Water in pipes

We need to keep two quantities apart. One is how fast the water moves: its speed $v$, in metres per second. The other is how much water passes each second: the volume flow rate $Q$, in cubic metres per second or litres per second.

Look at a pipe with the cross-section area $A$, full of water moving at the speed $v$. In a short time $\Delta t$, every bit of water moves forward the distance $v\,\Delta t$. So the water that crosses a chosen line across the pipe in that time is a slab with length $v\,\Delta t$ and area $A$, which has the volume $A v\,\Delta t$. Dividing by the time gives the flow rate:

One cubic metre is $1000$ litres, so $8\ \mathrm{L/s}$ is $0.008\ \mathrm{m^3/s}$.

Change the pipe’s radius and the water’s speed. The shaded slab is the water that crosses the dashed line in one second.

The Venturi pipe later in this lesson has an inlet radius of $6\ \mathrm{cm}$, so its area is $A_1 = \pi \times 0.06^2 = 0.01131\ \mathrm{m^2}$. With $Q = 0.008\ \mathrm{m^3/s}$, the speed there is $v_1 = Q/A_1 = 0.008/0.01131 = 0.707\ \mathrm{m/s}$.

Now let the pipe narrow. Water hardly squeezes: under ordinary pressures its volume stays the same, so it cannot pile up anywhere inside a full pipe. Once the flow is steady, the same volume must leave through the narrow part each second as enters through the wide part. So $Q$ is the same at every cross-section, and by chapter 1:

A pipe’s cross-section is a circle, with the area $\pi r^2$. If the narrow part’s radius is $0.55$ times the inlet’s, its area is $0.55^2 = 0.3025$ times the inlet’s area, because $\pi$ is the same and each radius is squared. The water must therefore move $1/0.3025 = 3.31$ times as fast: $v_2 = 0.707 \times 3.31 = 2.34\ \mathrm{m/s}$. This comparison is between two sections of the same steady flow. In the experiment below a pump is assumed to maintain the chosen flow rate. A real tap does not necessarily keep the flow rate fixed when a thumb changes the outlet: the total flow can decrease. Continuity still says that, at each steady setting, the same volume passes every section each second.

Make the narrow part narrower. The two shaded slabs are the water that passes each dashed line in half a second: the same volume, but long and thin where the pipe is narrow.

In the narrow part the water moves faster, so its kinetic energy is larger. Something must do work on it. In a level pipe nothing pushes the water except the water around it, so the extra speed must come from a difference in pressure: the water behind pushes harder than the water ahead. The work and energy page gives us the rule we need: the work done on a body equals the change in its kinetic energy.

Consider all the water between a wide inlet and a narrow outlet during a short time $\Delta t$. In steady incompressible flow, the same small volume $\Delta V$ enters and leaves. The overlapping water in the middle has the same velocity distribution before and after; its kinetic energy does not change. What differs is an incoming volume moving at $v_1$ and an outgoing equal volume moving at $v_2$. Each has mass $\rho\,\Delta V$, so their energy difference is $\tfrac12\rho\,\Delta V(v_2^2-v_1^2)$. Now count the pressure work at the two ends. The inlet pressure pushes with $p_1A_1$ through $d_1$, where $A_1d_1=\Delta V$, supplying $p_1\Delta V$. The outlet pressure pushes against the motion with $p_2A_2$ through $d_2$, where $A_2d_2=\Delta V$, taking away $p_2\Delta V$. In our level, frictionless model, the net work equals that kinetic-energy increase:

In the second line we divided by $\Delta V$, and in the third we moved the terms so that each side belongs to one place. So along the pipe the combination $p + \tfrac{1}{2}\rho v^2$ stays the same. This is Bernoulli’s equation for a level pipe. Where the water moves faster, $\tfrac{1}{2}\rho v^2$ is larger, so $p$ must be smaller. The pressure is lower in the narrow part, and that difference is exactly what speeds the water up as it goes in.

