Assumptions in Bernoulli — steady, inviscid, incompressible, along streamline: Visual walkthrough

Step 1 — Draw the road: a streamline

WHAT. We pick one such curve and call the distance measured along it $s$ (metres). As a blob travels, $s$ increases. Its height above the ground is $h$ (metres, measured straight up). Its speed along the road is $v$ (metres per second).

WHY these three? Because a moving blob can only do three energy-related things: push (pressure), move (speed $v$ ), climb (height $h$ ). We track exactly those and nothing else.

PICTURE. The red curve is the streamline. The arrow shows the blob's velocity, always tangent to the road.

$s$ ::: distance travelled along the curved streamline (not straight-line distance).
$v$ ::: the blob's speed, always pointing along the road.
$h$ ::: the blob's height above a chosen floor.

Step 2 — Cut out one tiny blob and name its size

WHAT. Zoom in and slice out a tiny cylinder of fluid sitting on the road. It has:

length $d s$ along the road (a tiny step),
cross-sectional area $A$ (the size of its front and back faces),
density $ρ$ (kilograms of water per cubic metre).

WHY a cylinder? Because pressure pushes on flat faces, and a cylinder gives us one clean back face and one clean front face pointing along the road — the only direction we care about.

Its mass is $mass = kg per m^{3} ρ \times volume A d s$

where $A d s$ is (face area) × (length) = the blob's volume, and multiplying by density $ρ$ turns volume into kilograms.

PICTURE. The red cylinder is our parcel; black arrows mark the two faces where pressure will act.

Step 3 — The two forces along the road

We only ever care about forces along the road (the $s$ direction), because sideways forces just bend the road; they don't change speed along it.

Force A — Pressure. Pressure $P$ (in pascals = newtons per m²) pushes on a face with force $P \times A$ .

Back face pushes the blob forward: $+ P A$ .
Front face pushes the blob backward, but the pressure there is slightly different, $P + d P$ : force $- (P + d P) A$ .

Add them: $push from behind P A - push from front (P + d P) A = - A d P$

Notice the whole $P A$ cancels — only the change $d P$ survives. So the net pressure force is $- A d P$ : if pressure grows as you go forward ( $d P > 0$ ), the net push is backward (minus sign).

Force B — Gravity. The blob's weight is $mass \times g = ρ A d s g$ , pointing straight down. Only the part along the road slows it or speeds it. That part is $- ρ A d s g \frac{d h}{d s}$ Here $\frac{d h}{d s}$ is "how much height you gain per step of road" — the steepness. On a flat stretch $d h / d s = 0$ (gravity does nothing along the road); on a vertical climb $d h / d s = 1$ (all of gravity fights you). The minus sign says: climbing ( $d h > 0$ ) pulls you back.

PICTURE. Red pressure arrows on the two faces; black weight arrow split into "along road" and "sideways" pieces.

Step 4 — Newton's law needs an acceleration: the two ways a blob speeds up

WHAT. Newton says $mass \times acceleration = force$ . We have mass and force; we need the blob's acceleration $a$ . But a moving blob's acceleration is subtle — it is the material derivative: $a = \frac{D v}{D t} = clock term \frac{\partial v}{\partial t} + travel term v \frac{\partial v}{\partial s}$

WHY two pieces? A blob can gain speed for two totally different reasons:

Clock term $\frac{\partial v}{\partial t}$ — you stand still at one spot and watch the whole flow crank up over time (someone opens a valve). Even a motionless blob would speed up.
Travel term $v \frac{\partial v}{\partial s}$ — nothing changes with time, but the blob drives into a narrower section where water is naturally faster. The $v$ out front is "how fast you cover road," and $\partial v / \partial s$ is "how much the speed grows per metre of road." Multiplied together = speed gained per second by travelling.

PICTURE. Left panel: same spot, flow rising with time (clock). Right panel: frozen-in-time pipe that narrows, so moving right means going faster (travel).

Step 5 — Assemble Newton's law for the blob

WHAT. Put mass × acceleration = net force: $mass ρ A d s (\frac{\partial v}{\partial t} + v \frac{\partial v}{\partial s}) = net force from Step 3 - A d P - ρ A g d h$

WHY divide by $A$ ? Area appears in every term, so it carries no physics — kill it. Divide through by $A$ : $ρ d s (\frac{\partial v}{\partial t} + v \frac{\partial v}{\partial s}) = - d P - ρ g d h$

PICTURE. The full force balance drawn on the blob: inertia on the left, pressure + gravity on the right, area crossed out.

This equation is still fully general — it is just Newton's law for a fluid blob on a streamline. No assumption made yet. Watch each assumption now delete a piece.

Step 6 — Assumption 1: STEADY kills the clock term

WHAT. Steady means: at any fixed point, nothing changes with time. So $\frac{\partial v}{\partial t} = 0$ . The clock term vanishes: $ρ d s (\frac{\partial v}{\partial t} + v \frac{\partial v}{\partial s}) = - d P - ρ g d h$

Notice $d s \cdot \frac{\partial v}{\partial s} = d v$ (the change in speed over that little step), so the left side tidies to $ρ v d v$ : $ρ v d v = - d P - ρ g d h$

WHY it matters. If someone opens a valve (Step 4 left panel), the clock term is not zero, and this line is a lie — the blob is being shoved by time-changes that $P$ , $v$ , $h$ at a point cannot bookkeep. That is the "tap turned on suddenly" case.

