Navier-Stokes equations — derivation from Newton's second law for fluid: Worked examples

This page is a drill ground for the parent Navier–Stokes note. There we built the equation from $F = m a$ . Here we use it in every situation it can throw at you — steady and unsteady, viscous and inviscid, at rest and gravity-driven, plus a word problem and an exam twist.

Before we begin, one reminder of the master sentence we are always plugging into:

Every symbol, so nobody is lost from line one:

Definition The alphabet of this page

$ρ$ (rho) — the fluid's density, mass packed into each cubic metre, units $kg / m^{3}$ .
$u$ — the velocity field: at every point in space and time it tells you which way and how fast the fluid there is moving. $u_{x}$ is its component along the $x$ -axis.
$P$ — pressure, the inward push per unit area on any surface, units $Pa = N / m^{2}$ .
$μ$ (mu) — dynamic viscosity, how "sticky/thick" the fluid is, units $Pa \cdot s$ (see Viscosity & Newton's law of viscosity).
$g$ — gravity as a vector (it has a direction, straight down), magnitude written $g = 9.8 m/s^{2}$ . Whenever we pick an axis we must project $g$ onto it: the component along axis $x$ is $g_{x} = g \cdot \overset{x}{^}$ . So $g$ (arrow) and $g$ (number) are different objects — always check which one a formula wants.
$\nabla$ (nabla) — the "slope in every direction" operator; $\nabla P$ is the vector pointing toward fastest pressure rise. $\nabla^{2}$ applies it twice (curvature).
$\frac{D}{D t}$ — the Material derivative, acceleration measured following the moving fluid: $\frac{D u}{D t} = \frac{\partial u}{\partial t} + (u \cdot \nabla) u$ . It appears explicitly in Ex 5 (pure convection) and Ex 6 (pure local term).

The scenario matrix

Every problem you can meet is one (or a blend) of these cells. The examples below are labelled with the cell they hit.

Cell	Case class	What is switched on / off	Example
C1	At rest ( $u = 0$ )	only pressure + gravity	Ex 1
C2	Steady viscous, pressure-driven	$\partial_{t} = 0$ , $μ \neq = 0$ , gradient drives	Ex 2
C3	Steady viscous, gravity-driven	$μ \neq = 0$ , $ρ g$ drives (falling film)	Ex 3
C4	Inviscid limit ( $μ = 0$ )	drop viscous term → Euler	Ex 4
C5	Convective term dominant	$(u \cdot \nabla) u \neq = 0$ , narrowing pipe	Ex 5
C6	Unsteady (transient start-up)	$\partial_{t} \neq = 0$	Ex 6
C7	Sign / direction check	which way does the sign of $\partial_{x} P$ push?	Ex 7
C8	Degenerate / limiting numbers	$μ \to 0$ , $h \to 0$ , $G \to 0$	Ex 8
C9	Real-world word problem	put numbers to Poiseuille flow rate	Ex 9
C10	Exam twist (dimensionless)	build Reynolds number from NS terms	Ex 10
C11	All terms live at once	unsteady + convective + viscous + free surface	Ex 11

Ex 1 — Fluid at rest (Cell C1)

Worked example A tank of water, 3 m deep. Find the gauge pressure at the bottom. (

$ρ = 1000 kg/m^{3}$ , $g = 9.8$ .) Forecast: guess the number before reading on. Deeper → more weight of water above → higher pressure. Roughly $1 0^{4}$ Pa per metre. What do you get for 3 m?

Step 1. Set $u = 0$ everywhere. Why this step? "At rest" is exactly $u = 0$ ; this kills the entire left side and the whole viscous term (both need a moving field). $0 = - \nabla P + ρ g$

Step 2. Take the vertical component. Let $z$ point up, so $g = (0, 0, - g)$ , i.e. its $z$ -projection is $g_{z} = - g$ . Why this step? $g$ is a vector; to read one scalar equation we project it onto $z$ . Nothing varies horizontally, so $\partial_{x} P = \partial_{y} P = 0$ and $P$ depends on $z$ alone — the partial becomes an ordinary derivative. $0 = - \frac{\partial P}{\partial z} - ρ g \Rightarrow \frac{d P}{d z} = - ρ g$

Step 3. Integrate from the surface down. Why this step? $d P / d z$ is constant, so pressure grows linearly with depth $h$ . $P = P_{surface} + ρ g h$

Verify: $P_{gauge} = ρ g h = 1000 \times 9.8 \times 3 = 29400 Pa$ . Units: $\frac{kg}{m ^{3}} \cdot \frac{m}{s ^{2}} \cdot m = \frac{kg}{m \cdot s ^{2}} = Pa$ ✓. This is exactly Hydrostatic pressure.

