Capillarity in Odd Shapes

No fountain from a short tube

In a clean glass tube of radius \(0.1\,\text{mm}\), water rises \(14\,\text{cm}\). Now take a tube of the same bore that stands only \(7\,\text{cm}\) above the water. Will the water climb to the top and spill over in a little fountain that never stops? It does not. The water reaches the top of the tube and stops there, as Fig 6.1 shows.

At the top of the short tube the water surface cannot go higher, so its shape changes instead. The meniscus flattens. A flatter surface has a larger radius \(R\), so the pressure step across it, \(2T/R\), is smaller. It flattens just enough for that step to hold the shorter column: \(2T/R = \rho g h'\), where \(h'\) is the length of the tube above the water. In the full tube, \(2T/r = \rho g h\). Dividing the two gives \(R h' = r h\).

For the \(7\,\text{cm}\) tube, \(R = r h/h' = 0.1 \times 14/7 = 0.2\,\text{mm}\), twice the radius of the tube. The water meets the glass at a larger apparent angle, with \(\cos\theta' = h'/h = 0.5\), so \(\theta' = 60^\circ\).

A tilted tube

Tilt a capillary at an angle \(\alpha\) to the vertical, keeping its foot in the water. The pressure just under the meniscus is still \(P_0 - 2T/r\), because the tube and the liquid have not changed. And pressure changes only with vertical depth, as the chapter on pressure in fluids showed. So the top of the column must stand at the same vertical height \(h\) as in an upright tube. Along the slant, the column is longer: its length is \(h/\cos\alpha\).

Tilt the tube so far that its top is less than \(h\) above the water. Then the liquid fills it to the end and stops, like the short tube. The meniscus flattens instead of overflowing.

A tube in a lift

Put the capillary in a lift that accelerates upward at \(a\). As the chapter on pressure in fluids showed, the liquid then behaves as if gravity were \(g + a\). Jurin's law becomes \(h = 2T\cos\theta/[\rho (g + a) r]\). The water rises less. Accelerating downward at \(a\), the effective gravity is \(g - a\), and the water rises more.

For the \(0.1\,\text{mm}\) tube, going up at \(10\,\text{m/s}^2\) doubles the effective gravity to \(20\,\text{m/s}^2\), and halves the rise to \(7\,\text{cm}\). Going down at \(5\,\text{m/s}^2\), \(g - a = 5\,\text{m/s}^2\) and the rise doubles to \(28\,\text{cm}\). In free fall, \(g - a = 0\): nothing pulls the column down, and the water climbs until it fills the tube, however long. On the Moon, where \(g\) is a sixth of its value on Earth, the rise would be six times larger, \(84\,\text{cm}\).

Between two plates

Stand two flat glass plates in water, parallel and a small gap \(d\) apart. Water rises between them. Take a width \(w\) of the plates. The water surface touches the glass along two lines of length \(w\), one on each plate, and each line pulls up with \(wT\cos\theta\). The water lifted is a thin sheet of width \(w\), thickness \(d\) and height \(h\).

Compare this with a round tube whose diameter equals the gap, so its radius is \(d/2\). The tube lifts water to \(2T\cos\theta/[\rho g (d/2)] = 4T\cos\theta/(\rho g d)\), twice as high as the plates. In the tube the water touches glass all round a circle. Between the plates it touches only two straight lines, as the small drawings seen from above in Fig 6.3 show. For a gap of \(0.2\,\text{mm}\), water rises \(7\,\text{cm}\) between plates and \(14\,\text{cm}\) in the tube.

The gap between two tubes

Put a thin glass tube of outer radius \(r_1\) inside a wider tube of inner radius \(r_2\), on the same axis. Dip both in water. Water climbs the ring-shaped gap. It touches the inner tube along a circle \(2\pi r_1\) and the outer tube along a circle \(2\pi r_2\). The pull is \(2\pi(r_1 + r_2)T\cos\theta\), and the weight of the ring of water is \(\pi(r_2^2 - r_1^2)h\rho g\).

Write \(r_2^2 - r_1^2 = (r_2 - r_1)(r_2 + r_1)\), and the factor \(r_1 + r_2\) cancels. What is left is \(h = 2T\cos\theta/[\rho g (r_2 - r_1)]\), the formula for plates with \(d = r_2 - r_1\). Tubes of radii \(0.5\,\text{mm}\) and \(0.7\,\text{mm}\) make a gap of \(0.2\,\text{mm}\), and water rises \(7\,\text{cm}\) between them, just as between plates \(0.2\,\text{mm}\) apart. A narrow ring is two plates bent round.

Plates that meet at an angle

Now let the two plates touch along one vertical edge and open at a small angle \(\alpha\), in radians, like a book opened a crack. At a distance \(x\) from the joined edge, the gap is \(d = \alpha x\). Each narrow strip of the gap behaves like two parallel plates of that gap. So the water rises to \(h = 2T\cos\theta/(\rho g \alpha x)\) there.

The product \(h x = 2T\cos\theta/(\rho g \alpha)\) is the same at every \(x\). The edge of the water is therefore a rectangular hyperbola: high near the joined edge, low far from it. For \(\alpha = 0.01\) rad, \(h x\) \(= 0.14/(10^4 \times 0.01)\) \(= 1.4 \times 10^{-3}\,\text{m}^2\), or \(14\,\text{cm}^2\). At \(2\,\text{cm}\) from the edge, the gap is \(0.2\,\text{mm}\) and the water stands \(7\,\text{cm}\) high. At \(4\,\text{cm}\) the gap has doubled, and the height has halved, to \(3.5\,\text{cm}\).

Wicks, blotting paper and soil

The diya of the opening works this way. A cotton wick is a bundle of twisted threads, and the gaps between the threads are narrow channels of every shape. Oil wets cotton, so it climbs these channels, as water climbs between plates. The narrower the gaps, the higher the oil can climb, so a tightly twisted wick feeds the flame from deep in the lamp. Blotting paper, a bath towel and a sugar cube dipped in tea lift liquid through the gaps between their fibres or grains.