Watering a garden with a hose, the water falls in a lazy arc a short way from the nozzle. Press a thumb over most of the nozzle, and a thin, fast jet shoots out to the far end of the garden. The tap was not touched. The same water now leaves much faster.
How much flows each second
Stand beside a pipe and ask how much water passes a given cross-section each second. Let the area of the section be \(A\) and let the water cross it at speed \(v\), at right angles to the section. In a short time \(\Delta t\), the water that was at the section moves on by \(v\,\Delta t\). All the water between the section and that new position has passed through. It fills a cylinder of volume \(A\,v\,\Delta t\).
\[ Q = \dfrac{A\,v\,\Delta t}{\Delta t} = Av \]
This is the volume flow rate. Its unit is m³/s. Smaller flows are given in litres per second, with \(1\,\text{L} = 10^{-3}\,\text{m}^3\). The mass flow rate, the mass that crosses the section each second, is \(\rho Q = \rho A v\).
A tap with a nozzle of area \(1\,\text{cm}^2\) fills a glass of \(0.5\,\text{L}\) in \(5\,\text{s}\). Find the volume flow rate and the speed of the water leaving the nozzle.
The volume flow rate is \(Q = 0.5 \times 10^{-3}\,\text{m}^3/5\,\text{s} = 10^{-4}\,\text{m}^3/\text{s}\), or \(0.1\,\text{L/s}\).
The nozzle area is \(1\,\text{cm}^2 = 10^{-4}\,\text{m}^2\). So \(v = Q/A = 10^{-4}/10^{-4} = 1\,\text{m/s}\).
A glass and a watch are enough to measure the speed of a stream.
A nozzle of area \(2\,\text{cm}^2\) fills a \(1\,\text{L}\) bottle in \(10\,\text{s}\). Find the speed of the water leaving the nozzle, in m/s.
Water flows at \(2\,\text{m/s}\) through a pipe of cross-section \(5\,\text{cm}^2\). Take the density of water as \(1000\,\text{kg/m}^3\). Find the mass of water that passes a section each second, in kg/s.
What goes in must come out
Take a tube of flow, or a real pipe, with a section of area \(A_1\) at one place and \(A_2\) further on. No fluid crosses the side of the tube. In steady flow nothing inside the tube changes with time, so the mass between the two sections stays fixed. Every kilogram that enters through \(A_1\) must therefore leave through \(A_2\) in the same time.
\[ \rho_1 A_1 v_1\,\Delta t = \rho_2 A_2 v_2\,\Delta t \]
Why. in the time \(\Delta t\) the mass that enters through \(A_1\) equals the mass that leaves through \(A_2\). None piles up inside, and none crosses the sides.
\[ \rho_1 A_1 v_1 = \rho_2 A_2 v_2 \]
Why. cancel \(\Delta t\). This form holds for any fluid, a gas included.
\[ A_1 v_1 = A_2 v_2 \]
Read it. for an incompressible fluid \(\rho_1 = \rho_2\). The volume flow rate is the same at every section.
In the steady flow of an incompressible fluid, the volume flow rate \(Av\) is the same at every cross-section of a tube of flow. The speed is inversely proportional to the area.
In Fig 2.1 the wide part carries \(Q = 4 \times 10^{-4} \times 1 = 4 \times 10^{-4}\,\text{m}^3/\text{s}\), or \(0.4\,\text{L/s}\). The narrow part has half the area, so \(v_2 = A_1 v_1/A_2 = 2\,\text{m/s}\). The slices of the second chip show why. Equal volumes must pass every section each second. A slice in the narrow part is twice as long as one in the wide part, so it must move twice as far each second.
For a round pipe the area is \(\pi d^2/4\), so the speed goes as \(1/d^2\). A pipe of half the diameter has a quarter of the area, and the water in it runs four times as fast. The same rule works for a river, with the area of its cross-section, width times depth.
A river in the hills is \(60\,\text{m}\) wide and \(3\,\text{m}\) deep, and it flows at \(1\,\text{m/s}\). Downstream it passes through a gorge only \(20\,\text{m}\) wide, where it is \(4.5\,\text{m}\) deep. Find the volume flow rate and the speed of the water in the gorge.
Treat each part of the river as a section of a wide pipe. The areas are \(A_1 = 60 \times 3 = 180\,\text{m}^2\) and \(A_2 = 20 \times 4.5 = 90\,\text{m}^2\).
\(Q = A_1 v_1\), and \(A_1 v_1 = A_2 v_2\).
