Physics · Mechanics

Bernoulli’s equation for ideal flow

Q: What does the Bernoulli equation say intuitively?

Under ideal conditions, $p+\frac12\rho v^2+\rho gh$ stays constant along a streamline. The terms represent static pressure and kinetic and elevation contributions to the mechanical balance per volume. At equal elevation, increased speed must therefore accompany lower static pressure. Elevation changes can outweigh that effect: faster water at a lower point may still have higher pressure. Different streamlines do not generally share one constant. The compact equation assumes steady incompressible flow without significant viscosity or machine work. It is a balance, not proof of those assumptions and not a complete description of the flow field around an obstacle.

Q: What is the Torricelli formula and how does it follow from Bernoulli?

Torricelli gives ideal outlet speed $v=\sqrt{2g\Delta h}$ for a small outlet below the open surface of a large tank. Both locations have the same ambient pressure. Surface motion is approximately slow and its change is treated quasi-steadily. Without significant losses, the elevation contribution per mass becomes the kinetic contribution. Five metres of elevation difference with $\frac{9.81 m}{s^{2}}$ gives about $\frac{9.90 m}{s}$, the same ideal speed magnitude as free fall through that height. A real jet can differ because of losses. Its volume flow also requires effective jet area; contraction means that area cannot simply be equated to the opening area.

Bernoulli relates static pressure, speed and elevation along a streamline under ideal flow conditions; at equal elevation higher speed accompanies lower static pressure.

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Formula

Calculate

p + \frac{1}{2} ρ v^{2} + ρ g h = const.

p in Pa · ρ in kg/m³ · v in m/s · $g=9.81\,\mathrm{m\,s^{-2}}$ · h in m

Variables & units

Variables & units
Symbol	Meaning	Unit
$p$	Static pressure in the fluid	Pa
$ρ$	Density of the fluid	kg/m³
$v$	Flow speed	m/s
$g$	Gravitational acceleration $(\frac{9.81 m}{s^{2}})$	m/s²
$h$	Height above the reference level	m

LaTeX

p + \frac{1}{2} \rho v^2 + \rho g h = \text{const.}

QUANTA STUDYquanta-study.de

Diagram: a horizontal pipe with a constriction, a short arrow v₁ in the wide part and an arrow v₂ four times as long in the constriction; the standpipe above the wide part (H₁) is filled higher than the one above the constriction (H₂). — Steady, incompressible, ideally lossless circular-pipe flow at a common axis elevation. Halving diameter quadruples speed; the lower standpipe level indicates a smaller piezometric head and lower axis pressure.

Bernoulli equation calculator

Find the pressure or speed at a second point on the same streamline, for ideal flow without friction and with absolute pressures.

Find

Absolute pressure at point 1 p₁

Absolute pressure at point 1 including air pressure, not gauge pressure, e.g. 120 kPa in the wide pipe.

Constant fluid density ρ

Density of the flowing liquid, e.g. 1000 kg/m³ for water.

Flow speed at point 1

Flow speed at point 1, e.g. 1.5 m/s in the wide pipe.

Flow speed at point 2

Flow speed at point 2, e.g. 4.5 m/s in a narrow section with a third of the area.

Gravitational acceleration g

Gravitational acceleration, 9.81 m/s² on Earth.

m/s²

Height at point 1 (shared zero) h₁

Height of point 1 above a zero level you choose, negative below it, e.g. 0 m.

Height at point 2 (shared zero) h₂

Height of point 2 above the same zero level, e.g. 0 m for a horizontal pipe.

Filled with example values. Type your own and the result updates straight away.

Result: Absolute pressure at point 2 p₂

p₂ = 111 kPa

With p₁ = 120 kPa, ρ = 1000 kg/m³, v₁ = 1.5 m/s, v₂ = 4.5 m/s, g = 9.81 m/s², h₁ = 0 m and h₂ = 0 m, p₂ = 111 kPa.

