Thuy Luc Va Cap Thoat Nuoc Dang Khoa 2323

7/31/2019 Thuy Luc Va Cap Thoat Nuoc Dang Khoa 2323

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3/102

1./ nh ngha mon hoc:-C lu chat la mot mon khoa hoc nghien cu cac quy luat

chuyen ong va ng yencua lu chat va cac qua trnh tng tac lccua no len cac vat the khac.

2./ oi tng nghien cu:Lu chat gom: cht lng, cht kh va plasma

*Tnh chat :

- Lc lien ket phan t yeu -> co hnh dang cua vat cha no.- Tnh chay c -> khong chu lc cat va lc keo- Tnh lien tuc


4/102

3./ Phng phap nghien cu:

- Cac nh luat C hoc cua Newton va cac nh luat ve baotoan va chuyen hoa trong c hoccac phng trnh mo ta trang thai giai u, p

- Phng phap giai:+ phng phap giai tch

+ phng phap thc nghiem


5/102

1./ Khoi lng rieng:

-Khoi lng rieng la khoi lng cua mot n v the tch lu chat.

Th nguyen: [] = ML-3n v: kg/m3

- Trong lng rieng : la lc tac dung cua trong trng len khoi lngcua mot n v the tch chat o. = gTh nguyen: [] = ML-3n v: kgf/m3hay N/m3

-Ty trong: ty so gia trong lng rieng cua mot chat vi trong lngrieng cua nc n ieu kien chuan = /n

V du: Ty trong cua thuy ngan 200C la 13,6

V

m

V

lim0


6/102

2./ Kh ly tng:Phng trnh trang thai cua kh ly tng:

p = RT+p la ap suat tuyet oi (N/m2= pascal= J/m3)+ la khoi lng rieng (kg/m3)+T la nhiet o tuyet oi (o Kelvin 0K)+ R la hang so, phu thuoc chat kh+ M la phan t khoi cua chat kh

V du: Mot bnh co the tch 0,2m3, cha 0,5kg Nitrogen. Nhiet otrong bnh la 200C. Xac nh ap suat kh trong bnh?Giai:Gia thiet kh Nitrogen la kh ly tng. Hang so kh ly tngcua Nitrogen la R= 0,2968kJ/kg.K.

Ap suat tuyet oi trong bnh la:

kPaKxKkg

kJx

m

kgRTp 218)20273(

.2968,0

2,0

5,03


7/102

3./ Tnh nen c:Suat an hoi ac trng cho tnh nen c cua lu chat.

- oi vi chat long:

Nc 200C co En= 2,2x109N/m2Lu chat c xem la khong nen c khi khoi lng rieng

thay oi khong ang ke ( = const). Chat long thng c xemla khong nen c trong hau het cac bai toan ky thuat.V du: Mot xilanh cha 0,1 lt nc 200C. Neu ep piston e thetch giam 1% th ap suat trong xilanh tang len bao nhieu?Giai: 200C, suat an hoi cua nc En= 2,2x109N/m2

The tch giam 1%: dV/V = -1/100Vay ap suat tang: dP = -EndV/V = 2,2x109x10-2 = 2,2x107 N/m2= 2,2x107Pa


8/102

- oi vi chat kh:+ Neu kh ly tng va qua trnh nen ang nhiet (T = const)T phng trnh p = RT => p/ = const

hay pV = const+ Neu qua trnh nen ang entropi (qua trnh nen khong ma sat vakhong co s trao oi nhiet): p/pk = const

k = cp/cv cp nhiet dung ang apR =

c

p c

vc

v nhiet dung ang tch

Van toc truyen am trong lu chat:

oi vi kh ly tng trong qua trnh nen ang entropi:

V du: khong kh 15,50C vi k =1,4; R = 287 m2/s2K=> van toc truyen am trong khong kh la c= 340,5m/s.

Nc 200C co E = 2,2GN/m2 va =998,2kg/m3

=> c =1484 m/s

E

d

dpc

kRTkpc


9/102

V du: Mot bnh bang thep co the tch V = 0,2m3 cha ay nc ieu kien chuan. Tm gia tang ap suat nc trong bnh sau khi nen

them vao V = 2lt nc cung ieu kien trong 2 trng hp:1/. Bnh c xem nh tuyet oi cng;2/. Bnh dan n. The tch bnh gia tang = 0,01%/at cho moi at apsuat gia tang.

Giai:1/. Bnh tuyet oi cng:Khoi nc ban au c xet la: Vb+ V = 0,202m3

The tch nc sau khi nen la the tch bnh Vb = 0,2 m3

Vay o gia tang ap suat la:'

')()(

VV

VVVE

V

VEp

b

bb

atPaxxx 2221018,2202,0

002,0102,2 79


10/102

2/.Neu bnh dan.Sau khi nen, the tch khoi nc la the tch bnh a dan

Vbs = Vb[1+p]Bien thien the tch nc sau khi nen la:V = Vbs - (Vb+ V) = Vb[1+p] (Vb+ V)

= pVb V

Vay:Suy ra

=68,9 at.

'

'

VV

VpVEp

b

b

'

'

VVEV

EVp

bb

atatPa

PaE 4

9

10.24,2/98100

10.2,2

)202,0)10.24,2()2,0(0001,0

)10.24,2()002,0(343

43

matxmx

atxm


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4./ Tnh nht:Lu chat khong co kha nang chu

lc cat, khi co lc nay tac dung, no sechay va xuat hien lc ma sat ben trong.

Chia lu chat thanh cac lp song song,ng suat ma sat gia cac lp do s chuyen ong tng oi gia chung

phu thuoc vao gradient van toc du/dy.ac trng cho ma sat gia cac phan t lu chat trong chuyen ong=> nh luat ma sat nht Newton

: ng suat tiep, n v N/m2=Pa

: o nht ong lc hoc, th nguyen [] = FTL-2, n v N.s/m2du/dy: suat bien dang hay bien thien van toc theo phng thang goc

vi chuyen ong

dy

du

du

dyu+du

u

x

y u


12/102

*Lu chat Newton la cac lu chat co

ng suat tiep t le thuan vi suat bien dang.

