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- Ch¬ng 13. Nh÷ng Trêng hîp nghiªn cøu
13.1. Tæng quan
Mét vµi trêng hîp nghiªn cøu giíi thiÖu trong ch¬ng nµy ®Ó minh häa c¸c kü
thuËt m« t¶ trong c¸c ch¬ng tríc ®îc sö dông vµ kÕt hîp ®Ó gi¶i quyÕt nh÷ng vÊn
®Ò thùc hµnh nh thÕ nµo. Ba vÝ dô sÏ tiªu biÓu cho c¸c lo¹i øng dông thêng xuÊt
hiÖn nhÊt. Nh÷ng vÝ dô ®· ®îc ®¬n gi¶n vµ lý tëng hãa ®Ó minh häa nh÷ng ®iÓm
chÝnh kÌm theo, kh«ng ®a ra nh÷ng phøc t¹p ®Æc trng t¹i tuyÕn thêng xuÊt hiÖn
trong nghiªn cøu thùc tÕ mµ cã thÓ ®ßi hái sù khÐo lÐo ®¸ng kÓ ®Ó vît qua. ViÖc sö
dông c¸c nghiªn cøu thùc tÕ lµm cho viÖc m« t¶ mçi trêng hîp l©u h¬n mµ l¹i kh«ng
®a ra thªm sù bæ sung s©u s¾c nµo.
Nh÷ng tÝnh to¸n trong c¸c nghiªn cøu ®îc tham chiÕu tíi nh÷ng ph¬ng tr×nh
vµ h×nh vÏ trong c¸c ch¬ng næi bËt. Tuy nhiªn, ph¬ng ph¸p thùc hiÖn ®¬n gi¶n
nhÊt ®èi víi nhiÒu tÝnh to¸n lµ b»ng gãi phÇn mÒm SandCalc vµ nh÷ng bíc cÇn
thiÕt ®Ó thùc hiÖn ®iÒu nµy ®îc m« t¶. Trong tÊt c¶ c¸c trêng hîp nghiªn cøu, ®Ó
®¬n gi¶n níc ®îc lÊy ë nhiÖt ®é 10oC vµ ®é muèi 35 o/oo, vµ mËt ®é trÇm tÝch 2650
kgm-3 (gi¸ trÞ ngÇm ®Þnh trong SandCalc). Nh÷ng gi¸ trÞ kh¸c cã thÓ dÔ dµng sö dông
th«ng qua c¸c menu 'Edit- Bed Material' vµ 'Edit- Water’ cña SandCalc, sö dông
'Derive’.
Mçi trêng hîp nghiªn cøu bao gåm mét tËp hîp nhá nh÷ng bíc ®îc ph¸c th¶o
trong 'Quy tr×nh Tæng qu¸t' (môc 1.5), bæ sung thªm nh÷ng quy tr×nh trêng hîp ®Æc
trng.
13.2. Sù æn ®Þnh cña viÖc chèng xãi
Mét dµn s¶n xuÊt dÇu sÏ ®îc l¾p ®Æt trong mét khu vùc khai th¸c dÇu ngoµi
kh¬i cã ®é s©u 20 m t¹i mét vÞ trÝ cã ®¸y c¸t mÞn. T¸m ch©n ®Õ cña dµn c¾m s©u vµo
®¸y, cã mÆt c¾t ngang h×nh trßn, ®êng kÝnh 2 m. §Ó xãi kh«ng xuÊt hiÖn, mét líp
phñ b¶o vÖ b»ng ®¸ sÏ ®îc ®Æt xung quanh c¸c ch©n. Ph¶i tÝnh to¸n kÝch thíc ®¸
cÇn thiÕt ®Ó chèng l¹i nh÷ng h h¹i ®¸ng kÓ víi chu kú lÆp 50 n¨m (vÝ dô).
