This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Prof. Daniel J. Bodony AE311, Fall 2011 Solutions to HW #7 Problem 1 Consider the laminar flow of a fluid layer falling down a plane inclined at an angle with the horizontal under the influence of gravity. If h is the thickness of the layer in the fully developed stage, show that the velocity distribution is u = g sin 2 ( h 2 y 2 ) where the x-axis points along the free surface and the y-axis towards the plane. Show that the volume flow rate per unit width is Q = gh 3 sin 3 and the frictional stress on the wall is o = gh sin . Answer 1. Let the x- y axes be oriented so that the x-axis lies on the free surface, with + x pointing downhill and the + y-axis pointing normal into the plane. Assume the fluid flowing down the wall is Newtonian, incompressible, steady ( / t = 0), and fully-developed ( / x = 0). Then the continuity equation implies u = u x + v y = so that v / y = 0. Thus v = constant which, when evaluated at the wall at y = h , implies v 0. The y-momentum shows that p / y = g cos which integrates to p = g cos y + f ( x ) for arbitrary function f . At the free surface, p = p atm is independent of x so f ( x ) = p atm . The x-momentum equation then simplifies to 2 u y 2 = g sin ....
View Full Document
- Fall '08