In short

For an ideal fluid (incompressible, non-viscous, irrotational, in steady flow), the sum

P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{constant along a streamline}

holds at every point on a given streamline. This is Bernoulli's equation. Reading the three terms: P is the pressure (energy per unit volume stored in the fluid), \tfrac{1}{2}\rho v^2 is the kinetic energy per unit volume, and \rho g h is the gravitational potential energy per unit volume. The headline consequence is where the flow is fast, the pressure is low. The derivation is a one-page work–energy argument applied to a small parcel of fluid that enters a flow tube slow and wide at one end and leaves fast and narrow at the other. Bernoulli's equation plus the equation of continuity (A_1 v_1 = A_2 v_2) solve almost every ideal-fluid problem you will meet at school: Venturi meters, atomisers, the lift on a plane wing, water flowing out of a tank (Torricelli), and the shape of a jet from a tap.

Stand on the platform of a Mumbai local and watch a fast train rush past at thirty metres per second. Before the train arrives, the air between you and the track is still. The moment the train is next to you, the air in the narrow gap between your body and the train's side is moving almost as fast as the train itself. The pressure in that fast-moving air is lower than the pressure behind you, and that pressure difference pushes you forward — toward the train. The yellow safety line on the platform is painted at the distance where that pull becomes dangerous. Every Indian Railways safety sign warning you to stand well back from the edge is, in effect, a warning about Bernoulli's equation.

This is what Bernoulli's principle says, in one sentence: a fluid moving fast has low pressure; a fluid moving slowly has high pressure. Put a little more carefully, the pressure, the kinetic energy, and the gravitational potential energy of an ideal fluid along a streamline trade off in a single conserved combination. The conservation law is an old friend in new clothes — it is the work-energy theorem, applied to a parcel of fluid instead of a block of wood on a frictionless slope.

The idea is remarkably general. The lift on an aeroplane wing, the suction of a household perfume atomiser, the accelerating spray from a garden-hose nozzle, the swing of a cricket ball as the seam cuts the air, the rise of water in a chimney when wind blows across the top, the way a hurricane lifts a tin roof — all of these are Bernoulli in different costumes. By the end of this article, you will derive the equation from scratch, read what each term means, explore the Venturi effect interactively, and know exactly which situations it can and cannot be trusted on.

Setting up the problem — one flow tube, two sections

Bernoulli's equation is a statement about what happens to a parcel of fluid as it moves along a streamline. To derive it, pick a flow tube — a bundle of streamlines — in a steady, incompressible, non-viscous, irrotational flow. (These are the same ideal-fluid assumptions you met in the equation of continuity. They are worth listing explicitly because Bernoulli fails, sometimes spectacularly, when any of them is violated.)

Consider two cross-sections of the flow tube:

In a short time \Delta t, the fluid advances through the tube. A thin slab of fluid that was just behind section 1 moves forward to just beyond section 1, and a thin slab that was just behind section 2 moves forward to just beyond section 2. Everything in between shuffles along by the same amount — same as the way a queue shifts when the person at the front of the line steps forward.

A flow tube with two cross-sections at different heights and areas A curved flow tube slopes upward from left to right. The left entrance is wide (area A1) and at lower height h1; fluid enters with speed v1 under pressure P1. The right exit is narrow (area A2) and at greater height h2; fluid leaves with speed v2 under pressure P2. Thin slabs of fluid of length v1 dt and v2 dt are shaded at each end to show the work done on and by the external fluid during the time interval dt. $A_1$ $A_2$ $v_1,\ P_1$ $v_2,\ P_2$ reference $h=0$ $h_1$ $h_2$ slab enters: length $v_1\Delta t$ slab leaves: length $v_2\Delta t$
A flow tube slopes from a wide, low entrance (section 1) up to a narrow, elevated exit (section 2). In a short time $\Delta t$, a slab of fluid of length $v_1 \Delta t$ enters at the left while a slab of length $v_2 \Delta t$ leaves at the right (red shading). The mass of the two slabs is equal by continuity.

Two quantities are crucial.

