Notes for the 2nd revised edition of TRANSPORT PHENOMENA by R. B. Bird, W. E. Stewart, and E. N. Lightfoot by R. B. Bird 9 Aug 2011 These "Notes," begun in 2009, are intended for use by students and instructors to fill in the missing steps in some of the derivations in the text. Comments and corrections will be greatly appreciated. Also, if there are places in the text where you feel additional explanation is needed, it would be appreciated if you would let us know. Many of these notes involve the Leibniz formula for differentiating integrals, the hyperbolic functions, the error function, Taylor series, and the gamma functions—all topics for which many undergraduates have received inadequate instruction. These topics are all reviewed in Appendix C. Pages in the text for which "Notes" have been prepared: p. 5 p. 18 p. 26 p. 35 p. 50 p. 51 p. 52 p. 54 p. 55 p. 58 p. 59 p. 78 p. 81 p. 82(i) p. 82(ii) p. 86(i) p. 86(ii) Conservation laws in binary collisions Evaluation of stress-tensor components Dimensional consistency in fluid dynamics Convective momentum flux Flow in tubes with elliptical cross section Flow of kinetic energy in tubes Conduits with circular and triangular cross sections Velocity distribution in annular flow Limiting cases of annular flow Flow of immiscible fluids Flow around a sphere Normal stresses at solid surfaces Equation of change for mechanical energy Proof that ( :v ) is positive for Newtonian fluids Conservation equation for angular momentum Vector identity needed for the Bernoulli equation Another way to look at the Bernoulli equation 1 p. 90(i) p. 90(ii) p. 117 p. 121 p. 123 p. 125 p. 139 p. 154 p. 170 p. 198 p. 199 p. 202 p. 218 p. 241 p. 242 p. 248 p. 250 p. 255 p. 259 p. 275 p. 286 p. 299 p.309 p. 315 p. 337 p. 341 p. 343 p. 346 p. 352 p. 375 p. 377 p. 379 p. 386 p. 388 p. 413 p. 415 p. 454 p. 494 p. 529 p. 534 p. 535 p. 547(i) Tangential flow between cylinders—pressure distribution Torque balance Slope of the complementary error function Fluid motion near an oscillating plate Equation for the stream function Alternative method of getting Stokes' law The Falkner-Skan equation Average velocity in turbulent flow in tubes Turbulent flow in a circular jet Derivation of macroscopic balances Efflux from a spherical tank The lawn sprinkler Unsteady flow from a cylindrical tank The Weissenberg-Rabinowitsch equation Power-law flow in circular tubes Viscoelastic flow near an oscillating plate The corotational Maxwell model Polymer flow analyzed with a FENE-P model The Casson equation Dimensional consistency in heat transfer Enthalpy of an ideal monatomic gas Temperature profile in flow with viscous heating Checking the cooling fin solution Temperature profile in tube flow Alternative equation of change for temperature The equation of change for entropy Tangential annular flow with viscous heating Temperature profile for transpiration cooling Stationary shock wave velocity distribution One-dimensional time-dependent heat conduction Two solutions for the slab heating problem Unsteady heat conduction with sinusoidal heating Steady-state potential flow of heat in solids Boundary layer flow with heat transfer Turbulent flow in tubes with heat transfer Turbulent flow in circular jets with heat transfer Derivation of macroscopic energy balance Planck's radiation law and Wien's displacement law Dimensional consistency in diffusion Binary formulas from multicomponent formulas Two formulations of Fick's law of diffusion Diffusion through a stagnant gas film 2 p. 547(ii) p. 555 p. 557 p. 563(i) p. 563(ii) p. 565 p. 584 p. 585 p. 589 p. 590 p. 591 p. 615 p. 622 p. 626 p. 628 p. 692 p. 766 p. 767 p. 768 p. 769 Taylor expansion of diffusion problem result Diffusion with chemical reaction Simplifying Eq. 18.4-18 Diffusion with solid dissolution Evaluation of mass flux from concentration profiles Verification of solution of diffusion problem Form of diffusion equation in molar units Diffusion with convection and chemical reaction Energy equation for multicomponent mixtures Simplification of the combined energy flux e Euler's theorem for homogeneous functions Time-dependent evaporation of a liquid Diffusion with time-dependent interfacial area Diffusion with chemical reaction Forced convection from flat plate Interaction of phase resistances Driving force in multicomponent diffusion Simplification of multicomponent diffusion result Relating Maxwell-Stefan and Fick diffusivities Illustrating interrelations between diffusivities 3 Note to p. 5 Section 0.3 is important for emphasizing some of the basic concepts and definitions. Here we work through some of the missing steps in Section 0.3, going from Eq. 0.3-3 to Eq. 0.3-4, and from Eq. 0.3-5 to Eq. 0.3-6. (a) In Eq. 0.3-3, replace rA1 by rA + R A1 and make analogous replacements for rA2 , rB1 , and rB2 . We also let mA1 = mA2 = 12 mA . With these substitutions, we then get: 1 2 ( ) ( ) ( ) ( ) 1 1 1 mA rA + R A1 + 2 mA rA + R A2 + 2 mB rB + R B1 + 2 mB rB + R B2 = 1 2 ) + 1 m ( r + R ) + 1 m ( r + R ) + 1 m ( r + R ) (1) mA ( rA + R A1 A A A2 B1 B2 2 2 B B 2 B B But, according to the drawing in Fig. 0.3-2, R A1 = R A2 and, analogously, R B1 = RB2 , so that mA rA + mB rB = mA rA + mB rB (2) This is the law of conservation of momentum in terms of the molecular masses and velocities. (b) We start with Eq. 0.3-5, and replace rA1 by rA + R A1 as above. We also let mA1 = mA2 = 12 mA . Then Eq. 0.3-5 becomes: 1 1 2 2 ( ( ) ( mA ( rA rA ) + 2 rA R A1 + R A1 R A1 )) R + 12 12 mA ( rA rA ) + 2 rA R A 2 + R + A A2 A2 ( ( ( ) ( ) ( + 12 12 mB ( rB rB ) + 2 rB R B1 + R B1 R B1 )) ) R + + 12 12 mB ( rB rB ) + 2 rB R B2 + R B2 B2 B = 1 1 2 2 ( ( ) ( ) ( R mA ( rA rA ) + 2 (rA R A1 ) + R A1 A1 )) 4 R + + 12 12 mA ( rA rA ) + 2 rA R A 2 + R A2 A2 A ( ( ) ( ( ) ( + R R + 12 12 mB ( rB rB ) + 2 rB R B1 B1 B1 )) ) R + + 12 12 mB ( rB rB ) + 2 rB R B2 + R B2 B2 B ( ) ( ) (3) The single-underlined terms just exactly cancel the doubly underlined terms in the following line, because R A1 = R A2 and also R B1 = RB2 . Hence we get R ) + 1 m ( R R ) + mA ( rA rA ) + 12 mA1 ( R A1 A1 A2 A2 A2 A 2 1 1 1 R ) + m (R R ) + + m ( r r ) + m ( R 1 2 = + B 2 1 mA 2 1 mB 2 B B B1 2 B1 B1 B2 2 B2 B2 B ) + m (R R ) + R (rA rA ) + mA1 (R A1 A1 A2 A2 A2 A 1 1 ) + B (rB rB ) + 2 mB1 (R B1 R B1 ) + 2 mB2 (R B2 R B2 1 2 1 2 (4) In the first line of the equation above, the terms have the following significance: Term 1 is the kinetic energy of molecule A in a fixed coordinate system; Term 2 is the kinetic energy of atom A 1 in a coordinate system fixed at the center of mass of molecule A; Term 3 is the kinetic theory of atom A2 in a coordinate system fixed at the center 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 atoms in 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. 0.36. This discussion of the collision between two diatomic molecules is interesting, for several reasons. It shows how the idea of "internal energy" arises in a very simple system. We encounter this concept later in §11.1 where the terms "kinetic energy" and "internal energy" are used in connection with a fluid regarded as a continuum. When the fluid is regarded as a continuum, it may be difficult to understand how one goes about splitting the energy of a fluid into kinetic and internal energy, and how to define the latter. In considering the collision between two diatomic molecules, however, the splitting is quite straightforward. 5 Another point is that, having seen the need for splitting the energy into two parts, one might be led to ask: why don't we need to split the momentum into two parts in a similar way? Here again, for the collision of diatomic molecules, the need for dividing the momentum 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 diatomic molecules is relatively straightforward. However, when applying them to a moving fluid, some notational problems arise associated with the necessity of dealing with fluid bodies in three dimensions. 6 Note to p. 18 In Fig. 3B.2 there is shown a duct with cross-section of an equilateral triangle. The height of the triangular cross-section is H, and the side length is 2H 3 . We want to evaluate the viscous stress tensor components for the incompressible flow in the z-direction, for which the velocity in the z-direction is given as a function of x and y in Eq. 3B.2-1: vz ( x, y ) = (P 0 P L ) 4 μ LH ( y H )( 3x 2 y 2 ) (1) and vx = 0 and v y = 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 modified pressure between the ends of the duct. What are the stresses at the surface y = H according to Eq. 1.2-6? For the velocity distribution given above, the nonzero components are yz = zy and xz = zx . From Eq. 1.2-6, we get: yz = μ = vz (P P L ) 3x 2 y y3 3Hx 2 + Hy 2 = μ 0 y 4 μ LH y ( (P 0 P L ) 4LH ( 3x 2 3y 2 + 2Hy ) (2) vz (P P L ) 3x 2 y y3 3Hx2 + Hy2 = μ 0 x 4 μ LH x (P P L ) 6x y H = 0 ( ) 4LH ( xz = μ ( ) ) ) (3) At the surface y = H: yz y= H = (P 0 P L ) 4LH ( 3x 2 ) H 2 ; xz y=H =0 (4,5) 7 Note to p. 26 It is very important to make a habit of checking equations for dimensional consistency. Show that the following equations in the text are dimensionally consistent: Eq. 1.4-14, Eq. 1.5-11, and Eq. 1.7-2. Do this by replacing the symbols in the formulas by the dimensions corresponding to the symbols in the table beginning on p. 872. Omit any numerical factors that appear. (a) (b) (c) M Lt = M Lt = ML2 ( M ) 2 (T ) tT ( ) L2 () 1 ML2 moles t L3 moles (1) (2) M M M L L M M M L L 2 = 2 + 3 t t = 2 + 2 + 3 t t Lt Lt L Lt Lt L (3) In each case, dimensional consistency is found. In (c) the unit tensor is a dimensionless quantity. 8 Note to p. 35 In Fig. 3B.7 we show the flow into a slot of width 2B. Far away from 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 the dyadic product of v with v. The components of vv in Cartesian coordinates 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 the velocity given in Eqs. 3B.7-2 to 4, we get ( vv )xx 1 2w = vx vx = + W 2 ( vv )xy 1 2w = vx v y = + W 2 (x 2 +y 2 ) 2 ) 4 x5 y (x ( vv )xz = vx vz = 0 2 +y 4 (1) (2) (3) ( vv )yx 1 2w = v y vx = + W 2 ( vv )yy 1 2w = vy vy = + W 2 ( vv )yz = v y vz = 0 ( vv )zx = vz vx = 0 ( vv )zy = vz v y = 0 ( vv )zz = vz vz = 0 x6 x5 y (x 2 +y 2 ) 4 x4 y2 (x 2 +y 2 ) 4 (4) (5) (6) (7) (8) (9) b. For a plane perpendicular to the x axis, the vector n is x . The components of the momentum flux are then 9 1 2w x vv x = vx v x = + W 2 1 2w x vv y = v x v y = + W 2 x vv z = vx vz = 0 (x x6 2 +y 2 ) 2 ) 4 x5 y (x 2 +y 4 (10) (11) (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 of surface perpendicular to the x axis, and x vv z is the amount of z momentum flowing per unit time through a unit of surface perpendicular 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. 10 Note to p. 50 Information about the flow in tubes of elliptical cross-section, and comparison with the flow in a circular tube. Label the semi-major axis of the ellipse as a and the semi-minor axis of the ellipse as b. The cross-sectional area of the elliptical tube is A = ab . The velocity distribution for laminar axial flow in a tube of elliptical cross-section is vz 2 2 P1 P 2 ) a 2 b 2 x ( y 1 = ( ) 2 μ L a2 + b2 a (1) b as given by J. Happel and H. Brenner, Low Reynolds Number Hydrodynamics, Prentice-Hall, Englewood Cliffs, NJ (1965), p. 38. When a = b = R, this velocity distribution reduces to Eq. 2.3-18 of BSL2e, since x 2 + y 2 = r 2 . To get the mass rate of flow, we integrate the velocity distribution over the cross section, thus P1 P 2 ) a 2 b 2 ( w = 4 ab ( 2μ L a + b 2 2 ) 1 0 0 1 2 (1 2 ) 2 dd Here the dimensionless variables = x a and = y b have been introduced. Integration over then gives P1 P 2 ) a3 b3 ( w = 4 ( 2 μ L a2 + b 2 ) ( ) 2 1 3 0 1 3 0 1 3 3 2 (P1 P 2 ) a b 1 = 4 1 2 0 2 2 3 2μ L a + b ( ) ( ) 3 2 1 2 d d (2) The integral may be found in H. B. Dwight, Tables of Integrals and Other Mathematical Data, Macmillan, New York, 4th edition (1961), Formula 855.41, and the result is 11 () () 3 3 5 1 2 (P1 P 2 ) a b 2 2 w = 4 2 ( 3 ) 3 2 μ L a2 + b 2 ( ) 3 3 2 (P1 P 2 ) a b = 4 3 2μ L a2 + b2 ( = (P1 P 2 ) a3 b3 ( 4 μL a2 + b2 ) ( )( ) ( ) ( ) 3 2 1 2 1 2 1 2 2 2! (3) ) () since 12 = . When a = b = R, this mass rate of flow expression reduces to Eq. 2.3-21 of BSL-2e. We can now compare the mass rates of flow for the circular tube and the elliptical tube of the same cross sectional area, as follows Circular tube: Elliptical tube: w = well = ( ) (4) ( ) (5) (P1 P 2 ) A 2 2 8μL (P1 P 2 ) A2 2 ab ( 4 μ L a2 + b 2 ) so that well = 2 1 + 2 w (6) where = b a 1 . Thus for tubes of the same cross-sectional area, the ratio well w decreases from unity (both tubes circular) to values less than unity as becomes smaller. 12 Note to p. 51 On this page several quantities are obtained from the velocity distribution of Eq. 2.3-18. We could also ask ourselves: how much kinetic energy is flowing per unit time in the axial laminar flow of a fluid in a circular tube? The volume rate of flow through an element of cross section rdrd is vz rdrd . The kinetic energy per unit volume of the fluid is vz2 , since the only nonzero velocity component is vz . Therefore the total amount of kinetic energy per unit volume flowing through the tube is: 1 2 0 0 ( 12 vz ) vz rdrd = 2 0 ( 12 vz ) vz rdr 2 R R 2 2 1 1 ( ) 3 3 2 = 2 12 R 2 0 vz3 d = R 2 vmax 0 1 d (1) In the second line, we have introduced the dimensionless coordinate = r R and the maximum velocity vmax = (P 0 P L ) R 2 4 μ L . Now all that remains is to evaluate the integral: 1 1 3 3 1 2 4 6 2 3 3 R 2 vmax 0 1 3 + 3 d = R vmax 2 4 + 6 8 4 6 + 4 1 1 3 3 = R 2 vmax = R 2 vmax 8 8 1 1 2 = R 2 vmax vmax 2 4 ( ( ) ) (2) The last form suggests a volume rate of flow multiplied by a kinetic energy per unit volume, both quantities evaluated for the maximum velocity. 13 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 crosssection of height H is described in Problem 3B.2. Both tubes have the same length, L. We want to compare these two flow problems. a. Compare the mass rates of flow for the two tubes when their cross-sectional areas are the same. b. Compare the mass rates of flow for the two tubes when the perimeters of their cross sections are the same. a. For flow in circular and triangular tubes we have for the mass flow rates (see Eq. 2.3-21 and Eq. 3B.2(b)): (P 0 P L ) R 4 w = 8μL w = 3 (P 0 P L ) H 4 320 μ L (1,2) In Eq. 1, R is the tube radius; in Eq. 2, H is the height of the triangular cross section, and 2H 3 is the length of a side of the triangle. To make the comparison, we need to express the flow rates in terms of the cross-sectional areas. Since for circular tubes A = R 2 , and for 1 2 equilateral triangular tubes, A = H , 3 w = ( ) (P 0 P L ) A 2 2 8μL ; w = 3 3 (P 0 P L ) A 2 180 μ L (3,4) Therefore w 3 3 (16 ) = 8 = 0.726 w 180 (5) b. The perimeters of the two tubes are P = 2 R for circular tubes, and P = 2 3H for triangular tubes. Therefore 14 (P 0 P L ) ( P 2 ) 8μ L 4 w = w = ( 3 (P 0 P L ) P 2 3 180 μ L ) 4 (6,7) Taking the ratio, as before w ( 8)( 2 ) = 0.265 3 = 4 w 2 3 (180 ) 4 ( ) (8) For the square cross section (see Problem 3B.3), the ratios corresponding to Eqs. 5 and 8 may be found to be w = 0.884 (same cross-sectional areas) w (9) w = 0.545 (same perimeters) w (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 a multiplied by 12 3 .] 