.37c, we obtain

(4.37d)

Substituting into Eq. 4.36 gives

(4.38)

The acceleration of the particle relative to an observer in the inertial XYZ system

is then

or

(4.39)

Both xyz and are measured relative to xyz, so the same caution observed in devel-

oping Eq. 4.37d applies. Thus

(4.40a)dV

dt i

du

dt j

d

dt k

dw

dt V a V

xyz

XYZ

xyz xyz xyz

r

r

r

r r

r

= + + + × = + ×ˆ ˆ ˆv

r

r r

V

r r

r

r r

a adV

dt

d

dt r

XYZ rf

xyz

XYZ

= +

+ ×( )

r

rr r

r r

adV

dt

dV

dt

dV

dt

d

dt r XYZ

XYZ rf xyz

XYZ

= = +

+ ×( )

r r rr r

V V V r XYZ rf xyz= + + ×

dr

dt V r xyz

r

r

r r

= + ×

r r

× = + +r x di

dt y

dj

dt z

dk

dt

ˆ ˆ ˆ

r

V i

dx

dt j

dy

dt k

dz

dt xyz = + +ˆ ˆ ˆ

dr

dt

d

dt xi yj zk i

dx

dt x

di

dt j

dy

dt y

dj

dt k

dz

dt z

dk

dt

r

= + + = + + + + +( ˆ ˆ ˆ) ˆˆ

ˆˆ

ˆˆ

k ˆ ji

r

r r

r

r

V rf

r

r r

r

r

r

V dX

dt

dR

dt

dr

dt V

dr

dt XYZ rf = = + = +

r r

r

X R r = +r

X

r

r i x jy kz= + +ˆ ˆ ˆ .




and

or

(4.40b)

Substituting Eqs. 4.40a and 4.40b into Eq. 4.39, we obtain

(4.41)

Equation 4.41 relates the acceleration of a fluid particle as measured in the twoframes (the inertial frame XYZ and the noninertial frame xyz). From your study of dy-

namics you will be familiar with each of the terms in the equation. They are

: Absolute rectilinear acceleration of a particle relative to fixed reference

frame XYZ.

: Absolute rectilinear acceleration of origin of moving reference frame

xyz relative to fixed frame XYZ.

: Rectilinear acceleration of a particle relative to moving reference

frame xyz (this acceleration would be that “seen” by an observer on

moving frame xyz; .

: Coriolis acceleration due to motion of the particle within moving

frame xyz.

: Centripetal acceleration due to rotation of moving frame xyz.

: Tangential acceleration due to angular acceleration of moving refer-ence frame xyz.

Substituting , as given by Eq. 4.41, into Eq. 4.35, we obtain

or

(4.42a)

But

(4.42b)

Combining Eqs. 4.42a and 4.42b, we obtain

r

r r

r

r r r r r

r

F a V r r dmdP

dt rf xyz

M

xyz− + × + × × + × =

∫ [ ( ) ˙ ]2

(system)system

r

v

r

r

a dmdV

dt

dmd

dt

V dmdP

dt

xyz xyz

xyz M M

xyz

M

xyz=

= =

∫ ∫ ∫ (system)(system) (system) system

r

r r

r

r r r r r r

F a V r r dm a dmrf xyz M

xyz M

− + × + × × + × =∫ ∫ [ ](system)

2 ( ) ˙( )

system

r

r r r

r

r r r r r

F a a V r r dmrf xyz xyz M

system(system)

[ ]= + + × + × × + ×∫ 2 ( ) ˙

r

a XYZ

.r r

× r

r r r

× ×( )r

2r

r

× V xyz

r

r

a dV dt xyz xyz xyz= / )

r

a xyz

r

arf

r

a XYZ

r r r r

r

r r r r r

a a a V r r XYZ rf xyz xyz= + + × + × × + ×2 ( ) ˙

d

dt r r V r xyz( ) ˙ ( )

r r r r r

r

r r r

× = × + × + × ×

= × + × + ×r r r

r

r r˙ ( ) r V r xyz

d

dt r

d

dt r

dr

dt ( )

r r

r

r r

r

× = × + ×

xyz




or

(4.43)

Equation 4.43 is a statement of Newton’s second law for a system. The systemderivative, d / dt , represents the rate of change of momentum, , of the systemmeasured relative to xyz, as seen by an observer in xyz. This system derivative can be

related to control volume variables through Eq. 4.25,

(4.25)

To obtain the control volume formulation, we set N ϭ , and ϭ . Then Eqs.

4.25 and 4.43 may be combined to give

(4.44)

Equation 4.44 is the most general control volume formulation of Newton’s sec-ond law. Comparing the momentum equation for a control volume moving with arbi-

trary acceleration, Eq. 4.44, with that for a control volume moving with rectilinear

acceleration, Eq. 4.33, we see that the only difference is the presence of three addi-

tional terms on the left side of Eq. 4.44. These terms result from the angular motion

of noninertial reference frame xyz. In dynamics these terms are often referred to as

“fictitious” forces that arise due to inertia effects present when we use a noninertial xyz coordinate system: the Coriolis force due to particle motion within the xyz coor-

dinates, and centripetal and tangential forces due to the xyz coordinate system’s rota-

tional motion, respectively. As we should expect, the general form, Eq. 4.44, reduces

to the rectilinear acceleration form, Eq. 4.33, when the angular terms are zero, and to

the inertial control volume form, Eq. 4.26, when all of the terms for the control vol-

ume motion ( , and ) are zero.

The precautions concerning the use of Eqs. 4.26 and 4.33 also apply to the useof Eq. 4.44. Before attempting to apply this equation, one must draw the boundaries

of the control volume and label appropriate coordinate directions. For a control vol-

ume moving with arbitrary acceleration, one must label a coordinate system xyz on

the control volume and an inertial reference frame XYZ .

EXAMPLE 4.13 Velocity in Fixed and Noninertial Reference Frames

A reference frame, xyz, moves arbitrarily with respect to a fixed frame, XYZ . A

particle moves with velocity xyz ϭ (dx / dt ) ϩ (dy / dt ) ϩ (dz / dt ) , relative to frame

xyz. Show that the absolute velocity of the particle is given by

r r r

r r

V V V r XYZ rf xyz= + + ×

k ˆ jir

V

r

r

r

ar ,

=∂∂

+ ⋅∫ ∫ t V d V V V dA xyz xyz xyz

r r r r

CV CS

r r

r r

r

r r r r r

F F a V r r dV S B rf xyz+ − + × + × × + ×∫ [ ( )˙

]2 CV

r

V xyz

r

P xyz

dN

dt t dV V dA xyz

=

∂∂

+ ⋅∫ ∫ system CV CS

r r

r

P xyz

r

P xyz

r r

r r

r

r r r r r

r

F F a V r r dV dP

dt S B rf xyz

xyz

V + − + × + × × + × =

∫ [ ( ) ˙ ]2

(system)system





GIVEN: Fixed and noninertial frames as shown.

X

r

ω

X

R

z

y

x

Particle

Y

Z

FIND: XYZ in terms of xyz, , , and rf .

SOLUTION:From the geometry of the sketch, so

Since

we have

or

The problem now is to evaluate d / dt , d / dt, and d / dt that result from the angular motion of frame xyz. To

evaluate these derivatives, we must consider the rotation of each unit vector caused by the three compo-

nents of the angular velocity, , of frame xyz.

Consider the unit vector . It will rotate in the xy plane because of z, as follows:i

r

k ˆ ji

dr

dt

V x di

dt

ydj

dt

zdk

dt

xyz

r

r

= + + +ˆ ˆ ˆ

dr

dt

dx

dt i

dy

dt j

dz

dt k x

di

dt y

dj

dt z

dk

dt

r

= + + + + +ˆ ˆ ˆˆ ˆ ˆ

r

r xi yj zk = + +ˆ ˆ ˆ

r

r r

r

r

r

V dX dt

dRdt

dr dt

V dr dt

XYZ rf = = + = +

r r

r

X R r = + ,

r

V r

r r

r

V r

V

y(t + ∆t )

i (t + ∆t )

x (t + ∆t )ω z

y(t )

x (t )i (t )

^

i (t + ∆t )^

i (t + ∆t )^

^i (t )^

i (t )^

∆θ –

Now from the diagram

But for small angles cos ⌬ Ϸ 1 Ϫ [(⌬ )2 /2] and sin ⌬ Ϸ ⌬ , so

ˆ( ) ˆ( ) ( ) ˆ ( )( )

( ˆ) ( ) ˆ ˆi t t i t j i j i+ − = + − = −

⌬ ⌬⌬

⌬⌬

1 12

12

2

ˆ( ) ˆ( ) ( ) sin ˆ ( )( cos )( ˆ)i t t i t j i+ − = + − −⌬ ⌬ ⌬1 1 1




In the limit as ⌬t : 0, since ⌬ ϭ z ⌬t ,

Similarly, will rotate in the xz plane because of y.i

di

dt

i t t i t

t

t jt

i

t

di

dt j

z

z

t t

z z

z

ˆlim

ˆ( ) ˆ( )lim

( ) ˆ ˆ

ˆ ˆ

due to

due to

=+ −

=−

=

→ →

⌬ ⌬

⌬

⌬

⌬⌬

⌬0 0

12

x (t )

z(t )

x (t + ∆t )

z(t + ∆t )

∆θ

ω y

Enlarged sketch

i (t + ∆t )^

i (t + ∆t )^

i (t + ∆t ) –^

i (t )^

i (t )^

i (t )^

Then from the diagram

For small angles

In the limit as ⌬t : 0, since ⌬ ϭ y ⌬t ,

Rotation in the yz plane because of x does not affect . Combining terms,

By similar reasoning,

Thus

x di

dt y

dj

dt z

dk

dt z y i x z j y x k y z z x x y

ˆ ˆ ˆ( )ˆ ( ) ˆ ( ) ˆ+ + = − + − + −

dj

dt k i

dk

dt i j x z y x

ˆˆ ˆ and

ˆˆ ˆ= − = −

di

dt j k z y

ˆˆ ˆ= −

i

di

dt

i t t i t

t

t k t

i

t

di

dt k

yt t

y y

y

y

ˆlim

ˆ( ) ˆ( )lim

( ) ˆ ˆ

ˆˆ

due to

due to

=+ −

=− −

= −

→ →

⌬ ⌬

⌬

⌬

⌬⌬

⌬0

12

ˆ( ) ˆ( ) ( ) ( ˆ) ( )( )

( ˆ) ( ) ˆ ˆi t t i t k i k i+ − = − + − = − −

⌬ ⌬

⌬⌬

⌬1 1

21

2

2

ˆ( ) ˆ( ) ( ) sin ( ˆ) ( )( cos )( ˆ)i t t i t k i+ − = − + − −⌬ ⌬ ⌬1 1 1



4-7 THE ANGULAR-MOMENTUM PRINCIPLE S-13

But

Combining these results, we obtainr r r

r r

r

V V V r V XYZ rf xyz XYZ = + + × ←

r r

× = = − + − + −r

i j k

x y z

z y i x z j y x k x y z y z z x x y

ˆ ˆ ˆ

( )ˆ ( ) ˆ ( ) ˆ

4-7 THE ANGULAR-MOMENTUM PRINCIPLE (continued)

Equation for Rotating Control Volume

In problems involving rotating components, such as the rotating sprinkler of Example

Problem 4.14, it is often convenient to express all fluid velocities relative to the rotat-

ing component. The most convenient control volume is a noninertial one that rotateswith the component. In this section we develop a formulation of the angular-momen-tum principle for a noninertial control volume rotating about an axis fixed in space.

Inertial and noninertial reference frames were related in Section 4-6. Figure 4.5

showed the notation used. For a system in an inertial frame,

(4.3a)

The angular momentum of a system in general motion must be specified relative toan inertial reference frame. Using the notation of Fig. 4.5,

With ϭ 0 the xyz frame is restricted to rotation within XYZ , and the equation be-

comes

so that

Since the mass of a system is constant,

or

(4.47)

From the analysis of Section 4-6,

(4.36)r r

r

V V dr

dt XYZ rf = +

r

r

r

r

r

T dr

dt V r

dV

dt dm XYZ

XYZ

M system

(system)= × + ×

∫

r

r

r

T d

dt r V dm XYZ

M system

(system)( )= ×∫

r

r

r

T d

dt r V dm XYZ

M system

(system)= ×∫

r

r

r

r

r

H r V dm r V dV XYZ M

XYZ V

system(system)

= × = ×∫ ∫ (system)

r

R

r r

r

r r

r

r

H R r V dm R r V dV XYZ M

XYZ V

system(system) (system)

( ) ( )= + × = + ×∫ ∫

r

r

T dH

dt system

system

=




With xyz restricted to pure rotation, rf ϭ 0. The first term under the integral on theright side of Eq. 4.47 is then

Thus Eq. 4.47 reduces to

(4.48)

From Eq. 4.41 with rf ϭ 0 (since xyz does not translate),

Substituting into Eq. 4.48, we obtain

or

(4.49)

We can write the last term as

(4.50)

The torque on the system is given by

(4.3c)

The relation between the system and control volume formulations is

(4.25)

where

Setting N equal to xyz)system and ϭ ϫ xyz yields

(4.51)

Combining Eqs. 4.49, 4.50, 4.51, and 4.3c, we obtain

dH

dt t r V dV r V V d A

xyz xyz xyz xyz

r

r

r

r

r r r

=

∂∂

× + × ⋅∫ ∫ system

CV

CS

r

V r

r r

H

N dm M

system = ∫ (system)

dN

dt t dV V dA xyz

=

∂∂

+ ⋅∫ ∫ system CV CS

r r

r

r

r

r r

r

T r F r g dm T s M

system shaftsystem

= × + × +∫ ( )

r

r

r

r

r

r dV

dt dm

d

dt r V dm

dH

dt

xyz

M xyz

M xyz

xyz×

= ×

=

∫ ∫ (system) system

system xyz

( )

= × = ×

∫ ∫

r r r

r

r a dm r dV

dt dm xyz

M

xyz

M xyz

(system) (system)

r

r r

r

r r r r r

T r V r r dm xyz M

system(system)

[ ( ) ˙ ]− × × + × × + ×∫ 2

r

r r r

r

r r r r r

T r a V r r dm xyz xyz M

system = × + × + × × + ×∫ [ ( ) ˙ ](system)

2

r r rr

r r r r r

a a V r r XYZ xyz xyz= + × + × × + ×2 ( ) ˙

a

r

r

r

r r

T r dV

dt dm r a dm

XYZ

M XYZ

M system

(system) (system)= × = ×∫ ∫

dr

dt

dr

dt

r r

× = 0

r

V




Since the system and control volume coincided at t 0,

(4.52)

Equation 4.52 is the formulation of the angular-momentum principle for a (noniner-

tial) control volume rotating about an axis fixed in space. All fluid velocities in Eq.

4.52 are evaluated relative to the control volume. Comparing Eq. 4.52 with Eq. 4.46

(for inertial XYZ cooordinates) we see that the noninertial (rotating) xyz coordinateshave an extra “moment” term on the left side that includes three components. As we

discussed following Eq. 4.44, these components arise because of “fictitious” forces: the

Coriolis force because of fluid particle motion within the xyz coordinates, and centripetal

and tangential forces because of the xyz coordinates’ rotational motion, respectively.

Equation 4.52 reduces to Eq. 4.46 when the control volume is not in motion (when

and are zero). Even though we have the extra term to evaluate, Eq. 4.52 is sometimes

simpler to use than Eq. 4.44 because a problem that is unsteady in XYZ coordinates be-comes steady state in xyz coordinates, as we will see in Example Problem 4.15.

