CoHocChatLuu

8/14/2019 CoHocChatLuu

1/42

NHP MN: C HOC CHT LUCHNG 1. CAC KHAI NIM V

CHT LU.

1.NH NGHA.Cac trang thai long va kh goi cac cht lu chung trai ngc

vi trang thai rn.-S khac bit gia cht long va cht kh.

Cht kh chim toan b th tch ma no c cha. Con chtlong th khng (v du: Bnh ng kh va bnh ng nc).

-Ranh gii gia cht long va cht kh t sai lch v ln cua(khi lng th tch) n* (mt ring hay mt hat). Cht long

ln hn khoang 1000 ln)

A:M|MM

*n;V

M[ AA

== vogadro.

iu nay cho thy: Khi lng th tch cang tng, th cac phn t

cang gn va cac lc tng tac phn t trong cht long rt quan trong.-S khac bit gia cht long va cht rn.+ D chay, ly dang cha no lam hnh dang.+ Co th cu tao lai sau khi rai ra (rot ra).Hin tng lun khac bit gia cht long va cht rn c giai

thch bi tnh di ng rt ln cua cac phn t trong trang thai long.Mt s khac bit na la vn tc cac im cua cht rn c

tnh theo theo cng thc:( ) ( ) MPMVPV +=

rrrr

Con i vi cht long vn nay rt tinh t khi cht long

chuyn ng.

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2/42

2.M HNHCUA CHT LU.-Theo kch thc v m: Cht lu la mi trnglin tuc; ngi

ta thng ly chiu dai c trng L quan sat kch thc v m

c ap t cho vn nghin cu.-Theo kch thc vi m: Cht lu la khng lin tuc no gm cac

phn t ang xao ng nhit lin tuc.-Theo kch thc trung m: La kch thc trung gian gia v

m va vi m. Cht lu vn la mi trng lin tuc+ Vi quan im nay cht lu c ct ra bng cac t bao phn

t hay phn t cht lu = hat cht lu (c cha rt ln s phn t).

-Vn tc cua hat cht lu tp trung tai im M thi im tbng gia tr trung bnh cua cac vn tc cua cac phn t c cha.

Kt lun: Kch thchat cua cht lu la trung m,no cho phep kt hp vaohato, nhng ai lng vm m ta cht lu nh

mt mi trng lin tuc.3.AP SUT CUA CHT LU.

1.nh ngha.Ap sut P(M) tai 1 im M.

Trong cht lu c xac nhc bi

( ) ndsMPFdr

r

=

ds: phn t din tch baoquanh im M

r

: phap tuyn i vi dsnP(M): ai lng v hng.

2


3/42

: Lc b mt tai im MFdr

2.iu kin bin.Goi P1 va P2 la ap sut 2 bn

cua phn t ds. Mt phn t thtch dV=hds, dm: phn t khilng. Theo phng trnh c bancua LH:

( ) dsnPP 1221dVfadm V.r

r

r

+= v h v cung be

21 PP0dV,dm ==

mt phn cach hai cht lu ap sut la lin tuc.

4.TNH NHT. phan anh chuyn ng cua cac cht lu thc. T thc

nghim ta a ra: lc ct (trt) hay goi lc nht trong chuyn ngmt chiu c th hin nh sau:

( )dientichS

etyvVSyVF x

:,

r

rr

r

==

: goi la nht; la hng s c trng cua cht lu

Co th nguyn la: [ML-1T-1]: (Kg/m.s) (N.s/m2) Pa.s1Pa = 1N/m2 Trong (SI) Pl = Pa.s (poisenille)Tnh nht la tnh cht cua cht lu chng lai s dch chuyn.

Tt ca cac loai cht lu thc u co tnh nht nht nh, th hin

di dang ma sat trong khi co s di chuyn tng i gia cac phnt cht lu.

Cac cht lu rt nht th co chng sc lai s di chuyn rtln.V du nh du m, nht...

Tnh nht c trng cho chay cua cht lu.

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4/42

5.PHN BIT DONG CHAY TNG VA DONGCHAY RI. S REYNOLDS.

1.Thc nghim cua dong chay cht lu thc.*Th nghim cua Reynolds.Dung bnh cha nc ni

vi ng thuy tinh. Khi m khoavoi nc co th chay vao ngvi cac vn tc khac nhau. Ncmu i t lo ng mu qua ngdn vao ng th nghim. Vi vntc nho, dong mau trong ngkhng b hoa tan vi nc xungquanh va co dang mt ng ch thng.

-Dong chay trong trng hp nay la dong chay tng. Khi tngvn tc trong ng, dong nc mau luc u co dang song, sau o hunh bin mt, hoa tan trn b mt ct va nhum u khp cht ncxung quanh.

-Chuyn ng cua cht lu tr nn hn loan, cac phn t ncc nhum mau bay i moi pha va va cham vi cac phn t khacva vi thanh ng: chuyn ng nay c goi la chuyn ng ri.

c trng c ban cua dong ri la: tn tai thanh phn vn tcngang so vi phng chuyn ng cua dong chay.

*Kt lun: Dong chay tng nu cac ng dong trt trnnhau, cac phn t lun gi phng song song; dong chay tng xay rakhi vn tc rt nho. Con ngc lai, vi vn tc ln ta co dong chayri ( khng n nh va cu truc ri loan).

