Baigang CLC 2010

Tom tat bai giang C lu chat-TS. Huynh cong Hoai-HBKtpHCM 1

C LU CHAT

FLUID MECHANICS

TOM TAT BAI GIANG

Giang vien:Huynh cong Hoai H Bach Khoa Tp HCMM.E (AIT,Thailand), Ph.D (INPT,Phap)(http://www4.hcmut.edu.vn/~hchoai/baigiang)

Tai lieu tham khao1. Bai gi ng C lu Chat

Gao trnh c lu chat - Bo mon C lu Chat3. B 4.Thuy lc ai cng Nguyen Tai, Ta ngoc Cau5.Thuy lc ( Tap I) Nguyen van Tao , Nguyen canh Cam6. Theory and application of fluid mechanics. Subramanya.K . Mc.Graw Hill 1993 7. Fluid mechanics John Doughlas, Janusz M. Gasiorek , John A. Swaffiield. Fourth edition, Prentice Hall, 20018. E-book : Fundamentals of fluid mechanics

9. E-book : Shaums interactive Fluid mechanics Giles R.V et al.10. Website:

Chean Chin Ngo, Kurt Gramoll

Thi gian giang day ly thuyet va bai tap: 42 tietHnh thc thi: Trac nghiem , c mang tai lieu vao phong thiThi gia hoc ky (20% iem) : 8 cau ly thuyet va 4 cau toan (45 phut) Thi cuoi hoc ky (80% iem ) : 12 cau ly thuyet va 8 cau toan (90 phut)

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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 :

www4.hcmut.edu.vn/~hchoai/baigiang

- Ky thuat giao thong , thiet ke che tao may bay, tau thuy, xe hi

- Xay dng thuy li, cau ng , cap thoat nc

- Thiet b thuy lc, bm, tua bin

- Moi trng, kh tng thuy van

- Y khoa, tim nhan tao

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

Nghien cu dong khi qua xe ang chuyen ong

Lc nang cua may bay

Lc can len tau thuy



+ 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

Nghien cu xoi l trong songNghien cu gio tac dung len nha cao tang

Nghien cu dong chay qua day cap cau treo

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

May bm Con oi



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

D bao bao D bao bao Katrina

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

o huyet ap



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

KgN (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/m3khong kh : 1,228 kg/m3 khong kh : 12,07 N/m3

- T trong : = /nc = /ncV du : nc = 1, thuy ngan = 13,6

2.2 TNH NHT CUA LU CHAT (Viscosity)Quan sat mot dong chay :

=Cong th c Newton ( ch dung cho dong chay tang)

Trong o :

: ng suat ma sat (N/m2) : he so nht ong lcu : van toc, phu thuoc vao yn v cua :

n v chuan : Pa.S

Ngoai ra : poise , 1 poise = 0,1

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 = 1cm2 /s = 10-4 m2/s



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 nht tang

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

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

Chat long Newton va phi NewtonHau het cac loai lu chat thong thng nh nc, xang, dau eu thoa man cong thc Newton (1) , tuy nhien co mot so chat long (hac n, nha nong chay, dau tho ..) khong tuan theo cong thc Newton c goi la chat long phi Newton, hoac oi vi chat long thong 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 th c: co ma sat

o he so nht

2.3 TNH NEN CUA LU CHAT :Moduyn an hoi: = hay =

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

oi vi chat kh ly tng : p = RTVi : p : ap suat tuyet oi (N/m2 )

: khoi lng riengR : 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 khong ang 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

K : moduyn h iV : th tch ban u



===

=

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 o V 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.

Mat canh may bm b xam thc do hien tng kh thc xay ra

Trong mot so ieu kien cu the, hien tng khthc (cavitation) xay ra khi ap suat chat long nhohn bao hoa



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 suat ma 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) 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 sat gia cac phan t tren AB se la:

CAU HOI TRAC NGHIEM



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 trong ieu 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 khc) Luon luon giam cho tat ca cac loai lu chatd) 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



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 tangc) Se giam 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

V du 1: Mot thang may trt tren 2 tam phang co kch thc nh hnh ve. Xac nh lc ma khi thang may chuyen ong vi van toc Vo = 0,5 m/s. Biet dau boi trn co o nht ong lc =9.10-2 Ns/m2

t=0,5cmb= 2 m

b= 2 m

h = 3 m

=

==

=

== 9.10-2. 0,5/0,005*2*2*3

= 108 N



V du 2: ng ong co ng knh d, dai l, dan dau vi he so nhn , khoi lng rieng . Dau chuyen ong theo quy luat sau:u=ady-ay2 (a>0; 0


Xy lanh ngoaiChat

long

L

V du 4: o nht cua lu chat co the o bang cach dung mot xy lanh o o nht nh hnh ve. Dung cu nay bao gom mot xy lanh ben ngoai co nh va xy lanh ben trong quay vi van toc quay . Neu biet ngau lc can thiet e quay vi van toc th o nht cua chat long c nam gia hai xy lanh c xac nh. Hay thiet lap mot phng trnh lien he gia , , , L va Ro va Rl . Bo qua anh hng ay va cho s phan bo chat long van toc gia hai xy lanh latuyen tnh.

=

==

=

Giai :

=

=

Phan bo van toc tuyen tnh :

= U

Ri

0y

y= 0 u = 0, y = Ro-Ri u = U



CHNG 2 : TNH HOC LU CHAT

I. KHAI NIEM-Tnh tuyet oi : can bang bi duy nhat la trong lcTnh 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:AFp

rr =Ap suat thuy tnh tai mot iem

AFlimp

A

rr

=

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 lc2.3 Th nguyen cua ap suat

Th nguyen cua ap suat = = =

n v cua ap suat: N/m2 ( Pa) : n v chuan dung e tnh toanKgf / cm2 , at , m nc, mm Hg

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

r


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

Ap suat kh tri : 98100 N/m2

Ap suattuyet oi

Ap suat d(tng oi)

Ap suatchan khong

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



III. PHNG TRNH VI PHAN C BAN CUA CHAT LONG CAN BANG

+p

x

z

ydx

dy

dz

AB

cD

EF

GH

Hnh 2.2

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

Lc tac dung len khoi hnh hop theo phngX la :

Lc khoi : dx dy dz FxLc mat :

Tong lc phng X: dx dy dz Fx + p dy dz - (p+ dx) dy dz = 0

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

phng y => Fy - = 0 (2.4)

phng z => Fz - = 0 (2.5)

Viet di dang vector (2.6)= r

rLc khoi n v (Fx, Fy, Fz)

p dy dz (p+ dx) dy dz

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

Thay vao =

= p = - gz + C

-->

+ =

pB

pAh

A

B

ZB

ZA

Mat chuan

Ap dung cho 2 iem A va B : +=+

suy ra: +=

p + gz = constChat long, khong nen=constant

= r

= =

= =

+=Bai tap : 1) 2.12 2) 2.13 3) 2.14 4) 2.15 5) 2.17



Chat kh la kh ly tng: =

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):

= +Goi p0 la ap suat ng vi z=0:

aRg

aRg

T

pCCTp ==

aRg

TazTpp

=Phng trnh kh tnh:

= Chat kh, nen c constant =

= = =

= Tch phan

= +

V. NG DUNG PHNG TRNH THUY TNH 5.1 Ap ke

*Ap ke tuyet oi: o ap suat tuyet oi=

* Ap ke o chenh. +=+

+=+T (a) va (b) ta suy ra: ( ) ( ) ( )+=++ ma pM = pN

* Ap ke o chenh co 2 chat long

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

=

+

+

+=+ +=+

=

=

+

+



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 oi5.3 Bieu o phan bo ap suat

