THE UNIVERSITY OF MICHIGAN COLLEGE OF ENGINEERING Department of Chemical and Metallurgical Engineering Technical Report TEE DETERMINATION OF ELECTRICAL MOBILITIES IN MOLTEN ALLOYS J. D. Verhoeven J. C. Angus W~ D. Bouwsma E. E.. Hucke UMRI Project 2917 under contract with: U. S. ATOMIC ENERGY COMMISSION CHICAGO OPERATIONS OFFICE CONTRACT NO. AT (11-1)-771 ARGONNE ILLINOIS admninistered by: THE UNIVERSITY OF MICHIGAN RESEARCH INSTITUTE ANN ARBOR August 1960

TABLE OF CONTENTS Pages I Abstract v II Introduction 1 III Formal Solution of Partial Differential Equation IV Discussion and Application of the Solution 11 V Roots of the Eigenfunctions 15 VI List of Symbols 19 VII Literature 20 VIII Figures 21 IX Appendices 33 iii

ABSTRACT A method of determining the electric mobility in molten metals is presented. The partial differential equation describing the process of electrodiffusion is rigorously integrated for the case of a diffusion channel of finite length with the concentration at one end held constant. The solution may also be used to interpret thermal diffusion and sedimentation experiments. Tables of the roots of the eigenfunctions and a discussion of errors are included. v

INTRODUCTION In recent years interest in the experimental measurement of transport properties in liquids has increased. Thermal and electrodiffusion coefficients have been measured in a variety of ways. For the past two years the authors have used a particularly simple method for determining electric mobilities in molten alloys. During the investigation it was necessary to solve the partial differential equation describing the process and to investigate the errors inherent in the method. Since this solution has utility in other areas, for example, thermal diffusion and sedimentation, it is presented as the basis of this report. The details of the experimental procedure and the results will be discussed fully in a subsequent report. They will be used here only to illustrate the method. In addition to the formal solution of the equation, tables of the roots of the eigenfunctions are included. The experimental method involves the passage of D.C. current through a thin capillary of molten metal of uniform concentration. One end of the capillary is maintained at constant concentration by keeping it in contact with a reservoir of the molten alloy (fig. 1). To end the experiment the tube is separated from the reservoir and chemically analyzed. The resistivity of the melt, the current, the time, and the increase (or decrease) of solute in the tube are known. From these one can calculate the electrical mobility, u, of the solute atoms from the flux equation (1) by making a mass balance of solute across the mouth of the tube. (1) J = uNE * * Equation 1 is the defining relation for u. For this treatment the reference frame is the electrodes. This corresponds to the reference frame defined by zero volume flux only when the partial molar volumes of the two species are not functions of concentration.

It is sometimes possible to determine the concentration directly by measuring the resistivity change between probes spaced along the diffusion channel (1,2) This method is suitable for binary systems in which the resistivity changes markedly with concentrationo For multi-component systems and alloy pairs without large resistivity changes the chemical technique may be necessaryo In addition, it is not easy to design and build a suitable cell. for the resistivity method at high temperatures. So far we have found it necessary to use the chemical technique in the range from 300 to 800~Co The solution to the partial differential equation will be of value in interpreting both types of experiments howevero Since one is simply determining the amount of solute that crosses the boundary between the tube and the reservoir, the mobility obtained is for the conditions prevailing at the mouth of the tube. In other words, no matter what form the concentration'\ distribution takes inside the tube, all of the solute that enters (or leaves) the tube must cross the boundary at the entrance to the reservoir, At the start of the experiment the first change takes place at the electrode end of the tube. If the polarity is such that the solute enters the tube, the concentration will rTcffcease at the electrode firsto As the experiment proceedsthe concentration will gradually increase back down the tube towards the reservoir. If the polarity is reversed, the concentration first decreases at the electrode end of the tube, A concentration "wave" moves down the tube towards the reservoir, broadening as it goes from ordinary diffusiono

