Chuong 3-Co So Dong Luc Hoc.ppt

1

THY LC I CNG Chng 3: C s ng lc hc cht lng

PGS. TS. Nguyn Thng 1

TRNG I HC BCH KHOA TP. HCM

Khoa KTXD - B mn KTTNN

Ging vin: PGS. TS. NGUYN THNG

E-mail: [email protected] or [email protected]

Web: http://www4.hcmut.edu.vn/~nguyenthong/

Tl. (08) 38 691 592 - 098 99 66 719


PGS. TS. Nguyn Thng 2

NI DUNG MN HC Chng 1. ac tnh chat long.

Chng 2. Thuy tnh hoc.

Chng 3. C s ong lc hoc chat long.

Chng 4. o ac dong chay.

Chng 5. Ton that nang lng.

Chng 6. Dong chay co ap trong mang li ong.

Chng 7. Lc tac dung len vat can.

Chng 8. Dong chay on nh eu trong kenh.

Chng 9(*)

. Dong chay on nh khong eu trong

kenh.

Chng 10(*)

. ap tran.

(*) : Trng hp mon Thuy lc m rong


PGS. TS. Nguyn Thng

MUC CH

Nghien cu cac quy luat cua chat long khi no chuyen ong:

- Phng trnh lien tuc.

- Phng trnh Bernoulli.

S tng tac cua nc chuyen ong vi vat can co nh:

- Phng trnh ong lng (Ch.5).

3


PGS. TS. Nguyn Thng

MOT SO KHAI NIEM

Xet mot moi trng chat long ang chuyen ong:

Quy ao : Quy ao la ng i cua mot phan t chat long trong khong gian.

ng dong: La ng cong (tng tng) tai mot thi iem cho trc, i qua cac phan t

chat long co vect van toc la tiep tuyen cua

ng cong o.

4

a b c

d

e

ng dong Va


PGS. TS. Nguyn Thng

CHU Y

Hai ng dong KHNG BAO GI CT NHAU

5

ng dong

SAI !!!!

V1

V2

A (1)

(2)


PGS. TS. Nguyn Thng

MOT SO KHAI NIEM

Dong nguyen to: Trong khong gian cha ay

chat long chuyen ong, lay mot ng cong

kn co dien tch vi phan ds, tat ca cac ng

dong i qua cac iem tren ng cong nay

tao thanh mot mat co dang ong goi la

DNG NGUYN T .

6

ds

Dong nguyn t

ng dong

2


PGS. TS. Nguyn Thng

MOT SO KHAI NIEM

Dong chay: Tap hp vo so cac dong nguyen to i

qua din tch hu hn S DONG CHAY.

B mt bao quanh dng chy NG DNG

Mat cat t: Mat thang goc cac ng dong.

7 Mat cat t phang

1

1

3

3

2

2 Mat cat t cong

ng dong


PGS. TS. Nguyn Thng

MT CT T

DIEN TCH T

ABCD

, ABC

(xet mat cat thang goc vi dong

chay)

8


PGS. TS. Nguyn Thng

ON

- Mat cat t : = (b+mh)h hnh thang.

- He so mai doc m: m=cotg()

9

b

m h

A

B C

D

A

B

C

DIN TCH T


PGS. TS. Nguyn Thng

CHU VI T

Xt mt ct t chiu di tip xc gia cht lng (nc) v lng dn GI L CHU VI T

CHU VI T: LABCD

; LABC

( : khi)

BAN KNH THUY LC :R= ABCD

/

10


PGS. TS. Nguyn Thng

- Chu vi t : hnh thang

- Ban knh thuy lc: R = /

- He so mai doc m: m=cotg()

2m1h2b

11

b

m h

A

B C

D

A

B

C

LABCD or LABC


PGS. TS. Nguyn Thng

V D

Bn knh thy lc cu ng ng trn ng knh d chy y:

R = (d2/4)/d = d/4

12

3


PGS. TS. Nguyn Thng

13

PHN LOI

DONG CHAY

Theo trang thai

Theo ap sut


PGS. TS. Nguyn Thng

14

PHN LOI

THEO

TRNG THI C 3 dang


PGS. TS. Nguyn Thng

DONG CHAY ON NH & EU

(V, h, Q, hang so theo thi gian & khong gian ).

15

V=hs.

h=hs.

i%=tg()

Q=hs.


