BaiGiangCoLuu Chat

8/8/2019 BaiGiangCoLuu Chat

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Tom tat bai giang C lu chat- TS Huynh cong Hoai H Bach Khoa tp

C LU CHATGiang vien: Huynh cong Hoai H Bach Khoa Tp HCM

M.E(AIT,Thailand), Ph.D (INPT,Phap)(http://www.hcmut.edu.vn/~hchoai/baigiang)

1. Gao trnh c lu chat - Bo mon C lu Chat2. Bi tp clu cht BMClu Cht -L song Giang - Nguyn th Phng3.Thuy lc ai cng Nguyen Tai, Ta ngoc Cau4.Thuy lc ( Tap I) Nguyen van Tao , Nguyen canh Cam5.Fundamentals of Fluid mechanics Phillip M. Berhart, Richard J. Gross, John I.Hochstein. Second edition, Addison wesley Publising Company Inc. 19856.Applied Fluid Mechanics- Robert L. Mott , Fourth edition , Macmillian PublishingCompany, 19907. Fluid mechanics John Doughlas, Janusz M. Gasiorek , John A. Swaffiield.Fourth edition, Prentice Hall, 20018. E-book : Fundamentals of fluid mechanics by Bruce R. Munson, Donald F.Young, Theodore H.Okiishi , John Wiley & Sons Inc. 2006

9. E-book : Shaums interactive Fluid mechanics Giles R.V et al.10. Website: https://ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=fl

Chean Chin Ngo, Kurt Gramoll11. Website : http://www.engin.umich.edu

Thi gian giang day: 42 tiet

Hnh thc thi: Trac nghiem

Gia hoc ky (20%) : 8 cau ly thuyet va 4 cau toan (45 phut)

Cuoi hoc ky (80%) : 12 cau ly thuyet va 8 cau toan (90 phut)

ted with FinePrint - purchase at www.fineprint.com


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Chng 1: M AU

I.GII THIEU MON HOC C LU CHAT- oi tng nghien cu : Lu chat : chat long va chat kh

- Pham vi nghien cu : - nghien cu cac qui luat cua chat long va chat kh khi nong yen va chuyen ong .

Tai sao phai nghien cu c lu chat ?Kien thc c ban cua mon CLC ng dung trong nhieu lnh vc :

+ Nghien cu thiet ke cac phng tien van chuyen : xe hi, tau thuy, may bay, hoa tien..

+ ng dung trong lnh vc xay dng nh cap, thoat nc, cong trnh thuy li (cong, e,ho cha, nha may thuy ien ..), tnh toan thiet ke cau, nha cao tang

+ Tnh toan thiet ke cac thiet b thuy lc : may bm, tua bin, quat gio, may nen..

+ ng dung trong kh tng thuy van : d bao bao, lu lut , ..

+ ng dung trong y khoa: mo phong tuan hoan mau trong c the, tnh toan thiet kecac may tr tim nhan tao..

II. CAC TNH CHAT VAT LY C BAN CUA LU CHAT:

2.1 KHOI LNG TRONG LNG

Khoi lng (KL) la mot ai lng khong thay oi theo khong gian

Trong lng (TL) = KL x g (gia toc trong trng ) => thay oi theo g

n v :

Khoi lng Trong lng

Kg N (kgm/s2)

Kgf (9,81 N) (kilogam lc)

Tf (1000 Kgf) (Tan lc)

- Khoi lng rieng () Trong lng rieng ()

n v : kg/m3 N/m3

V du : nc : 1000 kg/m3 nc : 9810 N/m3

khong kh : 1,228 kg/m3 khong kh : 12,07 N/m3

- T trong : = /nc = /nc

V du : nc = 1, thuy ngan = 13,6ted with FinePrint - purchase at www.fineprint.com


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2.2 TNH NHT CUA LU CHAT (Viscosity)

Quan sat mot dong chay :

dy

du =

Newton

Trong o :

: ng suat ma sat (N/m2)

: he so nht ong lc

u : van toc, phu thuoc vao y

n v cua :

n v chuan :ms

kg2

m

NsPa.S

Ngoai ra : poise , 1 poise = 0,1ms

kg

Ngoai he so ong lc, ngi ta con s dung he so nht ong hoc , c nh ngha

=

n v : m2/s hay stoke , 1 stoke = 1cm 2/s = 10-4 m2/s

y

u(y)

u

y

dy

Tnh chat cua he so nht:

He so nht phu thuoc vao nhiet o :

Chat long: khi nhiet o tang he so nht giam

Chat kh: khi nhiet o tang he so nhttang

He so nht phu thuoc vao ap suat:Chat long: ap suat tang he so nht tang

Chat kh : he so nht khong thay oikhi ap suat thay oi

Chat long Newton va phi Newton

Hau het cac loai lu chat thong thng nh nc, xang, dau eu thoa man cong thcNewton (1) , tuy nhien co mot so chat long (hac n, nha nong chay, dau tho ..) khongtuan theo cong thc Newton c goi la chat long phi Newton, hoac oi vi chat longthong thng khi chay trang thai chay roi cung khong tuan theo cong thc Newton.

Lu chat ly tng va lu chat thc

Lu chat ly tng: khong co ma sat

Lu chat ly tng: co ma sat



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2.3 TNH NEN CUA LU CHAT :

Moduyn an hoi:dV

dpVK = hay

d

dpK=

( v M = V => dM = dV +Vd = 0)Knc = 2,2 109 N/m2

oi vi chat kh ly tng : p = RT

Vi : p : ap suat tuyet oi (N/m2 )

: khoi lng rieng

R : hang so kh, phu thuoc vao loai kh

T : nhiet o tuyet oi ( nhiet o Kelvin , 0o C = 273 o Kelvin)

- Hau het cac loai chat long rat kho nen nen c xem nh la lu chat khong nen- Mot dong kh chuyen ong vi van toc nho th s thay oi khoi lng rieng khongang ke nen van c xem la lu chat khong nen.

- Khi dong kh chuyen ong vi van toc ln hn 0,3 lan van toc am thanh(khoang 100 m/s) thi mi xem la lu chat nen c

T phng trnh tren pV = const p : ap suat tuyet oi va V : thetch

2.4 AP SUAT HI BAO HOA:

Trong mot khong gian kn, khi cac phan t chat long boc hi at en trang

thai bao hoa tao ra mot ap suat trong khoang khong gian kn o c goi laap suat hi bao hoa.

Ap suat hi bao hoa tang theo nhiet oV du 32,20C, pbao hoa cua nc la 0,048at

1000C, pbao hoa cua nc la 1atKhi ap suat chat long Ap suat hi bao hoa chat long bat au soi (hoa kh).

V du co the cho nc soi 32,20C neu ha ap suat xuong con 0,048at.

