The Cause of the Production of Air Bubbles in a Supersound Beam of Water Citation Henny, George Christian (1922) The Cause of the Production of Air Bubbles in a Supersound Beam of Water. Master's thesis, California Institute of Technology. doi:10.7907/MJYY-XA48. https://resolver.caltech.edu/CaltechETD:etd-02092005-084424 Abstract No abstract. Item Type: Thesis (Master's thesis) Subject Keywords: Physics Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Physics Thesis Availability: Public (worldwide access) Research Advisor(s): Unknown, Unknown Thesis Committee: Unknown, Unknown Defense Date: 1 January 1922 Record Number: CaltechETD:etd-02092005-084424 Persistent URL: https://resolver.caltech.edu/CaltechETD:etd-02092005-084424 DOI: 10.7907/MJYY-XA48 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 570 Collection: CaltechTHESIS Deposited By: Imported from ETD-db Deposited On: 09 Feb 2005 Last Modified: 03 Oct 2019 23:00 Thesis Files Preview PDF (Henny_gc_1922.pdf) - Final Version See Usage Policy. 2MB Repository Staff Only: item control page

THESIS The Cause of the Production of Air Bubbles In @ Supersound Beam in Water by George C. Henny Presented in partial fulfillment of the requirements for the degree of Master of Science in Physics CALIFORNIA INSTITUTE OF TECHNOLOGY 1922 re sont te at .when standing waves of @ supersound beam are produced in water containing dissolved air, smill bubbles of the air are liberated frora the water. The exact cause for the liberation of these bubbles wag unknown and it is this problem which was undertaken and is dis- cussed here. The supersound is produced by miking use of the piezoelectric property (see apnendix) of quartz. A thin sheet of quartz is cut from 4 large clear crystal in the vroper manner and cemented between “between two blocks of steel, as indicated in the figure. When a high frequency voltage is applied scross the two steel blocks the quartz expands and contracts in _resonaence with the current and the steel pie. 1 blocks are set into high frequency vibra- tion. If the oottom block is in contact with water a supersound beam is projected downward. The apparatus to produce the high frequency current to excite the projectors was a standard 20 K W Poulsen Ind uclance are With inductance and large variable sheet metal and alr condenser. The are was run by & 500 volt D. G. generator, the apparatus being comected up as in Fig. 2. Al- ; + though the high frecuency current scross the condenser plates is not of the sine wave form, that taken from a small portion of the inductance is very neurly so, and was considered as such. In order to measure the frequency,a frequency meter con- sisting of an inductance, variable capacity, and a neon tube, was set. up, as in Pig. 35, and calibrated. At Neon resonance the induced voltage across the ¢ => inductance becomes & maximum and the neon Inductance tube glows a bright reddish color. Pig. 3 The enparatus used in the place research was devised after some preliminary experiments. In the first, it was found that bubbles would not form in boiled water, no ratter how intense the sound beam. This immediately suggests that the degree of saturation has a direct connection with the ease with whieh the bubb les form ond shows that the bubbles form from the dissolved air and not in any way by the decomposition of the water. For a given intensity of the sound beam more bubbles will form for hignly saturated water than for water with little dissolved air, this leads to the method of using water of different known saturations and finding the threshold value of the intensity of the sound beam thet will just cause the bubbles to start to appear. Now, since the saturation point of a gas in a liquid depends upon the temperature, the above method way be more readily accomplished by simply varying the temperature instead of changing the whole solution in the apparatus. But the saturation point of a gas in & liquid also depends on the pressure. Thus, instead of doing either of the above things, we may vary the pressure instead. As a matter of fact, in the experiments 411 three methods were used in connection with each other. The apparatus was set up as in Fig. 4. | Here P is a quartz steel pro- jector cemented so as to make 4 water tight joint onto the top af the beaker B, which is filled with water seturated at a known temperature and pressure. ¢ is a copper container through which the water is circulated by 4 small pump P. CG is either heated by a burner or codled by sale ice and serves for the tempera- Circulation > Put p ture regulation of the water. This