Capillary Rise in a Flattened Tube: 04-BS-7, May 2014, Q1

Question 1 — Capillary Rise in a Flattened Tube (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
Flat side length	2 mm
Gap between flats (= semicircle diameter)	1 mm
Total cross-section width	3 mm
Surface tension, σ	0.0728 N/m
Contact angle, θ	0°

Find. The capillary rise h of water in the tube.

Fig. Q1 — flattened-tube cross-section: a 2×1 mm rectangle capped by two semicircles of radius 0.5 mm (a "stadium" section). Water rises inside it by capillarity.

Approach. For a non-circular tube the reference-equation form of capillary rise, h = (σcosθ/ρg)×(perimeter/area), replaces the usual 4σ/(ρgD) for a round capillary.

Cross-section perimeter and area. The section is a 2×1 mm rectangle capped by two semicircles of radius 0.5 mm (their combined diameter of 1 mm matches the 1 mm gap, and their combined arc is one full circle): $$P = 2(2\text{{ mm}}) + \pi(1\text{{ mm}}) = 4 + 3.1416 = 7.1416\text{{ mm}}$$ $$A = (2\text{{ mm}})(1\text{{ mm}}) + \pi(0.5\text{{ mm}})^2 = 2 + 0.7854 = 2.7854\text{{ mm}}^2$$
Apply the capillary-rise reference equation. $$h = \frac{{\sigma\cos\theta}}{{\rho g}}\cdot\frac{{P}}{{A}} = \frac{{0.0728\times\cos 0^\circ}}{{1000\times 9.81}}\times\frac{{7.1416\times10^{{-3}}\text{{ m}}}}{{2.7854\times10^{{-6}}\text{{ m}}^2}}$$ $$h = (7.421\times10^{{-6}}\text{{ m}}^2)(2564\text{{ m}}^{{-1}}) = \boxed{{0.01903\text{{ m}} = 19.0\text{{ mm}}}}$$

Quantity	Result
Perimeter / area ratio	2564 m⁻¹
Capillary rise, h	19.0 mm

Topic: Capillarity in a non-circular tube.

Key relations: h = (σcosθ/ρg)×(perimeter/area); for a round tube this reduces to the familiar h = 4σcosθ/(ρgD) since P/A = 4/D there.

Why this works: the meniscus pulls upward with a force proportional to the wetted perimeter (σ·P·cosθ), while the weight of the risen column is proportional to the cross-sectional area (ρg·A·h). Equating these forces and solving for h gives the P/A form directly — it applies to any constant cross-section, not just circles.

Common pitfall: using the round-tube formula with an equivalent diameter instead of computing the true P/A ratio; also forgetting that two semicircular ends combine into one full circle's worth of arc, not two half-circles counted separately in area only.

Source: White, Fluid Mechanics, 8th ed., §2.9 (surface tension and capillarity).

Question 1 of 13: Capillary Rise in a Flattened Tube

Question 1 — Capillary Rise in a Flattened Tube (5 marks)