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5/42

2.S Reynolds.S chuyn t tng sang ri i vi cac dong chay c xet thc

hin bng:

-Vn tc trung bnh V cua cht lu: la thng s ta thy ro rangtrong th nghim trn.- nht cua cht lu. Ta hin nhin thy dong ri kho thc

hin vi du so vi nc.-ng knh ng D: Nu ng knh ng nho cho ta dong chay

tng hn ng co ng knh ln.-Khi lng th tch cua cht lu: Thng s nay khng anh

hng; nhng khi lng th tch lun co trong phng trnh tintrin. S khng th nguyn c goi s Reynolds, ky hiu nh sau:

=

VDR e

Thc nghim cho thy = 103 Kgm-3, = 10-3plnu V = 2,5cm/s va Re=300 : dongchay tng

nu V = 1,2m/s va Re = 14000 :dong chay ri.Kt lun: S Reynolds Re

2000: dong chay tngRe > 2000: dong chay ri

6.DONG CHAY CUA CHT LU LY TNG.Trong c hoc cht lu giam nhe vic giai mt s bai toan,khai nim v cht lu ly tng c s dung rng rai. Cht lu lytng c hiu la cht lu gia nh co tnh dch chuyn tuyt i,tc la hoan toan khng nht, cung nh khng nen tuyt i, khng

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6/42

dan n khi nhit thay i va tuyt i, khng co kha nng chnglai lc ct. n gian v tnh toan ta thng cht lu ly tng lamm hnh cho cht lu thc.

7.CAC C TRNG CUA DONG CHAYCHT LU.

1.Quy ao.Chuyn ng cua hat cht lu c tao thanh bi tp hp cac

im cua khng gian va thi gian khi no i qua la ( )tRr

co phng

trnh sau:( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )

dtt,tZ,tY,tXV

dZ

t,tZ,tY,tXV

dY

t,tZ,tY,tXV

dX

Zyx

===

2.ng dong. thi im t0 a cho, ng dong la ng cong ma tai o vec

t vn tc tip tuyn vi mi im co phng trnh:

),,,(),,,(),,,( 000 tzyxv

dz

tzyxv

dy

tzyxv

dx

zyx==

3.ng phat xa (anh du). thi im a cho,toan b cac hat i qua im nay u c

"anh du" va tao thanh mt ng cong goi la ng phat xa.

4.Dong chay dng.

Trng vn tc ( )rvrr

khng phu thuc tng minh thi gian t(i vi dong chay nay 3 ng trn trung nhau).

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7/42

CHNG II. NG HOC CHT LU.

1.M TA CHUYN NG THEO LAGRANGE.Chung ta nghin cu mt cht lu theo v m; chuyn ngcht lu trong 1 h qui chiu c goi la dong chay.

Nghin cu dong chay cht lu, ma m ta chuyn ng mi hatring bit cua cht lu, c xac nh trc.Trong khi bit qu ao

cua mi hat (t( )tR ir

( )0R ir

vi t=0), ta theo doi qua trnh chuynng cua no va tip tuc cho tt ca cac hat cua cht lu. M ta nay goi

la m ta theo Lagrange.V du: 1.Ngi cu ca.

2.Giao thng trn ng t.

Kt lun: Chuyn ng cht lu c m ta hoan toan bng sbit cac qu ao cua mi hat c anh du (nh trc) i cuacht lu.Con vn tc cua cac hat c xac nh bi:

( )tR ir

( )( )

( )( )t,tRVdt

tRdtV i

ii

rrr

r==

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8/42

( )( )

( )( )t,tRVdt

tRdtV

rrr

r==

Vi ( )tR ir

v tr thi gian t cua hat ma ban u co v tr ( )0R ir

thi im u t=0.Cac vn tc nay ch phu thuc ro rang vao thi gian va cac toa

ban u cua hat, tc la ( )tRr

.Ta thong dung ky hiu X(t),Y(t), Z(t) lam bin LagrangeV du ap dung. Cho dong chay m ta theo Lagrange:

( ) ( )

( )

=

+=

i,0i

i,0i

YtY

bt1XtX

( )constb = -Toa ban u cua hati khi t=0

i,0

i,0

Y

X

Xac nh vn tc ( )tVr

cua cac hat va tm ( )( )?t,tRV irr

Giai

( )( ) ( )( )

dt

tRdtVt,tRV iii

rrrr

==

( )( )

xix

i

i ebXedt

tdXtV

rrr

,0==

( ) ( )( ) x0ii ebXt,tRVtVrrrr

== , ma X0,i= bt

tXi

+1

)(

Vy ( )( )( )

xi

i ebbt1

tXt,tRV

rrr

+=

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9/42

2.M TA CHUYN NG THEO LE.

1.Khai nim.

Chung ta ng tai 1 im cua khng gian va xem xet (nghincu) qua trnh tin trin (bin i) mt ai lng v m nao o cuacht lu theo thi gian goi la m ta le.

V du: 1-Vn tc cac hat tai 1 v tr c nh.

2-Vn tc cac t tai 1 v tr c nh.

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10/42

2.Tnh c lp cua cac toa khng gian va thi gian.Trng vn tc ( ) ( ) ( )t,z,y,xVt,MVt,rV

rrrr== phu thuc khng

gian va thi gian: rr va t hay (x,y, z va t) la cac bin c lp.

3.Ta nh ngha m ta le cua cht lu.Chuyn ng cua cht lu c m ta hoan toan bi bit cacvn tc cac hat cua cht lu i qua 1 im M khng gian cho trc thi gian t.

-Cac toa khng gian va thi gian la cac bin c lp.-M ta nay dung m ta qua trnh tin trin (bin i) cua

cac ai lng c trng khac cua cht lu theo thi gian.