Nh÷ng ®iÒu kiÖn sãng cña c¬n b·o 50 n¨m t¹i vÞ trÝ ®îc chØ râ lµ Hs = 7 m. HiÖu
øng dßng ch¶y thñy triÒu bá qua. Gi¶ thiÕt r»ng thiÖt h¹i ®¸ng kÓ sÏ xuÊt hiÖn nÕu
1/3 nh÷ng sãng cao nhÊt trong mét b¶n ghi sãng ®· cho cã kh¶ n¨ng dÞch chuyÓn c¸c
khèi ®¸. Nh vËy ®iÒu kiÖn tíi h¹n t¬ng øng (theo ®Þnh nghÜa Hs) víi nh÷ng sãng
®¬n víi ®é cao thiÕt kÕ H = 7 m. Chu kú sãng t¬ng øng ®îc lÊy lµ chu kú ®Ønh Tp,
cña phæ. Sö dông nh÷ng ph¬ng tr×nh (49) vµ (48), hoÆc SandCalc-Edit-Waves-
178
- Derive víi h = 20 m (trong Edit-Water), Hs = 7 m cho ta Tz = 9,3 s vµ Tp = 11,9 s. Sö
dông lý thuyÕt sãng tuyÕn tÝnh cã thÓ tÝnh to¸n biªn ®é vËn tèc quü ®¹o ®¸y tõ
®êng cong ®¬n ®iÖu trong h×nh 14, hoÆc b»ng viÖc sö dông SandCalc
Hydrodynamics-Orbital Velocity-Monochromatic víi ®Çu vµo lµ ®é s©u níc = 20 m,
®é cao sãng = 7 m, chu kú = 11,9 s ®Ó cã ®îc vËn tèc quü ®¹o sãng = 1,98 ms-1. Thùc
nghiÖm bëi Sumer vµ nnk. (1992) cho thÊy r»ng øng suÊt trît t¹i ®¸y do chuyÓn
®éng sãng xung quanh mét h×nh trô trßn th¼ng ®øng cã ®êng kÝnh D cã hÖ sè
khuÕch ®¹i, do vËn tèc dßng ch¶y t¨ng xung quanh h×nh trô theo mét hµm cña sè
Keulegan -Carpenter KC. Trong trêng hîp ®ang xÐt, víi KC =Uw T / D = 1,98 x 11,9
/ 2,0 = 11,9, hä nhËn ®îc hÖ sè khuÕch ®¹i øng suÊt trît t¹i ®¸y lµ 2,2. §iÒu nµy cã
nghÜa lµ (xem ph¬ng tr×nh (57)) hÖ sè khuÕch ®¹i lµm vËn tèc quü ®¹o t¨ng (2,2)1/2
= 1,48. Nh vËy vËn tèc quü ®¹o hiÖu qu¶ gÇn ch©n dµn lµ 1,48 x 1,98 = 2,94 ms-1.
§o¸n tríc r»ng kÝch thíc ®¸ yªu cÇu sÏ lín h¬n 10 mm, ®êng kÝnh tíi h¹n dcr
cã thÓ nhËn ®îc tõ ph¬ng tr×nh (79), víi Uw = 2,94 ms-1, T = 11,9 s, g = 9,81 ms- l vµ
s = 2,58:
97,9 x 2,943,08
0,62m .
d cr
11,91,08 9,812,58 12,08
Nh vËy ®¸ cã ®êng kÝnh 0,62 m (khèi lîng xÊp xØ = x 0,623 x 2650/6 = 0,34
tÊn) sÏ kh«ng chÞu ®îc h háng ®¸ng kÓ víi ®iÒu kiÖn b·o cã chu kú lÆp hiÕm h¬n 50
n¨m. Trªn thùc tÕ, v× hÖ sè vËn tèc xung quanh mét h×nh trô trßn gi¶m nhanh theo
kho¶ng c¸ch kÓ tõ h×nh trô, mét lo¹i ®¸ h¬i nhá h¬n cã lÏ ®ñ.
