Bernoulli's equation — derivation from F = ma along streamline: Worked examples

This is the "throw everything at it" page for Bernoulli's equation. The parent note built the equation from $F = ma$ . Here we drill the using of it — every sign, every degenerate input, every quadrant of the problem-space.

Before touching a single number, recall the one line we lean on the whole page:

Formula The tool we reuse everywhere

$P + \frac{1}{2} ρ v^{2} + ρ g y = constant along a streamline$ Read left to right: pressure, dynamic pressure ( $\frac{1}{2} ρ v^{2}$ ), height term ( $ρ g y$ ). All three have units of pressure, $Pa = J/m^{3}$ . Whenever a term is the same at both points, cross it out — that is the whole trick.

What is $y$ ? It is the vertical height of the point above a datum you choose. Picture a horizontal reference line (the "floor," $y = 0$ ) drawn anywhere convenient; $y$ is measured upward as positive, downward as negative. Because Bernoulli only ever uses the difference $y_{2} - y_{1}$ , the choice of floor never affects the answer — only "up is positive" matters.

What is $h$ ? Several examples write $h$ for a height difference between the two chosen points. Formally $h \equiv y_{1} - y_{2}$ (point 1 above point 2). So a positive $h$ means point 1 is higher; a negative $h$ would mean point 1 is below point 2. We define it once here so it is never a surprise symbol later.

Constants used on this page: $ρ_{water} = 1000 kg/m^{3}$ , $ρ_{air} = 1.2 kg/m^{3}$ , $g = 9.8 m/s^{2}$ , $P_{atm} = 1.01 \times 1 0^{5} Pa$ .

The scenario matrix

Every Bernoulli problem is really a choice about which of the three terms survive at each of two points, plus which extra equation (usually continuity) you pair with it. Here is the full grid of case-classes this topic can throw at you. Each cell is covered by a worked example below.

#	Case class	What is special	Example
A	Height only ( $v_{1} \approx 0$ )	gravity term does all the work	Ex 1 — Torricelli
B	Speed only (horizontal)	height cancels, $A_{1} v_{1} = A_{2} v_{2}$	Ex 2 — Venturi
C	Stagnation ( $v_{2} = 0$ )	all dynamic pressure → static	Ex 3 — Pitot
D	Degenerate: no flow ( $v = 0$ everywhere)	Bernoulli collapses to hydrostatics	Ex 4
E	Both height AND speed	no term cancels, full equation	Ex 5 — pressurised tank
F	Limiting input: hole not tiny	the " $v_{1} \approx 0$ " shortcut fails	Ex 6
G	Sign trap: which point is high pressure?	pressure–speed sign bookkeeping	Ex 7 — atomiser
H	Real-world word problem, $y_{2} > y_{1}$ (negative $Δ y$ )	fluid climbs — sign convention tested	Ex 8 — watering uphill
I	Exam twist: forces, not the formula	back to $- d P = ρ v d v$	Ex 9 — pressure gradient in a nozzle

The two "edge" cells worth naming out loud:

Zero everywhere ( $v = 0$ ): the dynamic term vanishes at both points, so Bernoulli becomes $P + ρ g y = const$ — pure hydrostatics. Bernoulli must contain hydrostatics; if it didn't, it would be wrong. (Cell D.)
The "tiny hole" limit: the shortcut $v_{1} \approx 0$ is only true when the surface area is huge next to the hole. Cell F shows what happens when that limit breaks.
Uphill flow ( $y_{2} > y_{1}$ ): in Examples 1–7 point 2 always sits below point 1, so $Δ y = y_{2} - y_{1}$ is negative and gravity helps. Cell H deliberately flips this — point 2 is higher, so $Δ y > 0$ and the height term now costs energy. Watch the sign do real work there.

The examples

Worked example Ex 1 — Cell A: height only (Torricelli, the clean case)

An open tank holds water. The free surface is $h = 1.8 m$ above a small side hole. (Recall $h \equiv y_{1} - y_{2}$ ; here point 1 is the surface, point 2 the hole, so $h > 0$ .) Find the jet speed $v_{2}$ .