V du: Ap sut P(M,t); nhit T(M,t)-Quan im le m ta trang thai cht lu khi chuyn ng

bng cach kt hp cac trng v du trng vn tc, ap sut, nhit .Phn bit vi cach vit Lagrange,ta co x, y,z va Rr

rr ; Vv

rr

4. V du: (Biu din trng hp vn tc bng le).Khi nghin cu chuyn ng cua mt cht lu:tn tai mt ai

lng cho phep m ta dong chay v du nh:-mc nc trong ng.-lu lng tthoat ra.Cac ai lng nay cho phep m ta v m qua trnh chuyn

ng cua cht lu.Theo le chung ta tm ( )t,Mv

rtai moi im M cua cht lu

va cn phai xac nh toa cua M ma khng mu thun (tranh chp)

vi ai lng trc o.Kt lun Khi m ta chuyn ng cua cht lu bng le. No

tn tai:-1 bin xac nh trang thai cua cht lu.-1 bin cho phep nh mc le cua im M.

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11/42

Chov tr cua M bi cao z trn truc zz,h(t):toa phuc thuc thi gian t va cao cht lu

trong ng

z:toa le cua im MTh vn tc ( t,MV )r

theo le c cho bi biu thc

( ) ( )zetht,Mv

r&r =

Chu y:

S cn thit dung hai khai nim h va z v h(t) biu din caocua cht lu, con z la cao im M.

S phu thuc cua ( )t,Mvr

la ham cua:

-toa khng gian zer

-thi gian qua ( )th&

5.Tnh duy nht vn tc cua mt hat cht lu.-Theo le: Ta bit vn tc cua hat v tr M va thi gian t:

( ) ( )t,rvt,Mvrrr

=

-Theo Lagrang: Cn phai bit hat c anh du, ma qu aocua no i qua v tr M thi gian t ( )( )tRr

rr= khi t=0, ( )0R

r

Trong khi cho hat nay i qua tai rr thi gian t( )( tRr )

rr= . Vn tc cua no thi gian t la:

( )( )

( )( t,tRVd

)t

tRdtV

rrr

r== va ta co: ( )( ) ( )t,rvt,tRV rr

rr=

Vy va( ) ( t,rvt,Mvrrr

= ) ( )( )t,tRVrr

cn phai ng nht

Nhng x ly toan hoc rt khac nhau:-Theo Lagrang u tin cac hat cht lu c theo doi trong qua

trnh dch chuyn ma chung ta a vn tc vao.

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12/42

-Theo le, u tin cac v tr khng gian ma chung ta atrng vn tc cua chung vao, phu thuc khng gian va thi gian(cac bin c lp).

thi im t, tai v tr M: ( )( ) ( )zLog t,rvt,tRVrrrr

= Theo Lagrange tai v tr OMr =

r thi gian t cn phai tm hat

ma quy ao ( )tRr

i qua v tr M thi gian t.

( )tRrrr

= V du:

( ) ( )

( )

=

+=

0ii

0ii

YtY

bt1XtX

Theo trc ta co: ( ) ( )( )( )

xi

i ebbt1

tXt,tRVtV

rrrr

+==

Chuyn t Lagrange sang le, c thc hin trong khi noirng hat th i i qua im co hoanh x theo t nu x= Xi(t) nn

( ) xebbt1

xt,rv

rrr

+=

Con theo le, ta co th tnh nh sau:

,)(),( xetxtrvr

&rr

= ma = = X.

( )x t.

( )X t.

X i,o b v

Xi,o= bttX

+1)(

nn bbt

txb

bt

tXtx

+=

+=

1

)(

1

)()(&

vy ( ) xebbt1

xt,rv rrr+

=

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13/42

III.ao ham toan phn.1.Y ngha vt ly cua mt bin i toan phn.

Xet chuyn ng ri cua ngi du co vn tc thng ng

zevv

rr

= (v


14/42

T y: vdtdz

dTdTdu = , hay TdtgradvdTdu

r=

Ngoai ra, ta co:

vdt

dTdu = , ma co th vit

tng t Tgrad.vdt

dTdu r=

S bin i nay, trnhbay s bin i cc b cuanhit nhn theo ngidu xet ging nh 1 hat. Noc goi la s bin i toanphn hay ao ham toan phncua nhit c vit

t

T

Dt

DT

( 0t

T=

, trong

trng hp chung ta quan tm) hayz

T

ao ham ring i vi

cao z (ai lng bng trong trng hp trnh bay). Nh vy, ngi nhay du cm trong tay nhit k va quan sat

nhit bin i theo thi gian, o bng ao ham toan phnDt

DT.

Trong v du c xet, ngi nhay du quan sat mt s bin i c

goi i lu( convective). TgradvDt

DT r=

Nu dng s ri va quan sat trong ch khng dng (khngn nh) mt s bin thin cuc b theo nhit xut hin :

t

T

Dt

DT

=

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15/42

Trong trng hp chung ta co:

Tgradvt

Tgradvt

T

Dt

DT

+

=+

=rr

2.Y ngha vt ly cua bin i toan phn i vi cht lu.V du trn cho chung ta nm c khai nim ao ham toan

phn.Chung ta tng tng mt ngi ngi trn 1 hat cht lu.Cac bin cac ai lng c o la cac bin i toan phn.

i vi hat nay, ao ham 1 ai lng v hng (vec t Gr

),

c vitDt

GD;

Dt

DTr

( ) ( )dt t,Mgdtt;MdMgDtDg ++=r

vi ( )dtt,MvMd rr

=

( ) ( )dt

t,MGdtt;MdMG

Dt

GDrrrr

++= vi ( )dtt,MvMd

rr=

9


16/42

Bai tp:1.Cho dong chay theo Lagrange di dang:

( ) ( )

( )

=

+=

0

0

YtY

bt1XtX(vi b=const)

Tm gia tc cua 1 hat trc tip va s dung theo le2.Cho trng vn tc vi truc OZ thng ng, hng ln.

