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29 457 9 ASTIA rc7.l1 ON THE SOLENESS OF SOLUTION OF BASIC PROBLEM CONCERNING FREE THERMAl, CONEC(TION OF LIQUID By uisia B I. 0. Sevralk 295 796 r
ID-TT-62-1613i1 1 UNEDITED ROUGH DRAFT TRANSLATION ON THE SOLENESS OF SOLUTION OF BASIC PROBLEM CONCERNING FREE THERMAL CONECTION OF LIQUID BY: I. G. Sevruk English Pages: 6 SOURCE: Russian Periodical, Izvestiya ysshikh Uchebnykh Zavedeniy, Matematika (Kazan), Nr. 4, 1958, pp. 218-221 s/140-58-0-4-l THIS TRANSLATION IS A RENDITION OF THE ORIGI. NAL FOREIGN TEXT WITHOUT ANY ANALYTICAL OR EDITORIAL COMMENT. STATEMENTS OR THEORIES PREPARED BY: ADOCATED OR IMPLIED ARE THOSE OF THE SOURCE AND DO NOT NECESSARILY REFLECT THE POSITION TRANSLATION SERICES BRANCH OR OPINION OF THE FOREIGN TECHNOLOGY DI. FOREIGN TECHNOLOGY DIISION ISION. WP-AFB, OHIO. FTD-TT- 62-1613/1 Date 18 Jan 1963 B
On the Soleness of Solution of Basic Problem concerning Free hermal Convection of Liquid. by I* GSevruk 1 Assuming that in a certain volume, limited by a closed surface S is situated in state of thermal convective movement a viscous, mechanically incompressible liquids and assumng that on the boundary of the zone and at the initial moment of tim is given the distribution of velocity and temperature of the liquid. It is necessary to find the distribution of them values by the volume of the liquid at a given mmnt of time, Assuming the existence of a solution for the mentioned problem, we will ove its soleness, For this we employ the D*Ye*Dolidzell] method. Equations of free thermal convectior have the form of [2] 3 Squatins offreet a T-- ( vvt) = ZAT, ) dt l div v =-0, IjO where v- velocity of the liquid, p-read from hydrostatic pressure of the liquid, T6 temperature of the liquid, read from a certain constant mean value TO, (TO). density of the liquid, g-gravity acceleration,*-,, y - coefficients of kinenstie viscosity of thermal expansion and heat conduction of the liquid. The sought for values should satisfy the given conditionsa o-- (r), To =,(r)
(the sign a designates the boundary, and 0 - the initial value of corresponding value ). 29 We will assume, that there are two solutions for problem (1) - (5)s Then functions (, PI, T) H (,, P2, T). (.) U,-, q =pi-p ' o-t-t2 T2 will satisfy equations Ot au + o + () 2+ { r ) -A - 7 at+ vo+ ut2 divu = 0(..q and conditionsa U,--o0, es --o, jo UO=O, O O- To prove the theory of soleness it is necessary to, show that u = 0 and 0 P 0. For this purpose equation (7) is nmltiplied scalarly by u, equation (8)-by 0 we will combine same and then integate by the volume of liquid; we will obtain J'(u + o8) d'+flu(vlv)u+o(viv)od+fu(uq)v ' 2 d= =-fui-dv+f(ua+oao)d-f Ou(?g+ T 2 )d.(') P f Taking into consideration the continuum equal;ion and the boundary values of the functions u and (9, we transform the integrals included in (12) with the- aid of the Gauu-Ostrogradskiy and Grin formulas. In consequence we will obtain T-TT-62-i613/i 2
f UdTU dt / dt 2 2 d t v fp1u ±O(+0(iv) dp(,17)(u+ L) d= 1 Jdlv Iv, (U2 + 61)J d= 0, fv, q d- fdiv zi d-o, J'(UtU + YomO) d - f v (rot u) 2 + y (grad 0)' d. 710
Sabstituting the found values of integrals in (12) we will obtain d= -2 fv(rot g)' +(grad)d- 2 (,) v. 2 d - - 2 Jou(Sg+vT.,)d. Since v, y> 0, then the first ccmponent of the right part (14) is negtive and, consequently d 2fu (uv) 2d -2f (U 4T 2 W. v In the initial moment of time the function K(t) equals zero, and further on it /-be natia. Consequently in a certain interval of time 0( t - T dk - >' 0. In such a case, by ccuparing the absolute values of left and right parts of the inequality (15) for 04- t 4 -Z we obtain d<<2fu (U) 2 d + 2 fou(pg+t 2 )dj Evaluating the -ngnitude of the integrals, included in (16), we obtain fu(u)ad1'mfu~d<mk, where M - mean value 5 g) (1 - crosscut of direction u); fou(pg± T 2 )d ANflOluld,<Nf(U2 +63)d=.'-N.K, f ---_2 where N - mean value \ l * Taking into consideration the obtained evaluations, we will have f or 0 = t 4-- - I where A n 2M * No al 04 It is apparent, that the most rapid changes in function K with time in the inter- will be determined by condition dk. A K, LT dt i.eowhen K = conat.exp, Adto Henceconsidering (11) and (13) we obtain for 0 t I K a 0 and consequamtly FTh-TT-U-1613/1 3
Next we. will do as follows, We will break up the values t >0 into two classes, To the first class we will refer all t of the interval 0/ t ',. for which K = 0. The remaining t, for which K + 0, it is assumed are forming the second class. It is apparentothat the first class is net emptye Such a breakdown, according to the Dedekind theorem, is justified and is done by the number -, which appears to be maximm in first class. In such an instance for all tl*( 9 K(t)a0 and for any jexist such t o, 0 &. to-.t _ for which K(t') 0. On the other hand, it is evident from the proven facts with consideration of the continuity of function K(t) and condition K(') - 0, that for, a certain E 1 and all to satisfying inequality 0 Z t - U lok(t) = 0, which contradicts the made assumption. Consequeatly all values t are included in the first class and therefoe u O 0H0 O ;ka Bcex t>o. (1/7.) Aa to the pressure then as shown by equation (7) it is possible to synonymously determire he pressure gradient only. 3e The mentioned in point 2 proof of the soleness theory in the solution of the basic internal problem of free thermal convection of liquid can also be expanded for tk case of an external problem~kj. This can be realized by an ordinary maximum transition from the finite zone into the infinite, requiring the fulfillment of conditions of dampint at infinity8 L where alpha > 2, r - I r1+av J' Ir1+QT JI Ir'pl, Ir'+9rotvtI, Jrt+-gradT J- O (17c 1 ) (i=-1, 2), distance from a certain aasumed as origin of coordinates, point of heater. The author expresses thanks to B"1eGagyev and S.3,ihl'nk for the valuable information and a number, of critical notes. Su.mitted Febr.e819J8a m-t'-62-1613/i 4
The A.Mofo.Gkiy Form State Universi1ty Iiteratur., 1. D*Ye*Dlidze* Dolady Akadeuii Nauk SSM1, vol.96,no.39 1954 2. L.DLandau and Ye*M.Lifshitse Mchanics of Continuous bnidiaqgittl1954 FTD-TT-62-163A/ 5
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