Equation of Continuity

In short

For an ideal fluid — incompressible, non-viscous, irrotational, in steady flow — the volume crossing any cross-section of a tube per second is the same at every cross-section. This is the equation of continuity:

A_1 v_1 = A_2 v_2,

where A is the cross-sectional area and v is the flow speed at that section. The shared quantity Q = A v is the volume flow rate (m³/s). The derivation is a single line of conservation of mass applied to a small fluid parcel that does not pile up. Two useful consequences: fluid speeds up where the pipe narrows (pinching a garden hose), and streamlines — lines everywhere tangent to the fluid velocity — crowd together where the flow is fast. The principle generalises to compressible fluids by replacing volume flow rate with mass flow rate \rho A v.

Hold a running garden hose at arm's length and pinch the tip with your thumb until the opening is a quarter of its original area. The water, which was flowing out gently, suddenly shoots four times as fast. You have not turned up the tap. You have not added any energy. You simply squeezed the end of the pipe — and the water knew to speed up by exactly the right factor.

That "knew" is the whole interest of fluid dynamics. No single water molecule can see the whole pipe and plan its speed. Each molecule is responding only to the molecules right next to it. Yet the collective behaviour is exactly what a conservation law demands: the same volume of water has to pass every cross-section of the hose per second, because mass is not being created or destroyed anywhere in between. Narrow the opening by a factor of four and the water has to travel four times as fast through it, or else water would be piling up somewhere — and it isn't.

This is the equation of continuity. It is the first real result in fluid dynamics, and it is remarkably general. It explains why the Ganga speeds up through the gorges at Rishikesh and slows down across the plains at Kanpur. It explains why the Delhi Metro crowds thicken where the platform narrows near the stairs. It explains why a column of water falling from a tap thins as it falls and why a bonfire's smoke column spreads as it rises. And it is the foundation on which the next two ideas in fluid dynamics — Bernoulli's equation and Torricelli's law — are built.

The setup — an ideal fluid in steady flow

Before the derivation, a careful statement of the assumptions. Real fluids are complicated: water has a small viscosity, air is slightly compressible, and real flows are turbulent at high speeds. To get a clean result, you isolate the cleanest case and name the simplifications out loud. This is exactly the approach you saw in solid mechanics with Hooke's law — strip the real material down to an idealisation, derive the law there, and then name what changes if the idealisations fail.

An ideal fluid has four properties.

Incompressible. The fluid's density \rho does not change with pressure. Water at normal pressures is an excellent approximation. Air at low speeds (below roughly one-third the speed of sound) is a reasonable approximation too. Diesel fuel and blood also qualify.
Non-viscous. There is no internal friction between fluid layers. In reality all fluids have some viscosity (honey has a lot, water has a little, air has very little), but for many problems — especially over short distances — viscosity is negligible.
Irrotational. Individual fluid parcels do not spin as they move. Vortices (whirlpools, eddies, the spiral down a drain) are rotational flows; the ideal-fluid framework does not handle them directly.
Steady flow. At each point in space, the fluid velocity \vec{v} is constant in time. (It can differ from point to point — the fluid goes faster at one place than another — but at any given point, the velocity doesn't change as time passes.) The opposite is unsteady or turbulent flow, where the velocity at a fixed point fluctuates from moment to moment.

Assumptions. Throughout this article, unless stated otherwise, the fluid is incompressible, non-viscous, irrotational, and the flow is steady. The pipes or channels through which it flows have rigid walls.

Streamlines

In steady flow, you can draw curves called streamlines — lines that are everywhere tangent to the fluid velocity at each point. A streamline traces the path a tiny cork would follow if released into the flow.

Two essential properties follow.

Streamlines do not cross. If they did, at the crossing point the fluid would have two different velocities simultaneously, which is impossible for a single-valued velocity field.
The fluid stays on its streamline. Because the velocity at each point is fixed in time, a tiny cork released at a point always sees the same velocity at that point and always moves in the same direction — tracing out the streamline.

A flow tube (or stream tube) is a bundle of streamlines forming a tube-shaped surface. No fluid crosses the side wall of a flow tube (by definition of streamlines being tangent to the flow). Fluid only enters and exits through the two end cross-sections.