PGS. TS. Nguyn Thng

DONG CHAY ON NH KHONG EU

(V, h, Q, ap suat hang so theo thi gian va oi theo khong gian)

16

h1

h2

i%=tg()

Q=hs.

h1

Q=hs.


PGS. TS. Nguyn Thng

DONG CHAY KHONG ON NH

(V, h, Q, ap suat, thay oi theo khong gian & thi gian)

17

V h1(t)

h2(t)

Mc nc thay oi

theo trieu

Q (t)

h1 (t)


PGS. TS. Nguyn Thng

PHN LOI THEO

P SUT

Dong chy CO ap.

Dong chy KHNG ap.

18

4


PGS. TS. Nguyn Thng

DONG CHAY CO P Dong chy KHNG co mt thoang tip xuc khng khi Vi du: Dong chy trong h thng cp nc.

19


PGS. TS. Nguyn Thng

DONG CHAY KHNG P Dong chy CO mt thoang tip xuc khng khi Vi du: Dong chy trong sng, rach thin nhin.

20


PGS. TS. Nguyn Thng

PHNG TRNH LIN TC DONG CHAY N NH

C s Nguyn l bo ton khi lng ca vt cht.

21


PGS. TS. Nguyn Thng

22

CO AP KHONG AP

1

1

2

2 1

1

2

2

Q Q

V1

V2

1

1 1

1

2

2

2

2

dx1

dx2

t

(t0+dt)

(t0+dt) t

Q=V1S

1=V

2S

2

Q

(1-1)&

(2-2)

Mat

cat

t

S1

S2


PGS. TS. Nguyn Thng

T nguyen ly bao toan khoi lng cua nc:

ay la phng trnh lien tuc cua chat long chay

on nh (eu hoac khong eu)

23

1 1 2 2 1 1 2 2S dx S dx S Vdt S V dt

2211 VSVS

Q ra khoi m/c (2-2) Q vao m/c (1-1)


PGS. TS. Nguyn Thng

ng dng phng trnh lin tc

Bc 1: Chn 2 mt ct t 1-1 v 2-2.

Bc 2: p dung phng trnh lin tuc Lu lng vo m/c t 1-1 BNG lu lng ra khi m/c t 2-2:

j

j

i

i

ravaoQQ )()(

)22()11(

24

5


PGS. TS. Nguyn Thng

Bi tp ng dng phng trnh lin tc

Bai tap 1: Mot ng ong tron co D thay oi vi D

1=0.25m va D

2=0.5m. Van toc dong chay

vao D1 la 1m/s. Tnh van toc au ra.

Bai tap 2: Mot kenh hnh thang co b=5m, he so mai doc m=1. Tai mat cat t 1-1 kenh co

chieu sau 1,5m va van toc dong chay la 1m/s.

Tnh chieu sau tai mat cat 2 khi van toc dong

chay tai mat cat nay la 1,4m/s. Cho biet ay

la dong chay on nh khong eu.

25


PGS. TS. Nguyn Thng

Bai tap 3: Tnh van toc ha cua mat thoang

bnh trong he thong sau:

26

V1=?

A B

S1=0,5m

2

Nhan xet V1 khi S

1 rat ln so vi t/dien ong AB V

1?

B cha

ng ng tron co D=0,2m

V2=5m/s


PGS. TS. Nguyn Thng

Bai tap 4: Ngi ta muon co van

toc tia nc tai au ra cua mot

voi phun nc cha chay tang 16

lan so vi van toc au vao ong.

Anh (Ch) cho biet ng knh cuoi

ong phai tang (hoac giam) bao

nhieu lan so vi ng knh au

ong.

27


PGS. TS. Nguyn Thng

Bai tap 5: Cho ng ong tron re nhanh nh

hnh ve. Xac nh van toc nc V3.

28

V1=3m/s,

D1=100mm

V2=5m/s,

D2=50mm

V3=?,

D3=75mm

Nut

1

1

2

2


PGS. TS. Nguyn Thng

NANG LNG DONG CHAY TAI MAT CAT T (E)

(CHAY CO AP)

(1-1 & 2-2 : mat cat t)

Q=V1S

1 =V

2S

2

29

2

2 p

1

p2

Q V

2

V1

1

1 o o

Z2

Z1

Mat chuan

S1

S2


PGS. TS. Nguyn Thng

Nang lng dong chay E tai mat cat T bat ky (chay co ap):

30

E = Z + pd/g+V

2/2g (mH

2O)

Th nng

p nng (tai tm m/c t)

ng nng

6


PGS. TS. Nguyn Thng

NANG LNG DONG CHAY TAI MAT CAT T

(CHAY KHONG AP)

31

2

2

Q p

1

1

1

z1

z2

V1 V

2 h

2

h1

o o Mat chuan

ay kenh

Mat thoang

z0

h1/2


PGS. TS. Nguyn Thng

Nang lng dong chay E tai mat cat bat ky (chay khong ap):

32

E = z + h + V2/2g (mH

2O)

The nang

Ap nang

ong nang


PGS. TS. Nguyn Thng

Mt chun o-o: La mt nam ngang bt k.

