baigiangthuyluc_2009-ts. huynh cong hoai

8/7/2019 BaigiangThuyLuc_2009-Ts. Huynh Cong Hoai

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Tom tat bai giang - TS Huynh cong Hoai HBK tp HCM 1

THUY LC(HYDRAULICS)

TS. Huynh cong HoaiBo mon C Lu Chat - Khoa Ky thuat Xay dng H Bach Khoa tp HCMwww4.hcmut.edu.vn/~hchoai/baigiang

NN An, NT Bay, LS Giang, HC Hoai, NT Phng, LV Dc, Giao trnhThuy lc , Lu hanh noi bo HBK tp HCM, 2005

2. Nguyen canh Cam va cac tac gia Thuy lc tap II, NXB DH va THCN,1978

Nguyen canh Cam va cac tac gia Bai tap Thuy lc tap II, NXB DH vaTHCN, 1978

French R.H Open channel Hydraulics. McGra-Hill, Singapore 1986

Koupitas C.G. Elements of Computation Hydraulics . Pentics Pres, 1983

6. Haestad press. Computer Application Hydraulic Engineering , 2002

TAI LIEU THAM KHAO


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Chng:

1 DONG CHAY EU TRONG KENH H1.1 KHAI NIEM CHUNG

Khi dong chay eu xay ra th:- Chieu sau, dien tch t va bieu o phan bo van toc tai cac mat cat doctheo dong chay khong oi .

- ng dong, mat thoang, ng nang va ay kenh song song vi nhau.

Dong chay eu Dong khong euDong chay eu co ap Dong chay eu khong ap ( kenh h)

ieu kien can e co dong chay eu- Hnh dang mat cat t khong oi (kenh lang tru)- o doc khong oi (i = const)- o nham khong oi ( n = const)

1.2 CONG THC CHEZY VA MANNINGChezy (1769)

RiCV

C= Rn

1 61

Manning

iARn

Q 321

321 ARn

K iKQ

Kc goi la modul lu lng

Cong thc tnh toan dien tch t va chu vi t hnh thang

m = cotg : he so mai doc

: dien tch t

212 mhb P = chu vi t


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1.3 XAC NH HE SO NHAM

Cac yeu to anh hng en he so nham nh sau

o nham be mat Lp phu thc vat Hnh dang mat cat kenh

Vat can Tuyen kenh S boi xoi Mc nc va lu lng

1.3.1 Trng hp mat cat kenh n gian

Phng phap SCS (soil Conversation Service Method)Phng phap dung bangPhng phap dung hnh anhPhng phap dung bieu o lu toc

)95,0(78,6

)1( 61

x

hxn

h: Chieu sau dong chay

x =8,0

2,0

U

U U0,2: Van toc tai v tr 2/10 cua chieu sau hay 0,8 h tnh t ay,U0,8: Van toc tai v tr 8/10 cua chieu sau hay 0,2 h tnh t ay

Phng phap cong thc thc nghiem

Simons va Sentruk (1976): n = 0,047d1/6d: ng knh hat cua long kenh (mm).

1.3.2 Trng hp mat cat kenh phc tap

Cox(1973)

A

An

n

N

i

ii

e

1

A1: Dien tch t cua tng dien tch n gian

A: Dien tch t cua toan bo mat cat.

1.4 TNH TOAN DONG EU:

1.4.1. Bai toan kiem tra

a. Xac nh lu lng :

n1

n2

n

3

A2

A1 A3

Biet : A, i, n iARn

Q 321

b. Xac nh o sau h :

Biet : i, n, Q, hnh dang mat cat kenh

iARn

Q 321

h

Th dan -> h


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oi vi mat cat hnh tron co the dung bieu o

Modul lu lng:i

QAR

nK 3/2

1

Modul lu lng khi

chay ngap :

Tnh t so : K/Kng

T : K/Kng h/D h

1.4.2 Bai toan thiet ke

Dung bieu o

Neu kenh co cung ieu kien : i, n, mat cat co hnh dang li nhat ve thuy lc la :

a. Mat cat co li nhat ve thuy lc

- Co cung dien tch t A nhng cho lu lng ln nhat- Cung chay vi lu lng nhng co dien tch t A nho nhathoac

iARn

Q 321

T Mat cat co R ln hay co Pmin se la matcat co li nhat ve thuy lc

Nh vay trong tat ca cac loai mat cat, mat cat hnh tron la mat cat co linhat ve thuy lc

b. Mat cat hnh thang co li nhat ve thuy lc

Neu cac mat cat hnh thang cung mot dien tch t A, cung mai doc m, th matcat hnh thang nao co chu vi t nho nhat se la mat cat co li nhat ve thuy lc.

T so gia b/h e co mat cat co li nhat ve thuy lc c xac nh nhsau:


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c. Thiet kekenh

- Xac nh lu lng Q ( ma, nhu cau xa nc )- Xac nh o nham n ( loai vat lieu long kenh..)

- Xac nh o doc i ( phu thuoc a hnh ..)

- Xac nh hnh dang mat cat phu thuoc yeu cau thiet ke ( hnh tron, hnhthang, hnh ch nhat . )

- Xac nh kch thc kenh :

+ Mat cat ch nhat : xac nh b va h , phai cho b e tmh hoac ngc lai, hoac dung ieu kien b/h cua mat catco li nhat ve thuy lc+ Mat cat hnh thang : xac nh m da vao ieu kien onnh mai doc. Xac nh b va h nh trng hp mat cat

hnh ch nhat+ Mat cat hnh tron : xac nh ng knh D da vao tso o sau h/D cho phep trong cong

- Kiem tra van toc trong kenh phai thoa man : VKL < V < VKX

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3

o th dung e tnh toan cong tron

h/D

A/Ang

B/D

K/Kng

V/Vng

R/Rng

P/Png


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Cau 1: Cau nao sau ay ung:a) Dong eu ch co the xay ra trong kenh lang tru.

b) Trong kenh lang tru ch xay ra dong eu.c) Dong khong eu ch xay ra trong song thien nhien.d) Trong kenh co dien tch mat cat t khong oi th luon luon co dong eu

Cau 2: Dong chay eu trong kenh h co:

a) ng nang, ng mat nc va ay kenh song song nhau.

b) Dien tch mat cat t va bieu o phan bo van toc doc theo dong chay khong oi.c) Ap suat tren mat thoang la ap suat kh tri.

d) Ca ba cau tren eu ung.

