H.N. Abramson - The dynamic behavior of liquids in moving containers. With applications to space vehicle technology (798543), страница 47
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Theequations include six degrees of freedom, andthe parameters from the hydrodynamic solutionare matched with the parameters of both anequivalent pendulum system and an equivalentspring-mass system. Digital procedures forevaluating the parameters of the mechanicalmodels are given in reference 6.36.aLl&GUExcept for liquid helium a t temperaturesclose to absolute zero, all fluids e-xhibit a cert,ainamount of resistance to motion; that is, theyare viscous. Consequently, because of viscousfriction, a small portion of the energy of asloshing liquid is dissipated during each cycleof the motion, and energy must be supplied tot.hefluid to maintain a constant slosh amplitude.However, even considering the energy dissipation due to free surface effects, the totaldissipation or damping is so small in a cleantank (unbaffled with smoot,h walls) that practically no limit is placed on the slosh amplitudea t resonance.
Thus, the damping is usuallyincreased artificially by introducing baffles orother flow obstructions in the tank, as discussedin chapter 3. Although the damping in suchcases may still be quite small, an equivalentmechanical model must account for it throughsome sort of energy dissipation mechanism.The exact form of the damping of a sloshingliquid in a moving tank is extremely difficult tdodetermine, but, because the damping is small,i t seems reasonable to assume that it can berepresented adequately by equivalent linearviscous damping. The damping coefficient isthen determined experimentally, or obtainedfrom the extensive data presented in chapter 4.Figure 6.23 shows one proposed model forsloshing in a cylindrical tank in which this typeof damping has been included (ref.
6.25).The equation of motion for the nth sloshingmass is now216THE DYNAMIC BEHAVIOR OF LIQUIDSThe force exerted on the tank by the nth mass is-Kn~n-cnyn, so that the total lateral reactionforce can be expressed exactly as beforeThen by carrying through the analysis as before,one finds that-rnTGo2n-1For these calculations, it was assumed thatn=-yz=y from figure 6.24. Since the correlation is very good, one can conclude that themechanical model adequately duplicates thesloshing forces.I t can be seen from equation (6.41) that forvery large damping, the slosh force predictedby the model is equal toThat is, the free surface motion is damped outentirely, and the fluid behaves as a rigid body;this is the expected result for a very viscousliquid.rnnhn("l~)2+g14(6.41)m~ 1- ( w / @ ~ ) ~ + ~ ~ Y ~ w / w~al200s?k~where yo and o are both of the-form e"'.I t isnow nssumed that rn, and h, are the same forthe damped liquid as they were for the undamped liquid.
This leaves only yn to beevaluated, and this must be done experimentally. I t should be noted that the same7 , is used in both the translational part of theforce and the pitching part, but this assumptionhas been verified experimentally (ref. 6.25).Figure 6.24 shows a typical plot of anexperimentally determined damping factor.Using this damping factor, the calculated forceusing only the first two sloshing masses isshown and compared to actual slosh forces infigure 6.25; the phase angle of the force withrespect to the displacement is also shown.=8:1000(a)J dlgTheoretical-mechanical modelcr,= Y, -awl001,0024Yo Id, or equivalent Yo Id (pitching).0008.W32(b)FIGURE6.25.-ComparieonFIGURE6.24.-Comparison of damping factorn in pitchingand trannlation in a cylindrical tank witb conical-ringbaffler (ref.
6.25).we dlgof m u ~ u r s dand calculatedforce reeponm in a cylindrical tank witb conical-ringbafBes (ref. 6.25). (a) Translation force rsrponre; (b)pitching force rerponse.217ANALYTICAL REPRESENTATION OF LATERAL SLOSHINGBy following the same procedure as before,the pitching moment can be written for thismodel asI t is now specified that the moment of inertia of- a rigid body of fluid equal in mass and shape tothe actual fluid isI,.~=I.,+I;+~~+Emnh:(6.45)n-1The pitching moment acting on the tank is now-nhn(~l~n)'+gl4and the equation of motion of the disk is-m~%5l-(u/~~)~+2iy~~/u~As yn+- in this equation, the moment reducestoM=-k{~~+m&:+f:n-1m.h:)Now by assuming yo, yn, (PO, and 4 all vary ase'*', equations (6.39), (6.45), (6.46), and (6.47)can be combined to yield(6.44)But from equations (6.31) and (6.33), this isthe same as the moment due to an ideal inviscidliquid in a capped container.
However, itseems that the moment should be the same asthat of an equal mass of rigid fluid, since thefluid's resistance to shearing motion has theoretically approached infinity. In other words,I. should be modified somehow to account forthe damping. This is a relatively minor point,since the damping is actually quite small (notinfinite) and, furthermore, the moment aboutthe center of gravity usually contributes onlyslightly to the total moment when the centerof rotstion is more than 1 diameter from thecenter of gravity of the liquid. But to becomplete, the model is now modified so that itapproaches the right asymptotic limit.The proposed modification, shown in figure6.26, was originally developed by Bauer (refs.6.29 and 6.30).
