Bernoulli Equation Calculator

Compute pressure, velocity, or elevation between two points in an incompressible fluid using Bernoulli’s equation. Includes optional head loss and full energy-head breakdown.

Fluid mechanics Engineering Step-by-step

Bernoulli’s Equation Solver

Point 1
Point 2

Energy loss due to friction, fittings, etc.

Bernoulli’s equation – definition

Bernoulli’s equation is an energy balance for a flowing fluid. For steady, incompressible, inviscid flow along a streamline, the sum of pressure head, velocity head, and elevation head is constant (plus any pump/turbine or head loss terms).

General form (between points 1 and 2 with head loss):

\[ \frac{p_1}{\rho g} + \frac{v_1^2}{2g} + z_1 = \frac{p_2}{\rho g} + \frac{v_2^2}{2g} + z_2 + h_L \]

  • \(p\) – pressure [Pa, kPa, bar, psi]
  • \(\rho\) – density [kg/m³, slug/ft³]
  • \(v\) – flow speed [m/s, ft/s]
  • \(z\) – elevation above reference [m, ft]
  • \(g\) – gravitational acceleration [m/s², ft/s²]
  • \(h_L\) – head loss between 1 and 2 [m, ft]

Rearranged forms used by the calculator

Solve for pressure at point 2 (p₂)

Starting from the Bernoulli equation and solving for \(p_2\):

\[ p_2 = p_1 + \rho\left( \frac{v_1^2 - v_2^2}{2} + g(z_1 - z_2) \right) - \rho g h_L \]

Solve for velocity at point 2 (v₂)

Rearranging for \(v_2\):

\[ \frac{v_2^2}{2g} = \frac{p_1 - p_2}{\rho g} + \frac{v_1^2}{2g} + z_1 - z_2 - h_L \]

\[ v_2 = \sqrt{ 2g\left[ \frac{p_1 - p_2}{\rho g} + \frac{v_1^2}{2g} + z_1 - z_2 - h_L \right] } \]

Solve for elevation at point 2 (z₂)

\[ z_2 = \frac{p_1 - p_2}{\rho g} + \frac{v_1^2 - v_2^2}{2g} + z_1 - h_L \]

Worked example

Water flows from a large tank (point 1) through a pipe to an outlet (point 2) 5 m higher. Assume:

  • \(\rho = 1000\ \text{kg/m}^3\)
  • \(g = 9.81\ \text{m/s}^2\)
  • \(p_1 = p_2 =\) atmospheric (gauge = 0)
  • \(v_1 \approx 0\) (large tank, negligible velocity)
  • \(z_1 = 0\ \text{m}\), \(z_2 = 5\ \text{m}\)
  • Head loss \(h_L = 1\ \text{m}\)

Using the velocity form:

\[ \frac{v_2^2}{2g} = 0 + 0 + (0 - 5) - 1 = -6\ \text{m} \]

This gives a negative value, which is not physical. It tells us that with 5 m elevation gain and 1 m of head loss, there is not enough energy to sustain flow with zero pressure difference. In practice, either \(p_1 > p_2\) (pressurized tank) or the elevation difference/head loss must be smaller. The calculator will flag such situations as “no real solution”.

Assumptions and limitations

  • Steady flow: Conditions do not change with time.
  • Incompressible fluid: Density is constant (good for liquids, approximate for low-speed gases).
  • Inviscid / low friction: Viscous effects are small or lumped into the head-loss term \(h_L\).
  • Along a streamline: The equation strictly applies along a single streamline; for uniform sections it is often applied between cross-sections.
  • No pumps/turbines: This version does not include pump head added or turbine head extracted between 1 and 2.

Frequently asked questions

What is Bernoulli’s principle in simple terms?

Bernoulli’s principle states that where a fluid flows faster, its static pressure tends to be lower (if elevation and total energy remain the same). It is a consequence of energy conservation in a flowing fluid.

Is Bernoulli’s equation the same as the energy equation?

Bernoulli’s equation is a simplified form of the mechanical energy equation for fluids. The full energy equation includes pump/turbine terms, shaft work, and detailed loss terms. Bernoulli’s equation assumes no shaft work and groups losses into a single head-loss term.

How do I estimate head loss hL?

Head loss can be estimated using correlations such as the Darcy–Weisbach equation with a friction factor (Moody chart) plus minor-loss coefficients for fittings, valves, and entrances/exits. Many engineering handbooks and standards tabulate these coefficients.