PICTURE. The general equation with the clock term struck out in red, and the "before/after" flow snapshots shown identical (that's what steady looks like).

Step 7 — Assumption 2: INVISCID (the force we never drew)

WHAT. Look back at Step 3. We drew pressure and gravity — but no friction. Real water rubs against walls and against itself (viscosity), producing a shear force along the side of the blob, roughly $- τ (side area)$ .

WHY it's an assumption. By never writing that term, we silently assumed it is zero — that the fluid is inviscid. If it isn't, that missing force steals mechanical energy and turns it into heat, so our "constant" quietly leaks away downstream. This is exactly the rough-pipe example: same $v$ , same $h$ , yet pressure falls.

PICTURE. The blob with a red shear-drag arrow on its side surface — the arrow we left out — labelled "energy → heat."

Step 8 — Assumption 3: INCOMPRESSIBLE lets us integrate cleanly

WHAT. We now have $ρ v d v = - d P - ρ g d h$ . To turn this "tiny-step" statement into a whole-journey statement we integrate along the road (add up every step from point 1 to point 2): $\int ρ v d v + \int d P + \int ρ g d h = 0$

$\int d P = P$ — this one is always fine, no matter what the density does.
$\int ρ v d v = \frac{1}{2} ρ v^{2}$ — kinetic energy per volume, but only if we can slide $ρ$ outside the integral.
$\int ρ g d h = ρ g h$ — potential energy per volume, again only if $ρ$ comes out of the integral.

WHY incompressible — the real reason. The catch is not the $\int d P$ term (that always gives $P$ ). The catch is the two terms with $ρ$ inside them. If the density changes along the road, you cannot pull $ρ$ out front — $\int ρ v d v$ and $\int ρ g d h$ no longer collapse to the tidy $\frac{1}{2} ρ v^{2}$ and $ρ g h$ , and an extra internal-energy term appears. Assuming incompressible ( $ρ$ constant) is exactly the licence to factor $ρ$ out of both integrals.

For a gas, $ρ$ rides up and down with pressure, so this breaks. Rule of thumb: OK while Mach number $< 0.3$ .

Assuming $ρ$ constant and adding up: $P + \frac{1}{2} ρ v^{2} + ρ g h = const$

PICTURE. Left: liquid blob keeps its size along the road (constant $ρ$ ) → $ρ$ factors out of both energy integrals. Right: gas blob squeezed smaller in a high-pressure region (variable $ρ$ ) → $ρ$ trapped inside the integral.

Step 9 — Assumption 4: ALONG ONE STREAMLINE (and the degenerate multi-road case)

WHAT. Every single integral in Step 8 was taken along one road $s$ . So the constant we got is guaranteed the same only for points on that same streamline. A different road can carry a different constant.

WHY / degenerate case. In a draining vortex, inner streamlines spin faster than outer ones. Comparing an inner point to an outer point with one constant is nonsense — unless the flow is irrotational, a special condition that makes the constant the same everywhere.

PICTURE. Two separate red streamlines, each labelled with its own constant $C_{1} \neq = C_{2}$ ; a dashed arrow across them marked "not allowed unless irrotational."

Recall Quick self-test

On which of these can you equate Bernoulli directly? ::: Only two points lying on the same streamline (or anywhere, if the flow is irrotational).

The one-picture summary

The single figure above stacks the whole story: blob → forces → Newton with material derivative → strike the clock term (steady) → note the missing shear (inviscid) → factor $ρ$ out of the energy integrals (incompressible) → keep it on one road (along a streamline) → Bernoulli.

Recall Feynman retelling — the whole walk in plain words

Imagine a marble-sized blob of water sliding down a curved slide. Two things push it along the slide: pressure (the water behind shoves harder than the water in front, so only the difference in pressure counts) and gravity (only the part pointing down-slope helps or fights, set by how steep the slide is).

Newton says push equals mass times acceleration. But a moving blob accelerates two ways: because the whole river is speeding up over time (the clock), or because it slides into a faster stretch (travelling into a narrow part). Both together are the "material derivative."

Now we make four promises. Steady: the river isn't changing with time — the clock term dies. Inviscid: we pretend there's no friction — we simply never drew that drag arrow, so we're assuming it's zero. Incompressible: the blob never gets squished, so its density stays put — that lets us pull $ρ$ out of the kinetic and potential integrals and add everything up cleanly. Along one streamline: we only ever added things up along the one road the blob is on, so our final answer is a promise about that road only.

Add up kinetic energy ( $\frac{1}{2} ρ v^{2}$ ), pushing energy ( $P$ ), and climbing energy ( $ρ g h$ ) — their sum stays constant. Break any one of the four promises and the sum starts to drift: friction bleeds it into heat, unsteadiness sneaks in a hidden shove, compression stores it inside the gas, and hopping to another road lands you on a different number entirely.

Visual walkthrough — Assumptions in Bernoulli — steady, inviscid, incompressible, along streamline