
Optional unit · institution-dependent · Interactive lesson
Fluid Flow & Continuity
Before asking what pushes a fluid, you have to say how much of it is moving. Here you build that bookkeeping: a flow pattern that holds still, a flow rate counted through a surface, and the one conservation law that fixes the speed wherever a pipe narrows, widens or splits.
UPI-15.06 · Lesson 6 of 9 in this unit. Position is not a completion score.Optional unit · institution-dependent. Your university may place fluids elsewhere or omit it from Physics I. Optional work is not a required completion gate.
Build the model
What flows in must flow out: conservation of mass for a moving fluid.
The key idea
The continuity equation says that in steady flow the same mass crosses every cross-section of a stream tube each second, which for an incompressible fluid means the volume flow rate Q = Av is the same at every section.
Example. Water moving at 0.60 m s⁻¹ through a cross-section of 12 cm² enters a section of 4.0 cm²: Q = (12 × 10⁻⁴ m²)(0.60 m s⁻¹) = 7.2 × 10⁻⁴ m³ s⁻¹ = 0.72 L s⁻¹ at both sections, so the speed in the narrow part is 1.8 m s⁻¹.
The complete physical model and assumptions
A flowing fluid is described not by chasing parcels but by recording the velocity at each point in space. When that record does not change with time the flow is steady, parcels follow fixed curves called streamlines, and a bundle of streamlines forms a stream tube whose side walls no fluid crosses. Now count. In a time dt the fluid crossing a section of area A at speed v fills a slab of volume A v dt, so the volume flow rate is Q = Av and the mass flow rate is ρAv. In steady flow the mass held between two sections of a tube cannot change, so whatever enters each second must leave each second: ρ₁A₁v₁ = ρ₂A₂v₂. Liquids, and gases moving well below the speed of sound, hardly change density, so ρ cancels and A₁v₁ = A₂v₂: the fluid must speed up where the passage narrows, and no force has been mentioned. That is both the power and the limit of continuity. It is pure kinematics, so it survives viscosity and turbulence provided v means the average over the section; but it cannot tell you the pressure, its constant-density form fails for gas flows approaching a few tenths of the speed of sound, and the two-section balance fails whenever fluid is piling up or draining from storage in between.
1Describe the pattern, not the parcels
Following every parcel of water through a pipe is hopeless, so fluid mechanics watches fixed points instead and records the velocity there. If the velocity at every point stays the same from one moment to the next, the flow is steady. Steady does not mean that nothing accelerates: a parcel entering a nozzle speeds up because it moves to a place where the velocity is larger, while the pattern itself never changes. A streamline is a curve tangent to the velocity all along its length, and in steady flow it is also the path a parcel follows. Two streamlines can never cross, because fluid at the crossing point would need two velocities at once. Wrap a closed loop of streamlines around some fluid and you have a stream tube: the velocity is tangent to its walls, so nothing leaks through them. A stream tube is a pipe drawn by the flow itself, and everything below applies to it as well as to real pipes.
2A flow rate is a slab of fluid per second
Take a flat cross-section of area A where the fluid moves at speed v perpendicular to it. In a short time dt every parcel advances v dt, so the fluid that crosses fills a slab of volume dV = A v dt. Divide by dt: Q = dV/dt = Av, in m³ s⁻¹. A tap that fills a 1.5 L jug in 12 s delivers Q = 1.5 × 10⁻³ m³ ÷ 12 s = 1.25 × 10⁻⁴ m³ s⁻¹; through a spout of area 0.80 cm² that requires v = Q/A = (1.25 × 10⁻⁴) ÷ (0.80 × 10⁻⁴) = 1.6 m s⁻¹. Multiply by the density for the mass flow rate: dm/dt = ρQ = 0.125 kg s⁻¹ here. Because Q is a derivative, the volume delivered is its integral. A valve opened gradually so that Q(t) = (2.0 × 10⁻⁵ m³ s⁻²) t delivers ∫₀¹⁰ Q dt = ½(2.0 × 10⁻⁵)(10)² = 1.0 × 10⁻³ m³ in its first 10 s: one litre, half of what the final flow rate would have delivered in the same time.