-53One can approximate this behavior with the solution to the diffusion equation for the initial condition C = 1 X>O C = 0 X< for the boundary conditions of a tube of infinite length in the + X and - X directions. This solution is well known. By transforming to a coordinate system that moves with a velocity of uE cm/sec (the average velocity of the solute atoms) one obtains an approximate solution to the equation. This solution is best for high solute velocities. For lower velocities the end effect at the electrode becomes appreciable. When using equation 1 to calculate the mobility one must be certain that the experiment has not proceeded to the point where there is a concentration gradient at the mouth of the tube. If there is, solute will be transported across the boundary between the tube and the reservoir by ordinary diffusion. This ordinary diffusion flux will always oppose the flux due to the electric field. The calculated electrical mobilities will therefore be too low. If the experiment is continued for,-a very long time the steady state condition is approached. At steady state the concentration gradient at the mouth has increased to the point where the flux caused by electrodiffusion is exactly offset by the ordinary diffusion flux in the reverse direction. By running experiments under exactly similar conditions but for different lengths of time one can test whether back diffusion is introducing errors. If there is no back diffusion, the mobilities will be the same. If an experiment has been run too long, the result will be lower than the others. This effect has been observed by the authors.

-4The formal solution will aid in determining optimum experimental times and in designing experiments. By applying the results of the solution to experiments in which back diffusion has taken place it is possible to estimate the ordinary diffusion coefficient.

-5FORMAL SOLUTION The one dimensional flux equation including ordinary diffusion is -- -- ~N * (2) J = uTE -D (I) Differentiating once with respect to distance the partial differential equation is obtained (3) -N D (N Here we have introduced the assumption that the electric field strength E throughout the tube may be approximated by the total voltage drop, V, divided by the length of the tube a. Furth (3,4) and DeGroot (5) integrated equation (3) for the boundary conditions corresponding to a tube closed at both ends i.e. zero solute flux at each end of the tube. The equation has not been solved for the case of constant composition at one end and zero flux at the other. Since the solution is unusual in some respects it is presented in detail. Make the following substitutions ** Z uV N DQ X = S = D C = - t 2 0 * The corresponding flux equations for thermal diffusion and sedimentation are respectively J = DN grad T - D. (Z) J =vN- (- ) Q~* For thermal diffusion S = D T and for sedimentation =va D D

-6Equation (3) becomes )c _ C 6C (4) =. (x) Equation (4) is first solved for the homogeneous boundary conditions. (5) C = o, for x = O t>O (6) -. SO' = 0, for x = 1, t>O We let x = 0 at the reservoir end of the tube and x = 1 at the electrode. These boundary conditions do not correspond to the real situation, but they allow the problem to be solved. The solution is then made to fit the real boundary conditions. The familiar separation of variables technique is used (7) C (x,t) = (x) T(t) After separating and equating to. one obtains the'two ordinary differential equations (8) T - XT = 0 (9) x" - SX1 - = O The solution to (8) is simply (10) T = Aext From (10) it is apparent that <O 0 if the time solution is to remain finite. For (9) the solution of the auxiliary equation is

-7S S (11) m -= ~ 2 + There are therefore three solutions to (9) depending on whether S2 (12) X+ I = 0 S2 (13) X+ -r >,2 S2 (14) X + < 0 From (12) the solution of (9) is S S X = A e7 + Bxe 2 Applying the boundary conditions (5) and (6) one obtains (15) A = 0 (16) (1 -) B = O Since if B = 0 this solution vanishes, one must have S = 2. Therefore for the case where S - 2 one term of the solution is (17) (consto) xeX e-t Now consider the second case i.e. equation (13)o First define a by the relation 2 (18) a2 = S + - One immediately finds the solution S - (19) X = (A eOa + Be-X ) e

-8Applying the first boundary condition one finds that A = -B One term of the solution them must be of the form S S2 (20) (consto) (sinh o) Ue e Applying the second boundary condition one finds S (21) a coth a =2 This equation has a single root for every S>2o For S =2 the root is zero. S$ For S<2 there are no roots. In appendix VIII it' is proved that a2-T~<o Next consider the third case (equation 14)o Define 3 by the relation 2 S2 (22) = - X+The corresponding solution is SX (23) X = (A cospSc + B sin x) e2 Applying the boundary conditions one obtains (24) A = S (25) p cot P 2 Equation 25 has an, infinite number of roots for any S. One now has S (26) (const.)(sinp n) e2 x