CAU HOI TRACNGHIEM:

Cau 4: Trong dong chay eu:

a) Lc ma sat can bang vi lc trong trng chieu len phng chuyen ong.

b) Lc ma sat can bang vi lc quan tnh.

c) Lc gay nen s chuyen ong la lc trong trng chieu len phng chuyen ong.

d) a va c eu ung.

Cau 3: Trong kenh co mat cat hnh tron ng knh D:

a) Van toc trung bnh at gia tr cc ai khi chieu rong mat thoang B = 0,90D.

b) Van toc trung bnh at gia tr cc ai khi chieu rong mat thoang B = 0,78D.

c) Van toc trung bnh at gia tr cc ai khi chieu rong mat thoang B = 0,46D.

d) Van toc trung bnh at gia tr cc ai khi chieu rong mat thoang B = 0,25D.

Ve nha suy luan ???


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Cau 5: Trong kenh lang tru co lu lng khong oi:a) o sau dong eu tang khi o doc i giam.b) o sau dong eu khong oi o doc i tang.

c) o sau dong eu tang khi o doc i tang.d) Ca 3 cau tren eu sai.

Cau 6: Mat cat kenh co li nhat ve mat thuy lc :

a) Co the ap dung oi vi kenh co nhieu loai mat cat khac nhau.

b) at c lu lng cc ai neu gi dien tch mat cat t la hang so.c) at c dien tch mat cat t toi thieu neu gi lu lng la hang so.

d)Ca ba cau tren eu ung.


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Ch ng:

DONG ON NH KHONG EU BIEN OI DAN TRONG KENH H

2.1 CAC KHAI NIEM2.1.1 Nang lng rieng cua mat cat:

Nang lng toan phan Eg

Vha

g

VpzE

2cos

2

22

hh

a

Mat chuan nam ngang

Mat thoang

ay kenh

0 0

g

VhaE

2

2 o doc ay kenh nho cos = 1

Nang lng rieng cua mat catE0

vi mat chuan nam ngang i

qua iem thap nhat cua matcat o.

2

22

022 gA

Qh

g

VhE

Ta co the phan 2 loai chuyen ong khong eu trong kenh:

- Chuyen ong khong eu bien oi dan.- Chuyen ong khong eu bien oi gap.

E0min

hcr

E0

h

E0

h

Bien thien cua E0 theo h

Q = const

o

2

22

022 gAQh

gVhE

ng cong E0 = f(h)

Khi h E0 E0 hng phan giac th nhat E0 = h,la 1 ng tiem can

Khi h 0 E0

Truc hoanh E0 la 1 ng tiem can

2.1.3 o sau phan gii ( hcr):o sau phan gii hcr la o sau e cho nang lng rieng cua mat cat o at gia tr cc tieu.

00

crhhdh

dE

dh

dA

Ag

Q

gA

Qh

dh

d

dh

dE3

2

2

20 2

21

2

dA/dh = B

3

20 1

gA

BQ

dh

dE

phng trnh tnho sau phan gii: g

Q

B

A

cr

cr23

Trong o : Acr va la dien tch mat cat t , Bcr be rong mat thoang tnh vi o sau phan gii hcr.


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Kenh hnh ch nhat: v A = bh va B = bnen

3

2

32

2

g

q

gb

Qhcr

q = Q/b: lu lng tren 1 n v be rong kenh goi la lu lng n v

Kenh tam giac can: v A = mh2 va B = 2mh nen

Kenh hnh thang: cong thc gan ung

52

22

gm

Qhcr

crCNNN

cr hh

2105,0

31

b

mhcrCNN

32

2

gb

QhcrCN

trong o

Kenh hnh tron: ta co the ap dung cong thc gan ung25,02

26,0

01,1

g

Q

dhcr

vi ieu kien 85,002,0 d

hcr

2.1.4 So Froude

lctrong tnhquanlcsotvilet3

2

2

gABQFr

- He so sa cha ong nang. B - Chieu rong mat thoang

NB

gAC van toc truyen song nhieu ong nho trong nc tnh

so Froude the hien t so gia van toc trung bnh cua dong chay va van toc truyen song.

2.1.5 o doc phan gii

o doc phan gii icr la o doc cua mot kenh lang tru, ng vi mot lu lng chotrc, o sau dong chay eu trong kenh h0 bang vi o sau phan gii hcr.

Xac nh icr crcrcrcr iRACiRACQ 000

g

iRCA

B

A

g

Q

B

A crcrcrcr

cr

cr

cr

cr

2323

Ngoai ra

crcr

cr

crcrcr

cr

cr BC

gP

BRC

gA

i 22 suy ra

-Neu i< icr th h0 > hcr.-Neu i >icr th h0 < hcr.-Neu i = icr th h0 = hcr.


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h 0

00

h

E

00

h

E

Trangthai chay

Phan biet theo

o sauh

SoFroude

Vantoc

Em h > hcr Fr < 1 V < C

Phangii h = hcr Fr = 1 V = C

Xiet h < hcr Fr > 1 V > C 00

h

E

Y ngha vat ly trang thai chay

Vi C van toc truyen song trong nc tnh:B

gAC

B : be rong mat thoang va A dien tch t

Fr=0Nc tnh

Fr 1Chay xiet

2.1.6.Cac trang thai chay

h 0

00

h

E

00

h

E

Type of flow Way to determine

Depthh

Froudenumber

velocity

Subcriticalflow h > hcr Fr < 1 V < C

Criticalflow h = hcr Fr = 1 V = C

Super-critical flow h < hcr Fr > 1 V > C 00

h

E

2.1.6. Types of flow:

Subcritical flow :is the most common in nature and is relatively deep and slowmoving.

