The Continuity Equation
- State the continuity equation for incompressible flow
- Relate cross-sectional area to flow speed, A₁v₁ = A₂v₂
- Predict how flow speed changes in pipes of varying width
Flow rate is conserved
For an incompressible fluid (like water) flowing steadily through a pipe, the volume flow rate — the volume passing any cross-section per second — must be the same everywhere, because fluid cannot pile up or vanish. That flow rate equals the cross-sectional area times the flow speed, Av. What goes in one end comes out the other.
Narrow means fast
Since Av stays constant, a smaller area forces a larger speed: A₁v₁ = A₂v₂. Where a pipe narrows, the fluid must speed up to carry the same volume through the tighter opening; where it widens, the flow slows. This is why water shoots out faster when you cover part of a hose nozzle with your thumb.
Water flows at 2 m/s through a pipe of cross-sectional area 0.1 m². The pipe narrows to 0.05 m². Find the new flow speed.
- 1.Apply continuity: A₁v₁ = A₂v₂.
- 2.Substitute the knowns: 0.1 × 2 = 0.05 × v₂ → 0.2 = 0.05 × v₂.
- 3.Solve: v₂ = 0.2 ÷ 0.05.
- 4.Result: v₂ = 4 m/s — the flow doubled when the area halved.
Area and speed trade off inversely. If a pipe’s area drops to one-third, the speed triples. Check that your answer moves the opposite way from the area change.
Water flows through a pipe that narrows to a smaller cross-sectional area. In the narrow section, the water’s speed:
Water moves at 3 m/s through a pipe of area 0.2 m². It then flows into a wider section of area 0.6 m². What is the new speed?
Continuity assumes an incompressible fluid and steady flow. On the AP exam it is almost always paired with Bernoulli’s equation: first use continuity to get the speeds, then feed them into Bernoulli for the pressures.
Answer the 2 checkpoints as you read.
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