Xac nh bi0z

0x

vgtv

uvv

+=

==

r

3.Ta xet 1 dong chay cht lu gia mt y=0 va mt v han do daong X=asint. Ta co trng vn tc:

( ) ( ) ( ) xxKy et,yvekytcoseat,z,y,xv rrr == .

Tm gia tc cua hat.

Lu y: 0Dt

DP,0

Dt

D==

Ngha la: hat cht lu co khi lng khng i, th tch cua nonhng thay i theo thi gian; tng t i vi ap sut.

3.ao ham toan phn mt ai lng v hng g.Khi m ta ng hoc cac dong chay, chung ta a xet t quan

im Lagrange n quan im le trong khi quan tm n trngvn tc cua cht lu. Chung ta tm cach biu din ao ham toan phncua mt ai lng v hng. Chung ta bit rng

( ) ( )( )Lagole t,tRVt,rvrrrr

= , y ( )tRr

biu th qu ao cua hat i qua

im M thi gian t. Nh vy ta co:( ) ( ) M0tRtr ==

rr

Xet 1 ai lng v hng ( )t,rgr

: g = : khi lng th tch.g =P : ap sut.

10


17/42

Cn tm =Dt

Dg?

trong dt, hat dch chuyn t im M(X,Y,Z) ti

( )dZZ,dYY,dXX'M +++ vi dtvdZ;dtvdY;dtvdXzyx === hay dtvMd r

r= ai lng g bin thin Dg.

dtt

g

z

gdY

y

gdX

x

gDg

+

+

+

=

T y dtt

gv

z

gv

y

gv

x

gDg zyx

+

+

+

= ao ham toan

phn

ggradvt

ggradvtg

zgv

ygv

xgv

tg

DtDg zyx +=+=+++=

rr

Trong o:

- gradvr

(s hang) ao ham i lu (convective): no ch ra tnhkhng ng nht cua g.

-t

:ao ham cuc b, no ch ra tnh khng thng xuyn cua g.

Vy ta co th vit: ao ham toan phn cua khilng th tich

dgrav

tDt

D rr+

=

4.ao ham toan phn cua ai lng vec t Gr

zzyyxx eGeGeGG

rrrr++=

,eDt

DG

eDt

DG

eDt

DG

Dt

GDz

z

y

y

x

x rrrr

++= vix

xx

Ggradvt

G

Dt

DG r

+

=

vy ( )zzyyxx eGeGeGgradvtDt

GD rrrrr

++

+

=

Toan t ( )z

vy

vx

vgradv zyx

+

+

=r

trong toa cac.

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18/42

Tng t trc y: GgradvtDt

GD rrr

+

=

Trong toa tru: Gz

v

r

1v

t

v

tDt

GDzx

rr

+

+

+

=

Trong toa cu: Gsinr

1v

r

1vv

tDt

GD

r

r

rr

+

+

+

=

5.Ap dung: Gia tc cua hat.

( )vgrad.vt

v

Dt

aDa

rrrr

r+

==

( ) ( ) vvrot2vgradvgrad.v2

rrrr

+

=

V du: Cho dong chay hai chiu, trng vn tc c xac nh trongvung x>0; y>0 la ( )( )ky,kxt,Mv

rtc la:

( ) yx ekyekxt,rvrrrr

+= Hay tnh gia tc cua hat theo le va Lagrange:*Theo le:

( ) xkkxy

kyx

kxtDt

Dva 2xx =

+

==

( ) ykkyy

kyx

kxtDt

Dva 2

y

y =

+

==

Vy M0ka 2=r *Theo Lagrange: Quy ao c tm bi:

( )

( )

=

=

=

= kt

0

kt

0

eYY

eXX

tkYdt

dY

tkXdt

dX

( ) ( )tkXekXdt

dXtV

kt

0x ===

12


19/42

( ) ( )tkYekYdt

dYtV kt0y ===

( )( )tXkeXk

dt

tdVa 2kt0

2xx ===

( )( )tYkeYk

dt

tdVa 2kt0

2y

y ===

yx etYetXtrtRtRkarrrrrr

)()()()();(2 +===

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20/42

CHNG 3.S BAO TOAN KHILNG.

1.LU LNG KHI .1.nh ngha.

Ly mt mt nh

hng trong mt h quichiu.

Sr

Goi la khi lng

phn t i qua mttrong thi gian .

m

Sr

tTa goi Dm la lu lngkhi cua cht lu i qua

mt Sr

sao cho.tDmm =

Dm c xac nh nh sau: ( ) ttPvldNdSSd ,,rrrr

== Th tch phn t: ( ) ( ) tdSNtPvtSdtPvd .,,

rrrr==

Khi lng : ( ) ( ) ( ) mtdSNtPvtPSdtPS ==rr

,,,

Lu lng khi phn t:( ) ( ) ( ) ( ) dSNtPvtPSdtPvtPdDm

rrrr,,,, ==

Lu lng khi i qua 1 mt xac nh trc.( ) ( ) ( ) ( ) dSNt,Pvt,PSdt,Pvt,PD

SS

m

rrrr& ==

(khngong)

(khngong)

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21/42

( ) ( )= dSNt,Pvt,PDmrr

S(ongkn)t:

goi la

mt th tch cuadong khi lng

rr

( tPjv ,rr

= )

( )=S

m dSNtPjD ,

( )

=

Sm

dSNtPjDrr

,khong

on

ong

2.Mt kim soat va mt c bit.*Mt kim soat la 1 mt c nh trong 1 h qui chiu, mt nay

nh mt th tch kim soat*Mt c bit la 1 mt

trn o c xp t mtcach lin tuc cac hat cuacht lu.