Supercritical flow :is less common and is characterised by a very fast, relativelyshallow flowHowever, both may occur in the same channel at the same discharge

The ways to determine the types of flow

Specific energy curve E0 = f(h)

E0min

hcr

E0

h

E0

h

Q = const

o

Supercriticalflow

Subcritical flow


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2.1.7. Y ngha dong chay em va xiet

(ii) Dong chay qua ca cong

Cacong

Ca ieukhien mcnc

Mat nc sau cong

hcr

Chay em

Dong chay em qua cong

Ca congchuyen

ong xuong

hcr

Chay xiet

Mc nc conh Ncnhay

Dong chay xiet qua cong

Vi C van toc truyen song trong nc tnh:B

gAC

B : be rong mat thoang va A dien tch t

Fr=0

Nc tnh

Fr 1Chay xiet

(i) Lan truyen song trong dong chay

2.2 PHNG TRNH VI PHAN C BAN CUA DONG ON NH, KHONG EUBIEN OI DAN TRONG KENH H

0ds

dz

dhl

a

h

z

ng matnc

ng nang

V

Mat chuan

g

Vha

g

VpzE

22

22

g

V

ds

d

ds

dhi

g

V

ds

d

ds

dh

ds

da

ds

dEJ

22

22

Xem qui luat ton that doc ng cua dongkhong eu = dong eu=> J c tnh theo cong thc Chezy:

2

2

22

2

2

2

K

Q

RCA

Q

RC

VJ

ds

dA

gA

Q

gA

Q

ds

d

g

V

ds

d3

2

2

22

22

A = f{s,h(s)}ds

dhB

s

A

ds

dh

h

A

s

A

ds

dA

ds

dhB

s

A

gA

Q

g

V

ds

d3

22

2

ds

dhB

s

A

gA

Q

ds

dhi

RCA

Q3

2

22

2

3

2

2

22

2

1

.1

gA

BQs

A

gA

RC

RCA

Qi

dsdh

lang tru, A/s = 0 23

2

22

2

11

Fr

Ji

gA

BQ

RCA

Qi

ds

dh


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2.3 CAC DANG NG MAT NC TRONG KENH LANG TRU

2.3.1 Trng hp kenh co o doc thuan i > 0

Moun lu lng K RK = K(h) = CA

JKQ

ng vi o sau dong eu h0

ng vi o sau dong khong eu h

0000 RACK iKQ 0

K = CA R

iFr

KK

ds

dh2

22

01

1

2

3

2

22

2

11

Fr

Ji

gA

BQ

RCA

Qi

ds

dh

a. Trng hp kenh lai: 0 i icr

NK

K

aI

bI

0 < i < icr

B

hcr

h0

Mc nc tren khu aI :

hcr 0

ng nc dang

h ---> K ---> ts ---> 1

Fr2 ---> 0 ms ---> 1---> i

ng mat nc nam ngang

h ---> hoK ---> Ko ts ---> 0

Fr2 < 1 ms > 0--->0

ng mat nc tiem can vi ng N-N


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NK

K

aI

bI

0 < i < icr

B

hcr

h0Mc nc tren khu bI :

hcr 0

ng nc ha

h ---> hcrK < K0 ts < 0

Fr2 --->1 ms ---> 0+---> -

ng mat nc thang goc vi K-K

h ---> hoK ---> Ko ts ---> 0

Fr2 < 1 ms > 0--->0

ng mat nc tiem can vi ng N-N

NK

K

aI

bI

0 < i < icr

B

hcr

h0

Mc nc tren khu cI :

h < hcr < h0

K < K0 ts < 0K2o / K2 > 1

Fr2 > 1 ms < 0

ng nc dang

h ---> hcrK < K0 ts < 0

Fr2 --->1 ms ---> 0----> +

ng mat nc thang goc vi K-K


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b. Trng hp kenh doc: 0 i cr i

Mc nc tren khu aII :

h0 0

ng nc dang

h ---> K ---> ts ---> 1

Fr2 ---> 0 ms ---> 1---> i

ng mat nc nam ngang

h ---> hcrK > Ko ts ---> 0

Fr2 ---> 1 ms > 0--->

ng mat nc thang goc ng K-K

aII

bII

cII

0


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16/57


hcr

i icr

Nc nhay

Nc nhay

cII

bII

Cac th du ve ng mat nc


17/57


hcr

KK

i= 0

Co

hcr

KK

i< 0

C

Cac th du ve ng mat nc

Cau 4. Mot kenh co o doc i > icr, o sau nc trong kenh h < h0.a) o sau nc giam doc theo chieu dai kenh.b) Nang lng rieng cua mat cat tang doc theo chieu dai kenh.c) Nang lng rieng cua mat cat giam doc theo chieu dai kenh.

d) Ca 2 cau a) va c) eu ung.

Cau 3. Mot kenh co o doc i>icr, o sau nc trong kenh h > h0. Dong chay trongkenh trang thai:

a) Luon chay xiet b) Chay xiet neu h < hcr.c) Luon chay em d) Chay em neu h > hcr

Cau 1. Mot kenh co o doc i > icr, so Froude Fr > 1. Dong chay trong kenh trang thai:a) Chay xiet b) Chay em.

c) Chay xiet neu h < h0 d) Chay xiet neu h > hcr

Cau 2. o sau phan gii trong kenh:a) Nho hn o sau dong eu khi o doc kenh i > icr.b) Bang o sau dong eu khi o doc kenh i = icr.c) Ln hn o sau dong eu khi o doc kenh i < icr.d) Ca 3 cau tren eu ung.

