# 4 as we have seen in videos shown in class when a

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4. As we have seen in videos shown in class, when a glass capillary tube is placed in water or alcohol, the liquid rises up the tube to a height that is determined by the coefficient of surface tension and the tube radius. To analyze the dynamics of that process, assume that the flow in the tube is “quasisteady” (sometimes called “pseudosteady”). In other words, assume that even though the pressure gradient driving the flow decreases as the fluid rises, the flow is always approximately governed by the Hagen- Poiseuille equation, v z = dh dt = R 2 8 μ dP dz .
If the fluid moves slowly enough, then the dynamic pressure difference between the top and bottom of the liquid in the tube is a combination of the Young-Laplace pressure difference and the hydrostatic contribution, i.e. Δ P = 2 γ R − ρ gh(t) . Use this pressure difference and the quasisteady assumption to derive an equation that gives the time t required for the liquid height to become h(t). Let H be the final height of the liquid in the capillary.
8 μ R 2 2 γ R ρ g ( ) 2 ln 2 γ R = C and 8 μ R 2 h ρ g + 2 γ R ρ g ( ) 2 ln 1 ρ ghR 2 γ = t but the final height H was derived in class to be H = 2 γ R ρ g so t = 8 μ H ρ gR 2 h H + ln 1 h H . Comment on the driving force for flow in capillary rise The driving force for flow up the capillary tube is correctly stated in the original problem, but some students were wondering how it is obtained. The following is an explanation. Since the flow in this problem is vertically upward, gravitational effects are not trivially dismissed as in the derivation of the Hagen-Poiseuille equation for a horizontal tube. One way to account for gravity is to work in terms of the dynamic pressure, so that hydrostatic variations in pressure are removed. With the dynamic pressure P, the driving force for flow is accounted for in a single term that combines pressure and gravity, P = p − ρ g so for g pointing vertically down, P z = p z + ρ g and P = p + ρ gz + C , where the fact that C is a constant and not a function of r or q may be concluded from the components of the Navier-Stokes equations in the r- and q - directions. In the capillary tube, the pressure at z=0 is p 0 and the pressure just below the meniscus is p 0 -2 g /R, but the pressure drop driving flow is smaller than 2 g /R because the hydrostatic effect augments the pressure at z=0 by an amount that does not contribute to flow. The flow is contributed by the difference in P, or Δ P = P z = 0 P z = h = p