TS. Nguyen Th Bay - HBK tp HCM -Bai Giang CLC THE LU 1CHNGDong chay co the /thoa .k. (1) 0x y y x= 0yuxuxy=rot(u)=0dong chay phang, lu chat ly tng khong nen c chuyen ong on nh Gii han: I. CAC KHAI NIEM C BAN1. Ham the van toc:Ta nh ngha ham sao cho: = = = =r1u ;ru hayyu ;xur y xTrng vect u la trng co the khi: BAds urch phu thuoc vao hai v tr A va B. Ta co: B ABABABA) 1 ( thoa tontaiyBAxBAd) dyydxx( ds u ) dy u dx u ( ds u = = + = + = r rch phu thuoc vao gia tr ham the tai A va B.Ro rang t chng minh tren, BAds urVay:(1)ABnuunus0 dy u dx u 0 dy x= + = 2. Phng trnh ng ang the:3. Y ngha hamthe van toc:A B AB = = BAs ABds u la lu so van toc4. Tnh chat ham the:T ptr lien tuc, ta co: 0y x0y y x x0yuxu2222yx= + = + =+Ham the thoa phng trnh LaplaceTS. Nguyen Th Bay - HBK tp HCM -Bai Giang CLC THE LU 25. Hamdong:Khi dong chay lu chat khong nen c ton tai, th cac thanh phan van toc cua nothoa ptr lien tuc : ru ;r1u hayxu ;yu / 0yuxur y xyx = = = = =+ goi la ham dong. Nh vay ton tai trong moi dong chay,con ch ton tai trong dong chay the.6. Hamdong trong the phang:V la dong chay the nen:0y x0y y x x0yuxu2222xy= + = =Vay trong dong the th ham thoa ptr Laplace.7. ng dong va ptr:T ptr ng dong: 0 d 0 dxxdyy0 dx u dy uy x= = + = xyOnnxnydxdyds(-dx=ds.sin)Nh vay tren cung mot ng dong th gia tr la hang so.8. Y ngha hamdong:Ta co: = = = = + = + = = =BAA BBABAy xBAy xBAy y x xBABAn ABd dxxdyydx u dy uds sin u ds cos u ds n u ds n u ds n u ds u qr rVay:A B ABq =9. S trc giao gia ho cac ng dong va ng ang the: 0 ) u ( u ) u ( uy y x xx y y x= + = + Suy ra ho cac ng dong va cac ng ang the trc giao vi nhau.10. Cong the lu:......2 12 1+ + = + + = 11. Bieu dien dong the:vi z = x+iy = ei.The phc f(z): + = i ) z ( fNh vay:dydidxdiu udzdfy x+= =e bieu dien dong chay the, ta co the bieu dien rieng tng ham dong va ham the, tacung co the ket hp ham dong vi ham the thanh mot ham the phc nh sau:: TS. Nguyen Th Bay - HBK tp HCM -Bai Giang CLC THE LU 3II. CAC V DU VE THE LU xOy=0=1=2=3=-1=-2=-3=0=1=2=3=-3=-2=-1V01.Chuyen ong thang eu: t xa vocc ti, hp vi phng ngang mot goc.ux=V0cos; uy= V0sind = uxdy - uydx = V0ycos - V0xsin + CChon:=0 la ng qua goc toa oC=0.Vay: = V0ycos - V0xsinTng t: = V0xcos + V0ysinBieu dien bang ham the phc: F(z) = +i = (V0xcos + V0ysin) + i(V0ycos - V0xsin)=x(V0cos- iV0sin)+yi(V0cos - iV0sin)=az vi:a=(V0cos -iV0sin) la so phc; z=x+iy la bien phc.2. iem nguon, iem hut: vi lu lng q tam at tai goc toa o.(q>0:iem nguon; q0: xoay dngGhi chu:>0:xoay dng ngc chieu kim ong ho; = < = = =dng . iem . 0 Ru 4dng . iem . 1 Ru 4dng . iem . 2 Ru 4Ru 4sinR 2sin u 2 0 u00000Phan bo ap suat tren mat tru :2up2up2trtr20+ =+vi2120Rsin u u + = = =2020202tr20Ru 2 sin 2 12u )uu1 (2u pdtrGia s p=paLc tac dung tren mat tru:020tr yU d . sin R p F = = dPhng x: Fx=0Phng y:---Lc nang JukovsLu y :0 d . sin20n= TS. Nguyen Th Bay - HBK tp HCM -Bai Giang CLC THE LU 8/2Ru0=2/2Ru0=3/2Ru0=1FyCactrnghp xoay>0y| | /2Ru0=3StagnationPointrFy| | /2Ru0=1StagnationPointyr| | /2Ru0=2StagnationPointyrCactrnghp xoay< 0TS. Nguyen Th Bay - HBK tp HCM -Bai Giang CLC THE LU 9V du 1:Chuyen ong the cua chat long hai chieu tren mat phang nam ngang xoy viham the van toc =0,04x3+axy2+by3,x,y tnh bang m, tnh bangm2/s.1. Tm a, b.2. Tm o chenh ap suat gia hai iem A(0,0)va B(3,4),bietb khoi lngrieng long bang 1300kg/m3Giai:T ham the van