Nh÷ng yÕu tè bæ sung cã thÓ yªu cÇu cho nghiªn cøu ®Æc trng bao gåm:
• tÝnh to¸n dcr cho mét møc ®é chÆt chÏ h¬n cña thiÖt h¹i chÊp nhËn ®îc, vÝ dô
thiÖt h¹i do 1 % c¸c sãng cao nhÊt. §é cao sãng vît 1 % (xÊp xØ 1,5 Hs) cã thÓ tÝnh
to¸n b»ng viÖc gi¶ thiÕt ph©n bè Rayleigh cho ®é cao sãng (xem CIRIA/CUR, 1991)
• xÐt ®Õn nh÷ng c«ng tr×nh kh¸c, trong vïng l©n cËn ch©n dµn nh ®êng èng
hoÆc sù cã mÆt t¹m thêi cña thiÕt bÞ kÝch hoÆc cÇn cÈu nöa ch×m. HÖ sè vËn tèc tÝnh
to¸n cho c¸c c«ng tr×nh kh¸c ®îc nh©n víi hÖ sè (1,48) cho ch©n dµn khi tÝnh to¸n
vËn tèc quü ®¹o hiÖu qu¶
• trong c¸c khu vùc cã dßng ch¶y m¹nh (vÝ dô phÝa Nam BiÓn B¾c), cã thÓ cÇn
xÐt ®Õn nh÷ng hiÖu øng kÕt hîp cña sãng vµ dßng ch¶y b»ng c¸ch tÝnh to¸n øng suÊt
trît t¹i ®¸y cùc ®¹i b»ng ph¬ng ph¸p m« t¶ trong môc 5.3. ViÖc tÝnh to¸n chu kú
lÆp cña øng suÊt lµ mét bµi to¸n vÒ x¸c suÊt kÕt hîp, ®îc m« t¶ trong môc 12.3
• mét líp läc cña vËt chÊt nhá h¬n thêng ®Æt n»m díi líp ®¸ chÝnh, ®Ó ng¨n
ngõa c¸t lät qua c¸c khe hë gi÷a nh÷ng t¶ng ®¸.
Híng dÉn Xãi ë nh÷ng c«ng tr×nh biÓn (Whitehouse, 1997) ®a ra m« t¶ chi tiÕt
nh÷ng ph¬ng ph¸p chèng xãi.
179
- 13.3. X©m thùc trÇm tÝch t¹i c«ng tr×nh lÊy níc
Mét c«ng tr×nh lÊy níc lµm m¸t cho mét nhµ m¸y ®iÖn ®îc ®Æt ngoµi kh¬i trªn
®¸y biÓn cã c¸t. Ph¶i tÝnh to¸n lîng trÇm tÝch hµng n¨m ®i vµo theo níc ®Ó ®Ò
phßng vµ chuÈn bÞ mét bÓ l¾ng trÇm tÝch nÕu cÇn thiÕt ng¨n trÇm tÝch ®i vµo c«ng
tr×nh. §é s©u níc trung b×nh t¹i vÞ trÝ c«ng tr×nh lÊy níc lµ 7 m, ®é cao cña t©m
c«ng tr×nh lÊy níc lµ 3 m, ®é lín thuû triÒu lµ 2 m vµ ®¸y thuéc lo¹i c¸t mÞn ®îc
chän läc tèt víi d10 = 0,07 mm, d50 = 0,12 mm vµ d90 = 0,18mm. Dßng ch¶y thñy triÒu
chiÕm u thÕ trong khu vùc víi vËn tèc dßng ch¶y cùc ®¹i khi triÒu yÕu lµ 0,4 ms-1 vµ
0,6 ms-1 khi triÒu cêng. Sãng cã ®é cao Hs = 0,5 m víi tÇn suÊt 50%, Hs = 1,0 m víi
tÇn suÊt 10% vµ Hs = 3 m víi tÇn suÊt 1%. Híng sãng trung b×nh lµ 45o so víi
híng dßng ch¶y.