Working · rearrange for p₂, substitute, evaluate

1. Rearrange for p₂
$p_2=p_1+\frac\rho2(v_1^2-v_2^2)+\rho g(h_1-h_2)$
2. Substitute values (in SI units)
- p₁ = 120 kPa = 120 000 Pa
$p_2=120000+\frac{1000}2\cdot\left(1.5^2-4.5^2\right)+1000\cdot\left(9.81\right)\left(0-0\right)=111000\,\mathrm{Pa}$
3. Result
p₂ = 111 kPa

This is an ideal model balance. Cavitation was not checked without temperature and vapour pressure; the calculation does not confirm real non-cavitating flow.

Quick tips

Continuity first: A₁ · v₁ = A₂ · v₂. If the narrow section has a third of the cross-sectional area, the water flows three times as fast there.
At equal height: the faster the flow, the lower the static pressure. Height counts as ρ · g · Δh, about 10 kPa per metre in water; at equal speed the lower point has the higher pressure.

Assumptions and limits

Steady incompressible lossless flow along the same streamline, no pump/turbine. Positive constant density, nonnegative speed magnitudes, common height reference, approximately uniform g. Both pressures absolute in Pa; do not mix with gauge pressure. Validity remains conditional on no cavitation: temperature and vapour pressure are not supplied or checked.

Derivation & background

For steady incompressible flow with negligible viscosity and no machine work, $p+\frac12\rho v^2+\rho gh$ is constant along one streamline. Different streamlines may have different constants. Include elevation change; faster flow at a lower elevation can still have higher pressure. Continuity gives $A_1v_1=A_2v_2$ in ideal pipe models. Real head loss and pump or turbine work require additional terms.

Exam blueprint

Validity range

Along one streamline: steady, incompressible, negligible viscosity, no pump or turbine work. Different streamlines do not generally share one constant. Higher speed lowers static pressure only with matching remaining balance conditions.

Derivation steps

Pressure work changes motion and elevation of a fluid volume; without other work or losses the mechanical balance is constant along a streamline.

1Choose two points on one streamline of steady incompressible flow; neglect viscosity and machine work.
2Transferred volume $\Delta V$ has mass $\rho\Delta V$ at constant density.
3Pressure work in minus out is $(p_1-p_2)\Delta V$ .
4Kinetic change is $\frac12\rho\Delta V(v_2^2-v_1^2)$ ; potential change is $\rho\Delta Vg(h_2-h_1)$ .
5Set pressure work equal to these changes and divide by $\Delta V$ .
6Rearrange to $p_1+\frac12\rho v_1^2+\rho gh_1=p_2+\frac12\rho v_2^2+\rho gh_2$ .
7Only at equal elevation does higher v imply lower p; add losses or machine terms when needed.

Rearrangements

Speed from pressure difference

v = \sqrt{\frac{2 (p_1 - p_2)}{\rho}}

Ideal stagnation measurement: $p_1$ is stagnation pressure at $v_1=0$ and $p_2$ static pressure at speed v; equal elevation on one streamline. Not a formula for arbitrary pipe pressure losses.

Efflux speed (Torricelli)

v = \sqrt{2 g h}

Large open tank, small outlet and common ambient pressure; h is elevation difference. Quasi-steady lossless discharge with negligible surface speed.

Dynamic pressure

q = \frac{1}{2} \rho v^2

The part that appears as pressure when the flow is brought to rest.

Task variant

A large open water tank has a small outlet 5 m below the surface. Equal ambient pressure, negligible losses and quasi-steady lowering of the surface. Find ideal outlet speed.

Approach: $v=\sqrt{2g\Delta h}$ .
With $g=9.81\,\mathrm{m\,s^{-2}}$ , obtain $v=\sqrt{98.1}\,\mathrm{m\,s^{-1}}\approx9.90\,\mathrm{m\,s^{-1}}$ .

Check: $\frac12v^2=g\Delta h$ .

Wind hits a wall at 30 m/s ( $\rho_{\mathrm{air}}=1.2\,\mathrm{kg/m^3}$ ). What is the dynamic pressure?

For ideal deceleration at a stagnation point on the same streamline and elevation, $q=\frac12\rho v^2=540\,\mathrm{Pa}$ .
This is pressure increase, not absolute pressure.
Relative to 101325 Pa it is about 0.53 percent; total wall force additionally needs pressure distribution and area.

