1

i hc quc gia h ni

Trng i hc khoa hc t nhin

Nguyn Mnh Hng, Nguyn Th so

M hnh tnh sng

vng ven b

H Ni - 2005

2

3

mc lc

M u

Chng 1

L thuyt c bn v trng sng trn vng bin su v ven b

1.1 Cc yu t sng, dng sng v phn loi trng sng 5

1.2 Cc l thuyt m phng trng sng, phm vi p dng i vi cc vng nc su

v ven b 8

1.3 Tc ng v tng tc ca trng sng vi cc qu trnh thu thch, ng lc ven

b 15

Chng 2

Bin i cc yu t sng khi truyn vo vng ven b

2.1 Tc , di v cc yu t khc ca chuyn ng sng vng ven b 19

2.2 Bin dng sng vng ven b 28

2.3 Khc x sng vng ven b 30

2.4 Nhiu x sng do vt cn 33

2.5 Kt hp sng khc x v nhiu x 36

2.6 Phn x sng 40

2.7 Sng 41

2.8 Tng tc gia sng v dng chy vng ven b 48

Chng 3

ng sut bc x sng v cc qu trnh do sng sinh ra

vng ven b

3.1 Cc thnh phn ng sut bc x sng 54

3.2 Mc nc dng v rt ti vng sng 57

3.3 Cc loi dng chy do sng vng ven b 59

3.4 L thuyt dng chy sng dc b 60

3.5 Lp bin sng 65

3.6 Sng di vng ven b 69

4

Chng 4

L thuyt ph sng p dng cho vng ven b

4.1 Ph sng trong vng bin c su gii hn 71

4.2 Bin i ph sng vng ven b 78

Chng 5

Cc m hnh tnh ton sng gi, sng lng vng ven b

5.1 Cc yu t to sng v iu kin kh tng hi vn nh hng n trng sng 80

5.2 Cc phng php tnh sng da trn cc mi tng quan l thuyt v thc

nghim gia cc yu t sng v cc yu t to sng. Quy phm tnh ton sng

ca Vit Nam 93

5.3 Cc m hnh tnh sng vng ven b da trn phng php gii phng trnh lan

truyn sng 103

Ti liu tham kho 123

5

M u

Gio trnh M hnh tnh sng vng ven b c bin son nh mt s k tip cun

gio trnh ng lc hc Bin phn 1 Sng bin [1] c bin son nm 1998 dnh

cho hc sinh Hi dng hc ti khoa Kh tng, Thu vn v Hi dng hc . y l mt

cun sch vit kh y cc kin thc c bn v trng sng, trong cp n c

trng sng vng khi v trng sng ven b, cc phng php tnh ton d bo sng

trn c s l thuyt v thc nghim. Tuy nhin do s pht trin rt nhanh ca cc

nghin cu l thuyt, thc nghim ca ngnh Hi dng hc ni chung v ng lc sng

bin ni ring, c bit ti khu vc ven b l ni tp trung mi hot ng kinh t, xy

dng, du lch ngh dng, nn trong khong t nhng nm 90 li y, nhiu l thuyt,

m hnh tnh ton trng sng mi c nghin cu v a vo p dng trong nghip

v hng ngy. Cun gio trnh ny c bin son nhm p ng c cc yu cu nng

cao, cp nht cc l thuyt, m hnh tnh sng vng ven b, v vi phng hng nng

cao trnh , k nng thc hnh tnh ton cho sinh vin. Mt s cc phn l thuyt c

bn v trng sng s c nhc li so vi gio trnh u, tuy nhin cc l thuyt v

phng trnh lan truyn sng trn vng bin c dc thoi, l thuyt bc x sng v cc

m hnh tnh sng theo phng php s l nhng phn hon ton mi v nhng nm va

qua cc sinh vin c truyn t tng phn.

Gio trnh gm 5 chng xp xp theo th t t l thuyt c bn n thc hnh v

cc m hnh tnh sng.

Chng I cp n l thuyt c bn v trng sng vng bin su v ven b do

PGS. TS. Nguyn Mnh Hng bin son.

Chng II vit v bin i cc yu t sng khi lan truyn vo vng ven b do

PGS. TS. Nguyn Mnh Hng bin son.

Chng III trnh by l thuyt ng xut bc x sng v cc qu trnh do sng sinh ra

vng ven b do TS. Nguyn Th So bin son.

Chng IV lin quan ti l thuyt ph sng p dng cho vng ven b do PGS. TS.

Nguyn Mnh Hng bin son.

Chng V l cc m hnh tnh ton sng gi, sng lng vng ven b do PGS. TS.

Nguyn Mnh Hng v TS. Nguyn Th So cng bin son.

Trong qu trnh bin son, cc tc gi c gng trnh by mt cch c ng cc

phn l thuyt v thc hnh, lin quan n trng sng vng ven b. ng thi cng

chn la cc thut ng chung nht trong nghin cu sng, trong nghin cu a hnh a

mo vng b nhm bc u thng nht cc thut ng chuyn mn trong ngnh Hi

dng. Tuy vy c th vn cn nhng vn b st, cn c b sung v cc thut ng

cn c thng nht. Chng ti bit n v nh gi cao cc pht hin v ng gp ca

ngi c v cc bn ng nghip.

6

Chng 1

l thuyt c bn v sng trn vng bin su v ven b

Sng bin l mt trong cc yu t ht sc quan trng i vi cc hot ng trn i

dng, sng tc ng ln tu thuyn, cng trnh v cc phng tin trn bin. i vi

vng ven b, sng li cng tr nn quan trng. Sng l yu t c bn quyt nh n a

hnh ng b, n vic thit k cc cng trnh cng, lung ra vo cng v cc cng trnh

bo v b bin. Sng to ra cc dng vn chuyn trm tch dc b v ngang b lm thay

i a hnh y.

Sng l qu trnh thay i mt nc tun hon gia cc nh v bng sng. Hng

truyn sng c xc nh l hng truyn ca cc sng n. M phng dng chuyn

ng ca mt nc khi c sng ht sc kh khn do cc sng n tc ng qua li ln

nhau. Cc sng truyn nhanh hn s ui kp cc sng truyn chm v c th kt hp

thnh mt sng. Nh vy cc sng i khi s tng ln hoc b mt i do s tng tc gia

chng. Sng gi khi ra khi vng gi thi s n nh dn v tr thnh cc sng u hn -

sng lng. Nng lng sng b tiu hao trong bn thn khi nc, trong qu trnh tng

tc gia cc sng v trong qu trnh sng . Khi truyn vo vng ven b nng lng

sng cn b mt mt do ma st y. vng st b, mt ngun nng lng rt ln ca

sng s tc ng n b bin. Ngoi ra nng lng sng cng c th chuyn thnh nhit

nng trong qu trnh trao i ri trong khi nc khi sng hoc di tc ng ca

ma st y. Trong khi nhit nng khng c nh hng g ln th c nng (sng , p lc

sng) li ht sc quan trng i vi b bin v cc cng trnh trn bin. Nh vy vic

thit k cc cng trnh bin ph thuc rt nhiu vo chnh xc ca cc tham s sng.

D bo, d tnh trng sng thng c thc hin cho cc sng n, sau s dng cc

dng phn b nhn c trng sng thc t.

Vic nm vng cc l thuyt c bn ca chuyn ng sng l thc s cn thit cho

nghin cu cc m hnh sng vng ven b, phc v cho cc cng tc lp k hoch, thit k

xy dng v qun l vng ven b ni ring v vng bin ni chung.

1.1 Cc yu t sng, dng sng v phn loi trng sng

1.1.1 Cc yu t sng bin

Dao ng tun hon ca mt nc qua v tr mc nc trung bnh gi l sng. M

phng mt nc chuyn ng c th thc hin di dng mt sng - sng n hoc mt

nc chuyn ng ca nhiu sng - sng hn tp. Sng hnh sin hoc sng iu ho l cc

th d v sng n v b mt ca n c th m phng qua hm sin hoc cosin. Mt sng

chuyn ng so vi mt im c nh gi l sng tin, hng m sng chuyn ng ti gi

l hng truyn sng. Nu mt nc ch n thun dao ng ln xung gi l sng ng.

Nu trong chuyn ng sng mt nc c m phng bng qu o khp kn hoc gn

khp kn i vi mi chu k sng gi l dao ng hoc ta dao ng. nh ngha cc yu

t sng c nu ti bng 1.1

7

Bng 1.1 Cc yu t sng

Cc yu t sng K hiu nh ngha

Chu k sng T Thi gian mt nh v mt bng sng i qua mt im c nh

Tn s sng f =1/T: S dao ng trong mt giy

Tc pha C =L/T: Tc chuyn ng ca mt sng

di (bc) sng L Chiu di ca hai nh hoc hai bng sng k tip

cao sng H Khong cch thng ng gia nh v bng sng k tip

su d Khong cch t y bin n mt nc trung bnh

Lin h gia tc truyn sng, chiu di sng v chu k sng:

T

LC (1.1)

L

dgLC

2tanh

2 (1.2)

L

dgLC

2tanh

22

L

dgCTC

2tanh

22 ;

L

dgTC

2tanh

2 (1.3)

Gi tr L

2 gi l s sng (k) -s bc sng trong mt chu trnh sng.

Gi tr T

2 gi l tn s vng ca sng - s chu k sng trong mt chu trnh sng.

T (1.1) v (1.3) ta c:

L

dgTL

2tanh

2

2

(1.4a)

Tnh gn ng

)4

tanh(2 2

22

gT

dgTL

(1.4b)

Cng thc (1.4b) thun tin trong s dng v c chnh xc ph hp vi cc tnh ton

k thut. Sai s cc i khong 5% khi 12

L

d.

1.1.2 Dng sng bin

Dng sng biu th hnh dng ca mt nc khi c sng. Trn thc t, ph thuc vo

cc iu kin khc nhau (v d vng nc su, nc nng, vng gi thi vv..) sng s c

cc dng khc nhau v tnh cht sng cng c th khc nhau (sng iu ho v khng

iu ho). Dng sng n gin nht l sng tuyn tnh, i khi cng c cc tn gi khc

nh sng Airy, sng hnh sin, sng Stokes bc mt. Phng trnh m t dng ca mt

8

nc t do khi c sng l mt hm ca thi gian t, khong cch x i vi sng hnh sin c

dng:

tkxHT

t

L

xH

T

t

L

xa

cos

2

22cos

2

22cos (1.5)

Phng trnh (1.5) m t chuyn ng ca sng tin theo hng tng ca trc x, nu sng

truyn theo hng ngc li ta c du dng trong ngoc. Khi T

t

L

x 22 tin ti cc gi

tr 0, /2, , 3/2 ta c tin ti H/2, 0, -H/2, v 0. Hnh 1 v s cc yu t sng i vi

dng sng tin hnh sin.

Hnh 1.1 Cc yu t sng i vi dng sng tin hnh sin

1.1.3 Phn loi sng bin

Sng trn bin c th phn loi theo ngun gc, bn cht hin tng, cao, su, t s gia bc sng v su vv..

a. Phn loi sng theo ngun gc, hin tng

Sng gi l sng chu nh hng ca gi sinh ra n, sng lng l sng vt ra ngoi vng tc ng ca gi, cng tng t nh vy c th xc nh cc loi sng theo ngun gc sinh ra n. Bng 2.1 trnh by phn loi sng theo ngun gc, hin tng.

Bng 1.2. Phn loi sng theo ngun gc, hin tng

Hin tng Nguyn nhn Chu k

Sng gi Lc ko ca gi n 15s

Sng lng Sng gi truyn i n 30s

Sng Seiche p v gi 2-40 pht

Sng Surf beat Nhm sng 1-5 pht

Sng cng hng trong cng Tsunami, Surf beat 2-40 pht

Tsunami ng t 5-60 pht

Thu triu Lc ht ca mt trng, mt tri 12-24 gi

Nc dng Lc ko ca gi, gim p 1-30 ngy

9

b. Phn loi sng theo cao

Theo cao sng, c th phn loi sng theo t s gia cao v di sng ( dc)

v cao sng vi su bin. Sng c gi l c cao v cng nh khi dc nh

H/L0 v t s gia cao sng vi su bin nh H/d0. Sng c cao hu hn khi

khng tho mn mt trong hai iu kin trn.

c. Phn loi sng theo vng sng truyn, pht sinh

Theo t s gia su vi di ca sng c th phn ra 3 vng sng lan truyn

hoc pht sinh.

