notes final corrd 8-9-11-dist

7/30/2019 Notes Final Corrd 8-9-11-Dist

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Notes for the 2nd revised edition ofTRANSPORT PHENOMENA

by

R. B. Bird, W. E. Stewart, and E. N. Lightfootby R. B. Bird

9 Aug 2011

These "Notes," begun in 2009, are intended for use by studentsand instructors to fill in the missing steps in some of the derivationsin the text. Comments and corrections will be greatly appreciated.Also, if there are places in the text where you feel additionalexplanation is needed, it would be appreciated if you would let usknow.

Many of these notes involve the Leibniz formula fordifferentiating integrals, the hyperbolic functions, the error function,Taylor series, and the gamma functionsall topics for which manyundergraduates have received inadequate instruction. These topicsare all reviewed in Appendix C.

Pages in the text for which "Notes" have been prepared:

p. 5 Conservation laws in binary collisionsp. 18 Evaluation of stress-tensor componentsp. 26 Dimensional consistency in fluid dynamicsp. 35 Convective momentum fluxp. 50 Flow in tubes with elliptical cross sectionp. 51 Flow of kinetic energy in tubesp. 52 Conduits with circular and triangular cross sectionsp. 54 Velocity distribution in annular flowp. 55 Limiting cases of annular flowp. 58 Flow of immiscible fluids

p. 59 Flow around a spherep. 78 Normal stresses at solid surfacesp. 81 Equation of change for mechanical energy

p. 82(i) Proof that :v( ) is positive for Newtonian fluidsp. 82(ii) Conservation equation for angular momentump. 86(i) Vector identity needed for the Bernoulli equationp. 86(ii) Another way to look at the Bernoulli equation


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p. 90(i) Tangential flow between cylinderspressure distributionp. 90(ii) Torque balancep. 117 Slope of the complementary error functionp. 121 Fluid motion near an oscillating platep. 123 Equation for the stream function

p. 125 Alternative method of getting Stokes' lawp. 139 The Falkner-Skan equationp. 154 Average velocity in turbulent flow in tubesp. 170 Turbulent flow in a circular jetp. 198 Derivation of macroscopic balancesp. 199 Efflux from a spherical tankp. 202 The lawn sprinklerp. 218 Unsteady flow from a cylindrical tankp. 241 The Weissenberg-Rabinowitsch equationp. 242 Power-law flow in circular tubesp. 248 Viscoelastic flow near an oscillating platep. 250 The corotational Maxwell modelp. 255 Polymer flow analyzed with a FENE-P modelp. 259 The Casson equationp. 275 Dimensional consistency in heat transferp. 286 Enthalpy of an ideal monatomic gasp. 299 Temperature profile in flow with viscous heatingp.309 Checking the cooling fin solutionp. 315 Temperature profile in tube flowp. 337 Alternative equation of change for temperaturep. 341 The equation of change for entropyp. 343 Tangential annular flow with viscous heatingp. 346 Temperature profile for transpiration coolingp. 352 Stationary shock wave velocity distributionp. 375 One-dimensional time-dependent heat conductionp. 377 Two solutions for the slab heating problemp. 379 Unsteady heat conduction with sinusoidal heatingp. 386 Steady-state potential flow of heat in solidsp. 388 Boundary layer flow with heat transferp. 413 Turbulent flow in tubes with heat transferp. 415 Turbulent flow in circular jets with heat transferp. 454 Derivation of macroscopic energy balancep. 494 Planck's radiation law and Wien's displacement lawp. 529 Dimensional consistency in diffusionp. 534 Binary formulas from multicomponent formulasp. 535 Two formulations of Fick's law of diffusionp. 547(i) Diffusion through a stagnant gas film


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p. 547(ii) Taylor expansion of diffusion problem resultp. 555 Diffusion with chemical reactionp. 557 Simplifying Eq. 18.4-18p. 563(i) Diffusion with solid dissolutionp. 563(ii) Evaluation of mass flux from concentration profiles

p. 565 Verification of solution of diffusion problemp. 584 Form of diffusion equation in molar unitsp. 585 Diffusion with convection and chemical reactionp. 589 Energy equation for multicomponent mixturesp. 590 Simplification of the combined energy flux ep. 591 Euler's theorem for homogeneous functionsp. 615 Time-dependent evaporation of a liquidp. 622 Diffusion with time-dependent interfacial areap. 626 Diffusion with chemical reactionp. 628 Forced convection from flat platep. 692 Interaction of phase resistancesp. 766 Driving force in multicomponent diffusionp. 767 Simplification of multicomponent diffusion resultp. 768 Relating Maxwell-Stefan and Fick diffusivitiesp. 769 Illustrating interrelations between diffusivities


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Note to p. 5

Section 0.3 is important for emphasizing some of the basicconcepts and definitions. Here we work through some of the missing

steps in Section 0.3, going from Eq. 0.3-3 to Eq. 0.3-4, and from Eq.0.3-5 to Eq. 0.3-6.

(a) In Eq. 0.3-3, replace rA1 by rA +RA1 and make analogous

replacements for rA2, rB1, and rB2 . We also let mA1 = mA2 =12mA . With

these substitutions, we then get:

12mA rA + RA1( ) + 12mA rA + RA2( ) + 12mB rB + RB1( ) + 12mB rB + RB2( ) (1)

But, according to the drawing in Fig. 0.3-2, RA1 = RA2 and,analogously, RB1 = RB2, so that

mA rA +mB rB = mA rA +mB rB (2)

This is the law of conservation of momentum in terms of the

molecular masses and velocities.

(b) We start with Eq. 0.3-5, and replace rA1 by rA + RA1 as above. We

also let mA1 = mA2 =12mA . Then Eq. 0.3-5 becomes:

1212mA rA rA( ) + 2 rA RA1( ) + RA1 RA1( )( )

+ 1212mA rA rA( ) + 2 rA RA2( ) + RA2 RA2( )

+ A

+ 1

212mB rB rB( ) + 2 rB RB1( ) + RB1 RB1( )( )

+ 1212mB rB rB( ) + 2 rB RB2( ) + RB2 RB2( )

+ B

= 12

12mA rA rA( ) + 2 rA RA1( ) + RA1 RA1( )( )

12mA rA + RA1( ) +

12mA rA + RA2( ) +

12mB rB + RB1( ) +

12mB rB + RB2( ) =


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+ 1212mA rA rA( ) + 2 rA RA2( ) + RA2 RA2( )

+ A

+ 1

212mB rB rB( ) + 2 rB RB1( ) + RB1 RB1( )( )

+ 1212mB rB rB( ) + 2 rB RB2( ) + RB2 RB2( ) + B (3)

The single-underlined terms just exactly cancel the doublyunderlined terms in the following line, because RA1 = RA2 and also

RB1 = RB2 . Hence we get

12mA rA rA( ) +

12mA1

RA1 RA1( ) +

12mA2

RA2 RA2( ) + A

+12mB rB rB( ) +

12 mB1

RB1

RB1( ) +

12mB2

RB2

RB2( ) + B

= 1

2mA rA rA( ) +

12mA1 RA1 RA1( ) +

12mA2 RA2 RA2( ) + A

+ 1

2mB rB rB( ) +

12mB1

RB1 RB1( ) +

12mB2

RB2 RB2( ) + B (4)

In the first line of the equation above, the terms have the followingsignificance: Term 1 is the kinetic energy of molecule A in a fixedcoordinate system; Term 2 is the kinetic energy of atom A1 in acoordinate system fixed at the center of mass of moleculeA; Term 3 isthe kinetic theory of atom A2 in a coordinate system fixed at thecenter of mass of molecule A ; Term 4 is the potential energy ofmoleculeA as a function of rA2 rA1 , the separation of the two atomsin moleculeA. The sum of terms 2 to 4 we call the "internal energy"

uA of molecules A, and Eq. 4 may be rewritten in the form of Eq. 0.3-6.

This discussion of the collision between two diatomicmolecules is interesting, for several reasons. It shows how the idea of"internal energy" arises in a very simple system. We encounter thisconcept later in 11.1 where the terms "kinetic energy" and "internalenergy" are used in connection with a fluid regarded as a continuum.When the fluid is regarded as a continuum, it may be difficult tounderstand how one goes about splitting the energy of a fluid intokinetic and internal energy, and how to define the latter. Inconsidering the collision between two diatomic molecules, however,the splitting is quite straightforward.


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Another point is that, having seen the need for splitting theenergy into two parts, one might be led to ask: why don't we need tosplit the momentum into two parts in a similar way? Here again, forthe collision of diatomic molecules, the need for dividing themomentum into two parts is not necessary.

The subject of transport phenomena is built up on the laws ofconservation of mass, momentum, angular momentum, and energy.The application of these laws to the system of two colliding diatomicmolecules is relatively straightforward. However, when applyingthem to a moving fluid, some notational problems arise associatedwith the necessity of dealing with fluid bodies in three dimensions.


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Note to p. 18

In Fig. 3B.2 there is shown a duct with cross-section of anequilateral triangle. The height of the triangular cross-section is H,

and the side length is 2H 3 . We want to evaluate the viscous stresstensor components for the incompressible flow in the z-direction, forwhich the velocity in the z-direction is given as a function ofx andyin Eq. 3B.2-1:

vz x, y( ) =

P0 PL( )4LH

y H( ) 3x2 y2( ) (1)

and vx = 0 and vy = 0 . Here L is the length of the duct (which goesfrom z = 0 to z = L ) and P0 PL is the difference in modifiedpressure between the ends of the duct. What are the stresses at thesurfacey = Haccording to Eq. 1.2-6?

For the velocity distribution given above, the nonzerocomponents are yz = zy and xz = zx . From Eq. 1.2-6, we get:

yz = vzy

= P0 PL( )

4LH

y

3x2y y3 3Hx2 + Hy2( )

=P0 PL( )

4LH3x2 3y2 + 2Hy( ) (2)

xz =

vzx

= P0 PL( )4LH

x3x2y y3 3Hx2 + Hy2( )

=P0 PL( )

4LH6x y H( )( ) (3)

At the surfacey = H:

yz y=H

=P0 PL( )

4LH3x2 H2( ) ; xz y=H = 0 (4,5)


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Note to p. 26

It is very important to make a habit of checking equations fordimensional consistency. Show that the following equations in the

text are dimensionally consistent: Eq. 1.4-14, Eq. 1.5-11, and Eq. 1.7-2.Do this by replacing the symbols in the formulas by the dimensionscorresponding to the symbols in the table beginning on p. 872. Omitany numerical factors that appear.

