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Bernoulli Equation Calculator

Calculate fluid velocity using Bernoulli's equation.

The Bernoulli Equation Calculator is a free science calculator. Calculate fluid velocity using Bernoulli's equation. Solve physics and science problems with exact formulas.
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What is the Bernoulli Equation?

The Bernoulli equation describes how pressure, velocity, and elevation relate in flowing fluids. It states that for an ideal fluid in steady flow, the sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume remains constant along a streamline. This principle explains why airplane wings generate lift, why shower curtains billow inward, and why water speeds up when you partially block a garden hose.

Picture water flowing through a horizontal pipe that narrows from 10 cm diameter to 5 cm. Continuity requires the same volume flow rate through both sections, so velocity must quadruple in the narrow section (area is 1/4). Bernoulli's equation tells us this velocity increase comes at the expense of pressure — the narrow section has lower pressure. This pressure difference is how Venturi meters measure flow rate and how carburetors mix fuel with air.

Daniel Bernoulli published this principle in 1738, connecting Newton's laws to fluid behavior. The equation applies to incompressible, inviscid (frictionless) flow along a streamline. While real fluids have viscosity and turbulence, Bernoulli's equation provides excellent approximations for water flow, low-speed aerodynamics, and many engineering applications where friction losses are small compared to pressure and velocity changes.

How it Works: Formulas Explained

The Bernoulli equation reads: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂. Each term has units of pressure (Pa = N/m²). P is static pressure, ½ρv² is dynamic pressure (kinetic energy per unit volume), and ρgh is hydrostatic pressure (potential energy per unit volume). The sum P + ½ρv² + ρgh is called total pressure or Bernoulli constant.

For horizontal flow where h₁ = h₂, the equation simplifies to P₁ + ½ρv₁² = P₂ + ½ρv₂². If velocity increases (v₂ > v₁), then pressure must decrease (P₂ < P₁) to keep the sum constant. This inverse relationship between speed and pressure seems counterintuitive — we expect fast-moving fluids to have higher pressure. But the pressure that matters is static pressure (pushing perpendicular to flow), not impact pressure.

The continuity equation A₁v₁ = A₂v₂ often works alongside Bernoulli's equation. For incompressible flow, volume flow rate is constant. A pipe narrowing from area A₁ to A₂ causes velocity to increase by the ratio A₁/A₂. Combine this with Bernoulli to find the pressure change. For a 2:1 area reduction, velocity doubles and pressure drops by ½ρ(v₂² - v₁²) = ½ρ(4v₁² - v₁²) = 1.5ρv₁².

Working through numbers: Water (ρ = 1000 kg/m³) flows horizontally at v₁ = 2 m/s with pressure P₁ = 200,000 Pa. The pipe narrows so v₂ = 6 m/s. Using P₁ + ½ρv₁² = P₂ + ½ρv₂²: 200,000 + 0.5×1000×4 = P₂ + 0.5×1000×36. So 200,000 + 2,000 = P₂ + 18,000. Therefore P₂ = 202,000 - 18,000 = 184,000 Pa. The pressure dropped 16,000 Pa (0.16 atm) as velocity tripled.

Step-by-Step Guide

  1. Identify two points along the streamline. Choose point 1 where you know most variables (pressure, velocity, elevation) and point 2 where you want to find an unknown. For a tank draining through a pipe, point 1 might be the water surface and point 2 the pipe exit. For an airplane wing, compare a point far upstream to a point on the wing surface.
  2. List known values at each point. Write down P, v, and h for both points. Mark the unknown you're solving for. Example: P₁ = 150 kPa, v₁ = 3 m/s, h₁ = 5 m; P₂ = 101 kPa (atmospheric), h₂ = 0 m, find v₂. Use consistent SI units: Pa for pressure, m/s for velocity, m for height, kg/m³ for density.
  3. Apply continuity if velocities are unknown. For pipe flow, A₁v₁ = A₂v₂. If you know one velocity and both areas, calculate the other. A 15 cm diameter pipe (A₁ = 0.0177 m²) feeding a 5 cm pipe (A₂ = 0.00196 m²) gives velocity ratio v₂/v₁ = A₁/A₂ = 9. If v₁ = 1 m/s, then v₂ = 9 m/s.
  4. Write the Bernoulli equation and simplify. P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂. Cancel any equal terms. If horizontal, h₁ = h₂ cancels. If both points are open to atmosphere, P₁ = P₂ cancels. If the tank is large, v₁ ≈ 0 (surface drops slowly). Each simplification reduces the algebra.
  5. Solve algebraically for the unknown. Rearrange to isolate your target variable. For velocity: v₂ = √[(2/ρ)(P₁ - P₂ + ½ρv₁² + ρg(h₁ - h₂))]. For pressure: P₂ = P₁ + ½ρ(v₁² - v₂²) + ρg(h₁ - h₂). Plug in numbers carefully, watching units and signs. Calculate step by step to catch errors.
  6. Check your answer for physical reasonableness. Velocity should increase where pipe narrows. Pressure should drop where velocity increases (horizontal flow). Pressure should increase with depth. If water flows from a tank, exit velocity v = √(2gh) — Torricelli's theorem. A 10 m head gives v = √(2×9.8×10) = 14 m/s, about 50 km/h.

