Bernoulli as Energy Conservation
- Read each term in Bernoulli's equation as an energy per unit volume
- Combine continuity and Bernoulli to analyze flow through a changing pipe
- State the assumptions behind Bernoulli's equation and identify where they fail
Continuity first: what goes in must come out
For an incompressible fluid in a pipe with no leaks, the volume flow rate Av is the same everywhere. Narrow the pipe and the fluid must speed up, in exact inverse proportion to the area. Note that area goes as the square of the radius, so halving a pipe's radius quarters its area and quadruples the speed — a factor students routinely get wrong by a factor of two.
Bernoulli is the work–energy theorem for a fluid
Bernoulli's equation looks arbitrary until you notice what each term is. Take the familiar energy statement ½mv² + mgh = constant and divide through by volume. Since m/V = ρ, ½mv² becomes ½ρv² — kinetic energy per unit volume — and mgh becomes ρgh, the potential energy per unit volume. The pressure term P is the work per unit volume done by the surrounding fluid pushing the parcel along. Bernoulli's equation is conservation of energy per unit volume, and reading it that way makes the terms impossible to forget.
Faster flow means lower pressure
Hold the height constant and Bernoulli's equation reduces to P + ½ρv² = constant: where the fluid moves faster, its pressure is lower. This is the counterintuitive core of the unit. It explains why a shower curtain is pulled inward by fast-moving air, why two ships sailing in parallel are drawn together, and how an atomizer lifts liquid up a tube. The energy view removes the mystery — speeding up costs kinetic energy, and that energy has to come from the pressure term.
Water flows at 2.0 m/s through a horizontal pipe of radius 0.040 m, which narrows to radius 0.020 m. Find the speed in the narrow section and the pressure drop. (ρ = 1000 kg/m³)
- 1.Continuity: A ∝ r², and the radius halves, so the area drops to a quarter and the speed quadruples: v₂ = 4 × 2.0 = 8.0 m/s.
- 2.Horizontal pipe, so the ρgh terms are equal and cancel: P₁ + ½ρv₁² = P₂ + ½ρv₂².
- 3.Pressure drop: P₁ − P₂ = ½ρ(v₂² − v₁²) = ½(1000)(64 − 4).
- 4.Evaluate: ½(1000)(60) = 30 000 Pa.
Torricelli: a drained tank is a falling object
Apply Bernoulli to a tank draining from a small hole a depth h below the surface. Both the surface and the hole are open to the atmosphere, so the pressure terms cancel, and the surface drops slowly enough that its kinetic term is negligible. What remains is ½ρv² = ρgh, giving v = √(2gh) — precisely the speed an object reaches falling freely from height h. The fluid does not "know" it is a fluid; the same energy conversion is at work.
The assumptions, and where they break
Bernoulli's equation assumes flow that is steady (not changing with time), incompressible (constant density), non-viscous (no internal friction) and traced along a single streamline. Real flows violate all four to some degree. Viscosity in a long pipe causes a steady pressure drop that Bernoulli predicts nothing about, which is why water pressure falls at the far end of a long hose; and turbulence destroys the streamline picture entirely. Stating these limits is frequently its own free-response part.
Continuity and Bernoulli almost always appear together, and continuity must come first. Bernoulli's equation has two unknown speeds in it until continuity supplies the relationship between them.
Water flows through a pipe whose radius decreases by half. The flow speed in the narrow section is:
In a horizontal pipe, where the fluid moves fastest the pressure is:
Water exits a small hole 5.0 m below the surface of a large open tank. What is its exit speed? (g = 10 m/s²)
Check that every term in a Bernoulli equation comes out in pascals before you solve. P, ½ρv² and ρgh must all have the same units, and a term that does not is the fastest way to catch a substitution error under time pressure.
Answer the 3 checkpoints as you read.
Sign in to save your progress