Ae311_fall_2011_hw_7_soln

# Ae311_fall_2011_hw_7_soln - Prof Daniel J Bodony AE311 Fall...

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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 θ 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 θ 2 ν ( 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 θ 3 ν and the frictional stress on the wall is τ o = ρ gh sin θ. 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 = − ρ g cos θ which integrates to p = − ρ g cos θ 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 θ ν ....
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Ae311_fall_2011_hw_7_soln - Prof Daniel J Bodony AE311 Fall...

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