Exercises — Fluid kinematics — Eulerian vs Lagrangian description

Fixed station dekhta hai jo bhi hawa ek single point se guzre → Eulerian ($\partial T/\partial t$ fixed $\mathbf{x}$ par). Drifting balloon hawa ke ek parcel ke saath ride karta hai → Lagrangian ($DT/Dt$, particle label fixed). Picture: station = bridge, balloon = fish.

• $\frac{\partial T}{\partial t}$ = local (unsteady) term — temperature field khud time mein har jagah badal raha hai (jaise ki subah hote hue suraj poori valley ko garam kar raha ho).
• $u\frac{\partial T}{\partial x}$ = convective (advective) term — parcel move karke aise hawaon mein jaata hai jo pehle wali jagah se zyada garam ya thandi hoti hain.

Jhooth. Steady ka matlab sirf yeh hai ki fixed points par $\partial /\partial t=0$; convective term $(\mathbf{v}\cdot \nabla )\mathbf{v}$ phir bhi nonzero ho sakta hai, isliye ek particle jo faster region mein slide karta hai genuinely accelerate karta hai.

KYA/KYUN: fixed thermometer sirf local term feel karta hai; parcel local + convective dono feel karta hai. Local: $\frac{\partial T}{\partial t}=6$ K/s. → fixed thermometer padhta hai $\boxed{6\text{K/s}}$. Convective: $u\frac{\partial T}{\partial x}=3\times 2=6$ K/s. $\frac{DT}{Dt}=6+6=\boxed{12\text{K/s}}$ Moving parcel do guna tez garam hota hai — woh garam hawaon mein bhi ride kar raha hai.

Steady ⇒ $\partial u/\partial t=0$. Sirf convective bachta hai: $a_x=u\frac{\partial u}{\partial x}=4(1+0.5x)\cdot 4(0.5)=8(1+0.5x)$ $x=2$ par: $a_x=8(1+1)=\boxed{16{\text{m/s}}^2}$. Stopwatch "no time-dependence" dikhata hai phir bhi particle accelerate karta hai kyunki woh faster fluid mein enter karta hai.

Steady ⇒ $\partial /\partial t$ drop karo. Use karo $a_x=u\partial u/\partial x+v\partial u/\partial y$ aur similarly $a_y$ ke liye. $\partial u/\partial x=2,\partial u/\partial y=0$, toh $a_x=(2x)(2)+(-2y)(0)=4x$. $(1,1)$ par: $a_x=4$ m/s². $\partial v/\partial x=0,\partial v/\partial y=-2$, toh $a_y=(2x)(0)+(-2y)(-2)=4y$. $(1,1)$ par: $a_y=4$ m/s². $\boxed{a_x=4{\text{m/s}}^2,a_y=4{\text{m/s}}^2}$

Material derivative ko rearrange karke: $\frac{\partial T}{\partial t}=\frac{DT}{Dt}-(\text{convective})=5-(-3)=\boxed{+8\text{K/s}}$. Field fixed points par $+8$ K/s se garam ho rahi hai, lekin parcel thandi hawaon mein drift kar raha hai ($-3$ K/s convective), toh net sirf $+5$ K/s milti hai. Do competing causes hain, aur material derivative dono ko bookkeep karta hai.

$a_x=u\partial u/\partial x=u_0(1+kx)\cdot u_{0}k=u_0^{2}k(1+kx)$. Yeh zero hoga jab $1+kx=0$, yani $\boxed{x=-1/k}$. Lekin usi $x$ par velocity bhi $u=u_0(1+kx)=0$ hai — particle momentarily rest par hai, isliye woh kisi naye region mein move nahi kar raha. Interpretation: convective acceleration ke liye dono chahiye — velocity bhi aur spatial gradient bhi; kisi ek ko khatam karo aur woh vanish ho jaata hai.

A: $\partial u/\partial t=0$, $a_x=u\partial u/\partial x=(3x)(3)=9x$. $x=2$ par: $a_x=18$ 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 deta hai $\boxed{18{\text{m/s}}^2}$. Unsteady case mein local push (+1) bhi add hota hai aur convective piece bhi badal jaata hai kyunki $u$ khud larger hai.

(a) Lagrangian velocity: $u=\frac{\partial x}{\partial t}{|}_a=2a{e}^{2t}$. Label eliminate karo $a=x{e}^{-2t}$ use karke: $u=2(x{e}^{-2t})e^{2t}=\boxed{2x}$ (b) Lagrangian: $a_x=\frac{{\partial }^{2}x}{\partial t^2}{|}_a=4a{e}^{2t}=4x$ ($a{e}^{2t}=x$ use karke). Toh $a_x=4x$. (c) Eulerian: $u=2x$ ke liye $\partial u/\partial t=0$, $\partial u/\partial x=2$, toh $a_x=0+(2x)(2)=4x$. ✅ $\boxed{a_x=4x\text{(dono routes agree karte hain)}}$ Lesson: material derivative exactly woh machinery hai jo Eulerian route ko honest Lagrangian answer reproduce karaati hai.

Local: $\partial \rho /\partial t={\rho }_0(-0.2)=-0.2{\rho }_0$. Convective: $u\partial \rho /\partial x=5\cdot {\rho }_0(0.1)=0.5{\rho }_0$. $\frac{D\rho }{Dt}=-0.2{\rho }_0+0.5{\rho }_0=\boxed{0.3{\rho }_0{\text{kg/m}}^3\text{/s}}$ (Specific $x$ se independent hai kyunki gradients constant hain.) Ek parcel ki density badh rahi hai even though field time mein thin ho raha hai, kyunki woh denser fluid mein drift kar raha hai. Yahi $D\rho /Dt$ Continuity equation ko feed karta hai.

Operator set up karo: $a_i=\frac{\partial u_i}{\partial t}+u\frac{\partial u_i}{\partial x}+v\frac{\partial u_i}{\partial y}$. Point par values: $u=1+1=2$, $v=-2+2=0$. $a_x$: $\partial u/\partial t=1$, $\partial u/\partial x=1$, $\partial u/\partial y=0$. $a_x=1+(2)(1)+(0)(0)=\boxed{3{\text{m/s}}^2}$ $a_y$: $\partial v/\partial t=2$, $\partial v/\partial x=0$, $\partial v/\partial y=-1$. $a_y=2+(2)(0)+(0)(-1)=\boxed{2{\text{m/s}}^2}$ Har term matter karta hai: local (${\partial }_t$) pieces aur convective transport cleanly combine ho jaate hain.

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 tumhe particle acceleration purely spatial gradient ke through engineer karne deta hai — exactly yahi ek nozzle bina kisi unsteadiness ke gas accelerate karta hai.

Gauge ek fixed point par measure karta hai; kyunki flow steady hai, $\partial /\partial t=0$ har jagah, toh uski reading constant hai — koi contradiction nahi. Lekin ek particle jo narrowing se flow karta hai use speed up karna padta hai (mass conservation throat mein higher velocity force karta hai), toh uska convective acceleration $(\mathbf{v}\cdot \nabla )\mathbf{v}\ne 0$ hai. Euler's equation is particle acceleration ko pressure gradient se relate karta hai, toh pressure pipe ke along drop hoti hai even though woh time mein har jagah constant hai. Dono ek saath sach hain kyunki woh different ledgers use karte hain: gauge Eulerian-in-time-fixed hai, acceleration Lagrangian hai. Material derivative translator hai, aur yahi exactly woh setting hai jise Reynolds transport theorem aur Steady vs unsteady flow formalize karte hain.