Suy lu n :Dung bieu o phan bo ap suat xac nh ap lc khi dien tch chu lchnh ch nhat co canh 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 Ghi chu: Ap lc tac dung len mat phang chnh la the tch bieu o phan bo ap suat, vaphng nay s dung thuan li khi mat chu lc la hnh ch nhat va co canh song songvi mat thoang

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 OX A ydA=

Do o

( )+==

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

( ) ( ) +=+==

( ) +=+= +=

+=



yc

xc

x

iem at : D ( yD va xD) cua FXac nh yD :

- Moment cua F oi vi truc OXMox = 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 = pdA y= ( hdA)y = y2 sin dA

Vay moment cua F oi vi truc OX la :

=: moment quan tnh cua A /v OX =

(2.13)=(2.12) va (2.13) : yc sin A. yD= sin IOX

Suy ra : (2.14)=Moment quan tnh /v truc ox co the tnh tmoment quan tnh /v truc i ngang qua trongtam C theo cong thc

+= += = (2.15)

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

=

+=

trng hp ap suat tren mat thoang p0=0

Toa o xD : khong can xac nh neu dien tch A co mot truc oi xng song song v 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 apsuat po 0

Bai tap : 1) 2.11 2) 2.22 3) 2.24 4) 2.25 5) 2.26 6) 2.27



6.2 Ap lc chat long len mat cong:

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

Lc tac dung len mat cong tong quat: ++=Trng hp ab // oy nen Fy = 0, tm Fx va FzAp lc dF tren dien tch vi phan dA : dF = p. dA

Chieu dF tren phng ox dFx = p. dA sin = p. dAxDo o Fx =

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

Tng t , chieu dF len phng oz:

do o Fz = 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 l c 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

= Fz =



Th d 2: cho mat cong nh hnh ve, co be rong b

(?) 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, Fyva Fz

Bai tap : 1) 2.30 2) 2.31 3) 2.33 4) 2.35 5) 2.38



6.3 Lc ay Archimede:

V

A B

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.

VII. S CAN BANG MOT VAT TRONG CHAT LONG:

Can bang on nh

7.1 Vat ngap hoan toan trong chat long :

C tren D

C d i D

Can bang khong on nh



7.2 Vat ngap mot phan trong chat long :

Tam nh khuynh M n m trong CDMD c xac nh : =

Iyy: moment quan tnh cua mat noi oi vi truc quay yyW : The tch vat chm trong chat long

C tren D Tam nh khuynh M nam ngoai CD

C tren D

Can bang on nh

Can bang khong on nh

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

Bai tap: 1) 2.39 2)2.42

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:

==

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

=>= => p = -ax + f(z)

==>=

va t tren =

= => f = -gz +C1thay f vao p = -ax - gz + C1Phng trnh mat ang ap: dp = 0 => =

++

Thay cac ga tr +=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 l i ngang qua trung iem M c

= 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.

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

Phng trnh phan bo ap suat

+= Phng trnh mat ang ap +=

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.

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



T

Hnh 1.2

11

Cau 1: Hay cho biet cac ap suat nao bang nhau trong th nghiem sau ay va mcchat long nao ngang nhau neu tat ca cac ong eu co ng knh giong nhau

Cau 2 Hay cho biet bieu o phan bo ap suat tuyet oi nao sau ay la ung:

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

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

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



Cau 3: 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

F2nc

dau

F3

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

Cau 4. 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.

paCau 5 Hay tm hieu vat ap lc W trong cac hnh sau

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



ph

Cau 8: Mot thung cha nc co mat thoang tiep xuc vi kh tri trc tren mot mat nghieng co goc hp vi mat phang ngang mot goc . Chon chieu chuyen ong la chieu dng va goi a la gia toc cua thung cha th mat thoang cua chat long se nam ngang khi: a) a = 0 b) a = gc) a = g d) Ca 3 eu sai



Cau 9. 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

Qua cau

A

Cau 10: Mot ong hnh ch U, mot au bt kn va mot aue h tiep xuc vi kh tri. Khi ng yen mc nc trongbnh nam ngang nh hnh ve. Neu bnh quay tron qua tructhang ng oi xng vi van toc quay th ap suat tai A sovi luc ng yen se la :a)Nho hnb) Ln hnc) Khong oid) Cha xac nh con phu thuoc vao van toc quay



ap



ong ho o ap suat


V D TNH TON XC NH P LC TRN MT PHNG p lc sinh ra do p sut, do xc nh p lc bt buc phi bit p sut trn b mt . Ty theo bi tan m p dng phng php dng biu phn b p sut hoc dng cng thc tng qut. 1.Khi mt phng c din tch hnh ch nht v c cnh song song vi mt thang nn s dng phng php biu phn b p sut xc nh lc v im t. Th d: Tm p lc tc dng trn mt hnh ch nht PQRS cnh a,b v nghing mt gc nh hnh v. Cho a, b, H1, H2 v .

P H1

H2 Q b

SR

Q

S R

P pQ

pR

a

b

D d

Mt PQRS l mt mt phng hnh ch nht, c cnh PQ song song vi mt thong nn ta s dng phng php biu phn b p sut xc nh p lc. Biu phn b p sut trn mt PQRS nh hnh (b), p sut ti P v Q l : pQ = H1 v pR = (H1 + H2) Do p lc tc dng : abHHab

ppF RQ

2)2(

221 +=+=

im t lc D nm trn trc i xng v cch y mt on d l :

32

33

2

21

21 aHHHHa

pppp

dRQ

RQ

++=+

+= Ghi ch: Phng php dng biu p sut rt thun li khi bi ton c nhiu yu t phc tp. Nh trn mt thong khng phi p sut kh tri m l p sut po v c nhiu loi cht lng 1, 2 nh hnh 2. Cch xc nh p lc khng c g thay i, ch cn tnh li p sut pQ v pR pQ = po + 1H1 v pR = po + 1H1 + 2H2 (N/m2)

abHHpabpp

F oRQ2

222

2211 ++=+= (N) v im t D cng xc nh tng t:

322

333

2

22110

22110 aHHpHHpa

pppp

dRQ

RQ

++++=+

+= (m) Ch : Cc trng hp trn u c th dng cng thc tng qut xc nh p lc v cho mt kt qu nh nhau.

P H1

H2 Q b

SR

Q

S R

P pQ

pR

a

1b

D d

po

2

2. Khi mt phng khng phi l mt ct hnh ch nht hoc l mt ct hnh ch nht nhng c cnh khng song song vi mt thong, khi xc nh p lc nn s dng cng thc tng qut. V d: Mt mt trn tm C bn knh r nm trn mt phng nghing mt gc nh hnh v. Bit khong cch t tm C n mt thong H v trng lng ring ca cht lng . Xc nh p lc tc dng trn mt trn.