* Lu chat phi Newton la cac lu chat co

ng suat tiep khong t le vi suat bien dang

o nht ong lc hoc : la mot ac tnh

cua lu chat lien quan en ng suat tiep va

chuyen ong cua lu chat o.Vi lu chatNewton, la hang so phu thuoc loai lu chat

* o nht cua chat long giam khi nhiet o

tang do cac phan t chat long nam sat nhau,hut lan nhau vi mot lc hut manh va lc can gia cac lp chat long

chuyen ong tng oi phu thuoc vao lc hut phan t nay. Khi nhiet

o tang, lc hut giam => giam lc can.


13/102

* o nht cua chat kh tang khi nhiet o tang do khoang cach giacac phan t kh ln, lc hut phan t khong ang ke. Khi cac lp phan t khchuyen ong tng oi, lc can sinh ra do s trao oi ong lng cua cac

phan t kh cac lp sat nhau.Cac phan t kh co the b mang i do chuyen ong hon loan t vung

co van toc mach ong nho sang vung co van toc mach ong ln va ngclai. Khi nhiet o tang th chuyen ong hon loan cang tang do o o nhttang.

Cong thc Sutherland (dung cho chat kh):(C,S la hang so)

Cong thc Andrade (dung cho chat long):

= DeB/T

(D, B la cac hang so)T : nhiet o tuyet oi

ST

CT

2/3


14/102

V du: Chat long Newton (he so nht 1,9152 Pa.s) chay gia haitam phang song song, vi van toc phan bo theo quy luat:

V la van toc trung bnh. Vi V = 0,6m/s va h = 0,51m. Tnh ngsuat tiep tac dung len tam di va tai iem gia.

Giai:ng suat tiep c tnh t cong thc:Ta co:

Tai iem gia: y=0, du/dy = 0=> =0

h

V

dy

du

hy

3

tam d i

2

123

hyVu

dydu

2

3

h

Vy

dy

du

2/759,6)51,0(

)/6,0(3).9152,1( mN

m

smPa tamdi


15/102

5./ Ap suat hi:Ap suat hi cua chat long la ap suat cuc bo cua phan hi tren

be mat tiep xuc vi chat long bao hoa mot nhiet o nhat nh.Tai mot so vung nao o trong dong chay neu ap suat tuyetoi nho hn gia tr ap suat hi, chat long se tao bot. Cac bot khnay khi v se gay ton hai en be mat cua thanh ran goi la hientng xam thc kh.

6./ Sc cang be mat va hien tng mao dan:Sc cang be mat la lc hut phan ttren mot n v chieu

dai cua be mat chat long.Th nguyen [ ] = F/L, n v: N/m (SI)

Chenh lech ap suat ben trong va ben ngoai cua vong trondien tch R2 la:p =pi pe =2/R

Lc tac dung len na hat chat long


16/102

Hien tng mao dan xuat hien trong cac ong nho, tai mat giao tiepran long kh, gay ra bi sc cang be mat:

V du: Xac nh ng knh nho nhat cua ong thuy tinh sach ( 00) saocho o dang cua nc 200C trong ong do hien tng mao dan khongqua 1mm.

Giai:T . Suy ra: R =2cos/R.Nc 200C co = 0,0728 N/m va =9789 N/m3.

V 00nen e co h = 1mm thng knh ong nho nhat la : D = 2R = 0,0298m

Rh

cos2

R

h

cos2

mmmN

mNR 0149,0

)10)(/9789(

)/0728,0(233


17/102

Lc tac dung ch co lc phan bo va c chia thanh 2 loai:

+Noi lc+Ngoai lc

Ngoai lc gom lc khoi va lc mat1./ Lc khoi:

La ngoai lc tac dung len moi phan t cua the tch lu chat va tyle vi khoi lng lu chat- Vector cng o lc khoi

V du:Trong lc:Lc quan tnh:Lc ly tam:

F

V

fF

V

lim0

gF aF

rF2

A

f

V, V


18/102

2./Lc mat:La ngoai lc tac dung len the tch lu chat thong qua bemat bao boc va ty le vi dien tch be mat.

- Vector ng suat

V du:

Ap suatng suat ma satTrang thai ng suat

ng suat tren mat bat ky:

S

f

S

lim0

f

A

S

zzzyzx

yzyyyx

xzxyxx

(ij = ji)

x

y z

zzyyxnnnnn


19/102


20/102

Trong chng nay, lu chat trang thai ng yen hay chuyen ong,nhng khong co chuyen ong tng oi gia cac phan t.- Khong co ng suat tiep.- Lc mat tac dung ch la ap lc

1./ Ap suat tai mot iem:Ap suat la lc phap tuyen tren mot n v dien tch

Gia tr ap suat tai 1 iem trong lu chat ng yen khong phu thuochng at cua dien tch chu lc

Px = Py =PzGia tr ap suat phu thuoc toa o iem: p = f(x,y,z)

dA

dF

A

Fp nn

A

0lim


21/102

Ap suat co gia tr dng tng ng vi ng suat phap nen (hngvao dien tch chu lc)

Th nguyen: [p] = FL-2n v:

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


22/102

2./ Phng trnh c ban tnh hoc lu chat:

Ap dung nh luat Newton II cho 1 phan t lu chat:

Hai loai lc thong thng:Lc khoi:

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

SB FdFdFd

zyxadVadmaFd

zyxFFd B

F

zyx

2x

pp

zyx

2x

p-p

yxz

2z

pp

yxz

2z

p-p

O

x

z

y z

yx


23/102

Lc mat ap lc:

