transport phenomena addictional notes

8/7/2019 transport phenomena addictional notes

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p. 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 tank p. 242 Power-law flow in circular tubesp. 248 Viscoelastic flow near an oscillating platep. 255 Polymer flow analyzed with a FENE-P modelp. 275 Dimensional consistency in heat transferp. 286 Enthalpy of an ideal monatomic gasp. 299 Temperature profile in flow with viscous heatingp. 315 Temperature profile in tube flowp. 337 Alternative equation of change for temperaturep. 346 Temperature profile for transpiration coolingp. 375 One-dimensional time-dependent heat conductionp. 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 filmp. 547(ii) Taylor expansion of diffusion problem resultp. 555 Diffusion with chemical reactionp. 563(i) Diffusion with solid dissolutionp. 563(ii) Evaluation of mass flux from concentration profilesp. 565 Verification of solution of diffusion problemp. 585 Diffusion with convection and chemical reactionp. 615 Time-dependent evaporation of a liquidp. 622 Diffusion with time-dependent interfacial areap. 692 Interaction of phase resistancesp. 767 Simplification of multicomponent diffusion resultp. 768 Relating Maxwell-Stefan and Fick diffusivitiesp. 769 Illustrating interrelations between diffusivities


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Note to p. 5

Section 0.3 is important for emphasizing some of the basicconcepts and definitions. Here we work through some of the 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 +R A1 and make analogousreplacements for rA2 , rB1 , and rB2 . We also let mA1 =mA2 = 12 mA . Withthese substitutions, we then get:

12 mA rA +R A1( )+ 12 mA rA +R A2( )+ 12 mB rB +RB1( )+ 12 mB rB +RB2( ) (1)

But, according to the drawing in Fig. 0.3-2, R A1 = R A2 and,analogously, R B1 = RB2 , so that

mArA +mBrB =mArA +mBrB (2)

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

molecular masses and velocities.

(b) We start with Eq. 0.3-5, and replace rA1 by rA +R A1 as above. Wealso let mA1 =mA2 = 12 mA . Then Eq. 0.3-5 becomes:

12

12 mA rA rA( )+2 rA R A1( )+ R A1 R A1( )( )

+12 12 mA rA rA( )+2 rA R A2( )+ R A2 R A2( )

+ A

+12 12 mB rB rB( )+2 rB R B1( )+ RB1 R B1( )( )

+12 12 mB rB rB( )+2 rB R B2( )+ RB2 R B2( )

+ B

=12

12 mA rA rA( )+2 rA R A1( )+ R A1 R A1( )( )

12 mA rA +R A1( )+ 12 mA rA +R A2( )+ 12 mB rB +RB1( )+ 12 mB rB +RB2( )=


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+12 12 mA rA rA( )+2 rA R A2( )+ R A2 R A2( )

+ A

+12 12 mB rB rB( )+2 rB R B1( )+ RB1 R B1( )( )

+12 12 mB rB rB( )+2 rB R B2( )+ RB2 R B2( ) + B (3)

The single-underlined terms just exactly cancel the doublyunderlined terms in the following line, because R A1 = R A2 and also

R B1 = RB2 . Hence we get

12 mA rA rA( )+ 12 mA1 R A1 R A1( )+ 12 mA2 R A2 R A2( )+ A +

12 mB rB rB( )+

12 mB1 R B1 RB1( )+

12 mB2 R B2 R B2( )+ B

= 12 mA rA rA( )+ 12 mA1 R A1 R A1( )+ 12 mA2 R A2 R A2( )+ A +12 mB rB rB( )+ 12 mB1 R B1 RB1( )+ 12 mB2 R B2 R B2( )+ 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 of molecule 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.


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

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


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Note to p. 18

In Fig. 3B.2 there is shown a duct with cross-section of anequilateral triangle. The height of the triangular cross-section is H ,and the side length is 2H 3 . We want to evaluate the viscous stresstensor components for the incompressible flow in the z-direction, forwhich the velocity in the z-direction is given as a function of x and yin Eq. 3B.2-1:

vz x, y( )=

P 0

P L( )4 LH

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

and vx =0 and v y =0. Here L is the length of the duct (which goesfrom 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 =

vz

y

= P

0P

L( )

4 LH y3x2 y y3 3Hx2 +Hy2( )

=

P 0

P L( )

4LH 3x2 3 y2 +2Hy( ) (2)

xz =

vzx

= P

0P L( )

4 LH x3x2 y y3 3Hx2 +Hy2( )

=

P 0

P L( )

4LH 6x y H ( )( ) (3)

At the surface y = H :

yz y=H

=P

0P

L( )4LH

3x2 H 2( ); xz y=H =0 (4,5)


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Note to p. 26

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

t 2T

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 tensoris a dimensionless quantity.