Bng 1.3 Phn loi sng theo vng sng truyn, pht sinh

Phn loi d/L 2d/L tanh(2d/L)

Nc su >1/2 > 1

Bin dng 1/25 - 1/2 1/4 - tanh(2d/L)

Nc nng

10

sng vi cc bc cao hn c l thuyt trocoit (Gerstner - 1802) m phng dng sng c

hnh trocoit ng vi sng c bin hu hn. L thuyt Stokes bc cao cng ng vi sng

c bin hu hn. L thuyt sng cnoidal c Korteweg v De Vries xut nm

1885, m phng dng sng gn vi thc t hn trong vng nc nng. Tuy nhin p dng

l thuyt ny trong cc tnh ton thc t rt kh v thng c tnh sn thnh cc

bng. i vi sng vng nc nng, thun tin hn khi s dng l thuyt sng solitary.

L thuyt sng tuyn tnh gi l l thuyt sng Stokes bc 1, cc l thuyt sng

Stokes bc cao c p dng cho vng ven b khi bin sng tr nn ng k so vi

di sng v su. Trong l thuyt sng tuyn tnh p dng cc gi nh sau:

- Cht lng ng nht v khng nn, do vy mt nc khng i,

- B qua sc cng mt ngoi,

- B qua tc ng ca lc Coriolis i vi trng sng,

- p sut trn mt nc c coi l ng nht v khng i,

- Cht lng c coi l l tng khng nht,

- Sng khng tng tc vi cc chuyn ng khc trong cht lng. Dng chy trong sng

khng xoy, do vy qu o ht nc trong chuyn ng sng s khng xoy (ch tnh n

cc thnh phn lc vung gc b qua cc thnh phn tip tuyn).

- y bin bng phng theo phng ngang v c nh, khng thm. iu ny c ngha l

tc thng ng ti y b trit tiu.

- Bin sng nh v dng sng bt bin theo thi gian v khng gian.

- Trng sng hai chiu sng c nh di v tn.

Gi nh khng xoy trong chuyn ng sng cho php chng ta p dng hm th tc

. Hm th tc l i lng v hng vi gradient ca n theo trc x v z ti tt c

cc im ca cht lng l vect tc .

x

U

;

zW

(1.7)

vi: U, W l cc thnh phn tc cht lng theo trc x v z.

Hm c n v l m2/s. Nh vy nu bit hm th tc (x,z,t) trn ton min, c th

xc nh cc thnh phn tc qu o U v W.

Gi nh cht lng khng nn c ngha l ch c mt hm dng duy nht l hm

trc giao ca hm th tc . Cc ng ng hm th v cc ng ng hm dng

vung gc vi nhau. Nh vy nu bit c th tm c hoc ngc li, s dng cc

biu thc sau:

zx

;

xz

(1.8)

Biu thc (1.8) gi l iu kin Cauchy-Riemann (Whitham 1974, Milne-Thompson

1976). C v tho mn phng trnh Laplac i vi dng chy trong cht lng l

tng (tham kho chng 2 ca gio trnh sng bin).

11

Vi cc gi nh nu trn, phng trnh m phng mt sng tuyn tnh sng hnh

sin, l mt hm ca thi gian t v khong cch truyn sng x c dng :

cos2

cos2

22cos

2

Htkx

H

T

t

L

xH

(1.9)

vi: - bin i cao mt nc so vi mc nc bin trung bnh khi lng sng,

H/2 - bin sng (a).

Biu thc (1.9) biu th s lan truyn ca sng tin, tun hon hnh sin, lan truyn

theo hng trng vi hng dng ca trc x. Khi sng lan truyn theo hng ngc li,

du tr trong biu thc pha sng c thay bng du cng. Khi pha sng t cc gi tr

0, /2, , 3/2 cc gi tr mt nc s l H/2, 0, H/2 v 0 tng ng.

Chng 2 mc (2.1) s cp n cc yu t ca trng sng khi truyn vo vng

ven b trn c s l thuyt sng tuyn tnh, ni dung ca phn ny s tp trung chi tit

vo cc yu t sng ng vi cc l thuyt sng bc cao. i vi cc l thuyt sng ny,

phng trnh m phng tng qut mt sng c dng:

ndLBadLBadLBaa nn cos,..3cos,2cos,cos 3

32

2 (1.10)

vi: a=H/2 i vi sng bc 1 v 2; a0.01) hay t s

gia cao sng v su ng k (H/d>0.1) th l thuyt sng tuyn tnh bin nh

khng cn m phng gn ng c trng sng vi chnh xc cn thit na. Trong

trng hp ny phi p dng l thuyt sng Stokes bc cao i vi sng ngn - khi di

sng nh hn su, hay phi p dng l thuyt sng solitary hoc sng cnoidal khi

di sng ln hn su.

a. L thuyt sng ngn

L thuyt sng ngn c p dng i vi cc sng Stokes bc cao. V d phng

trnh mt nc c sng Stokes bc hai c vit di dng:

tkxkhkhkHtkxH 2coscothcoth316

cos2

32

21 (1.11)

Hnh 1.2 a ra hai dng sng tuyn tnh (Stokes bc 1) v sng ngn (Stokes bc 2).

Trn hnh ny chng ta thy bng sng ngn tr nn bng hn so vi sng tuyn tnh,

trong khi sn sng li tr nn dc hn v nh sng vn cao hn. Dng sng ngn

ny thng quan trc thy trn bin trong cc trng hp sng truyn vo vng ven b

c su nh hoc sng chu tc ng ca gi mnh.

Trong phng trnh thnh phn tc sng ngn theo hng truyn sng x, ngoi

cc thnh phn tun hon nh i vi sng tuyn tnh, xut hin thnh phn vn

chuyn theo x biu th s vn chuyn khi lng nc cng nh nng lng sng theo

hng truyn sng qua mi chu k sng gi l dng chy Stokes.

12

Hnh 1.2 So snh sng Stokes bc mt (tuyn tnh) v sng ngn (Stokes bc 2)

b. L thuyt sng di

Ti vng st b, khi su nh hn rt nhiu so vi di sng, cn p dng l

thuyt sng di. Phng trnh lan truyn sng di c dng:

2

22

2

2

xC

t

(1.12)

vi: gdC

Nu l t s gia cao sng v su ( = H/d) v l t s gia su v di

sng (=d/L), ta c cc trng hp sau:

- < 2 hay UR=HL2/d3 2 hay UR=HL2/d3 >> 1

Phng trnh vi phn ca mt nc v tc ht nc trong chuyn ng sng i

vi trng hp ny s c tuyn tnh ho di dng:

0

Ud

xt

(1.15)

0

xg

x

UU

t

U (1.16)

Cc phng trnh trn m t qu trnh phn tn bin sng v tc pha ca sng

trong trng hp ny s l .)( dgC

- = 2 1 hay UR=HL2/d3 1

13

Cc phng trnh trn chuyn thnh dng phng trnh Boussinesq:

03

13

33

xdUd

xt

(1.17)

0

xg

x

UU

t

U (1.18)

Trong trng hp c bit, sng di truyn theo mt hng x cho trc nhn c

phng trnh Korteweg De Vries:

06

1

2

313

32

xd

xdxtgd

(1.19)

C hai dng sng di vng ven b da trn c s l thuyt sng nu trn l sng

solitary v sng cnoidal.

1.2.3 L thuyt sng solitary

Sng solitary l loi sng tin c mt nh v bng duy nht (nh bn thn tn gi

ca loi sng ny), do vy y khng phi loi sng tun hon (khng c chu k v di

sng) nh chng ta nghin cu trn. Cc c trng ca sng solitary c J.

Scott Russel ln u tin m t vo nm 1844. Nm 1872 Boussinesq a ra c s l

thuyt ca sng solitary. Phng trnh m t chuyn ng ca nh sng solitary nh

sau:

d

x

d

HhHs

4

3sec 2 (1.20)

Trong mt sng s l to thng ng ca mt bin khi c sng so vi mc nc

trung bnh khi lng sng, cch to ti nh sng (x=0; s =H) mt khong cch x.

Tc pha ca sng solitary c xc nh theo:

....

20

3

2

11

2

d

H

d

HgdCs (1.21)

Chng ta thy rng tc ny ln hn so vi tc pha ca sng tuyn tnh ti vng

nc nng (2.7). Cng thc (1.21) c th cho cc kt qu gn ng nh sau:

Hdgd

HgdCs

1 (1.22)

Khi sng solitary truyn vo vng ven b c su gim, cao sng s tng v n

mt su nht nh mt sng s tr nn khng n nh v sng s . S khng n nh

ca mt sng cng s t c khi tc ht nc trong chuyn ng sng tng ng

vi tc pha. ng thi gc ca mt nc ti nh sng cng b gii hn bi ch tiu

1200. S dng cc ch tiu trn McCowan (1894) chng minh bng l thuyt ch tiu

sng i vi sng solitary.

78.0)( max d

Hb (1.23)

14

Tng nng lng ca sng solitary bao gm hai thnh phn, th nng v ng nng

gn nh bng nhau. Tng nng lng cho mt n v di nh sng s l:

32/3

33

8d

d

HgEsol

(1.24)

Tc ngang v thng ng ca ca ht nc trong sng solitary c xc nh theo

cc biu thc sau:

2/cosh/cos/cosh)/cos(1

dMxdMz

dMxdMzNCU s

(1.25)

2/cosh/cos/sin)/sin(

dMxdMz

dMxdMzNCW s

(1.26)

vi M v N l cc hng s do Munk a ra nm 1949 (xem hnh 1.3).

Hnh 1.3 Cc hng s M, N trong cng thc tnh tc ht nc trong chuyn ng sng solitary

Sng solitary l sng chuyn ti, c ngha l cc ht nc trong chuyn ng sng loi ny ch chuyn ng duy nht v pha trc, khng tn ti cc pha chuyn ng v pha sau (nh i vi sng tuyn tnh). Gi s chng ta quan trc sng solitary ti mt im, khi nh sng cch v tr khong 10 ln su cc ht nc bt u chuyn ng theo hng truyn sng x v ln pha trn. Vn tc ca ht nc t gi tr cc i ti v tr quan trc khi nh sng i qua. Sau khi nh sng i qua, ht nc s chuyn ng tin i xung v t ti v tr ban u. Nh vy sng solitary s gy chuyn ng tnh ca khi nc theo hng truyn sng. Lu lng nc ny cho mt n v nh sng tng ng vi khi lng nc ca sng solitary trn mc nc trung bnh khi lng sng v c xc nh nh sau:

2/1

2

3

14

d

HddxQ (1.27)

Gn nh ton b khi lng nc tp trung ti khu vc gn nh sng. i vi sng H/d=0.40, 90% lng nc trn tp trung trong vc x = 2.7d v cng mt phn trm nu trn ca nng lng sng tp trung trong khu vc x = 1.7d. V gn nh ton b nng lng sng tp trung ti khu vc gn nh sng, sng solitary c th c p dng i vi

15

trng sng thc t khi truyn vo st b. Khu vc ngoi ra ca nh sng solitary khng ng vai tr quan trng, do vy c th coi trng sng thc t l tp hp mt chui cc sng Solitory c nh lin tip i qua mt im, b qua s tng tc ca cc sng ny ti ra cch xa cc nh. xc nh c di ca cc sng solitary n c trong chui sng sao cho ln hn di hiu dng ca sng solitary c th t c chnh xc cho php khi b qua s tng tc ca cc sng ny ti ra cch xa cc nh. T c th xc nh c chu k sng thc t T phi ln hn gi tr chu k sng hiu dng (Bagnold 1947):

g

d

MTeff

2 (1.28)

Khi tin vo gn b, do nh hng ca dc y bin s lm bin i cc yu t ca

sng solitary nh bin , tc , dng sng so vi cc tnh ton l thuyt. iu ny lm

gim kh nng vn dng l thuyt sng ny trong cc tnh ton sng vng ven b.

1.2.4 L thuyt sng cnoidal

Sng cnoidal c Korteweg v De Vries nghin cu nm 1985. Li gii tng

qut ca phng trnh (1.19) l phng trnh dao ng sng vi chu k T v di L:

,22T

t

L

xKHcn (1.29)

vi: K() - tch phn ton phn bc nht ca module ,

- cao ca mt sng so vi v tr bng sng ti v tr to ngang x,

hm cn(r) l Jacobian ca hm elliptic (r).