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

- vector gia toc cua phan t lu chat

01 apg

zyx

zyx

zyx

x

p-

2x

pp

2x

p-pdFSx

01

0

1

01

z

pF

y

p

F

x

pF

z

y

x

0)(

1

pgradF

g

a


24/102

3./ Lu chat tnh trong trng trong lc:Phan bo ap suat trong lu chat tnh tuyet oi

+ 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

( )

a./Lu chat c xem la khong nen:= constPhan bo ap suat thuy tnh:

dp =-gdz p+gz = const hay

goi la cot ap tnh: o cao mc chat long trong ong o ap

tnh t mat chuan

gp

gdzdp

constp

z

x

B

y

zzB

zAA

hAB=zB-zA

pB

pA

ABBA

B

B

A

A

hpp

pzpz

pz

A

B

hABzA

zB

Mat chuan


25/102

*Nhan xet:+ Mat ang ap: la be mat ma ap suat tai moi iem tren o bang hang

so Mat ang ap la mat phang nam ngang. Mat phan chia gia 2 lu chat la mat ang ap. Mat ang apcung la mat ang nhiet

+ o chenh ap suat gia 2 iem A, B trong 1 moi trng lu chat phuthuoc khoang cach thang ng gia 2 iem o.

*nh luat Pascal:Trong chat long ng yen, o tang ap suat c truyen i nguyenven en moi iem trong chat long


26/102

b./Lu chat nen c:S dung phng trnh kh ly tng p =RT cho trong phngtrnh (*)

dp = -gdz

Trong tang bnh lu: z11kmT = T0-Lz

Trong tang oi lu: z>11kmT = T1= -56.5

0C

S thay oi nhiet o theo o cao

dzRT

g

p

dp

RLg

zT

L

p

p/

00

-1

)(

1

11

zzRT

g

ep

p


27/102

1./ Ap suat tuyet oi (Pt):La gia tr o ap suat so vi

chuan la chan khong tuyet oi.2./ Ap suat d (Pd):

La gia tr o ap suat so vichuan la ap suat kh tri tai vtr o.

pt >pa : ap suat d dngpt


28/102

V du 1:Xac nh gia tr ap suat oc tren ap keneu biet:h1 =76cm, h2 = 86cm,

h3 =64cm, h4 =71cm

Giai:

pBpC =HghBCpDpE =HghDEpDpC =nhDCSuy 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

h2

h1h3

h4

pa

E

D

C

B

AKh

Nc

Hg(13,6)

Nc


29/102

V du 2: Nc chay trong ong t A-B. e o o chenh cot ap tnhngi ta dung ong o ap o chenh nh hnh ve. Xac nh o chenhcot 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 iemA-M, M-N, N-B:

Hay

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

Hg

N

N

Hg

M

M

n

NN

n

BB

n

M

M

n

A

A

pz

pz

pzpz

pz

pz

hpp HgNM

n

B

B

n

A

AAB

pz

pzH

A

B

a

b

Mh

N

(1)

(2)


30/102

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

n

Hg

n

NMNM

n

N

N

n

M

M

hh

ppZZ

pz

pz

)(

mxhHn

Hg

AB 82,8)16,13(7,01

KPamNxmm

xZzHpnBAABAB

467,89/9810)3,082,8(][

3


31/102

1./ Ap lc thuy tnh tren mot dien tch phang.Cho 1 tam phang, dien tch A nam chm trong chat long va nghieng

mot goc so vi be mat chat long.

p0

x

Be mat chat long

Khoi lngrieng =

dF

h

hCF

ydA

C

D

y

yC

yD

Op0


32/102

*o ln:Xet mot vi phan dien tch dA, ap lc tac dung thang gocvao dien tch va co gia tr:

=> dF = pdAp = p0 + h

Ap lc tac dung len toan bo dien tch:

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

pC la gia tr ap suat tai trong tam C cua tam phang

ApAh

yAp

AydA

dAypdAhppdAF

CC

C

A

AA A

Ap

Asin

sinp

)sin()(

0

0

0

00


33/102

*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=>F = hCA = sinyCA

dAyydAp

dAyyp

dAhpy

ypdAydFdMM

A A

A

A

AAA

xx

2

0

2

0

0

00

sin

)sin(

)(

(*))(sinsin

22

0 AyJdAyM CxCA

x


34/102

Moment tnh theo ap lc F:M0x =yD F (**)

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:

Ay

Jyy

F

AyJy

C

xCCD

CxCD

)(sin2

Ay

Je

C

xC

Ay

Jxx

C

xyC

CD


35/102

=>JxyC = 0 => xD = xCNen ch can xac nh yD la u.+ Xet trng hp

a ve bai toan tng ng e giai. Trong o:

*Tnh ap lc thuy tnh bang phng phap bieu o:- Bieu o ap lc: la bieu o bieu dien phan bo ap suat p/tren

dien tch phang.

00 p

C

hC

C

x

Be mat chat long

y

O

P0

hC x

Be mat chat long

y

O

h0

=>

00

ph

p0


36/102

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

o ln cua ap lc tren toan bo

dien tch A:

F i qua trong tam CV cua the tch V

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

F = bF i qua trong tam C cua dien tch bieu o phan bo ap lc :

Be mat chat long

x

dF

h

y

O

p0

p/

dA

VF

Vd

dAp

pdAdFF

A

A

AA

y

p0Be mat chat long

x

F

O

AC


37/102

V du 1:Cho 1 ca van hnh ch nhat co be rong b = 5m. Chu ap

lc nc thng lu nh hnh ve vi H = 2m. Hoi ap lc thuytnh 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 1m

x 5m

x2m

= 98100 (N )V du 2:Cho 1 tam phang hnh tam giac chm trong chat long co tytrong = 1.2, co cac kch thc nh sau:

h = 3m, b = 2mH = 1m, = 600p0 = 0.06at = 600 kgf/m2

Giai

O

CH


38/102

Thay p0 bang lp chat long co be day tng ng:

F = pCA = h01/2bh = 1.2*1000kgf/m3*2.366m*1/2*3m*2m

= 8.5x103kgf

mh

y

m

hHhh

m

CC

mmm

C

73.260sin

366.2

sin

366.2

60sin3

315.0sin

3

0

0

0

hC x

p0

y

O

h0

CH

mmkgfx

mkgfph 5.0

/10002.1

/6002

2

00

mxy

h

bhy

bh

Ay

Je

CC

C

C 182.073.218

3

18

236

223


39/102

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: A

x, A

y, A

z

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

- Ap lc tren toan bo dien tch A:

- Ta co:

hp npdAFd

.

