In short

Bernoulli's equation — P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{constant along a streamline} — is one compact statement that explains an entire shelf of phenomena. A venturi meter measures flow speed from the pressure drop across a constriction: v_1 = \sqrt{\tfrac{2(P_1 - P_2)}{\rho\left[(A_1/A_2)^2 - 1\right]}}. A pitot tube measures the airspeed of an aircraft: v = \sqrt{2(P_0 - P)/\rho}. Torricelli's theorem gives the speed of fluid leaving a hole a depth h below the free surface: v = \sqrt{2gh}, the same as a body in free fall. The Magnus effect explains a spinning cricket ball's swerve: rotation drags air along with it, raising flow speed on one side and lowering pressure there, so a sideways force pushes the ball toward the low-pressure side. Aeroplane wings generate lift primarily through a combination of air being deflected downward (Newton's third law) and a pressure difference that Bernoulli's equation tracks — the naive "longer path on top" argument is not the whole story and deserves more care than textbooks give it.

A tanker truck pulls in at the petrol pump behind Connaught Place and the attendant fits a flow meter to the nozzle. It is a quiet, boring instrument. It has no moving parts in the flow path. It contains no paddle wheels. Yet every drop of diesel pumped into your rickshaw, every litre filling a lorry's tank, is metred by the difference in pressure between two points in the pipe. That difference, and the speed it reveals, are given to you by one equation — Bernoulli's.

Fifty kilometres above that petrol pump, a commercial airliner is on approach into IGI. The cockpit display shows a number — the aircraft's airspeed, 250 knots. Nobody on board is timing how fast the plane moves across the ground. They cannot — the ground is eight kilometres below, and the wind up there is blowing at 60 km/h in a direction they do not know. Instead, a slender forward-facing tube on the nose of the aircraft measures the pressure at its tip, compares it with the pressure on its side, and turns that difference into a speed. Same equation.

And a thousand kilometres south, on a cricket ground in Chennai, Jasprit Bumrah releases a ball at 145 km/h with the seam angled slightly to the off side. The ball swings sharply away from the right-handed batter. Why? Because one side of the ball has air rushing past it faster than the other side. The low-pressure side is sucked inward, curving the flight.

Three of the most different-looking phenomena you can name — petrol metering, aircraft instrumentation, fast bowling — and they are all the same idea wearing three different hats. That is the point of this article.

The tool you will use everywhere

This article is built on one equation and one consequence. If you are sure about both, the rest is mechanics.

Bernoulli's equation. For an ideal fluid (incompressible, non-viscous, irrotational, in steady flow), along any streamline the sum of pressure, kinetic energy density, and gravitational potential energy density is constant:

P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{constant}.
The derivation, assumptions, and precise statement are in Bernoulli's Principle. This article treats the equation as known and focuses on what it predicts in real situations.

The one-sentence consequence. Where a fluid moves faster, its pressure is lower (at the same height). Three Bernoulli-style applications will re-use this consequence with different geometries; one (Torricelli) will use the full three-term version; one (the wing) will need a careful combination of Bernoulli's equation and Newton's third law.

The other tool you will re-use is the equation of continuity — A_1 v_1 = A_2 v_2 for an incompressible fluid in a flow tube of varying area. See Equation of Continuity for the derivation.

With those two equations you can explain every application in this article.

Application 1 — The venturi meter

A venturi meter is a pipe with a deliberately narrowed middle section (the throat) used to measure the flow speed of a fluid by reading off the pressure drop between the wide and narrow portions. It is the device that measures flow in petrol pumps, water mains, chemical plants, and the inlet manifolds of some older carburettor-based engines.

The geometry is this: a horizontal pipe of cross-sectional area A_1 narrows smoothly to area A_2 < A_1, then widens back. Two small pressure taps — thin side-pipes sticking out of the main pipe — measure the static pressure at sections 1 and 2. A U-tube manometer between them reads the pressure difference P_1 - P_2.

Venturi meter cross-section with wide inlet, narrow throat, and U-tube manometer A horizontal pipe wide on both ends and narrow in the middle. Fluid flows left to right. Two vertical tubes at the wide and narrow sections connect to a U-tube manometer, with the liquid on the wide side higher than the narrow side. $v_1$ $A_1$ section 1 $v_2$ $A_2$ throat $v_1$ high low manometer: $P_1 - P_2$ reads here
A venturi meter. The pipe narrows at the throat (section 2); by continuity the fluid speeds up there; by Bernoulli the pressure drops. A manometer between the two pressure taps reads the drop, from which the flow speed $v_1$ is computed.

Derivation of the venturi formula

Step 1. Apply the continuity equation to the two sections.