temperature is read con- stantly from the thermometer T, Fig. 4 whose bulb is right inside the beaker. The whole circulation system is pressure tight and « hydro- static head muy be applied by raising the glass container @ full of water up or down. This pressure is read directly from the callbrated water manometer tube i. | It was found that for a given sound beam from the projector P, the bubbles would start to form at a particular pressure as indi- cated by M. At a different temperature or intensity of the sound beam, this pressure was found to be different. Readings were taken using water saturated at various temperatures, using different temperatures aod using different intensities of the sound beam. It is evident that to do any quantitative calculations with the above mentioned kind of readings it is necessary to know the output of the gound projector under different conditions. To get the best efficiency the projector was always run at resonance, the intensity be- | ing controlled by varying the voltage delivered to the orojector as measured directly by the electrostatic voltmeter VY. Perhaps the best way to find the total output is to absorb the sound in a calorimeter and measure the rate of rise in temperature. Since only one prbjector was used throughout the experiments this could be carefully dene and a curve showing relations between output in water and the voltage at the projector was drawn, Fig. 4a. Knowing the voltage (at resonance ) as read by V, delivered to the projector, we can, by referring to the curve, find the total output, and this divided by the area of the projector gives the intensity of the beam. The water investigated should be saturated at a known temperature and pressure. This was done by placing a large jar full of distilled water in a carefully regulated water thermostat, and gontly pubbling several streams of air through it for at least 36 hours. While handling, the water was always kept below its saturation tempera- ture so as to prevent supersaturation and possible evolution of pert of the dissolved gases. The apparatus was filled with this water, the projector brought to resonance at some voltage and the hydrostatic head applied. This pressure was slowly diminished until tiny air bubbles just began to form under the projector. This value of the pressure is herein called the “threshold value". All readings were taken at this point, the temperature of the whole apparatus changed and the performure. "U'O PEE ON ‘MHOA MAN "'00''yaSS3 © T344nay if, 3 8 Si i 4 i t | t i Pee alt 7 | Ow 3 i a RE 1. 3 || 4+ 1 kB i T t ie : - i i 4 ees t i { + { ns . ‘ r a i t 1 as au Lid Tlittli i , i aa AEE } am > aan Sanman Thy i HEHE EEEEEEEEE j t ti aeR { = mem iI | ‘ tH ann rt i } i i t it 4 ‘ u < Dt ae revented. Many sets of duta were obtained in this way, although onlv a few are given as the others are similar and lead to the same conclusion. A typleel short run with the anparatus is indicated in the following table. Teble 1. Saturation temperature of water, 19.5° ¢. Projector: Quartz~steel cemented with K cement. Steel reactors exch 50 x 50 x 10 mm Quartz Griver, 50 x 50 x 10 mm. Run at resonance, 46,500 cycles per second. Temperature Total output Output per sq. dm. Threshold pressure for bubbles 22° 09 watts ~116 watts 1&5 2, per 8d. Cit. 25.5 5 9 © wlle e4uOg, " uM 40 3. 1 i -1le4 ui ous Be ii i tf 454 3.2 ii Lee 270 ze u uu ti 25 2, 75 ul ello if 155 &. ii i tt P05 3.1 u 124 ui 95 2. u u u 19.5 2.37 ul 2095 u 0 @. it tt i Another run tuken with the same projector using water saturated at 35° C gave the following results: Table 2. Temperature Total Output Output per eq. om. Threshold pressure of bubbles 355° 3.1 watts 2124 watts 109 2. Der 8d on, 35 #29 » 116 76 #. 32 2.8 " 112 ui 25 Ze "i tf ui 2g 3.1 4“ 124 i 35 2 EF u ui 27 2.9 "i 116 it . 0 Bs iu it Dy 34, 6 209 " 116 u W353 ge. ut M 37 2.9 iu ellé w 7& & M u u 40 2.9 u »116 u Lu7 2. " u 35 243 . 9 u . 