H

r

o

c

y

x yc

D

yD

Trc i xng // vi oy

i vi mt phng hnh dng bt k phi dng cng thc tng qut xc nh p lc. p sut ti trng tm C ca din tch cn tnh lc, trong trng hp ny l mt trn bn knh r: pc = H (N/m2) p lc tc dng ln mt trn: F = pcA = H. r2 (N) xc nh im im t D, cn phi xc nh h trc ta xoy. Trc ox phi nm ngang vi mt thong ( p sut bng khng) v cng mt phng cha din tch cn tnh p lc. Trc oy thng gc vi ox v cng nm trong mt phng cha din tch cn tnh p lc. Ni cch khc, mt phng xoy l mt phng cha din tch cn tnh p lc v c trc ox ngang vi mt thong. p dng cng thc xc nh v tr im t D:

H

rH

rHrH

AyIyyC

oCD 4

sinsin

sin

64/)2(sin

2

2

4

+=+=+= (m)

Hay im D nm di im C mt on l :

H

rAy

IyyC

oCD 4

sin2 == (m) V hnh trn c trc i xng song song vi trc oy nn v tr im D phi nm trn trc i xng ny, do d dng xc nh im v tr D. Ch : Trong trng hp mt thong c p sut po khc khng, cn phi qui i mt thong v v tr p sut bng khng. Khi po > 0 tng v tr mt thong cao thm mt on p0/ , v khi po < 0 gim v tr mt thong xung mt on p0/. Lc v tr ca trc ox v gc O ta xoy phi xc nh theo v tr mt thong mi. V d trong hnh sau v tr ta xoy mi v v tr mt thong qui i khi p sut po > 0.

H

r

O

c

y

x

yc

D

yD

po/

Mt thong qui i c p sut trn mt thong bng khng

po>0 po=0

V D XC NH P LC TRN MT CONG p lc trn mt cong thng xc nh trn 3 phng Fx, Fy v Fz, tuy nhin thng thng cc mt cong thng gp c ng sinh song song vi trc oy nn Fy = 0, do ch cn xc nh Fx, Fz. Fx = Lc tc dng ln hnh chiu ca mt cong trn phng thng gc vi ox. Fz = VVAL (VVAL : th tch vt p lc ) V d: Xc nh lc tc dng ln mt cong ABCD l mt na mt tr bn knh r v di b nh hnh v. Trng lng ring ca cht lng l .

H

Po = 0

r

A

B

D

C

B

Ax

pc

b

2r

D

A

C

o

y

x

z

Mt cong ABCD c ng sinh AB song song vi trc oy nn p lc theo phng Fy = 0. Ch cn xc nh p lc Fx v Fz. xc nh lc Fx cn tm hnh chiu ca mt cong ABCD ln phng thng gc vi ox ( phng thng ng). Hnh chiu ca mt cong ABCD ln phng thng gc vi ox l mt phng hnh ch nht ABCD. Lc Fx chnh l lc tc dng ln hnh ch nht ny. Fx = pc . Ax = (H+r) b.2r xc nh lc Fz cn phi tm th tch vt p lc ca mt cong ABCD. Trn mt cong ABCD p sut thay i hng tc dng, t AB n MN p sut hng ln v t MN n CD p sut hng xung di, do phi tch mt cong ra lm 2 phn xc nh vt p lc. Vt p lc mt cong ABMN gii hn bi mt cong ABMN v cc mt bn thng ng ta vo cc cnh ca mt cong AB, BN, NM, MA ri ko di ln cho n khi gp phn ni di ca mt thong. Do vt p lc mt cong ABMN bng din tch mt phng AMMA nhn vi b. Vval(AMBN) = din tch (AMMA) .b Nu chn chiu dng hng ln th vt p lc ca mt cong ABMN c du dng v cht lng di v p sut hng ln trn. Tng t vt p lc cho mt cong MNCD l th tch gii hn bi mt cong MNCD v cc mt bn thng ng ta vo cnh mt cong MN, NC, CD, DM ri ko di ln cho n khi gp phn ni di ca mt thong. Do vt p lc mt cong MNCD bng din tch mt phng AMMD nhn vi b. V chiu dng hng ln nn vt p lc ca mt cong ABMN c du m v cht lng trn v p sut hng xung. Do

Vval(MNCB) = - din tch (AMMD) . b

Tng hp vt p lc cho ton b mt cong ABCD Vval = din tch (AMMA) .b - din tch (AMMD) . b = [din tch (AMMA) - din tch (AMMD)] b [din tch (AMMA) - din tch (AMMD)] chnh l din tch na hnh trn AMD bn knh r. Do vt p lc cho ton b mt cong ABCD l : Vval= -[r2/2]. b V lc Fz : Fz = -[r2/2]. b Tng lc tc dng ln mt cong:

( ) ( )22222 2/)(2 rrHrbFFF zx ++=+= Ch : d dng tnh ton vt p lc, trong cc trng hp c Fy = 0, ch cn v mt cong trong mt phng. V d trong trng hp trn mt cong ABCD c v mt ng AMD v vt p lc c xc nh qua din tch. Vt p lc ca mt cong AM l (din tch AMMA ) x b v vt p lc ca mt cong MD l ( din tch AMMD)x b. Vt p lc ca mt cong AMB bng tng vt p lc ca mt cong AM v MD hay chnh l din tch AMD x b = ( r2. b/2) .

Po = 0

r

A

B

D

C

H

M

B

N

M

N A

A M

B N

D

B

NA

A M

B N

+

C

M

b b

-

+ Chiu dng

Po = 0

r

A

D

H

M

AM

+

-

A M MA

A

M

D

+ =M

- r

A

M

D

Phng v chiu ca lc tc dng ln mt cong ch c th xc nh theo tng trng hp c th. Trong v d trn phng v chiu c th xc nh nh sau:

Po = 0

A

D

H

O F Fx

Fz

V p sut lun thng gc vi mt cong nn p lc cng thng gc vi mt cong. Trong trng hp ny mt cong l mt na mt tr nn tng lc F phi i qua tm O ca mt tr v nghing mt gc c xc nh bng, tg = Fz/Fx


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.

( )t,rfr orr = (3.1)hay ( )t,z,y,xxx ooo= ( )t,z,y,xyy ooo= ( )t,z,y,xzz ooo=

Van toc =rrrr ==

dtdzu ;

dtdyu ;

dtdxu zyx ===

===

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)

Gia toc

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=u rrva cac thanh phan

( )= ( )= ( )=

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

Gia toc cua chuyen ong : =rr

tren phng x:

tren phng y:

tren phng z:

Gia toc oi lu Gia toccuc bo

+

++

=+

++

=+

++

=



II. MOT SO KHAI NIEM2.1 ng dong : ng cong i qua cac phan t chat long co cac vector van tocla tiep tuyen vi ng cong o.

Phng trnh vi phan cua ng dong

= =

+ Hai ng dong khong cat nhau+ Trong chuyen ong on nh , ng dong trung vi qui ao

Tnh ch t:

2.2 Dong nguyen to :Dong nguyen toXet dien tch dA, cac ng dong bao quanh chu vi

dien tch dA tao thanh mot ong dong, chat long dichuyen trong ong dong c goi la dong nguyen to

Lu chat di chuyen trong dong nguyen to th khongi ra khoi va lu chat ben ngoai cung khong i vaodong nguyen to

ng dong

II. MOT SO KHAI NIEM2.1 ng dong : ng cong i qua cac phan t chat long co cac vector van tocla tiep tuyen vi ng cong o.

Phng trnh vi phan cua ng dong

= =

+ Hai ng dong khong cat nhau+ Trong chuyen ong on nh , ng dong trung vi qui ao

Tnh ch t:

2.2 Dong nguyen to :Dong nguyen toXet dien tch dA, cac ng dong bao quanh chu vi

dien tch dA tao thanh mot ong dong, chat long dichuyen trong ong dong c goi la dong nguyen to

