Similarity — geometric, kinematic, dynamic; Reynolds similarity: Question bank

2.2.27 · D5 · Physics › Fluid Mechanics › Similarity — geometric, kinematic, dynamic; Reynolds similar

Shuru karne se pehle, ek anchor yaad rakho: similarity ka matlab hai dimensionless ratios, kabhi bhi raw speeds, sizes, ya forces match karna nahi. Neeche ke almost har trap us ek idea ki disguised violation hai.

Definition Is page par use hone wale symbols aur terms — pehle ek baar padho

Length-scale ratio $L_{r} = \frac{L _{m}}{L _{p}}$ — ek constant jo har model length $L_{m}$ ko corresponding prototype length $L_{p}$ se relate karta hai. " $1/10$ model" ka matlab hai $L_{r} = 1/10$ .
Geometric similarity — model aur prototype ki shape ek jaisi hoti hai (saari lengths ek single $L_{r}$ se related hoti hain).
Kinematic similarity — motion similar hoti hai: velocity fields ka streamline pattern ek jaisa hota hai, sirf magnitude scale alag hoti hai. Iske liye geometric similarity zaroori hai.
Dynamic similarity — forces similar hoti hain: model aur prototype mein har corresponding force pair ka ratio same hota hai. Yeh kinematic (aur isliye geometric) similarity imply karta hai.
Mach number $Ma = \frac{V}{c}$ (flow speed $V$ divided by sound speed $c$ ) — inertia/elastic force ratio; tab matter karta hai jab compressibility matter karti hai. Dimensionless-number recipe ke liye Re dekho.
Weber number $We = \frac{ρ V ^{2} L}{σ}$ (surface tension $σ$ ke saath) — inertia/surface-tension force ratio; droplets, bubbles, thin films ke liye matter karta hai.

Neeche, picture in teeno levels of similarity ko ek image mein pin karti hai taaki True/False items ka ek visual ho jis par lean kiya ja sake.

Figure — Similarity — geometric, kinematic, dynamic; Reynolds similarity

True or false — justify karo

Do geometrically similar bodies ka streamline pattern automatically similar hota hai.

False — geometric (shape) similarity zaroori hai lekin kinematic similarity ke liye kaafi nahi; tumhe matched motion bhi chahiye, jo ek submerged body ke liye matlab hai matched Re.

Agar do flows ka Reynolds number same ho, toh unhe geometrically similar hona chahiye.

False — equal Re dynamic similarity ke liye zaroori hai lekin sirf diye gaye geometric similarity ke case mein; do alag shape ke bodies ek Re share kar sakti hain aur phir bhi bilkul alag behave kar sakti hain.

Dynamic similarity (equal force ratios) kinematic similarity (matched streamline patterns) imply karta hai.

True — matched force ratios (Newton's law non-dimensionalised) velocity fields ko same pattern rakhne par majboor karte hain, isliye motion automatically similar hoti hai.

Kinematic similarity (matched motion) dynamic similarity (matched force ratios) imply karta hai.

False — ladder sirf neeche ki taraf chadhti hai: shape → motion → forces. Matched motion matched force ratios guarantee nahi karta (jaise gravity/inertia alag ho sakti hai).

Geometric similarity ke saath, Reynolds number match karne par model aur prototype mein drag coefficient

C_{D}

equal hota hai.

True — geometric similarity aur equal Re ke saath Navier–Stokes equation ek identical dimensionless equation ban jaati hai, isliye har dimensionless coefficient including $C_D$ match karta hai.

Reynolds number match karne par model aur prototype mein drag force equal hoti hai.

Generally False —

C_{D}

match karta hai, lekin

F_{D} = \frac{1}{2} ρ V^{2} A C_{D}

phir bhi actual

ρ, V, A

par depend karta hai; forces sirf us special same-fluid case mein equal hoti hain jahan higher model speed exactly smaller area ko cancel karta hai.

Same fluid ke saath tum hamesha Reynolds number aur Froude number simultaneously match kar sakte ho.

False — Re-matching demand karta hai

V \propto 1/ L

jabki Fr-matching demand karta hai

V \propto L

; ye do speed laws same fluid mein scaled model ke liye incompatible hain.

Reynolds number dimensionless hota hai.

True —

\frac{ρ V L}{μ}

ki units hain

\frac{( kg/m ^{3} ) ( m/s ) ( m )}{kg/(m\cdots)} = 1

, yahi reason hai ki ise alag-alag size ke flows mein equate kiya ja sakta hai.

Geometric similarity ke under, areas

L_{r}

se scale hote hain aur volumes

L_{r}^{2}

se.

False — har length

L_{r}

se scale hoti hai, isliye areas (length²)

L_{r}^{2}

se scale hote hain aur volumes (length³)

L_{r}^{3}

se; exponents mix up karna ek classic gadbad hai.

Error dhundo

"Model 30 m/s par chal raha hai, same jaise car, isliye flows similar hain."

Error yeh hai ki raw velocity match ki ja rahi hai Re ki jagah; same speed par ek chhota model ka chhota

ρ V L / μ

hoga, isliye woh ek alag flow regime mein hoga — bilkul similar nahi.

"Kyunki shapes identical hain, measured model drag prototype drag ke barabar hai."

Geometric similarity akela shape fix karta hai lekin force ratios nahi; dynamic (Re) similarity ke bina

C_{D}

values alag hoti hain, isliye drags simply equate nahi ki ja sakti.