Common mistakes

Confusing static pressure p and total pressure.

Static pressure is one balance term. Its sum with dynamic and elevation contributions stays constant along the same streamline only under Bernoulli assumptions.

Forgetting the continuity equation.

For steady equal-density flow with suitable profiles, $A_1\bar v_1=A_2\bar v_2$ supplies missing mean speeds. Then check the separate validity of the energy balance.

Applying the equation to strongly viscous flow (narrow long pipes).

Include head loss in an extended mechanical energy balance. A pressure comparison needs matching boundary conditions; real pressures are not universally below every ideal value regardless of pumps and inlet pressure.

Exam context

Typical: Venturi tube with continuity plus Bernoulli, Torricelli efflux and qualitative explanations (atomiser, aerodynamic lift) via the pressure drop at high speed.

Formula cluster

Fluid dynamics

Connects pressure and density with energy conservation of moving fluids.

p = F/A

Pressure

ρ = m/V

Density

F_A = ρ·V·g

Buoyant Force (Archimedes)

Worked example

Water of density $\rho=1000\,\mathrm{kg/m^3}$ flows ideally and horizontally from $p_1=200\,\mathrm{kPa}$ , $v_1=2\,\mathrm{m/s}$ to $v_2=8\,\mathrm{m/s}$ . Then $p_2=p_1+\frac12\rho(v_1^2-v_2^2)=170\,\mathrm{kPa}$ . Dynamic pressure increases by $30\,\mathrm{kPa}$ , balanced by an equal decrease in static pressure.

Applications

Airflow around wings, Venturi nozzles and carburettors, Pitot tube (aircraft speed measurement), water jet pump, atomisers

How do you calculate with Bernoulli Equation?

Here is how to work through a typical Bernoulli Equation (p + ½ρv² + ρgh = const.) task step by step:

1
Task
A large open water tank has a small outlet 5 m below the surface. Equal ambient pressure, negligible losses and quasi-steady lowering of the surface. Find ideal outlet speed.
Solution path
1. Approach: $v=\sqrt{2g\Delta h}$ .
2. With $g=9.81\,\mathrm{m\,s^{-2}}$ , obtain $v=\sqrt{98.1}\,\mathrm{m\,s^{-1}}\approx9.90\,\mathrm{m\,s^{-1}}$ .
Check: $\frac12v^2=g\Delta h$ .
2
Task
Wind hits a wall at 30 m/s ( $\rho_{\mathrm{air}}=1.2\,\mathrm{kg/m^3}$ ). What is the dynamic pressure?
Solution path
1. For ideal deceleration at a stagnation point on the same streamline and elevation, $q=\frac12\rho v^2=540\,\mathrm{Pa}$ .
2. This is pressure increase, not absolute pressure.
3. Relative to 101325 Pa it is about 0.53 percent; total wall force additionally needs pressure distribution and area.

Understand and apply the equation

Water can speed up in a constricted pipe while its static pressure falls. Bernoulli explains this relation through pressure work, kinetic energy and elevation. The assumptions belong to the calculation: a lower point can have higher pressure even while the water moves faster.

What does Bernoulli’s equation state, and when does it apply?

“Steady” means flow quantities at a fixed measurement point do not change with time. A streamline is an imagined line whose direction follows the local flow direction everywhere. Choose two points on the same such line; the ideal mechanical energy balance is written between them.[1]

The compact equation applies along a streamline in steady incompressible flow with negligible viscosity and no pump or turbine work between the chosen points. Density ρ and g are taken constant. p is static pressure, v local speed and h elevation above a shared reference level. The pressure article explains the choice of reference and resulting normal forces; static pressure is not a velocity vector.[1]

p+\frac12\rho v^2+\rho gh=C

All three terms have units of pascals, equivalent to energy per volume.[1]

C is constant along the chosen streamline. Different streamlines can generally have different constants; a common constant throughout a connected region additionally requires irrotational flow and appropriate boundary conditions. Bernoulli alone does not determine the full flow field around an obstacle.[1]

How does Bernoulli follow from pressure work and energy conservation?

