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Bernoulli Equation Calculator

Calculate fluid flow parameters using Bernoulli's principle

Fluid Properties

Point 1

Point 2

Bernoulli's Equation

P1 + (1/2)*rho*v1² + rho*g*h1 = P2 + (1/2)*rho*v2² + rho*g*h2

This equation describes the conservation of energy in fluid flow.

What Is Bernoulli's Equation?

Bernoulli's equation is one of the most important principles in fluid dynamics, expressing conservation of energy along a streamline in steady, incompressible, inviscid flow. Published by Daniel Bernoulli in 1738, it states that the sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume remains constant: P + ½ρv² + ρgh = constant.

In practical terms, Bernoulli's principle explains why airplane wings generate lift (faster air over the curved top creates lower pressure), why a curveball curves (spin creates velocity differences), why a chimney draws better in wind (moving air at the top reduces pressure), and how a Venturi meter measures flow rate (pressure drops where the pipe narrows and velocity increases). Every term in the equation has units of pressure (Pa, or N/m²), representing energy density.

Bernoulli's Equation

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

Where:

  • P= Static pressure (Pa or N/m²)
  • ρ= Fluid density (kg/m³). Water = 1000, air ≈ 1.225
  • v= Flow velocity (m/s)
  • g= Gravitational acceleration (9.81 m/s² on Earth)
  • h= Height above a reference level (m)
  • ½ρv²= Dynamic pressure — kinetic energy per unit volume
  • ρgh= Hydrostatic pressure — potential energy per unit volume

Assumptions and Limitations

Bernoulli's equation is powerful but has specific conditions for validity:

AssumptionWhat It MeansWhen It Fails
Steady flowVelocity at each point doesn't change with timePulsating pumps, gusting wind
Incompressible flowDensity ρ is constantGas flows above Mach 0.3, compressible fluids
Inviscid flowNo friction/viscosity lossesLong pipes, highly viscous oils, boundary layers
Along a streamlineEquation is valid along a single flow pathComparing different streamlines requires the same constant

How to Use This Calculator

Solve for pressure, velocity, or height at a second point on a streamline:

  1. Select Target Variable: Choose what to solve for — Pressure at Point 2, Velocity at Point 2, or Height at Point 2. The corresponding input field becomes disabled (grayed out) since the calculator solves for it.
  2. Set Fluid Properties: Enter fluid density (1000 kg/m³ for water, 1.225 for air at sea level) and gravitational acceleration (9.81 m/s² for Earth).
  3. Enter Point 1 Values: Provide the known pressure (Pa), velocity (m/s), and height (m) at the upstream point on the streamline.
  4. Enter Point 2 Values: Fill in the two known quantities at Point 2. The calculator computes the third using Bernoulli's equation.
  5. Review Result: The computed value appears with its units. If solving for velocity, the calculator ensures the term under the square root is non-negative.

Real-World Applications

Bernoulli's principle explains aircraft lift: air travels faster over the curved upper surface of a wing than the flatter lower surface. The higher velocity reduces pressure above the wing (per Bernoulli), creating a net upward force. While the full explanation of lift also involves Newton's third law (deflecting air downward), Bernoulli's contribution is essential — together they form the complete aerodynamic picture used in wing design.

In civil engineering, Bernoulli's equation governs water supply systems. Municipal water towers use height to create pressure — the ρgh term at the elevated tank drives flow through distribution pipes. Engineers use Bernoulli to ensure sufficient pressure at every tap while accounting for velocity changes and elevation differences. The same principle explains why dams generate more power with greater head (height difference) — the potential energy term ρgh converts to kinetic energy in the turbine.

Worked Examples

Water Pipe Pressure Drop

Problem:

Water (ρ = 1000 kg/m³) flows through a horizontal pipe. At point 1: P₁ = 300 kPa, v₁ = 2 m/s. At point 2, the pipe narrows and velocity increases to v₂ = 8 m/s. Find P₂.

Solution Steps:

  1. 1Given: P₁ = 300,000 Pa, v₁ = 2 m/s, v₂ = 8 m/s, h₁ = h₂ (horizontal, so ρgh terms cancel)
  2. 2Rearrange Bernoulli: P₂ = P₁ + ½ρ(v₁² - v₂²)
  3. 3Substitute: P₂ = 300,000 + ½ × 1000 × (4 - 64) = 300,000 + 500 × (-60)
  4. 4P₂ = 300,000 - 30,000 = 270,000 Pa = 270 kPa
  5. 5The pressure drops by 30 kPa — this is the Venturi effect

Result:

Pressure at point 2 = 270 kPa. Faster flow means lower pressure — a 30 kPa drop for a 4× velocity increase. This Venturi principle is used in carburetors, aspirators, and flow measurement devices.

Water Tower Supply Pressure

Problem:

A water tower has its water level at h₁ = 40 m. A house tap is at h₂ = 2 m. At the tower outlet, v₁ ≈ 0 m/s (large tank) and P₁ = atmospheric (101.3 kPa). If water exits the tap at v₂ = 5 m/s, what is the pressure at the tap?

