chuong2 clc

7/23/2019 Chuong2 Clc

1/45

PetroVietnam University

CHNG 2: TNH HOC LU CHAT

I. Khai niemII. Ap suat thuy tnh

III. Phng trnh vi phan c ban cua tnhhoc lu chat

IV. Lu chat tnh trong trng trong lc

V. Tnh tng oi


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I. Khai niem

Tnh hc lu chtnghien cu lu chat trang thai can bang, khong co chuyen

ong tng oi gia cac phan t.

Tnh tuyet oi

Tnh tng oi


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II. Ap suat thuy tnh

1) nh

ngha:

Ap suat thuy tnh la lc phap tuyen tac dung len mot n v dien tch

Ap suat thuy tnh trung bnh:

Ap suat thuy tnh tai mot iem:

2) Tnh chat:

Ap suat thuy tnh tac dung thang goc va hng vao trong dien tch

chu lc

Gia tr ap suat thuy tnh tai mot iem khong phu thuoc hng at cua

dien tch chu lc

0

limA

FpA

F

A

Fp

A


4/45


*C/minh: Xet s can bang cua 1 vi

phan the tch lu chat hnh lang tru

tam giac

Lc do px tac dung len mat ABCD

chieu len Ox:px. y.z

Lc do ps tac dung len mat BCEF chieu len Ox:

-ps.y.s.sin= -ps .y.s.z/s = -ps .y.z

F la lc khoi n v, lc khoi tac dung len phan t lu chat chieu len

Ox la:

Do lu chat can bang:px .y.z-ps .y.z+(1/2).Fx .x.y.z =0

px - ps + (1/2).Fx .x = 0. Khix -> 0 px= ps

Tng t cho phng z: pz= ps=> px= pz = ps

II. Ap suat thuy tnh (tt)

x

F x y z 1

2

s

z

x

ypx

ps

A

B

C

D

E

F

z

x

y

O


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3) Th nguyen va n v cua ap suat:

Th nguyen cua ap suat :

n v cua ap suat :

+ He SI: N/m2 = Pa

+ He khac:1at=1kgf/cm2=10m nc=735mmHg=98100Pa(N/m2)

4) Ap suat tuyet oi, ap suat d, ap suat chan khong:

a./ Ap suat

tuyet oi (pt

):

La gia tr o ap suat so vi chuan la chan khong tuyet oi. La gia

tr ap suat that, v du ap suat khong kh Pa= 98100 N/m2

2[ ][ ] .[ ]

Fp F LA


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b./ Ap suat d

(pd):

La gia tr o ap suat so vi chuan la ap suat kh tri (pa)tai v tro.

pd= pt pa

pt>pa: ap suat d dng

pt


7/45PetroVietnam University

Xet phan t lu chat, tong ngoai lc tac dung chieu len phng Ox:

Lc khoi:

- lc khoi tac dung len mot n v khoi lng lu chat.

Lc matap lc:

Ap dung nh luat Newton I cho 1 phan t lu chat can bang :

III. PTVP c ban cua tnh hoc lu chat

Bx Sxd F d F d F

Bx xd F F x y z

Sx

pdF p p

x

p

- x

y z x y z

x y z

0xp

d F F x y z x y z

x

( , , )x y z

F F F F

xx

pp

p

x

z

y z

yxO



III. PTVP c ban cua tnh hoc lu chat (tt)

Vay phng trnh c ban tnh hoc lu chat la:

hay

Neu lc khoi tac dung ch la trong lc, phng trnh c ban tnh

hoc lu chat tr thanh:

- vector gia toc trong trng

10

10

10

x

y

z

pFx

pF

y

pF

z

1( ) 0F grad p

1

0g p

g




+ Lu chat tnh so vi he truc gan lien vi trai at.

+ Lc khoi tac dung ch la trong lc+ He truc toa o vi truc z hng thang ng len tri

Lc khoi theo tng phng se la:Fx= Fy= 0; Fz= -g.

Thay vao:

1)Lu chat c xem la khong nen: = const

Phan bo ap suat thuy tnh:

dp = -gdz p +gz = const

hay

1( ) 0F grad p

p g

dpg

dz

pz const



- goi la cot ap tnh

* He qua: Mat ang ap la mat nam ngang

Nhieu lu chat khac nhau, khoi lng rieng khac nhau, khong tron lan

vao nhau th mat phan chia la cac mat ang ap.

o chenh ap suat gia 2 iem A, B trong 1 moi trng lu chat ch phu

thuoc khoang cach thang ng gia 2 iem o.

Ap suat cac iem trong moi trng kh xem nh bang nhau.