FdF

zz

yy

xx

dFF

dFF

dFF

zzz

yyy

xxx

pdAnpdAdF

pdAnpdAdF

pdAnpdAdF

.

.

.

h

z

y

pa

()

x

A

dA

Az

Ax

Fd

n

dW


40/102


41/102

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 =pa), 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, Fy c xac nh tren cac dien tch phang Ax, Ay la cachnh chieu ng tng ng cua A theo cac truc x,y. Con Fz ctnh 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

222

zyx FFFF


42/102

V du 3: Van phang hnh tron at trenmat phang nghieng 1 goc 600 nh hnhVe. Van co the quay quanh truc nam

Ngang qua tam C. Bo qua ma sat.Xac nh:a./ Ap lc tac dung len vanb./ Momen can tac dung e m van.Giai:

F = pCA

yDyC = 0,0866mMC = 0

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

Nxm

xmxmNxD

hC3

23

2

1012304

)4()10()/1081,9(

4

m

mxm

mxm

Ay

Iyy

C

xCCD 6,11

)4()60sin/10(

)2()4/(

60sin

1020

4

0


43/102

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

Giai:

Fx =pCxAxAx=RL,

Fx =9,81KNFZ =10,93KN

h=1mDau (0,8)FZF

FX

pa

2

Rhp dCx

4

2

dd

RRhWFz

048

11,1tg

x

z

F

F


44/102

3./ Lc ay Archimede:Xet vat co the tch V chm trong chat long

Xet vi phan the tch dV hnh lang tru thangng. No co 2 be mat tren va di tiep xucvi chat long va thanh phan ap lc trenphng truc z tac dung tren 2 be mat nayla: dFz1va dFz2.

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 baoboc the tch V:

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

z

p0

x

dAz

dWz1

dFz1

dFz2

dWz2

V

dFz = dFz2 - dfz1dV

()

V

zz dVdFF

dVdWdWdFdFdF zzz 1212

0 yx FF


45/102

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

agF

)(

0p1

-)ag(

z

x

agz

p

ax

p

Cx

ag

az

z

x

Cxg

az

g

agF

a

a

x

z

01

pF

g

a

a

x

z

a

a=az zx

Cp

z


46/102

2./ Chat long tnh trong thung c huyen 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:

01

pF

rgF

2

Crg

z 22

2

Cp

z

constgzrp 2

2

2

g

rgF

2

x

z

r2

g

z

p

rr

p

2

20p

1-)rg(


47/102

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 tac

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

Tren mat AB ap suat phan bo theo quy luat:Nen:

Cpz

KNxxabpp

F BA 6,23212

108101000

2

1m

A

B

x

z 0,5m 0,6m

E


48/102

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 nhchieu 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

Cgah

Ch

Cg

ah

g

rz

2)3(

0

2

C2

22

3

2

22

1

22

g2

a10-3hh

22

2

h1

z

h3

hh2

a 3a

r


49/102


50/102


51/102

1./ Phng phap Lagrange:

- Chuyen ong cua the tch lu chat c mo ta bi v tr cua cacphan t theo thi gian cua the tch:

- u iem: mo ta chuyen ong mot cach chi tiet.

- Khuyet iem: so lng phng trnh phai giai qua ln (3n); khongthe mo ta cung mot luc quy ao cua nhieu phan t.- Kha nang ap dung: phong th nghiem.

dt

dua

dt

dua

dt

dua

dt

dzu

dt

dyu

dt

dxu

t,z,y,xzz

t,z,y,xyy

t,z,y,xxx

zz

y

y

xx

z

y

x

000

000

000


52/102

2./ Phng phap Euler:

- Chuyen ong cua the tch lu chat c quan niem la trng van

toc va c mo ta bi mot ham van toc lien tuc theo khong gian vathi gian:

u iem: ch co 3 phng trnh.Khuyet iem: khong cho thay ro cau truc cua chuyen ong.Kha nang ap dung: tnh toan.

aoQuy

tocGia

tzyxuu

tzyxuu

tzyxuu

zz

yy

xx

,,,

,,,

,,,

99 101 103 105 107 109 111 113 115 117 119 121-3

-1

1

3

5

7

9

11

13

15

17

19

21

23

25

Thng 1

Vn tc trn b mt

0.1m/s

0.5m/s

Scale

0.05m/s

0.01m/s


53/102

1./ ng dong:ng dongLa ng cong vach ra trong lchat chuyen ong sao

cho vector van toc cua cac phan t lu chat chuyen ong tren otiep tuyen vi no.Co the thay oi theo thi gian.

Phng trnh

2./ Ong dong, dong chay.Ong dongla be mat dang ong tao bi vo so cac ng dong cung i

qua mot chu vi khep kn.

Dong chayla khoi lng lu chat chuyen ong ben trong ong dongV du: mat trong cua ng ong; be mat long song cung vi matthoang la cac ong dong .

zyx u

dz

u

dy

u

dx

su

u


54/102

3./ Mat cat t, chu vi t, ban knh thuy lc.

- Mat cat t(A) la mat cat ngang dong chay sao cho trc giao vi

cac ng dong va nam ben trong ong dong.- Chu vit(P) la phan chu vi cua mat cat ni dong chay tiep xuc

vi thanh ran (0).

- Ban knh thuy lc(R)

4./ Lu lng, van toc trung bnh mat cat.