EXAMPLE 4.15 Lawn Sprinkler: Analysis Using Rotating Control Volume

A small lawn sprinkler is shown in the

sketch below. At an inlet gage pres-sure of 20 kPa, the total volume flow

rate of water through the sprinkler is

7.5 liters per minute and it rotates at

30 rpm. The diameter of each jet is 4

mm. Calculate the jet speed relative to

each sprinkler nozzle. Evaluate thefriction torque at the sprinkler pivot.

r

r

r

r

r r

r

r r

r

r r r r r

r

r

r

r r r

r F r g dV T

V dV

t r V r V V dA

s

xyz

xyz xyz xyz

× + × +

− × × + × × + ×

=∂∂

× + × ⋅

∫ ∫

∫ ∫

shaftCV

CV

CV CS

[ ( ) ˙ ]r r r

dV

2

r

r

r r

r

r r

r

r r r r r

r

r

r

r r r

r F r g dm T

r V r r dm

t r V r V V dA

s

xyz

xyz xyz xyz

× + × +

− × × + × × ×

=∂∂

× + × ⋅

∫ ∫

∫ ∫

) +

dV

M

M

shaft(system)

(system)

CV CS

[ ( ˙ ]2

V relV rel

ω

R = 150 mm

Q = 7.5 L/min= 30 rpmω

α = 30°

psupply = 20 kPa (gage)


GIVEN: Small lawn sprinkler as shown.

FIND: (a) Jet speed relative to each nozzle.

(b) Friction torque at pivot.

SOLUTION:Apply continuity and angular momentum equa-

tions using rotating control volume enclosing

sprinkler arms.

V relV rel

ω

R = 150 mm

α = 30°

psupply = 20 kPa (gage)

CV

(Control volumerotates with

sprinkler arm)

Q = 7.5 L/minω = 30 rpm z r

T f




art to be come

(4.52)

Assumptions: (1) Steady flow relative to the rotating CV.

(2) Uniform flow at each section.

(3) ϭ constant.

From continuity

Consider terms in the angular-momentum equation separately. As in Example Problem 4.14, the only ex-

ternal torque acting on the CV is friction in the pivot. It opposes the motion, so

(1)

The second integral on the left of Eq. 4.52 is evaluated for flow within the CV. Let the velocity and

area within the sprinkler tubes be V CV and ACV, respectively. Then, for one side, the first term (a Coriolis

effect) is

(The flow in the bent portion of the tube has no r component of velocity, so it does not contribute to the

integral.)

From continuity, Q ϭ 2 V CV ACV, so for both sides the integral becomes

(2)

The second term in the integral (a moment generated by centripetal acceleration) is evaluated as

so it contributes no torque. (The force generated by centripetal acceleration is radial, so it generates no

moment.)

r r r r

r r dV re k k re dV

re k re dV re r e dV

r r

r r r

× × × = × × ×

= × × = × − =

∫ ∫ ∫ ∫

[ ( )] ˆ [ ˆ ( ˆ ˆ )]

ˆ [ ˆ ˆ ] ˆ ( ˆ )]

CV CV

CV CV

2 0

r r

r

r V dV R Qk xyz× × =∫ [ ] ˆ

CV2 2

r r

r

r V dV re k V e A

re V e A

V A r V A k

xyz r

R

r

r

R

R

× × = × ×

×

{ }

∫ ∫ ∫ ∫

[ ] ˆ [ ˆ ˆ ]

ˆ ˆ

ˆ ˆ

CVCV

CV

CV CV

2 2

2

2

0

0

0

CV

CV

CV CV one side

dr

= dr

= dr k = R2

r

T T k f shaft = − ˆ

V Q

A

Q

D

V V

rel jet jet

2

3 2

2

rel

rel

. L

min ( ) mm

m

1000 L

mm

m

min

60 s

. m/s

= =

= × × × × ×

= ←

2 2

4

1

2

7 5 4 1

4

10

4 97

2 2

6

∂∂t ͵ CV

dV ϩ C͵S

V xyz и dA ϭ 0

ϭ 0(1)

r ϫ F S ϩ

͵CV

r ϫ g dV ϩ T shaft Ϫ

͵CV

r ϫ [2 ϫ V xyz ϩ ϫ ( ϫ r ) ϩ ϫ r ] dV

∂∂t

ϭ 0(1)

C͵V

r ϫ V xyz dV ϩC͵V

r ϫ V xyz V xyz и dAϭ

ϭ 0(3)

Governing equations:




The integral on the right side of Eq. 4.52 is evaluated for flow crossing the control surface. For the

right arm of the sprinkler,

The velocity and radius vectors for flow in the left arm must be described in terms of the same unit vectors

used for the right arm. In the left sprinkler arm, the component has the same magnitude but opposite

sign, so it cancels. For the complete CV,

(3)

Combining terms (1), (2), and (3), we obtain

or

T f ϭ R (V rel cos ␣ Ϫ R) Q

From the data given,

Substituting gives

T

T T

f

f f

= ×°

−

×

× × × ×

= ⋅ ←

150 4 97 30

0 0718

2

mm m

s

0.471 m

s

999 kg

m

7.5 L

m

1000 L 60 s

N m

3

3

. cos

min

min N .s

kg . m

m

1000 mm

.

R = × × × × =30 150 2rev

min

rad

rev

min

60 s

m

1000 mm0.471m/s

mm

− − = −T k R Qk RV Q k f ˆ ˆ cos ˆ

rel ␣ 2

r

r r r

r V V dA RV Q k xyz xyz× ⋅ = −∫ ␣ relCV

cos ˆ

r

r r r

r V V dA e V e k V A

RV k eQ

xyz xyz r × ⋅ = × − + +

= − + −

∫ ␣ ␣

␣ ␣

R ˆ [cos ( ˆ ) sin ˆ]{ }

[cos ( ˆ) sin ( ˆ )]

re rel jet

rel

lCS

2

This problem illustrates use of the angular momentum princi-ple for a noninertial (rotating) control volume. Note that in

this approach, unlike the inertial control volume of Example

Problem 4.14, the fluid particle position vector and velocity

vector are not time-dependent. As we should expect, the re-sults agree using either an inertial or noninertial control volume.

r

V

r

r



S-18 CHAPTER 6 / INCOMPRESSIBLE INVISCID FLOW

6-6 UNSTEADY BERNOULLI EQUATION—INTEGRATION OF EULER’SEQUATION ALONG A STREAMLINE

It is not necessary to restrict the development of the Bernoulli equation to steady

flow. The purpose of this section is to develop the corresponding equation for un-

steady flow along a streamline and to illustrate its use.The momentum equation for frictionless flow (Eq. 6.1) can be written (with inthe negative z direction) as

(6.17)

Equation 6.17 is a vector equation. It can be converted to a scalar equation by taking

the dot product with d , where d is an element of distance along a streamline. Thus

(6.18)

Examining the terms in Eq. 6.18, we note that

Substituting into Eq. 6.18, we obtain

(6.19)

Integrating along a streamline from point 1 to point 2 yields

(6.20)

For incompressible flow, the density is constant. For this special case, Eq. 6.20

becomes

(6.21)

Restrictions: (1) Incompressible flow.

(2) Frictionless flow.

(3) Flow along a streamline.

To evaluate the integral term in Eq. 6.21, the variation in ∂V / ∂t must be known as afunction of s, the distance along the streamline measured from point 1. (For steady

flow, ∂V / ∂t ϭ 0, and Eq. 6.21 reduces to Eq. 6.8.) Equation 6.21 may be applied to

any flow in which the restrictions are compatible with the physical situation.

Application of Eq. 6.21 is illustrated in Example Problem 6.9.

p V gz

p V gz

V

t ds1 1

2

12 2

2

21

2

2 2 + + = + + +

∂∂∫

dp V V g z z

V

t ds

1

222

12

2 11

2

20∫ ∫ +

−+ − +

∂∂

=( )

V dV V

t ds

dpg dz+

∂∂

= − −

∂∂

=

∇ ⋅ =⋅ =

V

sds dV V

p ds dp

k ds dz z

(thechange in along )

(the change in pressurealong )ˆ (the change in along )

s

s

s

r

r

DV

Dt ds

DV

Dt ds V

V

sds

V

t ds p ds gk ds

r

r r r

⋅ = =∂∂

+∂∂

= − ∇ ⋅ − ⋅1

ˆ

r

sr

s

DV

Dt p gk

r

= − ∇ −1

ˆ

r

g



6-6 UNSTEADY BERNOULLI EQUATION— INTEGRATION OF EULER’S EQUATION ALONG A STREAMLINE S-19

EXAMPLE 6.9 Unsteady Bernoulli Equation

A long pipe is connected to a large reservoir that initially is filled with water to a

depth of 3 m. The pipe is 150 mm in diameter and 6 m long. Determine the flow ve-

locity leaving the pipe as a function of time after a cap is removed from its free end.

EXAMPLE PROBLEM 6.9

GIVEN: Pipe and large reservoir as

shown.

FIND: V 2(t ).

SOLUTION:Apply the Bernoulli equation to the un-

steady flow along a streamline from

point to point .

Governing equation:

Assumptions: (1) Incompressible flow.

(2) Frictionless flow.

(3) Flow along a streamline from to .

(4) p1 ϭ p2 ϭ patm.

(5) V 21 Ӎ 0.

(6) z2 ϭ 0.

(7) z1 ϭ h ϭ constant.

(8) Neglect velocity in reservoir, except for small region near the inlet to the tube.

Then

In view of assumption (8), the integral becomes

In the tube, V ϭ V 2 everywhere, so that

Substituting gives

Separating variables, we obtain

Integrating between limits V ϭ 0 at t ϭ 0 and V ϭ V 2 at t ϭ t ,

dV

gh V gh

V

gh

t

L

V V

2

1

2 2 220

1

0

22

−=

=∫ −tanh

dV

gh V

dt

L

2

22

2 2−=

ghV

LdV

dt

= +22

2

2

∂∂

= =∫ ∫ V

t ds

dV

dt ds L

dV

dt

L L

0

2

0

2

∂∂

≈∂∂∫ ∫ V

t ds

V

t ds

L

1

2

0

gz ghV V

t ds1 2

2

1

2

2= = +∂∂∫

21

p1 1 ∂V ∂t

ds1͵

ϩ ϩ gz1 ϭ

Ӎ 0(5) ϭ 0(6)

V 2

2 p2 2ϩ ϩ gz2 ϩV

2

2

2

21

1

2

D = 150 mmh = 3 m

L = 6 m

V 2

z

Flow




Since tanhϪ1(0) ϭ 0, we obtain

For the given conditions,

and

The result is then V 2 ϭ 7.67 tanh (0.639t ) m/s, as shown:

t

Lgh

t t

22

2

1

6

7 670 639= × × =

m

m

s

..

22 9 81 3

7 67gh = × × =.

.m

s

mm/s2

1

2 2 2 2 22

1 2 2

2gh

V

gh

t

L

V

gh

t

Lgh

V t tanh tanh

( )

−

= =

←

or

0

8

6

4

2

01 2 3 4 5

t (s)

V 2

( m / s )

V 2 = 7.67 tanh (0.639 t )

Notes: This problem illustrates use of the unsteady Bernoulli

equation.

Initially the head available at state is used to acceler-

ate the fluid in the pipe; eventually the head at state

equals the head at state .

This problem is somewhat unrealistic except for the initial

instants— the asymptotic flow condition actually corre-sponds to a turbulent flow!

1

1

*6.7 IRROTATIONAL FLOW

We have already discussed irrotational flows in Section 5-3. These are flows in which

the fluid particles do not rotate ( ϭ 0). We recall that the only stresses that can gen-

erate particle rotation are shear stresses; hence, inviscid flows (i.e., those with zero

shear stresses) will be irrotational, unless the particles were initially rotating. Using

Eq. 5.14, we obtain the irrotationality condition

(6.22)

leading to

(6.23)∂∂

−∂∂

=∂∂

−∂∂

=∂∂

−∂∂

=w

y z

u

z

w

x x

u

y

v v

0

∇ × =r

V 0

r

* Note that Section 5-2 contains background material needed for study of this section.

2



6-7 IRROTATIONAL FLOW S-21

In cylindrical coordinates, from Eq. 5.16, the irrotationality condition requires that

(6.24)

Bernoulli Equation Applied to Irrotational Flow

In Section 6-3, we integrated Euler’s equation along a streamline for steady, incom-

pressible, inviscid flow to obtain the Bernoulli equation

(6.8)

Equation 6.8 can be applied between any two points on the same streamline. In

general, the value of the constant will vary from streamline to streamline.

If, in addition to being inviscid, steady, and incompressible, the flow field is also

irrotational (i.e., the particles had no initial rotation), so that (Eq. 6.22),

we can show that Bernoulli’s equation can be applied between any two points in theflow. Then the value of the constant in Eq. 6.8 is the same for all streamlines. To il-

lustrate this, we start with Euler’s equation in vector form,

(6.9)

Using the vector identity

we see for irrotational flow, where , that

and Euler’s equation for irrotational flow can be written as

(6.25)

Consider a displacement in the flow field from position to position + d ; the

displacement d is an arbitrary infinitesimal displacement in any direction. Taking thedot product of d = dx + dy + dz with each of the terms in Eq. 6.25, we have

and hence

or

dpd V gdz

+ + =

1

20

2( )

1

2

2d V

dpgdz( ) = − −

1

2

12∇ ⋅ = − ∇ ⋅ − ⋅( ) ˆV dr p dr gk dr r r r

k ˆ jir

r

r

r

r r

r r

r

1

2

1

2

12∇ ⋅ = ∇ = − ∇ −( ) ( ) ˆr r

V V V p gk

( ) ( )r r r r

V V V V ⋅ ∇ = ∇ ⋅1

2

∇ × =r

V 0

( ) ( ) ( )r r r r r r

V V V V V V ⋅ ∇ = ∇ ⋅ − × ∇ ×1

2

( ) ˆr r

V V p gk ⋅ ∇ = − ∇ −1

∇ × =r

V 0

p V gz

+ + =

2

2constant

1 1 10

r

V V

z

V

z

V

r r

rV

r r

V z r z r ∂∂

−∂∂

=∂∂

−∂∂

=∂∂

−∂∂

=




Integrating this equation for incompressible flow gives

(6.26)

Since d was an arbitrary displacement, Eq. 6.26 is valid between any two points

(i.e., not just along a streamline) in a steady, incompressible, inviscid flow that is also

irrotational (see Example Problem 6.5).

Velocity Potential

We were introduced in Section 5-2 to the notion of the stream function for a two-

dimensional incompressible flow.

For irrotational flow we can introduce a companion function, the potential func-

tion , defined by

(6.27)

Why this definition? Because it guarantees that any continuous scalar function ( x , y,

z, t ) automatically satisfies the irrotationality condition (Eq. 6.22) because of a funda-mental indentity7:

(6.28)

The minus sign (used in most textbooks) is inserted simply so that decreases in the

flow direction (analogous to the temperature decreasing in the direction of heat flow

in heat conduction). Thus,

(6.29)

(You can check that the irrotationality condition, Eq. 6.22, is satisfied identically.)

In cylindrical coordinates,

(3.18)

From Eq. 6.27, then, in cylindrical coordinates

(6.30)

Because ϫ ≡ 0 for all , the velocity potential exists only for irrotational

flow.

