Chapter 9 · Mechanical Properties of Fluids

9.2 Streamline Flow and Bernoulli's Principle

In steady flow an incompressible fluid speeds up where its pipe narrows, $A_1v_1=A_2v_2$, and — by the work–energy theorem — its pressure drops where it speeds up: $p+\tfrac12\rho v^2+\rho gh$ stays constant along a streamline.

Put your thumb over the end of a garden hose and the water that was dribbling out shoots across the lawn. A river that is broad and lazy in the plains races through a narrow gorge. Hold two sheets of paper a few centimetres apart and blow between them: instead of being pushed apart, they swing together. And an airliner of several hundred tonnes is held up by nothing but air flowing past its wings. All of these are fluids in motion, and two conservation laws you already know — conservation of mass and the work–energy theorem — explain every one of them.

This section describes the simplest kind of motion, steady streamline flow, and draws the pattern of the flow with streamlines. Conservation of mass for an incompressible fluid gives the equation of continuity: where the pipe narrows, the fluid must go faster. Applying the work–energy theorem to a moving piece of fluid gives Bernoulli's principle: where the fluid goes faster, its pressure is lower. With these two results you will find the speed of a jet leaving a tank, measure flow with a Venturi meter, and see why narrowed arteries and aircraft wings behave as they do.

Streamline flow and the equation of continuity

Watch a slow, clear stream: a leaf dropped at a given spot always follows the same path. Each particle of water passing that spot has the same velocity as the one before it. Speed the stream up over rocks and the paths become irregular and swirling — eddies form and no two particles follow the same route.

📘 Definition 9.4 — Steady (streamline) flow and turbulent flow

The flow of a fluid is steady (or streamline) if the velocity of the fluid at each fixed point does not change with time: every particle that passes a given point has the same velocity there. The path followed by a fluid particle in steady flow is a streamline; the velocity at every point of a streamline is along its tangent. A bundle of streamlines through a small area forms a tube of flow.

If the velocity at a point changes erratically from moment to moment, with eddies and swirls, the flow is turbulent. Flow is streamline at low speeds and becomes turbulent above a certain critical speed, which is lower for wider pipes and for less viscous fluids.

Two streamlines can never cross: if they did, a particle arriving at the crossing point would have two different velocities there. So fluid never passes through the side wall of a tube of flow — a tube of flow behaves like a pipe. Where streamlines crowd together, the fluid is moving faster. (Engineers predict whether a flow will be turbulent from a dimensionless combination of speed, size, density and viscosity called the Reynolds number; you do not need it here.)

From now on we consider an ideal fluid: incompressible (constant density), non-viscous (no internal friction), in steady flow. Real liquids such as water come close in many situations; viscosity is the subject of Section 9.3.

📘 Definition 9.5 — Volume flow rate

The volume flow rate $Q$ through a cross-section is the volume of fluid passing it per second. If the fluid crosses an area $A$ normally with speed $v$, in time $\Delta t$ it fills a cylinder of length $v\,\Delta t$, so

$$Q=\frac{\Delta V}{\Delta t}=Av\qquad(\text{unit } \text{m}^3\text{/s}).$$

The mass flow rate is $\rho Q=\rho Av$ (unit $\text{kg/s}$). Note $1\ \text{L}=10^{-3}\ \text{m}^3$.

Now take a tube of flow (or a pipe) whose cross-section is $A_1$ at one place and $A_2$ further along. In steady flow nothing builds up between the two sections — the mass inside stays constant — so the mass entering per second at section 1 must leave per second at section 2: $\rho_1A_1v_1=\rho_2A_2v_2$. For an incompressible fluid the densities are equal.

📐 Result 9.5 — Equation of continuity

For the steady flow of an incompressible fluid through a pipe or a tube of flow, the volume flow rate is the same at every cross-section:

$$A_1v_1=A_2v_2,\qquad Av=\text{constant}.$$

The speed is inversely proportional to the cross-sectional area: the fluid flows fastest where the pipe is narrowest. This is conservation of mass.

✏️ Example 1 — Thumb over the hose
Water flows at $0.50\ \text{m/s}$ in a garden hose of internal radius $1.0\ \text{cm}$. (a) Find the volume flow rate in litres per second. (b) A nozzle at the end has a radius of $0.25\ \text{cm}$. Find the speed of the water leaving it. (c) How long does it take to fill a $10\ \text{L}$ bucket, with and without the nozzle?
(a) $Q=Av=\pi(0.010)^2(0.50)=1.57\times10^{-4}\ \text{m}^3\text{/s}=0.157\ \text{L/s}$.
(b) $v_2=v_1\dfrac{A_1}{A_2}=0.50\times\left(\dfrac{1.0}{0.25}\right)^2=0.50\times16=8.0\ \text{m/s}$.
(c) $t=\dfrac{10\ \text{L}}{0.157\ \text{L/s}}=64\ \text{s}$ in both cases — the nozzle changes the speed, not the flow rate (as long as the supply delivers the same flow).
Check Quarter the radius, one-sixteenth the area, sixteen times the speed ✓.
✏️ Example 2 — From the aorta to the capillaries
Blood leaves the heart through the aorta, of cross-sectional area $3.0\ \text{cm}^2$, at an average speed of $0.30\ \text{m/s}$. In the capillaries it moves at only about $0.50\ \text{mm/s}$. (a) Find the total cross-sectional area of all the capillaries carrying the flow. (b) A capillary has a radius of about $4.0\ \mu\text{m}$. Estimate the number of capillaries open at one time.
Idea All the blood that passes through the aorta must pass through the capillaries in parallel, so continuity applies to their total area: $A_\text{aorta}v_\text{aorta}=A_\text{cap}v_\text{cap}$.
(a) $A_\text{cap}=\dfrac{3.0\times10^{-4}\times0.30}{5.0\times10^{-4}}=0.18\ \text{m}^2$ — six hundred times the aorta's area.
(b) One capillary: $\pi(4.0\times10^{-6})^2=5.0\times10^{-11}\ \text{m}^2$. Number $\approx\dfrac{0.18}{5.0\times10^{-11}}=3.6\times10^9$.
Check The very slow flow in the capillaries is what the body needs: it gives oxygen and nutrients about a second to diffuse out through the thin walls. Continuity explains why the slowest flow is where the vessels are thinnest — their combined area is huge.

