Exercises — Navier-Stokes equations — derivation from Newton's second law for fluid

• $\rho \frac{\partial \vec{u}}{\partial t}$ — local (unsteady) inertia. Units $\text{kg}{\text{m}}^{-2}{\text{s}}^{-2}=\text{N}{\text{m}}^{-3}$ (force per volume).
• $\rho (\vec{u}\cdot \nabla )\vec{u}$ — convective inertia. Same units, $\text{N}{\text{m}}^{-3}$.
• $-\nabla P$ — net pressure force per volume. $\frac{\text{Pa}}{\text{m}}=\frac{\text{N}/{\text{m}}^2}{\text{m}}=\text{N}{\text{m}}^{-3}$. ✓
• $\mu {\nabla }^2\vec{u}$ — net viscous force per volume. $[\mu ]=\text{Pa⋅s}$, $[{\nabla }^2\vec{u}]={\text{s}}^{-1}{\text{m}}^{-1}$, product $=\text{N}{\text{m}}^{-3}$. ✓
• $\rho \vec{g}$ — gravity (weight) per volume. ${\text{kg m}}^{-3}\cdot {\text{m s}}^{-2}=\text{N}{\text{m}}^{-3}$. ✓

Every term is a force per unit volume — that is why they can be added.

Yes. The flow is steady, so the local term $\partial \vec{u}/\partial t$ is zero, but the leaf is carried into faster water. The acceleration is entirely the convective term $(\vec{u}\cdot \nabla )\vec{u}$. A fixed observer sees constant speed at their point; the particle still speeds up. See Material derivative.

Steady ⟹ $\partial u_x/\partial t=0$. Convective term in 1-D: $a=u_x\frac{\partial u_x}{\partial x}=(3x)(3)=9x.$ At $x=2$: $a=9(2)=\boxed{18{\text{m/s}}^2}$. Notice the acceleration is nonzero despite steady flow — the convective term did all the work.

With $\vec{u}=0$, all inertia and viscous terms vanish. NS collapses to $0=-\nabla P+\rho \vec{g}$, i.e. $\nabla P=\rho \vec{g}$. Taking depth $h$ downward: $\frac{dP}{dh}=\rho g\Rightarrow P=P_0+\rho gh.$ $P=1.01\times 10^5+(1000)(9.8)(3)=1.01\times 10^5+2.94\times 10^4=\boxed{1.304\times 10^5\text{Pa}}.$ This is exactly Hydrostatic pressure — NS contains it as the zero-velocity case.

$Re=\frac{\rho UL}{\mu }=\frac{(1000)(2)(0.05)}{1.0\times 10^{-3}}=\frac{100}{10^{-3}}=\boxed{1.0\times 10^5}.$ $Re$ is the ratio (inertial force)/(viscous force). Here $Re\gg 1$, so inertia (the $\rho (\vec{u}\cdot \nabla )\vec{u}$ term) dominates and the flow is turbulent. See Reynolds number.

Case (a) gives the Euler equation (inviscid flow); case (b) gives hydrostatics.

$[\mu ]=\text{Pa⋅s}$. $[{\nabla }^2\vec{u}]=\frac{\text{m/s}}{{\text{m}}^2}={\text{s}}^{-1}{\text{m}}^{-1}$. Product: $\text{Pa⋅s}\cdot {\text{s}}^{-1}{\text{m}}^{-1}=\text{Pa/m}=\text{N}{\text{m}}^{-3}$. $[-\nabla P]=\text{Pa/m}=\text{N}{\text{m}}^{-3}$. ✓ Identical — both forces per volume.

Net $x$-force from the two horizontal faces (each area $dxdz$): $d{F}_x=[\tau (y+dy)-\tau (y)]dxdz=\frac{\partial \tau }{\partial y}dydxdz=\mu \frac{{\partial }^2{u}_x}{\partial y^2}dV.$ Per volume: $\mu {\partial }^2{u}_x/\partial y^2$. Why not first derivative? $\tau$ itself is a stress, not a net force. A net force needs the stress to differ across the blob — that difference is one more spatial derivative. See Viscosity & Newton's law of viscosity.

