Problem 3.38 - Problem 3.38 [Difficulty :2] Fluid 1 Fluid 2...

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Problem 3.38 [Difficulty :2] Fluid 1 Fluid 2 σπ D cos θ ρ 1 g Δ h π D 2 /4 Given: Two fluids inside and outside a tube Find: (a) An expression for height Δ h (b) Height difference when D =0.040 in for water/mercury Assumptions: (1) Static, incompressible fluids (2) Neglect meniscus curvature for column height and volume calculations Solution: A free-body vertical force analysis for the section of fluid 1 height Δ h in the tube below the "free surface" of fluid 2 leads to F 0 = Δ p π D 2 4 ρ 1 g Δ h π D 2 4 π D σ cos θ () + = where Δ p is the pressure difference generated by fluid 2 over height Δ h , Δ p ρ 2 g Δ h = Hence Δ p π D 2 4 ρ 1 g Δ h π D 2 4 ρ 2 g Δ h π
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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