Hnh 1.4 Vng p dng cc loi l thuyt sng

Sng cnoidal l loi sng tun hon c nh nhn v bng rt bng, ph hp vi

trng sng pha ngoi vng sng . im yu ca l thuyt sng ny l ng dng cc

hm ton hc phc tp, rt kh p dng trong thc t. Hnh 1.4 v cc vng p dng cc

16

l thuyt sng. Sng cnoidal p dng khi H/L26. Hnh 1.5 v dng cc

sng Airy, Stokes, cnoidal v solitary

Hnh 1.5 Dng cc sng Airy, Stokes, Cnoidal v Solitary

1.3 tc ng v tng tc ca trng sng vi cc qu trnh thu

thch, ng lc ven b

1.3.1 Tc ng v tng tc ca trng sng vi cc qu trnh ven b

Khi truyn vo vng ven b sng s chuyn ti mt ngun nng lng ln. Ngun nng lng ny c th di dng sng b mt nhit nng do qu trnh ri trong chuyn ng ca cc ht nc khi sng , hoc nhit nng truyn cho y bin do ma st v thm. Ngoi ra ngun nng lng do sng sinh ra di tc ng c hc i vi y bin khi sng truyn vo vng c su nh, khi sng v khi sng tc ng n cc cng trnh trn bin s ng vai tr c bit quan trng do n tc ng n y bin, b bin v n cc cng trnh nhn to vng ven b. Sng l yu t c bn quyt nh n a hnh ng b, n vic thit k cc cng trnh cng, lung ra vo cng v cc cng trnh bo v b bin. Sng to ra cc dng vn chuyn trm tch dc b v ngang b lm thay i a hnh y. Ngoi cc cu trc vi m ca b bin lun gn lin vi cc c trng trng sng, ti bt c mt vng b bin no trn th gii, chng ta cn thy rng, ng lc sng quyt nh n cc dng b bin trn tt c cc vng bin h, chu tc ng trc tip ca trng sng vng bin khi, i dng. Lewis (1938) nhn xt rng b bin lun c xu th pht trin vung gc vi cc hng sng thnh hnh. Silvester v Ho (1972) a ra dng b bin cn bng kiu ng cong logarit hoc ng cong trng li lim ti cc vnh. Cc loi ng cong ny c hng theo hng tc ng ca trng sng lng thnh hnh t i dng truyn n. Sng v dng chy do sng cng l nguyn nhn to ra cc yu t b bin a phng nh cc mi nh ra pha sau cc o

17

chn cc hng sng chnh hoc cc tombolo ni cc o vi khu vc t lin pha sau, c o che chn.

i vi nc ta trng sng ng mt vai tr c bit quan trng trn sut hn 3000 km ng b bin. Ch sng trong gi ma v c bit trong bo quyt nh mi hot ng trn ton vng bin v c bit l ti cc vng ven b. Nn kinh t ca chng ta ch yu da vo nng nghip, tp trung vo hai khu vc chu th ng bng sng Cu Long v ng bng sng Hng. c im ca hai vng chu th ny l cc vng t thp, rt d b tc ng ca nc dng, sng. Ngoi ra i vi cc cng trnh khai thc du kh vng khi v ven b pha nam, trng sng cng l yu t quan trng bc nht, quyt nh n mc kinh ph u t xy dng cng trnh khai thc thm d v n sn lng khai thc hng nm. Cc vng xi l b nghim trng phn b hu nh trn ton di ven b pha ng nc ta nh vng Hi Hu, vng ca Thun An, vng G Cng, vng Gnh Ho v nguyn nhn ca xi l l nh hng ca trng sng. Trong khi , trng sng cng gy vn chuyn trm tch, sa bi ti cc cng, lung lch ra vo cng v ca sng, lm nh hng n giao thng ng thu nh khu vc ca Nam Triu, cng Hi Phng, khu vc ca nh An v lung ra vo ca dn n cng Cn Th vv..

C th thng k s b nh hng v tng tc ca sng bin i vi cc qu trnh thu thch ng lc ven b sau:

a. Trng sng lm thay i phn b nhit mui trong nc bin, thay i phn b cc yu t ho bin theo su vo theo khng gian.

b. Trng sng lm thay i cc c tnh quang hc ca nc bin, thay i mu sc, trong sut ca nc bin.

c. Trng sng lm thay i tc v hng truyn m trong nc bin.

d. Trng sng tc ng n cc cng trnh bin vng khi v ven b.

e. Trng sng tc ng n b bin, gy bin ng b bin: xi l v bi t.

f. Trng sng tc ng n y bin vng ven b, gy bin ng y bin, bi lp cc knh ra vo cng, ca sng.

g. Trng sng gy dng chy ven b v dng vn chuyn trm tch, l nguyn nhn gy tc ng ca cc cng trnh ven b n cc vng ln cn. To ra cc loi mi t, tombolo a phng.

Chnh v ngha quan trng ca trng sng i vi cc vng bin su v ven b nn vic nghin cu l thuyt v thc nghim v sng bin c mt lch s lu i nht so vi cc yu t hi dng hc khc. Lch s nghin cu sng bin c trnh by kh chi tit trong gio trnh [1].

1.3.2 Cc vng tc ng ca trng sng v cc yu t a mo ven b.

Cn thit phi bt u nghin cu sng vng ven b bng vic xc nh cc vng tc ng ca trng sng khi truyn t vng khi vo ven b, cc thut ng v c ch vt l ca qu trnh. Thng thng do trng sng c lin quan trc tip n cc yu t a hnh, a mo vng ven b do n sinh ra nn vic phn chia cc vng tc ng ca trng sng lun i i vi phn chia cc yu t a mo ven b (cc bar y bin, g sng, vch b bin vv..).

a. Vng tc ng ca trng sng

Trn hnh 1.6 v cc vng tc ng ca trng sng khi truyn t vng khi vo ven

b.

18

Hnh 1.6 Cc vng tc ng ca trng sng ven b

- Vng ngoi khi l vng t im sng ra khi,

- i sng (ngha rng) l vng t gii hn ngoi ca vng sng v gii hn pha

trong ca vng sng v b. i sng (ngha hp) l vng t im sng n gii hn

pha ngoi ca vng sng v b.

- Vng bin dng l vng k t khi sng bt u chu nh hng ca y (d 1/2L) n

im sng .

- im sng l v tr ti sng t cao cc i v bt u .

- im sng b nho l v tr ti sng b ph hu hon ton khi nh sng b xung

mt nc pha trc.

- Vng sng l khu vc t gii hn ngoi ca i sng v im sng b nho.

- Vng sng v b l vng c gii hn pha trong cng v pha b do sng dn ti v

khu vc xo trn mnh gia nc rt ra v sng dn vo b.

- Vng sng leo l vng bt u t v tr ti sng bt u b cun ln bi v v tr gii

hn trong cng v pha b.

b. Cc yu t a mo v trng sng ven b

Nh trong cc phn trn chng ta thy rng trng sng c lin quan trc tip n

cc yu t a mo ven b do sng to ra, do vy vic phn cc vng tc ng ca trng

sng thng i i vi thng k cc yu t a mo ven b. Hnh 1.7 nu cc yu t a

mo c trng vng ven b trn mt ct vung gc vi b.

- Bar ngm dc b, thng xut hin ti v tr sng v sng b nho do ti y l khu

vc hi t ca dng vn chuyn trm tch ngang b vi hai hng, pha ngoi bar l

hng t khi vo b cn pha trong bar dng ny c hng t b ra.

- Bng ca bar ngm dc b to thnh lung su dc b.

- Mt bi bin l khu vc dc v pha bin ca bi bin lung lun hng chu tc ng x

b ca sng.

19

- G sng l mp gia bi bng phng pha trong t lin v sn dc pha ngoi do sng

to ra.

- Vch b bin l sng thng ng ca b bin do xi l to ra.

- ng b l ng tc ng tng tc ca t lin v nc*.

- Bc ngm l vch thng ng ngm di mt nc.

- n ct l cc lung ct ngay st b bin do gi to ra

Hnh 1.7 Cc yu t a mo ven b

* nh ngha ng b theo khi nim nu trn p dng chung trong trng hp mc nc n nh, khng i.

Ti cc vng chu tc ng ca thu triu khi nim ng b bin c m rng thnh ng b bin biu kin.

y l ng gianh gii gia mc nc trung bnh khi triu cng v bi bin, c th xc nh gn ng cc

vng c cc loi cy, thm thc vt ven bin bng gii hn pha ngoi bin ca di cy, thm thc vt (Ellis,

1978); tham kho thm trong Cm nang Cng ngh Ven bin 2001.

20

Chng 2

Bin i cc yu t sng khi truyn vo vng ven b

2.1 Tc , di v cc yu t khc ca chuyn ng sng vng ven

b

2.1.1 Tc v di sng vng ven b

Trong l thuyt sng trochoid, khi xt quy lut bin i ca p sut sng ti mt bin

su ta c:

1cos)(2

1 2020

20 Ckgk

rr

p

(2.1)

vi: r0 - bn knh qu o sng trn mt bin,

- tn s vng ca sng T

2 ,

k - s sng L

k2

,

- pha sng = kx - t.

Ti mt bin, khi khng xt tc ng ca gi c th coi p sut sng khng thay i

v khng ph thuc vo pha sng. tho mn iu kin ny, thnh phn th hai trong

v phi ca (2.1) phi b trit tiu c ngha l:

02 kg (2.2)

hay

2

22

2

2

T

LgL

k

kg

k

Theo nh ngha cc yu t sng ta c T

LC t rt ra:

2

2 gLC ti vng nc su.

vng bin dng, biu thc quan h gia tc truyn sng vi di sng v su c

dng:

L

dgLC

2tanh

2 (2.3)

vi:d - su bin.

Biu thc (2.3) cng c gi l h thc phn tn, n ch ra rng cc sng c chu k

khc nhau s chuyn ng vi cc tc khc nhau. Nu sng bao gm tp hp cc sng

n khc nhau, cc sng n c chu k ln hn s chuyn ng nhanh hn.

T (2.3) v nh ngha cc yu t sng (C =L/T) s nhn c:

)2

tanh(2 L

dgLC

(2.4)

21

hay: )2

tanh(2

2

L

dgTL

(2.5)

- Xp x gn ng cc hm hypecbol

Cc vng nc su, bin dng v nc nng, trong ng lc sng c biu th qua t

s gia su v di sng (d/L) hay l su tng i trong chuyn ng sng. Cc

biu thc lin h gia tc sng, chu k sng v di sng (2.3, 2.4) ph thuc vo cc

hm hypecbol ca su tng i. Bng 2.1 a ra cc xp x gn ng cc hm

hypecbol trong cc vng khi sng truyn t vng nc su vo vng ven b.

Bng 2.1 Xp x gn ng cc hm hypecbol

Hm

Biu thc

Xp x gn ng cho

cc bin ln

ee

Xp x gn ng cho cc

bin nh

1;1 ee

sinh

cosh

tanh

2

ee

2

ee

ee

ee

e2

1

e2

1

1

1

Vng p dng Bin dng Nc su Nc nng

Chng ta s s dng cc k hiu C0, C, Cs v L0, L, Ls ch tc pha v di ca

sng vng nc su, vng bin dng v vng nc nng. i vi vng nc su, su

tng i d/L0 ln ( 12

tanh0

L

d). T (2.4) v (2.5) ta c:

20gT

C hay 2

2

0

gTL (2.6)

Trong vng nc nng, su tng i nh (ss L

d

L

d 22tanh ). T (2.3) ta c:

gdL

dgLC

s

ss

2

2 (2.7)

Da vo su tng i lp ra bng phn loi sng theo cc vng nc su, vng

bin dng v vng nc nng (bng 2.1).

2.1.2 Tc qu o v gia tc ht nc trong chuyn ng sng

Thnh phn ngang v thng ng ca tc ht nc c dng:

T

t

L

x

Ld

Ldz

L

gTHU

22cos

/2cosh

/2cosh

2 (2.8)

22

T

t

L

x

Ld

Ldz

L

gTHW

22sin

/2cosh

/2sinh

2 (2.9)

(2.8) v (2.9) l cc biu thc tc ca ht nc trong chuyn ng sng ti cc v tr

(d+z) so vi y. Tc ca ht nc l mt hm tun hon theo x v t. i vi mt gc

pha cho trc T

t

L

x

22 cc hm cosh v sinh s ph thuc vo z di dng lu tha,

biu th s gim tc theo hm lu tha khi xung su di mt nc. Tc ht nc

theo chiu ngang t cc i theo hng dng khi = 0, 2 v t cc i theo hng

m khi = , 3. Tc theo chiu thng ng t cc i theo hng dng khi = /2,

5/2 v ngc li t cc i theo hng m khi = 3/2, 7/2 (xem hnh 2.1).

Gia tc ht nc s nhn c bng cch ly o hm ca tc theo thi gian t:

T

t

L

x

Ld

Ldz

L

Hgax

22sin

/2cosh

/2cosh (2.10)

T

t

L

x

Ld

Ldz

L

Hgay

22cos

/2cosh

/2sinh (2.11)

Hnh 2.1 v tc v gia tc ca ht nc trong chuyn ng sng. T hnh 2.1 ta

thy cc ht nc pha trn mt nc trung bnh khi c sng chuyn ng theo hng

truyn sng v cc ht nc pha di truyn theo hng ngc li.