The pipe is horizontal and the fluid is incompressible, so A_1 v_1 = A_2 v_2, giving

v_2 = v_1 \cdot \frac{A_1}{A_2}. \tag{1}

Why: mass cannot pile up at the throat — whatever volume per second enters section 1 must leave section 2. The narrower section forces the fluid to speed up.

Step 2. Apply Bernoulli's equation to a streamline running from section 1 to section 2.

Both sections are at the same height, so the \rho g h terms cancel:

P_1 + \tfrac{1}{2}\rho v_1^2 = P_2 + \tfrac{1}{2}\rho v_2^2. \tag{2}

Why: along a streamline, the Bernoulli sum is constant. Dropping the height terms is allowed only because the pipe is horizontal — in an inclined venturi you would keep them.

Step 3. Rearrange to isolate the pressure drop.

P_1 - P_2 = \tfrac{1}{2}\rho \left(v_2^2 - v_1^2\right). \tag{3}

Why: the pressure drop equals half the density times the jump in squared speed. Both sides have dimensions of pressure, as they must.

Step 4. Substitute the continuity result v_2 = v_1 (A_1/A_2) into equation (3).

P_1 - P_2 = \tfrac{1}{2}\rho\left[v_1^2\left(\frac{A_1}{A_2}\right)^2 - v_1^2\right] = \tfrac{1}{2}\rho v_1^2\left[\left(\frac{A_1}{A_2}\right)^2 - 1\right]. \tag{4}

Why: factor v_1^2 out of the square bracket. All the unknowns except v_1 (which is what you want) are now collected on the right.

Step 5. Solve for the inlet speed v_1.

\boxed{\; v_1 = \sqrt{\frac{2(P_1 - P_2)}{\rho\left[\left(A_1/A_2\right)^2 - 1\right]}} \;} \tag{5}

Why: the inlet speed is determined by the measured pressure drop, the fluid density, and the fixed area ratio of the meter. The square root is real and positive because the bracket is always positive (A_1 > A_2).

That is the venturi formula. Every petrol-pump meter, every water-mains flow-meter, every carburettor inlet, is built around this one equation.

Reading the manometer

If the manometer is a simple U-tube with a heavier liquid of density \rho_m, the height difference \Delta h between the two arms gives

P_1 - P_2 = (\rho_m - \rho) g \, \Delta h,

which substitutes into (5) in place of the raw pressure drop. In a petrol-pump meter \rho_m is mercury (13\,600\text{ kg/m}^3) and \rho is petrol (~740\text{ kg/m}^3); the \Delta h is typically a few centimetres.

Numbers: a Kanpur municipal water main

A 20 cm diameter water main narrows at a venturi throat to 10 cm diameter. A differential manometer reads P_1 - P_2 = 4000\text{ Pa} (about 41 cm of water column). What is the flow speed in the main, and what is the volume flow rate?

A_1 = \pi (0.10)^2 = 3.14 \times 10^{-2}\text{ m}^2, \quad A_2 = \pi (0.05)^2 = 7.85 \times 10^{-3}\text{ m}^2.

Area ratio A_1/A_2 = 4, so (A_1/A_2)^2 - 1 = 15. With \rho = 1000\text{ kg/m}^3:

v_1 = \sqrt{\frac{2 \times 4000}{1000 \times 15}} = \sqrt{0.533} \approx 0.73\text{ m/s}.

Volume flow rate: Q = A_1 v_1 = 0.0314 \times 0.73 \approx 2.3 \times 10^{-2}\text{ m}^3/\text{s} \approx 23\text{ L/s}.

A city's water supply engineer can read \Delta h off a manometer installed once, on a holiday, and get Q any day thereafter without opening the pipe.

Explore the venturi — drag the area ratio

Interactive: venturi pressure drop versus area ratio A curve showing pressure drop P1 minus P2 as a function of the area ratio A1 over A2 for fixed inlet speed 1 m/s and density 1000 kg/m³. Drag the red dot along the x-axis to read values. area ratio $A_1/A_2$ pressure drop $P_1-P_2$ (Pa) 0 100 200 300 400 500 2 4 6 8 10 drag the red point along the axis
Pressure drop $P_1 - P_2$ across a venturi as a function of the area ratio $A_1/A_2$, with inlet speed fixed at 1 m/s and water ($\rho = 1000$ kg/m³). The growth is quadratic — double the area ratio and the pressure drop quadruples. A 10:1 venturi reads 49 500 Pa, plenty for a manometer.

Application 2 — The pitot tube

A pitot tube is the airspeed sensor on every commercial aircraft, and it is the most direct Bernoulli instrument possible. It measures the pressure difference between a point where the air has been brought to rest (the stagnation point) and a point where the air is flowing past undisturbed at speed v. From the difference, it infers v.