Nh÷ng hiÖu øng cña c¶ dßng ch¶y vµ sãng cã vÎ quan träng trong viÖc lµm cho
trÇm tÝch ®i vµo tr¹ng th¸i l¬ löng vµ trong mét n¨m mét ph¹m vi réng cña nh÷ng tæ
hîp sÏ xuÊt hiÖn. Víi môc ®Ých minh häa, sö dông mét gi¸ trÞ ®¬n cña sãng vµ dßng
ch¶y, tøc lµ trung b×nh dßng ch¶y cùc ®¹i cña thñy triÒu U = 1/2(0,4 + 0,6) = 0,5 ms-1
vµ ®é cao sãng cã tÇn suÊt 10% lµ Hs= 1,0m. Chu kú sãng kh«ng cho, v× vËy nhËn
®îc ®¸nh gi¸ nhê sö dông ph¬ng tr×nh (49), cho ta Tz= 3,51.
C«ng thøc sö dông cho ph©n bè nång ®é do kÕt hîp sãng vµ dßng ch¶y lµ ph¬ng
tr×nh (115 a-e), yªu cÇu ®Çu vµo lµ sãng ®¬n. Sãng ®¬n t¬ng ®¬ng ®èi víi môc ®Ých
nµy lµ H = Hs/21/2 = 0,707 ms-1 vµ T = Tp = 1,28Tz = 4,50 s. Biªn ®é vËn tèc quü ®¹o
®¸y nhËn ®îc tõ h×nh 14 (xem vÝ dô 4.2) lµ Uw = 0,225 ms-1. øng suÊt trît cùc ®¹i
vµ trung b×nh t¹i ®¸y nhËn ®îc b»ng viÖc sö dông ph¬ng ph¸p trong môc 5.3 víi
hÖ sè 'DATA 13' tõ b¶ng 9, vµ sö dông ®é nh¸m liªn quan ®Õn h¹t z0 = d50/12 = 0,01
mm, nªn ta cã m = 0,285 N m-2 vµ max = 0,481 Nm-2.
TiÕp theo, tÝnh to¸n kÝch thíc h¹t trung vÞ cña trÇm tÝch l¬ löng nhê sö dông
ph¬ng ph¸p Van Rijn, ph¬ng tr×nh (97). Ph¬ng ph¸p nµy, thËt ra chØ dù ®Þnh cho
dßng ch¶y, yªu cÇu øng suÊt trît ma s¸t líp ®Öm ®èi víi dßng ch¶y cã thÓ tÝnh to¸n
nhê sö dông ph¬ng tr×nh (34) ®Ó cã T0s = 0,228 Nm-2, ngìng øng suÊt trît t¹i ®¸y
cã thÓ tÝnh to¸n tõ ph¬ng tr×nh (76) ®Ó cho ta cr = 0,151 N m-2. Còng yªu cÇu d16 vµ
d84, cã thÓ tÝnh to¸n b»ng néi suy log chuÈn (môc 2.2) gi÷a nh÷ng gi¸ trÞ ®· biÕt d10,
d50, d90 ®Ó cã d16 = 0,0793 mm vµ d84 = 0,164 mm. Ph¬ng tr×nh (97) cho kÝch thíc
h¹t l¬ löng trung vÞ, d50,s = 0,106 mm. VËn tèc ch×m l¾ng cña kÝch thíc h¹t nµy do
ph¬ng tr×nh (102) ®a ra, ws = 0,00633 ms-1.
Nh÷ng ®¹i lîng trªn cã thÓ sö dông trong ph¬ng tr×nh (115), víi z = 3 m, cho
ta nång ®é t¹i t©m c«ng tr×nh lÊy níc lµ C(z) = 6,38 x 10-6 theo thÓ tÝch, hoÆc 16,9
mgl-1 theo khèi lîng.