-Cac im trn mtc dch chuyn vi vntc nh vn tc cua cht lu.Mt nay nh mt th tchc bit.

Mt nay b keo i bicht lu.

H qua: Mt nay b keo i vi vn tc cua cht lu. Khng comt s chuyn qua mt nay.

2


22/42

Khi lng M nm trong th tch c bit bt bin theo thigian.

Nh vy: 0

Dt

DM=

-ao ham toan phn cua 1 tch phn theo th tch. Ta tnh:

Dt

DG, vi ( )=

V

dtt,MgG

Nu : khi lng th tch th G: la khi lng

cua cht lu c cha trong th tch c bit.

( ) ( t,Mt,Mg = )

Theo nh ngha ao ham toan phn cua G laDt

DGc biu

th:

( ) ( )

t

dtMgdttMg

Dt

DG tvttv

+=

+ )()(:

,,

( ) ( ) ( ) ( )

t

dt,Mgdtt,Mgdt,Mgdtt,Mg

1 23 3 V VV V

++

+

=

( )( ) ( )

t

dt,Mgdtt,Mg

t

tdt

t,Mg

1 23 V VV

+

+

=

Ta nhn c

=Dt

DG ( )

dtt,Mg

V

+ S

dSNtPvtPgr

r

),(),(

3


23/42

Kt lun: S hang( )

dt

t,Mg

Vbin cuc b

S hang SdSNtPvtPg

rr),(),(

bin i lu

V du: Nu g = 1 G = V. Theo strgradski

( ) dvdivdSNtPvDt

DV

S V

== )(,rrr

Nu ( ) 0vdiv =r : th tch toan phn khng i

Khi lng th tch cung khng i; 0Dt

D=

o la tnh c trng cua dong chay khng nen c

4


24/42

2.CN BNG KHI LNG.1.Phng trnh tng quat trong mi trng khng co ngun.

Xet 1 th tch V c nh (Th tch kim soat) cua khng gian

c chim bi cht lu nh bi 1 mt ong S (mt kim soat) trong1 h qui chiu, N

r- phap tuyn ngoai.

Khi lng cua cht lu m(t) cha trong th tch V, c vit

( ) ( ) =V

dt,Mtm

Phn cht lu i vao va ra lin tuc c nh (gii han) bi mtc nh nn khi lng m(t) ch phu thuc vao thi gian t.

*i vi 1 phn t th tch d cha khi lng. S bin i( ) = dt,Mdm ( )dm trong thi gian t

( )( )

tdt

t,Mdm

=

Khi lng toan b cua cht lu trong th tch V c bini trong thi gian la:t

( )

= V tdt

t,M

m V ( c nh)

*Khi lng mi qua mt phng S cnh c nh bi thtch V trong thi giant.

S tng khilng nay phu hp vikhi lng cua cht lua i qua mt S tngoai vao trong vikhoang thi gian t

5


25/42

ngha la:tDm ram ,=

Biu th di dang

( ) ( ) =S

t.dSNt,Pvt,Pmr

r

ongc nh gii hanV

Vy ta co phng trnh bao toan khi lng dang tch phn( )

( ) ( ) 0,,,

=+

V S

dSNtPvtPdt

tM rr

ong (kn)

Cng thc ghi nh.S cn bng cua qua

trnh bin i khi lngcha trong th tch V cnh khng co ngun cth hin bng phng trnhbao toan khi lng dangtch phn.

( ) =

V

m ra,Ddt

t,M

hay( )

( ) ( ) 0dSNt,Pvt,Pdt

t,M

V S

=+

rr

c nh (ong)

6


26/42

2.Trng hp mi trng co ngun.Goi Dm,ngun: lu lng khi ngun

Ta co:( )

ra,m

V

Dd

t

t,M=

+ Dm,ngun

Hay:( )

( ) ( ) +

V S

dSNt,Pvt,Pdt

t,M rr=Dm,ngun

ong gii han V

3.Bao toan lu lng khi

*Ch n nh (dng): = 0t

Dm,ra = Dm,ngun

*Cht lu khng chu nen : =

0t

Dm,ra = Dm,ngun

Nu khng co ngun cung cp: Dm,ngun = 0

4.Bao toan lu lng th tch cua cht lu khng chu nen.

Nu th goi DConst=

v la lu lng th tch ta co:SdtPvDDDS

VVm

rr

== ),( hay ( )=S

v dSNt,PvDrr

khng ong (m) ong

Kt lun:Trong trng hp cht lu khng chu nen, s bao toan lu

lng khi dn n s bao toan lu lng th tch.

7


27/42

3.DANG CUC B (A PHNG) CN BNGKHI LNG.

1.Phng trnh tng quat.Trng hp khng ngun, ta co phng trnh tng quat sau:

( )( ) ( ) =+

V S

0dSNt,Pvt,Pdt

t,M rr

ong

Theo cng thc Ostrogradski ta co:( )

( ) ( )[ ] =

+

V

0dt,Mvt,Mdiv

t

t,M r

( )( ) ( )[ ] 0t,Mvt,Mdiv

t

t,M=+

r

Hay( )

( ) 0t,Mfdivt

t,M=+

r

goi la phng trnh bao toan

khi lng dang cuc b (a phng).

Mt khac ta co: ( ) vgradvdivvdivrrr

. +=

Th vao phng trnh trn ta chu y: DtD

grad.vt

=+

r

Ta co th nhn c phng trnh bao toan khi lng dangcuc b con goi phng trnh lin tuc co dang sau:

0vdivDt

D=+

r

Trong ch n nh ta co: ( ) 0vdiv =r

i vi cht lu khng chu nen v 0DtD = ta co 0vdiv =r

V du ap dung (xem v du 3 trang 45) va 4,5 Bai tp 3,4.

