THE UNIVERSITY OF MICHIGAN COLLEGE OF ENGINEERING Department of Mechanical Engineering Heat Transfer Laboratory Technical Report No. 4 BOUNDARY LAYER ANALYSIS OF TWO-PHASE (LIQUID-GAS) FLOW OVER A CIRCULAR CYLINDER AND OSCILLATING FLAT PLATE M'. E'.: Goldste' in WQn-'Je i Yatig J,' A.''ClaTirk ORA Pruoject 05065 under contract witho AERONAUTICAL RESEARCH LABORATORY, OAR AERONAUTICAL SYSTEMS DIVISION AIR FORCE SYSTEMS COMMAND CONTRACT NO, AF 33(657)-8368 WRIGHT-PATTERSON AIR FORCE BASE, OHIO administered through: OFFICE OF RESEARCH ADMINISTRATION ANN ARBOR August 1965

This report was also a dissertation submitted by the first author in partial fulfillment of the rerequirements for the degree of Doctor of Philosophy in The University of Michigan, 1965o

ACKNOWLEDGMENTS The authors wish to express their appreciation to Professors Vedat S. Arpaci and Stanley Jacobs, for their interest and cooperation in serving as members of the Doctoral Thesis Advisory Committee,

TABLE OF CONTENTS Page LIST OF ILLUSTRATIONS vi NOMENCLATURE xii ABSTRACT xxiii CHAPTER I. INTRODUCTION 1 II. ANALYSIS 5 A. General Considerations 5 B. Gas/Liquid Flow Over an Infinite Circular Cylinder 6 1. Formulation 6 2. Solution 47 C. Semi-Infinite Oscillating Flat Plate in a Streaming Gas/Liquid Flow with Oscillations Parallel to the Flow 69 1. Formulation 69 2. Solution 86 III. NUMERICAL PROCEDURES 118 IV. DISCUSSION AND RESULTS 25 A. Convergence of Solutions 125 B. Numerical Results 141 1. The Circular Cylinder 141 2. Oscillating Flat Plate 186 V. CONCLUSION 204 A, Cylinder 204 B. Oscillating Flat Plate 205 BIBLIOGRAPHY 207 iii

TABLE OF CONTENTS (Continued) Appendix Page I. CALCULATIONS FOR DETERMINING THE MINIMUM VALUES OF Xe AND GAS REYNOLDS NUMBERS AT WHICH DROP TRAJECTORIES ARE STRAIGHT LINES 211 IIo ESTIMATION OF PARAMETERS FOR AIR/WATER MIXTURES 215 IIIo REDUCTIONS OF THE FIRST OF EQUATIONS (23) 218 IV. REDUCTION OF THE SECOND OF EQUATIONS (23) 222 Vo REDUCTION OF THE FIRST OF EQUATIONS (24) 226 VIo REDUCTION OF THE THIRD OF EQUATIONS (24) 231 VII. REDUCTION OF Tw/V i//2PU = E25*2 235 VIII. REDUCTION OF u/Uo = (25*) - 2355 anT IX. REDUCTION OF THE FIRST OF EQUATIONS (84) 237 Xo REDUCTION OF THE ENERGY EQUATION FOR SMALL E TO A SERIES OF ORDINARY DIFFERENTIAL EQUATIONS 239 XI. REDUCTION OF THE FIRST OR EQUATIONS (84) 242 XIIo REDUCTION OF THE SECOND OF EQUATIONS (85) 244 XIII, SIMPLIFICATION OF FIRST- AND SECOND-ORDER MOMENTUM EQUATIONS 247 XIV, SIMPLIFICATION OF THE FIRST- AND SECOND-ORDER MOMENTUM BOUNDARY CONDITIONS 250 XVo REDUCTION OF THE SECOND OF EQUATIONS (84) 252 XVI. REDUCTION OF THE THIRD OF CONDITIONS (85) 254 XVII. SIMPLIFICATION OF THE FIRST- AND SECOND-ORDER ENERGY EQUATIONS 256 XVIII SIMPLIFICATION OF FIRST- AND SECOND-ORDER ENERGY BOUNDARY EQUATIONS 258 iv

TABLE OF CONTENTS (Concluded) Appendix Page XIX. REDUCTION OF FIRST-ORDER MOMENTUM PROBLEM TO A SYSTEM OF DIFFERENTIAL EQUATIONS 259 XX. REDUCTION OF FIRST-ORDER ENERGY PROBLEM TO A SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS 263 XXI. REDUCTION OF THE SECOND-ORDER MOMENTUM PROBLEM TO A SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS 266 XXII. REDUCTION OF THE SECOND-ORDER ENERGY PROBLEM TO A SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS 276 XXIII. REDUCTION OF THE FORMULAS FOR VELOCITY, SHEAR STRESS, AND NUSSELT NUMBER 284 XXIV. CALCULATION FOR EXPRESSING THE VELOCITY, FALL SHEAR STRESS, AND NUSSLET NUMBER EXPLICITLY IN TERMS OF THE SOLUTIONS TO THE ORDINARY DIFFERENTIAL EQUATION 287 XXV. COMPUTER PROGRAMS INVOLVING INTEGRATION OF DIFFERENTIAL EQUATIONS 296 XXVI. COMPUTER PROGRAM FOR CALCULATING FILM THICKNESS, LOCAL SKIN FRICTION, AND THE TERMS OF THE EXPANSIONS (52) 329 XXVII. REDUCTION OF EXPERIMENTAL DATA FROM REFERENCE (1) 335 V

LIST OF ILLUSTRATIONS Table Page I. Values of, J and Vp 180 II. Drag Coefficients for a Sphere (28) 307 Figure 1. Coordinates and control volume for liquid film on cylinder. 8 2. Coordinates for dynamical equations of liquid droplet. 31 3, Sketch for linear drop density. 32 4. Control volume for the determination of physical quantities at the surface of liquid film from the trajectory data. 44 5. Coordinates and control volume for liquid layer on flat plate. 71 6. Plot of AIn+l/AIn versus Rn/Rn_l for iterative procedure. 1,22 7-a. Convergence of function f5 with respect to number of subdivisions used for numerical calculations in cylinder problem. 126 7-b. Convergence of function F4 with respect to number of subdivisions used for numerical calculations in cylinder problem. 127 7-c. Convergence of series for liquid film thickness with respect to the number of terms retained in cylinder problem. 128 7-d. Convergence of series for liquid velocity with respect to the number of terms retained in cylinder problem. 129 7-e. Convergence of series for Nusselt number with respect to the number of terms retained in cylinder problem. 130 vi

I;ST OF ILLUSTRATIONS (Continued) Figure Page 8-a. Convergence of solution for ydf with respect to size of increment used in integrating drop trajectories. 132 8-b. Convergence of solution for pd with respect to size of increment used in integrating drop trajectories. 133 9-a. Convergence of function F6 with respect to number of subdivisions for cylinder problem with E<.l, 154 9-b. Convergence of function F6' with respect to number of subdivisions for cylinder problem with E<.1. 135 9-c. Convergence of solutions with respect to number of terms retained in expansion for cylinder problem with E<.l. 136 10-a. Convergence of function f15 with respect to number of subdivisions used for numerical calculations in flat plate problem. 157 10-b. Convergence of function f25 with respect to number of subdivisions used for numerical calculations in flat plate problem. 158 10-c. Convergence of function F24 with respect to number of subdivisions used for numerical calculations in flat plate problem. 139 10-d. Convergence of function F24 with respect to number of subdivisions used for numerical calculations in flat plate problem. 140 11-a. Local liquid film thickness for cylinder with E2 >.01. 142 11-b. Local liquid film thickness for cylinder with E2 >.01. 143 11-c. Local liquid film thickness for cylinder with E < 0.1, rd/Ro =.004 and 2RoUoo/vg = 103. 144 vii

LIST OF ILLUSTRATIONS (Continued) Figure Page 11-d. Local liquid film thickness for cylinder with E < 0.1, rd/Ro =.004 and 2RoUo/vg = 5x103. 145 ll-eo Local liquid film thickness for cylinder with E < 0.1, rd/Ro =.004 and 2RoUo/vg = 104. 146 11-fo Local liquid film thickness for cylinder with E < 0.1, rd/Ro =.001 and 2RoUo/vg = 105. 147 11-go Local liquid film thickness for cylinder with E < 0.1, rd/Ro =.001 and 2RoUo/vg = 104 148 11-h. Local liquid film thickness for cylinder with E < 0.1, rd/Ro =.001 and 2RoUoo/Vg = 5x103. 149 11-i. Local liquid film thickness for cylinder with E < 0o1, rd/Ro =.0004 and 2RoUo/vg = 105. 150 12, Velocity profile in liquid film at x/Ro = -5 for cylinder with E2 >.01. 151 13-a1. Local wall shear stress for cylinder with E2 >.01. 152 13-b. Local wall shear stress for cylinder with E < 0.1, rd/Ro = ~004 and 2RoUj/vg = 103. 155 13-c. Local wall shear stress for cylinder with E < 0.1, rd/Ro =.004 and 2RoUo/Vg = 5x103 154 13-do Local wall shear stress for cylinder with E < 0.1, rd/Ro = o004 and 2RoUO/vg = 10. 155 13-e. Local wall shear stress for cylinder with E < 0.1, rd/Ro =.001 and 2RoUo/vg = 5x102. 156 13-fo Local wall shear stress for cylinder with E < 0.1, rd/Ro =.001 and 2RoUJ/Vg = 104 157 1.3-g Local wall shear stress for cylinder with E < 0o1, rd/RO =.001 and 2RoU /vg = 105 158 15-ho Local wall shear stress for cylinder with E < 0.1, rd/RO =.0004 and 2RoHoo/vg = 10. 159 viii

LIST OF ILLUSTRATIONS (Continued) Figure Page 14-a. Local Nusselt number for cylinder with Pr = 5~0 and E > 0.01. 160 14-b. Local Nusselt number for cylinder with Pr = 13.4 and E > 0.01. 161 14-c. Local Nusselt number for cylinder with Pr = 5.0, E < 0.1, rd/Ro =.004 and 2RoUo/vg = 5x103 162 14-d. Local Nusselt number for cylinder with Pr = 10, E < 0.1, rd/Ro =.004 and 2RoU,/vg = 5x103. 163 14-e. Local Nusselt number for cylinder with Pr = 5, E < 0.1, rd/Ro =.004 and 2RoUo/Vg = 104 164 14-f. Local Nusselt number for cylinder with Pr = 10, E < 0.1, rd/Ro =.004 and 2RoUoo/vg = 104. 165 14-g. Local Nusselt number for cylinder with Pr = 10, E <.1, rd/Ro =.001 and 2RoUo/vg = 5x103. 166 14-h. Local Nusselt number for cylinder with Pr = 10, E <.1, rd/R =.001 and 2RoUoo/vg 104 167 14-io Local Nusselt number for cylinder with Pr = 10, E <.1, rd/Ro =.001 and 2RoUoo/Vg = lo. 168 15- Peak values of local Nusselt number for cylinder with Pr = 10. 169 16-ao Drop trajectories around cylinder with rd/Ro =.01 and 2RolU/Vg = 103. 170 16-b. Drop trajectories around cylinder with rd/Ro =.004 and 2RUoo/vg = 103 171 16-c. Drop trajectories around cylinder with rd/Ro =.004 and 2RoUo/vg = 5x103 172 16-d. Drop trajectories around cylinder with rd/Ro = o004 and 2RoUl/Vg = 104. 173 16-eo Drop trajectories around cylinder with rd/Ro =.004 and 2RoUoo/Vg = 25850 174 ix

LIST OF ILLUSTRATIONS (Continued) Figure Page 16-fo Drop trajectories around cylinder with rd/Ro =.001 and 2RoU,/Vg = 5x103. 175 l6-go Drop trajectories around cylinder with rd/Ro = 001 and 2Rojvg vg 104 176 16-h. Drop trajectories around cylinder with rd/Ro =.001 and 2RoW,/Vg 10 lO 177 16-io Drop trajectories around cylinder with rd/Ro =.0004 and 2Ro U/vg = 105o 178 17. Comparison of calculated Nusselt number with experimental data for cylinder. 185 18-ao Local liquid fiLm thickness versus volume fraction for steady flat plate. 187 18-bo Replot of Figure 18-a in terms of oscillating variableso 188 19. Local wall shear stress versus volume fraction for steady flat plate. 189 20o Local Nusselt number versus v'olume fraction for steady flat plate 190 21-a. local amplitude of liquid film thickness on oscillating flat plate 1.91 21Nbo Local phase lag of liquid, film thickness on oscillating flat plate. 192 22-ao Local amplitude of wall shear stress on oscillating flat plate. 193 22-bo Local phase angle of wall shear stress on oscillating flat plate. 194 235-a Local amplitude of Nusselt number on oscillating flat plate with Pr = 5.0 and 10Oo, 195 25-bo Local phase angle of Nusselt number on oscillating flat plate with Pr = 50 196 x

LIST OF ILLUSTRATIONS (Concluded) Figure Page 23-c. Local phase angle of Nusselt number on oscillating flat plate with Pr = 10. 197 24. Permanent alteration in liquid film thickness due to oscillation of flat plate. 198 25. Permanent alteration in wall shear stress due to oscillation of flat plate. 199 26-a. Permanent alteration in local Nusselt number due to oscillation of flat plate for Pr = 5. 200 26-b. Permanent alteration in local Nusselt number due to oscillation of flat plate for Pr = 10. 201 27. Minimum values of E2 and gas Reynolds number for a straight drop trajectory. 214 28. Values of drag coefficients for spheres used in calculating drop trajectories. 306 xi

NOMENCLATURE Notation for Cylinder Problem Latin Letters a = constant appearing in the expansion for 6* 0 An = function of the f2n-'s and their derivatives and the b2n's appearing in the ordinary differential equations for the momentum problem, n = 1,3,5,... b2n = constants appearing in the expansion for 6*; n = 0,1,2... Bn = function of F2n's and b2n's appearing in the ordinary differential equation for the energy problem; n = 0,2,4 CD = drag coefficient for liquid drops Cp = specific heat at constant pressure of liquid Cp = specific heat at constant pressure of gas D2 = partial differential operator in momentum equations 2 D D, rewritten in terms of x,y xy -2 2 D = dimensionless version of D D2 = dimensionless version of D 2 2 E = 2Xe RoU/v f2n-i = n = 1,2,..., function of r in expansion for \* F2n = n = 0,1,2..., function of q in expansion for 9* Gn = n = 0,2,4,..., function of f2n-l's and F's appearing in the differential equations for Fn Hn = n = 1,3,5,..., function of b2n s f2 _ s and their derivatives appearing in differential equations for f'S 2n-1 xii

NOMENCLATURE (Continued) i = unit vector in xo direction = unit vector in yo direction Xe * = -- V* Xe r 00 k = thermal conductivity of the liquid kg = thermal conductivity of the gas Kn = linear operator for energy equations L = linear operator for momentum equations n x Xe ~ X r XJ= _ - Vr d x n = unit outward drawn normal to the cylinder N' = number of liquid drops lying along a fluid line whose length is equal to the radius of the cylinder No = number of drops per unit volume Nu = Nusselt number of the liquid Nug = Nusselt number of the gas P = pressure in the liquid Pg = pressure in the gas 2 P = P/l./2 p U0 P = P/l/2 p UX P = Pg/ 1/2 PgU Pr Prandtl Number of liquid Prg Prandtl Number of gas qw = heat flux at cylinder wall xiii

NOMENCLATURE (Continued) qn = terms of expansion for Nu/ -] gqg = heat flux into the gas at boundary of liquid film r = radial coordinate rd = radius of drop rd = position vector of drop R = radius of cylinder 0 Re = 2RoUo/v s = surface of liquid film t = time T = temperature of liquid Tw = temperature of cylinder wall T = temperature of flow system at infinity t* = t U/Ro u = vp; velocity of liquid in film in asimuthal direction * un = terms in expansion for u/UO ug = xo component of gas velocity & u ug = ug/U UO = velocity of liquid and gas at infinity V(> = velocity of liquid in film in asimuthal direction Vr = velocity of liquid in film in radial direction Vg = y component of gas velocity v^g = Vg/U V'P = asimuthal component of drop velocity at edge of liquid film xiv

NOMENCLATURE (Continued) V = radial component of drop velocity at edge of liquid film 5V = vc/u, v -v 00 Vr Vr/U V = V + V xo yo Vxo= drop velocity in xo direction VY = drop velocity in yo direction X0 = O V = V /U yo Vyo/U + A A V = i U + j Vg v* -= vg/u V* = V/UO0 Vyf = drop velocity in yo direction at edge of liquid film Vxof = drop velocity in xo direction at edge of liquid film V0of Vyof/ V*of = VXof/U AV* = > -(V V - u)2 + (Vy - vg) xo g yp g x = c: distance along surface of cylinder divided by Ro xZ = x x = x xo = coordinate in direction of flow at infinity xd = x position of drop ~"o x xo/Ro xv

NOMENCLATURE (Continued) Xd = Xd/Ro Xd = xo position of drop at surface of liquid film f Xdf = Xdf/R Xe = volume fraction of liquid outside of film Xe = volume fraction of liquid at infinity 00 y = coordinate in radial direction: r 1 Ro - = y/2Xe y = y/2 01 0 y = coordinate perpendicular to flow at infinity Yd = Yo coordinate of drop Ydoo = y coordinate of drop at infinity Ydf = Yo coordinate of drop at surface of film ao *= Yd/Ro y~Ad = ydo/R Ydoo = Ydo/Ro Ydf = Ydf/Ro Z(x) = dimensionless gas shear stress term Greek Letters a2n = terms in expansion for 6J P2n = terms in expansion for & y = cos-. - [1- 24(Pgp ) ( sin 2x ) E2 ff 7n = n = 1,3,5,..., terms in series expansion of Z(x) xvi

NOMENCLATURE (Continued) 6 = thickness of liquid film 6 = 6(e)/2Ro Xeo 6 = 6(x)/2R0 XeOo A 6 12R U\ 2 6 = b/Ro = function arising from integration of stream function n = y/6 6e = angle between tangent to cylinder and tangent to surface of liquid film i* = T - Too Tw- Tc = viscosity of liqued g9 = viscosity of gas v = p/p; kinematic viscosity of liquid Vg = ktg/pg; kinematic viscosity of gas S -= x/Ro p = density of liquid pg = density of gas Trr = radial component of shear stress in liquid film Trcp = azimuthal and radial components of shear stress in liquid film qTrp = azimuthal component of shear stress in liquid film Tg = shear stress exerted by gas on surface of film Tw = shear stress exerted by liquid on surface of cylinder xvii

NOMENCLATURE (Continued) 2UIO R T = n = 1,3,5,..., terms in expansion for T 2 /l/2pU nw v 00 Tef {4 Z(x) -xe v vy eff = E p f (v V*C PJ cp = azimuthal coordinate da = azimuthal coordinate of drop at surface of film'Vt = stream function in liquid film ~* = /U RoXe oo o Xeoo = V/UOOROXeo'V: /UORoXeO Special notation Vo = ^1 - + a A^ A 0 vd = 1Xa + jaV* R= R Vo * d= R Vd Subscripts d = drop g = gas i = integer j = integer 0o - upstream conditions m = integer n = integer xviii

NOMENCLATURE (Continued) Notation for Oscillating Flat Plate Latin Letters a,ain,a2n, a = coefficients in expansion for 6* (constants) A = area of control surface A = area of control surface CoS. c.s. Bn)Bn = terms appearing in momentum boundary conditions function of f's, an mn mn CnCn = terms appearing in energy boundary conditions function of f's F's m,n m, n cp = specific heat at constant pressure of liquid cg = specific heat at constant pressure of gas elnene2n = terms appearing in expansion for T*; n = 0,1,2,.o. E,En = nonhomogeneous terms appearing in differential equations for F,F's 2n 2n fo fln'f2n'2n, = coefficients in expansions for ~* (functions of ~ ) FoF1inF2nFn = coefficients in expansions for e* (functions of T ) gln, gan,92n = terms of expansion of un GlnG2n,'2n = terms of expansion for gn hnhn = terms appearing in expansion for u* HnHn = nonhomogeneous terms appearing in the differential equations for f2n'I2n s k = thermal conductivity of liquid kg = thermal conductivity of gas xix

NOMENCLATURE (Continued) Kn = differential operator for energy equations Ln = differential operator for momentum equations Nux = Nusselt number Nu = Nusselt number of gas Pr = Prandtl number of liquid Pr = Prandtl number of gas g PPnn = nonhomogeneous terms appearing in momentum equations qw = heat flux at plate q = heat flux into gas at edge of film g q* = coefficient of E in expansion for local-Nusselt number n rnrn = terms appearing in expansion for T* n R nRn = terms appearing in expansion for q* n n n s' = upper surface of liquid film s" = vertical boundary of control surface t = time T = temperature Tw = temperature of plate TX = temperature at infinity u = x-component of velocity in film u* = coefficient of e in expansion for u/UO n U = U ( + - sing.0t) 2 U~ = velocity of liquid drops and gas at infinity xx

NOMENCLATURE (Continued) V = y-component of velocity in film (x) V x = x-component of velocity of edge of liquid film (y) V,( = y-component of velocity of edge of liquid film O.S V- = normal component of velocity of edge of liquid film c oS x1 = x-distance in x-y coordinates x = coordinate direction along plate in coordinate system fixed to plate Xe = volume fraction of liquid in free stream y = coordinate direction perpendicular to plate in coordinate system fixed to plate Z1,Z2 = arbitrary complex quantities reek Letters aCLn = arbitrary complex quantities with cn imaginary for n = 0,4,8,..., an real for n = 2,6,... an = 0 for n = 1,3,5,... 6 = thickness of liquid film 6 = 5 1I n * = bn = 0,1,2,.., coefficient of e in expansion for 5* n E — = dimensionless amplitude of oscillation ~n ~ = y/6 * _= (T - T,)/(Tw - To) On = coefficient of En in expansion for 0* X = equal to ~ 1 = viscosity of liquid xxi

NOMENCLATURE (Concluded) l~g ~ = viscosity of gas v = kinematic viscosity of liquid Vg = kinematic viscosity of gas ea= I COX' 1 eUoo p = density of liquid p = density of gas g T = uwt T = component of liquid stress tensor xy T = shear stress exerted by gas at edge of film g T = shear stress at plate v* = coefficient of e in expansion for Tw = slope of liquid film and phase angle = stream function,* = V~ CD UOcXe X e Jr'f ~ = coefficient of e in expansion for'* n XD = frequency of oscillation of plate Subscripts avg. = time average value g = gas w = surface of plate n = integer steady = steady state value = upstream conditions xxii

ABSTRACT An analysis is carried out of a two phase (gas/liquid) flow over a circular cylinder and over an oscillating flat plate. The flow is assumed two dimensional and gravity, vaporization of the liquid phase, and compressibility are ignored. The liquid is assumed to be in the form of small drops far upstream from the body. The liquid film which forms on the surface of the body due to drop impingement is analyzed extensively. In the case of the cylinder the analysis is started from the full incompressible Navier-Stokes and Energy equations in the film which are simplified by using dimensional arguments. The order of magnitude of the physical properties of the fluids is taken to be those of air/water mixtures. The analysis is carried out for two ranges of a parameter appearing in the problem. For low values of the parameter the liquid drop trajectories deviate appreciably from straight lines and are obtained numerically. For the flat plate the boundary layer approximations are assumed to hold a priori in the film. The plate is taken to be oscillating sinusoidally in its plane with small amplitude and small frequency; and the drops are assumed to move in straight line s After appropriate changes of variable the solutions to the governing equations and boundary conditions are carried out by a series expansion technique which results in a series of ordinary differential equations which have been solved numerically on a 7090 computer. The solutions are used to calculate velocity and temperature profiles in the film and also such physical quantities as local film thickness, local Nusselt Number, and local skin friction. The analysis shows that in general there is a significant increase in heat transfer as well as skin friction over what would be obtained from a single component gas flow. In the case of the flat plate only a very small permanent change in the heat transfer was found due to the oscillationso In the case of the cylinder a peaking in the heat transfer, film thickness and skin friction was found with respect to the parameter E2 (the product of the volume fraction of the liquid in the free stream, squared, and the diameter Reynolds number based on liquid properties)o The peaking occurs at a value of this parameter of about unity. The theoretical prediction compares favorably with the experimental results of Acrivos et al. (ARL-64-116). xxiiif

CHAPTER I INTRODUCTION For some years now there has been a growing interest in the artificial creation of two phase flows for the purpose of significantly increasing the attainable level of surface heat transfer rates from solid bodies. Up until recently much of the interest has come from the expanding nuclear power field with much emphasis on systems in which solid particles, such as graphite dust, have been added to a coolent flow. A large number of papers and reports have also appeared dealing with two-phase (gas-liquid) hydrodynamics and heat transfer phenomena. Among them are a large number on forced convective flows of steam/ water mixtures. Data have been presented for channels of circular5'' 5 rectangular and annular3'5 cross-section with the flow 26 8,15 vertically upwards, downwards and horizontal over a wide range of values of steam quality, mass velocity and heat flux. In ad25 dition details are also given of investigations with refrigerents5 11 22 l4 and with two-component systems1 both with and without appreciable vaporization of the liquid phase. A number of impirical correlations have been suggested56 26'27 and some analysis has been carried out.7 An analysis has also been carried out for two-phase solid/liquid flowso 1

2 However due to the fact that relatively recent Russian liquid injection tests in flow across a tube bank have yielded increases in the measured heat transfer rates of an order of magnitude9 Research programs have currently been undertaken to investigate the heat transfer characteristics of external gas/liquid flow systems. Tests have been conducted at the Marquandt Corporation on a system which consisted of an air stream containing water droplets flowing over a heated vertical cylinder. An intensification of the heat transfer coefficient from 2.5 to nine times was noted. An exploratory analytic investigation was carried out by Tifford on the boundary layer characteristics of gas/liquid spray systems, with particular emphasis on heat transfer 29 characteristics. Experiments on single component systems have shown that oscillations of the system may result in significantly increased heat transfer performance. Analytical two dimensional studies2 2l24 however have not confirmed these results and it is believed that this disagreement between analysis and experiment may be due to the fact that the increased heat transfer is due to three dimensional effects. Nevertheless it may be possible that the two dimensional effects of oscillations may be augmented considerably by the increased mass which results in a two component system. In any event the use of two-phase heat transfer systems with oscillations is strongly encouraged by the following two important factors:

5 1. For a given wall temperature, which is often limited by the physical properties of material, a higher heat-transfer rate can be obtained. 2. For a given heat-transfer surface area, which is often limited by the space and other considerations, a large increase in heat-transfer rate can be obtainedo It is anticipated that a detailed analytical study would provide significant information on the dependence of the augmentative effects upon the choice of operating conditions for such a gas/liquid heat transfer system and would aid in the understanding of the physical mechanisms involved, The present study has therefore been undertaken in order to make some progress toward developing such detailed analytical investigations. Since however the physical situation is quite complex and changes considerably with the geometry and operating conditions the study was limited to two specific systems each in a certain range of operating conditions. Thus the present work is to study the momentum and heat transfer in a two-component (gas/liquid) steady flow over a circular cylinder and over a flat plate oscillating sinusoidally in its plane. Analytical results are obtained through the use of the perturbation technique. These results include velocity and temperature profiles, local friction factors, and heat transfer coefficientso

4 Numerical computations are performed by means of the Runge-Cutta method for a gas/liquid system which corresponds to air-water mixturee For low values of Reynolds Numbers and small drop sizes it has been found that the liquid drop trajectories deviate appreciably from straight 50 lines. Analytical and numerical studies have been carried out to determine drop trajectories around bodies in connection with icing problems by several investigators. Tribus3 has calculated trajectories about cylinders and airfoils on a differential analyzer. Langmuir19 has considered the cylinder analytically, and Bergran4 and Guibert et al.17 have studied airfoils numerically. It was disclosed that the presence of the liquid in the gas stream caused a significant increase in both the heat transfer and friction. The oscillations (in the case of the flat plate) caused only very small time average changes. Theoretical predictions on heat transfer and friction for the steady two phase flow over a circular cylinder agree qualitatively with the qualitative results of Tifford.29

CHAPTER II ANALYSIS A. General Considerations The physical system to be studied consists of a two-phase (gas/liquid) flow over a heated surface. Two cases are to be treated: A steady flow over a circular cylinder and a steady flow over a flat plate oscillating in its own plane. The experiments conducted at Marquandt Corporation indicate that when a streaming gas containing liquid drops impinges on a solid body a liquid film is formed on the surface of the body and is caused to flow in the down stream direction. The following assumptions are imposed on the solution: (1) The liquid drops contained in the streaming gas are uniformly distributed in the gas and move with the velocity of the gas far upstream. (2) The flow is taken to be two dimensional and laminar. (3) Due to inertial effects some of the drops in the free stream splash on the surface of the liquid film. The incoming drops contribute their mass, momentum and heat content to the film. However, the unsteady effects due to random drop impingement are negligible. (4) There are a sufficient number of drops in the free stream so that only the time average effect of drop impingement need be considered. 5

6 (5) It is disclosed, from Marquadt's experiments that waves formed on the surface of the liquid layer and that under certain conditions splashing occured at this surface. These latter two effects are also neglected, (6) On top of the liquid film a gas boundary layer forms which joins the flow in the region of the body to the external flow field. The usual boundary layer assumptions apply in this region. (7) Surface tension effects on the surface of the liquid layer are neglected. (8) The effects of compressibility and heat generated by dissipation can be ignored. The effects of gravity are also neglected. (9) The ratio of the gas/liquid properties will be taken to be of the order of those of air and water. (10) All fluid properties will be taken as constant. (e.g., the liquid and the gas are separately assumed incompressible) (11) No appreciable vaporization occurs. B. Gas/Liquid Flow Over an Infinite Circular Cylinder 1. FORMULATION With these assumptions, the following governing differential equations and boundary conditions are obtained for the two component (gas/liquid) flow about the infinite circular cylinder for three regions of the flow field: (1) The liquid film; (2) the gas/liquid free stream

7 flow; (5) the gas/liquid boundary layer. a. Equations and Boundary Conditions for the Liquid Film Consider the element of volume shown in Figure 1 bounded by dcp, d6 and ds. The law of conservation of momentum about 0 yields: Trcp(Ro+b, c ) [ Ro+ ] dCP+Tpp( R o+B, c) [ R+6]d + Tg(0,cp)[Ro+5]cos.e ds+Pg(0,cp)[Ro+6]sin.8 ds = p[p(Ro+6,cp)]2[Ro+b] dS - pr(Ro+6,cp)vRp(Ro+6,cp)[Ro+]2 dcp - Xe(R6) p(R p)[+6[VRo+6bcp) pV(R+ [Ro+][V(R )sin.QVr(Ro+S,C)cos. ] ds where terms in the square of small quantities have been neglected. A force balance in the radial direction yields upon neglecting the squares of small quantities: - Trr(Ro+6,CP)[Ro+6]dcp + Trp(Ro+6,cp)ds-Pg(0,cp)cos.e ds + Tg(0,cp)sin.Qds = pvCp(Ro+,Cr)Vr(Ro+bcp) db - [(vr(Ro+65,C)]2[Ro+6]dcp - Xe(Ro+5,p) pVr(Ro+,cp) [Vp(Ro+5,cp)sin9 - - V ( R+&,cp)cos o] ds

8 Vr P / - T T'g ds Fiur 1. Crf Liquid cyl er Trr \Vr Um f 8 uo/ /d i I, /I I'x /Cylinder Figure 1. Coordinates and control volume for liquid film on cylinder.

9 And the law of conservation of mass is: PYv(Ro+6'qp)db - Pvr(Ro+6,))[Ro+b6]dc = Xe(o+p)[p(Ro+,)p[V( +6,cp)sin.9-Vr(Ro+6,cp)cos.Q]ds From elementary calculus we have the relations: (as)2.o)2 d 2 )2 (ds = o R+2 + (d 2(d)2 1 d6 Tan.p = d Ro+6 dcP ds cos.8 = [Ro+6]dcp ds sin.@ = d6 After dividing the conservation laws by dcp and using the above relations between the differentials one obtains: -Tr(R+)+T dcp +(Ro+b)T g + d -pY2 d6 PVr VR+6 6) - PXe d6 (R +6)v] d6 (Ro+d)V~ rp dp (R+6)Trr- (R+)Pg+ Tg at db 2 db V - Pv (RRo+6) - Pe Vrp r = R +6 r CP dcp r e PLdcp 0 <cp -(R +6)V Pd cp P(Ro+6) vr = PXe (R P+6d)VrU dI d5,

10 Using the last of these two relations to simplify the first two, one arrives at: d6 db - TPr(Ro+6) + Tp dP + (R+6)Tg + Pg d PXe(VcpV) [vc d - (Ro+6)Vr Trr(Ro+6) + T d6 Pg(Ro+5) + g d= r =R+6 - 0(E^6). ^ H- 6) Tg d - 0 1) d6 dP = pXe(vr-Vr) L[ d_ - (Ro+6)V e r r (P dcp Orj d6 V db P dCp (R ) = PXe [ d Vr( R+ Within the film the liquid is assumed to satisfy the incompressible Navier-Stokes equations and the continuity equation. These equations and the relations of the components of the stress tensor to the velocity 18 field are:;v v v2 r vCp r vCp 1 aP v --- - - — + rr r p r rp ar 2_ I 2v a6r 2 v v r + t + 1 Vr 2 Vr vr2 r2 acp2 rr r r2 2) vv +p aV + vrcp 1 aP + r ar r cp r pr ac ~/as^ r.1 ) av + 2 av _ o6 V2V 1 a VT 1 aVCp 2 6vr \a2 2 arS2 r ar r a r2J Vr 1 av + = 0 5) ar r ap r

11 6vr T = - P + 2 -Vr rr 6r Tr ar P/Ip =vr -+ br V Tp = ( I+ — -) Ii) Since the equation of continuity Eq. (3) may be rewritten as: - (rvr) + = 0 3*) Tr r CP it will be satisfied identically if one introduces the usu4l form of the stream function W for cylindrical coordinates. That is: 1 aJ4 av Vr = - p; = ar Now since: d* = - dr + 6 dcp ar acp we have: dj. _= dr +. dcp ds ar r=R+6 ds a r=RO+6 ds a |. d6 + a. dcp rr=Ro+6 ds I r=Ro+6 ds d ds = S. d ds + ds dcp =R ds dcp acp r=R+ r=R+

12 d* = al | db + a dcp r dp ~cp along s r=Ro+6 along s r=Ro+6 | p - (Ro+b)vr at r = Ro+ 6) along s Now the vanishing of the velocity at the wall of the cylinder requires that: Vr = v = 0 at r = Ro; 0 < cp 7) in order to satisfy this condition it is sufficient to take: Vr = ~0 at r = R; 0 c 7) Introducing (5) into (2), yields: 6* 6 2 2 2i 6 6p a'a Va~ a' r 1 a 4 a 1= aP ar rcp ap ar2 r acp ar p aC < 3 + +_ 1 34 2 6 2' 63'\ ar3 ar2 r rc2 r2 r2 2 r }r _ a2 _ 2+6 + 2*V2 1(a )2 1 aP ar cp2 acp aracp Sr r \p p r 2*) + 4 3* + r 3 + - + 2 a\ \ 6cp3 6r 2cp 6racp r cp Using (6) in the last of Eqs. (1), one obtains: a = PXe [, d V(Ro+)] for r = Ro+ aCP ~ ~dcp r

13 or integrating both sides with respect to A along s, gives: 1 cp=o = P Xe- CVr(Ro+) dCp; r=RO+6 0 but since A must be an odd function of cp, CP=o = 0 and one has: P= P e [ dp - Vro(R +i d a1t r= Ro+S 1*) o Now using Eqs. (4) in the first two of Eqs. (1) and substituting Eqs. (5) into the resulting equations, rewriting Eqs. (2*) and collecting Eqs. (l*) and (7*) yields: I a _ * a 1 a2 11 aP + V D ar -r r a/ r c r p r acP ar _ _ls- ( —% - - v -D r acpr r a, or r2 ocp2 r or p 8r r op r=RO+ 8 0 <' Dj2 - -r + - I r r rr r 6cp2 Vr -= 1.; V = a ^ ~ vr r _cp > V;r

14 2g -1d ( 2 VI 1 L2 P I -r dj P( L — (R o+ ) - T (R - ( ) 2 ( a) L- 0 d 6 dcp r 2cp d~ p < ) (Ro+)(Rg + d ( 5 =) - P ) + g d 9) = - PXe 2 (R+ +6 V] [V (Ro+6)V - Ro+6 0 dr r=R0 - = at 0 10)r o < o - - ox * (Ro+6) + V r]R + flow in the liquid film if the gas conditions (Pg and T) and the liquid = P VCP eRp 0 r=Ro Tr = 10) These equations i.e. Eqs. (8), (9), and (10) would completely determine the flow in the liquid film if the gas conditions (Pg and Tg) and the liquid drop conditions (Xe,VC and Vr) were known at the edge of the liquid film. Thus the hydrodynamic problem in the film is completely determined by these relations.* Since the hydrodynamic equations are decoupled from the energy equations by the assumption of constant density and viscosity, the flow problem can be solved without consideration of the energy and the temperature field within the film can be solved for once the velocity field is known. *Note that Eqs. (8) constitute a fourth-order system in r and that (9) and (10) specify five boundary conditions. But the thickness of the layer is unknown and the additional condition is necessary to determine this.

15 With the assumptions that (1) the effects due to kinetic energy changes across this volume are negligible, and (2) the temperature of the drops is not significantly reduced from that of the free stream when they pass through the gas boundary layer, the energy balance produces: k - (Ro+0,cp) R+] dcp - k F (Ro+,c) +] do + qg ds P - pCpT(Ro+6,cp)vp(Ro+5,cp) d5 + pCpT(Ro+b,cp)vr(Ro+ ) [+b, Ro+] dcp + Xe (Ro+5,cp) CpPT, [Vcp(Ro+S,cp)sin e - - Vr(Ro+5,cp)cos ~] ds Dividing through by dcp, using the relations given above between ds dcp and sine and the last of Eqs.(l) one gets: at rk (Ro+) - k(Ro +) a + Rg +O) + 2 r=Ro+5< 11) p -= XeCpP(Tc-T) dcp - (Ro + )V ] Under the above restrictions the energy equation is: 1 a a/ T a aT k )(2T + 1 62T + 1 T jr \cr a cp r) pC p rar2 r2 rr 12) with the boundary condition at the wall: T = at r = R; O < cp 13)

16 Thus Eqs. (11), (12) and (13) determine the temperature field inside the liquid film completely once qg and the velocity field are known. Now the order of magnitude of the various terms in the governing equations is to be investigated. The following new variables are introduced: r A y R - x 1; x; 6/R and Eqs. (8) through (13) become: 6 -1 -1 R2 1 2 a[(y+l) x -(y+l) 4 yy = - - (y+l) P +v D D -1 - -2 -1 (y+l) X - (Y+l) [X-y (y+l) tXX+(y+l) y] = ( P - v(y+l) D j 2 O < X 2 -a 1 (ly)_ +2 14) l +y 6y 6y (l+y)2 6x2 1 4 vcp = R ay V-l k -2 -1 (l+y)1 [yTx - VxTy = k y+(l+y) Tx + (l+yTy yx PC at y = 0 0 = 4 = 0; T = Tw 15) O < x *The subscript notation for partial derivatives is being used. Thus - y -, ax - - - a -- - etc. Y y ~x x yy y ay2

17 +A-1 2-1 A v { 1+5 ) - (1+6) [y+1) yT + 262 a C +1 ] } y=5<x< 2(^ 1+) a 0l y) in ) + 2 x t 16) R2 X[1 (l+)+V v - (1 i RX] Rp A R2 A S + Tg(+6) + -) d x - o qg [() y V x R XeLRpP1 -1+6) -T VJL x - ( l)rr] thf sr t e e diffe ly d g AAlTAr 2 A2 whether the parameter E2 2R / is less than or greater than 1. k (P-P1~6 ) - k (lxx()+*) Rx -(g ) IL') +x = X R (T-T)F(l+) V NoWe wanteres to make thefocused on the asymptotic behavior of these en such a way thations Forall thlarge coeRffiientods nmbf the parameters appearing inof the resultinguid in We want to make these equations dimensionless in such a way that equations are of order 1 in order to determine which terms can be considered negligably smallO But this requires that the terms be made eutosidereofrdnergbysal Bt tin ordequre tohatemn -whic termsca be cone

18 dimensionless in different ways depending upon whether E2 is less or greater than 1.* Therefore these two cases have to be treated separately. Case a) E2> 1 Under these circumstances the numerical results show that: A = O(UORoXe ) and y, = O(2Xe ) Therefore the dimensionless quantities:' ='V/U RoXe; x = x; y = y/2Xe; = b/2Xe = b/2RX; VP = Vp/UO Vr = Vr/U; P = P/1/2 pUo2 are introduced and Eqs. (14), (15) and (16) become a+,_ ^^ ol +2Xe0y) }- (1+2Xey) y y - 2(+2Xe y) Px + 1 D2 _Y -1 2 0< ( Xe Y) x a- (1+2Xeo) - {(1+2XeOoy)J L- -1 (l+2Xe) 2 4. 6 _2 + 2X-^ ^J 4^o = - - - - (1+2eooy) a D 2e yj Py 2' It can be verified from the numerical results that all the numerical results that all the dimensionless terms are of the order 1.