Ex 2 — Steady viscous, pressure-driven (Cell C2)

Worked example Water between two fixed plates at

$y = 0$ and $y = h = 2 mm$ , pushed by pressure gradient $G = - \partial P / \partial x = 500 Pa/m$ , viscosity $μ = 1 0^{- 3} Pa\cdots$ . Find the centreline velocity. Forecast: the profile is a parabola, zero at both walls (no-slip), fastest in the middle. Will the max speed be millimetres/s or metres/s?

What to see in the figure: the two thick horizontal slate lines are the fixed plates. The lavender curve bulging to the right is the velocity profile $u_{x} (y)$ — notice it is pinned to zero at both plates (no-slip) and fattest in the middle. Each short coral arrow is the local flow speed at that height: they grow from nothing at the walls to longest at the centre, tracing the parabola. The butter arrow at the bottom is the pressure push $G = - \partial P / \partial x$ that drives everything, and the mint dot marks $u_{m a x}$ at $y = h /2$ . The picture is the derivation: constant curvature (Step 1) forced by no-slip walls (Step 2) can only be a symmetric parabola.

Step 1. Steady ( $\partial_{t} = 0$ ), flow only along $x$ , depends only on $y$ . Because the field now reduces to the single-variable function $u_{x} (y)$ , we switch from $\partial^{2} / \partial y^{2}$ to the ordinary $d^{2} / d y^{2}$ . Why this step? every dropped term is literally zero — no time change, no cross-flow, so convection $(u \cdot \nabla) u = 0$ . Also, continuity $\partial u_{x} / \partial x = 0$ guarantees $u_{x}$ cannot depend on $x$ — that is what lets us write $u_{x} (y)$ alone. $0 = G + μ \frac{d ^{2} u _{x}}{d y ^{2}}$

Step 2. Solve the ODE with no-slip $u_{x} (0) = u_{x} (h) = 0$ . Why this step? Second derivative is constant $\Rightarrow$ integrate twice; the walls pin the constants. This is Poiseuille flow. $u_{x} (y) = \frac{G}{2 μ} y (h - y)$

Step 3. Max at $y = h /2$ . Why this step? Symmetric parabola peaks at the midpoint. $u_{m a x} = \frac{G h ^{2}}{8 μ}$

Verify: $u_{m a x} = \frac{500 \times ( 0.002 ) ^{2}}{8 \times 1 0 ^{- 3}} = \frac{500 \times 4 \times 1 0 ^{- 6}}{8 \times 1 0 ^{- 3}} = 0.25 m/s$ . Units: $\frac{( Pa/m ) \cdot m ^{2}}{Pa\cdots} = \frac{m}{s}$ ✓.

Ex 3 — Steady viscous, gravity-driven (Cell C3)

Worked example A thin film of oil (

$μ = 0.1 Pa\cdots$ , $ρ = 900$ ) flows down a vertical wall, film thickness $δ = 1 mm$ , open to air (zero shear) on the free surface. Find the surface velocity. Forecast: here gravity is the driver, not a pressure gradient. The free surface is fastest (nothing drags it), the wall is stopped. Half-parabola.

Step 1 — choose axes and project $g$ . Let $x$ point straight down the wall (the flow direction) and $y$ measured horizontally out from the wall. Because $x$ is chosen aligned with the downward vertical, the gravity vector projects fully onto $x$ : $g_{x} = g \cdot \overset{x}{^} = + g$ and $g_{y} = 0$ . Why this step? $g$ is a vector; the $x$ -momentum equation only wants its $x$ -component. By aligning $x$ with $g$ we get the clean $g_{x} = g$ , and the $y$ -equation just gives the (constant) hydrostatic pressure across the thin film, which we ignore. Steady, $u_{x} (y)$ only (continuity again forbids $x$ -dependence, so $\partial^{2} \to d^{2}$ ): $0 = ρ g_{x} + μ \frac{d ^{2} u _{x}}{d y ^{2}} = ρ g + μ \frac{d ^{2} u _{x}}{d y ^{2}} \Rightarrow \frac{d ^{2} u _{x}}{d y ^{2}} = - \frac{ρ g}{μ}$

Step 2 — free-surface boundary condition. No-slip at wall $u_{x} (0) = 0$ ; zero shear at the free surface $\frac{d u _{x}}{d y}_{y = δ} = 0$ . Why this step? Air can't grip the oil, so the viscous stress $μ d u / d y$ must vanish at the open top — this is the free-surface pressure/stress condition: at an interface with a much lighter fluid, tangential stress $\to 0$ and surface pressure $=$ atmospheric.