A flow tube with a wide entrance (area $A_1$, slow speed $v_1$) and a narrow exit (area $A_2$, fast speed $v_2$). The streamlines (red dashed) stay inside the tube — no fluid crosses the walls. The streamlines crowd together on the right, which is the visual signature that the fluid is moving faster there.

Volume flow rate

The volume flow rate Q at a given cross-section of a flow tube is the volume of fluid crossing that section per unit time. For a cross-section of area A where the fluid moves at uniform speed v perpendicular to the cross-section, consider the fluid that crosses the section in a short time \Delta t. In that time, each fluid parcel on the section moves forward by v \Delta t, so the volume that passed through is a slab of area A and length v \Delta t:

\Delta V = A \cdot v \Delta t.

Dividing by \Delta t:

\boxed{\; Q = A v \;}

with SI units of m³/s. This is the single most useful quantity in the article.

Why: volume equals area times the distance fluid travels in \Delta t. Dividing by \Delta t turns distance into speed. The result has the dimensions of volume per time.

Numbers to calibrate

A kitchen tap running moderately delivers Q \approx 2 \times 10^{-4} \text{ m}^3/\text{s} = 12 \text{ L/min}.
The Ganga in flood at Haridwar carries Q \approx 10{,}000 \text{ m}^3/\text{s} — ten million litres per second, or about a million kitchen taps.
Your aorta (the main artery out of your heart) at rest carries Q \approx 5 \text{ L/min} \approx 8 \times 10^{-5} \text{ m}^3/\text{s}.
A 1.5-kilowatt Havells ceiling fan moves roughly 200 m³/min \approx 3.3 \text{ m}^3/\text{s} of air.

Deriving the equation of continuity

Take a flow tube. Pick two cross-sections — call them section 1 (area A_1, flow speed v_1) and section 2 (area A_2, flow speed v_2). In a short time \Delta t, the volume of fluid entering at section 1 is

V_1 = A_1 v_1 \Delta t,

and the volume leaving at section 2 is

V_2 = A_2 v_2 \Delta t.

Because the fluid is incompressible and the flow is steady, the mass between sections 1 and 2 is constant in time — no fluid is piling up, none is disappearing, and no fluid crosses the side walls. The only way mass can stay constant is if the mass entering at 1 equals the mass leaving at 2 in each interval \Delta t.

Equating masses:

\rho V_1 = \rho V_2,

\rho A_1 v_1 \Delta t = \rho A_2 v_2 \Delta t.

The \rho (constant density, because incompressible) and the \Delta t cancel:

\boxed{\; A_1 v_1 = A_2 v_2 \;}

This is the equation of continuity. The product A v is the same at every cross-section of a given flow tube.

Why: the whole derivation is one sentence — mass in equals mass out, for an incompressible fluid in steady flow, applied to a region bounded by streamlines. The cross-section can be as wide as you like or as narrow, but whatever volume of fluid flows in at one end has to flow out the other.

Equation of continuity

For an incompressible fluid in steady flow through a tube of varying cross-section, the product of cross-sectional area and flow speed is constant:

A v = \text{constant along the flow tube}.

Equivalently, the volume flow rate Q = A v is the same at every cross-section.

Consequences to hold in your head

Narrow → fast. Where the tube narrows (A small), the fluid speeds up (v large).
Wide → slow. Where the tube widens (A large), the fluid slows down (v small).
Speed and area are inversely proportional. Halve the area, double the speed. Quarter the area, quadruple the speed.
Streamlines crowd where flow is fast. If a tube of streamlines has a smaller cross-section, all those streamlines are still carrying the same total flow rate through a smaller area — so the speed (and the streamline density) is higher.

Why a falling water column thins

Open your kitchen tap slowly until a steady column of water falls. Look at the column from the tap down to where it hits the sink. The column narrows as it falls. Why?