- Lu lng(Q) la the tch lu chat chuyen ong ngang qua mat cat

t trong mot n v thi gian.- Van toc trung bnh mat cat(V):

PAR

AQV

A

P

AdA u


55/102

1./Theo anh hng cua o nht: Chuyen ong cua lu chat ly tng ( = 0)

Chuyen ong cua lu chat thc ( 0)2./Theo anh hng cua khoi lng rieng: Chuyen ong cua lu chat khong nen c ( = const) Chuyen ong cua lu chat nen c ( = var)

3./Theo anh hng cua thi gian: Chuyen ong cua lu chat la on nh ( ) Chuyen ong cua lu chat la khong on nh ( )

4./Theo khong gian cua chuyen ong: Chuyen ong cua lu chat la 1 chieu (u 0; v = w = 0) Chuyen ong cua lu chat la 2 chieu (u 0; v 0; w= 0) Chuyen ong cua lu chat la 3 chieu (u 0; v 0; w 0)

0 t

0 t


56/102

5./Theo trang thai chay:

Chuyen ong tang: la trang thai chay ma o cac phan t

lu chat chuyen ong trt tren nhau thanh t tang, tng lp,khong xao tron lan nhau.

Chuyen ong roi: la trang thai chay ma o cac phan t luchat chuyen ong hon loan, cac lp lu chat xao tron vao

nhau. Th nghiem Reynolds

Mc mau

Tia mc


57/102

- Xet phan t lu chat chuyen ongtren quy ao cua no (dung bien

Lagrange), gia toc cua pt :

Trong bien Euler, van toc la ham theo khong gian va thi gian ->van toc u c tnh theo u0 bang chuoi Taylor:

Thay vao bieu thc gii han:

va thc hien phep tnh gii han:

0u

s

Quy ao

u

0000 ,,, zyxt

zzz

yyy

xxx

ttt

0

0

0

0

t

uu

dt

uda

t

0

0lim

zz

uy

y

ux

x

ut

t

uuu

0

t

z

z

u

t

y

y

u

t

x

x

u

t

ua

t

0lim

z

uu

y

uu

x

uu

t

ua zyx


58/102

1./ Phng trnh lien tuc.

- nh luat bao toan khoi lng: toc o gia tang cua khoi lng

cua mot he vat chat bang khoi lng chuyen ong vao he trong 1n v thi gian.

- Ap dung cho lu chat trong the tch kiem soat:

Klng lu chat trong the tch:

Klng lchat cong ra khoi the tch:

- Theo L bao toan:

- oi vi lu chat khong nen c, =const:

VdV

0z

u

y

u

x

u zyx

S

ndSu

0

S

n

V

dSudVt

0

ut

0udiv

V

S

uun

un.dS

n


59/102

2./Phng trnh lien tuc cho dong chay on nh cua lc khong nenc.

- Xet the tch kiem soat la oan dong chay gia hai mcat 1-1 va 2-2Trong trng hp lu chat khong nen c, chuyen ong on nh

ptrnh lien tuc di dang tch phan c rut gon con:

Chia dien tch bao boc S= A1+ A2+ SnTach tch phan thanh tong cua 3 tch phan:

Hai tch phan au cho lu lng ngang qua cac mcat 1-1 va 2-2, contch phan th 3 bang khong:

0S

ndSu2

21

1

Sn

A2

A1

un=0

n

n

u

u

0

21

nS

n

A

n

A

n dSudAudAu

021 QQ 21 QQ constQ


60/102

Xet pt lu chat. iem M0 c chonlam cc cua pt.

Gia s van toc tai M0 a biet, cauhoi la van toc tai iem M?

S dung chuoi Taylor, bo qua so hangvo cung nho bac cao, tphan van toc ux:

Cong va tr so hang vao ve phai cuabieu thc tren,sau o sap xep lai se thu c bieu thc:

zz

uyyux

xuuu xxxxx

0

x

z

y

M0

M

x

z

y

0u

u

z

x

uy

x

uzy

2

1

zx

u

z

uy

x

u

y

ux

x

uuu

zxyxx

xx

2

1

2

10

zxu

zuy

xu

yu zxyx

2

1

2

1

uuuuu 11


61/102

at:

Thanh phan van toc uxse c tnh bang cong thc:

Tng t:

Gia s mat trai va mat phai cua pt ch chuyen ong theo trucx vi van toc u0x va ux tng ng cua iem M0 va M. Do co s

chenh lech van toc, sau 1 n v thi gian, pt dai ra mot oanla: ux-u

0x

Do o toc o gian dai tng oi cua pt la:

j

i

i

jk

j

i

i

jk

i

ii x

uxu

xu

xu

xu

2

1

2

1 ;;

zyzyxuuyzyzxxx

0

xzxzyuuzxzxyyy

0

yxyxzuuxyxyzzz

0

x

z

M0

M

x

z

ux-u0x

xuuxx

0

hi 0 uuu 0 y Mux-u0x u


62/102

Khi x 0, ta co:

i- toc o gian dai tng oi cua pt theotruc xi.

va

Gia s mat tren va mat di cua pt ch chuyenong theo truc x vi van toc u0x va ux tng ngcua iem M0 va M. Do co s chenh lech van toc,sau 1 n v tgian, pt se b o nghieng vi goc:

Tng t, do co s chenh lech thanh phan vantoc tren phng y gia mat trai va mat phai mapt cung se b o nghieng vi goc:

xxxx

xu

xuu

0

yu

yuu xxx

01

x

y

M0

M

x

y

uy-u0y

2

uyu0y

x

y

M0

M

x

y

x 0x

1

ux

u0x

xu

yuu yyy

02


63/102

Neu ca 2 chuyen ong ong thi xuat hien, pt se b thay oinh hnh:

Trong 1 n v thi gian pt b bien dang mot goc:

k- toc o bdang goc cua pt quanh truc xk.

Trong 1 n v thi gian pt quay i mot goc:

k - toc o quay cua pt quanh truc xk.

nh ly Hemholm: Cong cua pt lu chat bao gom cong cua

vat ran (theo cc va quay quanh cc) va cong bien dang(bdang dai va bdang goc).