Irrotationality may be a valid assumption for those regions of a flow in which

viscous forces are negligible.8 (For example, such a region exists outside the bound-

ary layer in the flow over a wing surface, and can be analyzed to find the liftproduced by the wing.) The theory for irrotational flow is developed in terms of an

imaginary ideal fluid whose viscosity is identically zero. Since, in an irrotational

flow, the velocity field may be defined by the potential function , the theory is often

referred to as potential flow theory.

V r

V r

V z

r z= −∂∂

= −∂∂

= −∂∂

1

∇ =∂∂

+∂

∂+

∂∂

ˆ ˆ ˆer

er

k z

r

1

u x y

w z

= −∂∂

= −∂∂

= −∂∂

, ,v and

∇ × = −∇ × ∇ = − ≡r

V curl(grad ) 0

r

V = −∇

r

r

p V gz

+ + =

2

2constant

7 That ϫ ( ) ≡ 0 can easily be demonstrated by expanding into components.8 Examples of rotational and irrotational motion are shown in the NCFMF video Vorticity.




All real fluids possess viscosity, but there are many situations in which the as-

sumption of inviscid flow considerably simplifies the analysis and, at the same time,

gives meaningful results. Because of its utility and mathematical appeal, potentialflow has been studied extensively.9

Stream Function and Velocity Potential for Two-Dimensional, Irrotational,

Incompressible Flow: Laplace’s EquationFor a two-dimensional, incompressible, irrotational flow we have expressions for thevelocity components, u and v, in terms of both the stream function , and the veloc-

ity potential ,

(5.3)

(6.29)

Substituting for u and v from Eq. 5.3 into the irrotationality condition,

(6.23)

we obtain

(6.31)

Substituting for u and v from Eq. 6.29 into the continuity equation,

(5.4)

we obtain

(6.32)

Equations 6.31 and 6.32 are forms of Laplace’s equation—an equation that

arises in many areas of the physical sciences and engineering. Any function or

that satisfies Laplace’s equation represents a possible two-dimensional, incompress-

ible, irrotational flow field.

It is worth summarizing what we’ve learned. For a two-dimensional incompress-

ible flow, we can always define a stream function ; if the flow is also irrotational,

will satisfy Laplace’s equation. On the other hand, for an irrotational flow, we can al-

ways define a velocity potential ; if the flow is also incompressible, will satisfy

Laplace’s equation.In Section 5-2 we showed that the stream function is constant along any

streamline. For ϭ constant, d ϭ 0 and

d x

d x y

d y

=∂∂

+∂∂

= 0

∂∂

+∂∂

=2

2

2

20

x y

∂∂

+∂∂

=u

x y

v

0

∂∂

+∂∂

=2

2

2

20

x y

∂∂

− ∂∂

=v

x u y

0

u x y

= −∂∂

= −∂∂

v

u y x

=∂∂

= −∂∂

v

9 Anyone interested in a detailed study of potential flow theory may find [4–6] of interest.




The slope of a streamline—a line of constant — is given by

(6.33)

Along a line of constant , d ϭ 0 and

Consequently, the slope of a potential line—a line of constant — is given by

(6.34)

(The last equality of Eq. 6.34 follows from use of Eq. 6.29.)Comparing Eqs. 6.33 and 6.34, we see that the slope of a constant line at any

point is the negative reciprocal of the slope of the constant line at that point; lines

of constant and constant are orthogonal. This property of potential lines and

streamlines is useful in graphical analyses of flow fields.

Example 6.10 Velocity Potential

Consider the flow field given by ϭ ax 2 Ϫ ay2, where a ϭ 3 sϪ1. Show that the flow

is irrotational. Determine the velocity potential for this flow.

d y

d x

x

y

u = −

∂ ∂∂ ∂

= −

/

/ v

d x

d x y

d y = ∂∂

+ ∂∂

= 0

d y

d x

x

y u u

= −

∂ ∂∂ ∂

= −−

=

/

/

v v


GIVEN: Incompressible flow field with ϭ ax 2 Ϫ ay2, where a ϭ 3 sϪ1.

FIND: (a) Whether or not the flow is irrotational.

(b) The velocity potential for this flow.

SOLUTION:If the flow is irrotational, 2 ϭ 0. Checking for the given flow,

so that the flow is irrotational. As an alternative proof, we can compute the fluid particle rotation (in the xy

plane, the only component of rotation is z):

then

so

Once again, we conclude that the flow is irrotational. Because it is irrotational, must exist, and

u x y

= −∂∂

= −∂∂

and v

2 2 2 2 2 0 2 z

x

u

y x ax

yay a a z=

∂∂

−∂∂

=∂

∂− −

∂∂

− = − + = ← v

( ) ( )

u y ax ay ay x ax ay ax =∂

∂ − = − = −∂

∂ − = −( ) ( )2 2 2 2

2 2and v

2

z x

u

yu

y x =

∂∂

−∂∂

=∂∂

= −∂∂

v

vand

∇ =∂∂

−( ) +∂∂

−( ) = − =22

22 2

2

22 2 2 2 0

x ax ay

yax ay a a




Consequently, Integrating with respect to x gives ϭ 2axy ϩ f ( y),

where f ( y) is an arbitrary function of y. Then

Therefore, and f ϭ constant. Thus

We also can show that lines of constant and constant are orthogonal.

ϭ ax 2 Ϫ ay2 and ϭ 2axy

The slopes of lines of constant and constant are negative reciprocals. Therefore lines of constant are orthogonal to lines of constant .

For constant henced

= = = − = −

=, ;0 2 2ay dx ax dy

dy

dx

y

x c

For constant henced

= = = − =

=, ;0 2 2axdx ay dy

dy

dx

x

yc

= +←

2axy constant

− = − − ∂∂ = − − =2 2 2 0ax ax f y y

ax df dy

df dy

( ) , so

v = − = −∂∂

= −∂

∂+2 2ax

y yaxy f y

[ ( )]

u x

ay x

ay= −∂∂

= −∂∂

=

2 2 and .

This problem illustrates the relations among the stream func-

tion, velocity potential, and velocity field.

The stream function and velocity potential are

shown in the Excel workbook. By entering the equations

for and , other fields can be plotted.

Elementary Plane FlowsThe and functions for five elementary two-dimensional flows—a uniform flow, a

source, a sink, a vortex, and a doublet—are summarized in Table 6.1. The and

functions can be obtained from the velocity field for each elementary flow. (We saw

in Example Problem 6.10 that we can obtain from u and v.)A uniform flow of constant velocity parallel to the x axis satisfies the continuity

equation and the irrotationality condition identically. In Table 6.1 we have shown the and functions for a uniform flow in the positive x direction.

For a uniform flow of constant magnitude V , inclined at angle ␣ to the x axis,

ϭ (V cos ␣) y Ϫ (V sin ␣) x

ϭ Ϫ(V sin ␣) y Ϫ (V cos ␣) x

A simple source is a flow pattern in the xy plane in which flow is radially out-ward from the z axis and symmetrical in all directions. The strength, q, of the source

is the volume flow rate per unit depth. At any radius, r , from a source, the tangential

velocity, V , is zero; the radial velocity, V r , is the volume flow rate per unit depth, q,divided by the flow area per unit depth, 2 r . Thus V r ϭ q /2 r for a source. The

and functions for a source are shown in Table 6.1.

In a simple sink , flow is radially inward; a sink is a negative source. The and

functions for a sink shown in Table 6.1 are the negatives of the corresponding func-

tions for a source flow.




T a b l e

6 . 1

E l e m e n t a

r y P l a n e F l o w s

U n i f o r m F

l o w ( p o s i t i v e x

d i r e c t i o n )

S o u r c e F l o w ( f r o m o

r i g i n )

O r i g i n i s s i n g u l a r p o i n t

q i s v

o l u m e fl o w r a t e p e r u n i t d e p t h

⌫

ϭ

0 a r o u n d a n y c l o s e d c u r v e

S i n k

F l o w ( t o w a r d o r i g i n )

O r i g i n i s s i n g u l a r p o i n t

q i s v

o l u m e fl o w r a t e p e r u n i t d e p t h

⌫

ϭ

0 a r o u n d a n y c l o s e d c u r v e

V

q r

q

V

q

r

r =

−

=

−

=

=

2

2

0

2

l n

V

q r

q

V

q

r

r =

=

=

=

−

2

2

0

2

l n

u

U

U y U

x

=

=

=

=

−

=

v

0 0

⌫

a r o u n d a n y c l o s e d c u r v e

U

U

x

y

y

x

ψ

φ

= c 3

ψ

= – c 3

ψ

= c 2

ψ

= – c 2

ψ

= c 1

ψ

= – c 1

ψ

= 0

= k 2

φ = – k 2

φ = k 1

φ = – k 1

φ = 0

x

y

rθ

y x

φ = – k 1

φ = – k 2

ψ

=

c 1

ψ =

c 2

ψ

=

c 3

ψ

=

c 4 ψ

=

c 5

ψ =

c 6

ψ

=

c 7

ψ

=

0

y

x r

θ

y

x φ =

k 1 φ

=

k 2

ψ

= – c 1

ψ

= –

c 2

ψ

= –

c 3

ψ

= –

c 4

ψ

= –

c 5

ψ

= – c 6

ψ

= – c 7 ψ

= 0




T a b l e

6 . 1

E l e m e n t a

r y P l a n e F l o w s ( c o n t ’ d )

I r r o t a

t i o n a l V o r t e x ( c o u n t e r -

c l o c k w

i s e , c e n t e r a t o r i g i n )

O r i g i n

i s s i n g u l a r p o i n t

K i s s t r e n g t h o f t h e v o r t e x

⌫

ϭ K

a r o u n d a n y c l o s e d

c u r v e e n c l o s i n g o r i g i n

⌫

ϭ 0

a r o u n d a n y c l o s e d

c u r v e n o t e n c l o s i n g o r i g i n

D o u b l

e t ( c e n t e r a t o r i g i n )

O r i g i n

i s s i n g u l a r p o i n t

⌳

i s s t r e n g t h o f t h e d o u b l e t

⌫

ϭ 0

a r o u n d a n y c l o s e d c u r v e

V

r

r

V

r

r

r = −

=

−

=

−

=

−

⌳

⌳

⌳

⌳

2 2

c o s

s i n

s i n

c o s

V

K

r

V

Kr

K

r =

=

−

=

=

−

0

2

2

2

l n

y

x r θ

y x ψ

= –

c 1

ψ

= – c 3 ψ

= – c 4

φ = – k 1

φ = – k 2

φ = – k 3

φ = – k 4

φ = – k 5

φ = – k 6

φ = –

k 7

φ =

0

ψ

= – c 2

y

x r θ

ψ

= – c 2

ψ

= –

c 3

ψ

=

0

ψ

=

c 3

ψ

=

c 2

ψ

= – c 1

ψ

=

c 1

φ = – k 1

φ =

k 1

φ =

k 2

φ = – k 2

x

y




The origin of either a sink or a source is a singular point, since the radial veloc-

ity approaches infinity as the radius approaches zero. Thus, while an actual flow may

resemble a source or a sink for some values of r , sources and sinks have no exactphysical counterparts. The primary value of the concept of sources and sinks is that,

when combined with other elementary flows, they produce flow patterns that ade-

quately represent realistic flows.

A flow pattern in which the streamlines are concentric circles is a vortex; in a free(irrotational) vortex , fluid particles do not rotate as they translate in circular paths around

the vortex center. The velocity distribution in an irrotational vortex can be determinedfrom Euler’s equation and the Bernoulli equation. For irrotational flow, the Bernoulli

equation is valid between any two points in the flow field. For flow in a horizontal plane ,

The component of Euler’s equation normal to the streamline is

Combining these equations yields

From the last equality

Integrating this equation gives

The strength, K , of the vortex is defined as K ϭ 2 rV ; the dimensions of K are L2 / t

(volume flow rate per unit depth). The irrotational vortex is a reasonable approxima-tion to the flow field in a tornado (except in the region of the origin; the origin is a

singular point).

The final “elementary” flow listed in Table 6.1 is the doublet of strength ⌳. This

flow is produced mathematically by allowing a source and a sink of numerically

equal strengths to merge. In the limit, as the distance, ␦ s, between them approaches

zero, their strengths increase so the product q␦ s /2 tends to a finite value, ⌳, whichis termed the strength of the doublet.

Superposition of Elementary Plane Flows

We saw earlier that both and satisfy Laplace’s equation for flow that is both in-

compressible and irrotational. Since Laplace’s equation is a linear, homogeneouspartial differential equation, solutions may be superposed (added together) to

develop more complex and interesting patterns of flow. Thus if 1 and 2 satisfy

Laplace’s equation, then so does 3 ϭ 1 ϩ 2. The elementary plane flows are

the building blocks in this superposition process. There is one note of caution:While Laplace’s equation for the stream function, and the stream function-

velocity field equations (Eq. 5.3) are linear, the Bernoulli equation is not; hence, in

the superposition process we will have 3 ϭ 1 ϩ 2, u3 ϭ u1 ϩ u2, and v3 ϭ

V r = constant

V dr r dV + = 0

dpV

V

r dr

= − =dV 2

12

dp

dr

V

r =

1

dp V = − dV




v1 ϩ v2, but p3 ϶ p1 ϩ p2! We must use the Bernoulli equation, which is nonlinearin V , to find p3.

We can add together elementary flows to try and generate recognizable flow pat-

terns. The simplest superposition approach is called the direct method, in which we try

different combinations of elementary flows and see what kinds of flow patterns are pro-

duced. This sounds like a random process, but with a little experience it becomes a quite

logical process. For example, look at some of the classic examples listed in Table 6.2.The source and uniform flow combination makes sense—we would intuitively expect asource to partially push its way upstream, and to divert the flow around it. The source,

sink, and uniform flow (generating what is called a Rankine body) is also not surpris-

ing— the entire flow out of the source makes its way into the sink, leading to a closed

streamline. Any streamline can be interpreted as a solid surface because there is no fl ow

across it; we can therefore pretend that this closed streamline represents a solid. We

could easily generalize this source-sink approach to any number of sources and sinksdistributed along the x axis, and as long as the sum of the source and sink strengths

added up to zero, we would generate a closed streamline body shape. The doublet-

uniform flow (with or without a vortex) generates a very interesting result: flow over a

cylinder (with or without circulation)! We first saw the flow without circulation in

Fig. 2.12a. The flow with a clockwise vortex produces a top-to-bottom asymmetry. Thisis because in the region above the cylinder the velocities due to the uniform flow and

vortex are in the same overall direction and lead to a high velocity; below the cylinderthey are in opposite directions and therefore lead to a low velocity. As we have learned,

whenever velocities are high, streamlines will be close together, and vice versa—

explaining the pattern shown. More importantly, from the Bernoulli equation we know

that whenever the velocity is high the pressure will be low, and vice versa—hence, we

can anticipate that the cylinder with circulation will experience a net upward force (lift)

due to pressure. This approach, of looking at streamline patterns to see where we haveregions of high or low velocity and hence low or high pressure, is very useful. We will

examine these last two flows in Example Problems 6.11 and 6.12. The last example in

Table 6.2, the vortex pair, hints at a way to create flows that simulate the presence of a

wall or walls: for the y-axis to be a streamline (and thus a wall), simply make sure thatany objects (e.g., a source, a vortex) in the positive x quadrants have mirror-image ob-

jects in the negative x quadrants; the y axis will thus be a line of symmetry. For a flowpattern in a 90Њ corner, we need to place objects so that we have symmetry with respect

to both the x and y axes. For flow in a corner whose angle is a fraction of 90Њ (e.g., 30Њ),

we need to place objects in a radially symmetric fashion.

