Similarity — geometric, kinematic, dynamic; Reynolds similarity: Worked examples

2.2.27 · D3 · Physics › Fluid Mechanics › Similarity — geometric, kinematic, dynamic; Reynolds similar

Shuru karne se pehle, symbols ka ek plain-language refresher — taaki hum kabhi kuch unexplained na likhen.

Definition Har equation ke characters

$L$ — ek characteristic length: body ka ek representative size (car ki length, pipe ka diameter). Ek single ruler imagine karo jo object ke saath rakha ho.
$V$ — body se door flow speed, metres per second mein.
$ρ$ (rho) — density, ek cubic metre mein packed mass ( $kg / m^{3}$ ).
$μ$ (mu) — dynamic viscosity, fluid kitna "chipchipa/geela" hai.
$ν = μ / ρ$ (nu) — kinematic viscosity, density per unit mein chipchipaahat ( $m^{2} / s$ ).
$g$ — gravitational acceleration, gravity fluid ko kitna neeche kheenchti hai; Earth par $g = 9.81 m/s^{2}$ . Ek giri hui droplet ko ground ki taraf accelerate hote socho. Yeh tab aata hai jab gravity/waves matter karte hain.
$A$ — ek reference area: frontal (ya projected) area jo flow "dekhta hai," $m^{2}$ mein. Geometrically similar bodies ke liye $A \propto L^{2}$ , isliye $A_{p} / A_{m} = 1/ L_{r}^{2}$ .
$Q$ — volume flow rate (discharge): kitne cubic metres fluid per second pass karta hai ( $m^{3} / s$ ). Yeh speed times cross-section ke barabar hai, $Q = V \cdot A$ — hum ise Ex 8 mein use karte hain.
$σ$ (sigma) — surface tension: liquid surface ki "skin-force per unit length" ( $N/m$ ). Yeh tab hi matter karta hai jab model tiny ho aur capillary effects aayein; yeh neeche Weber number drive karta hai.
$c$ — fluid mein speed of sound ( $m/s$ ); yeh Mach number drive karta hai, neeche hamara compressibility warning.
Subscript $m$ = model (chhota test piece), subscript $p$ = prototype (full-size asli cheez).
$L_{r} = L_{m} / L_{p}$ — length-scale ratio. Ek $1/10$ model ka $L_{r} = 0.1$ hota hai.

Scenario matrix

Har similarity problem jo tum kabhi dekhoge woh in cells mein se kisi ek mein aata hai. Har row ek alag tarah ka question hai; aakhri column us example ka naam deta hai jo use clear karta hai. Neeche ki figure inhi aath cells ko map ke roop mein draw karti hai taaki tum dekh sako ki har example kahan baithta hai.

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

#	Case class	Ismein tricky kya hai	Kaunsa number	Example
1	Same fluid, model speed dhundho	size ghata toh speed badhti hai	Reynolds	Ex 1
2	Different fluid, model speed dhundho	$ν$ ratio rakhna zaroori hai	Reynolds	Ex 2
3	Force scaling, same fluid	forces equal nikalta hai (surprise)	Reynolds + $C_{D}$	Ex 3
4	Force scaling, different fluid	real $ρ, V, A$ plug karne padte hain	Reynolds + $C_{D}$	Ex 4
5	Free-surface / wave flow	Reynolds galat number hai	Froude (+Weber caveat)	Ex 5
6	Conflict: Re aur Fr dono chahiye	ek fluid se usually impossible	dominant choose karo	Ex 6
7	Degenerate: $L_{r} = 1$ (full scale)	limiting case, sab barabar	either	Ex 7
8	Exam twist: hidden regime	tumhe decide karna hai kaunsa number	pehle decide karo	Ex 8

Inke andar do "signs/limits" hain: $L_{r} < 1$ (model chhota — normal case) Reynolds ke under speeds upar jaati hain; limit $L_{r} \to 1$ (Ex 7) saare ratios ko $1$ par collapse kar deti hai. Hum dono hit karte hain.

Case 1 — Same fluid, model speed dhundho

Neeche ki picture dono hulls ko scale ke saath unki speeds ke saath dikhati hai — dekho ki chhote model ko kitna bada speed arrow chahiye.

Number pick karo. Submerged body, koi free surface nahi → viscous/inertia contest → Reynolds match karo. Yeh step kyun? Paani ke andar gehra submarine koi waves nahi banata; gravity/free-surface effects absent hain, isliye Re controlling ratio hai.
Re-matching likho aur fluid cancel karo. $\frac{ρ V _{m} L _{m}}{μ} = \frac{ρ V _{p} L _{p}}{μ} \Rightarrow V_{m} L_{m} = V_{p} L_{p}$ Yeh step kyun? Same fluid matlab identical $ρ$ aur $μ$ ; woh cancel ho jaate hain, clean product rule $V L = const$ bacha.
Solve karo. $V_{m} = V_{p} \frac{L _{p}}{L _{m}} = 8 \times \frac{60}{3} = 8 \times 20 = 160 m/s$ Yeh step kyun? Chhota body ( $L_{r} = 1/20$ ) ko $ρ V L / μ$ fixed rakhne ke liye $20 \times$ faster push karna padega.