Solution Steps:

  1. 1Given: P₁ = 101,300 Pa, v₁ = 0, h₁ = 40 m, h₂ = 2 m, v₂ = 5 m/s, ρ = 1000, g = 9.81
  2. 2Apply Bernoulli: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
  3. 3Substitute: 101,300 + 0 + 1000×9.81×40 = P₂ + ½×1000×25 + 1000×9.81×2
  4. 4Compute left: 101,300 + 392,400 = 493,700 Pa
  5. 5Compute right terms: ½×1000×25 = 12,500; ρgh₂ = 19,620
  6. 6P₂ = 493,700 - 12,500 - 19,620 = 461,580 Pa = 461.6 kPa
  7. 7Gauge pressure at tap: 461.6 - 101.3 = 360.3 kPa ≈ 3.6 atmospheres

Result:

Tap pressure = 461.6 kPa absolute (360.3 kPa gauge, about 3.6 atm). That's excellent water pressure — typical household systems target 275-410 kPa.

Aircraft Wing Pressure Difference

Problem:

Air (ρ = 1.225 kg/m³) flows at 70 m/s over the top of a wing and 60 m/s along the bottom. Both surfaces are at the same height. What is the pressure difference generating lift?

Solution Steps:

  1. 1h₁ = h₂ → ρgh terms cancel
  2. 2P_bottom - P_top = ½ρ(v_top² - v_bottom²)
  3. 3Substitute: ΔP = ½ × 1.225 × (70² - 60²) = 0.6125 × (4900 - 3600)
  4. 4ΔP = 0.6125 × 1300 = 796.25 Pa ≈ 0.8 kPa
  5. 5For a wing area of 50 m²: Lift force ≈ 796 × 50 = 39,800 N ≈ 4,060 kg of lift

Result:

Pressure difference = 796 Pa (0.8 kPa). Over a 50 m² wing, this generates about 40 kN of lift — enough to support a 4-ton aircraft. Real wings use angle of attack and circulation for additional lift beyond this simple Bernoulli estimate.

Tips & Best Practices

  • ✓For horizontal flow (h₁ = h₂), the ρgh terms cancel — useful for Venturi tubes, pipe constrictions, and low-speed aerodynamics
  • ✓Always use consistent units: pressure in Pa (not kPa or atm), velocity in m/s, height in m, density in kg/m³
  • ✓The calculator solves for one unknown at Point 2 — make sure you provide the other two values at Point 2
  • ✓Bernoulli applies along a single streamline — comparing Point 1 on one streamline to Point 2 on a different one may give incorrect results
  • ✓For real (viscous) fluids, Bernoulli gives the ideal case — actual systems have additional frictional pressure losses

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Frequently Asked Questions

This is energy conservation: along a streamline, the sum P + ½ρv² + ρgh is constant. If height is unchanged, an increase in kinetic energy (½ρv²) must be balanced by a decrease in pressure energy (P). Physically, fluid particles accelerate because of a pressure gradient — they move from high to low pressure, gaining speed as they go. The Venturi effect is a direct demonstration: constrict a pipe to increase velocity, and the pressure drops.
Static pressure (P) is the pressure you feel if you move with the fluid — it's the thermodynamic pressure. Dynamic pressure (½ρv²) is the kinetic energy per unit volume, representing the pressure equivalent of the fluid's motion. The sum of static and dynamic pressure (ignoring height effects) is the stagnation pressure — the pressure at a point where the fluid is brought to rest isentropically. A Pitot tube on an aircraft measures stagnation pressure and uses the difference from static pressure (via Bernoulli) to compute airspeed.
Yes, with the important caveat that gases are compressible. Bernoulli's classical form (P + ½ρv² + ρgh = constant) assumes incompressible flow, which is a good approximation for air at speeds below about Mach 0.3 (~100 m/s or 360 km/h). Above that, density changes become significant and you need the compressible Bernoulli equation, which accounts for variable density. The calculator's results for air flows should be treated as approximate for high-speed applications.
Bernoulli's equation assumes no energy is added or removed from the fluid. Pumps add mechanical energy (increasing total head), and turbines extract it. To account for these, engineers modify the equation to include pump head (hp) and turbine head (ht): P₁/(ρg) + v₁²/(2g) + h₁ + hp - ht = P₂/(ρg) + v₂²/(2g) + h₂. This extended form is the basis of hydraulic engineering design.
Bernoulli's equation IS conservation of energy per unit volume for a fluid element. The pressure term P represents flow work (energy transferred by pressure forces), ½ρv² is kinetic energy density, and ρgh is gravitational potential energy density. When you multiply all terms by volume, you get the standard energy equation: PV + ½mv² + mgh = constant. Bernoulli derived this by applying Newton's laws to fluid elements, showing the mechanical equivalent of the First Law of Thermodynamics for inviscid flow.

Sources & References

Last updated: 2026-06-06

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Editorial Note

MyCalcBuddy Editorial Team

This page is maintained as an educational calculator reference.

Source

Formula Source: University Physics

by Young & Freedman

UpdatedLast reviewed: May 2026
CheckedFormula checks are based on standard references and internal QA review.