Trong mot so ieu kien cu the, hien tng kh thc (cavitation) xay ra khi ap suat

chat long nho hn Pbao hoa



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2.5 SC CANG MAT NGOAI VA HIEN TNG MAO DAN

(Xem tai lieu tham khao)

Cau 2 Mot dong chay neu co bieu o phan bo van toc eu nh hnh ve th ng suatma sat gia cac phan t tren AB se la:

a) Nho nhat A b) Ln nhat A

c) Nh nhat B d) Ca 3 ieu sai

A

B

a) Ln nhat A b) Ln nhat B

c) eu bang nhau tat ca moi iem tren AB

d) eu bang khong tat ca moi iem tren AB

A

B

Cau 1. Mot dong chay co bieu o phan bo van toc tuyen tnh nh hnh ve th ng suat ma satgia cac phan t tren AB se la:

CAU HOI TRAC NGHIEM



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Cau 3 . Mot lu chat co moduyn an hoi nho th:

a ) Kho nen b) De nen c) Kha nang an hoi kem d) Ca b) va c) eu ung

Cau 5:. S ma sat gia cac phan t chat long khi chuyen ong phu thuoc vao:

a) S phan bo van toc trong dong chay b) Tnh chat cua chat long

c) Ap suat cua dong chay d) Ca a) va b)

Cau 4 Mot khoi kh ly tng co khoi lng Mo

ap suat po

. Neu ap suat tang en p1

> po

trongieu kien nhiet o khong oi th khoi lng cua khoi kh (M1) trong ieu kien ap suat p1 se la :a) M1 = Mo b) M1 > Mo c) M1 < Mod) Cha the biet v con phu thuoc vao moduyn an hoi ln hay nho

Cau 7 Khi giam nhiet o th s ma sat gia cac phan t lu chat ang chuyen ong:

a) Luon luon giam neu la chat long

b) Luon luon giam neu la chat kh

c) Luon luon giam cho tat ca cac loai lu chat

d) Ca 3 eu sai

Cau 6 : Mot khoi chat long co the tch khong oi, khi at tren mat at va tren mat trang th :a) Trong lng khong oi b) Trong lng rieng khong oic) T trong khong oi d) Ca a) va b) eu ung



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Cau 8 He so nht ong lc hoc cua mot lu chat th :a) Mot so co th nguyen b) Phu thuoc vao trang thai chayc) Phu thuoc vao nhiet o d) Ca a) va c) eu ung

Cau 9 Khoi lng rieng cua mot chat kh th :a) Thay oi khi gia toc trong trng thay oi b) Se tang khi ap suat tang

c) Se gia m khi ap suat tang neu la chat kh ly tng d) Ca a) va b) eu ung

Cau 10 Mot dong chay co bieu o phan bo van toc nh hnh ben.ng suat ma sat () tai cac iem A,B,C se la:

a) A < B < C b) C < A < Bc) B = C < A d) C < B < A

C

B

A

V du 1: ng ong co ng knh d, dai l, dan dau vi he so nhn , khoi lngrieng . Dau chuyen ong theo quy luat sau:

u=ady-ay2 (a>0; 0


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V du 2 :Mot khoi co khoi lng 10 kg trttren mat nghieng co goc 20oso vi matphang nam ngang. Xac nh van toc cuakhoi neu gia khoi va mat nghieng co boimot lp dau co o nht ong lc = 0,38Pa.s, day 0,1 mm. Cho dien tch tiep xucgia khoi va tam nghieng la 0,2 m2

u

20o

0,1mm

Giai :

W la trong lng cua khoi W sin 20o = o A

=

=

=t

u

dy

du

y

o 0

At

uW

o

= 20sin

smA

Wu

oo

/0442,0)2,0)(38,0(

)20)(sin81,9)(10)(0001,0(20sin===

Xy lanh

ngoaiChat

long

L

Ro

V du 3: o nht cua lu chat co the o bang cach dungmot xy lanh o o nht nh hnh ve. Dung cu nay bao gommot xy lanh ben ngoai co nh va xy lanh ben trong quay

vi van toc quay . Neu biet ngau lc can thiet e quayvi van toc th o nht cua chat long c nam gia haixy lanh c xac nh. Hay thiet lap mot phng trnh lienhe gia , , , L va Ro va Rl . Bo qua anh hng ayva cho s phan bo chat long van toc gia hai xy lanh latuyen tnh.

iRU =

)()(io

i

ioRR

R

RR

U

=

=

dRR

LRLdRRR

RdAdF

io

ii

io

i

)()(

2

===

Ngau lc d tao ra

dRR

LRRd

RR

LRdFRd

io

ii

io

ii

)()(

32

=

==

===

2

0

2

0

332

0 )()(d

RR

LRd

RR

LRd

io

i

io

i

)(

2 3

io

i

RR

LR

=

U

Ri

Giai :



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S thay oi he so nht theo nhiet o



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CHNG 2 : TNH HOC LU CHATI. KHAI NIEM

-Tnh tuyet oi : can bang bi duy nhat la trong lc- Tnh tng oi: can bang bi nhieu lc (trong lc , lc quan tnh, lc ly tam .)

II AP SUAT THUY TNH2.1 Ap suat thuy tnh -nh ngha

Ap suat thuy tnh trung bnh:A

Fp

r

r

=

Ap suat thuy tnh tai mot iemA

FlimpA

r

r

0=

2.2 Tnh chat

- Ap suat thuy tnh tac dung thang goc vi dien tch chu lc va hng vao dien tch ay

- Tr so ap suat khong phu thuoc vao hng cua dien tch chu lc

2.3 Th nguyen cua ap suat

Th nguyen cua ap suat pF

AFL ML T= = = 2 1 2

n v cua ap suat: N/m2 ( Pa) : n v chuan dung e tnh toan

Kgf / cm2 , at , m nc, mm Hg

1 at = 1Kgf / cm2 = 10m nc = 735 mm Hg = 9,81.104 Pa (N/m2)

Fr

A

2.4 Ap suat tuyet oi, ap suat d va ap suat chan khong.

0

Ap suat kh tri :98100 N/m2

Ap suat

tuyet oi

Ap suat d

(tng oi)

0

-98100 N/m2

Ap suat

chan khong

0

98200 N/m2 100 N/m2

98000 N/m2 - 100 N/m2 100 N/m2

Ap suat tuyet oi la gia tr ap suat that , v du ap suat cua khong kh Pa = 98100 N/m2

Ap suat d( ap suat tng oi) la ap suat c so sanh vi ap suat kh tripd = ptuyet oi - pa

Ap suat chan khong la ap suat con thieu can phai them vao cho bang ap suat kh tripck = pa - ptuyet oi = 98100 N/m2 - ptuyet oi = -pdu



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III. PHNG TRNH VI PHAN C BAN CUA CHAT LONG CANBANG

dxx

pp

+p

x

z

ydx dy

dz

AB

cD

E F

GH

Hnh 2.2

3.1 Phng trnh vi phan c ban:Khoi cht lng vi phan , canh dx, dy, dz,can bang , khoi lng rieng .

Lc tac dung len khoi hnh hop theo phng

X la :Lc khoi : d x d y d zFx

Lc mat :