Verify karo: Re match hona chahiye. $Re_{p} \propto V_{p} L_{p} = 8 \cdot 60 = 480$ ; $Re_{m} \propto V_{m} L_{m} = 160 \cdot 3 = 480$ . ✓ Units: dono sides par $(m/s) (m)$ — consistent. Forecast check: haan, bada — exactly $20 \times$ .

Case 2 — Different fluid, model speed dhundho

Number pick karo. Internal pipe flow, koi free surface nahi → phir se Reynolds.
Poora Re rakho, $ν$ cancel mat karo. $\frac{V _{m} D _{m}}{ν _{m}} = \frac{V _{p} D _{p}}{ν _{p}}$ Yeh step kyun? Fluids alag hain, isliye $ν_{m} \neq = ν_{p}$ — kinematic-viscosity ratio bachta hai aur carry karna zaroori hai.
$V_{m}$ ke liye solve karo. $V_{m} = V_{p} \frac{D _{p}}{D _{m}} \frac{ν _{m}}{ν _{p}} = 3 \times \frac{0.6}{0.15} \times \frac{1.5 \times 1 0 ^{- 5}}{1.0 \times 1 0 ^{- 6}}$ $= 3 \times 4 \times 15 = 180 m/s$ Yeh step kyun? $D_{p} / D_{m} = 4$ (size effect) times $ν_{m} / ν_{p} = 15$ (fluid effect) — dono speed upar push karte hain, aur yeh multiply hote hain.

Verify karo: $Re_{p} = \frac{3 \cdot 0.6}{1 0 ^{- 6}} = 1.8 \times 1 0^{6}$ ; $Re_{m} = \frac{180 \cdot 0.15}{1.5 \times 1 0 ^{- 5}} = 1.8 \times 1 0^{6}$ . ✓ Dono dimensionless aur equal hain.

Case 3 — Force scaling, same fluid (the surprise)

Equal drag coefficients use karo. Re match hone par, $C_{D} = \frac{F _{D}}{\frac{1}{2} ρ V ^{2} A}$ (upar define kiya) model aur prototype ke liye identical hai. Yeh step kyun? Matched Re matlab identical dimensionless solution → identical $C_{D}$ . Drag Coefficient dekho.
Coefficients barabar karo aur shared factor $\frac{1}{2} ρ$ se divide karo. $\frac{F _{D, p}}{\frac{1}{2} ρ V _{p}^{2} A _{p}} = \frac{F _{D, m}}{\frac{1}{2} ρ V _{m}^{2} A _{m}}$ Constant $\frac{1}{2}$ aur density $ρ$ (same fluid) dono sides par identical hain, isliye har term ko $\frac{1}{2} ρ$ se divide karne par yeh cleanly remove ho jaate hain: $\frac{F _{D, p}}{V _{p}^{2} A _{p}} = \frac{F _{D, m}}{V _{m}^{2} A _{m}} \Rightarrow F_{D, p} = F_{D, m} \frac{V _{p}^{2} A _{p}}{V _{m}^{2} A _{m}}$ Yeh step kyun? Same fluid → $ρ$ common hai; $\frac{1}{2}$ universal constant hai. Dono term-by-term division survive nahi karte, sirf speed aur area ratios bachte hain.
Same-fluid relations $V_{m} = V_{p} / L_{r}$ aur $A = L^{2}$ so $A_{p} / A_{m} = 1/ L_{r}^{2}$ insert karo. $\frac{V _{p}^{2}}{V _{m}^{2}} = L_{r}^{2}, \frac{A _{p}}{A _{m}} = \frac{1}{L _{r}^{2}}$ $F_{D, p} = F_{D, m} \cdot L_{r}^{2} \cdot \frac{1}{L _{r}^{2}} = F_{D, m} = 90 N$ Yeh step kyun? Zyada model speed (jo force badhata hai) exactly chhote model area (jo force ghataata hai) ko cancel kar deti hai. Dono $L_{r}^{2}$ factors annihilate ho jaate hain.

Verify karo: cancellation $L_{r}^{2} \cdot L_{r}^{- 2} = 1$ kisi bhi $L_{r}$ ke liye hai — isliye same-fluid Re-matched drag hamesha equal hota hai. Units: newtons in, newtons out. ✓ Forecast: haan, equal — counter-intuitive answer.