First, in time \Delta t, the volume of fluid that enters at section 1 is \Delta V_1 = A_1 v_1 \Delta t, and the volume leaving at section 2 is \Delta V_2 = A_2 v_2 \Delta t. Because the fluid is incompressible,

\Delta V_1 = \Delta V_2 \equiv \Delta V,

and the mass of fluid that enters equals the mass that leaves,

\Delta m = \rho \Delta V.

This is just the equation of continuity (A_1 v_1 = A_2 v_2) in another form. Why: mass is not created or destroyed in a steady, incompressible flow, so whatever mass enters at one end leaves at the other in the same time.

Second, because the flow is steady, the fluid between section 1 and section 2 has exactly the same energy configuration before and after the time interval \Delta t. The kinetic and potential energy stored in the middle region does not change. So the net change in the fluid's energy over the interval is the change that happens as the slab at section 1 gets "absorbed" into the flow and the slab at section 2 gets "emitted" — as if the middle were a black box, untouched. This is the crucial insight that makes the derivation tractable.

Deriving Bernoulli's equation

Apply the work-energy theorem to the mass \Delta m as it moves from being at section 1 (with speed v_1, height h_1) to being at section 2 (with speed v_2, height h_2). The theorem, you will remember from work and kinetic energy, says the total work done on a body equals its change in kinetic energy:

W_\text{net} = \Delta \text{KE}.

Three works act on the slab.

Work 1 — the fluid behind pushes the slab into section 1. The fluid just behind section 1 pushes forward on the entering slab with pressure P_1, over an area A_1, through a distance v_1 \Delta t.

W_1 = (P_1 A_1)(v_1 \Delta t) = P_1 \cdot (A_1 v_1 \Delta t) = P_1 \Delta V.

Why: work is force times distance. The force is pressure times area, F = P_1 A_1. The distance the slab travels is v_1 \Delta t. The product telescopes into pressure times volume — a useful pattern that recurs in all fluid energy arguments.

Work 2 — the slab at section 2 pushes the fluid ahead of it. The exiting slab at section 2 is doing work on the fluid beyond, so the work done on our parcel by that external fluid is negative:

W_2 = -(P_2 A_2)(v_2 \Delta t) = -P_2 \Delta V.

Why: at section 2, the pressure P_2 pushes backward on our slab (in the opposite direction to the flow), because the fluid in front is resisting being pushed out of the way. So this is a negative contribution to the work done on our parcel.

Work 3 — gravity does work as the slab rises. Gravity acts on \Delta m = \rho \Delta V over a height change of h_2 - h_1. Since the slab rises (if h_2 > h_1), gravity does negative work:

W_3 = -\Delta m \cdot g \cdot (h_2 - h_1) = -\rho \Delta V \cdot g (h_2 - h_1).

Why: the work done by gravity on a body that rises by \Delta h is -mg\Delta h — the minus sign because gravity points down while the displacement is up. Write \Delta m = \rho \Delta V so the volume factors appear in the same place as in W_1 and W_2.

The net work is

W_\text{net} = W_1 + W_2 + W_3 = P_1 \Delta V - P_2 \Delta V - \rho \Delta V \, g (h_2 - h_1). \tag{1}

Now the right-hand side of the theorem — the change in kinetic energy. Because the middle of the tube is undisturbed, the net change is just "slab at section 2 has KE of \tfrac{1}{2}\Delta m\, v_2^2" minus "slab at section 1 had KE of \tfrac{1}{2}\Delta m\, v_1^2":

\Delta \text{KE} = \tfrac{1}{2}\Delta m\, v_2^2 - \tfrac{1}{2}\Delta m\, v_1^2 = \tfrac{1}{2}\rho \Delta V\, (v_2^2 - v_1^2). \tag{2}

Why: the steady-state argument collapses the full energy change of the whole tube into just the difference between the entering and exiting slab. Everything in the middle is identical before and after \Delta t, so it cancels out.

Equate (1) and (2):

P_1 \Delta V - P_2 \Delta V - \rho \Delta V\, g(h_2 - h_1) = \tfrac{1}{2}\rho \Delta V (v_2^2 - v_1^2).

Divide by \Delta V — every term has it, and it is non-zero:

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

Collect section-1 quantities on the left and section-2 quantities on the right:

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

Why: rearrange so each side depends only on quantities at one cross-section. Because sections 1 and 2 were arbitrary points along the same streamline, the combination on each side has the same value at every point of the streamline — it is a conserved quantity.

