1

Notes for the 2nd revised edition ofTRANSPORT PHENOMENA

byR. B. Bird, W. E. Stewart, and E. N. Lightfoot

by R. B. Bird9 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 functions—all 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 fluidsp. 59 Flow around a spherep. 78 Normal stresses at solid surfacesp. 81 Equation of change for mechanical energyp. 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

2

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

3

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 profilesp. 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

4

Note to p. 5

Section 0.3 is important for emphasizing some of the basicconcepts and definitions. Here we work through some of the missingsteps 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 =12 mA . With

these substitutions, we then get:

12 mA � �rA + � �RA1( ) + 1

2 mA � �rA + � �RA2( ) + 12 mB � �rB + � �RB1( ) + 1

2 mB � �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 themolecular masses and velocities.

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

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

12

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

�RA1( )( )

+ 1

212 mA �rA � �rA( ) + 2 �rA � �RA2( ) + �RA2 �

�RA2( )�

���

�+ �A

+ 1

212 mB �rB � �rB( ) + 2 �rB �

�RB1( ) + �RB1 ��RB1( )( )

+ 1

212 mB �rB � �rB( ) + 2 �rB �

�RB2( ) + �RB2 ��RB2( )�

���

�+ �B

= 1

212 mA � �rA � ��rA( ) + 2 ��rA � �RA1( ) + � �RA1 �

� �RA1( )( )

12 mA �rA + �RA1( ) + 1

2 mA �rA + �RA2( ) + 12 mB �rB + �RB1( ) + 1

2 mB �rB + �RB2( ) =

5

+ 1

212 mA ��rA � � �rA( ) + 2 � �rA � � �RA2( ) + � �RA2 �

� �RA2( )�

���

+ ��A

+ 1

212 mB � �rB � ��rB( ) + 2 ��rB �

� �RB1( ) + � �RB1 �� �RB1( )( )

+ 1

212 mB ��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

12 mA �rA � �rA( ) + 1

2 mA1�RA1 �

�RA1( ) + 12 mA2

�RA2 ��RA2( ) + �A

+12 mB �rB � �rB( ) + 1

2 mB1�RB1 �

�RB1( ) + 12 mB2

�RB2 ��RB2( ) + �B

=12 mA ��rA � � �rA( ) + 1

2 mA1� �RA1 �

� �RA1( ) + 12 mA2

� �RA2 �� �RA2( ) + ��A

+12 mB ��rB � � �rB( ) + 1

2 mB1� �RB1 �

� �RB1( ) + 12 mB2

� �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 A 1 in acoordinate system fixed at the center of mass of molecule A; 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 ofmolecule A as a function of rA2 � rA1 , the separation of the two atomsin molecule A. 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.

6

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.

7

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 of x and yin Eq. 3B.2-1:

vz x, y( ) =

P 0 �P L( )4μLH

y � H( ) 3x2� y2( ) (1)

and vx = 0 and vy = 0 . Here L is the length of the duct (which goes

from z = 0 to z = L ) and P 0 �P L is the difference in modifiedpressure between the ends of the duct. What are the stresses at thesurface y = H according 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 = �μ

�vz

�y= �μ �

P 0 �P L( )4μLH

�

�y3x2 y � y3

� 3Hx2+ Hy2( )

=

P 0 �P L( )4LH

3x2� 3y2

+ 2Hy( ) (2)

� xz = �μ

�vz

�x= �μ �

P 0 �P L( )4μLH

�

�x3x2 y � y3

� 3Hx2+ Hy2( )

=

P 0 �P L( )4LH

6x y � H( )( ) (3)

At the surface y = H:

� yz y=H

=P 0 �P L( )

4LH3x2

� H 2( ) ; � xz y=H

= 0 (4,5)

8

Note to p. 26

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

MLt

�

���

��=

M( )ML2

t2T

�

��

�

�� T( )

L2( ) �( )(1)

(b)

MLt

�

���

��=

1moles

�

���

��ML2

t

�

��

�

��

L3

moles

�

��

�

��

(2)

(c)

M

Lt2

�

���

��=

M

Lt2

�

���

��+

M

L3

�

���

��Lt

�

���

��Lt

�

���

��=

M

Lt2

�

���

��+

M

Lt2

�

���

��+

M

L3

�

���

��Lt

�

���

��Lt

�

���

��

(3)

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

9

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 flux tensor 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= �vx vx = +

1�

2w�W

�

���

�

2x6

x2+ y2( )

4 (1)

�vv( )xy= �vx vy = +

1�

2w�W

�

���

�

2 x5 y

x2+ y2( )

4 (2)

�vv( )xz= �vxvz = 0 (3)

�vv( )yx= �vy vx = +

1�

2w�W

�

���

�

2 x5 y

x2+ y2( )

4 (4)

�vv( )yy= �vy vy = +

1�

2w�W

�

���

�

2 x4 y2

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

10

��x ��vv�� ��x= �vxvx = +

1�

2w�W

�

��

2x6

x2+ y2( )

4 (10)

��x ��vv�� ��y= �vxvy = +

1�

2w�W

�

��

2 x5 y

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�� ��y

is the amount of y momentum flowing per unit time through a unit ofsurface perpendicular to the x axis, and

��x ��vv�� ��z

is the amount of zmomentum 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.

11

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 axisof 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( )a2b2

2μL a2+ b2( )

1 �xa

�

���

��

2

�yb

�

���

��

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 x

2+ y2

= r2 .To get the mass rate of flow, we integrate the velocity

distribution over the cross section, thus

w = 4 �P1 �P2( )a2b2

�

2μL a2+ b2( )

� ab 1 � � 2��

2( )d�d�0

1��2

�0

1�

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

Integration over � then gives

w = 4 �P1 �P2( )a3b3

�

2μL a2+ b2( )

1� �2( )� � 13�

3��

0

1��2

0

1� d�

= 4 �23�

P1 �P2( ) a3b3�

2μL a2+ b2( )

1� � 2( )3 2

d�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

12

w = 4 �23�

P1 �P2( )a3b3�

2μL a2+ b2( )

� 52( )� 1

2( )2� 3( )

= 4 �23�

P1 �P2( ) a3b3�

2μL a2+ b2( )

�

32( ) 1

2( )� 12( )� 1

2( )2 � 2!

=� P1 �P2( ) a3b3

�

4μL a2+ b2( )

(3)

since � 1

2( ) = � . When a = b = R, this mass rate of flow expressionreduces 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( ) A2

�2( )�

8μL(4)

Elliptical tube:

well =� P1 �P2( ) A2

�2( ) ab

4μL a2+ b2( )

(5)

so that

well

w�

= 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 lessthan unity as � becomes smaller.

13

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 muchkinetic 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

12 �vz

2 , since the only nonzero velocity component is vz . Therefore thetotal amount of kinetic energy per unit volume flowing through thetube is:

12 �vz

2( )vzrdrd�0

R�0

2�� = 2� 1

2 �vz2( )vzrdr

0

R�

= 2� � 1

2 � �R2 vz

30

1� �d�

= ��R2vmax

3 1 � � 2( )0

1�

3�d� (1)

In the second line, we have introduced the dimensionless coordinate

� = r R and the maximum velocity vmax = P0 �PL( )R2 4μL . Now allthat remains is to evaluate the integral:

��R2vmax

3 1 � 3� 2+ 3�4

� �6( )0

1� �d� = ��R2vmax

3 12�

34+

36�

18

�

�

��

= ��R2vmax

3 4 � 6 + 4 � 18

�

���

=

18��R2vmax

3

=

14

�R2vmax( )12�vmax

2�

���

�(2)

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

14

Note to p. 52

The laminar flow in a circular tube with radius R is discussed in§2.3, and the laminar flow in tubes with equilateral triangular cross-section of height H is 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�=� P 0 �P L( )R4

�

8μL w

�=

3 P 0 �P L( )H 4�

320μL(1,2)

In Eq. 1, R is the tube radius; in Eq. 2, H is the height of the triangularcross 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 ofthe cross-sectional areas. Since for circular tubes

A� = �R2 , and for

equilateral triangular tubes, A

�=

1

3H 2 ,

w�=� P 0 �P L( ) A2

�2( )�

8μL;

w

�=

3 3 P 0 �P L( )A2�

180μL(3,4)

Therefore

w�

w�

=3 3 16( )

180�8� = 0.726 (5)

b. The perimeters of the two tubes are P� = 2�R for circular

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

15

w�=� P 0 �P L( ) P 2�( )

4�

8μL w

�=

3 P 0 �P L( ) P 2 3( )4�

180μL(6,7)

Taking the ratio, as before

w�

w�

=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 andLifshitz, Fluid Mechanics, Addison Wesley (1959); our H is their amultiplied by

12 3 .]

16

Note to p. 54

Let's check a few things about Fig. 2.4-1.a. There the graph shows the transport of z-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 =

P 0 �P L( )R2L

rR

�

��

��� �

2 Rr

�

��

���

�

�� =

P 0 �P L( )R2L

1 � �2 R

r�

��

��

2�

�

�

��

Rr

(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 of � rz 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):

dds

1� s2�

1�� 2

ln 1 �( )ln

1s

�

��

�

�� = 0 � 2s +

1�� 2

ln 1 �( )

�

��

�

1s

(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

17

rmax = R

1�� 2

2 ln 1 �( )(4)

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

s =12 (1 +� ) . Therefore we now have to prove that

1 �� 2

2 ln 1 �( )<

12

(1+� ) or

2 1 ��( )1+�( ) ln 1 �( )

< 1 (5a,b)

It is easy to calculate the left side of Eq. 5b as a function of � :

�

2 1 ��( )1+�( ) ln 1 �( )

�

2 1 ��( )1+�( ) ln 1 �( )

0.1 0.711 0.6 0.9800.2 0.829 0.7 0.9910.3 0.895 0.8 0.999+0.4 0.935 0.9 0.999+0.5 0.962

It is evident that for � > 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 �P L( )R2

4μL1�

rmax

R�

��

�

2

�1 �� 2

ln 1 �( )ln

Rrmax

�

��

�

�

�

��

�

�

��

=P 0 �P L( )R2

4μL1�

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.

18

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:

lim��0

1 �� 4( ) = 1 (1)

lim��0

1 �� 2( )2= 1 (2)

l im��0

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.

19

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

vz

I=

p0 � pL( )b2

2μ IL1b

2μ I

μI+ μ

II

�

��

�

��+

μI� μ

II

μI+ μ

II

�

��

�

��xb

�

���

���

xb

�

���

��

2

��

�

�

dx�b

0

�

=

p0 � pL( )b2

2μ IL2μ I

μI+ μ

II

�

��

��+

μI� μ

II

μI+ μ

II

�

��

��� � �

2�

��

�

��d�

�1

0

� where � = x b

=

p0 � pL( )b2

2μ IL2μ I

μI+ μ

II

�

�

��

12

μI� μ

II

μI+ μ

II

�

�

��

13

�

�

��

=p0 � pL( )b2

2μ IL

6 � 2μ I� 3 μ

I� μ

II( ) � 2 μI+ μ

II( )μ

I+ μ

II

�

�

��

�

�

��

=

p0 � pL( )b2

2μ IL7μ I

+ μII

μI+ μ

II

�

��

�

�� (1)

Similarly for Region II we have

vz

II=

p0 � pL( )b2

2μ IIL1b

2μ II

μI+ μ

II

�

�

+

μI� μ

II

μI+ μ

II

�

�

xb

�

�

�

xb

�

�

2�

��

�

�

��

dx0

b

=

p0 � pL( )b2

2μ IIL2μ II

μI+ μ

II

�

��

��+

μI� μ

II

μI+ μ

II

�

��

��� � �

2�

��

�

��d�

0

1

�

=

p0 � pL( )b2

2μ IIL2μ II

μI+ μ

II

�

�

�+

12

μI� μ

II

μI+ μ

II

�

�

��

13

�

�

��

=p0 � pL( )b2

2μ IIL

6 � 2μ II+ 3 μ

I� μ

II( ) � 2 μI+ μ

II( )μ

I+ μ

II

�

�

��

�

�

��

=

p0 � pL( )b2

2μ IILμ

I+ 7μ II

μI+ μ

II

�

��

�

�� (2)

20

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 = �2μ

vr

r= �2μv

�

1R

�

��

��+

32

Rr

�

��

��

2

�32

Rr

�

��

��

4�

�

��

�

�

��cos�

=

3μv�

R�

Rr

�

��

��

2

+Rr

�

��

��

4�

�

�

��cos� (1)

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

� r� = �μ r

rv�

r�

��

��+

1r vr

�

�

��

�

�� = �μv

�

1R

�

��

��Rr

�

��

���

32

Rr

�

��

��

2

�Rr

�

��

��

4�

�

��

�

�

��sin�

�μv

�

1R

�

��

��Rr

�

��

���

32

Rr

�

��

��

2

�12

Rr

�

��

��

4�

�

�

���sin�( )

= +

32μv

�

RRr

�

���

�

4

sin� = +

32μv

�

RRr

�

���

�

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 thenecessary 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:

21

F n( )

= �p0 + gRcos +32μv

�

Rcos

�

��0

�

�0

2�� cos�( )R2sin�d�d�

= 2� �p0 + �gRcos� +

32μv

�

Rcos�

�

��

�

�0

� cos�( )R2sin�d�

= 2�p0R2 cos�

0

�

� sin� d� + 2�R2�gR +

32μv

R�

�

�cos2

� sin� d�0

�

� (3)

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

F n( )

= 2�R2�gR +

32μv

�

R�

���

23

�

���

=

43�R3

�g + 2�μRv�

(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�32μv

�

Rsin�

�

���

�0

�

� sin�( )R2sin� d�

= 3�μRv

�sin3

�0

�

� d� = 3�μRv�

43

�

���

�= 4�μRv

�(5)

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

22

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 thenormal 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 = �2μ

vr

r= �2μv

1R

+32

Rr

�

��

��

2

� 312

�

��

��Rr

�

��

��

4�

�

��

�

�

��

cos�

=

3μv�

R�

Rr

�

��

��

2

+Rr

�

��

��

4�

�

�

��cos� (1)

b. Then we get ���

from Eq. B.1-16:

���= �2μ

1r�v

�

��+

vr

r�

��

�

= �2μv

�

1R

�Rr

�

��

��+

34

Rr

�

��

��

2

+14

Rr

�

��

��

4�

�

�

��cos�

�2μv

�

1R

Rr

�

��

���

32

Rr

�

��

��

2

+12

Rr

�

��

��

4�

�

�

��cos�

= �

2μv�

R�

34

Rr

�

��

��

2

+34

Rr

�

��

��

4�

�

�

��

cos�

= �

3μv�

2R�

Rr

�

��

��

2

+Rr

�

��

��

4�

�

�

��cos� (2)

c. And, finally, we get ���

from Eq. B.1-17:

23

���= �2μ

1r sin�

�v�

��+

vr + v�

cot�r

�

��

�

= �2μv�

1R

Rr

�

��

��1�

32

Rr+

12

Rr

�

��

��

3�

��

�� cos� +

Rr

�

��

���1+

34

Rr+

14

Rr

�

��

��

3�

��

�� cos�

�

�

�

��

= �

3μv�

2R�

Rr

�

��

��

2

+Rr

�

��

��

4�

�

�

��cos� (3)

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

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

�

�tv

�

���

�+ 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

25

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

=

�

�t12� v �v( )

�

�

���

12

v �v( )�

�t� + �v � v ��v[ ]( ) (5)

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

=

�

�t12� v v( )

�

���

+

12

v v( ) � �v( ) + �v v �v[ ]( ) (6)

Now the second and third term may be combined by using Eq. A.4-19 with s replaced by

12 v �v( ) and v replaced by �v :

=

�

�t12� v v( )

�

���

+ �

12� v v( )v

�

���

=

�

�t12�v2�

���

+ �

12�v2v

�

���

(7)

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

26

Note to p. 82 (i)

We want to verify that ���:�v( ) , when written for a Newtonianfluid, 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( ) = � 1

2 �: ��( ) = 12 μ �� � 2

3� �v( )��� �� :��( ) + 1

2� v( ) �:��( ) (1)

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

���:�v( ) = 1

2 μ �� :��( ) � 4

3� �v( )

2��

�+ � �v( )

2(2)

This is equivalent to:

���:v( ) = 1

2 μ �� � 2

3 v( )��� �� : �� � 2

3 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:

��� � 2

3� �v( )��� : �� � 2

3� �v( )��� ( )

=��� :��( ) � 4

3� �v( )

2� 4

3� �v( )

2+ 4

9� 3 � �v( )

2

=��� :��( ) � 4

3� �v( )

2(4)

which is the coefficient of 12 μ in Eq. 2.

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 formthe cross product, term by term:

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

r �

�

�t�v

�

��

�

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

= � ijkx j�

�xl

�vlvkl��

��

�

�

k�

j�

= � ijk

l�

k�

j�

�

�xl

�xjvlvk � � ijk�vlvkl�

k�

j� � jl

=

�

�xl

�vl� ijk xjl�

k�

j� vk � � ijk�vj vk

k�

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 (

vj vk ) 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:

28

r � �p[ ]i

= � ijkk�

j� xj

�

�xk

p = � ijkk�

j�

�

�xk

xj p � � ijkk�

j� � jk p (4)

The last term is zero, since it involves a double sum on a pair ofindices that appear symmetrically in one factor and antisymmetric inanother. The other term can be rearranged as follows:

�

�xl

� ijk xj p�kl{ }l�

k�

j� =

�

�xl

� ijk xj p�klk�

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:

�

�xl

r � p�{ }il=

l�

�

�xl

r � 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 thesame manner as in part (c) above:

r � � � ��[ ] �� i

= � ijk xjk�

j� � � �[ ]k

= � ijk xjk�

j�

�

�xl

� lkl� (7)

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

� ijk

l�

k�

j�

�

�xl

xj� lk � � ijkl�

k�

j� � lk

�

�xl

xj

=

�

�xl

� ijk xj� kl†

k�

j�

�

�

�

��

l� � ijk

l� � lk

k�

j� � jl

=

�

�xl

r � �†{ }

il�

l� � ijk� jk

k�

j�

=

�

�xl

r � �†{ }

li

†+ � ikj� jk

k�

j�

l�

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.