Because Laplace’s equation appears in many engineering and physics applica-

tions, it has been extensively studied. One approach is to use conformal mapping with

complex notation. It turns out that any continuous complex function f ( z) (where z ϭ x

ϩ iy, and i ϭ ) is a solution of Laplace’s equation, and can therefore representboth and . With this approach many elegant mathematical results have been de-

rived [7–10]. We mention only two: the circle theorem, which enables any given flow

[e.g., from a source at some point (a, b)] to be easily transformed to allow for the pres-ence of a cylinder at the origin; and the Schwarz-Christoffel theorem, which enables a

given flow to be transformed to allow for the presence of virtually unlimited stepwiselinear boundaries (e.g., the presence on the x axis of the silhouette of a building).

Much of this analytical work was done centuries ago, when it was called “hy-

drodynamics” instead of potential theory. A list of famous contributors includes

Bernoulli, Lagrange, d’Alembert, Cauchy, Rankine, and Euler [11]. As we discussed

in Section 2-6, the theory immediately ran into dif ficulties: In an ideal fluid flow no

−1




T a b l e

6 . 2

S u p e r p o s

i t i o n o f E l e m e n t a r y P l a n e F l o w s

S o u r c e a n d U n i f o r m F

l o w ( fl o w p a s t a h a l f - b o d y )

S o u r c e a n d S i n k ( e q u a l s t r e n g t h , s e p a r a t i o n d i s t a n c e o n x

a x i s

2 a )

S o u r c e ,

S i n k , a n d U n i f o r m F

l o w ( fl o w p a s t a R a n k i n e b o d

y )

=

−

q

r r

U r

2

2 1

l n

c o s

=

+

+

=

+

+

=

−

+

−

s o

s i

u f

q

r

q

r

U x

1

2

3

1

2

2

2

l n

1 n

=

−

+

q

U r

2

1

2

(

)

s i n

=

+

+

=

+

+

=

−

+

s o

s i

u f

q

q

U y

1

2

3

1

2

2

2

=

+

=

+

=

−

+

=

s o

s i

q

r

q

r

q

r r

1

2

1

2

2 1

2

2

2

l

l

l

n

n

n

=

+

=

+

=

−

=

−

s o

s i

q

q

q

1

2

1

2

1

2

2

2

2

(

)

=

+

=

+

=

−

−

=

−

−

s o

u f

q

r

U x

q

r

U r

1

2

2

2

l

l

n

n

c o s

=

+

=

+

=

+

=

+

s o

u f

q

U y

q

U r

1

2

2

2

s i n

P

P

P

y

x r

P V 1

V 2

V

θ

y

x

V 2 V

1

r 2

r 1 r

V

P

θ 1

θ 2

( a , 0

)

( – a ,

0 )

y

x

V 2 (

× 3 )

V 3

V 1 (

× 3 )

r 2

r 1

r

V

P

θ

θ

1

θ 2




T a b l e

6 . 2

S u p e r p o s

i t i o n o f E l e m e n t a r y P l a n e F l o w s ( c

o n t ’ d )

V o r t e x ( c l o c k w i s e ) a n d U n i f o r m F

l o w

D o u b l e t a n d U n i f o r m F

l o w ( fl o w p a s t a c y l i n d e r )

D o u b l e t , V o r t e x ( c l o c k

w i s e ) , a n d U n i f o r m F

l o w ( fl o w p a s t

a c y l i n d e r w i t h c i r c u l a t i o n )

( C o n t i n u e d )

a

U

K

a U

r

K

U r

U r

a r

K

=

<

=

−

+

−

=

−

+

+

⌳

⌳

;

c o s

c o s

c o s

4

2

1

2

2 2

=

+

+

=

+

+

=

−

+

−

d

u f

r

K

U x

v

1

2

3

2

⌳

c o s

=

−

+

+

=

−

+

⌳ s i n

n

s i n

s i n

n

r

K

r

U r

U r

a r

K

r

2

1

2

2 2

l

l

=

+

+

=

+

+

=

−

+

+

d

u f

r

K

r

U y

v

1

2

3

2

⌳

s i n

n l

=

−

+

=

−

+

U

r

U r

U r

a r

⌳

c o s

c o s

1

2 2

=

+

=

+

=

−

−

=

−

−

d

u f

r

U x

r

U r

1

2

⌳

⌳

c o s

c o s

c o s

=

−

=

−

=

U

r

U r

U r

a r

a

U

⌳

⌳

s i n

s i n

1

2 2

=

+

=

+

=

−

+

=

−

+

d

u f

r

U y

r

U r

1

2

⌳

⌳

s i n

s i n

s i n

=

+

=

+

=

−

=

−

v

u f

K

U x

K

U r

1

2

2

2

c o s

=

+

=

+

=

+

=

+

v

u f

K

r

U y

K

r

U r

1

2

2

2

l

l

n

n

s i n

P

P P

y

x V 1

V 2

P

r

V

θ

y

x

V 1

V 2

P

r

V

θ

y

x V 1

V 2

V 3

P

r

V

θ




T a b l e

6 . 2

S u p e r p o s

i t i o n o f E l e m e n t a r y P l a n e F l o w s ( c

o n t ’ d . )

S o u r c e a n d V o r t e x ( s p

i r a l v o r t e x )

S i n k a n d V o r t e x

V o r t e x P a i r ( e q u a l s t r e n g t h , o p p o s i t e r o t a t i o n , s e p a r a t i o n

d i s t a n c e o n x

a x i s ϭ

2 a )

=

+

=

+

=

−

+

=

−

v

v

1

2

1

2

1

2

2

1

2

2

2

K

K

K

(

)

=

+

=

+

=

−

+

=

v

v

1

2

1

2

1

2

2 1

2

2

2

1

K

r

K

r

K

r r

l

l

n

n

n

=

+

=

+

=

−

−

=

+

=

+

=

−

s i

s i

q

K

r

q

r

K

v v

1

2

1

2

2

2

2

2

l

l

n

n

=

+

=

+

=

−

=

+

=

+

=

−

−

s o s o

q

K

r

q

r

K

v v

1

2

1

2

2

2

2

2 l

l

n

n

PP

P

y

x

r V

P V 1

V 2

θ

y

x

r

V

P V 1

V 2 θ

y

x

r 2

r 1

1

V 1

V 2

P

V

θ

2

θ

( a ,

0 )

( – a ,

0 )




body experiences drag — the d’Alembert paradox of 1752 —a result completely

counter to experience. Prandtl, in 1904, resolved this discrepancy by describing how

real flows may be essentially inviscid almost everywhere, but there is always a“boundary layer” adjacent to the body. In this layer significant viscous effects occur,

and the no-slip condition is satisfied (in potential flow theory the no-slip condition is

not satisfied). Development of this concept, and the Wright brothers’ historic first hu-

man flight, led to rapid developments in aeronautics starting in the 1900s. We willstudy boundary layers in detail in Chapter 9, where we will see that their existence

leads to drag on bodies, and also affects on the lift of bodies.An alternative superposition approach is the inverse method in which distributions

of objects such as sources, sinks, and vortices are used to model a body [12]. It is called

inverse because the body shape is deduced based on a desired pressure distribution.

Both the direct and inverse methods, including three-dimensional space, are today

mostly analyzed using computer applications such as Fluent [13] and STAR-CD [14].

EXAMPLE 6.11 Flow over a Cylinder: Superposition of Doublet and Uniform Flow

For two-dimensional, incompressible, irrotational flow, the superposition of a doublet

and a uniform flow represents flow around a circular cylinder. Obtain the stream

function and velocity potential for this flow pattern. Find the velocity field, locatethe stagnation points and the cylinder surface, and obtain the surface pressure distri-

bution. Integrate the pressure distribution to obtain the drag and lift forces on the

circular cylinder.


GIVEN: Two-dimensional, incompressible, irrotational flow formed from superposition of a doublet and a

uniform flow.

FIND: (a) Stream function and velocity potential.

(b) Velocity field.

(c) Stagnation points.(d) Cylinder surface.

(e) Surface pressure distribution.

(f) Drag force on the circular cylinder.

(g) Lift force on the circular cylinder.

SOLUTION:Stream functions may be added because the flow field is incompressible and irrotational. Thus from Table

6.1, the stream function for the combination is

The velocity potential is

The corresponding velocity components are obtained using Eqs. 6.30 as

V r r

U

= −

∂∂

= − −1

2

⌳ sinsin

V r r

U r = −∂∂

= − +

⌳ cos

cos2

= + = − − ← d u f r

Ur ⌳ cos cos

= + = − + ← d u f r

Ur ⌳ sin

sin




The velocity field is

Stagnation points are where

Thus

Thus V ϭ 0 when ϭ 0, .

Stagnation points are (r , ) ϭ (a, 0), (a, ).

Note that V r ϭ 0 along r ϭ a, so this represents flow around a circular cylinder, as shown in Table 6.2. Flow is

irrotational, so the Bernoulli equation may be applied between any two points. Applying the equation between

a point far upstream and a point on the surface of the cylinder (neglecting elevation differences), we obtain

Thus,

Along the surface, r ϭ a, and

since ⌳ ϭ Ua2. Substituting yields

or

Drag is the force component parallel to the freestream flow

direction. The drag force is given by

since dA ϭ a d b, where b is the length of the cylinder normal

to the diagram. Substituting

F p ab d U ab d

p ab U ab U ab

F F

D

D D

= − + − −

= −

−

+

= ←

∞

∞

∫ ∫ 0

212

0

22 2

0

212

2

0

212

2 43

3

0

2

1 4

0

cos ( sin ) cos

sin sin sin

p p U = + −∞12

2 21 4 ( sin ),

F p dA pa d b D A

= − = −∫ ∫ cos cos

0

2

p p

U

−= − ←

∞1

22

21 4

sin

Pressure distribution

p p U U U − = − = −∞12

2 2 2 12

2 24 1 4 ( sin ) ( sin )

V V a

U U 2 22

22 2 2

4= = − −

= ⌳

sin sin

p p U V − = −∞12

2 2 ( )

p U p V ∞ + = +

2 2

2 2

Stagnation points←

V r

U U r

= − − = − +

⌳ ⌳sinsin sin

2 2

V r U

ar = = =0 when . Also,⌳

V r

U U r

r = − + = − ⌳ ⌳

cos cos cos 2 2

r

V V e V er r = + =ˆ ˆ 0

r r

V V e V er

U er

U e V r r r = + = − +

+ − −

← ˆ ˆ

coscos ˆ

sinsin ˆ

⌳ ⌳

2 2

a

p dA

U

p∞

θ




Lift is the force component normal to the freestream flow direction. (By convention, positive lift is an up-

ward force.) The lift force is given by

Substituting for p gives

F p ab d U ab d

p ab U ab U ab

F F

L

L L

= − − −

=

+

+ −

= ←

∞

∞

∫ ∫ 0

212

0

22 2

0

212

2

0

212

23

0

2

1 4

4

34

0

sin ( sin ) sin

cos coscos

cos

F p dA pa d b L A

= − = −∫ ∫ ( sin ) sin

0

2

This problem illustrates: How elementary plane flows can be combined to generate

interesting and useful flow patterns.

d’Alembert’s paradox, that potential flows over a body do

not generate drag.

The stream function and pressure distribution are plot-

ted in the Excel workbook.

Example 6.12 Flow over a Cylinder with Circulation: Superposition of Doublet,Uniform Flow, and Clockwise Free Vortex

For two-dimensional, incompressible, irrotational flow, the superposition of a dou-blet, a uniform flow, and a free vortex represents the flow around a circular cylinder

with circulation. Obtain the stream function and velocity potential for this flow pat-

tern, using a clockwise free vortex. Find the velocity field, locate the stagnation

points and the cylinder surface, and obtain the surface pressure distribution. Integratethe pressure distribution to obtain the drag and lift forces on the circular cylinder. Re-

late the lift force on the cylinder to the circulation of the free vortex.


GIVEN: Two-dimensional, incompressible, irrotational flow formed from superposition of a doublet, a

uniform flow, and a clockwise free vortex.

FIND: (a) Stream function and velocity potential.

(b) Velocity field.

(c) Stagnation points.

(d) Cylinder surface.

(e) Surface pressure distribution.

(f) Drag force on the circular cylinder.

(g) Lift force on the circular cylinder.

(h) Lift force in terms of circulation of the free vortex.

SOLUTION:Stream functions may be added because the flow field is incompressible and irrotational. From Table 6.1,

the stream function and velocity potential for a clockwise free vortex are




Using the results of Example Problem 6.11, the stream function for the combination is

The velocity potential for the combination is

The corresponding velocity components are obtained using Eqs. 6.30 as

(1)

(2)

The velocity field is

Stagnation points are located where . From Eq. 1,

Thus V r ϭ 0 when

The stagnation points are located on r ϭ a. Substituting into Eq. 2 with r ϭ a,

Thus V ϭ 0 along r ϭ a when

Stagnation points: r ϭ a

=−

←

−sin

Stagnation points

1

4

K

Ua

sin or sin

= − =−

−K

Ua

K

Ua4 4

1

V U K

a

= − −

22 sin

V a

U K

a

U U

K

a

= − − −

= − − −2

⌳

⌳

⌳

sinsin

sinsin

2 2

r aU

= =←

⌳ Cylinder surface

V r

U U r

r = − + = −

⌳ ⌳coscos cos

2 2

r

V V e V er r = + =ˆ ˆ 0

rr

V r

U er

U K

r e V r = − +

+ − − −

←

⌳ ⌳coscos ˆ

sinsin ˆ

2 2

r

V V e V er r = +ˆ ˆ

V r r U

K

r

= −

∂

∂ = − − −

1

22

⌳ sin

sin

V r r

U r = −∂∂

= − +

⌳ cos

cos2

= − − + ← ⌳ cos

cosr

Ur K

2

= + +d uf f v

= − + + ← ⌳ sin

sin nr Ur

K

r 2 l

= + +d u f f v

f f

K r

K v v

= =2 2

ln




As in Example Problem 6.11, V r ϭ 0 along r ϭ a, so this flow field once again represents flow

around a circular cylinder, as shown in Table 6.2. For K ϭ 0 the solution is identical to that of Example

Problem 6.11.

The presence of the free vortex (K Ͼ 0) moves the stagnation points below the center of the cylinder.

Thus the free vortex alters the vertical symmetry of the flow field. The flow field has two stagnation points

for a range of vortex strengths between K ϭ 0 and K ϭ 4 Ua.

A single stagnation point is located at ϭ Ϫ /2 when K ϭ 4 Ua.

Even with the free vortex present, the flow field is irrotational, so the Bernoulli equation may be ap-

plied between any two points. Applying the equation between a point far upstream and a point on the sur-

face of the cylinder we obtain

Thus, neglecting elevation differences,

Along the surface r ϭ a and V r ϭ 0, so

and

Thus

Drag is the force component parallel to the freestream flow direction. As in Example Problem 6.11,the drag force is given by

since dA ϭ a d b, where b is the length of the cylinder normal to the diagram.

