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• 1. PGS.TS. Nguyen Thi Bay, DHBK tp. HCM; www4.hcmut.edu.vn/~ntbayCHNGGii han:dong chay phang, lu chat ly tng khong nen c chuyen ong on nhI. CAC KHAI NIEM C BAN 1. Ham the van toc: Ta nh ngha ham sao cho: u x = ; u y = xTrng vect u la trng co the khi: Ta co:BBABA1 ; u = r r (1) u ds ch phu thuoc vao hai v tr A va B.A tontai thoa (1) uds = ( u x dx + u y dy )yhay u r =BB uds = (AA dx + dy ) y xBRo rang t chng minh tren, Vay:= d = A BB udsAAch phu thuoc vao gia tr ham the tai A va B.Dong chay co the /thoa .k. (1) u y u x =0 = 0 y y x x y x 2. Phng trnh ng ang the: d = 0 u x dx + u y dy = 0 3. Y ngha ham the van toc: AB = B A 4. Tnh chat ham the: u BAB = u s dsAla lu so van tocn un uAusu y2 2 + =0 + =0 2 + 2 =0 T ptr lien tuc, ta co: x y x x y y x y x rot(u)=0 Ham the thoa phng trnh Laplace THE LUU 1B

2. PGS.TS. Nguyen Thi Bay, DHBK tp. HCM; www4.hcmut.edu.vn/~ntbay5. Ham dong: Khi dong chay lu chat khong nen c ton tai, th cac thanh phan van toc cua no thoa ptr lien tuc : u x u y 1 + = 0 / u x = ; uy = x y y x goi la ham dong.hayur =r ; u = rNh vay ton tai trong moi dong chay, con ch ton tai trong dong chay the.6. Ham dong trong the phang: u y u x 2 2 V la dong chay the nen: =0 2 + 2 =0 =0 x y x x y y x y Vay trong dong the th ham thoa ptr Laplace.7. ng dong va ptr:T ptr ng dong: u x dy u y dx = 0 dy + dx = 0 d = 0 y xNh vay tren cung mot ng dong th gia tr la hang so. 8. Y ngha B m dong: ha B Ta co:Vay:y BBnyq AB = u n ds = unds = u x n x ds + u y n y ds = u x cos ds + u y sin ds AAAABB = u x dy u y dx = dy dx = d = B A y x A A Aq AB = B AdyBndxnx ds(-dx=ds.sin)O9. S trc giao gia ho cac ng dong va ng ang the: = u x ( u y ) + u y ( u x ) = 0 + x x y ySuy ra ho cac ng dong va cac ng ang the trc giao vi nhau. 10. Cong the lu: = 1 + 2 + ... = 1 + 2 + ...11. Bieu dien dong the: e bieu dien dong chay the, ta co the bieu dien rieng tng ham dong va ham the, ta cung co the ket hp ham dong vi ham the thanh mot ham the phc nh sau::The phc f(z): Nh vay:vi z = x+iy = ei .f ( z ) = + idf d d = u x iu y = +i dz dx dyTHE LUU 2x 3. PGS.TS. Nguyen Thi Bay, DHBK tp. HCM; www4.hcmut.edu.vn/~ntbayII. CAC V DU VE THE LU 1. Chuyen ong thang eu: t xa vo cc ti, hp vi phng ngang mot goc . ux = V0cos; uy = V0sin d = uxdy - uydx =3 = V0ycos - V0xsin + C =2 Chon:=0 la ng qua goc toa o =1 =0 C=0. =-1 Vay: = V0ycos - V0xsin =-2 Tng t: = V0xcos + V0ysin =-3y V0 O=-3x =3 =2 =1 =0 =-1 =-2Bieu 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 dng ngc chieu kim ong ho; 0: xoay dng4. Lng cc: la cap iem nguon + hut co cung lu lng qat cach nhau mot oan vo cung nho (cho 0 vi ieu kien qm0 , la moment lng cc). V du ta xet trng hp nam tren truc hoanh: Tm ham dong: y y q q arctg arctg ( n h ) = 2 2 x x+ 2 2 y y y x y x + x + x q q 2 2 2 2 arctg arctg = = 2 2 2 y y x2 + y2 4 1+ x + x 2 2 = n + h =Khi 0 t so trong dau arctg tien ti 0 nen ta co the viet: y x y x + q y 2 2 q = = 2 2 2 2 x2 x2 + y2 + y2 4 4 THE LUU 4 y m0 2 x 2 + y 2 5. PGS.TS. Nguyen Thi Bay, DHBK