19 D2 = (1+2XeY) (1+2Xe Y) + -2 (1+2Xe Y) X+2 a (1+2Xe]) YTX - VTY2] = P YY -1 -2 1 + 2Xe (1+2X Y) Ty + 4Xe (1+2Xe, ) T — 00 oo 00 00 00 XX I-2 sI r 2e — n (-1-( +2Xem_) l1+2X )] + 4 oo ~ Xe()(+ 26[ 1 +2Xey) + X R T( 1+2Xe.o) 2 (Pg-P) 5 E2rIf 2Vl F2y sz 1/2pUo 2 x Y LoXP - (l+2Xe ) V Xe eo xeXe 8oy=6~eoo Xe (7)x(1+2Xeoo) b-(1+2Xe F) ay [1+2Xe) ]a 2Xs- R + 2(1+2x _ ) a- L+2xe ] vpu g xe+ +2 ( l+2Xe ~ i ) = XeoE LXe- l+2Xeoo 5) 2 1/2pU0 + V][2xcp (1+2Xe -)V] X 0 * + e C e dv o - Vr((+2X X) dx T (1+2Xe 6) - 4Xe (1+2Xe ) T61 + 2Xeoo q (1+2Xe )2+ 4X e 00 0O k 2e0 g q (+2Xeoo + 4XeooB ]

20 Xe E Pr (T,-T) V 2X, - (1+2Xe )Vr i Xe (p e00x or y = O; 0 < x V= =; T =Tw "~" Ty Now taking the limit as XeO + 0 and retaining all terms relating to the two-phase region one gets: t-L - ~ tfly = - 2P- + - Xy xy xyy x E2 yyy at o < < 1a) 0<3014) *=-T- 3F-T- T y x Xy r E2 YY r r -^ = E2{ cx [2Vp - V^ r + Xe } 0 eoo ey 3 g e og x 0 < x x X x r 2XepRRo X I =+^00 = -) e 2 *dx T_ + k g ~Xeog PrE (TW0T)Vr at y =; 0< x ( = y =0; T = T 16a) Where the pressure term in the first of Eqs. (15a) has been eliminated by substituting the second equation into it. Case b) E2 1 Under these circumstances the numerical results show that: t = O(U RoXe) and 6, y = 0 ( ) 00 ~ ~ \ ---

21 Therefore the dimensionless quantities: = = -. y 2ROUo - = V/UoRoXe; x = x y; = A b ~ 2RoUoo =2 i v VpP/; V/ r = Vr/U P = P/1/2pU; Re -2Rs are introduced and Eqs. (14), (15) and (16) now become: -2 1, - = ( - 1 2+ E D2 4Xe { ( 1+ r ) Al[X Y) 4J - O < x -2 P~ _ 4x(1+ 2~ y )1a2 22 - 1 (l= (l+ -y) a — ( -- y (l+ 2 2 2~_ Rey Re ) (1-f -~- Y(1- -2 -]2 4

22 2 2 ( (1+ -5 1+ y) 2~ ( (l+ + 2 a 1+ 2 y) ] + /2 (1+ 2 6) | E 26 L (1+ > e /2PU)2 - O < x 8 ( -61() - [1+ ) y) 2t ) =eocE XeL Le0'tx (1+ - ) + Il4R Scp -(1+e 6)V 2 P 2" 2 Xe VXe.r -Vr(1+, b)] y Re (1+ -6) T)e + k) Re xj -1/2pU,2 o % oC~~, Xe 1/2pU"2 x-f ^ v~l ]-~ fQ. I 2, 2. - )(-I at y=O; O<x; r=; T =T 2 X 2 s L]./2pU-oo2 (<+ ~i+2_) + ~ e MT 2 " %2A 2 2 X_ + E.8) * (+ - y; ~ ~ at0 _RRe Ry Re 0' e 2 2 2 y y ~ w;

25 Now taking the limit as Re -+ 0 and Xe+ 0 and retaining all those terms relating to the two-phase region yields: E xy yy = - 2P Eyyy 0 < y < a = 0 14b) o <x- y (u - 1 YY Xe i/2pU2 X CE - 2V I Vr X e at 00 y = [ - 2 ]' 2 0 < X, P - P+ -T =0 15b) Xe 4 r - r (Td-T) Vx aty us=0 o;0 < = 0; T = TE. l 6b) by use ofr to ocd he t s ec Eq. (b) In order to proceed further it is necessary to bring in the phenomenon which occurs in the two-phase region. b. Consideration of the Two-Phase Region The two-phase region may be thought of as consisting of a boundary

24 layer (on top of the liquid film) and a free stream flow outside of the boundary layer.* However due to the presence of the liquid drops there is the possibility of shear stresses being exerted on the gas outside of the boundary layer even though the mean velocity gradients are not steep. In any case we can conclude that the gas and liquid interact with one another only in the following ways: BI 1) The gas and liquid can interact at the surface of the liquid film by exerting a mutual shearing stress on one another. This is characterized by the term Tg appearing above. BI 2) There is a pressure coupling between the gas and liquid at the surface of the liquid film. This is characterized by the term Pg. BI 3) There is a heat transfer by conduction at the surface of the film from the liquid to the gas (or the opposite direction). This is characterized by the term q g DI 1) The gas and the liquid drops can exert mutual frictional effects on one another. The drops due to their inertia tend to move in straight lines. The gas on the other hand tends to flow around the cylinder. A difference in velocity therefore develops between the gas and the liquid drops in the neighborhood of the cylinder, and as a result there is a mutual shearing stress between the two of them. *And, of course, the region of the wake and separated flow which will not be considered here.

25 DI 2) The gas can exert pressure forces on the liquid drops.* Thus BI 1) through BI 3) are interactions which occur at the surface of the liquid film. We may anticipate that under certain conditions these interactions will have only a small effect on hydrodynamic and heat transfer behavior of the film. Because of the large difference in density between the gas and the liquid we will expect that the interaction BI 1) will have only a small effect on the film. We further expect that the interaction BI 2) will be dominated by the momentum carried into the film by the liquid drops provided that the velocity of the drops and the volume fraction of the liquid in the free stream is sufficiently high. And finally we might anticipate that the interaction BI 3) will be dominated by the enthalpy flux into the film due to the drops. Thus for two given fluids we anticipate, for sufficiently high velocities and amount of liquid, that the presence of the gas will have a negligible effect on the film at least as far as the interactions BI 1) through BI 3) are concerned. At the other extreme the liquid film will be moving at a very low velocity, will have a relatively uniform pressure, and there will be only a small temperature drop across it relative to the gas (since the thermal conductivity of the liquid is much higher than for the gas). Under these conditions we will expect that the presence of the film has only a small effect on the gas boundary layer. It is further possible that these two domains will over lap. Thus there is a *Note that we have already neglected the heat transfer effects between the drops and the gasO

26 domain where BI 1) through BI 3) are negligible for the behavior of the film, a domain where BI 1) through BI 5) are negligible for the gas boundary layer, and a region where the coupling between the two is mutually weak (i.e., the above two domains overlap).* But of course the lower boundary of the domain where the gas has only a small effect on the film (BI 1) through BI 3)) will depend also upon the dynamics of the drops (i.e., DI 1) and DI 2) since if the drops are strongly deflected by the gas stream little momentum, mass, and enthalpy can be carried into the film by the drops. Therefore before proceeding further with the interactions which occur at the surface of the film we will consider the interactions which effect the drop trajectories (i.e., DI 1) and DI 2)). In order to do this we should divide DI 1) and DI 2) up a little differently. We distinguish now between the total drag force exerted on the drops by the gas due to the difference in velocity between them and the force exerted on the drops by the gas due to the pressure gradients in the gas. The drops will be taken to be spherical and relatively small. Then the equations of motion for a single drop can be written as**: 4/53rd p = - 4/35nr VPg Crp - (Vg) *Note that even if at the lower end of the domain where the gas influences the film there is a fairly large influence exerted on the gas by the liquid only a small error will be introduced if we replace the interaction on the film by that which would result if only the gas was present. **l | denotes the absolute value.

27 A A A A A drd V =iA-^^_. - i +jV;o +x 0 a; = i VXo + j V - 0 i o xo Yo Yo dt \ \A \ A +>. A rd i xd + j Yd Vg = i Ug + j vg - A -~ V i UO for rd oo where CD is of course a function of IV - V I g If we introduce the notation: Vd 1 _ ia + j' Xd then it follows from elementary mechanics that: dV =. V V dt d We introduce the dimensionless quantities: * * * * xo Xo/Ro; Y o/Ro; Xd Xd/Ro; = RoV d Ro = V V/U; Vxo V /Uo; V =g/U, * * Ug= Ug/U; P P/ PgU1U; g ug/OO g g/ 1 eUtc and we get: + t ^ P g 1 P I * -i/* -^ * v*.vv* —. -2 -v P - 3/8 cD |v -v I(v -v) 17) Vd VP o (rd/Ro) P g g Now the drag forces exerted by the gas on the drops must equal minus the drag force exerted by the drops on the gas. Therefore if No is the number of drops per unit volume then on the average (i.e., we

28 assume that there are a sufficient number of drops per unit volume so that only the average effect influences the overall gas motion) the force exerted by the drops on the gas per unit volume is: + 1/2 NoCDr rd PglV -V ( -Vg) in the region outside the gas boundary layer where the average velocity gradients are not steep the frictional effects which are due to these gradients can be neglected, and the equation of motion for the gas can be written as Pg V VVg + 2 V ( - V + /2 - - Now it is obvious that N is related to the volume fraction of the 0 liquid Xe by the following formula: No(4/)3tr = X, Using this and the dimensionless quantities already defined the equation of motion for the gas becomes: +.* *... CD +. -. -. +. V oV g = - 2 VPg + 3/8 Xe IV -Vg(V Vg) 18) (rd/Ro) From Eq. (17) it might be anticipated that under certain conditions the drop trajectories may be straight lines (i.e., their motion is not appreciably influenced by the motion of the gas). We are now is a position to determine the lower limit of the domain where the presence of the gas has a negligible effect on the liquid film.

29 In order to do this it is necessary to consider the ratio of properties of the gas and the liquid specifically. We will limit ourselves in what follows to air and water since these are the substances of principal interest. From the first of Eqs. (15a) and (15b) it can be concluded that the shear stress on the surface of the liquid film will be negligibly small if: Xe 2g 1Pg 2 o2 1o/ 2 x r pU - E [ p g pU 19) XeOO CP r XePUOP2 ELjV P_ Tg Vg 2 By substituting the second of Eqs. (15a) and (15b) into the first of Eqs. (14a) and(l4b), respectively, it can be concluded that the pressure forces exerted on the liquid film by the gas are negligibly small whenever: E >1; 2P = 2(Pg 2Xe T ) l/2pU = P 2X 7 2U R 2 [ / e2p Uoo 4 T 2 R 1/2p U ~2 << 1 20a) P 1/2p U 2 E A g Vg or whenever: 2A 2 LA2)/1/2 U E2 > 1; 2P = 2(Pg - T 6 g )/l/2pU = Pg E1Pg 0e0X~~Vg 2- - ^ - e (T X /_ /2pU ] << E 20b) Pr ~ _l/2pgU^ Y ^^^v l g / 2pgu And finally from the last of Eqs. (15a) and (15b) it can be concluded that the gas conduction effects are negligible whenever:

50 Xe * 2XeORo e (T - TVr >> qg k P E [ 2]C [ ug 1 r Pg g 1 (T-To) | n i — Cp P i E that is whenever: i N X g1 [fig g Fp< KEV -X 21) u Vg Pr _ e X These inequalities are seen to depend upon the quantities - Vr and V* which in turn depend upon the liquid drop trajectories. If the drop trajectories were straight lines we would have from Figures 1 and 2: Xe = Xe V = - cos =- cos. x 22) r VC = sin x = sin x We now proceed to investigate under what conditions Eq.(22) hold with sufficient accuracy. Now if the drops are uniformly distributed then the number of drops per unit length of line in the plain of the flow 1/5 and perpendicular to the velocity at infinity is N. And the number 1/3 N' lying along the line segment ab in Figure 3 is No Ro or since: No = Xe/(-)rd *(14) and (21).

Yo r Liquid drop Vyo go i - \ U:o _- g, rd R - xo Fige 2liCy inder Figure 2. Coordinates for dynamical equations of liquid droplet.

32 Liquid drops 0000 0 00000 00000000000 000boOOQ00000 0-0 0-0 -0-0-0 0-0 0 0 0,00 00 0 \\ ooo oo oooo ys\\ 000 v 0 0 0 0 0 o0 0000 / a Cylinder N'= number of drops lying along this line. Figure 3. Sketch for linear drop density.

33 we have: xl/3 e A N' = 24)/3 rd \3 } Ro Now the number N' must be large enough so that the assumption that the unsteady effects due to drop impingement on the liquid film can be neglected, is valid. This fixes the minimum value of N'. From Eq. (17) one can see that if the dimensionless quantities are of the order unity then the particle trajectories will be straight if Pg/p and 3/8CD 1 Pg (rd/RO) p are both much less than 1. But if the drops did not in(rd/Ro) P fluence the gas appreciably then V*P* would certainly be of order 1 and og even if the presence of the drops could influence the gas pressure field by a factor of 100 the ratio Pg/p for air/water mixtures is of the order of 10-3 so that this term is almost certainly negligible. On the other hand the coefficient of 3/8 CD 1 Pg might con(rd/Ro) P ceivably be small almost everywhere if the liquid drops actually caused the gas to follow their trajectories. This is what we might expect if there were a large mass of liquid present. But with a large amount of liquid present (i.e,, large Xe) we could take (rd/Ro) large and still ( A) 1 Pg satisfy (23)and with large (rd/RO), /C - will certainly (rd/RO) p be small. We are interested rather in the other extreme when Xe is small. It is in this case that 23) requires that (rd/Ro) be small and therefore 3/8CD ( 1/ will be large not only from the (rd/RO) in the denominator but also from the fact that CD increases considerably(r/R the denominator but also from the fact that CD increases considerably

34 with a decrease in the drop Reynolds number and therefore (for a fixed Reynolds number based on the cylinder diameter) with decreasing rd/R. Now a comparison of Eqs. (17) and (18) show that the ratio of the drag force per unit volume acting on the gas to the drag force per unit volume acting on the drops is given by: XeP ) We anticipate (and the numerical results below verify) that this minimum value of X* is somewhat less than (pg/p)so that the gas velocity field at the minimum value of Xe is less affected by the liquid than the liquid velocity by the gas. Hence we conclude that at the minimum value of Xe for which the particle trajectories are straight lines the gas can not be affected much by the liquid if the liquid is not effected much by the gas. Therefore we anticipate the gas to behave as potential flow and 1 Pg the coefficient of 3/8CD 1 P/ g must be small. Of course the minimum (rd/Ro) P value of Xe depends also on the gas Reynolds number. That is for lower gas Reynolds numbers the minimum value of Xe must be larger in order to A 1 Pg satisfy (25) and still make 5/8CD (1R) - small. But the numerical (rd/R0) P results obtained below for air/water mixtures show that for reasonable values of (rd/Ro) the minimum value of Xe is always such for all gas Reynolds numbers that we can not expect to strong an effect on the gas *I.e., minimum value of Xe for which the particle trajectories are straight and there is a sufficient number of liquid drops.

35 by the liquid. In order to obtain numerical values for the minimum value of Xe the minimum value of E2 and the required rd/R for which the particle trajectories are straight lines for a given value of the gas Reynolds number we first choose a reasonable minimum value of N' of ten. And we have the following conditions: Drop Reynolds No. =( dx Gas Reynolds No. \Ro CD = Function of Drop Reynolds No.* 3/8 9 <.l 5/8 CD -g<.1 (rd/Ro) P X 1/3 N = > 10 \ ~r These relations serve to determine the desired minimum values and the results** are shown in Figure 7 for air/water mixtures. We have therefore shown that under reasonable restrictions we may expect the particle trajectories to be approximately straight. We will therefore impose this restriction temporarily on the analysis. And therefore require that Eqs.(22) hold. We now return to the consideration of the interaction which occurs at the surface of the liquid film and to what one might expect the minimum conditions to be for air/water mixtures for these interactions to have only a small effect on the liquid film. *See Schlichting, p. 16. **These were obtained from the computer program in Appendix I.

36 From Eq. (22) one gets: Xe - _ - -2 - V* V* = 2 cos X sin x = sin2X = sin2x Xe CP r 00 hence Xe. * -2- VpV = 0(1) Xe 00 and the inequality (19) becomes E >>] / 1/2 pg U0 19*) Now using the inequality (19*) in the inequalities (20a) and (20b) and remembering that the limit as Xe -+ 0 and Re oo is being taken the inequalities (20a) and (20b) become: E > 1 2P = 2 << 1 20a* 2 1/2 P g E2 < 1 2P = 2 U 2 << E 20b*) and finally the inequality (21) becomes: I r CLp P 4 E 21*)' Vg rrg one can now estimate the minimum value of E at which the effects of the gas on the liquid film are small for air/water mixtures by evaluating the terms in the inequalities (19*) through (21*) for a single component gas boundary layer (i.e., evaluating them from the relations for single

37 component flows over a cylinder). It should be noted that a low values of E the presence of the liquid does not modify the gas flow very much since the liquid in the film (which is thin) is moving at a low velocity so it is relatively stationary as far as the gas boundary layer is concerned. Low values of E2 also correspond to the range of parameters where the presence of the drops doesn't modify the potential flow very much. Hence the inequalities (19*) through (21*) become using the numerical values obtained in Appendix II: E > 2.2 x 10 2 19*A/W) E 2 > 1 2FPS5.2 x 10-3 20a*A/W) 3-3 E 2 1 2P&5.2 x 10 << E 20b*A/W) E ~ 1.2 x 103 21*A/W) Hence it can be concluded that for a value of E >.1 (or E > o0i) the terms in Eqs. (14a) through (16a) and Eqs. (14b) through (16b) which result from the presence of the gas can be neglected. Therefore using the relations (22) and the inequalities (19*A/W) through (21*A/W) we arrive at the following formulation: E2 > 1 (straight drop trajectories % conditions of Appendix I)

58 0 < 7"K< 5 T4- 1 -3 - - y xy'x yy E2 yyy /^, _ T 1< T~ ~ 14a*) O < x R T- T- = - y X X Y PrE2 YY Y = E- [2sin x - V-] cos x. s = sin x! 15a*) O<x y = PrE2 (T -T) cos x ~o <x <;_5 = 5-= 0; T =Tw 16a*).01 < < 1 (straight drop trajectories f conditions of Appendix I apply) O < x t~* - " Tww *, i\ j = = x - 14b*) oy xy x y E yy 0 < x < T -- O < f = [2 sin x - E ooss yy y n n t in x 15b*) 0 < x T^= EP (TB-T) cos x y r y = 0 J Z0 *= = 0; T= T 16b*) 0 <x y We now notice that Eqs. (14a*) through (16a*) and Eqs (14b*) through (16b*) are essentially the same. It is only necessary to use the relations y = yE; 6 = 6E to go from one to the other and thus there is a single formulation of the problem for E >.01 and the condition of straight particle trajectories. And therefore only the Eqs (14a*) through (16a*)

39 will be needed to solve the problem. Let us now consider the case when E2 <.01. For reasonable values of the Reynolds number this implies that Xe is small.* Hence there is only a small range of parameters for which one might expect the particle trajectories to be straight and stil-l have a sufficient number of drops to have a contuum. On the other hand, as argued above, one might expect that the presence of the liquid film does not influence the gas boundary layer very much and one may therefore evaluate the gas/liquid interaction terms in Eqs. (14a) through (14b) by assuming that the gas boundary layer and the pressure field in the region of the cylinder are the same as for single component gas flow about a cylinder. Also since we are considering the conditions when E <.1 we may take the limit as E + O. With one exception' In the energy relations the product EPr appears, and since the Prnadtl number is of order ten we must retain these terms. It seems rather unlikely that a film will form for values of E2 < 10 4 hence only the range.01 < E <.1 will be considered. From the inequality (21*A/W) one sees that one may also neglect the gas conduction term for this range of parameters. From Schlichting, p. 153 (28) we have for single component flow around a cylinder: T 2U0 - n(_) Un2n-) _ _ = 8 x - 72n-1 l/2p U 2 v (2n -1)' (2n-) *For air/water mixtures if the gas Reynolds number based on the cylinder diameter is 1000 than for E2 =.01, Xe = 103 larger Reynolds numbers and smaller values of E2 correspond to lower values of X.

40 t! Where the y.'s are the fi(O)'s there, and are listed in Table 9.1. It also follows from the general theory of potential flow that: -- P =P 2 sin. 2x Px l/2pgU2 And finally after noting as in the derivation of the inequalities (20a*) and (20b*) from (20a) and (20b) that P E E P Eqs. (14b) through P 1/2p UO2 (16b) become:.01 < E <.1 ^ 4^ /^ 4p ~ \ O < y < 6 - sin 2x 14c) 0 < x Tl = PrE j T - T) yy r yx xy 8Pg['~(L8pA Xe * if o = gZ Z(x) - 2 e V*V* yy Ep Uv xeo, cp r y = - J Xe Vr dx 15c) o Xer Y r Xe reoo y = 0 T = T 16c) y= 4f=V=o; T=T O<x 00 _ A (-1) ( 2n) yn- Z(x) x + (2n- 1) (2n-1) n=2 Of course V, Vr and Xe/X are still unknown. These must be obtained by integrating Eqs. (17) and (18) and using suitable boundary conditions.

41. 2 2 i.e., V + i xd + y2 - oo * A 2 2 V -i X + yO oo g o * A V.n = 0 x2 + y2 = R2 g 0 0 o A where n is the normal to the cylinder. This in general constitutes a difficult task. However for sufficiently small X the presence of the liquid drops don't influence the gas motion very much. By combining Eqs. (17) and (18) we get: - -4 V e V* ~ v Y* V* Vg = -2(1-Xe) e pV d V g o g Pg/p Hence for X <.1 Pg one would expect the gas flow to be essentially a P potential flow. Under these circumstances Eq. (17) can be solved for the components of the liquid velocity at the surface of the cylinder by replacing ug and vg by the corresponding values for potential* flow. Several numerical procedures have been developed for integrating this equation in connection with iceing problems on airfoils. See for example References.190 We now formulate the problem of drop trajectories for the case when the gas flow can be considered a potential flow to a sufficient * approximation i.e., when X <.1 pg/p. We first recall from potential flow theory that the velocity components outside of the boundary layer and away from the region of the wiake are given to a fair approximation by: *This approximation is probably good for larger values of X if the drops are large enough so that their trajectories are fairly straight.

42 *2 2 * Xd - Ug = 1 v = 2X g, *2 *2)2 (Xd + yd ) The situation is as shown in Figures 2 and 4. Next V P must be of o the order of unity and since pg/p is very small we are entirely justified in neglecting the pressure force term in Eq. (17). Now if one introduces the dimensionless time defined by: * t t -=-U 0o one may write V.\d V = V where the dot denotes differentiation.with respect to t. We also have: *)C.*.** yo = Yd Vo = Xd Equation (17) may now be rewritten in component form as: Y = V* Yd yo.* * x =V d xo.XO (ra/rR) p *c CD Pg(V*D P v* V vy0- (rd3/8), y- g

43 AV (VxO- +g) (Vy vg) ~*X t* *2 = 2Xd Yd2 (X2 + Yd2)2 This constitutes a system of four first order ordinary differential equations in Yd' x4d V0xo and V. We must therefore specify four Vyo boundary conditions. For our purposes we take these as: at t = 0 x - d Yd Ydo * *1 Xd d 0 These are the equations for an individual drop. We must now relate the solutions of these equations to the quantities at the edge of the liquid film which are of interest to us. In order to do this we consider the control surface in Figure 4 bounded on two sides by drop trajectories, on the third side by the liquid film and on the fourth side by a line perpendicular to the particle trajectories at a distance far enough from the cylinder so that they are parallel. Then the particles only cross the control surface at the latter two sides (i.e., they enter at the upstream side and leavet th the downstream side). Now from the previous developments we have that to the order of approximation of the quantities appearing in the liquid film equations the total flux of

Yo Vyof t7 -- v Drop trajectories Vr/ Y ydf xof Yd W i^\, from cylinder (assumed Control sur uniformly distributed.) Figure 4. Control volume for the determination of physical quantities at the surface of liquid film from the trajectory data.

45 liquid across the downstream surface (which is the flux of liquid into the film between cp = 0 and cp = cpd) is given by:* d XeVr R d(p 0 The flux through the downstream side is given by: Xe UOOYdd and since these must be equal to one another we have: CPd - Xe Vr Ro dp = Xe UOYd 0 cpd - (xe/ ) v dcp =y e/Xem r 0 differentiating both sides with respect to Yd we get: ~d d d * dPd i - e (Xe/x )vr dj = 1 hYdo r dyd (xe/, ) Vi d.P 00 *This is essentially equivalent to the approximation that the film is thin.

46 Next we note that: * -1 df (Pd = tan Ydf Xdf cos. 0ddcPd = dydf dydf dd = - Hence: X /x V - / dyjf -(Xe/X )V= x d / - dfX/ dyd and we may write from elementary analytic geometry at the edge of the liquid film * V, Sind + V cOsCPd vp = Xof sind + vof cosp 4\ and remembering that we have set x = cp, the results may be summarized as follows: -1 * rx 17*A. oC ) (Xe/ )V dx = r=e/Xe dr < e

47 - (Xe /Xe )Vr df / dy dyf _*= v x + Of f CP Xof f+ Yof Ydf We now notice that all the quantities on the right of the above relations can be obtained by integrating Eq. (17*) for different values. of yd 0 The quantities on the left are precisely those which are needed for the Eqs. (14c) through (16c). The calculations have been carried out numerically. These are discussed in Appendix XXV. We will consider here only the cases when (1) E >.01 and the drop trajectories are straight as formulated in (14a*) through (16a*) or equivalently in (14b*) through (16b*) (since these are the same equations), and when (2).01 < E <.1 and the external gas flow field is not influenced by the motion of the drops as formulated in (14c) through (16c) and (17*) and (17*A.C.). In the next section we therefore turn to the solution of Eqs. (14a*) through (16a*), and (14c) through (16c). 2. SOLUTION An examination of Eqs. (l4a*) through (16a*) (or equivalently Eqs. (14b*) through (16b*)) shows that the position of a boundary of the domain of the problem is unknown and must be determined by the problem. That is the thickness 5(x)is an unknown function. Since it is inconvenient to solve the problem in this form we will map the problem omain of the problem into one whose boundaries are known and in the process reintroduce

48 6(x) as a dependent variable (on the same footing as l). In order to accomplish this let us introduce the following variables: = 7Y/'7Y/ = / R0 _ -- v~ - x = x = x/R~ * _ - 6= = 6/E = 6/2RX * _ T - T Tw- TT Then we have: a x a + a a = a (a\ arl - ad - a a5 ~ a3 (ax)- at - dxs 71 a b2 1 62 1 a a d b ax5y" S aa 2 dx a) a Now using these in the first of Eqs. (14a*) we get: 1 * a6 2 1 d* arl ( 6^ 6TIm d6 6n 6Ti 6T a * a2* M1 d 4*a 1 a3II * -s * i a+ 1 rl qf- E3 a 6*2 S 6*3 dS *r T r ESS*3 3

49 which becomes after some rearrangement: * [r^ 22* _ si iT2 ] d6 / 2 1 _3 6 I 62 d E2 3 and wer get from the second of Eqs. (14a*): 6* r2 * _ a* 1 a KTl ae " Irl a J PrE2 ra|2 and finally since q = 1 when y = 6 and i = 0 when y = 0 Eqs. (l4a*) through (16a*) become: | ^ 6aTg T a62J ds T 2 E2 023 ) I o* ol a_ as c 1 = 1. ao 1 [LaO a " a J P.E2 a 12 a =E26*2c0 6 - 6i P 4 ] =n 1 * 4q = sin 24) 0 < * s*=- PrE2 G*5* cos S anI - n=0 = 0 o; e* = 25) o< E' The same results could also have been obtained from Eqs. (14b*) through (l6b*). The problem is now completely formulated except that we have not specified the boundary conditions in the I direction but

50 these follow from the symmetry of the problem. That is 6 and e must have even symmetry while f* must have odd symmetry. Because of these symmetries we seek a solution to the problem in the following form: 00 * [ E b 2n-1 27) 5- = fa( )1 -+ 2nW) (2n 28) e'= Fan) 1+ a^n s2 27) (2n)'. n=l If we substitute Eqs. (26) and (27) into the first Eq. (23) and equate coefficients of like powers of ~ after a lengthly calculation which is carried out in Appendix III we arrive at: 1 f + flf - fl2 = 0 a E2 29) Ln(fn) = (-1) n2 f + (n-2)f bn_ + Hn n = 3,5,7... Where: 1 d3 d2 d Ln --- + fl - (n+l)f + nf n aE2 d3 1 dr2 d a0E dj on And: H3 = 0

51 Hs = 20 Ef32 - f3f3) + b2(f3fl + 3f3f) H7 = i2(8ff3 - 3fsf3 - 5f3f3) - b2( 18fsf + 80[f32 - f3f3] -9 f5fi - 45frtf) - 10 b4( f3f + 4fff3 + 5f3f)] Upon substituting the expansion (26), (27), and (28) into the second of Eqs. (23) and equating coefficients of like powers of ~ we get after some calculation which is performed in Appendix IV: Kn(Fn) = Gn n = 0,2,4,6.... 30) Where: 1 da2 d K 2 -- fl - nfl n aoE Pr dri2 dri Go = G2 = (4f3 - b2fl)Fo G4 - 8(3f3F5 - 2f3F2) + 6b2(2fiF2 - flF') -(6f5 - 24b2f3 + b4fl)Fo G6 = 20(3f3F4 - 4f3F4) - 18(5fsFh - 2f5F2)

52 + 15b2(4fiF4 - flF4 - 16f3F2 + 24f3F2 ) + + 15b4(2fiF2 - flFk) + (8f7 - 90b2f5 + 60b4f3 -b6fl)Fo Next we get after substituting the expansions (26) and (27) into the first of Eqs. (24), noting that: 03 +5 7 sinS = - + -+..... 3' 5' 7r. 2 4 6 cos = 1 +- +. 2. 41. 6' and equating coefficients of like powers of t (see Appendix V for the calculations): 22 f' (1) + 2fl (1) - 4a = 0 aoE2 0 a0E n+l 2 (1) 2ff ( 2 1) + ) ba-f ( - 1) ~aoE2 nn n+l- 51) -2ao + fi (1) + An n = 3,5,7 where: A3 = - a 0

53 As = 152 2ao - fi(l)] + ao(lOb2 - 1) + 10(b2-l) 3b2ao + 2f'(l) - a A7 = 105(b2 - b4) L2a - fi(1) + 2ao(21 b2 - 7 4 4) 00o - 35b4 - 1) + 42(b2-l) Lao(5b4 - lOb2 + 1)- 3f5(l)] + - 70(b4 - 6b2 + 1) Lao(532-1) + 2f3(1)J j... O O 00 O O O.O. O....OO. O.O. O. * e e e e e o e e o ~ o ~ ~ e e e * e e e e e * e ~ ~ ~ ~ - ~ e ~ * o ~ ~ ~ ~ ~ ~ e - ~ - o * - e ~ Next we turn to the third of Eqs. (24). We substitute in the expansion (27) and (28) and equate coefficients of o. After the calculation performed in Appendix VI we arrive at:.2. F.(l) + Fo(l) = o aBE2 P + 0 aoEPr n2) - Fn (1) + F (1) + F(1) b + (-1) +B; n = 2,4,6,.. a E Pr Ln n n o r Where: B2 = 0 B4 =6 [b2F2(l) -F2(1) - b2Fo(l ) B6 = 15 [2F4(l) + b4F2(1) - F4(1) - 6b2F2(1) -b4 Fo(l) + F2(l) + b2 Fo(l)]

54 The second of Eqs. (24) yields upon comparison of: sio (_1)n 2n-1 sin = -X (2n-l)'. n=l with the expansion (26): fl (1) = 1 33) fn (1) = n = 3,5,7.... n+l And finally to satisfy the conditions (25) we must have dfn (0) f (0) = df =0; n = 1,3,5,7, n dr Fo(O) = 1 34) Fn(O) = 0 n = 2,4,6.... We have now completely formulated the problem in terms of a set of simultaneous ordinary differential equations with their appropriate boundary conditions. These equations can be solved successively, i.e., the second equation involves the solution to the first but not of the third, fourth, etc. We also note that the equations in fn are of the third order but that there are four boundary conditions. The fourth boundary condition is necessary to determine the ao, b2, b4, etc. which are to be determined by the problem. Before summarizing the results of this section we will derive explicit formulas for the velocity, shear stress, and heat transfer coefficients

55 in terms of the solutions to the equations derived above. We first recall that the velocity in the x direction (i.e., in the cp direction) vcp is given terms of the stream function in Eqs. (14) by: 1 a6 v = Ro If we now change the symbol from vcp to u (i.e., vcp = u) and introduce the new variables of this section (i.e., r*, r, *) we get: u _ = 1 f*' UOO 26* ar2 Also the wall shear stress in the cp direction is given by*: = -- \ w R2 ay2 y 0 Introducing the variables V, n, and 6 we get: iUORoXe 1 a2*? 2~ 2 *2 Tw 4R2 Xe2 S 2 or: 2URo \ Tw/' F7 1 a2,f* 56) 1/2PU 2 = E 5*2 62 6) _____1 p v) and fromv *Note that from the last of Eqo (4) Tr = M ( + - and rom (7) Vr = v. = 0 at r = Ro and therefore Vr = 0 at r = Ro hence Trp (Ro, ) = a I R a O R2 y2 O \r r R Y R2 O or Ir = R 0 y = 0 0y y = 0

56 Finally the heat flux at the wall qw is given by: aT k aT q = - k = - w ay r=Ro Ro ay y=O * * and substitution of,, and yields: k(Tw-Too) 1 a o qw = 2RoXe =* I _= or introducing the Nusselt number Nu = qw 2Ro/(T -Tw)k we get: Nu 1 a 2U R E 00 0 v rl=O Now upon substitution of the expansions (26) and (27) we get from Eq. (35) after a calculation performed in Appendix VII ~u -u3 u3 u 5 u7 58 u = uit -— ~ + _.~ 57 + -..... 8) Uo 3'. 5' 7'o Where: - 1 u.: = -- fi fo 2a 0 u3 = -- (4f3 + 3b2fi) 2ao u5= = - 6f5+ 40b2f3 + 5(6b2 -b4)f U7 = 2- 8f7 + 126b2f5 + 140(6ba - b4)f& + U7 6 = or

57 + 7(9Ob2 - 15b2b4 + b6)f f After substituting the expansions (26) and (27) into Eq. (36) the calculation of Appendix VIII yields: T 2V v Tw v T~ T T I /. 3' V 1/2pU e. 1 3 - 5 5 % 7Z7 +.. ) where: T71= - fl(O) T3 = [2 f (0) + 3b2f1()] 0 2 592 T5 = f5(0) + 4Ob2f3(O) + 5(9b2 - b4)f(0)]) a 2E.1 T7 a2E Lf 7(o0) + 126b2f (0) + 140( 9b - b)f ) + 7(180b - 45b2b4 + b6)fl(O) And finally the calculation of Appendix IX yields after substituting the expansions (27) and (28) into Eq. (37):

58 Nq *.i *.S * *..... 40) Nu + q2 2 + q4.4 q6 6 )+ 2 4) where: q = aE Ft (0) 02 a0E 0 q2 - 1 (0) - b2Fo(O) qoE q4 = k- [F4 (o) - 6b2F(0) + (6b2 -b4)Fo(0 q - 1 r q6 = -- E6 (0) - 15b2F4(0) + 15(6b - b4)F (0) 3 1 - (90b2 - 15b2b4 + b6)F6(0) We now summarize the results of this section: a fE ff' +_ f = 0; O Tl aoE 2 f'(l) + 2fi() - 4a = 0 s I flt.) ~41) f1(0) = fl(O) 0 f~(o): o

59 n+l n( n) = (-1) n- l ( n-1+ + Hn(f1',f2, f fl... fn_ 2...fn_ -2 -2 b2,...bn_3) n-5, 5,7.1 + f d n+1 n=n,5o,7.. -2a + f(l) aodb2,...bn f (l) = 1 fn(0) = 0 fn(1) = 0 42) K(Fn) Gn(fl* fn+l * F.**Fn2 aob2..bn) 1 d d K = oaPr + fl - nfl; 0 f < 1 n aE2P d2a d - 1, for o r for aoE2Pr Fn(1) + Fn(1) = e(n) to(l) + (-l))2]+ n=0,2,4,6... o _ + Bn [Fo(1)..., Fn_2(1), aob2,..o b2] } Fn(0) = 1 - (n) 0 n = 0 E(n) - 1 n 0o — 45) We now notice that the differential Eq. in (41) is of the third order but it contains the unknown parameter aoo There are thus three boundary

60 conditions plus a fourth condition to determine ao. Thus the problem for fi and ao is completely specified and can in principle be solved. The situation is exactly the same for each of the Eqs. in (42) and they thus determine the pair fn' bn-1 uniquely. It should be noted however that the solution to (41) and the solutions to the preceding Eqs. in (42) must first be found before any of the Eqs. (42) can be solved but this presents no problem since after the solution to (41) is determined the Eqs. of (42) can be solved successively. Finally the Eqs. of (43) are of second order and there are two boundary conditions. These depend of course on the solutions of Eqs. (41) and (42) and the solutions to the Eqs. (43) corresponding to lower value of n. Thus these equations can be successively solved once the solutions to (41) and (42) have been obtained. Having obtained the solutions to these equations the liquid film thickness (Eq. (27)) the liquid stream function (Eq. (26)) temperature (Eq, (28)) and velocity (Eq. (38)) can be computedo Also the local Nusselt number and shear stress can be obtained from Eqo (40) and (39), respectively. These equations have been solved numericallyo The actual calculations are discussed below. Before considering these solutions we turn to the case when o01 < E < ol and the presence of the liquid has no effect on the gas (i.e., there is potential flow upstream of the cylinder) Integrating the first of Eqs. (14c) and using the first two boundary conditions (16c) we get:

61 2 1 Pg. 3 +~'2 )'4 = 4- )g- sin 2xy3 + ( 44) 3 p Using this in the first and second of Eqs. (15c) we get: 4 Pg )y/ i, 8 Pg r,L) - g sin.2x 6 + 2 =(x) ()'p E p E Z() Xe * -2 -V* VV X r CP and: l x 2.1 P 9 3 x 2 1E Pg sin 2x 3 + (x)2 V* dx EP n~x6 + ( Xe) - r dx 0 or: 4 1 _ g sin.2x + Z() _- V} x+ XF Xe* XO + f e Vr dx = 0 45) Xe00 and: t~ f P g g X _ * * 2 Pg in-x (x) = - Z(x - V + in2x E p v Xe Next introducing the notation: 0 00 the solution to Eq. (45) can be written as3ZE x) e Vr Vef 6 - sin~ (2 cos. - 1) 46) 4(pg/X) V sx eo g 13 the solution to Eq. (45) can be written as ~ 4( PM) sin2x~ 3

62 where: 24 - 2 where:-l r 4P ) s in2 2x y = cos E2 T3 eff it should be noted that y + 4T and 7 + 2jt would also be solutions to (45) but y + 2t corresponds to negative values of 6 and 7 + 4n correspond to values of 6 which increase with increasing Teff which can be ruled out on physical grounds, Substitution of Eq. (44) into the second of Eqso (14c) yields: " =- - 2Pr - sin 2x y A-+ - P -- cos 2x y Oy yy r p 5 3 rp y + P E 2y G 3 y 8 3 r X y where we have introduced: * T - To Tw- To We next note that for air/water mixtures P, 10 Pg 13 x 10-3 P 2 Pr g ( 2.6 x 10 2 P Hence we will neglect the first two terms on the right of the energy equation and we have Q.= PrE f2_y Y 2 j - y, _~ 47)

65 Next we introduce the variables: = Y/ol = Y/5 e x and we have: - a a 1 d6 a and Eq. (47) now becomes: nn PrE 5 -2\ 62n Gt _2 8 - d} 48) O-V^^ ^Oe nT d I 48) Now if we define J(x) = J( ) by: X e * J - - V X r e.0 It follows from the equation preceding (45)e t f= Jdt +- 21 g (sin 2) b 3E p o Pg Now using this in Eq. (48) and neglecting terms of the order of 2Pr p we get: $ = PrE 5 2r( Jd - \ J 49) and from the last condition (15c) and (16c) we have:

64 9 = -PrE 6 J 8 at T = 1 50) e3 = 1 at T = 0 51) We now notice that 5 must be an even function of 5, J must be an even function of ~ and Jd~ must be an odd function of. We therefore set~ 00 tn =U X 2n 2n n=0 (2n)'. n=52) ~ 1^ (P2n 2n-1 5 Jd~ = 0 n=l (2n)'. and: 00 * Z F2n 2n 3) n=O (2n) where as usual we take'. = 1 Next differentiating the second of Eqs. (52) with respect to S and setting 5 = 0 we get upon comparison with the first Eq. (52) P2 = aC 54) Substitution of Eqs. (53) and (52) into Eq. (49) and equating the coefficients of like powers of 5 yields after the calculation performed in Appendix X: 0 < n < 1 Kk (F2k) = Gk - k = 0,1,2,... 55) OaO

65 Where: K - + r - 4kr k PE ao dqn2 d =0 k =0 = - 2 F' k = 1 2 0 2k-1 _, Gk n = rl' 0 2n-1) 2n P2(k-n+l) - (n )^Fn (?-2( kn n=l l -a2k F k > 1 2k 0 o In obtaining these equations the binomial coefficients which are written as () have been introduced and the relation m m! m).... _n/ (m-n)' n' has been used. Substitution of the expansion (52) and (53) into the boundary condition (50) yields after equating the coefficients of like powers of k n = 1; k ( )a2(k-n)Fn 56) and t u c i 5i2nt n=0 and the boundary condition (51) requires that

66 = 1 n =0 rI = 0 F2n 57) 0= n > 0O Now the heat flux at the wall qw is given by: k aT qw R7 6Y y =0 and substitution of e, T, and 6 yields: k(Tw - T) 2RoUg 1 a* 2R~ l n = 0 and the normalized Nusselt number is: Nu 1 a6* V or substituting in the expansion (53) yields 00 F (0) u 2n1 2 2UORo n- O (2n)' Next the wall shear stress can be written as in the derivations of Eqs. (36): T = 2 Ro2 y2 Iy = 0 and introducing r and y we get: *C.f. derivation of Eq. (37).

67 2U00R0 W V a = E 1/j2pUOO2 ay and using Eqs. (44) and (46) we have 2UR Tw, _ _ v = -2E T + 4 P 6 sin 2x /2p eff. *p Next we have: u= I. _ Ro ay and introducing r and y we get: u _E 6a UO 2 ay using Eqs. (44) and (46) we arrive at: *n (On pC LO u -= E Teff Y + Pg (sin.2x) (26 - y)y Uoo00 P and finally summarizing the results of this section we have: %^ E Teff. -: =E T.) (2 cos. - - 1) 4(pg/p) (sin.2x) 3 7 = cos jE2 (sin.22x)Ij y=COS-l {- ( 2i- 2x. E2 Te3ff = ET eff Y + LE (sin.2x) (26 - )y 2U R 00 0 1/ - = 2E Tff+ 4 sin.2x *C.f. derivation of Eq. (35).