Nh÷ng tÝnh to¸n nµy cã thÓ thùc hiÖn b»ng c¸ch kh¸c nhê sö dông SandCalcc
nh sau:
h Edit-Water, cho h = 7 m
d16, d84 Edit-Bed Material-Derive theo d10, d50, d90
H, T Edit-Waves-Derive, víi Hs =1,0 m
180
- Uw Hydrodynamics-Waves-Orbital velocity-Monochromatic
m, max Hydrodynamics-Waves + Currents-Total Shear-stress-Soulsby
Hydrodynamics-Currents-Skin-friction-Soulsby
0s
Sediments-Threshold-Bed shear-stress-Soulsby
0s
d50 Sediments-Suspension-Suspended Grain size-Van Rijn
ws Sediments-Suspension-Settling velocity-Soulsby
C(z) Sediments-Suspension-Concentration-Waves + Currents
Lu ý r»ng:
(a) kÝch thíc h¹t l¬ löng ph¶i th« h¬n mét chót nÕu nh÷ng hiÖu øng sãng ®îc
kÓ ®Õn trong tÝnh to¸n, vÝ dô nÕu d50,s = d50,b = 0,12 mm, nã lµm gi¶m nång ®é chØ ®Õn
1,9 mgl-1
(b) ph¬ng tr×nh sö dông ®Ó tÝnh to¸n ph©n bè nång ®é ®îc thiÕt kÕ cho ®¸y
ph¼ng, ®iÒu kiÖn dßng trÇm tÝch s¸t ®¸y, trong khi sö dông ®¸y cã vÎ h¬i gîn c¸t cho
sãng vµ dßng ch¶y. §¸y gîn c¸t lµm t¨ng nång ®é t¹i z = 3 m.
§Ó tÝnh to¸n x©m thùc trÇm tÝch hµng n¨m, cÇn bæ sung nh÷ng bíc sau:
• thùc hiÖn nh÷ng tÝnh to¸n trªn cho mçi tæ hîp ®iÒu kiÖn sãng vµ dßng ch¶y
thñy triÒu xuÊt hiÖn trong mét n¨m ®iÓn h×nh, céng nång ®é lÊy träng sè theo tÇn sè
xuÊt hiÖn cña chóng, sö dông c¸ch tiÕp cËn x¸c suÊt m« t¶ trong môc 12.3, nhËn
®îc nång ®é trung b×nh n¨m
• nh©n nång ®é trÇm tÝch trung b×nh hµng n¨m víi thÓ tÝch níc ®i qua c«ng
tr×nh lÊy níc trong mét n¨m, ®Ó cã x©m thùc trÇm tÝch hµng n¨m.
Nh÷ng nh©n tè bæ sung cã thÓ cÇn xÐt ®Õn bao gåm:
• tÝnh to¸n rÊt nh¹y c¶m víi nhiÖt ®é níc, bëi v× ®é nhít gi¶m lµm cho vËn tèc
ch×m l¾ng lín h¬n; mét sù t¨ng nhiÖt ®é tõ l0oC ®Õn 30oC lµm gi¶m nång ®é t¹i
z = 3 m tõ 16,9 mgl-1 xuèng 0,20 mgl-1.
• nÕu trÇm tÝch ®¸y ®¬c cÊp phèi réng, cã thÓ bao gåm c¸c nhãm mÞn nh bïn
vµ/ hoÆc nh÷ng nhãm kh«ng th« kh«ng l¬ löng, th× ph¶i sö dông ph¬ng ph¸p ®Ò cËp
®Õn mét d¶i nhiÒu kÝch thíc h¹t
• nÕu chØ nh÷ng kÝch thíc h¹t th« h¬n kÝch thíc ®· cho liªn quan ®Õn sù vËn
hµnh cña nhµ m¸y, (vÝ dô nÕu trÇm tÝch rÊt mÞn ®i th¼ng vµo), th× cÇn ph©n chia
tÝnh to¸n víi c¸c kÝch thíc h¹t
• v× tÝnh nh¹y c¶m nµy, mét vµi ®o ®¹c nång ®é trÇm tÝch l¬ löng t¹i tuyÕn lµ qói
gi¸ cho nh÷ng môc ®Ých hiÖu chØnh vµ kiÓm ®Þnh.
Ngoµi nh÷ng nhµ m¸y ®iÖn, viÖc ng¨n ngõa x©m thùc trÇm tÝch lµ quan träng
®èi víi nh÷ng c«ng tr×nh lÊy níc cho c¸c nhµ m¸y läc dÇu vµ khö muèi.
