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'6&1 7r ( NATIONAL ADVISORY COMMITTEE FOR AERONAUTICS TECHNICAL MEMORANDUM 1334 THE EFFECT OF HIGH VISCOSITY ON THE FLOW AROUND A CYLINDER AND AROUND A SPHERE*L By F. Homann For the determination of the flow velocity one is accustomed to measure the impact pressure, i.e., the pressure intensity in front of an obstacle. In incompressible fluids the impact pressure is 7v2/2g (7, kg/m3, is specific weight and v, m/sec, is velocity) if the influence of viscosity can be neglected. Such an influence is appreci able, however, when the Reynolds number corresponding to impact tube radius is under about 100, and must consequently be considered, if the velocity determination is not to be faulty. The first investigations of this influence are included in the work of Miss M. Barker2. In the following pages, experiments will be reported which determine the inten sity of impact pressure on cylinders and spheres; furthermore a theory of the phenomenon will be developed which is in good agreement with the measurements. The research apparatus consists of an oil circulation in which the velocity of the oil can be varied from 0.5 centimeter per second to 30 centimeters per second with the help of a vanetype pump lying entirely in the oil. A Russian bearing oil and a mixture of this with fuel oil is used for the measurements. Figure 1 illustrates the test setup. In this is indicated: P, the pump; a, turning vanes; G, straightener; and V, the actual test section which possesses a breadth of 0.148 meter, a depth of 0.15 meter calculated from the oil surface, and a length of 0.74 meter. It was provided with wall ports A in three different places. E is an entrance section for the pump; D, a diffuser; the immersion heater T and the cooling coil K provide "Der Einfluss grosser Zihigkeit bei der Strbmung um den Zylinder und um die Kugel", ZAMM, vol. 16, no. 3, June 1936, p. 153164. 1The suggestion of the present work, which was prepared in the Kaiser Wilhelm Institute in G6ttingen, I obtained from Herr Professor Dr. Prandtl, to whom in this place I express my most heartfelt thanks for the energetic furthering of the work and the valuable suggestions given me for its completion. Another work in the same field is published in "Forshung a.d. Gebiet des Ingenieurwesens", 1936, vol. 7, no. 1. M. Barker Proc. of Royal Society, 1922, Vol. A.101, p. 435. NACA TM 1334 temperature regulation. For impact pressure measurement a cylinder which was provided with a port was built rigidly into the test section. The diameters of the cylinders used were 1 centimeter, 1.377 centimeters, 1.953 centimeters, and 2'centimeters. The holes had a diameter of 0.1 centimeter and 0.2 centimeter. Two corrections one because of wall effect, the other because of finite size of hole (which originated with A. Thom and first had to be checked for the measuring range under consideration here) were applied to the measurements, which are illus trated in figure 7. The solid curve represents the theory. More precise information on the test setup and the measurement technique are found in the work cited in footnote 1. In the case of the measurement of static pressure on a sphere, a sphere provided with a hole was affixed to a pitot tube, the sphere having one time a diameter of 0.8 centimeter, the other time 1.6 centi meters. The execution of the measurements was in the same manner as in the case of measurement of the Barker effect. The result is shown in figure 8. The