Real-World Examples

Example 1: Airplane wing lift. Air flows over a wing with different path lengths. Upper surface: v₂ = 70 m/s. Lower surface: v₁ = 60 m/s. Air density ρ = 1.2 kg/m³. Pressure difference: ΔP = ½ρ(v₂² - v₁²) = 0.5 × 1.2 × (4900 - 3600) = 0.6 × 1300 = 780 Pa. A wing with area 30 m² experiences lift force F = ΔP × A = 780 × 30 = 23,400 N — enough to lift a 2,380 kg aircraft. Actual lift is higher due to angle of attack and circulation effects.

Example 2: Venturi flow meter. A pipe with diameter D₁ = 10 cm narrows to D₂ = 5 cm at the throat. Water flows through with measured pressure drop ΔP = P₁ - P₂ = 15,000 Pa. Area ratio A₁/A₂ = (10/5)² = 4, so v₂ = 4v₁. Bernoulli gives: ΔP = ½ρ(v₂² - v₁²) = ½ρ(16v₁² - v₁²) = 7.5ρv₁². Solving: v₁ = √(ΔP/7.5ρ) = √(15000/7500) = √2 = 1.41 m/s. Flow rate Q = A₁v₁ = 0.00785 × 1.41 = 0.011 m³/s = 11 L/s.

Example 3: Water tower supply. A water tower stands 50 m above a house. The tank is open to atmosphere (P₁ = 101 kPa), and the house faucet is also open (P₂ = 101 kPa). The tank is large, so v₁ ≈ 0. Bernoulli simplifies to ρgh₁ = ½ρv₂². Exit velocity v₂ = √(2gh₁) = √(2×9.8×50) = √980 = 31.3 m/s. This is theoretical maximum — friction in pipes reduces actual velocity to perhaps 10-15 m/s, but pressure at the faucet would be about 5 bar when closed.

Example 4: Spray bottle mechanism. Squeezing a spray bottle forces air across a tube at v₂ = 25 m/s. Still air in the bottle has v₁ ≈ 0. Pressure difference: ΔP = ½ρv₂² = 0.5 × 1.2 × 625 = 375 Pa. This slight vacuum (0.0037 atm) draws liquid up the tube. The fast air stream then atomizes the liquid into droplets. Same principle operates in perfume atomizers, paint sprayers, and carburetors.

Example 5: Fire hose nozzle. A fire hose with diameter 6 cm feeds a nozzle with 2 cm exit. Area ratio is 9:1, so exit velocity is 9 times hose velocity. If water enters the nozzle at v₁ = 5 m/s and P₁ = 800 kPa, find exit conditions. v₂ = 45 m/s. Bernoulli (horizontal): P₂ = P₁ + ½ρ(v₁² - v₂²) = 800,000 + 500(25 - 2025) = 800,000 - 1,000,000 = -200,000 Pa gauge. Negative gauge pressure means the jet is at atmospheric pressure (P₂ = 0 gauge) — the calculation shows all pressure converted to kinetic energy.

Common Mistakes to Avoid

Applying Bernoulli across streamlines instead of along one. Bernoulli's equation holds along a single streamline in steady, inviscid flow. Different streamlines can have different Bernoulli constants. In a curved flow (like water around a bend), pressure varies across streamlines due to centripetal acceleration. Always compare points on the same streamline, or verify the flow is irrotational where Bernoulli applies everywhere.

Forgetting that Bernoulli assumes incompressible flow. The standard Bernoulli equation assumes constant density, valid for liquids and gases below about Mach 0.3 (100 m/s for air). At higher speeds, compressibility matters — density changes as pressure changes. Compressible flow requires the energy equation with enthalpy. For air at 50 m/s, the incompressible assumption introduces only ~2% error.

Ignoring friction losses in long pipes. Real fluids have viscosity, causing friction that dissipates energy as heat. The extended Bernoulli equation includes a head loss term: P₁/ρg + v₁²/2g + h₁ = P₂/ρg + v₂²/2g + h₂ + h_loss. For short, smooth pipes at moderate velocity, h_loss is negligible. For long pipelines or high velocities, use the Darcy-Weisbach equation to estimate losses.