3Conserve mass between two sections
Mark two sections of a pipe or stream tube and treat the space between them as a control volume. Mass enters through section 1 at the rate ρ₁A₁v₁ and leaves through section 2 at the rate ρ₂A₂v₂; nothing crosses the side walls. In steady flow the density at each interior point is fixed, so the mass stored in the control volume cannot change, and the two rates must be equal: ρ₁A₁v₁ = ρ₂A₂v₂. This is conservation of mass written per second, and in this form it handles gases. Air of density 1.20 kg m⁻³ enters a heater duct of constant area at 3.0 m s⁻¹ and leaves hot at 1.00 kg m⁻³. The area cancels, so v₂ = (1.20/1.00)(3.0) = 3.6 m s⁻¹. The volume flow rate has risen by 20% while the mass flow rate has not changed at all, which is why mass, not volume, is the thing to balance.
4Cancel the density and let area set the speed
Water shrinks by only about 0.005% for each extra atmosphere of pressure, and air moving slower than roughly 100 m s⁻¹ changes density by only a few per cent, so in most first-year problems ρ₁ = ρ₂ and continuity becomes A₁v₁ = A₂v₂: the volume flow rate Q is the same at every section. Speed is inversely proportional to area. For a round pipe A = πd²/4, so v₂ = v₁(d₁/d₂)². Water moving at 0.40 m s⁻¹ in a 25 mm pipe that necks down to 10 mm leaves at 0.40 × (2.5)² = 2.5 m s⁻¹, not the 1.0 m s⁻¹ obtained by scaling with the diameter. Streamline pictures carry the same information: if neighbouring streamlines bound tubes of equal flow, they must sit closer together where the fluid is faster, so crowded streamlines mean high speed.
5Branches, tanks and anything that stores fluid
A junction is a control volume with more than two openings. In steady incompressible flow the flow rates in must add up to the flow rates out: ΣQᵢₙ = ΣQₒᵤₜ. Flow rates add; speeds do not. A main carrying 0.90 L s⁻¹ that feeds two branches, one metered at 0.35 L s⁻¹, must send 0.55 L s⁻¹ down the other whatever the pipe sizes are, and only then do the areas turn those flow rates into speeds. If the totals do not balance, the difference is being stored: dV/dt = Qᵢₙ − Qₒᵤₜ. A tank with a free surface of 0.60 m² that receives 0.90 L s⁻¹ while 0.30 L s⁻¹ drains away gains 6.0 × 10⁻⁴ m³ every second, so its level rises at dh/dt = (6.0 × 10⁻⁴) ÷ 0.60 = 1.0 × 10⁻³ m s⁻¹, one millimetre per second. The pipe rule A₁v₁ = A₂v₂ is the special case in which the stored volume is constant.
6What v means in a real pipe, and where the model stops
Real fluid sticks to the wall, so the speed across a section is not uniform: zero at the wall, greatest on the axis. The flow rate is then the integral Q = ∫ v dA, which for a round pipe is ∫₀ᴿ v(r) 2πr dr, a sum over rings of area 2πr dr. Continuity survives untouched provided v in A₁v₁ = A₂v₂ is read as the mean speed Q/A. For the parabolic profile of slow, smooth pipe flow the mean is exactly half the centreline speed, so a probe on the axis overstates Q by a factor of two. Turbulent flow is unsteady in its detail, yet its time-averaged flow rates still balance. What does fail is the constant-density shortcut, for gas flows faster than a few tenths of the speed of sound or with strong heating, and the two-section balance whenever fluid accumulates in between. And continuity is silent about pressure: that takes the energy argument of Bernoulli's equation, which comes next.
Key equations
Read the units, assumptions and limits before substituting.
Volume flow rate
Q = dV/dt = A v
Turns a speed into a delivery rate, and back. If the flow meets the surface at an angle θ to its normal, only the component v cos θ carries fluid through.