Tong lc phng X: d x d y d zFx + p dy dz - (p+ dx) dy dz = 0xp

x

p

Fx - = 0 => Fx - = 0 (2.3)xp

1

phng y => Fy - = 0 (2.4)

y

p

1

phng z => Fz - = 0 (2.5)z

p

1

Viet di dang vector (2.6)01

= pgradF

r

Fr

Lc khoi n v (Fx, Fy, Fz)

x

p

p dy dz (p+ dx) dy dz

p

A. TNH TUYET OI (Trang thai tnh di anh hng cua trong lc)IV. PHNG TRNH THUY TNH:

Di anh hng trong lc lc khoi theo tng phng se la:

x

y

z

g

Hnh 2.3

Fx = Fy = 0 Fz = -g (2.7)

Thay vao

1 p0 0

x

=

1 pg 0

z

=

p = - gz + C

-->p + z = const

pz const

+ =

pB

pA

h

A

B

ZB

ZA

Mat chuan

Ap dung cho 2 iem A va B :

p gz p gzA A B B+ = +

suy ra: hpp AB +=

p + gz = constChat long,khong nen=constant

01

= pgradF

r

p0

x

=

p=p(y,z)

1 p0 0

y

=

p0

y

=

p=p(z)

pg

z

=



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Chat kh la kh ly tng: p RT=

Neu nhiet o thay oi theo o cao theo o cao: T=T0 az; a>0,

T0 la nhiet o ng vi o cao z=0 (thong thng la mc nc bien yen lang):

0

gLnp Ln(T az) Ln(C)

aR= +

Goi p0 la ap suat ng vi z=0:

aRg

aRg

T

pCCTp

0

000 ==

aRg

T

azTpp

=

0

00

Phng trnh kh tnh:

dpg

dz= Chat kh,

nen c constant

p

R T =

dp pg

dz RT= dp p g

dz RT=

dp gdz

p RT=

0

dp gdz

p R(T az)=

Tch phan

o 0

gLnp Ln(T ) Ln(C)

aR= +

V. NG DUNG PHNG TRNH THUY TNH5.1 Ap ke

*Ap ke tuyet oi: o ap suat tuyet oi

hgp Hga ..=h

Hg

P = 0

Hnh 2.4

*Ap ke o chenh.MMAA gzpgzp +=+ (a)

NNBB gzpgzp +=+ (b)

T (a) va (b) ta suy ra:( ) ( ) ( )NMNMBBAA zzgppgzpgzp +=++

ma pM = pN

* Ap ke o chenh co 2 chat long

Hnh 2.6

A

B

h1

1

2M

N

(a), (b), (c) ta suy ra:

1

1

12 hp

zp

zh BBA

A

=

+

+=

MMAA gzpgzp 11 +=+ (a)

NNBB gzpgzp 11 +=+ (b)

12ghpp NM = (c)

1hp

zp

zh BBA

A =

+

+=

A

B

M

N

kh

Hnh 2.5

h1



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5.2 nh luat Pascal.Khi ap suat tai mot iem trong moi trng chat long thay oi, th tat ca moi iem trong moitrng o cung thay oi mot ga tr tng ngAp dung nh luat Pascal: Nguyen ly hoat ong cua con oi

5.3 Bieu o phan bo ap suatp a

h

pA= h A

AA

p a

h

pA = h A

A

A

B

pB = h Bh B

Suy luan :Dung bieu o phan bo ap suat xac nh ap lc khi dien tch chu lchnh ch nhat co ay nam ngang:

Hnh 2.8a. Ap lc tren mat ay F = h (Lb) Hnh 2.8b. Ap lc tren mat ben F = h2b/2

L

b

h

F

L

b

h

F

D

D

(1623-1662)

po

y

o

VI. AP LC THUY TNH6.1 Ap lc thuy tnh len mot mat phang

Tren dien tch vi phan

Lc tac dung len toan bo dien tch

: : moment tnh cua dien tch A oi vi truc OXA ydA

AyydA CA

=Do o

( )AhpF c0 +=

ApF c=

Vay ap lc F tac dung len dien tch A bang ap suat tai trong tam (pc ) dien tch Anhan cho dien tch o.

F

c

hch

pdA y

xyc

c

A

( ) ( )dAypdAhppdAdF sin00 +=+==

( ) +=+= AA 00 ydAsindApdAsinypF

+= A0 ydAsinAp

AysinApF c0 +=



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iem at : D ( yD va xD) cua F trng hp ap suat tren mat thoang p0=0Xac nh yD :

- Moment cua F oi vi truc OX

Mox = F. yD = hcA. yD = yc sin A. yD (2.12)

Ngoai ra: monent cua dF tren dA oi vi truc OX la:

dMox = dF . y = h dA y = y2 sin dAVay moment cua F oi vi truc OX la :

= A2

ox dAsinyM

at : moment quan tnh cua A oi vi OX= Axx dAyI2

(2.13)oxox IM sin=(2.12) va (2.13) : yc sin A. yD= sin IOXSuy ra : (2.14)

Ay

Iy

c

oxD =

Moment quan tnh /v truc ox co the tnh t moment quan tnh /v truc i ngang qua trongtam C theo cong thc

Ay

AyIy

c

2cc

D

+=

Ay

Iyy

c

cCD +=

Ay

Iyy

c

cCD =

(2.15)

Ic luon luon dng, do o . Ngha la v tr D thap hn Cy yD C>

Toa o xD : khong can xac nh neu dien tch A co mot truc oi xng song song vi oy th D senam tren truc oi xng o

Suy luan : Hay tm cach xac nh iem at ap lc trong trng hp tren mat thoang co ap suat po0

po

hch

y

yc

y

o

A

dA

c

p

cyD

D

D

F

= A2dAysin

AyII2

ccox +=

a

b

c

6.2 Ap lc chat long len mat cong:

Xet mot mat cong abc co canh ab song song vi truc oy

dAx

dA

b

a

x

z

y

o

dFx

dAx

dF

dA

Lc tac dung len mat cong tong quat: 222 zyx FFFF ++=

Trng hp ab // oy nen Fy = 0, tm Fx va Fz

Ap lc dF tren dien tch vi phan dA : dF = p. dA

Chieu dF tren phng ox dFx = p. dA sin = p. dAx

Do o Fx = xA

xpdA

Fx : chnh la lc tac dung len hnh chieu cua abc tren phng thang goc vi truc ox (phng thang ng) hay noi cach khac la lc tren mat phang abc



15/50


Tng t , chieu dF len phng oz:

do o Fz = zA

zpdA

Trng hp ap suat tren mat thoang bang khong va goi h la khoang cach thang ngt dien tch vi phan dA en mat thoang th :

Fz = . WW: c goi la the tch vat ap lc ( the tch abbc)

nh ngha VAL: The tch vat ap lc la the tch gii han bi mat cong va cac matben thang ng ta vao mep mat cong roi keo dai len cho en khi gap mat thoanghay phan noi dai cua mat thoang.

dFz = p. dA cos = p. dAz

x

z

y

o

a

b

c

dAz

dA

b

a

dFxdAx

dF

dA

= zz Az

Az hdAhdA Fz =

dAz

dFzh

Po=0

A

B

D

C

2R

2R

A

B

D

C

2R

2R

Po

1

2

Th d 2

(?) Xem xet trng hp co nhieu loai chat long va tren mat thoangkhong phai ap suat kh tri .