8


28/42

Cho trng vn tc cua cht lu khng chu nen, chay ra lulng khi lng Dm bi mt ngun theo n v chiu cao h trungvi truc 0Z, bit rng cac hat thoat ra vung goc vi dong tc la:

( ) ( ) ret,rvt,rvrrr

= xem hnh ve

hay tnh gia tc cua hat cht lu.

9


29/42

CHNG 4.M TA NH HNH MT VAILOAI DONG CHAY.

1.CAC C TRNG VN TC CUA CHT LU.1.M ta cuc b.

i vi dong chay bt ky, s vn ng cua mt phn t th tchcht lu t hp ba dang cuc b c nhn thy ring re:

+s gian n (dilatation).+s quay (rotation).+s bin dang (deformation).i vi dong chay phng la s vn ng cua mt phn t din

tch

2.Trng vn tc va s gian n: vai tro cua toan t vdivr

Mt dong hcay ba chiu c gia thuyt la mi thanh phn vn tcch phu thuc vao toa tng ng M(x,y,z):

( , , , ) ( , ). ( , ). ( , ). x x y y zv x y z t v x t e v y t e v z t e= + +r r r

z

r

Trong thi gian t, cac vach cua mt phn t th tch dxdydz dch

chuyn trc giao vi chnh chung.Canh co chiu dai dx cua hnh lp phng tr thanh :

' ( , ) [ ( , ) ] 1 xx xv

dx x dx v x dx t t x v x t t dx t x

= + + + + = +

Tng t ta co :

' 1yv

dy dy t y

= + va ' 1

zvdz dz t z

= +

Nh vy th tch nguyn t a bin thanh () sao cho :

( ) ' ' 'yx z

vv vdx dy dz dxdydz t divv t

x y z

= + + =

r

1


30/42

Ngha la :( )

=

t

1vdivr

= zv

yv

xv zyx

++

V cuc b: th h s bin i tng i cua th tch trong nv thi gian bng div cua trng vn tc.

Trng vn tc cua cht lu cho ta thng tin v s gian n cua

no nh trung gian div cua vn tc (Nu 0vdiv =r

ta co 0Dt

D=

dong chay khng chu nen c).

3.Trng vn tc va s quay: vai tro cua toan t vrotr

.

cuc b, trng cac vn tc cua cht lu co th ng dang(ging nh) trng vn tc cac im thuc vt rn quay (co vec tquay ). S quay c bit nay (s xoay) cua cht lu tai mt im

M se tn tai nu

r

( ) 02vrot =r

r

.Cho ta bit s tn tai cac vung xoay.

Toan t

=

y

G

x

Gx

G

z

Gz

G

y

G

Grot

xy

zx

yz

r

xem v du chng minh : =rr

2vrot

2


31/42

2.Cac c trng cua dong chay.1.Dong chay dng.Mt dong chay i vi trng vn tc le cua cht lu khng

phu thuc thi gian t tng minh:( ) 0=

=t

vvoiMvv

r

rr

Trong dong chay dng, lu lng khi i qua moi tit din cua

ng dong nh nhau v ( ) :M0t

=

nn Dm,ra = Dm,ngun.

2.Dong chay khng chu nen c.Dong chay ma th tch cua cac hat cht lu c bao toantrong qua trnh chuyn ng goi dong chay khng chu nen.

Ta co: ( ) 0t,Mvdiv =r

tai moi ni.Trong dong chay khng chu nen lu lng khi i qua moi

tit din cua ng dong nh nhau: Dm,ra = Dm,ngun.

3.Dong chay xoay va khng xoay.Mt dong chay c goi khng xoay nu vec t xoay 0=

r

0vrot =

r

tai moi ni, ngc lai nu r

0 goi la dong chay xoay.

4.Dong chay phng khng chu nen c.

( )

( )

( )

=

=

0v

t,y,xv

t,y,xv

t,z,y,xv

z

y

xr

; vArot0vdivr

r

r

==

( )

( )

=

t,y,x

0

0

t,z,y,xAr

3


32/42

Khi o ( )[ ] [ ] zz egradet,y,xrotvrrr

==

( )

=

x

yt,z,y,xvr

goi la ham dong( t,y,x )tng ng vi mt ng dongcte=

Khi co dong chay phng khng chu nen c, th co th xacnh 1 ham goi ham dong sao cho.

( )[ ] ( ) zz et,y,xgradet,y,xrotvrrr

== Cac ng dong ( ) ctet,y,x 0 = ng nht vi cac ng

dong t0.

5.Dong chay th.Mt dong chay khng xoay c goi la dong th tai moi im

cua dong chay (tn tai sao cho gradv =r

, goi la th vn tc)

Th cua cac vn tc phai thoa man = gradvr

-Nu dong chay th khng chu nen c, th ham se tun

theo phng trnh Laplace: 0= (v ( ) 0graddiv,0vdiv ===r ).

4


33/42

CHNG 5.PHNG TRNH NG LC HOCA PHNG I VI DONG CHAY LY TNG.

1.ng sut trong cht lu.1.Lc mt (ng sut mt).