(a)M

Lt

=

M( )ML2

t2T

T( )

L2( ) ( )(1)

(b)

M

Lt

=

1

moles

ML2

t

L3

moles

(2)

(c)

M

Lt

2

=M

Lt

2

+M

L

3

L

t

L

t

=M

Lt

2

+M

Lt

2

+M

L

3

L

t

L

t

(3)

In each case, dimensional consistency is found. In (c) the unit tensor is a dimensionless quantity.


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Note to p. 35

In Fig. 3B.7 we show the flow into a slot of width 2B. Far awayfrom the slot, the velocity components are given by Eqs.3B.7-2 to 4.

The convective momentum fluxtensor is given by vv, where vv is thedyadic product of v with v. The components of vv in Cartesiancoordinates are shown in the last three columns of Table 1.7-1.

a. What are these components for the flow in Problem 3B.7?b. What is the convective momentum flux through a plane

perpendicular to the x axis?

a. When we make use of the Cartesian components of thevelocity given in Eqs. 3B.7-2 to 4, we get

vv( )xx = vxvx = +1

2w

W

2x6

x2 +y2( )4

(1)

vv( )xy = vxvy = +1

2w

W

2x5y

x2 +y2( )4

(2)

vv( )xz = vxvz = 0 (3)

vv( )yx = vyvx = + 1 2wW

2

x

5

yx2 +y2( )

4 (4)

vv( )yy = vyvy = +1

2w

W

2x4y2

x2 +y2( )4

(5)

vv( )

yz= vyvz = 0 (6)

vv( )zx = vzvx = 0 (7)

vv( )zy = vzvy = 0 (8)

vv( )zz = vzvz = 0 (9)

b. For a plane perpendicular to the x axis, the vector n is x .The components of the momentum flux are then


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x vv x = vxvx = +1

2w

W

2x6

x2 +y2( )4

(10)

x vv y = vxvy = +

1

2w

W

2x5y

x2 +y2( )4 (11)

x vv z = vxvz = 0 (12)

Then x vv x is the amount of x momentum flowing per unit time

through a unit area of surface perpendicular to the x axis, x vv yis the amount ofy momentum flowing per unit time through a unit of

surface perpendicular to the x axis, and x

vv z

is the amount of z

momentum flowing per unit time through a unit of surfaceperpendicular to the x axis.

Verify that the units of the expressions on the right sides of Eqs.10 and 11 do indeed have the units of momentum per area per time.


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Note to p. 50

Information about the flow in tubes of elliptical cross-section, andcomparison with the flow in a circular tube. Label the semi-major axis

of the ellipse as a and the semi-minor axis of the ellipse as b. Thecross-sectional area of the elliptical tube is A = ab .

The velocity distribution for laminar axial flow in a tube ofelliptical cross-section is

vz =P1 P2( )a

2b2

2L a2 + b2( )1

x

a

2

y

b

2

(1)

as given by J. Happel and H. Brenner, Low Reynolds NumberHydrodynamics, Prentice-Hall, Englewood Cliffs, NJ (1965), p. 38.When a = b = R, this velocity distribution reduces to Eq. 2.3-18 of BSL-

2e, since x2 +y2 = r2 .To get the mass rate of flow, we integrate the velocity

distribution over the cross section, thus

w = 4 P1 P2( )a

2b2

2L a2 + b2( ) ab 1 2 2( )dd0

12

01

Here the dimensionless variables = x a and =y b have beenintroduced.

Integration over then gives

w = 4 P1 P2( )a

3b3

2L a2 + b2( )1 2( ) 13

3

0

12

0

1

d

= 4 2

3P1 P2( )a3b3

2L a2 + b2( )1 2( )

3 2d

0

1

(2)

The integral may be found in H. B. Dwight, Tables of Integrals andOther Mathematical Data, Macmillan, New York, 4th edition (1961),Formula 855.41, and the result is


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w = 4 2

3P1 P2( )a

3b3

2L a2 + b2( ) 5

2( )12( )

2 3( )

= 4 23 P1 P2( )a

3

b3

2L a2 + b2( )

32( )

12( )

12( )

12( )

2 2!

= P1 P2( )a

3b3

4L a2 + b2( )(3)

since 12( ) = . When a = b = R, this mass rate of flow expression

reduces to Eq. 2.3-21 of BSL-2e.

We can now compare the mass rates of flow for the circulartube and the elliptical tube of the same cross sectional area, as follows

Circular tube:

w

= P1 P2( ) A

2

2( )8L

(4)

Elliptical tube:

well = P1 P2( ) A

2

2( ) ab4L a2 + b2( )

(5)

so that

wellw

= 2

1 + 2

(6)

where = b a 1 . Thus for tubes of the same cross-sectional area, theratio well w decreases from unity (both tubes circular) to values less

than unity as becomes smaller.


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Note to p. 51

On this page several quantities are obtained from the velocitydistribution of Eq. 2.3-18. We could also ask ourselves: how much

kinetic energy is flowing per unit time in the axial laminar flow of afluid in a circular tube?

The volume rate of flow through an element of cross section

rdrd is vzrdrd. The kinetic energy per unit volume of the fluid is

12vz

2, since the only nonzero velocity component is vz . Therefore the

total amount of kinetic energy per unit volume flowing through thetube is:

1

2 vz2

( )vzrdrd0R

02

= 21

2 vz2

( )vzrdr0R

= 2 1

2R2 vz

3

0

1

d= R2vmax

3 1 2( )01

3

d (1)

In the second line, we have introduced the dimensionless coordinate

= r R and the maximum velocity vmax = P0 PL( )R2 4L . Now all

that remains is to evaluate the integral:

R

2

vmax3

1 32

+ 34

6

( )01

d

=R

2

vmax3 1

2

3

4 +

3

6

1

8

= R2vmax

3 4 6 + 4 18

=1

8R2vmax

3

=

1

4R2vmax( )

1

2vmax

2

(2)

The last form suggests a volume rate of flow multiplied by a kineticenergy per unit volume, both quantities evaluated for the maximum

velocity.


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Note to p. 52

The laminar flow in a circular tube with radius R is discussed in2.3, and the laminar flow in tubes with equilateral triangular cross-

section of height His described in Problem 3B.2. Both tubes have thesame length, L. We want to compare these two flow problems.

a. Compare the mass rates of flow for the two tubes when theircross-sectional areas are the same.

b. Compare the mass rates of flow for the two tubes when theperimeters of their cross sections are the same.

a. For flow in circular and triangular tubes we have for themass flow rates (see Eq. 2.3-21 and Eq. 3B.2(b)):

w

= P0 PL( )R

4

8L w =

3 P0 PL( )H4

320L(1,2)

In Eq. 1, R is the tube radius; in Eq. 2, His the height of the triangular

cross section, and 2H 3 is the length of a side of the triangle. Tomake the comparison, we need to express the flow rates in terms of

the cross-sectional areas. Since for circular tubes A

= R2, and for

equilateral triangular tubes, A =1

3 H2

,

w

= P0 PL( ) A

2 2( )8L

;w =

3 3 P0 PL( )A2

180L(3,4)

Therefore

ww =

3 3 16( )180 8= 0.726 (5)

b. The perimeters of the two tubes are P

= 2R for circular

tubes, and P = 2 3H for triangular tubes. Therefore


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w

= P0 PL( ) P 2( )

4

8L w =

3 P0 PL( ) P 2 3( )4

180L(6,7)

Taking the ratio, as before

ww

=3

2 3( )4

180( )8( ) 2( )

4

= 0.265 (8)

For the square cross section (see Problem 3B.3), the ratioscorresponding to Eqs. 5 and 8 may be found to be

w

w

= 0.884 (same cross-sectional areas) (9)

w

w

= 0.545 (same perimeters) (10)

What, if any, conclusions can you draw from this problem?

[The triangular duct problem is discussed on p. 58 of Landau and

Lifshitz, Fluid Mechanics, Addison Wesley (1959); our H is their amultiplied by 1

23 .]


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Note to p. 54

Let's check a few things about Fig. 2.4-1.a. There the graph shows the transport ofz-momentum in the

positive r-direction. This quantity is negative when r < R, positivewhen r > R, and 0 when r = R, where is defined by Eq. 2.4-12.We need to verify that this graph is consistent with Eq. 2.4-13.

b. The figure also shows the velocity distribution for flow in anannulus, as given in Eq. 2.4-14. What is the location of the maximumvelocity? Show that the position of the maximum is nearer the innercylinder of the annulus.

c. What is the velocity at the maximum in the curve?

a.Eq. 2.4-13 may be rewritten as

rz =P0 PL( )R

2L

r

R

2R

r

=

P0 PL( )R2L

1 2R

r

2

R

r(1)

If r R < , then the bracket in Eq. 2b.6-1 is negative, whereas if

r R > , then the bracket in Eq. 2b.6-1 is positive. This is inagreement with the graph ofrz vs. r in Fig. 2.4-1.

b. To find the maximum of the expression in brackets in Eq. 2.4-14, asa function of r R = s, we have to differentiate [ ] with respect to s asfollows 9 (alternatively, one may set rz equal to zero):

d

ds1 s2

1 2

ln 1 ( )ln

1

s

= 0 2s +

1 2

ln 1 ( )

1

s(2)

Setting the right side equal to zero, and solving for s gives

smax = 1 2

2 ln 1 ( )(3)

Hence, choosing the plus sign (why?), we get for the location of themaximum in the velocity curve


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rmax = R1 2

2ln 1 ( )(4)

The half-way point between the inner and outer cylinders is given by

s = 1

2(1 +) . Therefore we now have to prove that

1 2

2ln 1 ( ) 0.8, the maximum is very close to being halfway between the two cylinders, and that is to be expected, inasmuchas the annular-slit flow approaches a flat-slit flow. Problem 2B.5 givesa discussion of the interrelation of the flow in a plane slit and theflow in a narrow annulus.

c. The maximum velocity is then

vz,max =P0 PL( )R

2

4L

1 rmax

R

2

1 2

ln 1

( )

lnR

rmax

=P0 PL( )R

2

4L1

1 2

2 ln 1 ( )

1 2

ln 1 ( )ln

2ln 1 ( )1 2

(6)

This result may also be written in terms of, as in Eq. 2.4-15.