68 00 Nu 1 Fn() 2n..v n=O |2U00RO = 0 (2n):. Where the F's are determined by: 2n 1 O < <; Kk(Fk) = Gk k 1 Vt 2 k\ = 1; -F" = ) fj k = 0,1,2,5,... PrE 2k = PqE= 1; P r Fk= 2 n/ 2(k-n)F n=O 2n r= 1 n = 0 n = O; F2n L= 0 n > 0O j-_ -e V* X x r eoo 6 2, - _ Xe Vr dx Xe o 00 00 ~j = ~2n ~2n i (2n). n=O Teff - E {z(x) + J E p v p G r -= -0c2 T2 F' Gk 0 k-1 [2k F2 2(k-n+l)- Fn 2(k-n - rj F' 2k o

69 K - 1 + 2 - - 4kn PEc d2 ddn 00 Z (-1) n 2) ".2n-1 Z(x) = x - E (2n - 1) 7 (2n-l) n=2 7i 73 75 77 7s 711 1.2326 0.7244 1.0520 2.036833.280140 67.637501 These formulas can now be used to calculate the desired physical quantities once J,, and VD are known as functions of x. These quantities have been obtained numerically (see Appendix XXV) and the results are tabulated for various values of the Reynolds number and drop size. The numerical calculations are described in Appendix XXVI. Now the problem of the oscillating flat plate will be discussed. C. Semi-Infinite Oscillating Flat Plate in a Streaming Gas/Liquid Flow with Oscillations Parallel to the Flow 1. FORMULATION Now a semi-infinite oscillating flat plate which is maintained at a constant temperature Tw in a streaming gas which contains liquid drops will be considered. The plate is taken to be parallel to the velocity of the gas far upstream from it and to be oscillating sinusoidally in its plane with a small amplitude U. -. Now we will use the flow model 2 described above in the section entitled "Preliminary Consideration" -and assume from the outset that the boundary layer assumptions hold with

70 sufficient accuracy within the liquid film. The boundary layer approximations will hold with good accuracy in the film if it is sufficiently thin. But since the liquid capture area of the film in this case is only proportional to its thickness, whereas in the case of the cylinder it is equal to the cylinder cross sectional area one can be-sure that the film will be thin. Next since the pressure field due to the presence of the plate is only a second order effect and the gas flow outside of the boundary layer is essentially parallel, it will further be assumed that the drop trajectories are straight lines and the volume fraction Xe of the liquid outside the film is everywhere the same. The liquid film will now be considered. We set up an x, y coordinate system which is fixed to the plate. The velocity of the free stream U(t) relative to these coordinates is then U, minus the velocity of oscillation of the plate which is taken as - 2 U. sin. wt Thus: U(t) = U (1 + 2 sin. ct) 59) 2 Attention is now focused upon the control volume of Figure 5 which is bounded by the liquid free surface S' the surface perpendicular to the plate S" and by the plate itself. Its cross sectional area is A (x) If the x and y components of the velocity of the surface S' are Vc C.S

y n V< 4 8( 70xg V as(x,,it) at _ ^.s^^Y t,ds ^ xx Xe r 0a XZXj ~uc~~t)^~ t dx~ =~ ~Liquid SI' I IFilm S 8 (X|,t) Ac.s. S /- J y (0 7xI)^- L_______ < --- X, __{ Plate xi~~~~~~~~~~ Figure 5. Coordinates and control volume for liquid layer on flat plate.

72 and V(Y), respectively, then they must be related to 6 by: C.S. v(y) + v(X) c.s. at c.s. ax and they must be related to the velocity of the control surface per1 pendicular to itself V by: c.s. V = - V() cos.cp + V() sincp C.S. c.s. c.s. (x) (X) s s c.s cos. + V sincp + s incp [v(X) - cotan. c + sincp C.S. \x 6xs but a5 a = cotancp dx and: I a' V - sincp 60) ~CS.1 a6 Thus V s is dependent only on a and independent of what motion we may apply to the control surface along its length which is as it should be. The law of conservation of mass for the control volume may be written as: *Note that: ds V(y) as +a as a65 a5 ax as -d- VS = -+ + — = dt c.s. ~at aS at ~at ax as at a a dx -_ + d at a dt

73 d F p dA + pX -U cosp - Vc ds' dt e C.Sj Ac.s. + p u ds" = 0 61) ST" And remembering that an accelerated coordinate system is being used the law of conservation of x-direction momentum may be written as: dt / p u dA + pX U -U coscp - V ds' dt e Cc.s. Acsf C.S. S' x + p u ds" = - T (x,O,t)dxl + p - dy x j y a j dt str o Ac.s. + uTg sincp ds' 62) s' where the pressure forces have been neglected since it is assumed that the boundary layer approximations apply. And finally an energy balance on the control volume is performed. In doing this, we will neglect a priori the effects due to kinetic energy changes, and assume that the temperature of the drops is not significantly reduced from that of the free stream when they pass through the gas boundary layer. With these approximations one obtains: -t rJ pCMT dA+ pXeCpT U cosp - V ds' A s' C. S.

74 x + pC Tu ds" = k dxl + J P o ay y=O ST' + qg ds' 65) S' We next notice that: a6 - = cotan.cp ax f6 2 2 ds' = 1 +( dx = 1 + cotan cp dx dx sincp dA = dy dxl ds" = dy and by using these and Eqs. (60) in Eqs. (61) through (63), one obtains: x 6(xl,t) x d f dydxl - X 1 U F dx1 dt e Lat ax 0 0 6 + u(x,y,t) dy = 0 61A) 0 x x PX e U [ + U 1a dxl + JT dxl o t ox x 5(xl,t) 5 dt u dy dx + p (x,y,t)dy 0 o 0

75 x 5(x1,t) x p/ / dU dy + T (xl0,t) dxl 62A) dt xy O O o x x pXeCpT I + U -1 dxl + dx Ie P J t asinxp O o0 6(x,t) x 6(xl,t) = pCp T(x,y,t) u(x,y,t)dy + pCp - T dy dxt o o ~ O O 0 x - k T 1 dx 65A) O ay y=O Now differentiating (61A) through (63A) with respect of x yields: 6 xt) (x,t) (xt) a dJ y -e st + U + J ay 0 ~ + u(x,5,t) 5 = 0 ax 6(x,t) PXeU a + U 6 +T = dy Le Lt jxJ g gt 0 6(x,t) 5(xt) + p + 2 u dy + pu2(x,5,t) x + TXy (xO0t) -p dU dy 0 o dt - + q (X p~~Xe CT Fqu~ I x)(ut) dY pX C T + U 61 + s pC...dy e p o rat ax sincp Jx 0 6(x,t) + pC T(x,b,t) u(x,b,t) -_ + pC - / T dy P ox P ot - k (x,O,t) ay

76 or performing the integrations with respect to t: 5(x,t) p 2 u(x,6,t) p X U -dy 6t x xJ e Lt ] /O o1 o o PXeUCT + U -- - pu t (x,,6t) + u(xt) + T sincp P x at p x ay /(Xt) 6(xt) 6(xt) S p 2 + u dy + p t) - P dU dy ax — Cdy + Txdy(XpC T P 00 _ t Lx P xt ox k (xt) (xt) dt ~~o pXe U - u(x,6,t) ba + U 6 + Tg = - k T (x, La t to e oa F+(xOt) xy [T0 - uxt Lud + u + - k (x,O,t) 5(~a x st) a

77 5(x,t) 6(x,t) 6(x,t) pCe u + aT dy + pCp T udy- pC T(xt) u dy F 6 Sx 6j p x P 6x In order to proceed further the continuity equation is now introduced, i.e.: 6u 6v 6x ay By using this in the last two integrals on the right of the last two equations and performing an integration by parts one obtains: PXe [u - u(xBtj [a + U L + Tg p1 [ U au + V -] dy + Txy(x,o,t) p dy 64) 0L t 6x 6yj y d pC Xe FT - T(X6)t] LT + U a]+ s 6(x,t) Cp L + u + v 1 dy - k (x t) 65) 0o Lt ax ayJ y Since the liquid in the film is assumed to satisfy the boundary layer approximations, (and since for flow over a flat plate it is consistent with these to neglect the pressure forces) since we are using an accelerated coordinate system, the governing differential equations become: pFL 4+ -u.~ + u u + 2U Lat ay dt ay2 pC u t v a = 67) P -_t _ y2

78 au +v = O 68) 6x ay and the shear stress T y(x,Ot) is given by: u Txy(X',ot) = - -69) y=O Using Eqs. (66) and (67), and (69) in Eqs. (64) and (65) and performing the integrations we obtain: PX [U - u t + U + = 7 T y =x TO s I_. l =6~~~~~~~~70) O<x XeCp [TO- T][L - + U a + = ka t ^ sincp y Next we introduce the steam function 4 defined by: u =; V = - 71) ay ox which satisfies equation (68) identically. Now the boundary conditions at the plate are: T = Tw; u = v = O; y = O; < x the last two of which will be satisfied by taking: = O; y O, 0 < x 72) ay Using (71) in Eq. (61A) gives: x 6(xlt) x d 00 dy dxl - Xe + U

79 6(x,t) r a: + / -dy = J aya 0 and performing the integrations and using the boundary condition (72) gives x x d 6 (xit)dxi - X / ad dxi - X 5(x t)U + V(x,6,t) = 0 dt e 6 t e o o or x at y = b; <x; = XeU + (X - i) f dx 73) Now using Eqs. (71) and (59) in Eqs. (66), (67), (70) and (75) yields: 62 +~ 6* 2 6* a2 UoEW a3 +_ - _ = cos ct + v atay ay axay ax ay2 2 ay3 O<y<bS 74) 0<x aT t IT T _aT k a2T at ay ax x ~ y pCp y2 = Xe6 Uo( + 2 sin cot) + (Xe - 1) dxl y= g - V a 2 75) 0<x P ay2 Xe - T] + Uoo (1 + - sin cot) x + Ci e J [t 2 xj pC sincp k aT pC-p ay

80 w F 0<xw; V = = These equations and boundary conditions would completely determine the problem of the liquid film if Tg and q were known. Before discussing these, the following dimensionless variables are introduced: y/6 LOX XeU Xe UooXe XeV 6* 64 77) T =W)t T - Tt Tw - T It can be verified from the results of the calculations which this analysis leads to, that the terms of the Eqs. (75) through (76) expressed in terms of these new variables are of order unity. Now employing the chain rule, we get: a,( a + (ax a ax \6x/ t 6T \Xy t Oa yt y,t a (as ) + a''x,y x',y

81 o (1) l a 1a * a ay It+ -or - - ax - 2XeUoo L6 as " 11 c_, -w A * 78)'6t taT 5* a- -| 78 ) a 1 a ay vnd a and: a2 1 2 _ aJ 1 a2* 1 a a axay 2 XeUAVXe 65* Ux + 5*2 U| s i l t j + 79) _a 4 e 1fj a2!1 a a a1 atay ~ VXe 5 aa[5 T * aT Il arl _ ya3 + _ CO 1 a3 ay3 vX vX 5*3 6y3 _ I Rearranging the second of Eqs. (75) one obtains; - )6 U + U00, ) 2 U~0 sin cot a + 2U 6\ y\t 6x 2 Lot ox __ o^~|,e2 2 1 v 2 "a a~ y + U (1 - cos 2cut) = - - 6x 6y 4 2 x X 6y2 - P~e -at y = pXe where the identity: sin2. wt = 1/2 (1 - cos. 2ct)

82 has been used. Using the relations (77) and (79) in this,one obtains: Xe a4 r* 1 i * at q = 1; ( - ) (X e + - ) + 6* -* e2 2 E a* aO* x6 as6 a * + - sin T (X ~' + -' 2 e aT a 28* a6 an + (1 - cos 2T) = 16 a *2 2 T UW7FF[Pg 1_ Xep U2 I V I e goo u gL J Now the order of magnitude of the last term on the right of this equation can be estimated by evaluating it for a single component gas boundary layer and evaluating the property ratios for a particular liquid/gas combination. Since the principle interest is in air/water mixtures, we will limit ourselves to these. Thus from Appendix II we have: PgUi_ U x [ ] = 1.1 x 10-3 pgU2 v v P This term will be taken to be negligible if it is less than 0.1 1 a56 since the term - > 1 for all values of X as shown by the numerical 2 ~ e calculations. And the above results shown for air/water mixtures will be less than 0.1 if: X > 1.1 x 10-2

83 We therefore limit ourselves to values of Xe greater than this, and neglect the gas shear stress term, and we have:* 5* (65 - Xef) (Xe 5 + 1/2 56) + - 5* 6*(Xe&* + 6) ep eX- 2 Tr o Xe E2 *2 n6sinT + 5 56 (1 - cos 2T) = $n 2 ~ 16............................... For l = 1; 0 < 80) Next using the relations (77) and (79) in the third of Eqs. (75), one obtains: at r = 1 6* 1 c 6(T-T (T - T,) X 6* + 1 ( +- sin T),, = - e T T aT 2 2 i XePr 6* 6* 1[(qg) /( / (gg Pg 1 TAT) Xe T kg/g kg/ L v p sin We again estimate the order of magnitude of the last term in this equation for air/water mixture by calculating its value from the relations for a single component gas boundary layero From Appendix II its value is then approximately: 1 4 1 3.75 x 10- (T-T,) Xe *The subscript notation is used here to indicate partial derivatives i.e., i- - _ a; -n. *; *- a2;; etc.

84 and thus it is negligible compared with the term: 1 a6* i2 a (T - T) -2 for X > 10 which is the minimum value to be considered. Hence this e term is neglected and we have at 1r =1 0 <: x + 1/2 (1 + sin. T) 7] e 6T 2 TJ = -e* 1 -] 81) XePr Using the relations (77) and (78) in the first of Eqs. (75) yields: * = 56 (1 + sin.T ) + 2 (X - 1) d S; at = 1; 2 e 0 T 0 0 < 82) And using the relations (77), (78), and (79) in the first and second of Eqs. (74) gives: X 2 2 1a* + 1 _* _2r __a a* a62j e L ari 2 M 6an an a an2 J F al+ aa* ae a* a6 - *2t*l - -irr if - 2- ri}- 2e ae arl aT aI a82) S* cos T + 2 OTs *Note that: sin.p = 1 - 1 2 l+ cot.c p +

ra~~~~~~~~~~~~~~~~~~~~~~~~~~~L, CC cS -t LC\ CO CO CO I ~~~~~~~~~~~~~~~~~~~~~~~I I II QJ I l~~~~~~~~~~~~~ I & ALf Cl) |-m-| ~~~~ ~~* I 0 CD CU cmI CO C) mt- ~ ~ ~ ~~ ___ * I 1 7 0 0 co R ( (0K fO i r — o> CD?"" ^o/Ooo ~cc 11 I-) /10 c* I1 o I 0.D /c0 /"O 0 O (10, *1~~~~~~~~~~~- aIJn O (U l0 *1 O cO t cm C-7 -^lP ),, fO I/C> Olc",O / 0 CO j AL CD * I * *ol c r- U) IC0 CMi -r-\ * C ( 0 ( 0 * 1) (I +; P *(0 ( ( U CO, +10 0' ( ^ ~ ~ C -- (0 l(0*> ^ * ^ * | *< C +-C.. ( <0K) II,- - C' *|n I I - ^~~~ 7bO^?^ [ * 4 *.1 +" (0* u' L.. U^ ** Coj r~cv 1o7 +o Cl)^^cv 1O 1 I ( 0(0 *1/ U ~ *H I c -+ * ) ^ 1 r- I I ~ \,, -- co * I ai I^^n O 00O 1- CO + col^ r- I fci C^ j H * *o PI ( | | I CD 0tA G). CQ lU +- -- ^OOr- * C HIOiC wC 1/' Q, + i,^ < y l* ^+ *' (01( 0' ( -^00(0 w- * IijH U)n ~rl hO ~te 0 IO +o + N rA ol CO + *CD H - *- 43)n rC^ i * Rj (|1 * U co rA / 0 17 R 10 1(0 rt V b)0 CM H + /t0I(10 51 S-1 (/0(f0 co 0 0 *10 CM (O O -~-I~ C) * *1 II * t (i-4 Ut * mal) O *c -1 O a —--- --- co I H CO~n *l CO CO *O C vJ * * Ft 0- 0 (01(0 C * O ~ ~ ~ ~ ~ ~~ C _______ _______ VI ~ 0 U U Ua CW I Ii aE-n ro ~~~VI H&f 0 V I R0 0

86 XePrO5 [ XeT + sin T) 6 -= - " 6i = i~ 1; 1 _ =o 86) 0< a l J Equations (84) through (86) completely determine 8, 6, and r within the liquid film. The solution to this boundary value problem is now considered. 2. SOLUTION First we remember that the amplitude of the oscillations is taken to be small and expand the solutions in powers of e. Thus: * * * 2 * = to + E1 + E 2 + 6 = 60 +.C +... 87) * * * e- + +* es + e +.. o...7 ( a +9 E8 + E +..... 0 1 2 We now substitute the first two of the assumed expansions (87) into the first of Eqs. (84), the coefficients of like powers of e are equated, and after a rather lengthly calculation carried out in Appendix XI, we get: o *2 1 1* * * 2 0 tTTi 20'0T t - o 0fll l0o - 6 2 (xfr* - 6 t r = 1 88-0) o OT71OT i' -'n OT6 2 -- *0 )or o ~ O0rl O', Xre

cO * -H *&In * 0 H *0 o * 0 * GO __o + * 0 *H *0 * 0 * O Z * 0F * 0 *C -> *0 * * H 0 n- I=- - F- - I * + O ~* 0.JJ -" aL ~0 CM *.n F * 0 *0 * + 0,0 * 0 GO.+ * 0 * 0 ~_F + 0 G-O + to ++F * * *0 H * 0 I- CM * CI * *0 - cm GO 0^ *0 *H * GO *0 *H * 0 (O o_ +N * G *H *+ * - |+ l1 * H tLA * *0 *CM -O CO v - H * H *0 F *H -.,0 *0 + * 0 GO + GO * H G O * HGOf * H GCO FH " GO + " " + GO:.CMr- 1- T:,: -* H + H *H CO G *0 o * 0 r-|I IC\H H r-1 IC\j Go *: * 0 *0 * H |I ClJ + + + o GO Ln'-* + +F * H *,r + 0 *CM GO -* H * H, ( GO -" GO * H * * 0 (1' [r- (DO XO P- r( * H tO * 0 F GO * 0 G0 - * CM * co *H o *..+ aLn * ~ G0 * 0 *0 G0 Ci * a L CJ * H ~!GO Go H- + *" -_ _.n + + + Uj1n I I 4", 0,, * 0 — *0 * 0 ) 0 a'A GO cO * 0 CO,._0 & * CM R 0 AA*0 AL * I' - 0 H -H * 0 H * 0n C * * 0 Cr) * *o *' aa0 0 H *0 *0 * 0 I" * * 0 * GO *0 * * *H o- *CM si-A *%H> *0 I U:) ~ * H WI I lI II ~ + I COJ G0 rI - I I + +

88 *2 COS T + 88-2) e o 0v cos T + -o * *n 882 2X 01 X 2~T e e Next substitution of the first two of the expansions (87) into the first condition (85) gives upon equating the coefficients of like powers of E: ~~: * = 6% + 2(Xe 1) 89-0) E.: * = 5 + - O sin T + (X 1) d a d 89-1): = 6 + sin 0 at Substitution of the first two of the expansions (87) into the second condition (85) and equating the coefficients of like powers of e gives after the calculation performed in Appendix XII: 0 * * +1 sin 2(X 1) 89-2) e: o ) L2 e T + ) + Substitution of the first two of the expansions (87) into the second condition (85) and equating the coefficients of like powers of ~ gives after the calculation performed in Appendix XII' E 0 5 -(- 2 5) (5* -Xeo)0-0) o e OT 2 o 0 = *0'rl rl

89 + + sin T + (XeOT + 2 orOT 2 1 * * 60 at - Xe + sin T] 90-1) n=1 2 ( * -= 1 * * * *+ 2 *+ 1 * * * 1 * * * ~6 (+ X (X e 5 + 5 6)) + 5 ( X) j6X +2 + ) + (e5 + - + C +- ( -x - oX e x e 0) e2 2 e OT 2o *1 *o * * o* +5 6 o - -(6 +o61 ) * (X oT - "2f 90-2) and the boundary conditions (86) show that one must have: ~ ** at 0: n = =; n = 1, 2..... 91) Now Eq. (88-0) with the boundary condition (89-0), (90-0), and (91) (with n = 0) will be satisfied by: )* * e = 2a fo(); 65 = 2aS 92) 0 00~~~~~~~~~~~~~~~~-X

90 since (88-0) through (90-0) involve the time T only implicitly. Substitution of (92) into (88-0) through (90-0) and (91) show that f and a must satisfy the following equations and boundary conditions: O < I1 "f U+ f f f" = 0 95) o2 0 0 2a Xe f = 1 = 1 ~: 94) 1. f!! 1 - f' 2 f = 1 - Xef 1= 0: fo = f' = 0 95) 0 This as can easily be seen represents the steady state solution. Now (92) is used to simplify the Eqs. (88-1) through (90-1) and (88-2) through (90-2). First substitutions of (92) into Eqs. (88-1) and (88-2) yields after a calculation performed in Appendix XIII: 1i - 1 * * t f(* E:...2aX 2 -f + f' (1n~ - 0 - " *1 + 2-2 + (f2 - f - * <f - 2^'2 ^ *f + -a g3 cos T 88-la) Xe i [:a - 2a2Xl TI~ ~+ f0(^ - ~ l - 0fo L 2a 2X e2 - 6- f2f 0* 2 ) 2 * + f + 25 * + (f2- f f ) * - f2 - 2o2(f rf )6 2T-r 0 00 2 0 2 O o T

91 6a 3 5* COS T *5 = 1 cos T - 45 5 ~l - ^lj I~ + Sn + i ( fo -r + f' rl o g - fo n) iTI 1- 0 f 0 p 1* * + 2 * + 252(f' + f" ) 6 5 + 2g' f + 22 ( 0 0 1 iT +n l r ) 6* 88-2a) Substitution of Eqs. (92) into Eqs. (89-1) and (89-2) yields: ef = 5r + a a sin T + 2 (Xe-l) - dg 89-la) T1 ~ 1dS <~89-1a) 2= 2' 2 = F5 + 5 sin T + 2 (X -1) a g- 89-2a) 2 2 2 e OT e And finally using Eq. (92) in Eqs. (90-1) and (90-2) gives after some calculation performed in Appendix XIV: 2a: 2 q r = (l-nXefWo )l 2 t (e S6+ *) + a S sin T + & X - Xe ~ + a S sin T 90-la) 2 ftaFat* -' (1-xft) (\% + /*) + 2X a~26* E2an = 2a (l eo) [a(6) qI =! ~ o 1 + 5 * (X 6* + + ( * + ) sin T + a(* - Xe*) [(b + 6*) + 2 X 2* + a sin T + 1 (2a) ( - Xe ) + (2a)2 [( + 26*) + 2 (1) e + ) i 2 2X *] sin T + (2a aC (1 - cos 2T) 90-2a) 1 | 0 *""

92 Next substitution of the expansions (87) into the second of Eqs. (84) and equating coefficients of like powers of E yields after a calculation performed in Appendix XV: 0 1 - ~E: ~ 2 + - 6* (9* 4 -, 9*) - n5 6* 9* OT 0 2 o0 ( T OTI )-orl O0 OT OT = 9* 96-0) XeP OTrlrl - tfr 9* - 1 9i*) + 1 5 (9*~e t( - \|~ 9* ) 1T 2 O 2 1 O o o o - n6* 6* 9* - n (66* + 6*6* )9* = ~2 15 12 ~^ 2 1 25 ~^ + ( O2 ) 0 OT 1 q 1 1 oI T 1T O _xe_ *nn 96-1) r e 2 * + 2T 6 ~ + (6 + 2b66) 9* +- (i f* + 9) +9 -e* - - 2 2oT' 1i o 2 on 2T1 o *~ *(* * *' -TI 0 22 0e 1 * O* ( + 656 + 6* + ) 2 OT 1 1T 0 2T O 1 OT + 56*6 ) 9* *6 5* 9 9 *6-2) o iT rl o OT 2n PoX 2nn 9 2 -- z.~ -09~0 0~0 0080~0 0 re 00~9QQ 0~0~0~~~U~~0a8~ oq arl 000o0 0 ~ *~ 00 00b~~0 8 0 0~0 0 0 0 6

95 After substituting the expansions (87) into the third condition (85), equating coefficients of like powers of e and performing the calculation of Appendix XVI, on obtains: (X + + sin T = - * 1 o e 1T 2 1s 4 o o o X Pr 1I 97-1) E|: 90 65 (Xe,6oT + 2 ) = ( Xe + ~ * C 1 * *e * 1 1 ~ t* { 5* 1 5* -: 0e 6o (X 6 +-6 ) +e 6) 6(X + +- ) + 0 = e 1T 2 O o z_ 2 e OT 2 0o ~ * i&1 1 1 + (Xe6+ 2 ) + (Xe+ + s t i + ~ {~ S + 3 [G Gg + 1 * } sin T = - >' e r 1 = 2 jo 1 o XePr 98-) 97-1) 2 6* (x5*+i61* 4* 56* 1 * 2 oeOT 2 i Zi OT - 5~ +60 (Xe6 i + - + e x oI e - OT +6 (Xe 2: O e 2T 2 2 64 ~ + 8 + 6* 6 jjsin T e r And finally from the first boundary condition of (86), one must have: F 8o: 9 1=l 98-0).,-q = o o ei = o 98-1)

94 2 * 2: Q = 0 98-2) 2 ~ ~ 0 e 0~ ~ ~ ~ ~ e e e ~ Using (92) to simplify (96-0) and (97-0), one obtains: 1 * * * * 2 ~ = 0o f' - f + 2 2aX Pr oll o o o o OT 1 * * = 2a XPr orl Since the time is involved only implicitly these equations will be satisfied along with (98-0) by taking: 9 = F(r ) And Fo is a solution of the following boundary value problem: 0 < F < 1 F" + f F' = 0 100) 2a2PrXe 0 0 r 1 1 F' + F = 0 101) 2a2PrXe r =0: Fo =1 102) Now Eqs. (99) and (92) are used to simplify Eqs. (96-1) and (96-2). This gives, after a calculation performed in Appendix XVII: 1 * 2 * * * p1: - - Q_- 2~ 3 +f e -f 2a2PrX le 1 ~

95 2aftj + 6* f0 + 2-2 6 ) F' 96-la) 2 1 * 2 * * e: 2 + f a - f 2a PrXe 2T'f 2T 0 2a 0o a - * — * 2 * 1 —^ "- a( + f s + 2r 6 F' - 2a2t _ a 2 2T 0 * * * * * * 22 - (- 6 6 ) F a 4 2 1 ii T O~ 1 1 T 12 O 1 lb 1 T 1lJ (~* + f 58' ^J - ( f" + 2~2q 1^) 9r 1T6-2a) And using Eqs. (99) and (92) to simplify Eqs. (97-1) and (97-2) yields, after a calculation performed in Appendix XVIII: Z 1 * 1 ++ * + 5'] + 1 F o sin T = 0 97-la) =1 2 Lt a er 2fl +e2] +] ~l 2a + ) + 2e o 6j 2 * 22] 1 2 e: + + t + 2a ) + 2Xae + - ~l + 2 * sin T + * F (* + t ) sin T aX P 97-2a) e r The first order equations now may be solved, i.e., Eq. (88-la) with the boundary conditions (89-la) (90-la) and (91) with (n = 1) for the momentum problem and Eq (96-la) with the boundary conditions (97 and (298-1) for the energy problem 1 1 * 1 1 * 5 ~) +- ~ sin T + — F (5 + 5) sin T 21 22 2a o 1 A = 0 97-2a) The first order equations now may be solved, i.e., Eq. (88-1a) with the boundary conditions (89-1a) (90-1a) and (91) with (n = 1) for the

96 Solutions are sought to these equations in the form: 00 >* = i 2a f n() eni 103) 1 ~ ~ ~ n103) n00 5* =Q2a a, e 104) n=l 61 = 2a:F ann ei T 105) lin n=0 whereQ- denotes the real part and f n, an and Fin are complex quantities. A bar denotes the complex conjugate. Thus the complex conjugate of a is iain an O Upon substituting Eqs. (103) and (104) in Eq. (88-1a) through (90-1a) and (91)(with n = 1) and equating coefficients of like powers of ~, it is found from the calculation of Appendix XIX that the f's and the in a's; n = 13.,5,... must satisfy: in Ln(f) + n-l)f,2 + fof1 a = - O < <! -< ~11 _2i (1 - n) fo + q fo ) a- in 0o iz(n-2) I(n-2) fn ain + 2i(l-6) (Xe-) a(n-)/ in In 2 in in /n =1 P f" = (1 -' ) l( n + 6^t1)an 2 n rJ=l ff2 f' -( ~Xft n + nla + 5,7 +..... 2a in ef in 2 in 2i (1- ) Xe a + a - Xefn 6 106) -0 nf = f'n = 0 Lin. in

97 Where: 3 2 _1 d_ _ d d it L = f 2a d3(n f + n f 107) n 2a2X dT3 O df o - o e and 6mn is the "Kroneka Delta" defined by: 5mn = {~ 1; m= n mmn _ 0 m n and the f's and a's have the following restrictions: in ln f = a = 0 for n = 2,4,6,8,... in in f, a = pure imaginary quantity for n = 1,5,9,13,... in in f,n a = real quantity for n = 3,7,11,... in 108 by substituting Eqs. (103), (104) and (105) into Eqs. (96-la), (97-la) and (98-1), and equating coefficients of like powers of 5, it is found from the calculation of Appendix XX that the F's; n = 0, 2, 4,... must satisfy: K (F = - F' F(n+l)f(n) + f a n- -o L i(n+l) o i(n+l) L2i(l-6on )a(n-l + 2i(l-on)Fi(2) n=0,2,4,6,..., 1 F' + F + F I(2+n)a( 109) 2a2XePr in in o 0 (n+l) Ti=1 t e2i X,( on) al(n-l) 2 on r=o IF = 0

98 Where: 1 d2 d K - fd -n f n 2a PrXe dr O d and the F's have the following restrictions: in F =0 for n = 1Y35,5,7... in F = pure imaginary quantity for n = 0,48,... in F = real quantity for n = 2,6,10,... in 110) Before proceeding to the second order equations it will be convenient to develop a certain relation between complex quantities. Let us consider two complex variables Z and Z and let us evaluate the product: (6(Z e ) (6LZ ei) we first note that: iT 1 iT i1 WZ ei = (Z ee + Z e) 22 1 - (Z e + Z e ) Hence: (tLZ ie ) ( 2z eiT) = 1 (Z e + Z e ) (Z e -iT 1 2iT -- - 2iT 2 4 i2 1 i 2 1 -- -- 1Z + 1 - - (Z Z )+ZZ)= - +(Z Z ) + + 4 1 2 21 2 1Z2 4 1 2 1 2

99 and: iT i T 1 Z Z e + Z Z ) 111) (6LZ leT) (6eZ2eIT) 2 Z 2e4 e ZZ) +ll) 2 2 12 Next keeping this result in mind inspection of the right hand sides of Eqs. (88-2a), (89-2a), and (90-2a), and of Eqs. (96-2a) and (97-2a) shows that one might anticipate the solutions of the second order equations to be of the form: 00 = I2a 2n( )e + ( r ) 1 112) n=l 00 6n = e2a a e + a 13) 2 ~ L2n 2n n=l 00 3 = Fn (/ Inn)i F2nk( )]' 114) n=0 Upon substituting the expansions (112), (113), (103), and (104) into Eqs. (88-2a), (89-2a), (90-2a) and (91) (with n = 2) and using the relation (111) to simplify, it is found that two types of terms appear; those with a time dependence of the form e2i and those which are independent of time It is therefore necessary to equate the coefficients of e -with each other and the terms which are independent of time with each other. These calculations are performed in Appendix XXI. Then (Appendix XXI) the coefficients of like powers of 5 are equated in each of these expressions. It is then found from these calculations that the f's, and the a2n' s for n = 3 5 o.. must satisfy:

100 O<T<I n~Ln(fn) = - (n-l)f' + H — ["n o o o0J 2n n f = a - - a + 4i(l-S ) (Xe-1) a2(n-2) 2n 2n 4 in in n 1 \i 1S f" = (1 - Xe ) (n+l) (a - a l + 2a2 an 0 o an 4 in ] = 1 4 14i (1-6 )a (a - X f ) + a n=1,3,5,7... in 2(n-2) 2 in e in 2n 115) - Xft - i [(n+2) a + 2i (1-6 )Xea(n-2 - n iinin in e in + C n 8 1f = f' = O L 2n 2n Where H -— (1-5 )a + 4i(l-5 ) n 2Xe in i(n-2) in L2(n-2) - (f' + f")a 1 - - (n-1)f' f' o 0 2(n-2) 2 11 in 11 ln + (1-6 )n f" f + r2(n-l)f'f + f" f + f f"] a in 11 in l 11 O 11 o 11 in 0 in 0 n 0 in a 1+ 1-0) [nfifn - (n1)ff n ] f n=1n3l5,5~- (1-5in) ~ foI + f ) a a(n) + (f) fl 116) - f' )a -P 11 i(n-2) n and PI 0 P3 0

101 5 3f f" - 2f' f3 + Lof + Ps 2 f 13 -13 13 o i +13 + 3f f 2 f ai + ft)a a+ 0 13 0 13 131 11 13 + (fl3 - f a } 13 13 11 allac ll C1 = (l-Xef) +a 2a (all - Xefl) 2 C3 - (1-Xef o1 a3 + iXe a ] + 11 a13 + (all - Xefll) F4al + 2 i Xe all + 2a (a - Xef ) For C5 (1 - Xf) a l a + i Xe a + a [a a1 + i Xe all] + all a,5} + (all - Xef1l) 6 a15 + + 2 i Xe ai3 + (a3 - Xef3) 4a3 +2 i Xeal + 2 all (a15 - Xef15) o o o 0 0 0 o n o o ooo eo0 0 0 0 0 o o0 0 0 0 o a o 0 0 a e o * o * o * o o a ooo And that for n = 2,4,6,...; a - f; for n = 2,4,6,. 118) 2lan sn

102 And comparison of Eqs. (108) with Eqs. (115) through (117) shows that a n f = real quantities for n = 1,5,... 2n 2n a, f = pure imaginary quantities for n = 3,7,... 2n 2n 119) The calculations also show that for n = 1,5,... the f's and the a's 2n 2n must satisfy: 0<1 tLn ): n-l )f + HO + v\ if = a - -a 2n 2n 4 in 1,f, = (1 - Xefo) (n+l) (a i - i ( 2 n eo 2n 4 in 2 in -f') a2 - Xe - {(n+2)aln - 2i (1 n=l,5,.. i n 2n e n: 5l )X a + in + C in) e i(n-2) 8 n =0 ~*f f' = 0 an an Where: 3 - Hn = 2- (1-b )ai(n2) - lfn (nl)fllfn 2Xe in in n in + (1-n )nf0 in + [2(n-l)f' f + + foff a ] i n + (l-n) [nfo f - (n-l)fo fn + ff]n ] | Pn=l,5,.. P1 = 0,121)

103 "~P 3f1 f ~2f' f + f P = 2 13 13 13 [ 13 + 3f"f + 2 f' f al + i + i ) a13 0 13 0 13 13 11 + r + 3f' ) alT 13 13 1 And: C (1-Xef') a a + 2a (a - X f' ) 1 0 2 11 11 e 11 = (1-Xefo) 5 ] + a13 ai X a la1 + i X all + (.ee 2 For n=r r 122) - Xef') [6 a15 + 2 i Xea13 + (a13 - Xefi3) a13 + 2 i Xeeal + (a15 - Xe fs5) 2all And for n / 1,5,o.. 2n 2n And~ a, f real an an

104 Next substituting the expansions (103) through (105) and (112) through (114) into Eqs. (96-2a), (97-2), and using the relation (111) to simplify, it is again found that the only type of time dependent terms which appear are those whose time dependence is of the form e2iT. Therefore the coefficients of eT are equated to one another and the time independent terms are equated to one another. After carrying out these operations in Appendix XXII, the coefficients of like powers of e are equated in each of the resulting expressions. From this is found that the F2n's for n = 0,2,4,...must satisfy: Kn(F2) = 4 i (l-6)F - F (n+l) f 0 < r < 1 + fo a( ) + 4 i (1-6 )a a + E 2~ (n+m) on 2(n-i) n K1a2PXF-n12n ~2+n) (arl n=0,2,4,.. ir --- F1 + F + Fo (2+n) (a - a 2a2Pe 2n 2n 2( n+ 4 n l(n+) i> nl = 1 <;. 124) + 4 i Xe(1-6on)a F +4 B 0 2(n4 n n T = F2n = 0 F2n Where: 1 1 -- all F fll - F' (fll + all fo) E~2 0 2 o 2 o E2 = - 2 allfls + 2 i all all T + al3fli F1 + - 2 F12 (f'i + fo aii) - F' (fi + fl al )

105 - o (3 3+ f a 3 + i al ( 2Fo - Fl ) 10 13 13 10 10 0^ 1 _ 1 r' E4 = - - 5all f15 + 4 i T all a13 + 3a13 f13 2 125) + a15 flj Fo + 4F14 (fn + a11 f ) -Fi4 (f1 + fo a1) + 2F12 (fl3 + f a13) - F11 (3f3 + al3f) -Fo (5f15 + foa15) + i all (2F12 - Fia2) + i a13 (2Flo - T Fld) * o e e o o.o o o o e o e o e* o* o e e ~ * ~ o* o o o e o o o e* o o ~ ~ eZ e o * o@ o e o o o * ~ o o * ~ * o ee *e o oo e e ~* ~ ~ ~ And: B 1 1 B~ = a1 F1o + 2 a11 a1 F~ B2 = 2 Fl1(4al3 + 2 i Xall) + all F12 + all ( a13 2 2 1 + i Xe all) F0 + all al3 F 0 + F14 all + al (2 a15 + i Xeal3) F + al3 (2 a13 + 126) i Xeall) F + 2 all a15 Fo 0 2 0 0..... 0ID. 0 Q. 0. * 0. 00.0 00 e a 00 0 a 0 0 0 0

1o6 And for n = 1,3,5,... we find that: F() - 0; for n = 1,3,5,... 127) 2n A comparison of Eqs. (108), (110) and (119) with Eqs. (124) through (126) shows that: F = real quantity for n = 0,4,8,... 2n 128) F = pure imaginary quantity for n = 2,6,... The calculation of Appendix XXII further shows that the F2n must satisfy for n = 0,4,... 0 1 (F) = - F' n+l)f + fa2(n) + En an 0 (n+) 2(n+l) n 1 V i - F' + F + Fo(2+n) (a(n - a(n )) 2a2[PX 2n 2n 2(n+}) n+l)) n =i1 for -F + B = n=0,4,o. 4 in n -) 129) rl =0 ~ a~[2n rl = O <Fa = O Where:, 1 r1 - 1 - - E = — F' all fll - F' (f1l + f all) o 2 o 2 o E4 =a - all f1 + 3a13 f13 + a15 fl Fo + - F4F14 (fii + alf o) - F14 (fll + foall) + 2F2(fi| 21L

107 + fo a13) - F12 (3f3 + fOal3) - Flr (5?15 + fo ) O<rll + i al (2F12 + 1 F12) + i a13 (2Fio - Tr Fl ) 130) o ~ < o ~ o e o e ~ e o o ~ ~l e ~ l e o o o o o o ~ e * o ~ * ~ e o ~ ~ o e ~ o ~ e e e o0 and: v - 1 - Bo = all Flo + 2 all all Fo I 1 - 1 a1 B4(6 + 2 i XF al3) + 2 F (4al3+ TI 1 * > 131) 2 i Xe all) + allFz4 + all ( as15 + i X a13) F. + a13s ( a13 + i Xeall) Fo + - a1 a1e F 2 And for n f 0,4,8,..o we find that: F2n(dt) 0- for n a 0,4,8, oo. 32) And the F ns are real otherwise. The problem has now been completely formulated in terms of a set of simultaneous ordinary differential equations with their appropriate boundary conditionso These equations can be solved successively, ioe., once fo and a are known one can solve for fll and all etc. Then once these are known we can solve for fl3 and al3, etc. Once the solutions to the momentum equations are known (fm' s, amn s), one can solve

108 successively in the same way for the solution to the energy equation (i.e., the Fmn's). The remarks at the end of Section II-B,, about the general nature of the equations and their solutions are also appropriate here. The discussion will not be repeated now. It now remains to derive explicit formulae for the velocity, shear stress and heat transfer coefficients in terms of the solutions to the equations derived above. From the first of Eqs. (71) one has: u =6 by And using the dimensionless quantities defined in (77) we obtain: u _ Xe 1 U Xe a* 133) Next the wall shear stress is given by: bu I a2 w = L by I=o y2 y=0 y=O And using the dimensionless quantities of (77) we have: w 2 A? "12 1 - = 6*2e 6 2 134) 2 pU T n=0 And finally the heat flux at the wall is given by: y = - k y 135)

109 and using (77) we get: Nu -q-x / - j U"OX k(Tk-TW) v Substitution of the expressions (87) into the relations (133), (134), and (136) yields after the calculation of Appendix XXIII: U * 2 * O = u + ~U*j + C U2 + 157) U0 * r... 1 12+8) (T* + ET + E T +... ) 18) Where: u2oe 0 2 2 6"o ~~Xp 6p Uo N u l * * 140) 2 | 2= 4X (qo + l 2 + + " ( 1 -) 0<~~< 1 6, 219) 619 Where~ u* X- *Ot LI e oil o 6* 0 u2 = - * - 50* 2 o = 5 0 - 5 j *2 1ln 6ia=~ oi 1 = Lm orv i 141)

110 * F2S 262 * 5i 24 ^ 2 = 2*'1- i8om - 6- torl-6 * 5 * _ = * o o oI 1 X' * e i o * ( _ 6* O2 6* 5 i 6* JI$ U2 - e o~] - - in - 6- o 0 nO Now using the relations (91), (99), (103), (104), (105), (112), (l13), and (114) with the conditions (108), (110), (119), (123), (127) calculations of Appendix XXIV: u = Xef' o o ul = gin n eT n=0 143) 00 2 = (g2 eiT + )n n=0 where for n = 0,2,4,...:

Ill gin = Xe(f(+) - f al(n+l)) g2 = Xe(f(n+i) fo a() - hn ) = Xe(f - f' a hn) g2n 2(n+l) o 2(n+l) g2n e^ 2(n+l) - (n+) - ) O < T < 1 and: 144) 1 o 2 h2 = 1 t13 fl1 + all f13 - 2 all a13 f h4 = a5is fll + al3 fl3 + all fl - fo(2al al5 + a13 a13} LA 1 ho 2 0 h - 1 all ( fil -f af)5 = l tl fli + al3f13 + anf15 - f0 (2a.a.5 + a3a13) and: hn = 0 for n = 13,5,... h = 0 for n 4 0,4,5,.. n

112 T = - f o a o 00 * a n iT T =er^-2e e n=O * 2T 145) T2 L (e 2 e2 2n)tn n=O Where: _ fInf -1 2fo I a en = l(n+) 0 (n+l) e f - 2f' " a - 2n 2 (n+l+) o a2(n+l) n e2n f2(n+l)- 2fo a2(n+)- r and: 146) r = all (2fl - 3fo all) TI =0 r2 = 1 2al3 fll + 2all fl3 - 6all als fJ r 2 fl- u f3 2 r4 = 2- 2a15 fl + 2a13 fl3 + 2al fl - 3f (2all a15 + a13 a13)] r0 = 1 x (2f1' -5 all)

113 r24 = - L2ais fjl + 2a13 f" 2a f - 3fo (2a911 a15 + a13 13) and: rn = 0 for n = 1,3,5,... r = 0 for n 0,4,8,.. n = 1 q* = Ft o 2a ~ o00 q. -L X G1 n eT 147) ql = c3 —2a, n n=0 q =SG e G 2 ^ 2a 2n 2n n=O Where: G =F' - a F in in i(n+1) 0 G = F' - a F1 - R 2n 2n 2(n+)F' o n G =F - a F' -R 2n 2n 2(n+l) o n and: RO = _aaniFo 11 aa FF 0 R - all F' all allFo al 0 2 1O

114 R4 = - [a15 F + a13 F12 + all F14 - (2all a15 + 2 10 al3 ai3)F ] R = R = a ll Flo - all all F R4 = al5 Flo + al3 F12 + all F14 - (2all al5 + a,3 a-3)Fo] al3 a13)ro] and: R =0 for n = 13,55 oRn = 0 for n 0,4,8,..e It is convenient to have explicit formulas for the amplitude and phase of the first order harmonics (ioe., the coefficient of e to the first power) of the various quantities defined above. To accomplish this we note that they all can be put into the form: 00 Z ^= X n n ei n=O

115 Where: an imaginary for n = O04,8,.. an real for n = 2,6,.... c = 0 for n = 13,5,7, **. and: X = 0 or 1 Then: 00 Z =- I j ei( +(P') n=0 n=O 00 = n e X In sin. (T + cp + n=0 00 = | O ~ en | sin. (T + cp' + ) n=O 00 Where: n 00 -l an g nn= zl - arg = tan. n=0 n=O

116 and: cp = - (cp - ) Hence: 00 00 1 00 00 1 =x nnj m = V4 V nmn+m izl = n g 5m = a |n m n=O m=O n=O m=O -X |a | 21 + [2 + 2^^i?4 + 12 2541 -? {ozj {+ l2 lol:.J + Hence the amplitude is given by: zi = |1 + 1/2 a + I 2 + 150) and: -1 t + +o 4 +. -1 rai 4 Cp' = tan i = tan 2 + ^ + _. and therefore: = - -tan1 [I _o + 4 4 +o. 151) 2 a2 11) Now comparison of (149) with (104), (143), (145)), and (147) shows that we may write: u = jul sino (7 - cP)

117 T1 = | Sl l sin (T - (P ) *q |q I sin. (T - cpq ) 11 -1 61 = | 61 | sin. (T Cp5 ) Where: 1*1l I~-~ I Jl rj g12 -- -I - 2g14 glol lull = lol l F1+ rgg1 g+ g1 + + m = _ — tan1 [gl 4 t+v+' 1 2 leOg+[12 ig2 j+ IT1u L I 10 + 1 1 2 9 + 10g 4 +1 = -+ -.. C1 2 g~2 Lie2 iel2 + i Ig* = 12a 1 i i i + Ll2 2eGl4eio 4 + o 2Ca 2 U -t12 1 r_ o 1E CP - - tan o el__+ 4 ++.+ 1 2 L2 ie G 12 + Gl + 1- [ G1+ 2G1 a10 1 ] + = 2alai,| { 1 + 1!2 1 G__12. o~ L 2 Qlo!L aI I l J -1 r 1Glo G14 1 CP- = - tan _ — - ~ +4. 2 [La13iG ial3 ii:. i lala 2al5al 4 laill la,,

CHAPTER III NUMERICAL PROCEDURES For both the flat plate problem and the cylinder problem with 2 E > o01, there is the task of solving successively a system of ordinary differential equations. That is, the equations under consideration are such that the solution to the n1th equation depends on that of the n-lst equation, n-2nd equation, etc., but not on that of the n+lst, n+2nd, etc. The momentum equations (which of course must be solved before the energy equation) are of the third order. They each involve however an arbitrary parameter which must be determined by an additional boundary condition imposed on each of the equations. Thus there are two boundary conditions at q=O (independent variable in the notation) and two at Tr=1. 2 In the case of the cylinder problem with E <.01, only the energy equation is formulated in terms of a system of consectutive ordinary differential equations. In all cases however, once the solutions to the momentum equations are obtained the consecutive ordinary differential equations resulting from the energy equation can be solved. These are of second order with two boundaryconditions; one at ==O and one at r=1. The Runge-Cutta method which has been used to solve all these equations is a numerical procedure which performs a marching solution 118

119 to a system of n simultaneous first order equations. That is, it calculates numerically the value of the solution at any desired number of points. In order to apply this method to equations of the type considered here, some additional procedures must be used with it. First the n'th order equations which are to be solved must be reduced to a system of n first order equations. This is easy to do and an illustrative example will further clarify the procedure. Example: Consider the third order differential equation: Xl" +ax"+bx' = f(x) and let us define: U1 = x u - xi U3 = X1 It then follows that ui = U2 U2 = U3 u3 = f(ul)-au3-bu2 which is the desired decomposition. Next for a differential equation of order n the first n-l derivitives must be specified at a single point (r=0 in this case) before the Runge-Cutta method can be used to calculate the solution to the equation.

