3-13 - 3-13. In Figure 3.13 we have illustrated a capillary...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
3-13. In Figure 3.13 we have illustrated a capillary tube that has just been immersed in a pool of water. The water is rising in the capillary so that the height of liquid in the tube is a function of time. Later, in a course on fluid mechanics, you will learn that the average velocity of the liquid, , can be represented by the equation v z ⟨⟩ N o 2 o gravitational capillary force force viscous force 8v 2r z h gh r µ σ−ρ = ±²³ ±´² ´³ (1) in which is the average velocity in the capillary tube. The surface tension σ , capillary radius , and fluid viscosity μ can all be treated as constants in addition to the fluid density ρ and the gravitational constant g. From Eq. 1 it is easy to deduce that the final height of the liquid is given by v z o r o 2 h g r = σρ (2) In this problem you are asked to determine the height h as a function of time for the initial condition given by I.C. h t = = 0 , 0 (3) You may find it convenient to work with a dimensionless height defined by Hh t h = () (4) and you may wish to verify your analysis by doing a simple experiment. This would require that
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 07/15/2010 for the course ECM 051 taught by Professor B.g.higgins during the Winter '10 term at UC Davis.

Page1 / 4

3-13 - 3-13. In Figure 3.13 we have illustrated a capillary...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online