Pitot-static tube with stagnation port facing flow and static ports on the side A horizontal L-shaped tube pointing into the oncoming air. The front opening faces the flow (stagnation port). Small side holes sample the static pressure. Two internal channels connect to a differential pressure sensor. ΔP stagnation port ($P_0$) static ports ($P$) $v$ free stream
A pitot-static tube. Air enters the front port and slows to rest inside: the pressure there is the **stagnation pressure** $P_0$. Small holes on the side sample the undisturbed **static pressure** $P$. A pressure sensor reads the difference $P_0 - P$.

Derivation of the pitot formula

Take two points on the same streamline: point A, far upstream in the undisturbed flow (speed v, pressure P), and point B, at the mouth of the stagnation port, where the air has been brought to rest (speed 0, pressure P_0). The streamline lies at constant height, so the \rho g h terms cancel.

Step 1. Write Bernoulli's equation between A and B.

P + \tfrac{1}{2}\rho v^2 + 0 = P_0 + \tfrac{1}{2}\rho \cdot 0^2 + 0.
P_0 = P + \tfrac{1}{2}\rho v^2. \tag{6}

Why: along the streamline from the free-stream point to the stagnation point, energy is conserved (no viscosity in the ideal-fluid model). At the stagnation point the kinetic energy density \tfrac{1}{2}\rho v^2 has been converted entirely into pressure.

Step 2. Solve for v.

\boxed{\; v = \sqrt{\frac{2(P_0 - P)}{\rho}} \;} \tag{7}

Why: the difference P_0 - P is called the dynamic pressure. Squeezing that out of the differential pressure sensor, dividing by half the air density, and taking the square root gives you the airspeed directly.

Why aircraft use this and not a speedometer

A car's speedometer measures ground speed (via a transmission encoder). For an aircraft that is useless, because what keeps the plane in the air is the speed of air over the wings, not speed over the ground. A plane flying at 200 knots ground speed into a 100-knot headwind has an airspeed of 300 knots — and 300 knots is what determines whether the wings generate enough lift. The pitot tube measures exactly that.

Numbers: a Boeing 787 at cruise

At 12 km altitude, air density is about \rho \approx 0.3\text{ kg/m}^3 (one-quarter the sea-level value). The aircraft cruises at true airspeed 250 m/s (about 900 km/h, or Mach 0.85). The pitot tube senses:

P_0 - P = \tfrac{1}{2} \times 0.3 \times 250^2 = 9375\text{ Pa}.

That's about 9% of sea-level atmospheric pressure — readily measurable with a small diaphragm transducer. The cockpit electronics convert this pressure difference into the indicated airspeed on the pilot's display.

A caveat for fast flight. Equation (7) assumes the air is incompressible. Above about Mach 0.3, compressibility corrections become necessary; the pitot tube still works but the formula is modified. At supersonic speeds, a shock wave forms ahead of the tube and the relationship changes again. This is one of the subtle reasons the instrument panel shows separate indicated and true airspeeds, and why pitot icing incidents have caused accidents — if the tube clogs, the sensor lies, and the autopilot can feed the crew the wrong speed.

Application 3 — Torricelli's theorem

Drill a small hole at depth h below the free surface of a large open tank of water. With what speed does the water jet out of the hole? The answer, discovered in 1643, is startling in its simplicity:

\boxed{\; v = \sqrt{2 g h}. \;}

That is the same speed an object would acquire falling freely from rest through a height h. The jet does not know it is fluid — it comes out with exactly the speed gravity would give a stone dropped from the free surface.

Torricelli's theorem — water jetting from a hole at depth h A tall rectangular tank of water with a small hole at depth h below the free surface. A parabolic jet shoots horizontally from the hole and curves down under gravity. Arrows label the free surface, depth h, and exit speed v. $v = \sqrt{2gh}$ $h$ free surface (atmospheric pressure $P_\text{atm}$) hole
Torricelli's setup. The tank's cross-section is much larger than the hole, so the free surface descends very slowly — treat it as stationary. Apply Bernoulli from the free surface (point 1) to the jet just outside the hole (point 2), and the exit speed falls straight out of the algebra.

Derivation

Let point 1 be on the free surface of the tank and point 2 be at the mouth of the hole (just outside). Both points are exposed to the atmosphere, so P_1 = P_2 = P_\text{atm}. The tank's cross-section is much larger than the hole's, so by continuity the free surface moves very slowly — set v_1 \approx 0. Point 1 is at height h above point 2.

Step 1. Write Bernoulli's equation between points 1 and 2.

P_1 + \tfrac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \tfrac{1}{2}\rho v_2^2 + \rho g h_2.

Step 2. Substitute the known values: P_1 = P_2, v_1 = 0, h_1 - h_2 = h.

P_\text{atm} + 0 + \rho g h = P_\text{atm} + \tfrac{1}{2}\rho v_2^2 + 0.