tp. HCM; www4.hcmut.edu.vn/~ntbay 2 2 Tm ham the van toc: = n + h = q ln x + + y 2 ln x + y 2 4 2 2 2 x + + y2 q 2 x 2 = q ln 1 + = ln 2 2 4 4 x + y 2 x + y2 2 2 x2 + ... va bo qua cac so hang bac cao vo cung be, ta co: 2 q m0 2 x x khi 0 = 2 2 2 2 x + y2 2 x y 2 Trien khai ln(1 + x) = x Vay tom lai, oi vi chuyen ong lng cc th: m0 m 0 sin y = 2 r 2 x 2 + y 2 m m cos x = 0 2 = 0 2 x + y 2 2 r=m 0 cos i sin m 0 cos 2 + sin 2 m 0 1 f (z) = = = 2 r 2 r (cos + i sin ) 2 z-q+q5. Dong chay quanh na co the: La chong nhap cua chuyen ong thang eu ngang (U0)+ nguon tai goc toa o (q) q q ln( x 2 + y 2 ) = u 0 r cos + ln r 4 2 q y q = u0y + arctg( ) = u 0 r sin + 2 x 2iem dng = u0x +iem dng A: u A = 0 u xA = 0; u yA = 0 q q 2x x = u 0 + 4 x 2 + y 2 = 0 x A = 2 u 0 = q 2 y = 0 yA = 0 y 4 x 2 + y 2 THE LUU 5A 6. PGS.TS. Nguyen Thi Bay, DHBK tp. HCM; www4.hcmut.edu.vn/~ntbay6. Dong chay quanh co the dang Rankin La to hp cua dong chuyen ong thang ngang eu (u0) + nguon (+q) + hut(-q). Trong o iem nguon va hut nam tren truc hoanh, cach nhau mot oan 2a hu han,u0 AB2aq (x + a)2 + y 2 = uo x + ln 4 ( x a ) 2 + y 2 = uoy ++q -qq y y arctg x + a arctg x a 2 Co hai iem dng A va B: q 2y 2y = (x + a) 2 + y 2 (x a) 2 + y 2 = 0 {y = 0 y 4 = u + q 2(x + a) 2(x a) = 0 0 x 4 (x + a) 2 + y 2 (x a) 2 + y 2 u x = 0 u=0 u y = 0 the y = 0 u + q 2 2 = 0 0 4 (x + a) (x a) q 4a aq u0 + + a2 2 = 0 x = 2 u 0 4 x a 7. Dong chay quanh tru tron (=0) Xet to hp cua chuyen ong thang eu, nam ngang (u0)+lng cc (m0) = uox + m0 m cos m0 x = u o r cos + 0 = u o r cos 1 + 2 2 2 x + y 2 r 2u 0 r 2 m sin m0 m0 y = uo y + = u o r sin 0 = u o r sin 1 2 2 2 x + y 2 r 2u 0 r 2 Do khong co s trao m 0 bang ng Thay ng oi lu chat gia r= R= tron tron 2 u 0 trong va ngoai ng dong =0 R2 = u o r cos 1 + 2 r R2 = u o r sin 1 2 r iem dng THE LUU 6 Xet ng dong =0 =0 va m02 u 0r=m0 2 u 0th ban chat dong chay van khong oiTa co hnh anh cua dong chay bao quanh tru tron. (tru khong xoay) 7. PGS.TS. Nguyen Thi Bay, DHBK tp. HCM; www4.hcmut.edu.vn/~ntbayTm phan bo van toc tren mat tru r=R: 1 = 2 u 0 sin u = r r = R = 2 u 0 R cos u = 0 r Tm hai iem dng tren mat tru: u = 0 = 0 va = co hai iem dng A. B trc va sau mat tru. Tm hai iem co gia tr van toc ln nhat tren mat tru:3 u = u max = ; = 2 2 u C = 2 u 0 ; u D = 2 u 0pA = pB = u02/2 uC = -2u0C BA DuD = 2u0pC = pD = -3u02/2 C, D nam tren va di mat tru co gia tr van toc ln nhat.Khao sat phan bo ap suat ren mat tru: Ap dung P.Tr NL tren ng dong =0 t iem xa vo cc en iem tren mat tru: 2 2 2 2 u 2 u 2 u 0 u tr u 0 4u 0 sin 2 0 d p + = p tr + tr (1 2 ) = (1 ) Gia s p=pa p tr = 2 2 2 2 u0 2 u0 u 2 u 2 pA = pB = 0 Tai A, B: p d = 0 (1 4 sin 2 ) tr 2 2 2 Tai C, D: p = p = 3u 0 D D Do bieu o phan bo ap suat oi xng qua ox lan oy nen 2 Nhan xet: tong lc tac dung len mat tru trong trng hp nay = 0 7. Chuyen ong quanh tru tron xoay (0): Dong eu Bao gom chuyen ong quanh tru tron + xoay t do ( +) R2 = u o r cos 1 + 2 + r 2 R2 = u o r sin 1 2 ln r r 2 Phan bo van toc tren mat tru :V r = R nen u r = 0; u = 2 u 0 sin +L ccXoay t