Bernoulli's principle

Continuity says the fluid speeds up in a narrow part of a pipe. But something must push it to make it accelerate: there must be a net force on each piece of fluid entering the narrow part, so the pressure behind it must be greater than the pressure in front. Pressure is higher where the fluid is slow and lower where it is fast. To make this precise, apply the work–energy theorem (Section 5.1) to a moving portion of fluid.

Consider ideal fluid flowing steadily through a pipe from region 1 (cross-section $A_1$, speed $v_1$, pressure $p_1$, height $h_1$) to region 2 ($A_2$, $v_2$, $p_2$, $h_2$). Follow the fluid between the two sections for a short time $\Delta t$. Its rear end moves $v_1\Delta t$ and its front end $v_2\Delta t$; by continuity the volume swept out at each end is the same, $\Delta V=A_1v_1\Delta t=A_2v_2\Delta t$, of mass $\Delta m=\rho\Delta V$.

There is no viscous friction, so by the work–energy theorem $(p_1-p_2)\Delta V-\rho\Delta Vg(h_2-h_1)=\tfrac12\rho\Delta V(v_2^2-v_1^2)$. Divide by $\Delta V$ and collect the terms for each region:

$$p_1+\tfrac12\rho v_1^2+\rho gh_1=p_2+\tfrac12\rho v_2^2+\rho gh_2 .$$
📐 Principle 9.6 — Bernoulli's principle

For the steady flow of an incompressible, non-viscous fluid, along a streamline

$$p+\tfrac12\rho v^2+\rho gh=\text{constant}.$$

Each term is an energy per unit volume ($\text{J/m}^3=\text{Pa}$): $p$ is the work done by pressure per unit volume, $\tfrac12\rho v^2$ the kinetic energy per unit volume, and $\rho gh$ the gravitational potential energy per unit volume. Bernoulli's principle is conservation of energy for a flowing fluid. At equal heights, where the speed is greater the pressure is lower.

For a fluid at rest ($v=0$), Bernoulli's principle gives $p_1+\rho gh_1=p_2+\rho gh_2$, which is exactly Result 9.2 for a depth $h_1-h_2$. The principle fails where friction is important (long thin pipes, honey, blood in small vessels), in turbulent flow, and for gases when their density changes appreciably (at speeds approaching the speed of sound). For most everyday flows of water, and of air below about $100\ \text{m/s}$, it works well.

🧭 Problem-solving strategy — continuity and Bernoulli
  1. Pick two points on the same streamline where you know most quantities (a free surface open to the air has $p=p_a$; a wide tank has $v\approx0$; a jet in open air has $p=p_a$).
  2. Use continuity, $A_1v_1=A_2v_2$, to relate the speeds.
  3. Write Bernoulli between the two points, measuring heights from one chosen level. Use either absolute pressures throughout or gauge pressures throughout.
  4. Solve, and check the sign: faster flow at the same height must have lower pressure.
✏️ Example 3 — Water up a narrowing pipe
Water flows at $2.0\ \text{m/s}$ through a pipe of diameter $4.0\ \text{cm}$, at an absolute pressure of $2.00\times10^5\ \text{Pa}$. The pipe narrows to a diameter of $2.0\ \text{cm}$ and rises $1.5\ \text{m}$. Find the speed and the pressure in the narrow, upper section.
Continuity Half the diameter, a quarter of the area: $v_2=4\times2.0=8.0\ \text{m/s}$.
Bernoulli $p_2=p_1+\tfrac12\rho(v_1^2-v_2^2)-\rho g(h_2-h_1)$.
Numbers $p_2=2.00\times10^5+500(4.0-64)-1000(9.8)(1.5)=2.00\times10^5-3.0\times10^4-1.47\times10^4=1.55\times10^5\ \text{Pa}$.
Check Both effects lower the pressure: the water is faster ($-30\ \text{kPa}$) and higher ($-14.7\ \text{kPa}$). Energy per cubic metre balances: $200\,000+2000+0=155\,300+32\,000+14\,700$ ✓.

Speed of efflux, the Venturi meter, blood flow and dynamic lift

Speed of efflux. A wide tank, open to the air, has a small hole a depth $h$ below the water surface. Follow a streamline from the surface (point 1) to the jet just outside the hole (point 2). At both points the pressure is atmospheric. The tank is so much wider than the hole that, by continuity, the surface sinks very slowly: $v_1\approx0$. Bernoulli gives $p_a+0+\rho gh=p_a+\tfrac12\rho v^2+0$.

