Mechanical Properties of Fluids · Lesson 3 of 8
Steady Flow, Continuity and Bernoulli’s Principle
“Derive mass continuity and Bernoulli’s equation, then apply their distinct roles and limits to steady flow.”
• Distinguish steady flow, streamline geometry, laminar flow and turbulent motion. • Derive mass continuity and apply it to varying areas and multiple outlets. • Derive Bernoulli’s equation from pressure work and mechanical-energy changes. • State its assumptions, interpret pressure columns and use a consistent pressure reference.
Steady does not mean the same speed everywhere
Turn a tap on gently and the jet is initially smooth. Increase the flow and the jet becomes irregular. Fluid dynamics studies such motion. Instead of following every molecule, we describe the velocity at each position. Flow is steady when the velocity at any fixed point does not change with time: successive fluid elements passing that point have the same velocity. That velocity can still differ from the velocity at another point. A particle can therefore accelerate as it moves through a steady, narrowing pipe.
A streamline is a curve whose tangent at each point has the direction of the local fluid velocity. In steady flow, the streamline pattern stays fixed and a moving fluid particle follows a streamline. At an ordinary point with nonzero velocity two distinct streamlines cannot cross: a crossing would assign two velocity directions to the same point at the same instant. The drawing is a map of a velocity field, not a set of physical strings inside the liquid.
Laminar motion has orderly layers that slide past one another. Adjacent layers can move at different speeds; equal speed everywhere is not required. At sufficiently large speeds for a given geometry and fluid, irregular eddies and fluctuations can develop: this is turbulent motion. River rapids around rocks are a familiar example. The critical speed depends on the fluid and the size and geometry of the flow; it is not one universal speed for all liquids and pipes.
Deriving continuity from conservation of mass
Imagine a tube whose side boundary is made from streamlines. Fluid does not cross that side boundary because velocity is tangent to it. In a short time Δt, fluid at a section of area A travels distance vΔt. The volume that crosses is AvΔt when velocity is approximately uniform over that section; for a nonuniform profile use the cross-sectional mean speed. Multiplying by density gives the transported mass ρAvΔt.
In steady flow no mass accumulates inside the tube segment. The mass entering during Δt must equal the mass leaving. Cancel the same time interval to obtain ρ₁A₁v₁ = ρ₂A₂v₂. If density is constant, it cancels too. The resulting volume flow rate Q = Av is then the same at every section of a single unbranched tube.
If a circular pipe’s diameter halves, its area becomes one quarter and its mean speed becomes four times as large for the same volume flow. Do not use a simple inverse-diameter rule. Closely spaced streamlines can show a small stream-tube area and higher speed when the drawn lines represent equal flow portions. Arbitrarily drawing extra lines does not physically increase velocity.
Problem
A spray-pump cylinder has cross-sectional area 8.0 cm². Liquid in it moves at 1.5 m/min and exits through 40 holes, each of diameter 1.0 mm. Find the exit speed assuming equal flow through all holes.
- 1.Convert the inlet speed: 1.5/60 = 0.025 m/s. Inlet area is 8.0 × 10⁻⁴ m², so Q = 2.0 × 10⁻⁵ m³/s.
- 2.Each hole has radius 0.50 mm = 5.0 × 10⁻⁴ m. Total outlet area is 40π(5.0 × 10⁻⁴)² = 3.14 × 10⁻⁵ m².
- 3.Continuity gives exit speed Q/A_total = 0.637 m/s, approximately 0.64 m/s. The outlet speed increases because the same volume per second crosses a smaller total area.
- 4.A real pump must supply the required pressure and overcome losses; continuity alone does not tell us that pressure.
Why a pressure difference is needed to change speed
Continuity determines how velocities at different sections are related. It does not by itself explain the force that produces the acceleration. In a horizontal ideal-fluid constriction, the surrounding fluid exerts a net pressure force that accelerates the flowing material. To relate that pressure change to speed and elevation, apply conservation of mechanical energy. This is Bernoulli’s principle.
The pressure-work derivation of Bernoulli’s equation
Choose two sections of a pipe along a streamline. Let their pressures be P₁ and P₂, their heights above a common datum be y₁ and y₂, and their speeds be v₁ and v₂. Assume steady incompressible flow with negligible viscous dissipation. During a short interval Δt, the same volume ΔV enters section 1 and leaves section 2. Most of the fluid between the sections is replaced by identical fluid in the same steady state. The net energy change can therefore be found by comparing the entering volume with the leaving volume.
At the inlet, the pressure force is P₁A₁ in the direction of motion. Its point of application moves v₁Δt, so work done on the fluid is P₁A₁v₁Δt = P₁ΔV. At the outlet, the downstream fluid pushes back against the flow. Work done on the chosen fluid there is −P₂A₂v₂Δt = −P₂ΔV. Thus the net pressure work is (P₁ − P₂)ΔV. The minus sign at the outlet is essential.
The transported mass is Δm = ρΔV. Its kinetic energy changes by one half this mass times v₂² − v₁². Its gravitational potential energy changes by the same mass times g(y₂ − y₁). Pressure work supplies both changes. Equivalently, if gravity is counted as work, W_pressure + W_gravity = ΔK; moving the gravity term to the other side gives W_pressure = ΔK + ΔU.