Comparing pressure distributions, the free vortex contributes only to the terms containing the factor

K . The contribution of these terms to the drag force is

(3)

Lift is the force component normal to the freestream flow direction. (Upward force is defined as posi-

tive lift.) The lift force is given by

F p dA pa d b A

L = − = −∫ ∫ sin sin

0

2

F U

K Ua

ab K U a

abF

D f

D

v

12

22

0

2

22 2 2

0

2

22 4

0

= +

=

←

sin sin

F

U

K

Ua

K

U aab d

D f v

12

2

2

2 2 20

2 2

4

= +

∫ sin cos

F p d A pa d b D A

= − = −∫ ∫ cos cos

0

2

p p U K

Ua

K

U a p= + − − −

←

∞1

21 4

2

4

2 22

2 2 2

sin sin

( )

V

U

K

Ua

K

U a

= + +2

22

2 2 24

2

4sin sin

V V U K

a

2 22

22

= = − −

sin

p p U V U U

V − = − = −

∞1

2

1

21

2 2 22

( )

p U gz

p V gz∞ + + = + +

2 2

2 2

p∞

V

θ

a

p

U




Comparing pressure distributions, the free vortex contributes only to the terms containing the factor K . The

contribution of these terms to the lift force is

The circulation is defined by Eq. 5.18 as

On the cylinder surface, r ϭ a, and so

Substituting into the expression for lift,

or the lift force per unit length of cylinder is

F

bU L = −

← ⌫

LF

b

F UKb U b U b L = = − = − ( )⌫ ⌫

⌫

⌫

= − −

⋅

= − −

= − ←

∫

∫ ∫

22

22

0

2

0

2

0

2

U K

ae a d e

Ua d K

d

K

sin ˆ ˆ

sin

Circulation

r

V V e= ˆ ,

⌫ ϵ ͵ V и ds

Thus, F UKb F L f L

v

=←

F

U

K

Ua

K

U aab d

K

Ua ab d K

U aab d

Kb

U

K b

U a

F

L

L

f v

12

2

2

2 2 20

2

2

2

2 2 20

2

0

2

2

0

2 2

2 2

0

2

2

4

2

4

2

2 4 4

= +

= +

= −

−

∫

∫ ∫

sin sin

sin sin

sincos

f f

U

Kb

U

Kb

U

v

12

2

2 2

2

2

=

=

This problem illustrates: Once again d’Alembert’s paradox, that potential flows do

not generate drag on a body.

That the lift per unit length is − U ⌫. It turns out that this

expression for lift is the same for all bodies in an ideal

fluid flow, regardless of shape!

The stream function and pressure distribution are plotted

in the Excel workbook.



9-3 LAMINAR FLAT-PLATE BOUNDARY LAYER: EXACT SOLUTION S-39

9-3 LAMINAR FLAT-PLATE BOUNDARY LAYER: EXACT SOLUTION

The solution for the laminar boundary layer on a horizontal flat plate was obtained by

Prandtl’s student H. Blasius [2] in 1908. For two-dimensional, steady, incompressible

flow with zero pressure gradient, the governing equations of motion (Eqs. 5.27) re-

duce to [3]

(9.3)

(9.4)

with boundary conditions

at y ϭ 0, u ϭ 0, v ϭ 0

at y ϭ ϱ, u ϭ U , ϭ 0(9.5)

Equations 9.3 and 9.4, with boundary conditions Eq. 9.5 are a set of nonlinear, cou-

pled, partial differential equations for the unknown velocity field u and v. To solve

them, Blasius reasoned that the velocity profile, u / U , should be similar for all values

of x when plotted versus a nondimensional distance from the wall; the boundary-

layer thickness, ␦ , was a natural choice for nondimensionalizing the distance fromthe wall. Thus the solution is of the form

(9.6)

Based on the solution of Stokes [4], Blasius reasoned that ␦ ϰ and set

(9.7)

We now introduce the stream function, , where

(5.4)

satisfies the continuity equation (Eq. 9.3) identically. Substituting for u and v into Eq.9.4 reduces the equation to one in which is the single dependent variable. Defining

a dimensionless stream function as

(9.8)

makes f ( ) the dependent variable and the independent variable in Eq. 9.4. With de-

fined by Eq. 9.8 and defined by Eq. 9.7, we can evaluate each of the terms in Eq. 9.4.The velocity components are given by

(9.9)u y y

xU df

d

U

x U

df

d =

∂∂

=∂∂

∂∂

= =

f

xU

( )

=

u y x

=∂∂

= −∂∂

and v

= y U x

x U

u

U g

y= ∝( ) where

␦

∂∂

u

y

uu

x

u

y

u

y

∂∂

+∂∂

=∂∂

v 2

2

∂∂

+∂∂

=u

x y

v

0



S-40 CHAPTER 9 / EXTERNAL INCOMPRESSIBLE VISCOUS FLOW

and

or

(9.10)

By differentiating the velocity components, it also can be shown that

and

Substituting these expressions into Eq. 9.4, we obtain

(9.11)

with boundary conditions:

(9.12)

The second-order partial differential equations governing the growth of the laminarboundary layer on a flat plate (Eqs. 9.3 and 9.4) have been transformed to a nonlin-

ear, third-order ordinary differential equation (Eq. 9.11) with boundary conditions

given by Eq. 9.12. It is not possible to solve Eq. 9.11 in closed form; Blasius solved it

using a power series expansion about ϭ 0 matched to an asymptotic expansion for

: ϱ . The same equation later was solved more precisely— again using numeri-

cal methods—by Howarth [5], who reported results to 5 decimal places. The numeri-

cal values of f , df / d , and d 2 f / d 2 in Table 9.1 were calculated with a personal com-puter using 4th-order Runge-Kutta numerical integration.

The velocity profile is obtained in dimensionless form by plotting u / U versus ,using values from Table 9.1. The resulting profile is sketched in Fig. 9.3b. Velocity

profiles measured experimentally are in excellent agreement with the analytical solu-tion. Profiles from all locations on a flat plate are similar; they collapse to a single

profile when plotted in nondimensional coordinates.From Table 9.1, we see that at ϭ 5.0, u / U ϭ 0.992. With the boundary-layer

thickness, ␦ , defined as the value of y for which u / U ϭ 0.99, Eq. 9.7 gives

(9.13)␦

≈ =5 0 5 0. .

U x

x

x Re

at

→ ∞ =,df

d 1

at

= = =0 0, f df

d

2 03

3

2

2

d f

d f

d f

d + =

∂

∂ =

2

2

2 3

3

u

y

U

x

d f

d

∂∂

= −

∂∂

=

u

x

U

x

d f

d

u

yU U x

d f

d

2

2

2

2

2

v = −

1

2

U

x df d

f

v = −∂∂

= −∂∂

+

= − −

+

x xU

f

x

U

x f xU

df

d x

U

x f

1

2

1

2

1 1

2




The wall shear stress may be expressed as

Then

(9.14)

and the wall shear stress coefficient, C f , is given by

(9.15)

Each of the results for boundary-layer thickness, ␦ , wall shear stress, w, andskin friction coefficient, C f , Eqs. 9.13 through 9.15, depends on the length Reynolds

number, Re x , to the one-half power. The boundary-layer thickness increases as x 1/2,

and the wall shear stress and skin friction coefficient vary as 1/ x 1/2. These results

characterize the behavior of the laminar boundary layer on a flat plate.

EXAMPLE 9.2 Laminar Boundary Layer on a Flat Plate: Exact Solution

Use the numerical results presented in Table 9.1 to evaluate the following quantities

for laminar boundary-layer flow on a flat plate:

(a) ␦ */ ␦ (for ϭ 5 and as : ϱ).

(b) v / U at the boundary-layer edge.

(c) Ratio of the slope of a streamline at the boundary-layer edge to the slope of ␦ versus x .

C U Re

f w

x

= =

12

2

0 664.

w x U U x

U

Re= =0 332

0 3322

.

.

w

y

u

yU U x

d f

d =

∂∂

=

= =0

2

2

0

Table 9.1 The Function f ( ) for the Laminar Boundary Layer along a Flat Plate atZero Incidence

f f Ј

0 0 0 0.33210.5 0.0415 0.1659 0.3309

1.0 0.1656 0.3298 0.3230

1.5 0.3701 0.4868 0.3026

2.0 0.6500 0.6298 0.2668

2.5 0.9963 0.7513 0.2174

3.0 1.3968 0.8460 0.1614

3.5 1.8377 0.9130 0.1078

4.0 2.3057 0.9555 0.0642

4.5 2.7901 0.9795 0.0340

5.0 3.2833 0.9915 0.0159

5.5 3.7806 0.9969 0.0066

6.0 4.2796 0.9990 0.0024

6.5 4.7793 0.9997 0.0008

7.0 5.2792 0.9999 0.0002

7.5 5.7792 1.0000 0.0001

8.0 6.2792 1.0000 0.0000

′ = f u

U = y

U

x



S-42 CHAPTER 9 / EXTERNAL INCOMPRESSIBLE VISCOUS FLOW

EXAMPLE PROBLEM 9.2

GIVEN: Numerical solution for laminar flat-plate boundary layer, Table 9.1.

FIND: (a) ␦ * / ␦ (for ϭ 5 and as : ϱ).

(b) v / U at boundary-layer edge.

(c) Ratio of the slope of a streamline at the boundary-layer edge to the slope of ␦ versus x .

SOLUTION:The displacement thickness is defined by Eq. 9.1 as

In order to use the Blasius exact solution to evaluate this integral, we need to convert it from one involving u and

y to one involving f Ј (ϭ u / U ) and variables. From Eq. 9.7,

Thus,

(1)

Note: Corresponding to the upper limit on y in Eq. 9.1, max ϭ ϱ, or max Ϸ 5.

From Eq. 9.13,

so if we divide each side of Eq. 1 by each side of Eq. 9.13, we obtain (with f Ј ϭ df / d )

Integrating gives

Evaluating at max ϭ 5, we obtain

The quantity Ϫ f ( ) becomes constant for Ͼ 7. Evaluating at max ϭ 8 gives

Thus, ␦ * :ϱ is 0.24 percent larger than ␦ * ϭ5и

From Eq. 9.10,

v

v

= −

= −

= −

1

2

1

2

1

2

U

x

df

d f

U Ux

df

d f

Re

df

d f

x

, so

␦

␦

*

( )→ ∞

←

␦

␦

*

( . . ) .= − =1

5

8 0 6 2792 0 344

␦

␦

␦

␦

*( . . ) .

*( )= − = =

←

1

55 0 3 2833 0 343 5

␦

␦

*( ) max= −[ ]

1

5 0 f

␦

␦

* max

= −

∫ 1

51

0

df

d d

␦

≈5

U x

␦

* ( ) ( )max max

= − = −∫ ∫ 1 10 0

f x

U d

x

U f d Ј Ј

= = = yU

x y

x

U dy d

x

U , so and

␦ ␦

* = −

≈ −

∫ ∫

∞1 1

00

u

U dy

u

U dy




Evaluating at the boundary-layer edge ( ϭ 5), we obtain

Thus v is only 0.84 percent of U at Re x ϭ 104, and only about 0.12 percent of U at Re x ϭ 5 ϫ 105.

The slope of a streamline at the boundary-layer edge is

The slope of the boundary-layer edge may be obtained from Eq. 9.13,

so

This result indicates that the slope of the streamlines is about 1/3 of the slope of the boundary layer

edge—the streamlines penetrate the boundary layer, as sketched below:

Thus.

..

streamline streamline

dy

dx

d

dx

d

dx

dy

dx

= =

←

0 84

2 50 336

␦ ␦

d

dx U x

Ux Re x

␦ = = =−

51

22 5

2 51 2 ..

␦

≈ =

55

U x

x

U

dy

dx u U Re x

= = ≈

streamline

.v v 0 84

v v

U Re Re Re U x x x

= − = ≈ =←

1

25 0 9915 3 2833

0 837 0 845[ ( . ) . ]

. .( )

U U

uδ

This problem illustrates use of numerical data from the Bla-

sius solution to obtain other information on a flat plate lami-

nar boundary layer, including the result that the edge of the

boundary layer is not a streamline.



S-44 CHAPTER 12 / COMPRESSIBLE FLOW

12-3 FLOW IN A CONSTANT-AREA DUCT WITH FRICTION (continued)

Isothermal Flow

Gas flow in long constant-area ducts, such as natural gas pipelines, is essentially

isothermal. Mach numbers in such flows are generally low, but significant pressure

changes can occur as a result of frictional effects acting over long duct lengths.Hence, such flows cannot be treated as incompressible. The assumption of isothermal

flow is much more appropriate.

For isothermal flow with friction (as opposed to the adiabatic flow with friction we

previously discussed), the heat transfer ␦ Q / dm is not zero. On the other hand, we have

the simplification that the temperature is constant everywhere. As for adiabatic flow,

we can start with our set of basic equations (Eqs. 12.1), describing one-dimensional

flow that is affected by area change, friction, heat transfer, and normal shocks,

1V 1 A1 ϭ 2V 2 A2 ϭ VA ϭ ϭ constant (12.1a)

R x ϩ p1 A1 Ϫ p2 A2 ϭ V 2 Ϫ V 1 (12.1b)

(12.1c)

(12.1d)

p ϭ RT (12.1e)

⌬h ϭ h2 Ϫ h1 ϭ c p⌬T ϭ c p(T 2 Ϫ T 1) (12.1f)

(12.1g)

We can simplify these equations by setting ⌬T ϭ 0, so T 1 ϭ T 2, and A1 ϭ A2 ϭ A. In

addition we recall from Section 12-1 that the combination, h ϩ V 2 /2 is the stagnationenthalpy, h0. Using these, our final set of equations (renumbered for convenience) is

(12.22a)

(12.22b)

(12.22c)

(12.22d)

(12.22e)

(12.22f)

Equations 12.22 can be used to analyze frictional isothermal flow in a channel of

constant area. For example, if we know conditions at section (i.e., p1, 1, T 1, s1,

h1, and V 1) we can use these equations to find conditions at some new section 2

1

⌬s s s Rp

p= − = −2 1

2

1

ln

p RT =

˙ ( )˙

CSm s s

T

Q

AdA2 1

1−

∫ Ն

qQ

dmh h

V V = = − =

−␦ 0 0

22

12

2 1 2

R p A p A mV mV x + − = −1 2 2 1˙ ˙

1 1 2 2V V V Gm

A= = = = =

˙constant

⌬s s s cT

T R

p

p p= − = −2 1

2

1

2

1

ln ln

˙ ( )˙

CSm s s

T

Q

AdA2 1

1−

∫ Ն

␦ Qdm

h V h V + + = +112

222

2 2

mm

m



12-3 FLOW IN A CONSTANT-AREA DUCT WITH FRICTION S-45

after the fluid has experienced a total friction force R x . We have five equations (not

including the constraint of Eq. 12.22d) and five unknowns ( p2, 2, s2, V 2, and the heattransfer q that was necessary to maintain isothermal conditions). As we have seen be-

fore, in practice this procedure is unwieldy—we once again have a set of nonlinear,

coupled algebraic equations to solve.