Case 4 — Force scaling, different fluid

Phir se equal $C_{D}$ — $\frac{1}{2}$ pehle ki tarah cancel hota hai, lekin ab $ρ$ cancel mat karo. $\frac{F _{D, p}}{\frac{1}{2} ρ _{p} V _{p}^{2} A _{p}} = \frac{F _{D, m}}{\frac{1}{2} ρ _{m} V _{m}^{2} A _{m}} \Rightarrow F_{D, p} = F_{D, m} \frac{ρ _{p} V _{p}^{2} A _{p}}{ρ _{m} V _{m}^{2} A _{m}}$ Yeh step kyun? Different fluids matlab $ρ_{m} \neq = ρ_{p}$ ; har factor genuine hai aur plug in karna zaroori hai. Sirf shared constant $\frac{1}{2}$ jaata hai.
Numbers plug karo. $F_{D, p} = 30 \times \frac{1000 \cdot 4 ^{2} \cdot 5.0}{1.2 \cdot 5 0 ^{2} \cdot 0.20} = 30 \times \frac{1000 \cdot 16 \cdot 5}{1.2 \cdot 2500 \cdot 0.20}$ $= 30 \times \frac{80000}{600} = 30 \times 133. \overline{3} = 4000 N$ Yeh step kyun? $ρ$ -ratio aur $A$ -ratio force upar push karte hain; $V$ -ratio neeche push karta hai — arithmetic winner decide karta hai.

Verify karo: $\frac{1000 \cdot 16 \cdot 5}{1.2 \cdot 2500 \cdot 0.20} = \frac{80000}{600} = 133.33$ , times $30 = 4000 N$ . ✓ Units newtons mein cancel hote hain kyunki $C_{D}$ dimensionless hai. Lesson: kabhi force equality assume mat karo jab fluids alag hon — woh shortcut sirf Ex 3 ka tha.

Case 5 — Free-surface flow (Reynolds yahan galat hai)

Figure ship ka bow wave dikhati hai: gravity woh force hai jo uthaye hue paani ko wapis neeche kheenchti hai, isi liye $g$ enter karta hai.

Number pick karo. Ship paani ko upar wave mein push karta hai; restoring force gravity hai ( $g$ ke zariye). Inertia/gravity ratio Froude Number hai — yeh free surfaces ke liye controlling number hai. Yeh step kyun? Free surface ke saath, gravity waves drag dominate karti hain; Re match karna us physics ko ignore karta jo actually matter karti hai.
Froude match karo. $\frac{V _{m}}{g L _{m}} = \frac{V _{p}}{g L _{p}} \Rightarrow V_{m} = V_{p} \frac{L _{m}}{L _{p}} = V_{p} L_{r}$ Yeh step kyun? $g$ tank aur sea mein same hai, isliye cancel ho jaata hai, $V \propto L$ bachta hai.
Solve karo. $V_{m} = 10 \times \frac{1.5}{150} = 10 \times 0.01 = 10 \times 0.1 = 1.0 m/s$ Yeh step kyun? $L_{r} = 1/100 = 1/10$ , isliye chhota model slower jaata hai, Reynolds case se ulta.

Verify karo: $Fr_{p} = \frac{10}{9.81 \cdot 150} = \frac{10}{38.35} = 0.2608$ ; $Fr_{m} = \frac{1.0}{9.81 \cdot 1.5} = \frac{1.0}{3.836} = 0.2608$ . ✓ Equal. Note karo ki size effect ka sign flip hua: Reynolds → faster, Froude → slower.

Case 6 — The conflict: kya dono match kar sakte hain?

Dono required speeds likho. $V_{m}^{Re} = V_{p} \frac{1}{L _{r}} = 10 \times 100 = 1000 m/s, V_{m}^{Fr} = V_{p} L_{r} = 1.0 m/s$ Yeh step kyun? Reynolds chahta hai $V \propto 1/ L$ ; Froude chahta hai $V \propto L$ — $L$ mein opposite trends.
Contradiction dekhne ke liye ratio lo. $\frac{V _{m}^{Re}}{V _{m}^{Fr}} = \frac{1/ L _{r}}{L _{r}} = L_{r}^{- 3/2} = 10 0^{3/2} = 1000$ Yeh step kyun? Ek single model speed $1 m/s$ aur $1000 m/s$ ek saath nahi ho sakti — factor $1000$ apart, isliye ek fluid se simultaneous matching impossible hai.
Resolve karo. Dominant force match karo: ship ke wave drag ke liye, woh Froude hai ( $V_{m} = 1.0 m/s$ ). Viscous (Re) part alag se Boundary Layer friction-line formulas se correct kiya jaata hai — yeh ship resistance ka Froude ka classic split hai.

Verify karo: $L_{r}^{- 3/2}$ ke saath $L_{r} = 1/100$ gives $(100)^{1.5} = 1000$ . ✓ Mismatch fast badhta hai: sirf $1/10$ model bhi $1 0^{1.5} \approx 31.6 \times$ deta hai.

