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Fluid Mechanics

Bernoulli Equation Calculator

Solve pressure, velocity, or elevation at one of two points along a streamline using the ideal Bernoulli equation for steady incompressible flow.

Formula p + 1/2 rho v^2 + rho g z = constantLast updated Aug 14, 2026

Bernoulli's equation relates pressure, velocity, and elevation along a streamline when flow is steady, incompressible, and idealized with negligible losses. This calculator uses the two-point form p1 + 1/2ρv1² + ρgz1 = p2 + 1/2ρv2² + ρgz2 so you can solve for pressure, speed, or elevation at either point once the other terms are known.

Calculation Bench
Unknown Point
Solve for
01

ρ · Mass density of the flowing fluid, assumed constant between the two points.

02

g · Local gravitational acceleration magnitude.

03

v2 · Flow speed magnitude at point 2.

04

z2 · Elevation of point 2 relative to the same datum used for point 1.

05

p1 · Static pressure at the first point on the streamline.

06

v1 · Flow speed magnitude at point 1.

07

z1 · Elevation of point 1 relative to a chosen datum.

Pressures may be entered as either gauge or absolute values, but both points must use the same reference. This tool applies ideal Bernoulli only: no friction loss, no pump/turbine work, and no compressibility effects.

Solution

Enter the required values to calculate pressure at point 2.

p2 = p1 + 1/2 ρ (v1² - v2²) + ρ g (z1 - z2)

Formula Sheet

p1=p2+12ρ(v22−v12)+ρg(z2−z1)p_1 = p_2 + \frac{1}{2}\rho\left(v_2^2-v_1^2\right)+\rho g\left(z_2-z_1\right)
v1=2(p2−p1)ρ+v22+2g(z2−z1)v_1 = \sqrt{\frac{2\left(p_2-p_1\right)}{\rho}+v_2^2+2g\left(z_2-z_1\right)}
z1=z2+p2−p1ρg+v22−v122gz_1 = z_2 + \frac{p_2-p_1}{\rho g}+\frac{v_2^2-v_1^2}{2g}
p2=p1+12ρ(v12−v22)+ρg(z1−z2)p_2 = p_1 + \frac{1}{2}\rho\left(v_1^2-v_2^2\right)+\rho g\left(z_1-z_2\right)
v2=2(p1−p2)ρ+v12+2g(z1−z2)v_2 = \sqrt{\frac{2\left(p_1-p_2\right)}{\rho}+v_1^2+2g\left(z_1-z_2\right)}
z2=z1+p1−p2ρg+v12−v222gz_2 = z_1 + \frac{p_1-p_2}{\rho g}+\frac{v_1^2-v_2^2}{2g}
  • ρDensity
  • gGravity
  • p1Pressure at Point 1
  • v1Velocity at Point 1
  • z1Elevation at Point 1
  • p2Pressure at Point 2
  • v2Velocity at Point 2
  • z2Elevation at Point 2

Variables & Units

SymbolVariableDescriptionCommon Units
ρDensityMass density of the flowing fluid, assumed constant between the two points.kg/m³, g/cm³, lb/ft³
gGravityLocal gravitational acceleration magnitude.m/s², ft/s², g
p1Pressure at Point 1Static pressure at the first point on the streamline.Pa, kPa, bar, psi
v1Velocity at Point 1Flow speed magnitude at point 1.m/s, ft/s, mph
z1Elevation at Point 1Elevation of point 1 relative to a chosen datum.m, ft, in
p2Pressure at Point 2Static pressure at the second point on the streamline.Pa, kPa, bar, psi
v2Velocity at Point 2Flow speed magnitude at point 2.m/s, ft/s, mph
z2Elevation at Point 2Elevation of point 2 relative to the same datum used for point 1.m, ft, in

How to Use This Calculator

  • 01Choose which point contains the unknown first. Then choose whether you want to solve for pressure, velocity, or elevation at that point.
  • 02Enter the fluid density and gravity, plus the known pressure, velocity, and elevation values at the two points. Elevation is measured relative to any consistent datum.
  • 03Use pressure units consistently as either gauge or absolute pressure at both points. The equation works with either, but you must not mix them.
  • 04Select Calculate to see the answer, the active formula, and a substitution line in coherent SI units.

How the Formula Works

Bernoulli's equation is a mechanical-energy balance for ideal fluid flow. Along a streamline, the sum of static pressure energy per unit volume, kinetic energy per unit volume, and gravitational potential energy per unit volume stays constant. In pressure form, that balance is written as p + 1/2ρv² + ρgz = constant.

Because the three terms trade off against one another, a fluid can lose static pressure as it speeds up, or recover pressure as it slows down, while elevation changes shift energy between pressure and gravity terms. This calculator applies the steady two-point form directly, without adding pumps, turbines, friction losses, or minor losses. If those effects matter, the ideal Bernoulli result becomes a starting point rather than a complete system model.

