# The driving force for flow up the capillary tube is

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The driving force for flow up the capillary tube is correctly stated in the original problem, but somestudents 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 inthe derivation of the Hagen-Poiseuille equation for a horizontal tube.One way to account for gravity isto work in terms of the dynamic pressure, so that hydrostatic variations in pressure are removed.Withthe dynamic pressure P, the driving force for flow is accounted for in a single term that combinespressure and gravity,P=p− ρgso for gpointing vertically down,Pz=pz+ρgandP=p+ρgz+C,where the fact that C is a constant and not a function of r orθmay be concluded from the componentsof the Navier-Stokes equations in the r- andθ- directions.In the capillary tube, the pressure at z=0 is p0and the pressure just below the meniscus is p0-2γ/R, but the pressure drop driving flow is smaller than2γ/R because the hydrostatic effect augments the pressure at z=0 by an amount that does not contributeto flow.The flow is contributed by the difference in P, or
ΔP=Pz=0Pz=h=pz=0pz=h− ρgh=p0p02γR− ρghorΔP=2γR− ρgh,where the constant C cancelled out.This driving force is reduced relative toΔp by the hydrostaticcontribution to the pressure p at z=0 from the column with height h.This is the driving force given inthe problem.
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