Assumptions in Bernoulli — steady, inviscid, incompressible, along streamline: Worked examples

We build everything from the same master statement (never used before it is earned):

Formula The full parcel equation (before we throw anything away)

$unsteady ρ d s \frac{\partial v}{\partial t} + convective ρ v d v = pressure - d P gravity - ρ g d h + friction d F_{visc} + added energy d F_{pump}$ Read it as Newton's 2nd law on a sliding fluid blob. $P$ = pressure (force per area, unit pascal $= N/m^{2}$ ), $v$ = speed, $h$ = height, $ρ$ = density (mass per volume, $kg/m^{3}$ ), $g = 9.8 m/s^{2}$ , $d s$ = a tiny step along the streamline, $A$ = the parcel's cross-sectional area ( $m^{2}$ ). Bernoulli drops three of the six terms: it demands $\partial v / \partial t = 0$ so the unsteady term vanishes, and it assumes no friction and no pump so $d F_{visc} = 0$ and $d F_{pump} = 0$ . What remains is the balance of convective, pressure and gravity terms — that is the whole of Bernoulli. Every failure below is one of those three dropped terms coming back to life.

Definition The five off-page tools, restated here so this page stands alone

Material derivative — the acceleration of a moving parcel has two parts: $\frac{D v}{D t} = \frac{\partial v}{\partial t} + v \frac{\partial v}{\partial s}$ . The first is time-change at a fixed spot, the second is speeding up by moving into a faster region. (Full build in Material Derivative.)
Continuity equation — mass in = mass out for incompressible flow, so $A_{1} v_{1} = A_{2} v_{2}$ : a narrower pipe forces faster flow. (See Continuity Equation.)
Viscosity — internal fluid friction; near a wall it drags the flow and turns mechanical energy into heat, described by Poiseuille flow in pipes.
Boundary layer — the ultra-thin skin of fluid right against a surface where speed falls from the free-stream value to zero; viscosity always dominates here even when the outer flow is inviscid. (See Boundary Layer.)
Mach number $M = v / c$ ( $c$ = speed of sound) — measures how compressible the flow acts; below $M < 0.3$ density change stays under ~4.5% and we treat the gas as incompressible. (See Mach Number and Compressibility.)

The scenario matrix

Every situation a Bernoulli problem can throw at you belongs to exactly one row below. Our examples (Ex1–Ex9) are tagged with the cell they cover.

Cell	Situation class	Which term wakes up?	Example
A	Ideal case — all four assumptions hold	none (Bernoulli exact)	Ex1
A′	Ideal case with a height change ( $h_{1} \neq = h_{2}$ )	gravity term $ρ g d h$ active	Ex2
B	Friction present (rough/long/viscous)	$d F_{visc}$	Ex3
C	Unsteady / start-up / pulsing	$\partial v / \partial t$	Ex4
D	Compressible gas (high Mach)	$ρ$ not constant in $\int d P / ρ$	Ex5
E	Two different streamlines	constant differs per streamline	Ex6
F	Energy added by a pump/fan	$d F_{pump}$	Ex7
G	Degenerate / zero input ( $v = 0$ , flat, equal area)	terms vanish → limiting form	Ex8
H	Exam twist — a "valid-looking" trap combining cells	spot the hidden broken assumption	Ex9

The idea of a streamline (a curve everywhere tangent to velocity, that no fluid crosses) and the Material Derivative (the two-part acceleration) are the two tools we lean on; both are defined in the parent.

Cell A — the ideal case (all assumptions hold)

Worked example Ex1 · Water speeding through a smooth nozzle (Cell A)

Water ( $ρ = 1000 kg/m^{3}$ ) flows steadily through a smooth, level nozzle. At the wide inlet $v_{1} = 2 m/s$ , $P_{1} = 300 kPa$ . The nozzle narrows so that at the outlet $v_{2} = 8 m/s$ . Find $P_{2}$ .

Forecast: The water speeds up 4×. Does the pressure go up or down, and by roughly how much — tens of kPa, or hundreds?

Check the license. Steady ✔, smooth ⇒ inviscid ✔, water ⇒ incompressible ✔, single tube ⇒ one streamline ✔. Why this step? Cell A means we may use plain Bernoulli only after confirming all four; skipping this is the mistake steel-manned in the parent.
Write Bernoulli, cancel the height term. Level pipe ⇒ $h_{1} = h_{2}$ , so $ρ g h$ drops out: $P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2}$ Why this step? With gravity gone, the only trade is pressure ⇄ kinetic energy.
Solve for $P_{2}$ . $P_{2} = P_{1} + \frac{1}{2} ρ (v_{1}^{2} - v_{2}^{2}) = 300000 + \frac{1}{2} (1000) (4 - 64)$ $P_{2} = 300000 - 30000 = 270000 Pa = 270 kPa$ Why this step? Faster water ⇒ lower pressure; the $30 kPa$ drop is the "kinetic tax".

