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Boundary Layer (Reorganization of the Lecture Notes from Professor Anthony Jacobi and Professor Nenad Miljkoic) Consider a steady flow of a Newtonian, Fourier-Biot fluid oer a flat surface with

1 Boundary Layer (Reorganization of the Lecture Notes from Professor Anthony Jacobi and Professor Nenad Miljkoic) Consider a steady flow of a Newtonian, Fourier-Biot fluid oer a flat surface with constant properties, incompressible, -D flow and no body force. Note: 1. A Newtonian fluid is a fluid in which the iscous stresses arising from its flow, at eery point, are linearly proportional to the local strain rate-the rate of change of its deformation oer time. That is equialent to saying that those forces are proportional to the rates of change of the fluid's elocity ector as one moes away from the point in question in arious directions.. A non-newtonian fluid is a fluid whose iscosity is ariable based on applied stress or force. Many polymer solutions and molten polymers are non-newtonian fluids, as are many commonly found substances such as ketchup, starch suspensions, paint, and shampoo. In a Newtonian fluid, the relation between the shear stress and the strain rate is linear, the constant of proportionality being the coefficient of iscosity. In a non-newtonian fluid, the relation between the shear stress and the strain rate is nonlinear, and can een be time-dependent. Therefore a constant coefficient of iscosity cannot be defined. Steady flow: t = -D flow: w =, z = Therefore for the continuity equation what we are left with is: u + y = -momentum ρ ( u t + u u + u y + w u z ) = ρf P 4 u [ + 3 μ 3 μ ( y + w z )] After cleaning up: [μ ( u + y ρ (u u + u y ) = P + μ( u + u y ) y + )] [μ ( w + z + u z )] Assume the effects of the wall to be confined to a region near the wall and call that region the boundary layer, ( ). Scaling analysis Gien u + =, U y Therefore ~ P = ~ U 1

2 Gien ρ (u u + u y ) = μ( u + u y ) U U U, U ~ν( U, U δ ) U δ since and can be neglected. Therefore Energy equation u u + u y = ν u y U ~ν U ~ ν U = 1 Re (when, Re > 1) ρc ( T T T + u + t y T (k = ) T + w ) + T ( P z T ) ( u + y + w z ) (k T y ) + y T (k + z ) z μ : iscous dissipation U : internal heat generation Here we first neglect both iscous dissipation and internal heat generation u + = as is gien by the continuity equation. y Cleaning up and we get: u T T + y = k ( T ρc + T y ) Now consider a temperature boundary layer ( ). Assume,,. + μ + U Let T = T s T and k = k = α ρc ρc Scaling analysis U T, Since, U T ~α ( T, T )

3 Goerning equation becomes: u T = α T y U ~ α Plug in ~ ν U We get ~ ν α = Pr (when Pr<.1 ) Therefore h~ k q = k T y y= q ~ k T h = q T Nu = h k ~ = Re Pr For ~, Nu ~ Re Pr For >, Nu ~Re 1 (Pr) 1 3, ~ Pr 1 3 for high Pr. As a summary, for laminar boundary layer with Re > 1 ~ Re Pr, Nu ~ Re Pr Pr 1, Nu ~ Re Pr Pr, Nu ~ Re Pr 1/3 (neglect iscous dissipation) Now let s consider iscous dissipation: = [( u ) + ( y ) + ( w z ) ] + ( u y + ) + ( w y + z ) + ( w + u z ) 3 ( u + y + w z ) Since -D, incompressible, constant properties, steady flow, no body force, no U = [( u ) + ( y ) ] + ( u y + ) ~ U, δ U δ, U δ (dominant), U, δ U Since ~U / Therefore for a boundary layer with iscous dissipation 3

Continuity equation: u + y = Momentum equation: u u + u = ν u y y Energy equation: u T + T y = α T y + ν C p ( u y ) Our results from before are unchanged for continuity and momentum.

4 Continuity equation: u + y = Momentum equation: u u + u = ν u y y Energy equation: u T + T y = α T y + ν C p ( u y ) Our results from before are unchanged for continuity and momentum. We consider three scenarios for energy. Case one cannot happen. If is that big and dissipation matters, we can t hae T at as kinetic energy is dissipated into heat. Case two: ~ = δ U T, δ U T ~ α T δ δ, ν U C p δ 1~ k 1 ρc p U δ, ν U C p δ U T 1~ δ [ μ ρu k C p μ, U C p T ν U ] Recall Pr = C pμ, Re k = U ν And define Ec = U (Eckert number) C p T From preious analysis δ ~Re 1~ 1 Pr, Ec Adection ~ Diffusion, Dissipation For Pr~1, if Ec 1, adection~diffusion and we neglect dissipation; if Ec~1, adection~diffusion~dissipation; if Ec 1, not possible for ~. Case three: U T, U T ~ α T δ, ν U T C p δ 1~ k 1 ρc p U δ, ν U T C p δ 1~ δ [ 1 Re Pr δ, Ec ] T Re U T 4