Follow a small fluid volume ΔV between two points on one streamline in steady flow. Constant density gives mass $\Delta m=\rho\Delta V$ . The inlet receives pressure work $p_1\Delta V$ while the outlet delivers $p_2\Delta V$ . The difference changes kinetic and gravitational potential energy when no other mechanical transfers occur.[1]

(p_1-p_2)\Delta V=\frac12\rho\Delta V(v_2^2-v_1^2)+\rho\Delta Vg(h_2-h_1)

Signs distinguish pressure work received from pressure work delivered.[1]

p_1+\frac12\rho v_1^2+\rho gh_1=p_2+\frac12\rho v_2^2+\rho gh_2

Divide by ΔV and group inlet and outlet terms.[1]

This derivation is a balance under the stated assumptions, not proof that a real pipe has no friction or density changes. With $v_1=v_2=0$ , it recovers hydrostatics: $p_2-p_1=\rho g(h_1-h_2)$ , so lower elevation has higher pressure.[1]

Which units and continuity condition do you need?

Q=A_1\overline v_1=A_2\overline v_2

For steady incompressible pipe flow, Q is volume flow rate and v̄ cross-section mean speed.[2]

Pointwise Bernoulli uses local v. Simple pipe models assume approximately uniform profiles so local and mean values match. Strongly nonuniform profiles require kinetic-energy correction factors in a section-averaged energy balance. Halving circular diameter quarters area and quadruples mean speed in the ideal model. The continuity equation for flow connects the two cross-sections before their speeds are substituted into Bernoulli’s equation.[1]

\begin{aligned}[\rho v^2]&=\frac{\mathrm{kg}}{\mathrm{m\,s^2}}=\mathrm{Pa}\\p_2&=p_1+\frac12\rho(v_1^2-v_2^2)+\rho g(h_1-h_2)\end{aligned}

Elevation has a sign; do not assess pressure and speed terms in isolation.[1]

Use the same reference for both pressures. Subtracting one common reference leaves differences unchanged. Vapour-pressure or cavitation checks require absolute pressure explicitly. A low mathematical $p_2$ does not automatically mean the assumed liquid model remains possible.[1]

Calculating a constriction: does faster always mean lower pressure?

Horizontal ideal pipe with a smaller section

Given: Water at $\rho=1000\,\mathrm{kg\,m^{-3}}$ ; $A_1=12\,\mathrm{cm^2}$ , $A_2=4\,\mathrm{cm^2}$ ; $v_1=1.5\,\mathrm{m\,s^{-1}}$ ; $p_1=120\,\mathrm{kPa}$ absolute. Equal elevation and uniform profiles; ideal steady flow without losses or machinery.
Find: Find volume flow rate, v₂ and $p_2$ .

Approach: For steady incompressible lossless flow at equal elevation, apply continuity and then Bernoulli.
$\begin{aligned}Q&=A_1v_1=A_2v_2\\p_1+\frac12\rho v_1^2&=p_2+\frac12\rho v_2^2\end{aligned}$
Apply continuity to both sections.
$v_2=v_1\frac{A_1}{A_2}=1.5\cdot3=4.5\,\mathrm{m/s}$
Find volume flow in SI units.
$Q=12\cdot10^{-4}\cdot1.5=0.0018\,\mathrm{m^3/s}$
Evaluate the Bernoulli pressure difference at equal elevation.
$p_2=120000+\frac12\cdot1000(1.5^2-4.5^2)=111000\,\mathrm{Pa}$

Result: $Q=1.8\,\mathrm{L\,s^{-1}}$ , $v_2=4.5\,\mathrm{m\,s^{-1}}$ , $p_2=111\,\mathrm{kPa}$ absolute.

Check: Dynamic pressure rises from 1125 to 10125 Pa; static pressure falls by the same 9000 Pa. Their sum stays 121125 Pa.

