Exercises — Reynolds transport theorem

Woh galat hain. Woh term surface flux hai — woh rate jis par property fluid ke move karne ki wajah se boundary cross karti hai. Plain-word names:

• Left, $\frac{d{B}_{sys}}{dt}$ = "change ki rate original lump of matter ke saath chalte hue" (Lagrangian).
• Storage, $\frac{d}{dt}{\int }_{CV}\rho bdV$ = "fixed box ke andar baithee cheez ki matra kitni tezi se change ho rahi hai".
• Surface, ${\oint }_{CS}\rho b(\vec{v}\cdot \hat{n})dA$ = "walls se cheez ka net outflow" (out minus in).

Colleague ne andar ka change (storage) ko wall cross karna (surface) se confuse kiya.

$b=\frac{dB}{dm}$ = property per unit mass.

• (a) Mass: $B=m$, isliye $b=\frac{dm}{dm}=1$.
• (b) Momentum: $B=m\vec{v}$, isliye $b=\vec{v}$.
• (c) Energy: $B=E$, isliye $b=e$ (specific energy, energy per kilogram).

RTT $b=1$ (mass) ke saath set up karo: $\frac{d{m}_{sys}}{dt}=0$, aur steady flow storage ko khatam kar deta hai: $0=\underset{=0\text{ (steady)}}{\underbrace{\frac{d}{dt}{\int }_{CV}\rho dV}}+{\oint }_{CS}\rho (\vec{v}\cdot \hat{n})dA.$ Do ports ke upar surface integral evaluate karo. Inlet pe fluid andar flow karta hai, outward normal ${\hat{n}}_1$ ke opposite, isliye $\vec{v}\cdot {\hat{n}}_1=-v_1$. Outlet pe fluid bahar flow karta hai, ${\hat{n}}_2$ ke saath, isliye $\vec{v}\cdot {\hat{n}}_2=+v_2$. Pipe walls kuch contribute nahi karti ($\vec{v}\cdot \hat{n}=0$ wahan). Isliye: $0=\rho (-v_1)A_1+\rho (+v_2)A_2\Rightarrow v_2=\frac{A_1}{A_2}{v}_1=\frac{0.02}{0.008}(3)=7.5\text{m/s}.$ Mass flow: $\dot{m}=\rho A_1{v}_1=1000(0.02)(3)=\boxed{60\text{kg/s}}$ (dono ports pe same, jaisa hona chahiye).

$b=1$, lekin ab NOT steady — tank ka content badh raha hai, isliye storage term rakhte hain: $0=\frac{d}{dt}{\int }_{CV}\rho dV+{\oint }_{CS}\rho (\vec{v}\cdot \hat{n})dA.$ Sirf ek port (inflow), $\vec{v}\cdot \hat{n}=-v$, isliye surface term $=-{\dot{m}}_{in}=-4$. Density constant hai, bahar nikalo: $\frac{d}{dt}\int \rho dV=\rho \frac{dV}{dt}$. Phir: $0=\rho \frac{dV}{dt}-{\dot{m}}_{in}\Rightarrow \frac{dV}{dt}=\frac{{\dot{m}}_{in}}{\rho }=\frac{4}{1000}=\boxed{0.004{\text{m}}^3/\text{s}}.$

RTT $b=\vec{v}$ (momentum) ke saath use karo, steady ⇒ no storage: $\sum \vec{F}={\oint }_{CS}\rho \vec{v}(\vec{v}\cdot \hat{n})dA.$ Mass flow: $\dot{m}=\rho Av=1000(1.5\times 10^{-3})(20)=30\text{kg/s}$.

Inlet ($\vec{v}=(20,0)$, andar flow karta hai isliye $\vec{v}\cdot \hat{n}=-v$): contribution $=\rho \vec{v}(\vec{v}\cdot \hat{n})A=\dot{m}(-1)(20,0)=(-600,0)$.

Outlet ($\vec{v}=(0,20)$, bahar flow karta hai isliye $\vec{v}\cdot \hat{n}=+v$): contribution $=\dot{m}(+1)(0,20)=(0,+600)$.

Flux sum karo (yeh water par $\sum \vec{F}$ ke barabar hai): $\sum F_x=-600+0=-600\text{N},\sum F_y=0+600=+600\text{N}.$ Toh vane water ko $\boxed{F_x=-600\text{N},F_y=+600\text{N}}$ ke saath push karta hai (yaani leftward-and-up, taaki uski rightward motion hataye aur upward motion de). Magnitude $\sqrt{600^2+600^2}\approx 848.5\text{N}$.

Integrand hai $\rho \vec{v}(\vec{v}\cdot \hat{n})$. Do cheezein sign carry karti hain:

• $\vec{v}$ khud $+x$ hai (momentum jo carry ho raha hai), factor $+20$.
• $(\vec{v}\cdot \hat{n})$ inlet par $-v$ hai (velocity outward normal ke opposite hai), factor $-20$.