Lu chat di chuyen trong dong nguyen to th khongi ra khoi va lu chat ben ngoai cung khong i vaodong nguyen to

ng dong



2.3 Dien tch t - Chu vi ut Ban knh thuy lcDien tch t la dien tch thang goc vi cac ng dong va cha chat long

Chu vi t phan tiep xuc vi chat long va thanh ran

Ban knh thuy lc : t so gia dien tch t va chu vi t

=+=++====

2.4 Lu lngThe tch chat long i qua mat cat t trong mot n v thi gian (m3/s)

Khi lu lng tnh theo khoi lng(kg/s)

= Am udAQNh n xet: T (3.5) cho thay lu lng chnh la the tch cua bieu o phan bo van toc

= AudAQ (3.5)

2.5 Van toc trung bnh:. =

Bieu o phan bovan toc

+==

III. PHAN LOAI CHUYEN ONG3.1 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)

= =

* 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 Reynolds

= = : He so nhtong hoc



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

r=4.2. ao ham cua mot tch phan khoi

= W dWX (Tch phan kh i)( )= (ao ham cua mot tch phan kh i)

W

( ) ( ) +++=

Tai thi iem t1 +=Thi iem t2 = +

Trong thi gian t, co s bien oi

( ) ( )++=( ) ( ) += ++( ) ( )

+

= ++

ao ham theo t

Trong o

= SCt dAn.utX rr = SAt dAn.utX rrvat

dAn.utdAn.utlim

tXX

lim SSt

AtCt

t

+=

rrrr

+= rr

( ) ( )=

++

== + SSS dAn.udAn.u rrrr Thay (3.8), (3.9) vao (3.7)

Do o (2)

44 344 21444 3444 21 +

==

++

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

W

+= rrrr

+= rr



V. PHNG TRNH LIEN TUC

Bao toan khoi lng =Ap dung phng phap the tch kiem soat

=

=+= SCVHT

dAn.utm

dtdm rr

=+ SW dAn.udWt rr

( ) =+ WW dWudivdWt r( ) =

+W dWudivt r ( ) =+ udivt

r

(PT lien tuc)

Bien oi Gauss

=

0=+ SW dAn.udWt rr

* Chat long khong nen c: =udiv r

Hay

+ + =

Trong toa o cc ( ) =++zuu

rru

rrz

r

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

=++= A SAS b dAn.udAn.udAn.udAn.u rrrrrrrr =+ SW dAn.udWt rr

=+ AA dAn.udAn.u rrrr

= ==

= = =Chat long khong nen c: = =

= const)

==

( ) =+ udivt

r

021 =+ mm QQ



V PHAN TCH CHUYEN ONG CUA LU CHAT:

Van toc quay: uRot rr =

zyx uuuzyx

kjirrr

=

=zu

yu yz

x

=xu

zu zx

y

=yu

xu xy

z

Chuyen

ong

1. Tnh tien

2. Quay

3. Bien dang

Mot chuyen ongkhong quay th :

x = y = y = 0

Van toc bien dang

+

+

+

+=

+

++

+

=

+

+

+

+

=

=r

Bai tap : 1) 3.6 2) 3.7 3) 3.9 4) 3.11 5) 3.12 6) 3.14

Thu du 1: Xac nh ng dong cua mot dong chay co : ux = 2y va uy = 4x

=

=

==

+=

=



Th du 2 : ng dung phng trnh lien tuc viet cac lien he gia cac lulng

V du :

Cau 3. Mot bm hng truc at trong mot ng ong nh hnh ve. ng knh ong khong oi. Khi bm hoat ong chong chong se quay va bm nc vao , luc o van toc :

*a) V1 = V2 V1V2b) V1 > V2c) V1 < V2d) V1 = V2 ch khi xem la chat long 1y tng

Cau1: Mot dong chay co van toc tren phng x va y nh sau :ux = 5x(1+y) va uy = 5y(-1+x)

ay la mot chuyen onga) Khong on nh , 1 chieu b) Khong on nh , 2 chieuc) Khong on nh , 3 chieu d) Ca 3 eu sai

Cau 2. Dong chay nao sau thuoc chuyen ong on nha) Dong chay trong cong thoat nc ma*b) Dong chay trong ong di ap lc cua may bmc) Dong chay trong kenh thien nhien d) Ca ba eu sai



Cau 4 Mot dong chay co khoi lng rieng i qua mat cong S vi lu toc . Khoi lng lu chat i qua mat S trong thi gian dt se la:

a) ( vect phap tuyen cua mat ds )

b) ( la hnh chieu cua len phng thang goc cua mat ds )

c) ( la hnh chieu cua len phng tiep tuyen cua mat ds )

*d) Ca a) va b) eu ung

rr r

r

r

Cau5 Chuyen ong nao sau ay la chuyen ong khong quay:a) Ux = 4y(1+2t) Uy = -4x(1+2t) b) Ux = 4y(1+2t) Uy = 4x(1+2t) c) Ux = 4y(1+2t) Uy = 4x(1-2t) *d) Ux = 4y (-1+2t) Uy = 4x(1-2t)

ng dong qua mot xe ang chuyen ong



Chng 4 ONG LC HOC LU CHAT

I. PHNG TRNH VI PHAN CHO CHAT LONG LY TNG CHUYEN ONG (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 =>

+ Vi : lc khoi n vp(x,y,z,t) : ap suatu (x,y,z,t) : van toca (x,y,z) : gia t c :kh i l ng rieng

Fr

vr =

rr =


Neu viet tren phng x th :

dtdu

xpF xx =

+

++

=

them vao ve phai

+

++

=

xuu

xu

uxuu

xu

u zzy

yz

zy

y

+

rr =

sau khi bien oi, ta co:tu

yu

xu

uxu

zuu

uuuxx

pF xxyyzxzzyx

x +

+

++

=

[ ]

tuuu

uxx

pF xzyyzx ++

=

va viet di dang vector = + + ruuuuur uuuurr r r

pt Euler dang Lam-Gromeko

Tng t tren phng y [ ] = + + tren phng z = + +



uxugradtugradpF rrrr ++

=

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

= uuuurr

ngha la : =

= =

va (x,y,z) c goi la ham theTh 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 =

+= Neu chat long khong nen c: = const th : += Thay vao phng trnh Lamb Gromeko :

++ = + ruuuur 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:

=r

Chuyen ong khong quay =r

+ + + = uuuur

=+++

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

=++

+ + = + ruuuur r r ( ) + + =

uuuur uuuur

+ + = uuuur uuuur

i u ki n: (i) L u ch t ly t ng, (ii) khong quay, (iii) on nh, (iv) khong nen c , (v) anh hngduy nhat la trong lc



2. Chuyen ong on nh, tch phan doc theo ung dong:

Chuyen ong on nh : = r

+ + = uuuur r r

+ + = uuuur uur uurr r

marr

+ + = uuuur uur =

++

=++Neu chat long khong nen c va ch chu anh hng duy nhat la trong lc th thay (4.12) chotren mot ng dong la

=++

rr

rv

Ph ng trnh Bernoulli

i u ki n: (i) L u ch t ly t ng, (ii) tren 1 ng dong , (iii) on nh, (iv) khong nen c , (v) anhhng duy nhat la trong lc

L y la m t oan tren ng dong. va tiep xuc nhau Nhan hai ve cho

v

3. Chuyen ong on nh tch phan doc theo ng xoay:

ng xoay la cong i qua cac iem co vector van toc xoay la tiep tuyen.