"Safest rehne ke liye, hume Re, Fr, Ma aur We sab ek saath match karne chahiye."

Zyada matching automatically behtar nahi hota — ye criteria (Reynolds, Froude, Mach

Ma = V / c

, Weber

We = ρ V^{2} L / σ

) contradictory speed/fluid requirements impose karte hain; tumhe instead dominant force identify karni chahiye aur sirf uska number match karna chahiye.

"Inertia force

ρ V L^{2}

se scale hoti hai, isliye Re

= ρ V L^{2} / μ

Inertia force

ρ V^{2} L^{2}

hai (

V

ka ek extra factor acceleration

\sim V^{2} / L

se); viscous

μ V L

se divide karne par correct

ρ V L / μ

milta hai.

"Waves wale ship ke liye Reynolds similarity use karo kyunki paani viscous hota hai."

Free surface ka matlab hai gravity/wave forces dominate karti hain, isliye Froude similarity govern karta hai; viscosity present hai lekin wave pattern ke liye controlling force ratio nahi hai.

"Kinematic viscosity

ν

hamesha Re-matching mein cancel ho jaati hai, isliye ignore karo."

ν = μ / ρ

sirf tab cancel hoti hai jab model aur prototype same fluid use karein; alag fluids ke saath pura

ν_{m} / ν_{p}

ratio Re equation mein rakhna padta hai.

Why questions

Dynamic similarity hierarchy ke top par kyun hoti hai?

Kyunki force ratios match karna Newton's second law ko identically non-dimensionalise karta hai, jo automatically matched motion (kinematic) produce karta hai — ise achieve karna neeche sab kuch free mein deliver karta hai.

Re match karna identical dimensionless solutions guarantee karne ke liye kyun kaafi hai?

Jab tum Navier–Stokes ko

L

V

ρ V^{2}

se non-dimensionalise karte ho, har coefficient vanish ho jaata hai siwa viscous term ke aage

1/ Re

ke, isliye equal Re ka matlab hai same equation aur hence same solution.

1/10 scale model ko same fluid mein prototype speed ke 10× par kyun test karna padta hai?

Equal

ρ, μ

ke saath Re-matching deta hai

V_{m} L_{m} = V_{p} L_{p}

, isliye

L

ko 10 se shrink karna (matlab

L_{r} = 1/10

) ratio fixed rakhne ke liye

V

ko 10 se badhne par majboor karta hai.

Engineers ordinary wind tunnels ki jagah water tunnels ya pressurised air tunnels kyun prefer karte hain?

Same-fluid Re-matching model speeds ko impractically high drive kar deta hai; ek denser ya more suitable fluid (higher

ρ

, lower

ν

) tumhe target Re ek reasonable, safer speed par hit karne deta hai.

Boundary Layer concept kyun matter karta hai is baat ke liye ki Reynolds similarity reliable hai ya nahi?

Boundary layer ka behaviour (laminar vs turbulent transition) Re par depend karta hai, isliye agar model ka Re prototype ke Re se bahut door hai, layers differ karti hain aur extrapolated

C_{D}

unreliable ho jaata hai.

Hum dimensionless numbers forces ke ratios lekar kyun banate hain, forces khud lekar nahi?

Raw forces mein units hoti hain aur scale ke saath change hoti hain, lekin unke ratios pure numbers hain jo sizes mein equate ki ja sakti hain — yeh Π-theorem ka idea hai ki physics dimensionless groups par depend karta hai.

Edge cases

Agar Re-matching se model velocity speed of sound se zyada aa jaaye, toh kya Reynolds similarity valid hai?

Nahi — jab flow sound speed ke paas aati hai, compressibility enter karta hai (Mach number

Ma = V / c

significant ho jaata hai) aur Re-matching akela physics capture nahi karta; tum low-speed regime chhhod chuke ho jise criterion assume karta hai.

"Similarity" ka kya hoga agar prototype flow turbulent hai lekin model ka Re use laminar flow mein le jaaye?

Ye alag regimes mein hain, isliye matched geometry ke bawajood dimensionless solutions diverge karte hain — yahi reason hai ki tumhe correct Re tak pahunchna hai, sirf ek proportional Re nahi.

Creeping (bahut low Re) flow mein jahan inertia negligible hai, kya Re match karna phir bhi matter karta hai?

Limit Re → 0 mein inertia term equation se bilkul drop ho jaata hai, isliye flow sirf viscous balance se govern hoti hai aur results almost Re-independent ho jaate hain — matching trivial ho jaati hai, key constraint nahi.

Agar do flows same fluid aur same size body use karein, toh konsa velocity Re matched rakhta hai?

L

aur

ν

identical hone par, Re-matching

V_{m} = V_{p}

force karta hai — compensate karne ke liye koi scale factor nahi hai, isliye same fluid mein identical size simply identical speed ka matlab hai.

Kya scale factor

L_{r} = 1

ke saath geometric similarity ho sakti hai lekin phir bhi dynamic similarity na ho?

Haan — same size (ek full replica,

L_{r} = 1

) phir bhi alag speed ya fluid par chal sakta hai, alag Re deta hai, isliye shapes match karte hain jabki force ratios nahi.

Kya similarity ke liye perfectly symmetric body chahiye, ya sirf har scale par same shape?

Sirf ek constant $L_{r}$ se scaled same shape zaroori hai; symmetry irrelevant hai — ek asymmetric body aur uski scaled copy geometrically similar hain jab tak har length same factor use kare.