Bernoulli’s Equation
- State Bernoulli’s principle relating speed and pressure
- Apply Bernoulli’s equation along a streamline
- Explain everyday lift effects with the speed–pressure trade-off
Faster flow, lower pressure
Bernoulli’s equation is energy conservation for a flowing fluid: along a streamline, P + ½ρv² + ρgh stays constant. The key consequence at a fixed height is that where a fluid moves faster, its pressure is lower, and vice versa. The fluid’s energy is shared between pressure, motion, and height, so boosting one term must lower another.
Lift, roofs, and curveballs
This speed–pressure trade-off explains a lot of everyday physics. Air flowing faster over the top of an airplane wing leaves lower pressure above than below, producing upward lift. Wind racing over a roof lowers the pressure above it, and the higher still-air pressure below can lift the roof off. A spinning ball drags air faster on one side, creating a sideways pressure difference that curves its path.
Water (ρ = 1000 kg/m³) flows through a horizontal pipe at 2 m/s in a wide part and 6 m/s in a narrow part. Find the pressure difference between the two sections.
- 1.Horizontal flow, so the ρgh terms cancel: P₁ + ½ρv₁² = P₂ + ½ρv₂².
- 2.The pressure drop is P₁ − P₂ = ½ρ(v₂² − v₁²).
- 3.Compute the speeds squared: 6² − 2² = 36 − 4 = 32.
- 4.So ΔP = ½ × 1000 × 32 = 500 × 32 = 16000 Pa.
Do not assume "faster fluid pushes harder." It is the opposite — faster flow means lower pressure. The intuition that speed equals force fails for fluids and is a classic trap.
Along a horizontal streamline, where a fluid flows faster, its pressure is:
Wind blows rapidly across the top of a flat roof while the air inside the house stays still. The pressure difference tends to:
Solve fluid-flow free-response in two steps: continuity (A₁v₁ = A₂v₂) gives the speeds, then Bernoulli (P + ½ρv² + ρgh = constant) gives the pressures. For horizontal pipes, drop the ρgh terms to simplify.
Answer the 2 checkpoints as you read.
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