AJ Designer

Bernoulli Theorem Calculator

P1 equals rho g times (P2 over rho g plus V2 squared minus V1 squared over 2g plus Z2 minus Z1 plus h)

Solution

—
Share:

Bernoulli’s Equation

Bernoulli’s equation balances pressure energy, kinetic energy, and potential energy between two points along a streamline in a steady, incompressible flow. The head loss term captures real-world friction and turbulence losses.

P₁ + ½ρV₁² + ρgZ₁ = P₂ + ½ρV₂² + ρgZ₂ + ρgh

How It Works

Bernoulli’s equation balances pressure energy, kinetic energy, and potential energy between two points along a streamline in a steady, incompressible flow. When one form of energy increases, the others decrease to keep the total constant. The head loss term captures real-world friction and turbulence losses that the ideal equation ignores.

Example Problem

Water (density 1,000 kg/m³) flows through a horizontal pipe. At point 1 the pressure is 200,000 Pa and velocity is 2 m/s. At point 2 the velocity is 4 m/s. Both elevations are equal and head loss is zero. What is the pressure at point 2?

  1. Identify the governing equation: P₁ + ½ρV₁² + ρgZ₁ = P₂ + ½ρV₂² + ρgZ₂ + ρgh — total energy per unit volume is conserved along a streamline.
  2. Drop irrelevant terms: the pipe is horizontal (Z₁ = Z₂, cancels) and frictionless (h = 0), leaving P₁ + ½ρV₁² = P₂ + ½ρV₂².
  3. Rearrange to isolate P₂: P₂ = P₁ + ½ρ(V₁² − V₂²).
  4. Substitute numbers: P₂ = 200,000 + 0.5 × 1,000 × (2² − 4²) = 200,000 + 500 × (4 − 16).
  5. Simplify the kinetic term: 500 × (−12) = −6,000 Pa — velocity rose, so pressure must fall by 6,000 Pa.
  6. Report the final pressure: P₂ = 200,000 − 6,000 = 194,000 Pa (about 193.5 kPa, a 3% drop).

When to Use Each Variable

  • Solve for Pressure (P1) — when you know the velocity, elevation, and downstream conditions and need the upstream pressure, e.g., determining pump discharge pressure in a piping system.
  • Solve for Velocity (V1) — when you know the pressures, elevations, and downstream velocity and need to find the upstream velocity, e.g., sizing a pipe to achieve a target flow rate.
  • Solve for Elevation (Z1) — when you know the pressures and velocities and need to find the required height difference, e.g., determining siphon peak height limitations.
  • Solve for Head Loss (h) — when you know conditions at both points and need to quantify friction and turbulence losses, e.g., evaluating pipe network efficiency.

Key Concepts

Bernoulli's equation is an energy conservation statement for fluid flow: the sum of pressure energy, kinetic energy, and potential energy remains constant along a streamline in ideal flow. The head loss term extends the equation to real fluids where friction and turbulence dissipate energy. Density must remain constant (incompressible flow), which is valid for most liquid flows and low-speed gas flows below Mach 0.3.

Applications

  • Pipe system design: calculating pressure drops and required pump head in water distribution networks
  • Venturi meters: measuring flow rate by relating pressure difference to velocity change in a constriction
  • Aircraft aerodynamics: explaining lift by relating airspeed differences above and below a wing to pressure differences
  • Hydraulic engineering: analyzing flow over weirs, through siphons, and in open channels
  • Medical devices: understanding blood flow velocity changes through arterial stenosis using pressure measurements

Common Mistakes

  • Forgetting to use consistent units — mixing psi with m/s or feet with pascals produces nonsensical results; convert everything to SI or a single consistent system first
  • Applying Bernoulli's equation to compressible high-speed gas flow — above Mach 0.3, density changes significantly and the incompressible assumption breaks down
  • Setting head loss to zero in real pipe systems — ignoring friction losses leads to undersized pumps and overestimated flow rates; always account for pipe friction and fitting losses
  • Comparing points that are not on the same streamline — Bernoulli's equation only holds along a single streamline in the absence of irrotational flow assumptions

Frequently Asked Questions

What does Bernoulli's principle explain in everyday life?

Bernoulli's principle — faster-moving fluid has lower pressure — explains why airplane wings generate lift, why shower curtains get sucked in when the water runs, why a spinning soccer ball curves (Magnus effect), why two parked semi-trucks pull toward each other as a third rushes past, and why the top of a tornado has dramatically lower pressure than the surrounding air. Anywhere fluid speeds up or slows down, a pressure change comes along for the ride.

Why does water speed up when a pipe narrows?

Because mass has to be conserved. The continuity equation A₁V₁ = A₂V₂ says that for an incompressible fluid, the volume per second flowing through any cross-section is the same. When the pipe area halves, the velocity must double to carry the same flow rate. Bernoulli's equation then predicts the pressure drop that accompanies this speed-up: pressure energy is literally converting into kinetic energy.

What is Bernoulli's equation used for?

It relates pressure, velocity, and elevation at two points in a fluid flow. Engineers use it to size pipes and pumps, design Venturi and orifice flow meters, predict pressure drops through nozzles and valves, analyze siphons and open-channel flow, and as a first-order model for airfoil lift. Coupled with continuity and head-loss correlations, it is the workhorse equation for most of civil and mechanical fluid engineering.

Does Bernoulli's equation work for gases?