Step 3. Integrate: $u_{x} (y) = \frac{ρ g}{μ} (δ y - \frac{y ^{2}}{2})$ . Why this step? First integration + zero-shear fixes the slope constant; second + no-slip fixes the offset. Surface speed at $y = δ$ : $u_{surf} = \frac{ρ g δ ^{2}}{2 μ}$

Verify: $u_{surf} = \frac{900 \times 9.8 \times ( 0.001 ) ^{2}}{2 \times 0.1} = \frac{900 \times 9.8 \times 1 0 ^{- 6}}{0.2} = 0.0441 m/s$ ✓.

Ex 4 — Inviscid limit → Euler (Cell C4)

Worked example A frictionless ideal fluid (

$μ = 0$ ) flows steadily. Show NS becomes Bernoulli-friendly Euler, and evaluate the pressure drop when speed doubles from $2$ to $4 m/s$ ( $ρ = 1000$ , horizontal, no gravity term along the streamline). Forecast: with no stickiness, only pressure and gravity remain. Faster flow → lower pressure. How much lower?

Step 1. Set $μ = 0$ . Why this step? The whole viscous term dies, leaving the Euler equation (inviscid flow): $ρ \frac{D u}{D t} = - \nabla P + ρ g$

Step 2. Steady, along a horizontal streamline, integrate → Bernoulli: $P + \frac{1}{2} ρ u^{2} = const$ . Why this step? $ρ (u \cdot \nabla) u$ integrates to the kinetic term.

Step 3. $Δ P = - \frac{1}{2} ρ (u_{2}^{2} - u_{1}^{2})$ . Why this step? Constant total: pressure must fall to pay for the extra speed.

Verify: $Δ P = - \frac{1}{2} \times 1000 \times (4^{2} - 2^{2}) = - \frac{1}{2} \times 1000 \times 12 = - 6000 Pa$ . Pressure drops by 6 kPa ✓.

Ex 5 — Convective term dominant (Cell C5)

Worked example Steady flow in a narrowing channel: velocity along

$x$ is $u_{x} (x) = u_{0} (1 + α x)$ with $u_{0} = 1 m/s$ , $α = 2 m^{- 1}$ . Find the acceleration of a fluid particle at $x = 0.5 m$ using the material derivative. Forecast: even though $\partial u / \partial t = 0$ (steady!), a particle speeds up by moving into faster water. So acceleration is not zero. Positive or negative?

Step 1. Write the material derivative and drop the local term. Why this step? The Material derivative is the true particle acceleration; steady flow zeroes $\partial u / \partial t$ , leaving only convection: $\frac{D u _{x}}{D t} = = 0 \frac{\partial u _{x}}{\partial t} + (u \cdot \nabla) u_{x} = u_{x} \frac{d u _{x}}{d x}$

Step 2. $\frac{d u _{x}}{d x} = u_{0} α$ , so $\frac{D u _{x}}{D t} = u_{x} (x) \cdot u_{0} α = u_{0} (1 + α x) \cdot u_{0} α = u_{0}^{2} α (1 + α x)$ . Why this step? Chain rule on the given profile — this is convective acceleration made concrete.

Verify: $\frac{D u _{x}}{D t} = 1^{2} \times 2 \times (1 + 2 \times 0.5) = 2 \times 2 = 4 m/s^{2}$ . Positive → accelerating into the narrows, matching the leaf-in-a-river picture ✓.

Ex 6 — Unsteady start-up (Cell C6)

Worked example Stokes' first problem: a fluid rests above a plate. At

$t = 0$ the plate suddenly moves. Near the plate the equation is $\frac{\partial u}{\partial t} = ν \frac{\partial ^{2} u}{\partial y ^{2}}$ with $ν = μ / ρ = 1 0^{- 6} m^{2} / s$ . How deep has motion "diffused" after $t = 4 s$ ? Forecast: viscosity spreads momentum like heat spreads. The reached depth grows like $ν t$ — not linearly. Estimate the penetration depth $δ \sim ν t$ .

Step 1. Keep only $\partial_{t}$ and the viscous term. Why this step? No pressure gradient, no gravity along the plate; here the flow depends on both $y$ and $t$ , so we genuinely keep partial derivatives (unlike Ex 2/3 which collapsed to one variable). The material derivative reduces to its local piece $\partial u / \partial t$ since $u$ has no $x$ -dependence. This is the diffusion equation.

Step 2. Dimensional balance: $\frac{u}{t} \sim ν \frac{u}{δ ^{2}} \Rightarrow δ \sim ν t$ . Why this step? Matching the two surviving terms' sizes reveals the growth law.

Verify: $δ = ν t = 1 0^{- 6} \times 4 = 4 \times 1 0^{- 6} = 2 \times 1 0^{- 3} m = 2 mm$ ✓. Slow diffusion — that's why viscous startup is sluggish.

Ex 7 — Sign / direction check (Cell C7)

Ex 8 — Degenerate / limiting numbers (Cell C8)

Worked example Take the Poiseuille result

$u_{m a x} = \frac{G h ^{2}}{8 μ}$ from Ex 2. Examine three limits: (a) $μ \to 0$ , (b) $h \to 0$ , (c) $G \to 0$ . What does each mean physically, and evaluate $u_{m a x}$ if $h$ is halved to $1 mm$ (keeping $G, μ$ from Ex 2)? Forecast: thinner gap → dramatically slower flow because $h$ appears squared. Halving $h$ should cut speed by a factor of 4, not 2.

Step 1 (a). $μ \to 0$ : $u_{m a x} \to \infty$ . Why this step? No friction means nothing resists the push — unphysical, telling us real fluids always keep some $μ$ .

Step 2 (b). $h \to 0$ : $u_{m a x} \to 0$ . Why this step? Squeeze the gap shut and both walls stop the fluid everywhere — flow vanishes.

Step 3 (c). $G \to 0$ : $u_{m a x} \to 0$ . Why this step? No driving gradient, no flow — consistent with Cell C1 (at rest).

Verify: halve $h$ to $0.001 m$ : $u_{m a x} = \frac{500 \times ( 0.001 ) ^{2}}{8 \times 1 0 ^{- 3}} = \frac{500 \times 1 0 ^{- 6}}{8 \times 1 0 ^{- 3}} = 0.0625 m/s$ . That is $0.25/4$ — exactly a quarter of Ex 2's answer ✓.

Ex 9 — Real-world word problem (Cell C9)

Worked example Blood (treat as

$μ = 3.5 \times 1 0^{- 3} Pa\cdots$ ) flows through a capillary-like round tube, radius $R = 1 mm$ , length $L = 0.1 m$ , pressure difference $Δ P = 200 Pa$ . Find the volume flow rate $Q$ . Forecast: the round-pipe Poiseuille law is $Q = \frac{π R ^{4} Δ P}{8 μL}$ . Notice $R^{4}$ — flow is extremely sensitive to radius. Expect a tiny number in $m^{3} / s$ .

Step 1. Same NS reduction as Ex 2 but in cylindrical coordinates gives $u (r) = \frac{Δ P}{4 μL} (R^{2} - r^{2})$ . Why this step? Round geometry replaces the flat-plate parabola with a paraboloid, still from $μ \nabla^{2} u = - \partial_{x} P$ .

Step 2. Integrate the speed over the cross-section: $Q = \int_{0}^{R} u (r) 2 π r d r = \frac{π R ^{4} Δ P}{8 μL}$ . Why this step? Flow rate is speed summed over area.

Verify: $Q = \frac{π ( 1 0 ^{- 3} ) ^{4} \times 200}{8 \times 3.5 \times 1 0 ^{- 3} \times 0.1} = \frac{π \times 1 0 ^{- 12} \times 200}{2.8 \times 1 0 ^{- 3}} \approx 2.244 \times 1 0^{- 7} m^{3} / s$ ✓ (about $0.22 mL/s$ ). See Poiseuille flow.

Ex 10 — Exam twist: build the Reynolds number (Cell C10)

Worked example By comparing the sizes of the convective term and the viscous term in NS, derive the dimensionless group that decides "inertia vs friction." Evaluate it for water in a pipe:

$U = 1 m/s$ , $L = 0.05 m$ , $ν = 1 0^{- 6} m^{2} / s$ . Forecast: the ratio you get is the Reynolds number $R e = U L / ν$ . Big $R e$ = inertia wins (turbulent-prone); small $R e$ = viscosity wins (smooth).