Since sections 1 and 2 were chosen arbitrarily, the same combination must hold at every cross-section. Write the conclusion compactly:

\boxed{\; P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{constant along a streamline} \;} \tag{Bernoulli}

This is Bernoulli's equation. It is the work-energy theorem for an ideal fluid, expressed as a conservation law.

Bernoulli's equation

For an incompressible, non-viscous fluid in steady, irrotational flow, the sum of pressure, kinetic energy density, and gravitational potential energy density is constant along any streamline:

P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{constant}.

Equivalently, for two points 1 and 2 on the same streamline,

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.

Reading the three terms

The beauty of Bernoulli's equation is that every term has units of energy per unit volume (pascal, Pa = J/m³). Each term is an energy density — a storage form the fluid can hold, and the law says the total across the three forms is conserved along a streamline.

Check the units: in SI, P is in Pa = N/m² = kg/(m·s²) = J/m³. The kinetic term has units (kg/m³)(m/s)² = kg/(m·s²) = J/m³. The potential term has (kg/m³)(m/s²)(m) = kg/(m·s²) = J/m³. All three are energy densities. This is dimensional consistency at its most honest — not a check bolted on after the derivation, but the statement that the derivation is an energy bookkeeping.

The headline consequence: fast flow, low pressure

Suppose the flow is horizontal (so h_1 = h_2 and the gravitational term drops out):

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

Rearrange:

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

If v_2 > v_1 (the flow speeds up from section 1 to section 2), then the right side is positive, so P_1 > P_2 — the pressure drops where the flow is fast. This is Bernoulli's principle in its cleanest form. The fluid has a fixed total of pressure plus kinetic energy; if kinetic energy goes up, pressure has to come down to compensate.

A Venturi tube with pressure tappings A horizontal pipe with a smooth narrow section in the middle. Two vertical glass tubes (manometers) rise from the wide sections, with water level near the top. A third manometer rises from the narrow section but its water level is noticeably lower, visualising the lower pressure where the flow is faster. An arrow at the inlet labels the flow direction. $P_1$ $P_2$ $P_3$ $v_1$ $v_2 > v_1$ $v_3 = v_1$ throat
A Venturi tube — a horizontal pipe with a constricted throat. By continuity, the flow accelerates ($v_2 > v_1$) through the throat. By Bernoulli, the pressure drops ($P_2 < P_1$), which you can see in the lowered water level of the middle manometer. Downstream the pipe widens again and the pressure recovers: $P_3 = P_1$, $v_3 = v_1$.

This is the Venturi effect — the single experimental demonstration that Bernoulli predicts counter-intuitively and correctly. Water flows through a horizontal pipe with a narrow throat. Continuity (from the last chapter) tells you the water speeds up at the throat. Bernoulli tells you the pressure drops at the throat. The manometer above the throat reads lower than the ones above the wider sections — visible proof, in a working Venturi meter in any school lab, that the equation is correct.

Why this feels strange at first

Intuition from static fluids says "where the pipe is narrow, the water is more squeezed, so pressure should be higher." This is wrong, and the wrongness is instructive. In a static fluid, nothing is moving — pressure is set by depth, not by geometry. In a moving fluid, the same parcel of fluid has to accelerate to get through the narrow section. What accelerates it? A net forward force on the parcel. What provides that force? A pressure difference — high pressure behind the parcel, low pressure in front of it. So the narrow region is lower pressure than the wide region, because the fluid has to be pushed into it by the wider region behind. The pressure drop is what causes the acceleration.

Explore the Venturi relationship

The interactive figure below shows the throat pressure drop as a function of the area ratio A_1/A_2 for water flowing at v_1 = 1 m/s. Drag the red dot to change the area ratio and watch the pressure drop grow.

