Bernoulli's Principle

Beginner

Let's build Bernoulli's idea one small step at a time.

Step 1: what goes in must come out. Water cannot pile up inside a full pipe, so every second the same amount of water must pass every point. Where the pipe gets narrower, the water has to hurry. That is why water spurts faster when you squeeze the end of a garden hose.

Step 2: pressure is a push. Pressure tells you how hard the water pushes on each bit of surface it touches. A thin glass tube standing on the pipe shows it: the harder the push, the higher the water climbs in the tube.

Step 3: speeding up needs a push from behind. A bit of water entering the narrow part speeds up, so something must push it forward. That push comes from the water behind it, so the pressure behind (in the wide part) must be higher than the pressure ahead (in the narrow part). Fast water, low pressure; slow water, high pressure. You can also see it as a trade: the water gains energy of motion and pays for it with pressure.

In the animation, the white dots are small bits of water. Follow the red one: it pulls away from its neighbours, races through the narrow waist, then slows down again. The water column above the waist stands lowest.

Check yourself: two boats moving fast side by side, close together, tend to drift towards each other. Why? (Answer: the water squeezed between them flows faster than the water outside, so its pressure is lower, and the higher pressure on the outer sides pushes the boats together.)

Intermediate

What flows, and what we measure

Two boats moving fast side by side, close together, tend to drift towards each other. This lesson follows a steady flow of a liquid or a gas: the velocity at each fixed point never changes, although the fluid keeps moving through it, and each parcel travels along a fixed line called a streamline. Water is treated as an ideal fluid, incompressible, with a constant density $\rho =1000kg/{\mathrm{m}}^3$, and frictionless, which is a good model for a short, smooth pipe.

Four quantities describe a pipe flow: the flow speed $v$ (m/s), averaged over a cross-section; the cross-sectional area $A$ (m²), with $A=\pi d^2/4$ for a round pipe of diameter $d$; the volume flow rate $Q$ (m³/s), the volume that passes a section each second; and the pressure $p$, the push of the water on each square metre of the wall, in pascals ($1\mathrm{Pa}=1N/{\mathrm{m}}^2$).

A thin vertical tube on the pipe, a standpipe, shows the pressure as a height: the water climbs until the weight of its column balances the pressure below, $p=\rho gh$. All pressures here are gauge pressures, measured above the atmosphere's 101 kPa, which cancels in every difference.

Continuity: what goes in must come out

In a time $\Delta t$ the water crossing a section moves forward by $v\Delta t$ and fills a cylinder of volume $Av\Delta t$. Dividing by the time gives the flow rate:
$Q=Av$

Water cannot pile up inside a full pipe or vanish from it, so the same $Q$ must pass every section, wide or narrow:
$A_1{v}_1=A_2{v}_2$

The speed is therefore inversely proportional to the area. Because the area of a round pipe grows with the square of its diameter, halving the diameter divides the area by 4 and makes the water four times faster, not twice as fast.

Bernoulli's equation: trading pressure for speed

A parcel entering a narrowing speeds up, so a net force must push it forward. Along a horizontal pipe that force can only come from the pressure of the surrounding water, so the pressure behind the parcel must be higher than the pressure ahead of it: fast flow goes with low pressure.

Energy makes this exact: the work that the pressure forces do on a parcel moving from high to low pressure becomes kinetic energy, or potential energy if the parcel climbs. Per unit volume these energies are $p$, $\frac{1}{2}\rho v^2$ and $\rho gh$, all in pascals ($1\mathrm{Pa}=1J/{\mathrm{m}}^3$), and along a streamline their sum stays the same:
$p+\frac{1}{2}\rho v^2+\rho gh=\mathrm{const}$

In a horizontal pipe the height terms cancel, so the pressure drop equals the gain in kinetic energy per unit volume. Two standpipes show it as a difference of heights:
$p_1-p_2=\frac{1}{2}\rho (v_2^2-v_1^2)$
$\Delta h=\frac{p_1-p_2}{\rho g}$

Height counts too. In a uniform pipe that climbs 10 m the speed cannot change, so the pressure falls by $\rho gh=1000\cdot 9.81\cdot 10\approx 98\mathrm{kPa}$, almost one atmosphere, which is why tall buildings need pumps.