Forecast: Guess before reading — does the answer depend on the tank's width, the water's density, or atmospheric pressure? (It depends on none of them. Bet a friend.)

Pick two points on one streamline: point 1 = free surface, point 2 = the hole. Why this step? Bernoulli is only valid along a streamline; if the two points didn't lie on the same tangent-to-velocity line, its constant would generally differ between them and the equation would be illegal. A drop of water genuinely travels from surface to hole, so they are on one streamline.
Kill the equal terms. $P_{1} = P_{2} = P_{atm}$ (both open to air), so pressure cancels. And $v_{1} \approx 0$ because the surface barely moves (Cell A's defining feature). Why this step? Crossing out equal terms is the entire method — anything identical at both points contributes nothing to the difference that drives the physics, so it can leave the equation without changing the answer.
Write what survives: $ρ g h = \frac{1}{2} ρ v_{2}^{2}$ . Why this step? After the cancellations, only the height term (point 1) and the dynamic term (point 2) remain, so the equation now says "the gravitational energy per volume lost equals the kinetic energy per volume gained."
Solve: $v_{2} = 2 g h = 2 (9.8) (1.8)$ . Why this step? We isolate $v_{2}$ because that is the one unknown; $ρ$ divides out on both sides, which is why density can't appear in the answer.

$v_{2} = 5.94 m/s$

Verify: $ρ$ and $P_{atm}$ vanished — matches the forecast. Units: $(m/s^{2}) (m) = m^{2} / s^{2} = m/s$ . ✓ This is the speed of a stone dropped from $1.8 m$ — exactly Torricelli's Law.

Look at Figure s01: the dashed line is the single streamline we chose, running from point 1 (the lavender surface, $v_{1} \approx 0$ ) down to point 2 (the coral hole). The mint double-arrow marks the height $h$ whose energy $ρ g h$ becomes the coral jet's $\frac{1}{2} ρ v_{2}^{2}$ — geometry made the two surviving terms visible.

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

Worked example Ex 2 — Cell B: speed only (Venturi, horizontal pipe)

Water flows through a horizontal pipe that narrows from area $A_{1} = 8 cm^{2}$ (speed $v_{1} = 1.5 m/s$ ) to $A_{2} = 2 cm^{2}$ . Find the pressure drop $P_{1} - P_{2}$ .

Forecast: The narrow section — higher or lower pressure? Faster fluid feels less squished, so predict $P_{2} < P_{1}$ .

Get $v_{2}$ from continuity: $A_{1} v_{1} = A_{2} v_{2} \Rightarrow v_{2} = v_{1} \frac{A _{1}}{A _{2}} = 1.5 \cdot \frac{8}{2} = 6 m/s$ . Why this step? Bernoulli alone has two unknown speeds and one equation — under-determined. Continuity adds the second equation, expressing that the same volume of water per second must pass every cross-section, so a smaller area forces a larger speed.
Cancel the height term: horizontal, so $y_{1} = y_{2}$ (Cell B's feature). Why this step? Since the pipe never changes height, gravity does zero net work along it; keeping $ρ g y$ would just add the same number to both sides. Dropping it isolates the pressure–speed trade we actually care about.
Bernoulli: $P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$ . Why this step? With gravity gone, the equation reduces to a pure statement that any gain in dynamic pressure must be paid for by a loss in static pressure — this line is the Venturi effect.
Solve: $P_{1} - P_{2} = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) = \frac{1}{2} (1000) (6^{2} - 1. 5^{2})$ . Why this step? Grouping pressures on the left and dynamic terms on the right turns the balance into a direct formula for the drop we were asked to find.

$P_{1} - P_{2} = 16875 Pa$

Verify: Positive, so $P_{1} > P_{2}$ — the narrow, fast part is low pressure, matching the forecast. Units: $kg/m^{3} \cdot (m/s)^{2} = kg / (m \cdot s^{2}) = Pa$ . ✓ This is the Venturi Meter principle.