Gravity accelerates the water as it falls, so the speed at the bottom of the visible column is greater than at the top. By continuity, the cross-section must shrink in inverse proportion. Specifically, if the water leaves the tap at speed v_0 and falls a height h, its speed at depth h is v = \sqrt{v_0^2 + 2gh} (from constant-acceleration kinematics). Continuity says

A_0 v_0 = A(h) \cdot \sqrt{v_0^2 + 2gh},

A(h) = \frac{A_0 v_0}{\sqrt{v_0^2 + 2gh}}.

The column thins exactly because the water speeds up.

Explore the area–speed inverse relationship

Drag the area ratio A_1/A_2 below to see how exit speed responds. Inlet speed is fixed at v_1 = 1 m/s.

The exit speed $v_2$ grows linearly with the area ratio $A_1/A_2$ (inlet speed fixed at 1 m/s). Doubling the ratio doubles the speed. At a ratio of 10, the water emerges ten times as fast as it entered.

Why pinching the hose works — the 4× story

Return to the garden hose with which this article began. Let the hose have internal radius r_1 = 1 \text{ cm} and let the water flow at v_1 = 0.5 \text{ m/s} along its length. Now pinch the end until its opening becomes an ellipse with the same perimeter but much smaller area — say the new area is a quarter of the old.

Continuity: A_1 v_1 = A_2 v_2, so

v_2 = v_1 \cdot \frac{A_1}{A_2} = 0.5 \cdot 4 = 2 \text{ m/s}.

The water emerges at four times the original speed — it goes from a dribble to a jet. This four-fold speed-up is purely a consequence of conservation of mass. You did not do any work on the water (you just held the hose). The tap is at the same setting. No energy was added. What changed is how the same volume rate Q distributes through a smaller cross-section — and the only way it can is if the speed rises in inverse proportion.

Note what did happen: the water now has more kinetic energy at the exit than before. Where did that kinetic energy come from? It came from a drop in pressure at the exit — the pressure at the jet is lower than the pressure in the main hose, and the difference drove the fluid's acceleration as it entered the pinched region. This is the subject of Bernoulli's equation, which sits next to the equation of continuity as the second pillar of fluid dynamics. For now, just know that continuity tells you the speed changes — Bernoulli's equation tells you the pressure also changes, and by exactly the right amount to account for the kinetic-energy change.

Applications around India

Rivers — the Ganga at Rishikesh vs at Kanpur

At Rishikesh, the Ganga is squeezed through a narrow gorge in the Himalayan foothills, with a cross-sectional area that might be 200 m² at the narrowest. Downstream at Kanpur, the river spreads across a broad plain with a wet cross-section of perhaps 4000 m² — twenty times larger. The same volumetric discharge (ignoring tributaries, which Ganga has plenty, but as a sketch) must pass through both. Continuity gives

\frac{v_\text{Rishikesh}}{v_\text{Kanpur}} = \frac{A_\text{Kanpur}}{A_\text{Rishikesh}} = 20.

A Ganga that rages at 10 m/s through the gorge at Rishikesh flows placidly at 0.5 m/s across the Kanpur plain. This is why white-water rafting happens at Rishikesh and not at Kanpur, and why ferries run on the Kanpur stretch but nobody tries rafting on a paddleboat in Shivpuri canyon.

A nozzle

Fire-hoses, garden spray nozzles, and jet engines all work on the same principle: accelerate the fluid by forcing it through a narrowing cross-section. A fire-hose pump delivers a modest volume flow rate, but the nozzle at the end — often with a 20:1 area ratio — multiplies the speed by 20, producing a jet that can reach the tenth floor of a building.

Blood flow — the aorta vs the capillaries

The aorta has a cross-sectional area of about 4 cm². The total cross-sectional area of all the body's capillaries combined is about 5000 cm² — 1250 times larger. (There are billions of them, each tiny, but collectively they add up to a huge area.) By continuity, blood flows through the aorta at about v = 30 \text{ cm/s} and through the capillaries at about v = 30/1250 \approx 0.024 \text{ cm/s} — less than a millimetre per second. The slowness is necessary: the capillaries are where oxygen and nutrients diffuse into tissues, and diffusion needs time. Your circulatory system uses the continuity equation as its design principle.