Why: the pressures cancel because both points are exposed to the atmosphere. The only non-zero terms are the potential-energy difference on the left and the kinetic-energy density on the right.

Step 3. Simplify.

\rho g h = \tfrac{1}{2}\rho v_2^2.
v_2 = \sqrt{2 g h}.

Why: cancel \rho (non-zero). This is exactly the kinematic result for a particle falling from rest through height h — v^2 = 2gh — which is no coincidence. Bernoulli's equation is energy conservation per unit volume; kinematics is energy conservation per unit mass. They had to agree.

That is Torricelli's theorem. It is arguably the oldest quantitative result in fluid dynamics, and the derivation is three lines of Bernoulli.

Horizontal range of the jet

Once the water leaves the hole it is a projectile — it accelerates at g downward, travelling at v = \sqrt{2gh} horizontally. If the hole is at height H above the floor, the jet travels horizontally for a time t = \sqrt{2H/g} (the time to fall the remaining H) and lands at horizontal distance

R = v \cdot t = \sqrt{2gh} \cdot \sqrt{2H/g} = 2\sqrt{hH}.

The range depends symmetrically on the depth of the hole below the water surface and its height above the floor. The maximum range for a tank of total water height H_0 occurs when h + H = H_0 and h = H = H_0/2, giving R_\text{max} = H_0. In a one-metre-tall water tank, the farthest a leak can squirt is one metre — achieved when the leak is exactly halfway down.

Numbers: the slow-draining tank

A 20 cm diameter overhead water tank in your flat in Indiranagar has 1.5 m of water standing above a leaking seal. The seal's area is 2 mm² — a pinhole. Time to drain?

Exit speed: v = \sqrt{2 \times 9.8 \times 1.5} = \sqrt{29.4} \approx 5.4\text{ m/s}.

Volume flow rate out: Q = A_\text{hole} \cdot v = 2 \times 10^{-6} \times 5.4 = 1.08 \times 10^{-5}\text{ m}^3/\text{s} \approx 40\text{ mL/s}.

That's enough to drain several litres an hour — a spend on your water bill but not an immediate flood.

Strictly, as the tank drains h decreases, so v decreases, so the drain rate slows. The total draining time integrates to t_\text{drain} = \frac{A_\text{tank}}{A_\text{hole}}\sqrt{\frac{2 H_0}{g}}, but that derivation belongs to the going-deeper section.

Application 4 — The Magnus effect (and the swerving cricket ball)

A ball thrown without spin travels along a parabolic trajectory determined purely by its initial velocity and gravity. A ball thrown with spin curves sideways. The sideways force on a spinning ball moving through air is the Magnus force, and it is the reason Jasprit Bumrah's slower ball seems to die in the air, the reason Dilip Doshi's left-arm orthodox would drift toward the slip cordon, and the reason a football player can curl a free-kick around the wall.

The physical picture

Picture a cricket ball moving right at speed v, spinning anti-clockwise as viewed from above (the ball's top surface moving backward with respect to the ball's direction of motion, bottom moving forward). Now switch to the ball's rest frame: air rushes past to the left at speed v. The spinning ball's rough surface drags air along with it — the top surface drags air forward (adding to the free-stream speed there), the bottom surface drags air backward (subtracting from the free-stream speed).

So on top of the ball, the air moves faster. On the bottom, the air moves slower.

Bernoulli: where air is faster, pressure is lower. So the pressure on top is less than the pressure below — and the pressure difference, integrated over the ball's surface, gives a net upward force in the ball's rest frame, which translates to an upward deflection in the ground frame.

Switch the spin direction and the force reverses. Tilt the spin axis sideways and the force tilts with it. That is the Magnus effect in one paragraph: a spinning ball generates a sideways force perpendicular to both its velocity and its spin axis.

Magnus effect on a spinning ball moving rightward A cricket ball moves right and spins anti-clockwise. Airflow lines above crowd closer than below, indicating faster flow above. An upward arrow labels the Magnus force. spin ball's velocity $v$ Magnus force faster flow above → lower pressure slower flow below → higher pressure
A cricket ball moving right, spinning anti-clockwise as viewed from above. The top of the ball co-rotates with the free-stream air, producing faster relative flow — and by Bernoulli, lower pressure. Below, the ball's surface drags air backward; the relative flow is slower and the pressure is higher. Net force: upward. (Swap the spin to invert the force.)

A rough estimate of the force

An order-of-magnitude formula for the Magnus force on a sphere of radius r moving at speed v through air of density \rho, spinning at angular velocity \omega, is

F \sim \rho \cdot v \cdot \omega r^3 \cdot C,

where C is a dimensionless coefficient of order 1 that depends on the roughness of the ball's surface. Plug in a cricket ball: r = 3.5\text{ cm}, v = 40\text{ m/s} (about 140 km/h), \omega = 100\text{ rad/s} (about 16 revolutions per second — typical for a seam bowler), \rho = 1.2\text{ kg/m}^3:

F \sim 1.2 \times 40 \times 100 \times (0.035)^3 \times 1 \approx 0.21\text{ N}.