Verify: Units: $ρ v^{2}$ has $kg/m^{3} \cdot m^{2} / s^{2} = kg/(m\cdots^{2}) = Pa$ ✔. Sanity: pressure fell (correct — speed rose), and the total $P + \frac{1}{2} ρ v^{2}$ equals $302 kPa$ at the inlet and $302 kPa$ at the outlet. ✔

Cell A′ — ideal case with a height change

Worked example Ex2 · Water rising to a rooftop tank (Cell A′)

Water ( $ρ = 1000 kg/m^{3}$ ) flows steadily and smoothly up a pipe. At the ground point $P_{1} = 400 kPa$ , $v_{1} = 3 m/s$ , $h_{1} = 0$ . It emerges at a rooftop nozzle $h_{2} = 12 m$ higher where the speed is $v_{2} = 3 m/s$ (same area). Find $P_{2}$ .

Forecast: Only the height changes here. Will the pressure at the top be higher or lower than at the ground, and by how many kPa?

Check the license. Steady ✔, smooth ⇒ inviscid ✔, water ⇒ incompressible ✔, one pipe ⇒ one streamline ✔. Why this step? All four hold, so Bernoulli is exact — but now the gravity term stays.
Keep the $ρ g h$ term. Since $h_{1} \neq = h_{2}$ we may not drop it: $P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g h_{1} = P_{2} + \frac{1}{2} ρ v_{2}^{2} + ρ g h_{2}$ Why this step? Cell A′ is the whole reason the gravity term exists in the parent's derivation; equal speeds make the kinetic terms cancel.
Solve for $P_{2}$ . With $v_{1} = v_{2}$ the kinetic terms cancel: $P_{2} = P_{1} - ρ g (h_{2} - h_{1}) = 400000 - 1000 \cdot 9.8 \cdot 12$ $P_{2} = 400000 - 117600 = 282400 Pa \approx 282.4 kPa$ Why this step? Lifting water 12 m costs pressure energy — the gravity tax, mirror image of the kinetic tax in Ex1.

Verify: $ρ g Δ h = 1000 \cdot 9.8 \cdot 12 = 117600 Pa$ ✔. Units: $kg/m^{3} \cdot m/s^{2} \cdot m = Pa$ ✔. Pressure dropped going up — correct, gravity opposes the rise.

Cell B — friction wakes up

Worked example Ex3 · The rusty 50 m pipe (Cell B)

Water flows through a 50 m rough horizontal pipe of constant area. $P_{1} = 200 kPa$ , $P_{2} = 150 kPa$ ; $v$ and $h$ are equal at both ends. Bernoulli predicts $P_{1} = P_{2}$ — what went wrong, and how much energy per volume was lost?

Forecast: Bernoulli insists the pressures are equal, yet they differ by 50 kPa. Which of the four terms did we illegally cross out?

Apply naïve Bernoulli. Constant area ⇒ $v_{1} = v_{2}$ (from the Continuity Equation: $A_{1} v_{1} = A_{2} v_{2}$ , so equal areas force equal speeds); horizontal ⇒ $h_{1} = h_{2}$ . So $P_{1} = P_{2}$ is predicted. Why this step? We show the prediction so the contradiction is undeniable.
Observe the contradiction. Reality: $Δ P = 50 kPa$ lost. Why this step? A prediction that fails by 50 kPa signals a missing term, not a math slip.
Identify the woken term. The rough wall exerts viscous shear (the fluid friction that drags flow near a wall); the crossed-out $d F_{visc}$ returns. The lost mechanical energy per unit volume equals the pressure drop, $Δ P = 50000 Pa$ , converted to heat. Why this step? In a level, equal-speed pipe there is nothing else for the pressure to become — it must be dissipation.

Verify: Head-loss form $P_{1} = P_{2} + Δ P_{loss}$ ⇒ $Δ P_{loss} = 200 - 150 = 50 kPa$ ✔. This is the friction inside the thin Boundary Layer near the wall — exactly where "inviscid" dies.