📐 Result 9.7 — Speed of efflux (Torricelli's law)

Liquid flows out of a small hole a depth $h$ below the free surface of a wide open tank with speed

$$v=\sqrt{2gh},$$

the speed a body would gain falling freely through the height $h$ (assuming an ideal fluid and a hole much smaller than the tank's cross-section). If the space above the liquid is closed and kept at pressure $P$, while the jet emerges into air at $p_a$,

$$v=\sqrt{2gh+\frac{2(P-p_a)}{\rho}}.$$

The jet speed does not depend on the density of the liquid (for an open tank) — mercury and water spurt out equally fast from the same depth — nor on the direction of the hole. If the hole is a height $y$ above the ground and the jet leaves horizontally, the water lands at a horizontal distance $R=v\sqrt{2y/g}=2\sqrt{hy}$ (projectile motion, Section 3.3). In practice the jet narrows just outside the hole and some energy is lost to friction, so the measured flow is roughly $60\%$ of $A\sqrt{2gh}$ for a sharp-edged hole.

✏️ Example 4 — A leaking tank on a stand
A water tank stands on a platform. A small hole of area $1.0\ \text{cm}^2$ in its side is $0.80\ \text{m}$ above the ground and $2.0\ \text{m}$ below the water surface. Find (a) the speed of the jet, (b) where it hits the ground, (c) the initial flow rate through the hole.
(a) $v=\sqrt{2gh}=\sqrt{2\times9.8\times2.0}=6.26\ \text{m/s}\approx6.3\ \text{m/s}$, horizontal.
(b) Time to fall $0.80\ \text{m}$: $t=\sqrt{\dfrac{2\times0.80}{9.8}}=0.404\ \text{s}$. Range $R=vt=6.26\times0.404=2.5\ \text{m}$ from the foot of the tank wall.
(c) $Q=Av=1.0\times10^{-4}\times6.26=6.3\times10^{-4}\ \text{m}^3\text{/s}$, about $0.63\ \text{L/s}$ (for an ideal hole).
Check $R=2\sqrt{hy}=2\sqrt{2.0\times0.80}=2.53\ \text{m}$ ✓. As the tank empties, $h$ falls, and the jet slows and lands closer: Interactive 2 starts from this example.

The Venturi meter. To measure the flow in a pipe of cross-section $A_1$, insert a short smooth constriction (the throat) of cross-section $A_2$ and measure the pressure difference $p_1-p_2$ between the wide part and the throat, with vertical open tubes or a U-tube manometer.

📐 Result 9.8 — Venturi meter

For an ideal fluid in a horizontal Venturi meter, continuity gives $v_2=\dfrac{A_1}{A_2}v_1$, and Bernoulli at equal heights gives $p_1-p_2=\tfrac12\rho(v_2^2-v_1^2)$. So the speed in the wide pipe is

$$v_1=\sqrt{\frac{2(p_1-p_2)}{\rho\left[(A_1/A_2)^2-1\right]}}$$

and the flow rate is $Q=A_1v_1$. If the pressure difference is read from open vertical tubes whose liquid levels differ by $\Delta h$, then $p_1-p_2=\rho g\,\Delta h$.

The same effect is used on purpose in many devices: the carburettor of a petrol engine (fast air through a throat lowers the pressure and draws fuel in), a Bunsen burner (the gas jet draws in air), a scent spray or paint sprayer, and the filter pump in a chemistry laboratory.

✏️ Example 5 — Measuring the flow in a water main
A horizontal Venturi meter has a pipe diameter of $10\ \text{cm}$ and a throat diameter of $5.0\ \text{cm}$. Open vertical tubes fitted at the wide part and at the throat show water levels that differ by $15\ \text{cm}$. Find the speed of the water in the wide pipe and at the throat, and the flow rate.
Set-up $\dfrac{A_1}{A_2}=\left(\dfrac{10}{5.0}\right)^2=4$, and $p_1-p_2=\rho g\Delta h$, so $v_1=\sqrt{\dfrac{2g\Delta h}{4^2-1}}$.
Speeds $v_1=\sqrt{\dfrac{2\times9.8\times0.15}{15}}=\sqrt{0.196}=0.44\ \text{m/s}$; at the throat $v_2=4v_1=1.8\ \text{m/s}$.
Flow rate $Q=A_1v_1=\pi(0.050)^2(0.443)=3.5\times10^{-3}\ \text{m}^3\text{/s}=3.5\ \text{L/s}$.
Check $\tfrac12\rho(v_2^2-v_1^2)=500(3.136-0.196)=1470\ \text{Pa}=\rho g(0.15)$ ✓. Interactive 1 with $v_1=0.44\ \text{m/s}$ and a $5.0\ \text{cm}$ throat shows this.
Interactive 1 — Venturi Pipe

Water flows from left to right through a horizontal pipe of diameter $10\ \text{cm}$ with a narrower throat. The dots move with the local speed (continuity), and the open tubes show the local gauge pressure as a water height $p/\rho g$ (Bernoulli). Narrow the throat or speed up the flow and watch the throat tube fall. If the throat pressure drops below atmospheric, the tube would suck in air.

0.44 m/s5.0 cm
speed at throat $v_2$—
height difference $\Delta h$—
$p_1-p_2$—
flow rate $Q$—

Blood flow. When fatty deposits narrow an artery (a stenosis), continuity forces the blood to speed up through the narrow part, and by Bernoulli's principle its pressure there drops. The pressure outside the artery is unchanged, so the narrowed section can be squeezed shut; flow stops, pressure recovers, the artery reopens, and the cycle repeats — a fluttering that a doctor can hear through a stethoscope. In an aneurysm (a ballooned artery) the opposite happens: the blood slows and its pressure rises, pushing the weakened wall outwards further.

Dynamic lift. A body moving through a fluid can feel a sideways force, called dynamic lift, when the flow is faster on one side than the other. The wing of an aircraft (an aerofoil) is shaped and tilted so that the streamlines crowd together above it: the air moves faster over the top, its pressure is lower, and the pressure difference times the wing area pushes the wing up. (The wing also deflects air downwards; the two descriptions are two views of the same flow.) A spinning ball drags air round with it: on one side this adds to the flow past the ball, on the other it opposes it, so the ball is pushed towards the faster side. This Magnus effect makes a spinning football or table-tennis ball swerve.

✏️ Example 6 — Lift on a light aircraft
At take-off, air flows over the upper surface of a light aircraft's wings at $70\ \text{m/s}$ and under the lower surface at $60\ \text{m/s}$. The wings have a total area of $16\ \text{m}^2$; take the density of air as $1.2\ \text{kg/m}^3$ and neglect the small difference in height between the surfaces. Find (a) the pressure difference across the wings, (b) the lift force, (c) the largest mass the wings can hold up at this speed.
(a) $p_\text{below}-p_\text{above}=\tfrac12\rho(v_\text{above}^2-v_\text{below}^2)=0.60\times(4900-3600)=780\ \text{Pa}$.
(b) $F=\Delta p\,A=780\times16=1.2\times10^4\ \text{N}$, upward.
(c) $m=\dfrac{F}{g}=\dfrac{12\,480}{9.8}=1.3\times10^3\ \text{kg}$.
Check $780\ \text{Pa}$ is less than $1\%$ of atmospheric pressure — a tiny pressure difference over a large area gives a large force. Note that $v_\text{above}^2-v_\text{below}^2\ne(v_\text{above}-v_\text{below})^2$: squaring the difference would give only $60\ \text{Pa}$.
Interactive 2 — Draining Tank (Torricelli)

A wide tank (area $0.50\ \text{m}^2$) on a stand leaks through a hole of area $1.0\ \text{cm}^2$. The jet leaves horizontally at $v=\sqrt{2gh}$ and follows a projectile path to the ground. Press Play to let the tank drain (time is speeded up): as the depth $h$ falls, the jet slows and lands closer. Try to find the hole height that gives the longest jet for a given tank.

2.00 m0.80 m
depth above hole $h$—
jet speed $\sqrt{2gh}$—
range $2\sqrt{hy}$—
time elapsed—
⚠️ Common mistakes

1. "Faster flow, higher pressure." The opposite: at the same height, faster flow means lower pressure. A fast jet hitting your hand pushes hard because it is stopped, not because its pressure is high.

2. Diameter versus area. In $A_1v_1=A_2v_2$, halving the diameter quarters the area and quadruples the speed.

3. Mixing gauge and absolute pressures in one Bernoulli equation. Choose one and keep to it; at a free surface or in an open jet, $p=p_a$ (gauge $0$).

4. $\tfrac12\rho(v_2-v_1)^2$. The kinetic term is $\tfrac12\rho(v_2^2-v_1^2)$, a difference of squares.

5. Using Bernoulli where it does not apply — across a pump or a turbine (energy is added or removed), in very viscous or turbulent flow, or between points on different streamlines of a flow with eddies.

6. Efflux depth. In $v=\sqrt{2gh}$, $h$ is the depth of the hole below the free surface, not its height above the ground.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — how long does a tank take to empty?

In JEE Main, NEET and JEE Advanced (as an application of continuity and Bernoulli's principle); not required by the CBSE syllabus.

A tank of cross-section $A$ drains through a small hole of area $a$ at its base. When the water is $h$ deep, the jet speed is $\sqrt{2gh}$ and, by continuity, the surface falls at speed $\dfrac{a}{A}\sqrt{2gh}$:

$$\frac{dh}{dt}=-\frac{a}{A}\sqrt{2gh}.$$

The surface slows as the tank empties. Rewriting, $\dfrac{dh}{\sqrt h}=-\dfrac{a}{A}\sqrt{2g}\,dt$; since $\dfrac{d}{dh}\left(2\sqrt h\right)=\dfrac{1}{\sqrt h}$ (you will meet this rule in mathematics), integrating from depth $H$ to $0$ gives the emptying time

$$t=\frac{A}{a}\sqrt{\frac{2H}{g}} .$$

It is proportional to $\sqrt H$: the second half of the depth takes longer than the first. The time to fall from $H$ to $H/4$ equals the time from $H/4$ to empty. Example. A tank of area $0.50\ \text{m}^2$ holding water $1.8\ \text{m}$ deep, with a $1.0\ \text{cm}^2$ hole: $t=\dfrac{0.50}{1.0\times10^{-4}}\sqrt{\dfrac{3.6}{9.8}}=3.0\times10^3\ \text{s}$, about $51$ minutes (ideal hole).

Exercises

Take $g=9.8\ \text{m/s}^2$, $\rho_\text{water}=1.00\times10^3\ \text{kg/m}^3$ and $\rho_\text{air}=1.2\ \text{kg/m}^3$; treat all fluids as ideal unless stated otherwise.

I
Streamlines and continuity.
  • 1 Water flows at $3.0\ \text{m/s}$ in a pipe of cross-section $20\ \text{cm}^2$ which narrows to $5.0\ \text{cm}^2$. Find the speed in the narrow part and the volume flow rate.
  • 2 A hose of internal diameter $2.0\ \text{cm}$ fills a $20\ \text{L}$ bucket in $50\ \text{s}$. Find the speed of the water in the hose, and in a nozzle of diameter $0.50\ \text{cm}$.
  • 3 A river $40\ \text{m}$ wide and $3.0\ \text{m}$ deep flows at $1.2\ \text{m/s}$. Downstream it passes through a gorge $10\ \text{m}$ wide and $6.0\ \text{m}$ deep. How fast does it flow there?
  • 4 Explain why the stream of water from a gently running tap becomes narrower as it falls.
  • 5 In part 4, the stream has a diameter of $1.2\ \text{cm}$ and a speed of $0.50\ \text{m/s}$ at the tap. Find its speed and diameter $20\ \text{cm}$ lower down.
  • 6 Distinguish between streamline and turbulent flow. Why can two streamlines never cross?
  • 7 Water flowing at $0.60\ \text{m/s}$ in a pipe of diameter $2.5\ \text{cm}$ divides equally into five pipes of diameter $1.0\ \text{cm}$. Find the speed in each small pipe.
1 $v_2=3.0\times\dfrac{20}{5.0}=12\ \text{m/s}$; $Q=20\times10^{-4}\times3.0=6.0\times10^{-3}\ \text{m}^3\text{/s}$.
2 $Q=\dfrac{0.020}{50}=4.0\times10^{-4}\ \text{m}^3\text{/s}$; hose $v=\dfrac{4.0\times10^{-4}}{\pi(0.010)^2}=1.3\ \text{m/s}$ ($1.27$); nozzle: $16$ times faster, $20\ \text{m/s}$.
3 $v=1.2\times\dfrac{40\times3.0}{10\times6.0}=2.4\ \text{m/s}$.
4 The water accelerates as it falls, and the volume flow rate $Av$ is the same at every level (continuity), so the cross-section $A$ must shrink as $v$ grows.
5 $v=\sqrt{0.50^2+2(9.8)(0.20)}=2.0\ \text{m/s}$ ($2.04$). $d\propto\dfrac{1}{\sqrt v}$: $d=1.2\sqrt{\dfrac{0.50}{2.04}}=0.59\ \text{cm}$.
6 In streamline flow the velocity at each point is constant in time and particles follow smooth fixed paths; in turbulent flow the velocity at a point fluctuates irregularly, with eddies. At a crossing a particle would have two velocities at once, impossible in steady flow.
7 $A_1v_1=5A_2v_2$: $v_2=0.60\times\dfrac{2.5^2}{5\times1.0^2}=0.75\ \text{m/s}$.
II
Bernoulli's principle.
  • 1 Water flows through a horizontal pipe. Where the speed is $1.5\ \text{m/s}$ the pressure is $1.20\times10^5\ \text{Pa}$. Find the pressure where the speed is $4.5\ \text{m/s}$.
  • 2 Show that each term of $p+\tfrac12\rho v^2+\rho gh$ has the dimensions of pressure, and interpret each as an energy per unit volume.
  • 3 Water enters a building at ground level through a pipe of diameter $4.0\ \text{cm}$ at $1.0\ \text{m/s}$ and an absolute pressure of $1.60\times10^5\ \text{Pa}$. Find the speed and pressure in a $2.0\ \text{cm}$ pipe on a floor $10\ \text{m}$ higher.
  • 4 Explain, using Bernoulli's principle, why a strong wind can lift the roof off a house.
  • 5 A wind of $30\ \text{m/s}$ blows over a flat roof of area $120\ \text{m}^2$; the air inside is still. Estimate the net upward force on the roof.
  • 6 Two boats moving side by side in the same direction, close together, tend to be drawn towards each other. Explain.
  • 7 Show that for a fluid at rest Bernoulli's principle reduces to the formula for the variation of pressure with depth.
1 $p_2=1.20\times10^5-500(4.5^2-1.5^2)=1.20\times10^5-9.0\times10^3=1.11\times10^5\ \text{Pa}$.
2 $[\rho v^2]=[\text{M L}^{-3}][\text{L}^2\text{T}^{-2}]=[\text{M L}^{-1}\text{T}^{-2}]$; $[\rho gh]=[\text{M L}^{-3}][\text{L T}^{-2}][\text{L}]=[\text{M L}^{-1}\text{T}^{-2}]$, the same as $[p]$. $\tfrac12\rho v^2=\dfrac{\tfrac12mv^2}{V}$ is kinetic energy per unit volume, $\rho gh=\dfrac{mgh}{V}$ potential energy per unit volume, and $p$ is the work done by pressure forces per unit volume pushed through.
3 $v_2=4\times1.0=4.0\ \text{m/s}$. $p_2=1.60\times10^5+500(1.0-16)-1000(9.8)(10)=1.60\times10^5-7.5\times10^3-9.8\times10^4=5.45\times10^4\ \text{Pa}$ (absolute).
4 Fast air over the roof has a lower pressure than the still air inside the house; the pressure difference over the large roof area gives a big upward force, which can exceed the roof's weight and fixings.
5 $\Delta p=\tfrac12(1.2)(30)^2=540\ \text{Pa}$; $F=540\times120=6.5\times10^4\ \text{N}$, upward.
6 Relative to the boats, water streams backwards between them through a narrowed channel, faster than on their outer sides. Its pressure between the boats is lower, so the higher pressure outside pushes them together.
7 With $v_1=v_2=0$: $p_1+\rho gh_1=p_2+\rho gh_2$, i.e. $p_2-p_1=\rho g(h_1-h_2)$: the pressure is larger at the lower point by $\rho g$ times the depth difference — Result 9.2.
III
Speed of efflux and the Venturi meter.
  • 1 Find the speed of water leaving a small hole $3.2\ \text{m}$ below the surface of an open tank.
  • 2 A hole of area $2.0\ \text{cm}^2$ is $5.0\ \text{m}$ below the water surface in a large open tank. Find the volume flow rate out of it.
  • 3 A closed tank contains water with air above it at a gauge pressure of $3.0\times10^5\ \text{Pa}$. Water escapes through a small outlet $2.0\ \text{m}$ below the water surface. Find the speed of efflux.
  • 4 Water stands $1.25\ \text{m}$ above a small hole in the side of a tank; the hole is $0.80\ \text{m}$ above the ground. Find where the jet lands.
  • 5 An open tank on the ground is filled with water to a height $H$. Show that a jet from a small hole at depth $h$ below the surface lands at a distance $R=2\sqrt{h(H-h)}$, that holes at depths $h$ and $H-h$ give the same range, and that the range is greatest for $h=\tfrac{H}{2}$. Find this greatest range for $H=2.0\ \text{m}$.
  • 6 In a horizontal Venturi meter the pipe has area $50\ \text{cm}^2$ and the throat $10\ \text{cm}^2$. The pressure difference for flowing water is $1.2\times10^4\ \text{Pa}$. Find the speeds in the pipe and the throat and the flow rate.
  • 7 A Venturi meter with pipe diameter $6.0\ \text{cm}$ and throat diameter $3.0\ \text{cm}$ carries water; its open tubes show levels differing by $8.0\ \text{cm}$. Find the flow rate in litres per second.
1 $v=\sqrt{2(9.8)(3.2)}=7.9\ \text{m/s}$.
2 $v=\sqrt{98}=9.9\ \text{m/s}$; $Q=2.0\times10^{-4}\times9.9=2.0\times10^{-3}\ \text{m}^3\text{/s}$ ($2.0\ \text{L/s}$).
3 $v=\sqrt{2(9.8)(2.0)+\dfrac{2(3.0\times10^5)}{1000}}=\sqrt{39.2+600}=25\ \text{m/s}$ — the trapped air does most of the work.
4 $v=\sqrt{2(9.8)(1.25)}=4.95\ \text{m/s}$; fall time $\sqrt{1.6/9.8}=0.404\ \text{s}$; $R=2.0\ \text{m}$ (check: $2\sqrt{1.25\times0.80}=2.0$).
5 $v=\sqrt{2gh}$, fall height $H-h$, time $\sqrt{2(H-h)/g}$, so $R=2\sqrt{h(H-h)}$. Replacing $h$ by $H-h$ leaves the product unchanged. $h(H-h)=\tfrac{H^2}{4}-\left(h-\tfrac H2\right)^2$ is greatest at $h=\tfrac H2$, giving $R_\text{max}=H=2.0\ \text{m}$.
6 $\dfrac{A_1}{A_2}=5$: $v_1=\sqrt{\dfrac{2(1.2\times10^4)}{1000(25-1)}}=1.0\ \text{m/s}$, $v_2=5.0\ \text{m/s}$, $Q=50\times10^{-4}\times1.0=5.0\times10^{-3}\ \text{m}^3\text{/s}$.
7 $\dfrac{A_1}{A_2}=4$: $v_1=\sqrt{\dfrac{2(9.8)(0.080)}{15}}=0.32\ \text{m/s}$; $Q=\pi(0.030)^2(0.323)=9.1\times10^{-4}\ \text{m}^3\text{/s}=0.91\ \text{L/s}$.
IV
Applications and harder problems.
  • 1 (Modelling) A $1200\ \text{kg}$ light aircraft has a wing area of $16\ \text{m}^2$. In level flight the air passes under the wings at $50\ \text{m/s}$. How fast must it pass over the top?
  • 2 (Modelling) Blood ($1.06\times10^3\ \text{kg/m}^3$) flows at $0.40\ \text{m/s}$ in an artery. A deposit reduces the cross-section to one-third. Find the blood speed in the narrowed part and the pressure drop there, in Pa and in mm Hg ($1\ \text{mm Hg}=133\ \text{Pa}$).
  • 3 In a scent spray, air is blown at $25\ \text{m/s}$ across the top of a thin tube dipping into a perfume of density $800\ \text{kg/m}^3$. To what height above the liquid surface can the pressure drop lift the perfume?
  • 4 A fire-hose nozzle points straight up and the water leaves at $20\ \text{m/s}$. How high does the water rise? What gauge pressure is needed in the wide hose just behind the nozzle, where the water moves slowly?
  • 5 A footballer kicks a ball with backspin. Using the Magnus effect, explain whether the ball rises above or dips below the path it would follow without spin.
  • 6 As an open tank drains, the depth of water above the hole falls from $1.8\ \text{m}$ to $0.45\ \text{m}$. By what factor does the speed of efflux change? By what factor does the jet's range on the floor change, if the hole stays $0.50\ \text{m}$ above the floor?
1 Needed $\Delta p=\dfrac{1200\times9.8}{16}=735\ \text{Pa}$. $v_\text{top}^2=50^2+\dfrac{2(735)}{1.2}=2500+1225$, $v_\text{top}=61\ \text{m/s}$.
2 $v_2=3\times0.40=1.2\ \text{m/s}$; $\Delta p=\tfrac12(1060)(1.44-0.16)=6.8\times10^2\ \text{Pa}$, about $5.1\ \text{mm Hg}$.
3 $\Delta p=\tfrac12(1.2)(25)^2=375\ \text{Pa}$; $h=\dfrac{375}{800\times9.8}=0.048\ \text{m}=4.8\ \text{cm}$.
4 $H=\dfrac{v^2}{2g}=\dfrac{400}{19.6}=20\ \text{m}$ ($20.4$). Bernoulli from hose to jet (ignoring the short height of the nozzle): $p_\text{gauge}=\tfrac12\rho v^2=2.0\times10^5\ \text{Pa}$.
5 With backspin the bottom of the ball moves forward through the air, against the oncoming air, and the top moves backward, with it: the air flows faster over the top. The pressure is lower above, so there is an upward lift and the ball floats above its no-spin path (topspin makes it dip).
6 $v\propto\sqrt h$: $\sqrt{\dfrac{0.45}{1.8}}=\tfrac12$ — the speed halves. With the hole height fixed the fall time is unchanged, so the range also halves.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks State Bernoulli's principle and derive it for the steady flow of an ideal fluid through a pipe of varying cross-section and height.

