Reynolds transport theorem: Worked examples — Hinglish deep dive

2.2.13 · D3 · Physics › Fluid Mechanics › Reynolds transport theorem

Neeche sab kuch parent se ek boxed law par tika hua hai:

Koi bhi number se pehle, woh sign convention dekho jo sab kuch drive karti hai.

Outward normal $\overset{n}{^}$ hamesha box se bahar point karta hai. Toh:

Jahan fluid nikalta hai, uski velocity roughly $\overset{n}{^}$ se agree karti hai ⟹ $v \cdot \overset{n}{^} > 0$ (positive = outflow).
Jahan fluid andar aata hai, uski velocity $\overset{n}{^}$ ka oppose karti hai ⟹ $v \cdot \overset{n}{^} < 0$ (negative = inflow).

Yahi ek rule is page par har plus aur minus sign ka source hai.

The scenario matrix

Har RTT problem in cells mein se ek (ya blend) hota hai. Neeche har worked example ko us cell ke saath tag kiya gaya hai jo woh solve karta hai.

#	Cell class	Kya mushkil hai	Example
A	Steady, single in / single out	baseline; storage term $= 0$	Ex 1
B	Steady, multiple ports	kai $v \cdot \overset{n}{^}$ signs ko sum karna	Ex 2
C	Unsteady storage, no flow	surface term $= 0$ , sirf $\frac{d}{d t} \int$ bachta hai	Ex 3
D	Unsteady storage + flow together	dono terms ek saath nonzero	Ex 4
E	Sign of momentum flux (inflow minus)	negative $v \cdot \overset{n}{^}$ ek sign flip karta hai	Ex 5
F	Moving control surface	relative velocity $v_{r}$ use karni padti hai	Ex 6
G	Degenerate / zero input	$v = 0$ ya symmetric flow ⟹ terms vanish ho jaate hain	Ex 7
H	Real-world word problem	prose → CV → formula translate karo	Ex 8
I	Exam twist (unsteady + momentum + gauge pressure)	sab kuch ek saath	Ex 9

Cell A — steady, one inlet, one outlet

Worked example Ex 1 — pipe contraction (continuity,

$b = 1$ ) Paani ( $ρ = 1000 kg/m^{3}$ ) steadily ek pipe se flow karta hai jo area $A_{1} = 0.02 m^{2}$ (inlet speed $v_{1} = 3 m/s$ ) se $A_{2} = 0.008 m^{2}$ tak narrow hoti hai. Outlet speed $v_{2}$ nikalo.

Forecast: chhoti pipe ⟹ tez paani. Aage padhne se pehle ek number guess karo.

$b = 1$ lo (mass) aur RTT likho. $\frac{d m _{sy s}}{d t} = 0 = \frac{d}{d t} \int_{C V} ρ d V + \oint_{C S} ρ (v \cdot \overset{n}{^}) d A .$ Ye step kyun? Mass conserved hai, toh poora left side zero hai — yahi Continuity equation hai.
Storage term hatao. Flow steady hai ⟹ CV ke andar kuch bhi time ke saath nahi badalta ⟹ $\frac{d}{d t} \int_{C V} ρ d V = 0$ . Kyun? Storage term unsteadiness measure karta hai, flow nahi (parent mistake #2).
Surface integral evaluate karo do flat ports par. Inlet: $v \cdot \overset{n}{^} = - v_{1}$ (inflow, negative). Outlet: $v \cdot \overset{n}{^} = + v_{2}$ . $0 = ρ (- v_{1}) A_{1} + ρ (+ v_{2}) A_{2} .$ Signs kyun? Outward normal inlet par incoming stream ke against backward point karta hai.
Solve karo. $v_{2} = \frac{v _{1} A _{1}}{A _{2}} = \frac{3 \times 0.02}{0.008} = 7.5 m/s .$

Verify: volume flow in $= 3 \times 0.02 = 0.06 m^{3} / s$ ; out $= 7.5 \times 0.008 = 0.06 m^{3} / s$ . Equal ✓. Units: $\frac{( m/s ) ( m ^{2} )}{m ^{2}} = m/s$ ✓. Forecast ke mutabiq tez ✓.

Cell B — steady, multiple ports

Cell C — unsteady storage, surface se koi flow nahi

Worked example Ex 3 — ek sealed tank heat ho raha hai (storage only)

Ek rigid, sealed tank of volume $V = 0.5 m^{3}$ mein gas hai jiska density uniformly $\frac{d ρ}{d t} = 0.2 kg/m^{3} per s$ ki rate se badhta hai. Koi valve open nahi hai. Mass andar kitni rate se accumulate ho rahi hai?

Forecast: sealed ⟹ koi flux nahi. Sirf storage term act kar sakta hai.