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� � ijk vj � � v[ ]k

k�

j� = vi

i� � ijkv j �klm

�

�xl

vmm�

l�

k�

j� (3)

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

= � ijk�klmvi

m�

l�

k�

j�

i� vj

�

�xl

vm

= � ijk� lmk vi

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:

= � il� jm �� im� jl( )

m� vivj

�

�xl

vml�

j�

i� (5)

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

31

= vivj

j�

i�

�

�xi

vj � vivjj�

i�

�

�xj

vi (6)

In the second summation, we replace i by j and j by i to get:

= vivj

j�

i�

�

�xi

vj � vj vij�

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.

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

�

DDt

12 v2

+ �( ) = � 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 torewrite Eq. (1) as

� v �� 1

2 v2( ) + � v ���( ) = � v ��p( ) (2)

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

v �� 1

2 v2( ) + v � g�h( ) +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

dds

12 v2( ) � g

dhds

�1�

dpds

= 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.

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 thesystem (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 systembefore 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. 3.2-9and 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 state—that 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 of b = 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 =

2R2�o

2

1 �� 2( )2

�

���

�

�

���

2

�1�

d�

34

=�

2R2�o

2

1�� 2( )2

�

�2�

2�+�

2

�3

��

�� d�

=�

2R2�o

2

1�� 2( )2

12

�2

�2��

2

�2

��

�

� 2 ln� + C

��

�

���

(3)

The integration constant C is determined by the mass conservationstatement

�0�d� =

�

1� ��d�

�

1� or

��1dp�d� = 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

12

�2

�2��

2

�2

�

��

�

� 2ln� + C

�

�

���

�

1� �d� = 0 (6)

Performing the integration gives

12� 2

�4

4�

12�

2 ln� � 212�

2 ln� �14�

2�

��

��+

12

C� 2�

�

�

�

1

= 0 (7)

This gives for the integration constant

C = �1 �

1+� 2

4� 2�

3� 2 ln�1�� 2 (8)

Hence Eq. (3) becomes

��1dp

p0

p� =

�2R2

�o2

1 �� 2( )2

12

�2

�2��

2

�2

�

�

�

�� � 2ln�

�1�1 +� 2

4� 2�

3� 2 ln�1�� 2

�

�����

�

�

�����

(9)

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 � p0

�0

=�

2R2�o

2

1�� 2( )2

12

�2

�2��

2

�2

�

�

�

�� � 2 ln�

�1 �1+� 2

4� 2�

3� 2 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 � p0

�0

=�

4R2�i

2

1�� 2( )2

12

�2�

1�

2

�

�

�� � 2 ln�

�1 �1+� 2

4�

1+ 2� 2

1�� 2ln�

�

�����

�

�

�����

(11)

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 torquebalance on a shell of thickness �r and height L. Then the torque at

radius r is equal to 2�rL ��r� r

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

is 2 r + �r( )L� r� 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:

ddr

r2� r�( ) = 0 or

ddr

r 2 rddr

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

ddr

1r

ddr

rv�( )

�

�

��= 0 or

d2v�

dr2+

1r

dv�

dr�

v�

r2= 0 (2)

We start by performing the differentiations in the largeparentheses in Eq. 1b:

r2 r

ddr

v�

r�

��

��

�

�

�

� = r2 r

1r

dv�

dr�

v�

r2

�

��

��

�

�

�

� = r2 dv

�

dr� rv

� (3)

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

ddr

r2 dv�

dr� rv

�

�

���

�= 2r

dv�

dr+ r2 d2v

�

dr2� r

dv�

dr� v

�

= r2 d2v

�

dr2+

1r

dv�

dr�

v�

r2

�

��

�

� = 0 (4)

37

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

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

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

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 � :

dd�

erfc��=0

=d

d�1 �

2

�e��

2d�

0

�

��

���

�=0

= �

2

�e��

2

�=0= �

2

�= �1.1287 (1)

Here we have used the Leibniz formula for differentiating integrals in§C.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.

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 z1i

for z2 : z2r and z2 i

for 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

z1r wr � z1iwi = z2r wr � z2iwi (2)

Rearranging gives:

wr z1r � z2r( ) = wi z1i � z2 i( ) (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)

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

=14

(7)

Hence

a2

=12

,�12

and a = ±

1

2,±

1

2i (8)

whence

b = ±

1

2,±

1

2

1i

(9)

and the two square roots of i are

i = a + bi = ±

1

21 + i( ) (10)

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

i = rei�= 1ei� 2 (11)

Then the square root of i is

i = ±ei� 4

= ± cos�

4+ isin

�

4�

���

��= ±

1

2+ i

1

2

�

���

��(12)

41

Note to p. 123

Here we show how Eq. (A) of Table 4.2-1 is obtained.

The velocity is of the form

v x, y( ) = ��xvx x, y( ) + � yvy x, y( ) (1)

The velocity components are related to the stream function accordingto the relations in the 3rd column.

The first term in Eq. (A) comes from the first term in Eq. 4.2-1,which is

�

�t� � v[ ] = �z

�

�t� � v[ ]z = � z

�

�t

�vy

�x��vx

�y

�

��

= �� z

�

�t�

2�

�x2+�

2�

�y2

�

��

�

� = z

�

�t�

2� (2)

The right side of Eq. (A) comes from the right side of Eq. 4.2-1:

��

2� � v[ ] = �

�2

�x2+

�2

�y2

�

��

� � � v[ ]

= z�

�2

�x2+

�2

�y2

�

��

�

�

�2

�x2+�

2

�y2

�

��

�

� = �z��

4 (3)

The second term on the left side of Eq. (A) can be taken care ofsimilarly:

� � � v � � � v[ ]�� ��

= � � � v � ��z�

2�

�

��

= � � � ��x vx + � yvy( ) � �� z�

2�

�

��

= � � � ��� yvx�

2 + �xvy�

2�

�

��

42

=

��x � y �z

�

�x�

�y�

�z

0 vx�2 0

�

�x � y � z

�

�x�

�y�

�z

vy�2 0 0

= +��z

�

�xvx�

2�( ) + �z

�

�yvy�

2�( )

= �� z

�vx

�x+�vy

�y

�

��

� �

2� + �z vx

�

�x�

2� + vy

�

�y�

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�

2�

�

�y�

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.

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

�vr

�r�

���

2

=92

v�

2

r2��1� �

�3 �

��cos �( )

2(1)

2

1r�v

�

��+

vr

r�

���

2

=2v

�

2

r2�1 + 3

4 ��1+ 1

4 ��3

���cos� + 1� 3

2 ��1+ 1

2 ��3

���cos�( )

2

=

98

v�

2

r2��

�1+ �

�3�

�cos�( )

2(2)

2

v�cot�r

+vr

r�

��

��

2

=2v

�

2

r2�1 + 3

4 ��1+ 1

4 ��3

��cos� + 1 � 3

2 ��1+ 1

2 ��3

��cos�( )

2

=

98

v�

2

r2��

�1+ �

�3�

�cos�( )

2(3)

r�

�rv�

r�

���

+

1r�vr

��

�

��

�

2

=�v

�

�r�

v�

r+

1r�vr

��

�

���

2

=v�

2

r2

� 34 �

�1+ 3

4 ��3

��sin� + 1� 3

4 ��1� 1

4 ��3

��sin�

� 1 � 32 �

�1+ 1

2 ��3

��sin�

�

�

��

�

��

2

=

v�

2

r2� 3

2 ��3sin�

��

2(4)

When the above results are combined, we get

� ��:�v( )r2

= μv�

2 274 �

�2� 27

2 ��4

+ 274 �

�6�

�cos2

� + μv�

2 94 �

�6�

�sin2

�

(5)

44

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

Fk v

�= �2μ ��:�v( )r2dr

R

�

�0

� sin�d�

= 2 � 2

3 μv�

2 274 �

�2� 27

2 ��4+ 27

4 ��6�

��dr + 2 � 4

3 μv�

2 94 �

�6��

�dr

R

�

R

�

(6)

since

cos2�

0

�

� sin� d� = 23 and

sin3

�0

�

� d� = 43 . Then, finally

Fk = 9μv

�R �

�2� 2��4

+ ��6�

��1

�

d� + 6μv�

R ��6�

��1

�

d�

= 9μv

�R ��

�1+ 2

3 ��3� 1

5 ��5�

�� 1

�

+ 6μv�

R � 15 �

�5��

� 1

�

= 9μv

�R 1� 2

3 +15

�� �� + 6μv�

R 15

�� ��

= 9�μv

�R 2

5�� �� + 6�μv

�R 1

5�� �� =

245 + 6

5( )�μv�

R

= 6�μv�

R (7)

which is Stokes' law.

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

ve x( ) =

2c2 � �

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

�vx

�x+ vy

�vx

�y= ve

dve

dx+�

�2vx

�y2(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

�

��

���

2

�x�y

�

��

� +

�

�x�

��

���

2

�y2

�

��

� = ve

dve

dx+�

�2

�y2��

�y

�

��

�(3)

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

��

�y�

2�

�x�y���

�x�

2�

�y2=

c'2�

2 � �

�

��

�

1

x 2�3�( ) 2��( )��

�3�

�y3 (4)

Next we want to rewrite this equation in terms of f and � :

� 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

We start by converting the various derivatives from � x, y( ) to

derivatives of f �( ) :

��

�y= �Ax1 2��( ) df

d���

�y= �ABx1 2��( ) f '

1

x 1��( ) 2��( )= �ABx� 2��( ) f ' (7)

�2�

�y2= �ABx� 2��( ) ��

�yf " = �AB2 x� 2��( )

x 1��( ) 2��( )f " = �AB2 1

x 1�2�( ) 2��( )f " (8)

�3�

�y3= �AB2 1

x 1�2�( ) 2��( )

��

�y���f = �AB3 1

x 2�3�( ) 2��( )���f (9)

��

�x= �A

ddx

x1 2��( )�

���

f � Ax1 2��( ) df

d���

�x

= �A

x1 2��( )

2 � �( )1x

f � Ax1 2��( )�

By 1 � �( )2 � �( )

x� 1��( ) 2��( )�1�

��

�

� f '

= �A

12 � �( )

x1 2��( )

xf + A

1� �( )2 � �( )

x1 2��( )

x� f ' (10)

�2�

�y�x= �A

12 � �( )

x1 2��( )

x��

�yf '+ A

1� �( )2 � �( )

x1 2��( )

x��

�yd

d�� f '( )

= �AB

12 � �( )

x1 2��( )

x 1��( ) 2��( )xf '+ AB

1� �( )2 � �( )

x1 2��( )

x 1��( ) 2��( )x� f "+ f '( ) (11)

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

c'2�

2 � �

�

��

�

1

x 2�3�( ) 2��( )+ �AB3 1

x 2�3�( ) 2��( )���f =

c'2

2 � �

1

x 2�3�( ) 2��( )� + ���f( )

(12)

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

��

�y�

2�

�x�y���

�x�

2�

�y2= A2B2 1

2 � �( )x� 2��( )x1 2��( )

x 1��( ) 2��( )x�f 2

47

�A2B2 1� �( )

2 � �( )x� 2��( )x1 2��( )

x 1��( ) 2��( )x� �f f "+ �f 2( )

�A2B2 1

2 � �( )x1 2��( )

x 1�2�( ) 2��( )xf ��f

+ A2B2 1 � �( )

2 � �( )x1 2��( )

x 1�2�( ) 2��( )x� �f f "

= c'2 �

2 � �x� 2�3�( ) 2��( )

�f 2� c'2 1

2 � �( )x� 2�3�( ) 2��( ) f ��f

= c'2 1

2 � �( )x� 2�3�( ) 2��( )

� �f 2� f ��f( ) (13)

Combining the results in Eqs. 12 and 13 gives

c'2 1

2 � �( )x� 2�3�( ) 2��( )

� �f 2� f ��f( ) = c'2 1

2 � �( )x� 2�3�( ) 2��( )

� + ���f( )

(14)or

� �f 2� 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.]

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 =vzrdrd�

0

R

�0

2

�

rdrd�0

R

�0

2

�= vz ,max

1 � r R( )�� ��1 7

rdrd�0

R

�0

2

�

rdrd�0

R

�0

2

�

= vz ,max

2R2 1� 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.

49

Note to p. 170

Here we fill in some of the missing steps following Eq. 5.6-18.Setting C1 = 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

12 F2( )� = � �F( )� � 2 �F (2)

Each term may now be integrated with respect to � to give

12 F2

= � �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

�

dFd�

= 2F + 12 F2 or

dF2F 1 + 1

4 F( )=

d��

(4)

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

� 1

2 ln1+ 1

4 F

F= ln� + ln C3 or

ln

1+ 14 F

F= ln� + ln C3 (5a,b)

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

50

F1+ 1

4 F= C3�( )

2(6)

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

4 F( ) C3�( )2

or

F =C3�( )

2

1� 14 C3�( )

2(7)

Case II (negative sign):

F = � 1+ 1

4 F( ) C3�( )2

or

F = �C3�( )

2

1+ 14 C3�( )

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 approaches

minus 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 J in Eq.5.6-2 gives:

J = 2 �vz2

0

�

rdr = 2�� t( )22C3

2( )2

1+ 14 C3( )

2��

��

40

�

d (9)

We now let 14 C3�( )

2= x2 so that

�d� = 4 C3

2( )xdx ; then

51

J = 32�� t( )2C32 xdx

1+ x2( )40

�

= 32�� t( )2C32 1

6 1 + x2( )3

�

�

���

�

���

0

�

=163��

t( )2C32 (10)

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

w = 2 �vz0

�

rdr = 2��

t( )

z

2C32( )

1+ 14 C3( )

2��

��

20

�

d � z2

= 4�� t( )�zC3

2�

4C3

2

xdx

1+ x2( )20

�

= 16��� t( )z �1

2 1+ x2( )

�

�

�

�

0

�

= 8�� t( )z (11)

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:

ddt

�dV � n ��vS( )dS =S t( )�V t( )� � n � �v( )dS

S 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

ddt

�dVV t( )� = � n �� v � vS( )( )dS

S 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.

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

ddt

mtot = � n � �v( )dS � n � �v( )dSS2�S1

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

ddt

mtot = +�1 vdS � �1 vdSS2�S1

� (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)

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

ddt

�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. If g does not change over the volume of the flow system,then it may be removed from the integral, which gives mtot . Then

ddt

Ptot = � 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

� �v2dS � u2 �u2u2�� ��S2� �v2dS

= +�1 v1

2 S1u1 � �2 v22 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 � ��[ ]dS

S t( )� = + u1 � ��� ��S1� dS � u2 � ��� ��S2

� pdS � n � �[ ]dSSf +Sm�

55

� �Ff�s

��( ) (12)

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

p( )+ Ff �s

��( )= Ff �s = �Fs� f ):

ddt

Ptot = �1 v12 S1u1 � �2 v2

2 S2u2 + p1S1u1 � p2S2u2 + Fs� f + mtotg (13)

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

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 thedifference 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 x

2= 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( )dz

0

h�

= Rz2� 1

3 z3( )0

h

= Rh2

� 13 h3( ) (3)

Thus the liquid volume is:

V = Rh2 1 �

13

hR

�

���

�(4)

We may check this result at three points where we know the result:Tank full: h = 2R V = 4

3 �R3

Tank half full: h = R V = 23 �R3

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

ddt

Rh2 1�13

hR

�

���

� = �

P2 �P3( )D4�

128μL(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, and

h3 is the distance upward from the datum plane (i.e., plane "3"),which is zero. Hence, Eq. 5 becomes

�

ddt

Rh2 1 �13

hR

�

���

= �gh + �gL( )D4

128μL(8)

c. Eq. 8 may be rewritten as

�

Rh + L

ddt

h2�

13

h3

R

�

��

�

=

�gD4

128μL� A (9)

(which defines the quantity A) or

58

�

Rh + L

2h �h2

R

�

��

�

��

dhdt

= 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

HdHdt

= A (11)

or

H � 2 R + L( ) + 2R + L( )L

1H

�

��

�

��

dHdt

= A (12)

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

12

H 2� 2 R + L( )H + 2R + L( )L ln H

�

��

�

��

2R+L

L

= At0

tefflux (13)

This gives

12

L2�

12

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

L2R + L

= Atefflux (14)or after some cancellations

tefflux =

1A

2R2+ 2RL � 2R + L( )L ln

2R + LL

�

��

�

�� (15)

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

textbook is obtained.

59

Note to p. 202

A lawn sprinkler has four arms of length L and cross-sectional area S.Water enters the sprinkler at the center at a mass flow rate w andsplits into four streams. It leaves the sprinkler at right angles to thesprinkler arms. It is desired to find the angular velocity � of thesprinkler, when there is a frictional torque of

Tf .

The flow velocity is each arm is v = w 4S , so that the velocity at theoutlet stream, relative to the sprinkler arm is (at plane "2")

v2 = �� y

w4�S

� L��

���

Therefore, Eq. 7.3-2 gives for the angular momentum balance atsteady state (here we take the arm to be projecting in the x direction,the fluid to be issuing in the y direction, and the torque to be in the zdirection)

0 � w L��x � �� y

w4�S

� L��

��

�

�

�

�

�� �Tf �z

the entering stream contributing nothing to the angular momentumbalance. The unit vectors may now be removed

Tf = wL

w4�S

� L��

���

Then, solving for � we get

� =

w4�SL

�Tf

wL2

for the angular velocity of the sprinkler.

[For more on this problem, see Surely You're Joking Mr. Feynman," byR. P. Feynman, Bantam Books, New York (1986), pp. 51-53.]

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:

2h

d2h

dt2= 2h

ddt

u( ) = 2h 12 u�1 2 du

dt�

���

��= hu�1 2 du

dhdhdt

= hdudh

(1)

from which Eq. 7.7-8 follows at once:

h

dudh

� 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( )hN�2

+2g

N � 2�

��

�

�� � N �1( ) ChN�1

+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:

2ghN � 2

� N �1( )2gh

N � 2+

N � 2N � 2

2gh = 0 (4)

C-terms: C N � 1( )hN�1� N �1( )ChN�1

= 0 (5)

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

dhdt

�

���

��

2

= ChN�1+

2ghN � 2

(6)

to get an equation for the constant of integration

61

2gH

R0

R�

���

��

4

= CHN�1+

2gHN � 2

(7)

When this is solved for C, we get (noting that N = R R0( ) as definedimmediately after Eq. 7.7-3):

C =

2g

HN�2

1N

�1

N � 2�

���

��= �

4g

HN�2

1N N � 2( )

�

��

�

�� (8)

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

When we take the square root of Eq. 7.7-9 and introduce thedimensionless liquid height � = h H , we then get Eq. 7.7-10:

d�dt

= ±2g

N � 2( )H� �

2N�

N�1�

���

�(9)

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

2g1

� � 2 N( )�N�1d�

0

1�0

tefflux�

�

NH2g

� N( ) (10)

Keep in mind that tefflux = NH 2g is the quasi-steady-state solutionin Eq. 7.7-3—the solution that we would expect to be valid when N isextremely 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

62

� N( ) =12

N � 2N

1

� � 2 N( )�N�1d�

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( ) =

12

N � 2N

1

�1 �

2N

�N�2�

���

�1 2

d�0

1� (12)

The quantity to the �12 power can then be expanded in a Taylor

series (see Eq. C.2-1) about 2 N( )�N�2= 0 to get, after integrating

term by term

� N( ) =

12

N � 2N

1

�1+

11!