120 In order to accomplish this the derivitives at Tj=O which are unknown are first guessed at. The solution is carried out to r=1 by the Runge-Cutta method. Now the desired solution fd must satisfy a condition of the form:,~f d * of ( n - i ) G(fd(l),fd(l),... fd (1)) = at this point. The calculated solution fc is then substituted into these relations and the errors R: R = G(fc(l),f(l)...fe(n-l)(l))-G(fd(), fd(l)...fd(n-)(1)) are noted. On the basis of these errors the initial guesses are modified. Since many equations had to be solved it was desirable to have an efficient scheme for iterating on the initial conditions. This was accomplished by incorporating a procedure which could remember how much of a percentage change in R was caused by the previous change in the initial conditions and using this information to calculate what would be the most suitable change in the initial conditions. For example, suppose it is known that the previous change in the initial conditions decreased R by only a small percentage, then it would be desirable to make the next change in the initial conditions in the same direction as the previous one (since it decreased R and that is what is required) but it should be much larger (since the previous change in the initial conditions only caused a small percentage change in R and we want R to go to zero as rapidly as possible). In order to make these ideas quantitative consider the equation:

121 AIN+1 1 CSCH.(R/RN_- 1) AIN CSCH.(l) which is sketched in Figure 6. Where AIN is the N'-th increment in the initial conditions, RN is the value of the error in the boundary condition after the initial conditions were incremented by AIN. RN_1 is the error that existed before this, and AIN+1 is the new value of the increment. First it is noted that if RN/RN_ is negative then that means that AIN was made too large and it should be decreased by a certain percentage which is less than one (otherwise the error would be increased over the previous one). Hence: AIN+1 AIN is the proper change for AI to decrease RN in this case. That this function accomplishes this, can be varified from Figure 6. Now also the more negative RN/RN_1 is the closer AIN+1 should be to AIN in order to bring the error back closer to zero. Next it should be noted that 0 < RN/RN-1 < 1 means that the previous change in AIN accomplished a reduction in RN and therefore the next change in AIN should be in the same direction as the previous one. But if RN/RN-1 is close to 1 then the change in AIN was not to effection in reducing RN and hence AIN+i should be much larger than AIN and similarily if RN/RN-1 is close to zero then it was very effective and AIN+1 need now only be a small change compared with AIN

122 AIntl/ n Asymptotes v —----' -— d1 --- ^ ------ R n/R n-, -I Figure 6. Plot of AIn+i/AIn versus Rn/Rn_- for iterative procedure.

123 in order to bring the error close to 0. Finally notice that if RN/RN+1 is greater than 1, then the change, AIN, was in the wrong direction and AIN+1 should be in the opposite direction; i.e., AIN+i/AIN should be negative. Note also that it is now necessary to change AIN by such an amount that it not only compensates for the previous AIN moving the error further away from 0 but that it also decreases RN. It should further by noted that RN/RN_- close to one means that AIN didn't change RN very much, and a large change in AIN+1 will be necessary and conversely. That this will all be accomplished by the given function can easily be varified from Figure 6. It has been found that this function works very well in practice and that it usually requires less than ten iterations to get the desired solution when it is employed. It is still necessary to discuss how the parameters appearing in the momentum equations are to be determined. It can easily be seen that the same method as is used for the initial conditions can be used here. That is, an initial guess can be made for the value of the parameters. Then, the error in the boundary conditions can be checked and a correction can be made using the same function as described above. In the case of these equations, however, iterations must be performed on both an initial condition and the values of a parameter in order to satisfy the two boundary conditions at -=1l. This presents no problem since for a given assumed value for the initial condition,

124 iterations can be performed on the parameter until one of the boundary conditions at r=1 is satisfied. When this is accomplished, the error in the other boundary condition can be noted, the function described above can be used to calculate a new value for the initial condition, and so on until both boundary conditions are satisfied. In the case of the energy equations only a single iteration loop need be performedo Thus the procedure for both the oscillating flat plate problem and the circular cylinder is to solve each of the equations successively using the iteration procedure described above and storing the solutions to be used in the higher order equations. The Runge-Cutta method is also used for integrating the equations for the drop trajectories, but since this is an initial value problem no special difficulties are presented here. There calculations are all performed on a 7090 computer using the "MAD" language. The individual programs are listed in Appendix XXV.

CHAPTER IV DISCUSSION AND RESULTS A. Convergence of Solutions The computer programs listed in the appendix were run on the IBM 7090 digital computer located at The University of Michigan Computing Center. For those programs (Appendix XXV) which involved the integration of differential equations by the Runge-Kutta method it was necessary to determine the proper number of subdivisions into which the domain should be divided in order to get accurate answers (see RungeKutta Subroutine in Michigan Executive System). This was accomplished experimentally on the machine by calculating the solutions using various numbers of subdivisions. The results of these experiments for the differential equations describing the liquid film on the cylinder with 2 E >,01 are shown in Figures 7-a and 7-b. It can be seen from these that increasing the number of subdivisions past 100 results in no significant change in the solution. Of course, a point will be reached when round off errors become excessive and the solutions will begin to diverge again. But this could not be demonstrated, since it would involve excessive machine time to reach this point. It should be noted that the higher order terms are tabulated in these figures since the lower order terms were found to reach a stable value at a much smaller number of subdivisions. 125

.034.033 in to.032 E2l.031.030 50- 100 150 NUMBER OF SUBDIVIONS Figure 7-a. Convergence of function f% with respect to number of subdivisions used for numerical calculations in cylinder problem.

.02 E2 = 1.~ # —,%~ ~ ~ ~ ~ ~ ~~~P = 10 to~~~~~~~~~~~~~~~ %... LL!.01 50~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~P 10 1 NUMBER OF SUBDIVISIONS Figure 7-b. Convergence o cton F o Subdivisions used for numeri c with respect to be use fo nuercal calculat~ions in cylinder Problem.

128 2.4 4 Terms 3 Terms 2.2.-^2 Terms I Term 1.8 1.6.2s ~~~~2 0.2.4.6.8 1.0 1.2 1.4 1.6 x/Ro Figure 7-c. Convergence of series for liquid film thickness with respect to the number of terms retained in cylinder problem.

129 I Term.22 - 4 Terms 3/ Terms.20 / / 2 Terms.18 -.16 -.14.12.10.08- y/8 1/2 E2=1.06.0402 0 1 1 1 1 I I I 0.2.4.6.8 1.0 1.2 1.4 1.6 x/Ro Figure 7-d. Convergence of series for liquid velocity with respect to the numer of terms retained in cylinder problem.

1530 I.'~ — — ^ ^^ —---— I Term 1.0.9.8 2 Terms 3Terms 4 Terms.7.6 0 z.4 *3 2 E =1.2..1 0.2.4.6.8 1.0 1.2 1.4 1.6 x/Ro Figure 7-e. Convergence of series for Nusselt number with respect to the number of terms retained in cylinder problem.

131 Similarly the results of experiments on the equations for drop trajectories are shown in Figures 8-a and 8-b, and those for the energy equation in the liquid film with E <.1 are shown in Figure 9-a. The results for the equations describing the liquid film on the oscillating flat plate are shown in Figures 10-a through 10-d. In the last two cases, again only the higher order terms are shown since the lower order ones become stable at a lower number of subdivisions. In the case of the flat plate, it can be seen that a hundred subdivisions will give very satisfactory answers. It is in general impossible to prove convergence of the expansions that have been imployed in obtaining solutions to the partial differential equations. In the case of the cylinder problem for both large and small E it is hoped that a convergent series was obtained. In the 31 case of the flat plate only an asymptotic series was sought, (a coordinate expansion plus a parameter expansion in Van Dyke's terminology). The best one can hope to do is to show that increasing the number of terms retained in the expansion has a small effect on the solution after a certain number of terms. This has been carried out for the expansions used in the cylinder problem (for both large and small E) and the results are shown in Figures 7-c through 7-e for E2>.01 and in Figure 9-b and 9-c for E <.1. Because of the extreme complexity of the solutions for the oscillating flat plate problem this could not reasonably be done.

152.666-.664 -.662.660 \.660C \e Rrd/Ro =.004.658 - Yd 500.656.654.652-.650.6500.10.20 STEP SIZE Figure 8-a. Convergence of solution for Ydf with respect to size of increment used in integrating drop trajectories.

153.728.726.724.722.720 8\ rd /Ro =.004 ~.~~718"~ \ ~2RoU //vg= 104 I~~.716 \YdoD* =.500.716 \.714.712,,ICQ I I.710 - | 0.1.2 STEP SIZE Figure 8-b. Convergence of solution for cpd with respect to size of increment used in integrating drop trajectories.

154 4.80 4.79 4.78 4.774.76 LD 4.75 4.74 E=.04 /rd/Ro=.004 4.73- 2R oU//vg =104 Pr =10 4.72 4.71 4.70 0 20 40 60 80 100 120 NUMBER OF SUBDIVISIONS of subdivision for cylinder problem with E <.

135 9.96 9.95 iO 9.94 -; / E=.04 9.93 / rd/R=.004 2RoUOD/z =0 4 9.92 Pr = 10 9.91 9.90' 9.89L —— ______ 0 20 40 60 80 100 120 NUMBER OF SUBDIVISIONS Figure 9-b. Convergence of function Fe, with respect to number of subdivisions for cylinder problem with E <.1.

136.93 2= 1/2 4 Terms E =.04 5 Terms rd/Ro=.004 / 2 Terms.92 -2RoUc/vg104 / Pr =10.91- _ / 3 Terms.90.89.88 —'= —--—.88 —--- ~I Term.87.86 85 I I I I I 0.2.4.6.8 1.0 1.2 X/Ro Figure 9-c. Convergence of solutions with respect to number of terms retained in expansion for cylinder problem with E <.1.

.0137 Xe=0.5.0136 \ in.0135.01 34' 0 0 20 40 60 80 100 120 NUMBER OF SUBDIVISIONS Figure 10-a. Convergence of function fl5 with respect to number of subdivisions used for n8merical calculations in flat plate problem.

.00642.00640.00638 \. \05 0 00636 6.00634.00632 o063O0 20~ —~0 120 00650L3~-0 40 6 NUMBER OF sUBDIVISIONS N ~0mBERgence ~j ~f~i~fsith respect to number ox function f25 ai flat plate 10-b. Convergence calculations Figure ue for numeric of ubdisionsuse -problem.

.0154.0152 -.0150.0148 V P~~~~~~~~P 10 - ~.0146 X.5.0144.0142.0140 ----- 0 20 40 60 80 100 120 NUMBER OF SUBDIVISIONS Figure 10-c. Convergence of function F24 with respect to number of subdivisions used for numerical calculations in flat plate problem.

.0031.0030.0029.0028 | Io0.0027 Pr10 Xe.05.0026 0 20 40 60 80 100 120 140 NUMBER OF SUBDIVISIONS Figure 10-d. Convergence of function F24 with respect to number of subdivisions used for numerical calculations in flat plate prob lem.

141 B. Numerical Results 1. THE CIRCULAR CYLINDER The solutions to Eqs. (41) through (43) have been obtained numerically by means of the computer program listed in Appendix XVW. The results are available for values of E2 between.01 and 100 and Prandtl numbers between 5 and 13. These results can be used to calculate A*, 6*, and Q* by using Eqs. (26) through (28). They have also been used to calculate the terms of Eqs. (38) through (40) for velocity, skin friction and Nusselt numbers respectively. The liquid film thickness has been plotted in Figures 11-a and 11-b for various values of E. The local wall shear stress is plotted in Figure 13-a and the local Nusselt number has been plotted in Figures 14-a, 14-b, and 15. Typical velocity profiles are plotted in Figure 12. For E < 0.1 it is first necessary to calculate the liquid drop trajectories before any of the physical quantities such as skin friction can be found in the film. The calculation of the drop trajectories has been carried out by using the computer program listed in Appendix XXV to integrate Eqs. (17*). This program also calculates the quantities of Eqs. (17*A.C.) which must be known to solve the governing equations for the film. Typical drop trajectories are shown in Figures 16-a through 16-i. The quantities of (17*A.C.) were tabulated for values of rd/Ro ranging 3 5 between 0.0004 and 0.01 and values of 2RoUJvg between 10 and 10.

22 2.0 1 1.8' 1.0 *E —.Iv t Z. q i o 1.2 4E.OZ01 - FigUre -.o1uid film thickness for cne i

2.2 E= O 2.0 E2: a2 x IE: 10 0, 1.6 -___ 60 1.4 1.21.0 _ —-— 1 *0.0.2 4 6.8 1.0 4 x/Ro Figure 11-b. Local liquid film thickness for cylinder with E2 >.01.

144 1.8 0d/R =.004 2 R /- o.103 /E'.08 1.6 4 14 /E =06 1.42 v4**%E =.04 8r 0 1.0 cV a.81 cv E.02 C O o-6 -- — L__ 04.2 0 0.2 ~~~~.6.8 1.2 X/R 12 Figure n1 c Fig z~. 3J c~. L o c a l 3Lqgu~ d P i th clc k n es s f o r c.y 2i n ce r w it 12 ~ ~ o.j,.oo4 an 2R3, "bun/de v

&O ZpunF RF; pptTin Toc P- T ~ >t a T — —' ssau-9'~ 4 r -J P' nb >E' 80 1133i v i 0P-TT anT t00' o~/p, - 9' 0 tj -=~~/~P, —-- 490 20=3 A _. 08 90'-3 1l 90-=3 1 GE ^tT

146 E =.08 2.4 2.2 E=.06 2.0 rd/Ro =.004 2RoUoo/zg 104 1.8 E=.04 1.61.4 K.2 E=.02 I. - 4.2 0.2 4.6.8 1.0 1.2 1.4 x/Rn Figure 11-e. Local liquid film thickness for cylinder with E < 0.1, rd/Ro =.004 and 2RoUo/vg = 104.

147 E =.08 2.0 E=.06 1.8 E=.04 1.6- rd/Ro =.001 2RoUo/Vg 105 I 0 1.4 1.2 Vu E=.02 0 1.0 60.84.2 0.2.4.6.8 1.0 1.2 x/Ro Figure 11-f. Local liquid film thickness for cylinder with E < 0.1, rd/Ro =.001 and 2RoU<o/vg = 105.

148 rd / Ro.001 IA4 2RoUo/g 104 1.2 X E=.08 1.2l8^ /0E=0E.06.04,y. —-6 —-.E=.02 4..20.2 4.6.8 x/Ro Figure 11-g. Local liquid film thickness for cylinder with E < 0.1, rd/Ro =.001 and 2RoU/g = 104.

149 1.2 E=.06 I N./ E=.04 0 60.6 - | -E=.02 4 —F-^ ^rd/Ro=.001 2RoU /g 5 x 103.2 0 I I I 0.2.4.6 x/Ro Figure 11-h. Local liquid film thickness for cylinder with E O.1, rd/%o =.001 and 2R0U,/Vg = 5xl03.

150 rd/R =.0004 2Ro U,/vg 105 1.4 E=.08 E=.06 1.2 E=.04.8 1.0 0 (%1.8 E E.02.6 10.4.2 0.2.4.6.8 1.0 x/Ro Figure 11-i. Local liquid film thickness for cylinder with E < 0.1, rd/Ro =.0004 and 2RoUn/Vg - 105.

2 2 2 22 E =.I E.I E =1 E = E I.0.8.6 6<o / f < xx/Ro=.5.2.1.2.3.4 Figure 2. Velocity profile in liquid film at for cylinder e 12 Velocity profile in liquid film at xE >.01. for cylinder with E2 >.01.

152 E = 1.0.3 E =10.1 1 / / \E.10 0 _ 0.2.4.6.8 1.0 1.2 1.4 x/Ro Figure 15-a. Local wall shear stress for cylinder with E2 >.01.

153 rd/Ro =.004 2RoUO/vlg 103.07 E=.08.06 06 ~E =.06 958 A.05 0 ^ /"~ E=.04E a 9 " ~uE.04.03 02.01 I I I I I 0.2.4.6.8 1.0 1.2 x/Ro Figure 13-b. Local wall shear stress for cylinder with E < 0.1, rd /R =.004 and 2RoU /vg = 103.

154 rd / Ro =.004.10 2 Ro U/vg 5 x 103 E =.08 E=.08.09.~08 - / E=.06.07.06 - E =.04. 06 0.2.4.6.8 1.0 1.2 1.4 cQ..05 E =.02.04.03.02.01 0.2 4.6.8 1.0 1.2 1.4 x/Ro Figure 135-c. Local wall shear stress for cylinder with E < 0.1, rs/Ro =.00% and 2%U,/vg = 5x103.

155 rd/Ro=.004.10 2 RoUco/ 104 E.08 E =.04.06: Qa.05-.0 6/ E=.02 0.2.4.6.8 1.0 1.2 1.4 x/ Ro Figure 13-d. Local wall shear stress for cylinder with E < 0.1, r /R =.004 and 2- U /Vg = 104. rd/Ro:.00 and 2RnUoo/Vg 1O4.

156 rd/Ro=.001 2 Ro Uct/vg 5x 103.05 - E=.08 0o&I_~ //,E =.06'3.- |CE=.02.02.01 0.2 4.6.8 x/Ro Figure 13-e. Local wall shear stress for cylinder with E < 0.1, rd/Ro =.001 and 2RoU /vg = 5x10

157 rd/Ro =.001.06 2 Ro Uo/g 104.05 E 0.8. 03AE=.06 0.02 F ) / E=.04.02,01 0.2.4.6.8 x/Ro Figure 13-f. Local wall shear stress for cylinder with E < 0.1, ra/o =.001 and 2RoU/vg = 104.

158.10 rd /Ro =.001 2RoU</vg =105.09 E =.08.08 / E=.06.07.06- f/ E=.04 Q,.05 -~ ^ _ /// E =.02 0.2.4.6.8 1.0 1.2 1.4 x /R Figure 15-g. Local wall shear stress for cylinder with E < 0.1, rd/Ro =.001 and 2RoIU/vg = 105.

159 r /R =.0004 2Ro UD/Vg = 105.06 E=.08.05/E=.06 *u o E=.04.03 // E=.02.02.01 0.2 4.6.8 1.0 1.2 x/Ro Figure 13-h. Local wall shear stress for cylinder with E < 0.1, rd/% =.0004 and 2RU/vg = 105.

160 Pr = 5.0.9 2.8.6- E =.I'8 1 \.7.6 01 \.1r 2!-o.4 Z E:.OI.3.2.1 0 t. I.I.. I I I.I 0.2.4.6.8 10 1.2 14 x/Ro Figure 14-a. Local Nusselt number for cylinder with Pr = 5.0 and E > 0.01.

~~~I.'~161 6.l 0 ^ ^ -E =. 10. 2 \.5 z E = 100.2 Pr =13.4 0.2.4.6.8.1.0 1.2 x/Ro Figure 14-b. Local Nusselt number for cylinder with Pr and E > 0.01.

*OTXS = ^A/n00fl pus to00 = ~a/P T'O > E'0' = d Tqq-TA capuTTlhO loj laqmwnu, qI-ssnN sa I'OmI *o-tiT asjLS O~/x Zl O'1 8' 9' 0 Z' 0 I\. ^ I0 I X I \^ \ Z0'=3 z 0 90=3 o 8 80'= 3 ~ O'S = d ~ 01 xg = B/mn ~OZ t00oo = O/ P 39T

163 E =.08 r /R =.004.5 ~ 2 Ro Uo /vg = 5x 103 Pr = 10 E=.06.4.2 9 E0 Q:: 0 1 I I I I I I 0.2.4.6.8 1.0 1.2 X/Ro Figure 14-d~ Local Nusselt niumber for cylinder with Pr = 10, E < 0.1, rd/Ro =.004 and 2RoUloo/Vg = 5xlO03

164 rd/Ro =.004 4.4 2Ro UWv/Yg =10 Pr=5 E =.08 E.3 N = _a z E=.04 \ \ 0 1- I I I I I I I 0.2 4.6 8 1.0 1.2 1.4 X/Ro Figure 14-e. Local Nusselt number for cylinder with Pr = 5, E < 0.1, rd/Ro =.004 and 2RoUo/vg = 104.

165 E=.08. "5 - rd/Ro =.004 2Ro Uc/vg =104 E =.06 \Pr = 10.4E =.04.39 0 3.2 z E=.02 0 0.2.4.6.8 1.0 1.2 1.4 X/Ro Figure 14-f. Local Nusselt number for cylinder with Pr = 10, E < 0.1, rd/%o =.004 and 2RoUo/vg = 104.

166.3 rd/R=.001 2 R o/2RoUO/vg=5x103 8'2 Pr =10 EE =.06 o0 E=.04 z.1i E=.02 0 I I I 0.2 4.6 X/Ro Figure 14-g. Local Nusselt number for cylinder with Pr = 10, E <.1, rd/Ro =.001 and 2RoUo/vg = 5x103.

167.5 rd / Ro =.001 2 Ro Uo/Vg =104 Pr 10 r4 3 E=.08 8.: E =.06 N.\ J' ^.2 E=.04.1 E=.02 0 l I I I I I 0.2.4.6.8 1.0 X/RQ Figure 14-h. Local Nusselt number for cylinder with P E <.1, rd/%o =.001 and 2RoUoo/vg = 104.

168 rd /Ro =.001.5- E =.08 2Ro Uc/yg = 10 Pr =10.4 E=.06 3- E=.04::: 8.3 \ 0 N.2 E=.02 \.1I 0 I I I I I I 0.2.4.6.8 1.0 1.2 X/RQ Figure 14-1. Local Nusselt number for cylinder with Pr = E <.1, rd/Ro =.001 and 2Rolv/vg = 105.

169 1.I 1.0.9.8 e2 1.5 2 E 2 = 1.0.~~~~~~~~~2~~2.7,~~~E-..75.2 X/Ro Figure 15. Peak values of local Nusselt number for cylinder with Pr = 10. w.5 z.4.3..2.1 0.2.4.6.8 1.0 1.2 1.4 X/Ro vith P, = 10,

____J~~~~~_C --------— ~ rd/ROI.01 Yo 2cRo// =103 x Ro 0 0 - 0 -"o~-o Figure 16-a. Drop trajectories around C linde /. and 2%URJvg 103* rd,/,.01

~~~x0 /R~ ~ ~~~~YoR rd /Ros.004 1 ^_____________ —---— ^ ^./2RQ U,/ig ~ |03 -"o~~o 0% ^ ^ ----- a^ ^,^ ^^ D X0 /Ro Figure 16...b. Drop trajectories around and 2R0 Um'vgWith rd/] o

rd/R =.004 1 o 3 I Mr 2Ro Uo/Vg = 5x 103 -I oo /Ro Figure 16-c. Drop trajectories around cylinder with rd/R.004 and 2RoU,/vg 5x103.

r/R.=004 2ROUCD/vg 10 Xc,/Ro Figure l6-d. Drop trajectories around cylinder with rd/% =.OO4 and 2BU/Vg = 104.

_______________________rd/ RO =.004 -I~~~~~~~~~~~~~~~~~~~~2ROUCO/ =25830 -x0/R0 Figure 16-e. Drop trajectories around cylinder with rd/Ro =.004 and 2%Ujvg = 25850.

000~ ~~~~~~~~YoR rd R o'001 0~~~~~~ 2RO U(Z/Vg=5x io3 XOx/R, Figure 16-f. Drop trajectod Clnder with and 2RJ/v, 3g 5 around in wt,. 5x]- 0 0 ~ ~ ~ ra%.o

rd /o =001 2Ro U,(,=rpg 104 Ptgurei6g. Xn IF? and ~Ii~U,/~= L0 tL ectortesarll - ^ /...%-,...^,.,.. ~..70 rd/Ro ~ 001

y0 /R0 rd/R 0.001 2Ro Uo/l/g =10 -^^oI -x/Ro Figure 16-h. Drop trajectories around cylinder with rd/o =.001 and 2%oTU/vg = 10.

_ Yo/F-o rd - /.0004 2 f U,/>9 Uo5 Fieglre L6_i - X, /Ro' — dr0go -*0004

179 Representative ones are given in Table I. An examination of Table I and Figure 16-e shows that for a value of the gas Reynolds number of 25830 and an rd/Ro = 0.004 that the drop trajectories are practically straight and over 90% of the liquid drops which are heading toward the cylinder far upstream from it are intercepted by it. From Figure 27 the minimum value of the Reynolds number at which drops whose size is given by rd/Ro = 0.004 will move in straight lines is seen to be 25830. Hence the dimensional arguments used in Chapter II to obtain a lower limit to the validity of the solution corresponding to straight drop trajectories is seen to be verified by the actual calculations. These results for the drop trajectories were used in the computer program of Appendix XXVI to calculate the film thickness and local skin friction which are plotted in Figures 11-c through 11-i, and 13-b through 13-h, respectively. This program also calculates the terms of the expansions (52) by fitting a curve to the tabulated values of the quantities on the left. The 2n' s and APn'S of these expansions so obtained were used in the computer program of Appendix XXV which calculated the solutions to the Eqso (55). The results of these calculations are available for 0.01 < 0.1 and for the same range of parameters as the drop trajectories, and have been used to calculate the local Nusselt numbers which are plotted in Figures 14-c through 14-i. An examination of the figures mentioned above shows that the physical quantities, film thickness, skin friction,

180 TABLE I VALUES OF, J AND V* x J nC.040000.038124.952072.040817 080000.076185.949260.0c81548.120000.114117.944851.122107.160000.151864.937727.162340.200000.189307.931038.203038.240000.226467.921882.242616.280000.263240.909226.281863.320000.299520.898311.321262 366000.335348.881802.359166.400000. 370531.867531.397 569.440000.40 5 -79.85128 1 435666.480000.438993.830402.471253.520000.472101.811292.507434.560000.504390.790568.543176.600000.535 11.767969 577405.643000.566343.740299.609813.680000.595814.715034.643422.720000.624278..690639.675504.760)00.651706.659024. 704781.800000 677934.630416.734667.840000.703023.598205.763147.880000.726849.566219 *790230.920o0,'749424.532821.815848,960 00.7 7093.496437.839644.00C)00. 790555,460724,863966 1.040000.809037.423519.886282 1.080000.826112.383390.905594 1.120000.841677.342649.925682 1.160000.855731.300477.943274 1.203000.868265.255979.960C06 1.240000.879254.205461.975866 1.280000.888656.142798.990105 1.320000.896417.059399 1.001174 1.360L00.902464 -.055795 1.(2i06724

181 and Nusselt number attain a maximum at a value of E near one. The peak values of the Nusselt number for a Prandtl number of 10 are shown in Figure 15. In general, it can be seen from the figures that the behavior of the film changes at this value of E, for example Figures 11-a and 11-b, show that the film thickness increases in a downstream direction when E is less than one, but decreases slightly in the downstream direction when E is greater than one. Figure 12 shows that there is also a change in the behavior of the velocity profiles in the neighborhood of E=l. For values of E greater than one there is an increased tendency for the velocity gradient to become very steep at the outer edge of the film. This may represent an unstable condition. The steeping is probably due to the fact that large values of E correspond to a large flux of momentum into the film. The flux becomes so rapid that the momentum does not have time to diffuse into the film near the stagnation point and there it has the effect of rapidly accelerating a small mass of fluid at the outer edge of the film. Since the moment-um eventually penetrates into the film some distance downstream from the stagnation point this could account for the decrease in thickness of the film, since more and more of the liquid in the film is being speeded up. Further, since the drops don't penetrate deeply into the film at large E this could account for the decrease in the local Nusselt number as E increases beyond one. This can be seen from Figures 14-a and 14-b.

182 On the other hand, for E < 1 the velocity profiles tend to become increasingly linear as E decreases, (Figure 12), indicating that inertial effects are unimportant. This tends to verify the dimensional reasoning employed in Chapter II to eliminate the inertial terms in the momentum equation for E < 0.1. It should be noted that the behavior of the solutions is generally the same for values of E < 0.01 and values of E > 0.01, (Figures 11-a through 11-i, 15-a through 13-b, and 14-a through 14-i), even though different terms were retained in the equations which describe these regimes. However, there is an increasing tendency for the liquid film thickness to increase suddenly at some downstream point with decreasing values of E. The dimensionless film thickness, however, remains of the order of unity for all values of E which justifies the dimensional reasoning of Chapter II. This rapid increase in film thickness occurs at the point where the droplets no longer impinge on the cylinder for E < 0.1 as can be seen by a comparison of Figures 11-c through 11-i with Figures 16-b through 16-i. This is probably due to the fact that at this point the momentum carried into the film by the drops rapidly falls to zero, and the gas shear stress is not capable of moving the heavy liquid and as a result the liquid tends to pile up. This makes it very likely that separation will tend to occur in the neighborhood of this point, or at least the film will tend to fall off on vertical cylinders.

183 It can be seen from Figures 14-c through 14-i that the heat transfer coefficient drops off rapidly in this region. For all values of E, (Figures 14-a through 14-i), the heat transfer coefficient is a maximum at the stagnation point and decreases in a downstream direction. As E becomes large the heat transfer coefficient tends to decrease less in the downstream direction, as can be seen from Figures 14-a and 14-b. A comparison of Figures 14-a with 14-b, 14-c with 14-d, and 14-e with 14-f shows that the Nusselt number increases with increasing Prandtl number as is usual. For values of E less than 0.1 the effects of drop size and gas Reynolds number based on the cylinder diameter have a strong influence on the heat transfer coefficient, friction factors, and film thickness since they influence the amount of liquid coming into the film. It can be seen from Figures 16-a through 16-i that the drop trajectories are more nearly straight at higher Reynolds neumbers and larger drop sizes. Thus as can be seen from Table I larger amounts of mass and momentum are carried into the film with larger drop sizes and higher Reynolds numbers. The influence on the heat transfer (Figures 14-c through 14-i) skin friction (Figures 13-b through 13-h) and film thickness (Figures 11-c through 11-h) is what would be expected. They all increase with increasing values of gas Reynolds number and drop size. It can be seen from these figures that the drop size has a strong effect.

184 1,9 29 As noted experimentally, and predicted quantitatively there is a definite increase in heat transfer with the two-phase mixture above that, which would be obtained from a pure gas flow (Figures 14-a through 14-i). No value for the drop diameter was reported for the experimental results of Reference (1) due to the difficulty of measuring this quantity. However the data of Reference (1) after being reduced in Appendix XXVII, are compared with the analytical results corresponding to rd/Ro = 0.004* in Figure 17 and it can be seen that the agreement is close and well within the experimental scatter. It can be seen from Figures 13-a through 13-h that the skin friction usually reaches a peak value somewhere around the 500 point on the cylinder. From Figure 13-a it can be seen that for values of E > 1 this peak moves so far to the rear that it does not show up in the analysis. On the other hand, when E < 0.1 the peak will move forward when the drop size and gas Reynolds numbers are small. A comparison of Figures 15-a through 15-h with Figures 14-a through 14-i shows that an increase in heat transfer is always accompanied by an increase in friction. Thus one must always pay the price of increased friction to obtain increased heat transfer. *The diameter of the cylinder in the experiments was 1.5 in. For rd/Ro = 0.004 this corresponds to a drop size of 152 microns. The manufacturer of the nozzle used in the experiments, reports a volumemedian drop diameter of 167 microns for the nozzle under different operating conditions than those of the experiment.

185 2RoUo/vg = 8x104 rd/Ro =.004 Pr =10 ~= ~10 ~A Experimental E.02 points of acrivos eta/. (I) 8 D 1 0 IsC.2 I I I I I I 0.2 4.6.8 1.0 1.2 X/Ro Figure 17. Comparison of calculated Nusselt number with experimental data for cylinder.

186 2. OSCILLATING FLAT PLATE The solutions to Eqs. (93) through (95), (100) through (102), (106), (109), (115), (120), (124), and (129) have been obtained numerically by using the computer program listed in Appendix XXV. The results are available for 0.05 < Xe < 0.2 and Pr = 5 and 10. These results can be used to calculate *t, b*, and Q* by using them in Eqs. (92), (99), (103) through (105), and (112) through (114), and then using these new quantities in the expansions (87). The solutions have been used to calculate the terms in the expansions (138) and (139) for wall shear stress and local Nusselt numbers. The first term of each of the expansions (87), and (137) through (139) represents the steady solution. The second term represents the harmonic components (i.e,, it goes as e ) and the third term is the sum of a part which represents the permanent alteration to the flow field due to the oscillations and at a time dependent part. The physical quantities for the steady solution are plotted in Figures 18-a through 20. The harmonic components of the physical quantities are plotted in Figures 21-a through 23-c, and the permanent alterations in the physical quantities due to the oscillations are plotted in Figures 24 through 26-b. Figures 18-a through 20 show that the steady (nonoscillating) components of the film thickness, skin friction, and Nusselt number increase with increasing Xe. Figure 20 shows that the steady component

2.5 K 2.4 2.3 co 2.2 2.1 2.0 I 0.05.10.15.20 Xe Figure 18-a. Local liquid film thickness versus volume fraction for steady flat plate.

1.0.6'/e 4- 005 0.1.2.3.4.5.6.7.8.9 Vwx / Xe Uo Figure 18-b. Replot of Figure 18-a in terms of oscillating variables.

189.34.32.30.28.2 6.24.22 x.14 K8.20 D.10.08.06.04 10 I I I0.05.10.15.20 Xe Figure 19. Local wall shear stress versus volume fraction for steady flat plate.

.24 Pr=10.2 Pr<= 5/ 0 0 I.I =I. I 0.05.10.15.20 Xe Figure 20. Local Nusselt number versus volume fraction for steady flat plate.

.3 Xe=.20 w Xe.15.2 Q) X e=.10 US 0 z~~~~~~~~~~~~~~~~~~~~~ Co 60 0.1.2 4.5.6.7.8.9 /wx/Xe U0 Figure 21-a. Local amplitude of liquid film thickness on oscillating flat plate.

^Xe=.20 a: 3.2 Xe=.15 7r Xe=.1 L 3.0 -- L' 2.6- U) 2.4en - 2.02 0.1.2.3.4.5.6.7.8.9 l/cux /Xe Uo Figure 21-b. Local phase lag of liquid film thickness on oscillating flat plate.

.43 Cu Xe=.20 Xe =15 "u ________ ~Xe =.l0 ________- ^" 0 w. <I _,.I I-A ________Xe=.05 0 0.1.2.3.4.5.6 7.8.9 Ywx/Xe Uo Figure 22-a. Local amplitude of wall shear stress on oscillating flat -plate.

w tJ 1 3.2 c7r w 3.0 I < 2V.8 L ^s^. ~ 2.6 - UW oI~ ~ 2.4 ~- ~~Xe= 20 < e, Xe=.05,.10 < 2.2 r Q. 2.0 i I i 0.1.2.3.4.5.6 7.8.9 WX / Xe Um Figure 22-b. Local phase angle of wall shear stress on oscillating flat plate.

195 Xe =.20.05 Xe =.15.04 /,8.0~3-X e =.10 z; I 0 I.02 z /7. / Xe.05.01 Pr = 5.0 -- Pr =10.0 0! I - I I I1 0.1.2.3.4.5.6.7.8.9 /WX/Xe U(o Figure 23-a. Local amplitude of Nusselt number on oscillating flat plate with Pr = 5.0 and 10.0.

196 46 Xe =.05.42.40.38:.36 m.34 Pr =5.32 z.30 /.28 en u].26z.24 W.22.20.18 eO. 16,j.14-.12.10 - Xe =.15.08.06'~.04 s~- ~Xe.20.02 - 0.1.2.3 4.5.6.7.8.9 VWx/XeUoo Figure 23-b. Local phase angle of Nusselt number on oscillating flat plate with Pr = 5.

LLa co f- ~2(+ 120 La~t Pr 10.14 d.^2.e.05 LiJ.10. 08. 06.OejL >^,' CO.04 ^ ^e' X 02 - ^^ ^^.2 Xe~~~~~~~~. ^-:^, ^^-~~xX, flat Plate vtt4 phas 10. ~0 &ge Of Nusselt number on Q ^^^latIng

-.07 Xe=.20 -.06 - I / ^^ ^^^X,=.15 -.05 - -.04 0oo -.02 -.01 0 0.1.2.3 4.5.6.7.8.9 1/ X/XeUoo Figure 24. Permanent alteration in liquid film thickness due to oscillation of flat plate.

030 cql.028_ PXe =.20 \J.026 (N 8 =>.024 c~.020 — _ Xe =15 \.018.016 8^ ~ 0 14 _________X e= O.10.012 ^n.010'.008,0 Xe=.05.006.004.002 0 L I I I I I I! I 0.1.2.3.4.5.6.7.8.9 W/Ux/Xe Uco Figure 25. Permanent alteration in wall shear stress due to oscillation of flat plate.

-.001 e05 x =.10 8 -.002 e > Xe=.15 >- -.003 < 0 20 W 0 -.004 z Pr = 5.0 r Z I I I I I I I.1.2.3.4.5.6.7.8.9 1/'Wx/XeU) Figure 26-a. Permanent alteration in local Nusselt number due to oscillation of flat plate for Pr = 5.

Xe=O.05 -.001 X -.002 e ~ -.003 e1 -.004 e-20;Q:S^~ |Pr =110.0 z l I I I I I.. l I I.1.2.3.4.5.6 7.8 9 Wx / XeUa Figure 26-b. Permanent alteration in local Nusselt number due to oscillation of flat plate for Pr = 10.