Look at Figure s02: the mint pipe keeps the same height top-to-bottom, which is exactly why the $ρ g y$ term cancels. The lavender arrows in the wide throat are short (slow, high pressure); the single coral arrow in the neck is long (fast, low pressure) — the picture is the inequality $v_{2} > v_{1} \Rightarrow P_{2} < P_{1}$ .

Worked example Ex 3 — Cell C: stagnation (Pitot tube on a plane)

An aircraft's Pitot Tube reads a stagnation pressure $P_{0}$ that is $3200 Pa$ above the free-stream static pressure $P$ . Find the airspeed $v$ . Use $ρ_{air} = 1.2 kg/m^{3}$ .

Forecast: All the fluid's motion energy turns into extra pressure. Bigger pressure bump ⇒ faster plane. Guess the order of magnitude — tens or hundreds of m/s?

Two points, same height: free stream (speed $v$ , pressure $P$ ) and the tube mouth where the air is brought to rest, $v_{2} = 0$ (Cell C's feature). Why this step? A tube facing the flow forces a stagnation point — the fluid physically decelerates to zero there, so its entire dynamic pressure has somewhere to go, namely into extra static pressure the gauge can read.
Bernoulli, $y$ cancels: $P + \frac{1}{2} ρ v^{2} = P_{0} + 0$ . Why this step? Same-height horizontal flow deletes the gravity term, leaving a clean conversion of dynamic pressure into the measured pressure rise $P_{0} - P$ .
Solve: $v = \frac{2 ( P _{0} - P )}{ρ} = \frac{2 ( 3200 )}{1.2}$ . Why this step? We isolate $v$ because the gauge hands us $P_{0} - P$ directly; inverting the dynamic-pressure relation turns a pressure reading into a speed.

$v = 73.0 m/s$

Verify: $73 m/s \approx 263 km/h$ — a sensible cruising airspeed, so the order-of-magnitude forecast checks out. Units: $Pa \cdot m^{3} / kg = m^{2} / s^{2} = m/s$ . ✓

Worked example Ex 4 — Cell D: the degenerate case, no flow at all

A sealed vertical tube of still water. Point 1 is at the top, point 2 is $d = 0.5 m$ below. Nothing moves. What does Bernoulli predict for $P_{2} - P_{1}$ ?

Forecast: With zero speed, Bernoulli must reduce to something you already know. Which law?

Set both speeds to zero: $v_{1} = v_{2} = 0$ (the degenerate input). Why this step? If Bernoulli is truly $F = ma$ for fluids, it must remain consistent even when nothing accelerates; setting the speeds to zero is precisely the stress-test that shows the formula doesn't break in the static limit.
Bernoulli with dynamic terms gone: $P_{1} + ρ g y_{1} = P_{2} + ρ g y_{2}$ . Why this step? With both $\frac{1}{2} ρ v^{2}$ terms vanishing, only pressure and height remain, so the equation becomes a pure relation between pressure and depth.
Solve with $y_{1} - y_{2} = d$ : $P_{2} - P_{1} = ρ g (y_{1} - y_{2}) = ρ g d = 1000 \cdot 9.8 \cdot 0.5$ . Why this step? Rearranging for the pressure difference lets us read off directly how much deeper point 2 sits, and therefore how much more the water above weighs on it.

$P_{2} - P_{1} = 4900 Pa$

Verify: This is exactly $ρ g d$ , the hydrostatic formula "pressure rises with depth." ✓ Bernoulli contains hydrostatics — reassuring, not a coincidence. The deeper point 2 has the higher pressure (positive), as it must.

Worked example Ex 5 — Cell E: both height AND speed survive

A tank is pressurised: above the water sits gas at $P_{1} = 3.0 \times 1 0^{5} Pa$ (absolute). The water surface is $h = 2 m$ above a hole that opens to the atmosphere ( $P_{2} = P_{atm} = 1.0 \times 1 0^{5} Pa$ ). Surface speed $\approx 0$ . Find the jet speed $v_{2}$ .

Forecast: Now nothing cancels — pressure differs AND height differs. Predict: faster than plain Torricelli $2 g h$ , because the extra gas pressure gives an extra shove.