12

2N�

N�2�

���

�+

12!

12

32

2N

�N�2�

���

�

2

�

�

�

�

��

d�0

1�

=12

N � 2N

2 +1

N N � 32( )

+3

4N 2 N � 74( )

+��

��

�

�� (13)

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

� N( ) = 1 �

1N

�12

1N 2

+ O1

N 3

�

���

���

�

�� +

12

1N 2

+ O1

N 3

�

���

���

�

�� +

38

O1

N 3

�

���

���

�

�� +�

= 1�

1N

+O1

N 3

�

���

��(14)

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

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�� vzrdr

0

R = �2��

dvz

dr�

���

��r2

2

�

��

�

�� dr

0

R = +�� �� r2dr

0

R (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�R

3��� rz

20

�R d� rz (2)

In this equation, the shear rate �� is to be regarded as a function of the

shear 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

dd�R

�R3 w

R3�

�

��

�

= �� R�R

2 (3)

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

64

how the wall shear rate �� R can be obtained by differentiating the

flow-rate vs pressure-drop data.We now put Eq. 3 into a different form. Differentiating the

product with respect to �R gives

w

R3��3�R

2+ �R

3 dd�R

w

R3�

�

��

�

�� = �� R�R

2 (4)

Then division by R

2 w �R3�( ) gives

3 +

d ln w R3�( )

d ln�R

= �� R1

w R3�

(5)

Then finally

� �� R( ) =�R

�� R

=�R

w �R3�

3 +d ln w �R3

�( )d ln�R

�

�

��

�

�1

(6)

Equation 6 gives the viscosity as a function of shear rate at the wallfrom experimental measurements of w and P0 �PL . If we assume

that � �R( ) at the wall is the same as � ��( ) throughout the tube, then

Eq. 6 gives � ��( ) . This assumption seems to be valid for typical

polymeric 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.

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-lawmodel. First we have to obtain the expression for the scalar

�� that

appears in Eq. 8.3-5:

�� = 1

2�� : ��( ) = 1

2�� ij �� ji

j=1

3

�i=1

3

� and ��� = �v + �v( )

†(1,2)

where i and j 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 of v that is nonzero is the z-component and since that is a function of r 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 �

dvz

dr�

���

�

n�1dvz

dr= �

dvz

dr�

���

�

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:

66

m �

dvz

dr�

� �

��

n

=P0 �PL

2L�

� �

��r or

�

dvz

dr=

P0 �PL

2mL�

� �

��

1 n

r1 n (7,8)

Integration of Eq. 8 gives

vz = �

P0 �PL

2mL�

� �

��

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 �PL

2mL�

� �

��

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�

rR

�

�

�

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�0

2�� rdrd�

= 2R2

�P0 �PL( )R

2mL

�

��

�

��

1 nR

1 n( ) +11 � 1 n( )+1( )0

1 d (13)

where � = r R . Performing the integration gives

67

w = 2R2

�P0 �PL( )R

2mL

�

��

�

��

1 nR

1 n( ) + 1

1 n( ) + 1

2 1 n( ) + 3��

�

��

�

��

= R3

�P0 �PL( )R

2mL

�

��

�

1 n1

1 n( ) + 3(14)

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

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

i�t to appear explicitly on the right side of theequation (to match a similar factor on the left side):

�� i�v0ei�t{ } =�

d2v0

dy2ei�t �0

�1

e�s �1 e� i�sds0

�

��

�

���

��(1)

Next, we perform the integration over s:

�� i�v0ei�t{ } =�d2v0

dy2ei�t �0

�1

1� 1 �1( ) � i�

�

�

�� e�s �1 e� i�s

0

��

�

��

�

��

��

=�

d2v0

dy2ei�t �0

1+ i�1�

�

��

�

� �

��

���

��(2)

We may now remove the real-operator sign from both sides, as wellas the common multiplier e

i�t , to get:

�i�v0

=d2v0

dy2

�0

1 + i�1�

�

��

�

or

d2v0

dy2=

�i� 1 + i�1�( )�0

�

��

�

� v0 (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 a

2 . 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,

69

vx y,t( ) =� v0 y( )ei�t{ } =� Ae+ �+ i�( )y

+ Be� �+ i�( )y( )ei�t{ } (5)

Now we know that the oscillatory disturbance will not propagatewith increasing amplitude as the distance from the wall increases, sothat A 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�( )yei�t{ } = v0e��y

� ei �t��y( ){ } = v0e��ycos �t � �y( )(6)

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

�i� 1 + i�1�( )�0

= � + i�( )2

(7)

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

�

2� �

2= �

��1�2

�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�

��

2�0

�

��

21�

2= �

��1�2

�0

(10)

or, after multiplying through by �2

�

4+��1�

2

�0

�2�

��

2�0

�

��

2

= 0 (11)

70

This can be regarded as a quadratic equation in �2 , which has the

solution:

�

2= �

��1�2

2�0

±12

��1�2

�0

�

��

2

+ 4��

2�0

�

��

2

= �

��1�2

2�0

1 � 1 +1

�12�

2

�

��

�

��= �

��

2�0

�1� � 1 + �12�

2��

��

(12)

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

� = ±

��

2�0

�1� � 1+ �12�

2��

��

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:

� = +

��

2�0

�1� + 1 + �12�

2��

��

1 2

(14)

� = +

��

2�0

�1� + 1+ �12�

2�

� �

�1 2

(15)

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

� =

��

2�0

1+ �12�

2� �1�

�

� �

1 2

(16)

� =

��

2�0

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.

71

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

72

Note to p. 250

The simplest nonlinear viscoelastic model for polymers is thecorotational Maxwell model, which is obtained by replacing the partialtime 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-rateviscosity, and � , the relaxation time.

Here we show how to get the viscosity, the first normal-stresscoefficient, 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 01 0 00 0 0

�

�

��

�

�

��

�v( )†=

0 1 00 0 00 0 0

�

�

��

�

�

�� (2,3)

��� =

0 1 01 0 00 0 0

�

�

�

�

��

��

�� =

0 �1 01 0 00 0 0

�

�

��

�

�� (4,5)

Since the stress tensor �� does not depend on position and time, in theJaumann derivative of �� , the substantial derivative will notcontribute and we are left with only

12 � � � � � ��{ } .

73

�� � { } =0 �1 01 0 00 0 0

�

�

��

�

�

��

xx xy xz

yx yy yz

zx zy zz

�

�

���

�

�

���

�� =

� yx � yy � yz

xx xy xz

0 0 0

�

�

���

�

�

���

�� (6)

��{ } =

xx xy xz

yx yy yz

zx zy zz

�

�

���

�

�

���

0 �1 01 0 00 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

D t=

� yx12 xx � yy( ) � 1

2 yz

12 xx � yy( ) yx

12 xz

� 12 yz

12 xz 0

�

�

����

�

�� (8)

Thus the corotational Maxwell model in matrix form is

� xx � xy � xz

� yx � yy � yz

� zx � zy � zz

�

�

���

�

+ �

�� yx12 � xx � � yy( ) � 1

2 � yz

12 � xx � � yy( ) � yx

12 � xz

� 12 � yz

12 � xz 0

�

�

����

�

��

= ��0

0 1 01 0 00 0 0

�

�

��

�

�� (9)

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

� xx � � yy( ) � 2� yx� �� = 0 (10)

� yy � � zz( ) + � yx� �� = 0 (11)

74

� yx +

12 � xx � � yy( )� �� = ��0 �� (12)

� xz �

12 � yz� �� = 0 (13)

yz +

12 xz� �� = 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 � � zz

we find

� yx = ��0 ��

1

1+ � ��( )2

or

�

�0

=1

1+ � ��( )2

(15,16)

� xx � � yy = ��0 ��

2� ��

1 + � ��( )2

or

�1

�0�=

2

1 + � ��( )2

(17,18)

� yy � � zz = +�0 ��

� ��

1+ � ��( )2

or

�2

�0�= �

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 shearstress goes through a maximum at

� �� = 1. The second normal stress

is 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 qualitativebehavior.

75

Note to p. 255

Since the FENE-P dumbbell model (a molecular model)describes many of the observed rheological phenomena of polymersolutions, 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 �� yz

�� zx �� zy �� zz

�

�

�

�

���

=

0 �� 00 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 ��

2� p,xy � p,yy � p,yz

� p,yy 0 0

� p,yz 0 0

�

�

���

���

= �nKT�H ��

0 1 01 0 00 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:

Z p,xx = 2 p,yx�H �� and

Z� p,yx = �nKT�H �� (3,4)

in which

Z = 1+ 3 b( ) 1 � tr ��p 3nKT( )�

���= 1+ 3 b( ) 1 � � p,xx 3nKT( )�

���

(5)

76

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:

�

N3S

= 2S and ��

3S= 1 +

3b

1 � N( ) (12,13)

Then N may 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

3 arcsinh p�3 2q( ) (16)

or, in terms of the original variables:

77

� p,yx

3nKT= �2

b + 354

�

���

�

+1 2

sinh 13 arcsinh

b + 354

�

���

�

�3 2b

108�H ��

�

��

��

�

��

��

��

�

�

��

(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-Newtonianviscosity 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 1

3 p�3 2q ����

��( ) = �2p1 2 1

3 p�3 2q ����

�� +�{ } = � 2

3 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 + 16 x3

+� arcsinh x = x � 1

6 x3+� (19,20)

In terms of the original variables, Eq. 18 is

� p,yx = �3nKT � 2

3

b + 354

�

��

�

�1b

108�H �� = �nkT

bb + 3

�

��

��H �� (21)

This corresponds to

� ��s = nkT�H

bb + 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 Appendix

C.5). Therefore

S = �2p1 2sinh 1

3 ln 2p�3 2q( )( ) = �2p1 2 12 exp 1

3 ln 2p�3 2q( )( ) = � 2q( )1 3

(23)

78

In the original variables, this becomes

� p,yx = �3nKT 2 �

b108

�H ���

��

�

1 3

= �nKT 54 �b

108�H ��

�

��

�

1 3

(24)

This corresponds to a "power-law function":

� ��s = nKT�H

b2

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 = � 23 p�1q (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, see“Teaching with FENE Dumbbells,” by R. B. Bird, Rheology Bulletin,January 2007.]

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)

� = μ0 +

0

��when � � � 0 (2)

in which �0 is the yield stress, and μ0 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 �PL

2L�

���

�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 is

defined 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

� �� = μ0 �� + �0 or � rz = μ0 �� + �0 (4a,b)

80

Combining Eqs. 3 and 4b and rearranging gives

P0 �PL( )r2L

= μ0 �� + � 0 (5)

Solving Eq. 5 for �� we get

�� =1μ0

P0 �PL( )r2L

� 2P0 �PL( )r�0

2L+ �0

�

�

��

�

�� 0

μ0

rr0

� 2rr0

+ 1�

��

�

(6)

Inserting �� = �dvz

> dr , we can get a differential equation for the

dimensionless velocity profile �>= μ0 r0�0( )vz

> as a function of the

dimensionless radial coordinate � = r r0

�

d�>

d�= � � 2 � + 1 (7)

Integration of this equation, and using the boundary condition that

�>= 0 at � = R r0 , we get for r0 � r � R

�>=

12

Rr0

�

��

��

2

1�rR

�

��

��

2�

��

�

���

43

Rr0

�

��

��

3 2

1 �rR

�

��

��

3 2�

��

�

��+

Rr0

1 �rR

�

��

�� (8)

and for 0 � r � r0

�<=

12

Rr0

�

��

�

�

2

�43

Rr0

�

��

�

�

3 2

+Rr0

�16

(9)

The mass rate of flow can then be obtained by using Eq. 2.3-21 afterdoing an integration by parts. The dashed underlined terms are zero.

81

w = 2� 1

2 vzr2

r=0

r=R� 1

2 r2 dvz

drdr

0

R�

�

��

� = �� r2 dvz

drdr + r2 dvz

drdr

r0

R�0

r0�

�

��

�

----------- --------------

= ���

r03�0

μ0

�

�

� �

2 d�>

d�d� =

1

R r0� + ��

r03� 0

μ0

�

�

� �

2� � 2 � + 1( )d�1

R r0�

= �r0

3� 0

μ0

�

��

�

��

14

Rr0

�

��

�

��

4

�1

��

�

�

���

47

Rr0

�

��

�

��

7 2

�1

��

�

�

��+

13

Rr0

�

��

�

��

3

�1

��

�

�

��

�

�

��

�

��

��

= �r0�0R4

μ0

�

��

�

� 1 �

167

r0

R�

���

�

1 2

+43

r0

R�

���

��

121

r0

R�

���

�

4

��

�

�

��

= P0 �PL( )R4

�

8μ0L

1�167

2L� 0

P0 �PL( )R+

43

2L�0

P0 �PL( )R

�121

2L�0

P0 �PL( )R�

��

�

�

4

�����

�

�

�����

(10)

When �0 0 and μ0 � μ , the Newtonian result is recovered.

82

Note to p. 275

As an exercise in identifying the dimensions of thermalquantities, we verify that the following equations are dimensionallyconsistent: Eq. 9.3-12, Eq. 9.8-6, and Eq. 9.8-8. We do this byreplacing the symbols in these equation by the correspondingdimensions, making use of the "Notation" table on pp. 872 et seq.

(a)

ML

t3T�

���

��=

M( )ML2

t2T

�

��

�

�� T( )

L2

ML2

t2T

�

��

�

��

M (1)

(b)

M

t3

�

���

��=

M

L3

�

���

��Lt

�

���

��

2Lt

�

���

��+

M

L3

�

���

��L2

t2

�

��

�

��

Lt

�

���

��+

M

Lt2

�

���

��Lt

�

���

��+

M

t3

�

���

�� (2)

(c)

L2

t2

�

��

�

�� =

L2

t2T

�

��

�

�� T( ) +

L3

M

�

��

�

��

M

Lt2

�

���

��+ T( )

L3

M

�

��

�

��

T( )M

Lt2

�

���

�� (3)

In each of these cases, each term has the same dimensions.

83

Note to p. 286

We illustrate the use of Eq. 9.8-8 for an ideal monatomic gas.For the ideal monatomic gas, the heat capacity at constant pressure isgiven by

Cp =

52

RM

(1)

as given two lines after Eq. 9.3-15.The ideal gas law equation of state is

pV =

RTM

(2)

Therefore, the bracket in Eq. 9.8-8 is

V �T�V�T

�

��

�

�

p

= V � T�

�TRTpM

�

��

�

�

�

��

�

�

p

= V �TR

pM

�

��

�

�= 0 (3)

Hence, the only term that survives is the heat-capacity term, which is:

H � Ho

=52

RM

T �To( ) (4)

84

Note to p. 299

In connection with Fig. 10.4-2, we want to find the location ofthe maximum in the temperature vs. distance curve. We will expressthe result in terms of the Brinkman number, Br. We also need toknow what happens when the Brinkman number goes to zero (i.e.,negligible viscous heating).

To simplify the discussion, we introduce the followingdimensionless variables:

� =

T � T0

Tb �T0

(temperature) � =

xb

(distance) (1)

so that Eq. 10.4-9 becomes:

� =

12

Br� 1� �( ) + � (2)

To get the location of the maximum of the temperature, wedifferentiate with respect to � and set the derivative d� d� equal tozero:

d�d�

=12

Br 1 � 2�( ) + 1 = 0 (3)

From this we get the location of the maximum in the temperaturecurve:

�max =

12+

1Br

(4)

When there is negligible viscous heating, Br � 0 , and, according toEq. 4, �max � � . This result is nonsense, since the system extends onlyto � = 1 . Therefore, Eq. 4 has to be restricted to �max � 1, and when

�max = 1, Br = 2. Alternatvely, Eq. 4 has to be limited to 2 � Br<� .

85

Note to p. 309

It's always a good idea to check the solutions to problems. Herewe verify that the expression in Eq. 10.7-13 for the dimensionlesstemperature in the cooling fin satisfies the differential equation in Eq.10.7-9, and the boundary conditions in Eqs. 10.7-10 and 11.

First calculate the derivatives from Eq. 10.7-13 and Eqs. C.5-10and 11:

d�d�

=sinh N 1 ��( )

cosh N�N( ) (1)

d2�

d� 2=

cosh N 1 ��( )cosh N

+N 2( ) (2)

When the expression for � (in Eq. 10.7-13) and its second derivative(in Eq. 2) are substituted into Eq. 10.7-9, an identity results. Therefore,Eq. 10.7-13 satisfies the differential equation.

The boundary condition at � = 0 is satisfied, since

� =cosh N 1 ��( )

cosh N�=0

= 1 (3)

and the boundary condition at � = 1 is also satisfied, since

d�d�

� =1

=sinh N 1 ��( )

cosh N�N( )

�=1

= 0 (4)

The hyperbolic sine is shown in Figure C.5-2.

86

Note to p. 315

Here we verify the determination of the constants of integra-tion, C0 , C1 , and C2 in Eq. 10.8-27.

C1 : Boundary condition 1 requires that � 0,�( ) = finite :

� 0,�( ) = C0� + C1ln 0 + C2 (1)

This can be satisfied only if C1 = 0 .

C0 : Boundary condition 2 requires that �� �� = 1 at � = 1 :

��

���=1

= C02�4

�4�3

16

�

��

�

�=1

=14

C0 (2)

from which it follows that C0 = 4 .

C2 : The dimensionless form of condition 4 is given in Eq. 10.8-25.Substitution of Eq. 10.8-27 into that equation gives:

� = 4� + 4

�2

4��

4

16

�

��

�

� + C2

�

�

���

0

1� 1� �2( )�d� (3)

Next we evaluate the integrals:

4� 1� �2( )�d� = 4� � � �

3( )d� = 4�12�

14

�

��

�0

1�0

1� = � (4)

4

�2

4��

4

16

�

��

��0

1� 1 � �2( )�d� = 4

�3

4�

5�5

16+�

7

16

�

��

�

��0

1� d� =

4

116

�5

6 �16+

18 �16

�

���

�=

724 � 4

(5)

87

C2 1� �2( )�d� =

0

1�

14

C2 (6)

Combining the results of Eq. 3 to 6, we find that C2 = �7 24 .