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Note to p. 51

On this page several quantities are obtained from the velocitydistribution of Eq. 2.3-18. We could also ask ourselves: how 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 0R

02

=2 12 vz

2

( )vzrdr0R

=2

12 R

2 vz3

0

1 d = R2vmax

3 1 2( )01

3 d (1)

In the second line, we have introduced the dimensionless coordinate

= r R and the maximum velocity vmax = P 0 P L( )R2 4 L. Now all

that remains is to evaluate the integral:

R

2

vmax3

1 3 2

+3 4 6

( )01

d

=

R

2

vmax3 1

234 +

36

18

= R2vmax3

4 6 +4 18

=

18

R2vmax3

= 1

4 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 maximum

velocity.


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Note to p. 52

The laminar flow in a circular tube with radius R is discussed in2.3, and the laminar flow in tubes with equilateral triangular cross-section of height 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 8 L

w = 3P

0P

L( )H 4 320 L

(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 of the cross-sectional areas. Since for circular tubes A = R2 , and for

equilateral triangular tubes, A =13 H

2 ,

w =

P 0 P L( ) A2 2( ) 8 L

; w = 3 3

P 0

P L( )A2

180 L(3,4)

Therefore

ww =

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

b. The perimeters of the two tubes are P =2 R for circulartubes, and P = 2 3H for triangular tubes. Therefore


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w =

P 0

P L( ) P 2 ( )

4

8 L w =

3 P 0 P L( ) P 2 3( )4

180 L(6,7)

Taking the ratio, as before

ww

= 32 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

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

ww

=0.545 (same perimeters) (10)

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

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

Lifshitz, Fluid Mechanics , Addison Wesley (1959); our H is their amultiplied by 12 3 .]


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Note to p. 54

Let's check a few things about Fig. 2.4-1.a. There the graph shows the transport 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( )R

2LrR

2 Rr

=

P 0

P L( )R

2L1 2 R

r

2Rr

(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 ( )ln1s

=0 2s + 1 2

ln 1 ( )

1s

(2)

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

smax =

1 2

2ln 1 ( ) (3)

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


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rmax =R

1 2

2ln 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

2ln 1 ( )< 1

2(1+ ) or

2 1 ( )1+ ( )ln 1 ( )

0.8, the maximum is very close to being half way 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 =

P 0

P L( )R2

4 L1

rmax

R

21 2

ln 1

( )ln R

rmax

=

P 0

P L( )R2

4 L1 1

2

2ln 1 ( )1 2

ln 1 ( )ln

2ln 1 ( )1 2

(6)

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


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Note to p. 55

It is good practice to check limiting cases whenever possible.For example, one should show that Eq. 2.4-17 becomes the Hagen-Poiseuille formula for tube flow (Eq. 2.3-21) in the limit that 0 .(See the comment in the paragraph that begins four lines after Eq. 2.4-14.)

Solution: We have to show that the bracket quantity in Eq. 2.3-21 becomes 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

01 ( )= (3)

Thus the bracket quantity becomes:

1 4( ) 1 2( )2

ln 1 ( )lim

0 11 =1 (4)

and the Hagen-Poiseuille formula is recovered.


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Note to p. 58

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

In region I, the average velocity is given by

vz

I = p0 pL( )b2

2 IL1b

2 I

I + II

+

I

II

I + II

xb

xb

2

dxb

0

= p0 pL( )b

2

2 IL2 I

I + II

+

I

II

I + II

2 d

1

0

where =x b

=p0 pL( )b2

2 IL

2 I

I + II

1

2

I

II

I + II

1

3

= p0 pL( )b

2

2 IL6 2 I 3 I II( ) 2 I + II( )

I + II

= p0 pL( )b

2

2 IL7 I + II

I + II(1)

Similarly for Region II we have

vz

II = p0 pL( )b2

2 IIL1b

2 II

I + II

+

I

II

I + II

xb

xb

2

dx0

b

= p0 pL( )b

2

2 IIL2 II

I + II

+

I

II

I + II

2 d

0

1

= p0 pL( )b

2

2 IIL2 II

I + II

+1

2

I

II

I + II

13

= p0 pL( )b

2

2 IIL6 2 II +3 I II( ) 2 I + II( )

I + II

= p0 pL( )b

2

2 IIL

I +7 II

I + II(2)


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Note to p. 59

The verification of some equations requires a lot of algebraicdetail that falls in the category of "straightforward but tedious."Nonetheless, such derivations should be done. Here are twoexamples:

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

rr = 2

vrr

= 2 v 1R

+

32

Rr

2 32

Rr

4

cos

= 3 v

RRr

2+ R

r

4cos (1)