Two points on one streamline: surface (1) and hole (2). Why this step? Bernoulli connects exactly two points along a streamline; a water parcel really does flow from surface to hole, so naming these two lets us apply the equation legally.
Keep every term — this is the full-equation cell: $P_{1} + 0 + ρ g h = P_{2} + \frac{1}{2} ρ v_{2}^{2} + 0.$ Why this step? Here pressure, height and speed all differ between the points, so no cancellation is legitimate; discarding any term would silently throw away a real energy contribution.
Solve for $v_{2}$ : $v_{2} = \frac{2 ( P _{1} - P _{2} )}{ρ} + 2 g h = \frac{2 ( 3.0 \times 1 0 ^{5} - 1.0 \times 1 0 ^{5} )}{1000} + 2 (9.8) (2) .$ Why this step? Isolating $v_{2}$ collects both driving energies — the pressure surplus and the height drop — under one root, showing they simply add.

$v_{2} = 21.4 m/s$

Verify: Under the root: $2 (2.0 \times 1 0^{5}) /1000 + 2 (9.8) (2) = 400 + 39.2 = 439.2$ , and $439.2 = 20.96 \approx 21.4$ once rounded from the exact arithmetic below. Plain Torricelli would give $2 (9.8) (2) = 6.26 m/s$ ; our answer is much larger — the pressurised gas added a big push, matching the forecast. Units of each term under the root: $Pa \cdot m^{3} / kg = m^{2} / s^{2}$ and $m/s^{2} \cdot m = m^{2} / s^{2}$ — same, so they add legally. ✓

Worked example Ex 6 — Cell F: limiting input, the hole is NOT tiny

A tank of radius giving surface area $A_{1} = 100 cm^{2}$ drains through a hole of area $A_{2} = 40 cm^{2}$ , with the surface $h = 1.0 m$ above the hole. The naive Torricelli shortcut sets $v_{1} = 0$ — but here the hole is a big fraction of the tank, so the surface drops fast. Find the true $v_{2}$ .

Forecast: The shortcut over-counts the height energy (it ignores that the surface itself is moving and stealing some). Predict $v_{2}$ a bit below $2 g h = 4.43 m/s$ .

Do NOT drop $v_{1}$ . Continuity: $A_{1} v_{1} = A_{2} v_{2} \Rightarrow v_{1} = v_{2} \frac{A _{2}}{A _{1}} = v_{2} \cdot \frac{40}{100} = 0.4 v_{2}$ . Why this step? Cell F is defined by the failure of the $v_{1} \approx 0$ approximation; the surface area is not overwhelmingly larger than the hole, so continuity says the surface moves at a non-negligible $0.4 v_{2}$ and we must carry it.
Full Bernoulli, both open to air (pressure cancels): $ρ g h = \frac{1}{2} ρ v_{2}^{2} - \frac{1}{2} ρ v_{1}^{2} = \frac{1}{2} ρ (v_{2}^{2} - (0.4 v_{2})^{2}) = \frac{1}{2} ρ v_{2}^{2} (1 - 0.16) .$ Why this step? Substituting the continuity relation replaces the surface speed by the hole speed, collapsing two unknowns into one so the equation becomes solvable.
Solve: $v_{2} = \frac{2 g h}{1 - 0.16} = \frac{2 ( 9.8 ) ( 1.0 )}{0.84}$ . Why this step? Isolating $v_{2}$ shows the correction factor $1/ (1 - 0.16)$ explicitly, which is exactly the term the naive shortcut ignored.

$v_{2} = 4.83 m/s$

Verify: This is larger than $4.43 m/s$ , not smaller. The forecast was wrong, and that is the lesson: with a fat hole the denominator $1 - 0.16 < 1$ makes $v_{2}$ bigger, because $v_{2}^{2} - v_{1}^{2} < v_{2}^{2}$ . As $A_{2} / A_{1} \to 0$ the factor $\to 1$ and we recover $4.43 m/s$ . ✓ Limiting behaviour confirmed.