Interactive: Venturi pressure drop versus area ratio A rising curve showing the pressure drop from the wide section to the throat as a function of the area ratio A1/A2 for water of density 1000 kg per cubic metre flowing at 1 m/s in the wide section. At area ratio 1 the drop is zero. At area ratio 6 the drop is 17500 pascals. area ratio $A_1/A_2$ pressure drop $P_1 - P_2$ (Pa) 0 5 000 10 000 15 000 2 3 4 5 6 drag the red point along the axis
The pressure drop $P_1 - P_2 = \tfrac{1}{2}\rho\,(v_2^2 - v_1^2) = 500\,(r^2 - 1)$ Pa (for $\rho = 1000$ kg/m³, $v_1 = 1$ m/s, area ratio $r = A_1/A_2$). The drop grows *quadratically* with the area ratio because the throat speed grows linearly with $r$ and the kinetic term goes as speed squared. At a 6:1 constriction, the pressure at the throat is 17.5 kPa lower than upstream — enough to hold up a 1.8-metre column of water.

Animated Venturi — watch the pressure trade off with speed

The animation below shows a fluid parcel (red) flowing through a Venturi at constant volume flow rate. The parcel's speed increases through the throat (you can see it move faster there) and decreases back to the inlet speed downstream. A live bar alongside tracks pressure: low in the throat, high at the wide sections.

Animated: a fluid parcel through a Venturi A horizontal pipe with a narrow throat between x=4 and x=7 metres. A red dot represents a fluid parcel flowing rightward. In the wide sections (0-4 and 7-12) it moves at 1 m/s. In the throat (4-7) it moves at 4 m/s (because the area ratio is 4). The dot's trail shows the varying speed. wide: $A_1$ throat: $A_2 = A_1/4$ wide: $A_3 = A_1$
Watch the parcel (red) speed up as it enters the throat and slow down as it exits. The parcel covers equal volumes of fluid in equal times — that is continuity. Its extra speed through the throat comes from the pressure drop predicted by Bernoulli. Click replay to watch again.

Special cases worth memorising

Horizontal flow — pressure trades with speed

For a strictly horizontal flow (h constant), Bernoulli reduces to

P + \tfrac{1}{2}\rho v^2 = \text{constant}.

This is the cleanest form, and it governs every horizontal-pipe problem — Venturi meters, aerofoils, atomisers, fast-train suction.

Still fluid — hydrostatic pressure

If the fluid is at rest everywhere (v = 0), Bernoulli reduces to

P + \rho g h = \text{constant},

which is P_1 + \rho g h_1 = P_2 + \rho g h_2, or equivalently P_1 - P_2 = \rho g (h_2 - h_1). This is exactly the hydrostatic equation you already met in pressure in fluids. Bernoulli includes hydrostatics as a special case.

Torricelli's law — speed of efflux from an open tank

Open a small hole of area a at depth H below the surface of a large open tank of water. The water at the surface (section 1, where the area is essentially infinite compared to a) moves so slowly that v_1 \approx 0. Take h_1 = H, h_2 = 0. Both the surface and the hole are exposed to atmospheric pressure, so P_1 = P_2 = P_\text{atm}.

Bernoulli gives

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

Cancel P_\text{atm} and solve:

v_2 = \sqrt{2gH}.

Why: the pressure terms cancel (atmosphere acts identically at both ends), and the surface's negligible kinetic energy plus its gravitational PE gets converted into kinetic energy at the hole. The result is exactly the speed a freely falling body would reach after falling through height H under gravity — because the fluid mechanics is, at bottom, the same conservation of energy.

This is Torricelli's law: water leaves a hole in a tank at the same speed it would have if it had fallen freely from the water surface to the hole. Drill a hole 1.25 m below the water level in a water tank on your terrace; the jet emerges at v = \sqrt{2 \times 9.8 \times 1.25} \approx 4.95 m/s — about 18 km/h, fast enough to shoot a visible arc across the roof.

Worked examples

Example 1: Water flowing through a Venturi meter in an Indian municipal pipeline

A Municipal Corporation pipeline carries water through a horizontal Venturi meter used to measure flow rate. The wider section has an internal diameter of 20 cm and the throat has a diameter of 10 cm. The pressure in the wider section is measured at P_1 = 250\,\text{kPa} (gauge) and in the throat at P_2 = 232\,\text{kPa} (gauge). Find the flow speed in the wider section and the volume flow rate. Density of water \rho = 1000 \text{ kg/m}^3.

