EAS4101_S10_HW8S - EAS 4101 S10 HW#8 solutions Problem 1 a...

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EAS 4101 S10 HW #8 solutions Problem 1: a) 0 U h x y Assumptions (1) Fully developed: 0 V x = JG (2) Steady: ( ) 0 t = (3) Inc. (4) “Navier Stokes”: 2 DV pF V Dt ρρ =−∇ + + ∇ μ J GJ G (5) ( ) 0 w z == “2-D” (6) 0 b f = JJG (7) p const = throughout b) Continuity : u x N () 0 0 v vv x y = += = ( ) constant 1 v = X -momentum: 1/1 HW#6 N 0 steady uu u tx = ∂∂ + N 0 fully developed u v y ρ = + N N N 2 2 6 7 0 1 and continuity x b pu F xx ρμ = =− + + N 2 2 fully developed u y + 2 2 0 u y = governing differential equation 1-D Laplace: 1-D, 2 nd order, homogeneous ODE, therefore flow must be generated via BCs!
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EAS 4101 S10 HW #8 solutions c) Boundary conditions: () 00 vy vy h == 0 , uy uy h U d) Applying B.C.’s to continuity: ( ) constant 1 v = 0 everywhere v ∴= e) Integrate and apply B.C.’s to x momentum: 2 2 0 u y = Integrate twice 12 uc yc =+ Apply boundary conditions 2 c = 0 01 1 U uy h U ch c h = = Solution: 0 y uy U h = “linear profile” f) Shear Stress 2/2 HW#6 0 yx U du dy h τμ μ ==≡ constant Constant shear flow g) Vorticity k i ijk j u x ωε = z v x ω = N 0 FD U u yh −= , constant z
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EAS 4101 S10 HW #8 solutions Problem 2: Given: Consider the incompressible viscous flow of air between two infinitely long parallel plates separated by a distance h. Both plates are stationary, but a constant pressure gradient exists in the flow direction. Solution: (a) h x y 1 p 2 p g constant dp dx = Assume: 1) Newtonian Fluid 2) Laminar flow 3) Stokes’ Assumption 4) Incompressible 5) Fully-Developed 6) 2-D 7) steady 8) negligible body forces. (b) The Navier Stokes equation for this problem are 0 V V t ρ ∇⋅ = K K () 7 b VV p f ρρ +⋅ = + KK 8 2 V μ +∇ K Conservation of mass for 2-D gives u x 5 0 v y += From this, v is determined to be a constant. As v is equal to zero at the plates due to the solid-wall boundary condition, v = 0 everywhere. (c) B.C.s: u(y=0) = u(y=h) = 0 [No-slip condition] (d) As stated previously, v = 0 everywhere . 3/3 HW#6
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EAS 4101 S10 HW #8 solutions (e) Stating conservation of momentum in the x-direction, u u x ρ 5 v + 2 . 2 cons mass up u yxx μ ∂∂∂ =− + 5 2 2 u y + or () 2 2 2 12 1 1 2 du dp dy dx dp uy y Cy C dx = =+ + Solving for the constants using the no slip boundary condition at each wall yields 2 1 0 2 C hd p C dx = So 2 2 1 22 1 2 1 2 dp h dp y y dx dx or dp y h y dx or p y y dx h μμ ⎛⎞ ⎜⎟ ⎝⎠ (f) 1 2 du dp h dp y dy dx dx du dy τμ = For the bottom plate (y = 0), 2 hdp dx τ For the top plate (y = h), a negative must be added as the direction into the velocity profile from the flow is in the negative direction. 2 dx 4/4 HW#6
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EAS 4101 S10 HW #8 solutions This makes sense, as in the case where the pressure gradient is negative, the flow will be in the positive x direction, and a positive shear will be exerted on the plates.
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This note was uploaded on 09/05/2011 for the course EAS 4101 taught by Professor Sheplak during the Spring '08 term at University of Florida.

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EAS4101_S10_HW8S - EAS 4101 S10 HW#8 solutions Problem 1 a...

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