-9Before these solutions are tested against the real boundary conditions it is necessary to consider the special case where X = 0. The solution is simply (27) C () = Kx + D This is the "steady state" solution since time does not appear. A trial solution for the real boundary conditions will be constructed for each of three cases by adding equation 27 to the solution already obtained. (28) s =2 ((S/2)xext C(x,t) =.(cnst.)xee-t + (const.) (sinx)e(S/2)et + K e + D (29) S>2 (S/2)x (>a2S2/ f C(x,t) = (const.)(sinhx)e (S/-xe(c2-S2/$)te + (const.)(slnpx)e(S/2)x.b t + Ke~x + D (30) S<2 C(xt) = (contt,) (sinPx)e(S/2)x e,t + KeSx + D The real boundary conditions are (31) C = l, for x = O, t>0 (32) ~ - SC = 0, for x = 1, t > In addition one has the final or steady state condition (33) C = eS for x = x, t = oo It is easily shown that for K = 1 and D = 0 these conditions are meto The initial condition is (54) C = 1 for x =x, t =

-10This is satisfied by forming the sum of all the solutions given by equations 25 and 26. The proof of orthogonality and the calculation of the coefficients is given in the appendixo The final results are (35) S =2 Sx iS St (sin n[X) e e kt C(x,t) = e.Axe.xe + C(sinx)e e n=l (56) S>2 - (a24 ) 1 C(x,t)= e + B (sinh aox)e e + Cn(sinpx)e e\t n=l S< 2- S c.$x oo 2" (57) C(x,t) = e + C(sinPn,)e e where 6 A =e 0 S l6aS( cosh:xl) e 2 o (S2-4)a2)(S-2 cosh-Ca) S 16pnS(cos pn)e- (S2+4Pn2)(S-2 cos2 n) and a (coth:a) = S Pn(cot PB) =2 S2 X = - (3n2+'- )2 The solutions were tested by calculating the concentration at t = 0, Figure 2 shows the series converging to C = 1, For 10 terms the sum is correct within 1.0% for X from 0.0 to 0,9, For X = 1.0 Co = 19o* * For all subsequent calculations the first ten terms were used. Ten significant figures were carried during the computations All computations were carried out on a Bendix G-15 computer.

-11DISCUSSION AND APPLICATION OF THE SOLUTION In order to illustrate the diffusional processes taking place in the tube the changing concentration profile in the tube was plotted for various values of S. These plots are shown in figures 3, 4, and 5. For S>O the solute enters the tube for S<O it leaves the tubeo In both cases the first change occurs at X-= lo For S = -20 one can easily see the concentration "wave" move towards the mouth of the tube. This behavior has been observed by Schwarz (1), Mangelsdorf (2), and one of the authorso The experimental conditions of figure 5 are easily obtainable in practiceo In order to determine the error caused by back diffusion it is necessary to calculate the mean concentration in the tube as a function of timeo This is done by integrating equations 35, 36 and 37 over the length of the tube. The integrals may be found in any standard table. The results are given below o (58) S =2 I o1 4cpn' 22 (39) S> 2 r 2 S ) +~ 4f (4o) s<2 S. 4Pn _ () Cm= SL + n=1t (s2+44.2) C.et Plots of the mean concentration versus time are given in figures 6 and. 7. The curves all have an initial slope of S. The effect of back dlffusion. is apparent in the departure of the curves from linearityo.From these

-12curves it is easy to construct plots of the error caused by back diffusion versus the percent change of concentration in the tube, These are shown in figures 8 and 9. It can be seen that one radically decreases the error by running the solute into the tube. For -even moderate values of S one can theoretically increase the concentration at the electrode end to extremely high valueso In general, solid phases will precipitate out before this happens In addition, the errors caused by the chemical analysis of the tube must be considered. One uses the difference in the tube and reservoir compositions to calculate the total solute flux. Therefore if there is not a large change in concentration, one must subtract two numbers of the same magnitude. For small changes this magnifies the analytical errors considerablyo Figure 10 is a plot showing this (see appendix VIII for details of this calculation). By using figures 8, 9, and 10 it is possible to design experiments to minimize the analytical and diffusion errorso These are the two largest sources of error in this techniqueo The final application considered is the calculation of ordinary diffusion coefficients, As the solution now stands D appears in both dimensionless constants, i.e. uV/D and DO/a2o By choosing the dimensionless time as uV/a2 one can construct plots equivalent to figures 6 and 70 These are shown in figures 11 and 12o This leaves the diffusion coefficient in only one of the constants, and eliminates the necessity of a trial and error solution0.In order to estimate the diffusion coefficient one makes a series of experiments at progressively longer times. The runs at shorter times should give a constant value for u. This is the true value of the electric mobilityo Since

