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

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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
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Unformatted text preview: gd cos We are given that in an unknown fluid whose density is 3 2 w , where w is the density of water, the capillary rise is 4 15 times that of water. Thus, we are given h = 4 w w gd cos and 4 15 h = 4 3 2 w gd cos So, combining these two equations, we have 4 15 4 w w gd cos = 4 3 2 w gd cos = = 2 5 w Since the surface tension of water at 68 o F is w = 0.0050 lb/ft, the surface tension of the unknown fluid is = 0 . 0020 lb / ft...
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

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