15 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 , positive when 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 an annulus, as given in Eq. 2.4-14. What is the location of the maximum velocity? Show that the position of the maximum is nearer the inner cylinder 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 2 R P L ) R r P0 PL )R ( 2 R 2 R 1 R r = r r 2L 2L (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 in agreement 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, as a function of r R = s , we have to differentiate [ ] with respect to s as follows 9 (alternatively, one may set rz equal to zero): 1 2 1 d 1 2 1 2 ln = 0 2s + 1 s ln (1 ) s ds ln (1 ) s (2) Setting the right side equal to zero, and solving for s gives smax 1 2 =± 2 ln (1 ) (3) Hence, choosing the plus sign (why?), we get for the location of the maximum in the velocity curve 16 rmax 12 =R 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 1 2 < (1 + ) 2 ln (1 ) 2 or 2 (1 ) (1 + ) ln (1 ) <1 (5a,b) It is easy to calculate the left side of Eq. 5b as a function of : 0.1 0.2 0.3 0.4 0.5 2 (1 ) (1 + ) ln (1 ) 0.711 0.829 0.895 0.935 0.962 0.6 0.7 0.8 0.9 2 (1 ) (1 + ) ln (1 ) 0.980 0.991 0.999+ 0.999+ It is evident that for > 0.8, the maximum is very close to being half way between the two cylinders, and that is to be expected, inasmuch as the annular-slit flow approaches a flat-slit flow. Problem 2B.5 gives a discussion of the interrelation of the flow in a plane slit and the flow in a narrow annulus. c. The maximum velocity is then vz,max 2 P 0 P L ) R 2 rmax ( 1 = R 1 2 ln ln (1 ) rmax R P 0 P L ) R2 2 ln (1 ) ( 1 2 1 2 1 = ln 2 4 μL 2 ln 1 ln 1 1 ( ) ( ) 4 μL (6) This result may also be written in terms of , as in Eq. 2.4-15. 17 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 HagenPoiseuille 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. 2.414.) 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 terms inside the bracket become: ( ) lim (1 ) lim 1 4 = 1 (1) 0 2 0 2 =1 l im (1 ) = (2) (3) 0 Thus the bracket quantity becomes: ) ( ) 2 1 2 lim 1 4 1 =1 0 1 ln (1 ) ( (4) and the Hagen-Poiseuille formula is recovered. 18 Note to p. 58 Let us verify that the average velocities in the two regions are given by Eqs. 2.5-20 and 21. In region I, the average velocity is given by v (p = 0 I z (p = 2 pL ) b2 1 0 2 μ I μ I μ II x x + dx b b μ I + μ II μ I + μ II b b 2μ I L pL ) b 2 2 μ I μ I μ II 2 + d 1 μ I + μ II μ I + μ II 2μ I L p0 pL ) b 2 2 μ I 1 μ I μ II 1 ( I = I II II 2 3 2μ I L μ + μ μ + μ I I II I II p0 pL ) b 2 6 2 μ 3 μ μ 2 μ + μ ( = 2μ I L μ I + μ II p0 pL ) b 2 7 μ I + μ II ( = I II 2μ I L μ +μ 0 0 ( ) ( where = x b ) (1) Similarly for Region II we have 2 pL ) b 2 1 b 2 μ II μ I μ II x x v + dx 2 μ II L b 0 μ I + μ II μ I + μ II b b p0 pL ) b 2 1 2 μ II μ I μ II ( 2 + = d 2 μ II L 0 μ I + μ II μ I + μ II 2 p0 pL ) b 2 μ II 1 μ I μ II 1 ( + = 2 μ II L μ I + μ II 2 μ I + μ II 3 II I II I II p0 pL ) b 2 6 2 μ + 3 μ μ 2 μ + μ ( = 2 μ II L μ I + μ II p0 pL ) b 2 μ I + 7 μ II ( = 2 μ II L μ I + μ II II z (p = 0 ( ) ( ) (2) 19 Note to p. 59 The verification of some equations requires a lot of algebraic detail that falls in the category of "straightforward but tedious." Nonetheless, such derivations should be done. Here are two examples: 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 2 4 vr 3 R 1 3 R rr = 2 μ = 2 μ v + cos R 2 r r 2 r 2 4 3 μ v R R = + cos r R r (1) And from Eq. B.1-11 and Eqs. 2.6-1 and 2, we find r v 1 vr = μ r + r r r 2 4 1 R 3 R R = μ v sin R r 2 r r 2 4 1 R 1 R 3 R μ v ( sin ) R r 2 r 2 r 4 4 3 μ v R 3 μ v R =+ sin = + sin 2 R r 2 R r (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 is a special case of the general result given in Example 3.1-1.) The pressure p on the surface of the sphere is given in Eq. 2.6-8. Therefore the z-component of the force acting normal to the surface of the sphere is: 20 3 μ v cos (cos ) R 2 sin d d R 3 μ v = 2 0 p0 + gRcos + cos ( cos ) R 2 sin d 2 R n F( ) = 2 0 0 p0 + gRcos + 2 3 μ v = 2 p0 R 2 0 cos sin d + 2 R 2 gR + cos2 sin d 0 2 R (3) The first integral is zero, and the second is 2/3, so that 3 μ v 2 4 n = R3 g + 2μ Rv F ( ) = 2 R 2 gR + 2 R 3 3 (4) To get the z-component of the tangential force acting on the sphere, we substitute Eq. 2.6-11 into Eq. 2.6-10 and integrate: 3 μ v F (t) = 2 0 sin ( sin ) R 2 sin d 2 R 4 = 3μ Rv 0 sin 3 d = 3μ Rv = 4μ Rv 3 (5) This is the "friction drag" result displayed in Eq. 2.6-12. 21 Note to p. 78 The general result for normal stresses at solid surfaces given in Eq. 3.1-6 is very important. We have already seen in Eq. 2.6-5 that the normal stresses for creeping flow around a sphere are exactly zero at the 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: 2 4 vr 1 3 R 1 R rr = 2 μ = 2μ v + 3 cos 2 r R 2 r r 2 4 3 μ v R R = + cos r R r (1) b. Then we get from Eq. B.1-16: 2 4 1 v vr 1 R 1 R 3 R = 2 μ + = 2μ v + + cos r 4 r R r 4 r r 2 4 1 R 1 R 3 R 2 μ v + cos 2 r R r 2 r 2 4 2 μ v 3 R 3 R = + cos 4 r R 4 r 2 4 3 μ v R R (2) = + cos r 2R r c. And, finally, we get from Eq. B.1-17: 22 1 v vr + v cot = 2 μ + r r sin 3 3 3 R 1 R 3 R 1 R 1 R R + + = 2μ v 1 cos + 1 + cos r 2 r 2 r 4 r 4 r R r 3 μ v = 2R R2 R4 + cos r r (3) When r = R, all of the normal stresses are zero, in agreement with Example 3.1-1. 23 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 "engineering Bernoulli equation" is also used. The derivation we work through here is an excellent exercise in using some of the "del" relations given in Appendix A (see particularly §A.4). We start by forming the dot product of the local velocity v with Eq. 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 in Example 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 v + ( v v ) + v [ v v ] + ( v v ) ( v ) t t ---------------------------- ( ) (4) In the first term, we differentiate the product with respect to t, and in the second term, we use Eq. A.4-24. In the second line, the dashed underlined terms then sum to zero by using the equation of 24 continuity; furthermore, the first term is split up into two terms, and the third term is rearranged, thus: = 1 1 ( v v ) ( v v ) + v [ v v ] 2 t t 2 ( ) (5) We again use the equation of continuity to rewrite the second term in Eq. 5: = 1 1 v v ( ) + 2 ( v v ) ( v ) + v [ v v ] t 2 ( ) (6) Now the second and third term may be combined by using Eq. A.419 with s replaced by 12 ( v v ) and v replaced by v : = 1 1 ( v v ) + ( v v ) v 2 t 2 = 1 2 1 2 v + v v 2 t 2 (7) Clearly knowing that the final result is Eq. 3.3-1 is very helpful in doing the last several steps. 25 Note to p. 82 (i) We want to verify that ( :v ) , when written for a Newtonian fluid, may be written as a sum of squares as shown in Eq. 3.3-3, and is hence positive. First define a tensor = v + ( v ) , and then ( :v ) may be written for Newtonian fluids (for which is symmetric) as: † ( :v ) = 12 ( : ) = 12 μ ( 23 ( v ) : ) + 12 ( v )(: ) (1) where Eq. 1.2-7 has been used. Next, since ( : ) = 2 ( v ) , we get ( :v ) = 12 μ ( : ) 43 ( v )2 + ( v )2 (2) This is equivalent to: ( :v ) = 12 μ ( 23 ( v ) : 23 ( v ) ) + ( v )2 (3) which is another way of writing Eq. 3.3-3. The last step may be seen to be true by expanding the double-dot product in Eq. 3 thus: ( ( v ) : 23 ( v ) ) 2 2 2 = ( : ) 43 ( v ) 43 ( v ) + 49 3 ( v ) 2 = ( : ) 43 ( v ) 2 3 which is the coefficient of 1 2 (4) μ in Eq. 2. 26 Note to p. 82 (ii) Eq. 3.4.1 is obtained by taking the cross product of the position vector with the equation of motion in Eq. 3.2-9. To do this, we form the cross product, term by term: a. The time-derivative term is straightforward; for the ith component: r t v = t [ r v ]i 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 i component and expand the expression in terms of its components: r [ vv ] = ijk x j vl v k i j k l xl = ijk x j vl vk ijk vl vk jl xl j k l j k l = j k l v x v v v xl l ijk j k j k ijk j k (2) In the second line, we have moved the x j inside the differentiation and subtracted off a compensating term. In the third line, we have rearranged the first term and performed the sum on l in the second term. It can be seen that the second term is zero (inasmuch as it involves a double sum on a pair of indices that appear symmetrically in one factor ( v j vk ) and antisymmetrically in another ( ijk ); see Exercise 5 on p. 815). Now we convert Eq. 2 back to bold-face notation: r [ vv ] = v [ r v ] i i (3) c. Next we examine the term containing the pressure: 27 [r p]i = ijk x j x j k p = ijk j k k x p ijk jk p xk j j k (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 in another. The other term can be rearranged as follows: x j k l l { } ijk x j p kl = l xl ijk x j p kl j k (5) On the right side, the quantity within the braces can be recognized as the cross product of a vector with a tensor (see text just after Eq. A.319). Therefore the above result in Eq. 5 can be written as: x {r p }il = x {r p }li = {r p } l l l † l † 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 the cross product as indicated above. d. The term containing the tensor can be treated in somewhat the same manner as in part (c) above: r [ ] = ijk x j [ ]k = ijk x j i x lk j k j k l (7) l Next we write this intermediate result as the sum of two terms: ijk x j k l x j lk ijk lk j l k l x xl j † x ijk j kl ijk lk jl l j k j k l = r † ijk jk = r † il l xl j k l xl = xl { } { } † li + ikj jk j k 28 { = 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, straightforward. [r g ]i , is When all the terms are collected, Eq. 3.4-1 results. 29 Note to p. 86(i) In Example 3.5-1 it was pointed out that, to derive the Bernoulli equation, 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 most efficiently done by making use of summations. First we write the dot product in terms of the components: ( v v [ v ]) = v v [ v ] i i (2) i Next we write the cross product operations using the ijk symbol: = vi ijk v j [ v ]k = v i ijk v j klm i j k i j k l m v xl m (3) Then we rearrange the expression and make use of the cyclic property of the permutation symbol, i.e., klm = mkl = lmk : = ijk klm vi v j i j k l m vm = ijk lmk vi v j v xl xl m i j k l m (4) We can now use Eq. A.2-7 to replace the sum on k of products of two permutation symbols: ( ) = il jm im jl vi v j i j l m v xl m (5) After doing the sums on l and m we get two terms: 30 = vi v j i j v j vi v j v xi x j i i j (6) In the second summation, we replace i by j and j by i to get: = vi v j i j v j v j vi v =0 xi xi j i j (7) It may now be seen that both terms are the same. Consequently their difference is zero. Therefore, we have proven the identity in Eq. 1. 31 Note to p. 86(ii) Did you notice some similarity between Eq. 3.5-11 and Eq. 3.32? In this "Note," we demonstrate the connection. If we assume steady state, inviscid flow (as we did in Example 3.5-1), then Eq. 3.3-2 can be rewritten, with the help of Eq. 3.5-4 and Eq. A.8-19, as D Dt ( 1 2 ) ˆ = ( v p ) v2 + (1) where the inviscid flow assumption has been used to get rid of the terms containing . Next we use the steady-state flow assumption to rewrite Eq. (1) as ( ) ( ) ˆ = ( v p) v 12 v 2 + v (2) ˆ = gh , where h is Then we write the potential energy term using the coordinate in the direction opposite to the direction of gravity. This gives ( v v ) + ( v gh) + 1 ( v p) = 0 2 1 2 (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 the coordinate along a streamline. Equation (3) may now be written as d ds 1 dp =0 ( v ) g dh ds ds 1 2 2 (4) which is the same as Eq. 3.5-11. Equation (4) may then be integrated from point "1" to point "2" along the streamline to get the Bernoulli equation. Thus it is seen that the Bernoulli equation can be obtained directly from the mechanical energy balance. 32 Note to p. 90(i) In Example 3.6-3, we give Eq. 3.6-20, but we don't obtain the pressure distribution. If one knows the pressure at one point in the system (after a steady state velocity distribution has been obtained), then it is possible to determine the constant of integration and obtain the pressure distribution. Here we attack a different problem. We now show how Eq. 3.620 can be used to get the pressure distribution in the space between the cylinders in terms of the pressure p0 that exists in the system before the outer cylinder is rotated. We start by temporarily relaxing the assumption of incompressibility. It can be shown that the equation of motion in Eq. 3.2-9 and Newton's law of viscosity as given in Eq. 1.2-7 are still valid for a compressible fluid, provided that one assumes that the viscosity is independent of the pressure. The important new feature that we have to inject into the development is the equation of state—that is, the density as a function of the pressure. We do this via a Taylor series for the density as a function of the Gibbs free energy, G = H TS = 0 + (G G0 ) + T 0 (1) in which the subscript zero refers to properties of the system when the fluid is at rest before the outer cylinder is rotated. Next we truncate the Taylor series after two terms and write p = 0 1 + b0 (G G0 ) = 0 1 + b0 p 1 dp 0 (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 the density is a known function of the pressure 2 v2 2 R 2 2o 1 1 d p0 dp = r dr = 2 2 1 p ( ) 33 2 R 2 2o 2 2 = + 3 d 2 2 2 1 ( ) 2 R 2 2o 1 2 2 2 ln = + C 2 2 2 1 2 2 ( (3) ) The integration constant C is determined by the mass conservation statement 1 1 0 d = d or 1 p p 0 1 dp d = 0 (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 11 2 2 2 ln + C d = 0 2 2 2 (6) Performing the integration gives 1 1 4 1 2 1 1 1 ln 2 2 ln 2 + C 2 = 0 2 2 4 2 2 4 2 (7) This gives for the integration constant C = 1 1+2 4 2 3 2 ln 1 2 (8) Hence Eq. (3) becomes 1 2 2 2 ln 2 R 2 2o 2 2 2 p 1 p0 dp = 2 1 + 2 3 2 ln 1 2 1 4 2 1 2 ( ) (9) 34 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 twocylinder system with the outer cylinder rotating 1 2 2 2 ln p p0 2 R 2 2o 2 2 2 = 2 0 1 + 2 3 2 ln 1 2 1 4 2 1 2 ( ) (10) A similar development for the system with the inner cylinder rotating with angular velocity i and the outer cylinder fixed, gives 1 2 1 2 ln p p0 4 R 2 2i 2 2 = 2 2 2 2 0 1 1 1 + 1 + 2 ln 4 1 2 ( ) (11) 35 Note to p. 90(ii) An alternative method of solving the problem in Example 3.6-3 is 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 at radius r is equal to 2 rL r , whereas the torque at radius r + r 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 after dividing by r and letting r go to zero: ( ) d 2 r r = 0 dr or d 2 d v r r =0 dr dr r (1a,b) Eq. B.1-11 has also been used. Here we show that this is equivalent to Eq. 3.6-21, which is d 1 d rv =0 ( ) dr r dr or d 2 v dr 2 + 1 dv v =0 r dr r 2 (2) We start by performing the differentiations in the large parentheses in Eq. 1b: d v 1 dv v dv r 2 r = r 2 r 2 = r 2 rv dr dr r r dr r (3) Next do the differentiation with respect to r (in Eq. 1b) and set the result equal to zero: 2 dv dv d 2 dv 2 d v = 2r r rv + r r v dr dr dr dr dr 2 2 1 dv v 2 d v =r 2 + 2 =0 dr r dr r (4) 36 Therefore either r 2 is zero, or the quantity in parentheses is zero. But r 2 cannot be zero, and hence Eq. 2—which came from Example 3.63—must be the same as Eq. 1b (from the torque balance). 