Before doing any calculations, we can see that the Ts diagram for this process

will be simply a horizontal line passing through state . To see in detail what hap-pens to the flow, in addition to Eqs. 12.22, we can develop property relations as func-

tions of the Mach number. For isothermal flow, c ϭ constant, so V 2 / V 1 ϭ M 2 / M 1, andfrom Eq. 12.22a we have

Combining with the ideal gas equation, Eq. 12.22e, we obtain

(12.23)

At each state we can relate the local temperature to its stagnation temperature usingEq. 11.20b,

(11.20b)

Applying this to states and , with the fact that T 1 ϭ T 2, we obtain

(12.24)

To determine the variation in Mach number along the duct length, it is necessaryto consider the differential momentum equation for flow with friction. The analysis

leading to Eq. 12.18 is valid for isothermal flow. Since T ϭ constant for isothermal

flow, then from Eq. 12.18, with dT ϭ 0,

and

(12.25)

Equation 12.25 shows (set dx ϭ

0) that the Mach number at which maximum length Lmax is reached is M ϭ 1/ . Since T is constant, then the friction factor, f ϭ f ( Re),

is also constant. Integration of Eq. 12.25 between the limits of M ϭ M at x ϭ 0 and

M ϭ 1/ at x ϭ Lmax, where Lmax is the distance beyond which the isothermal flow

may not proceed, gives

(12.26) f L

D

kM

kM kM

h

max ln=−

+1 2

22

k

k

f

Ddx

kM d M

kM h

=−( ) ( )1

2 2

4

f

D

kM d x

kM d M

M h

2 2 2

22

1

2=

−

( )

T

T

k M

k M

0

0

22

12

2

1

11

2

11

2

=+

−

+−

21

T

T

k M 0 2

11

2= +

−

p

p

V

V

M

M

2

1

2

1

1

2

1

2

= = =

2

1

1

2

1

2

= =V

V

M

M

1




The duct length, L, required for the flow Mach number to change from M 1 to M 2can be obtained from

(12.27)

The distribution of heat exchange along the duct required to maintain isothermalflow can be determined from the differential form of Eq. 12.22c as

or, since T ϭ constant,

Substituting for dM 2 from Eq. 12.25,

(12.28)

From Eq. 12.28 we note that as M : 1/ , then dq / dx : ϱ . Thus, an infinite rate

of heat exchange is required to maintain isothermal flow as the Mach number ap-proaches the limiting value. Hence, we conclude that isothermal acceleration of flow

in a constant-area duct is only physically possible for flow at low Mach number.

We summarize the set of Mach number –based equations (Eqs. 12.23, 12.24, and

12.27, respectively, renumbered) we can use for analysis of isothermal flow of an

ideal gas in a duct with friction:

(12.29a)

(12.29b)

(12.29c)

12-6 SUPERSONIC CHANNEL FLOW WITH SHOCKS (continued)

Supersonic Diffuser

Analysis of the effects of area change in isentropic flow (Section 12-2) showed that a

converging channel reduces the speed of a supersonic stream; a converging channel is a

f L

D

kM

kM

kM

kM

M

M h

=−

−−

+1 11

2

12

22

22

12

22

ln

T

T

k M

k M

0

0

22

12

2

1

11

2

11

2

=+

−

+−

p

p

V

V

M

M

2

1

2

1

1

2

1

2

= = =

k

dqc T k kM

k M kM

f

Ddx

p

h

=−

+−

−

04

2 2

1

2 11

21

( )

( )

dq c T k

dM c T k

k M

dM p p=

−

=

−

+−

1

2

1

2 11

2

2 0

2

2( )

dq dh c dT c d T k

M p p= = = +−

0 021

1

2

f L

D f

L L

D

f L

D

kM

kM

kM

kM

M

M

h h

h

=−

=

−

−

−

+

max max

ln

1 2

1 112

12

22

22

12

22



12-6 SUPERSONIC CHANNEL FLOW WITH SHOCKS S-47

2 Boundary layers develop rapidly in adverse pressure gradients, so viscous effects may be important or

even dominant. In the presence of thick boundary layers, supersonic flows in diffusers may form com-

plicated systems of oblique and normal shocks.

supersonic diffuser. Because flow speed decreases, pressure rises in the flow direction,

creating an adverse pressure gradient. Isentropic flow is not a completely accurate

model for flow with an adverse pressure gradient,2 but the isentropic flow model with a

normal shock may be used to demonstrate the basic features of supersonic diffusion.

For isentropic flow, a shock cannot stand in a stable position in a converging

passage; a shock may stand stably only in a diverging passage. Real flow near M ϭ 1

is unstable, so it is not possible to reduce a supersonic flow exactly to sonic speed.The minimum Mach number that can be reached at a throat is 1.2 to 1.3.

Thus in real supersonic diffusers, flow is decelerated to M Ϸ 1.3 in a converging

passage. Downstream from the throat section of minimum area, the flow is allowed toaccelerate to M Ϸ 1.4, where a normal shock takes place. At this Mach number, the

stagnation pressure loss (from Eq. 12.41b) is only about 4 percent. This small loss is

an acceptable compromise in exchange for flow stability.Figure 12.22 shows the idealized process of supersonic diffusion, in which flow

is isentropic except across a normal shock. The slight reduction in stagnation pres-

sure all takes place across the shock.

In the actual flow, additional losses in stagnation pressure occur during the super-

sonic and subsonic diffusion processes before and after the shock. Experimental data

must be used to predict the actual losses in supersonic and subsonic diffusers [3, 4].

Supersonic diffusion also is important for high-speed aircraft, where a super-sonic external freestream flow must be decelerated ef ficiently to subsonic speed.

Some diffusion can occur outside the inlet by means of a weak oblique shock system[5]. Variable geometry may be needed to accomplish ef ficient supersonic diffusion

within the inlet as the flight Mach number varies. Multi-dimensional compressibleflows are discussed in Section 12-7, and are treated in detail elsewhere [6, 7].

4

5

321

Flow

4

32

1

5

p1

p01

p04

p1*

T 1

T 0 = constant

T * = constant

T

s

Fig. 12.22 Schematic Ts diagram for flow in asupersonic diffuser with a normal shock.




Supersonic Wind Tunnel Operation

To build an ef ficient supersonic wind tunnel, it is necessary to understand shock be-havior and to control shock location. The basic physical phenomena are described by

Coles in the NCFMF video Channel Flow of a Compressible Fluid . In addition to

choking— sonic flow at a throat, with upstream flow independent of downstream condi-

tions—Coles discusses blocking and starting conditions for supersonic wind tunnels.A closed-circuit supersonic wind tunnel must have a converging-diverging nozzle

to accelerate flow to supersonic speed, followed by a test section of nearly constant area,

and then a supersonic diffuser with a second throat. The circuit must be completed by

compression machinery, coolers, and flow-control devices, as shown in Fig. 12.23 [9].

Consider the process of accelerating flow from rest to supersonic speed in the

test section. Soon after flow at the nozzle throat becomes sonic, a shock wave forms

in the divergence. The shock attains its maximum strength when it reaches the nozzle

exit plane. Consequently, to start the tunnel and achieve steady supersonic flow in thetest section, the shock must move through the second throat and into the subsonic dif-

fuser. When this occurs, we say the shock has been swallowed by the second throat.

Consequently, to start the tunnel, the supersonic diffuser throat must be larger than

the nozzle throat. The second throat must be large enough to exceed the critical area

for flow downstream from the strongest possible shock.

Blocking occurs when the second throat is not large enough to swallow the

shock. When the channel is blocked, flow is sonic at both throats and flow in the test

Fig. 12.23 Schematic view of NASA-Ames closed-circuit, high-speed wind tunnel with sup-porting facilities [9]. (Photo courtesy of NASA.)

A. Dry Air Storage Spheres G. Cooling Tower

B. Aftercooler H. Flow Diversion Valve

C. 3-Stage Axial Flow Fan I. Aftercooler

D. Drive Motors J. 11-Stage Axial Flow Compressor

E. Flow Diverson Valve K. 9- by 7-Foot Supersonic Test Section

F. 8- by 7-Foot Supersonic Test Section L. 11- by 11-Foot Transonic Test Section



12-6 SUPERSONIC CHANNEL FLOW WITH SHOCKS S-49

section is subsonic; flow in the test section cannot be controlled by varying condi-

tions downstream from the supersonic diffuser.

When the tunnel is running there is no shock in the nozzle or test section, so the en-ergy dissipation is much reduced. The second throat area may be reduced slightly during

running to improve the diffuser ef ficiency. The compressor pressure ratio may be

adjusted to move the shock in the subsonic diffuser to a lower Mach number. A combi-

nation of adjustable second throat and pressure ratio control may be used to achieveoptimum running conditions for the tunnel. Small differences in ef ficiency are important

when the tunnel drive system may consume more than half a million kilowatts [10]!

Supersonic Flow with Friction in a Constant-Area Channel

Flow in a constant-area channel with friction is dominated by viscous effects. Even

when the main flow is supersonic, the no-slip condition at the channel wall guarantees

subsonic flow near the wall. Consequently, supersonic flow in constant-area channels

may form complicated systems of oblique and normal shocks. However, the basic be-havior of adiabatic supersonic flow with friction in a constant-area channel is revealed

by considering the simpler case of normal-shock formation in Fanno-line flow.

Supersonic flow along the Fanno line becomes choked after only a short length of

duct, because at high speed the effects of friction are pronounced. Figure E.2 (Appendix

E) shows that the limiting value of Lmax / Dh is less than one; subsonic flows can have

much longer runs. Thus when choking results from friction and duct length is increasedfurther, the supersonic flow shocks down to subsonic to match downstream conditions.

The Ts diagrams in Figs. 12.24a through 12.24d illustrate what happens when

the length of constant-area duct, fed by a converging-diverging nozzle supplied from

a reservoir with constant stagnation conditions, is increased. Supersonic flow on the

Fanno line of Fig. 12.24a is choked by friction when the duct length is La. When

additional duct is added to produce Lb Ͼ La, Fig. 12.24b, a normal shock appears.Flow upstream from the shock does not change, because it is supersonic (no change

in downstream condition can affect the supersonic flow before the shock).

In Fig. 12.24b the shock is shown in an arbitrary position. The shock moves to-

ward the entrance of the constant-area channel (toward higher initial Mach number)

as more duct is added.

Flow remains on the same Fanno line as the shock is driven upstream to state

by adding duct length; thus the mass flow rate remains unchanged. The duct length,

Lc, which moves the shock into the channel entrance plane, Fig. 12.24c, may be cal-

culated directly using the methods of Section 12-3.

When duct length Lc is exceeded, the shock is driven back into the C-D nozzle,

Fig. 12.24d . The mass flow rate remains constant until the shock reaches the nozzle

throat. Only when more duct is added after the shock reaches the throat does the

mass flow rate decrease, and the flow move to a new Fanno line.If the shock position is known, flow properties at each section and the duct

length can be calculated directly. When length is specified and shock location is to be

determined, iteration is necessary.

Supersonic Flow with Heat Addition in a Constant-Area Channel

Supersonic flow with heat addition in a frictionless channel of constant area is shown

in Fig. 12.25a. Assume the channel is fed by a converging-diverging nozzle, supplied

from a reservoir with constant stagnation conditions, and flow is supersonic at state

. Heat addition causes state points to move up and to the right along the Rayleigh1

1

f




line. Fig. 12.25a illustrates the condition in which the heat addition is just suf ficient tochoke the flow. Flow is sonic at the exit, so pe ϭ p* and T e ϭ T *; the heat addition per

unit mass, T ds, is represented by the shaded area beneath the Rayleigh line.

A normal shock involves no heat addition, so T 0 is constant across a shock. Con-

sequently, a shock in the constant-area channel would not change the heat addition

sse

1∫

1

1

*

T 0 p0

p1

p*

p0* p01

T * = constant

T 0 = constant

T 1

La

1 *

T 0 p0

La

Lb

Shock

1 *

T 0 p0

La

Lc

1

1

y

y

x

x

*

T 0 p0

La

Ld

Shock

Process path

T

s

(a) Choked supersonic flow in channel.

1 p1

p*

p0* p01

T * = constant

T 0 = constant

T 1

T

s

(b) Choked flow in channel with shock.

M > 1

M < 1

M = 1

Shock

M = 1

M < 1

1 p1

p*

p0* p01

T * = constant

T 0 = constant

T 1

T 1 p1T

s

(c) Choked flow with shock in nozzle exitplane.

p*

p0* p0 p01

T * = constant

T 0 = constant

T

s

(d ) Choked flow with shock in nozzle; subsonicflow in channel.

Fig. 12.24 Schematic Ts diagrams for supersonic Fanno-line flows with normal shocks.



t e1

1

M > 1 M > 1

M t = 1 M e = 1 M 1 = 2

p01

T 01

t 87

M < 1 M < 1

M t < 1 M 8 = 1 M 7 = 0.35

p01

T 01

t e21

M < 1 M > 1

M t = 1 M e = 1

p01

T 01

4 5 63

M < 1

M < 1 M > 1

M t = 1 M 4 = 0.701 M 6 = 1

M 3 = 1.5 M 5 = 0.407

p01

T 01

p01T 01

T 1 p1

T 0e p0e pe

T e

00

400

800

1200

1 2

Nondimensional entropy, (s* – s)/ R

3

T e m p e r a t u r e ,

( K )

(a) Choked supersonic flow.

0

800

1600

2000

1 2


3


( K )

(c) Choked flow with shock in nozzle; samemass flow rate, but flow shifts to a newRayleigh line.

p01 p02T 01

T 1

p1

p2T 2

T 0e

p0e

pe

T e

00

400

800

1200

1 2


3


( K )

(b) Choked flow with shock at nozzle exitplane.

0

800

400 400

1200 1200

1600

2000

1 2


3


( K )

(d ) Subsonic flow throughout; decreased mass flowrate and flow shifted to another new Rayleighline.

p06

p6

T 5T 4

T 3

T t

T 6

T 06

T 04

p01

p02

p3

T 01

p08

p01T 01

p8

T 8

T 7T t

T 08

t

Fig. 12.25 Schematic Ts diagrams for supersonic Rayleigh-line flows with normal shocks.




required to change the flow state from the inlet condition to choking. When the shock

stands at the channel inlet, Fig. 12.25b, the heat addition needed to reach Mach one

at the exit is the same as in Fig. 12.25a; the shaded areas also must be identical.If more thermal energy is added to flow at the conditions shown in Fig. 12.25b,

the shock will be pushed from the entrance of the constant-area duct back into the di-

verging portion of the nozzle, where the Mach number is lower.

With a shock in the nozzle, conditions at the duct entrance are changed, and heataddition occurs along a different Rayleigh line, as shown in Fig. 12.25c. There is no

change in T 0 or T * across the shock (thus T 03ϭ T 04

and T 3* ϭ T *

4), but the Mach

number downstream changes. Additional subsonic diffusion occurs from state to

the nozzle exit (state ), thus moving the choked condition upward on the Ts plane,

allowing for increased heat addition on the new Rayleigh line. All of these changes

occur at the same mass flow rate, because nozzle throat conditions remain un-

changed.The Mach number immediately upstream from the shock (state ) is less than

M 1 of Fig. 12.25b; the corresponding temperature, T 3, is higher than T 1. Since the shock

strength is reduced, the entropy rise across the shock is less, (s4 Ϫ s3) Ͻ (s2 Ϫ s1). The

subsonic diffusion following the shock results in a lower Mach number and higher

temperature at the duct entrance. Thus M 5 Ͻ M 2 and T 5 Ͼ T 2.When the heat addition rate is increased enough to drive the shock to the nozzle

throat, a further increase in heat addition will result in a decrease in mass flow rate.

The Mach number at the channel inlet is reduced, M 7 Ͻ M 5, and the channel flow

shifts to another new Rayleigh line, as shown in Fig. 12.25d .

Thus for specified mass flow rate, there is a maximum rate of heat addition for

supersonic flow throughout. For higher rates of heat addition, a shock occurs in the

nozzle and flow is subsonic in the constant-area channel, but the exit flow remains

sonic. If the shock position is specified, the heat addition along the Rayleigh line canbe calculated directly. If the heat addition is specified but the shock position or mass

flow rate are unknown, iteration is required to obtain a solution.

Additional consideration of flow with shock waves is given in [11].

3

5

4



S-53

Appendix H

A BRIEF REVIEW OF MICROSOFT

EXCEL

INTRODUCTION

Microsoft Excel is one of several available spreadsheet applications. All of thesespreadsheets work in a very similar fashion, with minor variations in syntax andbuilt-in features, so that most of our discussion of Excel will be directly applicable toother spreadsheet applications.