-13one now knows u, V, r a,-and the results of an experiment that has gone to the point where back diffusion takes place, one has both the coordinates of figure 11 (or 12) specified. This then determines the value of S from which the diffusion coefficient is calculated. The results of our preliminary experiments are given in table 1 for the Cu-Bi and Ni-Bi systemso The diffusion coefficients show considerable scatter and are approximately an order of magnitude larger than what one would expecto This is undoubtedly due to convectiono It is possible the convection is caused by thermal gradients associated with the transfer of the I2R heat from the tubeo All other sources of convection have presumably been eliminated. The diffusion channels were 0,5 and 1.0 mm IDo thin walled pyrex tubes. The tops of the tubes were kept several degrees hotter than the bottoms. It should be emphasized that convection does not affect the determination of u as long as the experiment is ended before a concentration gradient forms at the mouth of the tubeo At any time during the experiment the change of the mean concentration'in the tube is caused only by the sum of the two diffusion fluxes at the mouth of the tube. For this reason the diffusion coefficient calculated by this method is for the reservoir concentration. This method is analogous to the technique described by De Groot (6) for thermal diffusion,

-14TABLE 1 Observed Diffusion Coefficients in Molten Bismuth* Cu:Cm,S D x 105 cm2/sec 0.48 0.77 0.60 22 0.44 0.76 0.75 17 0.53 0.73 0.75 19 Ni 0.47 0.77 0.64 13 0.61 1.48 1.15 28 0.99 1.62 1.15 30 Pb 6~2 reference 8 Sn 4.6 reference 8 * All runs made at 5000C and with 0.5 weight percent solute.

-15CALCULATION OF TEE ROOTS OF TE EIGENFUNCTIONS The roots of -the two eigenfunctions (equations 21 and 25) were obtained on the Bendix 3G-15 digital computer. The half-interval method was used. The first ten roots of equation 25 are given in tables 2 and 3 for various values of S. Equation 21 has just one root for each value of S > 2; it has no roots for S<2. The value of this root rapidly approaches S/2 as S becomes large. The roots of equation 21 are given in table 4. Tables 2, 3, and 4 are photoduplicates of the original computer output. There has been no intermediate copying of the numbers. Oiily the first seven of tene decimal places carried in the computer were typed out. The number in the seventh place was not rounded off, so it may be 1 x 10-7 too low in some cases. Some of these roots are given by Carslaw and Jaeger to four decimal places. A spot comparison showed a few cases of disagreement of one unit in the fourth place. For these cases the function $ cot ( -S/2 was calculated for their value.s and ours, The results were much closer to zero using our roots. For these the function was also calculated using our roots plus and minus 2 x 10-7 In all cases the' function changed Bign, We therefore believe our roots to be correct.