Tng t nh tren ng dong, nhan 2 ve , la mot vector vi phan tren ng xoay

+ + = uuuur uur uurr r

ma tren ng xoay rr + + =

uuuur uur =

++

=++Neu chat long khong nen c va ch chu anh hng duy nhat la trong lc th thay (4.12) chotren mot ng xoay la

=++ i u ki n: (i) L u ch t ly t ng, (ii) tren 1 ng

xoay , (iii) on nh, (iv) khong nen c , (v) anhhng duy nhat la trong lc



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

Ta co : rr = rrr +=

rrr +

+

= Thay vao pt Euler:

( ) rr +

+

=+ Nhan 2 ve cho ( )[ ] + +=+ rr

( ) =+

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

=

+

( ) =+

khi nhng ng dong thang song song th R hay =

+

=+

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

r

Vector n v tren phng s va nvr

vr =

v=

Thi du tr ng hp co ng dong song song

1

1

2

2

Tren mat cat 1-1 va 2-2 ap suat phan bo theo qui luat thuy tnh



CV

III. PHNG TRNH NANG LNGXet the tch kiem soat W, bao quanhdien tch A. ai lng nghien cu la nang lng

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

+=

++

+

= rr

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

+=

=

Khong co strao oi nhiet +=

P: cong suat cung cap cho he thong, nhiet lng them vao trong 1 n v thi gian

dw

=

S bien thien nang lng trong mot n v thi gian, ( ap dung cong thc tnh ao ham cua mot tch phan khoi)

Khong co s trao oi nhiet =

P, cong suat do lc tacdung tren dien tch A bao quanh the tch kiemsoat, gom

( ) ( ) += rrrv

( ) ( ) +=

++

+

rrrvrr Thay vao:

sap xep lai, -- > Phng trnh nang lng dang tong quat :

( ) =

+++

+

rrrr

Chuyen ong on nh: ( ) =

++ rrrr

CV

dA

++

+

= rr

A

Trong o a biet:

Cong suat do ap lc( ) rv

( ) rr Cong suat do lc ma sat



++=

++ rrrrrr

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

( ) =

+++

+++

++ rrrrrrrr

V1

p2V2

p1

Mat chuan

z1z2

A1 A2

( ) =

++ rrrr

Sb

( ) =

+++

+++

++ rrrrrrrr

Th mat A bao gom A1, A2, va Sb nen

c viet lai

Ma tren Sb , va vuong gocr r

( ) =

+++

++ rrrrrr

Xet tch phan

++=

++= rr

Trong o 1 , 2 c goila he so sa cha (hieuchnh) ong nang : =

dong chay roi : 1

= rrrr

++=

++ rrrrrr

Tng t taimat cat A2cung co

++=

++ rr

dong chay tang : >1

Hang so ng dongthang song song

++=

++ rrrrrr



=

+++

++ rr

++=++rr

Thay vao phng trnh nang lng:

at : =rr

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

+++=++

( ) =

+++

++ rrrrrr

Phng trnhnang lng chotrng hp

(i) Chuyen ong on nh, (ii) Khong trao oi nhiet, (iii) Tai matA1, , A2 ng dong thang song song, (iv) Chat long khong nenva (v) Lu lng tai mat cat A1 va A2 bang nhau.

Xem chat long khong nen (1= 2 = ) va Q1 = Q2= Q . Chia 2 ve cho gQ va chu yg=

V nang cua mot n v trong lng chat long

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 )+Nang lng cua dong chay tnh tren mot n v trong lng chat long++= ( 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)=

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

11

21

1 2z

gp

gV ++ 212

22

22 2

+++= fhzgp

gV

+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)

QHB (watt) QHT (watt)

+++=++



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

++= +=

IV. NG DUNG PHNG TRNH NG L C VA PT NANG LNG1. o lu toc ( ong Pitot )

Ap dung phng trnh Bernoulli cho iem 1 va 2, xem lu chat ly tng

++=++ ==

Ap dung tnh chat ap suat tnh hoc tuyet oi

+=+ +=+

Suy ra += =

=Thay vao tren [ ] ==

=

m/ch

, chatlong

kh

kh

khM

N

p2p1

p1

p1

Ong Pitot



Ongpitot

Ong Pitot c lap tren may bay e ovan toc cua may bay (1knot = 1,852km/h)

=

CI : H s ng PitotCI > 1, N u ch t l ng ly tng CI= 1

=

Thong thng k> 1 nen

=

e xet en anh hng cua s ma satcua dong kh, he so CI c them vaoe ieu chnh van toc tnh bang congthc tren

Hay giai thch tai sao CI > 1 khi xet en s ma sat cua dong kh ?

2. o lu lng ( ong Ventury )

(C


Van toc qua lo tai mat cat co hep :

=Lu lng qua lo : =

Va at he so lu lng CD = . Cv, th

Ap dung phng trnh nang lng cho 2 mat cat 1-1 va mat cat co hep c-c (tai ay ng dong song song)

+++=++ Chon mat chuan i qua tam lo. Ap suat tai mat thoang va mat cat co hep la ap suat kh tri , p1= pc = 0. Van toc V1tai mat thoang rat nho nen ong nang xem nh bang khong ( V12/2g = 0)

+=Neu bo qua ton that nang lng hf1-c = 0

=Neu xet en ton that nang lng

Vi Cv he so lu toc(Cv < 1)

Vi Ac tiet dien mat cat co hep

Neu he so co hep : = Ac = ALu lng qua lo : =

3. Xac nh lu lng qua mot lo

Ac

H

1 1

C

C

VcMat chuan A

4. Xac nh cong suat mot may bm

Cong suat phai cung cap cho dong chay P = QHbNeu bm co hieu suat la (%) th cong suat cua may bm Pb = QHb /

2

H

bm

Q

Mat chuan

1 1

2V2

Ap dung phng trnh nang lng cho 2 mat cat 1-1 va mat cat 2-2

+++=+++ Chon mat chuan i qua tam lo. Ap suat tai mat thoang va mat cat 2-2 la ap suat kh tri , p1= p2 = 0. Van toc V1 tai mat thoang rat nho nen ong nang xem nh bang khong ( V12/2g = 0)

++=Hb :la nang lng ma mot n v trong lng cua dong chay phai nhan

Bai tap : 1) 4.16 2)4.18 3) 4.19 4) 4.21 5) 4.23 6) 4.24 7) 4.30 8) 4.319) 4.42 10) 4.43



5. Xac nh chieu cao at may bm

e tranh hien tng kh thc, chieu cao at bm Zb < [Zb]

Ap dung phng trnh nang lng cho 2 mat cat 1-1 va mat cat 2-2 ( mat cat ngay trc ca vao may bm)

+++=++ Chon mat chuan ngang mat thoang be hut. Ap suat tai mat thoang va mat cat 2-2 la ap suat kh tri , p1= 0. Van toc V1 tai mat thoang rat nho nen ong nang xem nh bang khong ( V12/2g = 0). Ap suat p2 la ap suat chan khong cho phep [pck]trong buong bm e khong xuat hien hien tng kh thc. Do o chieu cao cho phep [Zb] cua bm c xac nh nh sau:

[ ] [ ] ++=

bm

Mat chuan1 1

2

2[Zb]

[ ] [ ] = C i tien may bm ly tam, e tang chieu cao at bm (xem bai giang trong lp)

Khi may bm ly tam at qua cao so vi mat nc be hut, bm se tao mot ap suat chan khong rat ln e hut nc do ose xuat hien hien tng kh thc trong bm gay h hong may bm. e tranh hien tng kh thc can phai xac nh chieu cao toi a at bm cho phep [Zb].