(?) Xem xet trng hp mot phan tam qua cau trong chatlong, xac nh Fx va Fz



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Th nghim : Ottovon Guericke (8.5.1654) ti Maydeburg, c

Dng 2 bn cu D = 37 cm, bt kn v ht kh p sut tuyt i trong qa cu bngkhng .

Cho 2 n nga ko vn khng tch bn cu ra c. Vy phi cn 1 lc bng bao

nhiu tch hai bn cu ra (xem lc dnh gia 2 bn cu khng ng k)

DF =? F =?

Chn khng p(tuyt i) = 0

Suy lun

6.3 Lc ay Archimede:

V1

V2

V

A B

m

n

+

Mot vat nam trong moi trng chat long se b mot lc ay thang ung t dilen tren va bang trong lng cua chat long ma vat o chiem cho.

(287-212 BC)



17/50


CD

CD

VII. S CAN BANG MOT VAT TRONG CHAT LONG:

Can bang on nh

7.1 Vat ngap hoan toan trong chat long :

D di C

FA

G

C

D

FA

G

CD

FA

G

D tren C

Can bang khong on nh

C : im t trng lng, D : im t lc y archimede

FA

G

7.2 Vat ngap mot phan trong chat long :

C

D

Tam nh khuynh M nm trong CD

MD c xac nh :W

IMD

yy=

Iyy: moment quan tnh cua mat noi oi vi truc quay yy

W : The tch vat chm trong chat long

M

D

C

D

M

D di C Tam nh khuynh M nam ngoai CD

D di C

G

D

C

D

FA

G

C

D

FA

G

FA

D

G

D

FA

Can bang on nh

Can bang khong on nh

D : iem at lc ay ArchimedeC : iem at trong tam vat



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VIII. TNH HOC TNG OI :8.1- Chat long trong bnh chuyen ong thang ngang vi gia toc khong oi

Xet chat long chuyen ong thang vi gia toc a, ap dung phng trnh vi phanc ban cua chat long can bang:

ax

p

x

pa

=

=

01 => p = -ax + f(y,z)

001

0 ==>=

y

p

y

p

=> p = -ax +f(z)

gz

p

z

pg

=

=>=

01

va t trenz

f

z

p

=

gz

f=

=> f= -gz +C1

thay f vao p = -ax - gz + C1

Phng trnh mat ang ap: dp = 0 => 0=

+

+

dzz

pdy

y

pdx

x

p

Thay cac ga tr vao rut ra czpypxp /,/,/ zCxg

a

z +=Mat ang ap la nhng mat phang nghieng song song vi mat thoang.

Khi xe chuyen ong va chat long khong b tran ra ngoai th mat thoang lc ny s i ngang quatrung iem M ca mt thong khi xe ng yn

a>0

x>0

z

M

01

= pgradF

r

8.2 Chat long trong bnh chuyen ong quay eu quanh truc thang ng

Xet chat long trong bnh chuyen ong quay vi goc khong oi.

h/2

h/2h

z

r

Chat long can bang vi lc khoi tren cac phngnh sau:

Phng trnh phan bo ap suat

1

22

2

1Cgzrp +=

Phng trnh mat ang ap 222

2

1Cr

gz +=

Mat ang ap la nhng mat paraboloid tron xoay

Chu y : Do the tch chat long trong bnh trc va sau khi quay bang nhau, nen khoangcach t mat thoang chat long khi bnh ng yen en nh va chan cua paraboloidbang nhau.

xFx2

= yFy2

= gFz =

Thay vao phng trnh c ban va viet lai trentoa o tru co:



19/50


20/50

Tom tat bai giang C lu chat- TS Huynh cong Hoai H Bach Khoa tp H

paCau 3 Cac v du ve vat ap lc Vval:

Pdu

w

Fz

PaPck

w

Fz

PaPck

Pa

w

Fz

w

pa

w

pdpd/

Fz

w

pck

papck/Fz

pa

w

Fz

pck

pa

pck/

w

Fz

pck

pa

pck/w1

w2

Fz1

Fz2

Cau 4: Cac thung tren hnh ve eu co ay tron va cung ng knh, cha nc vadau. Goi F1, F2 va F3 la lc tac dung tren ay thung. Ta co :

nc

dau

F1nc

dau

F2

nc

dau

F3

a) F1 > F2 > F3 b) F1 < F2 < F3 c) F1 = F2 = F3 d) F1 > F1 = F2

Cau 5. Trong th nghiem cua Toricelli ong dung mot ong nghiem up tren mot chau thuyngan va hut het khong kh trong ong ra th thay mc thuy ngan dang len trong ongnghiem 76 cm. Neu thay thuy ngan bang nc th mc nc trong ong nghiem se la :

a) Thap hn mc thuy nganb) Cao hn mc thuy nganc) Bang mc thuy ngand) Co the cao hn hoac thap tuy thuoc vao ng knh cua ong nghiem ln hay nho.



21/50


Cau 6. Mot thung nc co trong lng Wn va mot qua cau cotrong lng Wc . Neu goi W la tr so oc tren can khi bo quacau vao trong nc tha) W < Wn + Wc khi qua cau noi tren mat thoangb) W < Wn + Wc khi qua cau noi chm l lng nh hnh vec) W = Wn + Wc khi qua cau chm xuong ay bnhd) Ca 3 eu ung

W

Qua cau

A

Cau 7: Mot ong hnh ch U, mot au bt kn va mot au eh tiep xuc vi kh tri. Khi ng yen mc nc trong bnhnam ngang nh hnh ve. Neu bnh quay tron qua truc thangng oi xng vi van toc quay th ap suat tai A so viluc ng yen se la :a)Nho hnb) Ln hn

c) Khong oid) Cha xac nh con phu thuoc vao van toc quay

Archimede 287-212 BC



22/50


Pascal 1623-1662 , Phap

(1 ) (2 ) (4 )

Hay cho biet bieu o phan bo ap suat tuyet oi nao sau ay la ung:

a) 1 b) 2 c) 3 d) 4

(3)

(2 ) (3 ) (4 )(1 )

Hay cho biet bieu o phan bo ap suat dnao sau ay la ung:

a) 1 b) 2 c) 3 d) 4



23/50


pa Cac v du ve vat ap lc Vval:

Pdu

w

Fz

PaPck

w

Fz

PaPck

Pa

w

Fz

w

pa

w

pdpd/

Fz

w

pck

papck/Fz

pa

w

Fz

pck

pa

pck/

w

Fz

pck

pa

pck/w1

w2

Fz1

Fz2



24/50


Chng 3 ONG HOC LU CHAT

I HAI PHNG PHAP NGHIEN CU1.1 Phng phap Lagrange.(J.L de Lagrange, nha toan hoc ngi Phap,1736-1883)

Theo doi qua trnh chuyen ong cua cac phant chat long va nhng dien bien trong quatrnh di chuyen cua no.

x

y

z

zo

xo

yo

to

t

( )t,rfr orr

= (3.1)

hay ( )t,z,y,xxx ooo= ( )t,z,y,xyy ooo= ( )t,z,y,xzz ooo=

Van toc c xac nh:

dt

rdu =r

2

2

dt

rd

dt

uda

rr

r

==

dt

dzu;

dt

dyu;

dt

dxu zyx ===

2

2

z2

2y

y2

2x

x

dd=a;

dd=a;

dd=a

dt

z

dt

u

dt

y

dt

u

dt

x

dt

u z ===

Trong phng phap Lagrange , cac yeu to chuyen ong la mot ham co bien so la thi gian

