Bernoulli Equation Calculator
Solve Bernoulli's equation for pressure, velocity, or height between two points in a fluid flow.
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.
What is dynamic pressure?
Dynamic pressure (q) is the kinetic energy per unit volume of a flowing fluid: q = ½ρv². It represents the pressure increase that would occur if the fluid were brought to rest (stagnation). Total pressure = Static pressure + Dynamic pressure (for horizontal flow). Dynamic pressure is used in: aerodynamics (lift and drag calculations), pitot tube measurements (airspeed indicators), wind load calculations on structures, and flow measurement devices. For air at sea level (ρ = 1.225 kg/m³) at 100 km/h (27.8 m/s), dynamic pressure = 0.5 × 1.225 × 27.8² = 473 Pa.
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).
What is the head form of Bernoulli's equation and when is it preferred?
Dividing every term by ρg converts the equation to metres of fluid: P/(ρg) + v²/(2g) + h = total head. These are the pressure head, velocity head and elevation head. Hydraulic engineers prefer this form because pump curves, gauge readings and pipe elevations are already quoted in metres. For water at 200 kPa the pressure head is 200000 / (998 × 9.81) ≈ 20.4 m, and at 8 m/s the velocity head is 64 / 19.62 ≈ 3.26 m.
How does a pitot tube use Bernoulli's equation to measure velocity?
A pitot tube faces the flow so the fluid stagnates at its mouth, recording total (stagnation) pressure, while a static port reads static pressure. Bernoulli for horizontal flow gives P_total − P_static = ½ρv², so v = √(2ΔP/ρ). In air at 1.225 kg/m³ a differential of 473 Pa gives v = √(2 × 473 / 1.225) = √772 ≈ 27.8 m/s, about 100 km/h. Airspeed indicators are simply this equation calibrated in speed units.
How do you extend Bernoulli's equation for pumps, turbines and pipe friction?
The extended energy equation adds shaft work and losses: total head at point 1 plus pump head equals total head at point 2 plus turbine head plus head loss h_L. Friction loss in a straight pipe comes from Darcy-Weisbach, h_f = f × (L/D) × v²/(2g), with the friction factor f set by Reynolds number and relative roughness; bends and valves add minor losses as K × v²/(2g). Ignoring h_L is the usual reason a hand-calculated delivery pressure exceeds the measured one.
Why can the low pressure at a Venturi throat cause cavitation?
Bernoulli says the throat trades pressure energy for kinetic energy, so a fast enough constriction can drive the local static pressure down toward the liquid's vapour pressure, roughly 2.3 kPa absolute for water at 20 °C. Vapour bubbles then form and collapse violently downstream where pressure recovers, eroding metal and generating noise. The same mechanism limits pump suction conditions, which is why net positive suction head is specified for pump inlets.
Does Bernoulli's equation fully explain how a wing generates lift?
Only partly. It correctly links the faster flow over the curved upper surface to reduced pressure there, but the popular equal-transit-time claim that both air parcels must meet at the trailing edge is wrong; upper-surface air actually arrives sooner. Lift is better described by circulation and by the wing turning a mass of air downward, with Bernoulli describing the resulting pressure field rather than causing it. Flat plates and inverted wings also generate lift at angle of attack.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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