Dl).IHOC Qu6c GIA TP. H6 CHI MINH
TRUt
ehi~u Ozmidov e6 dl;\ng:
au
--fv
at
av
-+fu
at
= ~(A
az
= ~(A
az
Z
Z
au)
(3.1)
av)
(3.2)
az
az
cae di~u kic$nbi~n:
- Tl;\i m~t biin eho tr\tde tr\t{Jng\fng Stitt tie'p tuye'n gi6
tr~n ~t
biin:
Po
A au(z, t) z
az
- 'tx
A av(z, t) Po Z az
- 'ty
(3.3)
- Tl;\i day biin vdi di~u kic$ndinh:
u(-H)
= v(-H) = 0
(3.4)
va di~u kic$nban ddu:
v(z,t=O) = 0
u(z,t=O)= 0 va
trongd6:
u va v la cae thanh phdn ohm ngangeua v~nt&-edong ehay.
Sir d~ng ph\tdngpMp phdn ttl hii'uhl;\n,gia tri gdn dung
eua cae thanh pMn v4n t6e dong ehay u va Y d\t<;1e
xtp xi quanh
cae nut eua phdn tii':
-(1)
U
-(n)
U
U 2 <1>(1)
-- U1 <1>(1)+
1
2
-- UN.l(n)+ UN+1(n)
2
<1>(1)+ v <1>(1)
- --- v N""'1
n,.(n) + vN+1
(n)
2
y(l)
V
(n)
1
V
1
2
2
8
V.di c1>lk)va c1>r)la cae ham d~ng dng vdi ph~n ur (k):
c1>lk)
= zk+l-z
L
c1>~k)=
va
Z
-LZk
(3.5)
Ap d\lDg phtidng phap Galerkin. m3i pUn td' dti<;Jebi~u
di~n dtfdi ma tr~n:
L/
3
/3
[L//6
L/
16
LI
13
~
Uk
L/
13
~~I+A
zfulk
~
u
~
- A
J i Z oz
]{Uk+l
L/
L/
3
7316
[L//6
-A
-
~
Vk
jl
~
Vk-l
-A
-
L
(Uk+l-U\::)+ Lf
-A
.1::+1
zk
ov-I +A
zOZlk
L
Vk + Vk+l
3
6
L
3
UI:: + Uhl
3
(3.6)
6
Uk+Uk+l
L
)))
VI:: + VI::+l
6
(Vk+l-VI<) Lf
zk
-A
\ - iAz Ov
OZ 1<+1 zk
l
(
(
(
(Vk+l-Vk)-Lf
(
(Uhl-Uk) + Lf
zk
6
3
))!
San khi lien ke't ta't ea cae pUn tU'd~ ~o thanh m<}tma
tr~n loan eve cho ml,tng ltidi. ta dti<;Jchai
M phtidng
thAnh ph~n v~ t6e u va v rieng bi~t e6 dl,t
trlnh dnh cae
~hti san:
~
[A] U}={Bu}
va
[A] VJ={BJ
(3.7)
Dtfa c c ai€u ki~n bien vao M (3. ) va chung dti<;Jcgiiii
r
bhng sai philn tie'n theo thCligian dng vdi mM btide l~p.
3.1.2 Cae ke't qua cua bai loan mot chi€u
1. TrtiClngdng sua't tie'p tuye'n gi6 kMng d5i: ke't qua tinh
toaD nMn dti<;JedtiClngxoAn 6e Ekman dng v"i cae vi de] khae
nhau
2. TrtiClngh<;JptrtiClngdng sua't tie'p tuye'n gi6 thong d5i
phtidng va de]!dn bie'n thien di€u bOa: cae kh6i nti"e l~i loon xoay
htidng eung chu ky eua trtiClnggi6. eung chi€u kim d6ng h6 d BAc
Ban C~u va ngti<;Jcl~i ~i Nam Ban C~u.
9
B~u mut cac vecto dong chay cac t~ng ve Den mQt du'Iii ham d~ng:
V=Vij + i;m j
vdi
aj
= XmYi
- XiYm ; bj = Ym- Yi;Cj =Xi
-
Xmva L\la di~n tich
cua m3i ph~n tit.
Ap d',lUgph1.fdngpMp Galerkin cho cae ph1.fdngtrinh (3.8)
- (3.10) va san khi lien ktt cac ma tr~n c\tc bQd~ ~o ma tr~n toaD
c\te cho ~ng
lu'di, eu6i ding,
[P]{~~}={Q,}
M phu'dng
trlnh trd thiloh:
[P]{~~}={Q.r
[P]{~;}=pJ
(3.17)
Cae phu'dog moh (3.17) d1.f<.1e
gilii bhog sai pMo lito theo
thCfigiao.
3.4 BM toan m6 hlnh ba chi~u
3.4.1 He ph1.fdngmoh xua't pM!. cae di~u kien bien va di~u kien
ban d~u
H~ ph1.fdngtrinh xua't pMt co d!,\og:
f}z
av
av
av
av
at;
g 0
at
Ox
Oy
f}z
oy
Po Oy z
-+u-+v-+w-+fu=-g
(A au)
av
( )
+A Llu
Ou+uOu+vOu+wOu-fv=-got;-JL~lpldz+~
at
Ox Oy
oz
ox PoOxz
7-
I
jpdz+-
0
&
z&
A -
z&
£
+A Llv
£
(3.18)
Tru'CJngnhi~t - mu6i d1.f<.1e
eho tr1.fdctU'cae tili li~u. H~ s6
nhdt r6i th£ng dti'ng d1.f<.1e
tinh trong di~u ki~n bO qua stf trao d6i
khutch tao nhi~t mu6i:
-
A,
= £'.1(::J'
+
(:: r
(3.19)
12
vdi f=f~(I-Rf)
la kich thudc r6i va fo =K(H+Z{I-~{I+ ~)]
la kich thudc r6i phan !lng trung dnh vdi K = 0,4 la hAngs6
Karman.
