Bernoulli's equation — derivation from F = ma along streamline: Question bank

Nothing on this page needs a calculator — this is about why, not how much.

Symbol & assumption refresher (read this first)

Definition Every symbol used below, in plain words

$P$ = static pressure — the ordinary squeeze at a point, measured in pascals ( $Pa = N/m^{2} = J/m^{3}$ ).
$ρ$ (Greek "rho") = density — mass packed into each cubic metre ( $kg/m^{3}$ ); assumed constant here.
$v$ = flow speed of the fluid at a point ( $m/s$ ).
$g$ = gravitational acceleration, about $9.8 m/s^{2}$ downward.
$y$ = height of the point above some chosen reference level ( $m$ ).
$h$ = a height difference (e.g. depth of a hole below the water surface), $m$ .
$A$ = cross-sectional area of the pipe or blob ( $m^{2}$ ).
$s$ = distance measured along the streamline — you "walk" along the curved path of the flow and $s$ is how far you've gone; $d s$ is a tiny step along it. This is the coordinate that makes $v \frac{d v}{d s}$ meaningful.
$θ$ = angle the streamline makes with the horizontal, so a step $d s$ raises you by $d y = d s sin θ$ .

The items lean on these four idealisations, the force-vs-energy readings, Equation of Continuity, and the convective-acceleration trick. The two figures below are the scenarios that keep recurring in the questions — glance at them before you start.

Figure — Bernoulli's equation — derivation from F = ma along streamline

True or false — justify

Bernoulli's constant has the same value everywhere in a flowing fluid.

False in general. The sum

P + \frac{1}{2} ρ v^{2} + ρ g y

is constant along one streamline; a different streamline can carry a different constant, unless the whole flow happens to be irrotational.

"Steady flow" means every fluid parcel has zero acceleration.

False. Steady means the velocity at a fixed point doesn't change in time; a parcel drifting into a faster region still accelerates via convective acceleration

a = v \frac{d v}{d s}

, where

s

is distance along the streamline.

If a horizontal pipe narrows, the pressure in the narrow part is lower.

True. Equation of Continuity forces the fluid to speed up in the constriction (see the Venturi figure above), and Bernoulli then requires the static pressure

P

to drop so the total stays constant.

Water shoots out of a hole faster the deeper the hole is below the surface.

True. Torricelli gives

v = 2 g h

where

h

is the depth below the free surface (left figure above), so a deeper hole (bigger

h

) means a faster jet — Torricelli's Law.

Bernoulli's equation is a form of energy conservation.

True. Multiply the derived line

d P + ρ g d y + ρ v d v = 0

by a volume and it is the Work-Energy Theorem per unit volume: pressure-work plus gravity-work equals change in kinetic energy.

You can apply Bernoulli across a water pump.

False. A pump adds energy to the fluid, violating the "no energy added" assumption (idealisation 3); you must insert a pump-head term, otherwise the equation is simply wrong there.

Low pressure is what causes the fluid to move fast.

False as stated. Causation runs through

F = ma

: a pressure difference is the net force that produces the speed change (

- d P = ρ v d v

). Speed and low pressure appear together, but the difference is the cause.

The term

\frac{1}{2} ρ v^{2}

has units of pressure.

True. Every Bernoulli term is energy per unit volume, i.e.

J/m^{3} = Pa

— that is why they can be added to a static pressure

P

For a fluid at rest, Bernoulli's equation reduces to

P + ρ g y = constant

True. With

v = 0

everywhere the dynamic term vanishes and you recover Hydrostatic Pressure: pressure rises linearly with depth.

Bernoulli works for honey flowing slowly through a thin tube just as well as for water in a wide one.

False. Honey is highly viscous, so internal friction dissipates energy (breaks idealisation 3) — that regime is governed by Viscosity and Poiseuille Flow, not inviscid Bernoulli.

Spot the error

P_{1} = P_{2}

at both ends of the draining-tank streamline, so the two static pressures cancel and the water can't move."

The pressures cancel, yes, but gravity does not — the

ρ g h

height difference is the leftover force, and it is what accelerates the jet to

2 g h

(left figure).

"In the Venturi, area shrinks, so by

F = ma

the mass shrinks and the parcel slows down."

Wrong link. Equation of Continuity fixes volume flow

A v

constant, so smaller area

A

means larger speed

v

; the parcel speeds up, and its mass isn't what's conserved here.