V. PHNG TRNH ONG LNG :ai lng nghien cu trong the tch kiem soat la ong lng.

CV

ong lng n v : r=

== rvr ong lng trong the tch kiem soat la

Ap dung nh luat ve ong lng:

= rr

= rr

=+rvrrr

Phng trnh ong lng =+rvrrr



V1V2

C.V- Chuyen ong on nh

=+rvrrr

=++ rvrrvrrvrr =+ rvrrvrr

=+ rvrrvrr o1 , o2 la he so sa cha ong lng

==

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

= rrr L ra L vao Tong lc

r (Lc khoi :Trong lng, ) r (Lc mat :Ap lc, lc ma sat, ..)

Xet mot chuyen ong

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

Neu Q1 = Q2 = Q = rrr = rrr VI.NG DUNG PHNG TRNH ONG LNG

1. Lc cua tia nc tac dung tren mot tam phangnghieng goc , van toc va lu lng en V1va Q1. Xem trong lng tia nc khong ang ke

2. Lc cua dong chay tac dung len mot tam chan cobe rong bang 1 n v, lu lng q va o sau h1, h2 . B qua ma sat ay

V1Q1

F =?

F = Q1 V1sin

Fqqh1

h2

( )

=

Bai tap : 1) 4.34 2) 4. 36 3)4.38 4)4.415) 4.42



Cau 1 : Mot ng ong dan nc vi p1 va p2 la ap suat trong ong nho va ong ln tai cac v tr nh hnh ve. Neu boqua ton that nang lng ta co :

*a) p1 < p2b) p1 = p2c) p1 > p2d) Cha the ket luan c v con phu thuoc vao

trang thai chay trong ong

Cau 2: He so sa cha ong nang ( ) trong phng trnh nang lng dung e :a) ieu chnh so hang ong nang do s mat nang trong dong chay*b) ieu chnh so han ong nang do s dung van toc trung bnh thay cho van

toc thc cua dong chayc) ieu chnh so han ong nang do s dung ap suat d thay cho ap suat tuyet

oid) Ca 3 eu sai

Cau 3 Mot dong song mat cat hnh ch nhat co be rong b, chieu sau h va chay vi van toc trung bnh V. Nang lng cua mot n v trong lng chat long cua dong chay ( lay mat chuan ngang ay song va=1):a) Vbh (V2/2g + h) *b) V2/2g + h c) (V2/2g + h) d) V2/2g

vi la trong lng rieng cua nc

Cau 4: Trong mot chuyen ong nam ngang, gia tr u2/2g + p = const tai tat ca moi iem khi :

a) Chuyen ong chat long ly tng, khong quay, on nh hoac khong on nh*b) Chuyen ong chat long ly tng, khong quay va on nhc) Chuyen ong chat long ly tng, on nh ( co the quay hoac khong quay)d) Ca 3 eu sai

++=++

Cau 5: Tai hai iem 1 va 2 trong dong chay khong quay, e thoa man c phng trnh:

th ieu kien nao sau ay khong can:a) Chat long khong nenb) Anh hng duy nhat la trong lc*c) iem 1 va 2 phai nam tren mot ng dong d) Ca 3 eu can



Cau 6: Mot ng ong dan nc nh hnh ve. Goi cot nc nang lng H = V2/2g + p/ + z va Q la lu lng. Neu bo qua ton that nang lng th tai 3 mat cat 1, 2, 3 ta co:a. H1 = H2 = H3b. Q1H1 = Q2H2 = Q3H3c. Q1H1 > Q2H2 + Q3H3*d. Ca 3 eu sai

Cau 7: Phng trnh Euler mo ta chuyen ong cua chat long nh sau:

=+ vrrr

trong o r van toc, r lc khoi, p ap suat, khoi lng rieng. rse co thanh phan la : a) (0,0,-z) *b) (0,0,-g) c) (0,0,-) d) Ca 3 eu sai

Neu chat long chuyen ong ch b tac ong bi trong lc th vector

May bm ly tam




Cau tao bo phancai tien may bmly tam co the hutsau

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



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

=01

1

2

2

Mat chuan

z1z2

L

=+

=+

V i J = hd / L , d c n ng l ng ng suat tiep ty le bac nhat theo r

PT c ban co the viet ==

Phng trnh c ban cua dong eu

+++=++ = =

=++

=+

RLhd

=LhR d=

ro r

dALc tac dung tren phngdong chay ( phng s) :

PT N ngl ng (1-2)

=++ (1)

=++ (2)

(1) Va (2) ->



III.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

=

Phan bo van toc trong chay tang co dang Parabol

drr

r0

= =

= = CrdrJu +

= 2+= =

( )=

r0

Xet dien tnh vi phan hnh vanh khan ban knh r , chieu day dr

ro

Lu lng va van toc trung bnh:

Ton that doc ng

Thay J = hd/L

=

Vi Re = VD/ ( He so Reynolds)

=T

Suy ra hd

=r

dA

= ( ) =( ) = =

( )=

==

= =

=

sap xep lai=

Tren dien tnh vi phan hnh vanh khan ban knh r , chieu day dr

Lu lng qua hnh vanh khan vi phan



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: =

vi c goi la h s nht roi =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 ong

=

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

Neu xem t le tuyen tnh vi ban knh r :

= Th

=

Thay vao :

=

Thay vao (1) :

=

=

=

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

T (2)

=

at = ( van toc ma sat , m/s)

= =

= + CTai tam ong y = ro u = umax thay vao cho =

=

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

y

u

ro

o

o

ng cong logarit

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) khi ch y r i co the lay bang 1

0 < y ro



Xac nh he so ton that :Dong chay tang:

Dong chay roi:

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

Prandtl-Nicuradse:

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

0,251000,1 1,46D Re

Colebrook: = + 1 2,512lg

3,71.D ReChay roi thanh hoan toan nham (khu sc can bnh phng) (Re rat ln >4.106 = f( /D).

Prandtl-Nicuradse:

+=D,lg,Dlg 173214121

= 2

dL VhD 2g

i v i dong r i t ly thuyet khong the suy ra c tonthat doc ng. Dung phng phap phan tch th nguyenva th nghiem chng to c ton that doc ung co dang

=

( )=

Ton that doc ng trong dong chay roi:

= hd t l V1

hd t l V2

0,000 01

0,000 0050,000 007

0,000 05 0,000 1 0,000 2

0,000 4 0,000 6

0,001

0,002

0,004 0,006 0,008 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 0,09 0,1

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

thanh nham

Khu chay roithanh trn

Khu chuyen tiep

=/ D_

O TH MOODY

==



1.Ton that ng dai: Cong thc tnh ton that doc ung co dang

(Darcy)= 2

dL VhD 2g

= f(Re, /D) : he so ton that : He so nham tuyet oi (chieu cao cac mo nham )

thay D = 4R =

vi J = hd/L = va at = ( he so Chezy) = ( Cong thc Chezy)

lu lng == goi la module lu lng=He so Chezy C co the tnhtheo cong thc Manning : = ( n la h s nham, tra bang

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

=T cong th tnh lu lng ==

trong o

Neu tnh C bang cong thc Manning th : =Neu ong tron th :

=

V. TNH TOAN TON THAT CUA DONG CHAY TRONG ONG

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

gVh cc =

c 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 ngot

=

=

vi V1

vi V2

mieng ra cua ong: c=1

mieng vao cua ong: c=0,5

Hai cong thc tren c chng minh t ly thuyet

A1

V1

A2

V2

Tham khao: So tay thuy kh ong lc hoc ng dung-Hoang ba Ch va Trng ngoc Tuan



VI. CAC BAI TOAN TRONG NG ONG

1. ng ong mac noi tiep

BfABBB

AAA hzp

gVzp

gV

+++=++ 2222

++++

++=

++++++=

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

l1; d1; 1l2; d2; 2

l3; d3; 3

h

A A

B BMat chuan

V1V2 V3

+=

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

Goi HA va HB 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 :

Tng t, xet dong chay t A en B tren ong 2 va 3 ton that ong 2 va 3 la :

Nh vay

Neu bo qua ton that cuc bo : ==

va

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

= (i)

= (ii)

(iii)

Q

Q1?

