{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

p0168

# p0168 - gd cos φ We are given that in an unknown fluid...

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

70 CHAPTER 1. INTRODUCTION 1.68 Chapter 1, Problem 68 Problem: A round capillary tube of diameter d is inserted into a pan of water. A second tube, also of diameter d , is inserted into a pan of unknown liquid whose density is 3 2 ρ w , where ρ w is the density of water. If the liquid rises to a height h in water and 4 15 h in the unknown liquid, what is the surface tension of the unknown liquid? Assume the wetting angle is the same in both tubes and that T =68 o F. Solution: In general, the height to which a fluid rises in a circular capillary tube is h = 4 σ ρ
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: gd cos φ We are given that in an unknown fluid whose density is 3 2 ρ w , where ρ w is the density of water, the capillary rise is 4 15 times that of water. Thus, we are given ∆ h = 4 σ w ρ w gd cos φ and 4 15 ∆ h = 4 σ 3 2 ρ w gd cos φ So, combining these two equations, we have 4 15 4 σ w ρ w gd cos φ = 4 σ 3 2 ρ w gd cos φ = ⇒ σ = 2 5 σ w Since the surface tension of water at 68 o F is σ w = 0.0050 lb/ft, the surface tension of the unknown fluid is σ = 0 . 0020 lb / ft...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online