Capillary Rise Between Touching Glass Rods: 04-BS-7, December 2015, Q1

Question 1: Capillary Rise Between Touching Glass Rods (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Rod diameter d	1 mm (rods touching each other)
Surface tension σ	0.073 N/m
Wetting angle θ	0° (cosθ = 1)
Water density ρ	1000 kg/m³

Find. The capillary rise h above the free surface.

Fig. Q1 — rods touch their neighbours (closely packed square array); the shaded curved-square "pore" between any four adjacent rods is the meniscus cell that draws water up.

Approach. Use the general capillary-rise relation given on the paper's Reference Equations page, h = (σcosθ/ρg)×(perimeter/area), applied to the wetted perimeter and open area of ONE interstitial pore between four touching rods (this reduces to the familiar 2σcosθ/ρgr for a circular tube, so it is the correct generalization here).

Identify the pore geometry. Four touching rods of radius r = 0.5 mm have centres on a square of side 2r (since adjacent rods touch). The square cell minus the four quarter-circles of rod material (which together make one full circle of radius r) is the open pore water can climb through: $$A_{pore} = (2r)^2 - \pi r^2 = r^2(4-\pi)$$
Wetted perimeter of the pore. The meniscus contacts glass along the four quarter-arcs bounding the pore, which together total one full circumference: $$P_{pore} = 4\left(\frac{2\pi r}{4}\right) = 2\pi r$$
Substitute r = 0.0005 m: $$A_{pore} = (0.0005)^2(4-\pi) = 2.146\times10^{-7}\ \text{m}^2, \qquad P_{pore} = 2\pi(0.0005) = 3.1416\times10^{-3}\ \text{m}$$
Apply the capillary-rise formula. $$h = \frac{\sigma\cos\theta}{\rho g}\left(\frac{P_{pore}}{A_{pore}}\right) = \frac{0.073(1)}{1000(9.81)}\left(\frac{3.1416\times10^{-3}}{2.146\times10^{-7}}\right)$$ $$h = (7.441\times10^{-6})(14\,640) = \boxed{0.1089\ \text{m} = 108.9\ \text{mm}}$$

Quantity	Result
Capillary rise h	108.9 mm (0.109 m) above the free surface

Topic: Capillary rise in a non-circular interstitial space (hydraulic-radius generalization).

Key relations: circular tube h = 2σcosθ/ρgr; general form h = (σcosθ/ρg)×(perimeter/area), which reduces exactly to the circular case since perimeter/area = 2/r for a circle.

Why this works. Capillary rise balances the vertical component of surface-tension force acting along the wetted perimeter against the weight of the raised column; expressing this per unit cross-sectional area gives a formula that works for any meniscus shape, not just circular tubes — here the "tube" is the curved-boundary gap between four touching rods.

Common pitfall. Using the rod pitch (2r, the array spacing) as if it were a tube diameter in the circular formula overstates the open area and understates the rise; the pore is smaller and more sharply curved than a circular tube of the same pitch, which is why the rise (109 mm) is larger than 2σ/ρg(r) alone would suggest.

Source: White, Fluid Mechanics, Ch. 1 (surface tension and capillarity).

Question 1 of 13: Capillary Rise Between Touching Glass Rods

Question 1: Capillary Rise Between Touching Glass Rods (5 marks)