It works for low-speed gas flows where density changes are negligible — typically below Mach 0.3 (about 100 m/s in air). Above that, density varies significantly with pressure and the incompressible assumption breaks down; you need the compressible energy equation with enthalpy terms instead. For HVAC, low-speed wind engineering, and subsonic aircraft performance at low speeds, Bernoulli's equation is accurate enough.

How much head loss is typical in a pipe system?

Head loss varies widely. A 100 m run of 50 mm steel pipe carrying water at 2 m/s may lose 5–15 m of head depending on roughness and fittings. Rules of thumb: 1–5 m per 100 m for smooth plastic, 5–15 m per 100 m for old steel, plus 0.5–3 m per fitting (elbows, tees, valves). Use the Darcy-Weisbach equation with a Moody-chart friction factor for precise estimates.

What assumptions does Bernoulli's equation make?

The equation assumes (1) steady flow — velocities and pressures don't change in time, (2) incompressible flow — density is constant, (3) inviscid flow — no friction (the head-loss term patches this), (4) flow along a streamline — both points are on the same streamline, and (5) no energy added or removed (no pumps, turbines, or heat exchange) between the two points. For pumps/turbines, add an energy term explicitly.

What is the difference between static, dynamic, and total pressure?

Static pressure (P) is what a pressure gauge flush to the flow reads — the pressure the fluid exerts on its surroundings. Dynamic pressure (½ρV²) is the kinetic energy per unit volume — what a pressure gauge would read if you stagnated the flow. Total (or stagnation) pressure (P + ½ρV²) is their sum and remains constant along a streamline in frictionless flow. Pitot tubes exploit this: the difference between total and static pressure gives the flow velocity.

Bernoulli's Equation

Bernoulli's equation is a statement of conservation of energy along a streamline in a steady, incompressible, inviscid flow. With an extra head-loss term to handle real-world friction, it takes the form:

P₁ + ½ρV₁² + ρgZ₁ = P₂ + ½ρV₂² + ρgZ₂ + ρgh

Where:

  • P — static pressure in pascals (Pa), at points 1 and 2 along the streamline
  • ρ — fluid density in kg/m³ (water ≈ 1,000; air ≈ 1.225 at sea level)
  • V — flow velocity in m/s, at points 1 and 2
  • g — gravitational acceleration, 9.81 m/s² on Earth's surface
  • Z — elevation head in meters above a reference datum
  • h — head loss in meters due to friction, fittings, and turbulence between points 1 and 2

Each term has units of pressure (Pa = J/m³), so Bernoulli's equation is literally an energy balance: static pressure energy plus kinetic energy (½ρV²) plus gravitational potential energy (ρgZ) stays constant along a streamline, minus any energy dissipated as head loss.

Worked Examples

Aviation — Wing Lift via Pressure Difference

How much does the static pressure drop above a wing where air flows 20% faster than the free stream?

A Cessna wing cruises at V₁ = 60 m/s free-stream. Over the upper surface, the air accelerates to V₂ = 72 m/s. Air density is 1.225 kg/m³. Both points are at the same altitude (no Z term) and friction losses are negligible. Solve for P₂ given P₁ = 101,325 Pa.

  • P₂ = P₁ + ½ρ(V₁² − V₂²)
  • P₂ = 101,325 + 0.5 × 1.225 × (60² − 72²)
  • P₂ = 101,325 + 0.6125 × (3,600 − 5,184)
  • P₂ ≈ 100,355 Pa — a 970 Pa drop

That 970 Pa differential acts over every square meter of wing, so a 16 m² wing sees roughly 15,500 N of total lift from this one mechanism — the essence of Bernoulli-based airfoil theory.

Plumbing — Garden Hose Nozzle Pressure

What exit velocity does a squeezed hose nozzle produce at 300 kPa supply pressure?

A garden hose supplies water at P₁ = 300,000 Pa and V₁ = 1 m/s. At the nozzle exit the water discharges to atmosphere (P₂ = 101,325 Pa) at the same elevation (Z₁ = Z₂) with negligible head loss. ρ = 1,000 kg/m³. Solve for V₂.

  • V₂ = √[ 2(P₁ − P₂)/ρ + V₁² ]
  • V₂ = √[ 2 × (300,000 − 101,325) / 1,000 + 1² ]
  • V₂ = √[ 397.35 + 1 ] = √398.35
  • V₂ ≈ 19.96 m/s (≈ 45 mph)

Real hoses lose 20–40% of this theoretical velocity to friction and turbulence through the narrowing — include a head-loss term of 5–10 m of water to match field measurements.

Meteorology — Wind Funnelling Between Buildings

How much does pressure drop in a wind tunnel between two skyscrapers?

Wind approaches at 8 m/s through an open plaza (P₁ = 101,325 Pa). It accelerates to 20 m/s between two adjacent skyscrapers. Both points are at the same elevation. Air density is 1.225 kg/m³. Solve for the pressure between the buildings.

  • P₂ = P₁ + ½ρ(V₁² − V₂²)
  • P₂ = 101,325 + 0.5 × 1.225 × (64 − 400)
  • P₂ = 101,325 + 0.6125 × (−336)
  • P₂ ≈ 101,119 Pa — a 206 Pa drop

That 206 Pa is enough to pull doors open, whistle through gaps, and lift loose paper — the “urban canyon effect” that makes certain downtown corners notoriously gusty. City planners use CFD with Bernoulli's equation as a sanity check during design.

Related Calculators

Related Sites