Q2?

Q3 ?



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

l1; d1; n1 l2; d2; n2

l3; d3; n3

J

12

3

Z1

Z2

Matchuan

Hj

Hj

ng nang ga sng nang

Hj

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

Hj

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

Ton that tren ong 1 = ==

Ton that tren ong 2 = =Ton that tren ong 3 = = =

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



l1; d1; n1 l2; d2; n2

l3; d3; n3

J

12

3

Z1

Z2

Matchuan

Hj

Th du trng hp 1 xay ra, Q1 > Q3

Ton that tren ong 1 : = =

Ton that tren ong 2 = =

Ton that tren ong 3 = =

Q1 Q2

Q3

4. Mang ng ong kn:

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 i2. Trong mot vong kn, tong ton that phai bang khong

Qui 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



Phng phap Hardy Cross

Ap dung cho nhng cong thc tnh ton that doc dng co dang

Th du = 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 2Ton that nang lng tren ong i khi a ieu chnh la

hdi = 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

( ) =+=

vi k la so ong trong mot vong

=+ =

=

=

== k

i

xii

k

i

xii

Qmx

QmQ

1

1

1

Bai tap : 1) 8.13 2) 8.15 3) 8.22 4)8.23 5) 8.25 6) 8.26

Cau 1. Hai ong noi tiep co ng knh nh nhau nhng bang vat lieu khac nhau. Mot lu lng kha ln chay trong ong, ngi ta thay he so ton that () tren hai ong eu bang nhau, luc o trang thai chay trong ong la :* a) Chay roi thanh trn

b) Chay roi thanh nhamc) Chay roi thanh hoan toan nhamd) Ca ba eu sai

Cau 2. Hai ng ong cung loai vat lieu, ong 1 dai hn ong 2 nhng co ng knh ln hn ong 2. Neu cung mot lu lng chay tren hai ong th :a)o doc ng nang ong 1 ln hn ong 2 *b)o doc ng nang ong 1 nho hn ong 2 c)o doc ng nang ong 1 bang ong 2 d)o doc ng nang ong 1 bang ong 2 neu chay tang

Cau 3. He so ton that doc ng () c xac nh banga) Ly thuyetb) Thc nghiemc) Ket hp ly thuyet va thc nghiem*d) Ch xac nh bang ly thuyet cho trng hp chay tang con tat ca cac

trng hp khac c xac nh bang thc nghiem.

Cau hoi trac nghiem



ng nangCau 4: Mot he thong 3 bon cha vi ng nang nh hnh ve th lu lng Q, Q1, Q2, Q3 se la :

a) Q = Q1 + Q2 + Q3*b) Q + Q1 = Q2 + Q3c) Q + Q3 = Q1 + Q2d) Ca 3 eu sai

Cau 5. ng ong chay co ap nh hnh ve, neu ap suat pB > pA , th se:a) Chay t B ve A b) Chay t B ve A neu xem ma sat khong ang kec) Chay t A ve B

*d) Co the chay t A ve B hoac t B ve A

Cau 5: Mot he thong 3 bon cha nh hnh ve co cac ton that cuc bo nho khong ang ke. Neu tai J ngi tagan mot ong thong vi kh tri e quan sat mc nc trong ong. Hay chon cau ung sau:a) Neu mc nc B, bon 1 se chay ve bon 2 va 3b) Neu mc nc A, bon 1 va 2 se chay ve bon 3c) Neu mc nc B, bon 2 se chay ve bon 1 va 3d) Ca 3 eu sai

12

3

J

A B

Th d 1: M t b m nh hnh v cung c p cho dong ch y cong su t P = 25 KW vi lu lng Q=0,04 m3/s. Biet ong co chieu dai L = 30 m, ng knh D = 60 mm, va he so t n th t =0,016. Xac nh lu lng chay trong ong neu khong lap bm vao. Bo qua ton that cuc bo



V du 2: Mot he thong nh hnh ve, biet ng knh cac ong nh nhau D = 0,10 m va co he so ma sat =0,02. Chieu dai ng ong va mc nc trong thung cho tren hnh ve. Xac nh lu lng chay vao thung B cho hai trng hp ong va m van thung C.

Z = 15m

Z = 0

ng knh tat ca cac ong D = 0,10m



r khong phu thuoc vao ng i t A en B

Chng 6 THE LU

Chuy n ong the la mot chuyen ongkhong quay

e ieu kien tren thoa man, can co motham (x,y) sao cho

=r =

(x,y) : Ham the van toc

Ngoai ra

I. NH NGHA THE LU:

vr === r

= va


Trong toa o descarde

yu;

xu yx

==

Trong toa o cc

=

=r

u;r

ur

+ Phng trnh ng ang the: Khi cho = Const => ng ang the

II. MOT SO KHAI NIEM1. Ham the van toc ():

=+

PT Laplace

==

Ham so the thoa man Phng trnh Laplace

uru

u

x

y

r

0



xu;

yu yx

==

ru;

rur

== 1

trong toa o Descarte trong toa o cc

=+

yx

+ Trong chuyen ong the thoa man phng trnh Laplace

=

Luon luon co

So sanh vi pt lien tuc =+

Luon tm c (x,y)

Mot so tnh chat cua ham dong:

+Trong bat ky dong chay nao cung co the tm c ham dong

yux

= xuy =

T =

0=)u(Rot v

=+

yx

2. Ham dong () :

=

-Khi cho = C th ay chnh la phng trnh mot ng dong

x

y

o

(x,y) = C

S thay oi gia tr tai 2 iem(x,y) va iem (x+dx, y+dy) trenng = C

+

=

ma (x,y) = C

Phng trnh ng dong

Vay cac iem nay thoa man pt ng dong hay noi cach khac ng (x,y) = C lamot ng dong

= =+=

+=



+ Lu lng (q) i qua gia 2 ng dong A,B bang q = B - A

uyux

x

ydx

dy dq

A

B

+ d

dq = ux dy-uydx

=

dq = d

Nh vay

3. Moi quan he gia ham dong va ham the:

ng dong va cac ng ang the trc giao vi nhau

+

=

==

=

+

Li thuy ong

- ay la mot chuyen ong the v =r

Ham the : (x,y)

Ham dong : (x,y)

II.MOT SO CAC CHUYEN ONG THE N GIAN

=

= =1. Chuyen ong eu nam ngang

= =

= =

= +

=

= =

= =

= = =

= +=

4/Uo3/Uo2/Uo1/Uo

1/Uo2/Uo

3/Uo

4/Uo



Ve nha ??

