Derive an expression for the change in height h in a circular tube of a liquid with surface tension Υ and contact angle θ , as in Fig. E1.8.

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Accepted Answer

The vertical component of the ring surface-tension force at the interface in the tube must
balance the weight of the column of fluid of height h:
[latex]2 \pi R\Upsilon \cos heta =\gamma \pi R^2 h[/latex]
Solving for h, we have the desired result:
[latex]h=\frac{2\Upsilon \cos heta }{\gamma R}[/latex]
Thus the capillary height increases inversely with tube radius R and is positive if [latex] heta \lt 90°[/latex] (wetting liquid) and negative (capillary depression) if [latex] heta \gt 90°[/latex] Suppose that R = 1 mm. Then the capillary rise for a water–air–glass interface, [latex] heta \approx 0°[/latex] , [latex]\Upsilon =0.073 N/m[/latex] and [latex] ho =1000 kg/m^3[/latex] is
[latex]h=\frac{2(0.073 N/m)(\cos 0°)}{(1000 kg/m^3)(9.81 m/s^2)(0.001 m)} =0.015(N.s^2)/kg=0.015m=1.5cm[/latex]
For a mercury–air–glass interface, with [latex] heta =130°[/latex] ,[latex]\Upsilon =0.48N/m[/latex] and [latex] ho =13.600 kg/m^3[/latex] the capillary rise is
[latex]h=\frac{2(0.48)(\cos 130°)}{13,600(9.81)(0.001)} =-0.0046m =-0.46 cm[/latex]
When a small-diameter tube is used to make pressure measurements (Chap. 2), these capillary effects must be corrected for.

Question 1.8: Derive an expression for the change in height h in a circula...

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