Chung ta nh ranh bn trong cht lu bi 1 mt ong c nh . Cachat cht lu bn ngoai mt tac dung ln cac hat bn trong, tacdung nay xay ra ngn va ln cn mt . Gia s co 1 phn t mt dS

cua . Hp lc Fdr

cac lc tac dung bi cac hat bn ngoai ln bn

trong c phn tch thanh 2 thanh phn NFdr

va TFdr

hay

TN FdFdFd

rrr

+=Thanh phn c goi ap lc (ap sut) va c xac nh:NFd

r

( ) dSNt,MPFd Nrr

=

Thanh phn TdSn.vgradFd Trrr

= ; nu Tvvrr

= goi lc nht.

i vi cht lu ly tng, ta bo qua lc nht 0Fd T =r

2.Lc th tch (Lc khi).Xet 1 phn t th tch d cua cht lu chu tac dung cua cac lc

th tch (v du trong lc). Nhng tac dung nay lin quan ti tt ca cachat, nh vy no t l vi s lng hat hay phn t th tch .d

Vy: = dffd vrr

fdr

goi la lc th tch.

Trong trng trong lc vi gia tc trong trng gr

Mt th tch cua lc la: gfvrr

= Mt khac ta co: ddm =

Lc th tch co th vit: == dfdmffd mmrrr

mv ffrr

= ; vi lc trong trng th gfmrr

=

1


34/42

Kt lun: Vi phn t th tch d va khi lng dm cua cht lu chutac dung cac lc khi hay th tch c biu din nh sau:

== dfdmffd vmrrr

vi mv ffrr

=

i vi trong lc : gfvrr

= vi gfm rr

=

3.S tng ng cua lc th tch, khi lng.Ta ch xet trng hp lc ap sut.(ap lc)Theo trc y khi xet cac lc nguyn t tac dung ln 6 mt

cua hnh hp phn t th tch cht lu ta co:

zyx edxdydzz

P

edxdydzy

P

edxdydzx

P

Fd

rrrr

=

dmPgradd

PgradPdgrad

===

Lc nay ng nht 1 lc th tch. Pgradfv =r

Kt lun: S tng ng cua lc th tch (khi lng) cua cac lcap sut trn mt c biu din di dang:

Pgradfv =r

=

Pgradfm

r

Cac tng ng nay dung tnh hp lc hay mmen cua caclc ap t ln 1 phn t bao quanh bi cht lu.

Cac tng ng nay khng dung tnh cng cua cac lc ap

sut.

2


35/42

2.Phng trnh le va ap dung.Ch xet cho dong chay ly tng khng co lc nht.

1.Phng trnh.

Phng trnh c ban cua ng lc hoc i vi mt hat cht luco khi lng dm. Di tac dung cua hp lc Fd

rta co th vit:

dmfdfFdDt

vDdm mv

rrrr

===

mfDt

vD rr

= ,toan b

v ( ) vvrot2

vgrad

t

vvgrad.v

t

v

Dt

vD2

rrr

rrrr

+

+

=+

=

m

2

fv22

vgrad

t

v rrrr

=+

+

= ,toan b

= vrot

2

1 rr

(s dung cng thc: ( ) AArot2

AgradAgrad.A

2 rrrr+

=

v ( )

+

+

=

Az

Ay

Ax

zAz

yAy

xAxAgrad.A

rr

z

AxAz

y

AxAy

x

AxAx

+

+

=

z

AyAz

y

AyAy

x

AyAx

+

+

+

z

AzAz

y

AzAy

x

AzAx

+

+

+

( ) ( ) ( )

+

+

= zAz

yAy

xAx2

1 222

3


36/42

y

Ax

x

Ayx

Az

z

Ax

z

Ay

y

Az

Arot

=r

Ta phn bit cac lc trn n v khi lng va tng ng cualc khi lng ch do cac ap sut, khi o ta co th vit : mf

r

,toan b

=

Pgradfm

r

Vy ta nhn c phng trnh le:

( )

=+ Pgrad

fvgrad.vt

vm

rrrr

hay:

=+

+

Pgrad

fv22

vgrad

t

vm

2 rrrr

i vi toa cac:

=

+

+

+

=

+

+

+

=

+

+

+

z

P1f

z

vv

y

vv

x

vv

t

v

y

P1f

z

vv

y

vv

x

vv

t

v

x

P1f

z

vv

y

vv

x

vv

t

v

m,Zz

zz

yz

xz

m,y

y

z

y

y

y

x

y

m,xx

zx

yx

xx

2/ Tch phn phng trnh le: (chiu dai cua 1 ngdong)

Ta s dung Phng trnh le theo dang sau:

Pgrad1

fv22

vgrad

t

vm

2

=+

+

rrrr

4


37/42

nhn v hng ldr

: phn t cua ng dong v//ldrr

( ) ld.Pgradfld.v2ld2

vgradld.

t

vm

2 rrrrrrrr

=+

+

ta ( ) ( ) ld//vvld.v2 rrrrrrr = nn ( ) 0ld.v2 =

rrr

Thng thng lc khi: Pmm egradf =r

( epm goi la th nng)

i vi trng trong lc: gfmrr

= (gzgradfm =

r

) va gzepm =

Ta co: 0ld.Pgrad1ld.e2vgradld.

tv pm

2

=+

++

rrrr

( )( )

( ) ( )( ) =+

++

B

A

B

A

B

A

2

pm 0ld.t,M

t,MPgrad

2

t,Mvt,Meld.

t

t,Mv rrr

y la tch phn phng trnh le theo chiu dai ng dong.V du ap dung 3.

3.Cac phng trnh Bernoulli (Benuli)1.Tn tai th nng lin kt vi lc th tch.T phng trnh le nhn v hng ld

r

(phn t chiu dai naoo)

( ) ld.Pgradfld.v2ld.2

vgradld.

t

vm

2 rrrrrrrr

=+

+

Nu ( )pmm egradf =r

( ) 0ld.Pgradv2e2

vgrad

t

vpm

2

=

++

++

rrr

r

5


38/42

2.Cht lu ng cht khng chu nen hay dong chaybartrp.

a. Cht lu khng chu nen.

