What is Bernoulli's Principle?
Bernoulli's principle is a fundamental concept in fluid mechanics stating that in a steady fluid flow, an increase in fluid velocity is accompanied by a decrease in pressure or potential energy. Mathematically expressed as: P + ½ρv² + ρgh = constant, which is a statement of the conservation of energy for flowing fluids. Kalkulab's Bernoulli's Law Calculator is designed to help high school students (grades 11-12), civil and mechanical engineering students, and piping professionals understand fluid dynamics. This tool covers four main calculations: (1) flow rate from pipe cross-section area and velocity, (2) continuity equation for pipe diameter changes, (3) Bernoulli's equation to find pressure or velocity at two points, and (4) Torricelli's theorem for tank discharge velocity.
Bernoulli's Equation & Continuity
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂Formula: Q = A × v ; A₁v₁ = A₂v₂ ; v = √(2gh) for TorricelliVariables:
- PPressure (Pascal/Pa)Fluid pressure at a point(e.g.: 100 kPa)💡 Water pressure in a pipe
- ρDensity (kg/m³)Fluid density(e.g.: 1000 kg/m³ (water))💡 Density of water vs oil
- vVelocity (m/s)Fluid flow velocity(e.g.: 2 m/s)💡 Water velocity in a hose
- gGravity (m/s²)Earth's gravitational acceleration(e.g.: 9.8 m/s²)💡 Earth's gravitational constant
- hHeight (meters)Height from reference point(e.g.: 10 m)💡 Water tank height
- QFlow Rate (m³/s)Volume of fluid per unit time(e.g.: 0.02 m³/s)💡 Household faucet flow rate
- ACross-Sectional Area (m²)Flow area(e.g.: 0.00785 m² (10cm pipe))💡 Water pipe area
Conservation of Energy for Fluids
Bernoulli's equation states that the sum of static pressure (P), kinetic energy per volume (½ρv²), and potential energy per volume (ρgh) is constant along a streamline in steady, incompressible, inviscid flow.
- 1Static Pressure (P): Energy per volume from fluid pressure
- 2Kinetic Energy (½ρv²): Energy per volume from fluid motion
- 3Potential Energy (ρgh): Energy per volume from height
- 4Continuity: A₁v₁ = A₂v₂ - Constant flow rate (Q = Av)
Categories:
How to Use the KalkuLab Bernoulli's Law Calculator
This calculator has 4 main modes. Follow the steps for your calculation:
- 1
Select Calculation Mode
Choose: (1) Flow rate (Q = A×v), (2) Continuity (A₁v₁ = A₂v₂), (3) Bernoulli pressure (P + ½ρv² + ρgh = constant), or (4) Torricelli (v = √(2gh)).
- 2
Enter Known Values
Enter cross-section area (A), velocity (v), pressure (P), density (ρ), and height (h). Use SI units (m, m/s, Pa, kg/m³).
- 3
Identify Points 1 and 2
For Bernoulli's equation, mark two points in the fluid system—e.g., wide pipe vs. narrow pipe. Assume steady, incompressible flow.
- 4
Click Calculate
Get instant results in m³/s or L/s for flow rate, Pa or kPa for pressure.
- 5
Analyze Results
Pipe constriction increases velocity but decreases pressure (Venturi effect). Torricelli speed depends only on height, not fluid density.
💡 Tip:
- •Water density = 1000 kg/m³, oil ≈ 800–900, air ≈ 1.2
- •1 m³/s = 1000 L/s; household faucet ≈ 0.1–0.2 L/s
- •Smaller area (A₂ < A₁) means higher velocity (v₂ > v₁)
- •On a horizontal pipe (h₁ = h₂): P₁ + ½ρv₁² = P₂ + ½ρv₂²
- •Venturi effect: constriction lowers pressure—used in carburetors and sprayers
Examples
Example 1: Household Faucet Flow Rate
Faucet diameter 2 cm, water speed 1.5 m/s. Find flow rate.
- 1.d = 0.02 m, v = 1.5 m/s
- 2.A = π × 0.01² = 0.000314 m²
- 3.Q = 0.000471 m³/s = 0.471 L/s
About 28 liters per minute flows from the faucet.
Example 2: Venturi Effect in a Pipe
Water at 2 m/s, 150 kPa narrows to 6 m/s on a horizontal pipe (ρ = 1000 kg/m³). Find pressure in the narrow section.
- 1.P₂ = P₁ + ½ρ(v₁² − v₂²)
- 2.P₂ = 150000 + 500×(4−36) = 134000 Pa
Pressure drops as velocity rises—the classic Venturi effect.
Example 3: Continuity in a Hose
Hose diameter shrinks from 2 cm to 0.5 cm. Speed in main hose 1 m/s. Speed at tip?
- 1.v₂ = (A₁/A₂) × v₁ = 16 m/s
Pinching the hose makes water exit 16× faster.
Example 4: Torricelli Tank Drain
Tank height 5 m above outlet. Exit speed? (g = 9.8 m/s²)
- 1.v = √(2 × 9.8 × 5) = √98 ≈ 9.9 m/s
Exit speed equals free-fall from 5 m— independent of fluid type.
Example 5: Pressure at Two Heights
Point 1: h=10 m, P=200 kPa, v=1 m/s. Point 2: h=2 m, v=3 m/s. Find P₂ (ρ=1000, g=9.8).
- 1.P₂ = P₁ + ½ρ(v₁²−v₂²) + ρg(h₁−h₂)
- 2.P₂ ≈ 274.4 kPa
Lower elevation increases pressure despite higher velocity.