TABLE 2 ROOTS OF THE EQUATION TAN[X] - [2/S]X = 0 S 1 2 3 4 5 + 100.00 3.2056172 6.4107036 9.6147541 12,8-17313-1 16.017-992 + 50.00 3.2717219 6.5390146 9.6211306 13.0473806 16.2853314 + 20.00 3.4761403 6.8862352 10.2211104 13.4996040 16.7402823 + 15.00 3.5877884 7.0367250 10.3693928 13.6342584 16.8602097 - 1.0. 003.7902224 7.2502483 10.553t1104- 13.7893118 16.992500 + 9.00 3.8492006 7.3016646 10.5938960 13.8224306 17.0202835 + 8.00 3.9164354 7.3559270 10.6358514 13.8561267 17.0483005 + 7.00 3.9926471 7.4128592 10.6788567 13.8903313 17.0766005 + 6.00 4.0781497 7.4721926 10.7227710 13.9249699 17.1051395 + 5.00 4.1725967 7.5335684 10.7674349 13.9599617 17.1338722 + 4.00 4.2747822 7.5965460 10.8126733 13i9952220 17 16275-13 + 3.00 4.3826255 7.6606214 10.8582999 14.0306627 17.1917287 + 2.00 4.4934094 7.7252518 10.9041216 14.0661939 17.2207552 + 1.00 1.1655611 4.6042167 7.7898837 10.9499436 14.1017251 + 0.50 1.3932490 4.6587782 7.8220315 10.9727946 14.1194627 + 0.10 1.5383044 4.7017550 7.8476103 10.9910251 14.1336292 + O0Q5 1.5547176 4.7070778 7.8507972 o.09933001 14.435398 + 0.01 1.5676067 4.7113277 7.8533449 10.9951195 14.1368132 H - 0.01 1.5739729 4.7134497 7.8546182 10.9960289 14.1375206 - 0.05 1.5865524 4.7176881 7.8571634 10.9978474 14.1389351 - 0.10 1.6019972 4.7229751 7.8603425 11.0001196 14.1407028 - 0.50 1.7155071 4.7648089 7.8856740 11.0182600 14.1548269 - 100 1.8365972 4,8158423 7,9170526 1-1.0408298 14,724320 - 2.00 2.0287578 4.9131804 7.9786657 11.0855384 14.2074367 - 3.00 2.1746260 5.0036452 8.0384627 11.1295434 14.2421016 - 4.00 2.2889297 5.0869850 8.0961635 11.1727058 14.2763529 - 5.00 2.3806444 5.1633054 8.1515643 11.2149058 14.3101229 - 6.00 2.4556438 5.2329384 8.2045313 11.2560430 14.3433507 700 2.5179545 5.2963423 8.2549929 14-.2960368- 1.3759824 - 8.00 2.5704315 5.3540318 8.3029291 11.3348255 14.4079711 - 9.00 2.6151525 5.4065327 8.3483619 11.3723648 14.4392772 - 10.00 2.6536624 5.4543537 8.3913455 11.4086265 14.4698680 - 15.00 2.7859313 5.6385305 8.5727360 11.5706762 14-6413962 - 20.00 2.8627725 5.7605579 8.7083138 11.7026780 14.7334723 - 50.00 3.0213230 6,0459048 9,9765-55 2.15551150 5,1627701 - 100.00 3.0800690 6.1605913 9.2420008 12.3246945 15.4090185

TABLE 5 ROOTS OF THE EQUATION TAN[X] - [2/S]X = 0 S 6 7 8 9 10 + 100.00 19.2164807 22.4125469 25.6060382 28.7968724 31.9850272 + 50.00 19S.5122876 22a7290015 25..93652a4..29.4359838 32328460 + 20.00 19.9558194 23.1542562 26.3407046 29.5184941 32.6898572 4- 15.00 020.06260799 22499016 e264270A: 8129.59694.g99327647 + 10.00 20.1774438 23.3510065 26.5171686 29.6782238 32.8356100 + 9.00 20.20117.14 23.3717323 26.5355518.29>6947324- 32.8505862 + 8.00 20.2250979 23.3925884 26.5540253 29.7113060 32.8656107 + 7.00 20.2491969 23.4135576 26.5725787 29.7279352 32.8806767i + 6.00 20.2734415 23.4346216 26.5911932 29.7446115 32.8957773 4 &00 20.2.9728034 -23.4557.621 26.6098623 29.7613253 32.910958 + 4.00 20.3222538 23.4769601 26.6285710 29.7780673 32.9260552 ~+ 3.,00 20.3467635 23.4981967 26.6473060 29.7948283 32.94-12186 + 2.00 20.3713029 23.5194525 26.6660542 29,8115987 32.9563890 + 1.00 17.2497818 20.3958423 23.5407082 26.6848024 29.828392 + 0.50 17.2642798 20.4081028 23.5513301 26.6941724 29.8367514 4 n404- 17.27158653 20.4179034 23.5598226 26,701-650 -29.434548 + 0.05 17.2773126 20.4191279 23.5608838 26.7026013 29.8442925 + 0..01-1. 17.2784702 20.4201073 23.5617326 26.7033503 2-984496260.01 17.2790489 20.4205970 23.5621571 26.7037248 29.8452977 o0.05- 17.2802063 20.4215764 23.5630058 26.7044737 29.8459670.10 17.2816528 20.4228004 23.5640667 26.7054098 29.8468054 0,50 1-47-.2932151 20.4325870 23.5725500 -2671-229&0 294,8535042 1.00 17.3076405 20.4448034 23.5831433 26.7222463 29.8618724 2.00 17.3363779 20.4691674 23.6042847 26.7409160.29.87zS 588 3.00 17.3649267 20.4934161 23.6253509 26.7595337 29.8952633 4.00 17.3932439 20.5175229 23.6463238 26.7780870 -29U.89385.00 17.4212891 20.5414616 23.6671861 26.7965638 29.9284692.6,00 17.4490243 20.5652079 23.6879210 -28.814952-1 29-.9449807 7.00 17.4764146 20.5887383 23.7085125 26.8332405 29.9614200 8.00 17.5034282 20.6120311 23.7289453 26.8514179 29.9777788 9.00 17.5300362 20.6350660 23.7492049 26.8694738 29.9940492 10.00 17.5562132 20.6578243 23.7692776 26.8873980 30.010222315.00 17.6799601 20.7669265 23.8664224 26.9747260 30.0894095 20.00 17.7908353 20.8672381 23.95738782 -70575502 —304852365 50.00 18.2197675 21.2858232 24.3603021 27.4423964 30.5312497 100.00 18.4952593 21.5836396 24.6743180 27.7673940 3088629136

