Bernoulli Equation Calculator

Find fluid pressure using Bernoulli's equation, the energy-conservation law linking pressure, speed, and height along a streamline.

🚰 Bernoulli Equation Calculator
Pa
kg/m³
m/s
m
m/s
m
Pressure P₂
Pressure change (P₂−P₁)
Step-by-step working

🚰 What is the Bernoulli Equation Calculator?

This Bernoulli equation calculator finds the pressure at a second point along a streamline given conditions at a first point. Enter pressure, density, velocity, and height at point 1, plus velocity and height at point 2, and it returns the pressure at point 2.

P + ½ρv² + ρgh = constant expresses conservation of energy for an idealized, incompressible, frictionless fluid flow, linking pressure, speed, and height along a single streamline.

This calculator directly reproduces two classic results: the Venturi effect (pressure drops where a fluid speeds up) and Torricelli's law (draining speed from a tank), both verified numerically before this calculator was built.

This calculator is useful for fluid dynamics, aerodynamics, and mechanical/civil engineering students studying pressure-velocity relationships in flowing fluids.

📐 Formula

P₁ + ½ρv₁² + ρgh₁  =  P₂ + ½ρv₂² + ρgh₂
P = pressure, ρ = fluid density (constant)
v = flow velocity, h = height, g = 9.80665 m/s²
Example: Venturi constriction (P₁=200,000 Pa, v₁=2 m/s → v₂=8 m/s): P₂ = 170,000 Pa.

📖 How to Use This Calculator

Steps

1
Enter pressure, density, velocity, and height at point 1.
2
Enter velocity and height at point 2.
3
Read the pressure at point 2.

💡 Example Calculations

Example 1 - Venturi constriction

1
P₁=200,000 Pa, ρ=1000 kg/m³, v₁=2 m/s, h₁=0, v₂=8 m/s, h₂=0
2
P₂ = 170,000 Pa, a 30,000 Pa drop
3
The classic Venturi effect: faster flow, lower pressure
P₂ = 170,000 Pa
Try this example →

Example 2 - Tank draining (Torricelli's law check)

1
P₁=101,325 Pa (atmospheric), ρ=1000 kg/m³, v₁=0 (tank surface), h₁=5 m, v₂=9.90285 m/s (Torricelli speed), h₂=0
2
P₂ = 101,325 Pa, exactly atmospheric again
3
Confirms v=√(2gh) is the correct exit speed for an open tank
P₂ = 101,325 Pa
Try this example →

Example 3 - Airflow over a wing

1
P₁=101,325 Pa, ρ=1.225 kg/m³ (air), v₁=100 m/s (below wing), h₁=0, v₂=120 m/s (above wing), h₂=0
2
P₂ = 98,630 Pa, lower pressure above the wing
3
The pressure difference (2,695 Pa) contributes to lift
P₂ = 98,630 Pa
Try this example →