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Note to p. 55

It is good practice to check limiting cases whenever possible.For example, one should show that Eq. 2.4-17 becomes the Hagen-

Poiseuille formula for tube flow (Eq. 2.3-21) in the limit that 0 .(See the comment in the paragraph that begins four lines after Eq. 2.4-14.)

Solution: We have to show that the bracket quantity in Eq. 2.3-21becomes equal to unity when 0 . In this limit, the various termsinside the bracket become:

lim0

1 4( ) = 1 (1)

lim0

1 2( )2 = 1 (2)

l im0

1 ( ) = (3)

Thus the bracket quantity becomes:

1 4( ) 1 2( )

2

ln 1 ( )

lim

0 1 1

= 1 (4)

and the Hagen-Poiseuille formula is recovered.


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Note to p. 58

Let us verify that the average velocities in the two regions aregiven by Eqs. 2.5-20 and 21.

In region I, the average velocity is given by

vzI =

p0 pL( )b2

2IL

1

b

2I

I + II

+

I II

I + II

x

b

x

b

2

dx

b

0

=p0 pL( )b

2

2IL

2I

I + II

+

I II

I + II

2

d

1

0

where = x b

=

p0 pL( )b2

2IL

2I

I + II

1

2

I II

I + II

1

3

=p0 pL( )b

2

2IL

6 2I 3 I II( ) 2 I + II( )

I + II

=p0 pL( )b

2

2IL

7I + II

I + II

(1)

Similarly for Region II we have

vzII =

p0 pL( )b2

2IIL

1

b

2II

I + II

+

I II

I + II

x

b

x

b

2

dx

0

b

=p0 pL( )b

2

2IIL

2II

I + II

+

I II

I + II

2

d

0

1

=p0 pL( )b

2

2IIL

2II

I + II

+

1

2

I II

I + II

1

3

=p0 pL( )b2

2IIL

62II + 3 I II( ) 2 I + II( )

I + II

=p0 pL( )b

2

2IIL

I + 7II

I + II

(2)


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Note to p. 59

The verification of some equations requires a lot of algebraicdetail that falls in the category of "straightforward but tedious."

Nonetheless, such derivations should be done. Here are twoexamples:

a. Verify the expressions for the stress components rr and r in Eqs.2.6-5 and 2.6-6. From Eqs. B.1-8 and 2.6-1, we get

rr = 2vrr

= 2v1

R

+3

2

R

r

2

3

2

R

r

4

cos

=3vR

R

r

2

+R

r

4

cos (1)

And from Eq. B.1-11 and Eqs. 2.6-1 and 2, we find

r = rr

v

r

+1

r

vr

= v

1

R

R

r

3

2

R

r

2

R

r

4

sin

v 1R

Rr

32Rr

2

12Rr

4

sin( )

= +

3

2

v

R

R

r

4

sin= +

3

2

v

R

R

r

4

sin (2)

This is the "form drag" result given in Eq. 2.6-8.

b. Show how Eqs. 2.6-9 and 2.6-12 are obtained by doing the

necessary integrations.To get the normal force acting on the sphere (from Eq. 2.6-7),we start by noting that rr on the surface of the sphere is zero. (This isa special case of the general result given in Example 3.1-1.) Thepressure p on the surface of the sphere is given in Eq. 2.6-8. Thereforethe z-component of the force acting normal to the surface of thesphere is:


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F n( ) = p0 + gRcos+

3

2

vR

cos

0

02

cos( )R2sindd

= 2 p0 + gRcos+

3

2

vR cos

0

cos( )R2

sind

= 2p0R

2 cos0

sind+ 2R2 gR +3

2

vR

cos2sind0

(3)

The first integral is zero, and the second is 2/3, so that

F n( ) = 2R2 gR +

3

2

vR

2

3

=4

3R3g + 2Rv (4)

To get the z-component of the tangential force acting on thesphere, we substitute Eq. 2.6-11 into Eq. 2.6-10 and integrate:

F(t) = 2

3

2

vR

sin

0

sin( )R2sind

= 3Rv

sin3

0

d= 3Rv4

3

= 4Rv

(5)

This is the "friction drag" result displayed in Eq. 2.6-12.


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Note to p. 78

The general result for normal stresses at solid surfaces given inEq. 3.1-6 is very important. We have already seen in Eq. 2.6-5 that the

normal stresses for creeping flow around a sphere are exactly zero atthe sphere surface, r = R. Verify that Eq. 2.6-5 is correct for rr, ,

and by using Eqs. B.1-15, 16, and 17 and Eqs. 2.6-1, 2, and 3.

Solution:

a. First, we get rr from Eq. B.1-15:

rr =

2vr

r = 2v

1

R +

3

2

R

r

2

3

1

2

R

r

4

cos

=3vR

R

r

2

+R

r

4

cos (1)

b. Then we get

from Eq. B.1-16:

= 21

r

v

+vr

r

= 2v1

RR

r

+3

4

R

r

2

+1

4

R

r

4

cos

2v1

R

R

r

3

2

R

r

2

+1

2

R

r

4

cos

= 2vR

3

4

R

r

2

+3

4

R

r

4

cos

= 3v

2R

R

r

2

+R

r

4

cos (2)

c. And, finally, we get

from Eq. B.1-17:


23/168

23

= 21

r sin

v

+vr + v cot

r

= 2v

1

R

R

r

1

3

2

R

r+

1

2

R

r

3

cos+

R

r

1 +

3

4

R

r+

1

4

R

r

3

cos

= 3v2R

R

r

2

+R

r

4

cos (3)

When r = R, all of the normal stresses are zero, in agreement withExample 3.1-1.


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24

Note to p. 81

Here we give the details of the derivation of Eq. 3.3-1 from Eq.3.2-9. Although Eq. 3.3-1 itself is not much used, the integral of Eq.

3.3-1 over large flow systems is widely used. We call this the"macroscopic mechanical energy balance"; the term "engineeringBernoulli equation" is also used.

The derivation we work through here is an excellent exercise inusing some of the "del" relations given in Appendix A (seeparticularly A.4).

We start by forming the dot product of the local velocity v withEq. 3.2-9. The last term presents no problems:

v

g( ) = v

g( ) (1)

The term involving [ ] may be rearranged using Eq. A.4-29 inExample A.4-1:

v [ ]( ) = v[ ]( ) + :v( ) (2)

The term containing p may be similarly rearranged by using Eq.A.4-19:

v p( ) = pv( ) + p v( ) (3)

We now tackle the remaining two terms by first putting both ofthem on the left side of the equation:

v

t

v

+ v vv[ ]( )

= v t v + v v( ) t

+ v v v[ ]( ) + v v( ) v( ) (4)------------- ----------------

In the first term, we differentiate the product with respect to t, and inthe second term, we use Eq. A.4-24. In the second line, the dashedunderlined terms then sum to zero by using the equation of


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25

continuity; furthermore, the first term is split up into two terms, andthe third term is rearranged, thus:

=

t

1

2

v v( )

1

2

v v( )

t

+ v v v[ ]( ) (5)

We again use the equation of continuity to rewrite the second term inEq. 5:

=

t

1

2 v v( )

+1

2v v( ) v( ) + v v v[ ]( ) (6)

Now the second and third term may be combined by using Eq. A.4-

19 with s replaced by12 v v( ) and v replaced by v :

=

t

1

2 v v( )

+ 1

2 v v( )v

=

t

1

2v2

+ 1

2v2v

(7)

Clearly knowing that the final result is Eq. 3.3-1 is very helpful indoing the last several steps.


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26

Note to p. 82 (i)

We want to verify that :v( ), when written for a Newtonian

fluid, may be written as a sum of squares as shown in Eq. 3.3-3, and ishence positive.

First define a tensor v + v( ), and then :v( ) may be

written for Newtonian fluids (for which is symmetric) as:

:v( ) = 12 : ( ) =

12 v( ) :( ) + 12 v( ) :( ) (1)

where Eq. 1.2-7 has been used. Next, since :( ) = 2 v( ), we get

:v( ) = 12 :( ) v( )

2

+ v( )2

(2)

This is equivalent to:

:v( ) = 12 v( ) : v( ) ( ) + v( )

2(3)

which is another way of writing Eq. 3.3-3. The last step may be seento be true by expanding the double-dot product in Eq. 3 thus:

v( ) : v( ) ( )

= :( ) v( )

2 v( )

2+ 3 v( )

2

= :( ) v( )

2(4)

which is the coefficient of 12 in Eq. 2.


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27

Note to p. 82 (ii)

Eq. 3.4.1 is obtained by taking the cross product of the positionvector with the equation of motion in Eq. 3.2-9. To do this, we form

the cross product, term by term:

a. The time-derivative term is straightforward; for the ith component:

r

tv

i

=

tr v[ ]i (1)

because the position vector r is independent of the time t.

b. For the next term in the equation, we consider only the icomponent and expand the expression in terms of its components:

r vv[ ] i = ijkxj

xlvlvk

l

k

j

= ijk

l

k

j

xlxjvlvk ijkvlvk

l

k

j jl

=

xlv

l

ijk

xjl

k

j v

k

ijk

vj

vkk

j (2)

In the second line, we have moved the xj inside the differentiation

and subtracted off a compensating term. In the third line, we haverearranged the first term and performed the sum on l in the secondterm. It can be seen that the second term is zero (inasmuch as itinvolves a double sum on a pair of indices that appear symmetricallyin one factor (

vjvk ) and antisymmetrically in another (

ijk ); see

Exercise 5 on p. 815). Now we convert Eq. 2 back to bold-facenotation:

r vv[ ] i = v r v[ ] i (3)

c. Next we examine the term containing the pressure:


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28

r p[ ]i = ijk

k

j xj

xkp = ijk

k

j

xkxjp ijk

k

j jkp (4)

The last term is zero, since it involves a double sum on a pair of

indices that appear symmetrically in one factor and antisymmetric inanother. The other term can be rearranged as follows:

xlijkxjpkl{ }

l

k

j =

xlijkxjpkl

k

j

l (5)

On the right side, the quantity within the braces can be recognized asthe cross product of a vector with a tensor (see text just after Eq. A.3-19). Therefore the above result in Eq. 5 can be written as:

xlr p{ }il =

l

xlr p{ }li

=

l r p{ }

i

(6)

In order to write Eq. 6 as the divergence of a tensor, the indices mustbe as in Eq. A.4-13. This requires introducing the transpose of thecross product as indicated above.

d. The term containing the tensor can be treated in somewhat the

same manner as in part (c) above:

r [ ] i = ijkxj

k

j [ ]k = ijkxj

k

j

xllk

l (7)

Next we write this intermediate result as the sum of two terms:

ijk

l

k

j

xlxjlk ijk

l

k

j lk

xlxj

= xl

ijkxjkl

k

j

l ijk

l lk

k

j jl

=

xlr { }

il

l ijkjk

k

j

=

xlr { }

li

+ ikjjk

k

j

l


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29

= r { }

i

+ :[ ]i (8)

If the stress tensor is symmetric, the :[ ]i term vanishes, and may

be replaced by in the first term.

e. The ith component of the external force term, r g[ ]i, isstraightforward.