Cell C — unsteady start-up

Worked example Ex4 · Suddenly opened valve (Cell C)

A horizontal pipe of length $L = 10 m$ and cross-sectional area $A$ (any value — it cancels), full of still water, is opened. In the first instant the water column accelerates uniformly at $a = 4 m/s^{2}$ everywhere. What extra pressure difference between the two ends does this acceleration demand — a number plain Bernoulli cannot see?

Forecast: With $v$ still near zero everywhere, does the $\frac{1}{2} ρ v^{2}$ term explain the pressure drop that drives this acceleration? Guess yes/no.

Kill the wrong terms. At the first instant $v \approx 0$ , so the convective term $ρ v d v \approx 0$ . Why this step? It proves the driving force is not the kinetic term Bernoulli keeps.
Keep the unsteady term. Newton on the whole water column: its mass is $ρ A L$ (density × area × length) and the net force pushing it is the pressure difference acting on the end face of area $A$ , namely $Δ P \cdot A$ : $Δ P \cdot A = ρ A L a \Rightarrow Δ P = ρ L a (A cancels)$ Why this step? $\partial v / \partial t = a \neq = 0$ is exactly the steady assumption breaking; the area $A$ appears on both sides and drops out, so the answer is independent of pipe width.
Plug numbers. $Δ P = 1000 \cdot 10 \cdot 4 = 40000 Pa = 40 kPa$ Why this step? This "inertial / water-hammer" pressure is invisible to plain Bernoulli.

Verify: Units: $kg/m^{3} \cdot m \cdot m/s^{2} = kg/(m\cdots^{2}) = Pa$ ✔. Once flow settles ( $a \to 0$ ), $Δ P \to 0$ and Bernoulli returns — matches the parent's Example 2.

Cell D — compressible gas

Worked example Ex5 · Fast air where density changes (Cell D)

Air ( $ρ = 1.2 kg/m^{3}$ , speed of sound $c = 340 m/s$ ) accelerates in a duct from $v_{1} = 50 m/s$ to $v_{2} = 200 m/s$ . Is Bernoulli (incompressible form) trustworthy at each end? Use the Mach number $M = v / c$ (restated in the tools box: how compressible the flow acts).

Forecast: One end is safe, one is not. Which end breaks, and at what speed does the rule of thumb $M < 0.3$ get crossed?

Compute both Mach numbers. $M_{1} = \frac{50}{340} \approx 0.147, M_{2} = \frac{200}{340} \approx 0.588$ Why this step? $M$ measures how much the flow can compress the gas; $\int d P / ρ$ is only clean when $ρ$ is constant.
Estimate the actual density change. For a smooth (isentropic) gas flow the fractional density change is approximately $\frac{Δ ρ}{ρ} \approx \frac{1}{2} M^{2}$ . $At M_{1} : \frac{1}{2} (0.147)^{2} \approx 0.0108 (\approx 1%), At M_{2} : \frac{1}{2} (0.588)^{2} \approx 0.173 (\approx 17%)$ Why this step? This is the number behind the rule of thumb: at $M = 0.3$ , $\frac{1}{2} M^{2} \approx 0.045$ , i.e. about a $4.5%$ density change — the largest error we usually tolerate. At $M_{2}$ a 17% density swing is far too large to ignore.
Verdict. $M_{1}$ : ~1% density change ⇒ incompressible Bernoulli fine. $M_{2}$ : ~17% ⇒ the crossed-out variable- $ρ$ term wakes up; use compressible Bernoulli (see Mach Number and Compressibility). Why this step? Mixing the two ends with one incompressible constant would carry ~17% error at the outlet.

Verify: $50/340 = 0.1471$ , $200/340 = 0.5882$ ✔. $\frac{1}{2} (0.3)^{2} = 0.045$ ✔ (the ~4.5% threshold). Crossover speed $v = 0.3 \times 340 = 102 m/s$ — the outlet's 200 m/s is well past it.

Cell E — two streamlines, two constants

Worked example Ex6 · Free vortex, inner vs outer streamline (Cell E)

In a free vortex (draining sink) the speed is $v = k / r$ with $k = 0.6 m^{2} / s$ . Compare an inner streamline at $r_{1} = 0.1 m$ with an outer one at $r_{2} = 0.3 m$ , same height. Naïvely equating one Bernoulli constant, what pressure difference does it predict — and why is that number not automatically legitimate here?

Forecast: Inner water is faster. Does one Bernoulli constant let you compare these two points? Guess valid/invalid before step 1.

The figure below shows the two circular streamlines. Read it as the map of the whole argument: the red inner circle ( $r_{1} = 0.1 m$ ) and the blue outer circle ( $r_{2} = 0.3 m$ ) are two separate streamlines that never touch — that is the visual reason Bernoulli's constant is not shared between them. The yellow arrows are the local velocities: the red arrow is long (fast, $6 m/s$ ), the blue arrow short (slow, $2 m/s$ ), so the arrow lengths literally draw the $v = k / r$ law. When you follow either circle all the way round, you are moving along one streamline — that is where Bernoulli is guaranteed; jumping radially from red to blue is the illegal move we are testing.

Get the two speeds. $v_{1} = \frac{0.6}{0.1} = 6 m/s, v_{2} = \frac{0.6}{0.3} = 2 m/s$ Why this step? Different radii ⇒ different speeds ⇒ different kinetic terms — matching the two arrow lengths in the figure.
What naïve single-constant Bernoulli would say. Setting one constant across both: $P_{2} - P_{1} = \frac{1}{2} ρ (v_{1}^{2} - v_{2}^{2}) = \frac{1}{2} (1000) (36 - 4) = 16000 Pa$ Why this step? We compute it precisely to expose the trap.
Why crossing streamlines is only allowed if the flow is irrotational — and why this flow is. These are different streamlines (the concentric circles in the figure never touch), so Bernoulli's constant is only guaranteed along one of them. Crossing is legal only if the flow is irrotational (zero curl — no local spin of a tiny paddle wheel). For a 2-D flow with only a tangential speed $v_{θ} (r)$ , the curl (vorticity) is $ω = \frac{1}{r} \frac{d ( r v _{θ} )}{d r} .$ For the free vortex $v_{θ} = k / r$ , we have $r v_{θ} = k = const$ , so $\frac{d ( r v _{θ} )}{d r} = 0$ and hence $ω = 0$ everywhere except $r = 0$ (the core, where $k / r$ blows up). So the number does survive here. Why this step? It shows the condition that rescues the calculation — and warns that a forced (rigid-body) vortex $v_{θ} = Ω r$ gives $r v_{θ} = Ω r^{2}$ , $ω = 2Ω \neq = 0$ , so there the naïve number would be wrong.

Verify: $\frac{1}{2} (1000) (36 - 4) = 16000 Pa$ ✔; and $\frac{d}{d r} (r \cdot \frac{k}{r}) = \frac{d k}{d r} = 0$ ✔ confirms zero vorticity. Rule: never equate constants across streamlines until you have checked irrotationality (parent Example 4).

Cell F — a pump adds energy

Worked example Ex7 · Pump in the line (Cell F)

A pump sits in a level pipe and raises the water from $P_{1} = 100 kPa$ to $P_{2} = 250 kPa$ with no change in $v$ (equal areas either side, so continuity gives $v_{1} = v_{2}$ ) or $h$ . Plain Bernoulli says the constant can't change. What pump-head (in metres) accounts for the jump?

Forecast: Bernoulli forbids the pressure from rising with $v, h$ fixed — yet it rose 150 kPa. Which term did the pump switch on?

Naïve Bernoulli. With $v_{1} = v_{2}$ and $h_{1} = h_{2}$ , both the kinetic and gravity terms cancel, so plain Bernoulli predicts $P_{1} = P_{2}$ . Why this step? We show the prediction so the contradiction is undeniable — identical in spirit to the friction case (Ex3), but here energy is added, not lost.
Restore the pump term $d F_{pump}$ . Go back to the master parcel equation and keep the term Bernoulli crossed out. Integrating the added energy per unit volume across the pump gives a head-rise $H_{pump}$ , so the energy balance becomes $P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g h_{1} + ρ g H_{pump} = P_{2} + \frac{1}{2} ρ v_{2}^{2} + ρ g h_{2} .$ With $v_{1} = v_{2}$ and $h_{1} = h_{2}$ this collapses to $ρ g H_{pump} = P_{2} - P_{1} = 150000 Pa .$ Why this step? The $d F_{pump}$ term Bernoulli deleted is exactly this injected energy; the "constant" is supposed to jump across a pump.
Solve for the head. $H_{pump} = \frac{P _{2} - P _{1}}{ρ g} = \frac{150000}{1000 \cdot 9.8} \approx 15.31 m .$ Why this step? Head has units of metres — the equivalent height of water the pump's energy could lift.