Statement For steady flow of an incompressible, non-viscous fluid, $p+\tfrac12\rho v^2+\rho gh$ is constant along a streamline. [1]
Work In time $\Delta t$ the volume $\Delta V=A_1v_1\Delta t=A_2v_2\Delta t$ enters at section 1 and leaves at section 2 (continuity). Net work by pressure $(p_1-p_2)\Delta V$; work by gravity $-\rho\Delta Vg(h_2-h_1)$. [1]
Energy $\Delta K=\tfrac12\rho\Delta V(v_2^2-v_1^2)$. Work–energy theorem, then divide by $\Delta V$: $p_1+\tfrac12\rho v_1^2+\rho gh_1=p_2+\tfrac12\rho v_2^2+\rho gh_2$. [1]
Exam tip Draw the pipe with $A_1,v_1,p_1,h_1$ and $A_2,v_2,p_2,h_2$ marked and state the assumptions; examiners give credit for the labelled diagram.

E2. NEET Water flows through a horizontal pipe whose radius decreases from $2r$ to $r$. If the speed in the wider part is $v$, the pressure difference between the wide and narrow parts is (a) $\dfrac{3\rho v^2}{2}$ (b) $\dfrac{15\rho v^2}{2}$ (c) $8\rho v^2$ (d) $15\rho v^2$

Idea Radius halves, area quarters: $v_2=4v$. $\Delta p=\tfrac12\rho(16v^2-v^2)$.
Answer $\dfrac{15\rho v^2}{2}$: (b).
Exam tip (a) uses $v_2=2v$ (radius ratio instead of area ratio); (d) forgets the $\tfrac12$.

E3. JEE Main (Numerical value) A large open tank standing on level ground is filled with water to a height of $5.0\ \text{m}$. A small hole can be drilled anywhere in its vertical side. Find the greatest horizontal distance, in metres, from the tank at which the jet can strike the ground.

Set-up Hole at depth $h$: $v=\sqrt{2gh}$, height above ground $H-h$, so $R=2\sqrt{h(H-h)}$.
Maximum $h(H-h)$ is largest at $h=\tfrac H2=2.5\ \text{m}$, giving $R=2\sqrt{2.5\times2.5}=5.0\ \text{m}$.
Answer $5$ (m) — equal to the depth of water, independent of $g$.
Exam tip Holes at depths $h$ and $H-h$ give equal ranges — a quick way to spot pairs in MCQs.

E4. JEE Advanced (One or more correct) A wide open tank on the floor holds water to a height $H$. A small hole is made in its side at a depth $h$ below the surface. Which statements are correct? (A) The speed of efflux would be the same for mercury filled to the same level. (B) The jet's range on the floor is greatest when $h=\tfrac{H}{2}$. (C) Holes at depths $\tfrac H4$ and $\tfrac{3H}{4}$ give jets of equal range. (D) The speed of efflux equals the speed a stone gains falling freely through the height $H$.

(A) $v=\sqrt{2gh}$ contains no density (both the driving pressure $\rho gh$ and the inertia $\tfrac12\rho v^2$ scale with $\rho$) ✓.
(B), (C) $R=2\sqrt{h(H-h)}$ is greatest at $h=\tfrac H2$ ✓ and symmetric under $h\to H-h$: $\tfrac H4$ and $\tfrac{3H}4$ both give $R=\tfrac{\sqrt3}{2}H$ ✓.
(D) The fall height that matters is $h$, the depth of the hole below the surface, not $H$ ✗ (true only for a hole at the very bottom).
Answer (A), (B), (C).
Exam tip Check symmetric claims with numbers: $H=4$, $h=1$: $R=2\sqrt3$; $h=3$: $R=2\sqrt3$ ✓.