Horizontal Venturi meter with measured pressures A horizontal pipe shown in cross-section view with a wider section on the left of diameter 20 cm and a narrower throat section in the middle of diameter 10 cm, then widening again. Gauges at the wide section read 250 kPa and at the throat read 232 kPa. An arrow points right indicating flow direction. $P_1 = 250$ kPa $d_1 = 20$ cm $P_2 = 232$ kPa $d_2 = 10$ cm flow
A horizontal Venturi meter. The pressure difference between the wide section and the throat, combined with the continuity equation, tells you the flow speed.

Step 1. Compute the areas.

A_1 = \pi r_1^2 = \pi (0.10)^2 = 3.14 \times 10^{-2} \text{ m}^2.
A_2 = \pi r_2^2 = \pi (0.05)^2 = 7.85 \times 10^{-3} \text{ m}^2.

Why: areas of circular cross-sections, radius in metres. The ratio A_1/A_2 = 4 — halving the diameter quarters the area.

Step 2. Use continuity to express v_2 in terms of v_1.

A_1 v_1 = A_2 v_2 \quad \Longrightarrow \quad v_2 = \frac{A_1}{A_2} v_1 = 4 v_1.

Why: the equation of continuity gives the speed ratio from the area ratio. Now Bernoulli has only one unknown, v_1.

Step 3. Apply Bernoulli's equation (horizontal, so no \rho g h term).

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

Substitute v_2 = 4 v_1:

P_1 + \tfrac{1}{2}\rho v_1^2 = P_2 + \tfrac{1}{2}\rho (4 v_1)^2 = P_2 + 8 \rho v_1^2.
P_1 - P_2 = 8\rho v_1^2 - \tfrac{1}{2}\rho v_1^2 = \tfrac{15}{2}\rho v_1^2.

Why: 16 v_1^2 - v_1^2 = 15 v_1^2, and the factor of \tfrac{1}{2} gives \tfrac{15}{2}. Now solve for v_1 in terms of the known pressure difference.

Step 4. Solve for v_1.

v_1 = \sqrt{\frac{2(P_1 - P_2)}{15 \rho}} = \sqrt{\frac{2 \times 18{,}000}{15 \times 1000}} = \sqrt{2.4} \approx 1.55 \text{ m/s}.

Why: pressure difference P_1 - P_2 = 250 - 232 = 18 kPa = 18{,}000 Pa. The 15 comes from Step 3.

Step 5. Volume flow rate.

Q = A_1 v_1 = 3.14 \times 10^{-2} \times 1.55 \approx 4.87 \times 10^{-2} \text{ m}^3/\text{s} \approx 48.7 \text{ L/s}.

Why: Q = A v from continuity, applied at the wide section where you know both the area and the speed.

Result: The water in the wide section flows at about 1.55 m/s; the volume flow rate is 48.7 L/s, or nearly 2.9 cubic metres per minute — enough to fill a standard 1000-litre rooftop Sintex tank in about 20 seconds.

What this shows: A Venturi meter turns a pressure difference (easy to measure) into a flow rate (hard to measure directly). Every municipal water meter, every industrial flowmeter at a Reliance refinery, every fuel meter at an Indian Oil petrol pump uses some form of this idea.

Example 2: Torricelli — a leaking overhead tank in Delhi

Your Sintex overhead tank stands on the roof of a Delhi house. The water surface is H = 3.5 m above a small hole of area a = 1.0 \text{ cm}^2 near the bottom of the tank. Assuming the tank is much wider than the hole (so the surface is essentially stationary), find the speed of water emerging from the hole, the volume flow rate through the hole, and the time taken to drain 10 litres. Ignore viscosity and the slow drop of the water level during the 10 litres.

A water tank with a hole near the bottom A rectangular tank shown in cross-section. The water surface is near the top labelled v1 approximately 0. A distance H equals 3.5 m is marked from the surface to a small hole on the right side near the bottom. An arrow from the hole to the right labels v2 as the unknown exit speed. The pressure at both the surface and the hole is atmospheric. surface, $v_1 \approx 0$, $P_1 = P_\text{atm}$ $H = 3.5$ m $v_2 = ?$ hole area $a = 1.0$ cm²
A Sintex tank with a small leak. The fluid surface is essentially stationary because the tank is much wider than the hole; the exit speed is set by Torricelli's law.