Worked example: the Venturi tube in the animation

On the animation's scale, which is the same for the pipe and the columns, the pipe's radius is 24 cm at the inlet and 13 cm in the throat, and the water enters at $v_1=1.0m/s$ with a gauge pressure of 7.85 kPa, which holds the first column 0.80 m above the axis.

Continuity, Bernoulli's equation and the standpipe rule give:
$v_2=v_1\frac{A_1}{A_2}=1.0\cdot {\left(\frac{24}{13}\right)}^2\approx 3.41m/s$
$p_1-p_2=\frac{1}{2}\cdot 1000\cdot (3.41^2-1.0^2)\approx 5.31\mathrm{kPa}$
$\Delta h=\frac{5310}{1000\cdot 9.81}\approx 0.541\mathrm{m}$

The throat pressure is $7.85-5.31=2.54\mathrm{kPa}$, so its column stands only 0.26 m above the axis. These are the animation's numbers: the labels $v_1=1.0m/s$ and $v_2=3.4m/s$, the purple $\Delta h$ arrow below the dashed inlet level, and the readouts, where $v$ climbs from 1.00 to 3.41 m/s while $p$ falls from 7.85 to 2.54 kPa as the red parcel enters the throat and its blue velocity arrow grows 3.41 times.

Past the throat everything runs backwards, and in this ideal fluid the last column climbs back to 0.80 m. The flow rate, $Q=\pi \cdot 0.24^2\cdot 1.0\approx 0.181{\mathrm{m}}^3/s$ or 181 litres a second, is the same in the throat: $\pi \cdot 0.13^2\cdot 3.41\approx 0.181{\mathrm{m}}^3/s$.

Worked example: reading a Venturi meter

A Venturi meter is such a tube built into a pipeline to measure its flow. Suppose a water main 12 cm in diameter narrows to a throat of 6 cm, and the columns over the two sections differ by 30 cm. How fast does the water flow, and how much passes each second?

The pressure drop is $\rho g\Delta h=1000\cdot 9.81\cdot 0.30\approx 2.94\mathrm{kPa}$. Halving the diameter quarters the area, so $v_2=4v_1$, and Bernoulli's equation becomes $\rho g\Delta h=\frac{1}{2}\rho (16-1)v_1^2$. The density cancels, and in general
$v_1=\sqrt{\frac{2g\Delta h}{(A_1/A_2{)}^2-1}}=\sqrt{\frac{2\cdot 9.81\cdot 0.30}{15}}\approx 0.63m/s$

Then $v_2=4\cdot 0.63\approx 2.5m/s$ and $Q=\pi \cdot 0.06^2\cdot 0.63\approx 7.1\times 10^{-3}{\mathrm{m}}^3/s$, about 7 litres per second. Fed with the animation's 0.541 m and area ratio 3.41, the same formula returns its 1.0 m/s. Note that $\Delta h$ grows with the square of the flow: twice the flow would give 1.20 m, so each meter suits a limited range.

Reading the columns and the dots

Each column top stands $p/(\rho g)$ above the axis, a height called the pressure head. Add the velocity head $v^2/(2g)$, 0.05 m at the inlet and 0.59 m in the throat, and every section reaches the same total head, 0.85 m: this is Bernoulli's equation divided by $\rho g$. Along real pipelines friction makes this level slope down.

The white dots enter at a steady rate, so the gap between neighbours in a row is the speed times a fixed time: in the throat the gaps open up 3.41 times, and the three rows close in, because streamlines crowd where the flow is fast.

Common mistakes

• The low pressure in the throat does not suck the water in. The water is pushed from behind, by 7.85 kPa at the inlet against 2.54 kPa in the throat.
• Square the speeds before subtracting them: for the animation, $\frac{1}{2}\rho (v_2-v_1{)}^2$ gives 2.90 kPa instead of the correct 5.31 kPa.
• A narrowing does not raise the pressure inside it. A thumb on a hose raises the pressure behind it by holding back the flow, but for a given flow through a horizontal pipe the narrowest section has the lowest pressure.
• The pressure is not used up in the throat. In an ideal fluid it recovers fully, so the last column is back at 0.80 m; in a real pipe friction leaves it several centimetres lower.