Divide by the common nonzero ΔV and rearrange so that the section-1 terms stand on one side and section-2 terms on the other. The result states that a specific sum stays constant along the streamline. It is conservation of energy per unit volume, not an additional conservation law unrelated to mechanics.
| Term | Meaning | Equivalent unit |
|---|---|---|
| P | Static pressure; pressure-work contribution per volume | Pa = J/m³ |
| ρv²/2 | Kinetic energy per volume | J/m³ |
| ρgy | Gravitational potential energy per volume relative to a datum | J/m³ |
Using the equation without turning it into a slogan
At the same elevation along a suitable streamline, higher speed corresponds to lower static pressure. If the heights differ, the ρgy change must also be included: “faster always means lower pressure” is not a universal rule. In static fluid v = 0, Bernoulli reduces to P₁ + ρgy₁ = P₂ + ρgy₂, which reproduces the hydrostatic pressure difference. This is a useful consistency check on signs.
Gauge pressures may replace absolute pressures if both are measured relative to the same atmospheric pressure: subtracting a common constant from both sides leaves the equation unchanged. If the reference atmospheric pressures are different, those different subtractions do not cancel. Keep absolute pressures or include the difference explicitly.
Viscous drag converts organised mechanical energy into internal energy. In a real pipe an extra pressure drop is needed to maintain flow; the downstream pressure is lower than a loss-free prediction for otherwise comparable conditions. Turbulent rapids have substantial dissipation and fluctuating velocities, so the elementary steady Bernoulli formula cannot be applied directly as a complete description. Low viscosity does not justify ignoring every loss in every geometry.
A vertical open tube attached to a horizontal pipe reads the local pressure through its stationary liquid height. With the same liquid in both tubes, a taller column means higher static pressure. At the constriction continuity predicts greater speed; Bernoulli then predicts lower pressure and a shorter column. A picture with a taller constriction column is incompatible with those assumptions. Decide the speed relation first, then the pressure relation; do not infer everything from Bernoulli alone.
Problem
Water moves at 2.0 m/s through a section of area 6.0 cm². A second section has area 2.0 cm² at the same elevation. Find the second speed and pressure change. Use ρ = 1000 kg/m³.
- 1.Continuity gives v₂ = (6.0/2.0) × 2.0 = 6.0 m/s.
- 2.At equal height, Bernoulli gives P₁ − P₂ = ρ(v₂² − v₁²)/2 = 1000(36 − 4)/2 = 16000 Pa.
- 3.The pressure at the narrow section is 16 kPa lower. This pressure difference supplies the kinetic-energy increase per unit volume. Without knowing P₁, neither absolute pressure is fixed.
Quick Quiz
Quiz
Which statement describes steady flow?
Steadiness concerns time variation at a fixed position. A particle may accelerate while moving to a place with a different velocity.
For steady incompressible flow, a pipe diameter decreases to one half. The mean speed becomes what?
The area becomes one quarter; Av constant requires four times the mean speed.
Which law directly gives the speed increase at a constriction?
Continuity is mass conservation. Bernoulli then relates the pressure and energy changes under its own assumptions.
Which quantity has the same units as pressure?
ρv² has units kg/(m·s²), equivalent to N/m² or J/m³.
Which pressure-tube pattern is consistent with ideal horizontal flow through a constriction?
Continuity gives faster flow at the constriction; Bernoulli gives lower static pressure and therefore a lower open-tube column.
Practice Problems
- A steady incompressible flow passes through pipe sections of areas 12 cm² and 3 cm². Its speed in the wider section is 0.50 m/s. Calculate volume flow rate and narrow-section speed, and explain whether an individual particle accelerates.
- A pipe divides into two outlets. Water enters with Q = 3.0 × 10⁻⁴ m³/s. One outlet of area 2.0 cm² has mean speed 0.50 m/s. Find the other outlet’s flow rate and its mean speed if its area is 4.0 cm².
- Derive Bernoulli’s equation from inlet and outlet pressure work. Explain the negative outlet-work sign, why the transported volumes are equal and what changes if energy is dissipated.
- In ideal steady water flow, point 2 is 2.0 m above point 1. Speeds are 3.0 and 5.0 m/s at points 1 and 2 respectively. Find P₁ − P₂ for ρ = 1000 kg/m³ and g = 9.8 m/s². Explain both contributions.
- Sketch a horizontal pipe narrowing at its centre, with open liquid-pressure tubes at wide and narrow sections. Show the correct tube levels. Explain why applying the same construction to turbulent river rapids requires more than the elementary Bernoulli equation.
Key Takeaways
• Steady flow can have spatially changing velocity; streamlines are tangent to the velocity field. • Continuity conserves mass; Av is constant only when density is constant in an unbranched steady flow. • Pressure work changes kinetic and gravitational energy, giving Bernoulli’s equation under stated assumptions. • Compare heights, pressure references and losses before using the faster-flow/lower-pressure result.