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Continuity and Bernoulli

Intermediate Fluid Mechanics and Hydraulics Reviewed 24 Oct 2026 2 min read

Two equations do most of the work in hydraulics. The continuity equation expresses conservation of mass: for an incompressible fluid in steady flow, the discharge Q = A V is the same at every section of a pipe or channel. Where the pipe narrows, velocity rises; where it widens, velocity falls.

Bernoulli's equation expresses conservation of energy along a streamline for steady, frictionless, incompressible flow: pressure head plus velocity head plus elevation head is constant. Each term is an energy per unit weight measured in metres of fluid. Real flows lose energy to friction and turbulence, so practical forms add a head loss term between two sections, and pumps add head while turbines remove it. Plotting the energy grade line and hydraulic grade line along a pipeline shows at a glance where pressure is high or low and where it might drop below atmospheric.

Diagram: Indiacon original, not to scale.

Applications are everywhere. A venturimeter or orifice meter measures flow from the pressure difference between a wide and a narrow section. A Pitot tube measures velocity from the difference between stagnation and static pressure. Flow from an orifice in a tank follows Torricelli's result, V = √(2gh), reduced by a discharge coefficient. Siphons, spillways, nozzles and pump suction lines are all checked with Bernoulli, keeping pressure above vapour pressure to avoid cavitation.

Tip

Draw the energy and hydraulic grade lines to scale for any long pipeline; low points of pressure jump out.

Key formulas
  • Continuity: Q = A₁ V₁ = A₂ V₂
  • Bernoulli: p/γ + V²/2g + z = constant
  • With losses: p₁/γ + V₁²/2g + z₁ + h_p = p₂/γ + V₂²/2g + z₂ + h_L
  • Venturimeter: Q = Cd A₁ A₂ √(2 g Δh) / √(A₁² − A₂²)
Common mistake

Applying Bernoulli across a pump or a hydraulic jump without adding the pump head or the energy loss.

Solved example

Problem. A venturimeter with 300 mm inlet and 150 mm throat shows a head difference of 0.9 m of water. Take Cd = 0.97. Find the flow.

  1. A₁ = 0.07069 m², A₂ = 0.01767 m².
  2. √(2 × 9.81 × 0.9) = 4.202 m/s.
  3. √(A₁² − A₂²) = √(0.004997 − 0.000312) = 0.06845.
  4. Q = 0.97 × 0.07069 × 0.01767 × 4.202 / 0.06845 = 0.0744 m³/s ≈ 74 L/s.

Answer. About 74 L/s

Quick quiz

1What does the continuity equation conserve?

Mass (volume, for incompressible flow).

2What does each Bernoulli term represent?

Energy per unit weight: pressure, velocity and elevation head.

3What is the hydraulic grade line?

The line of pressure head plus elevation along the flow.

Code note

    Clauses and documents to read

    • ISO 5167 (flow measurement by differential pressure devices, cited by number)

    Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.

    References and further reading

    Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.

    • P. N. Modi, S. M. Seth, Hydraulics and Fluid Mechanics including Hydraulic Machines, Standard Book House, latest edition.
    • V. T. Chow, Open-Channel Hydraulics, McGraw Hill, 1959.
    • K. Subramanya, Flow in Open Channels, McGraw Hill Education (India), latest edition.

    Free and official sources

    All references for this subject →

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