Problem 3.40

# Problem 3.40 - F ∑ = 2 w ⋅ σ ⋅ cos θ ⋅ ρ g ⋅...

This preview shows page 1. Sign up to view the full content.

Problem 3.40 [Difficulty: 2] Water Given: Water in a tube or between parallel plates Find: Height Δ h for each system Solution: a) Tube: A free-body vertical force analysis for the section of water height Δ h above the "free surface" in the tube, as shown in the figure, leads to F 0 = π D σ cos θ () ρ g Δ h π D 2 4 = Assumption: Neglect meniscus curvature for column height and volume calculations Solving for Δ h Δ h 4 σ cos θ () ρ g D = b) Parallel Plates: A free-body vertical force analysis for the section of water height Δ h above the "free surface" between plates arbitrary width w (similar to the figure above), leads to
This is the end of the preview. Sign up to access the rest of the document.

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 ⋅ =...
View Full Document

## 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.

Ask a homework question - tutors are online