As before, yo is the tank displacement and fi the pitching angle about thecenter of gravity. A weightless disk withmoment of inertia Idis attached a t the centerof gravity by means of the dashpot arrangementas shown. The angle of rotation of the diskrelative to the tank is #. The force responseof this model is the same as that of the oneshown in figure 6.23, since the disk is weightless.fM=-io(w21:)Irkld-Idw212,+cl,w/wn)'+&dy2 4-(w/~,,)~+2iy~u/w.*n+"(+ m ~-mT1-mni-(u/~.)~+2iy.w/u~(6.48)is defined in equat,ion (6.45) ; it iswhere IrlgldalsoThe effective decrease in the moment of inertiabecause the fluid is not actually rigid is then(w2r,+ca, )I.,w212,Thus, as the damping goes to infinity, themoment of inertia of the liquid approaches the@1; is not necessarily equal to the previous lo.218THE DYNAMIC BEHAVIOR OF LIQUIDSFree surfaceFIGURE6.26.-Equivalent mechanical model with moment of inertia damping.value of a similar rigid body, as it should; that is,For zero damping, the moment of inertia ofthe fluid isIF=Irlgld-Ia(6.50)and IFis given in figure 6.14.
Thus, the valueof Idmay be computed. The value of cd mustbe determined experimentally; it is approximately equal to the damping of the liquid whenit completely fills a closed tank pitching aboutthe center of gravity of the liquid. cd and I,have been determined in this way in the testsreport,ed in reference 6.33. The results showthnt even in a tank with baffles distributedthroughout the liquid depth, the coupling ofthe disk Idto the tank is so small as to benegligible (i.e., cd=O). Thus the inertia of thedisk does not enter in any of the calculations.Tnking this into account in eq~iations(6.45)rtnd (6.50) sllows thatConsequently, the model shown in figure 6.26has no real advantage over the one shown infigure 6.23. I t should be pointed out, however,that the results given in reference 6.33 showthat IFfor a baffled tank is significantly largerthan IFin an unbaffled tank; this is causedprimarily by the additional liquid motion nearthe baffles and not by the small increase indamping.A study of mechanical model representationof force response in a baffled spherical tank hasbeen given in reference 6.15, as indicatedpreviously.
(For descriptions and discussionsof the baffle configurations, see ch. 4 (fig. 4.17) .)Figure 6.27 compares measured and calculatedforce response curves in a spherical tank withvertical baffles (data are also shown for thehorizontal baffle configuration). Two values ofdamping coefficient (the larger of which corresponds to measured data) were employed in themechanical model calculations for the half-fulltank and one for the three-quarter-full tank.The agreement is not nearly so good as thatobtained in the cylindrical tank, no doubt asa result of the strong nodinear effects presentin spherical tanks.The mechanical models for other types oftanks can be modified in a similar way to include damping. Likewise, the simple, onesloshing mass models presented in section 6.2may also be modified by including an appropriate damping factor; in this way, they can bemade to give a very good approximation tosloshing in the fundamental mode (refs.
6.5,6.10, and 6.11).6.5 M E C H A N I C A L MODELS FOR TANKS WITHFLEXIBLE WALLSAll of the mechanical models described sofar have been for rigid tanks. When tankelasticity is included, the problem of determining an equivalent model is greatly complicatedbecause of the coupling between wall motionand fluid motion.
(See ch. 9.) Certain exact'solutions to the flexible-wall problem exist inthe literature, but their application is by noANALYTICAL REPRESENTATION OF LATERAL SLOSHING200Phase angle(deg.100II--+Experiment, unbaffled tank--*-Experiment, vertical baffles----Dimensionlessforceamplitude---6.--\Experiment, horizontal baffles Theory, vertical baffles(mechanical model)IIII9101112139'I01112'1330012345678(a)Dimensionless frequency parameter200Phase angle(deg.Dimensionlessforceamplitudela)300'123d567B(b)Dimtnsionlm frequency parameterFIGURE6.27.-Comparison of measured and calculated force-response in a spherical tank with vertically oriented baffles(ref. 6.15).
(a) Half-full tank; (b) three-quarter-full tank.220THE DYNAMIC BEHAVIOR OF LIQUIDSmeans direct. Hence, it appears to be reasonable to try to adapt the rigid-wall models to theflexible-wall problem in an approximate fashion.This can be accomplished by the method outrlined by Lukens, Schmitt, and Broucek (ref.6.4).At low liquid levels, some of the mechanicalelements in the rigid-wall models are locatedabove the liquid surface. (See, for example,I, in table 6.1 for small hid.) This presents noproblem so long as the walls are rigid, b u t ifthe force distribution on a flexible wall is to becalculated more accurately, the forces due tothe mechanical elements should act a t the correct location.
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