Transient Rise of Water in a Capillary Tube
Authors: Housam Binous and Ahmed Bellagi
Transient Rise of Water in a Capillary Tube
Authors: Housam Binous and Ahmed Bellagi
Authors: Housam Binous and Ahmed Bellagi
The rise of water in a narrow capillary tube is governed by the combined effects of capillary force, gravity, and viscous resistance. Here, we use a time-dependent mass of the rising liquid column, which introduces a nonlinear inertial term in the equation of motion. The governing equation is derived directly from Newton’s second law of motion applied to the liquid column of height (h(t)) and cross-sectional area (A). This governing equation includes:1. Inertia of the rising column, which accounts for the increasing mass as the column grows.2. Viscous damping, representing the laminar flow resistance along the tube walls.3. Gravity, which opposes the rise.4. Capillary driving force, which pulls the liquid upward in a wetting tube (theta < 90 -- for water we assume perfect wetting or theta=0).Initial conditions are set to a small positive height () to avoid singularity and zero initial velocity ((t=0)=0).The equation is solved numerically using Mathematica’s NDSolve. Then,we plot the height and velocity of water column as a function of time, showing the rapid initial rise followed by asymptotic approach to the equilibrium height: 1.467 cm. For a 1 mm radius glass capillary and standard water properties at room temperature, the equilibrium height is on the order of 15 mm, consistent with classical capillary rise theory. Finally, we check our laminar regime assumption by plotting the Reynolds number versus time.
h(t=0)=
-6
10
h
t
rho=1000;(*waterdensity,kg/m^3*)mu=1*10^-3;(*waterdynamicviscosity,Pa·s*)sigma=0.072;(*surfacetensionofwater,N/mor72dynes/cm*)theta=0;(*contactangledegrees--assumeperfectwetting*)
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R=1*10^-3;(*tuberadius,m*)A=PiR^2;(*tubecross-sectionalarea*)
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g=9.81;(*gravity,m/s^2*)
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Fcap=2PiRsigmaCos[theta];(*capillaryforce---positiveforwetting*)(*Newton'ssecondlawofmotion*)capillaryODE=rhoA(h[t]h''[t]+(h'[t])^2)+8muA/R^2h[t]h'[t]+rhogAh[t]==Fcap;(*Initialconditions*)ics={h[0]==1*10^-6,h'[0]==0};(*smallinitialheighttoavoidsingularity*)tmax=2;(*timespanof0-2secondsissufficient--characteristictimescalerhoR^2/mu=1s*)
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sol=NDSolve[{capillaryODE,ics},h,{t,0,tmax}];(*solutionofODEusingNDSolve*)
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hfinal=h[tmax]/.sol[[1]];Print["Final water height (cm): ",100hfinal];
Final water height (cm): 1.46772
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(*plotofWaterriseinmm*)Plot[Evaluate[100h[t]/.sol],{t,0,tmax},Frame->True,FrameLabel->{"Time (s)","Height (cm)"},PlotStyle->Darker@Blue,PlotLabel->"Capillary Rise of Water",GridLines->Automatic,AspectRatio->1,PlotRange->All]
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(*plotofvelocityinm/s*)Plot[Evaluate[h'[t]/.sol],{t,0,tmax},Frame->True,FrameLabel->{"Time (s)","Velocity (m/s)"},PlotStyle->Darker@Red,PlotLabel->"Velocity",GridLines->Automatic,AspectRatio->1,PlotRange->All]
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(*plotofReynoldsnumber*)Plot[Evaluate[Abs[h'[t]]2Rrho/mu/.sol],{t,0,tmax},Frame->True,FrameLabel->{"Time (s)","Reynolds number"},PlotStyle->Darker@Green,PlotLabel->"Reynolds number",GridLines->Automatic,AspectRatio->1,PlotRange->All]
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