E : nng lng dong chay tai v tr mt ct.

Z : tung o tam m/c so vi mat chuan (co ap), the nang.

z : tung o ay kenh (chay khong ap), the nang (chay khong ap).

h : chieu sau dong chay khong ap, ap nang.

p/g : ap nang (chay co ap).

V2/2g: ong nang.

33


PGS. TS. Nguyn Thng

Gii thich

11

1a

11

1 hzg

2

hgp

Zg

pZ

34

O O Z1

p1 h1

z1

Dong chy c p

Dong chy khng p

1

1

pa


PGS. TS. Nguyn Thng

35

KHI NIM V

NG NNG LNG E

(Energy line)&

NG CT NC

O P TON PHN H

(Hydraulic head)


PGS. TS. Nguyn Thng

36

ng nng lng khi dong chy i t A B

z1

p1/g

V12/2g

O O

p/g

z

V2/2g

E = z + p/g + V2/2g

Q

A

B

E EA

EB

1

1

2

2

7


PGS. TS. Nguyn Thng

37

ng ct nc o ap ton phn H

z1

p1/g

V12/2g

O O

p/g

z

V2/2g

p2 /g

z2

H=z+p/g

1

1

2

2


PGS. TS. Nguyn Thng

NHN XT

Theo chiu dng chy:

ng nng lng E lun lun gim (nm ngang).

ng ct nc o p ton phn H c th tng hoc gim hoc nm ngang.

38


PGS. TS. Nguyn Thng

Bai tap 1: Trong bai tap 1 (trc), gia thiet ap

suat d tai au vao la 0,2at. Chon mat chuan

cach tam ve pha di la 2m. Tnh nang lng E

dong chay tai mat cat au vao.

Bai tap 2: Trong bai tap 2 (trc), gia thiet mat

chuan qua ay kenh m/c (1-1). Tnh nang lng

E1 dong chay tai mat cat (1-1).

Bai tap 3: Trong bai tap 3 (trc), gia thiet mat

chuan qua truc ong AB. Khoang cach t truc

ong en mat thoang la 6m. Tnh nang lng E

dong chay tai mat thoang va tai mat cat thang

goc ong AB tai B (ngoai khong kh).

(1at=g*104 = 9,81.10

4 N/m

2)

39


PGS. TS. Nguyn Thng

Bai tap 4: Cho s sau. Tnh nang lng

tai mat cat (1-1). Biet rang bm X co cot

nc bm la 20m. Ap suat d tai A la

pA=0.2at. Hay e ngh p/p xac nh ap

suat tai B (!).

40

V=4m/s 1

1

2

2

A B

O O V=4m/s

~ Bm X


PGS. TS. Nguyn Thng

41

PHNG TRNH BERNOULLI


PGS. TS. Nguyn Thng

PHNG TRNH BERNOULLI

C s: nh ly ong nang cua he thong

42

V2

V1

z1

z2

p2

p1

O O

1

1

2

2

Q

Mot chat iem khoi lng m &

van toc V ong nang = mV2/2

8


PGS. TS. Nguyn Thng

nh ly ong nang: o bien thien ong

nang cua mot he thong se BANG cong

cua tat ca cac ngoai lc tac dung len he

thong.

)dl*F(mV2

1mV

2

1 21

2

2

43

2mV2

1 ong nang ; Cong dl*F

bin thin ng nng


PGS. TS. Nguyn Thng

F lc tc dung

m khi lng

L.F0mV2

1 21

44

m

t=0 m V=V1 F

L

V=0

t=dt


PGS. TS. Nguyn Thng

Tai t=0 Xt cht lng g/h bi 1-1 & 2-2

Tai t=dt Khi cht lng s gii han bi 1-1 & 2-2.