There are several versions of Excel in widespread use (e.g., Excel XP, Excel

2000, Excel 98, even Excel 97 ); we will use Excel 2000, again with the understandingthat most current versions are 100% compatible, and that most menu items are un-changed from version to version.

In this brief review we will discuss such fundamental features as how to createformulas, how to format cells, how to create graphs, and so on. The review of thesefundamentals will presuppose that you are familiar with the Windows environment,and it will be brief—more detailed explanations are available from a number of

sources, e.g., [1], [2], as well as from Excel’s built-in Help. We will then go on toexplore use of Excel’s Goal Seek and Solver features, and more sophisticated, lesscommonly used features such as use of macros and custom functions using Excel’sbuilt-in Visual Basic for Applications (VBA).

INSTALLING EXCEL FEATURES

We assume that you already have Excel installed on your personal computer (PC)—it comes with many different versions of Microsoft Office. If not, you may purchase a

stand-alone copy or a version of Office that includes it (and note that your collegebookstore will have heavily discounted academic versions available!); simply followthe default instructions for installation.

Although spreadsheets were initially developed for use in business for suchthings as tabulating and analyzing financial data, they are now so computationallypowerful that engineers and scientists are increasingly using them. Because the ma-

jority of users are still people such as accountants, the standard or default installationof Excel does not include some features that engineers and scientists find useful. (In a



S-54 APPENDIX H / A BRIEF REVIEW OF MICROSOFT EXCEL

similar way default installation of the word processor Microsoft Word does not in-clude the equation-creating component, which in Word is called the Equation Editor .)As an engineer you will find useful some features of Excel’s Analysis Toolpak , Visual

Basic for Applications (VBA), and we will use its Solver , so we will install them. The Analysis Toolpak has some useful mathematical features such as regression analysis;VBA is used for creating custom functions; the Solver will be very useful for solving

equations, optimizing a function, and so on.To install these, use the following steps:

Quit all programs.

On the Windows Start menu, point to Settings, and then click Control Panel.

Double-click the Add/Remove Programs icon.

Next, do one of the following:o If you installed Excel with Office, click Office on the Install/Uninstall tab, and

then click Add/Remove.

o If you installed only Excel, click Excel on the Install/Uninstall tab, and thenclick Add/Remove.



INSTALLING EXCEL FEATURES S-55

A window similar to the one below will appear. Click on Add or Remove Features.

A window similar to the one below will appear. Click on the plus sign next toMicrosoft Excel for Windows to expand the tree structure, and then on the plussign next to Add-ins.




A window similar to the following one will appear. You can now select add-ins youwould like. Start by selecting Analysis Toolpak, and select Run from My Com-

puter. Follow the instructions for installing the add-in (you will probably need yourOffice or Excel CD). Repeat this last step for the Solver, and any other add-ins youlike (a useful one is Internet Assistant VBA).



EXCEL FUNDAMENTALS S-57

The add-ins should now be available the next time you use Excel; depending on theversion, you may first have to restart your computer.

To proceed with this brief review, you should have Excel running. It will look some-thing like:

Note that your collections of icons may be different. You can view, hide, or customizetoolbars (collections of icons) using the same procedure used for other Office

applications.

EXCEL FUNDAMENTALS

Excel is an application for manipulating and performing calculations on data that arein a tabular format. Originally developed for financial calculations that lend them-selves to this format, Excel is also an extremely powerful tool for performing engi-neering calculations, whether or not data are tabular.

Entering Data into Cells

To enter data (numeric or text), simply click on a cell or use the arrow keys on thekeyboard to maneuver to the cell of interest, and type in the data. To conclude the en-try, press the Enter key. We can make some comments here:

You can do only limited editing while entering data into a cell: You can use theBackspace key to delete, but cannot use the left arrow to move to earlier digits or let-ters (using this or any other arrow key leaves the cell).




Excel internally recognizes whether a cell’s contents are numeric or text, and bydefault left-aligns text and right-aligns numbers.

Spaces should not be used in entering a number (e.g., 123 456) because Excel willinterpret this as text. Commas are allowed (but are not necessary) (e.g., 123,456).

Excel will use a general number format for numerical data; if the number is too largeto fit in a cell it will change to scientific format, e.g., 1.23E45 (or an error format

######, which can be changed— see below for how to widen columns). The fontand many other formatting properties of cells can be done by selecting the cell orcells and then using the Formatting toolbar:

or by right-clicking and choosing the cell-formatting option. To select several con-tiguous cells, simply drag over them with the mouse (or hold down the Shift keywhile using the arrow keys); for noncontiguous cells, hold down the Ctrl key whileselecting with the mouse. If a text data entry is too large to fit into its cell it will ap-pear to spill into the cell to the right. If you have several cells with text data Excel

will display them as if they were a deck of cards. For example, the words Data ina cell typed into three adjacent cells looks like:

even though Excel still remembers all three cells’ contents. There are several ways towiden a cell to display the full contents—one is to select the cell or cells and click on menu item Format . . . Column . . . Autofit Selection.

Once data have been entered in a cell (i.e., after pressing Enter), to edit the datasimply double-click on them, or select the cell and press F2. You will now be in editmode and can use the left and right arrow keys, the Del and BackSpace key, etc., and

finish by pressing Enter.

ENTERING FORMULAS INTO CELLS

The real power of Excel is its computational abilities. This involves creating formulasin cells for automatic calculation. As an example, suppose we wish to compute anynumber to any (reasonable!) power, e.g., 23, 121.5, eϪ , etc. The crude way is to typein a new formula each time we have two new numbers. Let’s do this first. The keyrule in Excel is that you must use the equals sign ( ϭ ) at the beginning of a formula

so that Excel knows it’s a formula and not numeric or text data. Hence to compute 23,select a cell and type 2^3؍ and press Enter (the exponential is created with ^). The

result (8) will be displayed. The disadvantage of this formula is that we need toretype the formula for each new computation, e.g., for 121.5 we would need to type12^1.5؍ in a cell. A much better approach is to use formulas with cell references.

A cell reference is simply the cell coordinates of the cell. For example, type 2 incell B3 and 3 in cell C3; B3 and C3 are cell references. Now, in cell D3, instead of computing 23 by typing ,2^3؍ we can type B3^C3. Excel؍ interprets this formulato say “take the contents of cell B3 and raise it to the value of cell C3,” and displaysthe result. You should have the following:



ENTERING FORMULAS INTO CELLS S-59

We have a “live” worksheet—we can type new numbers into cells B3 and C3 andget, after pressing Enter, the result immediately (e.g., 12 in B3 and 1.5 in C3 shouldproduce 41.56922 in D3).

Typing formulas with cell references can get tedious and can produce errors, espe-cially when there are many such references. A very useful mouse technique is to startcreating a formula, and instead of typing in a cell reference, use the mouse to click onthe cell you wish to reference. As long as you don’t press Enter, Excel will remain inthis formula creation mode, and you can continue typing the rest of the formula, usingthe mouse to click on additional cells while building the formula. For example, we canselect cell D3 and replace its current contents by typing ,؍ then clicking on cell B3

with the mouse, then typing ^, then clicking on cell C3 with the mouse, and finallypressing Enter (you can also use the arrow keys for this kind of formula building).

This is obviously a trivial example, but very complex spreadsheets can be devel-oped in which we have multiple formulas with multiple cell references. In addition, aworkbook can have multiple worksheets, indicated by the tabs (and you can addmore, rename them, and so on):

A formula on one worksheet can contain references to cells on other worksheets. To

do this you would start creating the formula, and when you needed to reference a cellon another worksheet, still in formula creation mode, click on the appropriate tab,then the cell, and then continue typing the formula.

We can copy and paste cells using any of the standard Windows techniques suchas using menu items, right-clicking, etc., and Excel has additional methods [1], [2].When we copy and paste a cell containing a formula, we need to be careful about how

Excel interprets the pasting. Excel pastes the logic of the formula, not the specific cell

references. For example, if we copy and paste our cell D3 to cell D5, we get (depend-ing on what else you have done to the worksheet), 0 or #NUM!, or some other result,not 8! If you click on cell D5 to see the formula, you will see that instead of the origi-nal formula B3^C3؍ we have .B5^C5؍ Excel has copied the logic, which in thiscase is “take the contents of the cell two cells to the left and raise it to the power of the

contents of the cell one cell to the left”—which in this example doesn’t compute be-cause in cells B5 and C5 we currently have zeros, and 00 is undefined. You only beginto get a result that makes sense if you now type numbers into cells B5 and C5.

In many cases we want this kind of pasting to happen. For example, suppose wewish to compute the mass of air in a container when the air is pressurized over arange of pressures. The formula for this is

(H.1)m pV

RT =




where p is the pressure, is the container volume, T is the air temperature, and R ϭ 287 J/kg-K is the air gas constant. Suppose ϭ 1 m3 and T ϭ 20°C ϭ 293K,and p varies from 10 to 100 kPa in steps of 10 kPa.

Hence,

(H.2)

In this formula p is the pressure in Pa and m will be the mass in kg.We can start a new worksheet (e.g., by clicking on the top left icon on the tool-

bar or by using menu item File . . . New) and type in column headings for the pres-sure and mass, and enter the pressure data (and note that there are ways in Excel toautomatically fill in the p sequence, and that we’ve done some formatting, and addedborders):

m p= ×1287 293

V

V

We can now create Eq. H.2 in cell C4, for the mass due to the first pressure: The for-mula should be 1/287/293*B4. (Note that Excel؍ follows the conventional prece-dents of exponentiation, then multiplication/division, then addition/subtraction, etc.)We could then create similar formulas for the other cells; e.g., cell C5 should have1/287/293*B5. A better approach is simply to copy cell C4 into cells C5؍through C13. The result is (after a little formatting):




We have just introduced what is called the idea of a relative reference— the referenceis to a cell relative to the cell containing the formula. Sometimes we want instead anabsolute reference. This is a reference to a fixed cell, so that if the formula is copiedto another cell the reference does not change. An example in which this would be de-sired is as follows: Suppose we wanted to be able to repeat the above computationsfor a different temperature or container volume. We want to have cells containingthese so that we can easily change them. Let’s put them in cells E4 and F4 (with ti-tles— can you create “m3”?):




We now need the following formula in cell C4 (corresponding to Eq. H.1):B4*E4/287/F4. The problem now is that if we copy this formula to the other؍cells in the mass column, we will get errors (try it, then click Undo!): The formulalogic says that each cell should obtain the volume and temperature data two andthree cells to the immediate right, e.g., cell C5’s formula as pasted would be,B5*E5/287/F5. The reference to B5 (the pressure) is correct؍ but E5 and F5 refer

to blank cells! What we wish to have in cell C5 is ,B5*E4/287/F4؍ i.e., we wish tocopy and paste our formula, but not have the E4/287/F4 part change. To do this wemake these references absolute. Before copying our original cell C4, we edit it (e.g.,by double-clicking on it) and place $ signs as follows: .B4*$E$4/287/$F$4؍

Excel understands that any cell reference that has the $ sign in front of it is absolute,and does not change it when copied and pasted. Note that E4, for example, has two$ signs, so that neither E nor 4 will change (strictly speaking, in this example E doesnot change anyway, so we don’t really need to $ it). If this formula is copied andpasted to the mass column, we get the correct results: Each computation obtains thepressure from the cell to the left, and the volume and temperature from cells E4 andE5. We can now change the temperature to, say, 100°C ϭ 373K, and obtain new data(don’t worry about the circle and square shown below— we’ll discuss them shortly):

Note that Excel will automatically create the $ signs in a reference, if you press the

F4 key during either formula creation or editing.Our last comment on formula creation is that Excel has many mathematical

functions built in, but you need to use the correct syntax and spelling. For example,to create 2 , you must type 2*pi(). Many functions are obvious, e.g., to obtainsin( /4) type sin(pi()/4); others are less so (e.g., !). Note that trigonometricfunctions in Excel use arguments in radians by default.

You can always search for functions while creating a formula by clicking on thePaste Function icon (circled above) to get a useful menu:




Let’s see how we can now graph the data we computed.

Creating Graphs

Excel has powerful graphing features. There are a number of ways to access these,but the main decision is whether you select the data to graph first and then go intographing mode, or vice versa. Both approaches work, but beginners will probablyfind it easier to start graphing mode without first having selected the data. As in allWindows applications, there are several ways to perform any action; here the sim-plest is to click on the Graph Wizard icon (squared in the spreadsheet image shownabove). If we do this we get the first of four self-explanatory windows:

Most of the time engineers will plot x - y data using XY (Scatter) (not Line, whichforces data to appear uniformly on the x axis!). There is nothing to preview yet soclick Next to get:




Click in Data Range and then drag the entire window out of the way a little as neces-sary so that you can select the range to plot (or alternatively click on the little icon inData Range on the right to go directly to the worksheet); select the cells containingour data (cells B4 to C13); and proceed with the rest of the steps. We end up with agraph something like:

It is not too interesting! It can be improved a lot by customizing the trace, adding ti-tles, etc. To do this we can right-click on any graph feature (e.g., a data point) for amenu of options, and we can also use the Formatting toolbar, and end up with some-thing like:




We have created a basic x - y graph: Excel can create many sophisticated graph typessuch as bar charts, pie charts, radar plots, and surface plots. In addition, it can be usedto fit a curve to experimental data. We will not do this here, but it involves: plotting thedata on an x - y plot; right-clicking any point on the curve to get a menu that includesAdd Trendline . . . ; selecting this (leading to the window shown next); choosing the

type of curve (Linear, Logarithmic, etc.), and specifying options with the Optionstab; clicking OK and observing the results.

This concludes our introduction to Excel basics. As you use Excel you will dis-cover many features as the need arises, and the references have additional help. We

mention one final feature to illustrate the usefulness of some of these features. Weearlier copied and pasted a formula from a single cell to a column using conven-tional Windows copy and paste techniques. Excel has an additional convenientshortcut for this: Simply click once on the cell to be copied, then move the mouse(without clicking) over the cell’s lower right corner until the Excel pointer changesto a “fill handle” (؉); you can then press and drag down a column (or a block of cells) to immediately copy the cell contents. Even neater is that if you have two ormore cells with contents that make a sequence (e.g., 12, 24, or Jan, Feb) and se-lect them, then move the mouse to the lower right to get the fill handle (؉) and fi-




nally drag down or across, Excel will fill with the sequence (e.g., 36, 48, . . . etc.,or Mar, Apr, . . . etc.!).

MORE ADVANCED FEATURES

Excel has a number of more advanced features. We will consider several of them:Goal Seek , Solver , use of macros, and creation of custom functions.

Goal Seek and Solver

Many problems in engineering end in an equation or equations to be solved for anunknown or unknowns. In some cases the equation or equations are not directly solv-able, and we need to use numerical techniques to obtain a solution. For example, sup-pose we wish to solve the following equation for x :

(H.3)

We cannot solve explicitly for x . We could make guesses for x until we converged on

a solution, or we could use one of several classic methods to converge to a solution ina more systematic way. Most scientific calculators, and certainly all spreadsheets, in-cluding Excel, also have equation solution techniques built in.