Compare to the ball's weight mg = 0.16 \times 9.8 = 1.6\text{ N}. The Magnus force is about 13% of the ball's weight — not huge, but over the 0.5 seconds of flight from bowler to batsman, that force produces a lateral deflection of

\Delta x = \tfrac{1}{2} \cdot (F/m) \cdot t^2 = \tfrac{1}{2} \cdot (0.21/0.16) \cdot 0.5^2 \approx 0.16\text{ m} = 16\text{ cm}.

Sixteen centimetres of drift in half a second is exactly enough to turn an edge into a clean beat of the bat.

Why the seam matters — the subtle cricket case

The classical Magnus effect needs spin to work. But a cricket ball that does not spin still swings when the bowler holds the seam at an angle to the line of flight. Why? Because the air flows differently across the two sides of the ball depending on whether the surface is smooth or rough — the seam trips the boundary layer from laminar to turbulent on one side, and turbulent boundary layers separate later from the ball's surface than laminar ones.

That asymmetric separation is what produces conventional swing — the outswinger (pitched-up ball moving away from the right-handed batter) works because the seam is angled toward the slip cordon, the turbulent side is closer to slip, and the lower pressure on that side pulls the ball sideways. The analysis is not pure Bernoulli — viscosity and boundary-layer separation are essential — but the final story still has a pressure difference across the two sides of the ball at its heart. Swing bowling is Bernoulli's equation with extra ingredients. See Reynolds Number and Turbulence for why laminar and turbulent regimes behave so differently.

Application 5 — Aeroplane wings and the honest lift story

Ask most textbooks why an aeroplane flies and you will get the equal-transit-time argument: "The air over the top of the wing has to travel a longer path in the same time as the air beneath, so it moves faster, and by Bernoulli's equation its pressure is lower, which pushes the wing up."

That argument is wrong in a specific way, and the way it is wrong matters. Two reasons.

  1. There is no law requiring equal transit times. The molecule that goes over the top and the molecule that goes under the bottom meet different fluid histories. They do not need to arrive at the trailing edge simultaneously. Observations show the top-going molecule arrives sooner, not at the same time.
  2. Symmetric wings also generate lift when tilted upward. A flat plate at an angle of attack lifts. The "curved top is longer" story cannot explain this at all.

The honest story, in a padho-wiki-shaped paragraph, is this:

A wing generates lift by deflecting air downward. The wing's underside pushes air down directly; its top surface, because of its shape and angle, drags a sheet of air downward behind it too. By Newton's third law, the air pushes the wing up with an equal and opposite force. Bernoulli's equation is not the cause — it is a consistent accounting of the same physics: where the air is deflected and moves fastest (typically just above the top surface), the pressure drops, and the integrated pressure difference across the wing matches the upward momentum imparted to the air. Lift = downward momentum flux of air = integrated pressure difference across the wing. Both descriptions are correct, and they are equivalent.

Bernoulli is not wrong for wings. What is wrong is the equal-time argument used to invoke it. The pressure is genuinely lower above the wing — but because the flow curves downward around the wing's upper surface, not because the path is longer.

A quick sanity check. Aerobatic pilots fly upside-down all the time. By the naive "curved top" story they would fall out of the sky. They don't, because the wing's angle of attack — the tilt of the wing relative to the oncoming air — is the dominant factor in lift generation, not the aerofoil's top-to-bottom curvature.

ISRO rocket fairings — Bernoulli in launch aerodynamics

When a rocket like the PSLV lifts off from Sriharikota and accelerates past the speed of sound, the fairing (the nose cone covering the payload) experiences peak aerodynamic pressure — called max-Q — at around 11–13 km altitude, where the product \tfrac{1}{2}\rho v^2 peaks (air density falling fast with altitude, speed rising even faster). Designing the fairing to survive max-Q is one of the central aerodynamic problems for ISRO, and that \tfrac{1}{2}\rho v^2 you recognise is exactly the dynamic-pressure term from Bernoulli's equation.

Worked examples

Example 1: Venturi meter on a CNG pipeline

A compressed natural gas pipeline in a Mumbai industrial estate has inner diameter 8 cm and a venturi meter installed with a throat diameter of 4 cm. A differential pressure transducer reads P_1 - P_2 = 500 Pa. The CNG density at the pipeline conditions is \rho = 0.9 kg/m³. Find the flow speed in the main pipe and the volume flow rate. Treat CNG as incompressible (reasonable for low-speed pipeline flow).

