Bernoulli Equation Calculator
Compute bernoulli equation using validated scientific equations. See step-by-step derivations, unit analysis, and reference values.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator
Bernoulli Equation Calculator
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Formula: P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
Worked example — P₂ = 170.1 kPa | ΔP = 29.9 kPa | v₂ = 8 m/s
Formula
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
Bernoulli's equation states that the total mechanical energy per unit volume is constant along a streamline in steady, incompressible, inviscid flow. The three terms represent static pressure energy, kinetic (dynamic) energy, and gravitational potential energy. Given five of the six variables (P, v, h at two points), the sixth can be calculated.
Worked Examples
Example 1: Venturi Flow Meter
Problem:Water (ρ = 998 kg/m³) flows at 2 m/s in a 100mm pipe that narrows to 50mm. If inlet pressure is 200 kPa, find the pressure at the throat.
Solution:By continuity: A₁v₁ = A₂v₂ v₂ = v₁ × (D₁/D₂)² = 2 × (100/50)² = 8 m/s Bernoulli: P₂ = P₁ + ½ρ(v₁² - v₂²) P₂ = 200000 + 0.5 × 998 × (4 - 64) P₂ = 200000 - 29940 = 170,060 Pa = 170.1 kPa Pressure drop = 29.9 kPa
Result:P₂ = 170.1 kPa | ΔP = 29.9 kPa | v₂ = 8 m/s
Example 2: Water Tank Drainage (Torricelli)
Problem:A water tank has an opening 3m below the water surface. Both the surface and opening are at atmospheric pressure. Find the exit velocity.
Solution:P₁ = P₂ = atmospheric (cancel out) v₁ ≈ 0 (large tank surface) h₁ = 3m, h₂ = 0m Bernoulli: ρgh₁ = ½ρv₂² v₂ = √(2gh) = √(2 × 9.81 × 3) = 7.67 m/s This is Torricelli's theorem
Result:v₂ = 7.67 m/s | Equivalent to free fall from 3m
Frequently Asked Questions
What is Bernoulli's equation?
Bernoulli's equation is a fundamental principle in fluid dynamics that describes the conservation of energy along a streamline in a steady, incompressible, inviscid flow. It states: P + ½ρv² + ρgh = constant, where P is static pressure (Pa), ρ is fluid density (kg/m³), v is flow velocity (m/s), g is gravitational acceleration (9.81 m/s²), and h is elevation (m). The three terms represent pressure energy, kinetic energy, and potential energy per unit volume. The equation shows that as fluid velocity increases, pressure decreases (and vice versa), which explains airplane lift, Venturi effect, and many other phenomena.
What are the assumptions of Bernoulli's equation?
Bernoulli's equation requires several assumptions: (1) Steady flow — flow properties don't change with time at any point. (2) Incompressible fluid — density remains constant (valid for liquids and low-speed gases below Mach 0.3). (3) Inviscid flow — no friction losses (real flows have losses, accounted for by adding a head loss term). (4) Flow along a streamline — the equation connects two points on the same streamline. (5) No energy addition or removal — no pumps, fans, or turbines between the points. When these assumptions don't hold, modified forms (with loss terms, compressibility corrections, or shaft work) are used.
How does Bernoulli's equation explain the Venturi effect?
The Venturi effect occurs when fluid flows through a constriction (narrowing). By the continuity equation (A₁v₁ = A₂v₂), velocity must increase in the narrow section. Bernoulli's equation then dictates that pressure must decrease to conserve energy: P₁ + ½ρv₁² = P₂ + ½ρv₂². Since v₂ > v₁, then P₂ < P₁. This pressure drop is used in: Venturi flow meters, carburetors (drawing fuel into airflow), aspirators and ejectors, atomizers and spray nozzles. The Venturi effect is essentially the conversion of pressure energy to kinetic energy.
How does Bernoulli's equation relate to Torricelli's theorem?
Torricelli's theorem is a special case of Bernoulli's equation for liquid draining from a tank through an orifice. Applying Bernoulli between the liquid surface (v₁ ≈ 0, P₁ = P_atm) and the orifice (P₂ = P_atm): ρgh = ½ρv₂², giving v₂ = √(2gh). The exit velocity equals the velocity of an object in free fall from height h. For example, water in a tank with the hole 2m below the surface: v = √(2 × 9.81 × 2) = 6.26 m/s. This assumes an ideal fluid; real discharge is modified by a coefficient of discharge (typically 0.6-0.65 for a sharp-edged orifice).
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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