13.4. Båi lÊp luång dÉn n¹o vÐt
Mét luång dÉn ®Õn mét bÕn c¶ng sÏ ®îc n¹o vÐt xuyªn qua mét vïng c¸t n«ng,
vµ yªu cÇu tÝnh to¸n n¹o vÐt b¶o tr× hµng n¨m. Luång ®îc ®Æt th¼ng gãc víi ®êng
181
- bê, do vËy nh÷ng dßng ch¶y thñy triÒu b¨ng qua nã theo híng vu«ng gãc, híng
sãng u thÕ lµ däc theo kªnh. §é s©u níc trung b×nh t¹Þ b·i n«ng lµ 5m vµ kªnh
®îc n¹o vÐt tíi mét ®é s©u th«ng thuyÒn trung b×nh lµ 10 m víi m¸i h¬i dèc vµ
chiÒu réng 100m t¹i phÇn s©u nhÊt. §é lín thñy triÒu lµ 2 m, dßng ch¶y thñy triÒu
b»ng nhau khi triÒu lªn vµ triÒu xuèng víi vËn tèc cùc ®¹i 0,6 ms-1 khi triÒu yÕu vµ
1,0 ms-1 khi triÒu cêng. Sãng cã tÇn suÊt 50% lµ Hs = 1,0 m vµ víi tÇn suÊt 10% lµ
Hs = 2,5 m vµ 1% lµ Hs = 2,5 m. C¸t trªn níc n«ng ®îc chän läc tèt, víi d10 = 0,1
mm, d50 = 0,2 mm, d90 = 0,3 mm.
C¬ chÕ båi lÊp (ë d¹ng c¸c sè h¹ng ®¬n gi¶n) ®îc t¹o ra mét phÇn bëi dßng ch¶y,
vµ mét phÇn bëi sãng. Dßng ch¶y mang trÇm tÝch qua b·i n«ng vµ vµo trong luång,
t¹i ®ã dßng ch¶y sÏ chËm h¬n (do tÝnh liªn tôc cña khèi lîng) trong ®é s©u lín h¬n.
Dßng ch¶y chËm h¬n mang Ýt trÇm tÝch h¬n ®i sang m¸i bªn kia cña luång, do ®ã cã
sù båi tô rßng cña trÇm tÝch trong luång. Qu¸ tr×nh trÇm tÝch mang vµo nhiÒu h¬n lµ
mang ra khái luång xuÊt hiÖn c¶ khi triÒu xuèng lÉn triÒu lªn, vµ sù ®èi xøng cña
chóng cã nghÜa lµ kh«ng cã sù dÞch chuyÓn thùc tÕ cña luång. Sãng gi¶ thiÕt cã cïng
®é cao vµ chu kú trong luång còng nh ngoµi b·i n«ng (tøc lµ bá qua khóc x¹ ®Ó ®¬n
gi¶n sù m« t¶). VËn tèc quü ®¹o sãng lµm t¨ng møc ®é vËn chuyÓn trÇm tÝch, nhng
v× nh÷ng vËn tèc quü ®¹o trªn b·i n«ng lín h¬n so víi trong níc s©u h¬n cña luång,
vËn chuyÓn trÇm tÝch trªn b·i n«ng lín h¬n. Nh vËy sãng kh«ng chØ lµm t¨ng, mµ
cßn cñng cè c¬ chÕ båi lÊp do dßng ch¶y.
§Ó minh häa, sÏ tÝnh to¸n møc båi lÊp ®èi víi mét vËn tèc dßng ch¶y vµ ®é cao
sãng ®¬n. Nh÷ng gi¸ trÞ ®îc chän lµ vËn tèc dßng ch¶y lín nhÊt cña thñy triÒu
trung b×nh, U = 1/2 (0,6 + 1,0) = 0,80 ms-1 vµ ®é cao sãng Hs = 1,0 m víi tÇn suÊt
10%. Chu kú c¾t qua kh«ng ®· kh«ng cho, v× vËy nã ®îc tÝnh to¸n tõ ph¬ng tr×nh
(49) lµ Tz = 3,51 s. §é s©u níc lÊy theo mùc níc trung b×nh, h = 5 m bªn ngoµi, vµ
10 m ë trong kªnh, v× biªn ®é thñy triÒu t¬ng ®èi nhá.