To use Excel’s Goal Seek feature, we first need to convert Eq. H.3 into a func-tion, the root of which we need to find:

(H.4)

The problem then becomes, find the root (or roots) of f ( x ). While not necessary forusing Goal Seek , the best procedure for this is first to plot the function over a rangeof x values to see the locations of the roots. After some experimentation with x

ranges, we obtain, using Excel,

f x e x x

( ) = −2 2

e x x 2 2=



MORE ADVANCED FEATURES S-67

This dialog box is almost self-explanatory: Click on the Set cell: window, then on thecell we wish to find the root of (C23); click on the To value: window and type 0;click on the By changing cell: window and click on the cell we wish to vary to findthe root (in this example, B23); click OK. If Excel can find a root, it tells us so:

and we have:

It looks like we have roots at approximately x ϭϪ0.8 and x ϭ 1.4. We would like moreaccurate answers. To use Goal Seek , set up another table containing a single x and f ( x ):

Note that cell B23 contains 0.0, an arbitrary starting point, and cell C23 contains theformula .EXP(B23/2)-B23^2؍ Goal Seek is accessed by using menu item Tools

. . . Goal Seek . . . , which brings up the following dialog box:




We can repeat this process by typing in another guess value for x (try 1) to find thesecond solution ( x ϭ 1.4296). When using this method you sometimes have to tryseveral guess values in order to find all the roots (in this example there is a thirdroot . . . can you find it?)— and of course, as with all numerical methods, sometimesit simply fails.

Note that we can also use Goal Seek to find an x at which f ( x ) attains any given

value, e.g., when does f ( x ) ϭ 2?Goal Seek is quick and easy. A more powerful feature is Solver . Solver is ac-

cessed by using menu item Tools . . . Solver . . . , which brings up this dialog box:

As for Goal Seek , we can still select an x cell that we will allow Excel to vary inorder to make a formula-connected f ( x ) cell attain a certain value—but we can domuch more! We can

Find the minimum or maximum of f ( x ).

Find the roots (or any other value) of a function of several variables, e.g., f ( x , y), f ( x , y, z) etc.

Do all of the above while applying constraints, e.g., while x Ͼ 2 and y Ͻ x 2.

The window is fairly self-explanatory, so we will not go into the details here. We willdemonstrate its use by finding the maximum of our function f ( x ) in Eq. H.4, graphedearlier. In the Set Target Cell: window select cell C23; for the Equal To: selectMax; click on the By Changing Cells: window and click on cell B23; click Solve.We end up with:

Finally, let’s consider an example of two variables:

f ( x , y) ϭ x 2 Ϫ 6 x ϩ y2 Ϫ 5

Suppose we wish to find its minimum value. We can set up a new worksheet, withcells for x and y (guess values of 0), and a formula cell (in the worksheet below theformula is :(B3^2؊6*B3؉C3^2؊5؍




Then all we need to do is use Solver with the following information:

The solution is f ( x , y) has a minimum value of Ϫ14 when x ϭ 3 and y ϭ 0.

Macros

A macro is nothing more than a set of instructions for automating a sequence of ac-tions in Excel (or Word or any number of other applications). We will illustrate theuse of macros by considering a fairly complicated but ultimately very useful exam-ple: We will create a worksheet that can be easily used to numerically evaluate a defi-nite integral. The formula we will use to approximate the definite integral is a form of Simpson’s Rule:

(H.5)

where N (which must be even) is the number of segments of size that therange (b Ϫ a) is divided into, and x 0 ϭ a with x iϩ1 ϭ x i ϩ ⌬ x , leading to x N ϭ b.Within limits, the accuracy of the evaluation increases with N . As a good check, wewould like to have the worksheet automatically evaluate the integral for N ϭ 10, 20,40, and 60 segments.

Before we automate this task with a macro for use with an arbitrary integral,let’s see how we create such a worksheet entirely from scratch for a specific integral.Suppose we wish to numerically evaluate the following:

⌬ x

b a

N =−( )

f x f x f x N N + + ⋅ ⋅ ⋅ + + ]−( ) ( ) ( )2 44 1

I f x dx x

f x f x f x f x a

b

= ≈ + + +[∫ ( ) ( ) ( ) ( ) ( )⌬

34 2 40 1 2 3




(H.6)

Obviously we can easily evaluate this analytically and find I ϭ 2; we can use this re-sult to test whether or not our worksheet is working. The following worksheet showsthe calculations for N ϭ 10, and part of the calculations for the other N values:

I x dx = ∫ sin( )0

Let’s discuss in detail the construction of the calculations for N ϭ 10 (the others arevery similar). In cell C4 we compute the value of ⌬ x : The cell contains the formula,pi()/10. The first x value is generated by entering the lower integration limit (0)؍and the other x values are created with formulas that add ⌬ x to each previous x value;e.g., cell C8 has ,C7؉$C$4. The f ( x ) column contains the integrand formula؍ with

x assumed to be in the cell to the left; e.g., cell D7 has؍sin(C7). The w columncontains the “weightings” 1, 4, 2, 4, 2, . . . 4, 1 used in Eq. H.5, and the Prod . columncontains the products of the f ( x ) and w columns (e.g., cell F7 contains the formulaD7*E7). We now need to sum the؍ Prod. column, shown in cell F18 (it contains theformula ,(SUM(F7:F17)؍ and this is the series in brackets in Eq. H.5. Finally,the integral I is given in cell D21 (it contains the formula C4/3*F18). The value؍

we obtain for the integral, 2.00011, is very close to the exact result of 2.This entire procedure needs to be done for the other computations, for N ϭ 20,

40, and 60 (with obvious changes, such as the computation of ⌬ x ). Your four valuesof I for Eq. H.6 should agree closely.

We now wish to modify the workbook so we can evaluate other integrals (differ-ent integrands and limits).

First, we can tidy up the entire workbook by relegating these calculations to asecondary “Calculations” worksheet (double-click the tab to edit its name; click and




In this worksheet we enter the formula for the integrand, assuming x is to the left (in thepresent case ,(sin(C6)؍ and the lower and upper limits, as shown. The Calcu-lations sheet must be modified so that ⌬ x in four locations is now automaticallycalculated from the limits on the Results sheet (e.g., cell C4 now contains,((Results!D11-Results!D10)/10؍ and the first x value in each column isnow computed from the lower limit (these cells contain the formula .(Results!D10؍

The workbook is now almost finished. To use it to evaluate the integral, we mustcopy the formula for f ( x ) in cell C6 from the Results sheet and paste it to the four f ( x )columns in the Calculations sheet. Finally, the four cells in the Results sheet showingthe integration results contain formulas for picking up these results from the Calcula-tions sheet (e.g., cell F15 contains the formula Calculations!D21). If all goes؍according to plan, the four evaluations appear as shown above.

It took quite a bit of effort, but we now have a workbook for numerically evalu-ating definite integrals! The workbook is fairly easy to use to evaluate an integral. Weuse the following steps:

1. Enter the integrand formula in cell D6, assuming x is to the immediate left.

2. Enter the lower and upper limits.

3. Copy the formula in cell D6 to the four f ( x ) columns in the Calculations sheet, and returnto the Results sheet to view the answers.

drag the worksheet tab to move the worksheet), and create a user-friendly “Results”worksheet as shown next:




This procedure works: You can test whether your worksheet is working correctly bytrying the following integral:

You should get the following results (the exact result is e Ϫ 2 ϭ 0.71828):

10 segments: 0.7183020 segments: 0.7182840 segments: 0.7182860 segments: 0.71828

We are now (finally!) ready to see how we can use a macro to automate this proce-dure. Steps 1 and 2 are straightforward, but step 3, involving a lot of copying andpasting from one sheet to another, is tedious and prone to error. We can automate thisstep by creating some Visual Basic code. If you are familiar with Visual Basic youcan write your own code, but Excel has a built-in macro-recording feature that willgenerate the code for you. The idea is that you switch on macro recording, and thenperform exactly the actions you wish to record that accomplish a task, and finally

switch off recording. Let’s do this now.Click on menu item Tools . . . Macro . . . Record New Macro . . . to get the

following window:

I x e dx x = ∫ 2

0

You may type in a name if you wish (e.g., Integration) and even assign a shortcut key(such as Ctrl ϩ i), then click OK. You will now be in macro-record mode, as indi-cated at the lower left of the screen:

You should be careful now because every action will be recorded! Select and copycell D6 on the Results sheet (try Ctrl ϩ C); switch to the Calculations sheet, selectthe f ( x ) column for N ϭ 10, and paste (try Ctrl ϩ V); select each of the other f ( x )columns and paste; to end up at the Results sheet, click on it and select a conven-ient cell, e.g., the lower limit cell D10. Now stop recording, by selecting menu itemTools . . . Macro . . . Stop Recording. . . . We have our macro! Your code shouldlook something like the following (to see it, use menu item Tools . . . Macro . . .

Macros . . . and select your code for editing; close the window when you’re done):




The code is almost self-explanatory and shows the steps used to accomplish our task.If you make a procedural error you can simply stop recording and use menu itemTools . . . Macro . . . Macros . . . where you can delete the macro and start over.

How do we use this macro? Whenever we enter a new integrand in cell D6, we canrun the macro by using menu item Tools . . . Macro . . . Macros . . . and selecting andrunning our macro. The macro will do all the copying and pasting for us! A more con-venient procedure is to assign an accelerator keystroke combination to run the macro (thecode above runs whenever Ctrl ϩ i is typed). An even better idea is to create an objectsuch as a picture or WordArt piece of text (click menu item Insert . . . Object . . . andchoose WordArt ); we can then right-click on it and choose Assign Macro . . . and assignour macro to the object. Then, whenever you click on the object the macro will run!

Before finishing this example, we can protect the Results and Calculationssheets (after unlocking cells we wish to be able to copy and paste to, or enter datainto) so a future user is not able to wreak havoc with the formulas we have “hard-wired,” and hide the grid lines, etc., and end up with something like the sheet shown

next. To use this worksheet we use the following steps:1. Enter the integrand formula in cell D6, assuming x is to the immediate left.

2. Enter the lower and upper limits.

3. Click on the “Integrate!” icon (or type, if you created them, the shortcut keys, e.g., Ctrlϩ i).

The user does not need to worry about complicated copying and pasting procedures,and has a very convenient integrator! Most of the needed formulas are “hard-wired,”and the needed copying and pasting is done by our macro.

Sub Integrate()’’Integrate MacroMacro recorded 12/6/02 by Philip J. Pritchard’

’Keyboard Shortcut: Ctrl ϩ i’

Range(“D6”).SelectSelection.CopySheets(“Calculations”).SelectActiveWindow.SmaIlScroll ToRight: ϭϪ5Range(“D7:D17”).SelectActiveSheet.PasteRange(“J7:J27”).SelectActiveSheet.PasteRange(“P7:P47”).SelectActiveSheet.Paste

ActiveWindow.SmallScroll ToRight: ϭ8Range(“V7:V67”).SelectActiveSheet.PasteSheets(“Results”).SelectRange(“D10”).SelectApplication.CutCopyModeϭ False

End Sub




In this example we saw how to create a simple macro. The potential of macros isobvious—you can include automatic graph generation, file saving, etc.— in fact, virtu-ally any actions you can perform in Excel can be recorded and hence made into a macro.

Customized Functions

Excel comes with a plethora of useful built-in functions, including many mathemati-cal, logical, database, and statistical functions. To see the list of available functions,click the menu item Insert . . . Function . . . (or click on the Paste Function icon wediscussed earlier) to get a window like the following:




Note that we have scrolled down to the collection of Engineering functions, whichMicrosoft has apparently decided should contain not only obviously engineering-ori-ented functions such as CONVERT(number,from_unit,to_unit) (for performing unitsystem conversions such as lb to kg), but also such things as complex math, and ad-vanced functions such as Bessel’s functions— Excel’s Math & Trig. group does notcontain these last two items but does contain equally advanced functions such has hy-

perbolic functions! When searching for a function in Excel, it is a good idea to searchunder All functions so you do not accidentally fail to find a built-in function.

Even with this impressive collection, there will be occasions when you wish todefine your own functions. This can be easily done in Excel. We will illustrate themethod for defining a customized function by considering the following example.Suppose we wish to compute the friction factor f for pipe-flow problems:

(H.7a)

(H.7b)

where is the pipe relative roughness and Re is the Reynolds number of the flow. If Re Ͻ 2300, the computation is simple; for Re Ն 2300 we have a slight problem, be-cause f is implicit in Eq. H.7b. To solve Eq. H.7b for f we need to iterate—given

and Re, we guess a value for f (e.g., 1) on the right side, and compute f on the left.This value is fed back into the right side, and the computation is repeated, until con-vergence to a suitable accuracy is obtained.

Let’s define a function in Excel that computes f from given values of and Re, us-ing Eq. H.7a or Eq. H.7b as needed. (Note that we could just define a function thatsolves Eq. H.7b, and then use a cell containing an If function to decide which equation touse, but it is neater to define a function that incorporates both equations automatically.)

In Excel we use its built-in Visual Basic editor to write the code for a function.To access this, use menu item Tools . . . Macro . . . Visual Basic Editor, to get a

window looking something like:

12 0

3 7

2 512300

f Re f Re= − +

≥. log.

.(Turbulent flow)

f Re

Re= <64

2300 (Laminar flow)




Note that your window may be slightly different, showing more or fewer subwin-dows—you can make subwindows visible or not using menu item View. You may ormay not have a Module subwindow visible—if not, create a module (a storage area forcode) using menu item Insert . . . Module.

We now need to create code for computing f from Eq. H.7a or H.7b. A possibleflowchart for the logic of the computation is:

Note that we have two decision boxes: The first is obvious; the second involves iterat-

ing Eq. H.7b until a suitable convergence has been obtained. We will iterate Eq. H.7buntil the change in f is less than 0.1%.

We need to write Visual Basic code for this flowchart. In this introduction to Excel, we cannot discuss in detail the Visual Basic syntax and grammar we must useto accomplish the tasks described in the flowchart, but there are several texts avail-able, e.g., [3]. However, next we present the code we need, which you can now typeinto the module subwindow (and note that as we type the first line the Declarations

tab changes to f, our new function):




The code works as follows:

We first perform the test for Reynolds number.

If Re Ͻ 2300, we compute f from Eq. H.7a and go to the end of the function.

If Re Ն 2300, we go to the Else line and define the error (Er) and initial valuesfor f n (oldf) and ⌬ f (deltaf), the error between successive values of f . Be-tween the Do While and Loop lines we iterate to converge on an accurate value of

f . For each iteration:

o We compute a value for f nϩ1 (newf) from the current value of f n and Re and

(represented by Re and e, respectively). [Note that we multiply the log by0.434294482 because Visual Basic treats Log as a natural log, so we convertto base 10 by multiplying by log(l0)/ln(10).]

o Next we compute the change ⌬ f (deltaf) in the value of f , and update thevalue of f .

o Finally, we return to the Do While line to perform the accuracy test: If ⌬ f/f Ն0.1% we repeat the iteration; if ⌬ f/f Ͻ 0.1% we finish iterating and output thevalue of f .

We can now close the Visual Basic window and return to the spreadsheet. We now havea new function f that computes Eq. H.7a or H.7b as appropriate! We can test the func-tion for laminar ( ReϽ 2300) and turbulent ( ReՆ 2300) flows, e.g., something like:




The procedure we just outlined showed how to create a custom function in Excel. Wedemonstrated this with a function that was fairly complicated—Eqs. H.7a and H.7b,with two decision branches needed—but the procedure can obviously be used fordefining simple functions, or ones of unlimited complexity.

REFERENCES

1. Gottfried, B. S., Spreadsheet Tools for Engineers. New York: McGraw-Hill, 2000.

2. Bloch. S. C., Excel for Engineers and Scientists. New York: John Wiley, 1999.

3. McKeown, P. G., and C. A. Piercy, Learning to Program with Visual Basic, 2nd ed. NewYork: John Wiley, 2001.

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