Thanh pMn th!ng dd'ng cua dong chay w du'<;1c
dnh to'
phudng trinh li~n D:tc:
au + av + aw = 0
(3.20)
ax ay az
Phudng trlnh d6i vdi dao dQngmtfc nu'dc:
at:"
at:"
at:"
at
ax
ay
-+u-+v-=w
.
khl z=t:"
(3.21)
Bi~u ki~n bi~n:
T~i m~t biln:
Cho tru'dc d'ng sutt ti€p tuy€n gi6 ~i m~t biln t: va t~:
a =~
z az
Po
5
A ~
va
a
15
az
Po
A -2 =.2....
z
(3.22)
T~i da y biln:
Dng sutt tru'<;1t~i day biln dU<;1c
dnh theo c<>ngthd'c:
1~ = rpou~u2 +V2
va
't~ = rpo V~U2 +V2 (3.23)
Bi~u ki~n bi~n ~i biJ r4n va bi~n long trong m<>hlnb ba
chi~u du'<;1cpMt bilu tudng tI! nhu m<>hlnh hai chi~u.
3.4.2 Su' tach mMn kMn!: !:ian troD!: m<>blnh ba chi~u
Thanh ph~n DAmngang cua dong chay trong d<;lidu'dng c6
thl du<;1cxem 18 t6ng cua hai thanh ph~n:
U=Uh +uz
(3.24)
{ V=Vh +vz
Do d6, phu'dng trlnh chuylo dQng ding du<;1ctach ra thanh
cac phu'dogmoh bai chi~u va mQtchi~u tUdngd'ng:
13
f
auh = -u au -Vau +fv-g at;_JL~ 1;p/dz+At Au
at
ax
()y
ax Po ax z
auz
=a A au -w au
at a( Zaz)
(3.26)
az
avh=-uav -vav -fu-gat;_JL~
at
avz
(3.25)
ax
()y
ay
f
(3.27)
1;p/dz+AtAV
Po By z
=~ A av -w av
at a( zaz)
(3.28)
az
Cac ph\fdng trlnh (3.26) va (3.28) d\f(jc giii b~ng bai loan
mQt chi~u thing dltng va cac ph\fdng trlnh (3.25) va (3.27) d\f(jc
giiHb~ng bili toaD hai chi€u n~m ngang.
3.4.3 Phltdng phap bie'n d6i loa do sigma
Ph\fdng phap bie'n d6i tQa dQ sigma chI th~t s1f.pM h(jp
cho khu V1f.cc6 dQ d6c day bi~n nM. Tuy nhi~n, n6 cho mQt
phltdng phap tinh ddn gian khi th1f.Chi<%n
bAi loan ba chi€u. S1f.ap
d~ng ph\fdng phap bie'n d6i tQa dQ sigma vito cac bAi toaD nlt~c
DongvAkhu v1f.cven b() IAmQtphltdng phap tinh hi<%u
qua cho cac
mo hlnh ba chi€u.
Th1f.c hi<%nphep bie'n d6i
d,
tit b€ m~t bi~n z
=t; de'n day
bi~n z =- H c6 tQa dQkhong thlt nguy~n d =0 vA d =-I:
I
X =x
I
Y =y
(3.29)
0"I =- z-t;
H+t;
Sau khi th1f.chi<%nphep bie'n d6i sigma, ap d~ng phltdng
phap tach mi€n khong gian thAnhcac bai loan mQtva hai chi€u:
14
Ul =Uhl +Uol
{ Vi
(3.30)
=Vhl +Vol
Do d6. cae bili toaD mQt chi~u 1£,)thAnh:
a1uol
1
al
a1u/
1 a1ul
---att= (H+<;)2&/ ( Ao &/ ) -w ac/
a'vol =
ae
1
al
(H + <;)200'
(
A alv' -w' alv'
0
00' )
001
(3.31)
(3.32)
va cae bili toaD hai chi~u:
auI
-al u h' = -u , -auf -v 1-+fv
ae
axl
I
ay'
-g--al<;
axl
0 , 1
0
al I
~ dcrl + JL J Qx ~dcrl
cI ax
Po d
00
Ja
- g(H + <;)
Po
al Vh'
-=-u
ae
I (}V,
--v
&/
, (}vI
--fu
ayl
I
Po
Ja
~ dcrl +
",' ay
(3.33)
al <;
-g-ayl
0 I ,
- g(H + <;)
+ At !:1.u'
0
~
J
Qy
Po cI
I I
a00~
dcrl + At !:1.v' (3.34)
3.4.4 Ap dung phtfdng phap pMn tU'hU'uban vao hili toaD ba chi~u
Bili toaD ba chi~u Ia tc5ngh6ngva .:1t= 240scho bai loan ven b<1.Trong
m6 hlnh ba chi~u, bai loan thing dti'ngmQtchi~u va bai tmin DAm
ngang hai chi~u c6 bu'<1cdnh I~n 1u'<;1t
la .:1r= 25s va .:1t= 6OOs cho
bili loan Dam Bi~n E>6ng.E>6ivdi bai loan ba chi~u ven b(1 cae
bu'<1c dnh I~n 1u'6ngva vinh Thai Lan lu6n
c6 h~ th6ng dong chay xoay. Stf xu!t hi~n cae xmly d~u lien quaD
d6n stf tu'dng lac eua tru'<1nggi6 va y6u 16 dia hlnh cua khu vf!.cva
e6 th~ giai thich sf!.xu!t hi~n cae xoay ireD cd sObao roan th6ng
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