TRAC NGHIEM


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Nc nhay la mot hien tng xay ra khi dong chay i t chay xiet sang chay em.Hien tng nc nhay tao ra mot cuon xoay lam bien oi ot ngot t o sau chayxiet (h< hcr) sang o sau chay em (h > hcr).

Chay em

C

Nc nhay

aK K

1

1

2

2

h

h

ln

V2V1

S o nc nhay

lsn

Chay xiet

hhhcr

AB

Chng:

NC NHAY3.1 KHAI NIEM

Tai sao nc nhay xuat hien ?:E0(h)

Chcr

h

h

E0= EE0min E0 E0

h

Khao sat cho trng hp i = 0Mat chuan la ay kenh

K

h

V1

h

i= 0

T bieu o E(h) cho thay nanglng se tang t Emin en E khio sau tang t hcr en h.

Khong the xay ra v nanglng theo dong chay ch cothe giam khong the tang

Ncnhay

Nang lng rieng =Nang lng toan phan

ng dung nc nhay :

Nc nhay tao ra mot cuon xoay manh liet nen dong chay qua nc nhay se b tieu

hao nang lng kha ln.Trong xay dng dung nc nhay e tieu hao nang lng sau cong trnh e tranh xoi l.


19/57


Cac loai nc nhay

Fr = 11,7 : nc nhay song

Fr = 2.54,5 : nc nhay dao ong

Fr = 1,72,5 : nc nhay yeu

Fr = 4.59 : nc nhay on nh

Fr > 9 : nc nhay manh

Nc nhay song : < 5%Nc nhay yeu : 5% 15 %Nc nhay dao ong : 15% 45%Nc nhay on nh : 45% 70%Nc nhay manh : 70% 85%

Tieu hao nang lng

3.2 PHNG TRNH NC NHAY

A

C1

yC1

S o tnh nc nhay

G

B D

hP2

P1

h

K

C

yC2C2

T

Gia thiet:

- Mat cat trc va sau nc nhayng dong thang song song - - >phan bo ap suat theo qui luat thuytnh- Bo qua ma sat ay kenh

Ap dung nguyen ly ong lng cho the tch ABCD, chieu tren phng s:

S

V1 , V2 van toc trung bnh cua dong chay tai mat cat AB, CDTs : lc ma sat tren long kenh => 0Gs : trong lng khoi nc tren phng S => 0Rs : phan lc ay tren phng S => 0P1S = P1 va P2S =P2: ap lc nc tai h va hAp suat phan bo theo qui luat thuy tnh

11Cs1 AyP

22Cs2 AyP

00201 Vi

2112 AyAyVVQ 2C1C0

Phng trnh nc nhay


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3.3 HAM NC NHAY

(h)

E0(h)

CC1hcr

h

h

, E0E0minmin

Bien thien cua E0 va theo hEn

dA

B

yC

C

dh x

h

Bdh

dA

trong o:

(ycA) la moment tnh cua dien tch Aso vi truc x c xac nh:

Khi h bien thien mot ai lng dh th A bien thien mot ai lng dA

-> moment tnh cua mat cat mi (A+dA) /v mat thoang dAdhA)dhy( c

va vi

0ABgA

Q2

20

g

Q

B

A 203

o= th cc tieu cua ham ncnhay trung vi cc tieu cua hamnang lng rieng

h=hcr

3.4 TNH TOAN NC NHAY3.4.1. Chieu sau nc nhay:

Trng hp ac biet: Kenh hnh ch nhat:

T phng trnh nc nhay

Suy ra khi nc nhay xuat hien th ham nc nhay AygA

Qh

C

2

0

tai mat cat trc va sau nc nhay se bang nhau:

Do o : Neu biet h A1 yc1 1Gia thiet h A2 yc2 2 1= 2

no

stopyes

2

h

gh

qbAy

gA

Qh

220

c

20

2

h

h

hb

2

h

h

hb

23cr

23cr

1h

h

812

h

h

3cr


21/57


Cong thc gan ung mat cho cat bat ky:Khi h< 5 hcr mot cach gan ung chieu sau noi tiep co the xac nh theocong thc cua A. N. Rakhmanov

cr

2cr

h2,0h

h2,1

h

cr

2cr h2,0

h

h2,1h

3.4.2 Ton that nang lng nc nhay:

22

2221

21

222

211

21n

gA2Qh

gA2Qh

g2

Vh

g2

Vh

EEE

oi vi kenh ch nhat:

hh4

a

hh4

hhE

33

n

hha vi

3.4.3 Chieu dai nc nhay ( ln):oi vi kenh ch nhat:

Safranez (1934):

Bakhmetiev va Matzke (1936): hh5ln

Silvester (1965): 01,121n 1Frh75,9l


22/57


3.5 CAC DANG NC NHAY KHAC

3.5.1 Nc nhay ngap

Khi mat cat trc nc nhay hoan chnh b ngap th ta co nc nhay ngap.

hc

hhKK

Nc nhay ngap

hng hC

a

A B

BA

C

h2h

h

ng

h

h1Fr21

h

h

la so Froude ng vi o sau ha lu hh va Vh la van toc ha lu

Theo Smetana, chieu dai nc nhay ngap c tnh:

Chnn hh6l


23/57


Chng:

DONG CHAY QUA CONG TRNHPHAN I DONG CHAY QUA AP TRAN

ap tran la mot cong trnh ngan dong chay va cho dong chay qua nh ap.ap tran c dung e kiem soat mc nc va lu lng.

Co 3 loai ap tran thong dung

ap tran thanh mong

ap tran mat cat thc dung

ap tran nh rong

4.1 AP TRAN THANH MONG

H

0,67H

ap tran thanh mong

< 0,67H

4.1.1 Cong thc tnh lu lngAp dung phng trnh nang lng hoac dungphng phap phan tch th nguyen:

m : he so lu lng

b : be rong ap tran

Vo: Van toc tien gan

Vo

Neu thay

mo co the c tnh theo cong thc

Bazin

Vi pham vi : 0,2 m


24/57


4.2 AP TRAN MAT CAT THC DUNG

0,67h 2 3HH

P1 P


Cai tien cua ap tran mat cat thc dung

ap tran Creager -Ophixerop

Qui ao tia nc ri

ieu kien chay ngap:

hh > P

Tr so pgPz

phu thuoc vao (phu luc 4).