"Since the flow is steady,

\frac{d v}{d t} = 0

, so there's no acceleration term and Bernoulli should have no

\frac{1}{2} ρ v^{2}

The parcel's

\frac{d v}{d t} = v \frac{d v}{d s}

is nonzero even in steady flow (convective acceleration, with

s

= arc length along the streamline); that very term integrates into the

\frac{1}{2} ρ v^{2}

piece.

"Gravity is

m g

downward, so the gravity term in

F = ma

- ρ A d s g

You forgot to project onto the streamline. Only the component along the flow acts, giving

- ρ A g d y

where

d y = d s sin θ

is the rise (not the full step length

d s

), and

θ

is the streamline's tilt.

"A wing has lower pressure on top, and low pressure sucks upward, therefore Bernoulli creates energy for flight."

No energy is created. The pressure difference is a force doing work already accounted for in the flow's kinetic and potential energy — see Aerodynamic Lift; Bernoulli is bookkeeping, not a source.

"At the Pitot tube mouth the fluid stops, so its energy disappears."

The kinetic energy doesn't vanish, it converts to extra static pressure:

P_{0} - P = \frac{1}{2} ρ v^{2}

, which is exactly what the gauge reads — Pitot Tube.

Why questions

Why does the cross-sectional area

A

cancel out of the derivation?

Because pressure force, gravity component, and mass are all proportional to

A

(each face has area

A

, the volume is

A d s

), so it divides out — the result is a property of the streamline, not the blob's fatness.

Why must we integrate along a streamline rather than in any direction?

The

F = ma

we wrote used only forces and acceleration along the flow (that is what the coordinate

s

tracks); components across the streamline are balanced separately, so the tidy exact-differential form only holds following the streamline.

Why does the pressure term appear as a difference

- A d P

and not as

P A

A uniform pressure squeezes both faces equally and cancels; only the change

d P

across the parcel gives a net push, so the physics depends on the pressure gradient.

Why is "fast means low pressure" true for a horizontal pipe but not automatically for a rising one?

On a horizontal streamline the height term

ρ g y

is fixed, so

\frac{1}{2} ρ v^{2}

and

P

must trade off directly; if the fluid also climbs, the

ρ g y

term joins the bargain and can flip the pressure comparison.

Why does incompressibility matter for the integration step?

With

ρ

constant it comes out of the integrals cleanly, giving the neat

\frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})

and

ρ g (y_{2} - y_{1})

; if

ρ

varied you couldn't pull it out and the equation would need a compressible correction.

Why does the shower curtain get sucked inward toward the water?

Fast-moving air (dragged by the spray) inside the curtain has lower static pressure than the still room air outside, and the outside pressure pushes the curtain in — the same fast=low-pressure trade seen in the Venturi Meter (right figure).

Edge cases

What does Bernoulli say when

v = 0

everywhere (no flow)?

It collapses to

P + ρ g y = constant

, the hydrostatic pressure law — Bernoulli contains statics as a special case.

What happens to the exit speed formula

2 g h

as the hole rises to the surface (

h \to 0

The jet speed goes to zero: water level with the hole has no head

h

to push it, so nothing squirts out.

If the tank's hole is not small (comparable to the tank's cross-section), is

v_{2} = 2 g h

still exact?

No — we assumed

v_{1} \approx 0

using Equation of Continuity; with a large hole the surface drops noticeably fast,

v_{1}

isn't negligible, and

2 g h

becomes an overestimate.

For a Pitot tube in a stationary fluid (

v = 0

), what does it read?

P_{0} = P

exactly — with no flow there's no dynamic pressure to convert, so the gauge shows just the ambient static pressure and reports zero speed.

What if the streamline goes straight up with no speed change (

v_{1} = v_{2}

The dynamic terms cancel and you're left with

P_{1} - P_{2} = ρ g (y_{2} - y_{1})

, i.e. pure hydrostatics along a vertical column.

Does Bernoulli apply the instant you switch a tap on, before flow settles?

No — that's unsteady startup flow (

\partial / \partial t \neq = 0

), so the steadiness assumption (idealisation 1) fails and an extra time-dependent term is needed until the flow settles.

Recall One-line self-test

If you can state, for each trap above, which of the four idealisations or which derivation step it attacks, you've mastered the concept map — not just the algebra.

Item :: pick any trap, name the assumption it targets, then reveal

Steady / incompressible / inviscid / single-streamline, or the convective-acceleration step — every trap on this page maps to exactly one of these.