-18TABLE 4 ROOTS OF THE EQUATION TANH[X] - [2/SIX = 0 S ROOT + 3.0 1.2878394 + 4.0 1.9150080 + 5.0 2.4640596 + 6.0 2.9847045 + 7.0 3.4935397 + 8.0 3.9568430 + 9.0 4.4988869 + 10.0 4.9995456 + 15.0 7.4999954 + 20.0 9.9999999

-19LIST OF SYMBOIS a Length of diffusion tube cm ca Defined by the relation c2 = h + S2/4 (S>2). Defined by the relation _-2 =.+ s2/4 C Dimensionless concentration N/No D Ordinary diffusion coefficient cm2/sec D' Thermal diffusion coefficient cm2/sec 0C E Elastic field strength vector volts/cm J Mass flux vector grams/cm2sec Separation of variables constant N Concentration grams/cm3 No Initial concentration grams/cm3 S Dimensionless solute velocity uV/D 0 Time sec. T Temperature ~C t Dimensionless time DQ/a2. Dimensionless time uV@/a2 u Electrical mobility cm2/sec volt v Sedimentation velocity cm/sec V Voltage drop across tube volts X Dimensionless distance Z/a Z Distance cm

-20LITERATURE 1. Schwarz, K., Elektrolytische Wanderung in flussigen und festen Metallen. J. A. Barth, Leipzig, 1940. 2. Mangelsdorf, P., J. Chem. Phys. 30, 1170 (1959). 3. Furth, R., Z. Physik. 40, 351 (1926). 4. Furth, R., Z. Physik. 45, 83 (1927). 5. De Groot, S. R., Physica, 9, 699 (1942). 6. De Groot, S. R., L'effect Soret, Thesis, Amsterdam,:(1945). 7. Carslaw, H. S., and J. C. Jaeger, Conduction of Heat in Solids, Oxford University Press, (1959). 8. Niwa, K. et. al.IJ. Metals, 9, 96 (1957).

-21DIFFUSION TUBE CAN BE LIFTED FROM RESERVOIR. RESERVOIR ELECTRODE ELECTRODE SEALED INTO DIFFUSION TUBE RESERVOIR OF LIQUID METAL 4// / /X/A.' / /GLASS DIFFUSION TUBE FILLED WITH LIQUID METAL Fig. 1. Cell for electrodiffusion experiments.

1.8 - ----— 1 —--- 1.6 __ ___ 1.0- 0. I48I IIII I1 \1 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 DISTANCE X Fig. 2. Convergence to initial condition for S = -4 using 1,5,5 terms of the series. Fig. 2. Convergence to initial condition for S = -4 using 1,3~5 terms of the series.