095 i 0 B. i ul £17 ‘From these tables it is evident that the more saturated the water is the more easily the bubbles are driven out. The theory is that cue to rarefactions in the sound waves or standing waves, the pressure is momentarily released and a tiny bubble set free. ‘hen the accompanying concensation comss along the tendency is for the pubbles to redissolve and terhaps most of them do. $e, even with the best measurements we can make, we cannot expect an exact check. But, judging from the accuracy of our readings and the accuracy of the check we do get, it is possible to suy there is good evidence for the theory. | The method of attack will be to derive theoretically the pressure amplitude of the sound waves. It ig found that even at saturation a certain energy in a sound beam is necessary before bubbles form. The pressure amplitude of this wave plus the threshold pressure necessary for the bubbles to form (i. ¢., the hydrostatic head which will just prevent the bubbles from forming) when a more powerful beam is employed, should check with the pressure amplitude of the more powerful beam. The method of deriving the expression for the pressure amplitude is the classical one. I am indebted to Dr. H. A. lorentz for his kind help in this derivation. The work was later cheeked by him. Thefollowing letters will denote the same things through- out. Let x, y, and z = coordinates of any point considered u, V, and w = the components in the x, y, and z, directions of the velocities of the medium under consideration. These are functions of x, y, or 2, and t. ¢ = density of the medium at "any instant. | = Te i" under normal conditions ro = pressure of the medium at any instant. Po 1" oH a nder normal conditions. A =amplitude of vibration of the particles of the medium. Then pu mass of the medium passing unit cross seotionul area per- pendicular to X, per second. apu ., a . . . ‘ x change of miss of the medium with respect to x. But 4m increase in density oh, is due to an increase of mass in unit volume. Considering the propagation along the x axis we my write: ; ¢ r* ieavine the - (aa entering = (- $6 for unit volume. unit cube this equation indicates that the change in mass per unit volume, or the change in density, equals the mass leaving the section minus that enter- ing. Instead of considering a unit volume. or unit cube as above, We nay take &@ section of thigkness dx. We may then write | (1) G08 ax = - 8 ax or (1) 964 4 Pe =o We have considered propagation only along the axis of x. Considering tis general case, we have, considering the matter entering aud leaving the sides dx, dy, and dz (2) 9¢4 dx(dydz ) + ger dy(dxdz) + BY dnl axdy) =-$e dxdydz. ry LZ t dydz, axdz, and dxdy ure the areas of the faces of the cube. The quant i- ty dxdydz occurs in each expression and may be cancelled out. We then have (3) He + He + Mw 4 Zl =, This is called the equation of continuity and states thut the mutter is continuous and is neither created nor destroyed. The derivation of the seceleration of each particle in the sound beam is as follows: The definition of acceleration is the change in velocity in unit time. Hence to find the change we multiply the rate of change by the interval. if dt = change in t, or the small interval, and. u = velocity in any direction, | then the general change in velocity is time dt is ghav + + ga udt + $y vat +94 wat = dt Q = oe 38 U+ v5y + wet , But if the x axis te taken along u, then in the ouse of sound = 0 and Ww = 0, so we get as the expression for the acceleration of each particle (ape= 2 ude, Since the force mist equal the tass times the acceleration, we have (Ju. = ~2Pax on unit ¢ nere 22 ax = (4) (Get ug \pdx = -5Fix on unit area, where $7 dx = the force or pressure at point dx. ‘Now p = f (p) due to characteristics of medium (5) and Do =f(2 ) Considering u. as infinitely small and aA as infinitely small (6) If s = condensation (a variable) =i +s) An expression for the pressure is given by (7) BDFD ade (e -?,) = Dg t eit since from (6) rate x interval cs de 2 x= (vel003 eet red, & ) Now we can put ap a = (velocity of propagation squared, as later found) (9) With (7) end (8) p=, + zoo t eae "From (1) and (0) G2 = 48% 0 get (10) t+ 49% =0 or $4 +88 - = 0 (11) k= ae from (4) end (7). From (7) Differentiating (10) in respect to x we get 2 (13) Si Ber = 0 or a2 gu, + a2 BSc = 0. and from (11), differentiating in respect to t (14) ge etsy = 0. Subtracting (13) from (14) we get (15) gu ~ 23h = In the same way differentiating (1) in respect to t and (11) in respect to x we get (18) os ~ a2F =0 The solution of (15) is (19) w= FP (x @at) + fitx + at) (One wave travelling in one direc- tion and the other in the opposite direction a = velocity of propagation UW = velocity of particle insélf at instentt) (20) Then $i = - af! (x - at) raf (x + at) and (21) ou =+t+ae Po (a = at) + a2 f'tx + at) In the same way from (19) (22) gu =i (x - at) + f'(x + at) and (23) SA =P (x = at) +F"(x + at) Substituting (21) end(23) into (18) satisfies the equation showing (19) is # solution. (24) Ifa = F(x - at), then the value of u at time t and place x is also equal to the value of u at time t + Ot and place x + adt for from (oy) x + eat Ye (4 +4t) =x - at Thus a = velocity of propagation. From (&) page8 a «gf . Thus 1a velocity of propagution, a, is ziven by (25) a= ye = Let K = coefficient of compressibility = fraction of diminution of vol. increase in pressure fraction increase of density (26) or K= inerease of pressure (27) cae -o4 to (20 = Kp and from (25) (for very small changes) y 1 (29) Velocity of propagation = Ke =. Ixample, for water K= 49x 107° per atmosphere = 49 + 10712 per dyne. . sol? Then for water, since P= Ll 4a =fipe = 1430. meters per seomd. Experimental value is 1437 meters at 15° 6. Applying (29 to steel, we mst use for K pressure per unit qr’ea fraction of change of leneth Then ge= o and (30) & =)G = v ! Young's Modulus = E = For steel E = approximately 19.5 x 1pil P= 7.8 ca Ue 104! = 5 x 10° om./second ana @ = 7.8 = 5000 meters per second. Experimental value is epproximately 5000 for seft steel. Thus this derivation for the velocity of the propagation of sound in any medium agrees with experiment. But we are concerned with finding the pressure amplitude. This can be done as follows: p=pt+ @ sa® from (9) pageS | where p =the pressure at any instant bo = " " under normal conditions. s = the condensation. 11 =\9P = velocity of propagation of sound. - rom (18) 4 eg B= 0 or 24 + 22 = 0 (31) | - du = 22 2 But the velocity of « particle u is giveny (32) u=B coe n (t -2- q) where B is the muximum velocity of the particle (33) Bax of where A = amplitude of tioration =AxXxn and f = frequency q = some arbitrary phase angle. From (32) and (32) (34) oa =-BYsinn(t - 2-4) or ¢ (35) s=+ Boosn(t - 4-4) - $(x) But from (9) page& 2 = pg + fa s. Thus (36) PF Do + pa? B cop n(t +q) at x =o. From this we get the expression for the tressure (37), amplitude as Ph = Zab But the problem is not solved yet peeause to find the pressure amplitude the amplitude of vibration, A, must be known. The relution between A and B is given in (33) i. e., B=AX 2Tlf = An To find A we mst make use of the fuct that the energy of the beam depends on A. Thus we must desl with the equations of energy. | | fhe kinetic energy of a system of plane waves moving in the direotion of x, having volume Jjj ax dy dz, density? and velecity - of each element of volume =uat point (x, y, z) and at time t is 12 vy =e fps dxdydz where u = velocity at the point (x, y, 2) and at time t. Now the pot ent ial energy from unit mass in expanding . through a small renge from volume v to standard volume vg is to the second order terms (38) 4 (ptpg) (ov: Now Pp =o t Es since b= P (424 py definition, and | e phe, , aa sis by (6) page8 , O= Pll +s). Pi ip. ‘ dp wy V (39) Then p=py t+ “de where s is Fig. 5 increase in density nag’ is vate of change of ) with the denisty. Algo (40) vg - V = SVQ where vg and v are volumes. (41) 4 (p + Bg) (vor ¥) = 4 Vo -F BV +E PQ Vo mE POV or (42) (a9V9 = Dov) +-¥ WVo ~ HPV - EF DoVQ +E PV = Dylvo -v) + HP - Pollvg = Vv) (43) or = (vo - ¥) +4 Es"v 5 since © p= Po + Hs and vo - V = 8vy The first term D9 Vo 7 v) will disappear when there is no total change of volume. Since this is assumed, we have for the potential enerzy CY4) ws afff seaxayas where 5 = ue ¢ from (30) pagelo . From (19) page9 u=i(x - at) + f (x + at) and from (10) pageS gu. -2 Bhs F'(x - at) + fix + at) (45) Ip = f(x tat) - Fix -at) s=-2f(x + at) +x wat) (46) or aS = - f(x + at) + P(x - at) This is for a double wave or two systems of waves, travelling in opposite directions with con- Stunt velocity a. Taking just one part of the wave, i. e., a pro- ij gressive wave, since u = K(x + at) C47) has =u Thus we see, since H =a" that (48) T= for the progressive wave. Consider the harmonic motion of any particle. (49) s =A cosn(t -#+q) £ = amplitude of vibration. (50) ira =-4énsin (% -% + q) = velocity. Wow Kinetic energy = -em ué and 1 =f per unit volume (51) Then KE = yp an’ sin® nt - #+q)=7 (52) But W+ T= constant. Hence maximim T gives (53) total energy, or T max = zs? Aen per unit length where A= amplitude of vibration and n = 2x frequency. The energy current or rate of flow of energy is given by (54) = an Ay where T = period li weve length in om. A A Tv =4 in om. per second (55) Hence we get energy current = & pont per second. - 10 A = From (29) page, & = =VK é Corse 2,2 -4 ppy*fe So we eun write Hnergy current = ae Ke Atn*=Zzfin K Sinee in our experiments we had stationary waves, we should derive the equation for these. (56) Let y =A sin 2m B- A) be the equation for the displacement . given to the particles of the medium by 4 progressive wave. The re- versed wave must be one for which at some instant, say t +0, the dis- placements giver to the same series of particles will