V du : u = at2 + b (a, b la hang so)

x

z

yror

1.2 Phng phap Euler. ( L. Euler, nha toan hoc ngi Thuy S, 1707-1783)

Mo ta cac yeu to dong chay tai tng iem trong khong gian, do o cac thong so dong chayla mot ham theo vi tr va thi gian

( )t,z,y,xu=urr

va cac thanh phan

( )tzyxuu xx ,,,=( )tzyxuu yy ,,,= ( )tzyxuu zz ,,,=

Th du : ux = 5x(1+t) , uy = 5y(-1+t)

Gia toc cua chuyen ong :du

a dt=

r

r

tren phng x:x x x x

x x y z

u u u ua u u u

x y z t

= + + +

tren phng y: y y y yy x y zu u u u

a u u ux y z t

= + + +

tren phng z:

Gia toc oi lu Gia toc

cuc bo

z z z zz z y z

u u u ua u u u

x y z t

= + + +



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III. PHAN LOAI CHUYEN ONG

3.1 Phan loai theo ma sat:

Chuyen ong chat long ly tng, : khong co ma sat

3.2 Phan loai theo thi gian:

* Chuyen ong on nh:

u = u(x,y,z) a = a(x,y,z)

u

t

a

t= =0 0

* Chuyen ong khong on nh

u = u(x,y,z,t) a = a(x,y,z,t)

3.3 Phan loai theo khong gian

Dong chay 1 D, 2D va 3D (Dimension)

Chuyen ong chat long thc: co ma satChuyen ong tang

Chuyen ong roi

He so ReynoldsVD 4VR

Re = =

: He so nhtong hoc

IV. PHNG PHAP THE TCH KIEM SOAT VA AO HAM CUA MOT TCHPHAN KHOI

4.1.Phng phap the tch kiem soat:

= W dWX

W: the tch kiem soat

X : ai lng can nghien cu

K: ai lng n v ( ai lng X tren 1 n v khoi lng)

Th du : ai lng n v cua khoi lng K=1ai lng n v cua ong lng u

r

=

4.2. ao ham cua mot tch phan khoi

= W dWX (Tch phan khi)

( )= W dWdtd

dt

dX (ao ham cua mot tch phan khi)?

S

W udw

CV



27/50


) )221122 AtCtBtAtAtBt

XXXXXXX +++=

Tai thi iem t1 111 BtAtt XXX +=

Thi iem t2 X X Xt Bt Ct2 2 2= +

Trong thi gian t, co s bien oi

) )1122 BtAtCtBt

XXXXX ++=

( ) ( ) 2212 AtCttBAtBA XXXX += ++( ) ( )

t

XX

t

XX

t

X AtCttBAtBA

+

=

++ 2212

ao ham theo t

Trong o (1)

=2

2 SCt dAn.utX

rr

=1

2 SAt dAn.utX

rr

va

t

dAn.utdAn.utlim

t

XXlim

SS

t

AtCt

t

+=

1222

00

rrrr

+= SCVCV

dAnut

X

dt

dX rr.

( ) ( )

t

X

t

XX tBAtBAt

=

++

12

0lim (3.8)

== + SSS dAn.udAn.urrrr

21

(3.9)

(3.10)

Thay (3.8), (3.9) vao (3.7)

Do o (2)

44 344 21444 3444 21

)2(

0

)1(

)()(

00

2212 limlimlimt

XX

t

XX

t

X

dt

dX AtCt

t

tBAtBA

tt

+

=

=

++

(3.7)

CV: the tch kiem soat ( Control Volume) S: dien tch bao quanh the tch kiem soat

t2

S2

S1

CA

BW

t1

+=12

..SS

dAnudAnurrrr

3.5. PHNG TRNH LIEN TUC

Bao toan khoi lng dmdt

= 0

Ap dung phng phap the tch kiem soat

= W dWX

0=+= SCVHT

dAn.ut

m

dt

dm rr

0=+ SW dAn.udWtrr

( ) 0=+ WW dWudivdWtr

( ) 0=

+W dWudiv

t

r

( ) 0=+ udivt

r

K=1

(PT lien tuc)

Bien oi Gauss

CV

= W dWm

0=+

SW dAn.udWtrr



28/50


CV

* Chat long khong nen c: 0=udivr

Hay

u

x

u

y

u

z

x y z+ + = 0

Trong toa o cc ( ) 011

=++ z

uu

rrurr

z

r

*Trng hp lu chat chuyen ong on nh , chon the tch kiem soat bao quanh dong chay

021

=++= A SAS b dAn.udAn.udAn.udAn.urrrrrrrr

0=+ SW dAn.udWtrr

u2

n

u1

A1

A2

n

Sb

021

222111 =+ AA dAn.udAn.urrrr

Q Qm m1 2=

o

ConstAVAV == 222111

1 2= = Const Q Q1 2=Chat long khong nen c: V A V A Const1 1 2 2= =

( = const)

ConstQQ == 2211

( ) 0=+ udivt

r

021

=+mm

QQ

3.6 PHAN TCH CHUYEN ONG CUA LU CHAT:

Van toc quay: uRotrr

2

1=

zyx uuuzyx

kjirrr

2

1=

=

z

u

y

u yzx

2

1

=x

u

z

u zxy

2

1

=

y

u

x

uxy

z2

1

Chuyen

ong

1. Tnh tien

2. Quay

3. Bien dang

Mot chuyen ong khong quay th :

x = y = y = 0



29/50


Thu du: Xac nh ng dong cua mot dong chay co : ux = 2y va uy = 4x

yx u

dy

u

dx=

x

dy

y

dx

42 =

ydyxdx 24 =

ydyxdx =2

Cyx

+=

222

22

Cyx = 222



30/50


z

ox

y

dx

dy

dz

Chng 4 ONG LC HOC LU CHAT

I. PHNG TRNH VI PHAN CHO CHAT LONG LY TNG CHUYENONG (P.Tr EULER)

Xet mot khoi hnh hop vi phan dxdydz trong khoi chat long ly tng chuyen ong.

