Reynolds transport theorem: Question bank — Hinglish deep dive

2.2.13 · D5 · Physics › Fluid Mechanics › Reynolds transport theorem

Master statement ka reminder, taaki koi bhi symbol yahan anjana na lage:

Sahi hai ya galat — justify karo

Har line mein: decide karo true ya false, phir ek-line reason do.

Storage term

\frac{d}{d t} \int_{C V} ρ b d V

zero hoti hai jab bhi fluid CV se flow kar rahi ho.

False. Woh term time mein unsteadiness measure karti hai, flow nahi; yeh sirf tab zero hoti hai jab CV ke andar kuch bhi time ke saath change na ho (steady flow), chahe fluid stream karte rehti ho.

Ek moving control volume ke liye, flux term fluid ki ground velocity

v

use karta hai.

False. Ek moving surface ko cross karna uske relative motion par depend karta hai, isliye

v_{r} = v - v_{C S}

use karte hain; ek surface jo fluid ke saath drift kare, woh zero flux dekhti hai.

Ek control volume ko space ka ek rigid, fixed region hona chahiye.

False. CV fixed, translating, deforming, ya kisi object ke saath move karta hua ho sakta hai (ek moving CV bas relative-velocity correction

v_{r}

add karta hai).

Agar flow steady hai, to

B_{sy s}

(matter ko follow karte hue) change nahi ho sakta.

False. Storage zero hai, lekin surface flux phir bhi nonzero ho sakta hai, isliye

d B_{sy s} / d t

us net efflux ke barabar hoga — ek bend se steady flow phir bhi momentum change karta hai.

Storage term ka integrand

ρ b

hai,

b

nahi.

True. Store ki gayi quantity hai (property per mass)

\times

(mass), aur

d m = ρ d V

, isliye

b

ρ

se weight karna zaroori hai.

b = 1

RTT ko Continuity equation mein badal deta hai.

True. Mass ke liye

B = m

, isliye

b = d B / d m = 1

, aur

d m_{sy s} / d t = 0

mass-conservation statement deta hai.

RTT ka left side,

d B_{sy s} / d t

, ek Eulerian (control-volume) quantity hai.

False. Yeh matter ke ek fixed lump ko follow karta hai — yeh Lagrangian (system) side hai; RTT ka poora kaam hai ise Eulerian terms mein rewrite karna. Dekho Eulerian vs Lagrangian description.

RTT sirf fluids ke liye ek special trick hai.

False. Yeh kisi bhi continuum dwara carried kisi bhi extensive property ke liye ek purely kinematic transport identity hai; fluids bas iska sabse common ghar hai.

Steady, incompressible flow ke liye ek inlet aur ek outlet ke saath,

A_{in} v_{in} = A_{o u t} v_{o u t}

True.

ρ

constant aur zero storage ke saath Continuity deta hai

\sum (A v)_{o u t} = \sum (A v)_{in}

, isliye narrow pipe mein flow faster hoti hai.

Error dhundo

Har line mein ek subtly galat claim hai. Batao kya galat hai.

"Inlet par,

v \cdot \overset{n}{^} > 0

kyunki fluid andar aa rahi hai."

Galat sign hai: inlet par flow CV ke andar point karta hai, outward normal

\overset{n}{^}

ke opposite, isliye

v \cdot \overset{n}{^} < 0

; sirf outlets positive value dete hain.

"Poori fluid speed

∣ v ∣

determine karti hai ki ek patch

d A

se kitni property cross karti hai."

Sirf normal component

v \cdot \overset{n}{^}

surface ko pierce karta hai; boundary ke tangent velocity ussi ke saath slide karta hai aur kuch bhi cross nahi karta.

"Kyunki momentum RTT ke left side par

\sum F

appear karta hai, surface term ek force hona chahiye."

Surface term ek momentum flux hai (wo rate jis par momentum mass flow dwara CS cross karta hai), force nahi; forces woh hain jo us flux plus storage ko balance karte hain.

"Humne RTT CV ko ek point tak shrink karke derive kiya."

Humne time interval

d t \to 0

shrink kiya, volume nahi; CV finite size rakhta hai. Point tak shrink karna uski jagah Material derivative deta hai.

"Ek wall se takrate jet ke liye hum storage term zero set karte hain, isliye flow incompressible honi chahiye."

Storage term isliye drop hoti hai kyunki flow steady hai (time-independent), incompressible hone ki wajah se nahi; yeh alag assumptions hain.

"Kyunki same molecules system define karte hain,

B_{sy s} (t) = B_{C V} (t)

hamesha hold karta hai."

Yeh sirf us ek instant

t

par hold karta hai jahan humne system aur CV ko coincide karna choose kiya; ek instant baad system drift kar chuka hota hai aur equality toot jaati hai.

\oint_{C S}

ka matlab hai sirf inlet aur outlet par integrate karo."

Yeh poore closed control surface par integrate karta hai; solid walls sirf isliye zero contribute karti hain kyunki

v \cdot \overset{n}{^} = 0

wahan hota hai (wall se koi flow nahi).

Why questions

Hum derivation ki shuruaat system aur CV ko time

t

par coincide kyun karte hain?

Taaki

B_{sy s} (t) = B_{C V} (t)

exactly ho, jo humein us instant par "system" ko "CV" se swap karne deta hai aur ek matter-tracking rate ko ek region-based rate mein convert karta hai.