be equal and opposite. The equation (57) y' =A sin or ate) satisfies this condition and represents 14 the reverse wave. The resultant for the standing wave is then the sum of y and y'. Thus (58) Y= Asin ant - 5) + A sin 2m (a ++) Now from trigonometry . (59) Sin (A +2) = sine cog + cosy sin’? (60) Sin (¥ 7@) = sind cosW- cosd sing Using these we get (61) ¥ = (2 A cos eT sin nt A Since n Ser an | (62 Y= 2 an cos a4 eos nt and for K. &. = 4 me (63) K.E. = 2pa’n® cos (2x, cos” nt A é We may write (63) as (64) KE, = pwn? (1 - cos (cos? nt) to get the total energy per wave length we may integrate in respect to x from 0 to A and miltiply by 2 since T= ¥W. See (48), page 15. This gives (6 = oe re) 2 ra 5) B= eA nn“ since the avefage value of cos” nt is =| and the energy density Dg = ean? aA® n@ From (55) the energy density is $ ae or (66) Dy = 2 aPn® This we see is just double the energy of the progressive wave, as might have been suspected. Using (54) and mowing all the values but A, this amplitude may be calculated. The energy current from the projector was measured calorimetrically and is given in the curve Fig. 4a page Has This amplitude may be substituted in equation (37) and the pressure amplitude calculated. We may then “compare this with the experimental values and elther sustain or 15 disprove the theory for the formation of the bubbles. To show that the maximum pressure of two waves pro- gressing in opposite directions is the. sum of the two, we have the ' following: To find the intensity of the reflected wave we may con- sider the energy current which the transmitted wave will have. The pottom of the beaker of the apparatus was practically all in contact with air during the experiments. Since the density of the glass pottom is less than 3 times that of water the amplitude of vibration is nearly the same. The density of the glass is, however, about 2400 times that of the air. Now the energy current was, by equation (54) seen to be directly proportional to the density. Hence the energy of the reflected wave is, well within experimental error, equal to that of the incident wave and will be considered as such in the calculations. Equation (57), representing two waves of the same ampli- tude progressing in opposite directions may be written (67) Y=A sin n (+ -% }+asinn (t +* ) decause n = otf = 2 =- and f = a = frequency. T (68) Then Y =A sin nt cos ~ A cos nt sin“ +A sin nt cos 7X a CU Go + A cos nt sin @X (69) or Y=2 A sin nt cos LA (70) then -24- = “BAN. gin nt sin 24 Oo Now the change in pressure is given (since 3 = 2b) py (71) dp = — fo at OY | therefore on OX (72) dp = +o af t24n) cin nt sin 71k and the maximum change in pressure is (73) Pm = 2fo anA or twice that of the progressive wave. Thus the maximum variation in pressure for the two waves progressing in 16 opposite directions is additive. From (54), page 13, we have | (74) am = ae £5 end from (33) and (37) we have (75) Pm = (2 Aan and substituting (76) Pm = V2ak . In the case of water this gives (77) Pm = 535/E where BE = energy in ergs per second and Pm is in- dynes per square cm. Putting # in terms of watts and Pm in grams per squere centimeter, we have (78) Pm = 25221090 /10 R= 1724/ B. In the case of the stand- ing wave we find this to be twice as great or (79) Total Pm = 3448/5" This formula will now be used with the data in the tables already given and also those below. Table 3 Saturation Temperature of water 20° 6. Temperature Total Output Gutput Threshold pressure - per sq. cm. for bubbles 20° 2.0 watts 080 watte 0 #4. per sq. cm. 20° 3.12 " 124 205 4 " u li. 3° 3.12 u leu # 0 i i wo i] 17° 3.12 . 124 ctf 165 i iil ti ii 25° 3,12 if lek i 210 ui fi FI ib 30° 3.0 it .120 u 270 8 i u Using water saturated at 30° &. the following results were ob tained. Table 4 Temperature Total Output Output per Threshold pressure Bq. om. for bubbles 30° 2.10 wats «084 watte O @. per gq. om. 30° 2.75 ell 110 # 4 #4 22° Ce 75 u wll ti 0 39 it} i u 26° . 2.75 i il a 108 " Hi i “i 34° 3.00 u .1P ' 130 4 ft " ii 35° 2.75 " «ll u 200 § 4 ou fw Using water saturated at 40° Gc, 17 Table 5 Temperature Total output Output per Threshold pressure sq. Om. for bubbles 40° 2.65 watts 106 watts O g. per sq. om. 4O° 3.05 “ize ue ww 37° 3.05 u 122 uo. 6 fH it i i 38° 4.00 u .160 u jp 4 8 ) 8 From Table 3, page 16, using the readings at 20° the results are Total Pm = 3448/1235 = Total Pm for threshold value = 344/.08 =_976 grams per sq. Difference = 234% =O " Experimental value = _ 205 " i Error eg OW ii From Table 4, using the readings at 30° Total Pm = 344,411 = Total Pm for threshold value = 34480841000 * " Difference = . 