37 Note to p. 117 When you look at Fig. 4.1-2, you might wonder whether the slope at y = 0 is 1 . You can answer that question by differentiating Eq. 4.1-15 with respect to : d d 2 erfc = 1 d d =0 = 2 e 2 0 e 2 =0 = d =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 than minus 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 the Leibniz theorem, it is vital to make this distinction. 38 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 these two locations. (a) Let the real and imaginary parts of the arguments be: z1r and z1i for z1 : z2r and z2i for z2 : wr and wi for w: Then the expression { z1 w} = {z2 w} becomes { } { } ( z1r + iz1i ) ( wr + iw i ) = ( z2r + iz2i ) ( wr + iwi ) (1) Then, taking the real parts of both sides, we get z1r wr z1i wi = z2r wr z2i wi (2) Rearranging gives: wr ( z1r z2r ) = wi ( z1i z2i ) (3) Since w, and hence wr and wi , are arbitrary, the only way that this equation can be satisfied is if z1r = z2r and z1i = z2i (i.e., z1 = z2 ). (b) The square root of i will be some number in the complex plane, say a + bi , so that i = a + bi (4) where the real quantities a and b must be determined. When the left and right sides of Eq. 4 are squared, we get ( ) i = a 2 b 2 + 2abi (5) 39 To find a and b, we equate the real and imaginary parts of the left and right sides: a2 b2 = 0 and 2ab = 1 (6) Eliminating b between these two equations gives 1 4 a4 = (7) Hence a2 = 1 1 , 2 2 and a=± 1 2 ,± 1 2 i (8) whence b=± 1 2 1 1 2 i ,± (9) and the two square roots of i are i = a + bi = ± 1 2 (1 + i) (10) Alternatively, one can write i in polar form and use the fact that i has a unit length: i = re i = 1e i 2 (11) Then the square root of i is i = ±e i 4 1 1 = ± cos + i sin = ± +i 4 4 2 2 (12) 40 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 ) = x vx ( x, y ) + y v y ( x, y ) (1) The velocity components are related to the stream function according to the relations in the 3rd column. The first term in Eq. (A) comes from the first term in Eq. 4.2-1, which is v y vx v = v = [ ] z t [ ]z z t x y t 2 2 = z 2 + 2 = z 2 t x t y (2) The right side of Eq. (A) comes from the right side of Eq. 4.2-1: 2 2 [ v ] = 2 + 2 [ v ] y x 2 2 2 2 2 = z 2 + 2 2 + 2 = z 4 y x y x (3) The second term on the left side of Eq. (A) can be taken care of similarly: v [ v ] = v z 2 = x vx + y v y z 2 y vx 2 + x v y 2 = ( ) 41 = x y z x y z x y z x y z 0 v x 2 v y 2 0 0 0 ( ) vx 2 + z v y 2 x y v y 2 v + z vx 2 + v y 2 = z x + y y x x = + z ( ) (4) The first term is zero, because of the assumption of incompressibility (see statement just above Eq. 4.2-1). Then we get finally 2 2 = z + x y y x = x y 2 x 2 y = ( , 2 ( x, y ) ) (5) When the results in Eqs. 2, 3, and 5 are combined, we get same expression that is given in Eq. (A) of the table. 42 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 is convenient to introduce the dimensionless variable = r R . 2 ( 2 vr 9 v 1 3 cos 2 = 2 r 2r 2 ) 2 (1) ( 2v2 1 v vr 2 + = 2 1 + 34 1 + 14 3 cos + 1 32 1 + 12 3 cos r r r 2 ( 9 v 1 + 3 cos = 2 8r 2 ) 2 ) 2 (2) 2v2 v cot vr 2 + = 2 1 + 34 1 + 14 3 cos + 1 32 1 + 12 3 cos r r r ( 2 ( 9 v 1 + 3 cos = 2 8r 2 v 1 vr v v 1 vr r = + r r + r r r r ) 2 ) (3) 2 3 1 + 3 3 sin + 1 3 1 1 3 sin 4 4 4 4 = r 1 3 1 + 1 3 sin 2 2 2 v2 2 = v2 3 3 sin 2 2 r 2 (4) When the above results are combined, we get ( :v ) r 2 = μ v2 274 2 27 4 2 + 27 6 cos2 4 + μ v2 94 6 sin 2 (5) 43 2 Then the kinetic contribution to the force on the sphere is given by Fk v = 2 μ 0 R ( :v ) r = 2 23 μ v2 R 274 2 2 drsin d 27 4 2 + 27 6 dr 4 + 2 43 μ v2 R 94 6 dr (6) since 0 cos 2 sin d = 2 3 and 0 sin 3 d = 43 . Then, finally Fk = 9 μ v R 1 2 2 4 + 6 d + 6 μ v R 1 6 d 1 1 = 9 μ v R 1 + 32 3 15 5 + 6 μ v R 15 5 = 9 μ v R 1 + 2 3 1 + 5 6 μ v R 15 = 9μ v R 52 + 6μ v R 15 = = 6μ v R ( 24 5 + 6 5 ) μ v R (7) which is Stokes' law. 44 Note to p. 139 We show here how to get the Falkner-Skan equation in Eq. 4.4-35 for the flow near a corner. For this system, the external flow ve was found earlier (see Eqs. 4.3-42 and 43) to be ve ( x ) = 2c ( 2 ) 2 x c' x ( ) 2 (1) We then have to solve Eq. 4.4-11 to get the velocity distribution in the neighborhood of the wedge shown in Eq. 4.3-4: vx vx dve 2 vx vx + vy = ve + x y dx y 2 (2) This equation can be rewritten in terms of the stream function ( x, y ) , by using the expressions for the velocity components in the first row of Table 4.2-1: 2 2 dve 2 y xy + x y 2 = ve dx + y 2 y (3) Insertion of Eq. 1 into this equation then gives Eq. 4.4-32: 2 2 c' 2 3 1 = y xy x y 2 2 x( 23 ) ( 2 ) y 3 (4) Next we want to rewrite this equation in terms of f and : 1 2 1 2 ( x, y ) = c' ( 2 )x ( ) f ( ) Ax ( ) f ( ) ( x, y ) = y y c' B 1 2 1 2 ( 2 ) x( ) ( ) x( ) ( ) (5) (6) 45 We start by converting the various derivatives from ( x, y ) to derivatives of f ( ) : 1 1 2 df 1 2 2 = Ax ( ) = ABx ( ) f ' (1 ) ( 2 ) = ABx ( ) f ' (7) d y y x ( ) 1 2 x = ABx f " = AB (1 ) ( 2 ) f " = AB2 (1 2 ) ( 2 ) f " (8) 2 y y x x 3 1 1 (9) = AB 2 (1 2 ) ( 2 ) f = AB3 ( 23 ) ( 2 ) f 3 y y x x d 1 2 1 2 df = A x ( ) f Ax ( ) dx x d x 2 ( 2 ) 2 1 2 x ( )1 1 2 By (1 ) (1 ) ( 2 )1 = A f Ax ( ) x f' 2 (2 ) x ( ) 1 2 1 ) x1 ( 2 ) ( 1 x ( ) = A f +A f' (2 ) x (2 ) x (10) 1 2 1 ) x1 ( 2 ) d ( 2 1 x ( ) = A f '+ A ( f ') yx ( 2 ) x y ( 2 ) x y d 1 2 1 ) x1 ( 2 ) ( x ( ) 1 = AB f '+ AB ( f "+ f ') (11) ( 2 ) x(1 ) ( 2 ) x ( 2 ) x(1 ) (2 ) x The terms on the right side of Eq. 4.4-32 are then: c'2 c'2 1 1 1 3 2 ( 23 ) ( 2 ) + AB ( 23 ) ( 2 ) f = 2 ( 23 ) ( 2 ) ( + f ) x x x (12) Next we write down the terms on the left side of Eq. 4.4-32: 2 2 1 x ( )x ( ) 2 2 2 =A B f y xy x y 2 ( 2 ) x(1 ) ( 2 ) x 2 1 2 46 A B 2 2 (1 ) x (2 ) x1 ( 2 ) f f "+ f 2 ) ( 2 ) x(1 ) ( 2 ) x ( 1 x ( ) A B ff ( 2 ) x(12 ) (2 ) x 1 2 2 2 +A B 2 2 (1 ) x1 ( 2 ) f f " ( 2 ) x(1 2 ) ( 2 ) x 1 23 ) ( 2 ) 23 ) ( 2 ) x ( f 2 c' 2 x ( ff 2 (2 ) 1 23 ) ( 2 ) = c' 2 x ( f 2 ff (2 ) = c'2 ( ) (13) Combining the results in Eqs. 12 and 13 gives c' 2 ( ) 1 1 23 ) ( 2 ) 23 ) ( 2 ) x ( f 2 ff = c' 2 x ( ( + f ) (2 ) (2 ) (14) or f 2 ff = + 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 from c', and the quantity c' was not defined.] 47 Note to p. 154 To get the average flow velocity from Eq. 5.1-4, we integrate the velocity distribution over the circular tube cross-section: 2 vz = R 0 v z rdrd 0 2 0 R 0 = vz ,max rdrd 2 R2 R 0 2 = vz ,max 0 R 0 1 ( r R ) 2 R 0 0 1 ( r R ) 17 17 rdrd rdrd rdr = 2vz ,max [1 ] d 1 17 0 (1) where = r R . To evaluate the integral, we make a change of variable 1 = . Then v z = 2vz ,max 1 7 (1 )d 1 (2) 0 This can then be written as the sum of two integrals, which can be evaluated: v z = 2vz ,max ( ) 1 8 7 15 7 17 87 0 d 0 d = 2vz ,max 8 7 15 7 = 0.82vz ,max 1 1 0 (3) Since 0.82 is approximately 4/5, relation in Eq. 5.1-5 is verified. 48 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: FF = ( F + F ) 2F (1) where the primes indicate differentiation with respect to . Next we note that Eq. 1 may be put into the form 1 2 ( F ) = (F ) 2F 2 (2) Each term may now be integrated with respect to to give 1 2 F 2 = F 2F C2 (3) which is the same as Eq. 5.6-19. The constant C2 is zero according to Eq. 5.6-16, with a, b, and d set equal to zero. [Note: The comment about setting = ln does not seem to be helpful.] Eq. 3 is a separable, first-order differential equation dF = 2F + 12 F 2 or d dF d = 2F 1 + 14 F ( (4) ) Then, according to a table of integrals (e.g., Formula 101.1 of Dwight's Table of Integrals), Eq. 4 gives, on integration ln 1 2 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 to obtain 49 F 2 = C ( ) 3 1 + 14 F (6) Now either a "plus" sign or a "minus" sign may be inserted inside the absolute value bars. Since we have no reason to prefer one over the other, we consider the two cases separately: Case I (plus sign): ( ) F = + 1 + 14 F ( C3 ) 2 or 2 C3 ) ( F= 2 1 14 ( C3 ) (7) or 2 C3 ) ( F= 2 1 + 14 ( C3 ) (8) Case II (negative sign): ( ) F = 1 + 14 F ( C3 ) 2 When F in Eq. 7 is plotted against ( C3 ) , it tends toward 2 infinity as ( C3 ) = 4 is approached from below, and approaches minus infinity when approached from above. Hence Case I is physically unreasonable behavior for the stream function. 2 When F in Eq. 8 is plotted versus ( C3 ) , is it monotone 2 decreasing over the entire range of ( C3 ) . 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: 2 J = 2 0 v z2 rdr = 2 ( ) We now let t 2 1 4 0 ( 2C ) 2 3 1 + 1 4 2 (C3 ) 2 4 d (9) (C3 )2 = x 2 so that d = ( 4 C32 ) xdx ; then 50 J = 32 ( ) C32 0 xdx t 2 ( 1 + x2 ) 4 1 t)2 2 ( = 32 C3 2 6 1 + x ( ) = 16 ( t )2 C 2 3 3 3 0 (10) whence Eq. 5.6-23 follows at once. Similarly the mass rate of flow is w = 2 0 ( ) vz rdr = 2 z t = 4 ( ) zC32 t 4 3 = 8 ( ) z t C2 0 ( 2C ) 2 3 0 1 + 1 4 xdx (1 + x ) 2 2 (C3 )2 2 d z 2 1 t) ( = 16 z 2 1 + x2 ( ) 0 (11) 51 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 in Fig. 7.0-1 on p. 197: V (t) t dV = V (t) ( v ) dV (1) We now apply the 3-dimensional Leibniz formula (Eq. A.5-5) to the left side and the Gauss divergence theorem (Eq. A.5-2) to the right side: d dV S(t) ( n vS ) dS = S(t) (n v ) dS dt V (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 parts in the system. Eq. 2 may be rewritten as ( ) d dV = S(t) n ( v vS ) dS dt V (t ) (3) We note that the mathematical surface S (t ) defining the system consists of several parts that we identify as follows: •the "inlet" surface S1 (on which v S = 0) •the "outlet" surface S2 (on which v S = 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 v S the surface velocity. The surface integrals are then split into four parts corresponding to the four types of surfaces. 52 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 = v S . Therefore we are left with d m = S ( n v ) dS S ( n v ) dS 1 2 dt tot (4) We now introduce the vectors u1 and u 2 , which are unit vectors in the direction of flow at planes "1" and "2", respectively. Thus the n in the 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 constant over 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 d m = + 1 S vdS 1 S vdS 1 2 dt tot (5) Here v is the velocity in the direction of flow, which varies across the cross section. Therefore integrations over the surfaces S1 and S2 give d m = 1 v1 S1 2 v 2 S2 = w1 w2 dt tot (6) is the average value over the cross section; w1 and w2 are where the mass rates of flow at the inlet and outlet, respectively. Eq. 6 is just the same as Eq. 7.1-1 in the textbook, which was written down by using elementary arguments (i.e., common sense). Macroscopic Momentum Balance When the equation of motion of Eq. 3.3-9 is integrated over the volume of the flow system in Fig. 7.0-1, we get V (t) t v dV = V (t) [ vv ]dV V (t) pdV V (t) [ ]dV + V (t) gdV (7) 53 Next apply the Leibniz formula to the left side and the Gauss divergence theorem (or Eq. A.5-2) to the surface integrals on the right side to get d vdV S(t) v ( n v S ) dS = S(t) [ n vv ]dS S(t) npdS S(t) [ n ]dV dt V (t ) + V t gdV (8) () The integral on the left side is just the total momentum Ptot in the flow 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 d P = S t n ( v v S ) v dS S t npdS S t [n ]dS + mtot g () () () dt tot (9) We now consider the three surface integrals seriatim: The first integral is zero on fixed and moving surfaces, and v S is zero at the entry and exit planes, so that S(t ) n ( v v S ) v dS = + S u1 u1u1 v 2 dS S u 2 u 2 u 2 v 2 dS 1 2 = + 1 v12 S1u1 2 v22 S2 u 2 (10) Here it has been assumed that the flow at the inlet and outlet planes is parallel to the container wall. The second integral will contribute both at the inlet and outlet planes and also to the force on the various solid surfaces: S(t ) npdS = + S u1 pdS S u 2 pdS S 1 2 f +Sm npdS ) = p1S1u1 p2S2 u 2 F(f s p (11) the last contribution being the force exerted by the fluid on the solid surfaces by the pressure. Finally the third integral will be S(t ) [ n ] dS = + S u1 dS S u 2 pdS S +S [ n ] dS 1 2 f m 54 ) F(f s (12) Note that we have omitted the contributions at the entry and exit planes because they would normally be quite small compared to the pressure terms. Therefore we are left with just the force exerted by the fluid on the solid surfaces because of the viscous forces. When all the forces are combined we get for the macroscopic p) ) momentum balance (with F (f s + F (f s = F f s = Fs f ): d Ptot = 1 v12 S1 u1 2 v22 S2 u 2 + p1S1u1 p2S2 u 2 + Fs f + mtot g (13) dt This is the same as Eq. 7.2-11 (or 7.2-12) in the textbook, obtained by elementary reasoning. [Note: The derivation of the macroscopic mechanical energy balance is given in §7.8, the derivation of the macroscopic energy balance is given in the Note to p. 454.] 55 Note to p. 199 Here we work through all the details of Example 7.1-1, (a) explaining how to get Eq. 7.1-4, (b) giving the details of how the difference in the modified pressures is obtained, and (c) working through 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 the liquid level. We imagine that the sphere is generated by a circle in the xz-plane, with its center at z = R and x = 0. The tank is draining in the negative z-direction, and the exit from the sphere into the attached tube is located at z = 0. The sphere is created by rotating the generating circle around the z-axis. The generating circle has the equation: x2 + ( z R ) = R2 2 or x 2 = 2Rz z 2 (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 = x 2 dz = 2Rz z 2 dz (2) Then the total volume of the liquid is: V= h ( ) ( 2 2 1 3 0 2Rz z dz = Rz 3 z ) h 0 ( = Rh2 13 h3 ) (3) Thus the liquid volume is: 1 h V = Rh 2 1 3 R (4) We may check this result at three points where we know the result: h = 2R V = 43 R 3 Tank full: Tank half full: h=R V = 23 R3 56 Tank empty: h=0 V =0 b. Next we want to apply Eq. 7.1-2 to the system. No liquid is entering 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 be given by the Hagen-Poiseuille formula (Eq. 2.3-21). Hence (P 2 P 3 ) D4 d 1 h 2 Rh 1 = dt 3 R 128 μ L (5) To get the modified pressures we must specify a datum plane; we choose 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 d 1 h ( gh + gL ) D = Rh 2 1 128 μ L dt 3 R 4 (8) c. Eq. 8 may be rewritten as R d 2 1 h 3 gD4 h A = h + L dt 3 R 128 μ L (9) (which defines the quantity A) or 57 R h 2 dh 2h =A R dt h + L (10) Now we introduce a new variable H = h + L in order to facilitate the solution of the differential equation 2R ( H L ) ( H L ) dH =A dt H 2 (11) or 1 dH H 2 R + L + 2R + L L =A ( ) ( ) H dt (12) Now we integrate this equation and make use of the initial and final conditions: L tefflux 1 2 H 2 R + L H + 2R + L L ln H = At ( ) ( ) 2 0 2R+L (13) This gives 1 2 1 L 2 L ( 2R + L ) 2 ( R + L ) L + 2 ( R + L ) ( 2R + L ) + ( 2R + L ) L ln 2 2 2R + L = Atefflux (14) or after some cancellations tefflux = 1 2 2R + L 2R + 2RL 2R + L L ln ( ) A L (15) When L2 is factored out of the bracket expression, Eq. 7.1-8 of the textbook is obtained. 