Hnh 2.1 Tc qu o v gia tc ht nc trong chuyn ng sng

2.1.3 Qu o chuyn ng sng

Qu o ca cc ht nc trong chuyn ng sng thng l hnh trn (vng nc

su) v ellip (vng bin dng v nc nng). Tch phn (2.8) v (2.9) theo x v d ta nhn

c s dch chuyn theo phng ngang v phng thng ng.

T

t

L

x

Ld

Ldz

L

HgT

22sin

/2cosh

/2cosh

4

2

(2.12)

23

T

t

L

x

Ld

Ldz

L

HgT

22cos

/2cosh

/2sinh

4

2

(2.13)

Ta c : L

d

L

g

T

2tanh

222

suy ra:

T

t

L

x

Ld

Ldz

L

H

22sin

/2sinh

/2cosh (2.14)

T

t

L

x

Ld

Ldz

L

H

22cos

/2sinh

/2sinh (2.15)

Cc biu thc (2.14) v (2.15) c vit li di dng:

2

2

/2cosh

/2sinh22sin

Ldz

Ld

aT

t

L

x

2

2

/2sinh

/2sinh22cos

Ldz

Ld

aT

t

L

x

Cng cc v ca h phng trnh trn vi nhau ta c:

12

2

2

2

BA

(2.16)

y l phng trnh ellip vi bn knh trc ln A (ngang) v bn knh trc nh B

(thng ng):

Ld

LdzHA

/2sinh

/2cosh

2

(2.17)

Ld

LdzHB

/2sinh

/2sinh

2

(2.18)

Nh vy theo l thuyt sng tuyn tnh, ht nc trong chuyn ng sng to thnh

qu o khp kn - sau mt chu k sng ht nc s tr v trng thi ban u. Trn thc

t khng hon ton nh vy, ht nc khng to thnh mt qu o khp kn v iu

ny gy ra vn chuyn vt cht.

Theo (2.17), (2.18) vng nc su ta c A=B: qu o ht nc trong chuyn ng

sng to thnh hnh trn:

LzeH

BA /22

vi d/L>1/2 (2.19)

Vng nc nng:

d

LHA

22

d

dzHB

2 vi d/L

24

Bin dao ng sng gim vi hm m theo su. Ti vng nc su su z=

L0/2 ta c A= B= H/2e- = H/2(0.04) (bng khong 4% bin trn mt nc). Ht nc

chuyn ng nh nht (0) ti y v cc i trn mt nc, bng mt na cao sng.

Hnh 2.2 v qu o chuyn ng sng vng nc su v vng ven b.

Hnh 2.2 Qu o chuyn ng sng vng nc su v ven b

2.1.4 p sut sng

T phng trnh Bernoulli cho th vn tc trong chuyn ng sng ta c:

02

1 22

tWUgz

P

(2.21)

vi l th vn tc trong chuyn ng sng (z

Wx

U

; ). Trong (2.21) p sut bao

gm c p sut thu tnh (-gz).

Nu ch ch n bin ng p sut do sng ta s c:

P

gzP

gzPP

Thay vo (2.21) ta c:

02

1 22

tWU

P

(2.22)

vi H/L rt nh ta c:

25

t

P

(2.23)

vi:

T

t

L

x

Ld

LdzHC

22sin

/2sinh

/2cosh

2

Thay t

vo (2.21) ta c:

T

t

L

x

Ld

LdzHgP

22cos

/2cosh

/2cosh

2 (2.24)

vng nc su:

T

t

L

xe

HgP Ld

22cos

2/2 (2.25)

p sut gim theo su theo quy lut hm m (e kd).

Nh vy P s t l vi cao sng H. Da trn nguyn tc ny ngi ta thit k cc

my o sng theo nguyn l o p sut ti tng su. Mng cm ng p sut c t

tng st y. Lc cao sng trn mt bin s c tnh theo:

La

Ld

g

PH

/2cosh

/2cosh

vi: P - dao ng p sut o c,

a - cao ca mng o p so vi y.

2.1.5 Tc nhm sng

Trn thc t mt bin c sng bao gm nhiu sng c cao, chu k v pha khc

nhau, do vy xut hin tc nhm sng. Tc ca tng sng ring bit (tc pha) C

s khc vi tc ca nhm sng Cg. vng nc su hoc vng bin dng, tc ca

nhm sng s nh hn tc ca tng sng C > Cg. din gii tc nhm sng, xt s

tng tc gia hai sng hnh sin 1 v 2, c cng cao v chuyn ng theo cng mt

hng vi s khc nhau rt t v di sng v chu k. Phng trnh mt bin c dng:

2211

21

22cos

2

22cos

2 T

t

L

xH

T

t

L

xH (2.26)

Do L1 rt gn vi L2, vi mt khong x no tng ng vi thi gian t, hai sng ny

s trng pha nhau v cao sng tng cng s l 2H, v ngc li s c thi im khi hai

sng ny ngc pha nhau v cao mt nc tng cng s b trit tiu. Hnh 2.3 m t

qu o v ng bao ca tng hai sng nu trn. Phng trnh ng bao c dng:

t

TT

TTx

LL

LLHbao

21

12

21

12cos (2.27)

Tc chuyn ng ca ng bao l tc ca nhm sng:

26

nCLd

Ld

T

LCg

/4sinh

/41

2

1

(2.28)

Hnh 2. 3 Nhm sng v ng bao

vi:

Ld

Ldn

/4sinh

/41

2

1

vng nc su:

0/4sinh

/4

Ld

Ld

ta c : 00

2

1

2

1C

T

LCg (2.29)

vng nc nng:

1/4sinh

/4

Ld

Ld

ta c: gdCT

LCg (2.30)

vng nc nng, tt c cc sng u truyn vi mt tc bng nhau, ph thuc

vo su. ngoi khi hoc vng bin dng tc pha ln hn tc nhm. Tc

nhm sng rt quan trng v n biu th tc truyn nng lng ca sng.

2.1.6 Nng lng sng

Tng nng lng sng bao gm ng nng v th nng:

- ng nng c gy ra bi tc qu o ca ht nc trong chuyn ng sng.

- Th nng th hin phn nc pha trn bng sng.

Theo l thuyt tuyn tnh, th nng tng ng vi mc nc trung bnh khi lng

sng. Cc sng chuyn ng theo mt hng th cc thnh phn th nng v ng nng

bng nhau. Nng lng sng cho mi bc sng trn mt n v b rng ca nh sng l:

81616

222 LgHLgHLgHEEE PK

(2.31)

Tng nng lng trung bnh cho mt n v b mt bin - mt nng lng sng, l:

27

8

2gH

L

EE

(2.32)

Thng lng nng lng sng l nng lng sng truyn theo hng truyn sng, qua

mt mt phng vung gc vi hng truyn sng tnh t mt bin n y bin. Thng

lng nng lng trung bnh cho mt n v nh sng, truyn qua mt mt phng vung

gc vi hng truyn sng s c tnh theo:

gCEnCEP (2.33)

P cng c gi l lc sng.

- Ti vng nc su: 0002

1CEP

- Ti vng nc nng: CECEP g

Khi nh sng song song vi cc ng ng su ta c phng trnh cn bng nng

lng sng:

nCECnE 000 (2.34)

Do n0=1/2 suy ra:

nCECE 002

1 (2.35)

Khi nh sng khng song song vi ng ng su, biu thc (2.35) s khng ng v

cc sng s truyn vi cc tc khc nhau (hin tng khc x sng).

2.1.7 Cc phng php tnh di sng vng ven b

Do trong vng bin dng v nc nng, di sng khng th tch ring ra mt v

trong biu thc tnh (2.5), tnh c yu t ny cn thit phi s dng cc phng

php khc nhau:

a, Phng php tra bng:

S dng bng tnh sn di sng v cc tham s sng khc thng qua cc s liu u

vo l cao sng, di sng vng nc su v su ti im cn tnh.

b, Phng php lp:

Tnh di sng theo cc bc sau:

i

iL

dLL

2tanh01 (2.36)

vi i=1, 2, 3, .. Sau so snh gia Li+1 v Li s dng ngng sai s xc nh kt qu tnh.

c, Phng php lp ci tin:

i

iL

dLL

2

012

2tanh

(2.37)

3

2 21222

iii

LLL

(2.38)

vi i =1, 2, 3, ..

28

Sau cng so snh gia L2i+1 v L2i s dng ngng sai s xc nh kt qu tnh.

d, Phng php tnh gn ng:

)2

tanh()2

tanh(0

00L

dL

L

dLL

(2.39)

Cng thc trn thun tin trong s dng v c chnh xc ph hp vi cc tnh ton k

thut. Sai s cc i khong 5% khi 12

L

d.

e, Phng php tnh gn ng PADE

i

id

Akk 0 (2.40)

))))0675.0(0864.0(462.0(6522.0(1

1

00000

0

iiiii

idkdkdkdkdk

dkA

(2.41)

Bng 2.2 a ra cc kt qu tnh bc sng ti su d=50m vi chu k sng T=19

giy. Nu dng cng thc (2.39) ta c L = 401.0 m cho sai s +5.1%. Nu s dng bng

ta c T=19s, d= 50 m suy ra L0 =563.80 m v d/L0 =0.1310 hay L=381.6 m ng vi kt

qu tnh trn bng 2.2.

Bng 2.2 Kt qu tnh di sng theo cc phng php khc nhau

S ln lp

n

Cng thc lp (2.36)

Li (m)

Cng thc (2.37), (2-38)

L2i+2 (m)

0 563.8 378.1

1 285.2 382.0

2 431.6 381.6

3 339.2 381.6

4 410.9

5 362.9

6 394.2

7 373.4

8 387.0

9 378.0

10 384.0

11 380.1

12 382.6

13 380.9

14 382.0

15 381.3

16 381.8

17 381.5

29

2.2 Bin dng sng vng ven b

Khi sng truyn vo vng ven b, cc tham s sng s b bin i do tc ng ca y

bin, do cc sng ct ti y bin, do c im trm tch y bin v cc vt liu y

bin. y bin tc ng ln sng truyn vo vng ven b thng qua cc hiu ng bin

dng, khc x. Ngoi ra, cc cng trnh bin vng ven b s lm thay i cc yu t sng

bi cc qu trnh nhiu x v phn x.

Nu sng truyn thng gc vo vng ven b c cc ng ng su thng v song

song vi ng b, s thay i dng sng xy ra ch do s thay i su, s thay i

ny gi l bin dng sng. Di tc dng ca hiu ng bin dng, u tin cao sng

gim dn sau tng t t, ng thi dng ca sng vn i xng. Vo st b, khi su

gim mnh, cao sng s tng nhanh ng thi dng ca sng tr nn bt i xng:

sn pha trc tr ln dc hn v cui cng s b . nh gi cc yu t sng di tc

dng ca hiu ng bin dng sng ph thuc vo l thuyt m phng trng sng v cc

loi phng php tnh bin dng trng sng. C ba loi phng php tnh ton bin

dng sng l phng php dng nng lng, phng php nhiu ng v phng php

s. Bng 2.3 a ra cc phng php tnh bin dng sng [6]. Hnh (2.4) v h s bin

dng sng theo cc l thuyt sng khc nhau.

2.2.1 Phng php tnh bin dng sng trn c s nng lng sng

Khi su thay i, cao v di ca sng s thay i. Tuy nhin chu k sng s

khng thay i do s cc con sng khng i. Nu cho rng p sut khng i v b qua

nht ca nc, c th thy rng nng lng sng s c bo ton. Trong iu kin

thc t, i vi trng sng n nh, iu kin nng lng s c bo ton khi b qua

dng chy, dng vn chuyn vt cht v tiu tn nng lng. Dng nng lng sng i

vi l thuyt sng bin nh c xc nh theo.

d

x dzucF2 (2.42)

vi du biu th gi tr trung bnh theo chu k sng.

Dng nng lng vng trung gian i vi sng bin nh c tnh theo:

8/2CngHFx (2.43)

i vi vng nc su ta c: n=1/2, C= C0, H= H0

16/020CgHFx (2.44)

H s bin dng c xc nh bng t s gia cao sng ti im tnh v cao

sng vng nc su trong iu kin bo ton nng lng (Fx = const).

L

dnC

C

nH

HKs 2

tanh

1

2

1

2

1 0

0

(2.45)

30

H s bin dng Ks l mt hm ca L

d2 hay ca

0L

d. Khi

0L

d gim, u tin h s bin

dng Ks gim nh hn 1 sau tng mnh. Vi vng rt nng d/L0

31

Hnh 2.4 H s bin dng sng

i vi cc l thuyt sng khc nhau (sng bin hu hn, sng Stokes bc cao) h

s bin dng s c tnh theo cc cng thc khc nhau. H s bin dng sng xc nh

theo (2.45) da trn gi thit l dc y bin rt nh (c s ca phng php nng

lng). i vi y bin dc, bo ton nng lng b ph v v h s bin dng c xc

nh theo cc phng php khc nh phng php nhiu ng hoc phng php s.