E5. JEE Main NEET Air of density $1.2\ \text{kg/m}^3$ flows past the upper surface of a wing at $60\ \text{m/s}$ and past the lower surface at $50\ \text{m/s}$. If the wing area is $20\ \text{m}^2$, the lift on the wing is about (a) $1.2\times10^3\ \text{N}$ (b) $6.6\times10^3\ \text{N}$ (c) $1.3\times10^4\ \text{N}$ (d) $2.6\times10^4\ \text{N}$

Idea $\Delta p=\tfrac12\rho(v_\text{top}^2-v_\text{bottom}^2)=0.60\times(3600-2500)=660\ \text{Pa}$.
Answer $F=660\times20=1.32\times10^4\ \text{N}$: (c).
Exam tip (a) squares the speed difference, $0.60\times10^2\times20$; (d) drops the $\tfrac12$; (b) is half the right answer — the $\tfrac12$ counted twice.

E6. Board · Case study A water company fits a Venturi meter in a horizontal main of diameter $20\ \text{cm}$; the throat diameter is $10\ \text{cm}$. Open tubes at the main and at the throat show water levels that differ by $30\ \text{cm}$. (i) Explain why the level in the throat tube is lower. [1] (ii) Find the speed of the water in the main. [2] (iii) Find the volume of water delivered per second. [1]

(i) By continuity the water moves faster in the narrow throat; by Bernoulli's principle (same height) faster flow has lower pressure, so the water column it supports is shorter. [1]
(ii) $\dfrac{A_1}{A_2}=4$, $p_1-p_2=\rho g(0.30)$: $v_1=\sqrt{\dfrac{2(9.8)(0.30)}{4^2-1}}=\sqrt{0.392}=0.63\ \text{m/s}$. [2]
(iii) $Q=\pi(0.10)^2\times0.626=2.0\times10^{-2}\ \text{m}^3\text{/s}$, about $20$ litres per second. [1]
Exam tip Write $\left(\dfrac{A_1}{A_2}\right)^2-1=15$ explicitly — forgetting the "$-1$" (it comes from the $v_1^2$ term) is the commonest slip.
🔬 Research–Teaching NexusResearch-tutored · optional, about 45 minutes

Research question. Does a real leaking bottle obey Torricelli's law — and if not, by how much does it fall short?

Why it matters. Engineers who size the outlets of water tanks, irrigation channels, rain-water harvesting systems and dam spillways use $Q=C_dA\sqrt{2gh}$, where the discharge coefficient $C_d$ (less than 1) corrects the ideal result for the narrowing of the jet and for friction. Measuring $C_d$ for a simple hole is a classic first experiment in fluid mechanics laboratories around the world.

What to do. Use a large plastic bottle with straight sides. Make a clean round hole about $4\ \text{mm}$ across near the bottom and tape over it. Fill the bottle and stand it on a box at the edge of a tray or sink.

  1. Measure the hole's diameter and its height $y$ above the tray. Mark depth levels $h$ above the hole on the bottle with a marker every $2\ \text{cm}$.
  2. Remove the tape. As the water falls past each mark, record the range $R$ of the jet on the tray (a friend with a ruler, or a phone video). Compare with $R=2\sqrt{hy}$.
  3. Collect the water for $10\ \text{s}$ at a fixed level (top it up continuously) and measure its volume. Compare with $A\sqrt{2gh}$ and find $C_d$.
  4. Time how long the level takes to fall between two marks and compare with the emptying law in the extension box.
  5. Plot $R^2$ against $h$. Is it a straight line? What does its slope tell you?

Tools: Plastic bottle, ruler, measuring jug, phone stopwatch or slow-motion video, spreadsheet for the graphs.

Hand in: A table of $h$, measured $R$ and predicted $R$; your measured $C_d$ with an estimate of its uncertainty; the $R^2$–$h$ graph; and a paragraph explaining two physical reasons for the shortfall.

Going further. Try a hole made with a short tube pushed into it, or a hole with ragged edges. How does $C_d$ change? Find out what the "vena contracta" is and why it reduces the flow.

Instructor note: for a $4.0\ \text{mm}$ hole, $A=1.26\times10^{-5}\ \text{m}^2$; at $h=0.10\ \text{m}$, ideal $Q=A\sqrt{2gh}=1.76\times10^{-5}\ \text{m}^3\text{/s}=17.6\ \text{mL/s}$. Students typically find about $11$ mL/s, $C_d\approx0.6$ for a sharp-edged hole, $0.8$ or more for a short tube. The range usually falls short of $2\sqrt{hy}$ by only a few per cent, because $R$ depends on the jet speed (reduced by friction, $C_v\approx0.97$) and not on the jet's narrowing — a nice way to separate the two effects. $R^2=4yh$ gives a straight line of slope $4y$.

🧪 Self-check quiz

0/5 correct

Q1. Water flows steadily through a pipe. Where the diameter of the pipe is halved, the speed of the water becomes

Q2. In a horizontal pipe carrying an ideal fluid, the pressure is lowest where

Q3. Water leaves a small hole $5.0\ \text{m}$ below the surface of a large open tank with a speed of about

Q4. Which of these is not an assumption in deriving Bernoulli's principle?

Q5. The lift on an aircraft wing (in the Bernoulli picture) arises because

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