Q in m³ s⁻¹, A in m², v in m s⁻¹ measured perpendicular to A; 1 L s⁻¹ = 10⁻³ m³ s⁻¹Mass flow rate
dm/dt = ρ A v = ρ Q
This, not Q, is what is conserved. Reach for it whenever the density differs between sections, as in a heated or compressed gas.
dm/dt in kg s⁻¹, ρ in kg m⁻³; for water, 1 L s⁻¹ carries 1 kg s⁻¹Continuity for steady flow
ρ₁ A₁ v₁ = ρ₂ A₂ v₂
Steady flow keeps the mass stored between the sections constant, so inflow and outflow have to balance second by second.
sections 1 and 2 cut the same stream tube; each side is a mass flow rate in kg s⁻¹Incompressible continuity
A₁ v₁ = A₂ v₂, so v₂ = v₁ (d₁/d₂)² in round pipes
Speed is inversely proportional to area, so diameters enter squared: one third of the diameter means nine times the speed.
A in m², v in m s⁻¹; d is the internal diameter, in any unit common to both sectionsJunctions and storage
Σ Qᵢₙ − Σ Qₒᵤₜ = dV/dt of the stored fluid
Handles tees, manifolds and tanks. A tank with free-surface area Aₛ rises at dh/dt = (Qᵢₙ − Qₒᵤₜ)/Aₛ.
every Q in m³ s⁻¹; V in m³ is the fluid held between the openings; the right side is zero in steady flowNon-uniform speed profile
Q = ∫ v dA = ∫₀ᴿ v(r) 2πr dr, with mean speed v̄ = Q/A
Real flow is slower near the wall. Continuity still holds exactly once v is read as the mean speed v̄ rather than a single probe reading.
v(r) in m s⁻¹ at radius r in m; each ring has area 2πr dr in m²; R is the pipe radiusInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Narrow the outlet and watch its slab stretch while its shaded area stays equal to the inlet slab's; then make the outlet wider than the inlet and the flow slows. Raising v₁ lengthens both slabs together. This duct has fixed depth, so v goes as 1/w; in a round pipe it goes as 1/d².
Live results
- FLOW RATE Q = A₁v₁
- 2.40 L s⁻¹
- OUTLET SPEED v₂
- 1.60 m s⁻¹
- AREA RATIO A₁/A₂
- 2.00
- MASS FLOW ρQ (WATER)
- 2.40 kg s⁻¹
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Inlet width w₁: 60 mm; Outlet width w₂: 30 mm; Inlet speed v₁: 0.8 m s⁻¹
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The traffic-jam picture says a fluid must slow down where the pipe narrows, as cars do where a road loses a lane. Cars slow because they can bunch up: their density rises and the jam stores vehicles. An incompressible fluid in steady flow can do neither, so the same volume per second has to pass through the smaller area, and the only way to manage that is to move faster. The companion error is to believe the narrow section carries a smaller flow rate than the wide one. A constriction may well reduce Q for the whole pipe, because it takes more pressure to drive the flow, but at any instant Q is the same at every section: the speed differs from place to place, the flow rate does not. And when you scale speeds between round pipes the ratio is of areas, so the diameters enter squared.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A drip-irrigation feed line has an internal diameter of 13 mm. With its valve fully open it fills a 9.0 L bucket in 60 s. Find the volume flow rate in SI units, the mean speed of the water in the line, and the mass flow rate. Take ρ = 1000 kg m⁻³.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Flow rate is volume per unit time: Q = V/t = (9.0 × 10⁻³ m³) ÷ (60 s) = 1.5 × 10⁻⁴ m³ s⁻¹, which is 0.15 L s⁻¹.
- Cross-section of the line: A = πd²/4 = π(0.013 m)²/4 = 1.33 × 10⁻⁴ m².
- Q = Av, so the mean speed is v = Q/A = (1.5 × 10⁻⁴) ÷ (1.33 × 10⁻⁴) = 1.13 m s⁻¹, or 1.1 m s⁻¹ to the two figures the data support.
- Mass flow rate: dm/dt = ρQ = (1000 kg m⁻³)(1.5 × 10⁻⁴ m³ s⁻¹) = 0.15 kg s⁻¹. For water the number of litres per second and of kilograms per second agree.