202 of the Nusselt number increases with increasing Prandtl number. It can be seen from Figures 21-a and 22-a that the amplitudes of the harmonic components of the film thickness remain relatively constant with increasing frequency (or distance along the plate) and then begin to increase somewhat, at larger values of frequency. However, Figure 23-a shows that although the amplitude of the harmonic component of the Nusselt remains relatively constant with frequency it may either increase or decrease at higher values of frequency depending on the volume fraction of the liquid in the free stream but not on the Prandtl number. It can be seen from Figures 21-b, 22-b, 23-b, and 23-c that the phase lag of the harmonic components of film thickness, skin friction, and Nusselt number increases with increasing frequency and decreasing volume fraction of liquid in the free stream. The amplitudes of these quantities decrease with decreasing volume fraction of liquid (Figures 21-a, 22-a, and 23-a). Increasing Prandtl number increases both the amplitude and phase lag of the harmonic component of the Nusselt number (Figures 23-a through 23-c). Examination of Figures 24 through 26-b shows that the oscillations result in an extremely small permanent alteration of the film thickness, shear stress, and Nusselt number. This is due to the fact that for small values of Xe the inertial terms in the momentum equation become relatively unimportant and so the equations are practically linear and linear equations cannot describe a physical situation in which there

203 is a permanent alteration of any physical quantity. In this regard, one might anticipate that oscillating the cylinder would result in no permanent alterations of any physical quantities with E less than one, since the equations are also almost linear-here. Figure 25 shows that the permanent increase in the skin friction due to oscillations decreases with increasing frequency at the higher values of frequency and remains constant with change in frequency at the lower values (quasi-steady). It also increases with increasing amounts of liquid in the free stream (Xe). Figures 25-a and 25-b show that there is a slight permanent decrease in the Nusselt number due to the oscillation which remains relatively constant with frequency. At higher frequencies this decrease becomes more pronounced. The decrease in the Nusselt number becomes larger with increase in the Prandtl number, and with increase in Xe.

CHAPTER V CONCLUSION A. Cylinder Solutions are obtained for two ranges of the parameter E Xe 2 V In case of steady flow the film thickness, skin friction, and Nusselt number have a maximum at E near unity. When E increases from unity, the liquid film thickness decreases toward the downstream with a decrease in the local Nusselt number and an increase in the velocity gradient at the outer edge of the film. This tends to unstabilize the flow in the liquid film. As E decreases from unity, the inertia effects become less important as indicated by the velocity profile which tends to become linear. The film thickness tends to increase more abruptly in the downstream direction. Because of the accompanying decrease in velocity the gravity force will probably become important in the region immediately downstream from the point at which this abrupt increase occurs on vertical cylinderso Decreasing drop diameter and gas Reynolds number tend to move the point of sudden increase in film thickness toward the forward stagnation points This is accompanied by a decrease in the local Nusselt number, skin frictions and film thickness. 204

205 The analytical predictions compared favorably with experimental results which indicated a substantial increase in local heat transfere rate over that of a single phase flow. B. Oscillating Flat Plate The steady (nonoscillating) components of the film thickness, skin friction, and Nusselt number increase with increasing Xe. The amplitudes of the harmonic components of the film thickness remain relatively constant with increasing frequency (or distance along the plate) and then begin to increase somewhat, at larger values of frequency. The amplitude of the harmonic component of the Nusselt number remains relatively constant with frequency. It may either increase or decrease at higher values of frequency depending on the volume fraction of the liquid in the free stream but not on the Prandtl number. The phase lag of the harmonic components of film thickness, skin friction, and Nusselt number increase with increasing frequency and decreasing volume fraction of liquid in the free stream. The amplitudes of these quantities decrease with decreasing volume fraction of liquid. Increasing Prandtl number, increases both the amplitude and phase lag of the harmonic component of the Nusselt number. Oscillations result in an extremely small permanent alteration of the film thickness, shear stress, and Nusselt number. This is due to

206 the fact that for small values of Xe the inertial terms in the momentum equation become relatively unimportant. One may anticipate that oscillating the cylinder would result in no permanent alterations of any physical quantities with E less than one. The permanent increase in the skin friction due to oscillations decreases with increasing frequency at the higher values of frequency and remains constant with change in frequency at the lower values (quasi-steady). It also increases with increasing amounts of liquid in the free stream (Xe)e There is a slight permanent decrease in the Nusselt number due to the oscillation which remains relatively constant with frequency. At higher frequencies this decrease becomes more pronounced. The decrease in the Nusselt number becomes larger with increase in the Prandtl number, and with increase in Xe

BIBLIOGRAPHY 1. Acrivos, A., et al., "Research Investigation of Two-Component Heat Transfer," Aerospace Research Laboratories Report, ARL 64-116, July, 1964, under contract AF 33(657)-10930, Project 7063, Task 7063-01. Also film made during experiments at Marquardt Corporation. 2. Bellman, A., Perturbation Techniques in Mathematics, Physics, and Engineering, Holt, Rinehart, and Winston, Inc. (1964). 3. Bennett, J.A.R., Collier J.G., Pratl, H.R.C., and Thornton J.D., "Heat Transfer to Two-Phase Gas-Liquid Systems, Part I. SteamWater Mixtures in the Liquid-Dispersed Region in an Annulus." Trans. Instr. Chem. Enginrs., 39, p. 113 (1961), AERE R-3159. 4. Bergman, N.R., "A Method for Numerically Calculating the AreaDistribution of Water Impingement on the Leading Edge of an Airfoil in a Cloud," NACA TN1397, August, 1947. 5. Bertoletti, S., et al., "Heat Transfer and Pressure Drop with Steam-Water Spray." C.I.S.E. R. 36, Milan, Italy, April, 1961. 6. Cicchetti, Ao, et al., "Two-Phase Cooling Experiments Pressure Drop, Heat Transfer and Burnout Measurements." Engin. Nuclear 7 (6), p. 407 (1960). 7. Collier, J.G., Pulling, O.Jo, "Heat Transfer to Two-Phase GasLiquid Systems." United Kingdom Atomic Energy Authority Research Group Report, AERE-R 3809, Chem. Eng. Div., Gerkshire (1962). 8. Davis, E.J., and David, M.Mo, "Heat Transfer to High-Quality Steam-Water Mixtures Flowing in Horizontal Rectangular Duct." Canadian Journ. of Chem. Engin., June, 1961, p. 99. 9. Elperin, I.T., "Heat Transfer of Two-Phase Flow with a Bundle of Tubes," Inzhereerno-Fizicheski Zhurnal, Vol. IV, No. 8, pp. 30-35, August, 1961. 10. Goldmann, K., "Burnout In Turbulent Flow-A Droplet Diffusion Model," Journal of Heat Transfer, Transactions of A.S.M.E. series C., pp. 158-162, May, 1961. 11. Groothuis, Ho, and Hendal, W.P.C., "Heat Transfer in Two-Phase Flow." Chem, Engino Sci, 11, pp. 212-220 (1959). 207

208 BIBLIOGRAPHY (Continued) 12. Guibert, A., et al., "Determination of the Rate, the Area, and the Destribution of Impingement of Water Drops on Various Airfoils from Trajectories Obtained on the Differential Analyzer," NACA RM 9A05, February, 1949. 13. Hodgman, C,, ed., C.R.C. Standard Mathematical Tables, 12th Ed., Chemical Rubber Publishing Co. (1961). 14. Juyghe, Jo, et alO, "Heat Transfer by a Mixture of a Liquid and a Gas in Turbulent Forced Convection with Little Vaporization of the Liquid Phase." Comptes Rendus des Seances de 1' Academiedes Sciences, 252, pp. 3015-3017, May, 1961. 15. Isbin, H.S., et al,, "Heat Transfer to Steam-Water Flows." Proc. Heat Transfer and Fluid Mech. Inst., Stanford University Press, Stanford, U.S.A., p. 70 (1961). 16. Jahnke, E., Emde, F., Tables of Functions, Dover Publication (1943). 17. Kreith, F., Principles of Heat Transfer, International Text Book Co., Scranton (1958). 18. Landau-Lifshitz, Fluid Mechanics, Addison-Wesley Publishing Company, Inc. (1959) 19. Langmuir, I,, and Blodget, Ko, "A Mathematical Investigation of Water Droplet Trajectories," AAF TR 5418, February, 1946. 20. Lighthill, MoJo, "The Response of Laminar Skin Friction and Heat Transfer to Fluctuations in the Steam Velocity," Proc. Royo Soc., Series A., 224, pp. 1-23 (1954). 21o Lin, CC., "Motion in the Boundary-Layer with a Rapidly Oscillating Exterrna Flow," Proc. 9th Int. Cong. Apple Mecho Brussel, 4, ppo 155-167 (1956). 22. Perroud, P., le la Harpe, "Heat Transfer by Liquids Entrained in a Turbulent Gas Stream." CEA No. 1422 (1960)o 23. Rohsenow, WoH., and Choi, H., Heat, Mass, and Momentum Transfer, Prentice-Hall, Inc. (1961), 24. Rosenhead, Lo, Laminar Boundary Layers, Oxford Press (1963).

209 BIBLIOGRAPHY (Concluded) 25. Sach, P., and Long, R.A.K., "A Correlation for Heat Transfer in Stratified Two-Phase Flow with Vaporization." Int. Journal of Heat and Mass Transfer, 2 (3), pp. 222-230 (1961). 26. Sani, R.Lo, "Downward Boiling and Non-Boiling Heat Transfer in a Uniformly Heated Tube," University of California UCRL-9023, January, 1960. 27. Schrock, V.E., and Grossman, L.V., "Forced Convection Boiling Studies." Forced Convection Vaporization Project-Final Report 73308-UCX 2182, November, 1959, University of California at Berkeley. 28. Schlichting H., Boundary Layer Theory, McGraw-Hill Book Company, Inc., New York (1960). 29. Tifford, A.N., Ohio State University, "Exploratory Investigation of Laminar Boundary Layer Heat Transfer Characteristics of Gas Liquid-Spray Systems," Aerospace Research Laboratories Report, ARL 64-136, September, 1964. 30. Tribus, M., "A New Method for Calculating Water-Droplet Trajectories about Streamlined Bodies," University of Michigan, ERI Report on Project M992-E, December, 1951. 31. Van Dyke, Mo, Perturbation Methods in Fluid Mechanics, Academic Press (1964).

APPENDIXES 210

APPENDIX I CALCULATIONS FOR DETERMINING THE MINIMUM VALUES OF Xe AND GAS REYNOLDS NUMBERS AT WHICH DROP TRAJECTORIES ARE STRAIGHT LINES To find these minimum values we set: - C Pg-.1 (i.1) 8 (rd/Ro) P N' = 10 and we have: Xe (3 (I2) \ R For a given rd/Ro this determines the minimum value of Xe and since the drag coefficient is a function of the Reynolds number, based on the drop diametero (I.1) determines this and therefore the gas Reynolds number, and therefore (I.1) and (1.2) determine the minimum 2 value of E These calculations have been carried out using the computer program listed in this Appendix. The values of CD(J) corresponding to RED(J) used in this program are tabulated in Table II of Appendix XXV. NOMENCLATURE FOR COMPUTER PROGRAM CD(J) tabulated value of CD at RED(J) 3_ CD(J)pg/p EP(J)'~' 8 (rd/Ro) 211

ES minimum value of E2 ET desired value of EP(J),.1 N N'; = 10 R rd/Ro RC ~ value of RE(J) at N' = 10 and EP(J) =.1 (minimum value) RE(J) F -o = Ro x RED(J) v rd RED(J) - tabulated value of rdUo/vg RG ~ minimum value of RoU/vg XE value of Xe at N' = 10 212

$CCMPILE MAC, PRINT OBJECT 014456 07/26/65 11 33 28.5 PM MAD (01 MAY 1965 VERSION) PROGRAM LISTING........ LIMENSION RE(100),RED(100),EP( 10C),CD(lCC) *001 INTEGER JJMAX *002 READ AND PRINT DATA *3 THROUGH SW2, FOR R=C.CC1,C.C005,R.G.RMAX *004 THROUGH SWL, FOR J=1,1,J.G.JMAX *005 RE(J)=RED(J)*7.52/R *006 SWI EP(J)=CD(J)*0.003/(8.*R) *0070 XE=(1.6*N*R).P.3 *: 0 WHENEVER EP(JMAX).L.ET.AND. XE.L.C.C3C0 r00 RC=TA6.(FT,EP(l),RE(1),1,1,4,JMAX,S) *010 01 01 H ES=(XE.P.2)*RC0 01 01 RG=RC/7.52 *012 01 0 PRINT RESULTS R,RG,ES,RC,XE *013 0 OThERWISE *014 01 CCNTINU *015 01 01 SW2 f\C CF CONDITIONAL*016 0 ENC CF PROGRAM *017

MINIMUM VALUE OF GAS REYNOLDS NO. BASED ON CYLENDS DIAMETER FOR WHICH DROPS OF A GIVEN (rd/Ro) WILL MOVE IN STRAIGHT LINES MINIMUM VALUE OF Xe REQUIRED TO HAVE 10 DROPS ALONG A FLUID LINE OF LENGTH EQUAL TO THE CYLINDER DIAMETER FOR A GIVEN (rd/RO) o, o, o, N - o 0'1 0" straight drop trajectory. A CD 0 6, 0 0, 01 0 MINIMUM VALUE OF E FOR GIVEN (rd/Ro) Figure 27. Minimum value of E2 and gas Reynolds number for a straight drop trajectory.

APPENDIX II ESTIMATION OF PARAMETERS FOR AIR/WATER MIXTURES A. PROPERTY RATIOS* FOR AIR/WATER MIXTURES Pg.081 -3 P- = 62.4 = 1.3xlO P 62.4 Vg.145x103 = 7.52 v 1.93x10-5 Vg = / 7.52 2.74 V Cpg.25... - = 2.25 Cp 1 Prg 7 B. CYLINDER PROBLEM Evaluation of surface phenomena and pressure terms for single component gas boundary layer. From Schlichting, p. 153, Figure 9.6: Tg 2U Ro l/2PgUoo2 From Schlichting, p 320, Figure 14.13: Nug < 1 lReg For potential flow: Pg 1/2pgo - *Kreith "Elements of Heat Transfer." 215

216 Hence for air/water mixtures one obtains: pg Tg 2 UR /l/2pgui 4.5'- 2.2.741.3xo-3 _3 2.2x10 p2 — Ll 2 1.3xl0-32 5.2xl0-3 r Ng P _ V 1.0 0.25. 1.3xlO - 2.7 12R, Pr j 1 Cp P v.7 L,, vg -3 1.2x10 C. OSCILLATING FLAT PLATE PROBLEM Evaluation of surface phenomena for single component gas boundary layer. From Schlichting, p. 120..332 PgUo Vg From Schlichting p. 305. (T —) / U~ /gCpg) Nr x 1-gc" g (T\E1 k/ Ng g kg Prg Hence for air/water mixtures one has: Li+ ] pU 2'. *332.2.74 1.3xlO —31. lxlOP g 0Uox ^^j ^ ^- - w^^w-3^ v

217 F9 x 2 1 [ gCpg2 i 7Pg 1 -TS kg /A Vg kg a L v p Cp CP x'^2 [2.74 L- 13x 10O - 25 -3 *.75x10-4 L(.7) /3

APPENDIX III REDUCTION OF THE FIRST OF EQUATIONS (23) Substitutions of Eqs. (26) and (27) into the first of Eqs. (23) yields: a (+ b2 2 + b4 4 + b6 6 +..) f3 3+ 6 5 f! 7 i + ) 7 1"_ f3T 2+ f 5 I f - ^ ( f1"vf.. K f3;H.~ (. ^ 3" I 8 f frf+..) 84 f ~2 6 4 8 b 6 4 a (b3 t+b t3+ b 65+.) (f4,_ 4- " f3.3 6 f+ 5 5 - f? 7 +..) + 6 3(f3 5 f7 43+ b +36 f f38 6 3( f }f Collecting terms and rearranging yields: (a b2 2+ b4 34+ b6 5~,+ fl 4 6 4 ft5f 2 4.13 6 2! 4! + 4f32 fl - 6 7 ( f' +f f~5 f3 + f5 f3 1 / 1 + I f7fl +...-.ff1"~( f3"f1+ 2! f^f) 3( f5fl 7~~I 3 3 5' 7' + b2 f- +fb4 b6...<2 L+ b b6 ~ +..) ff12~2.2 f,1f4 24 +...... + 6,4 +(.-!f5rf1+j4- f32) -— 3 fl1 f55f3+ If f53 3 2v'!3 51 6' 3 J12'4'''1

219 +-6 f515 s 8 f7T"+ 6' f' 5!! ( b2 2 b 4 > J6,.) T2 -fifl) +4 f3(4 4fW+ 4 f3f 1 l3 3 +( f ff +4 f3'- I2 ff- fl- 4- If3 f "f ( 88 f'f 7 5! 12:'3! 5! 3.:2! 4! 7 K: 4.6,8 8 4-6 4- 6 8 7 +3+-),f ~) (2..* aE(2f+ flt23+ f, fl, f i t 2-f. f7 f" = f f+1t, _ b4 flt2 = E 5! f5b1 2 3: 5: 3: 8 f - f3 1 6 4 714' 6 Now equating the coefficients of like powers of we arrive at: bt2-f I 2 f12 - - f" = 6f5'f'+ f -f'f- f3f3-5flyf5 3 - 3 ( f ff) 6 faoE 2 0 5: +-5 f3fl - - b4(ff"+f2) 6 b4 frt2 = 6 f5 2 1 3! aoE2 "! 2 II =, 21 80 2 80 20 (f3sfl a'' 3 3 3

220 1 8 8.8, 4 - 6.8 8 f2 t - a- f f f f3+ - f7 1 aE2 7! 7! 5''3! 7! + 46 f3 4.6 f + 8 b2 ft' 44,2 +! 52! bff3! 6b6f! 322 6 4,4 6ff>+.(1. b 6,+tf 44 f1 - f"' 8f7'fl'+ 68f5f3'-f7 fl-6f5 f3-105f3 f5-7fl f7 f+f 41 4 3' - ) b 8f5l+ (fi3'2-9ffl-")0b f345fl f+ 5 4f3"f 20 4 aE2 8f72f1'+168f5'f3'-f7'f1-63f 5f3- 105f3"f5-7f111f7 - 5 b2(l8f5Tft+80f, 2-9f5lf1-240f "f3-45ffflf5)- 12 b4(f3"fl 20 4 +3f3fl +4f,'f3')+ 7 b6(flfl"+5fl'2) Now if we define the operator Ln by: Ln.. d... 3 I fd - (n-l)fl' d + n fi" n 3= 5,7 a E2 dr3 d7r2 d Then we get upon collecting the results 0aE2 f1 "+flf1t_-f12 = 0 IS(f3) = 32(f,2+flfl") L5(f5) = 8 (f3' 2-f3f31)+ b2(f3" fl+3f3fl") b4(flfl"+3f' 2) 3 3' Lr,(f7) = 2l(8f5'f3'-3f5"f3-5f3t"f5)-.7 (18f5'fl'+80[f3'2-3f3f3"] -3f5"fl-15fl"f5)- 325 b4(f3 fl+4fl'f3'+f3fl" )+ 8 b6(flfl"+5f1'2) 2 8

221 And finally if we define Hn, n=3,5,7,.... by: H3 = 0 Hs = 20 [ 4(f323f3 I (f3 f 3)+b2(f3 +f3f) ] 3 H7 = [ 12(8f5'f3'-3f 5"f3-5f3 tf3 )-b2(8f5fl + 80[f3'2-3f3f3 ] 9fs5"fl-45f "f 5) - 0b4(f3lsfl +4f 1f3 1+3f3f )] We obtain: n+l n-l Ln(fn) = (-1) n [ffl"+(n2)ft 2]bn_ +H for n +l5,7, for n = 3,5,7....

APPENDIX IV REDUCTION OF THE SECOND OF EQUATIONS (23) Substitution of the expansions (26), (27), and (28) into the second of Eqs. (23) yields: 8!.+.) (7 23+ FT +Fr ) - (o' +?r'2 1 T! 6 i: F +* ) f- 3 f132+ 465 f54 8*7 6 +)\ = p...E2 (o"+ 2! F2" + " —' F4'+ 6! F6.+7 * + If44+ -F6 T jJ+- f{-6 5^ 7.! 6 2.! 2 4 6 Pr 2.! F+ 6+ Collecting terms and rearranging gives: a(l b2 + b4 4 6 b6 +)~jfiFo'+ (F2fl 1+ 2T f3Fot 1 F2T fl) 2+(1 fl'F4- - f3TF2- 1 flF4'+ f3F2 - F 3 ( flF6- f3F+ 57 f5fF2- 6 flF6 4 6 1 + 24! f3F4'- 4! fF2'+ 6.! 7F = 2F 2 1 2 = PrE2 1'+ +2! 2 + F4! 4 6! F6"6+. Upon equating the coefficients of like powers of r one obtains: - flF' = 1 Fo 222

225 1 fl F2+2f3Fo- - flF2 2 Fo 1 1 a - F2" 2 F2" -2-f F2+fl F' F 4f3Fo -b 2f 1F aor 1'F4- 2 1 F 1, fl'F4- f3'F,- flF4'+fFI3F'- fsFot 6 4! 4 b2 1b + (F2f1'+2f3Fo f F2')+ 4 fF 22 1 1 2' —- ~ F4" aoPrE 4! 1 aoPrE2 F4"-4f 1F4+f1F4 24f3F2 -16f3' F26f5Fo +6b2(2fl F2 -flF2'+4f3F' )-b4flFo' 1 1 6 1 5 fl'F8 -- f3' F4+ f5F2- F 1 + 2 f3F4 I 3L f5F2 8' f7F'+ 2~T5 6 6.' 7 b2 ( L f-1F4 f3TF23 flF4+f3F2 4 f5Fo )+ 4 (Ff1+2f+3Fo', I f1F2 ) 4e 2 b I 1 1 - 6 flFo a 2P E2 6:F6 1 — F6"-6fF6+F6' 603F4 aoPrE2 -90f5F2' +8f7FFo 80f3'F4+56f5'F2 +15b2(4f F4-flF4l' -16f3'tF+24f3F2'

224 -6fsFo')+15b4(2f'F2-flF2'+4f3Fo ) -b6f lF Now if the operator Kn is defined by: 1 d2 d Kn - + f - f nfl n=0,2,4,... n aoPrE 2 dri2 2 Then one obtains upon collecting the results: KO(Fo) = O K2(F2) = (4f3-b2fl)Fo K4(F4) = 8(3f3F2'-2f3'F2)+6b2(2fl'F2flF2' )-(6f5-24b2f3+b4fl)F' K6(F6) = 20(3f3F4'4f3'F4)-l8(f5F2'2f5'F2)+15b2(4f'F4-ffF4' 16f3'F2 +24f3F2' )+15b4(2f'F2-fFF2 )+(8f7 -90b2f 5+60b4f3-b6f )F0 And finally if Gn; n 0,2,4,6,...is defined by: G O Go = 0 G2 = (4f3-b2fl)Fo' G4 = 8(3f3F2'2f3 F2) +6b2(2f F2-fl 1F2 ) (6f 524b2f3+b4f )Fo Ge = 20(3f3F4' 4f3 Fa ) -18( 5f s5F2 -2f 5'F2)

225 +15b2(4fl'F -f4 F4'-16f3sF2+24f3F2 )+15b2(2f 1F2 -flF2' )+(8f7-90b2f $+60b4f3-b6f )Fo' One obtains: Kn(Fn) = Gn for n=0,2,4,6,...

APPENDIX V REDUCTION OF THE FIRST OF EQUATIONS (24) Substitution of the expansions (26) and (27) into the first of Eqs. (24) and using the relations: t3 5 I7 SINO = ~ - - + — +o. 3' 5' 7' C2 =4 6 + COS.S l — + L 6 +~~~ 2! 4: 6: yields (if we denote in this appendix fl(l) by fl; fl'(1) by fl'; f3'(1) by f3s etc.): f 4 6 8 fllg, f3l?3 6 f51,e5 f7ttg7+o0 31.5! 7 + b e ~ 1 +'b 4 a- aoE (2 b2+ b4 -4 -1 b+ -) 9a o (i + 2 2 + b4 4 + +6 3 (l - 4 (b6-D f5tt 8 7>; ( 2 2460' 36 >+ a0E22ao (~+[~ r] - b2 + b4] 7+o.. ao~+ b, 74,~ 6 6. 8 7 _____ fb~1 f7. + 1(bg 17~t2b4 _b22) t4+(be 1 b4 + b2 )e + 44 6+ a=E2 (a2ao-fi)+ [2o 1 ) +a b f3 + 2ao 226

227 + 4) _ 6! f5 + [ao( 7! + t b2' b4.. 6$ ) 8 f7 4! 5! o 7! 2!:.5' 4:'3: 7! +..{ +, (b2-1) ( + ~ + (4 - _:.k? 2! + 22' 6! 6 2... + 2'!)i+ 1 = 4aoE2 2ao-fl']+.T [2ao(3b2-1)+4f3']t 1 _ + L 2a(l-1ob2+5b4)-6f5,1' 5~ [2ao(-l+21b2-35b4+7b6)+ 8f7,]]7 5! 72 +.. _1 + 2- (b2-1)6+ 2. (b4+1-6b2)4+ 6! (b6-1-15b4152)6 ~+t' = aoE2 [2ao-fl']+ 3! [3b2ao+(2f3'-ao) ] + 2- [5a(b4 -2b5)-(3f0'-an)]~5 27! 7! + 2! (b2-1)t +' 1 (b4+1-6b2)g4+ 1 (b6-1-15b4+15b2)~6+.. 3'7 = fl "- -f3 t + 15 f5_155 f7i. + 35 7' Equating coefficients of like powers of 5 one obtains: fi" = aoE [2ao-fi] 3! " 4- f3" = aoE2 "2 [3b2ao+(2f3'_-ao)]+ 2 (b2l) [2ao-f, ]j aoE2 f3" -- {b2ao+(2f3'-ao) + 2 (b2-1)(2ao-f')} 2 2 6 f2:L 5! = aoE [5ao(b4-2b2)-(3f5'-ao)] + (b2-1)[3b2ao+(2f3'-ao)] + 4 (2ao-fl')(b4+l6b2)1

228 aoE 2 F2 20 5s" = 2 {2 [5ao(b4-2b )-(5f5'-a) ]+ 3 (b2-.)[3b2ao+(2f3 -ao)] + (2aO-f,')(b4+l-6b2) 8 f7 aE [7ao(5b2-5b4+b6)+(4f7'-ao)]+! (b2-l)[5a(b4 -2b2)-(3f5'-ao)]+ 4!2: (b4+l-6b2)[3b2ao+(2f3'-ao)]+! (b6-1 -15b4+15b2)[2ao-f' ]} -E 2 2 1 f7" = - ~2 s ^ [7ao(3b2-5b4+be)+4f7T'-ao]+ 2- (b2-1)[Sao(b4 -2b2)+ao-3f5']+ 5 (b4-6b2+l)[ao(3b2-l)+2f3']+ 7 (b6-1-15b4 2 T 4 +15b2) [2ao-fi'] 2 + 2f7+ (4a-f')b- (2a-tfl')+ (b-b4)(a-f f/2a+ E (4 O-4 2-b4)(2afb-)b 4 a )6 ( f) + ao(21b2-35b41)+ 21 (b2-l)[a0(5b4-10b2+1)-5f5']+ (b4-6b2 2 2 +1) [ao(5b2-1)+2f3'] Collecting the results and simplifying yields: fl1 1/2aaE2 + 2fj'-4ao = 0 / a +2f3'+ 3 (4ao-fl)b,- 2 (2ao-fi)-ao = 0

229 2" + 2f5 - 5 (4ao -fl)b4- 5- (2aofi')+1Ob2(2ao- f')+ 2 a(Ob-) 1/2aoE 5 5 5 20 (b2-l) [b2ao+(2f3 -ao)] = 0 3 1/2aE2 +2f7'+ (4ao-fl')b6- 74 (2ao- )+4 (b2-b4)(2a-f) + I ao(21b2-55b4-l) 21 35 + L2 (b2-l)[ao(5b4-10b2+l)-3f5']+ (b4-6b2+l)[ao(3b2_-)+2f3'] = 0 If we define As = -ao A=5 2 (15b2[2ao-fl']+ao(1Ob2-1)+10(b2-1)[3b2ao+2f3'-ao]] 5 A7 - { 105(b2-b4)[2ao-f-1']+2ao(21b2-35b4-1)+42(b2-l-)[ao(5b4 -10b2+l)-3fs ]+70(b4-6b2+) [ao(3b2-1)+2f3' ] Then we have: 2 fl1 +2fl4ao = 0 aoE

230 n+l 2 1"2'+2n 2 - f i~f +2f I b- [ 4ao +f, aE2 "+2fn + {(-1) b [4ao-f ] 2aon+f +An for n = 35,57,...

APPENDIX VI REDUCTION OF THE THIRD OF EQUATIONS (24) Noting that: 2 54 6 COS. 2 = 1- T + 6' +' and substitution of the expansions (27) and (28) into the third of Equations (24) yields: F F 2 4 4 Fe e ba 2 b4 4 b+ 6 ) 4 6 1 2 F4t tF61 +'6 where we have written (l) as F F a ( ) as F' etc Combining terms and simplifying we obtain: [F + +- (F2+Fob2 + (F4+ - F2b2+F b4) l 6+ 6.16' 6'. I4 o6 + 6 (6+ 4 F4b2+ 4 F2b4+F7be) 9 o + ] (- 2 + ^ - 6 +.~ ( ~ +2 F + s 4+ F6 t 6 ) - aoE2Pr F2 4 4F i 61) Equating coefficients of like powers of 5 yields: F e' F0 = F aoE2Pr aE 2Pt F2 (F2+FOb2 FO aoE 2p 2.- 2r. 6 251

232 - aoE2P F2' = F2+Fob2-Fo 1 1 1 1 1 - aoE2 F4' 4 = (F4+6F2b2+Fob4)- 22' (F2+Fob2)+. Fo - 1 F4' = F4+Fob4+6F2b2-6(F2+Fob2)+Fo aoE Pr 1 1 1 a2p F6e' = - (F6+15b2F4+15F2b4+Fob6) aoE Pr 1 1 1 -4 (F4+6F2b2+Fob4)+ 124 (F2+Fob2)- F 1 - aP — F6' = F6+Fob6-Fo+15(b2F4+F2b4-F4-6F2b2-Fob4+F2+Fob2) Now if we define: B2 = 0 B4 6 [b2F2-F2-b2Fo ] B6 = 15 [b2F4+b4F2-F4-6b2F2-b4Fo+F2+b2Fo ] We get upon collecting results: 1 2ao P F'+F 0 1 ] aEP F'+F+F L +(1) + Bn n = 2,4,6,.00 aoE2?nr n n

APPENDIX VII -1 * REDUCTION OF u/U0 = (2 *) Substitution of the expansions (26) and (27) into this equation yields: u 1 f_ 4 3 6 f 8 7 U= 1 If, _ f4 f 3' 6 5-8 1. - a [1-'T - 2a L3 7+ f5 x8 7, 7 b2 b 4 b6 61 b2 2 b b 4 x + (4! 6.1 2h4! If we now s 4ao 5 8 7' U 2f 3 ni 4 7 7 U0035!5! 7.5 X b2 2+ b2 b4 4 b2 2b2b4 u+6- 3 We get upon equating coefficients of like powers of 4 U =' -' — ~l U1?~- 4 1 f3" + 6 If~5 8 2 52 2ao 2 255 2f we now set: U0 3.' 5" 7' uh~b s 2b' 22rb ee ia 1- ~~~~~23'3 + --- + ~~

234 _ _. U3 = b23 f i (5b2fll+4f3') 2ao 2 fL + 2 f3 ] 2ao (3b u3 = 1 (4f3'+3b2f') 2ao US l f-'6b2 4 6 55 = 2 4 (6b b4)+ 3 2! b2f3'+ f5] 2a- 4,-26be,2-b4)2! + U5 2 -- [6fs'+40b2f3i+5(6b-b4)fl ] 2ao a -- (90b2-15b2b4+b)+'4' f3 (6b2-b4)+ fs'b + 8 f7] 1 2 3 U7 = -- [8f7'+126b2f'+140 4o(6b2-b4)f3'+7(90b2-15b2b4+be)fi'] 2ao After collecting the results we get: U U= U 3 3 U5 5 U7 7+ ~1 _ 7+ tll = ---- fil 2ao 1 U3 =- (4f3'+5b2fl' 2ao 1 2 U; _ _ (6f5 1+40b2f3' +5(6b2-b4) flt 2ao 1 2 3 U7 = — 8f7 +l26b2f5 +l40(6b b4)f3'+7(90b2-15b2b4+b6)fl 2ao a 00 00 0.0.0 00 0 00 0..0 000 00 a0 00 00 aa a0 a0 a0 0 0.

APPENDIX VIII 2 1 ___ REDUCTION OF Tw f\ //2pUc, E*2= 2 a2 Substituting the expansions (26) and (27) into this relation yields: 2U0QRQ TA v 4,, 6 5 8 78 l/2pU, - aE [ 4L r.3+ f5 8 x —- [1 "tf- - f3_ 2b *7 f65-,T fTbf *oo b2 2 b4 4 b6 6+ -2 a2E [1 f 6 f5 8 7f x* =a2E L f 5" 3 5 f7 + 7 I. 2 ao-E 7 7 - (9b2-b4 - (lb-5b2b4+b6 1 4 ao 5' 7. -W2 /1/Ptm 10 4 T36 f5 5t 7 7. i- 2 apo coleting f, of 3ik p o 7 o j^ / 2 a -^23 4- 1 6 Now setting~ TT R23 7' U-pon collecting coefficients of like powers of 5 one obtains. 255

236 = II- f3 + - 1- a1 T3 = ti 6 2) a E a oE LE 4 f1 6 2" It 0O ao E a2E [6f5ob+l(b b4)fl1T] = 2E [f 5"+4ob2f3"+5(9bb2) f3 +1?, 1 i 8 2,~6 452 (9b3'4)f3' ~ k(o r12. == -1- - b2f 511+ 4 (9b2-b4)f3 + - (l80b2 -7 ao LTE 2 5 6 -45b2b4+b6 )f'i T: = --- - [8f7"+252b2f5s" +28o(9b2-b4)f3l+14(l80b2-45bab4+be)f4'] o.2 2 3 2E [ 4f7"+126b2f 5+140( 9b-b4 )f3 +7( 180b2-45b2b4+b6)f" ] aO E And collecting the results one has: ao E 2 T ------ [2f3 1+~b 2f I 3 a 2E 0 2 2 2 2bs"+5(bb4)!] 2- 2 2 3 "a 0 ^ [ 4fT"+126bEfs"+i4o(9'b2-b4 )f3r"+7( 80bE-45b bA+be,)fz"] a. E

APPENDIX IX REDUCTION OF Nu/ o- 1 * / v ES* a Substituting the expansions (27) and (28) into this relation yields: Nu _aoE F 22 Ft 4+ F6' 6 1 L b2 2 F - o + f' + - 6' 2.' 2UoR0 a0E L 2: 4 6I 2t -t 2 _2 3 + (6b2b4 (90b2-15b2b4+b6) +o3 4r 60 Now setting: Nu q2 2 q4 4- q6 6 2U~oR0 2! 4' 6 + V We get upon equating coefficients of like powers of 5~ qo a E a0E q [ f2['-b2Fo'] E qj 2'bFa1 12 L ^ 22 2' 0 E 1 4! f2b2 1 q&4 [F4'-6b2F2' (6b2-b4)Fo' 1 E _ 1 F rFs' br4 + F2 (6b-b) 1 (90b-15bab4b)Fo, 61 Eao L6- 2o'o4 2'okt 6 237

238 1 2 3 q6 = — [Lfs''l5b2F4'+l(6b2b4) F2''-(9Cob2l5b bb6+b)Fo] Ea -o And upon collecting the results we have~ 1, qo a.E o 1] 2 q = -- {F.' 6b2F2 +(6b 2b4)Fo } a,E aoE 1b 1b2 bao,::,b -.1-b a E 2? Q0 0 80-(600 0000 0F000 00 0 00~0~6 0 0 ~ B00 0 0 0 b 0 0

APPENDIX X REDUCTION OF THE ENERGY EQUATION FOR SMALL E TO A SERIES OF ORDINARY DIFFERENTIAL EQUATIONS Substitution of the expansion (52) and (53) into Eq. (49) gives: 00 00 00 F_ F 2n-1 m2m PE(2n) 2 (2n-1) Z (2m-1) n=O rn.-l m=l 00 00 2 Z n z 2m z (2n).' (2m)!. n=O m=O Next rearranging the sums on the right, we get: F__ 2k 1, 7 F2ny2(krn)S PrE(2k) - nl (2n-l)!(2k-2n-l); k=O k=2 n=l1 oo k C2 7 F2na2(k-n) 2k (2n) (2k-2n)' k-O n=O F 2nP2(k-n+l) 2k - 2ra ) ) ------------ 2 J LZ(22n —1) I ( 2k-2n1+l kl n=l 00 > k t 2 T F2ng2(k-n) 2k 21 _s (2n)(2k-2n) 1 k=O Next introducing by binomial coefficients () and using the relation: (m\ mf nJ (m-n)'n: we have: 2539

240 1 1 (2k \" (2n-1)! (2k-2n+l)1 (2k) 2nl I and: 1 _ 1 f2k> (2n) (2k-2n)' (2k) t 2n Using this in the above equation, we get: F2k 2k P rE(2k): k=O 00 f- 1 ~ f- /2k ^ 2k 2r, ( ), L F2n2(k-n+l) j (2k): 2k=l _n=l 2 ^ 1 (2k-'' 2r k (2k)! L F2An2(k-n) (2n k=0 n=0. Equating the coefficients of like powers of ~ gives: k -K 2 1 F2Lnl P2(k.n+l) (16ok) n=l k 2 (2k) - n / 2I F2tng2(k-n) n=O or: 1 ___ Fk + a2 aoF~ - 4kP2pF2 = PrE k-l 2(1 l-So-&k) (n- ) F2n2(k-n+l) n=l

241 n2 (_l-ok) 2n F2n 2(k-n) n=0 k-1 -I Y 2n-)n n=l 2 /'2k 7 2 ( ) t2 F21 n2(kn) - ( Eok) O2kFo Next using (54) and defining: 1 d2 2d Kk = - + - - 4kll PrEOo dTl drl -O k = Gk =- 12 oaFo'; k = 1 k-1 3 F n 2( F2nP2(k-n+l)- n) F2n a2(k-n) n= 1 2 o-atk 2 Fo'; k > 1 We get; Kk(Fk)) = -; k = o,1,2,53.. ~l

APPENDIX XI REDUCTION OF THE FIRST OF EQUATIONS (84) Substitution of the first two expansions (87) into the first of Eqs. (84) yields~ (5o*.+c.*ie2*+o.. )]2(Q*Tl+C $+c2.+o o)+ 1/2 (5o*+c i ol*+c22*+ e oo) L(t +,t er 2 t o* )(2 21*+l'-~~~)(+cS* 22, )] 1 /2 [(*e +ce25.e7-+ o ~o) ] (( +~'*Ti-,2a4+ -ooo )- ~t (<o*+~*,l'E-20*+o ) ( O ^^^-**.^ ( r ^ri T 2^') - +/.e + ( +~*+-~-2 )( CO+ T+ cos + );/2Xe (*rnTf + *2:By coliecting terms we get~ 0 20 2 +25o-0..t..'2*o,*')+~o'( +" ~ ~ C +. )+ 1/2 (S5=*-^*....)[*L (C*.*o-o*o)+ (9n l o~r]~r2~),4-,-u o o ]~ -- (650o+~1~+~c-2 +oo,)[~0,* 2+c2t1,i]0 + 2 *** ^2( o*2^24jt* )~. ] -.' +~ (' ^ +2o _r3)+ ~~o ] - [5ooT+ ( olT+OT5i _ 1)t(O2Tot4 2422 2 0 4-2r~a4.