When all the terms are collected, Eq. 3.4-1 results.


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30

Note to p. 86(i)

In Example 3.5-1 it was pointed out that, to derive the Bernoulliequation, we need the vector identity:

v v v[ ] ( ) = 0 (1)

To show that this relation is true, we use A.2 and Eq. A.4-10.The proof requires replacing the dot and cross operations by

their expressions in terms of vector components. This is mostefficiently done by making use of summations. First we write the dotproduct in terms of the components:

v v v[ ] ( ) = vi

i v v[ ] i (2)

Next we write the cross product operations using the ijk symbol:

= vi

i ijkvj v[ ]k

k

j = vi

i ijkvj klm

xlvm

m

l

k

j (3)

Then we rearrange the expression and make use of the cyclicproperty of the permutation symbol, i.e., klm = mkl = lmk :

= ijkklmvi

m

l

k

j

i vj

xl

vm= ijklmkvi

m

l

k

j

i vj

xl

vm (4)

We can now use Eq. A.2-7 to replace the sum on k of products of twopermutation symbols:

= iljm imjl( )m vivj

xlvm

lji (5)

After doing the sums on l and m we get two terms:


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31

= vivjj

i

xivj vivj

j

i

xjvi (6)

In the second summation, we replace i byj andj by i to get:

= vivj

j

i

xi

vj vjvij

i

xi

vj = 0 (7)

It may now be seen that both terms are the same. Consequently theirdifference is zero. Therefore, we have proven the identity in Eq. 1.


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32

Note to p. 86(ii)

Did you notice some similarity between Eq. 3.5-11 and Eq. 3.3-2? In this "Note," we demonstrate the connection.

If we assume steady state, inviscid flow (as we did in Example3.5-1), then Eq. 3.3-2 can be rewritten, with the help of Eq. 3.5-4 andEq. A.8-19, as

D

Dt12v2 + ( ) = v p( ) (1)

where the inviscid flow assumption has been used to get rid of theterms containing . Next we use the steady-state flow assumption to

rewrite Eq. (1) as

v 1

2v2( ) + v ( ) = v p( ) (2)

Then we write the potential energy term using =gh, where h isthe coordinate in the direction opposite to the direction of gravity.This gives

v 1

2 v

2

( ) + v gh( ) +1

v p( ) = 0 (3)

We can now divide each term by v and introduce the unit vector

s = v v, and then recognize that s ( ) = d ds, where s is thecoordinate along a streamline. Equation (3) may now be written as

d

ds12v2( ) g

dh

ds

1

dp

ds= 0 (4)

which is the same as Eq. 3.5-11. Equation (4) may then be integratedfrom point "1" to point "2" along the streamline to get the Bernoulliequation. Thus it is seen that the Bernoulli equation can be obtaineddirectly from the mechanical energy balance.


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33

Note to p. 90(i)

In Example 3.6-3, we give Eq. 3.6-20, but we don't obtain thepressure distribution. If one knows the pressure at one point in the

system (after a steady state velocity distribution has been obtained),then it is possible to determine the constant of integration and obtainthe pressure distribution.

Here we attack a different problem. We now show how Eq. 3.6-20 can be used to get the pressure distribution in the space betweenthe cylinders in terms of the pressure p0 that exists in the system

before the outer cylinder is rotated.We start by temporarily relaxing the assumption of incom-

pressibility. It can be shown that the equation of motion in Eq. 3.2-9

and Newton's law of viscosity as given in Eq. 1.2-7 are still valid for acompressible fluid, provided that one assumes that the viscosity isindependent of the pressure. The important new feature that we haveto inject into the development is the equation of statethat is, thedensity as a function of the pressure. We do this via a Taylor seriesfor the density as a function of the Gibbs free energy, G = H TS

= 0 +

T

0

G G0( ) + (1)

in which the subscript zero refers to properties of the system whenthe fluid is at rest before the outer cylinder is rotated. Next wetruncate the Taylor series after two terms and write

= 0 1 + b0 G G0( ) = 0 1 + b0

1dpp0

p

(2)

in which b0 is the value ofb = 1 ( ) G( )T = p( )T at = 0 .The integration of Eq. 3.6-20 may now be performed by

inserting the velocity distribution of Eq. 3.6-29 and assuming that thedensity is a known function of the pressure

1dp

p0

p

=v

2

r dr =

2R2o2

1 2( )2

2

1

d


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34

=2R2o

2

1 2( )2

2

2

+2

3

d

=

2R2o2

1 2( )21

2

2

2

2

2

2 ln+C

(3)

The integration constant C is determined by the mass conservationstatement

0d=

1

d1

or

1dpd= 0

p0

p

1

(4,5)

where Eq. (2) has been used. Next, substitution of Eq. (3) into Eq. (5)

gives a relation from which C can be determined

1

2

2

2

2

2

2ln+ C

1

d= 0 (6)

Performing the integration gives

1

22

4

4

1

2

2

ln 2

1

2

2

ln

1

4

2

+

1

2C

2

1

= 0 (7)

This gives for the integration constant

C = 1

1 +2

42

32 ln

1 2(8)

Hence Eq. (3) becomes

1dpp0

p

=2R2o

2

1 2( )2

1

2

2

22

2

2ln

1 1 +2

42

32 ln

1 2

(9)


35/168

35

For an incompressible fluid, the left side of Eq. (9) becomes simply

(p p0 ) 0, so that we get for the pressure distribution in the two-cylinder system with the outer cylinder rotating

p p00

=2R2o

2

1 2( )2

12

2

2 2

2

2 ln

1 1 +2

42

32 ln

1 2

(10)

A similar development for the system with the inner cylinderrotating with angular velocity i and the outer cylinder fixed, gives

p p00

=4R2i

2

1 2( )2

12

2 12

2 ln

1 1 +2

4

1 + 22

1 2ln

(11)


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36

Note to p. 90(ii)

An alternative method of solving the problem in Example 3.6-3is given here. Eq.3.6-21 may be set up by making a shell torque

balance on a shell of thickness r and height L. Then the torque atradius r is equal to 2rL

r r r, whereas the torque at radius r + r

is 2 r + r( )Lr r+r r + r( ) . The torque is (force per unit area) x(area) x lever arm. When these torques are equated, we get afterdividing by r and letting r go to zero:

d

dr

r2r( ) = 0 or

d

dr

r 2 rd

dr

v

r

= 0 (1a,b)

Eq. B.1-11 has also been used. Here we show that this is equivalent toEq. 3.6-21, which is

d

dr

1

r

d

drrv

( )

= 0 or

d2v

dr2+

1

r

dv

drv

r2= 0 (2)

We start by performing the differentiations in the large

parentheses in Eq. 1b:

r2 rd

dr

v

r

= r

2 r1

r

dv

drv

r2

= r

2 dvdr

rv (3)

Next do the differentiation with respect to r (in Eq. 1b) and set theresult equal to zero:

d

dr r2

dv

dr rv

= 2r

dv

dr + r2d2v

dr2 r

dv

dr v

= r2d2v

dr2+

1

r

dv

drv

r2

= 0 (4)


37/168

37

Therefore either r2 is zero, or the quantity in parentheses is zero. But

r2 cannot be zero, and hence Eq. 2which came from Example 3.6-

3must be the same as Eq. 1b (from the torque balance).


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38

Note to p. 117

When you look at Fig. 4.1-2, you might wonder whether theslope at y = 0 is 1 . You can answer that question by differentiatingEq. 4.1-15 with respect to :

d

derfc

=0

=d

d1

2

e

2

d0

=0

=

2

e

2

=0=

2

= 1.1287 (1)

Here we have used the Leibniz formula for differentiating integrals inC.3 and the definition of the complementary error function in C.6.As may be seen from Fig. 4.1-2, the slope is somewhat steeper thanminus 1.

Notice that in Eq. 4.1-14, as well as in Eq. 1 above, we used abar over the to make a distinction between the variable ofintegration and the upper limit on the integral. When applying theLeibniz theorem, it is vital to make this distinction.


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39

Note to p. 121

Example 4.1-3 should be straightforward, except possibly for(a) the line immediately after Eq. 4.1-49, and (b) the line following Eq.

4.1-53. Here we show how to obtain the expressions given in thesetwo locations.

(a) Let the real and imaginary parts of the arguments be:for z1 : z1r and z1ifor z2 : z2r and z2ifor w: wr and wi

Then the expression z1w{ } = z2w{ } becomes

z1r + iz1i( ) wr + iwi( ){ } = z2r + iz2 i( ) wr + iwi( ){ } (1)

Then, taking the real parts of both sides, we get

z1rwr z1iwi = z2rwr z2iwi (2)

Rearranging gives:

wr z1r z2r( ) = wi z1i z2i( ) (3)

Since w, and hence wr and wi, are arbitrary, the only way that this

equation can be satisfied is if z1r = z2r and z1i = z2i (i.e., z1 = z2 ).