Ham the : Trong toa o cc (r,)

Trong toa o descarte

Ham dong: Trong toa o cc (r,)(Trong toa o descarte)

+=

2. iem nguon va gieng

= =

( )= = = = +( )= = = = +

=r

Xet mot iem nguon co cng o q (m2/s)

T phng trnh lien tuc

= =

=

=

= =

=

( ) = + =

=

q

Li thuy ong:

ng the :

= Phng trnh vongtron tam O ban knh

or

1

ng dong : Phng trnh ng thangqua tam nghieng mot goc

23

1= 0

234

or

1 23

1

234

iem nguon iem hut : tng tiem nguon thay -q

5

67 8

= =

= =

=

= =



xoay dng

Toa o cc

Toa o Descarte

Toa o Descarte

Toa o cc

3. Xoay t doDong chay tren nhng ng tron ong tam, co van toc

ro

Ghi chu:>0: xoay dng ngc chieu kim ong ho;


- Khi 0 th q m0 ( mo: cng o cua lng cc) Lng cc c nh ngha

Lng cc

Trong toa o cc

- Ham dong : Tng t co

Trong toa o cc

x

y

o

+= yxx

xq

+= > yxx

xqlimi

+= 22

0

2 yxxm

rcosm

=

yxym+

=

rsinm

=

III CHONG CHAP CAC CHUYEN ONG THE

Cac the lu eu thoaman pt Laplace

Chong chap chuyen ong the se cho ra motchuyen ong the

1. Chuyen ong qua na co the

Dong eu + nguon

Ham the van toc

= d+ n = ++ (Toa o descartes) = d+ n = + (Toa o cc)

Ham dong= d+ n = + (Toa o descartes)

+(Toa cc)

q

Uo = d+ n =



+=Thanh phan van toc

+==

( )

===

iem dng (u = 0)

( ur = 0 va u = 0 )Nhng iem tren tren truc x : u = 0

va ur = 0 khi =+ =

=iem dng S

Phng trnh ng dong i ngangqua iem dng S

+===

=+

=

iem dng

S

=

Ket hp mot chuyen ong eu va mot iem nguon co the dung e mo ta dong chay baoquanh na co the

Naco the

q/(2Uo)

2. Chuyen ong bao quanh tru tron

x

y

o

Uo

Dong eu (Uo) + Lng cc (m0)

Ham the van toc

= d+ lc =

= d+ lc =Toa o cc

++ 22

0

2 yxxmxUo

rcosmcosrU

+20

0

Toa o descartes

+= r

mrUcos2

00

Ham dong

= d+ lc =Toa o descartes

= d+ lc =Toa o cc

220

0 2 yxymyU +

rsinmsinrU

20

0

= r

mrUsin2

00



= r

mrUsin2

00

T ham dong ng dong

ng dong vi = 00

20

0 =

r

mrUsin

== ksin 0 02

00 =

rmrU

=

ng dong la 1 vong tron tamO ban knh r

q

= 0= 0

or

Cac ng dong = 1, = 2, . Co dangnhu tren hnh ve

Uo

=1

Mot dong eu ket hp vi motlng cc co the dung e mo tadong chay bao quanh mot tru tronNeu tru tron co ban knh ro thlng cc co cng o mo la

==Nh vay dong chay bao quanh tru tron ban knhr0 co ham the van toc va ham dong

+= 2

201

rrcosrUo

= 2

201

rrsinrUo

q

Uo

= 0

= 2 =2

Thanh phan van toc

+=

=

=

+=

=

+=

+=

Tren be mat hnh tru ( r = r0)

= and =Van toc cc ai A and B ( = /2)

A

B

iem dng C va D ( = 0, )

CD

02Uu A = 02Uu B =0== DC uu



Ap suat phan bo tren mat truXet mot iem xa mat tru co van toc Uo, apsuat p0 and va mot iem tren mat tru van toc us, ap suat psAp dung pt Bernoulli

++=++ Bo qua s thay oi (z) va thay uS

+=+

( ) +=

q

Uo

= 0

= 2 =2

A

B

CD

Ap suat cc ai tai C va D

Neu p0 la ap suat kh tri po= 0

( ) = ==Ap suat cc tieu tai A va B ==

( )= 2200 4121 sinUppS200 2

1 UCpp pS =

Ap suat phan bo tren mat tru

Dong chay the

( )= 241 sinCpng mau o , oi xng

Dong chay co quay

Cp khong oi xng , thnghiem cho ng mau xanh

Quan sat dong chay the va dongchay co quay qua mot xe angchuyen ong



3. Dong chay bao quanh tru tron vi mot xoay t do

Xoay t do : =

=

Uo

r0

Dong chay bao quanh tru tron r0

+=

=

Dong chay bao quanh tru tron + Xoay t do

+

+=+=

=+=

Ham the van toc

Ham dong

Thanh phan van toc tren mat tru

== =

+== =

iem dng ( ur = 0 va u=0) tren mat tru

= +=

Van toc tren mat tru

Tai iem u = 0 =+ =

=

Khong co iem dng tren mat tru ( iem

dng nam ngoai mat tru)

Uo

r0



Ap suat tren mat tru

T pt Bernoulli equation, xet iem o xa mat tru vaiem s nam tren mat tru

++=++ Bo qua s thay oi (z) va thay us vao

+

+

=+

++=

Tong lc tac dung tren mat tru khi po = 0

== = =

Chu y: khi hnh tru co chieu dai L th lc Fy phai nhan cho L

(Kutta Jouskowky law)

Fy ps

dA = r0d x

y

r0Uo

Fyps

dA = r0d

x

y

r0d

y

+=

Bai tap : 1) 6.5 2) 6.6 3) 6.8 4) 6.10 5)6.11 6) 6.13

S phan bo ap suat tren mat tru khi Re ln

Dong chay qua mot canh



Cau hoi trac nghiem :

Cau 1: Nhng tnh chat nao sau ay ch ung cho the lu a. Luon luon tm c cac ham dong b. p/ + z = const trong toan dong chay*c. Luon luon tm c cac ham thed. Ca 3 eu ung

Cau 2 : Trong chuy n ng ch ng ch p m t chuy n ng song song va m t i m ngu n nh hnh v th v n t c theo ph ng ti p tuy n ur = 0 t i ( ch n cau ung nh t)

a) m g c t a O*b)Nh ng i m n m tren truc hoanhc)Nh ng i m n m trn tr c tung d) Khong co iem nao

Cau 3: e mot chuyen ong 2 chieu (x,z) la mot the lu can phai co ieu kien : = b) + =

c) = *d) Ca 3 eu sai

b)

Cau 4: Dong chay bao quanh tru tron xoay. Chieu cua lc nang F trong hnh nao sau ay ung:

F

1 2 3 4

Cau 5 : Khi mot khoi hnh tru chuyen ong quay quanh no va di chuyen thang trong khong kh th:

*a.Luon luon co t nhat 1 iem dngb. Luon luon co t nhat 2 iem dngc. Luon luon co t nhat 1 iem dng tren mat hnh tru d. Co the khong co iem dng



yUo=

A

B

xB xA

yA

yB

Xac nh l u l ng qua A-B ?

Th du 1:


TLTK_10.pdfCH1_10.pdfCH2_10.pdfTinhapluc.pdfch3_10.pdfCH4_10.pdfCh5_10.pdfCh6_10.pdf

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Baigang CLC 2010