Phương pháp xác định mức nước dâng cực đại có thể xảy ra do bão tại khu vực công trình đê biển

Storm surges often happen in short time, but may have great affects on coastal areas. High waves combines with high surge levels can cause significant failures to sea dikes due to wave overtopping andor even crest overflows. Therefore, the determination of the possible maximum storm surge level is necessary for sea dike design. The envelop of maximum storm surge levels can be determined from numerical model simulations. The lowering of water level in a typhoon may also have affect on the lower part of a dike. The numerical model can provide also this result for stability consideration of the outer slope and bottom protection for sea dikes.

A method for determining possible maximum storm surge at a sea dike

Bui Xuan Thong 1 and Nguyen Van Lai 2

Abstract: Storm surges often happen in short time, but may have great affects on coastal areas High waves

combines with high surge levels can cause significant failures to sea dikes due to wave overtopping and/or even crest overflows Therefore, the determination of the possible maximum storm surge level is necessary for sea dike design The envelop of maximum storm surge levels can be determined from numerical model simulations The lowering of water level in a typhoon may also have affect on the lower part of a dike The numerical model can provide also this result for stability consideration of the outer slope and bottom protection for sea dikes

1 2

-

-

1 Center for Oceanography, Vietnam Administration for Sea and Islands, Ministry of Natural Resources and Environment; Address: 8 Phao Dai Lang Street, Dong Da, Hanoi, Vietnam; E-mail: thongkttv@yahoo.com

2

Faculty of Hydrology and Water Resources, Water Resources University; E-mail: ngvanlai@wru.edu.vn

(

-

L ) 2

2.1 Hệ phương trình toán học cơ bản ỷ

-



















y

V x

U t



(1)

) ( 1 x b x s w a w x P h x g fV y U V x U U t U                        (2) ) ( 1 y b y s w a w y P h y g fU y V V x V U t V                        (3)

) ( 1 x b x s w x gh fV t U              (4) ) ( 1 y b y s w y gh fU t V              (5) 0          y V x U t  (6) U, V

f - -1 / 2

),  - ρw / ³), h - y s x s   , - các y b x b   , -

- , (s x,s y), (b x,b y

trong h

“ ”

Vn = 0

2.2 Phương pháp sai phân

4 pháp sai phân

x

U U x









(7)

y

V V

y









(8)









j t

j

t

, 1





(9)

                   ) )( ( 2 ) ( 4 , 1, , 1 1, 1 , 1 , 1, , , 1 , j i j j j i j j t t j i t j t j t j h h x t g V V V V t f U U    ( , )/ /2 )) , 1 ( ) , ( ( 2 1, , , , j j i t w t x s t x s w h h j i tU k j i j i t             (10)                    ) )( ( 2 ) ( 4 , 1, , 1 1, 1 , 1 , , , 1 , 1 , t j t j j j j i j j i j t j t j h h x t g U U U U t f V V      /2 ) ( 2 , , 1 , , 1 , , , j j t j w t y j s t y j s w h h tV k t             (11) sai phân (10 11

L

c – H

,

-

3

3 p d ng hình t nh cho to n b v nh c

B B B

300 km × là 350 k ×

:

+ /

+

+ V = 6 m/s +

+

+

2006)

p d ng t nh toán ph ng cho v ng ven b r ng át –Hải Ph ng km × 35 k

km×1 km

k k

âu là 15m

phân

-

-

+

d +

-

3.3 h n t t quả phân b trư ng nư c dâng do b o 2

dâng ( – –

- -

0 50 100 150 200 250 0 50 100 150 200 250 300 2 –

+ -

–

0 5 10 15 20 25 30 0 5 10 15 20 25 4 –

Trong quá

Hình 3 -9-2005

4

à C

4.1 hái niệ ư ng bao nư c dâng do b o

t

,

4 â d ng các ch bản theo t s họ b o

typhoon f

–

-

θ = 0°

t W = 30m/s; b 30km; t

/

–

W1 = 25 m/s, W2 = 30 m/s, W /

– θ = 60°

4 t quả ph ng nư c dâng c c i c th ra

–

-

–

–

5

á

C

à

– -2000 (Trang 366-383) L – L

L L

-290