144 we: uw Experimental value = 110 4 “i Error 34 8 i u u 1210 grams per sq. cm. om. i] 1144 grams per sq. om. ii From Table 2, pace 5, using the readings at 35° the results are: Total Pm = 3445//,1ll6 = 1173 grams per sq. cm. Total Pm for threshold value = 1064 24 u # i Difference = 109 " u " a Experimental value = 76 =O 4 Er ror 43 ul i ii 1g From Table 5, page 17, using the readings at 40° the results are: . Total Pm = 3448/,122 = 1205 grams per sq. om. Total Pm for threshold value = 3448106 =1124 “ 4% 4 4 Difference = 81 " ul " ul Experimental value = 4& = # ‘ow Error 33. C«*S os The results do not check accurately, but the nearly constant error suggests some experimental error common to 411 the readings. For instance, the output curve of the projector might be shightly in error. Taking the results of Table 2 as an example, we see if instead of 16 watts per sq. om. output we had .114 and if instead of .095 watts per sq. em. output for the threshold value we had .0965, the results would check exactly. This means an error in the curve of about 1.7 per cent, which is as close as might be expected taking all things into consideration. Using the results of Table 1, page 5, we get in the same manner, for the reading at 20.5° and 19.5° respectively: Total Pm = 5yus / 124 an 1214 grams per sa. om. Total Pm for threshold value = 3448095 = = 1064 nL a Difference = 150 4 ‘ow4 Experimental value = _ 95 ou Diserepancy = 55 =O vo Nd This latter difference is seen to be larger than any of the others ’ but according to the theory it should be so because the temperature at which the hydrostatic head (of 95 @. per sq. om.) was applied was above the saturation temperature of the water. 19 Above the saturation temperature the water is supersaturated, the air is more easily freed, and hence the pressure should be greater to suppress the formation of bubbles. considering the unsteadiness of the Poulsen arc at the time of the experiment, due partly to the fact that one person ran it and took all readings at the same time, the experimental results may, with a fair degree of certainty, be said to agree with the theoretical ones. The whole theory is, perhaps, not yet clear. From a glance at the tables it is easily seen that the more saturated the water is, the more readily the bubbles form. But another thing that may be noticed is that for a saturated solution a certain intensity ef the sound beam is necessary to free the air in bubbles. At saturation, any reduction in pressure should free some air. However, the reduction in pressure is always followed, one-half period later, by an increase in pressure. This tends to redissolve any air that may have been set free. This is what is done, up to the . threshold value. The larger the bubbles that form, the less readily they will dissolve because of the smaller ratio of surface to volume. Thue, when the pressure passes a certain value (herein called the “threshold” value) the bubbles become large enough during the half period so as not to dissolve when the compression passes. Since the maximum pressures have been shown to be additive (equation/3), it is legitimate to subtract the threshold value from the total maximum value and expect to get the small additional “experimental” hydro- static pressure which had to be applied to repress the bubbles. This is what was done and a check (within experimental error, of course) obtained. 20 APPENDIX PIEZO-ELECTRIC EFFECT AND THE SUPERSOUND PROJECTOR In 1580 the brothers J. and >. Curie in France dis~ covered that when certain crystals, ~ as, for example, quartz, tourmaline, boracite, and Rochelle salts, cut in the proper way, - are subjected to compression or tension, opposite and equal charges of electricity appear at the ends of the crystals. They found that the amount of electricity appearing at the end faces of the crystal is proportional to the force applied. But the reverse action, within certain limits, was also found to take place. That is, if at the proper places on the piece of quartz electric charges of opposite signs are applied, the erystal will lengthen. If the charges are reversed, the crystal will shorten. This change in length may correspond to that produced by a foree of several kilo- grams. This action will be discussed later. Let us further investi- gate piezo-electricity. | We find the pliezo-electricity property to be associated only with crystals. Crystals are distinguished by an orderly and al practically invariable space distribution of planes of cleavage. These planes, together with the planes of growth, are in