(b) The square root of i will be some number in the complexplane, say a + bi, so that

i = a + bi (4)where the real quantities a and b must be determined. When the leftand right sides of Eq. 4 are squared, we get

i = a2 b2( ) + 2abi (5)


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40

To find a and b, we equate the real and imaginary parts of the left andright sides:

a2 b2 = 0 and 2ab = 1 (6)

Eliminating b between these two equations gives

a4 =

1

4(7)

Hence

a2 =1

2,

1

2and

a = 1

2

, 1

2

i (8)

whence

b =

1

2,

1

2

1

i(9)

and the two square roots ofi are

i = a + bi = 1

2 1 + i( ) (10)

Alternatively, one can write i in polar form and use the fact thati has a unit length:

i = rei = 1e i 2 (11)

Then the square root ofi is

i = e i 4 = cos

4+ isin

4

= 12

+ i 12

(12)


41/168


42/168

42

=

x y z

x

y

z

0 vx2 0

x y z

x

y

z

vy2 0 0

= +z

x

vx2( ) + z

y

vy2( )

= zvxx

+vyy

2 + z vx

x

2 + vyy

2

(4)

The first term is zero, because of the assumption of incompressibility(see statement just above Eq. 4.2-1). Then we get finally

= z

y

x

2 +x

y

2

=

x

y

x

2y

2=

,2( ) x, y( )

(5)

When the results in Eqs. 2, 3, and 5 are combined, we get sameexpression that is given in Eq. (A) of the table.


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43

Note to p. 125

Here we want to fill in the details of getting Eq. 4.2-20 from Eqs.4.2-18 and 4.2-19. This is a "straightforward but tedious exercise."

We begin by evaluating each of the four squared terms in Eq.4.2-19 using the velocity components in Eqs. 4.2-13 and 14; it isconvenient to introduce the dimensionless variable = r R .

2

vrr

2

=9

2

v2

r21 3 cos ( )

2

(1)

2

1

r

v

+

vrr

2

=

2v2

r2 1 +3

4

1

+1

4

3

cos

+ 1 3

2

1

+1

2

3

cos

( )

2

=

9

8

v2

r21 + 3 cos( )

2

(2)

2v

cot

r+vrr

2

=2v

2

r21 + 3

41 + 1

43

cos+ 1 321 + 1

23

cos( )2

=9

8

v2

r2

1 + 3 cos( )2

(3)

rr

v

r

+1

r

vr

2

=v

rv

r+1

r

vr

2

=v

2

r2

341 + 3

43

sin+ 1341 1

43

sin

1 321 + 1

23

sin

2

=v

2

r2 32 3

sin 2

(4)

When the above results are combined, we get

:v( )r2 = v

2 2742 27

24 + 27

46

cos

2+ v

2 946

sin

2

(5)


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44

Then the kinetic contribution to the force on the sphere is given by

Fkv = 2 :v( )r

2drR

0

sind

= 2 23 v

2 2742 272

4 + 274 6 dr + 2

43v

2 946 drR

R(6)

since cos2 0

sind= 23 and sin3

0

d= 43 . Then, finally

Fk = 9vR

2 24 + 6 1 d+ 6vR 6 1

d

= 9vR

1 + 233 1

55 1

+ 6vR

155 1

= 9vR 1 23 + 15 + 6vR 15

= 9vR

25

+ 6vR15

=245

+ 65( )vR

= 6vR (7)

which is Stokes' law.


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45

Note to p. 139

We show here how to get the Falkner-Skan equation in Eq. 4.4-35 forthe flow near a corner. For this system, the external flow v

e

wasfound earlier (see Eqs. 4.3-42 and 43) to be

ve x( ) =

2c

2 x 2( ) c' x 2( ) (1)

We then have to solve Eq. 4.4-11 to get the velocity distribution in theneighborhood of the wedge shown in Eq. 4.3-4:

vx

vxx + vy

vxy = ve dv

edx +

2

vxy2

(2)

This equation can be rewritten in terms of the stream function

x, y( ), by using the expressions for the velocity components in thefirst row of Table 4.2-1:

y

2xy

+

x

2y2

= ve

dvedx

+2

y2y

(3)

Insertion of Eq. 1 into this equation then gives Eq. 4.4-32:

y

2xy

x

2y2

=c'2

2

1

x 23( ) 2( )

3y3

(4)

Next we want to rewrite this equation in terms offand :

x, y( ) = c' 2 ( )x1 2

( )f ( ) Ax1 2

( )f ( ) (5)

x, y( ) =c'

2 ( )y

x 1( ) 2( ) B

y

x 1( ) 2( )(6)


46/168

46

We start by converting the various derivatives from x, y( ) toderivatives off ( ) :

y = Ax

1 2( ) df

d

y = ABx

1 2( )

f '1

x 1( ) 2( ) = ABx 2( )

f ' (7)

2y2

= ABx 2( )yf " = AB2

x 2( )

x 1( ) 2( )f " = AB2

1

x 12( ) 2( )f " (8)

3

y3= AB2

1

x 12( ) 2( )

yf = AB3

1

x 23( ) 2( )f (9)

x= A

d

dxx1 2( )

f Ax1 2( )

df

d

x

= Ax1 2( )

2 ( )1

xf Ax1 2( )

By 1 ( )2 ( )

x 1( ) 2( )1

f '

= A

1

2 ( )x1 2( )

xf+ A

1 ( )2 ( )

x1 2( )

xf ' (10)

2yx

= A1

2 ( )x1 2( )

x

yf '+ A

1 ( )2 ( )

x1 2( )

x

y

d

df '( )

= AB

1

2 ( )x1 2( )

x 1( ) 2( )x f '+ AB1 ( )2 ( )

x1 2( )

x 1( ) 2( )xf "+ f '( ) (11)

The terms on the right side of Eq. 4.4-32 are then:

c'2

2

1

x 23( ) 2( )+ AB3

1

x 23( ) 2( )f =

c'2

2 1

x 23( ) 2( )+ f( )

(12)

Next we write down the terms on the left side of Eq. 4.4-32:

y

2

xy

x

2

y2= A2B2

1

2 ( )x 2( )x1 2( )

x 1( ) 2( )xf 2


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47

A2B2

1 ( )2 ( )

x 2( )x1 2( )

x 1( ) 2( )x f f "+ f 2( )

A2B21

2 ( )

x1 2( )

x12

( )2

( )x

f f

+ A2B2

1 ( )2 ( )

x1 2( )

x 12( ) 2( )x f f "

= c'2

2 x 23( ) 2( ) f 2 c'2

1

2 ( )x 23( ) 2( )f f

= c'2

1

2 ( )x 23( ) 2( ) f 2 f f( ) (13)

Combining the results in Eqs. 12 and 13 gives

c'2

1

2 ( )x 23( ) 2( ) f 2 f f( ) = c'2

1

2 ( )x 23( ) 2( ) + f( )

(14)or

f2 f f = + f (15)

which is the same as Eq. 4.4-35, the Falkner-Skan equation.

[Note: In earlier printings of the textbook, the prime was omitted fromc', and the quantity c' was not defined.]


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48

Note to p. 154

To get the average flow velocity from Eq. 5.1-4, we integrate thevelocity distribution over the circular tube cross-section:

vz =vzrdrd0

R

02

rdrd

0

R

02

= vz,max

1 r R( ) 1 7rdrd

0

R

02

rdrd

0

R

02

= vz,max

2

R21 r R( )

1 7rdr = 2vz,max0

R

1 [ ]1 7

0

1

d (1)

where = r R . To evaluate the integral, we make a change of variable

1 = . Then

vz = 2vz,max

1 7 1 ( )0

1

d (2)

This can then be written as the sum of two integrals, which can beevaluated:

vz = 2vz,max 1 7

0

1

d 8 7

0

1

d

( )= 2vz,max

8 7

8 7

15 7

15 7

0

1

= 0.82vz,max

(3)

Since 0.82 is approximately 4/5, relation in Eq. 5.1-5 is verified.


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49

Note to p. 170

Here we fill in some of the missing steps following Eq. 5.6-18.Setting C

1

= 0 in Eq. 5.6-18 and rewriting the equation, we get:

F F = F + F( ) 2 F (1)

where the primes indicate differentiation with respect to . Next wenote that Eq. 1 may be put into the form

12F2( ) = F( ) 2 F (2)

Each term may now be integrated with respect to to give

12F2 = F 2F C2 (3)

which is the same as Eq. 5.6-19. The constant C2 is zero according toEq. 5.6-16, with a, b, and d set equal to zero. [Note: The commentabout setting = ln does not seem to be helpful.]

Eq. 3 is a separable, first-order differential equation

dF

d= 2F + 1

2F2 or

dF

2F 1 + 14F( )

=d

(4)

Then, according to a table of integrals (e.g., Formula 101.1 ofDwight's Table of Integrals), Eq. 4 gives, on integration

12

ln1 + 14 F

F

= ln+ lnC3 or

ln1+ 1

4F

F

= ln+ lnC3 (5a,b)

Next take the antilog of the equation and then square the result toobtain


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50

F

1 + 14F

= C3( )2

(6)

Now either a "plus" sign or a "minus" sign may be inserted inside the

absolute value bars. Since we have no reason to prefer one over theother, we consider the two cases separately:

Case I (plus sign):

F = + 1 + 1

4F( ) C3( )

2or

F =C3( )

2

1 14C3( )

2(7)

Case II (negative sign):

F = 1 + 1

4F( ) C3( )

2or

F = C3( )2

1 + 14C3( )

2(8)

When F in Eq. 7 is plotted against C3( )2, it tends toward

infinity as C3( )2

= 4 is approached from below, and approachesminus infinity when approached from above. Hence Case I isphysically unreasonable behavior for the stream function.

When F in Eq. 8 is plotted versus C3( )2, is it monotone

decreasing over the entire range of C3( )2. Since this is physically

reasonable, we choose the solution in Case II (which agrees with Eq.5.6-20 in the textbook).

When Eq. 8 (or Eq. 5.6-20) is inserted into Eq. 5.6-12 and 13, Eqs.5.6-21 and 22 follow immediately.