Delhi Metro crowd flow analogy

A crowd of commuters moving through a Delhi Metro platform behaves (roughly) like a fluid. As a train arrives and commuters pour from the platform into the staircase, the same number of people per second that were on the platform must also be on the staircase — so the crowd density (people per m²) on the narrower staircase is higher than on the wider platform. That is the continuity equation applied to a "human fluid." It is why stampedes happen at narrow bottlenecks, not at wide open spaces.

Worked examples

Example 1: Garden hose with a nozzle

Water flows through a garden hose of internal radius r_1 = 1.2 \text{ cm} at speed v_1 = 0.80 \text{ m/s}. The hose ends in a nozzle of internal radius r_2 = 3.0 \text{ mm}. Find the volume flow rate Q and the exit speed v_2.

A garden hose of internal radius 1.2 cm tapers to a nozzle of internal radius 3 mm. The exit speed is set by continuity — whatever volume rate enters the hose must emerge from the nozzle.

Step 1. Compute the cross-sectional areas.

A_1 = \pi r_1^2 = \pi (0.012)^2 = 4.52 \times 10^{-4} \text{ m}^2.

A_2 = \pi r_2^2 = \pi (0.003)^2 = 2.83 \times 10^{-5} \text{ m}^2.

Why: circular cross-section, so A = \pi r^2. Convert radii to metres before squaring.

Step 2. Compute the volume flow rate.

Q = A_1 v_1 = 4.52 \times 10^{-4} \times 0.80 = 3.62 \times 10^{-4} \text{ m}^3/\text{s}.

Converting to litres: Q = 0.362 \text{ L/s} = 21.7 \text{ L/min} — about an average tap.

Why: volume flow rate is area times speed. Units: \text{m}^2 \cdot \text{m/s} = \text{m}^3/\text{s}.

Step 3. Use continuity to find the exit speed.

v_2 = \frac{Q}{A_2} = \frac{3.62 \times 10^{-4}}{2.83 \times 10^{-5}} = 12.8 \text{ m/s}.

Alternatively, via the area ratio:

v_2 = v_1 \cdot \frac{A_1}{A_2} = 0.80 \times \frac{4.52 \times 10^{-4}}{2.83 \times 10^{-5}} = 0.80 \times 16 = 12.8 \text{ m/s}.

Why: the area ratio A_1/A_2 = (r_1/r_2)^2 = (12/3)^2 = 16. The exit speed is 16 times the inlet speed — the hose-to-nozzle squeeze is a 16× speed multiplier.

Result: Volume flow rate Q = 3.62 \times 10^{-4} \text{ m}^3/\text{s} (21.7 L/min); exit speed v_2 = 12.8 \text{ m/s} — about 46 km/h.

What this shows: A modest 0.8 m/s flow in the hose becomes a 13 m/s jet at the nozzle. This is why a thumb over a garden hose can water the top of a 3-metre tree: continuity amplifies the speed, and the jet then rises h = v_2^2/(2g) \approx 8.4 m under gravity (minus air drag) before falling back.

Example 2: Three pipes meeting at a junction

Three pipes join at a T-junction. Water enters pipe 1 (radius 2 cm) at 1.0 m/s and also enters pipe 2 (radius 1.5 cm) at 0.8 m/s. Both streams exit through pipe 3 (radius 3 cm). Find the exit speed in pipe 3.

Two input pipes merge at a junction. The total volume flow rate exiting through pipe 3 equals the sum of the rates entering through pipes 1 and 2.

Step 1. Compute each input's volume flow rate.

Q_1 = \pi r_1^2 v_1 = \pi (0.02)^2 \times 1.0 = 1.257 \times 10^{-3} \text{ m}^3/\text{s}.

Q_2 = \pi r_2^2 v_2 = \pi (0.015)^2 \times 0.8 = 5.655 \times 10^{-4} \text{ m}^3/\text{s}.

Step 2. Apply conservation of mass at the junction.

The junction is a short region; in steady state no water accumulates there, so total flow in equals total flow out:

Q_3 = Q_1 + Q_2 = 1.257 \times 10^{-3} + 5.655 \times 10^{-4} = 1.822 \times 10^{-3} \text{ m}^3/\text{s}.