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

r = r r

v r

+

1r

vr

= v 1R

Rr

32

Rr

2 Rr

4

sin

v 1

R

Rr

32

Rr

2

12

Rr

4

sin ( )

= +3

2 vR

Rr

4

sin = +3

2 vR

Rr

4

sin (2)

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

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

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


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F n( ) = p0 + gRcos +

32

vR

cos 0 0

2 cos ( )R2sin d d

= 2 p0 + gRcos +

32

vR cos 0

cos ( )R2

sin d

= 2 p0R2 cos 0

sin d +2 R2 gR+ 32 vR

cos

2 sin d 0

(3)

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

F n( ) = 2 R2 gR+ 3

2 vR

23

=

43

R3 g +2 Rv (4)

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

F(t) =2 3

2 vR

sin

0

sin ( )R2sin d

=3 Rv sin 3 0

d =3 Rv 43

=4

Rv (5)

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


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= 2

1r sin

v + vr +v cot

r

= 2 v 1

R

R

r

1 3

2

R

r+1

2

R

r

3

cos +

R

r

1+ 3

4

R

r+1

4

R

r

3

cos

= 3 v

2RRr

2

+ Rr

4

cos (3)

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


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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 of them on the left side of the equation:

v

t v

+ v vv[ ]( )

= v t v + v v( ) t + v v v[ ]( )+ v v( ) v( ) (4)

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

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


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

=

t

1

2 v v( )

1

2v 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 by12 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.


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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

=t

r 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

vlvklk j

= ijk

lk j xl x jvlvk ijk vlvk

lk j jl

=

xl v

l ijk

x jlk jv

k ijk

v jv

kk j(2)

In the second line, we have moved the x j inside the differentiationand 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 ( v jvk ) and antisymmetrically in another ( ijk); seeExercise 5 on p. 815). Now we convert Eq. 2 back to bold-facenotation:

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

c. Next we examine the term containing the pressure:


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r p[ ]i = ijk

k jx j xk

p = ijkk j xk

x j p ijkk j

jkp (4)

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

xl ijkx jp kl{ }

lk j=

xl ijkx jp kl

k jl(5)

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

xlr p{ }il =

l xlr p{ }li

=l

r p{ }i

(6)

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

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

same manner as in part (c) above:

r [ ] i = ijkx jk j

[ ]k = ijkx jk j xl

lkl

(7)

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

ijk

lk j xlx j lk ijk

lk j lk xl

x j

=

xl ijkx j kl

k jl ijk

l lk

k j jl

=

xlr { }ill

ijk

jk

k j =

xlr { }li

+ ikj jk

k jl


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= r { }

i+ :[ ]i (8)

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

be replaced by in the first term.

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

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


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Note to p. 86

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[ ]( )= vii v v[ ] i (2)

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

= vi

i ijkv j v[ ]k

k j= vi

i ijkv j klm xl

vmmlk j

(3)

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

= ijk klmvi

mlk jiv j xl

vm = ijk lmkvi

mlk jiv j 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 viv j xl vml ji (5)

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


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= viv j

ji xiv j viv j

ji x jvi (6)

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

= viv j

ji xiv j v jvi

ji xiv j =0 (7)

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


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Note to p. 90

An alternative method of solving the problem in Example 3.6-2is given here. Eq.3.6-21 may be set up by making a shell torque balance on a shell of thickness r and height L. Then the torque atradius r is equal to 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:

d

drr 2 r ( )=0 or

d

drr 2 r d

dr

v

r

=0 (1a,b)

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

ddr

1r

ddr

rv ( ) =0 or

d2v dr2

+1r

dv dr

v r 2

=0 (2)

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

r 2 r d

drv r

=r

2 r 1r

dv dr

v r 2

=r

2 dv dr

rv (3)

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

ddr r 2

dv dr rv

=2r

dv dr +r 2

d2v

dr 2 rdv

dr v

= r 2 d

2v dr2

+1r

dv dr

v r 2

=0 (4)


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28

Therefore either r 2 is zero, or the quantity in parentheses is zero. But r 2 cannot be zero, and hence Eq. 2which came from Example 3.6-3must be the same as Eq. 1b (from the torque balance).


