Bernoulli’s equation says that along a streamline in steady, incompressible, frictionless flow, the sum p + ½·ρ·V² + ρ·g·z stays constant. Pressure, speed and height trade off against each other: where a fluid speeds up, its pressure drops. With the continuity equation, it solves flow meters, nozzles, siphons and tank drains.
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The equation and its terms
Each term is an energy per unit volume, in pascals. p is the static pressure, ½·ρ·V² the dynamic pressure, and ρ·g·z the hydrostatic pressure due to elevation. Dividing by ρ·g turns them into heads in metres: pressure head p/(ρg), velocity head V²/(2g) and elevation head z. Their sum is the total head, and plotting it along a pipe gives the energy grade line.
Continuity supplies the second equation. For an incompressible fluid the volume flow is the same at every section, Q = A₁·V₁ = A₂·V₂, so a smaller area means a higher speed.
When Bernoulli applies
The equation comes from integrating Newton’s second law along a streamline, and it assumes four things: steady flow, incompressible fluid, negligible friction, and no pump or turbine between the two points. Liquids, and gases below about Mach 0.3, are close enough to incompressible.
Friction matters in long pipes. There, add a head-loss term from the Darcy–Weisbach equation and a pump head if there is one; the result is the extended energy equation. Bernoulli works best over short distances where the flow speeds up, such as a nozzle, a venturi throat or a tank outlet.
Pressures at a free surface or a free jet are atmospheric. Use gauge pressure throughout, or absolute throughout, but do not mix them. Absolute pressure matters when the answer is compared with a vapour pressure, as in the siphon example.
Key formulas
Formulas used on this page, with the textbook each is cited from
Formula
Expression
Where
Cited from
Bernoulli equation
p₁ + ½·ρ·V₁² + ρ·g·z₁ = p₂ + ½·ρ·V₂² + ρ·g·z₂
p = pressure (Pa), ρ = density (kg/m³), V = velocity (m/s), z = elevation (m)
Munson et al., Fundamentals of Fluid Mechanics
Head form
p/(ρ·g) + V²/(2g) + z = constant along a streamline
each term in metres of the flowing fluid
White, Fluid Mechanics
Continuity (incompressible)
Q = A₁·V₁ = A₂·V₂
Q = volume flow (m³/s), A = flow area (m²)
White, Fluid Mechanics
Venturi meter flow
Q = C_d·A₂·√(2·Δp / (ρ·(1 − β⁴))), β = D₂/D₁
C_d = discharge coefficient, Δp = p₁ − p₂ (Pa)
Munson et al., Fundamentals of Fluid Mechanics
Torricelli’s result
V = √(2·g·h)
h = height of free surface above the outlet (m)
White, Fluid Mechanics
Worked examples
Example: Flow through a venturi meter
Given
A horizontal venturi meter in a cooling-water line has a 100 mm inlet and a 50 mm throat.
The differential pressure between inlet and throat reads 18 kPa. Take ρ = 998 kg/m³ and C_d = 0.98, a typical value for a machined venturi.
Find
The ideal throat velocity.
The actual volume flow rate.
Solution
Write Bernoulli from inlet to throat (z₁ = z₂) and continuity, V₁ = V₂·β².p₁ − p₂ = ½·ρ·V₂²·(1 − β⁴), β = 50 / 100 = 0.5
Solve for the throat velocity.V₂ = √(2 × 18 000 / (998 × (1 − 0.5⁴))) = 6.20 m/s
Ideal flow through the throat area A₂ = π × 0.05²/4 = 1.963 × 10⁻³ m².Q_ideal = 1.963 × 10⁻³ × 6.20 = 0.01218 m³/s = 12.2 L/s
Apply the discharge coefficient for the small friction loss.Q = 0.98 × 12.18 = 11.9 L/s ≈ 43.0 m³/h
Answer: The throat velocity is about 6.2 m/s and the flow is about 11.9 L/s (43 m³/h). The inlet velocity is only 1.55 m/s, so most of the 18 kPa reading is the throat’s velocity head.
Example: Speed and crest pressure in a siphon
Given
A 20 mm hose siphons water out of an open tank. The outlet is 3.0 m below the tank’s water surface, and the top of the hose is 1.5 m above it.
Take ρ = 998 kg/m³, atmospheric pressure 101.3 kPa, and the vapour pressure of water at 20 °C, 2.34 kPa. Friction in the hose is neglected.
Find
The outlet velocity and flow rate.
The absolute pressure at the crest.
The highest the crest could be before the water boils.
Solution
Bernoulli from the free surface (p = p_atm, V ≈ 0) to the free jet at the outlet (p = p_atm). Pressures cancel.V = √(2 × 9.81 × 3.0) = 7.67 m/s
Flow through the hose area π × 0.020²/4.Q = 7.67 × 3.14 × 10⁻⁴ = 2.41 × 10⁻³ m³/s = 2.4 L/s
Bernoulli from the surface to the crest. The hose has one diameter, so V at the crest is also 7.67 m/s.p_crest = 101 325 − 998 × 9.81 × 1.5 − ½ × 998 × 7.67² = 57 300 Pa abs
The crest pressure falls by ρg per metre of height. Set it equal to the vapour pressure to find the limit above the surface.z_max = (101 325 − 2340) / (998 × 9.81) − 3.0 = 7.1 m
Answer: Water leaves at about 7.7 m/s (2.4 L/s), and the crest sits at about 57 kPa absolute, well above the vapour pressure. In this ideal case the crest could rise to about 7.1 m above the surface; friction and dissolved air make the real limit lower.
Common mistakes
Applying Bernoulli across a pump or a long pipe. Add pump head and friction head loss with the energy equation. Plain Bernoulli assumes neither is present.
Mixing gauge and absolute pressure. Pick one for the whole problem. Use absolute when you compare with vapour pressure.
Ignoring the inlet velocity in a flow meter. The upstream velocity is not zero. The 1 − β⁴ term accounts for it.
Forgetting that free-surface and free-jet pressures are atmospheric. At an open tank surface and in a jet leaving into air, p = p_atm, which often cancels.
Using Bernoulli for fast gas flows. Above about Mach 0.3, density changes matter. Use compressible flow relations.
In industry
Frequently asked questions
Does Bernoulli’s equation explain lift on a wing?
It relates the lower pressure above a wing to the faster flow there, which is correct, but it does not by itself explain why the flow is faster. Full explanations also use the turning of the flow and momentum.
What is the difference between Bernoulli’s equation and the energy equation?
The energy equation adds pump head, turbine head and friction losses between two points. Bernoulli is the special case where all three are zero.
Can I use Bernoulli across different streamlines?
Only if the flow is irrotational. For most textbook pipe and tank problems the points lie on one streamline anyway.
What is stagnation pressure?
The pressure where flow is brought to rest, p₀ = p + ½·ρ·V². A pitot tube measures it, and the difference from static pressure gives the velocity.
Is there a calculator for pipe flow with friction?
Yes. Coplain’s pipe pressure drop calculator finds the Reynolds number, friction factor and pressure loss, which is the friction term Bernoulli leaves out.