Why: the continuity equation, extended to a junction, says mass in equals mass out. For two pipes feeding one, the output volume flow rate is the sum of the inputs.

Step 3. Find the exit speed.

A_3 = \pi r_3^2 = \pi (0.03)^2 = 2.827 \times 10^{-3} \text{ m}^2.

v_3 = \frac{Q_3}{A_3} = \frac{1.822 \times 10^{-3}}{2.827 \times 10^{-3}} = 0.645 \text{ m/s}.

Why: for the exit pipe, v_3 = Q_3/A_3. The answer lies between v_1 and v_2 because pipe 3 is wider than either input, so the sum of flows is spread over a larger area.

Result: v_3 \approx 0.65 m/s.

What this shows: The equation of continuity extends naturally to pipe networks. At any junction, sum of flow rates in equals sum of flow rates out — a conservation law that powers everything from home plumbing to Delhi's water supply grid.

Example 3: A narrowing river

A straight stretch of a small river is 40 m wide and 3 m deep. Downstream, it enters a gorge where the cross-section reduces to 12 m wide and 2 m deep. If the river flows at 1.2 m/s in the wide section, estimate the flow speed in the gorge. Assume the river is incompressible and the flow is steady (no rainfall, no tributaries in between).

Step 1. Compute the two cross-sectional areas.

Wide section: A_1 = 40 \times 3 = 120 \text{ m}^2.

Gorge: A_2 = 12 \times 2 = 24 \text{ m}^2.

Step 2. Apply continuity.

v_2 = v_1 \cdot \frac{A_1}{A_2} = 1.2 \times \frac{120}{24} = 1.2 \times 5 = 6.0 \text{ m/s}.

Why: the cross-section decreases by a factor of 5, so the speed must increase by the same factor to conserve volume flow rate.

Result: The river speeds up from 1.2 m/s to 6.0 m/s as it enters the gorge — fast enough to sweep away anything unwary.

What this shows: The equation of continuity works at all scales, from a garden hose to a Himalayan river. The only requirement is that the flow is incompressible and steady, and that no water is lost or gained between sections.

Common confusions

"The fluid speeds up where the pipe narrows because it's being squeezed." Slightly wrong framing. The fluid's density is not changing — it is incompressible, and the "squeeze" does not compress anything. What changes is the geometry through which a fixed volume rate must pass. Because Q = A v and Q is fixed, a smaller A forces a larger v. The speed-up is kinematic (geometric), not because anything is being compressed.
"If the pipe widens out a lot, the fluid stops flowing." No. A pipe twice as wide has the fluid flowing at half the speed — still flowing, just slower. Only if the cross-section became infinite would the speed approach zero. In a large tank fed by a pipe, the water in the tank does move, but so slowly (because A is huge) that it looks stationary.
"The equation of continuity is about the fluid, not about the shape of the pipe." It is about both. Continuity relates the speed at any cross-section to the cross-section's area, so the geometry of the pipe (or channel) matters. Same fluid, different pipe → different speeds.
"Incompressible means zero compressibility." In the real world, no fluid is perfectly incompressible — water compresses by about 5% under 1000 atmospheres. The assumption means that at the pressures relevant to your problem, the density change is negligible. For water at normal pressures and ordinary pipe flows, this is an excellent approximation. For a gas flowing near the speed of sound, it breaks down entirely, and you need the compressible form of the continuity equation: \rho_1 A_1 v_1 = \rho_2 A_2 v_2.
"Continuity fails if the flow is turbulent." The equation of continuity holds even for turbulent flow, as long as you interpret the velocities as time-averaged. What fails for turbulent flow is the assumption of streamlines (which become chaotic). The conservation of mass always holds; it is the streamline picture that is idealised.
"The equation of continuity requires a pipe." Not quite — it requires a flow tube, which can be an imaginary tube bounded by streamlines in a larger flow, not a physical pipe. The mass-balance argument only uses streamlines as the side walls.

