052307-132a-hw5 - μ ∂u ∂z v v v v z =0 = A cos ωt....

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C HEMICAL E NGINEERING 132A Professor Todd Squires Spring Quarter, 2007 Homework 5 Due Wednesday, May 30 Problem 1. p503, prob 2 Problem 2. p504, prob 50 Problem 3. p504, prob 51 Problem 4. Fluid Mechanics example. The equation for a ‘uni-directional’ ±uid ±ow (i.e. in the x -direction) above a ±at, solid boundary (at z = 0 , say) is ∂u ∂t = μ ρ 2 u ∂z 2 , (1) where ρ is the ±uid density and μ is the ±uid viscosity. For water, ρ = 1g/cm 3 and μ = 10 - 2 g cm 2 /s. Say there is so much ±uid above the plate that you can ignore the top boundary. Say we drive the bottom boundary from side to side with a given stress (force per unit area) – this imposes a boundary condition on the ±uid
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Unformatted text preview: μ ∂u ∂z v v v v z =0 = A cos ωt. (2) Assuming you have been shaking the wall for a while (i.e. so that transients have died out), solve for the ±uid ±ow. This is very similar to the problem we did in class. Look at your solution – how far do the shear waves (i.e. the ±uid ±ow) propagate into the ±uid? How does it depend upon frequency? What about the amplitude of motion – how does it depend upon frequency? Does this make sense to you intuitively?...
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This note was uploaded on 12/29/2011 for the course CHE 132a taught by Professor Gordon,m during the Fall '08 term at UCSB.

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