Venturi meter on a CNG pipeline Horizontal pipe narrowing from 8 cm to 4 cm diameter at the throat. Pressure taps labelled P1 on the wide section and P2 at the throat. Arrow showing CNG flow direction. $v_1$ 8 cm $v_2$ 4 cm $P_1$ $P_2$
CNG flows left to right. The 8 cm pipe narrows to a 4 cm throat at the venturi. Pressure taps $P_1$ and $P_2$ feed a differential transducer.

Step 1. Compute the cross-sectional areas.

Main: A_1 = \pi (0.04)^2 = 5.03 \times 10^{-3}\text{ m}^2. Throat: A_2 = \pi (0.02)^2 = 1.26 \times 10^{-3}\text{ m}^2. Area ratio: A_1/A_2 = (0.04/0.02)^2 = 4.

Why: for circular pipes, A \propto d^2, so halving the diameter quarters the area. The ratio A_1/A_2 = 4 is what enters the venturi formula.

Step 2. Apply the venturi formula.

v_1 = \sqrt{\frac{2(P_1 - P_2)}{\rho\left[(A_1/A_2)^2 - 1\right]}} = \sqrt{\frac{2 \times 500}{0.9 \times (16 - 1)}} = \sqrt{\frac{1000}{13.5}} = \sqrt{74.1}\text{ m/s}.
v_1 \approx 8.6\text{ m/s}.

Why: plug the area ratio into (A_1/A_2)^2 - 1 = 15 and compute. The answer is a fast flow — CNG pipelines do move at roughly that speed.

Step 3. Compute the volume flow rate.

Q = A_1 v_1 = 5.03 \times 10^{-3} \times 8.6 = 4.33 \times 10^{-2}\text{ m}^3/\text{s}.

Converting to litres per minute: Q = 43.3\text{ L/s} = 2600\text{ L/min}.

Step 4. Sanity check — the throat speed.

v_2 = v_1 \cdot A_1/A_2 = 8.6 \times 4 = 34.4\text{ m/s}.

Why: by continuity, the narrower throat has four times the inlet speed. The incompressibility assumption is reasonable if v_2 stays well below the speed of sound (\approx 430 m/s for CNG), which it does.

Result: v_1 \approx 8.6 m/s, Q \approx 43 L/s, v_2 \approx 34 m/s.

What this shows: A small pressure difference (500 Pa — about 2 mm of water column) is enough to meter an industrial-scale gas flow, provided the area ratio is large enough. The venturi is a high-gain instrument: tiny signal in, accurate flow-rate out.

Example 2: Torricelli — draining a rainwater harvesting tank

A rooftop rainwater harvesting tank in a Chennai apartment block has a rectangular cross-section of 3 m × 2 m and is filled to a depth of 1.8 m. A tap at the bottom of one side wall (effectively at the tank floor) has a bore of 4 cm diameter. How fast does water initially exit the tap, and how long does it take the tank to empty completely?

Rainwater tank draining through a tap at its base A rectangular water tank with free surface at height 1.8 m. A tap near the base, shown with a short spout. An arrow indicates water exit speed. $v$ $h = 1.8$ m free surface (atm) tank floor — 3 m × 2 m
The tank has cross-section 3 m × 2 m. The tap at the base has a bore of 4 cm diameter. The initial water height is 1.8 m.

Step 1. Compute the initial exit speed.

v_0 = \sqrt{2 g h} = \sqrt{2 \times 9.8 \times 1.8} = \sqrt{35.28} \approx 5.94\text{ m/s}.

Why: Torricelli's theorem directly. The same speed a stone would have after falling 1.8 m from rest.

Step 2. Set up the draining differential equation.

The tank's cross-section is A_\text{tank} = 3 \times 2 = 6\text{ m}^2. The tap's bore area is A_\text{hole} = \pi (0.02)^2 = 1.26 \times 10^{-3}\text{ m}^2.

Let y(t) be the height of water at time t (starting at y(0) = 1.8 m). The exit speed when the water height is y is v(y) = \sqrt{2gy}. The volume flow rate leaving is Q = A_\text{hole} \sqrt{2gy}. The rate of change of water height is

A_\text{tank} \frac{dy}{dt} = -A_\text{hole} \sqrt{2gy}.

Why: the top of the water level drops at whatever speed matches the rate of volume leaving through the hole, divided by the tank's cross-sectional area. The minus sign is because y decreases.

Step 3. Separate variables and integrate.

\frac{dy}{\sqrt{y}} = -\frac{A_\text{hole}}{A_\text{tank}} \sqrt{2g}\, dt.