Faster flow with higher static pressure

Given: Water at 1000 kg/m³; $g=9.81\,\mathrm{m\,s^{-2}}$ . On the same ideal streamline: $p_1=140\,\mathrm{kPa}$ absolute, $v_1=2\,\mathrm{m\,s^{-1}}$ , $h_1=2\,\mathrm m$ ; $v_2=3\,\mathrm{m\,s^{-1}}$ , $h_2=0\,\mathrm m$ .
Find: Find $p_2$ and explain the sign of its change.

Approach: Along the same ideal streamline, include both elevation and speed changes in the Bernoulli balance.
$p_2=p_1+\frac12\rho(v_1^2-v_2^2)+\rho g(h_1-h_2)$
Calculate the speed contribution.
$\Delta p_{\mathrm{speed}}=\frac12\cdot1000(2^2-3^2)=-2500\,\mathrm{Pa}$
Include the lower final elevation.
$\Delta p_{\mathrm{height}}=1000\cdot9.81(2-0)=19620\,\mathrm{Pa}$
Add both contributions to initial pressure.
$p_2=140000-2500+19620=157120\,\mathrm{Pa}$

Result: Although v₂ is higher, static pressure rises to 157.12 kPa. Decreasing elevation term outweighs the increase in kinetic term.

Check: Check both complete pressure sums: initially $(140000+2000+19620)\,\mathrm{Pa}=161620\,\mathrm{Pa}$ , finally $(157120+4500)\,\mathrm{Pa}=161620\,\mathrm{Pa}$ . The pressure increase from lower elevation outweighs the reduction from acceleration. “Faster means lower pressure” needs equal elevation here.

View diagram: Pressure is redistributed into kinetic energy

Pressure is redistributed into kinetic energy

QUANTA STUDYquanta-study.de

Curve from the first example: constant density 1000 kg/m³, one ideal streamline, equal height and no machine work. The gap to the dashed line is dynamic pressure. The curve shows possible states, not a time evolution.[1]

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How do Torricelli’s law and stagnation-pressure measurement follow?

For a large open tank with a small outlet, compare the free surface with a point in the free jet. Both have the same ambient pressure; surface speed is approximately zero. Elevation energy per volume becomes kinetic energy. Real outlets have losses and jet contraction; distinguish speed from volume flow.[1]

v_{\mathrm{jet}}\approx\sqrt{2g\Delta h}

Measure vertical difference between free surface and the chosen exit point.[1]

A stagnation point is a location where flow is ideally brought to $v=0$ . On the same streamline at equal elevation, stagnation minus static pressure is $\Delta p=\frac12\rho v^2$ , giving $v=\sqrt{\frac{2\Delta p}{\rho}}$ . A measuring tube facing the flow and a side static-pressure tap detect different pressures; the measurement arrangement belongs to the interpretation.[1]

What do friction, a pump or a turbine change?

\frac{p_1}{\rho g}+\frac{v_1^2}{2g}+h_1+H_{\mathrm{pump}}=\frac{p_2}{\rho g}+\frac{v_2^2}{2g}+h_2+H_{\mathrm{turbine}}+H_{\mathrm{loss}}

An extended section balance also needs kinetic-energy correction factors for nonuniform profiles.[1]

Head loss represents mechanical-energy reduction per unit weight, not destruction of total energy. An ideal horizontal pipe of equal area without machinery gives $p_1=p_2$ . A real pipe can still have frictional pressure drop. Do not use the simple equation to eliminate that loss. The energy equation with head loss includes friction and machine work explicitly as additional balance terms.[3]

For gases, check whether density variation is negligible. One speed threshold is not a universal guarantee: Mach number, pressure change and thermal state matter. Turbulence and time fluctuations may also require an averaged balance with additional terms.[1]

Explore further: Airfoils and limits of the short explanation

Bernoulli can relate a suitably established velocity field to pressure distribution, but alone does not determine that field around a wing. Angle of attack, geometry, flow conditions and air deflection matter. Neighbouring air parcels have no requirement to arrive together after travelling above and below a wing.[4]

Aerodynamic lift comes from the overall surface pressure and, where relevant, shear forces. Air momentum balance and pressure analysis can describe the same effect with consistent system boundaries and models. A bare “faster above” comparison is not a complete theory of flight.[4]

Five problems with solutions

Calculate · Foundation

An ideal horizontal water flow accelerates from 1 to 5 m/s. Find $p_2-p_1$ for $\rho=1000\,\mathrm{kg\,m^{-3}}$ .