Where you meet it

Venturi meters measure flows of water, fuel and gas. Carburettors and perfume atomisers use the low pressure of a fast air stream to draw liquid up a tube, and a laboratory water-jet pump sucks air out of a flask the same way. A Pitot tube on an aircraft compares the pressure of air brought to rest with the static pressure to find the airspeed, and two ships sailing close side by side are pushed together by the faster water between them.

The lesson rests on Fluid Pressure and Pascal's Principle, for the columns, and on Work, Energy and Power, for the conservation of energy. The friction it leaves out, viscosity, is the subject of Drag and Terminal Velocity.

Advanced

Streamlines, stream tubes and the moving parcel

This section sets up the language of fluid flow and uses it to answer a first question: why does water speed up in a narrowing pipe, and how large is its acceleration? The running example is the animation's Venturi tube. Water of density $\rho =1000kg/{\mathrm{m}}^3$ enters a wide pipe of radius 24 cm at 1.0 m/s, with a gauge pressure (pressure above atmospheric) of 7.85 kPa, enough to hold up a water column 0.80 m high. It then passes a throat of radius 13 cm and widens again.

A flow is described by its velocity field $\vec{v}(\vec{r},t)$: the velocity of whatever fluid is at the point $\vec{r}$ at the time $t$. The flow is steady when this field does not change with time at any fixed point, $\partial \vec{v}/\partial t=0$. A streamline is a line that is tangent to $\vec{v}$ at each of its points. In a steady flow the streamlines stay fixed, and each small parcel of fluid travels along one of them.

Legend

$A_1$	Cross-section of the stream tube at section 1, where it is wide.	${\mathrm{m}}^2$
$A_2$	Cross-section of the same tube at section 2, where it is narrow.	${\mathrm{m}}^2$
$v_1$	Speed of the fluid through section 1.	$m/s$
$v_2$	Speed through section 2: larger, since $A_2{v}_2=A_1{v}_1$.	$m/s$
$\vec{v}$	Velocity at a point of the wall, which is itself a streamline: tangent to the wall there.	$m/s$

Figure 1 shows a pipe that narrows. The thin blue lines are streamlines, and the velocity at any point is tangent to the streamline through it. The wall is a streamline as well, because no fluid crosses it: the arrow $\vec{v}$ at the dot on the wall lies along the wall. The two heavier lines bound a stream tube, a bundle of neighbouring streamlines. Since no fluid crosses a streamline, none crosses the side of the tube either: whatever enters it through section 1, of area $A_1$, must leave through section 2, of area $A_2$.

For a liquid the density is constant, so the volume that enters per second must equal the volume that leaves per second. This volume flow rate is $Q=Av$, with $v$ the average speed across the section, so
$A_1{v}_1=A_2{v}_2$
Because $A_2$ is smaller than $A_1$, the speed $v_2$ is larger than $v_1$, by the factor $A_1/A_2$.

Look at the streamlines in Figure 1: they crowd together where the tube is narrow. Equal volumes flow between neighbouring lines, so their spacing is a picture of the speed: close lines, fast fluid. The two arrows $v_1$ and $v_2$ are drawn in the ratio $A_1/A_2$. In the animation $A_1/A_2=(24/13{)}^2\approx 3.41$, so the water reaches 3.41 m/s in the throat. The local form of this law is $\nabla \cdot \vec{v}=0$: a fluid of constant density can neither pile up nor thin out anywhere.

Steady does not mean unaccelerated. The animation's field never changes, yet the red parcel speeds up from 1.0 to 3.41 m/s, because it moves into a region where the field is faster. The rate of change of the parcel's own velocity is the material derivative:
$\frac{D\vec{v}}{Dt}=\frac{\partial \vec{v}}{\partial t}+(\vec{v}\cdot \nabla )\vec{v}$
The first term is the change of the field at a fixed point; it is zero in a steady flow. The second, the convective term, is the change the parcel meets because it moves: in the time $dt$ it travels $\vec{v}dt$ to a place where the field is different.

On the pipe's axis the velocity points along the axis, call it $x$, so the convective term is simply $vdv/dx$. In the animation it reaches about 25 m/s², some $2.5g$, near the end of the narrowing, and the widening section brakes the parcel at up to 21 m/s².

Energy balance on a stream tube

Why should faster fluid be at lower pressure? To speed up, a parcel needs a net forward force, and in a horizontal pipe only the pressure can provide it: the pressure behind the parcel must exceed the pressure ahead. An exact energy balance turns that idea into Bernoulli's equation.

Legend

$p_1{A}_1$	Force with which the fluid behind pushes the entering slab forward: pressure times area.	$\mathrm{N}$
$p_2{A}_2$	Force with which the fluid ahead pushes back on the leaving slab.	$\mathrm{N}$
$v_{1}dt$	Length of the entering slab: the distance the fluid moves in $dt$ at section 1.	$\mathrm{m}$
$v_{2}dt$	Length of the leaving slab, longer since $v_2>v_1$.	$\mathrm{m}$
$dV$	Volume of each slab, the same at both sections.	${\mathrm{m}}^3$
$h_1$	Height of the tube's axis at section 1 above a reference level.	$\mathrm{m}$
$h_2$	Height of the tube's axis at section 2 above the same level.	$\mathrm{m}$

Figure 2 shows a stream tube that narrows and rises, from section 1 at the height $h_1$ to section 2 at the height $h_2$, both measured to the tube's axis. Follow the fluid in the tube for a short time $dt$. At section 1 a slab of length $v_{1}dt$ enters, with the mass $dM=\rho A_1{v}_{1}dt$ and the volume $dV=dM/\rho$. At section 2 a slab of length $v_{2}dt$ leaves. By continuity it has the same mass and the same volume, so it is the longer of the two, because the tube is thinner there.

Between the two slabs nothing has changed: the flow is steady, so every point still holds fluid moving with the same velocity as before. The net effect of the interval $dt$ is therefore that the slab $dV$ has been taken from section 1 and put at section 2, with the speed and the height of section 2. Its energy has grown by
$dM\left(\frac{1}{2}{v}_2^2-\frac{1}{2}{v}_1^2\right)+dMg(h_2-h_1)$

What paid for this gain? Not the pressure on the side of the tube: it acts perpendicular to the flow and does no work. The work is done at the two ends. The fluid behind pushes the entering slab with the force $p_1{A}_1$ through the distance $v_{1}dt$, which is the work $p_1{A}_1{v}_{1}dt=p_{1}dV$. At section 2 the fluid ahead pushes back with the force $p_2{A}_2$ while the leaving slab moves $v_{2}dt$ forward against it, which is the work $-p_{2}dV$. Without friction, the work–energy theorem sets the net work equal to the gain in energy:
$(p_1-p_2)dV=dM\left(\frac{1}{2}{v}_2^2-\frac{1}{2}{v}_1^2+g{h}_2-g{h}_1\right)$

Divide by $dV$, use $dM/dV=\rho$, and collect the terms of each section on its own side: $p_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=p_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$. Sections 1 and 2 were any two sections, so along the tube
$p+\frac{1}{2}\rho v^2+\rho gh=\mathrm{const}$

Every term is an energy per unit volume, measured in pascals (1 Pa = 1 J/m³). $\frac{1}{2}\rho v^2$ is kinetic energy and $\rho gh$ potential energy. The pressure term $p$ is the work needed to push a unit volume of fluid into a region that is already full, called flow work. In the horizontal Venturi tube $h$ is the same everywhere, and raising the speed from 1.0 to 3.41 m/s costs $\frac{1}{2}\rho (v_2^2-v_1^2)\approx 5.31\mathrm{kPa}$ of pressure.

In a gas the internal energy per unit mass, $u$, joins the balance, and the sum conserved per unit mass becomes $u+p/\rho +\frac{1}{2}{v}^2+gh$. Its first two terms together form the specific enthalpy. Without heat exchange or shaft work, this is the steady-flow energy equation of a nozzle. In a frictionless liquid $u$ stays constant and drops out.

Euler's equation: along and across the flow

The energy argument treats a whole stream tube at once. Newton's second law, applied to each small volume of fluid, gives the same result in local form, shows where the constant holds, and adds a second rule for the pressure across the streamlines.