A B

A B

A B AB

p pz z

p p h

pz

y

z

zB

zA

A

hAB=zB-zA

pB

pA

B

x




nhluat Pascal:Trong chat long ng yen, o tang ap suat

c truyen i nguyen ven en moi iem trong chat long.Vd: nguyen ly hoat ong cua con oi

2)Lu chat nen c (chat kh ): const

Chat kh la kh ly tng, s dung phng trnh kh ly tng

p =RT

dpg

dz

p

RT

dp p dp g

g dzdz RT p RT



Nu nhiet o thay oi theo o cao: T = T0 - az, a> 0, T0 la nhiet

o ng vi o cao z = 0

Goi p0 la ap suat ng vi z = 0

Phng trnh kh tnh

Khi chiu cao z>11km:

T = T1= -56.50C


0

0

ln ln( ) ln( )

tchphndp g g dz p T az C p R T az aR

0 0ln ln( ) lng

p T C

aR

0 0

/

1-

g RLp a

zp T

11

( )

1

gz z

RTpe

p

0

0

0

gaR

T azp p

T



3)ngdung phng trnh thuy tnh:

a)Ap ke:Ap ke tuyet oi Ap ke d

b) Bieu o phan bo ap lc:

Bieu o ap lc: la bieu o bieu dien phan bo ap suat p/tren dien

tch phang.


aph

Aph

hpa

pv

B

A

hA

pa

APA/= hA

pa

hA

pa

PA/= hA

PB/= hBhAhB

A

B



V du 1: Xac nh gia tr ap suat oc tren ap ke

neu biet: h1=76cm, h2= 86cm, h3 =64cm, h4=71cm

Giai:

pBpC=HghBC

pDpE=HghDE

pDpC=nhDC

Suy ra gia tr ap suat d oc c la:

pA=pE=nhA-B + HghB-C -nhC-D+ HghD-E=0 - n(h1+h2)+ Hgh1- nh3+ Hgh4

= n(-h1-h2+13,6h1-h3+13,6h4)

=17,732n =17,732x9810Pa=173,95KPa

Vd minh ha

h2

h1h3

h4

pa

E

D

C

B

A Kh

Nc

Hg(13,6)

Nc



V du 2: Nc chay trong ong t A-B. e o o chenh cot ap tnh ngi tadung ong o ap o chenh nh hnh ve. Xac nh o chenh cot ap tnh va o

chenh ap suat gia 2 iem A va B. Biet chat long (1) la nc nc =1000kg/m3 (2) la thuy ngan Hg= 13,6, h =0,7m, b-a = 0,3m Giai: Phng

trnh thuy tnh ap dung cho cac cap iem A-M, M-N, N-B:

Hay

o chenh cot ap tnh gia 2 iem A va B la

Vd minh ha

A M

A M

n n

B N

B N

n n

M N

M N

Hg Hg

p pz z

p pz z

p pz z

M N Hgp p h

A B

AB A B

n n

p pH z z

A

B

a

b

M

h

N

(1)

(2)



o chenh ap suat gia 2 iem A va B la:

Vd minh ha

( )

M N

M N

n n

M N

M N

n

Hg

n

p pz z

p pz z

hh

1 0, 7 (13, 6 1) 8,82Hg

AB

n

H h m

3

[ ]

(8,82 0,3 ) 9810 / 89, 467

AB AB A Bnp H z z x

m m x N m KPa



x

Be mat chat long

Trong lngrieng =

dFhhC

F

y dA

CD

y

yC

yD

Op0

1. Ap lc thuy tnh tren mot dien tch phang.

Cho 1 tam phang, dien tch A nam chm trong chat long vanghieng mot gocso vi be mat chat long.

VI. Lu chat tnh trong trng trong lc


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*o ln:

Xet mot vi phan dien tch dA, ap lc tacdung thang goc vao dien tch va co gia tr:

=>dF = pdA

p = p0+h

Ap lc tac dung len toan bo dien tch:

Vay: F = pCA = (p0 +hC)A

pCla gia tr ap suat tai trong tam C cua tam phang

VI. Lu chat tnh trong trng trong lc (tt)

0 0

0

0

0

( ) ( sin )

p sin ( A

sin A

p A

A A A

C

A A

C

C C

F pdA p h dA p y dA

A yd A yd A y

p A y

h A p A

vi : la moment tnh A / v Ox)

x

Be mat chat long

Trong lng rieng =

dFhhC

F

y dA

C

D

yyC

yD

p0O

e


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*iem at lc D:

- Vi phan moment cua ap lc oi vi truc quay Ox:

dM0x=ydF

- Moment cua ap lc tren toan bo dien tch A oi vi truc quay Ox:

+ Xet trng hp p0=0

0 0

0

2

0

2 2

0

( )

( sin )

sin (

x x

A A A

A

A

A A A

M dM ydF ypdA

y p h dA

p y y dA

p ydA y dA y dA

oxvi I la moment quan tnh cua A / v Ox)

2 2 2

0 sin sin ( ) ( ) (*)x xC C ox xC CA

M y dA I y A I I y A v


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=>F =hCA =sinyCA

Moment tnh theo ap lc F:

M0x = yDF (**)

T (*) va (**) . Suy ra iem at lc D

o lech tam:

Xet moment cua ap lc tren toan bo dien tch A oi vi truc quay Oy.

Chng minh tng t :

Neu be mat phang co dang oi xng: =>IxyC = 0 => xD= xCNen ch can xac nh yDla u.

2sin ( )xC C

D

xC

D C

C

I y Ay

F

I

y y y A

xC

C

Ie

y A

xyCD C

C

Ix xy A


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+ Xet trng hp:

a ve bai toan tng ng e giai.

Trong o:

* Tnh ap lc thuy tnh bang phng phap bieu o:

- Xet vi phan dien tch dA, tai trong tam:p = p0 +h => dF = p.dA


0

0

ph

0

p 0

C

hC

C

x

Be mat chat long

y

O

P0hC x

Be mat chat long

y

O

h0

=>

Be mat chat long

x

dFh

y

Op0

p/

dA


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o ln cua ap lc tren toan bo dien tch A:

F i qua trong tam CVcua the tch V

- Trong trng hp A la hnh ch nhat co canh song song mat thoang:

F =b

F i qua trong tam Ccua dien tch bieu o phan bo ap lc:

A A

A

A

F dF pdA

pdA

dV

F V

Be mat chat long

x

F

O

AC

y


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Vd minh ha

V du 1: Cho 1 ca van hnh ch nhat co be rong b = 5m. Chu ap lcnc thng lu nh hnh ve vi H = 2m. Hoi ap lc thuy tnh F tac

dung len van? GiaiAp lc thuy tnh tac dung len van:

F = pCA =hCA

hC

= H/2 = 2/2 = 1 (m)

=> F = 9810N/m3 x 1mx 5mx2m = 98100 (N )

V du 2: Cho 1 tam phang hnh tam giacchm trong chat long co ty trong= 1.2,

co cac kch thc nh sau:h = 3m, b = 2m

H = 1m,= 600

p0 = 0.06at = 600 kgf/m2 Xac nh o lech tam

O

CH

hC x

p0

y

O

h0

C

H


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Vd minh ha

Gii: Thay p0bang lp chat long co be day tng ng:

F = pCA =hC1/2bh = 1.2*1000kgf/m3*2.366m*1/2*3m*2m

= 8.5x103kgf

0

0

0

3sin 0.5 1 sin 60

3 3

2.366

2.3662.73

sin sin 60

mm m

C

m

CC

hh h H

m

hy m

2

0

0 2

600 /

0.51.2 1000 /

p kgf m

h mx kgf m

hC x

p0

y

O

h0

C

3 2 23

0.18218 18 2.73

362

C

C CC

I bh he m

bhy A y xy

i


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Vd minh ha

V du 3: Van phang hnh tron at trenmat phang nghieng 1 goc 600 nh hnh

Ve. Van co the quay quanh truc namNgang qua tam C. Bo qua ma sat.

Xac nh:

a./ Ap lc tac dung len van

b./ Momen can tac dung e m van.Giai:

yDyC = 0,0866m

MC= 0 M = Fx(yDyC ) = (1230x103N)(0,0866m) = 1,07x105 N.m

4

0 0 210 ( / 4) (2 ) 11,6

sin 60 (10 / sin 60 ) (4 )

xCD C

C

I m x my y my A m x m

2 23 3 3(4 )(9,81.10 / ) (10 ) 1230.10

4 4C C

D mF p A h N m m N

VI L h h l ( )


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2. Ap lc thuy tnh tren mot dien tch cong.

- Ap suat tren mat thoang bang ap suat kh tri

- Ba hnh chieu cua A: Ax, Ay, Az

- Xet vi phan dien tch dA, tai trong tam:

- Ap lc tren toan bo dien tch A:

- Ta co:


p h .dF pdA n

h

z

y

pa

()

x

A

dA

Az

Ax

Fd

n

dW

F dF

x x

y y

z z

F dF

F dF

F dF

..