45

V2

V1

z1

z2

p2

p1

O O

1

1

2

2

Q 1

1

2

2

S1

dx1

dx2

Khi lng cht lng m

P1

P2 = p

2S

2


PGS. TS. Nguyn Thng

bin thin ng nng ca h thng (khng k phn chung g/h bi 1-1 v 2-2

m khi lng cht lng gii han bi 1-1 & 1-1 (cng l g/h bi 2-2 & 2-2 )

]1[mV2

1mV

2

1 21

2

2

46


PGS. TS. Nguyn Thng

CNG CA CC NGOI LC

p lc tai cac m/c 1-1 & 2-2:

Th nng:

Ch : m = S1dx1 = S2dx2

]2[dxSpdxSp 222111

47

]3[mgzmgz 21

P2

P1


PGS. TS. Nguyn Thng

NH LY NG NNG

[1] = [2] + [3]

Chia 2 v cho mg (m= S1dx1= S2dx2 )

21

222111

2

1

2

2

mgzmgz

dxSpdxSpmV2

1mV

2

1

48

9


PGS. TS. Nguyn Thng

y l p/t Bernoulli trong trng hp l tng khng c mt (b sung) nng lng khi cht lng i t 1-1 n 2-2.

g2

V

g

pz

g2

V

g

pz

2

222

2

111

49

Nng lng mt 1-1

Nng lng mt 2-2


PGS. TS. Nguyn Thng

PHNG TRNH BERNOULLI

Xet dong chay i t m/c t (1-1) (2-2):

Nang lng dong chay tai m/cat t (1-1) + nang lng bo sung (qua bm) hoc b ly i

(qua tuabin) khi dong chay i t (1-1)(2-2) (neu co)

BANG

Nang lng dong chay tai m/cat t (2-2) + tong

ton that nang lng khi dong chay i t (1-1)

(2-2).

50


PGS. TS. Nguyn Thng

PHNG TRNH BERNOULLI

51

2 2

1 1 2 21 bs 2 w

p V p Vz E z h

g 2g g 2g

w d ch dh dh

d cdh ; dh

Ebs

: nang lng bo sung hoc b ly i (neu co), Ebs > 0 bm (Hbm =Ebs) & Ebs < 0 tuabin (Htuabin =abs(Ebs) )

: tong ton that nang lng khi

dong chay i t 1-1 en 2-2.

: ton that ng dai, cuc bo.

E1

E2


PGS. TS. Nguyn Thng

Nguyen ly: Dong chay luon di chuyen t ni co

nang lng ln en ni co nang lng be hn

nang lng m/c t sau NHO hn nang lng m/c t trc (ngoai tr trng hp co

bo sung nang lng tren oan ng i).

Quy c: Khi ap dung phng trnh Bernoulli

cho oan dong chay g/han bi 2 m/cat t (1-

1) & (2-2) ta quy c dong chay i t mat cat

t (1-1) mat cat t (2-2) khi viet phng trnh Bernoulli.

52


PGS. TS. Nguyn Thng

CAC BC AP DUNG

PHNG TRNH BERNOULLI

Bc 1: Chon 2 m/c t 1-1 & 2-2. Chon mat chuan o - o.

- Mat cat t la m/cat thang goc vi dong chay.

- Nen chon m/c t sao cho ap suat tai tam va van toc

trung bnh cua m/c biet cang nhieu cang tot.

- Mat chuan la mat nam ngang bat ky. Nen chon sao cho

z1, z2 >=0.

Bc 2: P/t Bernoulli ap dung cho oan dong chay gii han bi 1-1 & 2-2 (viet ay u p/trnh ban au):

53


PGS. TS. Nguyn Thng

Trng hp chy C P:

Trng hp chy KHNG P:

122

2

222

2

111 wbs h

g

VhzE

g

Vhz

54

122

2

222

2

111 wbs h

g

V

g

pZE

g

V

g

pZ

10


PGS. TS. Nguyn Thng

Bc 3: Khao sat cac so hang

trong pt. [1].

Xac nh cac gia tr trong p/t nh vao so lieu ban au va cac gia thiet

(neu co).

55


PGS. TS. Nguyn Thng

Bc 4: Thay cac gia tr vao pt. [1].

Giai p/t Bernoulli thu gon neu p/t con 1

an so. Neu p/t con 2 an so, viet bo

sung p/t lien tuc cho oan dong chay

va giai he p/t.

Ghi chu: Neu p/t Bernoulli thu gon con >2 an so Ban a chon sai cac m/c t. Tr ve Bc 1 va chon lai

m/c t hp ly hn !