5 As before ~Re 1~ 1 Pr, Ec For Pr~1 or Pr 1, adection~dissipation. Ec~1 u T = ν C p ( u y ) For Pr 1 might get diffusion. Friction and heat transfer We hae the model equations; they tell us about friction and heat transfer τ = μ ( u y ) y= q = h(t s T ) = k ( T y ) y= Only the gradient at the wall matters. τ = 1 L L τ Note (u ) + (u) y (ut) + (T) T u = u + T + T y y So we can write (assume no dissipation) L h = 1 L h u u u u u u u = u + u + u + = u + + = u + y y y y T T T T T + = u + + = u + y y y Continuity equation: u + y = Momentum equation: (u ) + (u) = ν u y y Energy equation: (ut) + (T) = α T y y Integrate continuity equation, use Leibnitz, go to δ: Integrate momentum equation: Integrate energy equation: d ( u dy) d( udy)/ + = d ( utdy) = d ( udy ) + u u = ν[( u y ) ( u y= y ) ] y= d ( u dy) + u = ν[ ( u y ) ] y= + T T = α[( T y ) ( T y= y ) ] y= 5

6 d ( utdy) + T = α[ ( T y ) ] y= Substitute from continuity equation into the other two equations and rearrange. Alternatiely, d ( d ( u(u u)dy) u(t T)dy) τ ρ = d ( = ν ( u y ) y= = α ( T y ) y= u(u u)dy) q = d ( u(t T )dy) ρc p To use these equatons Assume a elocity profile; Obtain 1 st order ordinary differential equation for δ(); Sole for δ() and use profile to get τ. There is no analytical solution aailable!!!! Assume u U = f ( y δ ) = f(ƞ) where ƞ = y δ Boundary conditions: f(ƞ) = a + bƞ + cƞ + dƞ 3 u(, ) = a = u(, δ) = U b + c + d = 1 ( u y ) = b + c + 3d = y=δ ( u y ) = c = (at y=, u u + u = ν u = ) y= y y Let u U = 3 (y δ ) 1 (y δ )3 Soling O.D.E. of δ(), we get δ() = 4.64 ( ν U ).5 δ = 4.64 Re τ s = μ ( u y ) =.33μU.5 Re y= C f = τ s ρu / C f =.646 Re =.33ρU Re 6

7 Assume θ = T T s = e + fƞ + gƞ + hƞ 3 T T s where ƞ = y/ Apply boundary conditions θ = 3 y 1 ( y 3 ) Energy in integral form: U d ( u U (1 θ)dy) = α ( θ y ) y= Substitute the profiles of u and θ, for a case where ~ or Pr~1 or Pr 1 Sole and we get = 4.53 Re 1 Nu = h = q θ = = 3 =.331Re k (T s T )k y y= Pr 1 3 (in accord with the results from scaling analysis) Pr 1 3 Similarity solutions to boundary layer equations (Through similarity (introduction of φ), we turn PDE to ODE so that we can sole.) Steady state, constant properties, -d, no body force in, neglect, no U, boundary layers, no P Define Continuity equation: u + y = Momentum equation: u u + u = ν u y y Energy equation: u T + T = α T y y u = φ φ, = y Continuity is satisfied, substitute this into momentum equation. φ φ y y φ φ y = ν 3 φ y 3 Out of the blue: Assume φ = H(ε)f(ƞ) ε = ƞ = y U ν, H = νuε 1 φ y = φ ε ε y + φ ƞ ƞ y = + νuε f U ν = Uf = u φ = φ ε ε + φ ƞ ƞ = Uν 1 (f ƞf ) = φ y = U U ν f 7

3 φ y 3 = U ν f φ y = 1 Uƞf Substitute into momentum and rearrange where f = df/dƞ f + 1 f f = (magically ) Boundary conditions: u(, ) = when y= ƞ = for > Therefore @ ƞ = f = Uf ƞ= = (, ) = Therefore

8 3 φ y 3 = U ν f φ y = 1 Uƞf Substitute into momentum and rearrange where f = df/dƞ f + 1 f f = (magically ) Boundary conditions: u(, ) = when y= ƞ = for > ƞ = f = Uf ƞ= = (, ) = = f = u(, ) = U Therefore f ƞ = 1 The nonlinear PDE modeling momentum is now ODE. Sole numerically. How do we determine what to assume? The boundary layer equations model an artificial thing. No natural length scale. We contried δ. In the boundary layers, y goes from to 1. δ Scale analysis showed δ ~ ν U Therefore y δ ~y U ν ƞ This is how we get our assumption! Group ariables together and turn PDE to ODE!!! Recall our discussion T = 1 α T t T(, ) = T i T(, t) = T i T(, t > ) = T s δ is the penetration depth. T = T s T i Scale: T δ ~ 1 α T t δ~ αt δ ~ αt Let ƞ = and T(, t) = f(ƞ) we hae f + 1 f ƞ = αt 8

9 Similarity steps: 1. Group ariables together;. Turn PDE to ODE. The thermal boundary layer u T + T y = α T y T(, ) = T s T(, y) = T T(, ) = T Take θ = T T s T T s u θ + θ y = ν/pr θ y θ(, ) =, θ(, y) = 1, θ(, ) = 1 Define θ o (ƞ) = θ(, y), ƞ = y U ν The transform will yield θ o + Pr fθ o =, θ o () =, θ o () = 1 Nu = θ o Re Sole numerically! The numerical solution is ery closely approimated with Nu Re =.564Pr 1/ Pr<.5 Nu Re =.336Pr 1/3.6<Pr<1 Nu Re =.339Pr 1/3 Pr>1 9

CHAPTER 4 BOUNDARY LAYER FLOW APPLICATION TO EXTERNAL FLOW

CHAPTER 4 BOUNDARY LAYER FLOW APPLICATION TO EXTERNAL FLOW 4.1 Introduction Boundary layer concept (Prandtl 1904): Eliminate selected terms in the governing equations Two key questions (1) What are the