2.3 Khc x sng vng ven b

Do tc truyn sng ph thuc vo su, trong vng bin dng, khi sng truyn

vo b s chu nh hng ca su. Nu hng sng cho gc vi ng ng su s to

ra gradient ca tc truyn sng dc theo nh sng. Gradient tc truyn sng ny

lm cho sng thay i hng ng thi cng lm cho cao sng thay i. Hin tng

sng thay i hng khi truyn cho gc vo vng b gi l khc x sng. Theo l thuyt

sng bin nh, tc pha ca sng s l mt hm ca di sng L v su d (2.3).

kdk

gC tanh (2.46)

cao ca mc nc c th vit di dng [6]:

xkxexa txi rrrr r

,)( (2.47)

vi a l bin sng (a = H/2 ; H l cao sng), xr

l vect v tr (x,y) v kr

l vect s

sng vi ln k v c cng hng vi hng truyn sng. Tn s gc ( =2/T trong

T l chu k sng) tho mn h thc phn tn:

32

kdgk tanh (2.48)

Biu thc trn duy tr s lan truyn sng trn y c dc bin i t t. V s sng

kr

gn nh khng bin i trong trng hp cc b ny, h thc kr

= cng gn nh

khng bin i v:

kr

= 0 (2.49)

vi = ( yx /,/ ).

Mt khc, t phn tch hnh hc n gin dn n biu thc biu th hng sng sau:

C

c

1 (2.50)

vi v l cc to dc theo tia sng v ng nh sng nh v trn hnh (2.5).

Tng ng ton hc gia biu thc (2.49) v (2.50) c din gii qua ta chuyn

i v qua vic s dng nh ngha ca vect s sng:

kr

=(kcosx,ksinx),k =| kr

| (2.51)

Hnh 2.5 H to tnh khc x sng

Bin ca sng khc x, a c xc nh trn c s l thuyt bo ton dng nng

lng:

.(E gCr

) = 0 (2.52)

vi: E = ga2/2= gH2/8 l mt nng lng sng,

gCr

= ( kr

/k)nC l vct tc nhm sng.

Cho rng nng lng sng khng truyn ngang cc tia sng (trong mt cp tia sng

nng lng c bo ton), biu thc (2.52) c th vit li di dng:

33

0)(

bEnC

(2.53)

C ngha l dc theo mt cp tia sng t vng nc su (n=1/2) vo vng ven b ta c:

bEnCCEb 0002

1 (2.54)

hay: sr KKbnC

Cb

H

H

a

a.

200

00

vi0b

bKr (2.55)

Trong b0 l khong cch gia hai tia sng vng nc su v b l khong cch gia

hai tia sng vng trung gian. Ks l h s bin dng nu 2.2 v Kr l h s khc x,

biu th hiu ng bin i khong cch gia cc tia sng khi truyn t khi vo b ln

cao sng.

Ta c c th a ra biu thc lin h gia b v (hng truyn sng so vi trc x):

b

b

1 (2.56)

Bng cch th t (2.50) vo (2.56) ta c:

011

2

2

2

2

C

C

b

b (2.57)

Trong h to - ta c:

0coscossin2sin

sincos

2

2

222

2

2

2

2

by

C

yx

C

x

C

b

y

C

x

CbC

(2.58)

C th gii phng trnh (2.53) lin kt vi (2.58) xc nh s bin i cao sng

dc theo tia sng. Trng hp c bit vi a hnh y ng nht, c cc ng ng su

song song vi trc Y, tch phn ca (2.50) v (2.56) cho nh lut Snell:

0

0sinsin

CC

(2.59)

H s khc x trong trng hp ny c dng:

cos

cos 0rK (2.60)

Phng php gii phng trnh vi phn tia sng c thc hin theo (2.50) v (2.58);

khi tnh ton khc x sng theo li vi cc nt c nh s dng gii s theo cc biu

thc (2.49) v (2.52).

Nu tn ti trng dng chy Ur

c tc ng nht t y bin ln mt th h thc

phn tn (2.48) s c thay th bng:

Ukrr

.* kdgk tanh* (2.61)

34

v biu thc (2.52) s tr thnh:

0/. * UCE grr

(2.62)

Biu thc (2.62) biu th rng tc ng sng E/* s c bo ton thay v cho nng

lng sng. Phng trnh chuyn ng ca nc di tc ng ca sng thng qua ng

sut bc x sng Sxx, Sxy v Syy s l:

0/. *

y

VS

x

V

y

US

x

USUCE yyxyxxg

rr (2.63)

vi U v V l cc thnh phn dng chy trn trc x, y ca vect dng chy trung bnh Ur

.

Biu thc (2.63) l phng trnh bo ton nng lng sng dng tng qut (Longuet-

Higgins v Stewart, 1961; Phillips, 1971).

Trong trng hp khc x i vi cc sng khng u, Karlsson (1969) a ra phng

trnh bo ton nng lng dng:

0,,.

UfSCfS g

r (2.64)

Vi S(f,) l hm mt ph; S(f,)dfd l phn nng lng sng trong di tn (f, f+df)

v di hng (, +d). Lng dng nng lng sng qua mt phng vung gc vi

hng sng c tnh trong thnh phn th hai ca (2.64):

cossin

y

c

x

cnU (2.65)

Trong trng hp ring i vi sng n sc, phng trnh (2.64) tr thnh (2.50) v

(2.52).

Khc x sng tc ng ln qu trnh bin i b bin v y bin. Xu th chung l ti

vng c a hnh y li khc x s to nn vng tp trung (hi t) nng lng sng, cn

ngc li ti cc vng a hnh y lm to nn vng phn tn (phn k) nng lng

sng. Kt qu s to nn dng chy do sng vn chuyn vt liu y t cc vng tp

trung nng lng n cc vng phn tn nng lng sng, san bng cc bin ng cho

a hnh y bin vng ven b. Hnh 2.6 v cc trng hp khc x sng vi cc loi a

hnh y khc nhau [4].

2.4 Nhiu x sng do vt cn

Khi sng truyn vo cc vng c bo v, v d nh pha sau ca chn sng, s

xy ra hin tng nhiu x. i vi sng bin nh truyn trong vng c su bin

i ng nht, cc gi tr th tc , hm phn b thng ng ca tc qu o sng

theo phng ngang F(d,z) tho mn cc iu kin bin tuyn tnh trn mt bin (bin

sng nh so vi di sng) v iu kin bin trn y bin (bng phng) c dng:

),,(),( tyxzdFi

g

(2.66)

kd

dzkzdF

cosh

)(cosh),(

(2.67)

35

Hnh 2.6 Cc trng hp khc x sng vng ven b

36

tieyx ),(

(2.68)

vi: - cao mt nc,

- cao mt nc dng s phc.

Lc phng trnh Laplace s chuyn thnh phng trnh Helmholtz i vi

:

022

k (2.69)

Phng trnh trn c p dng i vi sng vng nc su v sng di.

i vi nhiu x sng do chn sng c mt u khng gii hn, Penney v Price (1952)

nhn c li gii ca (2.69) da trn nh lut Sommerfeld i vi nhiu x tia sng.

H s nhiu x Kd l t s gia bin sng b nhiu x v bin sng u chn

sng (cha b nhiu x) trong h to cc r v (Hnh 2.7)

)cos()cos(

2sin

8

2sin

8

ikrikrd e

L

rIe

L

rIK (2.70)

vi:

dei

Ii

2

2

2

1)( (2.71)

Hay: 2

)()(

2

)()(1)(

SCi

SCI

(2.72)

vi C() v S() l tch phn Fresnel:

dSdC2

sin)(.2

cos)(2

00

2

(2.73)

Hnh 2.7 Sng nhiu x do vt cn

37

2.5 Kt hp sng khc x v nhiu x

Khi truyn vo vng bin dng v vng ven b cc qu trnh khc x v nhiu x

sng thng xy ra ng thi. C s tnh ton trng sng di tc dng ng thi ca

hai qu trnh trn c nu ra di y:

2.5.1 Phng trnh dc y thoi

Phng trnh Laplace ca th tc sng vi gi thuyt l dng chy khng xoy,

c vit di dng:

02

2

2

2

zxi

(2.74)

vi: xi - (i=1,2) l to ngang,

z - to thng ng.

Phng trnh (2.74) nhn vi mt hm F v ly tch phn theo chiu thng ng t

y ln mt bin s nhn c:

0

2

22 0

d

dzz

FF

(2.75)

Phng trnh (2.75) biu th tch phn gn ng bc nht nng lng sng i vi y

dc. iu kin bin ti y l thnh phn vung gc ca tc qu o ht nc s b

trit tiu:

i

ix

duw

(z = -d) (2.76)

iu kin bin ti mt bin, ng vi l thuyt sng tuyn tnh trn mt nc c th

tho mn iu kin bin sng nh hn rt nhiu so vi di sng. T c th b

qua cc thnh phn bc cao khi khai trin chui Taylor cho iu kin bin trn mt bin:

tz

(z = 0) (2.77)

0

g

t (z = 0) (2.78)

Loi t (2.77) v (2.78) ta c:

2

21

tgz

(z = 0) (2.79)

vi (xi,t) l cao mt nc.

Ly tch phn thnh phn ca (2.75) vi iu kin bin ti y (2.76) nhn c.

0 0

0

20

0..d d zd

z

F

zFdzFdzFkdzF

(2.80)

Nu sng l sng hnh sin theo thi gian, thnh phn th t trong (2.80) s trit tiu.

Nu p dng cc iu kin bin trn mt bin (2.77), (2.78) v h thc phn tn (2.48),

li gii ca (2.80) s c ly di dng (2.66) v (2.68). Tuy khng hon ton tho mn

38

vi iu kin y dc, nhng theo cc kt qa nghin cu theo phng php nhiu ng

ca Biesel (1952) cho thy, thm ch i vi sng xp x bc mt, hiu ng dc ca y

bin c th b qua v hiu ng ca dc y bin rt nh, loi tr ti cc tng rt st

y. Nh vy t (2.80) nhn c phng trnh dc y thoi (Berkhoff, 1972,

1976; Smith v Sprinks, 1975; Mei, 1983).

0).( 22

nnC (2.81)

Trong cc thnh phn cha cc hm m bc cao hn v cc o hm ca su d c

b qua. Nu n l hng s th (2.81) chuyn thnh biu thc Helmholtz (2.69). L thuyt

phng trnh dc thoi c p dng i vi khu vc y bin c dc ti 1/3.

Phng trnh chuyn ng sng (2.74) vi iu kin bin trn y (2.76) p dng cho

th tc ca sng trong khu vc bin c dng chy n nh. Tuy nhin ng vi iu kin

dng chy n nh ny, c c s ph hp vi biu thc phn tn (2.61) cn a thm

cc thnh phn tng tc trong iu kin bin trn mt bin (2.77) v (2.78) nh sau

(Longuet-Higgins v Stewart, 1961):

0).(

gU

t

r (z=0) (2.82)

0.

U

tz

r (z=0) (2.83)

Bng cch gi nh cc biu thc ring s nh:

=F(d,z)(x,y,t) (2.84)

chng ta nhn c:

01.).( 2*22*2

n

k

nU

t

r (2.85)

Phng trnh (2.85) biu th phng trnh dc thoi m rng trong trng hp c

s tn ti ng thi gia trng sng v trng dng chy (Booij, 1981 [6]).

2.5.2 Phng trnh dc thoi theo theo thi gian

Dng cht ch ca tch phn theo chiu thng ng ca phng trnh lin tc c

dng:

0.

Q

t

r (2.86)

vi: Qr

l vect ca ln dng chy cho mt n v b rng v c xc nh theo:

d

dzuQrr

(2.87)

vi: ur

l vect tc qu o do sng.

Phng trnh (2.81) v (2.86) biu th tn ti mt biu thc gn ng:

0)(Q 2

n

n

C

t

r

(2.88)

39

Biu thc trn c coi l tch phn thng ng ca phng trnh chuyn ng.

nhn c phng trnh trn trc tip bng cch ly tch phn phng trnh chuyn

ng s rt phc tp. Trong thc t, bng cch th cc gi tr ca th vn tc trong

phng trnh (2.66) vo (2.74) chng ta s nhn c:

0)(2

22

z

FF (2.89)

C th thy rng s khng tn ti li gii tho mn phng trnh trn vi mi gi tr

z trong vng bin c a hnh tu . Nu gi tr hm s bin i z c loi tr t (2.89)

thng qua tch phn thng ng, phng trnh nhn c s c li gii di dng trung

bnh. Theo mt phng php khc, bng cch loi Qr

khi phng trnh (2.86) v (2.88)

chng ta c:

0)(. 22

nn

C (2.90)

Phng trnh ny bao hm o hm bc hai ca su dng n. Mc d cc o

hm ny khng lm gim bc chnh xc ca phng trnh, chng c th gy kh khn v

mt k thut trong tnh ton theo phng php s cc vng c a hnh y phc tp.