58 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 and splits into four streams. It leaves the sprinkler at right angles to the sprinkler arms. It is desired to find the angular velocity of the sprinkler, when there is a frictional torque of Tf . The flow velocity is each arm is v = w 4 S , so that the velocity at the outlet stream, relative to the sprinkler arm is (at plane "2") w v2 = y L 4 S Therefore, Eq. 7.3-2 gives for the angular momentum balance at steady 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 z direction) w 0 w L x y L Tf z 4 S the entering stream contributing nothing to the angular momentum balance. The unit vectors may now be removed w Tf = wL L 4 S Then, solving for we get = Tf w 4 SL wL2 for the angular velocity of the sprinkler. [For more on this problem, see Surely You're Joking Mr. Feynman," by R. P. Feynman, Bantam Books, New York (1986), pp. 51-53.] 59 Note to p. 218 Here we work through the details from Eq. 7.7-8 to the end of the example. The first term of Eq. 7.7-5 may be transformed as follows: 2h d2h dt 2 = 2h d dt ( u ) = 2h 1 2 u 1 2 du du 1 2 du dh = h = hu dh dt dh dt (1) from which Eq. 7.7-8 follows at once: h du ( N 1) u + 2gh = 0 dh (2) To verify that Eq. 7.7-9 is the solution to Eq. 7.7-8, we substitute the former into the latter to get 2g 2gh ( N 1) Ch N 1 + + 2gh = 0 h C ( N 1) h N 2 + N 2 N 2 (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: C-terms: 2gh 2gh N2 ( N 1) + 2gh = 0 N2 N2 N2 C ( N 1) h N 1 ( N 1) ChN 1 = 0 (4) (5) We next apply the initial conditions (Eqs. 7.7-6 and 7) to the solution in Eq. 7.7-9 2 2gh dh N 1 + dt = Ch N2 (6) to get an equation for the constant of integration 60 4 2gH R 2gH 0 = CH N 1 + N2 R (7) When this is solved for C, we get (noting that N = ( R R0 ) as defined immediately after Eq. 7.7-3): C= 4g 2g 1 1 1 = H N 2 N ( N 2) H N2 N N 2 (8) Even though the factor containing ( N 2 ) would cause C to become infinite for N = 2 , this need cause no alarm, since N is going to be a large number, i.e., when the radius outlet hole is small compared to the radius of the tank. When we take the square root of Eq. 7.7-9 and introduce the dimensionless liquid height = h H , we then get Eq. 7.7-10: d =± dt 2g 2 N 1 ( N 2) H N (9) We then choose the minus sign, because we know that the height of the fluid will be decreasing with increasing time, and therefore d dt must be negative. To get the efflux time, we integrate Eq. 9 from t = 0 when = 1 (full tank) to t = tefflux when = 0 (empty tank): t tefflux = 0efflux dt = ( N 2) H 2g 1 0 1 ( 2 N ) N 1 d NH (N ) 2g (10) Keep in mind that tefflux = NH 2g is the quasi-steady-state solution in Eq. 7.7-3—the solution that we would expect to be valid when N is extremely large, i.e., for the case that the outlet hole is so small that the system is never far from steady state. The function ( N ) is then given by 61 (N ) = 1 1 N2 1 d N 0 ( 2 N ) N 1 2 (11) This integral can probably not be performed analytically. However, in the expression under the square-root sign, for large N, the first term will predominate, over the range of integration (from = 0 to = 1 ). Therefore we take the first term outside of the radical and write 1 N2 1 1 2 N2 (N ) = 1 2 N N 0 1 2 d (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 2 1 N2 1 1 1 1 2 N 2 1 1 3 2 N 2 (N ) = + d 1 + 2! 2 2 N 2 N 0 1! 2 N 1 N2 1 3 (13) = 2 + + + N 2 N N 32 4N 2 N 74 ( ) ( ) When this expression is expanded in a Taylor series around 1 N = 0 , we get 1 1 1 1 1 1 1 3 1 + ( N ) = 1 + O + O 3 + 8 O 3 + 2 3 2 2 N 2 N N N N N 1 1 (14) = 1 +O 3 N N in which O ( ) means "term of the order of ( )." Thus we have arrived at Eq. 7.7-13 at the end of the example. 62 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 this kind of device, one may have to extract the function from tube-flow data. Show how to get the relation for the non-Newtonian viscosity w vs pressure ( ) from experimental data on mass rate of flow drop P 0 P L for flow through circular tubes. We know from §2..3 that, for any kind of fluid rz = R r R , where R = (P 0 P L ) R 2L is the shear stress at the tube wall The mass rate of flow through the tube is dvz r 2 R = + r 2 dr dr dr 2 0 R R w = 2 0 vz rdr = 2 0 r = R. (1) the second form being obtained by an integration by parts. In the third form we have made the replacement = dvz dr . Next we change the variable of integration from r to rz w R 3 = 1 R3 R 0 rz2 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 different lengths and radii should collapse onto a single curve when plotted as w R3 vs R . If now we multiply Eq. 2 by R3 and then differentiate with respect to R , we get d d R 3 w 2 R R 3 = R R (3) where the Leibniz formula for differentiating an integral has been used (see §C.3). This is the Weissenberg-Rabinowitch equation. It tells 63 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 R2 + R3 d w = R R2 3 d R R ( (4) ) Then division by R2 w R 3 gives 3+ ( d ln w R 3 d ln R ) = 1 R (5) w R3 Then finally ( d ln w R 3 R R 3 + ( R ) = = R w R 3 d ln R ) 1 (6) Equation 6 gives the viscosity as a function of shear rate at the wall from experimental measurements of w and P 0 P L . 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 for suspensions of fibers, because of the change in the fluid microstructure near the wall. The above analysis is applicable only when there is no appreciable viscous heating. 64 Note to p. 242 By working through all the intermediate steps in Example 8.3-1 we get a better idea how to solve problems involving the power-law model. First we have to obtain the expression for the scalar that appears in Eq. 8.3-5: = 1 2 ( : ) = 1 2 3 3 ij ji and i=1 j=1 = v + ( v ) † (1,2) where i and j take on the values 1 = r, 2 = , and 3 = z, since we are dealing with cylindrical coordinates. The components of in cylindrical coordinates may be obtained from Eqs. (S) to (AA) in Table A.7-2. Since the only component of v that is nonzero is the zcomponent and since that is a function of r only, the only components of that we need are the rz- and zr-components, rz = ( v )rz + (v )zr = vz r + vr z = vz r + 0 zr = ( v )zr + ( v )rz = vr z + vz r = 0 + vz r (3) (4) Therefore = 1 2 ( : ) = ( dvz dr ) = ± dvz dr 2 (5) where the minus sign must be chosen, since dv z dr is negative. Then the shear stress rz will be dv rz = m z dr n1 dvz dvz = dr dr n (6) By going through the above procedure, it is guaranteed that, when we take the fractional powers of the quantities in parentheses (see Table 8.3-2 for sample values of n), we will not get any imaginary quantities. When Eq. 6 is substituted into Eq. 2.3-13, we then get: 65 n dvz P 0 P L dv P PL m z = 0 = r or dr 2mL dr 2L 1n r1 n (7,8) Integration of Eq. 8 gives P PL vz = 0 2mL 1n r( ) +C (1 n ) + 1 1 n +1 (9) Application of the no-slip condition requires that vz = 0 at r = R : P PL 0 = 0 2mL 1n R( ) +C (1 n ) + 1 1 n +1 (10) Subtraction of Eq. 10 from Eq. 9 eliminates the integration constant C and leads to P PL vz = 0 2mL 1n 1 n +1 1 n +1 r ( ) R( ) (1 n) + 1 (11) Rearrangement then gives (P P L ) R vz = 0 2mL 1n r (1 n )+1 R 1 (1 n) + 1 R (12) The mass rate of flow w is then obtained by integrating the velocity distribution over the tube cross-section: 2 w = 0 R 0 vz rdrd (P P L ) R = 2 R 2 0 2mL 1n ( ) 1 R 1 n +1 1 ( ) d 0 (1 n ) + 1 (13) where = r R . Performing the integration gives 66 (P P L ) R w = 2 R 0 2mL 1n 2 (P P L ) R = R 0 2mL 3 (1 n ) + 1 R (1 n) + 1 2 (1 n) + 3 1n 1 (1 n ) + 3 (14) This is the power-law analog of the Hagen-Poiseuille equation for Newtonian fluids. 67 Note to p. 248 Here we work through the missing steps in Example 8.4-2. In Eq. 8.4-20, we make the change of variable s = t t after the first line in order to get a factor e i t to appear explicitly on the right side of the equation (to match a similar factor on the left side): { 0 i t i v e } 2 0 d v i t 0 s 1 i s = 2 e 0 e e ds 1 dy (1) Next, we perform the integration over s: { 0 i t i v e } 2 0 s i s 1 d v e 1e = 2 e i t 0 1 (1 1 ) i dy d 2 v 0 i t 0 = 2 e 1 + i1 dy 0 (2) We may now remove the real-operator sign from both sides, as well as the common multiplier e i t , to get: d 2 v 0 i (1 + i1 ) 0 d 2 v 0 0 = or i v = v 0 dy 2 dy 2 1 + i1 0 (3a,b) Eq. 3b is a differential equation for the complex function v0 ( y ) . The equation is of the form of Eq. C.1-4, where [] is a2 . Since this is a complex quantity, we write it as ( + i ) , where and are real, so that we can write the solution as 2 + + i ) y + i ) y v0 = Ae ( + Be ( (4) Then according to Eq. 8.4-19, 68 { {( } ) } + +i ) y + i ) y vx ( y,t ) = v0 ( y ) e i t = Ae ( + Be ( e i t (5) Now we know that the oscillatory disturbance will not propagate with increasing amplitude as the distance from the wall increases, so that 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 wall will be v0 . Therefore B will have to be set equal to v0 . Therefore Eq. 5 can be rewritten as { } { } + i ) y i t i t y ) vx ( y,t ) = v0 e ( e = v0 e y e ( = v0 e y cos ( t y ) (6) Now we must find out what and are. To do this, we have to go back to Eq. 3: i (1 + i1 ) 2 = ( + i ) 0 (7) We next equate the real and imaginary parts of both sides: 1 2 = 0 2 2 and 2 = 0 (8, 9) This gives us two equations for the two unknowns and . We can eliminate between the two equations and get an equation for : 2 1 1 2 = 0 20 2 2 (10) or, after multiplying through by 2 2 1 2 2 + =0 0 20 4 (11) 69 This can be regarded as a quadratic equation in 2 , which has the solution: 2 1 2 1 1 2 2 = ± + 4 2 20 2 0 0 1 2 = 20 1 1 1 + 2 2 1 2 2 2 1 + = 1 1 20 (12) We now extract the square root of both sides to get: =± 12 2 2 1 + 1 20 1 (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 finally get: =+ 20 + 1 + 2 2 1 1 =+ 20 + 1 + 2 2 1 1 12 (14) 1 2 (15) However, we could just as well have chosen both of the signs to be "minus." Then we would have = 12 2 2 1 + 1 1 20 (16) = 1 2 2 2 1 + 1 1 20 (17) The quantity given in Eq. 16 is still real and positive. So, how do we 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 not satisfy Eq. 8, whereas the latter do. Therefore, we conclude that Eqs. 70 16 and 17 are the proper quantities to insert in Eq. 6. Thus Eqs. 8.4-24 and 25 of the textbook are correct. 71 Note to p. 250 The simplest nonlinear viscoelastic model for polymers is the corotational Maxwell model, which is obtained by replacing the partial time derivative in Eq. 8.4-3 by the Jaumann derivative of Eq. 8.5-2. This gives + D = 0 Dt (1) This equation contains just two parameters: 0 , the zero-shear-rate viscosity, and , the relaxation time. Here we show how to get the viscosity, the first normal-stress coefficient, and the second normal-stress coefficient for the model given in Eq. (1). The imposed flow is vx = y with v y = 0 and vz = 0 . In learning how to analyze and evaluate nonlinear viscoelastic models, 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. 0 0 0 v = 1 0 0 0 0 0 0 1 0 = 0 0 0 0 0 0 (v )† 0 1 0 = 1 0 0 0 0 0 0 1 0 = 1 0 0 0 0 0 (2,3) (4,5) Since the stress tensor does not depend on position and time, in the Jaumann derivative of , the substantial derivative will not contribute and we are left with only 12 { } . 72 0 1 0 xx { } = 1 0 0 yx 0 0 0 zx xx { } = yx zx xy yy zy xz yx yz = xx 0 zz xy xz 0 1 0 yz 1 0 0 = yy zz 0 0 0 zy xy yy zy yy xy 0 xx xy xz yz xz 0 0 0 0 (6) (7) Hence the Jaumann derivative is (keep in mind that the stress tensor is symmetric) ( ) 1 yx xx yy 12 yz 2 D 1 1 = yy yx 2 xz D t 2 xx 1 12 yz 0 2 xz Thus the corotational Maxwell model in matrix form is ( xx yx zx xy yy zy ) yx xz yz + 12 xx yy zz 12 yz ( 0 1 0 = 0 1 0 0 0 0 0 ) 1 2 ( xx yy yx 1 2 xz ) (8) 12 yz 1 2 xz 0 (9) From this matrix equation, we can write down the following algebraic equations ( ( ) )+ xx yy 2 yx = 0 yy zz =0 yx (10) (11) 73 yx + 1 2 ( xx ) yy = 0 xz 12 yz = 0 (12) (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 or 1 = 0 1 + ( )2 (15,16) 2 or 1 2 = 0 1 + ( )2 (17,18) 2 or 2 1 = 2 0 1 + ( ) (19,20) 1 1 + ( ) xx yy = 0 yy zz = +0 2 2 1 + ( ) 1 + ( ) We know that the shear stress keeps increasing as the velocity gradient increases, but Eq. 16 indicates otherwise. In fact, the shear stress goes through a maximum at = 1. The second normal stress is negative, and that is in agreement with measurements for polymeric liquids. However, generally one expects 2 1 to be in the range from 0.1 to 0.4 . Hence from examination of this one particular flow, it is evident that caution must be used when drawing any conclusions from this model. However, for a model with just two parameters, it seems to be promising for predicting qualitative behavior. 74 Note to p. 255 Since the FENE-P dumbbell model (a molecular model) describes many of the observed rheological phenomena of polymer solutions, it is important to understand how we go from the constitutive equation in Eqs. 8.6-2 to 4 to the expressions for the rheological properties. Here we show how to get the non-Newtonian viscosity 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 = yx zx xy yy zy xz 0 0 yz = 0 0 0 zz 0 0 0 (1) Then Eq. 8.6-4 may be written in matrix form as follows: p,xx Z p, yx p,zx p,xy p, yy p,zy 2 p,xy p, yy p, yz p,xz p, yz H p, yy 0 0 p,zz 0 0 p, yz 0 1 0 = nKT H 1 0 0 0 0 0 (2) From this we get at once that the only nonzero components of p are p,xx and p,xy = p, yx . Then from the matrix equation, Eq. 2, we can write down the two nonvanishing component equations as: Z p,xx = 2 p, yx H Z p, yx = nKT H and (3,4) in which ( ) ( ) Z = 1 + ( 3 b ) 1 tr p 3nKT = 1 + ( 3 b ) 1 p,xx 3nKT (5) 75 In order to make the algebraic manipulations somewhat easier, we introduce 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 ZS = 13 (9) (10) Z = 1 + ( 3 b ) (1 N ) (11) Use the second of these equations to eliminate Z and rewrite Eqs. 9 and 11 thus: N = 2S 3S and 3 = 1 + (1 N ) 3S b (12,13) Then N may be eliminated between these two equations to get a cubic equation for the dimensionless shear stress S (for method of solving cubic equations, see a mathematics handbook): S3 + b+3 b S+ =0 18 54 or S 3 + 3pS + 2q = 0 (14,15) Eqs. 14 and 15 serve to define p and q. Eq. 15 has three solutions, two imaginary solutions (of no interest here) and one real solution: S = 2p1 2 sinh ( 1 3 arcsinh p3 2 q ) (16) or, in terms of the original variables: 76 p, yx b + 3 = 2 54 3nKT +1 2 b + 3 3 2 b 1 sinh 3 arcsinh H 108 54 (17) From this result one can plot the non-Newtonian viscosity as a function 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 be obtained from Eq. 17 or 16. Very small shear rate corresponds to very small q, so that S = 2p1 2 sinh ( 1 3 ) p3 2 q = 2p1 2 { 1 3 } p3 2 q + = 23 p1 q (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 x 3 + arcsinh x = x 16 x 3 + (19,20) In terms of the original variables, Eq. 18 is p, yx b + 3 = 3nKT 54 1 2 3 b b H = nkT b + 3 H 108 (21) This corresponds to b s = nkT H b + 3 (22) in the limit of zero velocity gradient. In the limit of infinite velocity gradient, we make use of the fact that for large values of the argument, sinh x 12 exp x (see Appendix C.5). Therefore S = 2p1 2 sinh ( ln ( 2p q)) = 2p 1 3 3 2 ( ln ( 2p q)) = ( 2q) 12 1 exp 13 2 3 2 13 (23) 77 In the original variables, this becomes p, yx b = 3nKT 2 H 108 13 b = nKT 54 H 108 13 (24) This corresponds to a "power-law function": b 1 s = nKT H 2 13 (25) H 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 quite small, the cubic term in S may be omitted, and one gets then S = 32 p1 q (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 ) 13 (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. In addition, 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.] 