245 +8~ a0 tC 2 ) * *'* (t +C ++-r l [ +C ( 5 1 5 O r 7o +&^&|)*. K+e2eE * 24* c [^.2 TO +c (5o0T+151T+5*O T+o2T)+ ~ ] [ o1 + t+ 2 + * ~ ~ = ] - [T 2Xe +3c5o 55+ (3 So+5 2 )+...]cos T + - ( n+ +' 2*2rj +- ~ ~ ) By equating the coefficients of like powers of c one arrives at: 1.2 1.*,*.* * 1 *.i7 2 R**. * Xe ~ ~o = 2o 0orm 2 oo o o o - 2 *oS(o* -OOTOT% **1~~~~~~ 2 - ~OTO1 ) Oo o 1 *'' *k T *.* * *** * 1 * Xe T-' + C1T - 01 COS T — = - T T2+ + o tiXe *2 2Xe o o T 2 f2l, 0 5 TT+) 2 2 0 IT"1 Xe ~211]?+2Xe 2_ ^ b*(1f otr.,, 1^ +5 * - Tl-O l]'b Xo4OI*-5 * (5 o5 T0) ton +51 t6OT t oT-5o 2 ~ O t ( V l n +2 wr t) o t ~0 0 1 0 T+5 o20 -, + ^ Tlt+2 o0 Ot ~T ~

APPENDIX XII REDUCTION OF THE SECOND OF EQUATIONS (85) Substitution of the first two of the expansions (87) into the second of (85) yields~ ) +23* 1 2* 1 * ( * 2 52)+. (5-Xe*t )+C( -xeaX )+C2(b5Xet )+~ ] + (5+C51 ) ( 0+C5(+C-25+o2 + ) o(Xe'Tb)+C(Xe T +so ) +C2 XeSt0T 52t) +~ ~ (2 +C )2t+o o ) ( 07 ^t6? 2. ef- ~ / i-x-~3 X,2 /e 32/ 2 S 0o o )(- cos2T) - o2 - Xe-T0f e 0 0.1 * 6^)(Xe&St-L)+(Xe T+ J &^)(5-Xe^ ) t + < (XeSStr + 2o & )(~O-Xetl)+(xe + 2 51 e)( ] -Xe ) +( XeLoe oT &t)(6-Xe9 +~~ + 5 C *..) + ) [ Xe- r 4-5e ) e-7 + &+ c5(Xe2T&+.so)+(oo Xe- (o4 - 2 e 6o*^ 162 - 2e2 2 o~)( c- 5, C — X e+ So, c + C l oXeXo+o ) (Xe-;+.z~ $', ~eS (, )~~ (,{*j sS,

245 Xe ** T + 2 (b2 2-^ ^ — os +. sn 16 ( + — )(l-cos2T) + Ce. 1Ta+jC2~?~+e. And equating the coefficients of like powers of e yields: &o(Xe 6oT+ 0 eoe *X *1 * *2 * *(X + b (Xe )4 o eoT+- 2 oXee OT 2-2 Xe *50* Ooi sin = *T~'q or: (&- Xe ) I [ (Xet 1T+ 2 5 )+5(Xe ts + 0 ~ ) + &o(Xe &sT+ E o) 1 -Xe1t + sin T] + ( Xe )sin T = 1 2 4 50o Xe52T+.T 520)(5-Xe2o)+(Xe&Tr + 1 S )(5I-Xei 2)+(Xe5oT + 2 5b )(62-Xet2 +S? (Xe~IT~+ 2 bS)(o-xe)+(xebT+ + 2 &')(& -Xe i)]+ + (Xe T~ 2 S )(1

246 -Xek)~ I [2 26b^(Xe r+51 )+~2(X ET* -Xeeo')+ 2 2515o (Xe58oe 0o ) Xe 18 T +e.* \ Xe ** * ** i)oXe (&O0l+l51o * Xe ** ]ol sin T *2 + 5o (l.-cos2T) = or: (5o-Xei) F(e_2+ 2 02)+0s(Xe 6oT+ 2- +o R +&!(Xe~05T+ &oi)3 +(61-Xe) *] Xe0l'T + 2 5|)+5*(Xeo+ 5 50 ) + o o(XeS5oT+ 2 5o)(& -Xe2')+ 2 2510(XeboT+oO 0) + 2(Xe5l>T o22 2 + 6 (l-cos2T-) 2r +2 (

APPENDIX XIII SIMPLIFICATION OF FIRST- AND SECOND-ORDER MOMENTUM EQUATIONS Substitution of (92) into Eq. (90-1) gives: 4a2~S~ITl] a ($i~]*, 2a2 + *1 ka t3 n4Ti~ + a&(t,,2atfo+2afo l-2atfo l I 2 f2) 2 \ f l22 * -2afo% ^ 2]) 3 a (4a o0f2O4a f02a b01fo -afO s4at25* -8a2Sf, Lq-h,(2a) 51T 8a3 1 * --- -6 cos? +- X 2Xe Xe Rearranging we get: -~ - ^ 4La r 2a2( f 4l2 tf foT Xe Xe 2''qo - " o fofy.^ n f(lOf T "S2O I2 (2(V 12-f &)Ffo )*-2a fo 2a 3 ndSubst collection of (92) nto Eq (90 gives ~~2* 0~~ 24 2a Xe Substitution of (92) into Eqo (90-2) gives:

*lTjrt. + 2afo - 2l afor l ** -2atfo*) + - 1 (2atfo'1 +2af1/* - 2afo|ll * 22*2 * * -2afor ) 2-2 (4a20f 2 4a2tffo)-2a2S2f 2 -2565 61a"afo-2a.Tfo Z3rt2VT? -2EbJ ib0a1*Oas igoba b 1T~ 251 ( **2a)( *t z **. 2a,2 - 2 (2a) (t+4ati fo1 o )-t(blblT+2atT)2a fO 2 6a2a3 * 1 * - t (2a^6iTiTlri = - 1s cos T + 1 * Xe Xe Upon rearranging we get: 2aa~ r~ 1.2* ~ fIr~f~ 2a - -— L e + fl-2 T + f2 2 Ef -fof*2 6a2,~',. x.. + ~ ~. 6.;. cos T - a(,(ri'~z",]'-"].)-q -aS1( fotq,,t f t o 2-" f ^'o-f it — ). +(.f, *2. " +2a: 6 /+ ~o 2a 1 -4+2ab I And finally we geto

LAnA -n *C\j H 0 r - 9 + F -^+ ca- " -t - &qlJ CU _l - q &(n a.n *0 a O ^ C- -- + H + qH 1n I n' 01 570 q. c an I * H OJ-O - i < r -0 + O-' I'& 9 G-O -~O c c H * r \i a OJ + I0 + + I 0 H +?~ l, 1-00 t- G O ^: I \O GO -- HF OCd CO C\] C C + H + + +

APPENDIX XIV SIMPLIFICATION OF THE FIRST- AND SECOND-ORDER MOMENTUM BOUNDARY CONDITIONS Substituting Eqo (92) into Eq. (90-1) gives: 2a (l-Xefo) L2a (Xe>l + 1 * )+ a5* + or: 2 e 2 T.-( Xefo) I2Xe5'T+('+~ ) + 2a2 l i + atsin T ~ 2a*XeIl S-+ a asin T And use of (92) in. Egq (90-2) gives.: _ 1 ~ 1 F.2 * a — aT1IO aa (Xe 12/2Xe ) +.a 6* +- 2 iC i * + 2 -$ 2 r8a't * 2 S S )XXe2a2(*J2+5L)f2 +-Xea in T ( -+Xe iT 3 -Xe2a2~9'!. sin r -- +:2(i -cos2T) = ^-2 T 250

251 Rearranging we get: 2a (1-Xef ) E2a 2Xe2 T+a( S2+2)+ 1( + 2 ) + a(6*-Xe) 2Xe IT""e3"1r 2* +(&+) + (2a) (2-Xe*2) + a (1 + 1 ) (1.-Xefo)+(S-Xel ) +(S1 +2z ) + + 2Xe-51CJT sin T + 2 a3(1-cos 2T) = t 2 and: 2a(l-Xefo) La( X2+62)+2XeaS S2T + ~ * 1 * a /* *\ %l(Xe&!-br+ - SAt) + 2 (&01'+S1) sin T + 2 2+ a(&s-Xer.l) L( i^ +6)+2Xe2 ~lT+aTsin ] + 2 (2a)2 2(&-Xe42)+ (2a)2 ( +2)+ -Xe2 sin 81 22 *-c Xe2S j sin T + -- (2a) a_ (1-cos2T) = 8 92

APPENDIX XV REDUCTION OF THE SECOND OF EQUATIONS (84) Substitution of the expansions (87) into the second of Eqso (84) yields: *2. ** * (5o +2ebob +c2( 2+2260o 1+ ~ ) (OoT *+ +2@* +~ ) + 1 (5*+ C2*+ ) [ ( * +C1-T+C_90 1 22T+' 2 *- 2 - / * 2 +Eo++e )(+o.o )( +c ++ o I )- ] -(t (o+ C f_+c 2-+o. o ) (Oo l cTT _c 2~T+o~ ~ ~ o O O ) (=++,+ 0) (0or r'+. XePr Upon collecting terms one obtains: t (0S 2 0.So l..+~ {6l +26o02} ~ )(0T+CO.LT+'-O~2?*o ~ ~ ) (' 1 2. ~ )-, ) +X-..+o *1 1l +@0. - ^ (So'+c'sic+2s6+o-o )[ ( %Kiol,-io.~o)+~(:o+%ift~i-S -o -r 0] ) +C 2( o * + ~ ~ +- - *o * *2t *o S (lo* + c~20 2 ] + OE + C ) XePr ( T 2 ~ 252

c) Cd?-~~~~~~~*0 CH CD ) ~o O * O Ln o) *0Q [ i- *o *0 i) Jn * 0 0 CD iP-O 0 c 0 *0 0 an in HI -nI 8 *H *+H *CD c~l aII G *o 0 - r1 *H + CD *C1 *r-{ H-I C - I + O LnC\ W c o 0 o *H * j d~*O I F o —-. r *0 H- I1': + D) Ga( Oan I an *0 H T *~H' *F0 *- O 0 *0 C) o I @ *: 0 H * - - *H I + ()D 0 *r4 *H CH ^ - r - *Co -> Go CH P *0 *0 * C CD 0 i) *0 -o- 00 GO GO o A) o * o O *0 1 0n *o 0 0m F *o 0 * - C H 4 I * 0 f * C\ -H- 0 0D - 5p CGO — r-5GI 0 *-0 o (0 + H: %: 0 >u 0 31 * H- G O *H * 0 -+ - + 0 Ln. * 0 i -P CD *0 - = * 0 o Cld - *4 4 — *r *rl C + i * 0 CD I ( -- + bC GO c * r-H + R * 11 ci) * 0 0 0 G n * 0 cO r- IC\J' *0o *0 -> - *H Q> <uln aj (?j'-;)i - * r + + *+ O Cn + *0 W.-C: H: 0 ) *- J H 0* 0 0 * H' —H Q * 0 *H CD *H GO o0 * *0 Go *0 G o0 CH o CGO i.n ) @ ( C O O *0 CM * 0 N O 0 o *0 t0 4 lc *0 - JO r- JI c r D (. 60 GO!n A.uL n <I Cl 4 + if A!n + i+

APPENDIX XVI REDUCTION OF THE THIRD OF CONDITIONS (85) Substitution of the expansions (87) into the third of (85) yields (o +c*1~..oO ) (5b+Cb+c2b5 +..) [(XeSboT 1/2 o ) + e(XeST+ 1/2 1) + c2(Xe 5(-',- 1/2 51*)o o ] + c/4 (9*+c~g +oo)( )So+ c(i o + )( &o| + ef+,.o )sinoT 1., 2g* - ( os ~-,Lc.' o zq +o6ooe ) XePr (QOCO1+c- 92o) [ (Xe6oT,+ 1/2 o ) + ( *5 )+*(Xe*5T+ 1/2 5) )+6) + c2(S (Xe5 -- 1/2 )+t(XeS)+ bo(Xe+S 1/2 esI))-+. ] + E/4 (o - 5o (Xe8,5T+ o + And equating the coeffcnts of like powers of e ge And equating the coefficients of like powers of c'we get;*~* * * 1 *0 ~o5o(Xel6oT- 1/2 &O) o= X or XePr 2 254

255 +C~(xSs 1/2 "* ) + Q*C. C6O JC(X, I~~JC -t~ 1/~ 6~E;+2* Q**(XeSt + 1/2 &5) + ~* {(XeT + 1/2 b) + 1 0 *(XeS*T 1/2 * + 1 /2 5*)} + 1/4 T*& sin T = - - Qo3(Xeo + 1/2 b5o ) + e {51(XeboT+ 1/2 bo) + &(Xet~5+ 1/2 5*9) + e(&(Xe5&0T+ 1/2 o ) + &z(XeT -T 1/2 56t) + &s(Xe52T + 1/2 5*)} 4+ l b5^+ 9 [55* + 6*F*] ] sin T XePr o o o XePr 2

APPENDIX XVII SIMPLIFICATION OF THE FIRST- AND SECOND-ORDER ENERGY EQUATIONS Substitution of Eqs. (99) and (92) into Eq. (96-1) yields: 4a238'T+a ( 2a fOgj-2afo!,-ltl^Fo ) e r And collecting terms we get: 1.. 1T1 - lT+foel-fOQl 2a 2P rXe _- (- 1{_ +61 fo+2l& 1T)Fo 2a Substitution of Eqs. (99) and (92) into Eq. (96-2) gives: 4a e2+4aS 5! ~+a (~l~l+2a fo~2 2* 2* e *. e -2afoQi- tti-Fo92 ) + -2 b1 (2aIo0 -2afo Q1-Fo ) ~ afo5 to (5; 1 T +2aT z)Fo0 22abIT91 * p -no 12a. PXgna 2 22f 0g ea to - 2a P r e 256

iJ. ID -~CO ). 4Z 9 ^ r~ % t -- t -0,-'O O i - O0 tH 4- (0 1-,. -'-' O t- % 0,1 0) r H N C 1 H *- (0 _ O * rt tCH dC 0'- -' (-0 CU I,,^- - -O +? +N i0 +: liA I C X J (0 H 3 C: (1 Pr-, rpI-4,'-4 7D r 4 HOn CdM Ct CM ^-" 1c0 ( t C i +-? -> LU 9 W o3, 1 ix

APPENDIX XVIII SIMPLIFICATION OF FIRST- AND SECOND-ORDER ENERGY BOUNDARY CONDITIONS Using (99) and (92) in Eq. (97ol) we get: ~l 2ata+Fo 0s a + 2a (Xet* + 1 1 * + - Fo 2a 2a sin T = - G 4 Xe T XePr and: t {, -1. n -t + } + - Fo {(61*+tb*) + 2Xe 6T) + |2 ^a =JXePr (a + Fo 2 sin T = 0 Using (99) and (92) in Eq. (97o2) we get: @2* a a + 0* (aa* + 2a 2 Xe 61T + a 651 1. 2 2 - XePr or: _ L- ~*(6* K) + 2 + 2Xe26 ~a2PrXe J2Q + 02*1+ ~* 2a + + F _ 2a( b6Q + ) + + 2 F* (XelT- 1* -} 0 2a 2 a + sin T + Fo 2 (5* 1 ) si.n - 0 258

APPENDIX XIX REDUCTION OF FIRST-ORDER MOMENTUM PROBLEM TO A. SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS Substitution of the expansion (103) and (104) into Eq. (88-1a) gives: fin -+ nfofin - finf nfin-fofl 2'5n} 2a Xe 00 2 n~n 1' Xe or: L 1n f 2 1 2a — ~-fn - in + (n-) ffin-nf fLn+[(l-n)fc -fofj]ai n 2aXe 00 1(n-2) 0 0 -f~?I a, _/ ~~tn. Xe And equating coefficients of like powers of I gives: 2a 2 Xe 0; n=l, 2 in n ln+'(XIX.l) -a Xe 259

260 where Smn is the Kroneka Delta.* Substitution of the expansions (103) and (104) into Eq. (89-1a) gives: <' 1 =iZ - a2 + n+2 n=l n=l n=l or: oo 00 00 y t^an= aln - + 2(Xe_ i I(Xe) aI(n2) n=l n=l n=3 And equating the coefficients of like powers of ~ we get: &in; n=l,2 = 1; fin-amn = (XIX.2) 2ii (Xe-1)a (n-2))/ n; n=3,4,5,. And finally substitution of the expansions (103) and (104) into Eqg (90-la) gives: - (p 2 (Nef 17 (2iXeal.p i j2n+l]aln) n nl n=l or: 00 000 7 a fn n n ( LXef ) 2iXeal(n-2) n n=l -=3 + (n+l)ain i n___rl nn=l ^* 1.; m=n bmn = 0; m7n

261 00 (aln-Xefln) - -- n-~l~2 And equating the coefficients of like powers of ~ one obtains: (l-Xeft) (n+l)aln - 51: n=l,2' i 5in In + a-Xefin- 2 in ^1=1:.fin =2 2a2 (1-Xefo)[ (n+l)aln+2iXea1(n 2)]+ + ain-Xefn: n=3,4,5,.9 (xIx.3) The boundary condition (91) (with n=l) shows that: = 0; fin = fln = 0 (xi. 4) Examination of Eqs. (XIX.1) shows that fl2 and al2 satisfy homogeneous equations with homogeneous boundary conditions which will be satisfied with f 12 a12 = 0. Furthermore, this then makes the equations and boundary conditions for fL4, a14, fl6^ a16, etc,) homogeneous. Hence: fin = an = 0; for n=2,4,6,.. A further investigation shows that we must take fliX all; f15. aL5; flg9 aL9; o to be pure imaginary and f13, a13; f17, a17;o..to be real. Hence introducing the operator Ln, defined by: - d3 d2, d 2a2Xe — -- + ^n^ oW-fd + nfo L =- 2aSe d.!3 o drn2 (n )f~ dr ~

262 one has: Ln(fln) + [(n-1)f o+fofo]aln X= - 3n 0 0Xe - 2i(l-81n)[(fo+nf")a(n )f(n_)]; n=l.,3,5,7,9o, n=15,35,7, fin = aln - - bn + 2i(l-Sln)(Xe-l)al(n-2)/n 2 fin = (1-Xefo) (n+l)aln- bln+2i(l- in)Xeal(n-_2) 2a2 2 t i + ain - Xefin - - bin 2 =O (f in = fin 0 for n=2,4,6,8,. fin = aln = 0 for n=1i5,9,9. fin amn pure imaginary for n=357,12,,. fin, a1n' real

APPENDIX XX REDUCTION OF FIRST-ORDER ENERGY PROBLEM TO A SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS Substitution of the expansions (103) through (105) into Eq. (96-la) gives: I l ia f,+ FlFn C - n fo Fln t i 2a2PrXe Fn n f n=O n=l 00 00 n+2 n-i z-1 2i in ~ = - Ft (flnn+foaln+29i~ aln)t n=O n-l F 1 F + f Finl Z nf' Fin n n=0 > ] n=1 2i> Fl(n-2)Sn = -Fo [(n+l)fl(n+l)+foal(n+l)]n n=2 n-=O 00 nZ: 2 - 2iTF0 a-n (.1)t n=2 A.nd equating the coefficients of like powers of 5, we get: 2 2p Fn + foF-nfoF1n -Ft [ fi()fl(n+l)+foal(n+i)]+ O i; n = 0,1 =02iFyPai( n-i)+2iF,(n-2); n=2,54,.. (XX- 1) Substituton of the expansions (103) through (105) into Eq. (97-la) gives: 263

264 2a2XeP Fin + F1 i + Fo ((l+n)aln+ n=0 n=l,,n-1 i 2iXeain2}] - 1 Fo = 0 2 or: 00 r 00 I ja2er Fin + Fi + Fo (n+2)ai(n+i)J. n=0 n=0 00 + 2iXeFo ai(n-i) 2 Fo = 0 n=2 And equating coefficients of like powers of { we get: - onFo n=0O1 = 2.1'o Fln+Fn+Fo(n+2)a(n+i) 2 2a XePr 2iXeFoal(n.i) n=2,,4,. (XX,2) and the boundary condition (98.1) implies: = 0; Fin = 0: for n=0,1,2~,,. (XX,5) Since fz2, a12 = 0, inspection of Eqs. (XXol) through (XX53).shows that Fn 1 satisfies homogeneous equations and boundary conditionso Therefore we may take F11 = 0. And since, the Fin's depend only on the Fl(n.2)'s and the fl(n+l), fi(n-i), al(n+l), al(n-i) explicitly, we have: 0 < < 1 Fn = 0 ~ for n=1,397,59,7, A further inspection of Eqs. (XXol) through (XX.3) shows that the F1i must be pure imaginary quantities for n = 0,4,8..oand the Fin must be real for n = 2,6,10,..o

265 Thus, we have, if we define the operator Kn by: 1 d2 d Kn - ---- + - nf0 2a2PrXe d 2 d' 0 0 _ < 1: _Kn(fn) = -F[(n+l)fl(n+l)+foal(n+l)-2i(l-bon)Tnal(n-l n=0,2...o, + 2i(l-bon)Fl(n-2) r1=: F2XePr Fin + Fin+F ((2+n)ai(n+i)+2iXe(1-bon)ai(n-i) a2XePr i n - I bon = 0 l=0o: Fin = 0 Fln - O for n-1,3l5979~~~ Fin 0: for n=13,5,5,7,- Fin = pure imaginary quantity: for n=0O,4,8, o Fin = real quantity: for n=2,6,10,,..

APPENDIX XXI REDUCTION OF THE SECOND-ORDER MOMENTUM PROBLEM TO A SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS Substitution of the expansions (103), (104), (112), and (113) into Eq. (88-2a), using the relation (111) to simplify and then equat21T ing the terms which are coefficients of e with each other and the time independent terms with each other (since no other terms appear as can be easily seen by inspection) yields for the coefficients of 2iT e oo 1 — ts~n - ftofn + (n-l)f O2n-nfof2n+[ (l-n)f o -foo ]a2n 00 O0 4~it {f2(n_2)-(fo+'fo)a2(n_2)}tn 3 a L-* X ~2Xe 0-1 n=3 n=l + 1 / / {-nflnflm+nf inf m+f nf sm-a m (nifof n 2 L Lf - n n m=l n=l + fof ln-nfof!n-fof in) +2nfoa l.nf m ) + o00 o00 + i 2 {-2alnf m+( fo-f )alnalm+aaln( flm m=l n=l n+m _L f im ) And for the time independent terms: n= -2a Xe mn=i ~- X1 fnnlff nf6 266

267 00 + [(l-n)fo -fOf ]a 2nj = - antnn + 2Xe 1 00 00 2+ -. ) [ -nfinflm+nfilnf m+flnflm-alm(nfof in m=l n=l + t, i,1 -, T nf' m+n + flfo nnff nfo ) +2nfoa nfim } 00 o00 + i 7e 7 [ (-2flmaln+(rfo+fo)ajmaln+ m=l n=l - " -, m+n aln(IfIm+flm) }m+ And equating coefficients of like powers of - we get from the above two equations: 2afXe fl2 + fof n(n-l)fOf2n+nf f2n - [(n-l)fo +fofol]an+Hn (XXI.1) -[(n-l)f +fofao'] n + fn (XXI 2) Where for n=13,5,^7, 57. Hn - (l~Sln)al(n-2)+)4i(1-ln)[f2(n-2) 2Xe -(fo+fo)a2(n2] 2 fllf in-(n- 1 )f llf l in 4-( I 1?, 4 I I +( l-ln)nf Lfln+[ 2(n-1)fofi+fof ll+fof ]aln (XXI. )

268 (l-1)[nfOf lIn-(n l)fofin fofh L]ai1} -i(l-Fln) (no+fo )allal(n-2)+(nf'Ll-fil)al(n-2) }-Pn (XXI.3) Where: P1 = 0 P3 O0 P 1 (3fl3f s32fi3f13+[fofs3+ + 3f 13l-2fof3 ]a3 }+i{(Nf( a+fo )allal3+(~f'3 -fl3)al } for n=2,4,6,8,...: Hn 0 (xxi.4) follows from (108) and for: n=l-,5, ~.. Hn.. - 1 {f fln- (n- 1)f f 3X (1-b1n)ai(n-2)- 2 (f in-(-l)fIn 2Xe 2 + (1-61n)nf tfl n+[2(n -)fof1l+fofl+fof1 ]aln +( l-&n)[nf~f -(n-l)fOfin-f foff n )"a Pn (xxI.5) where: P1 0 P2 - 2 {2 -- ^ +[ -fofi 3+ Ps = ( (3f1s3f132f13f1[ffSffS

269 + 2fof13i]'3}+i( ((1f 1+3f l )als+ (fjs3+3f3 )ajjL} For n=3,7,. e.we have from (108) Hn = pure imaginary quantity. And since we are actually seeking the real part of Hn we set for n=3,7,.oo Hn: n=2,4,6,.. (xxI 6) and for n=2,4,6,...it follows from (110) that Hn 0 n=2,4,6,... (XXI.7) Next substitution of the expansions (105)9 (104), (112), and (115) into Eq. (89-2a), using the relation (111) to simplify, and 21T then equating the terms which are coefficients of e with each other and the time independent terms with each other, (since no other terms appear) gives for the coefficients e 2: 00 00 00 00 fng = ann gn 7f2n -_a2n - IT - aln+4i(Xe-1) n+2 n=l n=l n=l n=l or: 00oo 00 o f2nt = ({an- a aln} + 4i(Xe-1).n n. 2-n Z 2rj - Z3j n-l n=l n=5 and for the time independent terms: f aa 7 a2n t 4 al7n_ n=l n=l nl n=l

270 And equating the coefficients of like powers of | in these last two equations we get: f0 n=l,2, fan - a2n- a (XX I8) i (Xe-l)a2(n-2)/n n=394,5,97 f2n =- an n n=l,2,53,o (XXI 9) And finally substitution of the expansions (103), (104), (112), and (113) into Eqo (90-2a), using the relation (111) to simplify and then 2iT equating the terms which are coefficients of e with each other and the time independent terms with each other (since no other terms appear) gives for the coefficients of e:.- 00 2aS(l-Xefo) a {(n+)a 2n+4iXea 2n 21n n=l 00 00 Z V m - l n+m + a ain(iXeailmi+ 2 ajm 1) n=l m=l 00 _ 00 00 a - (n+l)an + a (a n n=l n=l m=l -Xef n) [ (m+l)alm+2iXe ] }n m 00 00 -- i (alnf) 1i + (2a) (a2nXef ) 2 n=l j nl 00 1 (2a) 2 >(n+2)atn+2iXea n12 ] n=l 00 8( 2a) 2) a2a-n=l &2) t E tlnt

271 Simplifying we get: 00 1 (2a)2 J (1-Xef6) [(n+l)(a2n - am)+ n=l 4i(l-$!n)(1- 2n)a2(n-2) n + 0 00oo -r (1-Xefo) 7 7aIn[(l-(in)(1-62n)iXeal(m-2) ml= n=l 00 00 + I m alm]n n. + 7 7 (ain-Xefin) [(mn+l)aim 2 m=l n=l i n+m-i + 2iXe(1-b m)(1-tam)ai(m-2) -2^ b flm 00 00 + 7 (a2n-Xeftn) 1 X [(n+2)aln+ L[ L [+a n=1 n=l _" 00 n=l or oo or: 2 (2a)2 {(1-Xefo) [(n+l)(a2n - an)+ 2 nKlL m=1l (an-Xef2n)- ~ [ (n+2)amn+2i(l-&ln)(1-&2n)Xea(n-2)]

272 1 + - m amj]+(ain-Xefin)[(n+l)aln+ 2 m 00 n+m- 1i 2iXe(l-6m)(l-a2m)a1(m-2)1 I ] = E fn n=l and equating the coefficients of like powers of g gives: 1 2 (2a)2 1-Xefo) (n+l)(a2n-. aln)+ + 4i(l-1n-62n)a2(n-2) ]- (aln-Xefin)+(a2n-Xef2n) - [(n+2)aln+2i(l-6bn-b2n)Xeal(n-2)]- 6n + Cnl = fn (XXI.10) ~~~~~~4 8 where: C1 = (1-Xefo) a1a + (anl-Xefll)2anl 2 C3 = (l-Xefo)all[ al3+iXeal] + 1 al3a)ll 0 2 2 + (a.i-Xefll)[ 4al3+2iXeall]+2all(al3-Xef3s) C5 = ( lXef) {ai [ 5 ais+iXeal3]+ (XXI 1l) a13 [ 2 als+iXeaul]+ 2 a.las}+ (all-Xefl )[ 6ai,52iXeasl3]+(al3-Xef3)[ 4a13 + 2iXeall]+2all(al5-Xef 5)

273 And from (108) we have Cn = 0 for n=2,4,6,... (XXI.12) And similarily one obtains for the time independent terms: (2a)2{(1-Xef)(n+l)(a2n - ln)- (in-n) + (a2nXef2' [(n+2)aln_2i(l 6ln-b2n)Xel(n-2) ] "+ 8 C} fn 2n (XXI.5) where: Cl = (1-Xef') I 2ma+2an(Em-Xe i1) C5 = (l-Xefo)(all[ 5 als+iXeal3]+ 2 2 a!3[ 3 al3+iXeal4-]+ alla5}+ (~13 -2 3 (aii-Xef1) [ 6a 5+2iXeai3 ]+(a:3sXef3) [ -a13 + 2iXeai]i+(C5s-Xefi5)2all (xxIol4) And as before inspection of (110) shows that C3s C_, are pure imaginary and since we are only interested in the real part of fen we can set: Cn = 0 for n=35,7,9. (XXI.15) and (108) shows further that Cn = 0: for n=2,4,6,.oe (XXIo16)

274 The boundary condition (21) shows that we must have: df2n at ~ = 0; f2f2 = 0 eJ, ~ (XXI17) ~J df 2n at n = 0; f2n = dn = 0 Inspection of Eqs. (XXI.1) through (XXI.17) and use of the condition (108) shows that aan, f2n satisfy a homogeneous differential equation with homogeneous boundary conditions for n=2 and therefore a22, f22 = 0. This then implies that the higher order f2n for n=4,6,...satisfy a homogeneous equation with homogeneous boundary conditions and hence we have: a2n, f2n - 0: for n=2,4,6,.. and similarly: a2n, fn =0: for n=2,4,6,..o But since we are only interested in the real part of fa2n a2n we see that these also satisfy homogeneous conditions for n=35,7,.. Hence one has: aan, f2n 0; for n=35,7,.. Thus we have upon introducing the operator Ln defined by (107)~ For n=1,3,5, ~ o: 0 < K< 1 Ln(f2n) = -[(n-l)f +jffo]a2n+Hn

275 2n = a2n - i ain+4i(1- in)(Xe-1) a2(n-2) 4 n rl= 1 <- f2n = (-Xefo)[(n+l)(a2n - an)+ 2a 4i(l-bn)a2(n_2) ]- (aln-Xefin)+a2n - Xefen - [(n+2)aln+2i(l.-6n)Xeal(n-2)] Xbin C4 + Cn 8 = 0 f n fn = 0 and: a2n, f2n( ) -0: for n=2,4,6,m.o For n=1,5, o0 < T < 1 Ln(f2n) - [(n-l)fo-:fof0]2n + H I f -? _ f2n = a2n - Taln l -1 2a f2n = (l-Xefo)(n-1+) (a2 in)- a (ln n -f-n) -2nXef2n [ (n+2)aln=2i(l-tln)XeaI(n2) ] 4 + -in + Cn _ 8 T = 0 f2n = f2n = 0 and~ asn, f2n() - 0; for n=2,4,6,.. and n=5,79o, Where Hn, Hn, Cn and Cn are given by (XXI.3), (XXI.5) (XXI.ll) and (XXIo 14) respectively.

APPENDIX XXII REDUCTION OF THE SECOND-ORDER ENERGY PROBLEM TO A SYSTEM OF ORDINARY DIFFERENTIAL EQUATIONS Substitution of the expansions (103) through (105), and (112) through (114) into Eq. (96-2a), using the relation (111) to simplify, and then equating the terms which are coefficients of e2iT with each other and the time independent terms with each other (since no other terms appear as can easily be varified by inspection) gives for the coefficient of e j n+2 oo r oo X =2aPXe F2n + fo F2n - fnF2n i 22 2n n=O n=O 00 00 Zs ~ n-1 n+1. F6(n f2n + fo a2r)n - 41 F ) a2n n=l n=l 0o 00 2 _ 1 jj m ainfmFV n+mi.2 Va0naFl Y n+m n=l m=1 m=l n=l 00 00 00 00 V V n+m+. 1., T nm + 21 ) ) ainFlm + | [fln+foaln] mF.m n=l m=O n=1 m=O 00 00 00 00 2 Z / [finn+foa1n]Fimn 1 1 ) f ainFimr n=l m=O n=l m=O or using (108), (110), (118) and rearranging: oo 00 ) ~Faeran + foF2nnfF2} = 4i> (1 o6n-6in)F2(n-2) n=O n-O 276

277 00 -Fo {(n+l)f2(n+l)+foa2(n+l)+4iv(1-8on)a2(n-l))]n n=O 00 00 F 1 Fo l + (malnflm+2i(1-1m)alna(m2)n+mn=l m=l 00 00,, n+m-1 + in —0 n. 1mFlm(fln+foaln)-Fim(flnn+foaln) ] 2 n=L m=O 00 00 + i In+mi ~+ i), [(2ainFlm-a alnFlm] n=l m=O And for the time independent terms: O0 CO~~~~~~~~~~~~~~00 )E T F ^nof F2n-nfF2n - -Fo (nf2n+foa2n) 0a 0p r e n=O n=l -'- PO y y a 2 -iF y y =o w 00 00 0oo 0 1a n m — n+m-2 7 F imnazmn i-nfim m + )lo n n=1 m-1 n-l mrn1 0 00 00 0 + 2ir j f lF1en m+l+ L 1 <7 > [411+ftF-, n]F m- n=l m=O n=l m=O 00 00 00 00 1_V V1 - - m+n- n+m+n -I f [ ifnn+foa in]Fim -i lainFlms n=l m=0 n=l m=O And using (108), (110), and (123) and rearranging we get:'e Fen +fon' Fl -F o - (n+l)f2(n+l)+foa2(n+l) n=O n=O o0 00 2- Fo n; {malnflm+2i( l-&lm)alnal(m_2) ] n=l m=l

278 00 00 1 V V /- t-fll.m+n- 1 + - {mFl(fln+foaln)-Flm(flnn+foaln) } 2 Lnfo L-i n=l m=0 00 00 + i E {i 2lnFim-a lnFlm }n+m+1 n=l rm=Q And upon equating the coefficients of like powers of 5 and using (108), (110), (118), and (123), to simplify, one obtains: 2-P-X F2n + foFn - nfoF2n = 4i(l-6on-&b)F2(n-2) 2a 2P rXe -Fo[ (n+l)f2(n+l)+foa2(n+l)+4i(1-on)a2(n-l) ]+En (XXII. 1) where for n = 0,2,4,... 1! E - F allfll - 1 Fl(fll+foall) 0 2 0 E2 = - 2 {3allfi3+2ilallall+al3fll}Fo +' (2Fl2(fli+foa) -F12(fli+foall)-Flo (3fl3+foal3)]+iajl(2Flo-TlFo) (XXIIo2) E4= - 1 { 5allfl5+-4ialla+al3 al3fl3+al5f l 1}F + - (4FL4(fll+allfo)-F14(fll+foall)+2Fi2(f13+foal3)-F12(fl35+foa13) 2 l -Fo (5f 15+foal5) +ia 1(2F12-IFn12)+ial3 (2Flo- F1o) and: En = 0; for n = 1,3,5,... (XXII.3)

279 1 ~r ~'' ~ a +Xer foF2n nf OFn = -Fo {(n+l)^2( +l)+ 2(f l) +En (XXII.4) where for n = 0,4,8,.. Eo = 1 Foallfll- F10(fli+foa ) 02 2 E 1 - - i 1 E4 = 2 (5aiifl15+3al3fl3+a5lsfl]Fo + {4F14(fllall (XXII.5) -F14 (ll+fall )+2F12(fl3+foal3 )-Fl2(53f3+foa 3) -Flo ( 5f 15 t _t +foal5) +iall( 2l2+1rFl2)+ial3(2Frlo-rFlo) and En = 0 for n#0,4,8,... since we are taking the real parts and E2, E6,..are pure imaginary and hence contribute nothing. Next substitution of the expansion (103) through (105) and (112) through (114) into Eq. (97-2a), using the relation (111) to simplify and then equating the terms which are coefficients of e2 with each other and the time independent terms with each other (since no other 2iT terms appear) gives for the coefficients of e: 00 00 00 1 -- Fzn + F2n + 1 Fin{(l+m)aim 2a2PrXe 2n 2 F + 2a ep Fie n=0 n=0 m=l + 2iXe2 alm)5 1

280 00 2 fl-1 + Fo ) {(l+n)a n + 4iXea 2nS2 n=l 00 00 + FO Z ain(iXe2alm + am)+m-2 n = m =l 00 00 -4 Fin -i Fo (l+n)aln-n n=0 n=l And using (108), (110), and (118) and rearranging we get: > -2a2rX F2n+F2n + [Fo((2+n)(a2(n+i)- - al(n+l))+4iXe n=0 n=O (1l-on)a2(n-1) ) 00 00 i Fnn 171 7n Fin] + - ~Flin {(l+m)alm+2iXeal^ 2}n+m-1 n=0 m=l 00 00 V +Y" m 2 n+F,2 + a, L 2 a m + iXealj 2) n=l mr= and upon equating the coefficients of like powers of t and using the relations (108), (110), and (118) we get: 2a rX F2n + F2n Fo {(2+n)(a2(n 4iXe( n)a2(n-l) i F= + Bn 0 (XXIIo7) where for n=0,2,4,..

281 Bo = alF1o + 1 allallFo 2 1 b B2 Fo (4al3+2iXeall) + 1- 2 allF12 2 2 B4 = Flo (6a5+2iXeai3) + F12(4al3 —2iXeal)+ allF'4 (XXII.8) + al ( a 5 ( +iXeal3)Fo + a13 al3+i3eall) F ( al5allFo and Bn = 0 for n=1,3,5,... (XXII9) and similarly for the time independent terms, we get: 1 t r e ~ _ - 2a2P Xe F2n + F2n + Fo((2+n)(a2(n+i)- al(n+l)}- Fin + Bn = 0 (XII o10) where for n=0,4..o Bo = an1Flo + 2 allallF B4 - Flo(6ais+2iXeal3)+ 2 F12(4al3+2iXeal) (XXII 11) 2 2 +L aL1F14 + a(lla ais+ixeai3)FO - a13(2 ai3+iXea1.)Fo -2~~~~ 2 -+ - ala 1Fo 2 and Bn O; for n = 1,5,5,3o9

282 but since we are taking real parts and B2, B6,..are pure imaginary, we can set them equal to zero and we have: Bn = 0 for n#0,4,8,... (XXII.12) and finally the boundary conditions (98-2) require that: (XXIIo13) =r O; F2n = Inspection of Eqs. (XXII.1) through (XXII.13) and use of Eqs. (108), (110), (118), and (123) shows that F21 satisfies a homogeneous differential equation with homogeneous boundary conditions for n=2 and therefore F21()-=Oo This then implies that the higher order F2n for n=3,55,7,-.satisfy homogeneous equations with homogeneous boundary conditions and hence: F2n(n)-0; for n = 1,3,59., and similarly: F2n( )-0; for n = 13,5,5,.o But since we are only interested in the real part of F2n we see that they also satisfy homogeneous conditions for n=2,6,.o.Hence we have: F2n( )=0; for n#0,4,8,., Thus the relations of this appendix may be summarized introducing the operator Kn (defined below Eqs. (109)) by: For n = 0,2,4 9 oo

283 0 < < 1 K(F2n) = 4i(l-bon)F2(n-^2)-Fo'[(ll)f2(n+l) + foa2(n+1)+4i(1-bon)la2(n-) ]+E'n 1-a =e 1 + F + F 2 n + Fo((2+n)(a2(n+l)- r al(n+l)) 2a2PrXe + 4iXe(1-6on)a2(n-l) - Fin + Bn = 0 r)= 0 Fan = 0 and ^F2,(r)- for n=1l3,5,-.. for n=0,4,.0. 0 ] _< 1 Kn(F2n) = -Fo ((n+l)f2(n-+)+foa2(n+l))+En 1 ~' ~ F2n + Fn + Fo(2+n)(a2(n+1)- al(n+l)) 2a2PXe 4 = 1 e - 4 Fin + Bn = 0 n O F,2n =-' O T = 0 F2n = 0 and F2n (n)-0; for n#0,4,8,... Where En, En, Bn and Bn are given by (XXII.2), (XXII.5), (XXII.8), and (XXII. 1), respectively.

APPENDIX XXIII REDUCTION OF THE FORMULAS FOR VELOCITY, SHEAR STRESS, AND NUSSELT NUMBER Using the expansions (87) in the Eq. (133) gives: U- = Xe[o]+EC +C22++.. * * ] [o+CS-1+E2 5+.o. ] Xe o *[ol+C*17+C22. o ]j l 1 + 2 - * *a2 Xe r * * 2 s1 1[% +~~zn+ 2 +*.. + r1 _ ~ - ~2.2 50 - 6' (o 5* where: * uX = X 1 e = Xe Ll+~1 51. *, 2 Xe. 2 5 U2 = [ +] - - G -; where: Ul - "Xe' o * Xe -. + * ol - L o * o * 284

285 Using the expansions (87) in Eq. (134) gives: Uo0x Xel/2pU"2 - 2 o 2 [-2 1~ 2^~a] -2 - *2 [2^^T,+ 22 +.. ]' -1 2C1 26- + 612 ] + 5 2 %1 +. ] +5 E T*2.. = ro2*2^. 2: 3 - r*2 [*07 ~*(~r * * - o2. * 2 r+. ]2 * 2)1~ =+ + - i + C T2 +0. T o 6*2 2* * -2 [ * +cEq 5 o [ 5o Bo 00 0 06 60 o 47-Ux 26 - 36 ~ X~ P r1 * 2* _( + q..+Dq +..... o~~-E o ~ o

286 where: q* -= i0 6* on o~ 6* 0 2 6*2 o0 0 0 0 0 6 O g2 Q K8- 0

APPENDIX XXIV CALCULATION FOR EXPRESSING THE VELOCITY, WALL SHEAR STRESS, AND NUSSELT NUMBER EXPLICITLY IN TERMS OF THE SOLUTIONS TO THE ORDINARY DIFFERENTIAL EQUATION Using 0* = 2a f fo and b6* = 2a in the Eqs. (140) we get: Uo Xef u^ ^- q [414 fo5*] U = 2X ['[ -f to* + (fotl* )] 2a_ 0 2a5 Since: 00 n iT t= = tee 2a fln e n-l 00 n,',- = =e 2a aln e n=l we setfln iT uj = e 9 nl e n=O and we have: 00 o00,g n = Xe (fin - fo aln)" n=O n=l 00 Xe (<f(n+i) f f al(n+ )) n=O and equating coefficients: 287

288 gin = Xe(fi(n+l) - f al(n+l)); for n = 0,2,4,6,... since fin = an = 0 for n = 0,2,4,6,.o Since: 00 4f2* =e 2a (f2n e + f2n)n n=l 00 52* = -ie 2a 7 (a2n e2T + a2n)~ n=l we set: oo u2* =e 7 (g2n e + g2n)~ n=O and we have, using (111) and equating the coefficients of e2lro 00 00 g2] n = Xe (f2n - f a2n)t e 7 7an a(f- fo a nm)+-2 n=l =l =l pg2n Xe(fV(nam) vf a,(nJ)=h ) n+ m -2 n=O n= l m=l 1 a m1fi -fam a) ho 2 all(fll foall) hs~~~~ ~~~~ Y! asfifi-^i^fi)

289 2 [a-3f l+alf13 - 2alla3fo] 1 r,? I h4= 2 [15(f 11-1Oa 1)1+a3(f13-fa13) Since: tin, al, f2n, a2n = 0; for n 0,2,4,... and similarily for the time independent part: g2n = Xe (f2(n+1)-foa2(n+)hn); n = 0,4,8,... o = fan, aen = 0; for n1l,5,9,... Collecting the results, nwe get: uo - Xefo; ul* = geygl igr1; u2 =- e> (g\e2 iT+s )n n=0 n=0 gin =Xe(f1(n+i)~fo al(n+l)) n = 0,2,4,... g2n = Xe(ff(n+l)-foa2(n+l)-hn) n = 0,2,4,... ho = - ai(fli-fall) ) h2 =-1 a{ai3fi1allf3-a2allal3fo) h4 {al5 + llfal3fl3+allflfo(2allal5+al3al3) 1', he

290 g2n = Xef2(n+)-fo a2(n+i)-hn) n=0,4,... -! I2 1, - ho = 2all(f i-foa- ) h4 -= {als5f l+sal3f l3+allfl5-fo(2alazs+5ls3al3) }) using the relations: ~o* = 2a fo and 60 2aS in Eqs. (141), we get: To* = f1 = a o 1.,, T^= [- / l~l~l - 2 fo * 2~2a a.22a2 L 2a~ a n=0 We get from using the expansion (103) and (104) in the second of the above equations: emn = fi(n+i)-2f" al(n+l)~ n=0,2,4,. since alnj fin n 0 for n = 0,2,4,...and setting: 00 T* = a (e2n e + e2n)n n=0 We get by using the expansions (112) and (113), equating first time dependent and time independent parts and then the coefficients of powers of 5

291 e2n f(n+)-2 o a2(n+l) n; n=0,2,4,,.. r^ =! all(2fl-53oali) ro 2 1I 11 I 11 r2 - [2al3fll+2almfl3-6alal13fo] r 1 - 1 1t ro = ( all(2fll-3foall) r4 - [2als5f=l+2 -23f13+2allfl5 3fo(2allal5+al3al3)] since aln, fin, a2n fn - 0; for n=0,2,4,6o..e and a2n, f2n - ~ for nl1,5,9,9 o And we have also used Eqso (11.1) 00 00 To* -= a o = ene T2* = <!e. (e2ne +e2nJ n=0 n=O etn f= f (n+)2foa(n+l) n=0~,2,4,,) e2n = f(nl+l2fa 2(2) -rn ln02, 4 ro= a:(2fLi-53faj) r, = 1 [2a1 2a3f 1+2a 6aia13fo] 2 r4 e 1 [2a(5f'n+2as3f13+2aifir5-3fo(2azja15+a13a13)] ean = ~2(n+z)-2foa2(n+z)-rn n=O 4, o.