Integrating from y = 1.8 at t = 0 to y = 0 at t = T:

\int_{1.8}^{0} y^{-1/2}\, dy = -\frac{A_\text{hole}}{A_\text{tank}} \sqrt{2g} \cdot T.
\left[2 \sqrt{y}\right]_{1.8}^{0} = -2 \sqrt{1.8} = -\frac{A_\text{hole}}{A_\text{tank}} \sqrt{2g} \cdot T.

Solving for T:

T = \frac{2 \sqrt{1.8}}{\frac{A_\text{hole}}{A_\text{tank}} \sqrt{2g}} = \frac{A_\text{tank}}{A_\text{hole}} \sqrt{\frac{2 \cdot 1.8}{g}} = \frac{A_\text{tank}}{A_\text{hole}} \sqrt{\frac{2h_0}{g}}.

Why: the time to drain is the tank-to-hole area ratio multiplied by the free-fall time from the initial height. The larger the tank relative to the hole, the longer it takes — as physical sense demands.

Step 4. Plug in numbers.

T = \frac{6}{1.26 \times 10^{-3}} \times \sqrt{\frac{2 \times 1.8}{9.8}} = 4762 \times 0.606 \approx 2885\text{ s} \approx 48\text{ min}.

Result: Initial exit speed v_0 \approx 5.9 m/s. The tank takes about 48 minutes to drain through the 4 cm tap.

What this shows: Torricelli's result is not just a snapshot — combined with simple area accounting, it gives the complete draining curve for a tank. The \sqrt{y} dependence of speed means the tank drains fastest at the start and slows asymptotically as the level falls, which anyone who has watched a bathtub empty has observed.

Example 3: A cricket ball and the Magnus deflection

A spin bowler delivers a ball (m = 0.16 kg, r = 3.5 cm) at v = 25 m/s with a top-spin angular velocity \omega = 80 rad/s. Assume the Magnus coefficient is C \approx 0.5 (appropriate for a cricket ball — smoother than a tennis ball, rougher than a billiard ball). Air density \rho = 1.2 kg/m³. The pitch length from the bowler's release point to the bat is 18 m. Estimate the downward deflection of the ball's trajectory from top-spin (in addition to gravity's contribution).

Top-spinning cricket ball delivered down a pitch A cricket ball travels along a horizontal pitch, spinning forward (top-spin). A curved trajectory drops below the straight-line path, showing the combined effect of gravity and the downward Magnus force. straight-line (no gravity, no spin) actual trajectory top-spin $v = 25$ m/s pitch 18 m — bowler to bat
A top-spinning cricket ball. Gravity pulls it down; the Magnus force from the spin adds to that downward component (top-spin → downward Magnus force on a ball moving horizontally). The trajectory dips below the straight-line path.

Step 1. Estimate the Magnus force.

Using the order-of-magnitude expression from the body of the article:

F = \rho\, v\, \omega\, r^3\, C = 1.2 \times 25 \times 80 \times (0.035)^3 \times 0.5.
(0.035)^3 = 4.29 \times 10^{-5}.
F = 1.2 \times 25 \times 80 \times 4.29 \times 10^{-5} \times 0.5 = 5.14 \times 10^{-2}\text{ N}.

Why: cubic dependence on radius is what makes the effect strong at cricket-ball size but negligible for raindrops. The \omega r^3 combination has units of m³/s (spin times volume), which multiplied by \rho v gives force.

Step 2. Compute the corresponding lateral acceleration.

a_\text{Magnus} = \frac{F}{m} = \frac{0.0514}{0.16} \approx 0.32\text{ m/s}^2.

Why: Newton's second law, applied to the Magnus force alone. This acceleration adds vectorially to g = 9.8\text{ m/s}^2 pulling the ball downward.

Step 3. Time of flight down the pitch.

t = \frac{L}{v} = \frac{18}{25} = 0.72\text{ s}.

Why: treat the ball's horizontal speed as approximately constant; drag decelerates it slightly but the first-order estimate uses initial speed.

Step 4. Magnus-induced deflection over the pitch length.

\Delta y_\text{Magnus} = \tfrac{1}{2}\, a_\text{Magnus}\, t^2 = \tfrac{1}{2} \times 0.32 \times (0.72)^2 \approx 0.083\text{ m} = 8.3\text{ cm}.

For comparison, gravity alone produces \Delta y_\text{gravity} = \tfrac{1}{2} g t^2 = 0.5 \times 9.8 \times 0.5184 = 2.54\text{ m} — except the ball is not falling freely the whole time (it has horizontal momentum and the pitch is only 18 m long, plus spin bowlers pitch short of the batter, letting the ball bounce). The additional drop from Magnus — the 8.3 cm on top of a gravity trajectory — is what makes the ball dip faster than the batter expected.

Result: The top-spin adds about 8 cm of extra drop over the 18 m pitch length, on top of the expected drop from gravity alone.

