《等温椭圆柱上的自由对流边界层》由会员分享，可在线阅读，更多相关《等温椭圆柱上的自由对流边界层（4页珍藏版）》请在金锄头文库上搜索。
1、X3322002M3vJOURNAL OF TAIYUAN UNIVERSITY OF TECHNOLOGYVol.33 No.2M ar. 2002cI|:1007-9432(2002) 02-0153-04等温椭圆柱上的自由对流边界层任辛喜(=Sv)K1:就水平放置的椭圆柱体温度骤然升高所产生的自由对流问题,选取椭圆坐标系,建立了与时间相关的边界层方程,应用傅里叶级数简化方程组的边界条件,并以此逼近得到精确解。同时,介绍了用中心差分格式建立的数值解法,为进一步计算各种流体物理量奠定了理论基础。1oM:自由对流;边界层;椭圆柱体;傅里叶级数ms|:O357.4DSM:Ap1HZ,K8,sZV
2、YVaMMesZbOstrach1NSVW%)b7,8,59hnb3InEMM5:b8,t= 0HYWV8T06Twb1F8HZ,Merkin2BZE%bTV,0 H 45H,.#VHermann3LTM(HVYV8_/Z_)bIngham4df,86BTw,mM,THWwMvtbK7S(t= 0),tTElliott5S1,zBbMerkinIngham9HH5,Howarth6Nigam7o8f%,9?CB6Cb31ZTfy$,P)EIngham4ZEMb1边界层方程n5yUSbUS,8abb$bTBauK8bt=0HY,8BWT06Tw,T0Tw(%MbLB(Tw- T0)= B$Tl,
3、B8n;7.kGr= gB$Ta3/M2v104109W,M8T,gFbN,Ostrach8,TV9,8#(Vk)9byN,HWM1V1HZaIn5bUS(N, H) ,( N, 0)AT_/, N= N0Vb8V(u,v)bs(Bga$T) 1/2L=s,F(Bg$T/a) 1/2,T= T- T0/$T(T8L=),HWS= tBg$T/a 1/2bs(u,v)VfW(N,H,t)VU:u = cosh N0 d- 1 5W5H, v = - cosh N0 d- 1 5W5N. ( 1), W= f/ Bga3$T 1/2, d= ( cosh2N- cos2H) 1/2b,e,-XZ/
4、:cosh- 2N0 d2 5F5S- 5( W, F)5(N,H) =Gr- 12 52F5N2 +52F5H2 + cosh- 1N0 cosh N5T5Nsin H+sinh N5T5Hcos H , ( 2)X:8S=53(2000)10Te:,3, 1961M13,=,Z_:#3N,w, 041004l: 2001-08-06cosh- 2N0 d2 5T5S- 5(W,T)5(N,H) =1(Gr) 12Pr52T5N2+52T5H2 , (3)52W5N2+52W5H2= - cosh- 2N0d2F. (4)Pr+,8T1qbpZ(2)j(4)HHq:N= N0,S0H,W=
5、5W5N= 0,T= 1; (5)N,S0H,F,T0; (6)N,S0H,5W5N0; (7)H= 0180,NN0,S0H,F= W= 5T5H= 0. (8)8*,8BS/Gr1/2 1/21Hby7,HUSa,N= kx+ N0, k= 2 SGr 14. (9)!NVT/M:W= 14Gr12k37 ,F= 14Gr12kX,T= 8. (10)|(9),( 10)VrT(2)j(4) ,cosh2N0 d- 2 52X5x2 + 2x5X5x - 2X=4S5X5S+ 4S2 575H 5X5x - 575x 5X5H cosh2N0 d- 2 -4cosh N0 d- 2 cos
6、h (kx + N0) 585x sin H+k sinh (kx + N0) cos H585H -k2 52X5H2 cosh2N0 d- 2; (11)cosh2N0 d- 2Pr- 1 5285x2 + 2x585x =4S585S + 4S2 575H 585x - 585H575x cosh2N0 d- 2 -k2Pr- 1 5285H2 cosh2N0 d- 2; (12)5275x2 + k2 5275H2 = - X cosh- 2N0 d2. (13)Hq( 8),=)Tbk|,L7 =n= 1f n(x, S)sin nH,X=n= 1gn(x,S) sin nH,8 =
7、 12h0(x, S) +n= 1hn(x,t)cos nH.(14)Z(11),( 12)(13),7k= 0,HZ,|(14),4S 55SGn= 52gn5x2+ 2K1x5gn5x - 2K1gn-2S2m= 1,mnn- mf n- m5gm5x -m= 1( m+ n)f m+ n5gm5x -m= 1mgm 5f n+ m5x -m= 1,mnmgm sgn(n- m)5f n- m5x +2Dn, 15h05x - 25hn- 15x -5hn+ 15x -K2 2x 5gn+ 25x - 2gn+ 2+sgn(n- 2) 2x 5gn- 25x - 2gn- 2 ,n1; (
8、15)4K1S5h05S= 1Pr52h05x2+ 2K1x5h05x -4S2m= 1m f m5hm5x + hm5f m5x ,4S 55SH n= 1Pr52hn5x2+ 2K1x5hn5x -2S2 nf n5h05x +m= 1(n+ m)f n+ m5hm5x +m= 1,mnn- mf n- m5hm5x +m= 1mhm5f n+ m5x +m= 1,mnsgn(m- n)mhm5f m- n5x -K2 2x 5hn+ 25x + 5hn- 25x ,n1;(16)52f n5x2= - Gn,n1. (17)K1= (1+ K2) /2, K2= ( 1- K2)/ 4,
9、7K= b/a,Gn= K1gn- K2gn+ 2+ sgn(n- 2)gn- 2,H n= K1hn- K2hn+ 2- hn- 2.sgn(i- j)VU(i- j)|,+Ysgn( 0)= 0.ijH,Di,j= 0;i= jH,Dij= 1.YL,(14)K,V7B|,b1/Snv+n0Bh,Ll|154v33n0,Z(15), ( 16)(17),/Svn0f(,bHHq(5) ,(6)#(7)e(n1):x= 0,S0H,f n= 5f n5x = 0; (18)x= 0,S0H,hn= 2Dn, 0; (19)x,S0H,gn(n1),hn(n0)0; (20)x,S0H,5f n5x0. (21)CollinsDennis4?C,BNsT9(21),LL=b,Z(17)HVx = 0x=|s,P( 18), ( 19)(20),