292,' 1- ]. —. 0 _ t_1 _ 0 o= -1 a ( -5a r4 = 2 [2alsfll+2l13f 3+22alfl 5-3fo(2(llal5+a13a13) Using the relations (91) and (99) in Eqs. (142) gives: 2a 4a n 24 00 2a F~ t S~jCF ~A' - i* = C:e 2a ainn el n=l 00 Qi* = ( )j Fin~neiT n=l we set: 00 q1* =Qle p Gin e n=O and we have 00 00 00 G F F= Z * GinS = i Fln 0 - 4a t n=O n=O n=l And equating coefficients of,: Gin Fin - al(n+i)Fo; n=0,2,4,... since aln= 0 for n=0, 2 4,..e; Fin O for n=15,3,59.o Since:

295 o00 n=l @2* = e ) [F2n e + F2n] n=O We set: 00 q2* -= ice 1 [G2n e2 + G2n] 2a n=2iT and we have, using (111) for the coefficients of e 00 00 00 v n o n V nn=O n=O n=l 00 00 ^ ^ ^, n+m-1 2; ) a ln Flm t n=l m=O 00 00 1 V n+m- 2 + F 2 / ),aln am n n=l m=l And equating the coefficients of like powers of 5 we get: G2n = F2n- a2(n+i))o - Rn n=O,0,4,... 1 n 1 Ro = [alF1o -allallF] 1 2 0 nR2 e - [al3F1o all a F12 - Fo alal3] R4 = I 2aLF1o + al3F12 + allF14 - F(2allal5+al3al3)] Since a=, a2n = 0 for n-0,2,,4,.and F1n' F2n = O for n=1,3,5o0

294 And similarly for the time independent parts: G2n = Fn - a2(n+l) Fo - Rn n=0,4, o. RO = 2 [alFlO - a —llalFo] R4 = 2 [al5Fio + a13F12 + allF14 - Fo(2alal15+al3a13)] Since: F2n - 0 for n#0,4,8,.o. a2n 0 for n1,5,9,... Collecting the results: o* = 2a F 00 /<? 1 fTn iT l* -= L Gl e 2a n=0 00 q2* = [Gn e + G2n] 2a n=0 Gin = Fin - ai(n+i)Fo n=0,2,4,..! G2n = F2n - a2(n+l)Fo - Rn n=0p294Yoo. Ro = [allFlo - allalFo] 2 R4 = 2 [al3Flo + allF2- 2Foallal3a] R4 = [alFlo + a3F + allF14 - F(2allals+al3al3)]

295 G2n = Fn - a2(n+)F-Rn n=0,4,... Ro = - [alFlo - allallFo] 2 BR4 = - [El SF + al3F12 + all 4+ - Fo(2a11al5+al3a13)] 2

APPENDIX XXV COMPUTER PROGRAMS INVOLVING INTEGRATION OF DIFFERENTIAL EQUATIONS All the computer programs which involve the integration of differential equations by the Runge-Kutta method are listed in this appendix along with the nomenclature which is necessary for their understanding and for connecting them with the equations derived above. Each program is divided into blocks whose functions are indicated by titles. NOTATION FOR COMPUTER PROGRAMS INVOLVING THE INTEGRATION OF DIFFERENTIAL EQUATIONS Y(1) calculated value of solution to differential equation Y(2) first derivitive of Y(1) Y(5) second derivitive of Y(1) Y(4) third derivitive of Y(1) F(1) (U(1) - (U(l), U(2) are used for F(1) and F(2) for (t ) cylinder problem with.01 < E <.1) Terms F(2) \(2)/ of the differential equations which have been rewritten as a series of first order equations F(3) of the form: Y(2) = F(l) Y(3) = F(2) Y(4) = F(5) (c.f. discussion in Chapter III) 296

297 CORRESPONDENCE BETWEEN NOMENCLA.TURE USED IN COMPUTER PROGRAM FOR CYLINDER PROBLEM WITH E2 >.01 AND NOMENCLATURE USED IN TEXT A.(O) a~ A(K) bk; k = 2,4,6,... B(K) Bk C(K) Ak DE ~ * ES E2 FR Tw 2 R U pU,2 H(K,I) Hk(r); K,k 3,5,7,... H(K,I) Gk(q); K,k 2,4,6,... NU(J) q 0 N /2R0UQ V P(I) PR ~ P r RE T UoRo/V TAU(J) Tj U(I) f l(r) UP(I) fl(r) UDP(I) f1l) UTP(I) fl (l) V(K,I) ~ fk((n VP(K,I) f(T) K,k = 5,5,7,... VDP(K,I) ~ fr(l )

298 V(K,I) Fk(q) VP(K,I) Ft( l)> k,K 0,2,4,6,... VDP(K,I) F(I) W(JI) ~$1 u j(Q) X~7 XE Xe XI e

$COMPILE MAC, PRINT OBJECT 999595 06/30/65 7 28 50.9 AM MAD (01 MAY 1965 VERSION) PROGRAM LISTING........oo. CYLENOER_ ROBLEM WITH E 5Q UAE __ REATER THAN 0.01 * * * * * * * SET UP DIMENSION Y(3),F(3),D(3) *001 EXECUTE SETRKJ<D._ ( 3,Y( ),F (.).,D, X STEP) *0.02 INTEGER J,K,MN,I,T,EPS,TS,TMAXMMAXNMAX *003 VECTOR VALUES FMI=$1H1/1H- *$ *004 VECTOR VALUES FM2=$1H,S4,54(1H-) *$ *005 VECTOR VALUES FM3=$1H,S4,F5.3,4F12.6 $ *006 VECTOR VALUES FM4=$1H, 4F12.6 *$ *007 CIMENS.ION P( 150Q1_U.UL5.0),lUP-(150),UDP(15C) hUTP(150),B(1C) C(11) *008 1,A(ll),TAU(ll),NU(11) *008 DIMENSION R1(150),R2(150) *009 DIMENSION V(1400,VS) *010 DIMENSION VP(1400,PS) *011 DIMENSION VDP(1400,D.S) *012 DIMENSION W( 1400QTW) *013 DIMENSION H(1400,HS) *014 VECTOR VALUES VS=2,1505150 *015 VECTOR VALUES PS=2,,10,150 *016 VECTOR VALUES OS=2,150,150 *017 VECTOR VALUES TW=2,1,150 *018 VECTOR VALUES HS=2,150,15Q *019 READ AND PRINT DATA *020 Z2=-1.00 *021 ES=(XE.P.2)*2*RE *022 YDP=4.0/ (ES.P.0.50) *023 TS=1.*024 T=l *025 SW16 WHENEVER T.E.I, EPS=2 *026 WHENEVER T.G.1, EPS=1 *027 CALC. OF CORRECTED VALUES OF CONSTANT PARAMETER FROM ERRORS IN BOUNDARY COND'S TA=AC *028 CA=C.C01 *029 CY='.001 *030 1R()= 1. *031 R1( 1)=1. *032 =l1 *033 SW2 1=CF1*(F1(M)/R1 (M-1)-i). *034 WHENEVEP. Ql..87.20,Q1=87.20 *035 WHENEVER 01.L.(-87.20) Q'1=-87.20 *036 CE=-CA*((2.35040/(EXP.(Q1)-EXP.(-Q1)))+i) *037 CA=DB *038 tI=M+1 *039 TA=TA+DA *040 AN=A(0) *041 AP=TA *042

CALCo OF CORR. VALUES OF INITIAL CONDOS FROM PREVIOUS ERRORS IN BOUNDARY COND S YX=YCP *043 SW1 R2(0)=1.5 *044 R2(1)=1.0 *045 N=1 *046 SW4 Q2=CF2(R2(N )/R2(N-1 )-1) *047 WHENEVER Q2.Go87.2G0Q2=87.20 *048 WHENEVER Q2.L.(-87.20),Q2=-87.20 *049 DX=-Y*( (2.35040/(EXP(Q2)-EXP. (-Q2) ) )+1) *050 DY=DX *051 N=N+1 *052 YX=YX+DY *053 WHENEVER TS.E.1 *054 Y(3)=YX *055 01 Y(2)=O *056 01 Y () =0 *057 01 OR WHENEVER TS.E.O *058 01 Y(2)=YX *059 01 Y(1)=EPS-1 *060 01 END OF CONDITIONAL *061 01 INTEGRATION OF DIFF. EQS SW3 I=0 *062 X=O *063 START 1=1+1 *064 P (I) =X 065 5J WHENEVER T.E.1 *066 0 U(I) =Y(1) *067 01 UP(I)=Y(2) *068 01 UDP(I)=Y( 3) *069 01 UTP (I)=F(3) *070 01 OTHERWISE *071 01 V(T, I )=Y(1) *072 01 VP(T,I)=Y(2) *073 01 WHENEVER TS.E.1, VDP(TI)=Y(3) *074 01 WHENEVER TS.E.0, VDP(T,I)=F(2) *075 01 END OF CONDITIONAL *076 01 SW5 wHENEVER TS.E.1 *077 F(1) =Y(2) *078 01 F(2) Y(3) *079 01 WHENEVER T.E.1, F(3)=-AP*(Y(1)*Y(3)-(Y(2).P.2))*ES *080 01 WHEtNEVER T.NE.l., F(3)=-AN*(U(I)*Y(3)-(T+I)*UP(I)*Y(2)+T*UDP(I *081 01 1 )*Y(1)-H(TI)-((Z2).P.((T+1)/2))*IT/(T+I.))*(U(I)*UDP(I)+(T-2 081 2 )*(UP( I ).P.2) )*AP)*ES 081 GR WHLENVER TS.E.0 *082 01 F(1)=Y(2) *083 01 wHENEVEi T.E..) *084 01 F(2)=-MU*(U( I )*Y(2)) *085 02 CT-RW I S t *086 02 F(2)=-MU*(U( I )Y(2)-T*UP(I)*Y(1)-(2-EPS)*(H(T,I)-((Z2.P.(T/2 *087 02 1 ) ) (T+2)*V(T+1, )+A(T)*U(I) )*VP(OI))) *087 tNO uF CONDITIONAL *088 02 NiC CF CONDITIONAL *089 01 CALC 9=RKCL Q.(() *090;HEN~iEVEf S.c.l.i,TRANiSF ER TO SW5 *091 WHENEVER I.E.G.SIZ, TRANSFER TO SW6 *092

TRANSFER TO START' *093 SELECT. OF WHICH BOUNDARY CUND. TEST TO SUBJECT SOL'N TO DEPENDING ON WHETHER MOM. OR ENERGY EQU'S ARE BEING SOLVED SW6 WHENEVER TS.E.1, TRANSFER TO SW7 *094 WHENEVER TS.E.O, TRANSFER TO SW8 *095 CALC. OF ERROR IN CONT. BOUNDARY COND. SW7 WHENEVER.ABS.(Y(1)-(1.C*EPS/(T+1.0))).G.CTOL.AND. N.L.NMA *096 1 X *096 R2(N)=Y(1)-(EPS /(T+I.)) *097 01 TRANSFER TO SW4 *098 01 CALC. OF ERROR IN MOM. BOUNDARY CON3. OTHERWISE *099 01 WHENEVER N.GE.NMAX, PRINT RESULTS R2(1)...R2(N) *100 01 A(T-1 )=AP *101 01 C( 1.)=-4.0*(A.P.2) *102 01 MBC = A*2*Y(2) +C(T)-(2-EPS)*((Z2).P.((T-1)/2) )*(2.*T/(T+I)) *103 01 1 *A*(4*A-UP(I+1))*AP-(2.*T/(T+I))*A*(2.*A-UP(I+1))*(2-EPS) *103 END OF CONDITIONAL *104 01 WHENEVER.ABS.(Y(3) + MBC*ES*0.5 ).G.MTOL.AND. M.L.MMAX *105 R1 (M)=Y( 3)+MBC*ES*O.5 *106 01 TRANSFER TO SW2 *107 01 OTHERWISE *108 01 WHENEVER M.GE.MMAX, PRINT RESULTS Rl(i)...Rl(M) *109 01 STORAGE OF SOL'S AT LAST GRID PT. k WHENEVER T.E.1 *110 01 U(I+1)=Y( 1) *111 02 UP(I+1)=Y(2) *112 02 UDP( I+1)=Y(3) *113 02 UTP( I+1 )=F(3) *114 02 CTHERWISE *115 02 V(T, 1+1)=Y(1) *116 02 VP(T, I+1 )=Y(2) *117 02 VDP(T, I+1)=Y(3) 118 02 END OF CONDITIONAL *119 02 END GF CONDITIONAL *120 01 CALC. OF TERMS TO BE USED IN HIGHER ORDER MOM. EQU'S THROUGH SN14,FOR J=1tl,J.G.(GRSIZ+l) *121 WHENEVER T.E.1, H(3,J)=O *122 01 WHENEVER T.E.3, H(5,J)=(80./3. )*((VP(3,J).P.2)-(V(3,J)*VDP(3, *123 01 1 J1 ) )+( 2./ 3. )*A(2)* (VDOP(3,J)*U(J) )+(3.*V(3,J)*UDP (J) ) )*123 SW14 WHENEVER T.E.5, H(7,J)=7.*(3.*(8o*VP(5,J)*VP(3,J) *124 01 1 -5.*VCDP3,J)*V(5,J) )-0. 75*A(2)*(6.*VP(5,J)*UP(J)+(80./3.)*( *124 2 (VP(3,J).P.2)-3.*V(3,J)*VDP(3,J) -3.*VDP(5,J)*U(J)-I5.*UDP(J *124 3 )*V(:,J) )-2.5*A(4)*(VOP(3,J)*U ( J )+4.*UP ( J ) *VP(3, J )+3.*V(3,J) *124 4 *UCP( J ) )-9.*VDP( 5J )*V( 3J ) ) *124 WHENEVER T.E.1, C(3)=-(A.P.2) *125 wHENEVER T.E.3, C(5)=A*(1O.*A(2)*(2.*A-UP(GRSIZ+1 I)+(2./3.)* *126 1 A*(1O*A(2I)-l.)-(20./3.)*(A(2)-L)*(3.*A(2)*A+2.*VP(3,GRSIZ+1) *126 2 -A) )*126 WHEiNEVER T.E.5, C(7)=O.5*A*((105./2.)*(A(2)-A(4) )*(2.*A-UP(G *127

1 RSIZ+1))+A*(21.*A(2)-35.*A(4)-l.)+21.*(A(2)-1.)*(A(5.*A(4)- *127 2 10.*A(2)+l.)-3.*VP(5,GRSIZ+1) )+35.*(A(4)-6.*A(2)+1.)*(A*(3.* *127 3 A(2)-1.)+2.*VP(3,GRSIZ+1)).) *127 THROUGH SW12,FOR J=l,1,JG.(GRSIZ+1l) *128 WHENEVER T.E.I1, W(1,J )=UP(J )/(2.*A) *129 01 WHENEVER T.E.3, W(3,J)=(4.*VP(3,J)+3.*A(2)*UP(J) )/(2.*A) *130 01 WHENEVER T.E.5, W(5,J)=(6.*VP(5,J)+40.*A(2)*VP(3,J)+5.*(6.*(A(2). *131 01 1 P.2)-A(4))*UP(J))/(2.*A) *131 SW12 WHENEVER T.E.7, W(7,J)=(8.*VP(7,J)+126.*A(2)*VP(5,J)+140.*(6.*(A( *132 01 1 2).P.2)-A(4))*VP(3,J) +7.*(90.*(A(2).P.3)-15.*A(2)*A(4)+A(6)) *132 2 *UP(J))/(2.*A) *132 WHENEVER T.E.1, TAU(1)=UDP(1)/(ES*(A.P.2)) *133 WHENEVER T.E.3, TAU(3 )=2.*(2.*VOP(3,1)+3.*A(2)*UOP(1))/(ES*(A.P.2 *134 1 )) *134 WHENEVER T.E.5, TAU 5)=2.*(3.*VDP(5,1)+40.*A(2)*VDP(391)+5.*(9.*( *135 1 A(2).P.2)-A(4))*UOP(l))/(ES*(A.P.2)) *135 WHENEVER T.E.7, TAU(7)=2.*(4.*VDP(7,1)+126.*A(2)*VDP(5,1 )+140.*(9 *136 1.*(A(2).P.2)-A(4))*VDP(3,1)+7.*(180.*(A(2).P.3)-45.*A(2)*A(4 *136 2 )+A(6))*UDP(1)~/(ES*(A.P.2)) *136 TEST LOC. TO DET. WHICH MOM. EQ. IS BEING SOLVED AND WHEN MOM. PROB. HAS BEEN COMP. SW20 T=T+2 *137 WHENEVER T.L.TMAX,TRANSFER TO SW16 *138 SET UP OF ENERGY PROB. TS=0 *139 T=O 140 r YX=YP *141 MU=A*ES*PR *142 EPS=2 *143 TMAX=TMAX-1 *144 TRANSFER TO SW1 *145 CALC. OF ERROR IN ENERGY B.C. SW8 EBC=Y(1)+(2-EPS)*(V(0,+11)*(A(T)+(Z2.P.(T/2)))+B(T)) *146 WHENEVER.ABS.((Y(2)/MU)+EBC).G.ETOL.AND. N.L.NMAX *147 R2(N)=(Y(2)/MU)+EBC *148 01 TRANSFER TO SW4 *149 01 GTHERWISE *150 01 WHENEVER N.GE.NMAX, PRINT RESULTS R2()...R2(N),T *151 01 STORAGE OF SOL'S AT LAST GRID PT. V(T,I+1)=Y(1) *152 01 VP(T, I i)=Y(2) *153 01 VDPIT,I.+I)=F(2) *154 01 ENC OF CONDITIONAL *155 01 CALC. OF TERMS TO BE USED IN HIGHER ORDER ENERGY EQU'S THROUGH SW15, FOR J=1,1,J.G.(GRSIZ+1) *156 WHENEVER T.E.O, H(2,J)=U *157 01 WHENEVER T.E.2, H(4,J)=8.*(3.*V(3,J)*VP(2,J)-2.*VP(3,J)*V(2, *158 01 1 J ))+6.*A(2)*(2.*UP(J)*V(2,J)-U(J)*VP(2,J) )+24. *A ( 2 )*V13J)*V *158 2 P(0,J) *158 SW15 WHENEVER T.E.4, H(6,J)=20D.*(3.*V(3,J)*VP(4,J)-4.*VP(3,J)*V(4 *159 01

1,J))-18.*(5.*V(5,J)*VP(2,J)-2.*VP(5,J)*V(2,J))+15.*A(2)(4.* *159 2 UP(J)*V(4,J)-U(J)*VP(4,J)-16.*VP(3,J)*V(2,J)+24.V(3,J)*VP(2 *159 3,J))+15.*A(4)*(2.*UP(J)*V(2IJ)-U(J)*VP(2,J))+30.*(-3.*A(2)*V *159 4 (5,J)+2.*A(4)V{(3,J))*VP(0,J) *159 WHENEVER T.E.O, B(2)=0 *160 WHENEVER T.E.2, 8(4)=6.*(V(2,GRSIZ+1)*(A )-)-A(2)*V(O,GRS *161 1 IZ+1)) *161 WHENEVER T.E.4, BI6)=15.*(V(4,GRSIZ+1)*A(2)+V(2,GRSIZ+1)*A(4 *162 1 )-V(4,GRSIZ+I)-b.*.. V(2,GRSIZ+I )A(2)-V(0,GRSI Z+ )*A(4)+V(2,GR *162 2 SIZ+1)+V(OGRSIZ+1)*A(2) ) 162 WHENEVER T.E.O, NU(0)=VP(O,1)/A *163 WHENEVER T.E.2, NU(2)=(VP(2,1)-A(2)*VP(C,1))/A *164 WHENEVER T.E.4, NU(4)=(VP(4,1)-6o*A(2)*VP(2,1)+(6.*(A(2.P.2)-A(4 *165 1 ))*VP(O,1))/A *165 WHENEVER T.E.6NU(6= ( VP( 6,1 )-15.*A( 2 ) *VP(4,1)+1 5.* ( 6. ( A (2 ) P.2 *166 1 )-A(4))*VP(2,1)-(90.*(A(2)..P.3)-15.*A(2)*4)A()+A(6))*VP(O,) *166 2 /A *166 2__A______________ ______________________________________ _166 TEST LOC. TO DET. WHICH ENERGY EQ. IS BEING SOLVED AND WHEN ENERGY PROB. IS COMPL. T=T+2 167 WHENEVER ETOL.L.0.001, ETOL=ETOL*(100.P.(T/2)) *168 YX=YP *169 EPS=1 *170 WHENEVER T.L.TMAX, TRANSFER TO SW1 *171 PRINT STATEMENTS AND MISC. CALC'S PRINT FORMAT FM1 172 PRINT FORMAT FM2 *173 THROUGH SW10, FOR K=1,,K.G.(GRSIZ+1, *174 SW10 PRINT FORMAT FM3, P(K),UIK),V(3,K),V(5,K),V(7,K) *175 01 PRINT FORMAT FMI *176 PRINT FORMAT FM2 *177 THROUGH SW17, FUR K=l,1,K.G.(GRSIZ+1) *178 SW17 PRINT FORMAT FM3, P(K),W(1,K),W(3,K),W(5,K),W(7,K) *179 01 PRINT FORMAT FM1 *180 PRINT FORMAT FM2 *181 THROUGH SW18, FUR K=1,1,K.G.(GRSIZ+1) *182 SW18 PRINT FORMAT FM3, P(K),UP(K),VP(3,K),VP(5,K)K)VP(7K)183 01 PRINT FORMAT FMi *184 PRINT FORMAT FM2 *185 THROUGH SW19, FOR K=1,1,K.G.(GRSIZ+1) *186 SW19 PRINT FORMAT FM3, P(K),P(K),VDP(3,K),VP(5,KVDP(7,K) *187 01 PRINT FORMAT FM1 *188 PRINT FORMAT FM2 *189 THROUGH SW21, FOR K=1,1,K.G.(GRSIZ+l) *190 SW21 PRINT FORMAT FM3, P(K),V(OK),V(2,K),V(4,K),V(6,K) *191 c PRIiT FORMAT FMI *192 P RINT FVi MAT FM4, A,A(2),A(4),A(6) *193 PRINT FORMAT FM4, TAU(1),TAU(3),TAU(5),TAU(7) *194 PRINwT FORMAT FM4, NU(C),oNU(2),NU(4),NU(6)'195 PRINT FORMAT FM2 *196 THROUGH SW30, FOR XI=.,.1,XI.G.1. 5 *197 CE=A*(.+(A(2)/2.)*(XI.P.2)+(A(4)/24.)* XI.P.4)+(A(6)/720.) *198 01 1 (XI.P.6)) *198 CE=D*(ES.P.C.5) *199 01 FR=TAU( )*XXI-(TAU(3)/6.)XI.P.3)+(TAU(5)/120.)*(XI.P.5)-(TA *200 01 1 U(7)/504-. )(XI.P.7) *200

FR=FR~(ES.P.3.5) *201 01 G=NU(O)+(NU(2)/2.)*(XI.P.2)+(NU(4)/24.)*(XI.P.4)+(NU(6)/720. 202 01 1 )*(XI.P.6) *202 G=O/(ES.P.O.5) *203 01 SW30 PRINT FORMAT FM4, XI,DEO,,FR *204 01 SWI1 END OF PROGRAM *205 0

505 The procedure used in the program for calculating drop trajectories is to use the Runge-Kutta method to integrate the equations of motion for a drop starting at particular values of y* until the doo drop almost reaches the cylinder or until it entirely misses the cylinder. If the drop is going to hit the cylinder the calculated values of the solution are extended by linear interpolation to the required values at the cylinder. These values are then used to calculate the quantities of Eqs. (17*A.C.) which are needed to describe the conditions in the liquid film. At each step of the integration the current value of the absolute value of the vector difference between the drop velocity and the gas velocity at the position of the drop is used to calculate an instantaneous Reynolds number based on the drop diameter. The drag coefficient corresponding to this Reynolds number is obtained by interpolation from the tabulated values of Table II, which are put into the program as "data" and were obtained from Figure 28 which is a smoothing of the values obtained from Reference 28. This value of the drag coefficient is then used in the equations of motion to calculate the new position of the drop.

306 10 CD.1 L, | i1, l,,I, i i L,I, l I, l, l, I 10 100 1000 Re Figure 28. Values of drag coefficients for spheres used in calculating drop trajectories.

307 TABLE II DRAG COEFFICIENTS FOR A SPHERE (28) Re CD 0.2 12 40 0.4 64.5 o.6 44.5 0,8 36.0 1 28,4 2 14.4 4 8,4 6 6,2 8 5.1 10 4o4 20 2,8 40 L,88 6o 1.50 80 1,27 100 1.12 200 0o79 400 0 59 6oo 0.53 800 0.49 1000 o o48 2000 0. 40

308 CORRESPONDENCE BETWEEN NOMENCLATURE USED IN COMPUTER PROGRAM FOR CALCULATING DROP TRAJECTORIES AND NOTATION USED IN TEXT CD1 CD H AV* MDOT J MF(N) J (tabulated value) MTOT ~ i P(I) t* PH cd PHI(K) e Tpd (tabulated table) R(I) I rdl RE1 2rdAV*/vg (drop Reynolds number) RINF 2RoUo/Vg T t* V1 V* V2'yo yo VPHI V VXF V* xof VYF ~ V* yof X1 Xd X2 d XBA.R x XF - xf

309 XF(K) x* (stored value) df YF f d "df YF(K) y* (stored value) df YINF(K) - yd doc

$COMPILE MAD, PRINT OBJECT 013991_ 07/23/65 _ 3 49 47.3 AM MAD (01 MAY 1965 VERSION) PROGRAM LISTING..o o.. o..ALC. OF DROP TRAJECTORIES SET UP VECTOR VALUES FM1=$1H1/1H-. *$ * 001 VECTOR VALUES FM2=$1H,S4,48(1H-) *$ *002 VECTOR VALUES FM3=$1H,S4,4F12.6 *$ *003 DIMENSION Y(4),F(4),D(4) *004 EXECUTE SETRKD.(4,Y(1),F(1)vD.,T,STEP) 005_ DIMENSION IMAGE(10000) *006 EXECUTE PLOT2.(IMAGE,O.,-2.5,2.08335,O.) *007 DIMENSION RED(100),CD(100),P(1000)X1(1000),X2(1000),V (1000),V2(1 *008 1 000),MF(1000),XF(1000),YF(1000),PHI(1000),VPHI(1000)YINF(10 *008 2 00),R(1000) *008 INTEGER JMAX,I,IMAX,K,L,KE,M,N *009 INTEGER SW *010 INTEGER IIKI,IP,KP *011 READ AND PRINT DATA *012 K=O *013 SW=O *014 1 KI=O *015 ~ ALPH=3.*ROH/(RR8. ) *016 INITIAL CONDITIONS SW1 K=K+1 WHENEVER KI.E.5, KI=O KI=KI+1 Y(1)=-XINF Y(2)=INCY*(K-1i Y(3)=1.0 Y(4)=0. I=0 T=O'INTERGRATION OF IFFERENTIAL EQUATIONS I1=0 *026 START I=I+1 *027 II=1+1. *028 P(I)=T *029 Xl(I)=Y(1) *030 X2(I)=Y(2) *031 VI(I)=Y(3) *032 V2(I)=Y(4) *033 R(I)=(Y(1).P.2+Y(2).P.2).P.0.5 *034 WHENEVER KI.E.KP.AND. II.E.IP *035 EXECUTE PLOT3.($0$Y(1) 1)(2),1) *036 01 11=O *037 01 OTHERWISE *038 01

CONTINUE *039 01 END OF CONDITIONAL *040 01 SW5 F(1)=Y(3) *041 F(2)=Y(4) *042 H1=Y(3)-1.0+( (Y(l).P.2-Y(2).P.2)/(Y(l).P.2+Y(2).P.2).P.2)) *043 H2=Y(4)+(2.*Y(l)*Y(2)/((Y(1,).P.2+Y(2).P.2).P.2)) *044 H=(H1.P.2.+H2.P.2.).P.0.5 *045 REL=RINF*(.ABS.H )*RR *046 WHENEVER REI.GE.1.0 *047 CD1=TAB.(RE1lRED(1)~CD(1)~,ll,4,JMAXTS) *048 01 OTHERWISE *049 01 CDl=24.*(1.+(3./16.)*RE1)/RE1 *050- 01 END OF CONDITIONAL *051 01 F(3)=-ALPH*CD1*H1*H *652 F(4)=-ALPH*CD1*H2*H *053 S=RKDEQ.(O) *054 WHENEVER S.E.1.0, TRANSFER TO SW5 *055 WHENEVER (Y(1),P.2+Y(2).P.2).LE.1.0.OR, Y(1).GE.0., TRANSFER TO SW6 *056 TRANSFER TO START -057 STORAGE OF SOL'S AT LAST GRID PT. SW6 P(I+1)=T *058 X1(I+1)=Y(l) *059 X2(I+1)=Y(2) *060 VI(I+1)=Y(3) *061 V2(I+1)=Y(4) *062 R(I+1)=(Y(1).P.2+Y(2).P.2).P.O.5 *063 -----— ~ —--— * —* —-— * —--— * —----—. —-. —------— ~-. —---—. —. —-r~ CALC. OF TANG. VEL. AND TOTAL MASS FLOWH WHENEVER R(I+1).LE.1.0 * 064 RdI)=(XlII).P.2+X2(I).P.2).P.0.5 *065 01 A=(X1(I-1)*X2(I)-XL(I.)*X2(1-1))/(Xl (I-1)-XL(I)) *066 01 B=(X2(I)-X2(I-1))/(X1(I)-X1(I-1)) *067 01 C=(B.P.2-A.P.2+1.).P.0.5 6068 01 XF=-(A*B+C)/(1.+(B.P.2)) *069 01 YF=A+B*XF *070 01 VXF=V1(I) *071 01 VYF=V2(I) *072 01TAPH=YF/(.ABS.XF) *073 01 PH=ATAN.(TAPH) *074 01 PHI(K)=PH *075 01 VPHI(K)=VXF*SIN.(PH)+VYF*COS.(PH) -076 01 YINF(K)=INCY*(K-1) *077 01 XF(K)=XF *078 01 YF(K)=YF *079 01 CALC. OF LAST DROP TO HIT CYLINDER OR WHENEVER SW.E.O *080 01 SW=1" *o8- 01 KE=K~082 01 END OF CONDITIONAL *083 01 DET. OF LAST DROP IN PATH OF CYLINDER WHENEVER X2(1).L.1.O, TRANSFER TO SW1 - *084 PLOT STATEMENT

PRINT FORMAT FM1 *085 PRINT COMMENT $ *086 1 DROP TRAJECTORIES $ 086 EXECUTE PLOT4.(0,LABEL) *087 CALC OF MASS FLUX MF(KE-2)=0. *088 MF(KE-1)=0. *089 THROUGH SW8, FOR N=1,1,N.G. (KE-2) *090 MF(N)=INCY/(PHI(N+1)-PHI(N)) *091 01 PRINT STATEMENTS PRINT RESULTS KE *092 KE=KE+1 *093 PRINT FORMAT FM1 *094 H PRINT FORMAT FM2 *095 r0 THROUGH SW9, FOR M=1,1,XBAR.G.PHI(KE-2) *-096 XBAR=IPH*M *097 01 MDOT=TAB.(XBAR,PHI(1),MF(1) 1,1,4,KE-2,T6) *098 01 VPHI=TAB.(XBAR,PHI{I),VPHI(l),1,1,4,KE-2,T7) *099 01 MTOT=TAB.(XBAR,PHI(1),YINF(1),1l9,4,KE-2,T7) *10 01 PUNCH FORMAT FM3, XBAR,MTOT,MDOT,VPHI *101 01 SW9 PRINT FORMAT FM3, XBAR,MTOT,MOOT,VPHI 102 01 END OF PROGRAM *103

531 CORRESPONDENCE BETWEEN NOMENCLATURE USED IN COMPUTER PROGRAM FOR CYLINDER PROBLEM WITH E < 0.1 AND NOMENCLATURE USED IN TEXT ALP(2*J), 2j BC.(I,J) ~) BET(2*J) ~ 2j DEL(I) ~'(x) (tabulated value) DELT 6 E E F.(N) N! G (i NU Nu P(I) n PD(J) x (tabulated value) PR PR RINF 2RUoo/Vg RR rd/Ro V(2*K,I) F2k()) VP(2*K,I) F2 ) X ~ Tl XI x

SCOMPILE MAD, PRINT OBJECT 013985 07/23/65 3 38 53.1 AM MAD (01 MAY 1965 VERSION) PROGRAM LISTING SOLUTION TO ENERGY EQUATION FOR SMALL'El SET UP INTEGER KKMAXIJN,NMAXJMAX,IJ,IM,IN -001 INTEGER II *002 INTEGER T,TMAX *003 INTEGER M *004 DIMENSION ALP(50),BET(50),R(200),P(150),NO(75) 005 DIMENSION DEL(200),PD(200) *006 DIMENSION Y(3),U(3),D(3) *007 EXECUTE SETRKD.(2,Y(1),U(1),DX,STEP) *008 VECTOR VALUES FM1=$1H1/1H-*$ *009 VECTOR VALUES FM2=$1H,S4,82(1H-)*$ *010 VECTOR VALUES FM3=$1H,S4,F5.3,S4,10F13.6 $ *0 VECTOR VALUES FM4=$1H,S4,13,2F20.6 *$ *012 VECTOR VALUES FM5=$1H,S-4,10OF13.6 *$ *013 VECTOR VALUES FM6=$IH,S4,F5.3,F12.6*$ *014 VECTOR VALUES FM7=$1H,S4,F5.3,13*$ *015 DIMENSION V(2000,VS) *016 DIMENSION VP(2000,PS) *017 H VECTOR VALUES VS=2,150,150 *018 VECTOR VALUES PS=2,150,150 09 READ AND PRINT DATA *020 DEFINITION OF F. INTERNAL FUNCTION (IJ) *021 ENTRY TO F. *022 FACT=1 *023 WHENEVER IJ.E.O, FUNCTION RETURN FACT *024 THROUGH LOOP, FOR II=1,1,11.G.IJ'025 LOOP FACT=FACT*II *026 FUNCTION RETURN FACT *027 END OF FUNCTION *2028 ----- ----- -------- ---------- ----— ~ —--- - - - - - - - - - - DEFINITION OF BC. INTERNAL FUNCTION BC.(IM,IN)=F.(IM)/(F.(IN)*F.(IM-IN)) *029 --------------------------------— ~ —-------- ------ INPUT FROM MOM. PROGRAM T=O *030 SW13 T=T+1, *031 READ FORMAT FM7, E,M *032 THROUGH SW10, FOR J=,lJ.G.M *033 SWI1 READ FORMAT FM6, PD(J),DEL(J) *034 THROUGH SW1, FOR J=l,1,J.G.JMAX *035 SWi READ FORMAT FM4, NO(J),ALP(2*J-2),BET(2*J) *036 CAL. OF INITIAL COND'S FROM ERROR IN —------------------------------------— N B.C. CALC. OF INITIAL C.OND'S FROM ERROR IN B.C.

K=O 0037 SW2 YX=1.0 *038 DY=0.001 *039 R(O)=1.5 040 R(1)=1.C *041 N=l *042 SW4 Q=CF*(R(N)/R(N-1)-1).043 WHENEVER Q.G.87.2,Q=87.2 *044 WHENEVER Q.L.(-87.2)~ Q=-87.2 *045 OX=-DY*((2.35040/(EXP.(Q)-EXP.(-Q)))+1) *046 DY=DX *047 N=N+1l 048 YX=YX+DY *049 WHENEVER K.E.0 *050 Y(1)=l. *051 01 OTHERWISE *052 01 Y(1)=0. *053 01 END OF CONDITIONAL *054 01 Y(2)=YX *055 INTEGRATION OF DIFF. EQ'S SW3 1=0 056 X=0. *057 START 1=1+1 *058 P(I)=X *059 V(2KtI)=Y(1) *060 VP(2*KI)=Y(2) *061 SW5 U(1)=Y(2) *062 1 WHENEVER K.E.O *063 G=O. *064 01 OR WHENEVER K.E.1 *065 01 G=-ALP(2)*(X.P.2.)~VP(0,I) *066 01 OR WHENEVER K.G.1 *067 01 G= -ALP(2*K)*(X.P.2.~*VP(0,1) *068 01 THROUGH SW6~ FOR J=ll,J,.G(K-1) *069 01 SW6 G=G+X~(2.*BC.(2*K,2*J-1)*V(2*JI)~BET(2(K-J+1))-BC.(2*K,2*J *070 01 01 1 )*X*VP(2JTI)*ALP(2*(K-J)).).070 END OF CONDITIONAL *071 01 U(2)=-PR~E (ALP(*)*((X.P.2.)*Y(2)-4.*K*XY{1)~-G) *072 CALC S=RKDEQ.(O) *073 WHENEVER S.E.1.O, TRANSFER TO SW5 *074 WHENEVER I.E.GRSIZ, TRANSFER TO SW7 *075 TRANSFER TO START *076 CALC. OF ERROR IN B.C. SW7 WHENEVER K.E.O *077 EBC=O. *078 01 OTHERWISE *079 01 EBC=O. *080 01 THROUGH SW8, FOR J=0,I,J.G.(K-1) *081 01 SW8 EBC=EBC-BC.(2*K,2*J)*ALP(2*(K-J) )*V(2*J,I+1) *082 01 01 END OF CONDITIONAL *083 01 WHENEVER.ABS.(Y(2)/(PR*E)+ALP(0)Y(1)-EBC).G.ETOL.AND. N.L.NMAX *084 R(N)=Y(2)/IPR*E)+ALP(O)*Y()1-E8C *085 01 TRANSFER TO SW4 *086 01 STORAGE OF SOL'S AT LAST GRID PT.