Consider a small cube of fluid. The pressure pushes inwards on each face, and the forces on opposite faces cancel except for the difference in pressure between them. Per unit volume the net pressure force is therefore $-\nabla p$, pointing from high towards low pressure. Adding gravity, $\rho \vec{g}$ per unit volume, and setting the total equal to the density times the material derivative gives Euler's equation:
$\rho \left(\frac{\partial \vec{v}}{\partial t}+(\vec{v}\cdot \nabla )\vec{v}\right)=-\nabla p+\rho \vec{g}$

Three steps turn it into Bernoulli's equation. First, in a steady flow $\partial \vec{v}/\partial t=0$. Second, a vector identity splits the convective term: $(\vec{v}\cdot \nabla )\vec{v}=\nabla (\frac{1}{2}{v}^2)-\vec{v}\times \vec{\Omega }$, where $\vec{\Omega }=\nabla \times \vec{v}$ is the vorticity, twice the local angular velocity with which the fluid spins. Third, gravity points down while $h$ is measured up, so for a constant density $\rho \vec{g}=-\nabla (\rho gh)$. Moving all the gradients to the left,
$\nabla \left(p+\frac{1}{2}\rho v^2+\rho gh\right)=\rho \vec{v}\times \vec{\Omega }$

The right-hand side is a cross product with $\vec{v}$, so it is perpendicular to $\vec{v}$. The gradient of the sum therefore has no component along the flow: the sum is constant along each streamline, as found with Figure 2, but different streamlines may carry different constants. If the flow is irrotational, $\vec{\Omega }=0$, the whole gradient vanishes and one constant holds everywhere. An ideal fluid that starts from rest, or from a uniform stream like the water entering the Venturi tube, stays irrotational, so in practice a single constant usually serves.

Across the flow, Euler's equation says something different. Where the streamlines curve with radius $R$, each parcel needs the centripetal acceleration $v^2/R$ towards the centre of curvature, and, gravity aside, only a pressure difference across the streamlines can supply it:
$\frac{\partial p}{\partial n}=\frac{\rho v^2}{R}$
Here $n$ is the distance across the streamlines, measured away from the centre of curvature. The pressure therefore rises away from the centre of a bend, so air curving over a wing has a lower pressure next to it than farther out. Where the streamlines are straight and parallel, $R$ is infinite and the pressure does not change across them, apart from the hydrostatic change with height. That is why a tap in the wall of a straight section reads the pressure on the axis.

Reading the pressure graph

The animation's graph shows Bernoulli's equation at work along the whole tube. Its slopes show Euler's equation as well, and explain why real Venturi tubes are shaped the way they are.

Legend

$x$	Position along the tube, in the direction of flow.	$\mathrm{m}$
$p$	Static pressure (gauge) on the tube's axis.	$\mathrm{Pa}$
$\frac{1}{2}\rho v^2$	Dynamic pressure: kinetic energy per unit volume.	$\mathrm{Pa}$
$p+\frac{1}{2}\rho v^2$	Their sum, the same all along the tube.	$\mathrm{Pa}$

Figure 3 redraws the animation's pressure graph under a sketch of the tube, with the position $x$ along the pipe running to the right. Section 1 is the wide inlet and section 2 the middle of the throat, and the graph follows the red parcel along the pipe. The static pressure $p$ (purple) falls from 7.85 to 2.54 kPa through the narrowing, stays flat along the throat and recovers in the widening section. The dynamic pressure $\frac{1}{2}\rho v^2$ (blue) does the opposite: it rises from 0.50 to 5.81 kPa and falls back. Their sum (green) stays flat at 8.35 kPa. The three curves share one scale, so whatever height the blue curve gains, the purple one loses. Expressed as the height of the water column it could hold up, $(p+\frac{1}{2}\rho v^2)/(\rho g)$, the sum is 0.851 m, called the total head.

Now look at the slopes. On the axis Euler's equation reads $-dp/dx=\rho vdv/dx=\rho a$, with $a$ the parcel's acceleration. So the downward slope of the purple curve is the net force per unit volume on the parcel, and it equals the upward slope of the blue curve. It is steepest, about 25 kPa per metre, where the parcel accelerates most, on the steep flanks of the curves in Figure 3. It is zero along the throat, where the speed is constant.

In the widening section the pressure rises along the flow: the purple curve climbs on the right. Such a rise is called an adverse pressure gradient, because it pushes against the motion. The ideal fluid simply slows down. In a real fluid the layer next to the wall has already been slowed by friction and may lack the kinetic energy to climb the pressure rise; it stops, and the flow breaks away from the wall. That is why Venturi tubes widen more gently than they narrow: here over 60 cm, against 50 cm for the narrowing.

Tanks, jets and Pitot tubes

Bernoulli's equation also links points at different heights, and points where the fluid is brought to rest. Each case gives a simple and useful result.

