FLUID MECHANICS. Lecture 7 Exact solutions
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1 FLID MECHANICS Lecture 7 Eact solutions 1
2 Scope o Lecture To present solutions or a ew representative laminar boundary layers where the boundary conditions enable eact analytical solutions to be obtained.
3 Solving the boundary-layer equations The boundary layer equations V + 0 y dp ρ ρ V e + + µ y d y Solution strategies: Similarity solutions: assuming sel-similar velocity proiles. The solutions are eact but such eact results can only be obtained in a limited number o cases. Approimate solutions: e.g. using momentum integral equ n with assumed velocity proile (e.g. nd Yr Fluids) Numerical solutions: inite-volume; inite element, CFD adopts numerical solutions; similarity solutions useul or checking accuracy. 3
4 SIMILARITY SOLTIONS Incompressible, isothermal laminar boundary layer over a lat plate at zero incidence (Blasius (1908)) V + 0 y + V ν y y const. V0 at y0; as y Partial dierential equations or and V with and y as independent variables. For a similarity solution to eist, we must be able to epress the dierential equations AND the boundary conditions in terms o a single dependent variable and a single independent variable. dp e d 0 4
5 THE SIMILARITY VARIABLE Condition o similarity F ( η) η is likely to be a unction o and y / / F(η) Strategy: to replace and y by η Similarity variable y η δ How to ind δ beore solving the equation? 5
6 THE SIMILARITY VARIABLE To ind the order o magnitude o δ δ δ 1 [Re L ] L L O( ) [Re ] ( ) L O δ δ L [Re ] O( ) δ [δ ] O Re Re ν [ δ ] ν O Similarity variable: y y η O[ δ ] ν / 6
7 NON-DIMENSIONAL STREAM FNCTION For incompressible D lows stream unction ψ eists. ψ V y ψ Strategy: To replace and V with ψ Deine a non-dimensional stream unction such that is a unction o η only. 7
8 NON-DIMENSIONAL STREAM FNCTION Find the order o magnitude o stream unction ψ ψ y [ ψ ] O[ δ ] [ δ ] ν O [ ψ ] O[ ν ] Non-dimensional stream unction ψ ν ψ ν 8
9 SIMILARITY SOLTIONS Convert the boundary layer equations V + 0 y + V ν y y Automatically satisied by ψ or d d d η dη '' + ''' 0 0 Substituting ψ V y ψ ψ ν so as to replace, V by Let y y η δ ν / Replace,, y y by derivatives WRT η 9
10 Some details o analysis Thus with: ψ ν ( η) η y ν we ind: ψ ψ η ( ) ν η y η y ν ψ 1 ν V ( η ) ; η y η y ν So inally: + 0 with b.c. s η 0: 0; η : 1 10
11 SIMILARITY SOLTIONS Substituting into the boundary layer equation + V ν y y or d d d 3 η + 3 d η '' + ''' rd order ordinary dierential wall ψ 0, wall 0, y,, hence ' ' ψ ν
12 VELOCITY PROFILES At η y ν 5, 0.99 Boundary layer thickness: δ ν 5 5 Re η y ν The tabulated Blasius velocity proile can be ound in many tetbooks 1
13 VELOCITY PROFILES V is much smaller than. η y ν V ν η y ν 13
14 DISPLACEMENT AND MOMENTN THICKNESS Boundary layer thickness y η η 5 δ ν / y ν Re Displacement thickness y 1 ( 1 ' ) ν δ dy dη η 0 0 Momentum thickness y ' ' ν θ 1 dy ( 1 ) dη η 0 0 δ, δ*, θ all grow with 1/ δ θ Re Re δ * /δ 0.344, θ/δ
15 DISPLACEMENT AND MOMENTN THICKNESS Typical distribution o δ, δ* and θ δ, δ*, θ(mm) m/s, ν m /s δ (m) δ θ 15
16 Shape actor: Other seul Results δ Re, θ H. 59 Re θ δ Wall shear stress: C τ µ ν Re w '' (0) ρ ρ y w 16
17 Short Problem The boundary layer over a thin aircrat wing can be treated as that over a lat plate. The speed o the aircrat is 100m/s and the chord length o the wing is 0.5m. At an altitude o 4000m, the density o air is 0.819kg/m 3 and the kinematic viscosity is m /s, Assuming the low over the wing is D and incompressible, Calculate the boundary-layer thickness at the trailing edge Estimate the surace riction stress at the trailing edge Will the boundary layer thickness and riction stress upstream be higher or lower compared to those at the trailing edge? What will be the boundary-layer thickness and the surace riction stress at the same chord location i the speed o the aircrat is doubled? 17
18 SOLTIONS The Reynolds number at 0.5m: Re ν 0 10 The boundary layer thickness: δ m 6 Re.5 10 Friction stress at trailing edge o the wing τ w 0.5ρuC ρu Re N / m 18
19 SOLTIONS Will the boundary layer thickness and riction stress upstream higher or lower compared to those at the trailing edge? δ is smaller upstream as δ is proportional to 1/ τ w is higher upstream as δ is thinner. I the speed o the aircrat is doubled, Re will be doubled since Re ν δ will be smaller as Re increases. δ τ w will be higher as the increase in u has a greater eect than the increase in Re τ w 0.5ρu Re 5 Re 19
20 Plane stagnation low Flows with pressure gradients can be sel-similar but it has to be a pressure gradient compatible with sel-similarity. See Schlichting and other advanced tetbooks on luid mechanics or eamples. Stagnation low provides one such eample where e 0 / L and Ve 0 y / L potential low) Note 1 dpe d e by Euler equ n. e ρ d d The boundary layer equation thus becomes: 0 + V + ν y L L y or Here primes denote di n wrt η y 0 ν L 0
21 Stagnation low results Note that the boundary layer has a constant thickness! However, the mass within the boundary layer increases continuously since the velocity rises linearly with distance rom the stagnation point. nlike the lat-plate boundary layer, the shear stress decreases continuously rom the wall. For this low H δ*/ θ.1 i.e. less than or the zero-pressuregradient boundary layer. 1
22 Asymptotic Suction Flow Sometimes it may be desirable to withdraw luid through the wall V V w I the suction is uniorm a point is reached where the boundary layer no longer grows with distance downstream and no urther change with occurs. Thus the continuity equation is just V/ y 0, i.e V V w The -momentum equation becomes: which is readily integrated to give d d Vw dy dy ν yv 1 ep( w ) ν A question or you: What is the skin riction coeicient?
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