88

Note to p. 337

In the textbook, we transformed Eq. 11.2-2 into an equation forthe enthalpy (Eq. 11.2-3) and then used an equilibrium thermo-dynamic formula to get the equation of energy in terms of

Cp and T,

given in Eq. 11.2-5 (and also Eq. (J) in Table 11.4-1). Specifically, weused the equation for H T, p( ) :

dH =�H�T

�

��

�

��

p

dT +�H�p

�

��

�

��

T

dp

= CpdT + V � T�V�T

�

��

�

p

�

�

��

�

��

dp (1)

appropriately rewritten for a "particle" of fluid moving with the localvelocity v.

Alternatively, we can begin with Eq. 11.2-2 and use anotherequilibrium thermodynamic formula to get the equation of energy interms of CV and T (see Eq. (I) in Table 11.4-1). Specifically, we use

the equation for U T,V( ) :

dU =

�U�T

�

��

�

V

dT +�U

�V

�

��

�

T

dV = CV dT + �p +T�p�T

�

��

� V

�

��

��dV (2)

Then we can use this equation to obtain Eq. (I) in Table 11.4-1.First we rewrite Eq. 2 for a fluid particle moving with the fluid,

and then we multiply by the density of the fluid. This gives:

�

DUDt

= �CVDTDt

+ � �p +T�p�T

�

��

�� V

�

�

��

DVDt

(3)

Since V = 1 , we may rewrite DV Dt as follows:

�

DVDt

= �DDt

1�

�

���

�= � �

1�

2

�

��

�

�

D�

Dt(4)

89

Then if we use Eq. (A) of Table 3.5-1, we may rewrite this last relationas:

�

DVDt

= � �1�

2

�

��

� �� � v( )( ) = � v( ) (5)

Now substitute this expression into Eq. 3, and then make use of Eq.11.2-2 to get:

�CV

DTDt

+ �p + T�p�T

�

��

� V

�

��

�

�� � �v( ) = � � �q( ) � p � �v( ) � �:�v( ) (6)

When the terms involving � �v( ) are moved to the right side, there issome cancellation and Eq. (I) of Table 11.4-1 is obtained:

�CV

DTDt

= � � �q( ) � T�p�T

�

��

�� V

� �v( ) � �:�v( ) (7)

[Note: See footnote 1 at the bottom of p.338 for a discussion of"incompressible fluids."]

90

Note to p. 341

Verify that Eq. (T) of Table 11.4-1 is the equation of change forentropy.

We can start with the thermodynamic expression

dU = TdS � pdV (1)

which is written for a small isolated mass of fluid that is stationary.For a small mass of fluid that is moving with the fluid, we can thenwrite:

DUDt

= TDSDt

� pDVDt

(2)

where we are now assuming that this expression can be appliedlocally. When this equation is multiplied by and then combinedwith Eq. (G) of the table we get:

� q( ) � p v( ) � ��:v( ) = �T

DSDt

� �pDVDt

(3)

Next, divide by T and replace V by 1 / to get

�

DSDt

= �1T

q( ) �1T

p v( ) �1T

�:v( ) +�pT

DDt

1�

�

��

�(4)

The second and fourth terms on the right side may be seen to cancelone another is one makes use of the equation of continuity in theform of Eq. (A) of Table 3.5-1, so that we get

�

DSDt

= �1T

q( ) �1T

�:v( ) (5)

When we make use of Eq. 3.5-4, Eq. 5 becomes

91

�

�t�S + � ��Sv( ) = �

1T

� �q( ) �1T

�:�v( ) (6)

This is equivalent to Eq. (T) in Table 11.4-1. Eq. (T) is written in aform that emphasizes the entropy flux q / T( ) and the entropyproduction (the terms in brackets:

�

�t�S = � � ��Sv( ) � � �

qT

�

��

�+ �

1T 2

q ��T( ) �1T

�:�v( )�

��

�

�� (7)

See Problem 11D.1 and §24.1 for more on this subject.

92

Note to p. 343

Verify that Eq. 11.4-14 gives the location of the maximumtemperature. We'd also like to know whether the maximum is nearerthe inner cylinder or the outer.

Differentiate � in Eq. 11.4-13 with respect to � and set thederivative equal to zero:

d�d�

= �1

� ln�+ N

2�

3 � 1�1�

2

�

���

�1

� ln�

�

�

�� = 0 (1)

Solving for 2 �2 gives:

2�

2 =1

N ln�+ 1 �

1�

2

�

���

1

ln�(2)

whence, the location of the maximum temperature rise is

�max =2ln 1 �( )

1 �2( ) � 1� 1 N( )

(3)

Now is �max greater than or less than 12 � +1( )? Clearly this cannot be

answered until N is known; but N depends on the geometry, thetemperature difference, the thermal conductivity, and the viscosity. Ifwe take N to be infinity, then we get:

� 12

13

14

12 � +1( ) 0.75 0.67 0.625

�max 0.67 0.51 0.42

This suggests that the maximum occurs nearer the inner wall, wherethe velocity gradient is larger.

93

Note to p. 346

Here we want to derive Eq. 11. 4-27 from Eq. 11.4-26 by follow-ing the instructions in the text. First we rewrite Eq. 11.4-26 in terms ofdimensionless variables:

� =

T � T1

T�� T1

= dimensionless temperature (1)

� =

rR

= dimensionless radial coordinate (2)

Then Eq. 11.4-26 becomes:

d�d�

=RR0

dd�

�2 d�

d��

���

�where

R0 =

wrCp

4�k(3a,b)

Now introduce the change of variable u = �2 d� d�( ) , and rewrite the

differential equation as:

u

�2=

RR0

dud�

(4)

This equation may be solved to get:

�

1�=

RR0

ln u +RR0

ln C1 or C1u = exp �

R0

R1�

�

���

�(5a,b)

where the constant of integration has been written as R R0( ) ln C1 .Reverting to the original variable � gives:

C1�

2 d�d�

= exp �R0

R1�

�

���

(6)

Further integration then gives:

94

C1� = �

�2exp �R0

R1�

�

��

��

1� d� + C2

= exp �

R0

Rv

�

���

�1

v dv + C2

=

e� R0 R( )v� e� R0 R( )

� R0 R( )+ C2

=

e� R0 R�( )� e� R0 R( )

� R0 R( )+ C2 (7)

The constants of integration may be determined from the boundaryconditions that: at � =� , � = 1 , and at � = 1 , � = 0 . From the secondof these boundary conditions it is evident that C2 = 0 . From the firstboundary condition, C1 may be found. Then the final expression forthe dimensionless temperature is:

� =

e� R0 R�( )� e� R0 R( )

e� R0 R�( )� e� R0 R( )

(8)

which agrees with Eq. 11.4-27.

95

Note to p. 352

Fill in the missing steps, showing how to arrive at Eqs. 11.4-67, 11.4-68, and 11.4-75.

We begin by multiplying Eq. 11.4-66 by vx 1v1 to get

43

μ

�1v1

vx

dvx

dx= vx

2+

pvx

�1v1

�CIIIvx

�1v1

(1)

Eq. 11.4-61 may be used to simplify the second term on the right sideto p , and this may be further rearranged by using ideal gasrelations, thus:

p�=

RTM

=

�Cp ��CV( )T

M= Cp � CV( )T =

� � 1�

�

���

CpT =

� � 1�

�

���

CI �

12 vx

2( )(2)

In the last step, use has been made of Eqs. 11.4-64 and 65, the secondof which should have been written as

CI = CpT1 +

12 v1

2 . Then we get Eq.11.4-67

43

μ

�1v1

vx

dvx

dx=

� +1�

�

���

vx

2+

� � 1�

�

���

CI �

CIIIvx

�1v1

(3)

To put this equation in dimensionless form, we insert the expressionsfor the constants of integration, multiply the equation by

3��1 4μv1 and then introduce the quantities defined in Eqs. 11.4-69,70, 71, 73 and 74, as follows:

�

d�d�

=3��1

4μv1

� + 12�

�

��

�vx

2� �1 v1

2+

RT1

M�

��

�vx

�1v1

+� �1�

�

��

�CpT1 +

12 v1

2( )�

��

�

��

96

= �Ma1

�

�Ma1

3�1

4μv1

�

�

��

� + 12�

�

���

��v1

2

2� v1

2+

RT1

M�

���

�� +

� �1�

�

���

��CpT1 +

12 v1

2( )

��

��

= �Ma12�� +1

�

���

�� +12�

�

���

�

2� 1+

RT1

Mv12

�

��

�

� +

� � 1�

�

���

�

CpT1

v12

+12

�

���

�

��

�

�

�

��

= �Ma1

2�

2�� +1

�

���

�1 +

1� Ma1

2

�

��

�

� + 2

� � 1� + 1

�

���

�1

� �11

Ma12+

12

�

��

�

�

�

�

�

��

(4)

The coefficient of � may be seen by Eq. 11.4-72 to be �2� and theconstant term is +� . Therefore, Eq. 4 finally becomes

�

d�d�

= �Ma1 � � 1( ) � ��( ) (5)

which is in agreement with Eq. 11.4-68. This equation is separable

�d�� � 1( ) � ��( )

= �Ma1d� =�� �Ma1 � � �0( ) (6)

where �0 is the constant of integration. The integral on the left side isto be found on p. 29 of Table of Integrals and Other Mathematical Data,H. B. Dwight, MacMillan, New York, 4th edition (1961). There wefind

xdxa + x( ) c + x( )� =

1a � c

a ln a + x � c ln c + x�� �� (7)

For the problem at hand a = �1 and c = �� , so that

�d�� � 1( ) � ��( )

=1

� � 1� �1ln �1+� +� ln ��+�� � = �Ma1 � � �0( ) (8)

In order to remove the absolute value signs, we note that � < � < 1, sothat

97

�1ln 1� �( ) +� ln � ��( )� � = � �1( )�Ma1 � � �0( ) (9)

so that finally

1� �( )

� ��( )�= exp 1 ��( )�Ma1 � � �0( )� � (10)

which is the same as Eq. 11.4-75.

98

Note to p. 375

a. First we'll verify the formula for differentiation of the errorfunction in Eq. C.6-2.

b. Then we'll show that Eq. 12.1-8 is a solution to Eq. 12.1-3 andthe associated boundary and initial conditions.

a. The differentiation of the error function is given by Eq. C.6-2:

ddx

erf u =2

e�u2

�dudx

(1)

where it is understood that u is a function of x. Now differentiate theerror function in Eq. C.6-1 with respect to x using the Leibniz formulaof §C.3:

ddx

erf u =d

dx2

e�u2

du0

u�

�

���

=

2

�

�xe�u2

du + e�u2 �u�x

� e�02 �0�x0

u�

�

��

� (2)

Since u is a dummy variable of integration, it is not a function of x,and therefore the first term is zero. The last term is also zero. Hencethe second term is the only term in the parenthesis that contributes tothe derivative of erf u, and that leads directly to Eq. 1. [Note: It isimportant to designate the dummy variable of integration, u , and theupper limit in the integral, u, by two different symbols. This exampleemphasizes the importance of this statement.]

b. Now turn to Eq. 12.1-8, where the left side is the dimension-less temperature difference � . Form the derivatives that appear inEq. 12.1-3:

��

�t= �

2

�e�y2 4�t y

4��

12

�

���

��t�3 2�

���

��(3)

y= �

2

�e�y2 4�t 1

4�t

�

���

��(4)

2

y2= �

2

�e�y2 4�t 1

4�t

�

���

���

2y4�t

�

���

��(5)

99

When the results in Eqs. 3 and 5 are inserted into Eq. 12.1-3 it is foundthat the equation yields an identity. Therefore, Eq. 12.1-8 does satisfythe differential equation.

Eq. 12.1-8 also satisfies the boundary conditions at y = 0 and

y = � , as may be seen from Fig. 4.1-2. At t = 0, Fig. 4.1-2 tells us that

� = 0 .

[Note: Here, and elsewhere we have made use of the Leibniz formulafor differentiating an integral. For additional information and anec-dotes regarding the Leibniz formula see R. P. Feynman, Surely You'reJoking Mr. Feynman, Bantam Books, New York (1986), p. 72 and p. 93.Professor Feynman was a Nobel Prize winner in physics.]

100

Note to p. 377

As pointed out in the footnote, there are two solutions to thisproblem, one that converges rapidly for long times (see Eq. 12.1-31):

= 2

�1( )n

n + 12( )�

exp � n + 12( )

2�

2�

�� �n=0

�

� cos n + 12( )�� (1)

and one that converges rapidly for small times:

= 1 � �1( )

nerfc

n + 12( ) � 1

2�

�+ erfc

n + 12( ) + 1

2�

�

�

�

n=0

�

� (2)

We want to verify that both of these solutions satisfy the partialdifferential equation (Eq. 12.1-14), the initial condition (Eq. 12.1-15),and the boundary conditions (Eq. 12.1-16).

a. Solution in Eq. 1:When Eq. 1 is substituted into Eq. 12.1-14, we get by

differentiating the exponential once with respect to � and the cosinetwice with respect to �

2

�1( )n

n + 12( )�

exp � n + 12( )

2�

2�

�� �n=0

�

� cos n + 12( )�� � � n + 1

2( )2�

2�

� �=

2

�1( )n

n + 12( )�

exp � n + 12( )

2�

2�

�� �n=0

�

� cos n + 12( )�� � � n + 1

2( )��

� n + 1

2( )��

�

(3)and this is clearly an identity, so that the partial differential equationis satisfied.

When � is set equal to zero and � = 1 in Eq. 1, we get

1 = 2

�1( )n

n + 12( )�n=0

�

� cos n + 12( )�� (4)

101

We have to prove that this is an identity. To do this, multiply bothsides by

cos m + 1

2( )� and integrate over � from –1 to +1:

cos m + 1

2( )��d� = 2�1( )

n

n + 12( )�n=0

�

� cos m + 12( )��

�1

+1��1

+1� cos n + 1

2( )��d�

(5)Then performing the integrations we get

1m + 1

2( )�sin m + 1

2( )���1

+1= 2

�1( )n

n + 12( )�n=0

�

� �mn cos2 n + 12( )��

�1

+1� d�

or

2m + 1

2( )sin m + 1

2( ) = 2�1( )

m

m + 12( )

(7)

This is an identity, since sin m + 1

2( ) = �1( )m

.When � is set equal to +1 or –1, we get � = 0 , since

cos m + 1

2( )� = 0 .

b. Solution in Eq. 2When Eq. 1 is substituted into Eq. 12.1-14, we get by

differentiating the complementary error functions once with respectto � and then twice with respect to �

�erfc

n + 12( ) � 1

2�

�= �

2

�exp �

n + 12( ) � 1

2�

�

�

��

�

��

2�

��

�

�

��

n + 12( ) � 1

2��

�� � 1

2( )� �3 2

(8)

�erfc

n + 12( ) � 1

2�

�= �

2

�exp �

n + 12( ) � 1

2�

�

�

��

�

��

2�

��

�

�

��

�12� ��

�(9)

2

�2

erfcn + 1

2( ) � 12�

�= �

2

�exp �

n + 12( ) � 1

2�

�

�

��

�

��

2�

��

�

�

��

102

� �2n + 1

2( ) � 12�

�

�

��

�

�

�

�

�

��

�12� ��

�

�12� ��

�(10)

Thus, the nth term satisfies the partial differential equation, andhence the entire series does.

At � = 0 , all of the complementary error functions are zero, sothat � = 1 , and the initial condition is satisfied.

At � = ±1, � = 0 as may be seen as follows. First consider

� = +1:

= 1 � erfc

0

�+ erfc

1

�

�

��

��� erfc

1

�+ erfc

2

�

�

��

��+ erfc

2

�+ erfc

3

�

�

��

��

�

�

�� ��

(11)Hence there is cancellation of between the nth and (n + 1)th terms,and, since erfc 0 = 1, the dimensionless temperature is zero. A similarargument can be made for � = �1.

103

Note to p. 379

In Example 12.1-3 going from Eqs 12.1-38 and 39 to Eq. 12.1-40presents a few problems. Therefore we go through the details.

Substitution of Eq. 12.1-38 into Eq. 12.1-39 gives:

�k T0 � T( ) = q0 e� � 2� ycos �t �

�

2�y

�

��

�

�� dy

y

�

� (1)

Bars have been added to the variable of integration to distinguish itfrom the lower limit on the integral. It is easier to perform theintegration if the cosine is converted into an exponential of a complexquantity, thus:

T � T0 =

q0

ke� � 2� y

� ei�t� i � 2� y{ }y

�

� dy

=

q0

k� ei�t e� � 2� y� i � 2� y

y

�

� dy{ }

=q0

k� ei�t e� 1+ i( ) � 2� y

� 1+ i( ) � 2�y

��

�

�

�

�

=q0

k� ei�t e� 1+ i( ) � 2� y

1 + i( ) � 2�

�

��

�

��

��

(2)

Next we remove from the braces that portion of the exponential thatis real; we also multiply numerator and denominator by 1 � i( ) . Thenwe have

T �T0 =q0

ke� � 2� y

�ei�te� i � 2� y 1� i( )

2 � 2�

�

��

�

��

��

=

q0

2k2��

e� � 2� y� ei�te� i � 2� y 1� i( ){ } (3)

104

In order to proceed, we need to rewrite 1 � i( ) in the form rei� . Thenwe find r and � as follows:

1� i = rei�= r cos� + isin�( ) (4)

so that equating real and imaginary parts gives

1 = r cos� and �1 = r sin� (5,6)

Taking the ratio of these two equations we get

r sin�r cos�

=�11

or tan� = �1 or � = 34 ,� 1

4 (7,8,9)

Since 1 � i( ) is in the 4th quadrant, the appropriate choice is � = � 14 .

Next we square both Eq. 5 and Eq. 6

1 = r2 cos2� and 1 = r2 sin2

� (10)

Adding the two equations then gives:

r2= 2 or r = ± 2 (11)

The plus sign must be chosen, since r must be non-negative.Therefore we have shown that

1� i = 2e� i 4 (12)

Returning now to Eq. 3, we get:

T �T0 =

q0

2k2��

e� � 2� y� 2ei�t� i � 2� y� i� 4{ }

=

q0

k�

�e� � 2� y cos �t � � 2� y � 1

4 �( ) (13)

This agrees with Eq. 12.1-40 in the textbook.