Case 7 — Degenerate limit $L_{r} \to 1$

Model speed. Ex 1 se, $V_{m} = V_{p} / L_{r} \to V_{p} /1 = V_{p}$ . Yeh step kyun? Agar model prototype hai, toh use prototype ki speed par run karna chahiye — ek zaroori consistency check.
Force ratio. Ex 3 se, $F_{D, p} / F_{D, m} = L_{r}^{2} \cdot L_{r}^{- 2} = 1$ sabhi $L_{r}$ ke liye, isliye $1$ hi rehta hai. Yeh step kyun? Same body, same fluid, same speed → identical drag, trivially.
Reynolds ratio. $\frac{Re _{m}}{Re _{p}} = \frac{V _{m} L _{m}}{V _{p} L _{p}} = \frac{( V _{p} / L _{r} ) ( L _{r} L _{p} )}{V _{p} L _{p}} = 1$ . Yeh step kyun? Re-matching is ratio ko $1$ force karta hai by construction — limit sirf confirm karta hai ki kuch toot nahi raha.

Verify karo: har ratio $L_{r} \to 1$ par $1$ ke barabar hai. ✓ Ek theory jo is degenerate limit mein fail ho woh galat hogi; hamari pass karti hai.

Case 8 — Exam twist: pehle regime decide karo

Physics diagnose karo. "Crest ke upar behta hai, hawa mein khula" = free surface, gravity-driven → Froude, Reynolds nahi. Yeh step kyun? Exam choice hide karta hai; "free surface / open" phrase tell hai. Submerged/pipe → Re; free surface → Fr.
Velocity ke liye Froude match karo. $V_{m} = V_{p} L_{r} = 5 \times 1/25 = 5 \times \frac{1}{5} = 1.0 m/s$ Yeh step kyun? Ship ki tarah same $L_{r}$ rule — gravity cancel hoti hai.
Discharge $Q = V \cdot A$ derive aur scale karo. Yaad raho $Q$ volume per second hai, aur speed times cross-section ke barabar hai. Ratio model-to-prototype lo aur dono scaling laws substitute karo: $\frac{Q _{m}}{Q _{p}} = \frac{V _{m} A _{m}}{V _{p} A _{p}} = = L_{r} \frac{V _{m}}{V _{p}} \cdot = L_{r}^{2} \frac{A _{m}}{A _{p}} = L_{r} \cdot L_{r}^{2} = L_{r}^{5/2}$ $= (\frac{1}{25})^{5/2} = \frac{1}{2 5 ^{2} 25} = \frac{1}{625 \cdot 5} = \frac{1}{3125}$ Yeh step kyun? Velocity $L_{r}$ se scale hoti hai (Froude) aur area $L_{r}^{2}$ se (geometry, kyunki $A \propto L^{2}$ ); unka product $L_{r}^{5/2}$ standard Froude discharge law hai.

Verify karo: $Fr_{p} = \frac{5}{9.81 \cdot 50} = \frac{5}{22.15} = 0.2257$ ; $Fr_{m} = \frac{1.0}{9.81 \cdot 2} = \frac{1.0}{4.429} = 0.2258$ . ✓ Aur $2 5^{5/2} = 3125$ , isliye $Q_{m} / Q_{p} = 1/3125$ . ✓

Recall

Recall Quick self-test

Gehra torpedo run karne ke liye kaunsa number? ::: Reynolds (submerged, koi free surface nahi). Waves wale breakwater ke liye kaunsa number? ::: Froude (free surface, gravity). Same fluid, Re matched, $L_{r} = 1/8$ — model speed factor? ::: $\times 8$ ( $V \propto 1/ L_{r}$ se). Same fluid, Re matched — drag force ratio prototype/model? ::: exactly $1$ . Froude match, $L_{r} = 1/25$ — velocity factor? ::: $\times 1/5$ ( $L_{r}$ se). Froude match, $L_{r} = 1/25$ — discharge ratio $Q_{m} / Q_{p}$ ? ::: $1/3125$ ( $L_{r}^{5/2}$ se). Re aur Fr ek saath kyun match nahi kar sakte (same fluid)? ::: woh $V \propto 1/ L$ vs $V \propto L$ demand karte hain — contradictory (do $Π$ -groups, ek knob). Fast water-tunnel model kab unfaithful ho jaata hai? ::: jab woh cavitate kare (ya gas mein, jab $Ma ≳ 0.3$ aur compressibility appear ho). Tiny scales par simple Froude scaling kab toot jaati hai? ::: jab model waves mm tak shrink ho jaayein aur surface tension (low Weber) unhe distort kare.

Yeh bhi dekho: Navier-Stokes Equations (kyun matched Re identical dimensionless equation deta hai), aur parent topic note.