Ta co nnconst=

=PgradPgrad

b.Dong chay Bartrp cua chtlu: nu ( )P=

Th tn taiham s ( ) ( )=P

Pu

duP

0

( ) Pgrad.dP

dPgrad

=

va )(1)(

PdP

Pd

=

nn)(Pdgra

Pdgra

rr

=

Vy: ( )[ ]PgradPgrad

fm

==r

Vy i vi cht lu khng chu nen.( )

0ld.v2P

e2

vgrad

t

t,Mvpm

2

=

+

+++

rrrr

(1)

i vi cht lu Bartrp.

( ) 0ld.v2Pe2

vgrad

t

vpm

2

=

+

+++

rrrr

(2)

c. Cac trng hp ring

*Dong chay dng: 0t

v=

r

va v//ldrr

(v qu ao ng

dong)

(1) 0P

e2

vB

A

pm

2

=

++ (1)

(2) ( ) 0Pe2

vB

A

pm

2

=

++ (2)

6


39/42

*Dong chay khng xoay: ( ) 0vrot2

1==

rr

v ( )( ) ( )

t

t,Mgrad

t

t,Mvgradv;0vrot

=

==

rrr

T phng trnh le (2) i vi dong chay bartrp( )

( ) constPe2

v

t

t,Mpm

2

=

+++

(3/) thi gian t cho toan

b cht lui vi dong chay khng xoay va dng cua cht lu khng

chu nen.

(3/) ( ) constPe2v pm

2

=

++ (4) bt ky thi im t trong

toan b cht lu

v du 3 trang 105 va v du 4 trang 109.Cho cht lu khng chu nen cha trong 2 nhanh cua ng hnh

ch U vi tit din . Chiu dai tng cng cua cht lu trong ng la

L. trang thai cn bng mc nc 2 nhanh ng bng nhau. Hayxac nh chu ky dao ng cua cht lu trong ng.

Giai.Tch phn phng trnh le:

( )( )

( ) ( )( ) =+

++

B

A

B

A

B

A

2

0ld.tt,M

t,MPgrad

2

t,Mvt,Meld.

t

t,Mv rrr

(x)

v ( ) ( )Ttzt,Mv r&r =

1) ld.Tzld.t

v rr&&

rr

=

(v TdSldrr

= ) dSz&&=

7


40/42

( ) ==

B

A

B

A

LzdSzld.t

t,Mv&&&&

rr

2) ( ) ( ) gz2gzt,MeB

A

B

A==

( )0

2

t,MvB

A

2

=

3) ( ) 0PP1

ld.Pgrad1

ld.Lgrad

AB

B

A

B

A

=

=

=

rr

(x)L

g2k0gz2LZ 2 ==+ &&

Vy chu ky: g2L2

K2T ==

Bai tp:1.Cho dong chay theo Lag:

( ) ( )

( )

=

+=

0

0

YtY

bt1XtX(b=const

Tm gia tc cua hat trc tip va theo le.

Giai.

( )( ) ( ) ( ) ( ) ( ) ( ) yx etYetXtR;dt

tRdtVt,tRV

rrrr

rrr+===

Vy ( )( ) ( )

x0yx ebXedt

tdYe

dt

tdXtV

rrrr=+= ( v

( )0

dt

tdY= )

Hay ( )( )( )

xx0 ebbt1

tXebXt,tRV

rrrr

+==

0dt

dVa

0dtdVa

y

y

xx

==

==

hay 0dt

Vda ==

rr

Theo le: ( )( )( )

xebbt1

txt,trv

rrr

+=

8


41/42

Theo cng thc ( ) vgrad.vt

vta

rrr

r+

=

v v theor xer nn:

( )( ) x

bt1xb

.bt1

xb

t

t1xk

x

vv

t

vta

+

++

+

=

+

=

( ) ( ) ( )0

bt1

b.

bt1

xb

bt1

xb2

2

=++

++

=

2.Cho trng vn tc: ( ) z0x0 evgteuvrrr

++=

0x uv =

0Z vgtv +=

Phng trnh qu ao: dtvgtdZ

u

dX

00

=+

=

( )

( )

++=

+=

00

2

00

Ztvgt2

1tZ

XtutX

Kh t( )

0

0

u

XtXt

= :

( )( )

0

0

00

2

0

0 Zu

XXv

u

XtXg

2

1tZ +

+

=

Phng trnh parabol.

Phng trnh ng dong :00 vgt

dZ

u

dx

+=

0vgt

dzudx;

uvgt

dxdz

000

0 =++=

constvgt

z

u

x

00

=+

9


42/42

constxu

vgtZ

0

0 ++

= la ng thng

vi goc nghing:0

0

u

vgt

dx

dztg

+==

Con ng phat xa khng bin ng, no gi nguyn tai moithi gian t. No c cu tao ng parabol xut phat t im (0,0).3.Ta co dong chay qua mt y=0 va mt do dao ng cua mt y=0:

vi trng vn tc:tsinaX =( ) ( ) ( )

xx

ky et,yveekytcosat,y,xvrrr

== Tm gia tc cua hat cht lu:

i vi y = 0. ( ) xx edtdXetcosat,y,xv rrr ==

Nh vy vn tc cua hat khi y = 0 bng vn tc cua mt daong.

( ) ( ) xky2 eekytsinavgrad.vt

v

Dt

vDa

rrrrr

r =+

==

( )( ) ( )

0

y

et,yv.0

x

et,yvt,yv xx =

+

+

rr

Vy ( ) xky2 ekytsineaa

rr=

yx v

dy

v

dx=

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