Exercises — Fluid kinematics — Eulerian vs Lagrangian description

The fixed station watches whatever air blows past a single point → Eulerian ($\partial T/\partial t$ at fixed $\mathbf{x}$). The drifting balloon rides with one parcel of air → Lagrangian ($DT/Dt$, particle label fixed). Picture: station = the bridge, balloon = the fish.

• $\frac{\partial T}{\partial t}$ = local (unsteady) term — the temperature field itself changing in time everywhere (e.g. the sun heating up the whole valley as morning passes).
• $u\frac{\partial T}{\partial x}$ = convective (advective) term — the parcel moving into air that is warmer or cooler than where it was.

False. Steady means only $\partial /\partial t=0$ at fixed points; the convective term $(\mathbf{v}\cdot \nabla )\mathbf{v}$ can still be nonzero, so a particle sliding into a faster region genuinely accelerates.

WHAT/WHY: the fixed thermometer feels only the local term; the parcel feels local + convective. Local: $\frac{\partial T}{\partial t}=6$ K/s. → fixed thermometer reads $\boxed{6\text{K/s}}$. Convective: $u\frac{\partial T}{\partial x}=3\times 2=6$ K/s. $\frac{DT}{Dt}=6+6=\boxed{12\text{K/s}}$ The moving parcel warms twice as fast — it also rides into warmer air.

Steady ⇒ $\partial u/\partial t=0$. Only convective survives: $a_x=u\frac{\partial u}{\partial x}=4(1+0.5x)\cdot 4(0.5)=8(1+0.5x)$ At $x=2$: $a_x=8(1+1)=\boxed{16{\text{m/s}}^2}$. Even with a stopwatch showing "no time-dependence," the particle accelerates because it enters faster fluid.

Steady ⇒ drop $\partial /\partial t$. Use $a_x=u\partial u/\partial x+v\partial u/\partial y$ and similarly for $a_y$. $\partial u/\partial x=2,\partial u/\partial y=0$, so $a_x=(2x)(2)+(-2y)(0)=4x$. At $(1,1)$: $a_x=4$ m/s². $\partial v/\partial x=0,\partial v/\partial y=-2$, so $a_y=(2x)(0)+(-2y)(-2)=4y$. At $(1,1)$: $a_y=4$ m/s². $\boxed{a_x=4{\text{m/s}}^2,a_y=4{\text{m/s}}^2}$

Rearranging the material derivative: $\frac{\partial T}{\partial t}=\frac{DT}{Dt}-(\text{convective})=5-(-3)=\boxed{+8\text{K/s}}$. The field is heating at $+8$ K/s at fixed points, but the parcel is drifting into colder air ($-3$ K/s convective), so it only nets $+5$ K/s. Two competing causes, and the material derivative bookkeeps both.

$a_x=u\partial u/\partial x=u_0(1+kx)\cdot u_{0}k=u_0^{2}k(1+kx)$. This is zero when $1+kx=0$, i.e. $\boxed{x=-1/k}$. But at that same $x$ the velocity $u=u_0(1+kx)=0$ too — the particle is momentarily at rest, so it isn't moving into any new region. Interpretation: convective acceleration needs both a velocity and a spatial gradient; kill either and it vanishes.

A: $\partial u/\partial t=0$, $a_x=u\partial u/\partial x=(3x)(3)=9x$. At $x=2$: $a_x=18$ m/s². B: $\partial u/\partial t=1$; $\partial u/\partial x=3$; $u=3x+t$. $a_x=1+(3x+t)(3)=1+3(3\cdot 2+1)=1+21=\boxed{22{\text{m/s}}^2}$ Field A gives $\boxed{18{\text{m/s}}^2}$. The unsteady case adds the local push (+1) and changes the convective piece because $u$ itself is larger.

(a) Lagrangian velocity: $u=\frac{\partial x}{\partial t}{|}_a=2a{e}^{2t}$. Eliminate the label using $a=x{e}^{-2t}$: $u=2(x{e}^{-2t})e^{2t}=\boxed{2x}$ (b) Lagrangian: $a_x=\frac{{\partial }^{2}x}{\partial t^2}{|}_a=4a{e}^{2t}=4x$ (after using $a{e}^{2t}=x$). So $a_x=4x$. (c) Eulerian: $u=2x$ has $\partial u/\partial t=0$, $\partial u/\partial x=2$, so $a_x=0+(2x)(2)=4x$. ✅ $\boxed{a_x=4x\text{(both routes agree)}}$ Lesson: the material derivative is precisely the machinery that makes the Eulerian route reproduce the honest Lagrangian answer.

Local: $\partial \rho /\partial t={\rho }_0(-0.2)=-0.2{\rho }_0$. Convective: $u\partial \rho /\partial x=5\cdot {\rho }_0(0.1)=0.5{\rho }_0$. $\frac{D\rho }{Dt}=-0.2{\rho }_0+0.5{\rho }_0=\boxed{0.3{\rho }_0{\text{kg/m}}^3\text{/s}}$ (Independent of the specific $x$ here since the gradients are constant.) A parcel's density is rising even though the field is thinning in time, because it drifts into denser fluid. This exact $D\rho /Dt$ is what feeds the Continuity equation.

Set up the operator: $a_i=\frac{\partial u_i}{\partial t}+u\frac{\partial u_i}{\partial x}+v\frac{\partial u_i}{\partial y}$. Values at the point: $u=1+1=2$, $v=-2+2=0$. $a_x$: $\partial u/\partial t=1$, $\partial u/\partial x=1$, $\partial u/\partial y=0$. $a_x=1+(2)(1)+(0)(0)=\boxed{3{\text{m/s}}^2}$ $a_y$: $\partial v/\partial t=2$, $\partial v/\partial x=0$, $\partial v/\partial y=-1$. $a_y=2+(2)(0)+(0)(-1)=\boxed{2{\text{m/s}}^2}$ Every term matters: the local (${\partial }_t$) pieces and the convective transport combine cleanly.

Steady, 1-D: $a_x=u\frac{\partial u}{\partial x}=(2+cx)(c)$. Require $a_x(1)=12$: $(2+c)(c)=12\Rightarrow c^2+2c-12=0\Rightarrow c=\frac{-2+\sqrt{4+48}}{2}=-1+\sqrt{13}$ $c=-1+\sqrt{13}\approx \boxed{2.606{\text{s}}^{-1}}$ Check: $u(1)=2+2.606=4.606$; $a_x=4.606\times 2.606\approx 12.0$ m/s². ✅ Lesson: convective acceleration lets you engineer particle acceleration purely through the spatial gradient — this is exactly how a nozzle accelerates gas without any unsteadiness.

The gauge measures at a fixed point; because the flow is steady, $\partial /\partial t=0$ everywhere, so its reading is constant — no contradiction. But a particle flowing through the narrowing must speed up (mass conservation forces higher velocity in the throat), so its convective acceleration $(\mathbf{v}\cdot \nabla )\mathbf{v}\ne 0$. Euler's equation relates this particle acceleration to the pressure gradient, so the pressure drops along the pipe even while it is constant in time at each spot. Both are true at once because they use different ledgers: the gauge is Eulerian-in-time-fixed, the acceleration is Lagrangian. The material derivative is the translator, and this is exactly the setting the Reynolds transport theorem and Steady vs unsteady flow formalize.