Vector quay van toc:

zxy

yu

xu

2

1

2

112

x

y

M0

M

x

y

1

M

2

(2- 1)/2

zxy

y

u

x

u

2

1

2

112

urotukji zyx

2

1

2

1

V d 1 Ch 3 h h h


64/102

V du 1: Cho vector van toc gom 3 thanh phan:ux = x2 + y2 + z2uy = xy + yz + z2

ux = -3xz + z2/2 + 4Tm vector van toc quay?Giai

yyy

u

x

u

zzx

u

z

u

zyz

u

y

u

xy

z

zx

y

yz

x

22

1

2

1

322

1

2

1

202

1

2

1

kyjzizy

)2/()2/5()2/(

V d Ch


65/102

V du: Chuyen ong co vector van toc:ux = ay + by2uy = uz =0

Vi a, b la hang soa./ Chuyen ong co quay khong?b./ Xac nh a, b e khong co bien dang goc.Giai:

chuyen ong quay a, b 0

khong co cap a, b nao e bien dang goc bang 0

022

1

2

1

bya

y

u

x

u xyz

0221

21

bya

yu

xu xy

zy


66/102


67/102


68/102

Nhac lai hng trnh lien tuc:Nguyen ly bao toan khoi lng: khoi lng cua 1 he thong luon khongoi.

;

* Chuyen ong on nh:

Tong khoi lng lu chat ra vao the tch kiem soat = 0* Lu chat khong nen c :

Tong the tch lu chat ra vao the tch kiem soat = 0

Q1 = Q2+Q3 Q1 = Q2=Q3

V1 A1 = V2 A2 + V3 A3 V1 A1 = V2 A2 = V3 A3

0

systemdtdM 0

CV CS system

SdudVtdt

dM

00

CS

m SduQt

0 CSm SduQconst

V1

V2

V3

CVV2

V1 V31

1

2

2

3

3


69/102


70/102


71/102

a ti phng trnh Navier-Stokes tren truc x:

Di dang vector:

oi vi lu chat khong nen c:

Lu y gia toc c tnh:

z

u

y

u

x

u

xz

u

y

u

x

u

x

pF

dt

du zyxxxx

x

x

3

11

2

2

2

2

2

2

uupgradFdt

ud

3

11 2

upgradFdt

ud

21

uut

u

z

uu

y

uu

x

uu

t

u

dt

udzyx


72/102

1. Phng trnh van tai nang lng:

nh luat bao toan nang lng (L th nhat cua nhiet ong lc hoc): Toc o

bien thien cua ong nang va noi nang bang tong cong c hoc cua ngoai lc vacac dong nang lng khac tren 1 n v thi gian

e: noi nang (kh ly tng: ; chat long khong nen: )

: dong nhiet rieng i vao qua be mat bao boc

nh luat truyen nhiet Fourier:

Bien oi:

Thu c:

S

e

n

S

n

VV

dSqdSudVuFdVeu

dt

d ..

2

2

eq

V

iij

jS

jiij

Sn dVudSnudSu

.

.TT.gradq

TuuFeudt

d jji

ij

j

j

j

11.

2

2

Vjj

V

jjS

en dTdVqdSq

TceV

cTe


73/102

2. Phng trnh van tai ong nang:

Ptrnh Navier di dang tensor:

Nhan ptrnh tren cho ui :

3. Phng trnh van tai noi nang: Tr ptrnh van tai nang lng cho ptrnh van tai ong nang:

S dung gia thiet Stokes va cho lu chat khong nen c: ij

ijj

j uTdt

de

11

zyx

Fdt

du zxyxxxx

x

1 ijji

i Fdtdu

1

i

ij

ji

i uFdt

du

1 ij

ij

i

ij

j

i

i

uuuFu

dt

d

11

2

2

2

2

2

i

j

j

i

xu

xuT

dtde


74/102

Phng trnh Euler dang Lambo-Gromeko:

Gia thiet: = const

Phng trnh Euler dang Lambo-Gromeko thanh:

pgradFuugradtu 122

2

UgradF

022

2

uup

Ugradt

u


75/102

1. Trng hp chuyen ong co the. Chuyen ong co the: va

Phng trnh Euler tren thanh:

Trong trng trong lc: U = - gz

(Tphan Lagrange)

oi vi chuyen ong on nh:

tCg

upz

tg

2

1 2

Cg

upz

2

2

gradu

0

02

2

up

Ugradgradt

tCup

Ut

2

2


76/102


77/102

3 Trng hp lu chat chuyen ong on nh, tphan theo phng vuonggoc vi ng dong.

Phng trnh Euler trong he toa o t nhien:

Lay vi phan chieu dai ng phap tuyen vi ng dong: Nhan vo hng no vi pt. Euler:

Khi R :

Trong trng trong lc: U = - gz

(Tphan Euler)

nd

nCp

z

pUgradn

R

u

s

u

t

u 22 2

nd

pUgradndn

R

u

s

u

t

u

22 2

n

pUddn

R

u

2

nCp

U


78/102

Yngha nang lng cua cac so hang tch phan. Xet pt Bernoulli. Qua trnh thiet lap qua cac bc:

Cac so hang:

Phng trnh Bernoulli la pt bao toan nang lng

0.22

2

sduupUgrad

t

u

02

2

upUd

CupU 2

2

Cg

upz

2

2

ngQuanglchatklng

1vtrenLc

lchatklng1vtrasinhCong

lchattlng1vcualngangN

congtrongoithaykhongnova lchatklng1vcualngangN

gu 22

pz tnh)ap(cotlchattlng1vcuanangThe

toc)vanap(cotlchattlng1vcuanangong

g

upz

2

2

lchattlng1vcuaphantoanlngangNphan)toanap(cot


79/102

Xet oan dong chay on nh nam gia 2mcat t 1-1 va 2-2.