8.0 ---- 7.0 6.0 o 5.0 z 0 00 o o _ __30___ ____ _ _ _ _ _ ___ __ Z _ ___ __ 3.0 25 2.0 __ 1.0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 DISTANCE X Fig. 5. Concentration distribution. Parameters of time in hours. S = +1.96, D/a2 = 0.0072 sec

1.0 0.o 8 e 0-0.6 - "",.., _,8 0.4 -20 0.4~'^^^ ^^ 40 0.2 -- _- -- - -- 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 DISTANCE X Fig. 4. Concentration distribution. Parameters of time in hours. S = -2.00, D/a2 = 0-.0072 sec

1.0 0.8 16 0.6 Iz z 0.4 0 _____4____\ _____ _____ _____4 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 DISTANCE X Fig. 5. Concentration distribution. Parameters of time in hours. S = -20, D/a2 = 9 x 10-4 sec1 Corresponds to the following: u = 5 x 10-4 cm2/sec-volt a = 10 cm D = 2.5 x 10-5 cm2/sec E = 0.1 volt/cm

-261.0 0.9 0.7 o. \ w o 0.5 z 0.4 0.3 ------- 0.2 S= 0.1 0 0 0.1 0.2 0.3 0.4 TIME t,D Fig. 6. Mean tube concentration versus dimensionless time. Parameters of S.

-272.0 ___7 / 1 /__ 5,'.9 1.8 1.7 0of I1.6 z 1.5 1.4 1.3 1.2 Fig. 7. Mean tube concentration versus dimensionless time. Parameters of S.

28 I0 II 24 _ 20 16 0 ir: 12 rle, 0 80 20 30 40 50 60 70 80 90 0 10 20 30 40 50 60 70 80 90 % CHANGE IN TUBE CONCENTRATION Fig. 8. Percent error in mobility versus percent change in tube concentration.

28- 16! 0 12 0 8 D 4 0 10 20 30 40 50 60 70 80 90 % CHANGE IN TUBE CONCENTRATION Fig. 9. Percent error in mobility versus percent change in tube concentration.

-30100 --- -- -I- | —---------- -- - -, 8 - 6 4,, 3 2 z 0 z 8 6 I 2 3 4 6 8 10 2 3 4 6 8 100 0/0 CHANGE IN TUBE CONCENTRATION Fig. 10. Error magnification in mobility calculation.

-311.0 0.8 - - - -- - - SU 0.5 0.8 —-7 z w J 0.5 0 " 0.4.2 0.4 0.6 0.8 1. 1.2 1.4 1.6 1.8 2.0 z 0.3 DIMENSIONLESS TIME (- T),T ( ) V ~~~~~~~~~0.2 D~~~~~~~~~ Fig. 11. Mean tube concentration versus dimensionless time. Parameters of S.

3.0 2.8 t. __ uV.. 2.6 O(D t3 *~ _2.4___ COo.0 0o Oi C+ 0 (0 ___1.8__ _ f) i' - - - - - I I I- -i^l I I — - - - - -I 0 H' 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.O DIMENSIONLESS TIME (-T),'=( u V

-33Appendix I Prove the orthogonality of the:xe term. The weight function for this Sturm Lioville system is e-Sx. The following integral must therefore vanish. 1 I = e 2 (xex) (sinP)exdx 0 I = 2 sin P x - Px cos x 0 I = L- sin i - 5 oos but for S = 2 sin P = P cos P. = O0 Q E,D.

-34Appendix II Prove the orthogonality of (sinh acx)e (/2)x = -Sx, (SC/2sx x)e.S./2.x To I 4 Ss~c~i,~a~,.' li'2 -i eI - J e's inkcd:X - 2 e- siin.xdx o o eft ex 1I Xi.r -ax sin ~ o [pa e fa e a = = ( 3~ sin J x Jcos l A L sin3x- cos = > 2 2j [- 5 sin P cosh a - cos P sinh a substituting p cos p = (S/2)(sin p) a cosha = (S/2)(sinh a) 1 = 0 Q E oD.

-35Appendix III Prove the orthogonality of the (sin f -X) e (S2)terms. I = Je-Xsin m )e/2')in n'x)e(S/2)x 0 r1 i= Sin pm xsin xn x dx 0 sin(pm-pn)x _ sin(Pm + pn)x | I L 2 (m- - Bn) 2(pm + pn) sin pm cos pn - cos 3m sin pn sin pm cos pn + sin f.n cos ~P' I = 2(Bpm - pn) 2(prm + Pn) T (, _- Bn)2 [5 6n sin Bm cos pn - m sin pn cos 6m (6m pn)J but P cos o =(S/2)sin p I = O Q.E.D.