Step 1. Write Bernoulli between the surface (1) and the hole (2).

Both points are in contact with the atmosphere, so P_1 = P_2 = P_\text{atm}. The surface is essentially stationary, so v_1 \approx 0. Take the hole as the reference height (h_2 = 0) and the surface at h_1 = H.

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

Why: apply Bernoulli between two points on one streamline — the streamline that starts at a point on the surface and ends at the hole. Cancelling atmospheric pressure and the zero-speed/zero-height terms isolates the one unknown, v_2.

Step 2. Solve for v_2.

\tfrac{1}{2} v_2^2 = gH \quad \Longrightarrow \quad v_2 = \sqrt{2gH} = \sqrt{2 \times 9.8 \times 3.5}.
v_2 = \sqrt{68.6} \approx 8.28 \text{ m/s}.

Why: this is Torricelli's law — the exit speed equals the speed a body would have after falling freely through height H. Notice that the hole's area a does not enter: the speed is set purely by the head H.

Step 3. Compute the volume flow rate through the hole.

Q = a \cdot v_2 = (1.0 \times 10^{-4} \text{ m}^2)(8.28 \text{ m/s}) = 8.28 \times 10^{-4} \text{ m}^3/\text{s}.

Converting: Q = 0.828 L/s.

Step 4. Time to drain 10 L.

t = \frac{10 \text{ L}}{0.828 \text{ L/s}} \approx 12.1 \text{ s}.

Why: volume divided by rate, as long as the head H stays roughly constant (which it will for a small hole and a short time). If you wanted a full tank emptying, you would need to integrate v_2 = \sqrt{2gH(t)} as the level drops.

Result: Water exits at about 8.3 m/s. The hole passes 0.83 L/s, and 10 litres drain in 12 seconds.

What this shows: Torricelli's law gives the free-fall speed at the bottom of any open tank — the basis of every reservoir spillway, every drip irrigation nozzle, and every ISRO liquid fuel rocket where the propellant flows out of a pressurised tank into the combustion chamber.

Example 3: A perfume atomiser — why the air stream lifts the scent

A simple perfume atomiser has a horizontal tube along which air is blown at v_\text{air} = 15 \text{ m/s} from a rubber bulb. A narrow vertical tube dips into the perfume reservoir and opens into the horizontal air stream. The perfume is at atmospheric pressure (P_\text{atm} = 1.013 \times 10^5 Pa). Density of air \rho_\text{air} = 1.2 \text{ kg/m}^3, density of perfume \rho_\text{perf} = 900 \text{ kg/m}^3. Find the pressure at the top of the vertical tube (where the air rushes past), and the maximum height to which perfume can be lifted up the vertical tube by this pressure difference.

Atomiser: a horizontal air stream above a vertical tube dipping into a perfume reservoir A horizontal pipe at the top with an arrow showing air blown rightward at 15 m/s. A narrow vertical tube descends from the middle of the horizontal pipe into a wider container of perfume below. Drops of perfume are shown being pulled up the tube and flung out the pipe. air at 15 m/s perfume reservoir droplets rise
A perfume atomiser. Fast-moving air above the vertical tube creates a low-pressure region that sucks perfume up the tube, where the air stream then atomises it into a fine spray.

Step 1. Bernoulli across the top of the vertical tube.

Far upstream of the atomiser, the air is at rest (v_0 \approx 0) and at atmospheric pressure P_\text{atm}. Above the vertical tube, the air rushes past at v_\text{air} = 15 m/s. The flow is horizontal (same height), so

P_\text{atm} + 0 = P_\text{top} + \tfrac{1}{2}\rho_\text{air} v_\text{air}^2.

Solve for P_\text{top}:

P_\text{top} = P_\text{atm} - \tfrac{1}{2}\rho_\text{air} v_\text{air}^2 = 1.013 \times 10^5 - \tfrac{1}{2}(1.2)(15)^2.
P_\text{top} = 1.013 \times 10^5 - 135 \approx 1.0116 \times 10^5 \text{ Pa}.

Why: the air streaming past the top of the tube has \tfrac{1}{2}\rho_\text{air} v_\text{air}^2 of kinetic energy per unit volume, so its pressure is lower than stagnant atmospheric pressure by exactly that amount. The pressure drop is 135 Pa — tiny compared to atmospheric, but plenty to move a thin column of liquid.