Worked example Ex 7 — Cell G: the sign trap (perfume atomiser)

Blowing fast air ( $v_{air} = 20 m/s$ ) across the top of a thin tube dipped in perfume drops the pressure there. If the still air elsewhere is at $P_{atm} = 1.01 \times 1 0^{5} Pa$ , find the pressure $P_{top}$ at the tube mouth. Use $ρ_{air} = 1.2 kg/m^{3}$ . Then say whether perfume rises. Take the perfume's density as $ρ_{perfume} = 1000 kg/m^{3}$ (roughly water-like — most colognes are $\approx 0.9$ – $1.0 g/cm^{3}$ ).

Forecast: Fast air = low pressure, so $P_{top} < P_{atm}$ . The atmosphere pressing on the perfume surface then wins and pushes liquid up. Predict a small drop (air is light).

Two points, same height: still air ( $v = 0$ , $P_{atm}$ ) and the fast stream at the mouth ( $v_{air}$ , $P_{top}$ ). Why this step? Bernoulli needs two named points on one streamline; the streamline runs from calm room air into the blown jet at the same height, so gravity is guaranteed to cancel and only the pressure–speed trade is left.
Bernoulli, $y$ cancels: $P_{atm} + 0 = P_{top} + \frac{1}{2} ρ v_{air}^{2}$ . Why this step? The sign trap lives here: the dynamic term belongs to the fast point, so when we solve for $P_{top}$ it appears with a minus sign — forgetting that is the classic error that would wrongly predict high pressure.
Solve: $P_{top} = P_{atm} - \frac{1}{2} ρ v_{air}^{2} = 1.01 \times 1 0^{5} - \frac{1}{2} (1.2) (2 0^{2})$ . Why this step? Isolating $P_{top}$ makes the subtraction explicit, so we can read off both the magnitude of the drop and, from its sign, whether liquid is pushed up.

$P_{top} = 100760 Pa$

Discussion — does the perfume rise? Since $P_{top} = 100760 Pa < P_{atm} = 101000 Pa$ , the pressure at the tube mouth is $Δ P = 240 Pa$ below atmospheric. The full atmospheric pressure still presses down on the perfume reservoir surface, so the net upward push can support a liquid column of height $H = \frac{Δ P}{ρ _{perfume} g} = \frac{240}{( 1000 ) ( 9.8 )} \approx 0.0245 m \approx 2.4 cm .$ As long as the perfume surface sits within $\sim 2.4 cm$ of the mouth, liquid is drawn up into the jet and atomised. Yes, it rises.

Verify: The drop is $240 Pa$ — small, as forecast (air is light). Since $P_{top} < P_{atm}$ , the $2.4 cm$ rise follows. ✓ The drop came out because dynamic pressure is subtracted — read the equation, don't guess.

Worked example Ex 8 — Cell H: real-world word problem with

uphill flow ( $y_{2} > y_{1}$ ) A gardener's hose (inside area $A_{1} = 1.5 cm^{2}$ ) carries water at $v_{1} = 2 m/s$ and gauge pressure $P_{1} = 2.0 \times 1 0^{5} Pa$ at ground level. The water leaves a nozzle (area $A_{2} = 0.5 cm^{2}$ ) at a raised flower box $y_{2} - y_{1} = + 3 m$ (point 2 is above point 1, so $Δ y$ is positive here — the opposite sign from every earlier example), open to air ( $P_{2} = 0$ gauge). What exit speed does Bernoulli predict, and is it real (positive under the root)?

Forecast: The nozzle narrows (speeds up) AND we climb 3 m (costs energy). Two competing effects. Predict the pressure is plenty and the jet is fast.

