H.N. Abramson - The dynamic behavior of liquids in moving containers. With applications to space vehicle technology (798543), страница 43
Текст из файла (страница 43)
For thispurpose, it is often advantageous to replacethe liquid, conceptually , by an equivalentmechanical system, because the equations ofmotion for point masses and rigid bodies areusually included more readily in the overalltransfer function of the missile than are theequations for a continuously deformablemedium such as liquid fuel.The main point that should be noted aboutthe equivalent mechanical models describedin this chapter is that the liquid cannot becompletely replaced by a single body of suitablemass and moment of inertia that is rigidlyattached to the tank. Instead, a system ofspring masses or pendulums must be included.These oscillating masses are necessary to duplicate the oscillating slosh forces arising from thewave action a t the free surface.
I n contrast,where there is no free surface (the liquiduZhukovskiicompletely Fills u, ~ ~ t p p econtainer),(refs. 6.1 and 6.2) has proved that the liquidcan always be replaced by an equivalentrigid body; moreover, the mass and moment ofinertia of this equivalent system are independent of both time and the particular type ofmotion being considered. However, Okhotsimskii (ref. 6.3) has shown that this result isnot true in general for a liquid with a freesurface.
Although his proof is too long to beconsidered here, a short heuristic argumentfollowing his outline is developed in the nextfew paragraphs. Besides showing that t,heequivalent mechanical model must consist ofmore than merely rigid bodies, except for a fewspecific types of container motion, the proofalso shows how the masses of the free surfacemotion analogs may be computed.Consider a cylindrical tank with its axisvertical (the z-axis of an r, 8, z coordinatesystem fixed in space), and filled to a depth hwith an ideal liquid.
Then, as shown inchapter 2, the small linearized liquid motionscan be calculated from a velocity potential,a , which is the solution of Laplace's equationThe boundary conditions are2- v,,--a t the container walls, where V n is the component of the velocity of the tank motion inthe direction normal to the wall (the n-diection) ; anda t the free surface.The potential can be broken down into twopartsa=*,+az(6.4)which satisfies equations (6.1) and (6.2) butnot (6.3), is the potential of a liquid that completely fills a capped container whose height is..Preceding page blank LeB199200THE DYNAMIC BEHAVIOR OF LIQUIDSequal to twice the actual liquid depth. a* isthe potential of a liquid in a stationary tankand whose free surface is acted on by theu n s t e a d ~pressures due to a,. That is, 9,satisfies equation (6.1) and the homogeneousboundan.
conditiorlThe forces and moments acting on the tankcan be computed by integrating the fluidpressures over the walls of the tank. Sincethe dynamic pressures are proportional to &,i.e., proportional to- .-a t the tank walls; it also satisfies the freesurfnce condition, equation (6.3), which nowtakes the form*2&+g z=f(t)=-@,-g(6.6)The motion of the cont,ainer can be decomposed in the usual way into a translation anda rotation (the tnnk is assumed to be a rigidbody). Since the problem of calculating thefluid motion is a linear one, the effect of thetranslation and the rotation can each be treatedseparately. Thus, only one of the two basictank nlotions, say the transliltion, need beconsidered.I t is assumed without proof thnt 9, can beexpressed aswhere rr, is the time-varying tank displacement.other words, the fluid motion due to iP1 isdirectly proportional to the tank motion; thisis a rensonnble result considering the physicnldefinition ofThe poter~tial,a*,can be written as111it can be seen that the forces and moments arepartly due to a rigid body type of liquid motion(the term proportional to so), and partly due toa more complicated type of motion (the termsThus, one can concludeproportional to g,).that only when every g, is directly proportionalto xo can the liquid be replaced entirely by a nequivalent rigid body (even in this case theequivalent mass may be a function of parameters such as frequency or amplitude); ingeneral, this is not true.