78 Note to p. 259 The Bingham fluid is not the only empirical equation with a yield stress. Closely related to the Bingham equation is the twoconstant Casson equation, proposed for pigment-oil suspensions, but widely used for blood1,2,3 = = μ0 + 0 when 0 (1) when 0 (2) in which 0 is the yield stress, and μ0 is a parameter with dimensions of viscosity. Derive the expression for the mass rate of flow of a Casson fluid through a circular tube of radius R, and length L. 1 N. Casson, Rheology of Disperse Systems, C. C. Mill, ed., Pergamon, London (1959), pp. 84-104. 2 M. M. Lih, Transport Phenomena in Medicine and Biology, Wiley, New York (1975), 378-38. 3 E. 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 any kind of fluid was derived in Chapter 2 and found to be P PL r rz = 0 2L (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 = (P 0 P L ) 2L r0 . We further note that = dvz dr for flow in circular tubes, according to the definition given just before Eq. 8.3-2. For the region r0 r R , we then get from Eq. 2 = μ0 + 0 or rz = μ0 + 0 (4a,b) 79 Combining Eqs. 3 and 4b and rearranging gives (P0 P L ) r = 2L μ0 + 0 (5) Solving Eq. 5 for we get = 1 (P 0 P L ) r 2 2L μ0 (P0 P L ) r 0 2L r r 0 2 + 1 + 0 r0 μ0 r0 (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 > = 2 +1 d (7) Integration of this equation, and using the boundary condition that > = 0 at = R r0 , we get for r0 r R 2 3 2 2 3 2 1 R r 4 R r R r = 1 1 + 1 2 r0 R 3 r0 R r0 R > (8) and for 0 r r0 2 1 R 4 R = 2 r0 3 r0 < 3 2 + R 1 r0 6 (9) The mass rate of flow can then be obtained by using Eq. 2.3-21 after doing an integration by parts. The dashed underlined terms are zero. 80 r=R dv dv dv r R R w = 2 12 v z r 2 0 12 r 2 z dr = 00 r 2 z dr + r r 2 z dr r=0 0 dr dr dr ----------------------- r03 0 R r0 2 d > r03 0 R r0 2 = d d = + μ 1 2 + 1 d μ0 1 0 ( ) 4 4 R 7 2 1 R 3 r03 0 1 R = 4 r 1 7 r 1 + 3 r 1 μ 0 0 0 0 12 4 r0 0 R 4 16 r0 4 r0 1 r0 = 1 7 R + 3 R 21 R μ 0 16 2L 0 2L 0 4 + 1 4 7 3 P P R P P R ( ) ( ) (P 0 P L ) R 0 L 0 L = 4 8 μ0 L 2L 0 1 21 P P R ( 0 L) (10) When 0 0 and μ0 μ , the Newtonian result is recovered. 81 Note to p. 275 As an exercise in identifying the dimensions of thermal quantities, we verify that the following equations are dimensionally consistent: Eq. 9.3-12, Eq. 9.8-6, and Eq. 9.8-8. We do this by replacing the symbols in these equation by the corresponding dimensions, making use of the "Notation" table on pp. 872 et seq. ML2 M ( ) 2 (T ) tT ML2 t 2T (a) ML 3 = tT (b) 2 M M L L M L2 L M L M 3 = 3 t t + 3 2 t + 2 t + 3 L L t Lt t t (c) L2 (1) M L3 M L2 L2 L3 M M 2 = 2 (T ) + M 2 + (T ) T 2 ( ) Lt t t T Lt (2) (3) In each of these cases, each term has the same dimensions. 82 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 is given by Ĉp = 5 R 2M (1) as given two lines after Eq. 9.3-15. The ideal gas law equation of state is pV̂ = RT M (2) Therefore, the bracket in Eq. 9.8-8 is RT V̂ R V̂ T = V̂ T = V̂ T T pM pM = 0 p T p (3) Hence, the only term that survives is the heat-capacity term, which is: Ĥ Ĥ o = ( 5 R T To 2M ) (4) 83 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 express the result in terms of the Brinkman number, Br. We also need to know what happens when the Brinkman number goes to zero (i.e., negligible viscous heating). To simplify the discussion, we introduce the following dimensionless variables: = T T0 (temperature) Tb T0 = x (distance) b (1) so that Eq. 10.4-9 becomes: = 1 Br (1 ) + 2 (2) To get the location of the maximum of the temperature, we differentiate with respect to and set the derivative d d equal to zero: d 1 = Br (1 2 ) + 1 = 0 d 2 (3) From this we get the location of the maximum in the temperature curve: max = 1 1 + 2 Br (4) When there is negligible viscous heating, Br 0 , and, according to Eq. 4, max . This result is nonsense, since the system extends only to = 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< . 84 Note to p. 309 It's always a good idea to check the solutions to problems. Here we verify that the expression in Eq. 10.7-13 for the dimensionless temperature 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.5-10 and 11: d sinh N (1 ) = ( N ) cosh N d d 2 d 2 = cosh N (1 ) cosh N (1) ( +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 =1 (3) =0 and the boundary condition at = 1 is also satisfied, since sinh N (1 ) d = ( N ) d =1 cosh N =0 (4) =1 The hyperbolic sine is shown in Figure C.5-2. 85 Note to p. 315 Here we verify the determination of the constants of integration, C0 , C1 , and C2 in Eq. 10.8-27. Boundary condition 1 requires that ( 0, ) = finite : C1 : ( 0, ) = C0 + C1 ln 0 + C2 (1) This can be satisfied only if C1 = 0 . Boundary condition 2 requires that = 1 at = 1 : C0 : 2 4 3 1 = C0 = C 16 =1 4 0 4 =1 (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: 1 2 4 = 0 4 + 4 + C2 1 2 d 4 16 ( ) (3) Next we evaluate the integrals: 1 1 1 2 3 4 1 d = 4 d = 4 0 0 2 4 = 1 ( ) ( ) 3 2 4 5 5 7 1 2 0 4 4 16 1 d = 4 0 4 16 + 16 d = 5 1 7 1 4 + = 16 6 16 8 16 24 4 1 ( (4) ) (5) 86 ( ) 1 1 C2 0 1 2 d = C2 4 (6) Combining the results of Eq. 3 to 6, we find that C2 = 7 24 . 87 Note to p. 337 In the textbook, we transformed Eq. 11.2-2 into an equation for the enthalpy (Eq. 11.2-3) and then used an equilibrium thermodynamic formula to get the equation of energy in terms of Ĉp and T, given in Eq. 11.2-5 (and also Eq. (J) in Table 11.4-1). Specifically, we used the equation for Ĥ (T, p ) : V̂ Ĥ Ĥ dĤ = dT + p dp = Ĉp dT + V̂ T T dp p T p T (1) appropriately rewritten for a "particle" of fluid moving with the local velocity v. Alternatively, we can begin with Eq. 11.2-2 and use another equilibrium thermodynamic formula to get the equation of energy in terms of ĈV and T (see Eq. (I) in Table 11.4-1). Specifically, we use ( ) the equation for Û T ,V̂ : Û Û p dÛ = dT + dV̂ = ĈV dT + p + T dV̂ T V̂ T V̂ V̂ T (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: DÛ DT p DV̂ = ĈV + p + T T V̂ Dt Dt Dt (3) Since V̂ = 1 , we may rewrite DV̂ Dt as follows: 1 D DV̂ D 1 = = 2 Dt Dt Dt (4) 88 Then if we use Eq. (A) of Table 3.5-1, we may rewrite this last relation as: 1 DV̂ = 2 ( v ) = ( v ) Dt ( ) (5) Now substitute this expression into Eq. 3, and then make use of Eq. 11.2-2 to get: ĈV DT p + p + T ( v ) = ( q ) p ( v ) ( :v ) T V̂ Dt (6) When the terms involving ( v ) are moved to the right side, there is some cancellation and Eq. (I) of Table 11.4-1 is obtained: ĈV DT p = ( q ) T ( v ) ( :v ) T V̂ Dt (7) [Note: See footnote 1 at the bottom of p.338 for a discussion of "incompressible fluids."] 89 Note to p. 341 Verify that Eq. (T) of Table 11.4-1 is the equation of change for entropy. We can start with the thermodynamic expression dÛ = TdŜ pdV̂ (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 then write: DÛ DŜ DV̂ =T p Dt Dt Dt (2) where we are now assuming that this expression can be applied locally. When this equation is multiplied by and then combined with Eq. (G) of the table we get: ( q ) p ( v ) ( :v ) = T DŜ DV̂ p Dt Dt (3) Next, divide by T and replace V̂ by 1 / to get p D 1 1 1 1 DŜ = ( q ) p ( v ) ( :v ) + T Dt T T T Dt (4) The second and fourth terms on the right side may be seen to cancel one another is one makes use of the equation of continuity in the form of Eq. (A) of Table 3.5-1, so that we get DŜ 1 1 = ( q ) ( :v ) Dt T T (5) When we make use of Eq. 3.5-4, Eq. 5 becomes 90 1 1 Ŝ + Ŝv = ( q ) ( :v ) t T T ( ) (6) This is equivalent to Eq. (T) in Table 11.4-1. Eq. (T) is written in a form that emphasizes the entropy flux (q / T ) and the entropy production (the terms in brackets: 1 q 1 Ŝ = Ŝv + 2 (q T ) ( :v ) T T t T ( ) (7) See Problem 11D.1 and §24.1 for more on this subject. 91 Note to p. 343 Verify that Eq. 11.4-14 gives the location of the maximum temperature. We'd also like to know whether the maximum is nearer the inner cylinder or the outer. Differentiate in Eq. 11.4-13 with respect to and set the derivative equal to zero: 2 d 1 1 1 = + N 3 1 2 =0 ln ln d (1) Solving for 2 2 gives: 2 1 1 1 = + 1 2 2 N ln ln (2) whence, the location of the maximum temperature rise is max = ( 2 ln (1 ) (3) ) 1 2 1 (1 N ) 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, the temperature difference, the thermal conductivity, and the viscosity. If we take N to be infinity, then we get: 1 2 1 2 ( + 1) max 1 3 1 4 0.75 0.67 0.625 0.67 0.51 0.42 This suggests that the maximum occurs nearer the inner wall, where the velocity gradient is larger. 92 Note to p. 346 Here we want to derive Eq. 11. 4-27 from Eq. 11.4-26 by following the instructions in the text. First we rewrite Eq. 11.4-26 in terms of dimensionless variables: T T1 = dimensionless temperature T T1 r = = dimensionless radial coordinate R = (1) (2) Then Eq. 11.4-26 becomes: d R d 2 d = d R0 d d where R0 = wr Ĉp 4 k (3a,b) Now introduce the change of variable u = 2 ( d d ) , and rewrite the differential equation as: u 2 = R du R0 d (4) This equation may be solved to get: 1 R R = ln u + ln C1 R0 R0 or R 1 C1 u = exp 0 R (5a,b) where the constant of integration has been written as ( R R0 ) ln C1 . Reverting to the original variable gives: C1 2 R 1 d = exp 0 d R (6) Further integration then gives: 93 C1 = 1 2 exp R0 1 d + C2 = R R R v R R e ( 0 ) e ( 0 ) = + C2 ( R0 R ) R R R R e ( 0 ) e ( 0 ) = + C2 ( R0 R ) v 1 R exp 0 R v dv + C2 (7) The constants of integration may be determined from the boundary conditions that: at = , = 1 , and at = 1 , = 0 . From the second of 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: R e ( 0 = R e ( 0 R ) R R e ( 0 ) R ) R R e ( 0 ) (8) which agrees with Eq. 11.4-27. 94 Note to p. 352 Fill in the missing steps, showing how to arrive at Eqs. 11.4-67, 11.468, and 11.4-75. We begin by multiplying Eq. 11.4-66 by vx 1 v1 to get pvx CIII vx dv 4 μ vx x = v x2 + 3 1 v1 dx 1 v1 1 v1 (1) Eq. 11.4-61 may be used to simplify the second term on the right side to p , and this may be further rearranged by using ideal gas relations, thus: ( ) ( C p C V T 1 1 p RT ĈpT = = = = Ĉp ĈV T = CI 12 vx2 M M (2) In the last step, use has been made of Eqs. 11.4-64 and 65, the second of which should have been written as CI = ĈpT1 + 12 v12 . Then we get Eq. 11.4-67 ) ( C v dv + 1 2 1 4 μ vx + CI III x vx x = dx 1 v1 3 1 v1 ) (3) To put this equation in dimensionless form, we insert the expressions for the constants of integration, multiply the equation by 31 4 μ v1 and then introduce the quantities defined in Eqs. 11.4-69, 70, 71, 73 and 74, as follows: d 31 = d 4 μ v1 + 1 2 1 2 RT1 v x 1 2 v + + T + v v Ĉ 1 1 x M 1 v1 p 1 2 1 2 ( ) 95 3 1 + 1 2 2 2 RT1 1 1 2 = Ma1 v + + T + v v Ĉ 1 1 p 1 2 1 M Ma1 4 μ v1 2 RT1 2 + 1 2 1 ĈpT1 1 = Ma1 1 + + v 2 + 2 2 + 1 2 Mv 1 1 2 1 1 1 1 1 1 + = Ma1 2 + 2 + + 1 1 Ma 2 2 (4) 2 + 1 Ma 1 1 ( The coefficient of may be seen by Eq. 11.4-72 to be constant term is + . Therefore, Eq. 4 finally becomes d = Ma1 ( 1) ( ) d ) 2 and the (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 is to be found on p. 29 of Table of Integrals and Other Mathematical Data, H. B. Dwight, MacMillan, New York, 4th edition (1961). There we find xdx 1 ( a + x ) ( c + x ) = a 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 1 ln 1+ + ln + = Ma1 ( 0 ) (8) In order to remove the absolute value signs, we note that < < 1 , so that 96 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. 97 Note to p. 375 a. First we'll verify the formula for differentiation of the error function in Eq. C.6-2. b. Then we'll show that Eq. 12.1-8 is a solution to Eq. 12.1-3 and the associated boundary and initial conditions. a. The differentiation of the error function is given by Eq. C.6-2: d 2 u 2 du erf u = e dx dx (1) where it is understood that u is a function of x. Now differentiate the error function in Eq. C.6-1 with respect to x using the Leibniz formula of §C.3: d d 2 erf u = dx dx u u 2 0 e 2 u u 2 u 2 u 0 2 0 du = e du + e e (2) 0 x x x x, Since u is a dummy variable of integration, it is not a function of and therefore the first term is zero. The last term is also zero. Hence the 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 is important to designate the dummy variable of integration, u , and the upper limit in the integral, u, by two different symbols. This example emphasizes the importance of this statement.] b. Now turn to Eq. 12.1-8, where the left side is the dimensionless temperature difference . Form the derivatives that appear in Eq. 12.1-3: y 1 3 2 t 4 2 2 y 2 4 t 1 = e y 4 t 2 2 y 2 4 t 1 2y = e y 2 4 t 4 t 2 y2 = e t 4 t (3) (4) (5) 98 When the results in Eqs. 3 and 5 are inserted into Eq. 12.1-3 it is found that the equation yields an identity. Therefore, Eq. 12.1-8 does satisfy the 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 formula for differentiating an integral. For additional information and anecdotes 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.] 99 Note to p. 377 As pointed out in the footnote, there are two solutions to this problem, one that converges rapidly for long times (see Eq. 12.1-31): ( 1)n = 2 n=0 (n + ) 1 2 ( exp n + 1 2 ) 2 ( ) 2 cos n + 12 (1) and one that converges rapidly for small times: ( ) ( ) n + 12 12 n + 12 + 12 = 1 ( 1) erfc + erfc n=0 n (2) We want to verify that both of these solutions satisfy the partial differential 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 cosine twice with respect to 2 n=0 2 n=0 ( 1)n ( ) 2 2 cos n + ( ) 2 2 cos n + (n + ) exp n + 12 (n + ) exp n + 12 1 2 ( 1)n 1 2 ( 1 2 ) (n + ) ( 1 2 ) (n + ) (n + ) 1 2 1 2 2 2 = 1 2 (3) and this is clearly an identity, so that the partial differential equation is satisfied. When is set equal to zero and = 1 in Eq. 1, we get 1 = 2 n=0 ( 1)n (n + ) 1 2 ( ) cos n + 12 (4) 100 We have to prove that this is an identity. To do this, multiply both sides by cos m + 12 and integrate over from –1 to +1: ( ( +1 1 cos m + 1 2 ) ( 1)n ) d = 2 (n + ) 1 1 2 n=0 +1 ( ) ( ) cos m + 12 cos n + 12 d (5) Then performing the integrations we get ( +1 ( 1) +1 cos2 n + 1 1 sin m + = 2 mn 1 2 1 1 m + 12 n=0 n + 2 ( ) n ) ( ( ) 1 2 ) d or ( ( 1) 2 1 sin m + = 2 2 m + 12 m + 12 ( ) ) m ( (7) ) ( ) This is an identity, since sin m + 12 = ( 1) . When is set equal to +1 or –1, we get ( cos m + 1 2 m = 0 , since ) = 0 . b. Solution in Eq. 2 When Eq. 1 is substituted into Eq. 12.1-14, we get by differentiating the complementary error functions once with respect to and then twice with respect to ( ) ( ) ( ) ( ) n + 1 1 2 n + 12 12 2 2 2 n + 1 1 1 3 2 erfc = exp 2 2 2 (8) 2 n + 1 1 1 n + 12 12 2 2 2 2 erfc = exp (9) n + 1 12 n + 12 12 2 2 2 2 erfc = exp 2 ( ) ( ( ) ( ) ) 101 ( ) n + 1 1 1 1 2 2 2 2 2 (10) Thus, the nth term satisfies the partial differential equation, and hence the entire series does. At = 0 , all of the complementary error functions are zero, so that = 1 , and the initial condition is satisfied. At = ±1 , = 0 as may be seen as follows. First consider = +1 : 1 2 0 1 2 3 erfc + erfc = 1 erfc + erfc + erfc + erfc (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 similar argument can be made for = 1 . 