❓ Frequently Asked Questions

What is Bernoulli's equation?+
Bernoulli's equation states that for an idealized (incompressible, frictionless) fluid flowing along a streamline, the sum of pressure energy, kinetic energy, and gravitational potential energy per unit volume stays constant: P + ½ρv² + ρgh = constant. It is one of the most widely used relations in fluid dynamics.
What is the formula for Bernoulli's equation?+
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂, where P is pressure, ρ is fluid density, v is flow velocity, g is gravitational acceleration, and h is height, evaluated at two points along the same streamline.
What is the Venturi effect and how does this calculator show it?+
The Venturi effect is the pressure drop that occurs when a fluid speeds up passing through a constriction, such as a narrowed section of pipe. Increasing v₂ relative to v₁ (with height unchanged) directly lowers the computed P₂, exactly reproducing the Venturi effect from the equation itself.
What is Torricelli's law and how does it relate to Bernoulli's equation?+
Torricelli's law gives the exit speed of fluid draining from a hole in an open tank: v = √(2gh), where h is the height of fluid above the hole. It is a special case of Bernoulli's equation where both the tank surface and the exit hole are open to the same atmospheric pressure, so the pressure terms cancel.
What assumptions does Bernoulli's equation make?+
The classic form assumes the flow is incompressible (constant density), inviscid (no friction/viscous losses), steady, and evaluated along a single streamline with no external energy added (like a pump) or removed (like a turbine) between the two points.
Why does pressure drop when a fluid speeds up?+
Because total energy per unit volume must stay constant, an increase in kinetic energy (from higher velocity) must be balanced by a decrease in pressure energy, assuming height stays the same. This is the same principle that generates lift on an aircraft wing, where air moving faster over the curved upper surface has lower pressure than the slower air below.
Does Bernoulli's equation apply to gases as well as liquids?+
Yes, as long as the flow speed is well below the speed of sound (so compressibility effects are small, generally below about Mach 0.3), Bernoulli's equation works well for gases like air, which is exactly why it explains aircraft lift and Venturi-effect air flow devices.
What happens if I set v1 and v2 both to zero?+
With both velocities zero, the equation reduces to simple hydrostatic pressure: P₂ = P₁ + ρg(h₁−h₂), the familiar "pressure increases with depth" relationship for a fluid at rest.
Why is real-world Bernoulli behavior only approximate?+
Real fluids have viscosity (friction), which continuously converts some flow energy into heat, so real pressure and velocity relationships deviate from the ideal Bernoulli prediction, especially over long pipe runs or near solid boundaries. Engineers add empirical loss terms (like the Darcy-Weisbach friction factor) to account for this in practice.
Can this calculator be used for airflow around a wing?+
Yes as a simplified illustration: entering the air density, and different speeds above and below a wing (with height difference typically negligible), shows the pressure difference that contributes to lift, though real wing aerodynamics also involves additional effects like circulation and viscous boundary layers.

What is Bernoulli's equation?

Bernoulli's equation states that for an idealized (incompressible, frictionless) fluid flowing along a streamline, the sum of pressure energy, kinetic energy, and gravitational potential energy per unit volume stays constant: P + ½ρv² + ρgh = constant. It is one of the most widely used relations in fluid dynamics.

What is the formula for Bernoulli's equation?

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂, where P is pressure, ρ is fluid density, v is flow velocity, g is gravitational acceleration, and h is height, evaluated at two points along the same streamline.

What is the Venturi effect and how does this calculator show it?

The Venturi effect is the pressure drop that occurs when a fluid speeds up passing through a constriction, such as a narrowed section of pipe. Increasing v₂ relative to v₁ (with height unchanged) directly lowers the computed P₂, exactly reproducing the Venturi effect from the equation itself.

What is Torricelli's law and how does it relate to Bernoulli's equation?

Torricelli's law gives the exit speed of fluid draining from a hole in an open tank: v = √(2gh), where h is the height of fluid above the hole. It is a special case of Bernoulli's equation where both the tank surface and the exit hole are open to the same atmospheric pressure, so the pressure terms cancel.

What assumptions does Bernoulli's equation make?

The classic form assumes the flow is incompressible (constant density), inviscid (no friction/viscous losses), steady, and evaluated along a single streamline with no external energy added (like a pump) or removed (like a turbine) between the two points.

Why does pressure drop when a fluid speeds up?

Because total energy per unit volume must stay constant, an increase in kinetic energy (from higher velocity) must be balanced by a decrease in pressure energy, assuming height stays the same. This is the same principle that generates lift on an aircraft wing, where air moving faster over the curved upper surface has lower pressure than the slower air below.

Does Bernoulli's equation apply to gases as well as liquids?

Yes, as long as the flow speed is well below the speed of sound (so compressibility effects are small, generally below about Mach 0.3), Bernoulli's equation works well for gases like air, which is exactly why it explains aircraft lift and Venturi-effect air flow devices.

What happens if I set v1 and v2 both to zero?

With both velocities zero, the equation reduces to simple hydrostatic pressure: P₂ = P₁ + ρg(h₁−h₂), the familiar 'pressure increases with depth' relationship for a fluid at rest.

Why is real-world Bernoulli behavior only approximate?

Real fluids have viscosity (friction), which continuously converts some flow energy into heat, so real pressure and velocity relationships deviate from the ideal Bernoulli prediction, especially over long pipe runs or near solid boundaries. Engineers add empirical loss terms (like the Darcy-Weisbach friction factor) to account for this in practice.

Can this calculator be used for airflow around a wing?

Yes as a simplified illustration: entering the air density, and different speeds above and below a wing (with height difference typically negligible), shows the pressure difference that contributes to lift, though real wing aerodynamics also involves additional effects like circulation and viscous boundary layers.