Then substitution of Eq. 5.6-21 into the expression for Jin Eq.5.6-2 gives:

J= 2 vz2

0

rdr = 2t( )2

2C32( )

2

1 + 14C3( )

2

40

d (9)

We now let 14C3( )

2= x2 so that d= 4 C3

2( )xdx ; then


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51

J= 32 t( )2C32 xdx

1+ x2( )40

= 32 t( )2C32 1

6 1 + x2( )3

0

=16

3

t( )2C32 (10)

whence Eq. 5.6-23 follows at once.Similarly the mass rate of flow is

w = 2 vz0

rdr = 2t( )

z

2C32( )

1+ 14 C3( )2 20

d z

2

= 4 t( )zC32

4

C32

xdx

1+ x2( )20

= 16

t( )z 1

2 1+ x2( )

0

= 8t( )z (11)


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52

Note to p. 198

Here we show how to obtain the macroscopic mass and momentumbalances from the corresponding equations of change.

Macroscopic Mass Balance

The equation of change for conservation of mass is given in Eq.3.1-4. We want to integrate this equation over the system pictured inFig. 7.0-1 on p. 197:

tdV= v( )dV

V t( )V t( ) (1)

We now apply the 3-dimensional Leibniz formula (Eq. A.5-5) to theleft side and the Gauss divergence theorem (Eq. A.5-2) to the rightside:

d

dtdV n vS( )dS =S t( )V t( ) n v( )dSS t( ) (2)

in which n is the outwardly directed unit vector on the surface S t( ) .

The surface is a function of time t, because there may be moving partsin the system. Eq. 2 may be rewritten as

d

dtdV

V t( ) = n v vS( )( )dSS t( ) (3)

We note that the mathematical surface S t( ) defining the systemconsists of several parts that we identify as follows:

the "inlet" surface S1 (on which vS= 0)

the "outlet" surface S2 (on which vS= 0)the "fixed" surface Sf (on which v = vS = 0 )

the "moving" surface Sm (on which v = vS 0 )with v being the fluid velocity and vS the surface velocity. Thesurface integrals are then split into four parts corresponding to thefour types of surfaces.


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53

The integral on the left side is the total mass mtot in the system.

The surface integrals over Sf and Sm are zero, because v = vS .

Therefore we are left with

ddtmtot = n v( )dS n v( )dSS2S1 (4)

We now introduce the vectors u1 and u2, which are unit vectors inthe direction of flow at planes "1" and "2", respectively. Thus the n inthe S1 integral will be u1, whereas the n in the S2 integral will be

+ u2 . Now we make the assumptions that (i) the density is a constantover the cross section, and (ii) that the velocity is always parallel to

the walls of the entry and exit tubes, so that v = u at plane "1" andv = u at plane "2". Then Eq. 4 becomes

d

dtmtot = +1 vdS 1 vdSS2S1 (5)

Here v is the velocity in the direction of flow, which varies across thecross section. Therefore integrations over the surfaces S1 and S2 give

ddt mtot =

1 v1 S1 2 v2 S2 = w1 w2 (6)

where is the average value over the cross section; w1 and w2 arethe mass rates of flow at the inlet and outlet, respectively. Eq. 6 is justthe same as Eq. 7.1-1 in the textbook, which was written down byusing elementary arguments (i.e., common sense).

Macroscopic Momentum Balance

When the equation of motion of Eq. 3.3-9 is integrated over thevolume of the flow system in Fig. 7.0-1, we get

t

v

V t( ) dV= vv[ ]V t( ) dV pdV [ ]V t( )V t( ) dV+ gdVV t( )

(7)


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54

Next apply the Leibniz formula to the left side and the Gaussdivergence theorem (or Eq. A.5-2) to the surface integrals on the rightside to get

d

dt vdV v n vS( )dS = S t( )V t( ) n vv[ ]S t( ) dS npdS n [ ]S t( )S t( ) dV+

gdVV t( ) (8)

The integral on the left side is just the total momentum Ptot in theflow system. Ifg does not change over the volume of the flow system,then it may be removed from the integral, which gives mtot . Then

d

dtPtot = n v vS( )v S t( ) dS npdS n [ ]S t( )S t( ) dS + mtotg (9)

We now consider the three surface integrals seriatim:The first integral is zero on fixed and moving surfaces, and vS

is zero at the entry and exit planes, so that

n v vS( )v S t( ) dS = + u1 u1u1 S1 v

2dS u2 u2u2 S2 v2dS

= +1 v12 S1u1 2 v2

2 S2u2 (10)

Here it has been assumed that the flow at the inlet and outlet planes isparallel to the container wall.

The second integral will contribute both at the inlet and outletplanes and also to the force on the various solid surfaces:

npdS

S t( ) = + u1S1 pdS u2S2 pdS npdSSf+Sm

= p1S1u1 p2S2u2 Ffs

p( ) (11)

the last contribution being the force exerted by the fluid on the solidsurfaces by the pressure.

Finally the third integral will be

n [ ]dSS t( ) = + u1 S1 dS u2 S2 pdS n [ ]dSSf+Sm


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55

Ffs

( ) (12)

Note that we have omitted the contributions at the entry and exitplanes because they would normally be quite small compared to the

pressure terms. Therefore we are left with just the force exerted bythe fluid on the solid surfaces because of the viscous forces.

When all the forces are combined we get for the macroscopic

momentum balance (with Ffsp( ) + Ffs

( ) = Ffs = Fsf ):

d

dtPtot = 1 v1

2 S1u1 2 v22 S2u2 + p1S1u1 p2S2u2 + Fsf +mtotg (13)

This is the same as Eq. 7.2-11 (or 7.2-12) in the textbook, obtained by

elementary reasoning.

[Note: The derivation of the macroscopic mechanical energy balance isgiven in 7.8, the derivation of the macroscopic energy balance is givenin the Note to p. 454.]


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56

Note to p. 199

Here we work through all the details of Example 7.1-1, (a)explaining how to get Eq. 7.1-4, (b) giving the details of how the

difference in the modified pressures is obtained, and (c) workingthrough the algebraic details of the remainder of the problem.

a. (This development was given by Professor L. E. Wedgewood,University of Illinois at Chicago)

We have to find the volume of liquid in the sphere below theliquid level. We imagine that the sphere is generated by a circle in thexz-plane, with its center at z = R and x = 0. The tank is draining in thenegative z-direction, and the exit from the sphere into the attachedtube is located at z = 0. The sphere is created by rotating thegenerating circle around the z-axis.

The generating circle has the equation:

x2 + z R( )

2= R2 or x2 = 2Rz z2 (1a,b)

Then we visualize the liquid volume as being made up of a stack ofthin circular disks of thickness dz, each with a volume of

dV= x2dz = 2Rz z2

( )dz (2)

Then the total volume of the liquid is:

V= 2Rz z2( )dz0

h

= Rz2 1

3z3( )

0

h

= Rh2 1

3h3( ) (3)

Thus the liquid volume is:

V= Rh2

1 1

3

h

R

(4)

We may check this result at three points where we know the result:

Tank full: h = 2R V=43R3

Tank half full: h = R V=23R3


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57

Tank empty: h = 0 V= 0

b. Next we want to apply Eq. 7.1-2 to the system. No liquid isentering at plane "1" so that w1 = 0, and, if the diameter of the exit

tube is sufficiently small that the flow in it is laminar, then w2 will begiven by the Hagen-Poiseuille formula (Eq. 2.3-21). Hence

d

dtRh2 1

1

3

h

R

=

P2 P3( )D4

128L(5)

To get the modified pressures we must specify a datum plane; wechoose this to be at the tube outlet (i.e., plane "3"), so that:

P2 = p2 + gh2 = gh + patm( ) + gL (6)

That is, p2 is the pressure due to the liquid in the sphere above plane

"2" plus the atmospheric pressure, and h2 is the height of plane "2"above the datum plane (i.e., plane "3"). Furthermore,

P3 = p3 + gh3 = patm + 0 (7)

since p3 is the pressure atmospheric pressure at the tube outlet, andh3 is the distance upward from the datum plane (i.e., plane "3"),which is zero. Hence, Eq. 5 becomes

d

dtRh2 1

1

3

h

R

= gh + gL( )D4

128L(8)

c. Eq. 8 may be rewritten as

R

h + Ld

dth2

1

3

h3

R

=gD4

128L A (9)

(which defines the quantityA) or


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58

R

h + L2h

h2

R

dh

dt= A (10)

Now we introduce a new variable H= h + L in order to facilitate thesolution of the differential equation

2R H L( ) H L( )2

H

dH

dt= A (11)

or

H 2 R + L( ) + 2R + L( )L1

H

dH

dt

= A (12)

Now we integrate this equation and make use of the initial and finalconditions:

1

2H2 2 R + L( )H+ 2R + L( )L lnH

2R+L

L

= At0

tefflux (13)

This gives

1

2L2

1

22R + L( )

2 2 R + L( )L + 2 R + L( ) 2R + L( ) + 2R + L( )L ln

L

2R + L

= Atefflux (14)or after some cancellations

tefflux =

1

A2R2 + 2RL 2R + L( )L ln

2R + LL

(15)

When L2 is factored out of the bracket expression, Eq. 7.1-8 of thetextbook is obtained.


59/168


60/168

60

Note to p. 218

Here we work through the details from Eq. 7.7-8 to the end of theexample.