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29

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

= dd

1 2

e2d0

=0

= 2

e

2

=0= 2

= 1.1287 (1)

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

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


30/119


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31

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

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

Eliminating b between these two equations gives

a4 = 1

4(7)

Hence

a2 = 12 , 1

2and

a = 1

2

, 1

2

i (8)

whence

b = 1

2 , 1

21i

(9)

and the two square roots of i are

i = a +bi= 12 1 +i( ) (10)

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

i = rei =1ei 2 (11)

Then the square root of i is

i = ei 4 = cos

4+ isin

4

=

12

+ i 12

(12)


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32

Note to p. 125

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

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

2

vrr

2

= 92

v2

r 2 1 3 cos ( )2 (1)

21r

v +

vrr

2

=2v2

r 2 1 +34

1

+14

3

cos

+ 132

1

+12

3

cos

( )2

= 9

8v2

r 2 1 + 3 cos ( )

2(2)

2

v cot r

+ vrr

2

= 2v2

r 21+ 34

1 + 14 3 cos + 1 32

1 + 12 3 cos ( )

2

= 9

8

v2

r2

1 + 3 cos ( )2

(3)

r

rv r

+

1r

vr

2

= v r

v r

+1r

vr

2

= v2

r 2

34

1 + 34 3 sin + 1 34

1 14

3 sin

1 32 1 + 12

3 sin

2

=v2

r 2 32 3sin

2

(4)

When the above results are combined, we get

: v( )r 2 = v2 274

2 272

4 + 274 6 cos2 + v2 94

6 sin2

(5)


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33

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

Fkv = 2 : v( )r 2drR 0

sin d

=2 2

3 v2 27

4 2 27

2 4 + 274 6 dr +2 43 v2 94 6 drR R

(6)since cos2 0

sin d = 23 and sin 3 0

d = 43 . Then, finally

Fk =9 v R 2 2 4 + 61 d +6 v R

61 d

=9 v R 1 + 23

3 15

5

1+6 v R 15

5

1

=9 v R 1 23 + 15 +6 v R 15

=9 v R 25 +6 v R 15 = 245 + 65( ) v R

=6 v R (7)

which is Stokes' law.


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34

Note to p. 139

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

ewas

found 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 vxx +v y vx y =ve dvedx +

2

vx 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

dvedx

+ 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

1x 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

yx 1 ( ) 2 ( )

(6)


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35

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

y = Ax1 2 ( ) df

d y = ABx1 2 ( )

f '1

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

f ' (7)

2

y2= ABx 2 ( )

y f " = 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 d

dxx1 2 ( )

f Ax

1 2 ( ) df d x

= A x

1 2 ( )

2 ( )1x

f Ax1 2 ( )By 1 ( )

2 ( )x 1 ( ) 2 ( ) 1

f '

= A 1

2 ( )x1 2 ( )

x f +A 1

( )2 ( )

x1 2 ( )

x f ' (10)

2

y x= A 1

2 ( )x1 2 ( )

x y f '+A 1

( )2 ( )

x1 2 ( )

x yd

d f '( )

= AB

12 ( )

x1 2 ( )

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

1x 2 3 ( ) 2 ( )

+ AB3 1x 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 12 ( )

x 2 ( )x1 2 ( )

x 1 ( ) 2 ( )x f 2


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36

A2B2

1 ( )2 ( )

x 2 ( )x1 2 ( )

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

A2B2 1

2 ( )

x1 2 ( )

x1 2

( )2

( )x f 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 ( )x2 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 ( )x2 3 ( ) 2 ( ) f 2 f f ( )=c' 2 12 ( )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.]


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37

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 02

rdrd 0

R 02

=vz ,max1 r R( ) 1 7 rdrd

0

R 02

rdrd 0

R 02

= vz ,max

2 R2

1 r R( ) 1 7 rdr = 2vz ,max0R 1 [ ]1 70

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 70

1

d 8 7

0

1

d

( )=2vz ,max

8 7

8 7

15 7

15 7

0

1

=0.82vz ,max

(3)

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


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38

Note to p. 170

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

1=0 in Eq. 5.6-18 and rewriting the equation, we get:

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

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

12 F

2( )= F( ) 2F (2)

Each term may now be integrated with respect to to give

12 F

2 = 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 dF

2F 1 + 14 F( )= d

(4)

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

12 ln

1+ 14 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


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39

F1+ 14 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+14 F( )C3 ( )

2or

F =

C3 ( )2

1 14 C3 ( )2 (7)

Case II (negative sign):

F = 1+14 F( )C3 ( )

2or

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 monotonedecreasing 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 vz20 rdr =2 t( )2 2C3

2( )2

1+ 14 C3 ( )2 40

d (9)

We now let 14 C3 ( )2 =x2 so that d = 4 C32( )xdx; then


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40

J =32 t( )2C32xdx

1+x2( )40

=32 t( )2C321

6 1+x2( )30

=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 ( )220 d z

2

= 4 t( ) zC324

C32

xdx

1+x2( )20 =16 t( )z 1

2 1+x2( )0

=8 t( )z (11)


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41

Note to p. 198

Here we show how to obtain the macroscopic mass and momentum balances 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( )dSS t( ) (2)

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

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

ddt

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

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

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

the "outlet" surface S2 (on which vS = 0)the "fixed" surface S f (on which v =vS =0)the "moving" surface Sm (on which v =vS 0)

with v being the fluid velocity and vS the surface velocity. Thesurface integrals are then split into four parts corresponding to thefour types of surfaces.