Answer. Q = 1.5 × 10⁻⁴ m³ s⁻¹ (0.15 L s⁻¹); v = 1.1 m s⁻¹; dm/dt = 0.15 kg s⁻¹.
MediumExample 2
In a cooling loop a supply main of internal diameter 50 mm carries water at a mean speed of 1.20 m s⁻¹ to a tee. One leg, of diameter 25 mm, feeds a heat exchanger, and a meter on it reads a mean speed of 1.80 m s⁻¹. The other leg is a bypass of diameter 30 mm. The flow is steady. Find the mean speed in the bypass and the fraction of the water that takes it.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Label the main 1, the exchanger leg 2 and the bypass 3. Flow rates add at a junction, speeds do not: Q₁ = Q₂ + Q₃, with each Q = (πd²/4)v.
- The factor π/4 is common to every term, so work with d²v: main 50² × 1.20 = 3000; exchanger leg 25² × 1.80 = 1125, both in mm² m s⁻¹.
- Bypass: d²v = 3000 − 1125 = 1875 mm² m s⁻¹, so v = 1875 ÷ 30² = 1875 ÷ 900 = 2.08 m s⁻¹.
- In SI units: Q₁ = (π/4)(0.050)²(1.20) = 2.36 × 10⁻³ m³ s⁻¹ and Q₃ = (π/4)(0.030)²(2.083) = 1.47 × 10⁻³ m³ s⁻¹.
- Fraction through the bypass: Q₃/Q₁ = 1875/3000 = 0.625. Check: the exchanger leg carries Q₂ = 8.84 × 10⁻⁴ m³ s⁻¹, and 0.884 + 1.473 = 2.357 ≈ 2.36 in units of 10⁻³ m³ s⁻¹.
Answer. The bypass runs at 2.08 m s⁻¹ and carries 62.5% of the flow, 1.47 × 10⁻³ m³ s⁻¹ of the main's 2.36 × 10⁻³ m³ s⁻¹.
HardExample 3
Oil flows slowly and smoothly through a round pipe of radius R = 12 mm with the speed profile v(r) = v₀(1 − r²/R²), where the centreline speed is v₀ = 0.60 m s⁻¹. Integrate over rings to find Q and the mean speed. The pipe feeds a nozzle of exit radius 4.0 mm, where the profile is nearly uniform: find the exit speed, and the error made by putting v₀ into A₁v₁ = A₂v₂.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- A ring of radius r and width dr has area dA = 2πr dr and carries dQ = v(r) 2πr dr, so Q = ∫₀ᴿ v₀(1 − r²/R²) 2πr dr.
- Integrate term by term: Q = 2πv₀ [r²/2 − r⁴/(4R²)] from 0 to R = 2πv₀ (R²/2 − R²/4) = ½πR²v₀.
- Numbers: Q = ½π(0.012 m)²(0.60 m s⁻¹) = 1.36 × 10⁻⁴ m³ s⁻¹, or 0.136 L s⁻¹.
- Mean speed: v̄ = Q/(πR²) = v₀/2 = 0.30 m s⁻¹. It is only half the centreline value because the fast fluid near the axis occupies little area, while the slow outer rings are large.
- Continuity with mean speeds: v₂ = 0.30 × (12/4.0)² = 0.30 × 9 = 2.7 m s⁻¹. Check: Q/A₂ = (1.36 × 10⁻⁴) ÷ (π × 0.0040²) = 2.7 m s⁻¹.
- Using v₀ = 0.60 m s⁻¹ as if it were the mean gives 0.60 × 9 = 5.4 m s⁻¹, twice the true value: a 100% overestimate of both the exit speed and Q.
Answer. Q = 1.36 × 10⁻⁴ m³ s⁻¹; mean speed 0.30 m s⁻¹ = v₀/2; exit speed 2.7 m s⁻¹. Treating the centreline reading v₀ as the mean would double every one of these.
Quick check
Select an answer and explain it before checking.
Review & continue
The continuity equation says that in steady flow the same mass crosses every cross-section of a stream tube each second, which for an incompressible fluid means the volume flow rate Q = Av is the same at every section.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics I completion tasks.