Confusing static, dynamic, and total pressure. Static pressure (P) is what you measure with a pressure tap perpendicular to flow. Dynamic pressure (½ρv²) represents kinetic energy. Total pressure (P + ½ρv²) is what a Pitot tube measures facing into the flow. Bernoulli says total pressure is constant (no losses). A common error is measuring static pressure and treating it as total pressure, or vice versa.

Pro Tips

Use Torricelli's theorem for tank draining. When a tank drains through a small hole, exit velocity is v = √(2gh), where h is the height of water above the hole. This is Bernoulli with P₁ = P₂ (both atmospheric) and v₁ ≈ 0. A 2-meter deep tank drains at v = √(2×9.8×2) = 6.26 m/s. The jet trajectory follows projectile motion — horizontal distance x = v√(2H/g) where H is height from hole to ground.

Recognize when elevation terms dominate. In tall buildings, the ρgh term often overwhelms velocity effects. Water pressure increases by about 10 kPa per meter of depth. A 100 m skyscraper has about 1000 kPa (10 bar) pressure difference between top and bottom floors. Velocity head ½ρv² for typical pipe velocities (1-3 m/s) is only 0.5-4.5 kPa — negligible for vertical pressure calculations.

Combine with momentum equation for forces. Bernoulli gives pressures and velocities, but not forces on objects. To find the force of water on a pipe bend or turbine blade, use the momentum equation: F = ṁ(v_out - v_in), where ṁ is mass flow rate. Combine Bernoulli (for velocities) with momentum (for forces) to analyze pumps, turbines, and pipe fittings.

Apply the stagnation pressure concept. When fluid decelerates to zero velocity (stagnates), all dynamic pressure converts to static pressure. A Pitot tube facing into airflow measures stagnation pressure P₀ = P + ½ρv². Subtract static pressure P to find dynamic pressure, then solve for velocity: v = √[2(P₀ - P)/ρ]. This is how aircraft airspeed indicators work.

FAQs

Energy conservation requires it. The Bernoulli equation is essentially conservation of energy for fluids. When a fluid accelerates, kinetic energy increases. Since total energy is fixed (no external work), this energy must come from somewhere — it comes from pressure energy. Think of pressure as stored energy that can convert to motion. Higher speed means more kinetic energy, leaving less as pressure.

Yes, for low-speed gas flow where density changes are negligible (below Mach 0.3, about 100 m/s for air). Use the gas density at your conditions. For compressible high-speed flow, you need the compressible Bernoulli equation or full energy equation with enthalpy. At Mach 0.1, incompressible Bernoulli is accurate within 0.25%. At Mach 0.5, error reaches about 6%.

Fast-moving water from the showerhead drags air with it, creating airflow inside the shower. This moving air has lower static pressure than still air outside (Bernoulli effect). The pressure difference pushes the curtain inward. The effect is stronger with hot water because rising warm air creates additional convection currents. Heavier curtains resist this pressure-driven motion better than lightweight ones.

Partially. A spinning ball drags air around it, creating different velocities on opposite sides. The side spinning into the airflow has higher relative velocity and lower pressure; the other side has lower velocity and higher pressure. This pressure difference creates a lateral force (Magnus effect). Bernoulli explains the pressure difference; the complete picture requires boundary layer analysis and wake dynamics.

Written and reviewed by the CalcToWork editorial team. Last updated: 2026-04-29.

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Frequently Asked Questions

Energy conservation requires it. The Bernoulli equation is essentially conservation of energy for fluids. When a fluid accelerates, kinetic energy increases. Since total energy is fixed (no external work), this energy must come from somewhere — it comes from pressure energy. Think of pressure as stored energy that can convert to motion. Higher speed means more kinetic energy, leaving less as pressure.
Yes, for low-speed gas flow where density changes are negligible (below Mach 0.3, about 100 m/s for air). Use the gas density at your conditions. For compressible high-speed flow, you need the compressible Bernoulli equation or full energy equation with enthalpy. At Mach 0.1, incompressible Bernoulli is accurate within 0.25%. At Mach 0.5, error reaches about 6%.
Fast-moving water from the showerhead drags air with it, creating airflow inside the shower. This moving air has lower static pressure than still air outside (Bernoulli effect). The pressure difference pushes the curtain inward. The effect is stronger with hot water because rising warm air creates additional convection currents. Heavier curtains resist this pressure-driven motion better than lightweight ones.
Partially. A spinning ball drags air around it, creating different velocities on opposite sides. The side spinning into the airflow has higher relative velocity and lower pressure; the other side has lower velocity and higher pressure. This pressure difference creates a lateral force (Magnus effect). Bernoulli explains the pressure difference; the complete picture requires boundary layer analysis and wake dynamics.