102 Note to p. 379 In Example 12.1-3 going from Eqs 12.1-38 and 39 to Eq. 12.1-40 presents 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 y e 2 y cos t y dy 2 (1) Bars have been added to the variable of integration to distinguish it from the lower limit on the integral. It is easier to perform the integration if the cosine is converted into an exponential of a complex quantity, thus: T T0 = q0 k 2 y y e { { e i ti 2 y }dy q0 2 y i 2 y e i t y e dy k q0 i t e (1+i ) 2 y = e k (1 + i ) 2 y q0 i t e (1+i ) 2 y = e k 1 + i 2 ( ) = } (2) Next we remove from the braces that portion of the exponential that is real; we also multiply numerator and denominator by (1 i ) . Then we have e i t e i 2 y (1 i ) 2 2 q 2 2 y i 2 y = 0 e e i t e (1 i ) 2k q T T0 = 0 e k 2 y { } (3) 103 In order to proceed, we need to rewrite (1 i ) in the form re i . Then we find r and as follows: 1 i = re i = r ( cos + i sin ) (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 1 = r cos 1 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 = r 2 cos2 1 = r 2 sin 2 and (10) Adding the two equations then gives: r2 = 2 r=± 2 or (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 2k { } 2 2 y i 2 yi 4 e 2e i t q 2 y = 0 e cos t 2 y 14 k ( ) (13) This agrees with Eq. 12.1-40 in the textbook. 104 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 = re i , then ln z = ln r + i = ln x 2 + y 2 + i arctan ( 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 sin Z 1 1 sin ( X + iY ) 1 = ln ln + i sin Z + 1 sin ( X + iY ) + 1 = 1 sin X cos iY + cos X sin iY 1 ln sin X cos iY + cos X sin iY + 1 = 1 sin X cosh Y + icos X sinh Y 1 ln sin X cosh Y + icos X sinh Y + 1 ( sin X cosh Y 1) ( sin X cosh Y + 1) + cos 2 X sinh 2 Y 1 +i cos X sinh Y (sin X cosh Y + 1) cos X sinh Y ( sin X cosh Y 1) = ln 2 2 sin X cosh Y + 1 + cos X sinh Y ( ) ( ) ( ) 2 2 2 2 1 sin X cosh Y 1 + cos X sinh Y + 2i cos X sinh Y = ln (sin X cosh Y + 1)2 + (cos X sinh Y )2 (2) In going from the third to the fourth line, we have multiplied the numerator and denominator by the complex conjugate of the denominator. The imaginary part of the expression in Eq. 2 is then 105 = 2cos X sinh Y 1 arctan 2 2 2 2 sin X cosh Y 1 + cos X sinh Y = 2 ( cos X sinh Y ) 2cos X sinh Y 1 1 arctan = arctan 2 2 1 ( cos X sinh Y )2 sinh Y cos X 2 tan A 1 arctan 2 1 tan A (3) The last expression serves to define A . But the quantity in parentheses is just tan 2A (see, e.g., formula 406.02 of Dwight's Tables of Inegrals and Other Mathematical Data, 4th edition), and the "angle whose tangent is tan 2A " is just 2A (i.e., arctan(tan2A ) = 2A. However, A = arctan ( cos X sinh Y ) so that, finally = cos X 2 cos x b 2 arctan arctan = sinh y b sinh Y (4) which is the result in Eq. 12.3-6. 106 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 the right side of the equals sign in Eq. 12.-4-4 (the second term is zero, because ve = v = a constant in this problem): 0 v x ( v vx ) dy = v2 ( x ) 0 1 vx v vx 1 d v (1) Here ( x ) is the velocity boundary-layer thickness, and = y ( x ) is the dimensionless coordinate in the y-direction. In the second integral we 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 assumed velocity profile into Eq. 1 gives: 0 v x ( v vx ) dy ( = v ( x ) ( 2 4 )( ) = v2 ( x ) 0 2 2 3 + 4 1 2 + 2 3 4 d 1 1 0 2 315+56718035 315 ( = v ( x )( 2 ) 2 3 + 9 4 4 5 4 6 + 47 8 d 2 = v2 ( x ) 1 43 12 + 95 32 74 + 12 19 )= 37 315 ) v2 ( x ) (2) Similarly, the integral appearing in Eq. 12.4-5 may be evaluated: 0 Ĉp vx (T T ) dy = Ĉp v (T T0 ) T ( x ) 0 vx v T T T T = Ĉp v (T T0 ) T ( x ) 0 vx v T T0 T0 T T T T T dT 0 0 1 1 0 dT 107 = Ĉp v (T T0 ) T ( x ) 0 1 vx v T0 T 1 T T 0 dT (3) in which T = y T ( x ) = y ( x ) , and is assumed to be independent of x. Then, inserting the postulated profiles for velocity and temperature into Eq. 3, we get: 0 Ĉp vx (T T ) dy ( = Ĉp v (T T0 ) T ( x ) 0 2T 2T3 3 + T4 4 1 = Ĉp v (T T0 ) T ( x ) ( 2 15 3 1 140 3 + 180 4 ) ) (1 2 T ) + 2T3 T4 dT (4) When the expressions in Eqs. 2 and 4 as well as Eqs. 12.4-6 to 9 are substituted into Eqs. 12.4 and 5, we get differential equations for the boundary-layer thicknesses as a function of the distance along the plate: 2 μ v 37 d = v2 315 dx 2k (T T0 ) T = ( 2 15 (5) ) 3 1 140 3 + 180 4 Ĉp v (T T0 ) d dx T (6) Eq. 5 for ( x ) may be solved as follows: 315 2 μ v d = ; 37 v2 dx 0 d = 630 μ 37 v x 0 dx ; = 1260 μ x 37 v (7,8,9) and Eq. 6 for T ( x ) may also be solved: 2k (T T0 ) T d T T = dx ( = ( 2 15 ) 3 1 140 3 + 180 4 Ĉp v (T T0 ) 1 2 15 ) 2k 3 1 140 3 + 180 4 Ĉp v d dx T (10) (11) 108 T 0 T d T = T = ( ( 2 2 15 3 1 140 3 + 180 4 4 2 15 3 1 140 3 + 180 4 ) ) v x 0 dx x v (12) (13) It remains to find = T ( x ) ( x ) as a function of the physical properties. Forming the ratio we get: = T = ( 4 ( 37 1260 ) 2 15 3 1 140 3 + 180 4 ) μ (14) Squaring both sides and collecting all the terms on the left side, we find: 35 1 k 1 3 5 6 2 3 1 140 + 180 = (15) = Pr= 15 315 Pr μ Ĉp Ĉp μ k For large Prandtl numbers, we can drop all but the lead term on the left side and get = 3 35 15 Pr 1 3 315 2 = 3 0.880Pr 1 3 = 0.958Pr 1 3 (16) As pointed out in the textbook, 0.958 may be replaced by 1 to fit the exact 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.] 109 Note to p. 413 The object here is to fill in the missing steps between Eq. 13.4-11 and Eq. 13.4-16. First we multiply Eq. 13.4-11 by t d ( ) d 1 + = C0 d d (1) Then first integration with respect to gives: t ( ) d 1 + = C0 0 d + C1 C0 I ( ) + C1 d (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: C0 I ( ) C1 d = + d 1 + (t ) 1 + (t ) ( ) ( (3) ) A second integration with respect to gives: = C0 0 () I ( ) t 1 + ( ) d + C1 0 ( 1 ) t 1 + ( ) d + C2 (4) in which we must remember that ( ) is now a function of . If next we consider the limit of the above expression as 0 , it may be seen that 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: t 110 () I ( , ) = C0 + C0 0 ( ) t 1 + ( ) d + C2 (5) Next, apply the boundary condition at = 1 : 1 = C0 () I ( ) t 1 + ( ) =1 = C0 I (1) C0 = I (1) or 1 (6) inasmuch as ( ) is zero at the wall. Next we want to get the driving force 0 b , which is the dimensionless wall temperature minus the dimensionless bulk temperature (defined in Eq. 10.8-33): t ( ) t 1 + ( ) I ( ) 1 = C0 0 0 0 I ( ) 1 0 b = C0 0 ( ) t 1 + ( ) d d C0 C0 1 I (1) 0 ( () I 1 ( ) t 1 + ( ) d d 1 0 d 1 d () ) 1 + ( ( ) ) d I t (7) In the second expression we have interchanged the order of integration. The second term in Eq. 7 may be rewritten: C0 1 I ( 1) 0 ( 1 0 d 0 d () ) ( ) I 1 + () (t ) d I C0 1 = d I (1) I t I (1 ) 0 1 + ( ) () ( ) 111 I ( ) C 1 = C0 0 d + 0 0 d t) t) ( ( I 1 ( ) 1+ 1+ 1 I ( ) ( 2 ) ( (8) ) The first term in Eq. 8 above just cancels the first term in Eq. 7, and hence we are left with: I ( ) C 1 0 b = 0 0 d = I (1) 1 + ( t ) 2 ( ) I ( ) I (1) d 0 t) ( 1+ 2 1 ( ) (9) This is in agreement with Eq. 13.4-16 in the textbook. 112 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 (t ) F (t ) 1 ( ) F F + = r z r z z Pr(t ) r r r (1) where the primes indicate differentiations with respect to . We next ( ) make the change of variables = r z and = ( ) w z , and further let ( r, z ) = ( , ) = f ( ) , so that t f 1 1 = + = + 0 = r z r r z (2) t f r ( ) f r = + = 2 + = 2 2 z z z z w z z (3) Substitution of these expressions into Eq. 1 gives t f 1 (t ) F f r f ( ) F F z z z z 2 2 z t ( ) 1 f 1 = (t ) 2 Pr z (4) Multiplication of the entire equation by z 2 ( ) gives t F f 1 1 f F f + F f + = ( ) [ ] Pr(t ) (5) 113 On the left side, the terms involving F f cancel, and the equation may be rewritten as follows: 1 1 f 1 d Ff ) = (t ) ( d Pr (6) This equation can be multiplied by Pr( ) and integrated once to give t Pr( ) Ff = t 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) t = Pr ( ) F 0 t d = Pr( ) 0 C32 1+ 1 4 (C ) 2 3 t d = Pr( ) ln 1 + 1 4 (C ) 2 3 2 (8) or, taking the antilogarithm of both sides f ( ) = 1 + f ( 0 ) 1 4 (C ) 2 3 t 2Pr( ) (9) When this result is compared with Eq. 5.6-21 for the velocity profile in a circular jet, and use is made of the definition in Eq. 13.5-8, we obtain finally max vz = vz ,max t Pr( ) (10) which is a rather simple, and apparently fairly satisfactory, result. 114 Note to p. 454 Verify that Eq. 15.1-1 can be obtained from Eq. 11.1-9 by the method described on p. 454. We start by integrating Eq. 11.1-9 over the volume of the flow system shown in Fig. 7.0-1: V (t) t ( 2 v 1 2 ( ( ) ˆ dV = + Û + V (t) 1 2 )) ˆ v dV v 2 + Û + ( ) V (t ) ( q )dV V ( t ) ( pv )dV V (t ) [ v ] dV (1) We next apply the Leibnitz formula to the left side of the equation and the Gauss divergence theorem to the right side: d dt V (t ) ( 1 2 ( ( ) ˆ dV n v 2 + Û + S(t ) ( ( = S(t ) n 1 2 1 2 ) ) ˆ v dS v 2 + Û + S )) ˆ v dS ( n q )dS v 2 + Û + S(t ) ( ) S( t ) ( n pv )dS S(t ) n [ v ] 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 left side can be combined with the first term on the right side. Thus we get ( ( ) ) d ˆ ( v v ) dS Ktot + Utot + tot ) = S t n 12 v 2 + Û + ( S () dt S t ( n q )dS S t ( n pv )dS S t n [ v ] 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 S1 and the outlet section S2 , the surface velocity v S is zero and collinear with the outwardly directed unit vector n. The fluid velocity vector v is assumed to point in the direction opposite to the n vector at the entry plane, and in the same direction as as the n vector at the exit 115 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 the potential energy are constant over the cross section. Then when the integration over the cross sectional area is performed we get: ( ( S(t ) n 1 2 ) ) ˆ ( v v S ) dS v 2 + Û + ˆ v S = 12 1 v13 S1 + 1Û1 v1 S1 + 1 1 1 1 ˆ v S 12 2 v23 S2 + 2Û 2 v2 S2 + 2 2 2 2 (4) The second term on the right side (the q-term) is the integral over all surfaces of the normal component of the heat flux vector and is thus the rate of total heat addition to the system, Q: S f +Sm +S1 +S2 (n q)dS = Q (5) It is assumed that the heat addition at surfaces S1 and S2 is usually be small compared to the heat added at the solid surfaces. The third term on the right (the p-term) has to be evaluated at all the surfaces. At the inlet and outlet planes, we will get S 1 +S2 (n pv )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 the entry and exit planes. On the solid surfaces we get S f +Sm (n pv )dS = Wmp (7) This term is the rate that pressure does work on the system at the moving 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 are considered, but it is assumed that these will be small compared to the 116 pressure terms in Eq. 6. On the solid surfaces there will be a contribution similar to that for the pressure forces on the moving surfaces: S f +Sm (n [ v ])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 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: ( ) d ˆ v S Ktot + Utot + tot ) = 12 1 v13 + 1Û1 v1 + 1 ( 1 1 1 dt ˆ v S +Q+ W 12 2 v 23 + 2Û 2 v2 + 2 2 2 2 m ( ( + p1 v1 S1 p2 v2 S2 ) ) (8) or, introducing the enthalpy ( ) d ˆ v S Ktot + Utot + tot ) = 12 1 v13 + 1 Ĥ1 v1 + 1 ( 1 1 1 dt ˆ v S +Q+W 12 2 v 23 + 2 Ĥ 2 v2 + 2 2 2 2 m ( ) (9) Either Eq. 8 or Eq. 9 is referred to as the unsteady-state macroscopic energy balance. 117 Note to p. 494 The derivations of the Stefan-Boltzmann law and Wien's law from the Planck black-body distribution law are quite important and therefore it is a good idea to understand all the intermediate steps in the development. a. The Stefan-Boltzmann law We integrate Planck's distribution law over all wavelengths: e q(b ) = 0 (e) qb d = 0 1 2 c 2 h d 5 e ch KT 1 (1) Next we make a change of variable x = ch KT . Then 5 KT ch 1 2 c h 5 d = 2 c h x 5 2 dx ch KT x 2 1 2 = 2 ( KT ) 4 2 3 c h ( x ) dx 3 (2) Hence the integral becomes 2 ( KT ) e q(b ) = 2 3 4 c h x3 0 ex 1 dx (3) Then expand the denominator of the integrand as a Taylor series about x = 0, to get: ( ) e x 1 = 1 + e x + e 2x + e 3x + 1 (4) Then a term by term integration of Eq. 3 gives: 2 ( KT ) c 2 h3 4 0 n=1 3 nx x e dx = 2 ( KT ) c 2 h3 4 1 2 ( KT ) 4 6 4 = c 2 h3 15 n=1 n 4 (5) 118 On p. 171 of Planck's book, The Theory of Radiation, Dover, New York (1959), which is a translation of Vorlesungen über die Theorie der Wärmestrahlung, 5th edition, Barth, Leipzig (1923), the StefanBoltzmann equation is obtained as shown above. However, Planck did not evaluate the summation in Eq. 5 exactly. Instead, he simply evaluated 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 Mathematical Data, Macmillan, New York, Fourth Edition (1961)), the integral over x in Eq. 3 may be found; see Formula 860.33 on p. 231. When Eq. 5 is compared with the Stefan-Boltzmann law for a e black body q( ) = T 4 , then we get the Stefan-Boltzmann constant: 2 5K4 = 15 c 2 h3 (6) which interrelates key constants from several different fields of physics. b. Wien's displacement law First rewrite Eq. 16.3-7 in terms of x thus: 2 c 2 h e) ( qb = 5 1 e ch KT 5 KT x 5 = 2 c h x ch e 1 1 2 (7) We can then differentiate this with respect to x to get 5 e 4 dq(b ) x5 e x 2 KT 5x = 2 c h ch e x 1 e x 1 dx ( ) 2 (8) Then setting the derivative equal to zero gives the value of x at which e the maximum in the q(b ) curve occurs: 119 5 xmax e xmax e xmax 1 =0 (9) whence ( xmax = 5 1 e xmax ) (10) from which one can find, by trial and error, xmax = 4.9651... , or maxT = 0.2884 cm K . This is Wien's displacement law. 120 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 equation instead of the mathematical symbols (and omit the numerical factors) we get: (a) L2 M t Lt ML2 t 2 T (T ) = 1 L (1) (b) moles moles L 2 = Lt L3 t (2) (c) M L M L2 1 3 t = 3 t L L L (3) In each case, the dimensions on the left side are the same as the dimensions on the right side. 121 Note to p. 534 It is important to know how to simplify multicomponent relations to their corresponding binary equations. We illustrate this procedure by showing how to get the binary formula in Eq. (Q') from the multicomponent formula in Eq. (Q) in Table 17.7-1. In the sum in Eq. (Q), the indices and can take on only the values 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 M2 M + xA ( MB MA ) xB (1) Next use Eqs. (M) and (J) of the table, written for a binary system: A = MA ( xA MA + xB MB ) 2 ( x A MA + xB MB ) + x A ( MB MA ) ( x A ) (2) Within the bracket, the terms xA MA cancel each other, and the remaining terms may be combined, since xA + xB = 1 . We are then left with: A = + MA MB ( xA MA + xB MB ) 2 xA (3) which is just Eq. (Q'). 122 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 analogous term in Eq. 17.7-4, with a multiplying factor: ( ) D AB A = ( cM ) D AB MA MB M 2 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 ( v A v ) = ( c A MA ) v A ( A v A + B v B ) = ( c A MA ) B ( v A v B ) = (c A MA ) ( xB MB M ) ( v A v B ) ( = ( c A MA ) ( MB M ) v A ( xA v A + xB v B ) = cA ( v A 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. In the 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) of Table 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 ( v A v ) JA = c A ( v A v * ) (3a,b) which are given in Eqs. (E) and (I) of Table 17.8-1. It remains, then, to verify that jA = JA ( MA MB M ) (4) This may be done by rewriting Eq. 4 as 123 jA = JA ( M ) ( MA M ) ( MB M ) = JA ( c ) ( A x A ) ( B xB ) (5) Here Eqs. (G) and (O) of Table 17.7-1 have been used. The result in Eq. 5 may also be written thus: JA jA = A B cx A xB (6) This important equation is also given in Eq. 17B.3-1. 