.

x x x

y y y

z z z

dF pdA n pdA

dF pdA n pdA

dF pdA n pdA

VI L h t t h t t t l (tt)


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Thanh phan ap lc tren truc toa o x

thanh phan ap lc tren truc x

chnh bang ap lc thuy tnh tren

dien tch phang Ax:

Tng t cho thanh phan ap lc tren truc toa o y:

Thanh phan ap lc tren truc toa o z:

(Wthe tch vat ap lchnh tru thang ng gii han ay la mat cong)

x Cx xF p A

.

x

x x x xS SF dF pdA n pdA

z

pa

x

hA

dA

Ax

Fd

n

dA

x

pp

dFx

y Cy yF p A

.

z z

z z z z

A A A

F dF pdA n hdA dW zF W



28/45



W - The tch vat ap lc la the tch hnh lang tru thang ng taobi dien tch cong A , co ng sinh trt tren chu vi cua A, giihan bi A va keo dai cho en khi gap mat t do (p =p

a

), hay matthoang keo dai cua chat long tac dung len dien tch cong o.

Xac nh ap lc thuy tnh tren dien tch cong A la xac nh 3thanh phan cua no tren 3 truc toa o. Trong o, 2 thanh phan namngang Fx, Fyc xac nh tren cac dien tch phang Ax, Ayla cachnh chieu ng tng ng cua A theo cac truc x, y. Con Fzc

tnh bang the tch vat ap lc.Vay:

Tr so ap lc d c tnh bang :

Trong trng hp dien tch cong phc tap (co hnh chieu chongchap) => chia no thanh cac phan n gian va tnh tng phan roicong lai

2 2 2

x y zF F F F

V d i h h


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Vd minh ha

V du 4: Xac nh ap lc do dau tac dung len mot van cung dang hnh tru co ban knh 0,5m, dai 2m nam di o sau h =1m.

Giai:

Fx =pCxAx, trong o Ax=RL,

Fx = 9,81 kNFy = 0

FZ= 10,93 kN

2

Rhp dCx

h=1mDau (0,8)

FZF

FX

pa

2

d d

4z

RF W Rh L

0tg 1,11 48z

x

F

F

KNFFFF zyx 69,14222



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3. Lc ay Archimede

Xet vat co the tch V chm trong chat long

Xet vi phan the tch dV hnh lang tru thang

ng. No co 2 be mat tren va di tiep xuc

vi chat long va thanh phan ap lc tren

phng truc z tac dung tren 2 be mat nayla:dFz1vadFz2.

Ap lc tong cong tac dung len dV theo truc z:

Thanh phan ap lc tren truc z tac dung len toan bo be mat bao bocthe tch V:

Tng t, tnh c 2 thanh phan ap lc tren 2 truc con lai:z

F V

z

p0

x

dAz

dWz1

dFz1

dFz2

dWz2

V

dFz= dFz2- dfz1

dV

()

z zV

F dF dV

2 1 2 1z z zdF dF dF dW dW dV

0x y

F F



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KL:Mot vat nam trong moi trng lu chat se b mot lc ay thangng t di len tren va bang trong lng cua chat long ma vat ochiem cho.

4. S can bang cua mot vat trong chat long

a)Vat ngap hoan toan

trong chat long

C: iem at tronglng

D: iem at lc ay

archimede



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VI. Lu chat tnh trong trng trong lc (tt)a)Vat ngap mot phan trong chat longC: iem at trong lng, D: iem at lc ay archimede

MD c xac nh: MD =Iyy/W Khi MD>CD can bang on nh.

Iyy: momen quan tnh cua mat noi oi vi truc quay yy, W: the tch vat chm

trong chat long.

Khi MD


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1. Chat long tnh trong thung chuyen ong thang vi gia toc khongoi.

Vector cng o lc khoi:

dp = - axdx -(g+az)dz

Phng trnh cua ap suat thuy tnh:

p +axx+(g+az)z = const

Mat ang ap:*Chuyen ong thang ngang: p +ax + gz = const;

Tren mat phang x=const:

*Chuyen ong thang ng: p +(g+a)z = const; z =C

V. Tnh hoc tng oi

F g a

1(g a) - p 0

( )

x

z

pa

x

pg a

z

pz C

x

z

az x C

g a

az x C

g

g

agF

a

a

x

z

10F p

g

a

a

x

z

a

a=az z

x

V T h h t i


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V. Tnh hoc tng oi

V Tnh hoc tng oi (tt)


35/45


V. Tnh hoc tng oi (tt)2. Chat long tnh trong thung chuyen ong quay tron eu.

Vector cng o lc khoi:

dp =r2dr -gdz

=> Phng trnh cua ap suat thuy tnh:

Mat ang ap:

->Ho cac mat cong paraboloid tron xoay

Tren mat tru r = const:

10F p

2

2 1(g r) - p 0

pr

r

pg

z

2F g r

22

2

2z r C

g

2

pz C

2 2

2 2

1

2 2

p r gz const p r gz C

g

rgF

2

x

z

r

2

2

2

x

y

z

F x

F y

F g



36/45


V. Tnh hoc tng oi (tt)



37/45


V. Tnh hoc tng oi (tt)

V d minh ha


38/45


Vd minh ha

V du 5: Mot bnh kn rong 2m, cha ay nc chuyen ong nhanhdan eu theo phng ngang, chieu nh hnh ve (gia toc a = 2m/s).