\(Q = 180 \times 1 = 180\,\text{m}^3/\text{s}\). Then \(v_2 = 180/90 = 2\,\text{m/s}\).
About \(180\,\text{m}^3\) of water pass each second, and the river races through the gorge at \(2\,\text{m/s}\), twice its speed upstream.
Water flows steadily through a pipe that narrows to half its radius. What happens to the speed of the water?
Water moves at \(0.5\,\text{m/s}\) in a hose of inner diameter \(4\,\text{cm}\). It leaves through a nozzle of diameter \(1\,\text{cm}\). Find its speed at the nozzle, in m/s.
The stream from a tap
Watch the smooth stream from a tap turned a little. It is thickest at the tap and thins as it falls. The reason is continuity. Once the water leaves the tap, it falls freely and speeds up. By the third equation of motion, after a fall \(h\) its speed is \(v = \sqrt{v_0^2 + 2gh}\), where \(v_0\) is the speed at the tap.
The same volume passes every level each second, so \(A_0 v_0 = A v\). A larger speed lower down needs a smaller area there.
\[ A = \dfrac{A_0 v_0}{v} = \dfrac{A_0 v_0}{\sqrt{v_0^2 + 2gh}} \]
A stream that leaves a tap of \(2\,\text{cm}^2\) at \(2\,\text{m/s}\) is moving at \(\sqrt{4 + 12} = 4\,\text{m/s}\) after falling \(60\,\text{cm}\). Its area there is \(2 \times 2/4 = 1\,\text{cm}^2\), half the area at the tap.
Try the slider above. At \(0.5\,\text{m/s}\) the water is moving at \(1.80\,\text{m/s}\) after \(15\,\text{cm}\), so the area shrinks to \(0.28\,\text{cm}^2\). At \(2\,\text{m/s}\) the speed grows only to \(2.65\,\text{m/s}\), and the area stays at \(0.76\,\text{cm}^2\). The fall adds the same \(2gh\) to \(v^2\) in both cases. That is a large change for a slow stream and a small one for a fast stream.
A thin stream does not thin for ever. A few tens of centimetres down, it breaks into a line of drops. The cause is surface tension, which pulls a long thin column of liquid into beads. The chapter on Surface Tension explains the force behind it.
Water leaves a tap of area \(2\,\text{cm}^2\) at \(1\,\text{m/s}\) and falls freely. Take \(g = 10\,\text{m/s}^2\). Find the area of the stream \(40\,\text{cm}\) below the tap, in cm².
Why does a smooth stream of water from a tap become thinner as it falls?
A kitchen tap turned fully fills a \(15\,\text{L}\) bucket in about a minute. Its mouth is about \(1.5\,\text{cm}\) across. Estimate the speed of the water leaving the tap, in m/s.
Benedetto Castelli, a pupil of Galileo, was asked to settle quarrels about the rivers and canals of Italy. In 1628 he published a book on the measurement of running water. Its central idea was that the same quantity of water passes every section of a river in the same time. So a river must run faster where its section is smaller. This is the equation of continuity, stated for rivers.
Now return to the garden hose. The tap sends roughly the same volume each second into the hose, and all of it must leave through the nozzle. Suppose the nozzle has an area of \(2\,\text{cm}^2\) and the water leaves at \(2\,\text{m/s}\). A thumb that covers three quarters of it leaves only \(0.5\,\text{cm}^2\). By continuity the water now leaves at about \(8\,\text{m/s}\), four times as fast. From the chapter on Motion in Two Dimensions, the range at \(45°\) is \(v^2/g\): \(0.4\,\text{m}\) before, \(6.4\,\text{m}\) after.
The thumb trick has a limit. A narrower opening also resists the flow, so the tap delivers a little less water each second. The speed at the nozzle then grows somewhat less than the ratio of the areas. Continuity is still exact; it is \(Q\) that changes.
- Scaling the speed with the diameter instead of its square.The area of a round pipe is \(\pi d^2/4\). Half the diameter gives four times the speed.
- Mixing units: cm² with m/s, or litres with m³.Put the area in m² and the speed in m/s; then \(Q\) is in m³/s, and 1 L/s is \(10^{-3}\,\text{m}^3/\text{s}\).
- Writing \(A_1v_1 = A_2v_2\) for a gas that is being squeezed.For a compressible fluid, conserve mass instead: \(\rho_1A_1v_1 = \rho_2A_2v_2\).