Bernoulli's Equation — P + ½ρv² + ρgh = constant

Bernoulli's equation explained: pressure, speed and height trade-offs in flowing fluids, key assumptions, and a worked pipe example.

What it means

Bernoulli's equation is energy conservation for flowing fluids, written per unit volume: pressure energy plus kinetic energy plus gravitational potential energy stays constant along a streamline. The famous consequence is the pressure–speed trade-off — where a fluid speeds up, its pressure drops. That single idea explains Venturi meters, carburetors, curveballs and (partially) wing lift. The equation fails where its assumptions fail: in viscous, turbulent or compressible flow, energy is lost to friction and the constant is no longer constant.

Variables

Assumptions

• Steady, incompressible, inviscid (frictionless) flow
• Flow along a single streamline
• No pumps or turbines adding or removing energy between the points

Derivation

Apply the work-energy theorem to a small fluid parcel moving along a streamline: pressure forces do work $P\Delta V$, gravity does work $-\rho g\Delta h\,\Delta V$, and the parcel's kinetic energy changes by $\tfrac{1}{2}\rho\Delta V\Delta(v^2)$. Dividing by $\Delta V$ gives Bernoulli's equation — conservation of energy per unit volume.

Worked example

Given:

• P₁ = 200 kPa
• v₁ = 1 m/s
• v₂ = 3 m/s
• ρ = 1000 kg/m³ (water)
• horizontal pipe, so h₁ = h₂

1. Write Bernoulli for horizontal flow: $P_1 + \tfrac{1}{2}\rho v_1^2 = P_2 + \tfrac{1}{2}\rho v_2^2$

2. Rearrange: $P_2 = P_1 + \tfrac{1}{2}\rho(v_1^2 - v_2^2)$

3. Substitute: $P_2 = 200{,}000 + 0.5 \times 1000 \times (1^2 - 3^2)$

4. Calculate: $P_2 = 200{,}000 + 500 \times (-8) = 200{,}000 - 4000$

5. Result: $P_2 = 196{,}000\ \mathrm{Pa} = 196\ \mathrm{kPa}$