Step 1 — size the convective (inertia) term. Replace each derivative by (typical change)/(typical length): a velocity of scale $U$ changing over a distance $L$ . Why this step? We only want the order of magnitude, so exact derivatives are unnecessary. $ρ (u \cdot \nabla) u \sim ρ U \cdot \frac{U}{L} = \frac{ρ U ^{2}}{L}$

Step 2 — size the viscous term. The viscous term is a second derivative, so it carries two length divisions: $μ \nabla^{2} u \sim μ U / L^{2}$ . Why this step? Same order-of-magnitude accounting, applied to $\nabla^{2}$ (curvature $\Rightarrow$ divide by $L$ twice). $μ \nabla^{2} u \sim μ \frac{U}{L ^{2}}$

Step 3 — form the ratio. Divide inertia by viscosity; the $U / L$ common piece cancels and $μ / ρ = ν$ (kinematic viscosity) appears. Why this step? A pure ratio of two forces must be dimensionless — that number is the Reynolds number. $R e = \frac{ρ U ^{2} / L}{μU / L ^{2}} = \frac{ρ U L}{μ} = \frac{U L}{ν}$

Verify: $R e = \frac{1 \times 0.05}{1 0 ^{- 6}} = 50000$ . Large → inertia dominates, flow likely turbulent ✓. Dimensionless check: $\frac{( m/s ) ( m )}{m ^{2} / s}$ = pure number ✓.

Ex 11 — All terms live at once (Cell C11)

Worked example Impulsively started pipe: at

$t = 0$ a long round pipe (radius $R$ ) is switched to a pressure gradient $G$ . Before the parabola settles, all NS terms can matter. Confirm which survive, and after steady state find the mean speed if $R = 1 mm$ , $G = 500 Pa/m$ , $μ = 1 0^{- 3} Pa\cdots$ . Forecast: at start-up the flow is unsteady ( $\partial_{t} \neq = 0$ ); if the pipe also narrows the parcel accelerates ( $(u \cdot \nabla) u \neq = 0$ ); friction is always present ( $μ \nabla^{2} u$ ); and the pipe outlet meets air (free-surface / atmospheric pressure condition). The final settled mean speed is half the centreline.

Step 1 — which terms are alive? During the transient, the $x$ -momentum equation keeps every term: $ρ (unsteady \frac{\partial u _{x}}{\partial t} + convective (u \cdot \nabla) u_{x}) = G + viscous μ \nabla^{2} u_{x}$ Why this step? This is the only example where none of the three left-block effects is switched off — it is the genuine "combined" cell promised in the matrix. In a straight uniform pipe continuity $\partial u_{x} / \partial x = 0$ kills convection; in a narrowing pipe it would survive.

Step 2 — settle to steady state. As $t \to \infty$ , $\partial_{t} \to 0$ ; in a straight pipe convection is off, so we recover steady Poiseuille $u (r) = \frac{G}{4 μ} (R^{2} - r^{2})$ with centreline $u_{m a x} = \frac{G R ^{2}}{4 μ}$ . Why this step? The transient dies away and the balance reduces to Cell C2/C9 — a healthy consistency check across cells.

Step 3 — mean speed. For a round pipe the area-average of a paraboloid is exactly half the peak: $\overset{u}{ˉ} = \frac{1}{2} u_{m a x} = \frac{G R ^{2}}{8 μ}$ . Why this step? Flow rate ÷ area; the paraboloid's mean is half its max.

Verify: $\overset{u}{ˉ} = \frac{500 \times ( 0.001 ) ^{2}}{8 \times 1 0 ^{- 3}} = \frac{500 \times 1 0 ^{- 6}}{8 \times 1 0 ^{- 3}} = 0.0625 m/s$ ✓. (Consistent with the free-surface exit at atmospheric pressure setting the downstream boundary; continuity fixed convection to zero here.)

Recall Which cell was hardest for you?

Cell C7 sign trap (force is $- \nabla P$ ) ::: fluid is pushed from high to low pressure, so a positive $\partial_{x} P$ gives a negative $x$ -force. Why does Ex 5 have acceleration despite steady flow? ::: convective term $(u \cdot \nabla) u$ in the material derivative — the parcel moves into faster fluid. In Ex 8, halving $h$ changes $u_{m a x}$ by what factor? ::: one quarter, because $u_{m a x} \propto h^{2}$ . When does the convective term survive in a pipe (Ex 11)? ::: only if the pipe narrows; continuity kills it in a straight uniform pipe.