Unka product negative hai: momentum $+x$ enter kar raha hai, aur "entering" ek negative outflow ki tarah likha jaata hai. Physically, inlet se momentum andar aana woh momentum hai jo CV gain karta hai, isliye water ki momentum steady rakhne ke liye vane ko equal-and-opposite force supply karna padta hai. Negative sign exactly wahi hai jo baad mein (via $\sum F=\text{outflow}-\text{inflow}$) reaction force dene ke liye flip hota hai.

Step 1 — continuity ($b=1$) se $v_2$ milta hai: $v_2=\frac{A_1}{A_2}{v}_1=\frac{0.01}{0.002}(2)=10\text{m/s},\dot{m}=\rho A_1{v}_1=1000(0.01)(2)=20\text{kg/s}.$ Step 2 — momentum ($b=\vec{v}$), steady. $x$-momentum equation: $\sum F_x={\oint }_{CS}\rho v_x(\vec{v}\cdot \hat{n})dA=\dot{m}{v}_2-\dot{m}{v}_1=\dot{m}(v_2-v_1).$ (Outlet $+\dot{m}{v}_2$; inlet $-\dot{m}{v}_1$ kyunki inflow minus carry karta hai.) $\sum F_x=20(10-2)=160\text{N}.$ Step 3 — real forces list karo jo $\sum F_x$ banate hain: inlet face par pressure fluid ko $+x$ direction mein push karta hai: $+p_1{A}_1$. Exit face par pressure atmospheric gauge $=0$: $+p_2{A}_2=0$. Plus wall reaction $R_x$: $\sum F_x=p_1{A}_1+p_2{A}_2+R_x=50000(0.01)+0+R_x=500+R_x.$ Step 4 — solve: $500+R_x=160\Rightarrow R_x=160-500=\boxed{-340\text{N}}$. Nozzle water ko $-x$ mein 340 N se pull karta hai; reaction mein water nozzle ko forward 340 N se pull karta hai (yahi woh thrust hai jo tum hose nozzle pakad ke feel karte ho).

Key idea: moving CV ke liye surface term relative velocity ${\vec{v}}_r=\vec{v}-{\vec{v}}_{CS}$ use karta hai, kyunki sirf surface ke relative motion hi actually surface cross karta hai. $v_r=v-v_{CS}=15-5=10\text{m/s}.$ Moving surface across mass flow $v_r$ use karta hai: ${\dot{m}}_r=\rho A{v}_r=1000(2\times 10^{-3})(10)=20\text{kg/s}.$ Momentum RTT (relative), cart frame mein steady, $b={\vec{v}}_r$: $\sum F_x=\oint \rho v_{r,x}({\vec{v}}_r\cdot \hat{n})dA.$ Inlet: relative velocity $+10$, inward cross karta hai $({\vec{v}}_r\cdot \hat{n}=-v_r)$ ⇒ term $={\dot{m}}_r(-1)(+10)=-200$. Outlet: $180^{\circ }$ deflect hua, relative velocity $-10$, outward cross karta hai $(+v_r)$ ⇒ term $={\dot{m}}_r(+1)(-10)=-200$. $\sum F_x=-200+(-200)=\boxed{-400\text{N}}.$ Vane water ko $-x$ mein 400 N se push karta hai; water cart ko forward (+x) 400 N se push karta hai.

Shuru karo ($b=1$, fixed CV): $0=\frac{\partial }{\partial t}{\int }_{CV}\rho dV+{\oint }_{CS}\rho (\vec{v}\cdot \hat{n})dA.$ (Time-derivative andar $\partial /\partial t$ ke roop mein jaati hai kyunki CV fixed hai.) Surface flux ko volume integral mein convert karo Divergence theorem ke saath, ${\oint }_{CS}\vec{F}\cdot \hat{n}dA={\int }_{CV}\nabla \cdot \vec{F}dV$, $\vec{F}=\rho \vec{v}$ use karke: ${\oint }_{CS}\rho (\vec{v}\cdot \hat{n})dA={\int }_{CV}\nabla \cdot (\rho \vec{v})dV.$ Ek volume integral mein combine karo: ${\int }_{CV}\left[\frac{\partial \rho }{\partial t}+\nabla \cdot (\rho \vec{v})\right]dV=0.$ Yeh CV ke har choice ke liye valid hai, chahe kitna bhi chhota ho, isliye bracket pointwise vanish hona chahiye: $\boxed{\frac{\partial \rho }{\partial t}+\nabla \cdot (\rho \vec{v})=0.}$ Yeh exactly point analogue hai, aur yeh Material derivative aur Continuity equation notes se connect karta hai.

L1 :::: terms ka naam lena, $b$ choose karna (Recognition) L2 :::: numbers plug karna, steady vs unsteady storage (Application) L3 :::: vector momentum, inflow-minus signs (Analysis) L4 :::: continuity + momentum + pressure forces combined (Synthesis) L5 :::: moving-CV relative velocity, differential form via divergence theorem (Mastery)