Unformatted text preview: F ∑ = 2 w ⋅ σ ⋅ cos θ ( ) ⋅ ρ g ⋅ Δ h ⋅ w ⋅ a ⋅ − = Solving for Δ h Δ h 2 σ ⋅ cos θ ( ) ⋅ ρ g ⋅ a ⋅ = For water σ = 72.8 mN/m and θ = 0 o (Table A.4), so a) Tube Δ h 4 0.0728 × N m ⋅ 999 kg m 3 ⋅ 9.81 × m s 2 ⋅ 0.005 × m ⋅ kg m ⋅ N s 2 ⋅ × = Δ h 5.94 10 3 − × m = Δ h 5.94 mm ⋅ = b) Parallel Plates Δ h 2 0.0728 × N m ⋅ 999 kg m 3 ⋅ 9.81 × m s 2 ⋅ 0.005 × m ⋅ kg m ⋅ N s 2 ⋅ × = Δ h 2.97 10 3 − × m = Δ h 2.97 mm ⋅ =...
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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.
 Fall '07
 Lear
 Fluid Mechanics

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