Cc phng trnh tng ng hon ton vi phng trnh (2.81) c a ra di

dng:

0)Q.(1 '

rn

nt

(2.91)

0t

Q 2'

C

r

(2.92)

vi 'Qr

c xc nh theo:

0

22

2'Q

d

dzuFn

gk

t

C

r (2.93)

Hay:

t

nn

.

C-QQ

2

2'

rr (2.94)

v 'Qr

khng hon ton trng lp chnh xc vi lng dng chy thc t Qr

.

Mc d cc phng trnh dc thoi theo thi gian c dn ra nh trn, chng

khng m phng c trng sng vng trung gian v cc chuyn ng sng tun hon

c gi nh trc trong phng trnh (2.68).

2.5.3 Cc phng trnh parabolic

Mc d phng trnh dc thoi (2.81) rt c ch cho m phng trng sng nhng

n c vit di dng elliptic, do vy tnh ton trng sng cn thit phi gii theo

phng php lp. Radder (1979) a ra xp x dng parabolic ca phng trnh dc

thoi bng cch cho rng sng truyn ch yu theo trc ca hng truyn sng v c th

b qua cc sng phn x:

40

)(2

)(2

1 22

2

2 ynC

yknC

iknC

xknCik

x

(2.95)

Vi l bin sng dng s phc, c tnh theo bin sng, a, v pha sng :

iae (2.96)

T cc phn thc v phn o ca (2.95) ta c

0221 2

2

2

2

k

xk

yy

anC

yanC

(2.97)

0)( 2222

a

ynC

yaknC

x

(2.98)

Xt vng b c ng ng su thng v song song theo trc y. Trc x theo hng

vung gc vi b, trng sng ban u cho gc vi cc ng ng su ng nht dc b:

0

y

a, .sin constk

y

(2.99)

vi: - l gc gia hng sng vi ng vung gc vi ng b.

Vi cc iu kin trn, phng trnh (2.97), (2.98) s c vit li dng:

cossin2

11 2 kk

x

(2.100)

0)( 2

nca

x (2.101)

Phng trnh (2.100) ch ng vi cc gc nh, vi ln n s cho kt qu ln hn

thc. T phng trnh (2.101) cho thy phng trnh parabolic (2.95) tnh n s bin

dng sng (anC-1/2) trong khi hiu ng khc x sng (a(cos)-1/2) b loi tr. Vic

khng tnh n ton vn cc hiu ng ny l do vic lm gn ng theo phng trnh

parabolic, gi nh rng sng truyn theo ch yu theo trc X. cao sng tng khng

gii hn cc im hi t tia sng, thng xy ra trong tnh ton theo phng php tia

sng ti cc vng c a hnh phc tp. Berkhoff, Booy, Radder (1982) v Hashimoto

(1982) cho rng phng php parabolic cho cc kt qu tnh sng kh hin thc, thm tr

trong trng hp a hnh kh phc tp to ra hiu ng gradient ngc li ca bin

sng ln hm pha sng . C th nhn c cc phng trnh parabolic gn ng ca

phng trnh dc thoi, v d nh:

ynC

ynCk

inCk

xnCkk

kki

x xx

xx

y

x

22

2

2

2

22

1

2 (2.102)

vi: kx, ky - cc thnh phn ca vect s sng theo trc x,y:

coskkxr

; sinkk yr

(2.103)

Cc phng trnh ring r ng vi (2.97), (2.98) c cho di dng:

41

0221 22

2

2

2

yxx kk

xk

yy

anC

yanC

(2.104)

0)( 2222

a

ynC

yaknC

xx

(2.105)

Vi cc iu kin ring (2.99), cc phng trnh trn c th c tch phn v nhn

c cc biu thc di y m phng chnh xc hiu ng khc x:

coskx

(2.106)

0)cos( 2

nCa

x (2.107)

2.6 Phn x sng

Sng s b phn x bi cng trnh trn bin hoc y bin dc. H s phn x Kp c

xc nh l t s gia cao sng phn x Hp vi cao sng truyn ti HI :

Ipp HHK / (2.108)

Miche tnh c gii hn ca dc sng c c phn x ton phn i vi y

dc:

2

max0

0 sin2

L

H (2.109)

vi l dc y bin.

Nu sng truyn ti c cao sng ln hn cao sng v phi ca phng trnh

(2.109), th nng lng ln hn nng lng ng vi dc gii hn ca phng trnh

(2.109) s b tiu hao qua hiu ng sng . Nh vy h s phn x i vi y s l:

max0

0

0

0

max0

0

0

0

0

0

max0

0

1L

H

L

H

L

H

L

H

L

H

L

H

Kp (2.110)

Phng trnh (2.110) cho cc kt qu tnh h s phn x cao hn thc khi h s phn

x Kp gn bng 1.

Battjes (1974) nhn c cng thc thc nghim cho h s phn x i vi y dc:

21.0 pK (2.111)

Vi c gi l tham s ng nht i sng c xc nh theo:

LH //tan (2.112)

Nu mt ct y phc tp, tan c th c xc nh t dc pha trc ca bi st

b bin. Madsen,1974 thc hin cc nghin cu l thuyt v h s phn x i vi

tng chn thm nc, tuy nhin i vi cc loi tng thng ng thm nc phc tp

42

nh dng cc chn sng dng tiu hu nng lng, cn c cc th nghim bng m

hnh. Goda (1985) a ra cc gi tr gn ng ca h s phn x i vi cc dng cng

trnh bin khc nhau.

2.7 Sng

Sng khi truyn vo b bin l mt qu trnh khc lit nht trong ng lc ven b.

S khc lit v c ch vt l l ch qu trnh sng tiu tn hu nh ton b nng

lng ca sng. Nng lng ny s to ra dng chy ngang b v dc b v c th dn

n vn chuyn trm tch lm bin ng y bin. Sng cng c th ngoi vng bin

su khi hnh dng ( dc) sng vt qua gii hn cho php.

Trong thi gian gn y, t c kt qu kh tt v nghin cu chuyn ng ca

cc ht cht lng trong i sng . Tuy nhin trong cc iu kin t nhin, s tng tc

gia cc ht nc trong i sng cn cn phi c nghin cu v vn cha c c

mt m hnh tng qut c th m phng c bin i ca ton b di ph ca trng

sng trong i sng . Mt trong cc kh khn l cha c c mt m hnh ton m t

y chuyn ng ca cht lng trong i sng khi m chuyn ng ny thng l

phi tuyn v ph thuc vo thi gian. Gia tc ca ht nc trong chuyn ng sng i

sng khng cn c coi l nh so vi gia tc trng trng, tc qu o ca ht nc

cng khng c coi l nh so vi tc pha. Cc qu trnh sng s c nghin cu

theo trnh t cc sng dc dn khi i vo b, c ch sng v bin i trng sng trong

i sng .

2.7.1 Qu trnh tng dc sng dn ti sng

T l thuyt sng Trochoid nhn c gii hn ca sng l gc nh sng t 1200

(hnh 2.8 [5]).

Hnh 2.8 Gc gii hn ca nh sng

V c ch vt l th gii hn trn biu th gii hn ca tc qu o so vi tc

pha ca trng sng. Chng ta cho rng nh sng c to bi hai ng thng l tip

tuyn ca mt nc cong trong chuyn ng sng trn thc t. Chuyn ng sng c

43

coi l chuyn ng n nh trong gii hn nu trn hnh 2.8. iu kin v tc qu o

v tc pha c hiu mt cch khc l tc qu o ti nh sng phi b trit tiu

(= 0). Th tc trong vng nh sng theo to cc (r,) trn hnh 2.8 c th c xp

x bng biu thc:

)sin(),( nBrr n (2.113)

vi B v n l cc h s hiu chnh, r v l to cc.

hiu chnh h s n vi thc t mt nc l mt ng dng ta c:

01

ru (2.114)

Hay: 0)cos(

n ti = 0 (2.115)

v: 2

0

n (2.116)

Cho rng p sut trn mt thong bng khng, phng trnh Bernoulli ca mt nc

ln cn nh sng s l:

0)(2

1 22 gzuur (2.117)

Th cc gi tr tc r

ur ;

r

u v to trn mt z = - r cos vo

(2.117) ta c:

constgrBn n 03222 cos

2

1 (2.118)

V v phi ca (2.118) khng i do vy gi tr m ca r phi bng 0:

2n3=0 hay n=3/2 (2.119)

Thay vo (2.116) ta c:

00 60 (2.120)

Thay B t (2.118) vo (II.113) ta c th tc di dng:

)3

2sin(

3

2),( 2/32/1 rgr (2.121)

Hay dng s phc (Longuet-Higgins, Fox, 1977):

2/312/1

3

2zg

iw (2.122)

Trong : w = + i v z1 = r exp(i).

Cc thnh phn ca tc qu o trong chuyn ng sng s c dng:

)2

3cos(

1

)2

3sin(

2/12/1

2/12/1

rgr

u

rgr

ur (2.123)

44

C hai thnh phn ca tc qu o s tin ti 0 khi r tin ti 0, iu ny tho mn

ch tiu ng lc ca hin tng sng . Gi nh th hai v nh sng c to bi hai

ng thng tip tuyn ca hai mt sn sng s c biu th qua vic xc nh mt

nc t do vi mt gc to cc 0 . Li gii (2.120) cho rng gc ca mt sng gim ti

gi tr ti hn 1200. y l trng hp ng vi cc sng dc nht. Khi r >0 nh sng s

tr ln t hn. xc nh dng chy ti hn chng ta xt to (r, ) vi im gc 0

khong cch l pha trn nh sng (hnh 2.9). Chng ta cn tm li gii theo vi r/l

s c dng dng chy gc Stokes; (2.122). Longuet- Higgins v Fox (1977) t cc kt

qu tnh ton nhn c mt nc t do s ct tim cn ca chng khong r/l =3.32

v sau t ti tim cn vi dao ng v hai pha rt chm. Gia hai ln ct tim cn

r/l =3.32 v r/l =68.5 gc cc i ca dc vt qu mt cht so vi 300 v gi tr tnh

ton s l 30.370. Hn na gia tc thng ng ca ht nc ti nh sng l 0.388g nhng

v tr xa nh r/l gia tc tin ti gi tr 1/2g tng ng vi dng chy gc Stokes.

chng minh gi tr ny chng ta xc nh cc thnh phn gia tc ht nc trong

h to chuyn ng sau:

r

uuu

r

u

r

uua

r

uu

r

u

r

uua

rr

rrrr

2

(2.124)

S dng phng trnh (2.123) tnh cc thnh phn ca phng trnh (2.124) v

tm c rng a = 0 v:

)2

3(cos)

2

3(cos

2

3)

2

3(sin

2

1 222 gggar (2.125)

Hay:

gar2

1 (2.126)

Hnh 2.9 Dng tim cn ca cc sng c dc cc i

45

Nh vy gia tc ht nc gn nh sng gim trc tip t nh sng tr xung vi gi

tr l (1/2)g. vng nc su, sng c dc cc i l 0.142, thng c biu th l

sng c cao cc i t 1/7 di. i vi sng su hu hn, khi y bin ng

vai tr quan trng i vi qu trnh sng , cc c trng v dc a phng ca mt

nc trong chuyn ng sng v ch tiu ng lc gii hn sng s hn ch s tng

cao sng vng ven b. Miche (1944) nhn c biu thc dc gii hn cho vng

bin dng:

)tanh(142.0max

kdL

H

(2.127)

Trong thc t, ch tiu ng lc gii hn sng thng c biu th qua t s ln

nht gia cao sng v su (H/d)max. T s ny gi l h s sng (d) v ph thuc

vo dc ca y bin v chu k sng. Hnh 2.10 v h s sng vi cc dc v chu

k khc nhau [4].

Hnh 2.10 H s sng vi cc dc y bin, chu k sng khc nhau

2.7.2 Cc dng sng vng ven b

C 4 dng sng di tc ng ca su khi sng truyn vo vng ven b:

- Sng dng st nh: nh sng tr ln khng n nh v b st v pha trc to

thnh mt lp bt trng.

- Sng dng b nho: nh sng b vn v pha trc v xung ngay ti chn

pha trc ca sng.

- Sng dng v chn: nh sng gi nguyn nhng sn pha trc b khi tin

vo bi st b bin.

- Sng dng st sn: nh sng gi nguyn nhng sn pha trc b vn ln

tng dc v xung.

46

Ngoi ra di tc ng ca gi, sng cng c th b di dng bc u. Hnh 2.11

a ra hnh nh ca bn dng sng nu trn.