specific arrangements in which planes and axes of symmetry are found. These arrangements possess different degrees of syametry; so crystallized matter may be divided on this basis into about thirty-two classes, differentiated by definite symmetry relationships, which are perhaps due to definite arrangement of the molecules or atoms. These classes are further grouped in several systems according to the defined axes to which these geometrical arrangements are naturally referred. In about twenty of these crystallographic classes no center of symmetry is found. These classes are technically designated as hemi-morphic, hemi-hedral and tetarto-hedral. In these classes in general, opposite ends of a line or direction, through the crystalline medium are not characterized by the same grouping of planes or association of physical properties. In other words, these directions are poler. So all matter whose crystalline ‘structure is lacking in a center of symmetry has at least one direction, and usually many, with two or more principal directions, along which the application of a fores, a8 pressure or tract ion, will produce electric polarization and the accumulation of electric charzes of opposite sign at opposite ends of the axes. In most eases this effect is exceedingly small. In a few substances, such as Rochelle salt, the effect produced by the application of force is very much greater for some directions than for others. | When there are two or more principal directions through & substance which are principal axes of piezo-electricity, it happens that the application of force along one direction may produce 22 perceptible polarization, not only at the extremities of that line, but at the extremities of another, also, This is the case with quart2, and plays a very important part in the action of the guartz projector. | Evenly developed quartz orystals look like hexagonal prisms terminated by hexagonal pyramids. ‘The axis running longi- tudinally from the top to the bottom pyramid is called the optical or orincipal axis of the crystal and is not an axis of piezo- electricity, so that force exerted in its direotion produces no piezo-electric effect. But the three horizontal directions which biseot the angles between the hprizontal crystal exes are the chief axes of piezo-electricity (see fig. 7). These axes emerge at the angles petween the prism faces. Force applied as pressure or tension upon planes cut perpendicular to any one of these intermediate directions develops electric polarization and the acoummlation of charges whose quantities are elven by the formula | q=DF where gq = electrical quantity of electrostatic units, F = epplied force measured in dynes, D =the piezo-electric constant, an empirical coefficient of the piezo-electric effect reduced from observation and experiment. (for quartz the Curies found for the chief directions D= 6.32 x 1097S e. S. u. of quantity per dyne.) But, now, if we compress or dilate a crystal plate out from quartz in a direction perpendicular to both the principal 23 crystallographic axis é and to any one of the piezo- electric axes; asa, Fig. &, thus applying the foree upon the Planes A,B,C,D, and E,F,G, Fig. 9, again electric polarization is produced with reference to the piezo-electric axis, 4), and 4 charge is set free and accumulated on the faces ACH and BDPG, which stand perpendicular to the electric exis @)+ (The electricity set free and thus accumulated is opposite in sign to that which would be produced by force Similerly applied directly upon these faces.) In this case, however, the quantity of elec- tricity thus set free is given by the formula -1 Q=DGF 5 =the constant piezo-electric (when, as before D = 6.32 x 10” coefficient found by the Curies for this direction also) where 1 and I are measured in centimeters. As a result of extended investigations with quartz and other piezo-electric substances, the Curies were able to state the following laws: 24 1. The quantities of electricity set free at the end of an axis due to a deformation are equal and of opposite sign. 2. The quantities of electricity set free at one of the ends of an axis ere equal and of opposite sign for two deformations inverse in sense to each other. 3. The quantities of electricity set free on each end are proportional to the variation in force applied. 4, For one and the same variation in force applied, exerted in the direction at whose extremities the electricity is collected, the quantities of electricity set free are independent on the dimensions of the crystal. 