The first term of Eq. 7.7-5 may be transformed as follows:

2hd2h

dt2= 2h

d

dtu( ) = 2h 12 u1 2

du

dt

= hu1 2du

dh

dh

dt= h

du

dh(1)

from which Eq. 7.7-8 follows at once:

hdu

dh N 1( )u + 2gh = 0 (2)

To verify that Eq. 7.7-9 is the solution to Eq. 7.7-8, we substitute theformer into the latter to get

h C N 1( )hN2 +

2g

N 2

N 1( ) Ch

N1 +2gh

N 2

+ 2gh = 0 (3)

where C is the constant of integration. We then see that the terms in"g" and the terms in "C" separately sum to zero:

g-terms:

2gh

N 2 N 1( )

2gh

N 2+N 2N 2

2gh = 0 (4)

C-terms: C N 1( )hN1 N 1( )ChN1 = 0 (5)

We next apply the initial conditions (Eqs. 7.7-6 and 7) to the solutionin Eq. 7.7-9

dhdt

2

= ChN1 + 2ghN 2 (6)

to get an equation for the constant of integration


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61

2gH

R0R

4

= CHN1 +2gH

N 2(7)

When this is solved for C, we get (noting that N= R R0

( ) as definedimmediately after Eq. 7.7-3):

C =2g

HN21

N

1

N 2

= 4g

HN21

N N 2( )

(8)

Even though the factor containing N 2( ) would cause C to becomeinfinite for N= 2, this need cause no alarm, since Nis going to be alarge number, i.e., when the radius outlet hole is small compared to

the radius of the tank.When we take the square root of Eq. 7.7-9 and introduce the

dimensionless liquid height = h H, we then get Eq. 7.7-10:

d

dt=

2g

N 2( )H

2

NN1

(9)

We then choose the minus sign, because we know that the height of

the fluid will be decreasing with increasing time, and therefore d dtmust be negative. To get the efflux time, we integrate Eq. 9 from t = 0when = 1 (full tank) to t = tefflux when = 0 (empty tank):

tefflux = dt =N 2( )H

2g

1

2 N( )N1d

0

1

0tefflux

NH

2g N( ) (10)

Keep in mind that tefflux = NH 2g is the quasi-steady-state solution

in Eq. 7.7-3the solution that we would expect to be valid when Nisextremely large, i.e., for the case that the outlet hole is so small thatthe system is never far from steady state.

The function N( ) is then given by


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62

N( ) =1

2

N 2N

1

2 N( )N1d

0

1

(11)

This integral can probably not be performed analytically. However,in the expression under the square-root sign, for large N, the firstterm will predominate, over the range of integration (from = 0 to

= 1 ). Therefore we take the first term outside of the radical andwrite

N( ) =1

2

N 2N

1

1

2

NN 2

1 2

d0

1

(12)

The quantity to the 12 power can then be expanded in a Taylorseries (see Eq. C.2-1) about 2 N( )N2 = 0 to get, after integratingterm by term

N( ) =1

2

N 2N

1

1 +

1

1!

1

2

2

NN2

+1

2!

1

2

3

2

2

NN2

2

d

0

1

=1

2

N 2

N

2 +1

N N 32( )+

3

4N2

N 74( )+

(13)

When this expression is expanded in a Taylor series around 1 N= 0,we get

N( ) = 1 1

N1

2

1

N2+ O

1

N3

+

1

2

1

N2+ O

1

N3

+

3

8O

1

N3

+

= 1

1

N+O

1

N3

(14)

in which O ( ) means "term of the order of ( )." Thus we have arrivedat Eq. 7.7-13 at the end of the example.


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63

Note to p. 241

In the text it is shown how to get the function ( ) from data

obtained with a cone-and-plate viscometer. In the absence of thiskind of device, one may have to extract the function from tube-flowdata. Show how to get the relation for the non-Newtonian viscosity

( ) from experimental data on mass rate of flow w vs pressuredrop P0 PL for flow through circular tubes.

We know from 2..3 that, for any kind of fluid rz = Rr R,

where R = P0 PL( )R 2L is the shear stress at the tube wall r = R .The mass rate of flow through the tube is

w = 2 vzrdr0R

= 2dvzdr

r2

2

dr

0

R

= + r2dr0R

(1)

the second form being obtained by an integration by parts. In thethird form we have made the replacement = dvz dr .

Next we change the variable of integration from r to rz

w

R3 =1

R3

rz

2

0

R

drz (2)

In this equation, the shear rate is to be regarded as a function of theshear stress. Equation 2 tells us that data taken in tubes of differentlengths and radii should collapse onto a single curve when plotted as

w R3 vs R . If now we multiply Eq. 2 by R

3 and then differentiate

with respect to R, we get

ddR

R3 wR3

= RR2 (3)

where the Leibniz formula for differentiating an integral has beenused (see C.3). This is the Weissenberg-Rabinowitch equation. It tells


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64

how the wall shear rate R can be obtained by differentiating theflow-rate vs pressure-drop data.

We now put Eq. 3 into a different form. Differentiating theproduct with respect to R gives

w

R33R

2 + R3 d

dR

w

R3

= RR

2 (4)

Then division by R2 w R3( ) gives

3 +d ln w R3( )

d lnR

= R1

w

R

3

(5)

Then finally

R( ) =RR

=R

w R33 +

d ln w R3( )d lnR

1

(6)

Equation 6 gives the viscosity as a function of shear rate at the wall

from experimental measurements of w andP

0 PL . If we assume

that R( ) at the wall is the same as ( ) throughout the tube, thenEq. 6 gives ( ) . This assumption seems to be valid for typicalpolymeric fluids. It would not, however, be expected to hold forsuspensions of fibers, because of the change in the fluidmicrostructure near the wall. The above analysis is applicable onlywhen there is no appreciable viscous heating.


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65

Note to p. 242

By working through all the intermediate steps in Example 8.3-1we get a better idea how to solve problems involving the power-law

model. First we have to obtain the expression for the scalar thatappears in Eq. 8.3-5:

= 12

: ( ) = 12 ij jij=1

3

i=1

3

and = v + v( )

(1,2)

where i andj take on the values 1 = r, 2 = , and 3 = z, since we aredealing with cylindrical coordinates. The components of in

cylindrical coordinates may be obtained from Eqs. (S) to (AA) inTable A.7-2. Since the only component ofv that is nonzero is the z-component and since that is a function ofr only, the only componentsof that we need are the rz- and zr-components,

rz = v( )rz + v( )zr = vz r + vr z = vz r + 0 (3)

zr = v( )zr + v( )rz = vr z + vz r = 0 + vz r (4)

Therefore

= 1

2 : ( ) = dvz dr( )

2= dvz dr (5)

where the minus sign must be chosen, since dvz dr is negative. Then

the shear stress rz will be

rz = m dvzdr

n1dvzdr

= dvzdr

n

(6)

By going through the above procedure, it is guaranteed that, whenwe take the fractional powers of the quantities in parentheses (seeTable 8.3-2 for sample values of n), we will not get any imaginaryquantities. When Eq. 6 is substituted into Eq. 2.3-13, we then get:


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66

m

dvzdr

n

=P0 PL

2L

r or

dvzdr

=P0 PL

2mL

1 n

r1 n (7,8)

Integration of Eq. 8 gives

vz =

P0 PL2mL

1 nr 1 n( )+1

1 n( ) + 1+ C (9)

Application of the no-slip condition requires that vz = 0 at r = R :

0 = P0 PL

2mL

1 nR 1 n( )+1

1 n( ) + 1+C (10)

Subtraction of Eq. 10 from Eq. 9 eliminates the integration constant Cand leads to

vz =

P0 PL2mL

1 nr 1 n( )+1 R 1 n( )+1

1 n( ) + 1(11)

Rearrangement then gives

vz =P0 PL( )R

2mL

1 nR

1 n( ) + 11

r

R

1 n( )+1

(12)

The mass rate of flow w is then obtained by integrating the velocitydistribution over the tube cross-section:

w = vz0

R

02

rdrd

= 2R2P0 PL( )R

2mL

1 n R

1 n( ) +11 1 n( )+1( )0

1

d (13)

where = r R . Performing the integration gives


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67

w = 2R2P0 PL( )R

2mL

1 nR

1 n( ) + 11 n( ) + 1

2 1 n( ) + 3

=R

3

P0 PL( )R2mL

1 n1

1 n( ) + 3 (14)

This is the power-law analog of the Hagen-Poiseuille equation forNewtonian fluids.


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68

Note to p. 248

Here we work through the missing steps in Example 8.4-2. InEq. 8.4-20, we make the change of variable s = t t after the first linein order to get a factor e it to appear explicitly on the right side of theequation (to match a similar factor on the left side):

iv0e it{ } = d2v0

dy2e it

01es 1 e isds

0

(1)

Next, we perform the integration over s:

iv0e it{ } = d2v0

dy2e it

01

1

1 1( ) i

es 1 e is

0

= d2v0

dy2e it

01 + i1

(2)

We may now remove the real-operator sign from both sides, as well

as the common multiplier e it, to get:

iv0 =

d2v0

dy20

1 + i1

or

d2v0

dy2=

i 1 + i1( )0

v

0 (3a,b)

Eq. 3b is a differential equation for the complex function v0 y( ) . The

equation is of the form of Eq. C.1-4, where [ ] is a2 . Since this is a

complex quantity, we write it as + i( )2, where and are real, so

that we can write the solution as

v0 = Ae+ + i( )y + Be + i( )y (4)

Then according to Eq. 8.4-19,


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69

vx y, t( ) = v

0 y( )e it{ } = Ae+ +i( )y + Be + i( )y( )e it{ } (5)

Now we know that the oscillatory disturbance will not propagatewith increasing amplitude as the distance from the wall increases, sothatA will have to be set equal to zero and will have to be positive.Also we know that the amplitude of the disturbance right at the wallwill be v0 . Therefore B will have to be set equal to v0 . Therefore Eq. 5can be rewritten as

vx y, t( ) = v0e

+ i( )ye it{ } = v0ey e i ty( ){ } = v0eycos t y( )(6)

Now we must find out what and are. To do this, we have to goback to Eq. 3:

i 1 + i1( )0

= + i( )2

(7)

We next equate the real and imaginary parts of both sides:

2

2

=

12

0 and 2=

0 (8, 9)

This gives us two equations for the two unknowns and . We caneliminate between the two equations and get an equation for :

2

20

21

2=

12

0(10)

or, after multiplying through by 2

4 +

12

02

20

2

= 0 (11)


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70

This can be regarded as a quadratic equation in 2, which has thesolution:

2

=

1

2

20 1

2

1

2

0

2

+ 4

20

2

= 1

2

201 1 +

1

122

=

201 1 + 1

22

(12)

We now extract the square root of both sides to get:

=

201 1+ 1

22

1 2

(13)

Because must be real and positive, the sign must be chosen to be+, and the sign must also be chosen to be "plus." Thus we finallyget:

= +

201+ 1 + 1

22

1 2

(14)

= +

201+ 1+ 122

1 2

(15)

However, we could just as well have chosen both of the signs to be"minus." Then we would have

=

201+ 1

22 1

1 2

(16)

=

20 1 + 12

2

1

1 2

(17)

The quantity given in Eq. 16 is still real and positive. So, how dowe make a choice between Eqs. 14 and 15 on the one hand, and Eqs.16 and 17 on the other? It is easy to show that the former do notsatisfy Eq. 8, whereas the latter do. Therefore, we conclude that Eqs.