Instead, as can beseen from equation (6.9), a series of spring-massor pendulum elements must be included in themechanical model, with the masses and thespring constants (or pendulum lengths) chosenin such a way that equation (6.9) is satisfied.'The inertial parameters of the equivalentmechanical model for sloshing suffer from thedisadvantage that they may be functions of thetype of tank motion under consideration.However, in addition to easier visualization,the mechanical model has the advantage thatthe slight damping normady present in sloshingcan be treated by adding linear dashpots to thespring-mass or pendulum elements.I t should be emphasized that the aboveresults apply only for rigid containers and linearliquid motions.
Very few results are availablefor mechanical models that include tankelasticity or nonlinear liquid motions.where ~ $ 2 . ~nre the ordinary sloshing mode6.9 SIMPLE MECHANICAL MODELSfunctions, ~ ~ then d g,(t) can be calculated withthe aid of equation (6.6). After sub~titut~ion,Models With O n e Sloshing Massequntion (6.6) can be simplified to giveAccording to the theoretical developments ofthe preceding section, a complete mechanicalanalogy for transverse sloshing must include anIn equntiorl (6.9) use hiis been 1111~deof theinfinitenumber of oscillating masses, one forfuct tllrit b@,/bz=O 11t the free slirfnce for thistype of tnnk motion. w , is the ~iaturalfre1 Equation (6.9) is identical in form, of course, t o thequency of the mth sloshing mode, i ~ n dA, isequation of forced motion of a spring-mass oscillator orIL constant.pendulum whose natural frequency is om.201ANALYTICAL REPRESENTATION OF LATERAL SLOSHINGeach of the infinity of normal sloshing modes.A detailed analysis, such as that to be given insection 6.3, shows that the size of each of thesependulum or spring-mass elements decreasesrapidly with increasing mode number.
Thus itis generally acceptable to include in the mechanical model only the pendulum or spring masscorresponding to the fundamental mode (atleast as long as the exciting frequency is notnear the natural frequencies of any of the highermodes). Furthermore, it is nearly impossibleto include more than one sloshing mass in thosecases for which no analyses exist and for whichthe inertial parameters of the mechanical modelmust be determined by a suitable experimentalprogram. Consequently, a number of simpleone sloshing mass models are described in thissection.The model parameters mtiy be calculateddirectly from the sloshing force and momentanalysis, if such an analysis is available;several examples of this kind of model aregiven later.
In all cases, the model should bechecked with experimental measurenlents, and,indeed, the model parameters can even bedetermined from these tests. I t appears worthwhile to describe how this may be done, withthe development following closely that given inreferences 6.4 and 6.5.Experimental Derivation of M o d e l ParametersFigure 6.1 shows a typical mechanical model,which in this instance uses a pendulum as thesloshing mass. The pendulum mass, ml, isassumed to have no centroidal moment ofinertial2but the fixed mass, m0, is assumed tohave a moment of inertia, Io. cp, is the pitchingangle, and xo is the translation of the tank.In order to correlate this model with theresults of the experimental sloshing tests, it isnecessary to have expressions for the force andmoment exerted on the tank. The totai iateraiforce of the mechanical model can be written aswhere TIis the tension in the pendulum rod.Since only linearized motions are considered,sin $ has been replaced by #.
For the sameI n other words, the pendulum bob is a point mass,and the pendulum rod is weightless.IFree surfacetFIGURE6.1.-&%ode1 with one slosh'ig mass.reason, TImay be replaced by mlg, where g isthe longitudinal acceleration of the tank. I nother wordsF= -moxo+moh,ii)-mlg$(6.10)The summation of moments about the centerof gravity of the liquid isThe pendulum angle, $, can be calculated bywriting down the equation of motion for thependulum (compare this equation with eq. (6.9))Other assumptions are sometimes made; twocommon ones are that the penduium is aiwsysperpendicular to the liquid surface and thatthe sum of mo and m, equals the total liquidmas33 The pendulum angle \L can be determined by theuse of the first assumption.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