OTHERWISE *087 01 PRINT RESULTS N *088 01 V(2*K,I+1)=Y(1) *089 01 VP(2*K,I+1)=Y(2) *090 01 END OF CONDITIONAL *091 01 K=K+1 *092 CALC. ANC PRINT-OUT OF NUSSELT NO. PRINT RESULTS ERR,RINF,PR *093 THROUGH SW12, FOR J=1,1,((J+1)*INXI).G.XIMAX *094 XI=J*INXI *095 01 DELT=TAB.(XI,PD(1),OEL(1),,1,4,M,T6) *096 01 NU=O *097 01 THROUGH SW11, FOR I=O,1,I.G.KMAX *098 01 SW11 NU=NU+VP(2*I,1)*(XIP.(2*I))/(F.(2*I)*DELT) *099 02 SW12 PRINT RESULTS XI,NU *100 01 PRINT STATEMENTS PRINT FORMAT FM 101 PRINT FORMAT FM2 *102 THROUGH SW9, FOR I=1,1,I.G.(GRSIZ+1) *103 SW9 PRINT FORMAT FM3, P(I),V(OtI),V(2,I),V(4,I),V(6,I),V(8,I) *104 01 PRINT FORMAT FMI 105 PRINT FORMAT FM2 *106 PRINT FORMAT FM5, VP(Ol1), VP(2,1),VP(41),VP(61),VP(81) 107 WHENEVER T.LE.TMAX, TRANSFER TO SW13 *108 END OF PROGRAM *109

317 Note: that all quantities which occur in program for the oscillating plate are either pure imaginary or real. The equations are written in this program so that the i = /-l does not appear. It is to be understood that all quantities calculated in this program which correspond to pure imaginary numbers should be multiplied by i to get'the quantities appearing in the text. CORRESPONDENCE BETWEEN NOMENCLATURE USED IN COMPUTER PROGRAM FOR OSCILLATING PLATE WITH NOMENCLATURE IN TEXT A.(O) a A(J) alj A2(J) aj A2H(J) a2j A.D1 ~ 61l AND Iq| A1l ~ qj I ARD1 ep c ARF1 ~ p ARN 1 Pq1 DELTA.(I,J) bi j El(J) ~ eLj E2(J) e2j EH(J)'e Gl(J) G:i

31.8 G2(J) G2j GH(J) G2~ H(I) nonhomogeneous part of differential equations for momentum problem, terms appearing in nonhomogeneous part of differential equations for energy problem. NU (N -N )/ V (uavg Nusteady/ P(I) o ~ PBl1 (b/x) u PB2, & PB3 To - PUoo PI /2 PR Pr R Rj, rj; j = 0,2,4,o.. depending on which equation is being solved IRH Rj, rj; j = 0,2,4,... depending on which equation is being solved RES nonhomogeneous part of differential equation for energy problem SG u* o SG1 g1o SG12 gl2 SG14 _ g.4 SG2 gao SG22 g22 SG24 g24 SGH g o

319 SGH4 g24 U(I) fro(n) UP(I) fo(n) UDP(I) f0o( V(J,I) fij ( ) VP(J,I) fij~() j,J = 1,5,5,... VDP(J,I) flj(-i) V(J,I) Fjj(7) VP(J,I) Flj(j) j,J = 0,2,4,... VDP(J,I) Flj (i) W(J,I) ~ f2j () WP(J,I) f~j( r) j,J = 1,3,5,. WDP(J,I) ~ f2j(n W(J,I) F2j ( ) WP(J,I) F2j(T) j,J = 0,2,4,.. WH(J,I) f2j(rQ) WHP(J,I) ~ FI)Fj jJ = 15. WHDP(J, I) ~ j (l) WH(J,) - F2j(n) WHP(J.,I) -,() 7 jJ = 0,4,... WHDP(J,I)~ F('r) | XE ~ Xe Z(I) Fo(T) ZP(I) Fo(T) ZDP(I) F"(i)

sCCMPILE MAC, PRINT UBJECT 014457 07/26/65 11 33 38.1 PM MAD (01 MAY 1965 VERSION) PROGRAM LISTING...... PROGRAM FOR OSC. FLAT PLATE SET UP DICENSION Y(3),F(3),D(3) *001 EXECUTE SETRKD.(3,Y(1),F(1),D,X,STEP) *002 INTEGER J,K,M,N,T,EPS,GRSIZ,TAU,MP,NP,I,Tl,T2,NFE,NSE,TMAX *003 1,MMAX,NMAX,TS *003 INTEGER CELTA. *004 1,T3 *004 VECTOR VALUES FM1=$1H1/1H- *$ *005 VECTOR VALUES FM2=$1H,S4,111(1H-)*$ *006 VECTCR VALUES FM3=$1H,S4,F5.3,S4,F1C.6,2(S4,3F1C.6), S4, 2F10.6*$ *007 VECTOR VALUES FM4=$1H, F10.6, 2(S4,3F1C.6), S4, 2F1C.6 *$ *008 CIMENSION P(15C),U(150),UP(150),UDP(15C),UTP(150, (11),C(11) *009 1,A(11),A2(11),A2H( 11 ),H(150),Z(150),ZP(150),ZDP(C) *009 CIMENSION E1(6),E2(6),G1(6),G2(6)G,EH(6),GH(6) 010 CIMENS IUiN R1(5C),R2(5C0) 011 CIMENSION V(1400,VS) *012 DIMENSION VP(1400,PS) +013 DIMENSION VDP(140CO,DS) *014 CIMENSION' W(14CO,WT) *015 LIMENSION WP(14tC,PT) *016 CIMENSION WDP(140CN,DT) 017 CIMENSION WH(1400,HT) *018 CDIM.ENSIOiN WHP(14CO,PH) *C19 CIMENSION WHCP(140C,DH) *020 VECTOR VALUES VS=2,15C,15C *021 VECTCR VALUES PS=2,15G,150 *022 VECTOR VALUES DS=2,15C,15C *023 VECTOR VALUES WT=2,15,15C5 *024 VECTOR VALUES PT=2,150,150 *025 VECTCR VALUES DT=2,15C,150 *026 VECTOR VALUES HT=2,150,15C *027 VECTOR VALUES PH=2,5150,15 *028 VECTOR VALUES DH=2,15,15t') *029 DEFINITION O'F bLTA. INTERNAL FUNCTION (MP,NP) *C30 ENTRY TO CELTA. *031 WHENEVER V P..E.NP *032 FUNCT IOC- RETURN 1 *033 01 OTHtRWISE *034 01 FUNCTICN RETURN C *035 01 ENC OF CCNOITIONAL *036 01 ENC OF FUNCTION *037 INTEGER CAT,CATM *C38 CAT=C *039 SW3I CAT=CAT+1 *C40

REAC AND PRINT DATA *041 Z2=-1.00 *042 TS=l *043 TAU=: *044 INDEX FOR MGM.EQU'S SW16 WHENEVER TAU.LE.T1 *045 T=2*TAU-1 *046 01 EPS=Z2.P.(TAU) *047 01 CR WHENEVER TAU.LE.T2 *048 01 T=2* (TAU-T )-i *049 01 EPS=Z2.P.(TAU-T1) *050 01 CR WHENEVER TAU.LE.T3 *051 01 T=4*(TAU-T2)-3 052 01 EPS=Z2.P.(TAU-T2) *053 01 END CF CCNDITIONAL *054 01 CALC. OF CCRRECTED VALUES OF CCNSTANT PARAMETER FROM ERRORS IN BCUNDARY COND' S TA=ANO *055 CA=O.C01 *056 CY=O.CC1 *057 R 1()=1.5 *058 Rl( 1 )=1.0 *059 y=1 *060 SW2 Cl=CF1-(R1(M)/R1(M-1 )-1) *061 WHENEVER Q1.G.87.2C,Q1=87.20 *062 WHENEVER Q1.L.(-87.20),Q1=-87.20 *063 CB=-CA*.((2.35040/(EXP.(Q1)-EXP.(-Q1)))+1) *064 DA=DB *065 i=M+l *066 (o TA=TA+DA *067 AN=A () *068 AP=TA *069 CALC. OF CORR. VALUES OF INITIAL CCOND'S FRON PREVICUS ERRORS IN RCUNDARY COND'S YX=YCP *0C70 SW1 R2(0)=1.5 *071 R2( 1)=1. *072 N= 1 *C73 SW4 i2=CF2*( 2(N)/R2(N-1)-1) *074 WHENEVER Q2.G.87.>2-,2=a-7.2C *075 wHENEVER Q2.L.(-87.2C),Q2=-87.2C *076 CX=-CY*((2.3 5<40/(EXP.(02)-EXP.(-Q2)))+1) *C77 CY=CX *078: = N+i *079 YX=YX+DY *080 vHENCVER TS.E.1 *081 Y(3)=YX *082 01 Y(2) = *083 01 Y(1)== *084 01 CR WHENEVER TS.E.O *085 01 Y(2)=YX *086 01 Y( l)=LELTA.(r,TAU) *087 01 END OF CCNDITIONAL *088 01 _- _ -_- _ -_-_- _ -_- _ -_- _ -_-_- _ -_- _ -_- _ -_-_- _ -_- _ -_- _ -_- _ -_-_- _ -_- _ -_- _ -_-_- _ -_- _ -_- _ -_-_

INTEGRATION OF DIFF. EQ'S SW3. I=O *089 X= 0*090 START 1=1+1 *091 P ( I ) =X *092 WHENEVER TAU.E.O.AND. TS.E.1 *093 U I )=Y( 1) *094 01 UP(I)=Y(2) *095 01 UDP(I,)=Y ( 3 ) *096 01 OR WHENEVER TAU.E.0.AND. TS.E.O *097 01 Z( I )=Y(1) *098 01 ZP( I)=Y(2) *099 01 ZDP( I)=F(2) *10 01 OR WHENEVER TAU.LE.T1 *101 01 V(T, I )=Y( 1) *102 01 VP(T, I)=Y(2) *103 01 WHENEVER TS.E.1, VDP(T, I )=Y(3) *104 01 WHENEVER TS.E.O, VOP(T,I )=F(2) *105 01 CR WHENEVER TAU.LE.T2 *106 01 W(T, I )=Y(1) *107 01 WP(T, I)=Y(2) *108 01 WHENEVER TS.E.1, WDP(T,1)=Y(3) *109 01 WHENEVER TS.E.O, WDP(T,I)=F(2) *110 01 CR WHENEVER TAU.LE.T3 *111 01 WH(T, I )=Y( 1) *112 01 WHP(T, I )=Y(2) *113 01 WHENEVER TS.E.1, WHOP(T,I)=Y(3) *114 01 WHENEVER TS.E.O, WHDP(T, I )=F(2) *115 01 o ENC OF CCNDITIONAL *116 01 WHENEVER-TS.E.1, UTP(I)=F(3) *117 WHENEVER TS.E.1 *118 F( 1 )=Y(2) *119 01 F(2)=Y(3) *120 01 WHENEVER TAU.E.O'*121 01 F(3)=-2.*XE* (AP.P.2)*Y(1)*Y(3) *122 02 CTHERWISE *123,02 F(3)=-2*(AN.P.2)*(U(I)*Y(3)-(T-l)*UP(I)*y(2)+T*UCP(I)*Y(1)+( *124 02 I (T-1)*(UP( ).P.2)+U ( I )*UDP( ))*AP-H(I)) I XE 124 END OF CONDITIONAL *125 02 OR WHENEVER TS.E.O *126 01 F(1)=Y(2) *127 01 WHENEVER TAU.E.-J *128 01 RH S=0 *129 02 OR WHENEVER TAU.LE.tt *130 02 RHS=-ZP( I )*( (T+1)*V(T+1,I )+U( I )*A(T+1 )-2.*(1-DELTA. (C,T) )EP *131 02 1 S*I*STEP*A(T-1) )-2.*( 1-DELTA. (, T ) )*EPS*V(T-2, I) 131 CR WHENEVER TAU.LE.T2 *132 02 RHS=-ZP( I)*((T+1)*W(T+1,I)+U(I )*A2(T+1)+4.*(1-CELTA.(,T))*E *133 02 1 PS*I*STLP*A2(T-1) )+4.*(1-DELTA.(C,T))*EPS*W(T-2, I )+h(I) *133 CR WHENEVER TAU.LE.T3 *134 02 RHS=-ZP( I)*( ( +l)*WH(T+1,I)+U(i)*A2H(T+l))+H(I) *135 02 END CF CONDITIONAL *136 02 F(2)=-MU*(U( I )*Y(2)-TUP( I )*Y( 1)-RHS) *137 01 END (IF CNDITItNAL *138 01 CALC S=RKLCQ. ( ) *139 WHENEVER S.E.l.'0, TRANSFiR TO SW5 140 "',FNF-VER^ I E.GGRSIZ,TRANSF-tR TO SW6 *141

TRANSFER TO START *142 SELECT. OF WHICH BOUNDARY COND. TEST IC SUBJECT SCL'N TO CEPENDING ON WHETHER MNU. UR ENERGY ECU'S ARE BEING SOLVED SW6 WHENEVER TS.E.1, TRANSFER TO SW7 *143 WHENEVER TS.E.O, TRANSFER TO SW8 *144 CALC. OF ERROR IN CONTINUITY BOUNDARY CONDITION SW7 WHENEVER TAU.E.O *145 IBC=1.O *146 01 OR WHENEVER TAU.LE.T1 *147 01 WHENEVER T.E.1 *148 01 I BC=AP-0.5 *149 02 CTHERW [SE *150 02 IBC= AP+2*(Z2.P.TAU)*( 1-XE)*A(T-2)/T *151 02 END OF CCNDITIONAL *152 02 CR WHENEVER TAU.LE.T2 *153 01 IBC=AP-O.25*EPS *A(T)+(4. *(1-DELTA.(1,T))*EPS*A2(T-2) *154 01 1 *(XE-1.)/(Ti*1.)) *154 CR WHENEVER TAU.LE.T3 *155 01 IBC=AP-O.25*A(T) *156 01 END OF CONDITIONAL *157 01 WHENEVER.ABS.(Y(1)-IBC).G.ITOL.AND.N.L.NMAX *158 R2(N)=Y( 1)-IBC *159 l01 TRANSFER TO SW4 *160 01 CALC. OF TERMS FOR MOM. BOUNDARY CCNO. ---------------------------------------------------------------------— p IN) CTHERWISE *161 01 WHENEVER TAU.E.O *162 01 A(0 ) =AP- *163 02 MBC=-(2.* Y(3)/((2.*A).P.2))+l.-XE*Y(2) *164 02 CR WHENEVER TAU.LE.T1 *165 02 MBC=-(2.* Y(3)/((2.*A).P.2))+(l 1.-XE*LP(1+1))* ((T+I)*AP-2.* *166 02 1 XC*EPS*(i-DELTA.(1,T) )*A(T-2)-DELTA.(I1,T)*.5)+AP-XE*Y(2 )-DEL *166 2 TA.(1,T)*.5 *166 CR WHENEVER TAU.E.(Ti+l) *167 02 MBC=-(2.* Y(3)/((2.*A).P.2))+(1.-XE*LP(I+1))*(2.*(AP+0.25 *168 02 1 *A( 1) )-':".5*(A( i1 ).P.2) )+(AP-XE*Y(2) )-(2.*A(Il)-.5)*(A( 1 )-XE*V *168 2 P(1,1+1) )+(3./4. )*A( 1)-( 1./.) *168 CR WHENEVER TAU.LE.T2 *169 02 MBC=-(2.* Y(3)/((2.*A).P.2) )+(I.-XE*LP(I+1) )* ((T+1)*(AP-U. *170 02 1 25*rPS *A(T ))+EPS*4.*XE*A2(T-2)-XE*A(1I)*A(T-2) +(T+1i)*0.5*E *170 2 PS*A( 1)*A(T))+AP-XE*Y(2)1+EPS*(2.*A(1)-.5)*(A(T)-XE*VP(T,I+1) *170 3 )+(A( 1)-XE*VP( 1, [+1 )*(CPS*(T+i)*A(T)-2.*XE*A (T-2))-L..25*( * 170 4 EPS* (T+2)*A(T)-2.*YE*AIT-2))+DCLTA.(5,T)*(A(3)*((3./2.)*A(3 *170 5 )-Xc*A(i))+(A(3)-XE*VP(3,T+l))*(4.*A.(3)-2.*XE*A(l) I-XE*UP(I+ *17C 6 1)*A(3)*((3./2.)*A(3)-X A(l1))) *170 CR WH E'hVCR TAU.E. ( T2+1I) *171 02 i32C=-(2.* Y(3)/( (2.*A).P.2) )+( I.-XE*LP (I+1) )* (2.*(AP-0.25 *172 02 I A(1))+".5*(A(1).P.2))+(AP-XE*Y(2))+(2.*A(1)-.5)*(A(1)-XE*V *172 2P( 1, + ) )-(3./4. )*A( i)+( 1./E. *172 CLi W i'LN:VR TAU.LF.T3 *173 02 P;.C:=-(2.:' Y(3)/(( 2.*A).'. 2) )+(1.-XE*LP(I1+1) )* ( (T+1)*(AP-O *174 02 1.25*A(T) )+X *:A(1 )*A(T-2) +. S*FPS*A ( 1)*A ( T)* (T+1))+AP-XE*Y(2) *174 +(2.*A( 1 )-.-5)*(A(T)-XL*VP(T,1+1))+(A(1)-XE*VP (I1, I+1))*(EPS*( *174 3 T+I )*A( T )+2.*Xs *A(T-2 ) )-.2S * ( (T+2) *A ( T )+2.*EPS*XE*A (rf-2) ) 174 q + ELT'. { 5, T ), (A( 3) * ( ( 3./2. )*A( )-XE*A ( 1 ) )+ (A (3 )-XE*vP ( 3, I+1 ) *174

5 )*(4.*A(3)-2.*XE*A(1~)-XE*UP( I+1)*A(3)*((3./2.)*A13)-XE*A(l) *174 6 )) *174 END OF CONDITIONAL *175 02 END OF CONDITIONAL *176 01 CALC. OF ERROR IN MOM. BOUNDARY CONDITION WHENEVER.ABS.MBC ~G.MTOL.AND. M.L.MMAX *177 RI(M)=MBC *178 01 TRANSFER TO SW2 *179 01 STORAGE OF SOL'S AT LAST GRID PT. CTHERWISE *180 01 WHENEVER TAU.E.3 *181 01 U(I+1)=Y(1) *182 02 UP(I+1)=Y(2) *183 02 UDP(I+1)=Y(3) *184 02 CR WHENEVER TAU.LE.Tl *185 02 A(T)=AP *186 02 Ei(T-l)=VDP(T,1.)-2.*UDP(1)*A(T) *187 02 V(T,I+1)=Y(1) *188 02 VP(T,1+1)=Y(2) *189 02 VDPIT,I+1)=Y(3) *190 02 CR WHENEVER TAU.LE.T2 *191 02 A2(IT)=AP *192 02 c2(T-l)=WDP(T,1)-2.*UDP(I)*A2(T)-R *193 02 *A(T,I+1)=Y(1) *194 02 wP(T,I+1)=Y(2) *195 02 WOP(T, I+1=Y(3) *196 02 CR WHENEVER TAU.LE.T3 197 02 AZH ( T )=AP *198 02 cH(T-l)=WHDP(T,I)-2.*UDP(1)*A2-I(T)-RH199 02 4H(T, 1+1 )=Y( 1)200 02 WHPIT,I+1)=Y(2) *201 02 WHDP(T,I+1)=Y(3) *202 02 END CF CONDITIONAL *203 02 UTP(I +1)=F(3) *204 01 END OF CONDITIONAL *205 01 CALC. OF TERMS TO 3E USED IN HIGHER ORDER MCN. ECU'S THROUGH SW14,FOR J=1,1,J.G.(GRSIZ+1) *206 WHENEVER TAU.E.0 *207 01 HIJ)=O *208 01 01 OR WHENEVER TAU.L.TI *209 01 01 H(J)=-2.C*(Z2.P.TAU)*((UP(J)+STEP*J*UDP(J)~*A(T)-VP(f,J)) *210 01 01 1 -DELTA.(1,T)*(l./XE) *210 GR WHENEVER TAU.E.T1 *211 01 01 H(J)=0.5*((UCP(J)* V(IlJ)+U(J)*VDP(1,J))*A(1)+VDP()l,J~*V(l,J *212 01 01 1 )1 *212 OR WHENEVER TAU.L.T2 *213 01 01 HJC) = 0.5*(Z2.P.(TAU-T1))*((2*(T+1)*UP(J)*VF(1,J)+UDP *214 01 01 I (J)* V(IJ)+U(J)*VDP(I,J))*A(T+2)-((T+1)*UP(J)*VP(T+2,J)-(T+ *214 2 2)*UDP(J)* V(T+2,J)-U(J)*VDP(T+2,J))*A(l)+(T+2)*vDP(lJ)*V(T *214 3 +2,J)+V(l,J)*VDP(T+2,J))+A(T)*(.STEP*J*VDP(lJ)-VP(l,J)+A(l)* *214 4 tSTEP*J*UDP(J)+UP(J)))-4*(Z2.P.(TAU-T1))*(WP(T,J~-(LP(J)+STE *214 5 P*J*UDP(J))*A2lTD)-DELTA.(3,T)*(C.5*(3*V(3,J)*VDP(3,J)-2*(VP *214 6 (3,J).P.2))-A(I)*(ST?~*J*VDPt3,J)-VP(3,J))-A(ll*A(3)*(STEP*J *214 7 *UDP(J)+UP(J))+).5*(O(J)*VDP(3,J)+ 3*UDP(J)*V(3,J)+2.*UP(J)* *214

8 VP(3,J))*A(3))-D.5*EPS*(T+l)*VP(1,J)*VP(T+2,J)-(3./(2.*XE))* *214 9 A(T) *214 OR WHENEVER TAU.E.T2 *215 01 01 H(J)=-O.5*((UDP(J)* V(1,J)+U(J)VDP(1,J) )*A(1)+VCP(1,J)*V(1, *216 01 01 1 J)) 216 OR WHENEVER TAU.L.T3 *217 01 01 H(J)=-3*A(T+2)/(2 E)- ((2.XE)-.5 2T+3)*UP(J)*VP(1,J)+U 218 01 01 I GP(J)* V(1,J)+U(J)*VDP(1,J) )*A(T+4)-( (T+3)*UP(J)VP(T+4J)-( *218 2 T+4)*UDP(J)* V(T+4,J)-U(J)*VOP(T+4,J))*A(l)-(T+3)* *218 3 VP(1,J)*VP(T+4,J)+(T+4)*VDP(1,J)*V( *218 4 T+4,J)+V1,J+4, +VJ)V 4,J) )+(A(1)*(STEP*J*VDP(3,J)+3*VP(3,J))- *218 5 A(3)*(STEP*J*VDP(1,J)+3.*VP(1,J)))-C.5*(3.*V(3,J)*VDP(3,J)-2 *218 6.*(VP( 3,J).P.2)+(U(J)VDP ( 3,J )+ 3.*LDP (J)V(3, J)+2.*P (J)*VP *218 7 (3,J))*A(3)) )218 SW14 END OF CCNDITIONAL *219 01 01 WHENEVER TAU.E.T1, R=-0.5*(2.*VOP(1,1)-3.*UDP(1)*A(1))*A(1) *220 WHENEVER TAU.E.(Tl+1), R=0.5*(2.*A( 3)*CP( 1,1)+2-.A(1)*VDP(3,1)-6 *221 1.*A( 1)*A(3)*UDP( 1) *221 WHENEVER TAU.E.(T1+2), R=-0.5*(2.*A(5)*VCP(1,1)-2.*A(3)*VDP(3,1)+ *222 1 2.*A(1)*VDP(5,1)-3.*UDP(1)*(2.*A(1)*A(5)-(A(3).P.2))) *222 WHENEVE" TAU.E.T2, RH=C.5*(2.*VDP(1,1)-3.*UDP(1)*A(1))*A(1) *223 WHENEVER TAU.E.(T2+1), RH=C.5*(2.*A(5)*VDP(1,1)+2.*A(3)*VDP(3,1)+ *224 1 2.*A( 1)*VDP(5 ( 5,1)-3.UP(1)*(2.*A(1)*A(5)+(A(3).P.2) ) ) 224 TEST LOC. TC DET. WHICH MOM. EQ. IS BEING SOLVED AND WHEN MOM. PROB. HAS BEEN COMP. SW20 TAU=TAU+l *225 WHENEVER TAU.LE.T3, TRANSFER TO SW16 *226 SET UP UF ENERGY PROB. ----------------------— n T=J *227 TAU= 3 *228 TS=0 *229'U=2.*(A.P.2)*PR*XE *230 YX=YP *231 TRANSFER TO SW1 *232 CALC. OF TERMS FOR ENERGY BOUNDARY CCNO. SW WHENEV ER TALJ.E.J *233 EPC=:J *234 01 CR WHENEVER TAU.LE.T1 *235 01 EiC=( (2+T)*A(T+1)-0.5*OELTA.(C, T)-2.*EPS*(1-[ELTA. (C,T) )XE *236 01 1 A(T-i))*Z(I+1) *236 OR WHENEVER TAU.E.(T1+1) *237 01 EBC=Z(T+1)*(2.*( A2(1 +C.25*A(1) ))-0.5*(A(1).P.2))-(A(1)-0.25) *238 01 1 *V(O, I+1) *238 CR WHENEVER TAU.E.(TI+2) *239 01 EBC=Z( 1+1)*(4.*(A2(3)-C.25*A(3) )+2.*A(1)*A(3)+XE*(4.*A2(1)-( *240 01 1 A(1).P.2)))+(A(1)-0.25)*V(2, I+1)+0.5*(4.*A(3)-2.*XE*A(1))*V( *240 2,1+1) *240 OR WHENEVER TAU.E. (T+3) *241 01 C=Z (I+l)*( 6.*(A2(5)+C. 2? A(5))-3.*A( 1)*A( )-XE*(4.*A2(3)+A *242 01 1 ( 1)* 3( ) )-( A( ) —.25)*V(4, I+l)-. 5*(6.*A(5)+2.*xE*A(3) )*V(O *242 2, I+1)+ 0. 3-(4.*A(3)-2.*X*a(. 1) )*V(2, I+1)+A(3) (1. 5*A(3)-XE*A( *242 3 1))LZ(1+1) *242 uR WHENEVER TAU.E.(T2+1) *243 01 t3C=Z( I+1 ) (2.*(A2H( I)-. 25.? A( 1) )+.5*(A(1 ).P.2) )+(A(l)-0.25 *244 01

1 )*V(30,I+1) 244 OR WHENEVER TAU.E.(T2+2) *245 01 EBC=Z( I+1)*6.*(A2H(5)-0.25*A(5))+3.*A(1)*A(5)+X:*A1l)*A(3)) *246 01 1 +(A(1)-0.25)*V(4, I+1)+.5*(6.*A(5)+2.*XE*A(3) )*V(0,I+ +0.5* *246 2 (4.*A(3)-2.*XE*A(1) )*V(2,1+1)+A(3)*(1.5*A(3)-XEA( 1) *ZI+1) *246 END OF CONDITIONAL *247 01 CALC. OF ERROR IN ENERGY B.C. WHENEVER.ABS.((Y(2)/MU)+YI1)+EBC).G.ETOL.AND. N.L.NMAX *248 R2(N)=(Y(2)/MU)+Y 1)+EBC *249 01 TRANSFER TO SW4 *250 01 STORAGE OF SOL'S AT LAST GRID PT. OTHERWISE *251 01 WHENEVER TAU.E.O *252 01 Z( I+1)=Y(1) *253 02 ZP(I+1)=Y(2) *254 02 ZCP(I+1)=F(2) *255 02 OR WHENEVER TAU.LE.T1 *256 02 G1(T)=VP(T,1)-ZP(1)*A(T+1) *257 02 V(T, I+1)=Y(1) *258 02 VP(T,I+1)=Y(2) *259 02 VDP(T, I+)=F(2) *260 02 OR WHENEVER TAU.LE.T2 *261 02 G2(T)=WP(T,1)-ZP( 1 )A2(T+1)-R 262 02 W(T, I+1)=Y( 1) 263 02 WP(T, I+)=Y(2) *264 02 WDP(T,I+1)=F(2) *265 02 GR WHENEVER TAU.LE.T3 *266 02 GH(T)=WHP(T, )-ZP 1)*A2H(T+1)-RH *267 02 WH(T,I+1)=Y(1) *268 02 WHP(T,I+1)=Y(2) *269 02 WHCP(T,1+1 )=F(2) *270 02 END OF CONDITIONAL *271 02 END OF CONDITIONAL *272 01 CALC. OF TERMS TO BE USED IN HIGHER ORDER ENERGY EQU'S THROUGH SW15, FOR J=1,1,J.G.(GRSIZ+1) *273 WHENEVER TAU.E.T1, H(J)=0D.5ZP(J)*A( 1)*V(I,J)+C.5VP(I,J)*(V *274 01 1 (1,J)+U(J)*A(1)) *274 WHENEVER TAU.E.(T1+1), H(J)=-0.5*ZP(J)*(3.*A(I)*V(3,J)-2.*ST *275 01 1 EP*J*(A( 1).P.2)+A(3)*A(1))+O.5*(2.*V(2,J)*(VP(1,J)+U(J)*A(1) *275 2 J)V(,J+U(J)A)-VP J,J)(3.VV(3,J)+UJ)*AA(3)))-P( *275 3 A(1)*(2.*V(O,J)-STEP*J*VP(C,J)) *275 WHENEVER TAU.E.(TI+2),H(J)=C.5*ZP(J)*(5.*A(l)*V(5,J)+4.*STEP*J.A( *276 01 1 1)*A(3)-3.*A(3)*V(3,J)+A(5)*V(1,J))-C.5*(4.*V(4,J)*(VP(1,J)+ +276 2 A(l)*UP(J) )-VP(4,J )(V(1,J)+A(1)*U(J))-2.*V(2,J)*(VP(3,J)+A( *276 3 3)*UP(J))+VP(2,J)*(3.*V(3,J)+A(3)*U(J))-VP(O,J)*15.*V(5,J)+A *276 4 (5)*U(J)))-A(1)*(2.*V(2,J)-STEP*J*VP(2,J) )-A(3)*(2.*V(O,J)-S 276 5 TEP*J*VP(O,J)) *276 WHENEVER TAU.E.T2, H(J)=-C.5*ZP(J)*A(1)*V(1,J)-C.5VP(O,J)*( *277 01 1 V(1,J)+A(l1)U(J) )277 SW15 WHENEVER TAU.E.(T2+1), H(J)=-O.5*ZP(JI*(5.*A(1)*V(5,J)+3.*A(3)*V( 278 01 1 3,(IIA(5)*V(1,J))+0.5-(4.*V(4,J)*(VP(1,J)+A(1)*UP(J))-VP(4,J *278 2 )*(V(1,J)+U(J)*A(1))+2.*V(2,J) (VP(3,J)+LP(J)*A 3)-VP(2,J)* 278 3 (3.*V(3,J)+U(J)*A(3) )-VP(0,J) 5.V5, (5. (5J)+U(J)*A(5) )+A(1)*(2 278 4.*V(2,J )+STEP*J*VP(2J ) )-A(3)*(2.*V(C,J)+STEP*J*VP(C,J)) 278

WHENEVER TAU.E.Tl, R=-C.5*A(1)*Gl(C) *279 WHENEVER TAU.E.(T1+1), R=C.5*(A(1)*G1(2)+A(3)*G1(O)) *280 WHENEVER TAU.E.(T1+2), R=-C.5*A(5)*G1(C)+A(1)*G1(4)-A(3)*GI(2) ) *281 WHENEVER TAU.E.T2, RH=C.5*A (1)*GI(O) *282 WHENEVER TAU.E.(T2+1), RH=0.5*(A(5)*Gl(C)+A(1)*GI(4)+A(3)*GI(2)) *283 TEST LOC. TO DET. WHICH ENERGY EQ. IS BEING SOLVED AND WHEN ENERGY PROB. IS COMPL. TAU=TAU+i *284 YX=YP *285 WHENEVER TAU.LE.T1 *286 T=2*TAU-2 *287 01 EPS=Z2.P.TAU *288 01 CR WHENEVER TAU.LE.T2 *289 01 T=2* ( TAU-T )-2 *29C 01 EPS=Z2.P. (TAU-T1) *291 01 GR WHENEVER TAU.LE.T3 *292 01 T=4* ( TAU-T2 )-4 *293 01 EPS=Z2.P.(TAU-T2) *294 01 END OF CONDITIONAL *295 01 WHENEVER TAU.LE.T3, TRANSFER TO SW1 *296 PRINT STATEMENTS AND MISC. CALC'S PRINT FORMAT FM1 *297 PRINT FORMAT FM2 *298 THROUGH SWIG, FOR K=I,1,K.G.(GRSIZ+l) *299 SWIO PRINT FORMAT FM3, P(K),U(K),VII,K),V(3,K),V(5,K),W(i,K),W(3,K),W(5,K) *300 01 I,WH(1,K),WH(5,K) *300 PRINT FORMAT FM1 *301 R) PRINT FORMAT FM2 *302 THROUGH SW9,FOR K=1,I,K.G.(GRSIZ+I) *303 SG=UP(K) *304 01 SG1=VP(I,K)-UP(K)*A(1) *305 01 SG12=VP(3,K)-UP(K)*A(3) *306 01 SG14=VP(5,K)-UP(K)*A( 5 ) *307 01 SG2=WP(1,K)-UP(K )*A2( 1 )+G.5*A() *SG1 *308 01 SG22=WP(3,K)-UP(K)*A2(3)-0.5*(A(1)*SG12+A(3)*SG1) *309 01 SG24=WP(5,K)-UP(K)*A2(5)+0.5*(A(5)*SG1+A ( 1 ) *SG14-A(3)*SG12) *310 01 SGH=WHP(1,K)-UP(K)*A2H(1)-C.5*A(1)*SGl *311 01 SGH4=WHP(5,K )-UP(K)*A2H( 5)-C.5*(A( 5 )*SG1+A(1)*SG14+A(3) *SG12 *312 01 1) *312 SW9 PRINT FORMAT FM3, P(K),SG,SG1,SG12,SG14,SG2,SG22,SG24,SGH,SGH4 *313 01 PRINT FORMAT FM1 *314 PRINT FORMAT FM2 *315 THROUGH SW12, FOR K=I,1,K.G.(GRSIZ+1) *316 SW12 PRINT FORMAT FM3, P(K),UP(K),VP(1,K),VP(3,K),VP(5,K),WP(1,K),WP(3,K) *317 01 1 WP(5,K),WHP( 1,K ),WHP( 5,K) *317 PRINT FORMAT FM1 *318 PRINT FORMAT FM2 *319 rHROUGH SW13, FOR K=1,1,K.G.(GRSIZ+l) *320 SW13 PRINT FORMAT FM3, P(K),Z(K),V(C,K),V(2,K),V(4,K),W(0,K),W(2,K),W(4,K) *321 01 1,WH(j,K),WH(4,K) *321 PRINT FORMAT FMI *322 PRINT FO{RMAT FM4, A(I),A(1),A(3),A(5),A2(1),A2(3),A2(5),A2H(1),A2H(5) *323 PRINT FORMAT FM4, UDP( 1),El(-),Ei(2),E 1(4),E2(C),E212),E2(4),EH(O),E) *324 1 (4) *324 PRINT FORMAT FM4, ZP(1),Gl(O),Gl(2),GI(4),G2(I),G2(2),G2(4),GH(O),GH( *325 1 4) *325

PRIST FORMAT FM2 *326 PI=1. 5708 *327 THROUGH SW3C, FOR XI=.1,.'',XI.G.C.9 *328 ACI=2.*A*XI*(.ABS.A( 1) )*(1.+(C'.5*( (A(3).P.2)+2.*A(5)*A(1) )/( *329 01 I A(1).P.2))*(XI.P.4))*(XE.P..5,) *329 AFI=(.ABS.E1(O) )*( 1.+(C.5*( ((1(2).P.2)+2.*EI(4)*E1 (C))/(Ei(O *330 01 1 ).P.2) )*(XI.P.4) )*XE/A *330 ANi=(.APS.G1(C) )*( 1.+(C.5*( (G1(2).P.2)+2.*Gl(4)*GI(L))/(Gl(0 *331 01 1 ).P.2) )*(XI.P.4) )*(XE.P.:.5)/(2.*A) *331 ARD1=PI-ATAN.((A(1)+A(5)*(XI.P.4))/(A(3)*(XIP.2 ) )) *332 01 ARFi=P [-ATAN.((E1(C)+Ei( 4)*(XI.P.4) )/(E1 (2)*(XI.P.2))) *333 01 ORN1=PI-ATAN.((G1(0)+G1(4)*(XI.P.4))/(G1(2)*(X[.P.2))) *334 01 NOU=(GH('' ) +GH(4)*(XI.P.2) )*(XE.P.C.5)/(2.*A) *335 01 SW3'i PRINT FORMAT FM4, XI,AD1,AF1,AN1,ARD1,AiRF1,ARN1,NU *336 01 PB1=2.*A*(XE.P.0.59) *337 P P 22.*A *338 Pe3=UDP( 1 ) *X/A *339 C PB4= ZP( 1)*(XE.P.0.5)/(2.*A) *340 PRINT FORMAT FM2 *341 PRINT FORMAT FM4, PB1,PB2,PPS3,P34 *342 WHENEVER CAT.LE.CATM, TRANSFER TO SW31 *343 Swil END OF PROGRAM *344

APPENDIX XXVI COMPUTER PROGRAM FOR CALCULATING FILM THICKNESS, LOCAL SKIN FRICTION, AND THE TERMS OF THE EXPANSIONS (52) This program calculates the liquid film thickness, local skin friction, and the terms of the expansions (52) from the numerical values of f2z, J, and V* obtained from the program for drop trajectories Once & is calculated, the product 21 b and J b are formed, and are tabulated. The expansions (52) are assumed to contain a finite number of terms and a system of linear equations is obtained by writing these expansions at various values of t which we denote by io Thus we have: m 6( i) _ r 2n,(U = n " 9,, i = 0,l,2,n.m L (2n)! 2 n-O This system of equations is written in matrix form and solved for the o2n's by inverting the matrix. Thus if we denote the matrix A by: 2n A.= i I j (2n) j the vector c by 329

330 o ) and the vector of by: a2 C = 0 And C6 A.A C CORRESPONDENCE BETWEEN NOTATION FOR COMPUTER PROGRAM FOR CAILCUJLATING FILM THICKNESS-, LOCAL SKIN FRICTION, AND TERMS OF THE EXPANSION (52), AND NOT.ATIONS USED IN THE TEXT AG(2K-1) e% DEL(M) b DR pg/p E' E Fo (MN) N0

331 FF(M) F / 2 pU 2 GAM 7 MDOT ~ P(M) x R Teff RINF 2RUoo /vg RR rd/Ro VR Fl zX ~ xv Z.(x) Z(x)

$COMPILE MAD, PRINT OBJECT 013997 07/23/65 3 54 38,9 AM MAD (01 MAY 1965 VERSION) PROGRAM LISTING * **~ * * SET UP INTEGER 1 *001 INTEGER TITMAX *002 INTEGER IMAX,SM 003 INTEGER MK,NIP *004 DIMENSION Q(8),D(8),AG(20) *005 DIMENSION C(100),D(100) *006 DIMENSION FF(200),A(200),8(200),ALP(200),8ET(200),CV(200),P(200) *007 DIMENSION XBAR(500),MDOT(500),J(500),VPHI (500) *008 DIMENSION DEL(200) 009 DIMENSION C1(4000~CS).010 DIMENSION C2(4000,CT) 011 VECTOR VALUES CS=2,1,60 *012 VECTOR VALUES CT=2,1,60 *013 VECTOR VALUES FM1=$.lH1/1H-$ *014 VECTOR VALUES FM2=$1H S4,53(1H-)*$ *015 o VECTOR VALUES FM3=$lH S4,F5.3,2F12.6*$ *016 VECTOR VALUES FM4=$1H S4,13,2F20.6*$.017 VECTOR VALUES FM5=$1H,S44F12.6*$ *018 VECTOR VALUES FM6=$IH S4,F5.3,F12.6*$ *019 VECTOR VALUES FM7=$1H,S4,F5.3,I3*$ *020 READ AND PRINT DATA *021 THROUGH SW9, FOR K=l,1rK.G.KMAX *022 SW9 READ FORMAT FM5, XBAR(K),MDOT(K),J(K),VPHI(K) *023 AG(1)=1.2326 *024 AG(3)=0.7244 *025 AG(5)=1.0320 *026 AG(7)=0,5792+7.*0.L829+(70.*0.0076/3.) *027 AG(9)=0.5399+12.*0.1520+1126.*0.0572/5.)+84.*0.0607-280.*.0 *028 1308 *028 AG(11)=0.5100+(55.*0.1323/3.)+66.~0.0742+220.0.0806+462.*O. *029 1 1164-1540.*,0.1796+(15400.*0.0516/3.) *029 THROUGH SW4, FOR K=1,1,K.G.6.030 SW4 PRINT RESULTS AG(2*K-1).031 -----------------------------—; —----------------- DEFINITION OF F. INTERNAL FUNCTION F.(IJ)=GAMMA.(IJ+1) *032 DEFINITION OF Z. INTERNAL FUNCTION (IX) *033 ENTRY TO Z. *034 SZ=AG(1)*IX *035 THROUGH SW5, FOR K=2,1,K.G.6 *036 SW5 SZ=SZ-(((-I.).P.K)*(2*K)*AG(2~K-I)(IX.P.(2*K-1))/F.(2.*K-1. *037 1 )) 037 FUNCTION RETURN SZ *038

END OF FUNCTION *039 CALC. OF DELTA AND RELATED QUANTITIES. T=O *040 E=O. *041 SW10 T=T+1 *042 E=E+O.02 *043 K1=DR/E *044 K2=4.*Kl*VR *045 M=0 *046 SW1 M=M+l *047 X=M*INCX *048 N=IP*M *049 P(M)=X *050 Q =(4./3.)*K1*SIN.(2*X) *051 R =K2*Z.(X)+J(N)*VPHI(N) *052 S=MDOT(N) *053 WHENEVER (.ABS.(1.-(27./2.)*({Q.P.2)*S/(R.P.3)))).G.1. *054 M=M-1 *055 O0 TRANSFER TO SW15 *056 01 OTHERWISE *057 01:ONTINUE *058 01 END OF CONDITIONAL *059 01 GAM=ARCCOS.(-(1.-(27,/2o)*((Q.P.2)*S/(R.P.3)))) *060 DEL(M)=(R/(3.*Q))*(2.*COS.(GAM/3.)-l.) *061 CALCo OF LOCAL SKIN FRICTION FF(M)=2.*E*R +3.*E*Q *DEL(M) 062 CALC. OF ALPHA SUB. 2J AND BETA SUB. 2J __- ____________________________ —---------— ___-_-_ —__-___ C(M)=DEL(M)*J(N) *063 D(M)=DEL(M).*MDOT(N) *064 WHENEVER X.L.XMAX, TRANSFER TO SW1 065 THROUGH SW3, FOR K=l,1,K.G.IMAX *066 WHENEVER K.E.1 *067 01 X=P(I) *068 01 01 A(1)=C(1) *069 01 01 B(1)=D(1) *070 01 01 OR WHENEVER K.E.IMAX *071 01 01 X=P(M) *072 01 01 A(K)=C(M) *073 01 01 B(K)=D(M) *074 01 01 OTHERWISE'075 01 01 X=X+(P(M)-P(1))/(IMAX-1.4 *076 01 01 A(K)=TAB.(X,P(1),C(l)91,,4 t4M,T6) *077 01 01 B(K)=TAB.(X,P(1),D(1),l,l,4,M,T6) *078 01 01 END OF CONDITIONAL *079 01 01 THROUGH SW3, FOR I=1,1,I.G.IMAX *080 01 C1(K I)=(X.P.(2*I-2)) *081 02 SW3 C2(K,I)=(X.P.(2*I-1)) *082 02 S1=SLE.(IIMAX,60,Cl(1,1),ALP(I),A(1),CV,O) *083 PRINT RESULTS S1 *084 S2=SLE.(IMAX,60,C2(1,1),BET(1),B8(),CV,O) *085 PRINT RESULTS S2 *086 PRINT STATEMENTS ____________________________________________________________

PRINT RESULTS RINF *087 PRINT RESULTS RR *088 PRINT RESULTS E,M *089 PUNCH FORMAT FM7, E,M *090 PRINT FORMAT FM1 *091 PRINT FORMAT FM2 *092 THROUGH SW7, FOR K=117,K.G.M *093 PUNCH FORMAT FM6, P(K),OEL(K) *094 01 SW7 PRINT FORMAT FM3, P(K),ODEL(K),FF(K) *095 01 PRINT FORMAT FM1 *096 PRINT FORMAT FM2 *097 THROUGH SW8, FOR K=l, 1,K.G.IMAX'098 ALP(K)=ALP(K)*F.(2.*K.-2.) *099 01 BET(K)=BET(K)*F.(2.*K-1.) *100 01 PUNCH FORMAT FM4, K,ALP(K),BET(K) *101 01, SW8 PRINT FORMAT FM4, K,ALP(K),BET(K) *102 01 WHENEVER T.LE.TMAX, TRANSFER TO SWIG *103 ~ END END OF PROGRAM *104

APPENDIX XXVII REDUCTION OF EXPERIMENTAL DATA FROM REFERENCE (1) From Reference 1 we shall consider the data corresponding to the SC-6 nozzle with an air velocity of 110 ft/sec and a water flow rate of 0.083 lb/sec, given in Table II-B, page 48. The reported value of 2RoU/vg was 8x104. The measured total value of flow using a drop collecting device and performing a numerical integration was 0.061 lb/sec. The actual total flow obtained from their rotameters was 0.083 lb/sec. The measured value of the flow impinging on the cylinder using the drop collecting device was 0.002 lb/sec. We therefore take as a corrected value for the total flow impinging on the cylinder 0.002 x (0.083/0.061) or 0.00272 lb/sec. On page 24, Reference 1, is reported that for an air velocity of 110 ft/sec the drop -velocity was 85 ft/sec. From Reference 23, Table E-2, v = 0.045 ft2/hr at the drop temperature of 57~. The diameter of their cylinder was 1.5 in. Hence: 2ROU 1.5 5600 x 85 = 8.30xl05 V 12.o46 Since the cylinder is 2 in. in length, the mean liquid flux impinging on the cylinder actual flow impinging on the cylinder area of cylinder 335

336.00272 x 144 =.13 lb/sec ft 1.5x2 Hence the effective mass of liquid per unit volume in the free stream is: 13 1-.52x103 lb/ft3 85 and the volume fraction of liquid in the free stream, Xe, is Xe = -5 x 103 = 2.44x105 62.4 Hence: | V Roo A ~8 e o8.0x1o = 910 E = Xe 2 U.022 v From Reference 23, Table E-2, at a wall temperature of 78~, k =.35 Btu/hr-ft~F. Hence: -2RO.5.. —- = 5394 k____ 12xo35x910 V Table IITB (from Reference 1) Position h average 4 v Btu/ft2 hr-~F 0~ 451 0.178 30~ 372 0.147 60~ 270.01065 330~ 448 Oo 168

UNIVERSITY OF MICHIGAN 3 9015 03126 3596