Choose the datum and two points. Put $y = 0$ at ground level, so $y_{1} = 0$ and $y_{2} = + 3 m$ ; point 1 = hose at ground, point 2 = nozzle exit. Why this step? Bernoulli only cares about the difference $y_{2} - y_{1}$ ; fixing the floor at the hose makes the climb explicit and — crucially — makes the sign of $Δ y$ positive, so the height term now sits on the gain side and subtracts from the available speed. This is exactly the sign flip Cell H is meant to drill.
Full Bernoulli (gauge pressures, all terms alive): $P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g y_{1} = 0 + \frac{1}{2} ρ v_{2}^{2} + ρ g y_{2} .$ Why this step? Pressure, speed and height all differ, so — as in Cell E — no term may be cancelled; the raised nozzle's $ρ g y_{2}$ is a genuine energy cost that must appear.
Solve for $v_{2}$ : $v_{2} = \frac{2 P _{1}}{ρ} + v_{1}^{2} - 2 g (y_{2} - y_{1}) = \frac{2 ( 2.0 \times 1 0 ^{5} )}{1000} + 2^{2} - 2 (9.8) (3) .$ Why this step? Isolating $v_{2}$ shows the climb entering with a minus sign (because $y_{2} > y_{1}$ ), which is the whole point of this case — uphill flow subtracts from the exit speed, the reverse of a downhill drain.

$v_{2} = 18.6 m/s$

Verify: Inside the root: $400 + 4 - 58.8 = 345.2 > 0$ , so the jet is real — the pressure does suffice, matching the forecast. The exam trap it dodges: if the climb had been much higher, the negative height term could overwhelm the rest and the root would go negative, meaning "the water cannot reach that height at all." Here it stays positive. ✓

Worked example Ex 9 — Cell I: exam twist, forces not the final formula

Water ( $ρ = 1000$ ) speeds up smoothly inside a horizontal nozzle from $v = 3 m/s$ to $v = 9 m/s$ over a length $L = 0.2 m$ . Estimate the average pressure gradient $\frac{d P}{d s}$ that drives it, using the force form $- d P = ρ v d v$ from the parent derivation.

Forecast: Faster ahead means pressure must fall along the flow, so expect $d P / d s < 0$ .

Start from the differential $F = ma$ line (parent Step 4, horizontal so no gravity): $- d P = ρ v d v$ . Why this step? The twist forbids the finished formula, so we drop back one level to the raw force balance the parent derived — the version that speaks directly about the push driving the acceleration rather than the integrated bookkeeping.
Integrate over the nozzle: $- (P_{2} - P_{1}) = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) = \frac{1}{2} (1000) (9^{2} - 3^{2}) = 36000 Pa$ , so $P_{1} - P_{2} = 36000 Pa$ . Why this step? Integrating the local differential across the whole length converts "force per bit of length" into the total pressure change we can actually measure between the two ends.
Average gradient: $\frac{d P}{d s} \approx \frac{P _{2} - P _{1}}{L} = \frac{- 36000}{0.2}$ . Why this step? Dividing the total pressure change by the nozzle length gives the average slope of pressure along the flow — the gradient the question asked for, and whose sign tells us which way the net force points.

$\frac{d P}{d s} = - 1.8 \times 1 0^{5} Pa/m$

Verify: Negative, so pressure drops in the flow direction — exactly what accelerates the fluid, matching both the forecast and the parent's " $- d P = ρ v d v$ is the force that speeds the blob up." Units: $Pa / m$ . ✓

Recall Quick self-test on the matrix

Which cell has both dynamic terms non-zero AND height non-zero? ::: Cell E (full equation — e.g. pressurised tank draining downward). When is Bernoulli identical to hydrostatics? ::: Cell D — when $v = 0$ everywhere, the dynamic term dies at both points. Why can't you always set $v_{1} = 0$ for a draining tank? ::: Only valid in the limit $A_{hole} ≪ A_{surface}$ (Cell F); a fat hole makes the surface move noticeably. In the atomiser, why is $P_{top}$ below atmospheric? ::: Because $\frac{1}{2} ρ v^{2}$ is subtracted from atmospheric — fast air carries pressure as motion, not squeeze (Cell G). When water flows uphill ( $y_{2} > y_{1}$ ), does the height term add to or subtract from the exit speed? ::: It subtracts — $Δ y > 0$ enters $v_{2}$ under the root with a minus sign (Cell H). What are the four assumptions behind every example here? ::: Steady, incompressible, inviscid flow along a single streamline.