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42

The integral on the left side is the total mass mtot in the system.The surface integrals over S f 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 u 1 and u 2 , 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

+u 2 . Now we make the assumptions that (i) the density is a constantover the cross section, and (ii) that the velocity is always parallel to

the walls of the entry and exit tubes, so that v = u at plane "1" 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 + gdV V t( )

(7)


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44

F f 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 macroscopicmomentum balance (with F f s

p( ) +F f s( ) =F f s = Fs f ):

ddt

Ptot = 1 v12 S1u 1 2 v22 S2u 2 +p1S1u 1 p2S2u 2 +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 balanceisgiven in 7.8, the derivation of the macroscopic energy balanceis givenin the Note to p. 454.]


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45

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 x2 =2Rz z2 (1a,b)

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

dV = x2dz = 2Rz z2

( )dz (2)

Then the total volume of the liquid is:

V = 2Rz z2( )dz0

h = Rz2 13 z

3( )0

h

= Rh2 1

3 h3( ) (3)

Thus the liquid volume is:

V = Rh2

113

hR

(4)

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

43 R

3

Tank half full: h =R V =23 R

3


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46

Tank empty: h =0 V =0

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

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

ddt

Rh2 1 13

hR

=

P 2 P 3( )D4 128 L

(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:

P 2 = 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,

P

3 = 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( )D4128 L

(8)

c. Eq. 8 may be rewritten as

Rh +L

ddt

h2 13

h3

R

= gD

4

128 LA (9)

(which defines the quantity A) or


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47

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

H dH dt

=A (11)

or

H 2 R +L( )+ 2R +L( )L 1

H

dH

dt

=A (12)

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

12

H 2 2 R +L( )H + 2R +L( )Lln H 2R+L

L

=At 0tefflux (13)

This gives

12

L2 12

2R +L( )2 2 R +L( )L+2 R +L( ) 2R +L( )+ 2R +L( )Lln L2R +L

=Atefflux (14)or after some cancellations

tefflux =

1A

2R2 +2RL 2R +L( )Lln 2R +LL

(15)

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


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48

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 thatappears in Eq. 8.3-5:

= 12 :( )= 12 ij ji

j=1

3

i=1

3and = 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

= 12 :( )= dvz dr( )

2 = dvz dr (5)

where the minus sign must be chosen, since dvz dr is negative. Thenthe shear stress rz will be

rz = mdvz

dr

n 1 dvz

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:


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49

m

dvzdr

n

=P

0P

L

2L

r or

dvzdr

=P

0P

L

2mL

1 n

r1 n (7,8)

Integration of Eq. 8 gives

vz =

P 0

P L

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 = P

0P L

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 =

P 0

P L

2mL

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

1 n( )+1 (11)

Rearrangement then gives

vz =

P 0

P L( )R

2mL

1 nR

1 n( )+11 r

R

1 n( )+1

(12)

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

w = vz0R 0

2 rdrd

= 2 R2

P 0

P L( )R2mL

1 n R1 n( )+1 1

1 n( )+1( )01 d (13)

where = r R . Performing the integration gives


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50

w =2 R2

P 0

P L( )R

2mL

1 nR

1 n( )+11 n( )+1

2 1 n( )+3

= R

3

P 0

P L( )R

2mL

1 n1

1 n( )+3 (14)

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


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51

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

i v0ei t{ }= d

2v0

dy2ei t 0

1e s 1 e i sds0 (1)

Next, we perform the integration over s:

i v0ei t{ }= d2v0

dy2ei t 0

1

11 1( ) i

e s 1 e i s

0

= d

2v0

dy2ei t 0

1+ i 1

(2)

We may now remove the real-operator sign from both sides, as wellas the common multiplier ei t , to get:

i v0 = d

2v0

dy20

1 + i 1

or

d2v0

dy2= i 1 +i 1 ( )

0v0 (3a,b)

Eq. 3b is a differential equation for the complex function v0 y( ). Theequation is of the form of Eq. C.1-4, where [ ]is a2 . Since this is acomplex quantity, we write it as + i ( )2 , where and are real, so

that we can write the solution as

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

Then according to Eq. 8.4-19,


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52

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 go back 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)


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53

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

2

= 1

2

2 0 12

1

2

0

2

+4

2 0

2

= 1 2

2 01 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 21 2

(15)