Verify: $150000/ (1000 \cdot 9.8) = 15.306 m$ ✔. Units: $Pa / (kg/m^{3} \cdot m/s^{2}) = m$ ✔. Sanity: pressure rose across the pump — correct, since a pump inputs energy (matches the parent's fourth mistake).

Cell G — degenerate / zero inputs

Worked example Ex8 · The static tank (Cell G, zero-velocity limit)

A large open tank of water is perfectly still ( $v = 0$ everywhere). A point is $d = 3 m$ below the free surface (where $P_{atm} = 101 kPa$ ). What does Bernoulli give, and does it collapse to a familiar law?

Forecast: With every speed zero, the $\frac{1}{2} ρ v^{2}$ term dies. What famous formula is Bernoulli's limiting form?

Zero out the kinetic term. $v = 0$ at both points ⇒ $\frac{1}{2} ρ v^{2} = 0$ . Why this step? The degenerate input $v = 0$ removes one whole term — the limiting case that tests whether Bernoulli stays consistent.
What survives is hydrostatics. Take the surface point (height $h_{surface} = d$ , pressure $P_{atm}$ ) and the deep point (height $0$ , pressure $P_{deep}$ ): $P_{atm} + ρ g d = P_{deep} + ρ g (0)$ $\Rightarrow P_{deep} = P_{atm} + ρ g d .$ Why this step? Bernoulli must reduce to $P = P_{0} + ρ g d$ when nothing moves — a consistency check on the whole theory.
Plug numbers. $P_{deep} = 101000 + 1000 \cdot 9.8 \cdot 3 = 130400 Pa \approx 130.4 kPa .$ Why this step? Confirms the pressure-with-depth law is just Bernoulli's still-water shadow.

Verify: $101000 + 1000 \cdot 9.8 \cdot 3 = 130400 Pa$ ✔. Removing the $v$ term is the "degenerate cell" — Bernoulli gracefully becomes hydrostatics, so no assumption is violated; the kinetic term is simply zero. ✔

Cell H — the exam trap

Worked example Ex9 · The "looks valid" wing question (Cell H, hidden failure)

Air ( $ρ = 1.2 kg/m^{3}$ ) at $30 m/s$ flows over a wing; on top it reaches $45 m/s$ . (a) Predict the pressure drop with Bernoulli. (b) The exam then asks for the pressure on the skin itself, inside the thin surface layer — is your answer still valid?

Forecast: Part (a) is clean Cell A. Part (b) is the twist. Which assumption silently fails right at the metal surface?

Check Mach for compressibility. $M = 45/340 \approx 0.132 < 0.3$ ⇒ incompressible OK. Why this step? Rules out Cell D so we don't waste the compressible correction from Mach Number and Compressibility.
Part (a): Bernoulli pressure drop. $Δ P = \frac{1}{2} ρ (v_{top}^{2} - v_{\infty}^{2}) = \frac{1}{2} (1.2) (4 5^{2} - 3 0^{2})$ $= \frac{1}{2} (1.2) (2025 - 900) = 675 Pa .$ Why this step? Faster top surface ⇒ lower pressure ⇒ lift; valid outside the surface layer.
Part (b): spot the hidden failure. Right at the skin, velocity drops to zero through the Boundary Layer (the ultra-thin viscous skin restated in the tools box), and viscous shear dominates — the inviscid assumption dies. Bernoulli's $675 Pa$ applies just outside the layer, not at the metal. Why this step? The exam trap is applying a valid answer in an invalid region — a different streamline family where a crossed-out term (friction) has woken up.

Verify: $\frac{1}{2} (1.2) (2025 - 900) = 675 Pa$ ✔. $45/340 = 0.1324 < 0.3$ ✔. Lesson: always ask where on the flow you are before trusting Bernoulli — see Bernoulli's Equation.

Recall Self-test — name the failing assumption

Rough 50 m pipe, equal $v$ and $h$ , pressure drops ::: Inviscid (friction → heat). Cell B. Valve just opened, water accelerating everywhere ::: Steady (unsteady term $\partial v / \partial t$ ). Cell C. Air duct where $M = 0.6$ ::: Incompressible ( $ρ$ varies ~17%). Cell D. Comparing inner and outer streamlines of a vortex ::: Along a streamline (need irrotational to cross). Cell E. Pump raises pressure with $v, h$ fixed ::: None of the four — energy added, constant jumps. Cell F. Water flowing up 12 m, all four assumptions hold ::: None fail — keep the gravity term. Cell A′. Still tank, all speeds zero ::: None fail — Bernoulli reduces to hydrostatics. Cell G.