124 Note to p. 547 (i) Starting with the concentration profile for species A in Eq. 18.211, derive the subsequent results up through Eq. 18.2-16. First obtain the expression for xB,avg : xB,avg xB1 0 ( xB2 xB1 ) d 1 = 1 0 d 1 xB2 xB1 ) ( = ln ( xB2 xB1 ) 0 (1) ( ) where we have used the integral a x dx = a x ln a + C . Therefore xB,avg xB1 = ( xB2 xB1 ) 1 ln ( xB2 xB1 ) or xB,avg = xB2 xB1 ( xB )ln ln ( xB2 xB1 ) (2a,b) Hence the rate of evaporation of A at the gas-liquid interface is: NA z= z1 = cD AB dxB xB1 dz = z= z1 cD AB dxB xB1 d =0 d cD AB d ( xB xB1 ) = dz z2 z1 d (3) =0 Then, using the derivative da x dx = a x ln a , we get: NA z= z1 cD AB = z2 z1 x x cD AB xB2 B2 ln B2 = ln z2 z1 xB1 xB1 xB1 =0 (4) Then, multiplying the numerator and denominator by ( xB2 xB1 ) : NA z= z1 = cD AB ( xB2 xB1 ) cD AB ( xA1 xA2 ) = z2 z1 ( xB )ln z2 z1 (1 xA )ln (5) Next we obtain the solution for very small values of xA1 and xA2 : 125 NA z= z1 cD AB ( xA1 x A2 ) ln (1 xA2 ) (1 xA1 ) = z2 z1 (1 xA2 ) (1 xA1 ) = cD AB ln (1 xA2 ) ln (1 x A1 ) z2 z1 = cD AB 2 1 3 1 2 1 3 xA2 1 xA2 x + x + x + x A2 A1 A1 2 3 2 3 A1 z2 z1 ( ) ( ( ) ) ( ) 2 2 3 3 x A1 xA2 cD AB ( xA1 xA2 ) 1 xA1 xA 2 1 = 1+ 2 +3 + z2 z1 x x x x ( A1 A 2 ) ( A1 A2 ) = cD AB ( xA1 xA2 ) z2 z1 ( ) 2 2 1 + 1 ( xA1 + xA2 ) + 1 x A1 + x x + x + A1 A2 A2 2 3 (6) The Taylor expansion in Eq. C.2-3 has been used for expanding the logarithms in line 2. 126 Note to p. 547 (ii) a. The result in Eq. 18.2-11 may be written as follows: xB xB2 = xB1 xB1 = in which z z1 z2 z1 (1,2) We want to expand this result in a Taylor series in to get a result of the form xB = 1 ( ) + xB1 (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 so small that it can be neglected with respect to unity). We can then show 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 xB2 = xB1 xB1 =0 d xB2 + d xB1 ( 0) + =0 x = 1 + 1 ln B2 1 + xB1 (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) + 13 ( x 1) 2 3 1 4 ( x 1) 4 + (5) Therefore, 127 2 3 x xB x x 1 1 B2 B2 B2 = 1 + 1 1 + 1 + + xB1 3 xB1 2 xB1 xB1 (6) If xB2 is only slightly greater than xB1 (which would be the case if species A is present only in a small amount), then we need retain only the first term inside the bracket. The final result is then: x xB = 1 + B2 1 xB1 xB1 (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: d 2 xA dz 2 =0 (8) Since xA + xB = 1 , the same differential equation (with A replaced by B) is valid for species B. Furthermore, since z and are related by the linear expression given in Eq. (2) we may write: d 2 xB d 2 =0 (9) This equation may be integrated to give: x xB = 1 + B2 1 xB1 xB1 (10) in agreement with Eq. 7. 128 Note to p. 555 We wish to verify that Eq. 18.4-9 does satisfy Eq. 18.4-7 by substituting into the differential equation. Differentiate Eq. 18.4-9 with respect to , using Eq. C.5-10: d sinh (1 ) ( ) = d 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 expression in Eq. 18.4-9 into Eq. 18.4-7, it is seen that an equality is obtained. 129 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 common denominator N= 2 cosh + (V S ) cosh sinh 1 sinh cosh + (V S ) sinh (1) Next make use of Eq. C.5-5 to get N= 2 sinh + (V S ) cosh sinh sinh cosh + (V S ) sinh (2) Finally cancel the sinh appearing in the numerator and denominator to get sinh + (V S ) cosh N = cosh + (V S ) sinh (3) As (V S ) 0 , N tanh , and as (V S ) , N coth . 130 Note to p. 563 (i) Here we verify that Eq. 18.6-8 is a solution to the differential equation given in Eq. 18.6-2. Because of the definition of f just above Eq. 18.6-6, f = cA c A0 must satisfy Eq. 18.6-2. Therefore, we calculate the derivatives using the Leibniz formula given in §C.3: f 1 = 4 exp 3 z z 3 ( ) () = ( )( ) ( ) 1 4 3 exp 3 13 yz 4 3 ( a 9D AB ) 13 (1) f 1 1 f = 4 exp 3 = 4 exp 3 y z y y 3 3 ( ) () = 2 f y 2 = ( ) ( ) 1 4 3 exp 3 ( a 9D AB z ) ( ) ( ) 1 4 3 ( ) () exp 3 ( a 9D AB z ) 13 13 (2) ( 3 ) ( a 9D z ) 13 2 AB (3) Then, substituting these expressions into Eq. 18.6-2, we get: ( ) ay + 13 yz 4 3 ( a 9D AB ) 13 ( ) = D AB +3 2 ( a 9D AB z ) 23 If both sides are multiplied by ( 9D AB z a ) , and replace 13 y 2 ( a 9D AB z ) 23 (4) 2 by , we get ay 2 = 3D AB y 2 ( a 9D AB z ) 3z (5) which is an identity. This concludes the proof. 131 Note to p. 563 (ii) The concentration profiles are given by Eq. 18.6-8. cA = c A0 3 e d ( ) 4 3 (1) where = y ( a 9D AB z ) . We show how to get the next two equations from this result. Notice in Eq. 1 the bar over , which is used to make a distinction between the dummy variable of integration and the dimensionless distance. This distinction is vital when we apply the Leibniz rule (Eq. C.3-2) for differentiation of an integral to get the local molar flux at the wall is then: 13 N Ay y=0 = D AB c A y = y=0 D AB c A0 d 3 d e dy 43 d y=0 (2) () Note that only the third term in Eq. C.3-2 contributes to the derivative (i.e., the term involving the lower limit of the integral): N Ay y=0 13 13 D AB c A0 3 a D AB c A0 a e (3) = = 9D AB z 43 43 9D AB z =0 () ( ) () Then the total molar flow across the surface of width W and length L is given by the integral over that surface: W WA = 0 L 0 N Ay D c a dzdx = AB4 A0 y=0 9D () 3 D c a = AB4 A0 9D () 3 AB z2 3 13 W 2 3 L 0 AB 13 L W 0 z 1 3 dz 2D AB c A0 WL a = 9D L 4 43 AB 3 () 13 (4) 132 In the last step, Eq. C.4-4 can be used to replace agree with Eq. 18.6-10. 4 3 ( ) by ( ) to 4 3 7 3 133 Note to p. 565 Let us verify that Eq. 18.7-9 satisfies the differential equation in Eq. 18.7-6. To simplify the problem, it is a good idea to introduce dimensionless variables: r = ; R c = A ; c AR = k1aR 2 DA (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 d 2 d = 2 ; 2 d d = 1 sinh sinh (4,5) First we use Eq. 5 to evaluate the derivative (see also §C.5): d 1 sinh cosh = 2 + d sinh sinh (6) Multiplication by 2 then gives: 2 d cosh sinh = + sinh sinh d (7) Further differentiation gives: cosh cosh d 2 d 2 sinh = + + sinh sinh sinh d d (8) Division by 2 gives: 1 d 2 d 1 sinh = 2 2 sinh d d (9) 134 But the right side of the equation is just , according to Eq. 5. We have therefore shown that the differential equation is satisfied. 135 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 c + x = ( N ) + R t t (1) Next we use Eq. 19.1-12 to obtain c x + x t N cv * + ) R = ( N ) + R ( =1 (2) Then we go to Eq. 17.8-2 to find N = J + c v * , so that (2) becomes N x c x ( cv *) + ( c v *) = J + R x R t =1 ------------------------------ ( ) (3) Next we have to show that the two dashed-underlined terms on the left 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 a scalar 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 xA RB = xB RA xA RB (5) because for a binary system N in the upper limit of the sum is 2, there being just two components, A and B. 136 Note to p. 585 The equation to be solved is in Example 19.1-1 is dc A d 2 cA v0 = D AB k1c A dz dz 2 (1) which is of the form of Eq. C.1-7a. We know that the concentration of A will decrease with increasing distance from the porous plug, so we assume that it will have the form c A = exp ( az ) , where a is a positive constant. When this is substituted into the Eq. 1, we get (after canceling exp ( az ) from each term): v0 a = D AB a2 k1 or D AB a2 + v0 a k1 = 0 (2) This quadratic equation can be solved for a to get: a= v0 ± v02 + 4D AB k1 (3) 2D AB To insure that a is positive, we must choose the plus sign. Hence a is given by: ( ) v a = 1 + 1 + 4D AB k1 v02 0 2D AB (4) and the concentration profile is: ( ) v z c A = exp +1 1 + 4D AB k1 v02 0 2D AB (5) which is the result in Eq. 19.1-20. 137 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 and then perform mathematical operations on the new left side. First we use Eq. 3.5-4 to rewrite the new left side as: DĤ + ( q ) = Ĥ + vĤ + ( q ) Dt t ( ) (1) For a multicomponent mixture, the heat flux vector q may be written as described in Fn. (a) of Table 19.2-4, and then neglect the x contribution q( ) . Then Eq. 1 may be rewritten as N H N DĤ + ( q ) = c H + vĤ ( kT ) + j t =1 Dt =1 M ( ) (2) Here we have also used the relation Eq. 19.3-9 to rewrite the first term on the right side. Next we combine the second and fourth terms on the right side to get N DĤ N + ( q ) = c H + vcH + J H ( kT ) Dt t =1 =1 N N N = c H + v c H + J H ( kT ) (3) t =1 =1 =1 Then, since c v + J = N from Table 17.8-1, Eqs. (G) and (H), N N DĤ + ( q ) = c H + N H ( kT ) =1 t =1 Dt (4) Combining this with Eq. (E) then gives Eq. (H). 138 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 to get the second line of Eq. 19.3-4: ( ) e = Û + v v kT + 1 2 2 (a) (b) N H M =1 (c) j + pv + [ v ] (d) (e) (1) (f) We next move terms (c) and (d) to the left, and terms (b) and (f) to the right: e = kT + N H M =1 (c) (d) ( ) j + Û + pV̂ v + 12 v 2 v + [ v ] (a) (e) (b) (2) (f) where used has been made of the expression V̂ = 1 to get term (e). We next combine terms (a) and (e) to introduce the enthalpy: e = kT + N H M =1 (c) (d) j + Ĥv + 12 v 2 v + [ v ] (a+e) (b) (3) (f) Next we make use of the relation Ĥ = cH . 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 = H n = x H . Hence term (a+e) can be enthalpy per mole is H rewritten so that (3) becomes e = kT + (c) N N H j + M c H v + 12 v 2 v + [ v ] =1 =1 (d) (a+e) (b) (f) (4) Then we use Eq. H of Table17.7-1, Eq. S of Table 17.8-1, and Eq. P of Table 17.8-1 to rewrite parts of the summands in term (d+a+e) thus: 139 j j + v n + c v = = = N M M M (5) When this is substituted into (4) we get e = kT + (c) N H N + 12 v 2 v + [ v ] (6) =1 (d+a+e) (b) (f) Then, for systems in which terms (b) and (f) may be neglected, we finally arrive at Eq. 19.3-6. 140 Note to p. 591 The theorem of Euler (pronounced "Oiler") for homogeneous functions is used in order to get Eq. 19.3-9. We here give a proof of Euler's theorem. First we give a definition of a homogeneous function. A function of n variables is said to be "homogeneous of degree k" if f ( x1 , x2 , x3 , xn ) = k f ( x1 , x 2 , x3 ,xn ) (1) That is, if in the function f ( x1 , x 2 , 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 ) f ( x2 ) f ( xn ) + + = k k1 f ( x1 , x2 , x3 ,xn ) ( x1 ) ( x2 ) ( xn ) (2) where, on the left side, it is understood that by f we mean the function f ( x1 , x2 , x3 , xn ) . It is understood that the function f has continuous first partial derivatives. We now perform the differentiations on the left side to get f f f x1 + x2 + xn = k k1 f ( x1 , x2 , x3 ,xn ) ( x1 ) ( x2 ) ( xn ) (3) Next, we set = 1, to get x1 f f f + x2 + xn = kf ( x1 , x2 , x3 ,xn ) x1 x2 xn (4) which is Euler's theorem. Here, the functionality of f is exactly the same on both sides of the equation. Since the enthalpy is a homogeneous function of degree "1" Eq. 19.3-9 follows directly. 141 H =H n ( ) ,T ,p n n (5) or n H =H (6) This result is frequently used in discussions of mixtures. An example is the relation immediately after Eq. 17C.1-3, where, for a binary mixture nAVA + nBVB = V or, when the entire equation is divided by V, c AVA + cBVB = 1 Another example of Euler's equation is in going from Eq. 1 to Eq. 2 in the Note to p. 589. 142 Note to p. 615 We want to verify that Eqs. 20.1-16 and 17 are a solution to 20.19 and 10. We make use of §C.6 on the error function. First we find the first and second derivatives of X with respect to Z: X= 1 erf ( Z ) (1) 1 + erf dX 1 2 ( Z )2 = e dZ 1 + erf d2X dZ2 = (2) 1 2 ( Z )2 2 ( Z ) e 1 + erf (3) When the derivatives in Eqs. 1, 2, and 3 are substituted into Eq. 20.19, it is seen that the latter equation is satisfied. Next we substitute the first derivative from Eq. 2 into Eq. 20.110 to get: 1 2 ( Z )2 1 x = A0 e 2 1 xA0 1 + erf Z=0 2 x e = A0 1 xA0 (1 + erf ) (4) which agrees with Eq. 20.1-17. 143 Note to p. 622 Here we give a more detailed discussion of the development between 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 dg =0 d (1) where g = c A c A0 and = z ( t ) , which has the solution (by analogy with Example 4.1-1) cA z = 1 erf cA0 (2) with given by Eq. 20.1-70. To get Eq. 20.1-72, we differentiate the concentration profile: N Az0 = D AB c A z z=0 z = D AB c A0 erf 2 t z 4D AB 0 S ( t ) S (t ) dt z=0 (3) Use Eq. C.6-2 to differentiate the error function, and get N Az0 = +D AB c A0 = c A0 1 2 2 2 t 4D AB 0 S ( t ) S (t ) dt D AB 1 t t 0 S ( t ) t S ( t ) dt 2 1 2 (4) The total number of moles of A that have crossed the mass-transfer surface S(t) at time t is then given by: MA = S ( t ) c A0 0 1 erf ( z / ) dz = S ( t ) c A0 0 erfc ( z / )dz (5) 144 The complementary error function "erfc (...)" is defined in §C.6. We now insert this function into Eq. 5: MA = S ( t ) c A0 2 0 2 z / e d dz = S (t ) c A0 2 2 e d 0 0 ( z ) d (6) In the second form, the order of integrations over and z has been interchanged. Now the integral over z can be performed to give 2 MA = S ( t ) c A0 2 0 e d = S ( t ) cA0 1 2 2 (7) Inserting the expression for from Eq. 20.1-70, we then get MA = S ( t ) c A0 1 4D AB 0 S ( t ) S ( t ) dt = c A0 2 t 4D AB 0 S ( t ) t 2 dt (8) which is Eq. 20.1-73. Another expression can be obtained from integrating Eq. 20.172: MA = 0 S ( t ) N Az0 ( t )dt t 4D AB = c A0 S( t ) t 0 0 S ( t ) S ( t ) t dt 2 (9) dt It is relatively easy to show that Eqs. 8 and 9 are the same. We first note that the S ( t ) in the denominator can be removed from the integral, so that Eq. 9 can be rewritten as MA = c A0 4D AB S ( t ) t 0 2 0 S ( t ) t dt 2 (10) dt 145 Then we make use of the fact that the square root in the denominator contains no t and hence can be removed from the integral over t to give MA = c A0 4D AB = c A0 4D AB = c A0 4D AB 0 S ( t ) t 2 () dt 2 0 S t dt t 0 S ( t ) dt 2 t 0 S ( t ) t 2 dt (11) This is the same as Eq. 8. Proving that the two expressions for MA are the same is a stronger statement than that the two expressions give the same results for dMA dt . This example is also a good illustration of the importance of using a symbol for the dummy variable of integration that is different from that used as one of the limits in the integral. 146 Note to p. 626 We want to verify that the limiting solutions for slow and fast reactions in Eqs. 20.2-16 to 18 and 20.2-19 are correct. (a) Slow reactions We 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: 1 4 d 3 = + 3 + 2 Sc 3 d (1) Then, using the expansion in Eq. 20.2-16, we have ( ) = Sc 1 3 1 + a1 + a2 2 + ( ( ) (1 + 3a + ( 3a + 3a ) (2) ) + ) 2 = Sc 2 3 1 + 2a1 + a12 + 2a2 2 + (3) 3 = Sc 1 (4) 2 1 1 2 2 Substitution of these three expressions into Eq. 1 gives 1= ( ( ) ) ( ( ) ) 4 3a1 + 2 3a12 + 3a2 2 + + 1 + 3a1 + 3a12 + 3a2 2 + 3 + Sc1 3 1 + 2a1 + a12 + 2a2 2 + ( ( ) ) (5) We now equate terms in the same powers of : Zeroth power of : 1=1 First power of : 0 = 4a1 + 3a1 + Sc1 3 a1 = 17 Sc1 3 or Second power of : 0= or (6) 8 3 (a 3 1 ) (7) ( ) + 4a1 a2 2 + a13 + 4a1 a2 2 + 2Sc1 3 a1 2 ( ) 0 = 11 a12 + a2 + 2Sc1 3 a1 147 ( or 0 = 11 or 3 a2 = + 539 Sc 2 3 1 49 ) ( Sc 2 3 + a2 + 2Sc1 3 17 Sc1 3 ) (8) Thus we have obtained the first two coefficients in the expansion of Eq. 20.2-16, and hence the first few terms in the expression for as a function of . (b) Fast reactions When 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 1 4 = ( 3m ) K 3 3m + K 3 3m + K 2 2m+1 Sc 3 (10) The ratio of the last term to either of the first two terms is proportional to m+1 , which is a positive quantity. For large , the last term is the dominant term on the right side. Therefore Eq. 1 becomes 1 = 2 Sc (11) and therefore the solution to Eq. 1 for this case is = ( Sc ) 1 2 (12) in agreement with Eq. 20.2-19. 