105

Note to p. 386

The derivation of Eq. 12.3-6 from Eq. 12.3-5 is given here. Firstwe note that, if z = x + iy = rei� , then

ln z = ln r + i� = ln x2+ y2

+ iarctan y x( ) (1)

Now we have to resolve Eq. 12.3-5 into its real and imaginary parts.We introduce the abbreviated notation Z = �z b , X = �x b , and

Y = � y b . Then

w =

1�

lnsin Z � 1sin Z + 1

�

��

�

��=

1�

lnsin X + iY( ) � 1

sin X + iY( ) + 1

�

��

�

�� � � + i�

=

1

lnsin X cos iY + cosXsin iY� 1sin X cos iY + cosXsin iY + 1

�

��

�

�

=

1

lnsin X coshY + icosXsinhY � 1sin X coshY + icosXsinhY + 1

�

��

�

�

=1

ln

sin XcoshY �1( ) sin XcoshY + 1( ) + cos2 Xsinh2 Y

+icosXsinh Y sin X coshY + 1( ) � cosXsinhY sin XcoshY �1( )

sin X coshY + 1( )2+ cosXsinhY( )

2

�

�

�����

�

�

=1

lnsin2 Xcosh2 Y �1 + cos2 Xsinh2 Y( ) + 2icosXsinhY

sin XcoshY +1( )2+ cosXsinhY( )

2

�

�

��

�

�

(2)

In going from the third to the fourth line, we have multiplied thenumerator and denominator by the complex conjugate of thedenominator.

The imaginary part of the expression in Eq. 2 is then

106

� =

1�

arctan 2cosXsinhY

sin2 X cosh2 Y � 1+ cos2 Xsinh2 Y

�

��

�

�

=

1

arctan2cosXsinhY

sinh2 Y � cos2 X

�

��

�

�

=1

arctan2 cosX sinhY( )

1� cosX sinhY( )2

�

���

�

�

�

1

arctan2tan A

1� tan2 A

�

��

�

� (3)

The last expression serves to define A . But the quantity inparentheses is just tan 2A (see, e.g., formula 406.02 of Dwight's Tablesof Inegrals and Other Mathematical Data, 4th edition), and the "anglewhose tangent is tan 2A " is just 2A (i.e., arctan(tan2A ) = 2A.However, A = arctan cosX sinhY( ) so that, finally

=

2

arctancosXsinhY

�

��

�

�=

2

arctancosx b

sinh y b

�

��

�

�(4)

which is the result in Eq. 12.3-6.

107

Note to p. 388

Here we work through the missing steps to get the result in Eq.12.4-16. We begin by evaluating the integral in the first term on theright side of the equals sign in Eq. 12.-4-4 (the second term is zero,because ve = v

�= a constant in this problem):

�vx v

�� vx( )dy = �v

�

2� x( )

vx

v�

0

10

�

1�vx

v�

�

��

�� d� (1)

Here � x( ) is the velocity boundary-layer thickness, and � = y � x( ) isthe dimensionless coordinate in the y-direction. In the second integralwe have changed the upper limit to "1" because 1� vx v

�( ) in Eq.12.4-6 and 7 is zero beyond � = 1 . Then substituting the assumedvelocity profile into Eq. 1 gives:

�vx0

�

� v�� vx( )dy

= �v

�

2� x( ) 2� � 2�3

+�4( ) 1� 2� + 2�3

��4( )d�0

1�

= �v

�

2� x( ) 2� � 4�2

� 2�3+ 9�4

� 4�5� 4�6

+ 4�7��

8( )d�0

1�

= �v

�

2� x( ) 1� 4

3 �12 +

95 �

23 �

47 +

12 �

19( )

= �v

�

2� x( ) �315+567�180�35

315( ) =37

315 �v�

2� x( ) (2)

Similarly, the integral appearing in Eq. 12.4-5 may beevaluated:

�Cp0

�

� vx T��T( )dy

= �Cpv

�T�� T0( )�T x( )

vx

v�

T�� T

T��T0

�

��

��0

1 d�T

= �Cpv

�T��T0( )�T x( )

vx

v�

T��T0

T��T0

�T0 �TT0 �T

�

�

��

��0

1 d�T

108

= �Cpv

�T��T0( )�T x( )

vx

v�

1�T0 �TT0 �T

�

�

��

��0

1 d�T (3)

in which �T = y �T x( ) = y �� x( ) , and � is assumed to be inde-pendent of x. Then, inserting the postulated profiles for velocity andtemperature into Eq. 3, we get:

�Cp0

�

� vx T��T( )dy

= �Cpv

�T��T0( )�T x( ) 2�T� � 2�T

3�

3+�T

4�

4( )0

1�

1 � 2�T + 2�T3��T

4( )d�T

= �Cpv

�T�� T0( )�T x( ) 2

15 � � 3140 �

3+ 1

180 �4( ) (4)

When the expressions in Eqs. 2 and 4 as well as Eqs. 12.4-6 to 9 aresubstituted into Eqs. 12.4 and 5, we get differential equations for theboundary-layer thicknesses as a function of the distance along theplate:

2μv�

�=

37315

�v�

2 ddx

� (5)

2k T�� T0( )

�T

= 215 � � 3

140 �3+ 1

180 �4( )�Cpv

�T��T0( )

ddx

�T (6)

Eq. 5 for � x( ) may be solved as follows:

�

ddx

� =31537

2μv�

�v�

2;

�d�

0

�

� =63037

μ

�v�

dx0

x� ;

� =

126037

μx�v

�

�

��

�

�

(7,8,9)and Eq. 6 for �T x( ) may also be solved:

2k T�� T0( )

�T

= 215 � � 3

140 �3+ 1

180 �4( )�Cpv

�T��T0( )

ddx

�T (10)

�Tddx

�T =1

215 � � 3

140 �3+ 1

180 �4( )

2k

�Cpv�

(11)

109

�T d�T =2

215 � � 3

140 �3+ 1

180 �4( )

�

v�

dx0

x�0

�T� (12)

�T =4

215 � � 3

140 �3+ 1

180 �4( )

�xv�

�

��

� (13)

It remains to find � = �T x( ) � x( ) as a function of the physicalproperties. Forming the ratio we get:

� =�T

�=

4 37 1260( )2

15 � � 3140 �

3+ 1

180 �4( )

��

μ

�

��

�(14)

Squaring both sides and collecting all the � terms on the left side, wefind:

2

15 �3� 3

140 �5+ 1

180 �6=

35315

1Pr

Pr=

μ

k

Cp

=1

Cpμ k

�

���

�

���

(15)

For large Prandtl numbers, we can drop all but the lead term on theleft side and get

� = 35

315152

3 Pr�1 3= 0.8803 Pr�1 3

= 0.958Pr�1 3 (16)

As pointed out in the textbook, 0.958 may be replaced by 1 to fit theexact curve in Eq. 12b.2-15 within 5%. This replacement leads to Eq.12.4-16 in the textbook.

[Note: In earlier printings of the textbook, the second integrals in Eqs.12.4-10 and 11 had an upper limit of infinity rather than 1.]

110

Note to p. 413

The object here is to fill in the missing steps between Eq. 13.4-11and Eq. 13.4-16. First we multiply Eq. 13.4-11 by �

dd�

� 1+�

t( )

�

�

��

��

d�d�

�

��

�� = C0� (1)

Then first integration with respect to � gives:

� 1 +

�t( )

�

�

��

��

d�d�

= C0 �0

�

d� + C1 � C0I �( ) + C1 (2)

where � �( ) is the dimensionless velocity defined just after Eq. 13.4-6,

and the abbreviation I �( ) is introduced. Eq. 2 may now be rewritten:

d�d�

=C0I �( )

� 1+ �t( )�( )�

���

+C1

� 1 + �t( )�( )�

���

(3)

A second integration with respect to � gives:

� = C0

I �( )� 1+ �

t( )�( )�

���

0

�

� d� + C11

� 1 + �t( )�( )�

���

d� + C20

�

� (4)

in which we must remember that �t( ) is now a function of � . If next

we consider the limit of the above expression as � 0 , it may be seenthat the first term goes to zero and the second term goes to infinity(and therefore violates B. C. 1); therefore C1 must be taken to be zero.Hence the expression for the dimensionless temperature becomes:

111

� ,�( ) = C0� + C0

I �( )� 1+ �

t( )�( )�

���

d�0

�

+ C2 (5)

Next, apply the boundary condition at = 1 :

1 = C0

I �( )� 1 + �

t( )�( )�

��� � =1

= C0I 1( ) or C0 = I 1( )�� ���1

(6)

inasmuch as �t( ) is zero at the wall. Next we want to get the driving

force �0 ��b , which is the dimensionless wall temperature minusthe dimensionless bulk temperature (defined in Eq. 10.8-33):

�0 ��b = C0

I �( )

� 1+ �t( )�( )�

�

d�0

1� � C0

��I �( )

� 1+ �t( )�( )�

�

d�d�0

�

�0

1�

��d�0

1�

= C0

I �( )

� 1 + �t( )�( )�

���

d�0

1 �

C0

I 1( )�d�

�

1( )

I �( )� 1 + �

t( )�( )�

���

d�0

1 (7)

In the second expression we have interchanged the order of integra-tion. The second term in Eq. 7 may be rewritten:

�C0

I 1( )�d� � �d�

0

�

0

1( )

I �( )� 1+ �

t( )�( )�

���

d�0

1

= �C0

I 1( )I 1( ) � I �( )��

��0

1�

I �( )� 1+ �

t( )�( )�

���

d�

112

= �C0

I �( )

� 1+ �t( )�( )�

���

0

1� d� +

C0

I 1( )

I �( )�� ��2

� 1+ �t( )�( )�

���

0

1� d� (8)

The first term in Eq. 8 above just cancels the first term in Eq. 7, andhence we are left with:

�0 ��b =C0

I 1( )

I �( )�� ��2

� 1 + �t( )�( )�

���

0

1 d� =

I �( ) I 1( )�� ��2

� 1 + �t( )�( )�

���

0

1 d� (9)

This is in agreement with Eq. 13.4-16 in the textbook.

113

Note to p. 415

Some additional material is given here on §13.5.Perform the indicated substitutions into Eq. 13.5-1 to get for the

partial differential equation for :

�t( )

zF�� �F

�

��

����

�r+ �

�t( )

z�F�

�

��

����

�z=

�t( )

Pr t( )

1r�

�rr��

�r�

��

��(1)

where the primes indicate differentiations with respect to . We next

make the change of variables = r z and � = ��

t( ) w( )z , and further

let r, z( ) =

�

� ,�( ) = f �( ) � , so that

r=�

�

�

r+�

�

�

r=�

��1z+�

��0 =

�f�

�1z

(2)

��

�z=��

�

��

��

�z+��

�

��

��

�z=��

�

���

r

z2

�

��

�+��

�

��

� t( )

w

�

��

�

=

�f�

�r

z2

�

���

��

f

�2

�

z�

���

�

(3)Substitution of these expressions into Eq. 1 gives

�t( )

zF�� �F

�

���

� �f

�

1z

�

��

�

�� �

�t( )

z�F�

�

��

�

�

�f�

�r

z2

�

���

� �

f

�2

�

z�

���

� �

��

�

��

=

�t( )

Pr t( )

1�

�

����f��

�

��

�

1�z2

(4)

Multiplication of the entire equation by z2� �

t( ) gives

F�� �F

�

��

��f[ ] + �F( ) �f +

f�

�

�

�� =

1

Pr t( )

1�

�

����f��

�

��

�(5)

114

On the left side, the terms involving �F �f cancel, and the equationmay be rewritten as follows:

1

dd

Ff( ) =1

Pr t( )

1

�

��f�

�

���

�(6)

This equation can be multiplied by Pr t( ) and integrated once to give

Pr t( )Ff =

�f�

+ C (7)

According to Eq. 5.6-20, F = 0 at = 0 , which means that C = 0. Afurther integration from 0 to then yields

lnf �( )f 0( )

= Pr t( ) F

�0

�

� d� = �Pr t( ) C32�

1+ 14 C3�( )

2d�

0

�

� = �Pr t( )ln 1+ 1

4 C3�( )2�

���

��

2

(8)or, taking the antilogarithm of both sides

f �( )f 0( )

= 1+ 14 C3�( )

2�

���

��

�2Pr t( )

(9)

When this result is compared with Eq. 5.6-21 for the velocity profilein a circular jet, and use is made of the definition in Eq. 13.5-8, weobtain finally

max

=vz

vz,max

�

��

�

��

Pr t( )

(10)

which is a rather simple, and apparently fairly satisfactory, result.

115

Note to p. 454

Verify that Eq. 15.1-1 can be obtained from Eq. 11.1-9 by the methoddescribed on p. 454.

We start by integrating Eq. 11.1-9 over the volume of the flowsystem shown in Fig. 7.0-1:

�

�t12 �v2

+ �U + ��( )�

��

�dV = � � � 1

2 �v2+ �U + ��( )v( )dV

V t( )�V t( )�

� � q( )V t( ) dV � � pv( )V t( ) dV � � �� v[ ]( )V t( ) dV (1)

We next apply the Leibnitz formula to the left side of the equationand the Gauss divergence theorem to the right side:

ddt

12 �v2

+ �U + ��( )dV � n 12 �v2

+ �U + ��( )vS( )S t( )V t( ) dS

= � n 1

2 �v2+ �U + ��( )v( )S t( ) dS

� n �q( )S t( )� dS

� n � pv( )S t( )� dS � n � �� �v[ ]( )S t( )� dS (2)

The integral in the first term on the left side is the total energy(kinetic + internal + potential energy). The second term on the leftside can be combined with the first term on the right side. Thus weget

ddt

Ktot +Utot +�tot( ) = � n 12 �v2

+ �U + ��( ) v � vS( )( )S t( ) dS

� n �q( )S t( )� dS � n � pv( )S t( )� dS � n � �� �v[ ]( )S t( )� dS (3)

We now analyze the terms on the right side seriatim:The first term can be seen to contribute nothing on the fixed

surface Sf and the moving surface Sm . At the inlet cross section S1

and the outlet section S2 , the surface velocity vS is zero and collinearwith the outwardly directed unit vector n. The fluid velocity vector vis assumed to point in the direction opposite to the n vector at theentry plane, and in the same direction as as the n vector at the exit

116

plane; therefore, at the entry n �v( ) = �v , and at the exit n �v( ) = +v .We make the further assumption that the internal energy and thepotential energy are constant over the cross section. Then when theintegration over the cross sectional area is performed we get:

� n 1

2 �v2+ �U + ��( ) v � vS( )( )S t( ) dS

= 1

2 �1 v13 S1 + �1U1 v1 S1 + �1�1 v1 S1

� 1

2 �2 v23 S2 + �2U2 v2 S2 + �2�2 v2 S2 (4)

The second term on the right side (the q-term) is the integralover all surfaces of the normal component of the heat flux vector andis thus the rate of total heat addition to the system, Q:

� n �q( )Sf +Sm +S1 +S2� dS = Q (5)

It is assumed that the heat addition at surfaces S1 and S2 is usually besmall compared to the heat added at the solid surfaces.

The third term on the right (the p-term) has to be evaluated atall the surfaces. At the inlet and outlet planes, we will get

� n � pv( )S1+S2� dS = p1 v1 S1 � p2 v2 S2 (6)

by the same arguments leading to the internal energy terms in Eq. 4.These terms represent the rate of doing work on the system at theentry and exit planes. On the solid surfaces we get

� n � pv( )Sf +Sm� dS = Wm

p (7)

This term is the rate that pressure does work on the system at themoving surfaces Sm ; there is no work done at the fixed surfaces

Sf ,

inasmuch as the rate of doing work is a force times a velocity, and at

Sf the surface velocity is zero.

The fourth term on the right (the �� -term) is evaluated similarlyto the p-term. First the contributions at surfaces S1 and S2 areconsidered, but it is assumed that these will be small compared to the

117

pressure terms in Eq. 6. On the solid surfaces there will be acontribution similar to that for the pressure forces on the movingsurfaces:

� n � �� �v[ ]( )Sf +Sm� dS = Wm

� (8)

and, here again, the contribution at Sf will be zero. The contributions

from Eqs. 7 and 8 will be added to give Wmp+ Wm

��= Wm , the total

work done on the system through the moving surfaces.When all the contributions in Eqs. 4 through 8 are added up we

get Eq. 15.1-1 of the textbook:

ddt

Ktot +Utot +�tot( ) = 12 �1 v1

3+ �1U1 v1 + �1�1 v1( )S1

� 1

2 �2 v23

+ �2U2 v2 + �2�2 v2( )S2 +Q + Wm

+ p1 v1 S1 � p2 v2 S2( ) (8)

or, introducing the enthalpy

ddt

Ktot +Utot +�tot( ) = 12 �1 v1

3+ �1H1 v1 + �1�1 v1( )S1

� 1

2 �2 v23

+ �2H2 v2 + �2�2 v2( )S2 +Q + Wm (9)

Either Eq. 8 or Eq. 9 is referred to as the unsteady-state macroscopicenergy balance.

118

Note to p. 494

The derivations of the Stefan-Boltzmann law and Wien's lawfrom the Planck black-body distribution law are quite important andtherefore it is a good idea to understand all the intermediate steps inthe development.

a. The Stefan-Boltzmann lawWe integrate Planck's distribution law over all wavelengths:

qb

e( )= qb�

e( )0

�

d� =2�c2h

�5

1ech �KT

�10

�

d� (1)

Next we make a change of variable x = ch �KT . Then

2�c2h

1�

5d� = 2�c2h

KTch

�

��

��

5

x5

chKT

�

��

���

1x2

�

��

��dx

=

2 KT( )4

c2h3�x3( )dx (2)

Hence the integral becomes

qbe( )=

2 KT( )4

c2h3

x3

ex� 10

�

� dx (3)

Then expand the denominator of the integrand as a Taylor seriesabout x = 0, to get:

ex

� 1 = 1+ e�x+ e�2x

+ e�3x+�( ) � 1 (4)

Then a term by term integration of Eq. 3 gives:

2 KT( )4

c2h3x3e�nx

0

�

dxn=1

�

� =2 KT( )

4

c2h36

1n4

n=1

�

� =2 KT( )

4

c2h3

4

15

�

��

�� (5)

119

On p. 171 of Planck's book, The Theory of Radiation, Dover, New York(1959), which is a translation of Vorlesungen über die Theorie derWärmestrahlung, 5th edition, Barth, Leipzig (1923), the Stefan-Boltzmann equation is obtained as shown above. However, Planckdid not evaluate the summation in Eq. 5 exactly. Instead, he simplyevaluated the sum numerically as:

� = 1+

124

+134

+144

+�= 1.0823 (5)

Nowadays, even in a small integral table (such as H. B. Dwight,Tables of Integrals and Other Mathematical Data, Macmillan, New York,Fourth Edition (1961)), the integral over x in Eq. 3 may be found; seeFormula 860.33 on p. 231.