Legend

$H$	Depth of the free surface above the hole.	$\mathrm{m}$
$A_T$	Area of the tank's free surface.	${\mathrm{m}}^2$
$a$	Area of the hole.	${\mathrm{m}}^2$
$v$	Speed of the jet just outside the hole.	$m/s$
$dH/dt$	Rate of change of the level: negative, since the level sinks.	$m/s$

Figure 4 shows a wide tank draining through a small hole. Follow the dashed streamline from point 1, on the free surface, to point 2, in the jet just outside the hole, a depth $H$ lower. At both ends the pressure is atmospheric: at 1 the surface is open to the air, and at 2 the streamlines of the jet are straight and parallel, so the pressure inside it equals that of the air around it. The surface is almost still, because it is much wider than the hole. So the pressure terms cancel, the kinetic term at 1 is negligible, and the potential energy $\rho gH$ lost per unit volume all becomes the kinetic energy $\frac{1}{2}\rho v^2$ of the jet:
$v=\sqrt{2gH}\text{(Torricelli)}$

The jet leaves as fast as a stone dropped from the height $H$. This is the $v=\sqrt{2gh}$ of the formula list, with the depth written $H$ because it changes as the tank drains.

How long does the tank take to empty? If the level sinks slowly, the flow is almost steady at every moment (quasi-steady), and Torricelli's result holds at the current depth. Continuity then links the two ends of the streamline: the free surface, of area $A_T$, sinks at the rate $dH/dt$, which is negative, while the jet leaves the hole, of area $a$, at the speed $v$:
$A_T\frac{dH}{dt}=-av=-a\sqrt{2gH}$
Since $A_T/a$ is large, the surface crawls while the jet shoots out, which is why Figure 4 cannot be to scale. Separate the variables, $dH/\sqrt{H}=-(a/A_T)\sqrt{2g}dt$, and integrate from the initial depth $H_0$ down to 0. The left side gives $-2\sqrt{H_0}$, so $2\sqrt{H_0}=(a/A_T)\sqrt{2g}t$, and the tank empties in
$t=\frac{A_T}{a}\sqrt{\frac{2H_0}{g}}$

A barrel of 0.25 m² filled 0.80 m deep, draining through a 1 cm² hole, has $A_T/a=2500$ and empties in $2500\cdot \sqrt{1.6/9.81}\approx 1.0\times 10^3\mathrm{s}$, about 17 minutes. The level drops fast at first and slowly at the end, since the jet speed goes as $\sqrt{H}$.

A Pitot tube measures speed with a stagnation point. Its open end faces the flow, and the fluid arriving there is brought to rest, $v=0$; at the same height the pressure at the mouth is therefore $p+\frac{1}{2}\rho v^2$, the stagnation pressure. Small holes in the side of the tube, which the flow runs past in parallel, read the static $p$. The difference is $\frac{1}{2}\rho v^2$, from which $v$ follows.

On the animation's axis the open end would read the green 8.35 kPa everywhere, while the wall taps read the purple curve. An aircraft at 60 m/s in air of density 1.2 kg/m³ sees a difference of $\frac{1}{2}\cdot 1.2\cdot 60^2=2.16\mathrm{kPa}$.

Worked problem: when the throat sucks and boils

Keep the animation's tube and its inlet gauge pressure of 7.85 kPa, but turn up the flow. At what inlet speed does the throat reach atmospheric pressure, and at what speed does the water boil there? Take 101.3 kPa for the atmosphere and 2.34 kPa for the vapour pressure of water at 20 °C.

The tube is horizontal, so Bernoulli's equation between the inlet and the throat is $p_1+\frac{1}{2}\rho v_1^2=p_2+\frac{1}{2}\rho v_2^2$. Continuity gives $v_2=(A_1/A_2)v_1$, with $A_1/A_2=3.41$. Substituting $v_2$ gives the throat gauge pressure
$p_2=p_1-\frac{1}{2}\rho v_1^2\left[{\left(\frac{A_1}{A_2}\right)}^2-1\right]=7.85-5.31v_1^2$
with $p_2$ in kPa and $v_1$ in m/s; the 5.31 is $\frac{1}{2}\rho (3.41^2-1)$ in these units.