What this shows: Eight centimetres is the difference between a ball the batter plays cleanly off the front foot and one that thuds into the pad, plumb LBW. The Magnus effect is not a curiosity; it is the working mechanism of every ball that dips or rises mysteriously in a spinner's delivery.

Common confusions

If you can work out exit speeds from venturi, pitot, and Torricelli formulae and can explain Magnus qualitatively, you have the working content. What follows is the cleaner derivations and the edge cases.

The coefficient of discharge — why Torricelli's prediction overshoots

The formula v = \sqrt{2gh} is an ideal-fluid result. In practice, the jet emerging from a real orifice has a smaller effective area than the hole (the streamlines contract as they approach the opening, forming what is called the vena contracta), and some energy is lost to viscosity at the hole's edges. Both effects reduce the actual discharge rate. The correction is bundled into a single empirical coefficient of discharge:

Q_\text{actual} = C_d \cdot A_\text{hole} \sqrt{2gh}.

For a sharp-edged circular hole, C_d \approx 0.61 (a combination of the vena contracta's area reduction — about 0.62 — and a small velocity-correction). A rounded-edge orifice has C_d \approx 0.98. The choice between these in engineering practice is between a cheap sharp hole (with a known, characterised loss) and a more expensive rounded one (with near-ideal performance).

The venturi's pressure recovery — why the meter is "cheap"

Downstream of the throat, the venturi widens back to the full pipe diameter. If the widening is gradual (typically a 7° diffuser angle), the fluid slows back down and the pressure recovers almost to its upstream value — the only loss is a small viscous dissipation. Compare an orifice plate (a thin plate with a hole, mounted across the pipe) which also measures flow from pressure drop but dissipates most of the pressure drop permanently. The venturi is the "efficient" flow meter because it wastes very little pumping energy — the pressure drop across the meter that is permanent can be as little as 10% of the drop measured at the throat.

Pitot tube compressibility correction

For high-speed flow (Mach numbers above 0.3), air cannot be treated as incompressible and equation (7) is modified. The isentropic compressible-flow version is

\frac{P_0}{P} = \left(1 + \frac{\gamma - 1}{2} M^2\right)^{\gamma/(\gamma-1)},

where M = v/c is the Mach number, c is the local speed of sound, and \gamma = c_p/c_v \approx 1.4 for air. For M \ll 1 this reduces to equation (7) plus small corrections; for M approaching 1, the correction is large. Commercial airliners, which cruise at M \approx 0.8, need this compressibility-corrected calibration.

Why Bernoulli alone is not enough for a viscous boundary layer

The derivations in this article assume a non-viscous fluid. Real air has a small but non-zero viscosity, and near a solid surface (the wall of a pipe, the skin of an aeroplane, the surface of a cricket ball) a thin boundary layer forms where viscosity dominates and Bernoulli's equation fails. Outside the boundary layer — the bulk flow — Bernoulli is excellent; inside, you need the Navier–Stokes equations. The boundary-layer picture is critical for understanding cricket-ball swing (the seam trips one boundary layer from laminar to turbulent, changing its separation behaviour), aeroplane stall (the boundary layer separates abruptly at high angle of attack), and the Reynolds-number transition between laminar and turbulent pipe flow. The next article, Viscosity and Stokes' Law, opens up the viscous story.

The Kutta-Joukowski lift theorem — the honest aerofoil story

For a two-dimensional aerofoil in steady flow, the lift per unit span is

L' = \rho\, v\, \Gamma,

where \Gamma is the circulation around the aerofoil — the line integral of the velocity around any closed curve enclosing the wing. The Magnus effect on a spinning cylinder is the same formula applied to the cylinder's circulation (driven by the ball dragging air with its spin). For a wing, \Gamma is set by the Kutta condition — the requirement that the flow leaves smoothly from the sharp trailing edge. The result is the lift coefficient C_L that engineers use in design. This is the honest "why does a wing lift" story, and it involves circulation, not path length.

A subtle historical note

Bernoulli's equation was derived by Daniel Bernoulli in 1738, using energy arguments on a piston-driven fluid element. Torricelli's theorem predates Bernoulli by nearly a century — Evangelista Torricelli stated it in 1643 as an empirical observation, before the machinery of energy conservation existed. That Bernoulli's equation reproduces Torricelli's theorem in a few lines is a testament to how powerful the energy principle is once you have it.

In the Indian tradition, the Sulba Sutras and later Aryabhatan astronomy used water clocks (ghatika-yantra) whose draining rate was calibrated empirically by the principle that halving the head height approximately reduces the drain rate by \sqrt{2} — Torricelli's theorem in all but name. The modern quantitative treatment belongs to the 17th and 18th centuries, but the physical fact was operational in Indian timekeeping long before.

Where this leads next