H

P

Z

hh

P1

4.2.1 Cac trang thai chay (i) chay t do (ii) chay ngap

H

P1P

hh

H

P

Z

hh

Chay t do Chay ngap


25/57


4.2.2 Cong thc tnh lu lng

Trong thc te do chieu rong ap ln Be rong ap b c chia thanh nhieu nhp

mo tru gia va mo ben

dong chay se b co hep ngang

: He so co hep ben do anh hng cua tru gia va mo ben

mb: He so co hep do mo ben

mb: He so co hep do mo ben

n: So nhp ap

b: Be rong moi nhp mb = 1 mb = 0,7 mt = 0,8 mt = 0,45 mt = 0,25

He so co hep do mo tru va mo ben

m : he so lu lng

m = mtc.hd.H

mtc: He so lu lng tieu chuan

ap tran loai Creager mtc = 0,48 0,5ap tran hnh a giac mtc = 0,3 0,45 phu luc .4.3

H: He so ieu chnh do cot nc tran Hkhac vi cotnc thiet ke (Htk).H > Htk : ap co chan khong H> 1H = Htk : H= 1H < Htk : ap khong co chan khong H< 1

phu luc 4.4

hd: He so ieu chnh do thay oi hnh dang ap so vi hnhdang tieu chuan

phu luc 4.5


26/57


4.3 AP TRAN NH RONG

3H 8HH0

P1

Hh

1

1

4.3.1 Cac trang thai chay(i) chay t do(ii) chay ngap

Chay ngap

Hh

P

Z1

K K hh

H0

P1

Hh

1

1

Chay t do

H0

P1

Z2

hn

Cac trang thai chay qua ap tran nh rong

ieu kien chay ngaphh > P

Chay ngap

Hh

P

Z1

K Khh

H0

P1

Z2

hn

4.3.2 Cong thc tnh lu lng

Trng hp chay khong ngap

Viet phng trnh nang lng cho 2 mat cat 0-0 va 1-1

H0

P1

Hh

1

1

Khi ca ap hnh ch nhat:


27/57


Ta co the bien oi a ve dang nh sau:

at

at

m: he so lu lng cua ap tran nh rong phu luc 4.6

: he so lu toc phu luc 4.7

T Neu biet m va co the suy ra k (k1 va k2)Co k suy ra h (o sau tren nh ap)

phu luc 4.7

Chu y: k1 cho h ng vi dong chay xiet tren nh ap

k2 cho h ng vi dong chay em tren nh ap

Chng minh tng t

Trng hp chay ngap:

Trng hp ca ap hnh ch nhat:

Thong thng mc nc hn gan bang h , do o co the tnh :

Chay ngap

Hh

P

Z1

K Khh

H0

P1

Z2

hn

n: he so lu toc phu luc 4.7 phu luc 4.6


28/57


29/57


Chay ngap

Gia thiet rang ap suat tren mat cat co hep phan bo theo quy luat tnh

a hc hng

Chay ngap

Nc chay ngap

H0 Hhh

Viet phng trnh nang lng cho hai matcat 0-0 va c-c

Vc = )(2ngo hHg

Q = Vc A = .b a )(2 ngo hHg

hng : xac nh theo cong thc nc nhay ngap trong chng 3

Gan ung co the lay hng = hh, cong thc tnh Q co the viet

Khi o m cong a kha nho hn so vi o sau hh, th xuat hien nc ngap lang hng = hh

4.5 CONG NGAMc xay dng qua e, ap hoac di ng co mat cat khep knCong ngam thng co mat cat hnh tron hoac hnh ch nhat

4.5.1 Trang thai chay trong cong ngam

Chay khong ap

- Chay ban ap

- Chay co ap

NN

KK

KK

Mc nc thng ha luthap hn nh cong

Mc nc thng lu caohn nh congMc nc ha lu thap hnnh cong

Mc nc thng halucao hn nh cong


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4.5.2 Cong thc tnh toan

a. Chay khong ap:

Chieu dai cong L < 8H

Tnh toan nh chay quaap tran nh rong

b. Chay ban ap:

Tnh toan nh chay qua

ap cong lo thien (h)

Lvao = 1,4 a

Khoang cach t ca cong en mat cat cohep co the c xac nh theo cong thcthc nghiem sau Lvao

a hc

L

c. Chay co ap:

Tnh toan nh chay quamot ong ngan co ap

Q = c.A. ogZ2

Zo = Z+g

Vo

2

2

Zh > Zd- D/2 Z= Ztl - Zh

Z= Ztl (Zd-D/2)

ZVo Zd

Ztl

Zh

A: Tiet dien ngang cong

Z: Chenh lech mc nc thng ha lu

c: He so lu toc qua cong

c: He so ton that cuc boL: Chieu dai cong

R: Ban knh thuy lc mat catthang ng congC: He so Chezy

D

Zh


31/57


CAU HOI TRAC NGHIEM

Cau 1. Dong chay qua ap tran thc dung, che o chay ngap, co:a) Mc nc ha lu cao hn ngng ap tranb) Dong chay qua ap tran la chay emc) Mc nc ha lu anh hng ti lu lng qua ap trand) Ca 3 cau eu ung

Cau 2. Dong chay qua cong h (lo thien), che o chay t do, co:a) Nc nhay ngapb) Nc nhay t doc) Dong chay qua ca cong che o chay emd) Ca 3 cau eu sai.

Cau 3. Ap suat tren be mat ap tran Creager:a) Bang ap suat khong kh.b) La ap suat chan khong khi cot nc tren ngng tran cao hn cot

nc thiet ke.c) Ln hn ap suat khong kh khi cot nc tren ngng tran cao hn cot

nc thiet ke.d) Ca 3 cau eu sai.