If you can derive A_1 v_1 = A_2 v_2 from conservation of mass and apply it to pipes and rivers, you have the working content of this article. What follows is for readers who want the differential form of continuity, the extension to compressible fluids, and the link to Bernoulli's equation that comes next.

The differential form

The algebraic form A v = \text{constant} works for one-dimensional flow in a pipe. For three-dimensional flow — air around an aircraft wing, water swirling in a river — you need the differential form.

Consider a fluid of density \rho(\vec{r}, t) moving with velocity \vec{v}(\vec{r}, t). Pick an arbitrary volume V bounded by a surface S. Conservation of mass says

\frac{d}{dt}\int_V \rho \, dV = -\oint_S \rho \vec{v} \cdot d\vec{A},

which reads: "the rate of change of mass inside V equals the negative of the mass flux out through the boundary." The - sign is because d\vec{A} points outward.

Apply the divergence theorem to the right-hand side:

\oint_S \rho \vec{v} \cdot d\vec{A} = \int_V \nabla \cdot (\rho \vec{v}) \, dV.

Substituting and moving everything under one integral:

\int_V \left[\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{v})\right] dV = 0.

Since this holds for every volume V, the integrand must vanish:

\boxed{\; \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{v}) = 0 \;}

This is the continuity equation in its differential form, one of the fundamental equations of fluid mechanics.

For an incompressible fluid, \rho is constant in space and time, so \partial \rho/\partial t = 0 and \rho factors out of the divergence:

\nabla \cdot \vec{v} = 0.

The fluid's velocity field is divergence-free. In one dimension, this reduces to d(Av)/dx = 0 along a flow tube — the algebraic form.

Compressible fluids

For gas dynamics (especially near the speed of sound), density varies and cannot be factored out. The one-dimensional steady form becomes

\rho_1 A_1 v_1 = \rho_2 A_2 v_2,

the mass flow rate \dot{m} = \rho A v is conserved, not the volume flow rate. This is the version used in rocket engines, jet nozzles (supersonic flows through converging-diverging nozzles), and wind tunnels. Interestingly, in supersonic flow (gas moving faster than the local sound speed), narrowing the nozzle slows the gas down — the opposite of subsonic behaviour. This counter-intuitive fact underlies the design of rocket nozzles (ISRO's PSLV and GSLV engines use converging-diverging nozzles for exactly this reason).

Continuity plus Bernoulli

The equation of continuity is one half of a pair. The other half is Bernoulli's equation, which combines conservation of energy with the continuity result to relate pressure, speed, and elevation along a streamline:

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

Here is how continuity enters: given the area changes along a pipe, continuity tells you how v changes. Then Bernoulli tells you how P changes in response. Together they answer the question at the end of this article's hose story: "Where did the kinetic energy at the nozzle come from?" Answer: a drop in pressure, as Bernoulli's equation quantifies exactly.

Streamline crowding as a visual diagnostic

In a streamline diagram, wherever the streamlines crowd together, the fluid is moving faster. This is because the volume flow rate between any two adjacent streamlines is the same — and volume rate is area times speed, so a smaller cross-section between streamlines means higher speed. This is why aircraft wings are drawn with dense streamlines above and sparser streamlines below (indicating faster flow above, and hence by Bernoulli, lower pressure above — which is the lift force).

A subtle historical note

The algebraic form of the continuity equation — mass in equals mass out — was stated by Leonardo da Vinci in his notebooks around 1500 in the context of riverbed erosion, long before the full mathematical framework of fluid mechanics existed. Aryabhata's calculations for the water clocks (ghatika-yantra) used by Indian astronomers implicitly assumed the same principle: the time for water to drain depended on the aperture area and the flow speed, and the rate was consistent with what we now call the continuity equation. The modern differential form is due to Leonhard Euler (1757) and sits at the centre of the Euler equations of fluid flow.

Where this leads next

Pressure in Fluids — the hydrostatics foundation.
Pascal's Law and Hydraulic Machines — the other great idea that follows from the pressure-depth law.
Archimedes' Principle and Buoyancy — what happens when a body is submerged in a fluid at rest.
Newton's Second Law — the conservation-of-momentum law that underpins Bernoulli's equation, the next step after continuity.