56


PGS. TS. Nguyn Thng

Bai tap 1: Tnh ap suat tai B (N/m2 v at) mat

cat (2-2). Biet rang tai A co pA=1.4at. ng

ong co DA=0,2m & D

B=0,4m va dan lu lng

nc 25l/s. Bo qua ton that nang lng khi

dong chay i t A en B. Lay g=9,81m/s2.

57

h=3m 1

2

Q

(Chu y: 1at =1kgf/cm2 =g(N/m2)

= 9,81(N/cm2)= 9,81.104 (N/m2)

VA

VB

2

1

B

A


PGS. TS. Nguyn Thng

Bai tap 2: Tnh ap suat tai B (N/m2, at).

Biet tai A co pd

=1,2at. ng ong co

DA=0,25m, D

B=0,1m va dan lu lng

nc 40l/s. Bo qua ton that nang lng

khi dong chay i t A B.

58

Chu y: 1at =1(kgf/cm2)= g(N/cm

2)

= 9,81(N/cm2)= 9,81.10

4 (N/m

2)

A

B H=5m

Q


PGS. TS. Nguyn Thng

Bai tap 3: Cho dong chay co ap nh s

o sau. Tnh ap suat nc tai v tr B.

Bo qua ton that nang lng.

59

A

B pA=400kN/m2

pB= ???

DAB

=0,2m H=20m


PGS. TS. Nguyn Thng

Bai tap 4: Cho dong chay co ap qua tuabin X

nh hnh ve. Biet ap suat tai A la 0.5at, tai

B la 20at. Van toc dong chay trong ong la

7m/s. Ong co tiet dien 0.5m2. Bo qua ton

that nang lng. Tnh Htuabin

, t o tnh

cong suat cua tuabin P=gQHtuabin

(watt)

vi Q la lu lng qua ong, =0.85 hieu

suat.

60

A

B

ZB=5m

ZA=0m

X

11


PGS. TS. Nguyn Thng

Bai tap 5: Mt bm X hoat ng theo s sau. Bm c Hb=25m. Bm ang hoat ng vi lu lng 10l/s. ng ng c d=0.12m. Mt nng lng cuc b qua ng ht . Tnh p sut (at) tai v tr ngay trc v sau bm (gi thit cao nh nhau). Gi thit nc 270C v s bc hi

chuyn sang th kh

nu pt < 3mH2O.

Nhn xt ?

61

H=6m

X

g2/V5 2


PGS. TS. Nguyn Thng

Bi tp 6: Tinh lu lng nc t bnh 1 sang bnh 2. Gi thit mt nng lng dng chy i t 12 l:

Cac bnh co tit din rt ln so vi t/din ng.

g2

V.

d

L10.5h

22

w

62

1 2

L=40m

d=0.1m

H=4m

A B

V


PGS. TS. Nguyn Thng

Bi tp 7: Tinh lu lng chy ra khi ng AB. Gi thit mt nng lng t bnh ra khi ng l:

Bnh co tit din rt ln so vi tit din ng.

g2

V.

d

L10.4h

22

w

63

1 L=30m d=0.1m

H=6m

A B V


PGS. TS. Nguyn Thng

Bi tp 8: Mt nh may (NM) thy in co s lm vic nh hnh v. B qua tn tht nng lng khi dong chy i t h n NM. QNM=32m

3/s. Tinh cng sut (MW) ca NM. Bit hiu sut NM l =0,8. Ly g=10m/s2.

64

Tuabin X

Zh=650m

ZNM=25m

QNM

H cha

NM


PGS. TS. Nguyn Thng

Bai tap 9: Ap suat tai B la 0,4at va tai A la 0,2at. ng ong co D hang so va dan lu lng

Q=10l/s. Bo qua ton that nang lng khi dong

chay i t A en B. Tnh cong suat P cua bm

X. Hieu suat bm =0,8. Lay g=10m/s2

[P=gQHb/ (watt)] .

65

A

B Q

VA

VB

X H=15m


PGS. TS. Nguyn Thng

Bi tp 10: Tinh ap sut tai B (N/m2). Bit rng ap sut d tai A l 0,2at. DA=DB=0,2m. Bm X co ct nc bm l 20m. B qua tn tht nng lng khi dong chy i t A n B. Lu lng bm l 40l/s. Tnh cng sut bm vi =0.8.

66

A

B Q

VA

VB

X H=14m

12


PGS. TS. Nguyn Thng

67

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Chuong 3-Co So Dong Luc Hoc.ppt