5. For one and the same variation in pressure, exerted normal to the direction alone which we collect the electricity, the quantities of electricity set free are proportional to the ratio of the length to the thickness, they are independent of the third dimension, and these lead, together with the phenomena of pyro- electricity, to the supplementary generalization. 6. For one and the same crystal the quantities of elec- tricity set free are proportional to: the intensity of the deformation at each point. 7. For one and the same intensity of deformation at each point, the quantities of electricity set free sre pro- portional to the surface upon which the electricity is collected end are independent of the third dimension of the crystal. These laws plainly teach that when we are limited to a given total quantity of foree and to a given volume of the cerystal- lized substance also, when we collect the charge upon Dianes per- pendicular to a piezo-electric exis which is parallel to the direction of this applied force, we will obtain a given quantity of liberated electricity; dependent only upon the total amount of the foree applied; but in this case we way alter the dimensions of the erystal plate but in so doing, if we increase the collecting area, we decrease the foree per unit of areas applied, and so the intensity of deformation at each point in equivalent ratio, and vice versa. 25 Since the piezo-electric action is reciprocal, we see we may obtain a greater force per unit area for a.given charge of eleotricity by reducing the size of the plane perpendicular toa piezo-electric axis upon which are the metallic electrodes. Hoaw- ever, if we are interested in producing the largest force with the least quartz, regardless of the charge to produce it, the best way is to cut the crystal perpendicular to a piezo-electric axis into thin (2 or 3 mm.) sheets. We have now seen something of how the piezo-electric action works and may proceed to see how it is applied in supersound projectors. Sinee a crystal of quartz, for example, when cut in the right way, will lengthen or shorten when cpposite charges are brought to the opposite ends of a piezo-electric axis, it will lengthen and shorten alternately if the charges alternate in sign. Thus, if an alternating current is applied to the crystal, a vibration in the crystal of the same frequency as that of the current will take place. ‘This is found to be the case even when very high frequencies are employed. It is found that as the frequency is gradually increased & point is reached where this frequency and the natural mechenical frequency of vibration of the crystal come into resonance. At this point the amplitude of vibration is very much grester than at any other point. In fact, the amplitude may build up to such an extent that the vibration breaks the orystal. The end of the vibrat- ing crystal, being in contact with the air, causes successive rare- factions. and condensations to be propagated outwards. These constitute supersound in sir. “However , since the amplitude of vibration is so | small and the air is so easily compressible, very little energy is 26 dissipated in the form of sound waves. If one face of the vibrating eryetal is placed in contact with a large body of water, supersound waves are sent outward from the projector face. A erent deal of ‘energy may be communicated to the water in this way, although the amplitude of vibration is very small. This amplitude may be only one-fourth that of the wavelength of sodium light, and, partly due to the high frequency (say 50,000 per second) the energy current of the sound beam would be about 13 x 10! ergs. In nearly all of our work the projectors were made of quartz crystals which were used as the drivers, although we are now trying tourmaline. Quartz sheets were cut from the crystals so that a piezo-electric axis was perpendicular to the surface. Since the sheets were cut thin (due to the scarcity of clear quartz) their resonant fibration frequency is very high. In order to bring this down to more desirable limits it was found possible to cement 4 piece of steel to each surface of the quartz slab in such a way that the steel formed part of the YZ vibrating system. These pieces of steel are called reactors, and Stee! : Cement —___§_ys their effect is to lower the Quvarlz ———> Steele_____s natural frequency of the projector. Equal thicknesses of steel are Fig. 6 always cemented to the faces of the quartz, as shovm in Fig. 6. This is done so as to have the quartz at the center of the vibrating system. This gives the maximum amplitude of vibration at resonance and puts the least strain on the cement.