solid curve corresponds to the theory. In order to arrive at a clearer picture of the viscous flow around a cylinder or a sphere, the case of viscous flow against a plate was next calculated. The differential equation appearing in the two dimensional case has already been solved by Hiemenz3 and will be sketched once more for the sake of a better understanding of the final form. The solution will than be used in the flow around the cylinder. After this, the threedimensional flow against a plate in a fluid jet will be treated, to be used on the sphere. VISCOUS FLOW IN THE VICINITY OF A STAGNATION POINT (TWODIMENSIONAL CASE) Potentialflow theory gives for the velocity components in the neighborhood of a stagnation point, for the case of flow perpendicular to a plane wall (fig. 2): U = ax V = ay 3Hiemenz Dissertation, Dingl. Polyt. Journal, V.326 (1911), No. 2126. NACA TM 1334 3 The pressure is found from the Bernoulli equation to be S p (U2 + V2) pa22 2) Po 2 2 With consideration of viscosity, Hiemenz3 makes the formulation: u = f(x) v = yf'(x) (1) po p a2 (x) (2) The continuity equation is fulfilled; the boundary conditions read: for x = 0 (that is, at the wall): u = v = O, f = f' = O for x = m: v = V, f' = a In equation (2) if po signifies the pressure at the stagnation point, then F(o) = 0. From the equations of motion 6u u 1 vp .2 vfu 62u u u+ v + W+ ax ay P ax a2 y2 u+v + + 6x dy P 6y Yx2 y 2] one obtains as determining equations for f(x) and F(x) ff' = F' Vf" (3) f,2 ff" = a2 + Vf'' (4) with the boundary conditions above. Equation (4) has already been integrated by Heimenz. make the coefficients equal to unity, he sets f(x) = AO(k) NACA TM 1334 In order to S= ax From comparison of the coefficients, it follows that A = va a = f With this, equation (4) becomes 0,2 0", = 1 + 0"' The new boundary conditions read for 0 = 0: 0 = = 0 for w = m: 0' = 1 The behavior of 0 and the first two derivatives is shown in figure 4. We need the pressure difference between the stagnation point and the pressure for x = a% For x = m: 0' = 1 f = a 0 = 0.647 From integration of equation (3), F is determined to be a2 1 3 F = Vf' +1 2 2 If one now forms (P p) minus (p p'), as given by the Bernoulli equation, one obtains: (P0 P) (Po' P') a (F() y2) (f2 y }2f2) NACA TM 1334 For the stagnation streamline, for which y = 0: (Po P) (P' P')= p2 F P 2 2 2 If one puts in for F the previously obtained value, there results (Po p = (2vff + f 2 2 f2)= pvfm For x = mo, f = a; therefore, one obtains as a final formula: (6a) (Po P) (Po' P') = pva VISCOUS FLOW AT A STAGNATION POINT ROTATIONALLYY SYMMETRIC CASE) For the solution of the differential equation arising, all the expressions, such as the equations of motion, the velocity components, etc., were reduced to cylindrical coordinates. If z, r and p are the coordinates (fig. 3), then corresponding to the twodimensional case, there will apply: v = f(z) Vr = (z) pa2 2 P p F Z) r2) The continuity equation is again fulfilled; v = O, since we are dealing with a rotationally symmetric process. These expressions stem from the frictionless problem of a fluid jet against a plate, where Vr = ar r Vz = 2az p p a2 (4z2 r2) o 2 6 NACA TM 1334 The quantity 2az in the frictionless case is replaced by f(Z) in the viscous case. In the case at hand the equations of motion read: avr r aVz Vr + vr vz VZ Z v z Z z _ 8+ Vr P 6r r2 1 6r r 6r r2 +z2 r) oz2 (9) =1  P 6z 6z2V 1 S r St 2 +r r z2 / Substituting equations (7) and (8) in equation (9) gives: 1 f,2 ff"' = 2a2 + vf''' 2 2 ff" a2 2 (10) (11) F' Vf"'' The boundary conditions read for z = 0: f = f' = 0 for z = o: f' = 2a If one finds from equation f from equation (10), one can therewith determine (11). One next substitutes the transformation f(z) = Ao(() t = az (12) into equation (10), in order to make the two coefficients equal to unity. This yields: a 2A2~'2 a2A2"' = 2a2 + 3AO''' 2 From equating the coefficients: 1 2 2 2 3 aA = 2a = Va3A 2 NACA TM 1334 A = 2v a = (13) From equation (10) with equations (12) and equations (13) there results the final differential equation: 0"' + 20" 0'2 + 1 = 0 (14) with the boundary conditions: I for = 0: 0 = = 0 for t = o: = 1 The differential equation (14), just as Hiemenz', is no longer elementarily integrable. Its solution was obtained, accordingly, through a power series development from zero: 0 = ao + al + a 2+ + agn (15) By the method of undetermined coefficients, ai can be determined. Since, however, one boundary condition lies at infinity, one coefficient remains undetermined; and in fact it turns out to be a2. From the recursion formulas = ao + al1 + a 22 + a33 + = al + 2a2E + 3a3 2 + 0" = 2a2 + 2 x 3a 3 + 3 x 4a2 + . result as coefficients: NACA TM 1334 ao = al = 0 a2 = for the present undetermined a3 = 0.166667 a4 = 0 a5 = 0 a6 = 0.555556 x 102 a2 a7 = 0.396825 x 103 ag =0 a9 = 0.440917 x 103 a22 al0 = 0.793651 x 104 a2 all = 0.360750 x 105 a12 = 0.374111 X 104 a 3 al3 = 0.114597 X 104 a22 a14 = 0.115735 x 105 a2 ai5 = 0.301482 x 105 a4 0.385784 x 107 a16 = 0.134896 x 105 a23 a17 = 0.211005 X 106 a22 a18 = 0.224141 x 106 a25 + 0.157758 x 107 a2 a19 = 0.135546 x 106 a24 0.415153 x 109 a20 = 0.316633 x 107 a23 a21 = 0.152798 x 107 a26 0.295658 x 108 a22 a22 = 0.119505 x 107 a25 + 0.199390 x 109 a2 a23 = 0.371665 x 108 a24 0.433457 x 1011 a24 = 0.956242 x 109 a27 + 0.554360 x 109 a23 a25 = 0.943031 x 109 a26 0.462914 x 1011 a22 NACA TM 1334 9 In order now to be able to determine a2, a second series develop ment from infinity was set up, which was adjusted to the boundary condi tion for 0 at infinity. To this end one sets 0 = 0 + 1 (16) in which 0i corresponds to a small quantity, which one can neglect in the following expressions when it appears squared. 0o is the solution for 5 = m. 0' = 0o + 0 = 1 + ' since for e = m: 00' = 0' = 1 0" = 0" "'' = 1'" (16a) The boundary condition reads for = c: 1' = 0 Furthermore, 0o= " The integration constant is omitted, since in the following calculation it comes in again automatically. If one substitutes the above values into equation (10), one obtains 1"' + 2(o + 0101) (Oo2 + 20o'0'1 ,+ + 1 = 0 (17) Or if one neglects the squared terms in 0: 01''' + 2l"'' 201' = 0 (18) 10 NACA 'M 1334 To solve this differential equation one sets Q = 1 With this, equation (18) gives 0" + 2W0' 20 = 0 (19) A special, not identically vanishing solution of equation (19) is: Ol= E If 02 is an additional solution, then 01 2 1 ID = e a _2 t02 2 = e 2 02 2 1e 2 = 0 This equation is directly solvable. Its solution is O 1en TI 02 = E f J , The general solution of equation (19) is then: = + = + 1 e2 S= Clz + C2"2 = CiE + C2 21 e2 dT] go T1 NACA TM 1334 Since for t = i' = 0 = 0, then C1 = 0. Therefore S' = D = C2 J 'J 01=) Se'2 dT q2 = C2 e2 2 Ic 0 da 1 = 3 + C2 e2 2q' I eu du dq C32 = C C2 e' d'q iJ 0  2C2f The double integral becomes, according to Blasius , dq eu2 du = 2C2 2 Se 2 d9 e"12 dTi With this, equation (20) becomes: i C3 C02 1 e2 2 2 d e' d 2 e2 di 2 f O o 2 PR e2 d'i = 2 0 2 e d2] J/ (22) d = 2 FU fo 