Steady, unidirectional ⟹ inertia terms vanish, the $x$-momentum NS becomes $0=-\frac{\partial P}{\partial x}+\mu \frac{d^2{u}_x}{d{y}^2}=G+\mu \frac{d^2{u}_x}{d{y}^2}\Rightarrow \frac{d^2{u}_x}{d{y}^2}=-\frac{G}{\mu }.$ Integrate twice: $u_x=-\frac{G}{2\mu }{y}^2+C_{1}y+C_2.$ Apply $u_x(0)=0\Rightarrow C_2=0$. Apply $u_x(h)=0\Rightarrow C_1=\frac{Gh}{2\mu }$. So $\boxed{u_x(y)=\frac{G}{2\mu }y(h-y)}.$ Maximum where $d{u}_x/dy=\frac{G}{2\mu }(h-2y)=0\Rightarrow y=h/2$ (the centre), with $u_{\max }=\frac{G}{2\mu }\cdot \frac{h}{2}\cdot \frac{h}{2}=\frac{G{h}^2}{8\mu }.$ The parabola is the fingerprint of viscous pressure-driven flow. See Poiseuille flow.

$u_{\max }=\frac{G{h}^2}{8\mu }=\frac{(200)(0.01{)}^2}{8(10^{-3})}=\frac{200\cdot 10^{-4}}{8\times 10^{-3}}=\frac{0.02}{0.008}=\boxed{2.5\text{m/s}}.$ At $y=h/4=0.0025$: $u_x=\frac{G}{2\mu }y(h-y)=\frac{200}{2\times 10^{-3}}(0.0025)(0.0075)=(10^5)(1.875\times 10^{-5})=\boxed{1.875\text{m/s}}.$

Along the flow ($z$, downward), NS gives $0=\mu \frac{d^2{u}_z}{d{x}^2}+\rho g$, so $\frac{d^2{u}_z}{d{x}^2}=-\frac{\rho g}{\mu }.$ Integrate: $\frac{d{u}_z}{dx}=-\frac{\rho g}{\mu }x+C_1$. Zero-shear at $x=h$: $0=-\frac{\rho g}{\mu }h+C_1\Rightarrow C_1=\frac{\rho gh}{\mu }$. Integrate again: $u_z=-\frac{\rho g}{2\mu }{x}^2+\frac{\rho gh}{\mu }x+C_2$. No-slip $u_z(0)=0\Rightarrow C_2=0$. So $\boxed{u_z(x)=\frac{\rho g}{\mu }\left(hx-\frac{x^2}{2}\right)}.$ Maximum at the free surface $x=h$: $u_{\max }=\frac{\rho g}{\mu }\left(h^2-\frac{h^2}{2}\right)=\frac{\rho g{h}^2}{2\mu }.$

$u_{\max }=\frac{\rho g{h}^2}{2\mu }=\frac{(1000)(9.8)(0.002{)}^2}{2(0.1)}=\frac{(1000)(9.8)(4\times 10^{-6})}{0.2}=\frac{0.0392}{0.2}=\boxed{0.196\text{m/s}}.$

$\frac{\rho U^2/L}{\mu U/L^2}=\frac{\rho U^2{L}^2}{L\mu U}=\frac{\rho UL}{\mu }=Re.$ Numerically: $\frac{(1000)(2^2)/0.05}{(10^{-3})(2)/0.05^2}=\frac{80000}{0.8}=\boxed{1.0\times 10^5}=Re$. ✓ This is why $Re$ measures inertia-to-viscosity — it drops straight out of comparing two NS terms.

Every term is a force per volume ::: so they add; check units are ${\text{N m}}^{-3}$. Steady flow can still accelerate ::: via the convective term $(\vec{u}\cdot \nabla )\vec{u}$. Viscous force is a second derivative ::: it's the difference of stress across the blob. Second-order ODE ::: needs two boundary conditions. Free surface ::: zero shear, not no-slip.