Worked Example 01

Pressure at point 2 in a horizontal pipe contraction

Known

  • Density (ρ): 1000 kg/m³
  • Gravity (g): 9.81 m/s²
  • Pressure at Point 1 (p1): 200 kPa
  • Velocity at Point 1 (v1): 2 m/s
  • Elevation at Point 1 (z1): 5 m
  • Velocity at Point 2 (v2): 5 m/s
  • Elevation at Point 2 (z2): 5 m

Formula

p2 = p1 + 1/2 ρ (v1² - v2²) + ρ g (z1 - z2)

Substitution

p2 = 200,000 + 1/2 × 1000 × (2² - 5²) + 1000 × 9.81 × (5 - 5)

Result

p2 = 189.5 kPa

Because the fluid speeds up while elevation stays the same, static pressure drops by 10.5 kPa between the two points in the ideal Bernoulli model.

Worked Example 02

Velocity at point 2 from pressure and elevation changes

Known

  • Density (ρ): 1000 kg/m³
  • Gravity (g): 9.81 m/s²
  • Pressure at Point 1 (p1): 150 kPa
  • Velocity at Point 1 (v1): 3 m/s
  • Elevation at Point 1 (z1): 10 m
  • Pressure at Point 2 (p2): 100 kPa
  • Elevation at Point 2 (z2): 12 m

Formula

v2 = √(2 (p1 - p2) / ρ + v1² + 2 g (z1 - z2))

Substitution

v2 = √(2 × (150,000 - 100,000) / 1000 + 3² + 2 × 9.81 × (10 - 12))

Result

v2 ≈ 8.35 m/s

The pressure drop contributes strongly to acceleration, while the higher elevation at point 2 offsets part of that gain.

Worked Example 03

Elevation at point 2 from Bernoulli balance

Known

  • Density (ρ): 1000 kg/m³
  • Gravity (g): 9.81 m/s²
  • Pressure at Point 1 (p1): 180 kPa
  • Velocity at Point 1 (v1): 4 m/s
  • Elevation at Point 1 (z1): 12 m
  • Pressure at Point 2 (p2): 150 kPa
  • Velocity at Point 2 (v2): 7 m/s

Formula

z2 = z1 + (p1 - p2) / (ρ g) + (v1² - v2²) / (2 g)

Substitution

z2 = 12 + (180,000 - 150,000)/(1000 × 9.81) + (4² - 7²)/(2 × 9.81)

Result

z2 ≈ 13.38 m

Some of the pressure energy difference appears as elevation rise, while the higher velocity at point 2 consumes part of that available head.

Applications

  • 01Checking pressure recovery or pressure drop between two sections of idealized flow
  • 02Estimating jet or constriction velocity from pressure and elevation changes
  • 03Introductory fluid-mechanics problems involving venturi, nozzle, and streamline energy balance concepts

Assumptions

  • 01Flow is steady, incompressible, and evaluated along a streamline.
  • 02Viscous losses, shaft work, and heat transfer effects are neglected.
  • 03Pressure, velocity, and elevation are interpreted as static pressure, speed magnitude, and geometric elevation at the two points.

Where This Model Stops

  • 01Not valid as a complete model when friction losses, pumps, turbines, cavitation, compressibility, or strong streamline curvature effects are important.
  • 02Velocity solutions require the implied squared speed to stay non-negative; otherwise the stated combination of pressures and elevations is not achievable under the ideal Bernoulli assumptions.
  • 03The page uses the pressure form of Bernoulli's equation and does not directly compute flow rate, pipe diameter, or continuity constraints.

References

  1. [1]
    Lecture 9: Bernoulli Equation for Potential Flow

    MIT Marine Hydrodynamics

    Summarizes the streamline Bernoulli relation for steady ideal flow and its assumptions.

  2. [2]
    Lecture 25: Energy (Bernoulli) Principle

    MIT OpenCourseWare

    Covers the Bernoulli energy principle and head interpretation used in undergraduate fluid mechanics.

Frequently Asked Questions

Can I use gauge pressure instead of absolute pressure?

Yes, as long as both pressure inputs use the same reference. Because Bernoulli's equation depends on pressure differences between the two points, consistent gauge pressures work just as well as consistent absolute pressures.

Why might the calculator reject a velocity solution?

If the pressure difference, elevation change, and known opposite-point speed imply a negative squared velocity, then that state cannot exist under the ideal Bernoulli assumptions. In a real system, losses, pumps, or other effects may need to be modeled explicitly.

What is the head form of Bernoulli's equation?

Dividing the pressure-form equation by ρg gives the head form: p/(ρg) + v²/(2g) + z = constant. The three terms are pressure head, velocity head, and elevation head. This calculator solves the equivalent pressure form directly.

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