Show hint

Elevation terms are equal.

Learning goal: Find pressure change from speed change.

Show solution and calculation

Approach: Both points lie on the same ideal streamline at equal elevation.
Thus $p_2-p_1=\frac12\rho(v_1^2-v_2^2)$ .
Substitution gives $-12\,\mathrm{kPa}$ .

Check: dynamic pressure increases from 0.5 to 12.5 kPa, requiring the same 12 kPa fall in static pressure. An absolute pressure cannot be found without initial pressure.

p_2-p_1=\frac12(1000)(1^2-5^2)\,\mathrm{Pa}=-12000\,\mathrm{Pa}

Transfer · Application

An ideal outlet is 1.8 m below the open surface of a large tank. Find jet speed with $g=9.81\,\mathrm{m\,s^{-2}}$ .

Show hint

Use equal ambient pressure and negligible surface speed.

Learning goal: Apply Bernoulli to a tank outlet.

Show solution and calculation

Approach: Both locations have the same ambient pressure and the large free surface is approximately stationary.
Elevation energy becomes kinetic energy: $v=\sqrt{2g\Delta h}$ .
This gives about 5.94 m/s.

Check: $\frac{v^2}{2}\approx17.658\,\mathrm{J\,kg^{-1}}=g\Delta h$ . This is ideal outlet speed, not a volume flow including losses.

v=\sqrt{2\cdot9.81\cdot1.8}\,\mathrm{m\,s^{-1}}\approx5.94\,\mathrm{m\,s^{-1}}

Rearrange · Application

An ideal stagnation-pressure measurement in fluid at $\rho=800\,\mathrm{kg\,m^{-3}}$ gives $\Delta p=1600\,\mathrm{Pa}$ at equal elevation. Find v.

Show hint

Rearrange $\Delta p=\frac12\rho v^2$ for v.

Learning goal: Rearrange the stagnation-to-static pressure difference to find speed.

Show solution and calculation

Approach: Ideal stagnation at equal elevation gives $\Delta p=\frac12\rho v^2$ .
Solving for speed gives 2.0 m/s.

Check: $\frac12(800\,\mathrm{kg\,m^{-3}})(2\,\mathrm{m\,s^{-1}})^2=1600\,\mathrm{Pa}$ . This pressure difference is stagnation minus static pressure, not an arbitrary pipe pressure loss.

v=\sqrt{\frac{2\Delta p}{\rho}}=\sqrt{\frac{2\cdot1600}{800}}\,\mathrm{m\,s^{-1}}=2.0\,\mathrm{m\,s^{-1}}

Interpret · Application

Water flows steadily through a horizontal constant-area pipe. The two measurement sections have identical velocity profiles and equal mean speed but $p_1-p_2=6000\,\mathrm{Pa}$ , with no pump or turbine. What do ideal Bernoulli and the real loss balance imply? Find loss head for $\rho=1000\,\mathrm{kg\,m^{-3}}$ , $g=9.81\,\mathrm{m\,s^{-2}}$ .

Show hint

Equal elevations and speeds cancel; the remainder requires a loss term.

Learning goal: Interpret pressure loss despite unchanged speed.

Show solution and calculation

Approach: lossless Bernoulli would require $p_1=p_2$ .
The measurement contradicts the lossless model, not energy conservation.
Including loss head gives $H_{\mathrm{loss}}=\frac{p_1-p_2}{\rho g}\approx0.611621\,\mathrm m$ .

Check: $\rho gH_{\mathrm{loss}}$ reconstructs 6000 Pa. Dissipation in real flow is a possible cause; two pressure measurements do not locate it spatially.

H_{\mathrm{loss}}=\frac{6000\,\mathrm{Pa}}{1000\,\mathrm{kg\,m^{-3}}\cdot9.81\,\mathrm{m\,s^{-2}}}\approx0.611621\,\mathrm m

[1][3]

Error analysis · Challenge

In steady incompressible inviscid but rotational flow, A and B lie on different streamlines. Elevations are equal; $v_A=2\,\mathrm{m\,s^{-1}}$ , $v_B=4\,\mathrm{m\,s^{-1}}$ , $\rho=1000\,\mathrm{kg\,m^{-3}}$ . A solution claims $p_B-p_A=-6000\,\mathrm{Pa}$ without further information. Is this value determined? State the missing assumption.

Show hint

Start with separate constants CA and CB.

Learning goal: Recognize that Bernoulli’s constant is streamline-specific.

Show solution and calculation

Approach: each streamline initially has its own constant.
The pressure difference includes $C_B-C_A$ in addition to the kinetic term. −6000 Pa follows only if $C_A=C_B$ .
A connected irrotational region with suitable conditions could justify equality, or boundary data could establish it; neither is supplied.

Check: evaluating squared speeds does not remove the unknown constant difference.

p_B-p_A=(C_B-C_A)+\frac12\rho(v_A^2-v_B^2)=(C_B-C_A)-6000\,\mathrm{Pa}

[1]

Key points

Choose two points on the same streamline and check steady, incompressible, lossless assumptions. Balance static pressure, speed and elevation together. Continuity can supply missing speeds; losses and machinery require additional terms.

Sources for this explanation

Quanta · Updated October 1, 2026

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This explanation, 2 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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Change history

October 1, 2026Made Bernoulli conditions, energy derivation, units, constriction, elevation counterexample, Torricelli and real losses accessible through questions; retained flow calculations and linked pressure reference, continuity and loss accounting in context.
October 1, 2026Clarified the entry and learning transitions; checked and refined existing exercise solutions, typeset mathematical prose and added an original quantitative article figure.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

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Quanta. Bernoulli Equation. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/bernoulli-gleichung

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Frequently asked questions about Bernoulli Equation

What does the Bernoulli equation say intuitively?+

Under ideal conditions, $p+\frac12\rho v^2+\rho gh$ stays constant along a streamline. The terms represent static pressure and kinetic and elevation contributions to the mechanical balance per volume. At equal elevation, increased speed must therefore accompany lower static pressure. Elevation changes can outweigh that effect: faster water at a lower point may still have higher pressure. Different streamlines do not generally share one constant. The compact equation assumes steady incompressible flow without significant viscosity or machine work. It is a balance, not proof of those assumptions and not a complete description of the flow field around an obstacle.

How do you solve a typical Bernoulli problem?+

Choose two points on the same ideal streamline and first check density, losses and elevations. At equal height, $p_2=p_1+\frac12\rho(v_1^2-v_2^2)$ . Continuity supplies missing mean pipe velocities only with suitable profile assumptions. For example, water at $\frac{1000 kg}{m^{3}}$ and initial pressure 200 kPa accelerates from 2 to $\frac{8 m}{s}$ . The speed term reduces pressure by 30 kPa, leaving 170 kPa. Both pressures use the same reference. Check: the static-pressure fall balances the dynamic-pressure increase. Different elevations require an additional elevation term; real pipe losses must not be mistaken for acceleration.

What is the Torricelli formula and how does it follow from Bernoulli?+

Torricelli gives ideal outlet speed $v=\sqrt{2g\Delta h}$ for a small outlet below the open surface of a large tank. Both locations have the same ambient pressure. Surface motion is approximately slow and its change is treated quasi-steadily. Without significant losses, the elevation contribution per mass becomes the kinetic contribution. Five metres of elevation difference with $\frac{9.81 m}{s^{2}}$ gives about $\frac{9.90 m}{s}$ , the same ideal speed magnitude as free fall through that height. A real jet can differ because of losses. Its volume flow also requires effective jet area; contraction means that area cannot simply be equated to the opening area.

Does Bernoulli explain why aeroplanes fly?+

Bernoulli can connect a suitably determined velocity field to static pressure, but does not alone determine flow around a wing. Geometry, angle of attack, flow conditions and separation must be considered. The familiar assertion that neighbouring air parcels must reach the trailing edge simultaneously above and below the wing is not a valid boundary condition. Aerodynamic lift comes from the total surface forces; alternatively, analyse the deflection and momentum change of surrounding air. Both approaches need consistent system boundaries and assumptions. A simple “faster above means lower pressure” comparison replaces neither determination of the velocity field nor integration of forces on a real wing.

What are the limits of the Bernoulli equation?+

The compact equation applies along a streamline for steady incompressible flow without significant viscosity or pump and turbine work. Real pipe friction reduces mechanical energy; extended balances add head loss. For gases, check density change, Mach number and thermal conditions; one threshold in metres per second does not guarantee incompressibility. Different streamlines do not generally share the same constant. Turbulence or time fluctuations may require additional terms in averaged balances. Using section-mean pipe speeds also needs suitable profiles or kinetic-energy correction factors. A formally calculated pressure below vapour pressure may invalidate the assumed single-phase liquid model. Revisit assumptions instead of simply accepting that numerical result.

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Which formula describes Bernoulli Equation?

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How do you rearrange Bernoulli Equation for Speed from pressure difference?

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Which common mistake happens with Bernoulli Equation?

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Subject references and further reading

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Related formulas

E_{kin} = \frac{1}{2} m v^2

Kinetic Energy

\frac{p_1}{\rho g}+\alpha_1\frac{v_1^2}{2g}+z_1+h_p=\frac{p_2}{\rho g}+\alpha_2\frac{v_2^2}{2g}+z_2+h_t+h_L

Extended Bernoulli energy equation with head loss

\rho_1 A_1 v_1 = \rho_2 A_2 v_2

Continuity equation for flow

More formulas: Mechanics

\sum \vec F_{\mathrm{ext}} = \vec F_{\mathrm{res}} = m \vec a

Newton's Second Law

F_G = G \cdot \frac{m_1 \cdot m_2}{r^2}

Newton's Law of Universal Gravitation

E_{pot} = m \cdot g \cdot h

Potential Energy (Gravitational)

F = k \cdot x

Hooke's Law (Spring Force)

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Create a curated FSRS exam set for p + ½ρv² + ρgh = const.: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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Bernoulli’s equation for ideal flow

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How do you calculate with Bernoulli Equation?

What does Bernoulli’s equation state, and when does it apply?

How does Bernoulli follow from pressure work and energy conservation?

Which units and continuity condition do you need?

Calculating a constriction: does faster always mean lower pressure?

Horizontal ideal pipe with a smaller section

Faster flow with higher static pressure

Pressure is redistributed into kinetic energy

How do Torricelli’s law and stagnation-pressure measurement follow?

What do friction, a pump or a turbine change?

Explore further: Airfoils and limits of the short explanation

Five problems with solutions

An ideal horizontal water flow accelerates from 1 to 5 m/s. Find p2−p1p_2-p_1p2​−p1​ for ρ=1000 kg m−3\rho=1000\,\mathrm{kg\,m^{-3}}ρ=1000kgm−3.

An ideal outlet is 1.8 m below the open surface of a large tank. Find jet speed with g=9.81 m s−2g=9.81\,\mathrm{m\,s^{-2}}g=9.81ms−2.

An ideal stagnation-pressure measurement in fluid at ρ=800 kg m−3\rho=800\,\mathrm{kg\,m^{-3}}ρ=800kgm−3 gives Δp=1600 Pa\Delta p=1600\,\mathrm{Pa}Δp=1600Pa at equal elevation. Find v.

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More formulas: Mechanics

Retain Bernoulli Equation for exams

An ideal horizontal water flow accelerates from 1 to 5 m/s. Find $p_2-p_1$ for $\rho=1000\,\mathrm{kg\,m^{-3}}$ .

An ideal outlet is 1.8 m below the open surface of a large tank. Find jet speed with $g=9.81\,\mathrm{m\,s^{-2}}$ .

An ideal stagnation-pressure measurement in fluid at $\rho=800\,\mathrm{kg\,m^{-3}}$ gives $\Delta p=1600\,\mathrm{Pa}$ at equal elevation. Find v.