Tong lc tac ong tren khoi hnh hop vi phan =>

p p dx

x

+ p

Vi : lc khoi n vp(x,y,z,t) : ap suatu (x,y,z,t) : van toc

Fr

dt

udpgrad

1F

v

r

=

Neu viet tren phng x th :

dt

du

x

pF xx =

1

dt

dt

t

u

dt

dz

z

u

dt

dy

y

u

dt

dx

x

u

x

pF xxxxx

+

+

+

=

1

them vao ve phai

t

u

z

uu

y

uu

x

uu

x

pF xxz

xy

xxx

+

+

+

=

1

x

uu

x

uu

x

uu

x

uu zz

y

yz

z

y

y

+

dt

udpgradF

r

r

=

1

sau khi bien oi, ta co:t

u

y

u

x

uu

x

u

z

uu

uuu

xx

pF xx

y

yzx

z

zyx

x

+

+

++

=

2

1222

[ ]t

uuu

u

xx

pF xzyyzx

++

=

222

12

va viet di dang vector 21 u uF gradp grad 2 u

t 2

= + +

r

uuuuur uuuurr

r r

pt Euler dang Lam-Gromeko

Tng t tren phng y[ ]

2y

y x z z x

u1 p uF u 2 u 2

y y 2 t

= + +

tren phng z2

zz y x x y

1 p u uF u 2 u 2

z z 2 t

= + +



31/50


uxu

gradt

ugradpF

rr

r

r

22

12

++

=

II.TCH PHAN PHNG TRNH CHUYEN ONGLc co the: Lc khoi n v F la lc co the khi co the tm c mot ham (x,y,z) sao cho

F grad uuuur

ngha la :x

Fx

=

yFy

=

zF

z

=

va (x,y,z) c goi la ham the

Th du : Lc khoi n v la trong lc la mot lc co the vi : (x,y,z) = gz

Ham ap suat: (x,y,z) goi la ham ap suatkhi

pgradgrad

1=

Cdp

+=

Neu chat long khong nen c: = const th : Cp

+=

Thay vao phng trnh Lamb Gromeko :

2u ugrad 2 xu

2 t

+ + = +

r

uuuur

r r

1. Trng hp chuyen ong khong quay (chuyen ong the):

Mot chuyen ong khong quay luon luon tm c mot ham the van toc (x,y,z,t) sao cho:

gradu =r

chuyen ong khong quay 0=r

2ugrad 0

t 2

+ + + =

uuuur

Cu

t=+++

2

2

chuyen ong on nh, khong nen c va ch chu anh hng duy nhat la trong lc

Cup

gz =++2

2

2u ugrad 2 xu

2 t

+ + = +

r

uuuur

r r

( )2u

grad grad2 t

+ + =

uuuur uuuur

2ugrad grad

2 t

+ + =

uuuur uuuur



32/50


33/50


4. Chuyen ong on nh, tch phan theo phng phap tuyen vi ung dong

Ta co : rr

.uu =dt

du

dt

du

dt

ud

r

r

r

+= .

nR

uu

s

uu

t

u

dt

ud rrr

+

+

= .

Thay vao pt Euler: ( ) nRuu

suu

tugrad

rr

+

+=+.

Nhan 2 ve cho dn ( )[ ] dnnR

uu

s

uu

t

udngrad ...

+

+

=+rr

( )[ ] dnR

udngrad

2

. =+

Neu chat long chuyen ong on nh, khong nen c va ch chu anh hng duy nhat la tronglc th cho tren phng phap tuyen cua ng dong la

R

up

gzn

2

=

+

( )R

u

n

2

=+

khi nhng ng dong thang song song th R hay 0=

+

pgz

nC

pgz =+

ap suat phan bo theo qui luat thuy tnh tren phng thang goc vi ng dong

R

s ,o

n

ur

Vector n v tren phng s va nnvr

,

W

CV

III. PHNG TRNH NANG LNG

Xet the tch kiem soat W, bao quanh

dien tch A.

u

n

dA

ai lng nghien cu la nang lng

X = E

Nang lng n vK = u2/2 + gz (ong nang + the nang)

+=

W

dWgzu

E

2

2

dAnugzu

dWgzu

tdt

dE

AW

++

+

=rr

22

22

Theo nh luat I nhiet ong lc hoc,s bien thien nang lng trong mot he thong cua cacphan t chat long trong mot n v thi gian (dE/dt) , bang cong suat cung cap cho he thongcong vi nhiet lng them vao he thong trong mot n v thi gian

dt

QdP

dt

dE~

+=

= W dWX

Khong co strao oi nhiet dt

QdP

dt

dE~

+=



34/50


Khong co s trao oi nhiet Pdt

dE=

P do lc tac dung trendien tch A bao quanhthe tch kiem soat gom

nr ap suat-p

r

ng suat do ma sat

( ) ( ) dAuudAnpPAA +=

rrrv

( ) ( ) +=

++

+

AAAWdAuudAnpdAnugz

udWgz

u

t

rrrvrr

22

22

Thay vao:

sap xep lai :

( ) dAudAnugzpudWgzut AAW =

+++

+

rrrr

22

22

Chuyen ong on nh: ( ) dAudAnugzpu

AA =

++

rrrr

2

2

W

CV

u

n

dA

dAnugzu

gzu

tdt

dE

AW

++

+

=rr

22

22

++=

++=

++ dAnugz

pdAnu

udAnugz

pdAnu

udAnugz

pu rrrrrrrrrr

222

222

Trng hp chon the tch kiem soat la mot oan dong chay tai mat cat A1 va A2 co ngdong song song :

( ) dAudAnugzpu

dAnugzpu

dAnugzpu

ASbAA =

+++

+++

++rrrrrrrr

222

2

22

222

11

121

V1

p2V2

p1

Mat chuan

z1z2

A1 A2

( ) dAudAnugzpu

AA =

++

rrrr

2

2

Hang so ng thangsong song



35/50


QgzpV

dAnugzpu

++=

++ 22

22rr

++=

++ dAnugz

pdAnu

VdAnugz

pu rrrrrr

22

22

Trong o c goi lahe so sa cha (hieuchnh) ong nang :

= dAVu

A 3

31 oi vi dong chay roi : 1

=

+++

++ dAuQgz

pVQgz

pV rr

2

2

2

221

1

2

11

22

Q

dAuz

p

g

Vz

p

g

V

++=++

rr

22

2

221

1

2

11

22

Thay vao :

at :Q

dAuh Af

=rr

21 Goi hf1-2 la ton that nang lng cho trong dong chay

2122

2

221

1

2

11

22+++=++ fhz

p

g

Vz

p

g

V

++=

++ dAnugz

pdAnu

udAnugz

pu rrrrrr

22

22

= dAnuV

dAnuu

rrrr

22

22

V nang cua mot n v trong lng chat long

g

V

2

2

1

p

Y ngha cac so hang:

ong nang cua mot n v trong lng chat long.

Ap nang cua mot n v trong lng chat long.

Z

The nang ( cot nc o ap )zp

+

Nang lng cua dong chay tnh tren mot n v trong lng chat longzg

p

g

VE ++=

2

2

( Cot nc nang lng , m)

Nang lng toan dong tai mot mat cat trong mot n v thi gian(Cong suat cua dong chay, Watt)QEP =

Phng trnh nang lng khi co may bm hoac tua bin:

11

21

1 2z

g

p

g

V+

+ 212

22

22 2

++

+= fhzg

p

g

V+HB -HT

HB Nang lng may bm cung cap cho mot n v trong lng chat long (m)

HT Nang lng tua bin lay t mot n v trong lng chat long (m)

P = QHB (watt)

P = QHT (watt)

2122

2

221

1

2

11

22+++=++ fhz

p

g

Vz

p

g

V



36/50


ng nang va ng cot nc o ap ( chat long ly tng)

ng cotnc o ap

ng nang

Mat chuan

iem dng

ng nang ng cotnc o ap

zp

g

VH ++=

2

2

zp

Hp +=

IV. NG DUNG PHNG TRNH NANG LNG

1. o lu lng ( ong Ventury )

Th nghiem ong Venturytrong phong th nghiem

Ong Ventury trongng dung thc te



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(C


38/50


V. PHNG TRNH ONG LNG :

ai lng nghien cu trong the tch kiem soat la ong lng.