The throat gauge pressure is zero, so the throat is at atmospheric pressure, when $v_1=\sqrt{7.85/5.31}\approx 1.22m/s$. At higher speeds the throat is below atmospheric pressure, so a tube opened there draws air or liquid in instead of holding up a column. This is how a water-jet pump and a carburettor work.

Cavitation, boiling at room temperature, begins when the absolute throat pressure falls to the vapour pressure. Absolute pressure is gauge pressure plus atmospheric, so this happens at the gauge pressure $2.34-101.3\approx -99.0\mathrm{kPa}$. Setting $7.85-5.31v_1^2=-99.0$ gives
$v_1=\sqrt{\frac{7.85+99.0}{5.31}}\approx 4.49m/s,v_2=3.41\cdot 4.49\approx 15.3m/s$

The pipe then carries $\pi \cdot 0.24^2\cdot 4.49\approx 0.81{\mathrm{m}}^3/s$, four and a half times the animation's flow. In practice bubbles form a little earlier, around tiny pockets of gas and in turbulent eddies. Swept into the widening section, where the pressure recovers, they collapse violently, and these repeated collapses are what pit pump impellers and ship propellers.

Limits of validity

Bernoulli's equation rests on three idealisations: no friction, constant density and steady flow. Each fails somewhere, and the paragraphs below take them in turn and show where.

Viscosity, the fluid's internal friction, is the first to fail. The Reynolds number $\rho vd/\eta$ compares the fluid's inertia with its viscous forces; here $d$ is the pipe's diameter and $\eta$ the viscosity of water, about $1.0\times 10^{-3}\mathrm{Pa}\cdot \mathrm{s}$. For the animation it is $1000\cdot 1.0\cdot 0.48/(1.0\times 10^{-3})\approx 5\times 10^5$, far above the few thousand at which pipe flow turns turbulent, so the real flow would be turbulent.

Yet friction hardly matters in a short, smooth narrowing, and Venturi meters agree with the ideal formula to within a few per cent. Across the whole tube, though, a real Venturi permanently loses 5 to 20 % of its pressure drop, so the column after the widening would stand some 3 to 11 cm below the first.

Incompressibility is excellent for liquids. The animation's 5.3 kPa drop changes the density of water by the fraction $\Delta p/K\approx 2.4\times 10^{-6}$, where $K\approx 2.2\mathrm{GPa}$ is its bulk modulus, the pressure needed per unit fractional change of volume. Gases qualify while the flow is well below the speed of sound. At Mach 0.3, three tenths of the speed of sound or about 100 m/s in air, the density changes by about 4.5 %, and the incompressible formula predicts a Pitot reading about 2 % too small.

Steadiness fails when a valve shuts quickly. The term $\partial \vec{v}/\partial t$ then dominates, and a pressure wave, called water hammer, runs along the pipe. Stopping the animation's 1.0 m/s flow abruptly would raise the pressure by about $\rho c\Delta v\approx 1000\cdot 1400\cdot 1.0\approx 1.4\mathrm{MPa}$, where $c$ is the speed of the wave in the water-filled pipe and $\Delta v$ the sudden change of speed.

Where this leads

Bernoulli's equation is the work–energy theorem of Work, Energy and Power, written per unit volume. Torricelli's jet leaves with the speed of Free Fall and traces the parabola of Projectile Motion, and $\partial p/\partial n=\rho v^2/R$ is the centripetal force of Circular Motion applied to a fluid. For gases the enthalpy form leads to the First Law of Thermodynamics and Heat Engines, and the speed of sound that bounds the incompressible model returns in Sound and the Doppler Effect.

Lift is often misexplained. A popular claim says that the air over a wing must rejoin the air beneath it at the trailing edge, and so must move faster along the longer upper path. This is false: the upper air arrives first. Moreover, a steady ideal flow without circulation exerts no net force on a body at all.

What a wing does is set up a circulation $\Gamma =\oint \vec{v}\cdot d\vec{l}$, the integral of the velocity around a loop that encloses the wing: a net swirl added to the oncoming stream. Its value is fixed by the sharp trailing edge, which the flow must leave smoothly. In a stream of speed $U$ the lift per unit span is then $\rho U\Gamma$, and Bernoulli's equation turns the faster flow above the wing and the slower flow below into the pressure difference that carries it. The viscosity that selects that circulation, and causes drag, is the subject of Drag and Terminal Velocity.