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Unformatted text preview: 5 Laminar External Flow 5.1 The boundary layer A boundary layer is an easy physical concept to grasp; it is the result of the noslip boundary condition as fluid flows over the surface of an object. Because of viscosity, the fluid directly adjacent to the object is sheared, and this effect is confined to a relatively thin layer parallel to the surface, i.e., the boundary layer. A boundary layer is also a mathematical concept, and boundary layer behavior can occur whenever the higherorder derivatives in a differential equation are multiplied by relatively small parameters. This statement is more arcane, yet it is important in terms of gaining insight into why physical boundary layers are thin, and what solution methods can be used to solve for the transport of momentum and heat across the boundary layers. A simple example both physical and mathematical of boundary layer behavior can be seen in a 1D convectivediffusive problem. Say a uniform, 1D flow enters a control volume with temperature T . At some distance L downstream, the flow is brought to a new temperature T 1 . All the while the density and the velocity u stay constant. The boundary at x = L might represent, for example, a porous surface which is maintained at T 1 . The steady 1D energy equation for this situation is UC dT dx = k d 2 T dx 2 (1) Denote nondimensional variables as T = T- T 1 T- T 1 , x = x L (2) and the DE becomes d T d x = 1 Pe L d 2 T d x 2 (3) T (0) = 1 , T (1) = 0 (4) with Pe L = uL = C uL k (5) The solution to this problem is T ( x,Pe L ) = e Pe L- e Pe L x e Pe L- 1 (6) A plot of the results is shown below. 0.2 0.4 0.6 0.8 1 x/L 0.2 0.4 0.6 0.8 1 (T-T 1 )/(T-T 1 ) Pe L = 0.1 Pe L = 1 Pe L = 10 Pe L = 100 Note that as Pe L increases, the temperature distribution goes from being linear to increasingly piled up into a narrow region adjacent to the x = L boundary. Indeed, for Pe L = 100 the temperature is basically 1 uniform at T = 1 except for a small region of steep gradient at the outflow surface. This region of steep gradient would be considered a boundary layer. The 2 nd order derivative, for this case, is multipled by the small parameter 1 / 100 = 0 . 01. This might lead one to assume that the 2 nd derivative term could be neglected from the DE and it could, through most of the flow region. Getting rid of this term would result in d T d x = 0- T = constant = 1 (7) where the constant is evaluated from the first BC. However, the problem has two boundary conditions, and the zerotemperature condition must be maintained at the outflow wall. A firstorder DE cannot accommodate two boundary conditions, so somewhere the secondorder derivative term must become significant. It will become significant in the neighborhood of the T = 0 surface....
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- Fall '10
- Heat Transfer