Xet 1 ng dong trong oan dong chay.Neu cho rang lu chat la ly tng, ptrnhBernoulli cho ng dong:

Phng trnh tren the hien tnh bao toan.Neu lu chat la thc th:

Bay gi xet 1 dong chay nguyen to. Nang lng cua no bien oi theo ptrnh:

Nh vay cho toan bo dong chay, nang lng cua no se bien oi theo ptrnh:

1

1

2

2

dQ

dQ

dQ

Q

g

upz

g

upz

22

2

222

2

111

fhg

upz

g

upz

22

2

222

2

111

dQhdQg

upzdQg

upz f

22

2

222

2

111

lchat)tlngv1cuanlngthatton:fh(

Q

f

AAAA

dQhdQg

udQ

pzdQ

g

udQ

pz

221122

2

222

2

111


80/102

Thc hien cac tch phan:

Thay vao cho ket qua:

1. ieu kien ap dung pt Bernoulli cho dong chay: ; =const; Tai hai mcat ap dung pt, dong chay phai la bien oi cham. Trong oan dong chay gia 2 mcat, khong co nhap lu hoac tach lu.

2. Neu trong oan dong chay gia 2 mcat viet pt co turbine, may bm:

Qp

zdQp

z

A

chamboilachaydongAtmcattai:kienieu

Qg

VdQ

g

u

A

22

22

nang,hchnhhso: 10,105,11

3

A

dAV

u

A

QhdQh fQ

f

lchattlngv1cualngnangthatton:fhap)cotthat(ton

fhg

Vpz

g

Vpz

22

2

222

2

111

0 t gF

BTff HHhh


81/102

1. Phng trnh bien thien ong lng.

Nguyen ly bien thien ong lng: toc o bien thien cua ong

lng cua mot he vat chat bang vector tong ngoai lc tac dunglen he.Ap dung cho lu chat trong the tch kiem soat:

Bien oi:

oi vi dong chay on nh, ptrnh bien thienong lng la:

RdVudt

d

V

V

S

uun

un.dS

n

RdSuudVut

S

n

V

RdSuuS

n


82/102

2. Ptrnh bien thien lng cho dchay on nh cua lchat khong nen c.Xet the tch kiem soat la oan dong chay gia hai mcat 1-1 va 2-2

Chia dien tch bao boc S= A1+ A2+ SnPtrnh bien thien ong lng thanh:

Tch phan th 3 bang khong con hai tch

phan au c viet lai thanh:

Cac tch phan nay c thc hien:

Thay vao cho ket qua:

RdSuudSuudSuu

nS

n

A

n

A

n

21

2

21

1

Sn

A2

A1

un=0

n

n

u

u

RdQudQuAA

21

QVdQu

A

1122 VVQR

lng,hchnhhso: 05,102,11

2

A dAV

u

A

111222 VQVQR


83/102

1.1 Hai trang thai chay. Chay tang: ReD 2300

Chay roi: ReD > 2300

1.2 Mo hnh Boussinesq Phan tch Reynolds:

( - van toc trung bnh thi gian;uvan toc mach ong) Mo hnh Boussinesq:

Van toc tnh toan la van toc trung bnh thi gian. Lu chat trong chuyen ong roi co o nht la o nht hieu dung:

(to nht roi) Mo hnh Prandtl (1925)

uuu

u

t

u

(Chay roi)

u

t

u

(Chay tang)

u

u

teff

dy

dult2 tronxaodaichieu-yl

1. CAC KHAI NIEM (2/2)



84/102

1.3 Lp mong chay tang.

> -> che o chay thanh trn thuy lc

-> che o chay thanh nham thuy lc

(Loi roi)

(Lp mong chay tang)

2. PHNG TRNH C BAN CUA DONG CHAY EU (1/2)


85/102

2.1 Phng trnh c ban. Ngoai lc tac dung tren phng

chuyen ong: Gs= lAsin- trong lc

P1- P2= (p1- p2)A ap lc

Fms=0lP lc msat tren vo ong

Ptrnh bthien lng tren phng s:

Ptrnh Bernoulli cho oan dong chay t mc 1-1 -> mc 2-2:

T (1) va (2) =>

112221 VVQFPPG mss

(1)lR

pz

pz

022

11

P1

P2

G

Gs

s

1

1

2

2

l

V1

V2

0

lsin

z1

z2

0 0

(2)fhp

zp

z

2

21

1fh

g

V

pz

g

V

pz

22

222

2

211

1

RJ 0 lcthuydoco lhJ f

2. PHNG TRNH C BAN CUA DONG CHAY EU (2/2)


86/102

2.2 Li giai.Xet mat tru ban knh r, ptrnh c ban cua dong eu:

a) Chay tang.

b) Chay roi.

Xet mat tru ban knh r sat thanh ong, rR0:

Tch phan cho ket qua:

Jr

dr

du

2 drdu 220

4rR

Ju

const

t

0

2

2

0

dy

duy

y

u

dy

du 1*

Eyu

u ln*

0* u

r

R0y

2rR Jr2

tocvanbophan Lp mong chay tang

ng cong Parabol

ng cong Logarit

y

3. TON THAT COT AP DOC NG (1/4)


87/102

3.1 Cong thc Darcy. T phng trnh c ban cua dong eu rut ra:

ng suat ma sat c xac nh bang thc nghiem:

Thay 0 t (2) vao (1), rut ra:

- he so ton that cot ap doc ng hoac he so ma sat ng ong c xacnh bang thc nghiem vi:

lR

hl

0

,,,,0 VDf DfV

Re,2

0

g

V

R

lhl

24

2

g

V

D

lhl

2

2

hoac cho ong tron

(1)

(2)

Df Re,



88/102

Th nghiem Nikurade (1933):

Cac cong thc thc nghiem

- Chay tang (ReD < 2300):

- Chay roi (ReD > 4000):

DRe

64

DRe

51.271,3

log21

25.0

Re

10046.11.0

D

(Colebrook-1939)

(Altsun-?)