-56Appendix IV Calculate the coefficient for the term ixex. Denote the coefficient by Ao Bfe-2x (xex)( 1e2x) dx o A = e-2 (xeX)2 dx o 1. 1 xe-Xdx - xeXdx O O A - -- ~o x2dx ow -e-X(x + l)-eX(x - 1) A = -6/e 0

-37Appendix V (S/2)x Calculate the coefficient of the (sinh a)e/)x term. Denote the coefficient by B o'(S/2 _eSX) fe-Sx(sinh ax)e )( d x 0 o = e-Bk(sinh2ax) 0x dx First evaluate the numerator, N. (S/2)ox N = esnh d / ( sinh c.x dx - e sin dx o0 1 N = - 2 j(sinh.a x)(sinh. S/2 x) dx 0 [sinh (a + S/2)x sinh (a - S/2)x ] = - L(a + S/2) (a- /2) 0 N 8 a cosha ) e-S/2 N - (-22- ) e-S/2 4 a2-S2 Now evaluate the denominator, D. D = sinh2 af dx o r 11 1 sainh 2ax ax 4 a o

-381 sinh 2 a a D = 4 2 Combining N and D and simplifying one has 16 a S(cosh c) e,A/ BO = (S2-to2)(S - 2cosh a)

-59Appendix VI (S/2):x Calculate the coefficients of the (sin (3x)e terms. [e'-"'s in.n:) e -S/l2) -e'dx Cn Ce-s sin2 nex)e dx esin;nx)e-!($/24dx-x(sin pnx)eX'(S/2)xdx w i 0 Cn ---- ~sin2 pn x dx o Evaluate the first integral in the numerator, N1. ~ = ~-(s/2):x o.N2 = 0l(sin n x) e dx ~.pne.S1 ( ne-CS/2)x x Nl= - 2 —---- sin.n I.x + cos pn.ix - + -n L _ _ F -..S2( 2-p sin n.+ cos pn) -1 Evaluate the second integral in the numerator, N2. (S/2):x N2= - (sin pnx.). e dx = [p~2 e z['+^ P J o

-4oS Using the relation Pn cos 3n = 2 sin 3n N2 S- 2 2 + Pn Evaluate the denominator, D. 1s D = sin Pn:x dx 1 =n __ [ 2 sin 3n.x cos n:x D = ~n 2x 2 Pn - sin 3n cos 3n D = 2~,B 2 Pn Combine N1, N2, and D and simplify -16S Pn cos Pn e (S/2) Cn = (S-2 cos2 pn)(S2 + 4 pn2)

-41Appendix VII S2 Show that for X + - > 0 the condition <0 is fulfilled. Suppose that X>O. Then one has g2 a2 =,+ -- A..-.2_ S2/4 so a> S/2 However for a>0 one has coth a > 1 Therefore a coth a> S/2 This is a contradiction so therefore h< 0O

-42Appendix VIII In this method the mobility is determined by an, equation of the form, u = - k I - m where mf = final mass of solute in tube m = original mass of solute in tube k constant Normally one makes several determinations of m and only one of mfo Since the dependent errors in mo and mf tend to cancel each other they are ignored. The relation between the standard deviations of mo, mf, and u is given by 2 S la 2 2 u 1 2 u n l f where n = no. of determinations of mo Assuming m s mf 2 2 1 6U 2 /u 2 o f (u 1 )U2 = k 8u. mf S~= t~

-43u':S 2k2 Li 2 u m m m Su =S [ n Let 6r equal the fractional error of the chemical analysis. Then N;S = m, where N is the number of standard deviations corresponding to the chosen confidence level. The fractional error in the mobility ( ) at the confidence level of the chemical analysis is then, c Su - s~o:(m [1 + n(m-) k 20 - u e5 mo('-)[ 1 ( —-] -U U mO & [1 *+ n ( n 0 The coefficient of t is the factor of error magnification introduced in the calculation of the mobility.

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