When Eq. 5 is compared with the Stefan-Boltzmann law for a

black body qe( )= �T4 , then we get the Stefan-Boltzmann constant:

� =

215

�5

K4

c2h3 (6)

which interrelates key constants from several different fields ofphysics.

b. Wien's displacement lawFirst rewrite Eq. 16.3-7 in terms of x thus:

qb

e( )=

2�c2h

�5

1ech �KT

� 1= 2�c2h

KTch

�

���

�

5x5

ex�1

(7)

We can then differentiate this with respect to x to get

dqbe( )

dx= 2c2h

KTch

�

���

�

55x4

ex�1

�x5ex

ex�1( )

2

�

�

��

�

�

(8)

Then setting the derivative equal to zero gives the value of x at which

the maximum in the qbe( ) curve occurs:

120

5 �

xmaxexmax

exmax � 1= 0 (9)

whence

xmax = 5 1� e�xmax( ) (10)

from which one can find, by trial and error, xmax = 4.9651... , or

�maxT = 0.2884cm K . This is Wien's displacement law.

121

Note to p. 529

We want to verify that Eq. 17.4-4, Eq. (I) of Table 17.8-1, and Eq.(E) of Table 17.8-2 are dimensionally consistent.

If we put the dimensions of the quantities into the equationinstead of the mathematical symbols (and omit the numerical factors)we get:

(a)

L2

t

�

��

�

��

MLt

�

���

��

ML2

t2T

�

��

�

�� T( )

=1L

(1)

(b)

molesL2t

�

���

��=

molesL3

�

���

��Lt

�

���

��(2)

(c)

M

L3

�

���

��Lt

�

���

��=

M

L3

�

���

��L2

t

�

��

�

��

1L

�

���

��(3)

In each case, the dimensions on the left side are the same as thedimensions on the right side.

122

Note to p. 534

It is important to know how to simplify multicomponentrelations to their corresponding binary equations. We illustrate thisprocedure by showing how to get the binary formula in Eq. (Q') fromthe multicomponent formula in Eq. (Q) in Table 17.7-1.

In the sum in Eq. (Q), the indices � and � can take on only thevalues of A and B in a binary system. Then in the sum, � must be B.Therefore, Eq. (Q') becomes for a binary system:

��A = �

MA

M2M + xA MB � MA( )�� ���xB (1)

Next use Eqs. (M) and (J) of the table, written for a binary system:

��A = �MA

xA MA + xBMB( )2

xA MA + xBMB( ) + xA MB � MA( )�� �� ��xA( )

(2)Within the bracket, the terms xA MA cancel each other, and theremaining terms may be combined, since xA + xB = 1 . We are then leftwith:

��A = +MA MB

xA MA + xBMB( )2�xA (3)

which is just Eq. (Q').

123

Note to p. 535

Here we want to verify that Eq. 17.7-4 can be derived from Eq.17.7-3 using the relations in Tables 17.7-1 and 2.

First we transform the last term in Eq. 17.7-3 into the analogousterm in Eq. 17.7-4, with a multiplying factor:

��D AB�A = � cM( )D AB MA MB M2( )xA

= �cD AB�xA � MA MB M( ) (1)

Here, Eq. (F) of Table 17.7-1 was used, as well as Eq. ( �Q ).Next we transform the mass concentration times the diffusion

velocity as follows:

�A vA � v( ) = cA MA( ) vA � �AvA +�BvB( )�� ��

= cAMA( )�B vA � vB( ) = cA MA( ) xBMB M( ) vA � vB( )

= cAMA( ) MB M( ) vA � xAvA + xBvB( )( )

= cA vA � v *( ) � MA MB M( ) (2)

In this development, we end up with the same factor MA MB M( )appearing. In the first step above, we used Eq. (B) of Table 17.7-2. Inthe second step we used Eq. (K) of Table 17.7-1. In the third step Eq.(O) of Table 17.7-1 was used. In the fourth step we used Eq. (J) ofTable 17.7-1, and in the last step Eq. (C) of Table 17.7-2.

The rest of the proof makes use of the definitions

jA = �A vA � v( ) JA�= cA vA � v *( ) (3a,b)

which are given in Eqs. (E) and (I) of Table 17.8-1. It remains, then, toverify that

jA = JA�� MA MB M( ) (4)

This may be done by rewriting Eq. 4 as

124

jA = JA�� M( ) MA M( ) MB M( ) = JA

�� � c( ) �A xA( ) �B xB( ) (5)

Here Eqs. (G) and (O) of Table 17.7-1 have been used. The result inEq. 5 may also be written thus:

jA

��A�B

=JA�

cxAxB

(6)

This important equation is also given in Eq. 17B.3-1.

125

Note to p. 547 (i)

Starting with the concentration profile for species A in Eq. 18.2-11, derive the subsequent results up through Eq. 18.2-16.

First obtain the expression for xB,avg :

xB,avg

xB1

=xB2 xB1( )

�d�

0

1�

d�0

1�

=xB2 xB1( )

�

ln xB2 xB1( )0

1

(1)

where we have used the integral

ax� dx = ax ln a( ) + C . Therefore

xB,avg

xB1

=xB2 xB1( ) � 1

ln xB2 xB1( )or

xB,avg =

xB2 � xB1

ln xB2 xB1( )� xB( )ln

(2a,b)

Hence the rate of evaporation of A at the gas-liquid interface is:

NA z=z1=

cD AB

xB1

dxB

dz z=z1

=cD AB

xB1

dxB

d��=0

d�dz

=cD AB

z2 � z1

d xB xB1( )d�

� =0

(3)

Then, using the derivative dax dx = ax ln a , we get:

NA z=z1=

cD AB

z2 � z1

xB2

xB1

�

��

�

��

�

lnxB2

xB1

�

��

�

��

�

�

�

��� =0

=cD AB

z2 � z1

lnxB2

xB1

�

��

�

�� (4)

Then, multiplying the numerator and denominator by xB2 � xB1( ) :

NA z=z1

=cD AB

z2 � z1

xB2 � xB1( )xB( )ln

=cD AB

z2 � z1

xA1 � xA2( )1� xA( )ln

(5)

Next we obtain the solution for very small values of xA1 and xA2 :

126

NA z=z1

=cD AB

z2 � z1

xA1 � xA2( ) ln 1 � xA2( ) 1� xA1( )�� �

1� xA2( ) � 1� xA1( )

=

cD AB

z2 � z1

ln 1� xA2( ) � ln 1� xA1( )�� �

=

cD AB

z2 � z1

�xA2 �12 xA2

2� 1

3 xA23

��( ) + xA1 +12 xA1

2+ 1

3 xA13

��( )��

�

=cD AB xA1 � xA2( )

z2 � z1

1 + 12

xA12

� xA22( )

xA1 � xA2( )+ 1

3

xA13

� xA23( )

xA1 � xA2( )+�

�

�

��

�

�

��

=

cD AB xA1 � xA2( )z2 � z1

1 + 12 xA1 + xA2( ) + 1

3 xA12

+ xA1xA2 + xA22( ) +��

� �

(6)

The Taylor expansion in Eq. C.2-3 has been used for expanding thelogarithms in line 2.

127

Note to p. 547 (ii)

a. The result in Eq. 18.2-11 may be written as follows:

xB

xB1

=xB2

xB1

�

��

�

��

�

in which � =

z � z1

z2 � z1

(1,2)

We want to expand this result in a Taylor series in � to get a result ofthe form

xB

xB1

= 1� � � �( )� + � � � (3)

b. Next we rework the problem in §18.2 by omitting the xA

term in the denominator of Eq. 18.2-1 (that is, assume that xA is sosmall that it can be neglected with respect to unity). We can thenshow that this gives the first two terms of the series in Eq. 3.

a. From Eq. C2.1 we get by expanding about � =0:

xB

xB1

=xB2

xB1

�

��

�

�

�=0

+d

d�xB2

xB1

�

��

�

�

�=0

� � 0( ) + � � �

= 1+ 1 � ln

xB2

xB1

�

��

�

�� �1

�

�

���� + � � � (4)

Now from Eq. C.2-3, with (1 + x) replaced by x, and x replaced by(1 – x), we have, for 0 < x � 2

ln x = x � 1( ) � 12 x � 1( )

2+ 1

3 x �1( )3� 1

4 x �1( )4+ � � � (5)

Therefore,

128

xB

xB1

= 1+xB2

xB1

� 1�

��

�� �

12

xB2

xB1

� 1�

��

��

2

+13

xB2

xB1

�1�

��

��

3

+��

�

�

��� + � � � (6)

If xB2 is only slightly greater than xB1 (which would be the case ifspecies A is present only in a small amount), then we need retainonly the first term inside the bracket. The final result is then:

xB

xB1

= 1+xB2

xB1

� 1�

��

�

� � (7)

b. When xA can be neglected with respect to unity in Eq. 18.2-1,the differential equation for xA as a function of z is:

d2xA

dz2= 0 (8)

Since xA + xB = 1 , the same differential equation (with A replaced byB) is valid for species B. Furthermore, since z and � are related by thelinear expression given in Eq. (2) we may write:

d2xB

d� 2= 0 (9)

This equation may be integrated to give:

xB

xB1

= 1+xB2

xB1

� 1�

��

�

� � (10)

in agreement with Eq. 7.

129

Note to p. 555

We wish to verify that Eq. 18.4-9 does satisfy Eq. 18.4-7 bysubstituting into the differential equation.

Differentiate Eq. 18.4-9 with respect to � , using Eq. C.5-10:

d�d�

=sinh 1��( )� � �( )

cosh(1)

A second differentiation then gives, using Eq. C.5-11:

d2�

d� 2=

cosh 1��( )� � +2( )

cosh(2)

By substituting the second derivative from Eq. 2 and the expressionin Eq. 18.4-9 into Eq. 18.4-7, it is seen that an equality is obtained.

130

Note to p. 557

Here we show that Eq. 18.4-18 can be rearranged into a simpler form.First we put the expression in the large parentheses over a commondenominator

�

N =

sinh

cosh2 + V S�( ) cosh sinh � 1

cosh + V S�( ) sinh

�

��

�

(1)

Next make use of Eq. C.5-5 to get

�

N =�

sinh�

sinh2� + V S�( )� cosh� sinh�

cosh� + V S�( )� sinh�

�

��

�

� (2)

Finally cancel the sinh� appearing in the numerator anddenominator to get

�

N = �sinh� + V S�( )� cosh�

cosh� + V S�( )� sinh�

�

��

�

� (3)

As V S�( ) 0 , �

N � tanh� , and as V S�( )� � , �

N � coth� .

131

Note to p. 563 (i)

Here we verify that Eq. 18.6-8 is a solution to the differentialequation given in Eq. 18.6-2.

Because of the definition of f just above Eq. 18.6-6, f = cA cA0

must satisfy Eq. 18.6-2. Therefore, we calculate the derivatives usingthe Leibniz formula given in §C.3:

�f�z

= �1

� 43( )

exp ��3( )

��

�z

= �

1� 4

3( )exp ��

3( ) � 13( ) yz�4 3 a 9D AB( )

1 3(1)

�f�y

= �1

� 43( )

exp ��3( )��

�z�f�y

= �1

� 43( )

exp ��3( )��

�y

= �

1� 4

3( )exp ��

3( ) a 9D ABz( )1 3

(2)

�2 f

�y2= �

1� 4

3( )exp ��

3( ) a 9D ABz( )1 3

�3�2( ) a 9D ABz( )1 3

(3)

Then, substituting these expressions into Eq. 18.6-2, we get:

ay + 1

3( ) yz�4 3 a 9D AB( )1 3

= D AB +3�2( ) a 9D ABz( )2 3

(4)

If both sides are multiplied by 9D ABz a( )1 3

, and replace �2 by

y2 a 9D ABz( )

2 3, we get

ay2

3z= 3D AB y2 a 9D ABz( ) (5)

which is an identity. This concludes the proof.

132

Note to p. 563 (ii)

The concentration profiles are given by Eq. 18.6-8.

cA =

cA0

� 43( )

e��3

�

�

d� (1)

where � = y a 9D ABz( )1 3

. We show how to get the next two equationsfrom this result.

Notice in Eq. 1 the bar over � , which is used to make adistinction between the dummy variable of integration and thedimensionless distance. This distinction is vital when we apply theLeibniz rule (Eq. C.3-2) for differentiation of an integral to get thelocal molar flux at the wall is then:

NAy y=0= �D AB

cA

yy=0

= �D ABcA0

� 43( )

dd�

e��3

�

�

� �d�dy

�

�

�

y=0

(2)

Note that only the third term in Eq. C.3-2 contributes to thederivative (i.e., the term involving the lower limit of the integral):

NAy y=0= �

D ABcA0

� 43( )

�e��3

( ) � a9D ABz

�

��

�

�

1 3

�

�

�

���=0

=D ABcA0

� 43( )

a9D ABz

�

��

�

�

1 3

(3)

Then the total molar flow across the surface of width W andlength L is given by the integral over that surface:

WA = NAy y=0

dzdx =0

L�0

W�

D ABcA0

� 43( )

a9D AB

�

��

�

1 3

W z�1 3dz0

L�

=

D ABcA0

� 43( )

a9D AB

�

��

�

��

1 3

Wz2 3

0

L

23

=2D ABcA0WL

43 �

43( )

a9D ABL

�

��

�

��

1 3

(4)

133

In the last step, Eq. C.4-4 can be used to replace 43 �

43( ) by

� 7

3( ) toagree with Eq. 18.6-10.

134

Note to p. 565

Let us verify that Eq. 18.7-9 satisfies the differential equation inEq. 18.7-6. To simplify the problem, it is a good idea to introducedimensionless variables:

=

rR

; � =

cA

cAR

; � =

��k1aR2

D A

(1,2,3)

Then the differential equation in Eq. 18.7-6 and the solution in Eq.18.7-9 may be rewritten as:

1�

2

dd�

�2 d�

d��

���

= �

2� ;

� =

1�

sinh��sinh�

(4,5)

First we use Eq. 5 to evaluate the derivative (see also §C.5):

d�d�

= �1�

2

sinh��sinh�

+�

�

cosh��sinh�

(6)

Multiplication by �2 then gives:

�

2 d�d�

= �sinh��sinh�

+ ��cosh��sinh�

(7)

Further differentiation gives:

dd�

�2 d�

d��

��

�= ��

cosh��sinh�

+ �cosh��sinh�

+ �2�

sinh��sinh�

(8)

Division by �2 gives:

1�

2

dd�

�2 d�

d��

���

= �

2 1�

sinh��sinh�

(9)

135

But the right side of the equation is just � , according to Eq. 5. Wehave therefore shown that the differential equation is satisfied.

136

Note to p. 584

Here we show how to get Eq. 19.1-15 from the preceding equations.Start with Eq. 19.1-10, and let c� = cx

� , so that we get

c�x

�

�t+ x

�

�c�t

= � N�( ) + R

� (1)

Next we use Eq. 19.1-12 to obtain

c�x

�

�t+ x

�� � � cv *( ) + R

��=1

N

�

�

�

� = � � �N

�( ) + R� (2)

Then we go to Eq. 17.8-2 to find N�= J

�

�+ c

�v * , so that (2) becomes

cx

�

t� x

�� cv *( ) + � c

�v *( ) = � � J

�

�( ) + R�� x

�R

��=1

N

� (3)

------------------------------

Next we have to show that the two dashed-underlined terms on theleft side of (3) are the same as the c v * ��x

�( ) in Eq. 19.1-15:

�x�

� � cv *( ) + � � c�

v *( ) = �x�

� � cv *( ) + � � x�

cv *( )

= �x�

� � cv *( ) + x�� � cv *( ) + cv * ��x

�( )

= c v * ��x�( ) (4)

Here we have made use of Eq. A.4-19 to differentiate the product of ascalar with a vector.

The last two terms in Eq. 19.1-17 for a binary system are obtained thus:

RA � xA RA + RB( ) = 1� xA( )RA � xARB = xBRA � xARB (5)

because for a binary system N in the upper limit of the sum is 2, therebeing just two components, A and B.

137

Note to p. 585

The equation to be solved is in Example 19.1-1 is

v0

dcA

dz= D AB

d2cA

dz2� ���k1 cA (1)

which is of the form of Eq. C.1-7a. We know that the concentration ofA will decrease with increasing distance from the porous plug, so weassume that it will have the form cA = exp �az( ) , where a is a positiveconstant. When this is substituted into the Eq. 1, we get (aftercanceling exp �az( ) from each term):

�v0a = D ABa2� k1

��� or D AB a2+ v0a � k1

��� = 0 (2)

This quadratic equation can be solved for a to get:

a =

�v0 ± v02+ 4D ABk1

���

2D AB

(3)

To insure that a is positive, we must choose the plus sign. Hence a isgiven by:

a = �1 + 1 + 4D ABk1

��� v02( )�

�

�

v0

2D AB

�

��

�

� (4)

and the concentration profile is:

cA = exp +1� 1+ 4D ABk1

��� v02( )�

�

�

��

v0z2D AB

�

��

�

��

�

�

��

� (5)

which is the result in Eq. 19.1-20.

138

Note to p. 589

It is desired to show how to obtain Eq. (H) of Table 19.2-4 from Eq.(E).

We move the � �q( ) term to the left side of the equation andthen perform mathematical operations on the new left side. First weuse Eq. 3.5-4 to rewrite the new left side as:

DHDt

+ � �q( ) =�

�tH + � �vH( ) + � �q( ) (1)

For a multicomponent mixture, the heat flux vector q may be writtenas described in Fn. (a) of Table 19.2-4, and then neglect the

contribution qx( ) . Then Eq. 1 may be rewritten as

�

DHDt

+ � �q( ) =�

�tc�

H�

�=1

N

� + � ��vH( ) � � � k�T( ) + � �H

�

M�

j�

�=1

N

��

��

�

� (2)

Here we have also used the relation Eq. 19.3-9 to rewrite the firstterm on the right side. Next we combine the second and fourth termson the right side to get

�

DHDt

+ � �q( ) =�

�tc�

H�

�=1

N

� + � � vc �H + J�

H�

�=1

N

��

�

�

��

�

��

�

�� � � � k�T( )

=

�

�tc�

H�

�=1

N

� + v c�

H�

�=1

N

� + J�

H�

�=1

N

��

�

�

��

�

��

�

�� � kT( ) (3)

Then, since c�v + J�= N

� from Table 17.8-1, Eqs. (G) and (H),

�

DHDt

+ � �q( ) =�

�tc�

H�

�=1

N

� + � � N�

H�

�=1

N

��

���

�� � � k�T( ) (4)

Combining this with Eq. (E) then gives Eq. (H).