toc = 0,04x3+ axy2+ by3ta co:2y2 2xby 3 axy 2yu ; ay x 12 , 0xu + = = + = =Cac thanh phan van toc phai thoa phng trnh div(u)=0 nen:0 by 6 x ) a 2 24 , 0 ( 0 by 6 ax 2 x 24 , 0 0yuxuyx= + + = + + =+V div(u)=0 ung vi moi iem nen the (x=0; y=1) vao ta c b = 0(x=1; y=0) vao ta c a = -0,12V ay la chuyen ong the nen p.tr Ber ung cho hai iem bat ky A va B, ta co: = += + 2) u u () p p (2u p2u p2A2BB A2B B2A A uA=0; uB= ((0,12*32-0,12*42)2+(-0,24*3*4)2)1/2= 3 m/s22ABm / KN 85 , 52) 3 ( 1300p = = V du 2:Giai:) x y (21) y , x (2 2 = xyDong chay the uon congmot goc 900vi ham the van tocc cho nh sau: (x,y tnh bang m).Tm lu lng phang quang thang noihai iem A(1,1) va B(2,2) yyu ; xxuy x= = = =) y ( C yx x y uxy+ = = = const xy const ) y ( C x ) y ( ' C x uyx+ = = = + = s / m 3 1 * 1 2 * 2 q2A B AB= = = TS. Nguyen Th Bay - HBK tp HCM -Bai Giang CLC THE LU 10Ap suat d tren mat tru bang:) sin 4 1 (2u p220 =dtrFydF 0 Rd ) cos( ) sin 4 1 (2u) cos( pds dF F02200 0x x= = = = = = = = = 0 02200220y02200 0y yd ) sin( 3 )) (cos( d ( cos 4 (2u Rd ) sin( ) 3 cos 4 (2u RFRd ) sin( )) cos 1 ( 4 1 (2u) sin( pds dF F3u R 53433432u Rcos34cos 32u RF20200320y= + = = Giai:V du 3:Gio thoi qua mai leu dang ban tru R=3m viV=20m/s,khong kh co khoi lng riengbang 1,16 kg/m3. Tm lc nang tac dung len1m be dai leu. e tm lc nang Fy tac dung len 1mbe dai leu,tren ban tru ta chonmot viphandien tch ds, tm lc dF tac dung len ds, sau o chieu dF len phng y dFy. Va tchphan (dFy) tren toan ban truN 2320 Fy = Giai:V du 4:Mot xilanh hnh tru tron di chuyen trongnc vi van toc u0khong oi o sau10m.Tm u0e tren be mat xilanh khongxay ra hien tng kh thc , biet nc 200CA BCDuC= -2u0uD= 2u0pA= pB= u02/2pC= pD= -3u02/2200C ap suat hi bao hoa cua nc : pbh= 0,25m nce tren be mat xi lanh khong xay ra hien tng kh thcth ptrut> pbh= 0,25m nc ptruck- 9,75m ncAp suat d nho nhat tren mat tru (neu tru di chuyen tren mat thoang ), nh ta abiet, tai v tr C va D, ba bang:pC= pD= -3u02/2Suy ra, van toc toi a ma tru co the di chuyen c e khong co hien tng kh thcxay ra tren mat tru phai giai t bat p.tr : pC= pD= 10n-3u02/2 Vay neu tru di chuyen o sau 10m th :Ptrud= 10n-3u02/2 > - 9,75 n 3u02/2 < 19,75 n u0< 11,365 m/sTS. Nguyen Th Bay - HBK tp HCM -Bai Giang CLC THE LU 11Giai:V du 5:Hai na xilanh c noi vi nhau va at trong trngchay eu co the nh hnh ve. Ngi ta khoet 1 lo nho taiv tr goc e cho khong co lc tac dung len hai moi noi. Gia thiet rang ap suat ben trong xi lanh bang ap suat benngoai xi lanh tai lo khoet. Xac nh goce cho khong co lc tac dung len hai moi noi th tong lc Fxtac dung len moi namat tru phai bang khong. Dobieu o ap suat tren mat tru phan bo oi xng quatruc ox,nen ta ch can xettong lc Fxtren mat tr. Ta xet tren mat tru t 0 en/2:Ap suat d tren mat tru: ) sin 4 1 (2u p220 =dtrTren mat tru ta chon vi phan ds, goi dFnla lc tac dung len ds t ben ngoai mattru, ta co:dFn=pds dFnx= - pdscos = -pRcosd6R usin34sin2R uRd cos ) sin 4 1 (2uF202 /03202 /0220nx= = = 0 /2dFdFxds[ ] R p sin R p Rd cos p ds p F2 /02 /02 /0tx = = = = Goi pla ap suat tai lo khoet, ta co: ) sin 4 1 (2up220 =d) sin 4 1 (2R uF22otx = Ta co:Fnx+Ftx=0Suy ra: 31sin31sin34sin 4) sin 4 1 (2R u6R uF F2 222o2otx nx= = = = =026 , 35 = Nhan xet:Lc Fnx>0hng theo chieu dnglc Ftxt ben trong mat tru phai hng theochieu am. Nh vay, ap suat tai lo khoet phai la ap suat chan khong0 /2 FnxFtx