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

=

2 01+ 12 2 1

1 2(16)

=

2 0 1 + 1

2 2

1 1 2

(17)

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


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54

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


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55

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 0

0 0 00 0 0

(1)

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

Z p,xx p,xy p,xz

p,yx p,yy p,yz

p,zx p,zy p,zz

H

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)


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56

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

Dimensionless shear stress: S =

p,yx3nKT (6)

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

Dimensionless shear rate: = H (8)

Then Eqs. 3, 4, and 5 become:

ZN =2S (9) ZS =

13 (10)

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

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

N 3S

=2S and 3S

=1 + 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+3

18S + 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 13 arcsinh p

3 2q( ) (16)

or, in terms of the original variables:


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57

p,yx

3nKT = 2 b+3

54

+1 2

sinh 13 arcsinhb+354

3 2 b

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-Newtonian

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

S = 2p1 2sinh 13 p

3 2q( )= 2p1 2 13 p 3 2q +{ }= 23 p 1q(18)

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

sinh x = x + 16 x3 + arcsinh x =x 16 x

3 + (19,20)

In terms of the original variables, Eq. 18 is

p,yx = 3nKT 2

3

b+354

1 b

108 H = nkT

b

b+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 AppendixC.5). Therefore

S = 2p1 2sinh 13 ln 2p

3 2q( )( )= 2p1 2 12 exp 13 ln 2p 3 2q( )( )= 2q( )1 3(23)


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58

In the original variables, this becomes

p,yx = 3nKT 2

b108

H

1 3

= nKT 54 b108

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, seeTeaching with FENE Dumbbells, by R. B. Bird, Rheology Bulletin, January 2007.]


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59

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)

MLt3T

=

M( ) ML2

t2T

T ( )

L2

ML2

t2T

M(1)

(b)

Mt3

=

ML3

Lt

2 Lt

+

ML3

L2t2

Lt

+

MLt2

Lt

+

Mt3

(2)

(c)

L2

t2

= L

2

t 2T

T ( )+ L

3

M

MLt2

+ T ( )

L3

M

T ( ) MLt2

(3)

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


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60

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

R M

(1)

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

p V = RT

M(2)

Therefore, the bracket in Eq. 9.8-8 is

V T V T

p= V T

T RT pM

p= V T R

pM

=0 (3)

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

H H o = 5

2R

MT T o( ) (4)


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61

Note to p. 299

In connection with Fig. 10.4-2, we want to find the location of the 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 T 0

T b T 0(temperature)

= x

b(distance) (1)

so that Eq. 10.4-9 becomes:

= 1

2Br 1 ( )+ (2)

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

dd = 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< .


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62

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:

dd

= 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

dd =1

= sinh N 1 ( )

cosh N N ( )

=1=0 (4)

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


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63

Note to p. 315

Here we verify the determination of the constants of integra-tion, C

0 , C

1 , and C

2in Eq. 10.8-27.

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

0 , ( )=C0 +C1ln0 +C2 (1)

This can be satisfied only if C1 =0 .

C0 : Boundary condition 2 requires that =1at =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

+C20

1 1 2( ) d (3)

Next we evaluate the integrals:

4 1 2( ) d =4 3( )d =4 12

14

0

1 01 = (4)

4

2

4 4

16

0

1 1 2( ) d =4 3

45 5

16+

7

1601 d =

4 1

165

6 16+ 1

8 16

=

724 4

(5)


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64

C2 1

2( ) d =01 14 C2 (6)

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


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65

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( ):

d H =

H T

pdT +

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 another

equilibrium 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 ( ):

d U =

U

T

V dT +

U

V

T d V = CV dT + p +T

p

T

V d V (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:

D U Dt

= CV DT Dt

+ p +T pT

V

D V Dt

(3)

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

D V Dt

= DDt

1

=

1 2

D Dt

(4)


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67

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:

dd

= 1 ln

+N 2 3

1 1 2

1 ln

=0 (1)

Solving for 2 2 gives:

2 2

= 1N 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 than12

+1( )? Clearly this cannot beanswered until N is known; but N depends on the geometry, thetemperature difference, the thermal conductivity, and the viscosity. If we 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.


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68

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 of dimensionless variables:

= T T 1

T T 1= dimensionless temperature (1)

= r

R= dimensionless radial coordinate (2)

Then Eq. 11.4-26 becomes:

dd

= RR0

dd

2 dd

where R0 =

wr Cp4 k

(3a,b)

Now introduce the change of variable u = 2 d d ( ) , and rewrite thedifferential equation as:

u 2

= RR0

dud

(4)

This equation may be solved to get:

1 = R

R0ln u + R

R0ln C1 or

C1u =exp

R0R

1

(5a,b)

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

C1

2 dd

=exp R0R

1

(6)

Further integration then gives:


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69

C1 = 2exp

R0R

1

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 first boundary condition, C1 may be found. Then the final expression for

the dimensionless temperature is:

= e

R0 R ( ) e R0 R( )

e R0 R ( ) e R0 R( )(8)

which agrees with Eq. 11.4-27.