W

CV

ong lng n v : ur

=

== WW dWudWKrv

ong lng trong the tch kiem soat la

Ap dung nh luat ve ong lng:

= Fdt

Kd rr

= FdWudtd

w

r

r

hay

=+

FdAnuudWu

t Aw

r

vrrr

)(

Phng trnh ong lng =+

FdAnuudWt

u

Aw

r

vrr

r

)()(

V1

V2C.V

A1A2

Sb

- Chuyen ong on nh

=+

FdAnuudWt

u

Aw

r

vrr

r

)()(

=++ FdAnuudAnuudAnuuSbAA

r

vrrvrrvrr

)()()(21

=+ FdAnuudAnuuAA

r

vrrvrr

21

)()(

=+ FdAnuVdAnuVA

o

A

o

r

vr

r

vr

r

2

22

1

11 )()(

o1 , o2 la he so sa cha ong lng

AV

dAu

AV

dAu

AAo 2

2

2

2 ==

Vi chuyen ong tang trong ong : o =4/3va roi th o = 1,02 -1,05=+ FQVQV oo

rrr

)()( 222111

= FVQVQ oorrr

111222

L ra L vao Tong lc mF

r

(Lc khoi :Trong lng, )

SFr

(Lc mat :Ap lc, lc ma sat, ..)

Xet mot chuyen ong

- The tch kiem soat la mot oan dong chay:- Khong nen c



39/50


40/50


I.Cau tao bo phan cai tien

Source: HT Hung-NHD Kha-NT Hai-LH Dng

2Q

1Q

Q



41/50


22

1

1

bomh

1Q

2Q

Q



42/50


Chng 5 DONG CHAY EU TRONG ONG

I. HAI TRANG THAI CHAY

Th nghiem Reynolds:

1. Chay tang : Khi van toc nho , Re = VD/ < Regh

Qua o:

2. Chay roi : Khi van toc ln , Re = VD/ > Re gh

Trong th nghiem nhan thay:

Tang RoiRegh(tren)

Tang RoiRegh(di) =2300

II. PHNG TRNH C BAN CHO DONG EU TRONG ONGTrong ong xet oan vi phan dong chay eu hnh tru co dien tch dA nh hnh ve:

F2=p2dA

F1=p1dA

G

Gsin

s

=max

=0

1

1

2

2

Mat chuan

z1z2

L

Lc tac dung tren phng dong chay ( phng s) : 0sin 21 =+ msFFFG

0)(

2121 =+

PLdApdAp

L

zzLdA

at J = hd/ L

ng suat tiep ty le bac nhat theo r

T pt c ban co the viet0

max

0

max2 r

rhay

rJ ==

Phng trnh c ban cua dong eu (5.1)

R

Lpz

pz

=++ )()( 22

11

JR =

2/Jr =

0)()( 221

1 =++R

Lpz

pz

0)( 2121 =+dA

PLppzz



43/50


II.PHAN BO VAN TOC TRONG DONG CHAY TANG

Tai r=r0 ta co u=0, suy ra

Tai r=0 ta co u=umax

= =

2 22 0

max 0 max 20

J r ru r u u4 r

hay

=

2

2

max 1or

ruu

Phan bo van toc trong chay tang co dang Parabol

ou

r

drr

parabol

r

r0

dr

du =

2

rJ =

2

rJ

dr

du = dr

rJdu

2= = dr

rJu

2

C4

rJu

2

+

=

4

2

0rJC=

( )224

rrJ

u o =

Newton

PTCB

ro

Lu lng va van toc trung bnh:

Ton that doc ngThay J = hd/L

g

V

D

Lhd

2Re

64 2=

Vi Re = VD/ ( He so Reynolds)

L

hrV do

8

2

=T

Suyrahd

udAdQ = r

dA

rdru 2= ( ) rdrrrJdQ o

2

4

22 =

( ) =or

o rdrrrJ

Q0

22

42

4

8oJrQ

=

( )224

rrJ

uo

=

2

4

8o

o

r

Jr

A

QV

==

8

2

oJrV =2

maxuV =

8

2

oJrV =

sap xep lai2

8

o

dr

VLh

=

TS. Hunh cng Hoi BM Clu Cht - HBK tp HCM



44/50


oi vi dong chay roi, ng suat tiep phu thuoc chu yeu vao o chuyen ong hon loan cua cacphan t lu chat.

Theo gia thiet cua Prandtl:dy

du=

vi c goi la ng suat nht roidy

dul2=

y

u

ro

o

o

y : khoang cach t thanh en lp chat long ang xet

l :chieu dai xao tron

Prandtl: ng suat nht roi khong phu thuoc vao tnh nht cua lu chat.

Theo th nghiem cua Nikudrase, chieu dai xao tron l trong ong2/1

or

y1kyl

=

Vi k : hang so Karman ( k = 0,4)

Neu xem t le tuyen tnh vi ban knh r :

=

oo

r

y1 Th

=oo

1r

y

Thay vao :2/1

o

kyl

=

(1)

(2)

(3)

Thay vao (1) :2

22o

dy

duyk

=

dy

duyk

0

22

= (5.8)

ky

1

dy

du o

=

III. PHAN BO VAN TOC TRONG DONG CHAY ROI TRONG ONG

T (2)

ky

1

dy

du o

=

at

= o*u ( van toc ma sat)

ky

u

dy

du *=

y

dy

k

udu

*

=

yLnk

uu

*

= + C

Tai tam ong r = ro u = umax thay vao cho o*

max rLnk

uuC =

y

rLn

k

uuu o

*

max =

Phan bo lu toc trong trng hp chay roi co dang ng logarit

y

u

ro

o

o

Umax

ng conglogarit

Do o ta nhan thay s phan bo van toc trong trng hp chay roi tng oi ong eu gan vi vantoc trung bnh hn so vi trng hp chay tang. o cung la ly do tai sao cac he so sa cha ongnang () hay he so sa cha ong lng (o) co the lay bang 1




45/50


III. TNH TOAN TON THAT CUA DONG CHAY TRONG ONG1.Ton that ng dai: Dung phng phap phan tch th nguyen chng to ton that doc ung codang

(Darcy)= 2

dL VhD 2g

= f(Re, /D) : he so ton that

: He so nham tuyet oi (chieu cao cac mo nham )

thay D = 4RL

hR

gV d

8=

vi J = hd/L RJg

V

8= va at

gC

8= ( he so Chezy)

RJCV = ( Cong thc Chezy)

lu lng JKRJACQ == Vi module lu lng RACK =

He so Chezy C co the tnh theo cong thc Manning : 6/11

Rn

C= ( n la o nham)

Cong thc Manning ch dung khi dong chay roi thanh hoan toan nham

LK

Qh

2

2

d =T cong th tnh lu lng

2. Xac nh he so ton that :Dong chay tang: = 1d

64 h VRe

Dong chay roi:

Roi thanh trn thuy lc: (2300 < Re < 105 ) = f(Re).