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16 and 17 are the proper quantities to insert in Eq. 6. Thus Eqs. 8.4-24and 25 of the textbook are correct.


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72

Note to p. 250

The simplest nonlinear viscoelastic model for polymers is thecorotational Maxwell model, which is obtained by replacing the partial

time derivative in Eq. 8.4-3 by the Jaumann derivative of Eq. 8.5-2.This gives

+ D

Dt= 0 (1)

This equation contains just two parameters: 0, the zero-shear-rate

viscosity, and , the relaxation time.Here we show how to get the viscosity, the first normal-stress

coefficient, and the second normal-stress coefficient for the modelgiven in Eq. (1). The imposed flow is vx = y with vy = 0 and vz = 0 .

In learning how to analyze and evaluate nonlinear viscoelasticmodels, it is useful to display the various parts of the models in

matrix form. We start by getting v, v( ), , and for the flow

field being considered.

v =

0 0 0

1 0 00 0 0

v( )

=

0 1 0

0 0 00 0 0

(2,3)

=0 1 0

1 0 0

0 0 0

0 1 01 0 0

0 0 0

(4,5)

Since the stress tensor does not depend on position and time, in the

Jaumann derivative of, the substantial derivative will not

contribute and we are left with only 12

{ } .


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73

{ } =0 1 01 0 0

0 0 0

xx xy xz

yx yy yz

zx zy zz

=

yx yy yzxx xy xz

0 0 0

(6)

{ } =

xx xy xz

yx yy yz

zx zy zz

0 1 01 0 0

0 0 0

=

xy xx 0

yy xy 0

zy xz 0

(7)

Hence the Jaumann derivative is (keep in mind that the stress tensoris symmetric)

D

Dt=

yx12

xx yy( ) 12 yz12

xx yy( ) yx 12xz 1

2yz

12xz 0

(8)

Thus the corotational Maxwell model in matrix form is

xx xy xz

yx yy yz

zx zy zz

+

yx12

xx yy( ) 12 yz12

xx yy( ) yx 12 xz 1

2yz

12xz 0

= 0

0 1 0

1 0 0

0 0 0

(9)

From this matrix equation, we can write down the followingalgebraic equations

xx yy( ) 2yx= 0 (10)

yy zz( ) + yx= 0 (11)


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yx +

12

xx yy( )= 0 (12)

xz

12yz = 0 (13)

yz +

12xz = 0 (14)

From the last two equations we see that xz = yz = 0 . From solving

the first three simultaneous equations for yx, xx yy, and yy zzwe find

yx = 0 1

1 + ( )2

or

0=

1

1+ ( )2

(15,16)

xx yy = 0 2

1 + ( )2

or

10

= 21 + ( )

2(17,18)

yy zz = +0

1 + ( )2

or

20

= 1

1 + ( )2

(19,20)

We know that the shear stress keeps increasing as the velocitygradient increases, but Eq. 16 indicates otherwise. In fact, the shear

stress goes through a maximum at = 1. The second normal stressis negative, and that is in agreement with measurements forpolymeric liquids. However, generally one expects 2 1 to be inthe range from 0.1 to 0.4 . Hence from examination of this oneparticular flow, it is evident that caution must be used when drawingany conclusions from this model. However, for a model with just twoparameters, it seems to be promising for predicting qualitative

behavior.


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75

Note to p. 255

Since the FENE-P dumbbell model (a molecular model)describes many of the observed rheological phenomena of polymer

solutions, it is important to understand how we go from theconstitutive equation in Eqs. 8.6-2 to 4 to the expressions for therheological properties. Here we show how to get the non-Newtonianviscosity and the first normal-stress coefficient in steady shear flow.

For the steady shear flow, vx = y, the rate-of-strain tensor may be displayed as a matrix (see A.9) thus:

xx xy xz

yx

yy

yzzx zy zz

=

0 0

0 0 00 0 0

(1)

Then Eq. 8.6-4 may be written in matrix form as follows:

Z

p,xx p,xy p,xz

p,yx p,yy p,yz

p,zx p,zy p,zz

H

2p,xy p,yy p,yz

p,yy 0 0

p,yz 0 0

= nKTH0 1 0

1 0 0

0 0 0

(2)

From this we get at once that the only nonzero components of p are

p,xx and p,xy= p,yx . Then from the matrix equation, Eq. 2, we can

write down the two nonvanishing component equations as:

Zp,xx = 2p,yxH and

Zp,yx = nKTH (3,4)

in which

Z = 1 + 3 b( ) 1 tr p 3nKT( ) = 1 + 3 b( ) 1 p,xx 3nKT( ) (5)


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In order to make the algebraic manipulations somewhat easier, weintroduce the following dimensionless quantities:

Dimensionless shear stress:S =

p,yx

3nKT (6)

Dimensionless normal stress:N= p,xx 3nKT (7)

Dimensionless shear rate:= H (8)

Then Eqs. 3, 4, and 5 become:

ZN= 2S (9)

ZS = 13 (10)

Z = 1 + 3 b( ) 1 N( ) (11)

Use the second of these equations to eliminate Z and rewrite Eqs. 9and 11 thus:

N

3S= 2S and

3S= 1 +

3

b1 N( ) (12,13)

Then Nmay be eliminated between these two equations to get a cubicequation for the dimensionless shear stress S (for method of solvingcubic equations, see a mathematics handbook):

S3 +

b + 318

S +b

54= 0 or S

3 + 3pS + 2q = 0 (14,15)

Eqs. 14 and 15 serve to define p and q. Eq. 15 has three solutions, twoimaginary solutions (of no interest here) and one real solution:

S = 2p1 2sinh 1

3arcsinh p3 2q( ) (16)

or, in terms of the original variables:


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p,yx

3nKT= 2

b + 354

+1 2

sinh 13

arcsinhb + 354

3 2b

108H

(17)

From this result one can plot the non-Newtonian viscosity as afunction of the shear rate.If one wants only the limiting values of the non-Newtonian

viscosity at zero and infinite shear rates, this information can beobtained from Eq. 17 or 16. Very small shear rate corresponds to verysmall q, so that

S = 2p1 2sinh 13p3 2q ( ) = 2p

1 2 13p3 2q +{ } = 23 p

1q

(18)

if we keep only the terms linear in q in the Taylor series expansion ofthe hyperbolic sine and the arc hyperbolic sine functions:

sinh x = x + 1

6x3 +

arcsinh x = x 1

6x3 + (19,20)

In terms of the original variables, Eq. 18 is

p,yx = 3nKT2

3

b + 3

54

1b

108H = nkT

b

b + 3

H (21)

This corresponds to

s = nkTH

b

b + 3

(22)

in the limit of zero velocity gradient.In the limit of infinite velocity gradient, we make use of the fact

that for large values of the argument, sinh x 12 exp x (see AppendixC.5). Therefore

S = 2p1 2sinh 1

3ln 2p3 2q( )( ) = 2p1 2 12 exp 13 ln 2p3 2q( )( ) = 2q( )

1 3

(23)


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78

In the original variables, this becomes

p,yx = 3nKT 2 b

108H

1 3

= nKT 54 b

108H

1 3

(24)

This corresponds to a "power-law function":

s = nKTHb

2

1

H

1 3

(25)

in the infinite velocity gradient limit.Now the results in Eqs. 22 and 25 can be obtained directly from

Eq. 15 in another way. If the velocity gradient (and hence, q) is quitesmall, the cubic term in S may be omitted, and one gets then

S = 2

3p1q (26)

directly (see Eq. 18). Similarly, if the velocity gradient is quite large,then the linear term in S may be omitted, and one gets

S = 2q( )1 3

(27)

immediately (see Eq. 23).From Eq. 17, one can also find the molecular stretching as a

function of the shear rate, as described at the top of p. 255. Inaddition, from Eq. 3 the first normal stress coefficient can be obtained

1 =

2 s( )2

nKT(28)

this formula being predicted for the entire shear-rate range.

[Note: For more information on the FENE-P dumbbell model, seeTeaching with FENE Dumbbells, by R. B. Bird, Rheology Bulletin,

January 2007.]


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79

Note to p. 259

The Bingham fluid is not the only empirical equation with ayield stress. Closely related to the Bingham equation is the two-

constant Casson equation, proposed for pigment-oil suspensions, butwidely used for blood1,2,3

when 0 (1)

= 0 +0

when 0 (2)

in which 0 is the yield stress, and 0 is a parameter with dimensionsof viscosity. Derive the expression for the mass rate of flow of aCasson fluid through a circular tube of radius R, and length L.

1N. Casson, Rheology of Disperse Systems, C. C. Mill, ed., Pergamon, London(1959), pp. 84-104.2M. M. Lih, Transport Phenomena in Medicine and Biology, Wiley, New York (1975),378-38.3E. N. Lightfoot, Transport in Living Systems, Wiley, New York (1974), pp. 35, 430,438, 440.

The expression for the shear stress in a circular tube for anykind of fluid was derived in Chapter 2 and found to be

rz =

P0 PL2L

r (3)

At some radius r0, the yield stress will be exceeded. The region

0 r r0 will be a plug-flow region, which moves as a solid, and r0 isdefined by 0 = P0 PL( ) 2L( )r0 . We further note that = dvz drfor flow in circular tubes, according to the definition given just beforeEq. 8.3-2. For the region r0 r R, we then get from Eq. 2

= 0 + 0 or rz = 0 + 0 (4a,b)


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Combining Eqs. 3 and 4b and rearranging gives

P0 PL( )r

2L=

0

+ 0

(5)

Solving Eq. 5 for we get

=1

0

P0 PL( )r2L

2P0 PL( )r0

2L+ 0

0

0

r

r0 2

r

r0+ 1

(6)

Inserting = dvz>

dr, we can get a differential equation for thedimensionless velocity profile > = 0 r00( )vz

> as a function of the

dimensionless radial coordinate = r r0

d>

d

notes final corrd 8-9-11-dist

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