148 Note to p. 628 Example 20.2-2 is, for the most part easy to follow, except for the steps going from Eq. 20.2-30 to Eq. 20.2-34. Here we provide those missing steps. First, we assume will be a function of alone. Then the derivatives of with respect to x and y must be converted to derivatives with respect to , using Eq. 20.2-33: v 1 3 2 d 1 d d = = y x x 2 x = d 2 x 2 d d y y (1) d d v d = = = y d y x d 2 x d y x (2) 2 2 d v d 2 v d 2 = 2 = y d 2 x = 2 2 y x d 2 x d y (3) Then Eq.20.3-30 becomes d d 1 v0 ( x ) y d 2 = + dy v d 2 x v x 0 v d y v d 2 y 2 (4) Now multiplication by ( v ) ( y ) leads to 2 v ( x) y d v + 0 dy d x 0 v v (a) (b) (c) 2v x d 1 d 2 = d d 2 (d) (5) 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 x and . We shall show that term (c) contains only , and later (look ( ahead to Eq. 20.2-37) arguments are given that v0 ( x ) v may be set equal to a constant, K. )( 2v x ) 149 Let us, therefore, proceed to an analysis of term (c). We begin by replacing the integral over y by an integral over : y 2 x dy = d v x 0 v x 0 v = 2v x ( d ) v 0 2 x v x ( d ) 0 v (6) Then the derivative with respect to x is replaced by a derivative with respect to : ( d ) d + ) 2v x ( y dy = 2v x x 0 v = v 0 2v x v v 0 When Eq. 7 is multiplied by ( y x 0 v dy v 1 3 2 2 x d y d x v v d 0 2 2 (7) 2v x d d , we then get ) 2v x d d d = 0 v d + v d d d ( c1 ) ( c2 ) (8) Thus, term c 2 cancels term a, and the remaining term appears in the final equation for : (K ( x) 0 ) d 1 d 2 v d = d 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 calculated from 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- 150 43, which is obtained by setting d d = ( ) in Eq. 9 and then solving the first-order differential equation for ( ) . One obtains d + f = 0 d (10) the solution of which is ( d = = C1exp 0 fd d ) (11) A further integration gives ( ( ) ) ( , , K ) = C1 0 exp 0 f , K d d + C2 (12) Then the constants of integration are obtained from the boundary conditions in Eqs. 20.2-35 and 36, with the final result for the profiles ( ( , , K ) = exp ( ( ) ) f ( , K ) d ) d 0 exp 0 f , K d d 0 (13) 0 which is the same as Eq. 20.2-43. 151 Note to p. 692 We begin by formulating the problem as in Problem 12.1-4: Solid Cs 1 2 Cs = 2 Cs = finite at = 0 Cs = Cl at = 1 Cs = 1 at = 0 Liquid (1) Cs = NCl (5) =1 (2) (3) (4) Taking the Laplace transform of the problem we get Solid 1 d 2 dCs pCs 1 = 2 d d Liquid (6) Cs = finite at = 0 (7) at = 1 (8) Cs = Cl dCs d = NCl (9) =1 The solution to the homogeneous equation corresponding to Eq. 6 is the complementary function Cs,cf = K1 K cosh p + 2 sinh p (10) The particular integral of Eq. 6 is, by inspection Cs,pi = 1 p (11) The complete solution to Eq. 6 is then Cs = K1 K 1 cosh p + 2 sinh p + p (12) 152 The boundary condition in Eq. 7 requires that K1 = 0 . The boundary condition in Eq. 8 requires that Cs =1 = K 2 sinh p + 1 = Cl p (13) and combination of Eqs. 9 and 12 gives ( NCl = K 2 sinh p + p cosh p ) (14) Elimination of Cl between Eqs. 13 and 14 leads to an expression for the integration constant K 2 , and hence also to Cs : Cs = 1 N p p sinh p (15) p cosh p + ( N 1) sinh p Next, the Laplace transform of the total amount of A within the sphere is MA 4 R 3 c0 = = 1 0 Cs 1 N 2 3p p 2 d = 0 p 1 N 2 3p p 0 x sinh x p p cosh p + ( N 1) sinh p x sinh x p cosh p + ( N 1) p dx x cosh x sinh x 1 N = 2 3p p p cosh p + ( N 1) sinh p 0 = p 1 N p cosh p + ( N 1) sinh p Nsinh p 2 3p p p cosh p + ( N 1) p 1 N = 1 3p p 2 = dx Nsinh p p cosh p + ( N 1) p 1 N p cosh p + ( N 1) sinh p Nsinh p 2 3p p p cosh p + ( N 1) sinh p 153 1 N = 1 3p p 2 p cosh p + ( N 1) sinh p Nsinh p 1 N N2 = + 3p p 2 p 2 p coth p + ( N 1) (16) Now we take the inverse Laplace transform of the above; the transforms of the first two terms may be found in an elementary table of transforms: 1 2 1 = N + N L 2 4 R 3 c 0 3 p MA ( p coth p + ( N 1) 1 (17) ) To get the inverse transform of the last term, we can use the Heaviside partial fractions expansion theorem for repeated roots, which is: If f ( p ) = N ( p ) D ( p ) with D ( p ) = ( p a1 ) N ( p ) is a polynomial of degree less than i k , then f (t ) = n mk m1 ( p a2 )m ( p an )m 2 ( m ) 1 , and j n , ai ak for kl ( ak ) t mk l e akt k=1 l=1 ( mk l ) ! ( l 1) ! D ( p) d l1 N ( p ) with kl ( p ) = l1 and . D p = ( ) k mk dp Dk ( p ) ( p ak ) The contribution from the factor p 2 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 154 11 ( 0 ) = N2 p coth p + ( N 1) p=0 N2 = = N 1 + ( N 1) (19) d N2 12 (0 ) = dp p coth p + ( N 1) p=0 ( ( d dp ) p coth p + ( N 1) = N 2 2 p coth p + ( N 1) ( ) ) p=0 = 1 3 (20) The contributions in Eqs. 19 and 20 just exactly cancel the first two terms in Eq. 17. The contribution from the remaining factor in L1 { } is N ( p ) N ( ak ) + a e k =L = 3 4 R c0 D ( p ) k= 2 D' ( ak ) MA ( ) =N 2 1 k= 2 p2 ( 1 2 (21) e + ak ) ( ) p coth p + ( N 1) p= ak ------------------------------ p1 2 coth p 12 csch 2 p + 2p where the prime on D indicates differentiation with respect to p, and the ak are the zeros of the denominator in the braces in Eq. 17 (except for the double zero from p 2 , which we have already taken into account). This means that the dashed underlined term in Eq. 21 may be omitted. We know on physical grounds that the quantity M(t) must be a decreasing 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 N ( p ) 2 sinh 2 p 2 e k L == 2N k= 2 p 2 p 1 2 sinh p cosh p 1 D ( p ) p= k2 1 ( ) 155 = 2N 2 k= 2 = 2N 2 i 2 sin 2 k i k ( )( i sin 4 k k= 2 ( ) k 2 cos k ik ) sin 2 k e k 2 k3 ( k sin k cos k ) e k (22) Therefore, the total amount of A within the sphere is at any time t is: MA ( t ) 4 R 3 c0 3 ( = 6N 2 Bn exp n2D AB t / R 2 n=1 ) (23) where the n are determined from n cot n + ( N 1) = 0 (24) and the Bn are Bn = N 2 sin 2 n n3 ( n sin n cos 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 2 n cot n ) sin 2 n n2 ( 1 Bn 3 = 4 = n ( n sin n cos n ) n (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 Wärme156 übertragung, Springer-Verlag, Berlin, 3rd edition (1961), pp. 55-62. 157 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 driving force at the bottom of the page. When Eq. 24.1-2 is applied to a moving element of fluid, we write DŜ DV̂ N G D DÛ =T p + Dt Dt =1 M Dt Dt (1) in which N DÛ = ( q ) ( :v ) + ( j g ) Dt =1 Table 19.2-4, Eq. D (2) [Note footnote b to Eq. D!!] 1 D 1 DV̂ D 1 = = = + ( v ) Dt Dt 2 Dt D = ( j ) + r Dt Table 19.2-3, Eq. A (3) Table 19.2-3, Eq. B (4) Substitution of these expressions into Eq. 1 gives N p DŜ 1 = ( q ) ( :v ) + ( j g ) + ( v ) Dt T T =1 ( 1 N G ( j ) + r T =1 M ) N G 1 1 1 N G = ( q) ( j ) T (:v ) T M r T =1 M =1 1 N + ( j g ) T =1 158 N G 1 1 = q j q 2 T + T T =1 M j =1 N 1 G 1 1 1 N G g ( :v ) r M T T T T M =1 (5) Comparison of Eq. 5 with Eq. 24.1-1 gives the entropy flux vector and the entropy production rate (Eqs. 24.1-3 and 4): N G 1 s = q j T =1 M 1 N gS = q 2 T j T =1 (6) 1 G 1 1 1 N G g ( :v ) r T =1 M T M T T (7) h Next we rewrite Eq. 6 by replacing q by q( ) + N ( H =1 ) M j --that is, we subtract off the heat flux associated with the diffusion of the chemical species. This gives us then for the entropy flux (Eq. 24.1-5): 1 ( h) N S s = q j T =1 M (8) The entropy flux is now written as the sum of two terms: the first term is the entropy flux associated with heat flow, and the second term is that connected with diffusion of the chemical species. (For h more on the reasons for replacing q by q( ) + N ( H =1 ) M j , see S. R. de Groot and P. Mazur, Non-Equilibrium Thermodynamics, NorthHolland, Amsterdam (1962), pp. 24-25.) In order to rewrite the entropy production term, we make use of 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): 159 G p N 1 Ud + Vd n d = 0 T T =1 T (9) where n is the number of moles of species . Division by V and doing some elementary manipulations gives us the form of the Gibbs-Duhem equation that we need: N 1 G H 1 1 T + p =0 T M T 2 T M =1 =1 N (10) Next, in Eq. 7, we add four terms inside the bracket in the second term on the right side in such a way that the term is not changed: 1 gS = q 2 T T N j =1 H 1 1 G 1 p + 2 T T T M M T N H 1 1 1 g 2 T g + T T =1 M T 1 1 N G :v ( ) T M r T =1 (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: h 1 gS = q( ) 2 T T N j =1 H 1 1 G 1 p + 2 T T M T M T N 1 1 g g + T =1 T 1 1 N G ( :v ) r T T =1 M (12) or 160 ( ) h TgS = q( ) lnT j 1 1 N G cRTd T ( :v ) T M r =1 =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 GibbsDuhem equation, and the fourth and fifth terms clearly combine to sum to zero. The above development has led us to the expression for the generalized driving forces for diffusion: G cRTd = c T + c H lnT p g + T N g (14) =1 To get the alternative form for d given in the second line of Eq. 24.18, we follow the discussion of Ref. 5 on p. 766. The quantities G , H , and p are functions of the state of a fluid element, which may be described by the temperature T, the pressure p, 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: G c T = c G c G lnT T = c = c N 1 G =1 N 1 G =1 G G T + c p c G lnT T p x x + c x x c TS lnT + c V p c G lnT (15) 161 When Eq. 15 is combined with Eq. 14, it is seen that the lnT terms exactly cancel, and the p terms may be combined. Furthermore, when the relation dG = RTdlna is used, we get finally N 1 lna cRTd = c RT =1 lnx x + ( ) p g + T ,p,x N g =1 (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 fraction gradients, the pressure gradient, and the external forces. Equation 16 is in agreement with Eq. 7.8 of Ref. 5 at the bottom of p. 766, and also with Eq. 11.1-27 of Molecular Theory of Gases and Liquids, 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 be misleading. Here and elsewhere (Eqs. 24.2-8, 24.2-9, 24.2-10, 24.4-1, and 24.5-4) the abbreviated notation lna =1 lnx N 1 x T ,p lna (17) T ,p,x 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 of Transport Phenomena, Oxford University Press (1998), Eq. 11.6-2. Deen, however, states that his treatment is for dilute mixtures. Both of these authors, however, appear to have their summations going from 1 to N (instead of 1 to (N – 1). The discussion in Multicomponent Mass Transfer, by R. Taylor and R. Krishna, Wiley, New York (1993), also uses essentially the notation 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 from some notational differences, the same as Eq. 16 above). 162 In Advanced Transport Phenomena, by J. C. Slattery, Cambridge University Press (1999), p. 450, Eq. 8.4.3-4, gives a result that is consistent with Eq. 16 above, although the notation is considerably different from ours. There is also a discussion of multicomponent systems in Transport Phenomena Fundamentals, by J. L. Plawsky, pp. 66-69 (heat flux) and pp. 69-73 (mass fluxes); however, no derivations of the expressions are given. Thermodynamics of Irreversible Processes, by G. D. C. Kuiken, Wiley, New York (1994), has a discussion of the driving force d on p. 183. 163 Note to p. 767 Just as a check, we show that Fick's law for a binary system may be obatained by simplifying Eq. 24.2-3. This equation is, for a binary system with the two components labeled "A" and "B": jA = + A ( D AAd A + D ABd B ) (1) Use Eqs. (A) and (C) of Table 24.2-1 and the equation d A + d B = 0 (see line 2 on p. 767), to eliminate the generalized Fick diffusivities in favor of the Maxwell-Stefan diffusivities: B2 jA = + A DABd A A B DABd A x A xB x A xB (2) This may be rewritten, using the fact that A + B = 1 , to give jA = A B D d xA xB AB A (3) Next use the second line of Eq. 24.1-8, omitting the terms for thermal diffusion, pressure diffusion, and forced diffusion. This yields: jA = ln aA A B DAB ( xA ln aA ) = A B DAB x A xA xB xA xB ln xA (4) Then, to obtain the molar flux of species "A" we use Eq. 17B.3-1 to get the relation J*A = ( cxA xB A B ) jA . Finally, combine this equation with Eq. 4 and introduce the binary diffusivity D AB from Eqs. 24.3-2 and 4, to get: J*A = cD AB xA (5) This is the form of Fick's (first) law given in Eq. B of Table 17.8-2 for a binary system. 164 Note to p. 768 Here we show how to derive the generalized Maxwell-Stefan relations (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 + DT lnT , in order to avoid having to carry along the thermal diffusion term throughout the derivation. Then the generalized Fick equation, Eq. 24.2-3, for species may be written as N j# = D d =1 where j# = 0, (1) d = 0 , D = D , and D = 0 (see above Eq. 24.2-3). Next, write a similar equation for species and form the difference of the two equations; then multiply both sides of the equation by the quantity x x D to get x x j# j# x x = D D (D N =1 ) D d (2) Now we sum both sides over the index x x j# x x j# N D = D D D =1 ( )d (3) If the bracket quantity on the right side is set equal to , we see that we get N x x j# j# N d = d = D d = d (0) = d =1 =1 ( ) 165 x x j# j# = d D or (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, also lead to the generalized Maxwell-Stefan equation. If Eq. 5 is multiplied by 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 bracket expression 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 by defining 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 equation for = : x x D (B ) = ( ) (6) Next we perform some operations on the relation (D ) D + D = 0 or ( B ) D +D = 0 = 0: (7) 166 ( ) Then we multiply Eq. 6 by B1 and sum on to get: ( ) x x = B1 D (8) ( ) Multiplying the second form of Eq. 7 by B1 and summing on gives ( ) = D B1 (9) Then multiplying the reciprocal of Eq. 8 by Eq. 9, and replacing the index 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 ) = (adjB ) (det B ) 1 (11) where adjB is the matrix adjoint to B and det B is the determinant of the matrix B . The matrix adjB is the transpose of the matrix of cofactors (or "signed minors") of B . 167 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 a ternary system: D12 = x1 x2 D12 ( adjB1 )22 + D13 ( adjB1 )32 1 2 (adjB1 )22 + (adjB1 )32 (1) The 2 x 2 matrix ( B1 ) may be displayed thus: (B1 )22 (B1 ) = ( B1 )32 (B1 )23 D22 + D12 = (B1 )33 D32 + D12 D 23 + D13 D + D 33 (2) 13 The adjoint matrix is the transpose of the matrix of cofactors (a cofactor is a "signed minor"): ( adj B1 )22 (adj B1 ) = ( adj B1 )32 (adj B1 )23 (B1 )33 (B1 )23 = (adj B1 )33 (B1 )32 (B1 )22 D + D13 = 33 D32 D12 D 23 D13 D 22 + D12 (3) We are now ready to substitute into Eq. 24.2-7: D12 = x1 x2 D12 ( D33 + D13 ) + D13 (D32 D12 ) 1 2 ( D33 + D13 ) + (D32 D12 ) D12D33 + D13D32 D D + D + D 12 33 13 32 = x1 x2 1 2 = x1 x2 +D12D33 D13D 23 1 2 D12 + D 33 D13 D 23 (4) In the last step, we have made use of the symmetry of the D . 168