Neu tai E co 1 lo nho. Xac nh ap suat tai A va B va ap lc tacdung len mat ng AB.

Giai:

pE= pa= 0P +axx +gz =const

pA+ axxA+gzA= pE+ axxE+gzEpA= pE+ax(xE - xA) = 0 +1000x2x0,5 =1000 Pa

pA+ axxA+gzA= pB+ axxB+gzBpB= pA+ghAB= 1000 +1000x9,81x1 =10810 Pa

Tren mat AB ap suat phan bo theo quy luat:

Nen:

pz C

1000 108101 2 23, 6

2 2

A Bp pF ab x x KN

1m

A

B

x

z0,5m 0,6m

E

V d minh ha


39/45


Vd minh ha

V du 6: Ba ong nho cung ng knh cao H = 1m noi vi nhau nhhnh ve, cha nc en o cao h = 0,5m. Biet a =0,4m. Xac nh chieu

cao nc trong 3 ong neu 3 ong quay eu quanh truc z vi van toc =2rad/s

The tch chat long khong oi:

h1+ h2+ h3 = 3h

h1= 0,424m; h2 =0,391m; h3= 0,685m

2

2 210 a

h 3h -2g

2 2

1

2 2

2

2 2

3

2

C 02

(3 )

2

ah C

gr

z h Cg

ah C

g

h1

z

h3

hh2

a 3a

r

V d minh ha


40/45


Vd minh ha

V du 7: Cho mot xe co kch thc H = 3m

, L = 5m, b = 2m, cha nc en chieu cao

h = 2,5m, chuyen ong nhanh dan eu theo

phng ngang vi gia toc a nh hnh ve. Hoi:

1) amax e nc khong tran ra ngoai ?

2) Tnh ap lc nc len thanh sau xeGiai:

T phng trnh mat thoang:

Ap lc len thanh sau cua xe:

Psau= b = 9810N/m3 x(0,5x3mx3m)x2m = 88290 N

az x C

g

max 1

5

atg

g

x

z

g

a

a

L

Hh

0,5

0,5

2

max

11,962 /

5a g m s

H

H

V d minh ha


41/45


Vd minh haV du 8: Cho mot bnh h co kch thcR = 3m, H = 4m, cha nc en chieu

cao h = 3,1m, chuyen ong quay tronxung quanh truc cua bnh vi van tocnh hnh ve. Hoi:

1)max e nc khong tran ra ngoai ?

2) Tnh ap lc nc len thanh bnhGiai:T phng trnh mat thoang:

Tai A:

Tai B:H1= C


pysau

= 2R = 9810N/m3 x(0,5x4mx4m)x2x3m = 470880 N

2 2

max

2

RH C

g

2 2max

1 4 2 0, 9 2, 22

RH H x m

g

2 2

2

rz C

g

g

rgF

2

x

z

r

2

R

h H

A

B

H1

0,9

0,9

max 2,19 /rad s H

H

V d minh ha


42/45


Vd minh haV du 8: Cho mot bnh h co kch thcR = 3m, H = 4m, cha nc en chieu

cao h = 3,1m, chuyen ong quay tronxung quanh truc cua bnh vi van tocnh hnh ve. Hoi:

1)max e nc khong tran ra ngoai ?

2) Tnh ap lc nc len thanh bnhGiai:T phng trnh mat thoang:

Tai A:

Tai B:H1= C


pysau

= 2R = 9810N/m3 x(0,5x4mx4m)x2x3m = 470880 N

2 2

max

2

RH C

g

2 2max

1 4 2 0, 9 2, 22

RH H x m

g

2 2

2

rz C

g

g

rgF

2

x

z

r

2

R

h H

A

B

H1

0,9

0,9

max 2,19 /rad s H

H

Cu hi trc nghim


43/45


Cuhi trc nghim

Cu hi trc nghim


44/45


Cuhi trc nghim

Cu hi trc nghim


45/45

Cuhi trc nghim

chuong2 clc

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