mech7220-laminar-bl-flow

# mech7220-laminar-bl-flow - 5 Laminar External Flow 5.1 The...

This preview shows pages 1–3. Sign up to view the full content.

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 no–slip 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 higher–order 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 1–D convective–diffusive problem. Say a uniform, 1–D flow enters a control volume with temperature T 0 . At some distance L downstream, the flow is brought to a new temperature T 1 . All the while the density ρ 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 1–D energy equation for this situation is ρUC dT dx = k d 2 T dx 2 (1) Denote non–dimensional variables as T = T - T 1 T 0 - 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 = u L α = ρ C u L k (5) The solution to this problem is T ( x, Pe L ) = e P e L - e P e L x e P e L - 1 (6) A plot of the results is shown below. 0 0.2 0.4 0.6 0.8 1 x/L 0 0.2 0.4 0.6 0.8 1 (T-T 1 )/(T 0 -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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 zero–temperature condition must be maintained at the outflow wall. A first–order DE cannot accommodate two boundary conditions, so somewhere the second–order derivative term must become significant. It will become significant in the neighborhood of the T = 0 surface. The thickness of the boundary layer, for this simple problem, can be determined by reformulating the problem. Reverse the coordinate direction, so that x runs from the T = 0 surface, and define a scaled (or stretched) coordinate ˜ x via ˜ x = x Pe L (8) The scaled DE becomes - d T d ˜ x = d 2 T d ˜ x 2 (9) T (0) = 0 , T x
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern