Bernoulli's Equation

Where a fluid speeds up, its pressure drops.

12 minUpdated 21 October 2026Bernoulli's equationwork–energy theorempressure and speed

Lesson at a glanceShow the key points

Bernoulli's equation is the work–energy theorem applied to a moving slab of ideal fluid in steady flow.
P₁ + ρgh₁ + ½ρv₁² = P₂ + ρgh₂ + ½ρv₂² (13.10), i.e. P + ρgh + ½ρv² is constant along the flow (13.11).
At the same height, a faster-moving region has lower pressure.
Combined with continuity, it gives pressure differences in pipes of changing cross-section.

Blow between two hanging sheets of paper and they swing together, not apart. Fast-moving fluid is at lower pressure. Bernoulli’s equation makes this precise, and it is nothing more than the work–energy theorem applied to a flowing liquid.

Setting up

Take steady, irrotational flow of an ideal (incompressible, nonviscous) liquid of density ρ through a tube whose cross-section and height vary. Follow the liquid between sections A and B. At A: area A₁, speed v₁, pressure P₁, height h₁. At B: A₂, v₂, P₂, h₂. In a short time Δt this liquid moves forward so that it lies between A′ and B′, where $AA' = v_1\Delta t$ and $BB' = v_2\Delta t$ . By continuity the slices AA′ and BB′ have the same volume and the same mass:

\Delta m = \rho A_1v_1\Delta t = \rho A_2v_2\Delta t

Work done on the liquid

The liquid behind pushes forward with P₁A₁ through a distance v₁Δt: $W_1 = P_1A_1v_1\Delta t = P_1\Delta m/\rho$ .
The liquid ahead pushes back with P₂A₂: $W_2 = -P_2\Delta m/\rho$ .
Gravity: the net effect is as if the slice Δm were lifted from h₁ to h₂, so $W_3 = \Delta m\,g h_1 - \Delta m\,g h_2$ .
The walls push perpendicular to the flow and do no work.

Change in kinetic energy

The flow is steady, so the liquid between A′ and B (common to both moments) has the same kinetic energy before and after. The change is just that of the slice moving from AA′ to BB′: $\tfrac12\Delta m\,v_2^2 - \tfrac12\Delta m\,v_1^2$ .

Setting total work equal to the change in kinetic energy and dividing by Δm/ρ:

Bernoulli's equation(13.10, 13.11)

P_1 + \rho g h_1 + \tfrac12\rho v_1^2 = P_2 + \rho g h_2 + \tfrac12\rho v_2^2, \qquad P + \rho g h + \tfrac12\rho v^2 = \text{constant}

Interactive

Bernoulli in a pipe

Water flows left to right. The open vertical tubes rise to a height P/ρg above each point (gauge pressure). Where the pipe narrows the water speeds up and the pressure falls; where it climbs, pressure falls too. P + ρgh + ½ρv² stays the same, as the stacked bars show. The level difference between the tubes is exactly what a venturi meter reads.

point 1

point 2

ρgh

½ρv²

v_2

2.00m/s

P_2

13.50kPa (gauge)

tube level difference15.3cm

Pressure in the wide part (gauge)

P_1

15.0kPa

Speed in the wide part

v_1

1.0m/s

Area ratio

A_1/A_2

2.0

Rise of the narrow part

h_2

0.00m

Worked example

Pressure drop at a constriction

Water flows through a horizontal pipe of cross-section 8.0 cm² at 0.50 m/s; the pipe narrows to 2.0 cm². Find the pressure difference between the wide and narrow parts.

Worked example

Pumping water uphill

Water flows steadily through a uniform pipe that climbs 5.0 m. How much lower is the pressure at the top?

What to remember

1Bernoulli's equation is the work–energy theorem for steady flow of an ideal liquid.
2 $P + \rho gh + \tfrac12\rho v^2$ is constant along the flow.
3At equal height, faster flow means lower pressure; continuity tells you where the flow is faster.

Check yourself

Water speeds up from 1 m/s to 3 m/s in a horizontal pipe. The pressure drops by…

Which assumption is NOT needed for Bernoulli's equation as derived here?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

WatchFluids in motionCrash Course Physics #15 PlayFluid Pressure and FlowPhET · CU Boulder ReadBernoulli's equationOpenStax University Physics §14.6 ReadBernoulli equationHyperPhysics ReadBernoulli's principleWikipedia

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Bernoulli's Equation

Where a fluid speeds up, its pressure drops.

12 minUpdated 21 October 2026Bernoulli's equationwork–energy theorempressure and speed

Lesson at a glanceShow the key points

Bernoulli's equation is the work–energy theorem applied to a moving slab of ideal fluid in steady flow.
P₁ + ρgh₁ + ½ρv₁² = P₂ + ρgh₂ + ½ρv₂² (13.10), i.e. P + ρgh + ½ρv² is constant along the flow (13.11).
At the same height, a faster-moving region has lower pressure.
Combined with continuity, it gives pressure differences in pipes of changing cross-section.

Setting up

AA' = v_1\Delta t

and

BB' = v_2\Delta t

. By continuity the slices AA′ and BB′ have the same volume and the same mass:

\Delta m = \rho A_1v_1\Delta t = \rho A_2v_2\Delta t

Work done on the liquid

The liquid behind pushes forward with P₁A₁ through a distance v₁Δt:

W_1 = P_1A_1v_1\Delta t = P_1\Delta m/\rho

The liquid ahead pushes back with P₂A₂:

W_2 = -P_2\Delta m/\rho

Gravity: the net effect is as if the slice Δm were lifted from h₁ to h₂, so

W_3 = \Delta m\,g h_1 - \Delta m\,g h_2

The walls push perpendicular to the flow and do no work.

Change in kinetic energy

The flow is steady, so the liquid between A′ and B (common to both moments) has the same kinetic energy before and after. The change is just that of the slice moving from AA′ to BB′:

\tfrac12\Delta m\,v_2^2 - \tfrac12\Delta m\,v_1^2

Setting total work equal to the change in kinetic energy and dividing by Δm/ρ:

Bernoulli's equation(13.10, 13.11)

P_1 + \rho g h_1 + \tfrac12\rho v_1^2 = P_2 + \rho g h_2 + \tfrac12\rho v_2^2, \qquad P + \rho g h + \tfrac12\rho v^2 = \text{constant}

Interactive

Bernoulli in a pipe

point 1

point 2

ρgh

½ρv²

v_2

2.00m/s

P_2

13.50kPa (gauge)

tube level difference15.3cm

Pressure in the wide part (gauge)

P_1

15.0kPa

Speed in the wide part

v_1

1.0m/s

Area ratio

A_1/A_2

2.0

Rise of the narrow part

h_2

0.00m

Worked example

Pressure drop at a constriction

Water flows through a horizontal pipe of cross-section 8.0 cm² at 0.50 m/s; the pipe narrows to 2.0 cm². Find the pressure difference between the wide and narrow parts.

Worked example

Pumping water uphill

Water flows steadily through a uniform pipe that climbs 5.0 m. How much lower is the pressure at the top?

What to remember

1Bernoulli's equation is the work–energy theorem for steady flow of an ideal liquid.
2 $P + \rho gh + \tfrac12\rho v^2$ is constant along the flow.
3At equal height, faster flow means lower pressure; continuity tells you where the flow is faster.

Check yourself

Water speeds up from 1 m/s to 3 m/s in a horizontal pipe. The pressure drops by…

Which assumption is NOT needed for Bernoulli's equation as derived here?

Setting up#

Work done on the liquid#

Change in kinetic energy#

Check yourself

Keep exploring

Setting up#

Work done on the liquid#

Change in kinetic energy#

Check yourself

Keep exploring

Setting up

Work done on the liquid

Change in kinetic energy

Setting up

Work done on the liquid

Change in kinetic energy