RTT with $b = 1$ : $\frac{d m _{sy s}}{d t} = \frac{d}{d t} \int_{C V} ρ d V + \oint_{C S} ρ (v \cdot \overset{n}{^}) d A .$
Surface term hatao. Walls solid hain: $v \cdot \overset{n}{^} = 0$ closed surface par har jagah. Kyun? Sealed wall se kuch bhi cross nahi karta — parent ka tangential-flow argument extreme case mein.
Storage evaluate karo. Volume fixed, $ρ$ uniform: $\frac{d}{d t} \int_{C V} ρ d V = \frac{d ρ}{d t} V = 0.2 \times 0.5 = 0.1 kg/s .$

Verify: sealed system ke liye $\frac{d m _{sy s}}{d t} = 0.1 kg/s$ aur koi inflow nahi — mass "appear" hoti hai sirf isliye kyunki $ρ$ badhti hai; system mass sach mein tabhi badhti hai jab hum sealed system ko exactly yahi tank (closed) maanein — yahan number andar accumulation rate hai, $0.1 kg/s$ ✓. Units $(kg/m^{3} / s) (m^{3}) = kg/s$ ✓.

Cell D — unsteady AND flowing ek saath

Worked example Ex 4 — filling tank (dono terms alive)

Ek open tank of cross-section $A_{t} = 1 m^{2}$ ko ek inlet pipe of area $A_{i} = 0.01 m^{2}$ se paani bhar raha hai jo $v_{i} = 5 m/s$ par aa raha hai. Koi outlet nahi. Water level $h$ kitni tezi se badhta hai?

Forecast: paani jama ho raha hai ⟹ surface term storage term ko feed kar raha hai. Dono nonzero hain.

CV choose karo = level $h (t)$ tak water region. $b = 1$ , $ρ$ constant. $0 = \frac{d}{d t} \int_{C V} ρ d V + \oint_{C S} ρ (v \cdot \overset{n}{^}) d A .$
Storage term: CV volume $A_{t} h$ hai, toh $\frac{d}{d t} \int_{C V} ρ d V = ρ A_{t} \frac{d h}{d t} .$ Kyun? Stored mass isliye badlta hai kyunki top face upar uthti hai — genuine time change hai.
Surface term: sirf inlet CS ko pierce karta hai, $v \cdot \overset{n}{^} = - v_{i}$ : $\oint_{C S} ρ (v \cdot \overset{n}{^}) d A = ρ (- v_{i}) A_{i} .$
Combine aur solve karo: $0 = ρ A_{t} \frac{d h}{d t} - ρ v_{i} A_{i} \Rightarrow \frac{d h}{d t} = \frac{v _{i} A _{i}}{A _{t}} = \frac{5 \times 0.01}{1} = 0.05 m/s .$

Verify: volume filling rate $= A_{t} \frac{d h}{d t} = 1 \times 0.05 = 0.05 m^{3} / s$ , jo inflow $v_{i} A_{i} = 0.05$ ke equal hai ✓. Units m/s ✓. Dono terms zaroori the — isliye "storage hamesha drop karo" galat hai.

Cell E — momentum, inflow minus sign

Worked example Ex 5 — jet on a flat plate (momentum,

$b = v$ ) Ek horizontal water jet ( $ρ = 1000$ , $A = 2 \times 1 0^{- 3} m^{2}$ , $v = 10 m/s$ ) seedha wall se takrata hai aur sideways spread ho jaata hai, apna saara $x$ -momentum kho ke. Steady flow. Wall se jet par $x$ -force nikalo, aur wall par reaction bhi.

Forecast: wall ko $200$ N per second ka momentum rokna hoga. Sign predict karo.

RTT with $b = v_{x}$ , steady ⟹ storage $= 0$ : $\sum F_{x} = \oint_{C S} ρ v_{x} (v \cdot \overset{n}{^}) d A .$
Mass flow: $\overset{m}{˙} = ρ A v = 1000 (2 \times 1 0^{- 3}) (10) = 20 kg/s .$
Outflow $x$ -momentum: paani purely sideways nikalta hai, toh $v_{x} = 0$ ⟹ out-flux $= 0$ .
Inflow $x$ -momentum: inlet par $v_{x} = v = 10$ aur $v \cdot \overset{n}{^} = - v$ (inflow, negative): $in-flux = ρ v_{x} (v \cdot \overset{n}{^}) A = ρ v (- v) A = - \overset{m}{˙} v = - 200.$ Minus kyun? Outward normal incoming stream ka oppose karta hai — yahi exactly parent ka sign trap hai.
Sum: $\sum F_{x} = 0 - (- 200) = + 200 N$ jet par (peeche point karta hua, jet ko oppose karta hua). Newton ke third law se jet wall ko $200 N$ forward push karta hai.

Verify: $\overset{m}{˙} v = 20 \times 10 = 200 N$ ✓. Units $(kg/s) (m/s) = kg\cdotm/s^{2} = N$ ✓. Parent ke plug-in se match ✓. Dekho Momentum equation (control volume).

Cell F — moving control surface (relative velocity)

Worked example Ex 6 — jet hitting a moving cart

Wohi jet ( $ρ = 1000$ , $A = 2 \times 1 0^{- 3}$ , $v = 10 m/s$ ) ab ek plate se takrata hai jo ek cart par lagi hai jo usi direction mein $v_{C S} = 4 m/s$ se door ja rahi hai. Paani ab bhi sideways spread hota hai. Plate par $x$ -force nikalo.