Tríc hÕt, tÝnh to¸n suÊt vËn chuyÓn trÇm tÝch qin, do kÕt hîp sãng vµ dßng ch¶y
trªn níc n«ng, sö dông c«ng thøc Soulsby - Van Rijn (ph¬ng tr×nh (136)). §¸y gi¶
thiÕt cã gîn c¸t víi z0 = 6 mm, vµ hÖ sè ma s¸t chØ do dßng ch¶y CD ®îc tÝnh to¸n tõ
ph¬ng tr×nh (37) víi ®Çu vµo h = 5 m, z0 = 0,006 m, cho ta CD = 0,00488.
VËn tèc quü ®¹o sãng RMS nhËn ®îc tõ h×nh 14 nhê sö dông ®êng cong
JONSWAP (xem vÝ dô 4.3) víi ®Çu vµo Hs = 1,0m, Tz= 3,51 s, h = 5 m, cho ta Urms =
0,208 ms-1.
Ngìng vËn tèc dßng ch¶y ®îc tÝnh to¸n tõ ph¬ng tr×nh (71) víi h = 5 m, d50 =
0,2 mm, d90 = 0,3 mm, cho ta Ucr = 0,391 ms-1.
SuÊt vËn chuyÓn trÇm tÝch ®îc tÝnh to¸n tõ ph¬ng tr×nh (136) víi h = 5 m, U =
0,8 ms-1, CD = 0,00488, Urms = 0,208 ms-1, Ucr = 0,391 ms-1, J = 0 (®¸y n»m ngang), cho
ta qt = l,838 x 10-4 m2s-1.
B©y giê tÝnh to¸n suÊt vËn chuyÓn trÇm tÝch trong r·nh, cã h = 10 m. VËn tèc
dßng ch¶y qua r·nh cã thÓ tÝnh to¸n tõ ph¬ng tr×nh liªn tôc : U1h1 = U2h2, cho U =
0,8 x 5/10 = 0,4 ms-1. Mét lo¹t tÝnh to¸n gièng nh trªn níc n«ng, nhng sö dông h
182
- = 10 m, U = 0,4 ms-1 cho CD = 0,00388, Urms =0,083 ms-1, Ucr = 0,416 ms-1 vµ qt = 5,28
x 10-8 m2s-1.
Nh÷ng tÝnh to¸n trªn, tríc hÕt qua níc n«ng vµ sau ®ã trong r·nh, cã thÓ
thùc hiÖn b»ng c¸ch kh¸c nhê sö dông SandCalc nh sau.
h Edit-Water, ®Æt h = 5 m
Tz Edit-Waves-Derive, víi Hs = 1,0m
CD Hydrodynamics-Currents-Total shear-stress-Logarithmic
Urms Hydrodynamics-Waves-Orbital velocity-Spectrum
Ucr Sediments-Threshold-Current-Van Rijn
qt Sediments-Total load-Waves & Currents-Soulsby/ Van Rijn
Trë l¹i Edit-Water, ®Æt h = 10 m, vµ lÆp l¹i tÝnh to¸n.
SuÊt vËn chuyÓn trÇm tÝch rßng lµ 1,838 x 10-4 x 5,28 x 10-8 = 1,837 x 10-4 (vËn
chuyÓn ra khái r·nh rÊt nhá trong trêng hîp nµy). TrÇm tÝch l¾ng ®äng trªn 100 m
chiÒu réng cña r·nh, víi tèc ®é båi lÊp trung b×nh trªn giê, kÓ c¶ kh«ng gian xèp cña
trÇm tÝch ®· l¾ng ®äng víi e = 0,40, lµ:
(1,837 x 10-4 x 3600)/(100(1 - 0,40)) = 11 mm/h
§iÒu nµy chØ x¶y ra khi dßng ch¶y thñy triÒu lín nhÊt, vµ tèc ®é båi lÊp vµo thêi
®iÓm thñy triÒu kh¸c sÏ Ýt h¬n nhiÒu, v× vËy tæng sè lÊy trung b×nh qua mét chu kú
thñy triÒu sÏ nhá h¬n.