Step 2. Perfume column that balances the pressure difference.

At the perfume surface in the reservoir, the pressure is P_\text{atm}. At the top of the vertical tube, the pressure is P_\text{top}, which is 135 Pa lower. The extra atmospheric pressure on the reservoir surface pushes perfume up the tube until the weight of the raised perfume column balances the deficit:

\rho_\text{perf} \, g\, h = P_\text{atm} - P_\text{top} = \tfrac{1}{2}\rho_\text{air} v_\text{air}^2.

Solve for h:

h = \frac{\rho_\text{air} v_\text{air}^2}{2 \rho_\text{perf} g} = \frac{1.2 \times 225}{2 \times 900 \times 9.8} = \frac{270}{17{,}640} \approx 1.53 \times 10^{-2} \text{ m}.

Why: hydrostatic balance inside the vertical tube — the extra atmospheric push on the reservoir lifts the liquid by a height whose weight-per-area equals the pressure deficit. Notice the elegant way the perfume's density appears in the denominator: denser liquids rise less under the same suction.

Result: The pressure at the top of the vertical tube drops to about 101,165 Pa (135 Pa below atmospheric). Perfume rises roughly h \approx 1.5 cm up the tube — which is more than enough, since the tube is only a few centimetres tall.

What this shows: The same principle powers a car's carburettor (petrol lifted into an air stream by Bernoulli suction), a paint spray gun, a Bunsen burner's air inlet, and the humble pichkari water gun on Holi. Fast-moving fluid at one port, a slower-moving fluid at another port, a connecting channel — and the pressure difference does the work.

Common confusions

Where Bernoulli fails

Bernoulli's equation is an ideal-fluid result. It fails, quantifiably, whenever its four assumptions break:

  1. Viscosity matters — long pipes, thick fluids, low flow speeds. Replace with Poiseuille's law or the engineering Darcy–Weisbach equation.

  2. Turbulence — high Reynolds number flows (swirling smoke, rapids, breaking waves). Streamlines stop existing in the ordinary sense.

  3. Compressibility — gas flows near or above the speed of sound. Use the compressible-flow form where kinetic energy is traded against internal energy as well.

  4. Unsteady flow — water hammer, surges, transient flows. Extra terms appear in the equation from the time-dependent velocity field.

Knowing when to trust Bernoulli is as important as knowing the equation itself.

If you have the equation, can read the three terms, and can solve Venturi, Torricelli, and atomiser problems, you have the working content of this article. What follows is the alternative derivation using Euler's equation, the general form of Bernoulli for rotating and unsteady flows, and the link to the Magnus effect that explains cricket-ball swing.

Derivation from Euler's equation

The work-energy derivation you saw above uses integral quantities (slabs of volume \Delta V) and is the shortest route to the result. There is a differential derivation that shows Bernoulli's equation as a direct consequence of Newton's second law applied pointwise to a fluid element.

Take a small fluid element of density \rho moving with velocity \vec{v}(\vec{r}, t) under the influence of gravity and a pressure gradient. Newton's second law per unit volume reads

\rho \frac{D\vec{v}}{Dt} = -\nabla P - \rho g \hat{z},

where D/Dt = \partial/\partial t + \vec{v}\cdot\nabla is the material derivative — the rate of change following a fluid element. This is Euler's equation for inviscid flow.

For steady flow (\partial \vec{v}/\partial t = 0), and using the vector identity (\vec{v}\cdot\nabla)\vec{v} = \nabla(\tfrac{1}{2}v^2) - \vec{v}\times(\nabla\times\vec{v}), Euler's equation becomes

\rho \nabla(\tfrac{1}{2}v^2) - \rho \vec{v}\times(\nabla\times\vec{v}) = -\nabla P - \rho g \nabla z.

Project onto the flow direction \hat{t} (tangent to a streamline). The term \vec{v}\times(\nabla\times\vec{v}) is perpendicular to \vec{v}, so its projection onto \hat{t} vanishes. What remains is

\hat{t}\cdot \nabla\left(P + \tfrac{1}{2}\rho v^2 + \rho g z\right) = 0.