2 e dq 1 4Blasius Dissertation, Z.F.M.u. Phys. 1908, p. 1. and since e'2 d f 2 JC 2C2 f TI (20) +  Now: 2f 0 (21) e2 d' = 2 O o .2 12 NACA TM 1334 If one substitutes equation (22) in equation (21), one can calculate pointwise, since 2 ' Ff e72 dq is tabulated. Therefore 1 C3 = e2 C2 2 C2 = e2 C2 e 2 d j  + t 7  t {A f eT2 [n o dT 1. = e2 1d 0 Jfo Herewith 0, 0', and 0" are determined for the development at infinity, as a comparison with equations (16) and equation (16a) shows. In both developments a2, C2, and C3 appear as unknowns. If one now combines both solutions at the point t = to, and determines that the value of the function and the first two derivatives of the series development at zero are equal to the corresponding values that one obtains from the development at infinity, then three determining ( NACA TM 1334 H 0 0 u a + a ) E + 0 0 0  (U I a, a F + : CM 0 0 I a) 0 cu 0 Cd Cu1 + *H I 0, I 'H 0 'II rd I OJ 0 0 C) I 0 a 0 N 0 I o 0 up od ru *M O ri C 01 II ri +U rd a) I 01 0 II (,1 Cd I' CQ x CM + I *n 01 rLO crd CU + * *I 1 a0 4, rd d a, 0 c rj 0= "a 0 0 aL D a Cd o a) 0 m  0 0 + Ur a 4C o c O i 0 H U Ul In U 3 : . 0 OO NACA TM 1334 Thereby a2 is determined accurately coefficients of the power series this a2 a3 a6 a7 ag a10 all 12 al 3 a13 al4 a15 to at least five places. yields: = 0.658619 = 0.166667 = 0.365900 x 102 = 0.396825 x 103 = 0.191261 x 103 = 0.522714 x 104 = 0.360750 x 105 = 0.106882 x 104 = 0.497098 x 105 = 0.762253 x 106 = 0.605859 x 106 For the al6 a17 a18 a19 a20 a21 a22 a23 a24 a25 a = a1 = a = a = a = 0 The values for ', and are shown more accurately how ever, in Table I. In this case, 0' is calculated accurately to two decimal places, 0' to two, and 0 to three. With this the differ ential equation (14) is solved. From integration of equation (ll), one obtains a F = Vf + f2 =2av( + 2) (23) 2 2 = 0.385391 x 106 = 0.958673 x 107 = 0.381678 x 107 = 0.259200 x 107 = 0.904605 x 108 = 0.252966 x 108 = 0.161233 x 108 = 0.703675 x 109 = 0.209783 x 109 = 0.970520 x 1010 NACA 1 1334 As in the plane case, one uses again the pressure difference between the stagnation pressure and the pressure for z = c. For equation (23) is equal to zero; for E = 0' = 1 f' = 2a If one now forms again (po P) the Bernoulli equation, one obtains: 0 = t 0.557611 minus (po' P'), as given by (P P) (P' P') = P( Vf' + I f"2 f.2) = pVf' As a final formula one obtains (P P) (Po' P') = 2pva STAGNATION PRESSURE ON A CYLINDER For the stagnation streamline the tion gives NavierStokes differential equa du 1 dP 82u u +1 v=  + ax P 6x ax2 (25) Figure 12 shows the variation of u on this streamline. The different behavior of u at the stagnation point from potential flow is explained by the influence of viscosity. If one integrates between the boundary R and , one gets UR2 2  2 2 + (P R P)= v 6 + . / OX 10 fR w2u a y2 Figure 13 shows that R vu =0 ax m t = 0, (24) likewise uR = 0, and we want to identify PR with po ceding calculation. One obtains, therefore (P P) 2 NACA TM 1334 of the pre 2 dx 8y2 (26) r u dx " oy2 00 6 is calculated approximately in that for u the value corresponding to the potential flow is put in. The contribution of the boundary layer to the integral is, in the case of not too small Reynolds number, small in comparison. As potential function $ of the flow around the cylinder, one obtains 0 = Uo x + 2 and with it: 82u By2 2R2 r4 8x2R2 r6 r6 S8y2R2 r6 48x2y2R2) r8 For y = 0, therefore, along the stagnation streamline: S2 ay 6UoR2 x4 R 2 0oo oy dx = 2pUo dx  