§Ó tÝnh to¸n yªu cÇu n¹o vÐt b¶o tr× hµng n¨m, cÇn thiÕt bæ sung nh÷ng bíc
sau:
• thùc hiÖn tÝnh to¸n nh trªn cho mçi tæ hîp ®iÒu kiÖn sãng vµ dßng ch¶y thñy
triÒu xuÊt hiÖn trong mét n¨m ®iÓn h×nh, vµ lÊy tæng suÊt vËn chuyÓn trÇm tÝch cã
träng sè theo tÇn sè xuÊt hiÖn cña chóng, sö dông c¸ch tiÕp cËn x¸c suÊt m« t¶ trong
môc 12.3, nhËn ®îc tèc ®é båi lÊp trung b×nh n¨m
• thùc hiÖn nh÷ng tÝnh to¸n t¹i mét sè ®iÓm ®¹i diÖn däc theo kªnh, vµ céng
lîng trÇm tÝch cho tõng ®o¹n.
Nh÷ng yÕu tè bæ sung cã thÓ quan träng trong tÝnh to¸n båi lÊp luång bao gåm:
• khóc x¹ sãng trªn luång; díi nh÷ng ®iÒu kiÖn nhÊt ®Þnh, cã thÓ g©y ra 'néi
ph¶n x¹ toµn bé' nh÷ng sãng ®Õn xiªn
• sù lÖch nh÷ng dßng ch¶y ®Õn xiªn mét gãc, do hiÖu øng chËm l¹i khi dßng ch¶y
c¾t qua luång
• m¸i rÊt dèc cã thÓ g©y ra mét khu vùc bãng, hoÆc khu vùc hoµn lu ngîc, ®èi
víi dßng ch¶y
• trÇm tÝch l¬ löng cã thÓ mét phÇn ®îc mang ngang qua luång mµ kh«ng l¾ng
®äng, ®Æc biÖt ®èi víi nh÷ng luång hÑp, nh÷ng trÇm tÝch mÞn, vµ nh÷ng dßng ch¶y
nhanh, dÉn ®Õn sù gi¶m ‘hiÖu øng bÉy'
• khi luång ®Çy lªn víi trÇm tÝch, vµ trë thµnh n«ng h¬n, hiÖu øng bÉy cña nã
gi¶m, dÉn ®Õn sù gi¶m tèc ®é båi lÊp theo hµm mò
183
- • nh÷ng luång dÉn thêng ®Çy bïn, còng nh c¸t, v× vËy yªu cÇu kü thuËt tÝnh
to¸n kh¸c nhau.
Nh÷ng nguyªn lý tæng quan bao hµm trong båi lÊp luång ®îc bµn luËn bëi Lean
(1980), vµ nh÷ng kü thuËt m« h×nh sè phøc t¹p cho tÝnh to¸n båi lÊp tiªn tiÕn h¬n
®îc m« t¶ bëi Van Rijn (1985) vµ Miles (1995).
Nh÷ng ph¬ng ph¸p luËn m« t¶ trong trêng hîp nghiªn cøu nµy ®èi víi båi lÊp
nh÷ng luång tµu réng cã thÓ ¸p dông cho sù båi lÊp c¸c hè ®µo réng kh¸c (vÝ dô
nh÷ng èng tuynen ch×m trong níc vµ nh÷ng èng lÊy níc lµm m¸t), vµ cho nh÷ng
r·nh hÑp t¹m thêi ®Ó ch«n nh÷ng èng dÉn, nh÷ng nguån th¶i vµ c¸p ngÇm. Nh÷ng
nguyªn lý tæng quan cã thÓ øng dông theo híng ngîc l¹i ®èi víi sù ph¸t t¸n c¸c
®èng ®Êt vµ c¸c ®èng trÇm tÝch díi níc.
184
- 185
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