That is: along every streamline, the gradient of P + \tfrac{1}{2}\rho v^2 + \rho g z is zero, so the sum is constant along a streamline. This is Bernoulli's equation, rederived from Euler. The strength of this derivation is that it shows why the "along a streamline" qualification appears — it comes from projecting onto the flow direction.

For irrotational flow (\nabla\times\vec{v} = 0), the \vec{v}\times(\nabla\times\vec{v}) term vanishes everywhere, not just along the flow. So the sum is constant throughout the fluid — one Bernoulli constant for the whole flow field. This is why irrotational is an extra assumption sometimes listed with Bernoulli: without it, the constant can differ between streamlines.

Unsteady Bernoulli

If the flow is time-dependent, Euler's equation picks up an extra term and Bernoulli becomes

\int \frac{\partial \vec{v}}{\partial t} \cdot d\vec{\ell} + \frac{P}{\rho} + \tfrac{1}{2}v^2 + g z = \text{constant along a streamline}.

The integral is along the streamline and accounts for the acceleration of the fluid at each point. This form is important for water hammer in pipes (when you suddenly close a tap, the water's momentum creates a pressure spike that can burst pipes), for oscillating flows in pulsatile blood pumps, and for wave propagation in liquids.

Compressible Bernoulli

For a compressible fluid (typical in gas dynamics), the pressure–volume work is no longer P\,\Delta V but \int P\,dV along the process. If the gas is also adiabatic (no heat exchange with surroundings), the result is

\frac{\gamma}{\gamma - 1}\frac{P}{\rho} + \tfrac{1}{2}v^2 + g z = \text{constant},

where \gamma = C_p/C_v is the ratio of specific heats. This is the compressible-flow Bernoulli equation, used in the design of rocket nozzles, jet engines, and wind-tunnel diffusers.

An interesting consequence: for an adiabatic air flow, the "stagnation temperature" (temperature the air would have if brought to rest) is always higher than the moving-air temperature, by exactly \tfrac{1}{2}v^2/C_p. A supersonic aircraft's skin can heat up to hundreds of degrees Celsius just from the air being compressed at the leading edge — a direct consequence of compressible Bernoulli.

The Magnus effect — why a cricket ball swings

When a spinning ball moves through air, the air on one side is sped up (the side where the spin direction matches the air-flow direction relative to the ball) and slowed on the other. By Bernoulli, the pressure is lower on the fast side and higher on the slow side. The net force is perpendicular to the ball's motion and points from the slow side toward the fast side. This is the Magnus effect.

For a cricket ball bowled with seam orientation and backspin (as in Jasprit Bumrah's outswinger), the seam trips turbulence on one side of the ball but leaves the other side in laminar flow. The turbulent side keeps the air attached longer around the ball (it does not separate as early), which creates more air deflection on that side. The pressure asymmetry makes the ball curve sideways in flight — the famous swing. Bernoulli's equation is the quantitative engine here: low pressure on the fast-air side produces a sideways force of the right order of magnitude to match the observed curvature (roughly 30 cm over a 20 m flight for a well-bowled outswinger).

Bernoulli and Torricelli's law as one idea

The speed of efflux from an open tank, v = \sqrt{2gH}, is exactly the speed a freely-falling body would reach after falling through height H. This is not a coincidence — Bernoulli's equation is conservation of energy, and the speed of efflux is the fluid-mechanics analogue of "PE converts to KE." Every drop of water at the surface of the tank has the potential energy it will have at the hole's level, and it emerges with a kinetic energy exactly equal to that drop in height. In this sense, Torricelli's law is older than Bernoulli's equation (Torricelli 1643; Bernoulli 1738), but it is really the first theorem of fluid dynamics expressed before the general equation existed to contain it.

An Indian footnote

Varahamihira's Bṛhat Saṃhitā (6th century CE) contains detailed observations of water's behaviour in fountains, pressurised pots, and siphons — including the observation that water flowing out of a narrow spout moves faster than water in the wide pot above. This is Bernoulli's principle stated qualitatively, over a thousand years before Daniel Bernoulli wrote it down mathematically. The quantitative derivation had to wait for Newton's laws, the work-energy theorem, and the idealisation of a fluid as a continuum.

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