Blasius: = 14tr

0,316Re

Prandtl-Nicuradse: = tr

1 2lg(Re ) 0,8

Roi thanh nham thuy lc: ( Re > 105 ) = f(Re, /D).

Antersun: = +

0,251000,1 1,46D Re

Colebrook: = +

1 2,512lg

3,71.D Re

Chay roi thanh hoan toan nham (khu sc can bnh phng) (Re rat ln >4.106 ) = f( /D).

Prandtl-Nicuradse:

+

=

D17,3lg(214,1

Dlg2

1



46/50


0,000 011 2 3 4 5 7

x1031 2 3 4 5 7

x1041 2 3 4 5 7

x1051 2 3 4 5 7

x1061 2 3 4 5 7

x1071

x108

0,000 005

0,000 007

0,000 050,000 10,000 2

0,000 40,000 6

0,001

0,002

0,0040,0060,0080,010,0150,02

0.030,040,05

0,0080,0090,01

0,015

0,02

0,025

0,03

0,04

0,05

0,06

0,070,080,090,1

Khu chay roi thanh nham hoan toan (Khu sc can bnh phng)KhuChay tangKhu chay roi

thanh nham

Khu chay roithanh trn

Khu chuyen tiep

Re = vD/

=/D_

O TH MOODY

3.Ton that cuc bo: Tnh theo cong thc thc nghiem Weisbach:

g

V

h cc 2

2

=

la he so ton that cuc bo (phu thuoc vao tng dang ton that)

V la van toc dong chay tai v tr sau khi xay ra ton that

M rong ot ngot1

1

2

2

V1

V22

2

11

=

A

A

2

1

2 1

=

A

A

vi V1

vi V2

mieng ra cua ong: c=1

mieng vao cua ong: c=0,5




47/50


IV. CAC BAI TOAN TRONG NG ONG1. Phan biet ng ong dai, ngan hc5%hd : ong ngan h f = hd + hc

h f = hd

2. ng ong mac noi tiep

l1; d1; 1 l2; d2; 2

l3; d3; 3

H

A A

B BMat chuan

BfAB

BB

A

AA hzp

g

Vz

p

g

V+++=++

22

22

++++

++=

g2

V

g2

V

g2

V

g2

V

g2

V

3d

3l

g2

V

2d

l

g2

V

d

lH

23

ra

22

th

21

mr

21

vao

23

3

222

2

21

1

11

++++++= 2

3

ra22

th21

mr21

vao23

322

22211

11

2

A1

A1

A1

A1

A1

3d3l

A1

2dl

A1

dl

g2QH

Trong o A1, A2 , A3 la tiet dien ong 1, 2, va 3. Q chay trong ong neu bietcac thong so con lai

3. ng ong mac song song (bo qua ton that cuc bo).

A B

l1, d1, n1

l2, d2, n2

L3, d3, n3

Goi EA va EB la nang lng tai A va B.

Neu xet dong chay i t A en B tren ong 1 , ta co ton that tren ong so 1 la : hf1 = EA EB

Tng t, xet dong chay t A en B tren ong 2 va 3 ton that ong 2 va 3 la : hf2 = EA EBhf3 = EA EBNh vay hf1 = hf2 = hf3

Neu bo qua ton that cuc bo : 3d2d1d hhh ==

hay

va

T 3 phng trnh (i), (ii), (iii) Q1, Q2 va Q3

222

22

121

21 l

K

Ql

K

Q= (i)

323

23

222

22 l

K

Ql

K

Q= (ii)

Q=Q1 + Q2 + Q3(iii)



48/50


4. ng ong noi 3 ho cha (bo qua ton that cuc bo).

l1; d1; n1 l2;

d2; n

2

l3; d3; n3

J

12

3

Z1

Z2

Matchuan

Ej

Ej

ng nang ga sng nang

Ej

ng nangChay t J ve 2Chay t 2 ve J

Khong chay tren ong 2

ng nang gas

l1; d1; n1 l2; d2; n2

l3; d3; n3

J

12

3

Z1

Z2

Matchuan

Ej

Gia s cao trnh nang lng tai J ,Ej ngang vi mc nc trong bon 2

Ton that tren ong 1 121

21

1d lK

Qh = => 12

1

21

21 lK

Qzz =

1

2111

l

zzKQ

=


22

2d lK

Qh = => 22

2

2

20 lK

Q= Q2 = 0


23

3d lK

Qh = => 32

3

2

32 l

K

Qz =

3l

zKQ 233 =

Q1 > Q3 => trong ong 2 dong chay i t J ve bon 2Q1 < Q3 => trong ong 2 dong chay i t bon 2 ve JQ1 = Q3 => trong ong 2 khong co dong chay

Cach xac nh chieu dong chay tren ong 2



49/50


l1; d1; n1 l2; d2; n2

l3; d3; n3

J

12

3

Z1

Z2

Matchuan

Ej

Th du trng hp 1 xay ra, Q1 > Q3

Ton that tren ong 1 : 121

21

j1 lK

QEz = =>

1

j111

l

EzKQ

=


22

2j lK

QzE = =>

2

2j22

l

zEKQ

=

Ton that tren ong 332

3

23

j lK

QE = =>

3

j33

l

EKQ =

Q1 = Q2 + Q3

Ej , Q1 , Q2 , Q3

Q1 Q2

Q3

5. Mang ng ong kn:

Q

QL

QD

QB

A

B

C D

E

F

G

H

I

J

K

L

i ii iii

viiviiv

Lu lng trong tng ong c xac nh da vao 2 ieu kien cua dong chay trong mang kn nh sau1. Tai mot nut lu lng en phai bang lu lng i

2. Trong mot vong kn, tong ton that phai bang khongQui c dong chay theo chieu tnh toan ton that lay dau dng (+)va dong chay ngc chieu tnh toan ton that lay dau am (-)

Bc 1: T phan phoi lu lng tren tng ong sao cho thoa man ieu kien 1Bc tnh toan

Bc 2: ieu chnh lai lu lng tng ong sao cho thoa man ieu kien 2

Ap dung phng phap Hardy Cross



50/50


Phng phap Hardy Cross

Ap dung cho nhng cong thc tnh ton that doc dng co dang hd = m Qx.

Th du lK

Qh

2

2

d = m = x =l/K2 2

Goi Qi la lu lng t phan phoi c tren ong i ( cha thoa man ieu kien 2)

Q la lu lng can ieu chnh trong mot vong e thoa man ieu kien 2

Ton that nang lng tren ong i khi a ieu chnh lahdi = mi (Qi +Q)x hdi = mi (Qi x +xQ Qx-1 + .)

Gan ung hdi = mi (Qi x +xQ Qx-1)

Trong mot vong kn, tong ton that phai bang khong

( ) 0QQxQmk

1i

1xi

xii

=+=

vi k la so ong trong mot vong

01

1

1

=+ =

=

k

i

x

ii

k

i

x

ii QmQxQm

=

==k

i

x

ii

k

i

x

ii

Qmx

Qm

Q

1

1

1

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