Those equations are a 2 1 0 a 2 b 2 c 2 3 0 a 2 b 2 0

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Those equations are a 2 + 1 = 0 a 2 + b 2 + c 2 3 = 0 a 2 b 2 = 0 and a 3 + 1 = 0 a 3 + b 3 + c 3 = 0 a 3 b 3 2 = 0 . So a 2 = 1 b 2 = 1 c 2 = 1 and a 3 = 1 b 3 = 1 c 3 = 0 . The Pi groups are then Π 1 = D h Π 2 = ρ Uh μ and
Π 3 = γ μ U . Note that Π 2 is a Reynolds number and Π 3 is the inverse of the Capillary number. c) Doing as suggested means setting up exactly the same problem as was solved for part “b”, except that the velocity V replaces U. Also, h has been removed from the list of dimensional variables. The only remaining dimensional length scale is D. We therefore obtain two dimensionless groups, numbered in a way that corresponds to part “b”, as Π 2 = ρ VD μ = ρ UD 2 μ h and Π 3 = γ μ V = γ h μ UD . Physically the velocity V represents the velocity variation over the particle length scale. d) If the mean diameter D is independent of the density r , then Π 3 must be independent of Π 2 . The performance of the emulsifier, as described by D, must then depend only on one group, so Π 3 = γ h μ UD = K where K is a constant, and D = γ h K μ U . This result implies that higher surface tensions and larger gap heights h yield larger oil droplets, and higher viscosities and velocities yield smaller droplets, under the conditions where the density becomes unimportant. 3. In the figure below, the embankment surface consists of two linear sections, both submerged under water. Above the water is air at atmospheric pressure. In computing the force components, include the effect of the atmospheric pressure in your answer. The width of the embankment in the y- direction is W. a) (20 pts.) Derive an expression for the horizontal, or x-directed component of force on the embankment. b) (20 pts.) Derive an expression for the vertical, or z-directed compononent of force on the embankment.
Solution

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