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70

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 u

2 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 = ddx

2

e u2du0

u = 2

xe u

2du +e u

2 ux

e 02 0

x0u

(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 to

the 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 y

2 4 t 14 t

(4)

2

y2= 2

e y

2 4 t 14 t

2 y4 t

(5)


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71

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 =0and 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're Joking Mr. Feynman , Bantam Books, New York (1986), p. 72 and p. 93.Professor Feynman was a Nobel Prize winner in physics.]


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72

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 T 0 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 T 0 =

q0k

e 2 y ei t i 2 y{ } y dy

= q0

kei t e 2 y i 2 y y dy{ }

= q0k

ei t e1+i( ) 2 y

1+ i( ) 2 y

= q0

kei 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 T 0 = q0k e

2 y ei t

ei 2 y

1 i( )2 2

= q0

2k2

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


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73

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 , 14 (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 =r

2 sin 2 (10)

Adding the two equations then gives:

r 2 = 2 or

r = 2 (11)

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

1 i = 2ei 4 (12)

Returning now to Eq. 3, we get:

T T 0 =

q02k

2 e

2 y

2ei t i 2 y i 4

{ } = q0

k e 2 y cos t 2 y 14 ( ) (13)

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


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74

Note to p. 386

The derivation of Eq. 12.3-6 from Eq. 12.3-5 is given here. First

we 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 ln

sin Z 1sin Z +1

=

1 ln

sin X +iY ( ) 1sin X +iY ( )+1

+i

= 1

ln

sin Xcos iY +cosXsin iY 1sin Xcos iY +cosXsin iY +1

= 1

ln

sin Xcosh Y + icosXsinh Y 1sin Xcosh Y + icosXsinh Y +1

= 1 ln

sin Xcosh Y 1( ) sin Xcosh Y +1( )+cos2 Xsinh2 Y +icosXsinh Y sin Xcosh Y +1( ) cosXsinh Y sin Xcosh Y 1( )

sin Xcosh Y +1( )2 + cosXsinh Y ( )2

= 1

ln

sin 2 Xcosh 2 Y 1 +cos2 Xsinh 2 Y ( )+2icosXsinh Y sin Xcosh Y +1( )2 + cosXsinh Y ( )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


75/119


76/119

76

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 = v2 x( )

vxv0

1 0 1vxv

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

= v2 x( ) 2 2 3 + 4( )1 2 +2 3 4( )d0

1 = v

2 x( ) 2 4 2 2 3 +9 4 4 5 4 6 +4 7 8( )d01

= v2 x( )1 43 12 + 95 23 47 + 12 19( )

= v2 x( ) 315+567 180 35315( ) =

37315 v

2 x( ) (2)

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

Cp0 vx T T ( )dy

= Cpv T T 0( ) T x( )

vx

vT T T T 0

0

1

d T

= Cpv T T 0( ) T x( )

vxv

T T 0T T 0

T 0 T T 0 T

0

1 d T


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77

= Cpv T T 0( ) T x( )

vxv

1T 0 T T 0 T

0

1 d T (3)

in whichT

= 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 T 0( ) T x( ) 2 T 2 T 3 3 + T 4 4( )0

1 1 2 T +2 T 3

T 4( )d T

= Cpv T T 0( ) T x( ) 215 3140 3 + 1180 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 the boundary-layer thicknesses as a function of the distance along theplate:

2 v

= 37

315 v2 d

dx (5)

2k T T 0( ) T

= 215

3

140

3 + 1180

4

( ) C

pv T T

0( ) d

dx T

(6)

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

ddx

= 31537

2 v v2

; d 0 = 63037

vdx0

x ; = 126037 x v

(7,8,9)

and Eq. 6 for T x( )may also be solved:

2k T T 0( ) T

= 215 31403 + 1180

4( ) Cpv T T 0( ) ddx T (10)

T

ddx

T =1

215

3140

3 + 1180 4( )2k

Cpv(11)


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78

T d T =

22

153

1403 + 1180 4( )

vdx0

x 0 T (12)

T =

4215 3140 3 + 1180 4( )

x

v

(13)

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

=

T

= 4 37 1260( )2

153

1403 + 1180 4( )

(14)

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

2

153 3

1405 + 1180

6 = 35315

1Pr

Pr=

k Cp

= 1 Cp k

(15)

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

=35315

152

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.]