We assumed that no energy is lost to friction on the way, which is a good approximation over a short, smooth stretch; chapter 5 shows what changes when it is not. If the pipe also climbs by $h_2 - h_1$, gravity takes away the work $\rho\,\Delta V g\,(h_2 - h_1)$ from the parcel. Including it and dividing by $\Delta V$ again adds a gravity term: along the pipe, $p + \tfrac{1}{2}\rho v^2 + \rho g h$ stays the same.

For the Venturi pipe of chapter 2, use the unrounded speeds from $v=Q/A$ to obtain $p_1-p_2=\tfrac12\rho(v_2^2-v_1^2)\approx2484\,\mathrm{Pa}$. Substituting the displayed rounded speeds $2.34$ and $0.707\,\mathrm{m/s}$ instead gives about $2488\,\mathrm{Pa}$; the small difference is rounding, not new physics.

Here is the pipe. Two thin upright tubes stand on it, one over the wide part and one over the narrow part. Water rises in each tube until the weight of its column balances the pressure in the pipe below. From chapter 2 of the pressure and buoyancy lesson, a column of height $h$ presses with $\rho g h$, so the height is $h = p/(\rho g)$. A pressure difference of $2484\ \mathrm{Pa}$ therefore shows as a height difference of $2484/(1000 \times 9.81) = 0.253\ \mathrm{m}$, about $25\ \mathrm{cm}$.

Press Play. The gold rings are marks in the water: watch them spread apart where the water speeds up. Change the flow and the narrowing, and compare the readouts with the formulas above. The pipe and the tubes are drawn larger than their real size so that you can see them.

Real water is a little sticky. Layers of water sliding past each other rub, a little like two hands rubbed together, and the rubbing turns some energy into heat. This stickiness is called viscosity and written $\mu$ (the Greek letter mu). Honey has a large viscosity; water has a small one, about $0.001\ \mathrm{Pa\,s}$, which we write as $1\ \mathrm{mPa\,s}$.

Look at a long, straight pipe of constant width. By chapter 2 the speed is the same everywhere along it, so by Bernoulli’s equation the pressure would be the same too: water without stickiness would keep flowing with no push at all. Real water does not. The water touching the wall does not move, the layer next to it moves slowly, and the middle moves fastest. The layers rub, and the flow would die away unless a pressure difference $\Delta p$ between the two ends kept pushing it.

How much water flows? Here we assume a straight circular pipe, a liquid whose viscosity stays constant, and fully developed laminar flow: smooth layers with a velocity profile that no longer changes along the pipe. We also neglect entry and exit effects. In this steady flow the push from the pressure difference balances the rubbing. Double $\Delta p$ and the push doubles; the rubbing grows in proportion to how fast the layers slide past each other, so the speeds, and with them $Q$, double too. A liquid twice as viscous rubs twice as hard at the same speeds, so at the same push it flows half as fast. For a pipe with radius $R$ and length $L$ these combine into Poiseuille’s law, derived step by step in the box below:

The $R^4$ is the striking part: a pipe twice as wide carries $2^4 = 16$ times as much water for the same pressure difference. In the pipe below, $R = 6\ \mathrm{mm}$, $L = 1\ \mathrm{m}$, $\Delta p = 20\ \mathrm{Pa}$ and $\mu = 1\ \mathrm{mPa\,s}$, so $Q = \frac{\pi \times 20 \times 0.006^4}{8 \times 0.001 \times 1} = 1.02 \times 10^{-5}\ \mathrm{m^3/s}$, about $10.2\ \mathrm{mL/s}$. The average speed is $Q/(\pi R^2) = 0.090\ \mathrm{m/s}$, and the middle moves twice as fast as the average.

Press Play. The gold dye sheet starts flat and is stretched into a curved shape: fastest in the middle, still at the wall. Change $\Delta p$ and $\mu$ and compare the flow with the formula.