do1 R 2 < 4 Ru 0 2.iem.dng suy ra: sin = = 4Ru 0 1.iem.dng u = 0 2 u 0 sin = 2 R 4 Ru 0 > 4Ru 0.iem.dng 0 Phan bo ap suat tren mat tru : 2 2 u u 1 vi u = 2u0 sin + p + 0 = p tr + tr 2 2 R 2 2 2 2 2 u 0 u 0 u tr Gia s p=pa p d = 1 2 sin (1 2 ) = tr 2 u0 2 2 Ru 0 Lc tac dung tren mat tru: Lu y : Phng x: Fx =0 2 2 Phng y: d n --Lc nang Jukovs Fy = p tr R sin .d = U 0 sin .d =0 0THE LUU 70 8. PGS.TS. Nguyen Thi Bay, DHBK tp. HCM; www4.hcmut.edu.vn/~ntbayCac trng hp xoay >0/2Ru0=1 iem dng/2Ru0=2 iem dngFy/2Ru0=3 iem dngCac trng hp xoay < 0y Fyy r rStagnation iem dng Pointy| | /2Ru0=1Stagnation iem dng Point | | /2Ru0=2 r Stagnation iePoint g m dn| | /2Ru0=3 THE LUU 8 9. PGS.TS. Nguyen Thi Bay, DHBK tp. HCM; www4.hcmut.edu.vn/~ntbayV du 1: Chuyen ong the cua chat long hai chieu tren mat phang nam ngang xoy vi ham the van toc = 0,04x3 + axy2 + by3 , x,y tnh bang m, tnh bang m2/s. 1. Tm a, b. 2. Tm o chenh ap suat gia hai iem A(0,0) va B(3,4), bietb khoi lng rieng long bang 1300kg/m3 Giai: T ham the van toc = 0,04x3 + axy2 + by3 ta co: ux = = 0,12x 2 + ay 2 ; uy = = 2axy + 3by 2 x y Cac thanh phan van toc phai thoa phng trnh div(u)=0 nen: u x u y + = 0 0,24x + 2ax + 6by = 0 (0,24 + 2a )x + 6by = 0 x y 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 = -0,12 uA=0;uB = ((0,12*32 -0,12*42)2+(-0,24*3*4)2)1/2 = 3 m/sV ay la chuyen ong the nen p.tr Ber ung cho hai iem bat ky A va B, ta co: 2 2 2 pA u2 pB uB (uB u2 ) A A + = + (pA pB ) = p AB = 1300 ( 3 ) = 5,85 KN / m 2 2 2 2 2V du 2: Dong chay the uon cong mot goc 900 vi ham the van toc c cho nh sau: 1 ( x, y ) = ( y 2 x 2 ) 2 (x,y tnh bang m).Tm lu lng phang qua ng thang noi hai iem A(1,1) va B(2,2)yxGiai:y(phi=70)25 y(phi=60) ux = = x x ; uy = = y y20y(phi=50) y(phi=40)15y(phi=30) = uy = yx x = yx + C(y) = u x x + C'(y) = x y10y(phi=20) y(phi=10)5y(phi=0)0 -30-20-10y(phi=-10)0 -5102030 y(phi=-20) y(phi=-30) C(y) = const = xy + const qAB = B A = 2 * 2 + 1*1 = 3m2 / s THE LUU 9 10. PGS.TS. Nguyen Thi Bay, DHBK tp. HCM; www4.hcmut.edu.vn/~ntbayV du 3: Fy dF Gio thoi qua mai leu dang ban tru R=3m vi V=20m/s, khong kh co khoi lng rieng bang 1,16 kg/m3 . Tm lc nang tac dung len 1m be dai leu. Giai: e tm lc nang Fy tac dung len 1m be dai leu, tren ban tru ta chon mot vi phan dien tch ds, tm lc dF tac dung len ds, sau o chieu dF len phng y dFy. Va tch phan (dFy) tren toan ban tru 2 u 0 d p tr = (1 4 sin 2 ) Ap suat d tren mat tru bang: 2 2 u 0 (1 4 sin 2 ) cos( )Rd = 0 2 0 0 0 2 u 0 Fy = dFy = pds sin( ) = (1 4(1 cos 2 )) sin( )Rd 2 0 0 0 Fx = dFx = pds cos( ) = 2 2 Ru 0 Ru 0 2 2 Fy = (4 cos 3) sin( )d = (4 cos ( d(cos()) 3 sin( )d 2 0 2 0 0 2 2 2 R u 0 4 Ru 0 4 4 5Ru 0 3 Fy = 3 cos 3 cos = 2 3 + 3 3 3 = 3 2 0 Fy = 2320NV du 4: Mot xi lanh hnh tru tron di chuyen trong nc vi van toc u0 khong oi o sau 10m. Tm u0 e tren be mat xi lanh khong xay ra hien tng kh thc , biet nc 200C Giai: 200C ap suat hi bao hoa cua nc : pbh = 0,25m nc e tren be mat xi lanh khong xay ra hien tng kh thc th ptru t > pbh = 0,25m nc ptru ck < 9,75m nchaypA = pB = u02/2 u