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79

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( )

dd

=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:

dd

= C0I ( ) 1+ t( ) ( )

+ C1 1 + t( ) ( )

(3)

A second integration with respect to gives:

=C0I ( )

1+ t( ) ( )0 d +C1 1

1 + t( ) ( )d +C20

(4)

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


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80

, ( )=C0 +C0I ( )

1+ t( ) ( )d 0

+C2 (5)

Next, apply the boundary condition at =1:

1 =C0I ( )

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 drivingforce

0 b , which is the dimensionless wall temperature minus

the dimensionless bulk temperature (defined in Eq. 10.8-33):

0 b =C0I ( )

1+ t( ) ( )d 0

1 C0

I ( ) 1+ t( ) ( )

d d 0 0

1

d 01

=C0 I ( ) 1 + t( ) ( )

d 01 C0I 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:

C0I 1( )

d d 0 0

1 ( ) I ( ) 1+ t( )

( )

d 01

= C0I 1( )

I 1( ) I ( )01

I ( ) 1+ t( ) ( )

d


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81

= C0I ( )

1+ t( ) ( )01 d + C0I 1( )

I ( ) 2

1+ t( ) ( )01 d (8)

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

0 b =C0I 1( )

I ( ) 2

1 + t( ) ( )01 d =

I ( ) I 1( ) 2

1 + t( ) ( )01 d (9)

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


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82

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( )

zF

z

= t( )

Pr t( )1r r

rr

(1)

where the primes indicate differentiations with respect to . We next

make the change of variables = r z and = t( ) w( )z , and furtherlet r , z( )= , ( )= f ( ) , so that

r=

r+

r=

1z

+

0 = f

1z

(2)

z

=

z+

z=

rz2

+

t( )

w

= f

rz2

f 2

z

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

t( )

zF

F

f

1z

t( )

zF

f

rz2

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

= 1Pr t( )

1

f

(5)


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

1

d

d 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. A

further integration from 0 to then yields

ln f ( ) f 0( )

=Pr t( ) F 0

d = Pr t( ) C32

1+ 14 C3 ( )2 d 0

= Pr t( )ln 1 + 14 C3 ( )

2 2

(8)or, taking the antilogarithm of both sides

f ( ) f 0( )

= 1+ 14

C3

( )

2 2Prt( )

(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.


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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 v

2 + U + ( ) dV =

12 v

2 + 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 equation

and the Gauss divergence theorem to the right side:

ddt

12 v

2 + U + ( )dV n 12 v2 + U + ( )vS( )S t( ) V t( ) dS

= n 12 v

2 + 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

K tot +U tot + 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 S f and the moving surface Sm . At the inlet cross section S1and 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


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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 12 v

2 + U + ( )v vS( )( )S t( ) dS = 12 1 v1

3 S1 + 1 U 1 v1 S1 + 1 1 v1 S1

12 2 v2

3 S2 + 2 U 2 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( )S f +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( )S f +Sm dS =W m

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

S f the surface velocity is zero.The fourth term on the right (the -term) is evaluated similarly

to the p-term. First the contributions at surfaces S1 and S2 areconsidered, but it is assumed that these will be small compared to the


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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[ ]( )S f +Sm dS =W m (8)and, here again, the contribution at S f will be zero. The contributions

from Eqs. 7 and 8 will be added to give W mp +W m =W m , 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 K tot +U tot + tot( )=

12 1 v1

3

+ 1 U 1 v1 + 1 1 v1( )S1

12 2 v2

3 + 2 U 2 v2 + 2 2 v2( )S2 +Q +W m + p1 v1 S1 p2 v2 S2( ) (8)

or, introducing the enthalpy

d

dt

K tot +U tot + tot( )= 12 1 v13 + 1 H 1 v1 + 1 1 v1( )S1

12 2 v2

3 + 2 H 2 v2 + 2 2 v2( )S2 +Q +W m (9)Either Eq. 8 or Eq. 9 is referred to as the unsteady-state macroscopicenergy balance.


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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

51

ech KT 10 d (1)

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

2 c2h 1

5d =2 c2h KT

ch

5

x5 chKT

1x2

dx

= 2

KT ( )4

c2h3x3( )dx (2)

Hence the integral becomes

qbe( ) =

2 KT ( )4

c2h3x3

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 nx0 dx

n=1= 2

KT ( )4

c2h36 1

n4n=1= 2

KT ( )4

c2h3 4

15

(5)


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

=1+ 1

24+ 1

34+ 1

44+ =1.0823 (5)

Nowadays, even in a small integral table (such as H. B. Dwight,Tables of Integrals and Other Mathemati

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transport phenomena addictional notes