Dng sng c th c xc nh bng tham s:

tan

2/1

0

00

L

H (2.128)

Sng dng v chn st sn: 3.3< 0

Sng dng b nho: 0.5

47

2.7.3 Bin i trng sng trong vng sng

S bin i trng sng trong vng sng , hay ni cch khc bin i nng lng

sng trong vng sng l nguyn nhn c bn ca nhiu qu trnh ven b: hon lu

gn b, sng leo, vn chuyn trm tch vv... Cc m hnh tnh m phng qu trnh bin

i lan truyn sng trong vng ny c chia thnh 3 loi:

- Da trn c s cao sng gii hn,

- Cc m hnh lan truyn Bore,

- M phng tng qut s bin i theo khng gian cc tham s sng (nng lng

tc ng sng).

Theo loi cui, phn b cao sng c tnh theo phng trnh cn bng nng lng:

)()(

xx

ECg

(2.129)

Trong : - l nng lng tiu hao cho mt n v din tch mt bin do ma st y, do

dng ri pht sinh khi sng vv...

Cc nghin cu ca Horikawa v Kno (1966) v o c thc t cho thy cao sng

s t c gi tr n nh sau khi sng u tin trn y ng nht.

hHstab (2.130)

Xt bin i trng sng sau khi sng (hnh 2.12).

Hnh 2.12 S bin i sng trong i sng

Theo hnh 2.12 nng lng tiu hao gia mt ct AA v BB l:

stabggg ECECdx

EC)(

)(

(2.131)

vi: (ECg)stab - dng nng lng ca sng n nh.

K - h s tiu tn, khng th nguyn.

48

Trong phng trnh (2.129) p dng gdCg vng nc nng ta c:

)()()(

)( 2/32 xdxGxdx

xG

(2.132)

trong :

)()()( 2/12 xdxHxG

Trong trng hp n gin nht khi sng trong vng sng c a hnh l tng,

ta c cc li gii sau:

a. su khng bin i:

2/1

22

2

exp

d

x

d

H

d

H

b

(2.133)

x = 0 : H/d = (H/d)b im sng .

b. dc khng i:

Nu su bin i tuyn tnh theo x, c ngha l:

xdxd b )( (2.134)

Ta c:

2/12

)1(

b

r

bb d

d

d

d

H

H (2.135)

vi:

2

1,

2

5

22

mr

H

d

b

(2.136)

Trong trng hp c bit, nu K/ =5/2; phng trnh (2.135) c dng:

2/1

1 ln1

bbb d

d

d

d

H

H (2.137)

trong :

2

21

2

5

bH

d

(2.138)

Ch rng nu K=0, phng trnh (2.135) tr thnh biu thc m t nh lut Green.

Cc cng thc tnh nu trn ph thuc rt nhiu vo cc h s K v .

S dng phng php bnh phng ti thiu c th xc nh c cc gi tr K v

ng vi cc dc khc nhau (bng 2.4 [5])

49

Bng 2.4 H s K v ng vi cc dc y bin khc nhau

dc K

1/80 0.200 0.350

1/65 0.115 0.355

1/30 0.275 0.475

i vi y bin c dng tu , vic lan truyn sng trong vng sng c tnh n

cc hin tng bin dng, sng v sng ti to cha c c cc kt qu gii tch.

Trong cc tnh ton s cn tnh ti mc nc bin i trong vng sng da trn c s

l thuyt ng sut bc x sng (chng III).

2.8 Tng tc gia sng v dng chy vng ven b

S tng tc gia sng v dng chy ng vai tr quan trng trong hu ht cc khu

vc ven b c bit l cc vng ca sng, lung ra vo cng, lch triu. Vic tnh n

qu trnh tng tc ny trong tnh sng cn thit cho thit k cng trnh, tnh ton vn

chuyn trm tch, hng hi, giao thng ven bin.

2.8.1 Truyn sng trn dng chy ng nht.

Gi nh trng sng tin n nh, truyn vo vng b c su d vi mt gc

ngc li vi trc x. Trng dng chy )0,0,( 1UU

song song vi trc x. H thc phn tn

c dng:

)tanh()( 2/1 kdgkUk

(2.139)

Hay:

cos||Uk (2.140)

Trong :

2/1)]tanh([|,| kdgkkk

(2.141)

Tn s l tn s quan trc trn c s ca h c nh cn l tn s thc t trn c s h

chuyn ng vi tc

U , cho rng:

2/1* )tanh(,cos||

, kdkdd

Ub

g

da

(2.142)

Lc phng trnh (2.140) c th vit li di dng:

*)( kdba (2.143)

Phng php n gin nht gii (2.143) i vi

k l xc nh vng ct nhau ca

mt phng:

kdba )( (2.144)

50

vi s bin i b mt:

* (2.145)

Trong khng gian (kd,) hng quay c tnh ngc li vi trc . Nu

U = 0 hay

= 900 th b=0 v = . Nh vy nu vct

k vung gc vi

U th dng chy khng nh

hng n li gii. Vi hng bt k khc, li gii s l ng cong trn hnh 2.13. Trn

mt ct phng xuyn tm ca hnh 2.13 c th phn bit bn li gii khc nhau ti cc

im A, B, C, D. im A, trong =A > 0 v (kd)A>0 tng ng vi sng c thnh phn

k cng hng vi dng chy. Tn s quan trc s ln hn tn s thc t A. Th =A

vo (2.140) v chia cho s sng kA ta c:

||)()()(

UkCkC AAAaA ti = 0

0 (2.146)

vi: CA - tn s pha tng ng vi tn s A ngha l CA=A/(kA)

CA(a) - tc pha tuyt i ca ngi quan trc ng trn h khng chuyn ng.

Hnh 2.13 Li gii ca h thc phn tn khi sng truyn trn trng dng chy

Ti im B v C chng ta c A >0 v kd

51

|)(||)(||| BBBg kCkCU

(2.149)

Trong khi ti im C:

|)(||||)(| BBBg kCUkC

(2.150)

Sng trong tng hp ny truyn ngc li vi dng chy c ngha l nh ca n

truyn v pha ngun ca dng chy cn nng lng li b cun v pha cui dng chy.

im D tng ng vi sng vi D

52

Tc nhm tuyt i:

UCC g

a

g

)(

Hng ca tia sng:

coscos

sinsintan

g

g

CU

CU

(2.157)

Khi U=0 v = hng ca tia sng s trng vi hng vung gc vi nh sng.

Hnh 2.14 S vect dng chy v hng sng

2.8.3 Truyn sng trn dng chy thay i theo chiu ngang

Nu sng truyn trn dng chy vi trng tc thay i th s thay i dng m men s dn n s trao i nng lng gia sng v dng chy. n gin ho chng ta

gi nh rng )0,0),(( 1 xUUU

tng ng vi trng dng chy trong cc knh hoc

sng, vi tc thay i theo su d(x).

Gi thuyt l dng chy n nh ta c:

kdgkUk tanh).( 212 (2.158)

vi cc gi tr U1 v d, ch cn xc nh k trong phng trnh (2.158). Cho rng i vi trng hp n gin nht l vng nc su:

constUCk )( 1 (2.159)

Gi tr K c ly bng 0 l tn s sng im c U1 =0 c ngha l 0 = (gk0)1/2 v

Co=(gk0)1/2. Phng trnh (2.159) c th c vit li di dng:

00

1

0

2

0

C

U

C

C

C

C (2.160)

vi li gii l:

2/1

0

1

0

41

2

1

2

1

C

U

C

C (2.161)

53

Li gii (2.161) cho thy tc ca dng chy ngc U1=(1/4)C0 = (-1/2)C =- Cg tng

ng vi gii hn ng lc. Nh vy, nu tc dng chy ln hn v ngc li vi tc

nhm a phng ca trng sng, nng lng s khng truyn c theo hng ngc

dng chy.

Cc kt qu nghin cu chi tit v truyn sng trn nn dng chy bin i theo chiu

ngang c th tm c trong cc cng trnh nghin cu ca Kirby (1988).

2.8.4 Truyn sng trn nn dng chy bin i theo su

Chng ta nghin cu tng tc gia sng v trng dng chy c c bin i theo

thi gian ln hn nhiu so vi chu k sng do vy c th coi nh dng chy n nh. Cng

tng t nh vy, bin thin dng chy theo chiu ngang c kch c ln hn nhiu so vi

bc sng. Tuy nhin trong trng hp dng chy bin i theo chiu thng ng th cc

gi nh trn khng ph hp. Kh khn nht a ra cc li gii gii tch l phng

trnh xut pht khng th gii chnh xc i vi s sng tng hp v tn s sng tr khi

h s xoy c gi nh khng i. S n gin nht ca profile thng ng ca tc

dng chy U1(z) c dng nh hnh 2.15 . U1 trt ti y v phn b ng dng theo

chiu thng ng. Chuyn ng c gi nh l khng xoy v c th p dng l thuyt

sng n nh khng xoy. Xt s tng tc gia sng tuyn tnh tun hon vi dng chy

trt n nh vi profile tu trong khng gian 2 chiu (hnh 2.15), bin i mc nc

(x,z) theo hnh sin c ngha l:

)cos(),( tkxatx (2.162)

v trng dng chy tng cng (uT(x,z,t), wT (x,z,t)) c th c vit di dng:

)sin()(),,(

)cos()()(),,( 1

tkxzwtzxw

tkxzuzUtzxu

T

T

(2.163)

Tc U1(z) tng ng vi dng chy khi khng c sng.

Hnh 2.15 H toa bin i dng chy theo su

54

T l thuyt n nh ng lc, a ra phng trnh di dng thun tin m

phng tng tc gia sng v dng chy thng qua tc thng ng w.

01

12

1

2

2

2

w

dz

Ud

kU

kk

dz

wd

(2.164)

Cc iu kin trn thch hp vi w(z) l:

w(z) =0 ti z = -d (2.165)

0)()( 2112

1 wgkdz

dUwkUk

dz

dwkU ti z = 0 (2.166)

)()( 1kUazw ti z = 0 (2.167)

iu kin (2.166) biu th h thc phn tn cn (2.167) l iu kin ng lc ca mt

thong. Cc phng trnh (2.164) - (2.167) i vi k v w(z) c gii vi gi thit rng

,a,k v U1(z) bit. Trong trng hp ring, t s gia tc qu o ngang v thng

ng trong sng u(z) v w(z) c th nhn c t phng trnh lin tc:

dz

dw

kzu

1)( (2.168)

Tuy nhin cc phng trnh (2.164) - (2.167) khng th gii bng phng php gii

tch cho cc s sng k v tn s tu , tr khi dng chy ph thuc vo su (bin i

tuyn tnh theo su). Trong trng hp tng qut, cn gii s cc phng trnh (2.164)

- (2.167). Cc kt qu tnh ton c so snh vi s liu thc nghim v c s ph hp

kh tt. Cc nghin cu tng tc gia sng v dng chy i vi dng chy xoy cn rt

t v khng th p dng i vi phn b xoy tu v sng c bin hu hn.

55

Chng 3

ng sut bc x sng

v cc qu trnh do sng sinh ra vng ven b

3.1. Cc thnh phn ng sut bc x sng

3.1.1. Sng vung gc vi b

Cc sng mt sn sinh ng lng M theo hng lan truyn sng. i lng ny xc

nh bng:

C

Ekd

gTgHkd

THkdaM coth

2

8

1coth

2

8

1coth

2

1 222

(3.1)

Trong : - mt nc, H - cao sng, a = H/2 - bin sng, T - chu k sng, =

2/T - tn s sng, k = 2/L - s sng, L - di sng, d - su, E - nng lng trung

bnh trn mt n v din tch b mt.

Khi sng ang lan truyn b chn li bi vt cn nh chn sng, s xy ra phn x

ti mt vt cn v nh vy hng ng lng b thay i. iu c ngha l sng to ra

mt lc thu ng c ln bng sut bin i ng lng. Lc ny lin quan n ng

sut gi l ng sut bc x, cc ng sut ny tng ng vi dng ng lng v hng

v xc nh bng dng ng lng d do chuyn ng sng to nn.

Mc d trong thc t mi cht lng u nht v nn c, nhng trong nhiu trng

hp s hng nht c th b qua so vi s hng p sut v gia tc. Trng hp ny ta c

chuyn ng khng nht, khng xoy v gi l chuyn ng th. Gi thit sng lan

truyn theo hng x, dng th i vi chuyn ng sng c th m t bng phng trnh

Euler. Lc , cn bng ng lng theo hng lan truyn sng x l:

x

P

z

UW

y

UV

x

UU

t

U

)()()()( (3.2)

Trong : U, V, W l nhng vn tc tc thi theo hng x (vung gc vi b), y (song song

b) v z (thng ng).

Phng trnh (3.2) c th vit li nh sau:

xF

z

UW

y

UV

x

UUP

t

U )()()()( .

