Ideal fluid flow: use continuity and an explicit energy balance

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The same volume can move at different speeds along one tube. Calculate those speeds first, then account for pressure, elevation, and the assumptions behind Bernoulli’s equation.

About 20 minutes, including the worked examples and checks.

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Keep volume flow separate from local speed

A fluid can move faster through a narrow section while the same volume passes every section each second. The speed describes motion at a location; the volume flow rate describes how much fluid crosses a section per time. AAMC’s fluid scope includes continuity, Bernoulli’s equation, constrictions, and pressure-based flow measurements.¹

Volume flow rate is Q = ΔV/Δt. For a uniform velocity perpendicular to a section, Q = Av; when the profile varies, v in this relation means the cross-sectional average speed. Useful units include m³/s and mL/s, with 1 mL = 10⁻⁶ m³. Under steady incompressible flow, with no leaks, branches, or accumulation between two sections, A₁v₁ = A₂v₂. This is conservation of mass expressed for constant density.²

A branch changes the bookkeeping: incoming volume flow equals the sum of outgoing flows under those same conditions. If density changes substantially, mass flow rather than volume flow is the conserved quantity. A claim that “flow is the same everywhere” must specify which quantity and which flow path it concerns.

Use the energy relation only under its stated conditions

For steady incompressible flow with negligible viscous loss, Bernoulli’s relation along a streamline is p + ½ρv² + ρgz = constant. Here p is static pressure, ρ is density, v is speed, and z is height measured upward. The comparison must also exclude a pump or turbine adding or removing mechanical energy between the selected points.³

Each term has pressure units, equivalently energy per volume: 1 Pa = 1 J/m³. The pressure term represents the contribution of pressure work to the fluid’s mechanical-energy balance; the other terms describe kinetic and gravitational contributions. All pressures can be absolute, or they can share one common gauge reference that cancels between the points.

The simple section calculations below additionally idealize each section’s velocity as uniform, so its streamline speed and cross-sectional average coincide. Steady flow does not mean that a moving fluid element never accelerates. An element can speed up as it enters a constriction even though the speed at each fixed location is constant over time.

Worked application: a constriction followed by a rise

All tube quantities and data below are original illustrative values. Water of density 1000 kg/m³ flows through the unbranched tube. Treat the flow as ideal under the conditions just stated and use g = 10 m/s². At A, the cross-sectional area is 4.0 cm², speed is 1.0 m/s, and gauge pressure is 12.0 kPa. B has half that area at the same height. C returns to A’s area but is 0.10 m higher.

First find Q = (4.0 × 10⁻⁴ m²)(1.0 m/s) = 4.0 × 10⁻⁴ m³/s = 400 mL/s. Continuity then gives v_B = 2.0 m/s and v_C = 1.0 m/s. The constriction changes local speed; it does not consume half the volume flow.

Between A and B, the height term cancels. The kinetic contribution rises from ½(1000)(1.0²) = 500 Pa to ½(1000)(2.0²) = 2000 Pa. Pressure must fall by 1500 Pa, giving p_B = 10.5 kPa. This is the ideal pressure decrease associated with a Venturi constriction at unchanged height.³

Between A and C, the speeds are equal, so the kinetic terms cancel. The rise in gravitational contribution is (1000)(10)(0.10) = 1000 Pa. Thus p_C = 11.0 kPa. A raised section requires accounting for height even when its speed matches the starting section.⁴

Pressure ordering depends on more than speed

The worked pressures satisfy p_A > p_C > p_B. Comparing A with B isolates the speed change because their heights match. Comparing B with C requires both the speed and height terms. Raising C farther could make its pressure lower than B’s even though C has the lower speed. “Faster means lower pressure” needs the rest of the energy comparison held appropriate.

In an ideal tube that returns to the same area and height, the pressure can return to its earlier value. A persistent pressure loss in a real tube calls for an energy-loss term; it cannot be explained by merely asserting that fluid has already passed a narrow point.

A pitot comparison measures a local kinetic contribution

A pitot-static comparison uses a stagnation point, where the approaching flow is brought to rest, and a nearby static-pressure reading. At the same height with negligible loss, p_stagnation − p_static = ½ρv². The difference gives v = √[2Δp/ρ]. It is not a comparison between two arbitrary static-pressure taps.³

For the worked flow at B, the dynamic contribution is 2.0 kPa. A local stagnation reading would therefore be 12.5 kPa gauge, compared with 10.5 kPa static. Using Δp = 2000 Pa and ρ = 1000 kg/m³ returns v = 2.0 m/s. The inlet static pressure of 12.0 kPa is not interchangeable with this local stagnation pressure.

Know what a failed ideal prediction leaves unresolved

In a long horizontal tube of constant area, steady incompressible flow gives equal average speeds at its ends. The loss-free Bernoulli relation would then give equal end pressures. A measured pressure decrease can instead be required to overcome viscous resistance while maintaining the flow. Including dissipation changes the energy relation without violating continuity.⁵

For a new problem, calculate the conserved flow first when area and speed data permit it. Then compare kinetic, height, and pressure terms under a named model. If the measurements disagree, inspect the neglected losses, added energy, density changes, and measurement conditions before changing a conservation principle.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

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From forces to fluid flow

Lesson 5 of 6.

• Previous in From forces to fluid flow: Pressure, buoyancy, and the role of fluid interfaces
• Next in From forces to fluid flow: Viscous flow: explain pressure loss and radius dependence
• Return to this path’s overview

Related lessons

• Work, energy, and power: choose the useful conservation model
• Pressure, buoyancy, and the role of fluid interfaces
• Viscous flow: explain pressure loss and radius dependence

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Sources

1. AAMC: C/P Content Category 4B

2. OpenStax University Physics 1: Fluid Dynamics

3. OpenStax University Physics 1: Bernoulli’s Equation

4. OpenStax College Physics 2e: General Applications of Bernoulli’s Equation

5. OpenStax University Physics 1: Viscosity and Turbulence