Ek patch se cross karne wali fluid ka swept-volume

(v \cdot \overset{n}{^}) d t d A

kyun hota hai,

∣ v ∣ d t d A

nahi?

Fluid ek slanted cylinder sweep karti hai; uski height $d A$ ke perpendicular sirf normal velocity component times

d t

hoti hai, isliye tangential motion surface se volume add nahi karta.

Ek single closed surface integral automatically inflow aur outflow dono ko kyun handle karta hai?

Outward-normal sign convention

v \cdot \overset{n}{^}

ko inlets par negative aur outlets par positive banata hai, isliye ek

\oint_{C S}

alag bookkeeping ke bina inflow subtract aur outflow add karta hai.

Momentum-flux term velocity mein quadratic kyun hai jabki mass-flux term linear hai?

Momentum flux hai

ρ v (v \cdot \overset{n}{^})

— ek

v

momentum-per-mass hai (

b = v

) aur doosra

v \cdot \overset{n}{^}

hai woh rate jis par mass cross karta hai — isliye dono velocities appear hoti hain,

v^{2}

dependence deta hai.

Divergence theorem RTT ko ek differential (point) law mein kyun convert kar sakta hai?

Yeh surface flux

\oint_{C S} ρ b v \cdot \overset{n}{^} d A

ko ek volume integral

\int_{C V} \nabla \cdot (ρ b v) d V

mein rewrite karta hai; integrands ko equate karne par (kisi bhi CV ke liye valid) local conservation equation milta hai.

Bernoulli energy RTT ka restricted case kyun hai, ek fresh law nahi?

Yeh steady, incompressible, inviscid, along-a-streamline conditions ke under

b = e

case hai; inhe hata do aur tum general energy control-volume balance par wapas aa jaate ho.

Hum extensive property

B

aur uske intensive partner

b

mein distinguish kyun karte hain?

RTT CV par

ρ b

integrate karta hai, isliye ise per-unit-mass form

b

chahiye; total

B

feed karna

ρ d V

ke andar already existing mass ko double-count kar deta.

Edge cases

Behaviour ke har quadrant — pin down karo kya hota hai jab inputs zero, extremes, ya degenerate ho jaate hain.

Ek CV jiska surface har jagah exactly local fluid ke saath move kare.

Relative velocity

v_{r} = v - v_{C S} = 0

surface par hogi, isliye flux term zero hai aur CV system (Lagrangian) ban jaata hai — koi bhi matter kabhi ise cross nahi karta.

CV ke andar fluid bilkul still hai (

v = 0

) lekin walls heat ho rahi hain.

Flux term zero hai, phir bhi energy ke liye (

b = e

) storage term

\frac{d}{d t} \int ρ e d V

nonzero hai — property purely unsteady storage se change ho sakti hai bina kisi flow ke.

Ek closed rigid tank sealed walls ke saath, koi inlet ya outlet nahi.

v \cdot \overset{n}{^} = 0

poore surface par hai, isliye flux zero hai aur

d B_{sy s} / d t

sirf storage term ke barabar hai — RTT reduce ho jaata hai "CV is the system."

Flow parallel patches se equal

\overset{m}{˙}

lekin opposite normal directions se enter aur leave kare.

Mass ke liye (

b = 1

) fluxes cancel hote hain (

\sum ρ b v \cdot \overset{n}{^} = 0

); momentum ke liye (

b = v

) inhe zaroor nahi cancel hona chahiye, kyunki

v

direction mein differ kar sakta hai chahe speeds match karein — isi tarah bends force generate karte hain.

Surface ka ek patch jahan flow tangentially graze kare (

v ⊥ \overset{n}{^}

v \cdot \overset{n}{^} = 0

wahan hoga, isliye woh patch zero flux contribute karta hai chahe fluid fast ho — boundary ke parallel motion kuch bhi cross nahi karta.

Derivation mein limit

d t \to 0

Overlap-region terms

\frac{d}{d t} B_{C V}

mein collapse ho jaate hain aur thin inflow/outflow slabs (regions I aur III) surface integral ban jaate hain — yahi limit discrete bookkeeping ko exact theorem mein badal deta hai.

Density

ρ \to

constant (incompressible) continuity RTT mein.

ρ

dono integrals se bahar aa jaata hai; steadiness storage ko khatam kar deti hai, sirf geometric

\sum (A v)_{o u t} = \sum (A v)_{in}

bachta hai — volume, sirf mass nahi, conserved hota hai.

Ek outlet jahan

v \cdot \overset{n}{^} = 0

momentarily ho (flow reversing sign).

Woh instant wahan zero flux contribute karta hai; jaise hi yeh zero cross karta hai, patch inlet aur outlet ke beech switch karta hai, aur

v \cdot \overset{n}{^}

ka sign continuously iske saath flip karta hai.

Connections

Parent: RTT topic note
Continuity equation — $b = 1$ traps yahan rehte hain.
Momentum equation (control volume) — $b = v$ sign traps.
Eulerian vs Lagrangian description — left-vs-right-side confusion.
Material derivative — "Spot the error" mein point-limit contrast.
Divergence theorem — "Why questions" mein surface-to-volume conversion.
Bernoulli equation — restricted energy case.