R Herewith equation (26) gives S ) 2 (p p) = 2 2pvUo R (27a)  Uo NACA TM 1334 17 or P p 4 p = 1 + (27) pUo2/2 Re If one substitutes 7 = pg, then formula (27) reads Po p 4 o 1 +  (27b) Uo2/2g Re where g = 9.81 meters per second2 In order to be able to accomplish a comparison of test results with theory, the "displacement thickness" (see Tolmien: Hdb. d Experimental physik, v. 4, 1st part, p. 262, "Grenzschichttheorie (Boundary Layer Theory)") on the cylinder must yet be considered in the calculation. Solution of the differential equation (5) yields (fig. 5): E* = ax* = nb* = 0.647 where 5* is the displacement thickness. Therefore 0.647 = 5*" (28) If one compares the flow in the region nearest the stagnation point for the cylinder and for the flow against a plate, one obtains from equa tions (27a) and (6a) 2pVUo 2Uo pva 2v a R If one substitutes this value in equation (28), one obtains for the displacement thickness 0.647 5 0 = *Re RV R R V/Re The dependence of 5*/R on Re is shown in figure 6. NACA TM 1334 Now since in the radius R, the actual tion (27b) is altered test results Re is formed from the cylinder effective radius is therefore (R + 8*), and equa to: Po 4v yUo2g 1 + + 7Uo2/2g Uo(R + 6*) With this one obtains cylinder as a final rule for the stagnation pressure on the P p 4 7Uo 1 + 7Uo2/2g Re t 0.457+e (29) In figure 7 the solid curve again gives the theory, which agrees very well with the practice. STAGNATION PRESSURE ON A SPHERE Corresponding to a cylinder, for the stagnation streamline of a sphere 8u 1 6p U +  6x P ax a2u  + ox2 va ( 6y If one integrates again over x from o to R: pU 2 Po P) = 2 2 + pv / Sco Vy2 2 d 68z2 (30) since HR Lm udx = 0, 6x2 UR = O The integral on the right side of equation (30) one again solves by NACA TM 1334 substituting for u the value for the potential flow. function of this flow is 0 = Uo For potential flow it is further true that x2 3x The potential 62u 3+ 0 2 az 82u 3y2 Therefore R( 2 Jo\Ia r u d.x R ax2 X IR If one substitutes in the last formula the value for one obtains f, aRf 2^ = oby 6u/3x given by 0, ,+ udx = 3Uo 2 R z / Herewith equation (30) becomes 2 pUo 3pVUo p p + 0 2 R (31a) (31) Po P 6+ PUo2/2 Re or with 7 = pg (31b) Po P 6 2, = 1 + R YU 2 29g or z2 20 NACA TM 1334 If here one also puts the displacement thickness into the calculation, one gets (since in the rotationally symmetric case E* = 0.5576): 0.5576 = B7i/v Comparison of equation (24) and equation (31a) yields a 3 0.5576 D= 5* 2R V 2VR (32) 6* 0.455 R %fRe It appears that the displacement thickness for a sphere and a cylinder are equal within I percent, although the displacement thickness in the case of plane flow against a plate is different from the corresponding threedimensional flow. If one considers the displacement thickness in equation (31b), one obtains as a final stagnation pressure formula for a sphere p p 0 = + 6(33) 7Uo2/2g Re + 0.455JFc The solid curve in figure 8 corresponds to the theory; the agreement with test results is again satisfactory. From the final stagnation pressure formula the dependence of the numerical factor c on Re can be determined, if osne sets Po P C pop 7Uo2/2g Re For the sphere there results 6 Re Re + 0.