Nhng thnh phn bn v phi phng trnh (3.2) c th xem nh lc hiu qu tc

ng ln th tch cht lng c cc cnh x, y, z, lc ny sn sinh s thay i ng

lng cc b theo thi gian ca th tch ang xt. Gi thit su nc khng i, tch

phn nhng thnh phn phng trnh (3.2) theo chu k sng T v theo ton b su (-d

n ) v tr i p sut cht lng thu tnh khi khng c sng, c th nhn c cc lc

hng ngang, trung bnh thi gian trn b rng n v:

d

T

d

XX dzPdzdtUUPT

S 00

)(1

(3.3)

56

T

d

XY dzdtUVT

S0

)(1

(3.4)

Trong : P0 - p sut thu tnh ca cht lng ti su z.

V sng lan truyn vung gc vi b (V = 0) ta c SXY = 0.

Hnh 3.1 ng sut bc x sng

Tng t i vi hng y cho ta:

d

T

d

YY dzPdzdtVVPT

S 00

)(1

. (3.5)

Nhng lc biu th bng phng trnh (3.3-3.5) gi l ng sut bc x, v l nhng

lc trn b rng n v (N/m). C th lin h cc ng sut bc x ny vi cc tham s

sng bng cch ly tch phn. S dng cc biu thc ca l thuyt sng tuyn tnh ta c:

EngHnSXX )2/12(8

1)2/12( 2 (3.6)

EngHnSYY )2/1(8

1)2/1( 2 (3.7)

0XYS (3.8)

Trong : n = Cg/C- t l tc nhm v tc pha, H - cao sng,E - mt nng lng sng trung bnh.

Khi sng lan truyn trong mt min c su khng i v gi thit khng c tn

tht nng lng ( cao sng khng i), ta c SXX khng i, SYY khng i v SXX/x =

0, SYY/y = 0, c ngha l ng lng khng i. Nhng gradient SXX, SYY v SXY theo cc

hng x, y c th nguyn l lc trn n v din tch*. Chng th hin ngoi lc tc ng

trn din tch n v ln mt phn t cht lng cao h:

x

SXXXX

,

y

SYYYY

,

x

SXYXY

. (3.9)

57

V d:

Tnh ton gi tr SXX v SYY cho mt sng lan truyn t nc su (h= 5 m, h = 150 m, T =

12 s, = 1000 kg/ m3, g = 9,81 m/ s2) ti b.

ng sut bc x su

nc

cao

sng

Bc

sng

H s Nng lng

SXX SYY

d (m) H (m) L (m) n E (N/m) (N/m) (N/m)

150 5,0 225 0,502 30656 15451 61

100 4,91 225 0,521 29561 16022 620

60 4,59 220 0,611 25834 20150 2868

25 4,56 170 0,799 25497 27996 7623

10 5,07 120 0,919 31480 42120 13190

3.1.2. Sng truyn di mt gc vi b

ng sut bc x sng trong trng hp ny, thun tin cho cc tnh ton cc qu

trnh ng lc ven b, c xc nh theo h to ca ng b (x vung gc v y song

song vi b) v c k hiu bng Sxx, Sxy, Syy. Cc thnh phn ng sut bc x da trn

h to ca ng b c tnh t cc thnh phn ng sut bc x da trn h trc to

ca trng sng SXX, SXY, SYY . chuyn i c th s dng s Mohz (xem hnh

3.2). Cc thnh phn ng sut bc x sng theo h trc to ca ng b c dng sau:

2cos22

YYXXYYXXxx

SSSSS

2cos22

YYXXYYXXyy

SSSSS

(3.10)

2sin2

YYXXxy

SSS

Hay:

EnnSxx )cos2/1(2 (3.11)

EnSS yxxy sincos (3.12)

EnnS yy )sin2/1(2 (3.13)

* Cc gradient ng sut bc x sng XX, XY, YY, trong thc t chnh l ng sut bc x sng vi th nguyn l

lc trn n v din tch (Van Rjin 1989). Tuy nhin hin nay trong cc sch chuyn mn u coi cc lc bc x

SXX, SXY, SYY, l ng sut bc x sng nn trong gio trnh ny chp nhn cc quy c trn [7].

58

Trong : - gc gia hng lan truyn sng v hng x vung gc vi b, = 0o i vi

sng vung gc vi b. Nhng lc Sxx v Syy l lc php tuyn. Sxy l lc tip tuyn.

Hnh 3.2 S Mohz chuyn i ng sut bc x sng sang h to ng b bin

3.2. Mc nc dng v rt ti vng sng

Sng tc dng mt lc ln khi cht lng m trong chng lan truyn. iu ny to

ra mt dng khi lng v mt dng ng lng rng, dn ti nhng bin i su nc

trung bnh (dng v rt), khi c gradient cao sng hng ngang. Khi sng tip cn b

di mt gc, s pht sinh dng chy dc b trong vng sng . Hin tng ny c th

gii thch bng khi nim ng sut bc x ca Longuet - Higgins v Stewart (1964) nh

ni trn. Dng ng lng rng v dng khi lng rng l nhng hiu ng phi tuyn

bi v lin quan n s hng H2, c th nhn c cc ng sut ny bng cch p dng l

thuyt sng tuyn tnh.

3.2.1. Nc rt do sng trong sng khng

Nhng phng trnh (3.6) n (3.13) hp l trong trng hp dc y thay i dn

dn. Cn bng lc theo hng x cho ta:

0'

)'( dx

dhhdgxx (3.14)

Trong : d - su nc tnh, h' - bin i mc nc so vi mc nc tnh, xx = Sxx/x

gradient ng sut bc x.

Phng trnh (3.14) cho thy rng gradient ng sut bc x ngang cn bng vi

gradient p sut thu tnh do s bin i mc nc trung bnh.

59

Hnh 3.3 Bin i mc nc v dng chy do sng

Gi thit cn bng dng nng lng khng c nhng hiu ng tiu tn d( E nC)/dx = 0 v gi thit h'

60

22 )'(16

3

2

3hdgESxx (3.17)

ng sut bc x gim theo hng vo b do mc nc gim, dn ti mt nc trung

bnh tng, ph hp vi phng trnh (3.14). Thay phng trnh (3.16) vo phng trnh

(3.14) ta c:

0)')'(8

3( 2 hhd

dx

d . (3.18)

Nh vy: consthhd ')'(8

3 2 hoc KhH '8

3 . (3.19)

H s K c th xc nh t phng trnh (3.16). Thay phng trnh (3.16) vo phng

trnh (3.19):

KHH brbr 16

1

8

3

Hoc: KHbr 16

5 cho ta: HHh br

8

3

16

5' (3.20)

Trong : Hbr - cao sng ti ng sng , H - cao sng, - h s sng .

Gi tr ln nht l h'max = 5/6Hbr i vi d = 0. chnh lch tng cng ca mc nc

trung bnh trn vng sng l:

brbrbrbr dHHHh2

8

3

8

3

16

1

16

5' . (3.21)

Phng trnh (3.20) cng c th biu th nh sau:

2

2

8

31

8

3

6

5

'

dHh

br

(3.22)

Phng trnh trn cho ta dng mc nc tuyn tnh trong vng sng trong trng

hp y phng dc.

Nc dng do sng l mt hin tng lin quan n hot ng ca sng trong mt

thi gian thit lp nhng iu kin cn bng. Nhng nhm sng ln vn chuyn

lng nc tng i ln v pha ng b, gy ra nc dng, nhng mt t nc ny c

th chy ngc li trong nhng khong thi gian tng i yn tnh gia nhng nhm

sng.

3.3 Cc loi dng chy do sng vng ven b

Trong i dng tn ti nhng dng chy c hng v vn tc hu nh khng i

sut c nm. Chng thng do gi sinh ra v c phn thnh dng chy tri v dng

chy gradient, hoc dng chy mt , dng chy m v dng chy lnh, tu theo c ch

pht sinh ra chng. Nhng dng chy ny t nh hng n vng ven b. Mt loi dng

chy khc do chuyn ng ca thy triu sinh ra gi l dng triu. Dng triu b tc ng

mnh ca y bin v hnh dng ng b.

61

khu vc gn b thng tn ti dng chy do sng, thng c gi l dng chy ven

b. Khi sng truyn vo vng nc nng ven b, do bin i ca a hnh y v ng

b, sng b khc x, phn x, bin dng. Di tc ng ca ma st y, xy ra tiu tn

nng lng sng, ng thi vi hin tng sng dn mt khi lng nc vo b to

ra cc ng sut khng ng u gy ra cc dng chy. Loa dng chy do sng ny c

nghin cu cch y khng lu, v theo D. W. Johnson (1919) th c th chia ra hai loi:

dng chy dc b v dng sng di hay dng tch b ngm khi p dng tnh ton vn

chuyn trm tch.

Nhng ngi dn nh c, ngi cu h v nhng ngi sng ven bin nhn thy c

nhng dng chy kh mnh hng t b ra thng ngoi khi. Do vy vo nm 1941

Shepard, Emery v La Fond gi y l dng tch b (cn gi l dng gin on), chng

a nc bin do sng mang vo b tr li bin. Nhng dng ny ch yu trn mt,

khc vi dng sng di nm di y. V sau, nm 1950 Shepard v Inman t cc quan

trc hin trng thit lp h thng dng chy gn b nh c m t trn hnh 3.4.

Trong s 3 loi dng chy do sng: dng dc b, dng tch b v dng sng di th dng

dc b c nghin cu nhiu nht cng nh d quan trc nht v n thng xuyn hin

din v thng mt quy m khng gian kh ln. Hn na, dng chy dc b ng vai

tr ch o trong vic vn chuyn trm tch v bin i a mo b, do ta s nghin

cu k hn mc di y.

3.4 L thuyt dng chy sng dc b

Sng ng vai tr ch o trong vic to ra cc dng chy chuyn ng n nh nh

dng chy dc b, dng sng di, dng gin on. Khi sng v trong vng sng , chng

gim ng lng, gy ra ng sut bc x. Thnh phn ngang b ca ng sut bc x y

nc vo b v to ra s dng mc nc, mc nc tng v pha b so vi mc nc tnh.

dc mt nc do n gy ra cn bng vi gradien ngang b ca thnh phn ng sut

bc x vung gc vi b. i vi sng n xin mt gc vi b, cn c thnh phn dc b

ca ng sut bc x, gradient ca n to nn dng chy dc b bn trong vng sng

(v ngay st bn ngoi), cn bng vi ma st y.

3.4.1 M u

C mt lot cc tham s tc ng ln dng chy sng dc b, n gin chng ta gi nh rng trng sng n nh, hai chiu truyn cho gc vi ng b. Trong vng sng h s sng c coi l khng i. Bi bin c coi l thng, di v tn, c cc ng ng su song song, dc y va phi v y khng thm. Dng chy dc b tnh c trong iu kin b qua lc tc ng ca gi, lc Coriolis, lc tc ng ca thu triu, lc cn ca y ngoi vng sng v tng tc gia sng v dng chy. Vi cc gi nh nu trn, phng trnh cn bng lc i vi dng chy dc b s l:

Lc tc ng + Ma st y + Trao i ri ngang = 0

Trong h to nu trn hnh 3.3 ta c:

0.

, dx

dd

dx

dS xyyb

xy . (3.23)

62

Trong : dx

dSxy l ng sut bc s sng; yb, l ng sut ma st y v

dx

dd xy. l ng

sut trao i ri ngang.

Sau khi tnh c cc thnh phn trong (3.23) c th ly tch phn cho ton vng

sng v nhn c phn b dng chy dc b trong vng sng .

Hnh 3.4 H thng dng chy gn b

3.4.2 Bn ngoi vng sng

T l thuyt khc x, vi gi nh y bin nu trn ta thy rng sin/C = const, ta

c:

dx

Fd

CnCE

dx

d

Cdx

dS xxy

sin)cos(

sin (3.24)

Trong : cosncEF x - dng nng lng sng theo hng x.

Gi thit rng dng nng lng khng i bn ngoi khu vc sng (tiu tn bi ma

st y khng ng k), ta c:

63

0dx

dSxy. (3.25)

Nh vy, lc tc ng bng khng v khng c dng chy pht sinh theo hng dc

b.

3.4.3 Bn trong vng sng

Dng nng lng xF khng phi l hng s do tiu tn nng lng bi sng . V

d xF /dx < 0 ( xF gim theo hng x dng), trong khi gradient ng sut bc x tc ng

theo hng y dng i vi lan truyn sng nh trn hnh 3.2, gradient ng sut bc x

c xc nh nh sau:

dx

Fd

Cdx

dS xxy sin . (3.26)

Cho rng cos 1, n 1, C (gd)0,5 v H d (b qua nc dng do sng) trong nc

nng, ta c:

dx

ddKgd

dx

CEd

Cdx

dSxy )()(

16