Cau 4. Trong cong thc tnh lu lng qua cong ngam khi chay co ap, heso lu lu toc c tnh

, trong o ton that doc ung chay trong cong c tnh vi ieu kiendong chay trong cong la:a) Chay roi thanh trn thuy lc b) Chay roi thanh nham thuy lcc) Chay roi thanh hoan toan nham d) Cho tat ca trang thai chay

Cau 10. Neu cung mot o sau H trc nh ap va cung be rong b, dong chayqua ap la chay t do th loai ap cho lu lng ln nhat la :a) ap tran thanh mong b) ap tran mat cat thc dungc) ap tran nh rong d) Ca 3 eu bang nhau


32/57


ap tran thanh mong

ap tran thanh mong hnh tam giac


33/57



ap tran Tr An


34/57


35/57


Cong lo thien

Cong ngam mat cat hnh tron


36/57


Cong ngam mat cat hnh hop

Cong ngam mat cat hnh hop


37/57


CHNG

NOI TIEP VA TIEU NANG

PHAN 1. NOI TIEP DONG CHAY HA LU CONG TRNH

Dong chay t thng lu qua ap tran hay quacong c noi tiep vi kenh dan ha lu banghai hnh thc

Noi tiep chay ay

Noi tiep chay mat

Noi tiep chay ay Noi tiep chay mat

Van toc ln xuat hien ay Van toc ln xuat hien be mat

5.1 NOI TIEP CHAY AY5.1.1 Cac dang noi tiep chay ay

5.1.2 Cong thc tnh toan

Nc nhay phong xa

hhHng

hc

c

c

Nc nhay ngap

Viet phng trnh nang lng cho 2 mat cat 0-0 va c-c.

c

c

hhh

0

0

H

P

V0

Mat chuan

fcc

c hg

VhE

g

VPH

22

2

0

20

Ton that nang lng

Mat cat hnh ch nhat cc hEgbhQ 02 hc

th danhc So sanh hh

Nhay ngapNhay t do


38/57


5.2. NOI TIEP CHAY MAT

Noi tiep chay mat thng gap trong ieu kien cong trnh co bac thang

Noi tiep chay mat

Noi tiep chay mat ay khong ngapMc nc ha lu dang len

Noi tiep chay mat ngap

Mc nc ha lu tiep tuc dang cao

Noi tiep chay mat ay ngapMc nc ha lu dang cao

PHAN 2. TIEU NANG HA LU CONG TRNH

Khi noi tiep chay ay thnggay xoi l tren mot oan daisau cong trnh

Cong trnhtieu nang

a ncnhay taichan congtrnh

Giamchieu daioan chayxiet

Noi tiep chay ay

Be tieu nang

Tng tieu nang

Be tng ket hp

oan chay xiet


39/57


5.3 BE TIEU NANG

Phng phap tnh toan

Nguyen tac tnh toan

o sau d sao cho h2 la o sau sau ncnhay cua hc

hc: xac nh t phan tnh toan dong chayay

o sau sau nc nhay khi co nc nhay tai matcat co hep

Dong chay t be qua kenh ha lu c xemnh dong chay qua mot ap tran nh rong

Vo2/2g

H

EP

hc

h2

H02

d

h

h

V2

V22/2g

gV

bhg

Qz

h2'2

22

2

2

d= h2 - z - hh

= 0,95 1 ( he so lu toc qua dap tran )

he so an toan: = 1,05 - 1,1.

5.4 TNG TIEU NANGNguyen tac tnh toanChieu cao tng sao cho h2 la o sau saunc nhay cua hcPhng phap tnh toan

hc: xac nh t phan tnh toan dong chayay

Lu lng chay qua tng tieu nang ctnh nh qua ap tran mat cat thc dung

o sau sau nc nhay khi co nc nhay tai mat

cat co hep

23022/

tn HgbmQ

mt :he so lu lng khi chay qua tng tieu nang mt= 0,4 0,42.n :he so ngap

32

022

/

tn gbm

QH

ma

C= h2 - H2

(= 1,05 - 1,1)

Vo2/2g

H

E0 P

hc

h2H02

c

V2/2g

hhH2


40/57


41/57


CHNG

DONG THAM QUA CONG TRNH AT

6.1 MOT SO KHAI NIEM VA NH NGHA

6.1.1 o rong (n)T le phan tram the tch gia phan rongva toan the phan at a

Wo : The tch lo rong

W : The tch at a bao gom ca phan rong va phan ran

6.1.2 Van toc tham thc va van toc tham trung bnh (V):

Van toc tham thc te la van toc tham qua cac khe rong cua at a

Van toc tham trung bnh la van toc tham c xem nh tham qua ca phan atva phan khe rong cua at a

: the tch nc tham trong thi gian tA : Dien tch mat cat ca phan rong va phan at

6.1.3 Cot nc o ap (thuy lc) :

Mat chuan

p/

z

hA

6.1.4 He so tham (k) ( o dan thuy lc)

Dong tham V : nho V2/2g

Cot nc o ap cot nc nang lng

Lu lng tham trem mot n v tiet dien ngang cua dong tham khi chu tac ongbi mot n v cot nc thuy lc tren mot n v chieu dai tham ( ngha la co moto doc thuy lc bang mot n v).

Loai at He so tham (k)(cm/s)

Set 10-9 - 10-6

Bui, bui cha cat 10-6 - 10-4

Cat tuyen chon tot 10-3 - 10-1

oi vi at khong ong chat , d hng th k thay oi theo tng iem va tai mot iemth kx ky kz


42/57


6.2 NH LUAT DARCY (Henry Darcy ,1856 )Khi dong tham on nh, lu lng tham t le vi o doc cot nc o ap(hydraulic gradient) va dien tch tham A

h

h1 V

L

Mat chuan

h2

A

V la van toc tham (van toc Darcy)

Tong quat : van toc tham tai tng iemtrong mien tham

ieu kien ng dung nh luat Darcy

Vi : Re = 5

6.3 CONG THC DUPUIT FORCHERHEIMERTrong trng hp tham khong ap vi o doc nho, cac ng dong tren mot mat catt c xem song song th cot nc o ap h = p/+z la hang so tai cac iem trenmot mat cat .