89/102

- o th Moody (1944):

0,000 011 2 3 4 5 7

x10 31 2 3 4 5 7

x10 41 2 3 4 5 7

x1051 2 3 4 5 7

x10 61 2 3 4 5 7

x10 71

x108

0,000 0050,000 007

0,000 05

0,000 1

0,000 2

0,000 40,000 6

0,001

0,002

0,004

0,0060,008

0,01

0,015

0,02

0.03

0,04

0,05

0,008

0,009

0,01

0,015

0,02

0,025

0,03

0,04

0,05

0,06

0,07

0,08

0,090,1

Khu chay r oi thanh nham hoan toan (K hu sc can bnh thng)Kh u

Ch ay tangKhu ch ay roithanh nham

Khu ch ay roi

thanh t rn

Khu chuyen t iep

Re =VD/

D

OTHMOODY



90/102

3.2 Cong thc Chezy. Cong thc Chezy:

So sanh vi cong thc Darcy:

So Chezy thng c tnh theo cong thc Manning:

Cac cong thc suy dien t Chezy:

gC

8

JKRJACQ

RJCV (C - So Chezy)

6/11 Rn

C (n - he so nham Manning)

RACK

lRC

Vl

K

Qhl 2

2

2

2

(Kmodule lu lng)

4. TON THAT COT AP CUC BO (1/1)


91/102

4.1 Khai niem.

Trong oan lm:

4.2 Cong thc Darcy - Weisbach

g

Vhcb

2

2

(- he so ton that cot ap cuc bo)

lm (2050)D

PP

hcbE

E

t

dy

du

dy

dueff fh

5. TNH TOAN THUY LC NG ONG (1/4)


92/102

5.1 Gii thieu. Cac phng trnh, cong thc c ban:

Ptrnh Bernoulli cho dong chay Ptrnh lien tuc

Cac cong thc tnh ton that cot ap (ton that cot ap doc ng va cuc bo)

Cac gia thiet:

lm


93/102

5.2 Cac bai toan.a. ng ong ngan ve mat thuy lc.

Ch xet ng ong n gian Xem bai toan tong quat. Ptrnh Bernoulli

t mcat 1-1 ti mcat 2-2:

a ti:

vi

T ptrnh tren neu cho Q se tnh c H, hoac ngc lai neu cho H se tnhc Q

g

VkH d

2

2

2

fhg

V

pz

g

V

pz

22

2

222

2

111

12

2

2

2

4

1

2

11

1

1

d

l

d

d

d

lk

d1, l1, 1

d2, l2, 2H1

2 (Vd2)1 1

2 200

V2

Q



94/102

b. ng ong dai ve mat thuy lc.

b1. ng ong n gian

Xem bai toan tong quat. PtrnhBernoulli t mcat 1-1 ti mcat 2-2:

a ti:

T ptrnh tren neu cho trc 2 trong so 3 thong so Q, H va HB, se tnh cthong so con lai.

2

2

2

2

1

12

K

l

K

lQHHB

Bf

222

2

211

1 Hhg2

V

pz

g2

V

pz

d1, l1, n1

d2, l2, n21 1

V2

QH

B

2

2



95/102

b2. ng ong gan noi tiep

Trong tnh toan c thay the bang 1 ong tng ng vi:

b3. ng ong gan song song

Trong tnh toan c thay the bang 1 ong tng ng vi:

i i

i

T

T

Kl

Kl

22

A BA B

1 2 3 T

QQ

A B

1

2

3Q

A B

TQ

i i

i

T

T

lK

lK

Ch 6 CHUYEN ONG CO THE 2 CHIEU


96/102

Chng 6. CHUYEN ONG CO THE 2 CHIEU

1. Cac khai niem2. Cac chuyen ong the phang n gian3. Chong chap cac chuyen ong the



97/102

1.1 Chuyen ong co the. n: Cong cua lu chat c goi la co the khi ton tai mot ham sao cho:

- ham the van toc; ng cong (x,y) = const ng ang the Tnh chat:

Phng trnh:

1.2 Ham dong. n: Ham (x,y) sao cho c goi la ham dong.

ng cong (x,y) = const ng dong

Tnh chat:

Phng trnh:

gradu

02

1 urot

0

xuyu yx ;

2112 q

0



98/102

1.3 Ham the phc. Ham dong va han the co tnh trc giao do:

=> mo ta bang ham the phc:

Cac ai lng:

1.4 Tnh chong chat.

0

yyxx

izf

phctocvan yxiuyxuzV yx ,,

hplientocvan yxiuyxudz

zdfzV yx ,,

yxuyxuyxu

yxyxyx

yxyxyxzfzfzf

,,,

,,,

,,,

21

21

21

21

2. CAC CHUYEN ONG THE PHANG N GIAN (1/2)


99/102

2.1 Chuyen ong thang eu.

U0 van toc dong chay

2.2 iem nguon va gieng

q - lu lng n v

zUzf 0

xU0

yU0

2

ln2

ln2

q

rq

zq

zf

2. CAC CHUYEN ONG THE PHANG N GIAN (2/2)


100/102

2.3 Xoay t do.

- lu so van toc

2.4 Lng cc.

m- moment cua lng cc

r

z

i

zf

ln2

;2

ln

2

2222;

yx

ym

yx

xm

z

mzf

3. CHONG CHAP CAC CHUYEN ONG THE (1/2)


101/102

3.1 Chuyen ong bao ban vat.

(dong thang eu + nguon)

3.2 Chuyen ong bao vat Rankine.

(dong thang eu + nguon + gieng)

2;ln

2

ln

2

00

0

qyUr

qxU

zq

zUzf

azazqzUzf

ln2

0

3. CHONG CHAP CAC CHUYEN ONG THE (2/2)


102/102

3.3 Chuyen ong bao tru tron.

(dong thang eu + lng cc)

3.4 Chuyen ong bao tru tron co lu so van toc

(dong bao tru tron + xoay t do)

2

2

02

2

0

2

0

1sin;1cosr

RrU

r

RrU

z

RzUzf

Alemberd'lynghch 0xP

ziz

RzUzf ln2

2

0

nanglc UP

Py

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