(34) Re + 0.455V1ie NACA TM 1334 From Stokes' calculation one obtains for small Re: C = 3. In figure 9 is drawn log Re as abscissa, e as ordinate. In the region from about Re = 0.1 to Re = 1 the course of e is essentially dif ferent, since Stokes' law describes an approximation for very small Reynolds number and the above law is an approximation for large Reynolds number. For the cylinder one obtains in the same fashion e= 4Re (35) Re + 0.457vRe According to Lamb5, for small Re, for which the validity of the formula extends to about Re = 0.5: Re Re (36) 1.309 ZnRe In figure 10 is again shown the dependence of e on log Re. Within the accuracy of measurement the test results here also confirm equa tion (29). With the help of the flow against a plate it is now also possible to establish approximately the course of u, 6u/6x, and from this p, on the stagnation streamline. A single curve was assumed in which, inside the displacement thickness 6*, the magnitudes as given by the flow against a plate were used. From the displacement thickness on, which had a value of 0.0455 in the foregoing case for Re = 100, the potential flow was calculated. To explain the transition from viscous to potential flow, I would like to go through the calculation of u as an example. The solution of the viscous problem ul/uo has as asymptote the tangent to the curve u2/uo, which was determined from potential theory, at the point 5* = 0.0455 centimeter. In figure 11 this tangent is labelled t. The difference k between the asymptote t and ul at the point xo gives the deviation of viscous flow from potential flow at this point. Therefore to the value ul at the point xo was added the proper k. With the help of this procedure one obtains point wise the transition from ul to u2. 5Lamb "Hydrodynamics" (2nd Edition 1931; German Edition by E. Helly, p. 696, par. 343). NACA TM 1334 au/6x was determined correspondingly; the pressure p was found from the equations of motion to be, in the case of the sphere: Po P Ux2 1 bux 6 (37) 1+ (37) Uo2/2g 400 200 6x Uo(R + x)4 Instead of 6 in the last term of the preceding equation (37), in the case of the cylinder one gets the factor 4. Figures 12, 13, and 14 are the results; by way of comparison the corresponding curves for the cylinder and the sphere are shown on one sheet. The curves are true, as already said, for Re = 100, in which R = 0.01 meter; Uo = 1 meter; v = 0.0001 kilogram x second per meter2 was assumed. Naturally the last curves give only an approximation, which can be made essentially better through a second approximation; yet this task in the framework of the foregoing work would lead too far. SUMMARY In the foregoing work the stagnation pressure increase on cylinders and spheres brought about through the influence of large viscosity, was reported on. For the threedimensional problem, hence the flow around a sphere, a differential equation was set up which corresponded to that of Heimenz, who had already solved the twodimensional case. The solution was ascertained likewise through an approximate method. The solutions for the two and for the threedimensional case were used for the flow around the cylinder and sphere respectively; the formulas so obtained for the stagnation pressure increase stood in good agreement with the reported test results. Finally, a procedure to determine the velocity and pressure variation, as well as the variation of 6u/ox on the stagnation streamline was shown and used on the practical case of Re = 100. Translated by D. C. Ipsen University of California Berkeley, California NACA TM 1334 TABLE I 0 0 0 1.3172 1.4 0.8546 0.9476 0.1697 .1 .0064 .1267 1.2172 1.5 .9502 .9635 .1301 .2 .0250 .2434 1.1173 1.6 1.0472 .9762 .0895 .3 .0548 .3502 1.0181 1.7 1.1453 .9863 .0622 .4 .0974 .4471 .9200 1.8 1.2424 .9905 .o418 .5 .1439 .5343 .8235 1.9 1.3436 .9935 .0276 .6 .2012 .6129 .7298 2.0 1.4430 .9962 .0180 .7 .2659 .6804 .6400 2.1 1.5413 .9979 .0115 .8 .3370 .7400 .5548 2.2 1.6409 .9986 .0073 .9 .4137 .7847 .4742 2.3 1.7416 .9991 .0042 1.0 .4951 .8352 .4015 2.4 1.8417 .9995 .0027 1.1 .5805 .8712 .3351 2.5 1.9420 .9997 .0016 1.2 .6653 .9025 .2760 2.6 2.0423 .9999 .0009 1.3 .7608 .9247 .2241 NACA TM 1334 Figure 1. Test tunnel. Figure 2. Streamline picture for flow against a plate (twodimensional). Figure 3. NACA TM 1334 0 0.5 I I5 2.0 25 Figure 4. 