Cong thc Darcy tr thanh cong thcDupuit - Forcherheimer

Vi: h :o sau dong thamV :van toc c xem phan boeu tren mat cat

Do o :

h

Tang khong tham

Tang thamu

s

Khi ay tang khong tham namngang, chon lam mat chuan thh chnh la o sau mc ncngam ( z = 0)

Mat chuan


43/57


6.4 CHUYEN ONG ON NH CUA DONG THAM VAO GIENG NC.6.4.1. Gieng phun: Tham co ap

R r

ro

h

s

Q

Tang khong tham

Tang tham ncb

Hng cot nc

o ap

oTang khong tham

ho

Xet mot mat tru ban knh rong truc vi thanh gieng

Lu lng tham qua mattru ban knh r, cao b

Van toc

Tch phan t ro en r ng vi h t ho en h

Tch phan t ro en R ng vi h t ho en H

Goi : R ban knh anh hng, H chieu cot nc o ap khi cha bm

S = H ho: chieu sau hut nc

6.4.2 Gieng thng

Xet mat tru tron ong tamvi thanh gieng, ban knh r

Van toc

Lu lng dong thamqua mat tru

Tch phan t ro en r ng vi h t ho en h

Tch phan t ro en R ng vi h t ho en H

S = H - ho

Khi S/2H0

R

ro

h

s

Q!

Tang tham nc

H

oTang khong tham

ng bao hoa

ho

S


44/57


R c xac nh :Theo tng loai at ma lay gan ung nh sau:

- oi vi at c hat trung bnh R=250m500m

- oi vi at hat to R=700m1000m

Cong thc kinh nghiem

(i-cac d )

6.5 THAM QUA AP AT

H m

A

B

C

A

O

H

F

E

a0

h

x

1

m1

LoTang khong tham

m = cotg

m1 = cotg 1

Bien oi mai nghieng AC thanh thang ngAO, sao cho lu lng qua AC giong nh AO

Mikhailop

Dupuit-Forcherheimer:

V

Tch phan t 0 en x tng ng t H en h

k

Lay goc toa o tai O


45/57


Hm

A

B

C

A

O

H

F

E

a0

h

x

1

m1

LoTang khong tham

V

L

tai x = H+L, h = ao

Can xac nh ao

h

Xac nh v tr ng bao hoa

chieu dai cua dai nguyen to

m1 z

m1 zDarcy

Lu lng tham qua MN

Lu lng tham nay cung bang lu lng tham i qua ap at

Giai phng trnh bac 2 cho a0:

Co ao , xac nh c q va ng bao hoa

dzz

a0

M

N

m1

ao


46/57


6.6 THAM CO AP QUA AP

Tang khong tham

Tang tham

H1

H2

e xac nh van toc tham (u) , cot nc o ap (h), da vao phng trnh tham Darcy nh sau:

x

z

Phng trnhlien tuc

h(x,z)

at ong chat va ang hng:

V D TNH TOAN TH M CO AP B NG PH NG PHAP SAI PHAN H U H NMot ap betong co be rong 60m, thng lu cha nc co o sau 10m, ha lu khong conc. ap c xay tren mot tang at tham nc ong chat day 80m, vi he so tham lak= 10-4m/s. Mot hang c c ong gia ay ap sau 35m e giam lu lng thamqua ap. Xac nh lu lng tham qua ay ap neu xem dong tham ch anh hng trongpham vi 105m ve pha thng lu ap va 135m ve pha ha lu ap.

A B C

D

E F G

HI

10 m

Kch thc , v tr ap va li tnh toan


47/57


Lu lng tham trong 1 ngay em la: q = 39,007 m3/m ngay em

A B C

D

E F G

HI

10 m

u1u2

u3u4u5

ng ang cot nc o ap va van toc tham di hang c

Tnh toan chi tiet va chng trnh may tnh xem trong phu luc


48/57


CHNG

DONG KHONG ON NH TRONG KENH

I.KHAI NIEM Dong khong on nh la dong chay ma cac yeu to cua dong chay eu phuthuoc vao thi gian,

o sau : h = h(x,t)

Van toc : V = V(x,t)

Lu lng: Q = Q(x,t)

Dong khong on nh thng xuat hien

Song, kenh b anh hng thuy trieu

Kenh xa nha may thuy ien khi lu lng xa thay oi ot ngot

II.PHNG TRNH LIEN TUC:

h

Q

dX

S thay oi the tch trong oan dx trong thi gian dt

h

B

h

th bang lu lng ra lu lng vao trong thi gian o

- =

Phng trnh lien tuc


49/57


Cac dang khac :

III. PHNG TRNH ONG LNG:Cac gia thiet:

Bien oi cham, bo qua lc quan tnhTon that nang lng c tnh nh dong euPhan bo ap suat tren mat cat ng c xem theo qui luat thuy tnho doc nho


50/57


51/57


Ap dung phng trnh ong lng: S bien oi ong lng trong mot n v thigian trong mot the tch kiem soat th bang tong cac lc tac ong len the tch o

hQ

dX

S bien oi ong lng trong the tch kiemsoat gii han m/c 1-1 va 2-2 1

1

2

2

ong lng vao m/c 1-1

ong lng ra m/c 2-2

S gia tang ong lng trong the tch kiem soat:

Theo phng trnh ong lng

n gian va chia 2 ve cho va dx:

Cac dang cua phng trnh ong lng

Chu y:

Viet lai:

Hay

Dong eu

Dong khong euDong khong on nh


52/57


IV.PHNG PHAP NG AC TRNG SONG BIEN O NHO

Co phng trnh lien tuc

Phng trnh ong lng :

Van toc truyen song: Van toc lan truyen khi gay mot nhieu ong trongnc tnh, co o sau h

Neu song co bien o nho va vi mot so ga thiet sau

1. Kenh nam ngang : o doc i = 0

2. Khong co ma sat: o doc nang J = 0

3. Van toc dong chay nho va s bien oi van toc nho :

4. Tiet dien mat cat t A doc theo dong chay xem bang mot tietdien trung bnh khong oi Ao va

5. Be mat thoang B bang be rong trung bnh khong oi Bo

Mat cat ch nhat

He phng trnh lien tuc va ong lng co the viet lai thanh dang n gian:

at neu mat cat hnh ch nhat th

Nhan cho pt (1)

(1)

(2)

Nhan g cho pt (2)

Cong ve theo ve

Neu th :

(3)

(4)


53/57


Tr ve theo ve va neu

Y ngha phng trnh ng ac trng :

x

t

L0 x

x

M

L R

tM

Lt

R1

x

R

M

L

R t

L1

h0

0x

t

t

l

h0

L

R

Cach xac nh o sau (h) va

van toc (V) tai M(x,t)Tai M ve ng ac trngthuan Co+ (co o doc 1/Co)cat truc hoanh (t = 0) tai L

Tren ng MLcho :

Tng t ve ng ac trngnghch Co- (co o doc -1/Co),

MR:

(1)

(2)

T (1) va (2)

V hL,VL,hR,, Vra biet tai thiiem t = 0

VM, hM,

x

x

M


54/57


x

t M

Lt

R1 R

M

L

R t

L10 x

M

L l

xx

h0

R

Tuy nhien neu v tr M ganau kenh hoac cuoi kenh thphai can them ieu kien bien

V du iem M(x,t)

e xac nh VL , hL, tai Lta ve mot ng ac trngnghch Co-

Biet hR1 , VR1, biet t ieu kien au +VL

Hoac hL

hL

VL

ieu kien bien

VM, hM

(1)

(2)

(1)(2)

Biet hL1 , VL1, biet t ieu kien au +VR

Hoac hR

hR

VRieu kien bien

VM, hMT ng t M

Cac ieu kien au va ieu kien bien song bien o nho:

1.ieu kien ban au : V(0, x), h(0,x)2. ieu kien bien :

au kenh : V(t,0) hoac h(t,0)

Cuoi kenh : V(t,l) hoac h(t,l)

e giai toan song co bien o nho hay mot bai dong khong on nh thong thng, canthiet phai co :


55/57


V d : M t kenh ch nh t rong b = 10m, dai 1000 m noi vi mot ho cha. Tai thi iem t= 0, nc tnh vi o sau ho= 5m. Tai au kenh o sau thay oi theo thi gianh = ho + sin(2 t/T) vi T = 400. Xac nh V, h tai v tr x= 500 m thi iem t = 200 s.

co= 7 m/s

tL=128,57 shL=5,9 m

VL=1,26 m/s

hR1=5 m

VR1=0 m

xR1=900 m

VR=0 m/s

hL1

=5 m

VL1=0 m

xL1=100 m

hR=5 m

VM=1,26 m/s

hM=5,6 m

ho=5m

500 m 1000 m0

M200 s

L128,57 s

R1

R

L1

C+

0

t

x

ho

C+

C-

C-

100 m 900 m

V. PHNG PHAP NG AC TRNG CHO MOT KENH MAT CAT HNH CH

NHAT:Van toc truyen song trong kenh hnh ch nhat :

== >

do o

va

== >

Xet cho mot n v be rong kenh B = 1m => A = h = =>

Thay vao pt lien tuc va ong lng :

chia (1) cho C/g

nhan (2) cho g

(1)

(2)

cong hai ve :


56/57


hay

Neu th :

ng ac trng thuan ( C+) :

ng ac trng nghch ( C-) :

M

C+ C-

Chayem

M

L RL RChay xieti ra khoikenh

t

XL

M

C+ C-

0

Chu y

Nhng neu dong chay xiet (V>C) th ng ac trng nghch va thuan cung chieu nhau

Khi V va C eu dng, neu dong chay em (V < C) th ng ac trng nghch va

thuan ngc chieu

Tong quat ieu kien bien va ieu kien ban au cho bai toan dong khong on nh nh sau:

1.ieu kien ban au : V(0, x), h(0,x)

2.ieu kien bien :

Chay em :

Chay xiet:

+au kenh Dong chay i vao kenh: can 2 ieu kien bien V(0,t) va h(0,t)

Dong chay i ra kenh : khong can ieu kien bien+ Cuoi kenh : Dong chay i ra khoi kenh: khong can ieu kien bienDong chay i vao kenh can 2 ieu kien bien V(0,t) va h(0,t)

+ au kenh : ch can 1 ieu kien bien V(0,t) hoac h(0,t)+ Cuoi kenh : ch can 1 ieu kien bien V(0,t) hoac h(0,t)


57/57

VI. KHAI NIEM PHNG PHAP SAI PHAN HU HAN:

Xet mien tnh toan Xot c ri rac hoa nh hnh ve

i+1ii-1

n

n+1

t

x

0

t

x

Tai iem i va i+1 thi iem t ta co :

Neu bo cac so hang bac cao , suy ra

Tng t , neu tai iem i thi iem n va n+1 ta cung co

Thay vao trong phng trnh lien tuc va pt ong lng :

Van toc va o sau hin+1 , Vin+1hi , Vin

thi iem n thi iem n+1oi vi nhng iem nam tren bien, can phai bo sung them ieu kien bien mi xacnh c cac gia tr h va V

ieu kien on nh cua pp sai phan hien

ieu kien Courant - Friedrichs Lewy (CFL)

baigiangthuyluc_2009-ts. huynh cong hoai

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