139

Note to p. 590

Simplification of the expression for the combined energy flux e,given in the first line of Eq. 19.3-4. First we insert Eq. 19.3-3 for q toget the second line of Eq. 19.3-4:

e = � U + 1

2 v2( )v � k�T +H

�

M�

j�+ pv + � �v[ ]

�=1

N

� (1)

(a) (b) (c) (d) (e) (f)

We next move terms (c) and (d) to the left, and terms (b) and (f) to theright:

e = �k�T +

H�

M�

j�+ � U + pV( )

�=1

N

� v + 12 �v2v + � �v[ ] (2)

(c) (d) (a) (e) (b) (f)

where used has been made of the expression V = 1 to get term (e).We next combine terms (a) and (e) to introduce the enthalpy:

e = �k�T +

H�

M�

j�+ �H

�=1

N

� v + 12 �v2v + � �v[ ] (3)

(c) (d) (a+e) (b) (f)

Next we make use of the relation H = c �H . Then we can use the fact

that enthalpy is "homogeneous of degree 1" (see Example 19.3-1 on p.591 and in particular Eq. 19.3-9), H = �

�n�

H� . Therefore, the

enthalpy per mole is �H = H n =�

�x�

H� . Hence term (a+e) can be

rewritten so that (3) becomes

e = �k�T +

H�

M�

j�+ c

�H

��=1

N

��=1

N

� v + 12 �v2v + � �v[ ] (4)

(c) (d) (a+e) (b) (f)

Then we use Eq. H of Table17.7-1, Eq. S of Table 17.8-1, and Eq. P ofTable 17.8-1 to rewrite parts of the summands in term (d+a+e) thus:

140

j�

M�

+ c�

v =j�+ �

�v

M�

=n�

M�

= N�

(5)

When this is substituted into (4) we get

e = �k�T + H

�N

��=1

N

� + 12 �v2v + � �v[ ] (6)

(c) (d+a+e) (b) (f)

Then, for systems in which terms (b) and (f) may be neglected, wefinally arrive at Eq. 19.3-6.

141

Note to p. 591

The theorem of Euler (pronounced "Oiler") for homogeneousfunctions is used in order to get Eq. 19.3-9. We here give a proof ofEuler's theorem.

First we give a definition of a homogeneous function. A functionof n variables is said to be "homogeneous of degree k" if

f �x1 ,�x2 ,�x3 ,��xn( ) = �k f x1 , x2 , x3 ,�xn( ) (1)

That is, if in the function f x1 , x2 , x3 ,�xn( ) , we replace x1 by �x1 , x2

by �x2 , etc., then this will give a result that is the same as multiplying

the original function by �k .

If we differentiate both sides of Eq. 1 with respect to � , we get

�f� �x1( )

� �x1( )��

+�f

� �x2( )� �x2( )��

+��f

� �xn( )� �xn( )��

= k� k�1 f x1 , x2 , x3 ,�xn( )

(2)where, on the left side, it is understood that by f we mean thefunction

f �x1 ,�x2 ,�x3 ,��xn( ) . It is understood that the function fhas continuous first partial derivatives. We now perform thedifferentiations on the left side to get

�f� �x1( )

x1 +�f

� �x2( )x2 +�

�f� �xn( )

xn = k� k�1 f x1 , x2 , x3 ,�xn( ) (3)

Next, we set � = 1, to get

x1

�f�x1

+ x2

�f�x2

+�xn

�f�xn

= kf x1 , x2 , x3 ,�xn( ) (4)

which is Euler's theorem. Here, the functionality of f is exactly thesame on both sides of the equation. Since the enthalpy is ahomogeneous function of degree "1" Eq. 19.3-9 follows directly.

142

n�

�

��H�n

�

�

�

��

n� ���( ),T ,p

= H (5)

or

n�

�

� H�= H (6)

This result is frequently used in discussions of mixtures. An exampleis the relation immediately after Eq. 17C.1-3, where, for a binarymixture

nAVA + nBVB =V

or, when the entire equation is divided by V,

cAVA + cBVB = 1

Another example of Euler's equation is in going from Eq. 1 to Eq. 2 inthe Note to p. 589.

143

Note to p. 615

We want to verify that Eqs. 20.1-16 and 17 are a solution to 20.1-9 and 10. We make use of §C.6 on the error function.

First we find the first and second derivatives of X with respectto Z:

X =

1� erf Z ��( )1 + erf�

(1)

dXdZ

=�1

1+ erf�2

�e� Z��( )

2

(2)

d2X

dZ2=

�11 + erf�

2

�e� Z��( )

2

�2 Z ��( )�� �� (3)

When the derivatives in Eqs. 1, 2, and 3 are substituted into Eq. 20.1-9, it is seen that the latter equation is satisfied.

Next we substitute the first derivative from Eq. 2 into Eq. 20.1-10 to get:

� = �

12

xA0

1 � xA0

�

��

�

�

�11 + erf�

2

�e� Z��( )

2�

���

�Z=0

=

xA0

1 � xA0

�

��

�

�

e��2

1+ erf�( ) �(4)

which agrees with Eq. 20.1-17.

144

Note to p. 622

Here we give a more detailed discussion of the developmentbetween Eqs. 20.1-67 and Eq. 20.1-74.

When the bracket in Eq. 20.1-67 is set equal to unity, we get:

d2 g

d 2+ 2

dgd

= 0 (1)

where g = cA cA0 and � = z � t( ) , which has the solution (by analogywith Example 4.1-1)

cA

cA0

= 1 � erfz�

(2)

with � given by Eq. 20.1-70. To get Eq. 20.1-72, we differentiate theconcentration profile:

NAz0 = �D AB

�cA

�z z=0

= �D ABcA0 ��

�zerf

z

4D AB S t( ) S t( )��

�

2dt

0

t�

�

�

���

�

z=0

(3)Use Eq. C.6-2 to differentiate the error function, and get

NAz0 = +D ABcA0

2

4D AB S t( ) S t( )�

��

2dt

0

t�

�

���

�1 2

= cA0

D AB

t1t

S t( ) S t( )��

�

2dt

0

t�

�

��

�

�1 2

(4)

The total number of moles of A that have crossed the mass-transfersurface S(t) at time t is then given by:

MA = S t( )cA0 1 � erf z / �( )�� ��0

�

� dz = S t( )cA0 erfc z / �( )0

�

� dz (5)

145

The complementary error function "erfc (...)" is defined in §C.6. Wenow insert this function into Eq. 5:

MA = S t( )cA0

2

�e��

2

z /�

�

�0

�

� d�dz = S t( )cA0�2

�e��

2d z �( )d�0

�

�0

�

�

(6)

In the second form, the order of integrations over and z � has beeninterchanged. Now the integral over z � can be performed to give

MA = S t( )cA0�

2

�e��

2

0

�

�d� = S t( )cA0�2

��12

(7)

Inserting the expression for � from Eq. 20.1-70, we then get

MA = S t( )cA0

1

�4D AB S t( ) S t( )�

���

2dt

0

t�

= cA0

4D AB

�S t( )��

��

2dt

0

t�

(8)which is Eq. 20.1-73.

Another expression can be obtained from integrating Eq. 20.1-72:

MA = S t( )NAz0 ( t )dt

0

t�

= cA0

4D AB

�

S t( )

S t( ) S t( )��

��

2dt

0

t�

0

t� dt (9)

It is relatively easy to show that Eqs. 8 and 9 are the same. We firstnote that the S t( ) in the denominator can be removed from theintegral, so that Eq. 9 can be rewritten as

MA = cA0

4D AB

�

S t( )��

��

2

S t( )��

��

2dt

0

t�

0

t� dt (10)

146

Then we make use of the fact that the square root in the denominatorcontains no t and hence can be removed from the integral over t togive

MA

= cA0

4D AB

�

S t( )��

��

2dt

0

t�

S t( )��

��

2dt

0

t�

= cA0

4D AB

�S t( )��

��

2dt

0

t�

= cA0

4D AB

�S t( )��

��

2dt

0

t� (11)

This is the same as Eq. 8. Proving that the two expressions for MA arethe same is a stronger statement than that the two expressions givethe same results for dMA dt .

This example is also a good illustration of the importance ofusing a symbol for the dummy variable of integration that is differentfrom that used as one of the limits in the integral.

147

Note to p. 626

We want to verify that the limiting solutions for slow and fastreactions in Eqs. 20.2-16 to 18 and 20.2-19 are correct.

(a) Slow reactionsWe start by rewriting Eq. 20.2-13 in terms of dimensionless

variables. We do this by using the definition of � in Eq. 20.2-17:

1Sc

=43�

dd�

�3+ �

3+ ��

2 (1)

Then, using the expansion in Eq. 20.2-16, we have

� = Sc�1 3 1 + a1� + a2�

2+�( ) (2)

�

2= Sc�2 3 1+ 2a1� + a1

2+ 2a2( )�2

+�( ) (3)

�

3= Sc�1 1+ 3a1� + 3a1

2+ 3a2( )�2

+�( ) (4)

Substitution of these three expressions into Eq. 1 gives

1 =

43

3a1� + 2 3a12+ 3a2( )�2

+�( ) + 1+ 3a1� + 3a12+ 3a2( )�2

+�( )

+ Sc1 3

� 1 + 2a1� + a12+ 2a2( )� 2

+�( ) (5)

We now equate terms in the same powers of � :

Zeroth power of � : 1 = 1 (6)

First power of � : 0 = 4a1� + 3a1� + Sc1 3�

or a1 = � 17 Sc1 3 (7)

Second power of � : 0 = 8

3 a13+ 4a1a2( )�2

+ a13+ 4a1a2( )�2

+ 2Sc1 3 a1�2

or 0 = 11 a1

2+ a2( ) + 2Sc1 3 a1

148

or 0 = 11 1

49 Sc2 3+ a2( ) + 2Sc1 3

� 17 Sc1 3( )

or a2 = + 3539 Sc2 3 (8)

Thus we have obtained the first two coefficients in the expansion ofEq. 20.2-16, and hence the first few terms in the expression for � as afunction of � .

(b) Fast reactionsWhen the trial function for � is taken to be of the form

� = K�m m < 0 (9)

When this is substituted into Eq. 1, we get

1Sc

=43

3m( )K3�

3m+ K3

�3m

+ K 2�

2m+1 (10)

The ratio of the last term to either of the first two terms isproportional to �

�m+1 , which is a positive quantity. For large � , thelast term is the dominant term on the right side. Therefore Eq. 1becomes

1Sc

= ��2 (11)

and therefore the solution to Eq. 1 for this case is

� = Sc�( )�1 2

(12)

in agreement with Eq. 20.2-19.

149

Note to p. 628

Example 20.2-2 is, for the most part easy to follow, except for thesteps going from Eq. 20.2-30 to Eq. 20.2-34. Here we provide thosemissing steps.

First, we assume � will be a function of � alone. Then thederivatives of � with respect to x and y must be converted toderivatives with respect to � , using Eq. 20.2-33:

��

�x�

��

�� y

=d�d�

��

�x�

��

�� y

=d�d�

yv

2�

12

x�3 2�

��

��=

d�d�

�12�

x�

��

��(1)

�

y

�

�

�� x

=d�d�

�

y

�

�

�� x

=d�d�

v�

2 x=

d�d�

�

y

�

�

��(2)

2�

y2

�

�

��

x

=

yd�d�

v�

2 x

�

�

�� =

d2�

d�2

v�

2 x=

d2�

d�2

�

y

�

�

��

2

(3)

Then Eq.20.3-30 becomes

�v

d�d�

�12�

x�

���

�+

v0 x( )v

��

�x�vdy

0

y�

�

��

�

�

d�d�

�

y

�

��

�

�=

v�

d2�

d�2

�

y

�

��

�

�

2

(4)

Now multiplication by v�

�( ) y �( )2 leads to

��v

d�d�

� +v0 x( )

v

��

�x�vdy

0

y�

�

��

�

�

2v

x

d�d�

=1�

d2�

d�2 (5)

(a) (b) (c) (d)

Our attempt to use a combination of variables to get an equation for� �( ) has not been successful because of the appearance of the term(b), which contains both x and � . Also term (c) may contain both xand � . We shall show that term (c) contains only � , and later (look

ahead to Eq. 20.2-37) arguments are given that v0 x( ) v

�( ) 2v�

x �( )may be set equal to a constant, K.

150

Let us, therefore, proceed to an analysis of term (c). We beginby replacing the integral over y by an integral over � :

�

�

�x�vdy =

0

y� �

�

�x�vd�

2xv

0

�

��

��

�

�

= �

2v

x

�

��

�

� �vd�

0

�

�( ) � 2xv

�

�x�vd�

0

�

�( ) (6)

Then the derivative with respect to x is replaced by a derivative withrespect to � :

�

�

�x�vdy =

0

y� �

2v

x

�

��

�

� �vd�

0

�

�( ) � 2xv

dd�

�vd�0

�

��

���

�y

v

2�

12

x�3 2�

���

�

= �

2v

x

�

�

�� �vd�

0

�

( ) +

2v

x

�

�

���v� (7)

When Eq. 7 is multiplied by 2v�

x d� d� , we then get

�

�

�x�vdy

0

y�

�

���

�2v

x

d�d�

= � �vd�0

�

�( ) d�d�

+�v�d�d�

(8)

( c1 ) ( c2 )

Thus, term c2 cancels term a, and the remaining term appears in thefinal equation for � :

K x( ) � �vd�

0

�

( ) d�d�

=1�

d2�

d�2(9)

which is Eq. 20.2-34.When K is taken to be constant, and the quantity in parentheses

is set equal to � f , then the velocity profile �v may be calculatedfrom Eq. 20.2-39 with the associated boundary conditions in Eqs.20.2-40 to 42. The remaining profiles may be obtained from Eq. 20.2-

151

43, which is obtained by setting d� d�=� �( ) in Eq. 9 and then

solving the first-order differential equation for � �( ) . One obtains

d�d�

+ �f � = 0 (10)

the solution of which is

d�d�

= � = C1exp �� fd�0

�

�( ) (11)

A further integration gives

� �,�, K( ) = C1 exp �� f � , K( )d�0

�

( )d�0

�

+ C2 (12)

Then the constants of integration are obtained from the boundaryconditions in Eqs. 20.2-35 and 36, with the final result for the profiles

� �,�,K( ) =exp �� f � ,K( )d�0

�

�( )d�0

�

�

exp �� f � ,K( )d�0

�

�( )d�0

�(13)

which is the same as Eq. 20.2-43.

152

Note to p. 692

We begin by formulating the problem as in Problem 12.1-4:

Solid Liquid

�Cs

��=

1�

2

�

���

2 �Cs

��

�

���

(1)

��Cs

���=1

= NCl (5)

Cs = finite at � = 0 (2)

Cs = Cl at � = 1 (3)

Cs = 1 at � = 0 (4)

Taking the Laplace transform of the problem we get

Solid Liquid

pCs � 1 =

1�

2

dd�

�2 dCs

d�

�

��

�

� (6)

�

dCs

d��=1

= NCl (9)

Cs = finite at � = 0 (7)

Cs = Cl at � = 1 (8)

The solution to the homogeneous equation corresponding to Eq. 6 isthe complementary function

Cs,cf =

K1

�cosh p� +

K2

�sinh p� (10)

The particular integral of Eq. 6 is, by inspection

Cs,pi =

1p

(11)

The complete solution to Eq. 6 is then

Cs =

K1

�cosh p� +

K2

�sinh p� +

1p

(12)

153

The boundary condition in Eq. 7 requires that K1 = 0 . The boundarycondition in Eq. 8 requires that

Cs �=1

= K2sinh p +1p= Cl (13)

and combination of Eqs. 9 and 12 gives

NCl = �K2 �sinh p + p cosh p( ) (14)

Elimination of Cl between Eqs. 13 and 14 leads to an expression for

the integration constant K2 , and hence also to Cs :

Cs =

1p�

Np�

�

��

�

�

sinh p�

p cosh p + N � 1( )sinh p(15)

Next, the Laplace transform of the total amount of A within thesphere is

MA

4�R3c0

= Cs0

1� �

2d� =

13p

�N

p2

xsinh x

p cosh p + N � 1( )sinh p0

p� dx

=

13p

�N

p2

xsinh x

p cosh p + N � 1( ) p0

p� dx

=13p

�N

p2

xcosh x � sinh x

p cosh p + N � 1( )sinh p

�

��

�

��

0

p

=

13p

�N

p2

p cosh p + N � 1( )sinh p � Nsinh p

p cosh p + N � 1( ) p

�

��

�

��

=

13p

�N

p21�

Nsinh p

p cosh p + N � 1( ) p

�

��

�

��

=

13p

�N

p2

p cosh p + N � 1( )sinh p � Nsinh p

p cosh p + N � 1( )sinh p

�

��

�

��

154

=

13p

�N

p21�

Nsinh p

p cosh p + N � 1( )sinh p

�

��

�

��

=1

3p�

N

p2+

N 2

p2 p coth p + N �1( )��

�

(16)

Now we take the inverse Laplace transform of the above; thetransforms of the first two terms may be found in an elementary tableof transforms:

MA

4R3c0

=13� N� + N 2

L�1 1

p2 p coth p + N � 1( )( )

�

��

��

�

��

��

(17)

To get the inverse transform of the last term, we can use theHeaviside partial fractions expansion theorem for repeated roots,which is:

If f p( ) = N p( ) D p( ) with D p( ) = p � a1( )

m1 p � a2( )m2� p � an( )

mn ,

N p( ) is a polynomial of degree less than mj( ) � 1, and ai ak for

i k , then

f t( ) =

�kl ak( )mk � l( )! l � 1( )!l=1

mk

�k=1

n

� tmk � leakt

with �kl p( ) =

dl�1

dpl�1

N p( )Dk p( )

�

��

�

� and

Dk p( ) =D p( )

p � ak( )mk

.