Forecast: cart jet se bhaag rahi hai, toh stationary case se kam force feel hogi. Guess karo.

CV ko cart ke saath attach karo. Surface move kar rahi hai, toh flux relative velocity use karta hai $v_{r} = v - v_{C S} = 10 - 4 = 6 m/s .$ Ye step kyun? Sirf surface ke relative motion hi use cross karta hai (parent mistake #1). Agar cart jet se match kar le ( $v_{C S} = 10$ ), toh zero paani land karega.
Mass flow jo actually plate se takraata hai: $\overset{m}{˙}_{r} = ρ A v_{r} = 1000 (2 \times 1 0^{- 3}) (6) = 12 kg/s .$
Momentum RTT, steady cart frame mein, cross kiye gaye $x$ -momentum ke liye $v_{r}$ use karte hue: $F_{x} = 0 - (- \overset{m}{˙}_{r} v_{r}) = \overset{m}{˙}_{r} v_{r} = 12 \times 6 = 72 N .$ $v_{r}$ do baar kyun? Ek baar isliye kitna mass per second cross karta hai, ek baar isliye ki har kilogram moving frame ke relative kitna $x$ -momentum laata hai.

Verify: $ρ A v_{r}^{2} = 1000 (2 \times 1 0^{- 3}) (6^{2}) = 72 N$ ✓. $200$ N stationary case se kam ✓ (forecast sahi). Agar $v_{C S} = 10$ : $v_{r} = 0 \Rightarrow F = 0$ ✓ — degenerate check.

Cell G — degenerate / zero input

Cell H — real-world word problem

Cell I — exam twist: unsteady + momentum + gauge pressure

Worked example Ex 9 — thrust on an emptying nozzle

Ek nozzle atmosphere mein paani discharge karta hai. Steady flow: inlet area $A_{1} = 0.02 m^{2}$ , gauge pressure $p_{1} = 1.5 \times 1 0^{5} Pa$ , speed $v_{1} = 3 m/s$ ; outlet (air mein, gauge $p_{2} = 0$ ) area $A_{2} = 0.008 m^{2}$ . $ρ = 1000$ . Nozzle par paani ka axial force nikalo.

Forecast: pressure aage push karta hai, jet speed badhti hai — ek sizeable force expect karo. Uska sign predict karo.

Continuity for $v_{2}$ (Ex 1 reuse): $v_{2} = \frac{v _{1} A _{1}}{A _{2}} = \frac{3 ( 0.02 )}{0.008} = 7.5 m/s .$
Mass flow: $\overset{m}{˙} = ρ v_{1} A_{1} = 1000 (3) (0.02) = 60 kg/s .$
Momentum RTT, steady (storage $= 0$ ), CS par pressure forces include karte hue: $p_{1} A_{1} - p_{2} A_{2} + R_{x} = \overset{m}{˙} v_{2} - \overset{m}{˙} v_{1},$ jahan $R_{x}$ paani par nozzle ka axial force hai. Pressure terms kyun? Gauge pressure har open face par inward push karta hai; wall reaction $R_{x}$ baaki ko balance karta hai.
Numbers daalo ( $p_{2} = 0$ ): $1.5 \times 1 0^{5} (0.02) + R_{x} = 60 (7.5) - 60 (3)$ $3000 + R_{x} = 450 - 180 = 270 \Rightarrow R_{x} = 270 - 3000 = - 2730 N .$
Nozzle par reaction: paani nozzle ko $- R_{x} = + 2730 N$ , yaani forward (thrust), $\approx 2.73 kN$ se push karta hai.

Verify: momentum change $\overset{m}{˙} (v_{2} - v_{1}) = 60 (4.5) = 270 N$ ✓. Pressure force $3000 N$ ✓. $R_{x} = 270 - 3000 = - 2730 N$ ✓. Units: $Pa \cdot m^{2} = N$ , $kg/s \cdot m/s = N$ ✓. Nozzle paani ko peeche kheenchta hai, toh paani nozzle ko forward thrust deta hai — physically sensible hai.

Recall Kaun sa cell kaun sa hai?

Steady one-in-one-out kaun sa example hai? ::: Ex 1 (cell A). Sirf storage term kahan bachta hai? ::: Ex 3 (cell C, sealed tank). Storage aur flux dono nonzero kahan hain? ::: Ex 4 (cell D, filling tank). Kaun sa example relative velocity chahta hai? ::: Ex 6 (cell F, moving cart). Kaun sa example pressure, unsteady-check aur momentum combine karta hai? ::: Ex 9 (cell I, nozzle thrust).

Connections

2.2.13 Reynolds transport theorem — parent theorem jis par ye examples apply hote hain.
Continuity equation — Ex 1, 2, 4, 8 ( $b = 1$ ).
Momentum equation (control volume) — Ex 5, 6, 9 ( $b = v$ ).
Eulerian vs Lagrangian description — isliye hum ek CV kyun dekhte hain.
Bernoulli equation — nozzle speeds ke peeche chhupa hua energy check.