0, 0', 0 ". The curves drawn out illustrate the twodimensional solution, those not drawn out the threedimensional. L/ 0.647 Figure 5. R Figure 6. Momentum thickness c* on a cylinder = 0.457 and on a R . d 0 U.45D sphere = . R NACA TM 1334 o Test Point S P P 20 4 Re + 0.457'C Figure 7. Stagnation pressure on cylinder. o Test Point PO P ____ YUL Re + 0.455 Nf 29 PoP 2u 29 40 Re Figure 8. Stagnation pressure on sphere. POP 29 NACA TM 1334 o Test Point Curve I: e = 3 (stokes solution) Curve II: f 6Re Re +*o0.455 V a 8  6 f o 2 3 2 1 0 I 2 g Log Re Figure 9. e for sphere. o Test Point Curve I : E x Re +4 Re (Oseens' solution) 1.309 in Re Curve I: 'Re o.45  Ike + 0.457q~ Log Re Figure 10. e for cylinder.  NACA TM 1334 Figure 11. Illustration of interpolation for transition from viscous to potential flow. U 0 0 0.2 0.4 0.6 0 1.0 1.2 IA X R Figure 12. Velocity variation on stagnation streamline; cylinder: Curve I; sphere: Curve IL Re = 100. NACA TM 1334 u 150 dx Figure 13. 3U Variation of x on stagnation streamline; cylinder: Curve I; sphere: Curve II. Re = 100. POP Yru 2g Figure 14. Pressure variation on stagnation streamline; cylinder: Curve I; sphere: Curve II. Re = 100. NACALI gle 81152 1000 u = C E o 0 Q ,; EcN uwE F a o C i0 Q '0  Nfl U.* in E ,' LEa a 14U MF cgs i' s cn~ Sz .E" ES U D P 0 O U.<= < rn Z m w u,. CZ ~zL. L... n > U t~ i 2 E < L .n r. mr Z t u ;2 C O u NCE L40 Lo J _* z  >^r S3 m.* C Lcb5M> L ^B pj 2u0 N EM  ,a ~ls cV, c0 l 'C 0 'Oi O r cr C DCc 10.cu E. E wl 'a* uCLCUC)Z m E on c 2 a *'. Ln u . Q,, CL 0,  3~ CLC 0.L5C1 C IL 0~ 9. .C % U a 0 o Y' Cc 0L O E L 'A E g .ic 0.'L CuE lCPt Lb 'aLrL CL.O Cu cL Hu1,... CO.01C' JSL C^SS? 2tt~gai ^i. gc. El.9i LC a s5 L ' S' *i53 ; * ^~~ (L s'^ '"^ c' ,TI 3 0 L' ,3 i cnr cE o^ ^c  I it :Jm m g o I I' o Sa V > . 3 9 i C Z a. ^ ^ a ," N w * N* E C 47 C S a go ca x 0lU L B B U < w.0 E 0 U c <1 M S z N 0 bD, , r" 1. 0  aZ u CU0 5.Oo'0'.o 6 0 a 0~ (U JA kQ0 < * S2 R bw U0 ( Q D R N m3""20 u R wod&.4 C^%[raZ a? ^ z z E. 0! '0 faU~0 wu I ) w ^ 2S '. 4 0 1 0 L cU ~ ~ tu .sa a) M H z vm~ 5.0~00 T Lo 0 c ' ~jgg^ ^Ssg 0.Dg0,0 5.a k 0. L 5 f B 0 cu 0. (D 45 a n^ ot' ai or gpp z3 ' C.s a)s tsS '^gl^g0 W w be W w e~lssilgE. ( 0) .. C' C qc O '0. eS :a gM3 ES m.E to o o, 5 ~ 0 640.0o)~yag S.00. *U 0)E 0 0 cc CU 10 'U) 5. 0 0 0)U' 0 a) 4E' 0. 05 a. t E 5 001 0.0L '.4 v m f0 ok m P gw W.t 2 .4 A'C 'Ui U) L6 .4 g Q< 0 E ca Q . 0'Ea 0 1 C A. z zr P < .4 e :0 A~J I7 40'01 0.m0 LU0. r q o r 4)m r 0 bb > X CU ;, a M t :z 0 ) o 0 0 14 = U .=a DO M.~ 0..290.0 S.2 41  o"60 t' sSa^'3 0 CLB. =, 3 o o Fr Z! h > >> 01 0, E.l0 E = 3 j E 0 *S ok E  02 9 co c S 2 eo w o 'a' oIF. qS5 LU 0 0 0 .0 t:sc PIst ~ rL i aji h. u '" i~ i 025.00; U 0^ uoo 0 2  VU bo01 WU z 0 ar. e 2 uI. L 1 0 M 01  0'01. u S .l~g33, < S o S. E no 0. 020m ; i< Z 9 Im L z NC P ?? C'S C wu a 01..s. 0 .l .0 c cc >a k u w 0 w fn C M acn t < LU . > 4 gigillvi 4 w LU ~" 0~S (ULU4 00) h bCarai.<6Sa Cisi a ,Si 5~$~NE . *. 0 0.i cUco UL^ io i s Q 0.'V 0z 0 m 20 o