The contribution from the factor p2 is then (with a1 = 0 , m1 = 2, k=1)

�11 0( )2 � 1( )! 1 �1( )!

t +�12 0( )

2 � 2( )! 2 � 1( )!= �11 0( )� +�12 0( ) (18)

where

155

�11 0( )� =N 2

p coth p + N � 1( )p=0

� =N 2

1+ N � 1( )� = N� (19)

�12 0( ) =d

dpN 2

p coth p + N � 1( )

�

�

��

�

�

��

p=0

= �N 2d dp( ) p coth p + N � 1( )( )

p coth p + N �1( )( )2

�

�

���

�

�

���

p=0

= �

13

(20)

The contributions in Eqs. 19 and 20 just exactly cancel the first twoterms in Eq. 17.

The contribution from the remaining factor in L�1 { } is

MA �( )4�R3c0

=

L�1 N p( )

D p( )

���

�

��

��=

N ak( )D' ak( )k=2

�

� e+ ak� (21)

= N 2 e+ ak�

p2 12 p�1 2coth p � 1

2 csch2 p( ) + 2p p coth p + N �1( )( )��

�� p=ak

k=2

�

�

------------------------------

where the prime on D indicates differentiation with respect to p, andthe ak are the zeros of the denominator in the braces in Eq. 17 (except

for the double zero from p2 , which we have already taken into

account). This means that the dashed underlined term in Eq. 21 maybe omitted.

We know on physical grounds that the quantity M(t) must be adecreasing function of time. Therefore, the ak must be negative. This

can be guaranteed by setting ak equal to ��k2 where the �k are real

numbers. Then we have

L�1 N p( )

D p( )

��

��

��

��== 2N 2 sinh2 p

p2 p�1 2sinh p cosh p � 1( )��

�� p=��k

2k=2

�

� e��k

2�

156

= 2N 2 i2sin2�k � i�k

�k4( ) isin �k �cos�k � i�k( )k=2

�

� e��k

2�

= 2N 2 sin2�k

�k3( ) �k � sin�kcos�k( )k=2

�

� e��k

2� (22)

Therefore, the total amount of A within the sphere is at any time t is:

MA t( )43 �R3c0

= 6N 2 Bnn=1

�

� exp ��n2D ABt / R2( ) (23)

where the �n are determined from

�ncot�n + N �1( ) = 0 (24)

and the Bn are

Bn =

N 2sin2�n

�n3�n � sin �ncos�n( )

(25 )

For infinite N these last two expressions may be simplified. If, in Eq.24, N is infinite, sin�k must be zero, and therefore �k must be n� . If,in Eq. 25, N �

Bn

�ncot�n( )2

sin2�n

�n3�n � sin �ncos�n( )

=�n

2

�n4=

1

n�( )2

(26)

Thus we obtain the results in Eqs. 22.4-34 and 35.

Two references should have been cited here:E. N. Lightfoot, in Lectures in Transport Phenomena, AIChE, New

York (1969), pp. 59-60. H. Gröber, S. Erk, and U. Grigull, Die Grundgesetze der Wärme-

157

übertragung, Springer-Verlag, Berlin, 3rd edition (1961), pp. 55-62.

158

Note to p. 766

Here we work through the details of the development on p. 766,leading up to the expression for the generalized diffusional drivingforce at the bottom of the page.

When Eq. 24.1-2 is applied to a moving element of fluid, wewrite

DUDt

= TDSDt

� pDVDt

+G

�

M�

D��

Dt�=1

N

� (1)

in which

�

DUDt

= � � �q( ) � �:�v( ) + j��g

�( )�=1

N

� Table 19.2-4, Eq. D (2)

[Note footnote b to Eq. D!!]

DVDt

=DDt

1�

�

��

�= �

1�

2

D�

Dt= +

1�

� �v( ) Table 19.2-3, Eq. A (3)

�

D��

Dt= � � � j

�( ) + r� Table 19.2-3, Eq. B (4)

Substitution of these expressions into Eq. 1 gives

�

DSDt

=1T

� � �q( ) � �:�v( ) + j��g

�( )�=1

N

��

���

�+

pT

� �v( )

�

1T

G�

M�

� � � j�( ) + r

�( )�=1

N

�

= �

1T

q( ) �G

�

M�

j�( )

�=1

N

��

��

�� �

1T

�:v( ) �1T

G�

M�

r�

�=1

N

�

+

1T

j��g

�( )�=1

N

�

159

= � � �

1T

q �G

�

M�

j�

�=1

N

��

�

��

�

�

�� � q �

1T2

�T�

�

��+

j�� �

1T

G�

M�

�1T

g�

�

��

�

��

�

�

�

�

�=1

N

� �

1T

�:v( ) �1T

G�

M�

r�

�=1

N

� (5)

Comparison of Eq. 5 with Eq. 24.1-1 gives the entropy flux vector andthe entropy production rate (Eqs. 24.1-3 and 4):

s =

1T

q �G

�

M�

j�

�=1

N

��

��

�

(6)

gS = � q

1T 2

T�

��

��� j

�

1T

G�

M�

�1T

g�

�

��

�

��

�

��

��

�=1

N

� �1T

�:v( ) �1T

G�

M�

r�

�=1

N

�

(7)

Next we rewrite Eq. 6 by replacing q by q h( )

+ H�

M�( ) j

��=1

N

� --that is,

we subtract off the heat flux associated with the diffusion of thechemical species. This gives us then for the entropy flux (Eq. 24.1-5):

s =

1T

q h( )�

S�

M�

j�

�=1

N

��

��

�

(8)

The entropy flux is now written as the sum of two terms: the firstterm is the entropy flux associated with heat flow, and the secondterm is that connected with diffusion of the chemical species. (For

more on the reasons for replacing q by q h( )

+ H�

M�( ) j

��=1

N

� , see S. R.

de Groot and P. Mazur, Non-Equilibrium Thermodynamics, North-Holland, Amsterdam (1962), pp. 24-25.)

In order to rewrite the entropy production term, we make useof the Gibbs-Duhem relation in the entropy representation (see H. B.Callen, Thermodynamics and an Introduction to Thermostatics, Wiley,New York (1985), pp. 60-62):

160

Ud

1T

�

���

+Vd

pT

�

���

� n

�d

G�

T

�

��

�

�=1

N

� = 0 (9)

where n� is the number of moles of species � . Division by V anddoing some elementary manipulations gives us the form of theGibbs-Duhem equation that we need:

� �

�

H�

M��=1

N

�1

T 2T +

1Tp � �

�

1T

G�

M�

�

�

�� = 0

�=1

N

� (10)

Next, in Eq. 7, we add four terms inside the bracket in the secondterm on the right side in such a way that the term is not changed:

gS = � q �1

T 2�T

�

���

���

j�

��

�

���

1T

G�

M�

�1T��

��p + �

�

H�

M�

1T 2

�T

���

H�

M�

1T2

�T �1T��

g�+

1T��

���g�

�=1

N

�

�

�����

�

�

�����

�

�

����

�

�

����

�=1

N

�

�

1T

�:v( ) �1T

G�

M�

r�

�=1

N

� (11)

The terms containing H� clearly cancel one another, and the terms

containing �p and ����g� do not contribute to the sum, inasmuch as

��j�= 0 . Thus, Eq. 11 may now be rewritten as:

gS = � q h( )�

1T 2

�T�

���

���

j�

��

�

���

1T

G�

M�

�1T��

��p + �

�

H�

M�

1T2

�T

�1T��

g�+

1T��

���g�

�=1

N

�

�

�����

�

�

�����

�

�

����

�

�

����

�=1

N

�

�

1T

�:v( ) �1T

G�

M�

r�

�=1

N

� (12)

or

161

TgS = � q h( )

lnT( ) �j�

��

cRTd�

�

��

��

�=1

N

� �

1T

�:v( ) �1T

G�

M�

r�

�=1

N

�

(13)

The d� are the generalized driving forces for diffusion . Note that the

above development guarantees that ��d�= 0 , as is required, since

��j�= 0 . We can see that ��

d�= 0 , since the first three terms in the

bracket in the first line of Eq. 12 sum to zero according to the Gibbs-Duhem equation, and the fourth and fifth terms clearly combine tosum to zero.

The above development has led us to the expression for thegeneralized driving forces for diffusion:

cRTd

�= c

�T�

G�

T

�

��

�

� + c

�H

��lnT ��

��p � �

�g�+�

���g�

�=1

N

� (14)

To get the alternative form for d� given in the second line of Eq. 24.1-8, we follow the discussion of Ref. 5 on p. 766.

The quantities G� , H� , and p are functions of the state of a fluidelement, which may be described by the temperature T, the pressurep, and the set of (N – 1) mole fractions x� where � = 1, 2,3, ... ( N – 1).

Then the first term in Eq. 14 may be written by expanding �G� using

the chain rule of partial differentiation to give:

c�T�

G�

T

�

��

�

= c

��G

�� c

�G

��lnT

= c

�

�G�

�x�

�

��

�� �x

�+ c

�

�G�

�T

�

��

�� �T + c

��=1

N�1

��G

�

�p

�

��

���p � c

�G

��lnT

= c

�

�G�

�x�

�

��

��

�=1

N�1

� �x�� c

�TS

��lnT + c

�V��p � c

�G

��lnT (15)

162

When Eq. 15 is combined with Eq. 14, it is seen that the �lnT termsexactly cancel, and the �p terms may be combined. Furthermore,

when the relation dG�= RTdlna

� is used, we get finally

cRTd�= c

�RT

�lna�

�lnx�

�

�

��

T ,p,x

x�

�=1

N�1

+ ����

�( )p � ��

g�+�

���g�

�=1

N

(16)

in which �� = c�V� is the volume fraction of species � , and the

subscript x stands for all the x� except x� and xN . Thus, we see that

the driving forces d� include the contributions from the mole fractiongradients, the pressure gradient, and the external forces.

Equation 16 is in agreement with Eq. 7.8 of Ref. 5 at the bottomof p. 766, and also with Eq. 11.1-27 of Molecular Theory of Gases andLiquids, by Hirschfelder, Curtiss, and Bird, Wiley, New York (1964).Therefore, the first term of the second line of Eq. 24.1-8 of BSL may bemisleading. Here and elsewhere (Eqs. 24.2-8, 24.2-9, 24.2-10, 24.4-1,and 24.5-4) the abbreviated notation

�lna�

�lnx�

�

��

�

�

T ,p,x

�x�

�=1

N�1

� � �T ,plna� (17)

is used (this is just the chain rule applied to �lna� with the �p and

�T terms omitted). This notation has also been used by E. N.Lightfoot, Transport Phenomena and Living Systems, Wiley, New York(1974), Eqs. 1.2.7 and 12, on p. 163, and W. M. Deen, Analysis ofTransport Phenomena, Oxford University Press (1998), Eq. 11.6-2.Deen, however, states that his treatment is for dilute mixtures. Bothof these authors, however, appear to have their summations goingfrom 1 to N (instead of 1 to (N – 1).

The discussion in Multicomponent Mass Transfer, by R. Taylorand R. Krishna, Wiley, New York (1993), also uses essentially thenotation of Eq. 17 above, with the summation going from 1 to (N – 1)as may be seen on pp. 23, 24, and 29 (their Eq. 2.3.10 is, aside fromsome notational differences, the same as Eq. 16 above).

163

In Advanced Transport Phenomena, by J. C. Slattery, CambridgeUniversity Press (1999), p. 450, Eq. 8.4.3-4, gives a result that isconsistent with Eq. 16 above, although the notation is considerablydifferent from ours.

There is also a discussion of multicomponent systems inTransport Phenomena Fundamentals, by J. L. Plawsky, pp. 66-69 (heatflux) and pp. 69-73 (mass fluxes); however, no derivations of theexpressions are given.

Thermodynamics of Irreversible Processes, by G. D. C. Kuiken,Wiley, New York (1994), has a discussion of the driving force d� onp. 183.

164

Note to p. 767

Just as a check, we show that Fick's law for a binary systemmay be obatained by simplifying Eq. 24.2-3. This equation is, for abinary system with the two components labeled "A" and "B":

jA = +A DAAdA + DABdB( ) (1)

Use Eqs. (A) and (C) of Table 24.2-1 and the equation dA +dB = 0 (seeline 2 on p. 767), to eliminate the generalized Fick diffusivities infavor of the Maxwell-Stefan diffusivities:

jA = +�A �

�B2

xAxB

DABdA ��A�B

xAxB

DABdA

�

��

�

(2)

This may be rewritten, using the fact that �A +�B = 1, to give

jA = �

�A�B

xAxB

DABdA (3)

Next use the second line of Eq. 24.1-8, omitting the terms for thermaldiffusion, pressure diffusion, and forced diffusion. This yields:

jA = �

�A�B

xAxB

DAB xA� ln aA( ) = ��A�B

xAxB

DAB

� ln aA

� ln xA

�

�

�� �xA (4)

Then, to obtain the molar flux of species "A" we use Eq. 17B.3-1 to getthe relation JA

*= cxAxB �A�B( ) jA . Finally, combine this equation

with Eq. 4 and introduce the binary diffusivity D AB from Eqs. 24.3-2and 4, to get:

JA*= �cD AB�xA (5)

This is the form of Fick's (first) law given in Eq. B of Table 17.8-2 for abinary system.

165

Note to p. 768

Here we show how to derive the generalized Maxwell-Stefanrelations (Eq. 24.2-4) from the generalized Fick equations (Eq. 24.2-3).This proof was first given by H. J. Merk, Appl. Sci. Res., 73-99 (1959),§5, Eqs. 89 and 90.

We start by introducing the abbreviation j�#= j

�+ D

�

T�lnT , in

order to avoid having to carry along the thermal diffusion termthroughout the derivation. Then the generalized Fick equation, Eq.24.2-3, for species � may be written as

j�

#

��

= D��

d�

� =1

N

� (1)

where

j�

#

�

� = 0 ,

d�

�

� = 0 , D

��=D

��, and

�

��

� D��

= 0 (see above Eq.

24.2-3). Next, write a similar equation for species � and form thedifference of the two equations; then multiply both sides of theequation by the quantity x�

x�

D�� to get

x�

x�

D��

j�

#

��

�j�

#

��

�

��

�� =

x�

x�

D��

D��

�D��( )d�

� =1

N

� (2)

Now we sum both sides over the index �

x�

x�

D�����

�j�

#

��

�j�

#

��

�

�

�� =

x�

x�

D��

D�

�D� ( )

���

��

�

��

�

�

�� =1

N

� d

(3)

If the bracket quantity on the right side is set equal to ��

���

, wesee that we get

x�

x�

D����

j�

#

��

�j�

#

��

�

�

�� = �

����

�( )� =1

N

� d�= d

���

�d�= d

���

�0( )

� =1

N

� = d�

166

or

x�

x�

D�����

�j�

#

��

�j�

#

��

�

�

�� = d

�(4)

This is exactly the generalized Maxwell-Stefan equation for species (see Eq. 24.2-4). At the same time we get

x�

x�

D��

D��

�D��( ) =

���

� ���

�� (5)

From this we will get the relations between the D� and the D

� .One might wonder why we replace the bracket expression in

Eq. 3 by ��

���

rather than simply ��

, which would, after all, alsolead to the generalized Maxwell-Stefan equation. If Eq. 5 is multipliedby �

and summed over , one gets the identity 0 = 0, because of

�

� D�

= 0 . But if the � is not included on the right side of Eq. 5,

multiplication of the equation by � and summing over will give 0

= � . Similar arguments may be made for not setting the bracketexpression equal to

��

� A��

, where A is an arbitrary constant. Let us now return to 5 and get the relation between the Fick

and the Maxwell-Stefan multicomponent diffusivities. We start bydefining a matrix B with matrix elements

B�( )

� = �D

� +D

� ; the

matrix B is an N � 1( ) � N � 1( ) matrix, with the ( � = )-row and the(� = )-column missing. We now rewrite Eq. 5, omitting the equationfor � = :

x�

x�

D��

B�( )

� =

���

� ��� (� � � ) (6)

Next we perform some operations on the relation

��

�

� D��

= 0:

�

�D

� �D

� ( ) + ��

�

��

� D�

= 0 or

��

B�( )

��+

�

� D��

= 0 (7)

167

Then we multiply Eq. 6 by B�

�1( )�

and sum on to get:

x�

x

D�

= ���

B�

�1( )�

�

� (8)

Multiplying the second form of Eq. 7 by B�

�1( )�

and summing on

gives

�

= � D

��B�

�1( )�

�

� (9)

Then multiplying the reciprocal of Eq. 8 by Eq. 9, and replacing theindex � by the index � , gives Eq. 24.2-7 of the textbook:

D��

=x�

x�

���

�

D��

adjB�( )

��� ��

�

adjB�( )

��� ��

�(10)

In the last step we have also made use of the fact that

B�

�1( ) = adjB�( ) detB

�( ) (11)

where adjB� is the matrix adjoint to B� and detB

� is the determinantof the matrix B� . The matrix adjB

� is the transpose of the matrix of

cofactors (or "signed minors") of B� .

168

Note to p. 769

As a check on Eq. 24.2-7, we use it to get Eq. (B) of Table 24.2-2.We first write Eq. 24.2-7 for the specific case of the 1-2 pair of aternary system:

D12 =

x1x2

�1� 2

D12 adjB1( )22+D13 adjB1( )32

adjB1( )22+ adjB1( )32

(1)

The 2 x 2 matrix B1( ) may be displayed thus:

B1( ) =

B1( )22B1( )23

B1( )32B1( )33

�

��

�

�� =

�D22 +D12 �D23 +D13

�D32 +D12 �D33 +D13

�

��

�

��(2)

The adjoint matrix is the transpose of the matrix of cofactors (acofactor is a "signed minor"):

adj B1( ) =

adj B1( )22adj B1( )23

adj B1( )32adj B1( )33

�

��

�

�� =

B1( )33� B1( )23

� B1( )32B1( )22

�

��

�

��

=

�D33 +D13 D23 �D13

D32 �D12 �D22 +D12

�

��

�

��(3)

We are now ready to substitute into Eq. 24.2-7:

D12 =

x1x2

�1� 2

D12 �D33 +D13( ) +D13 D32 �D12( )�D33 +D13( ) + D32 �D12( )

=

x1x2

�1� 2

�D12D33 +D13D32

�D12 �D33 +D13 +D32

�

��

�

�

=

x1x2

�1� 2

+D12D33 �D13D23

D12 +D33 �D13 �D23

�

��

�

� (4)

In the last step, we have made use of the symmetry of the D

�� .