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Lecture_28_notes - ChE 37 4—Lecture 28—Boundary Layers...

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Unformatted text preview: ChE 37 4—Lecture 28—Boundary Layers Navier-Stokes Equations: — Complex: PDE, 3D, unsteady, nonlinear, 4 equations. — Solve by simplifying: Inviscid, laminar, reduce dimensions, steady state. Boundary Layer Method. — Split flow into two regions that are matched at the interface: 1 An outer region that is inviscid. Solve the resulting Euler Equations. - Many analytic solutions exist (especially in 2D) for complex geometries. - But does not apply near walls. 2 An inner boundary layer region in reduced dimensions and simplified by dropping terms. Boundary layer region. — No gravity, 2D, Steady state, thin. — Scale the governing equations to determine properties of the flow and the boundary layer equa- tions: Navier—Stokes equations (SS, no gravity), scale x,y with just L, 17 with U, and P with pUZ: 17- Vf)‘: —%VP + MVZU. Scale it to get: (17- V17)* = (VP)* + Rie(V217)*. -— High Re gives no viscous term which makes no sense. Instead, we need two scales, L and 6, the boundary layer thickness. — KEY RESULT: Need two scales, L and 6, boundary layers are thin. Continuity: 3—“ + Q = 0, scale to 3—“ * + M Q * 2 0. 81 (9y 8x 6U 3y — KEY RESULT: we}: 2 U6/L, and Uref << U. - Q E _ _lfl 32v 32” Y—Momentum. uaz + ’Uay — p 8y + 12%; + 1/3112. * ’k =0: * _ av * av _ L2 8P 1 3% L2 621) scaled~ (Max) + (Day) 6? (By) + Re (3:62) + 6 Re (all?) ' — KEY RESULT: g—l; = 0. (Pressure can vary along the length, but not through the boundary layer thickness). This is because the boundary layer is thin and the streamlines are nearly parallel. X—Momentum: See exam 2, problem 2, where we dropped one of the viscous terms. 3n 871V__1_§.IZ 827; u$+v3~y _ 96w +VW' — This and continuity are the laminar boundary layer equations. — KEY RESULT: ignore the % term (that is, we ignore diffusion of momentum in the downstream direction). — Note: —%% = U%. (Just difierentiate Solution proceedure: Solve U(X) for outer flow using Inviscid equations; Solve Boundary layer equations given U(x); Solve for wall stress, drag, etc. Bernoulli equation with respect to x). SEE POSTED SOLUTION OF THESE EQUATIONS FOR REFERENCE (not required). Boundary Layers apply to balls, wings, jets, wakes, mixing layers. As for pipe flow, we have laminar, transitional, and turbulent. Take Re : 5 x 105 as the cutoff between laminar and turbulent. Shear stress decreases with distance for laminar and turbulent, but wall stress (friction) is greater for turbulent than for laminar. 51%; 29-— Tsomam7 Lw/arr 10,4 ?rwim(~/ —- NE}, 27%. A P06} '5'? {fr-wt} nofilv‘acm gp‘ot § ” ’h‘fixfléercj lo gala go, 22:5,»! no“); Ql‘flx ctr ComP‘Q-m’z‘m/D‘ 0 CPD ‘Hme/ '{o 486 “ Hal) AQQUMPWM ' {7 9014;.“ Jaw, @ Kama/M7 L47“ OFFIWIL '— F1914 aég‘ump-p‘im '“ goldb‘w {334' )3 L, Arprowk U a 'g log.) (Ive/7 6L L2) \‘ A» ' Nita? HSNJV 7149‘ 3,1,, bwtlafl (.1 1,4 ,7 3411:? -t ‘ )1) $3 «3 1 "Vita/W CD Ody) F100 I '2. 2:31; Benawwu’ £7'x (75,5) M018)”, no /L w—p— 9‘ KW 4§rv$=~’-'~‘7 *7 U) M (7/32) «A f F J 17 my Aw» gamma 2 m, CD .__————' Jaimprfié/ Hg, In KL M” SD’“. '9 Ri‘IUCC 'H: bin/7’ Dre? TEAM ma r/(a‘IZ L\ T3 L ¥ (on-35 ..._——--v f f_______________—- EL )1 ~51;de . 7D 0671" Cfln+"nu)47 _ no —’ \/—— mam ? X, Mar-fl ' 40’ . 141;,“ W [7&11'77 £0 2 Origin». "U’;U€’-“ L4 LA.»-\i(1¢.:>m7}~/ [ 90!.u+;0v\ @ $0M 0v+MP10v3 mn- JVW 1““ 75¢, Jaw +0 a)!» We» M74 ow ‘ ’4’ (yr/(’5’ Vfat/ [\lecé [Jig/L L ,7 11".“ g LaM'.~6‘/], L9“) wIZ—i 6} no //0LJ gePafla-JJWI MM A9 1/”, Fan Pb”), W a. 7(4..wrvlw, (£9 354 \u UM Tam img RC >11 Rafi Cf '3 YT“) l v ""2 / —-\>1-A “ELEM 70172- 7 b J <1 é: O'ééLII/ fit Lflr";"M C g 0.017 7M5 43 . mm Q: 09“ 6% 9+“? if, ‘I 6'7 94% “MW D/Sflma Dun Flawl't7 (Law) [.Mua}w K“? may 6’»; laDL In,” P1539) (WM) {0 ‘6 ...
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