Similarity — geometric, kinematic, dynamic; Reynolds similarity

1. The three levels of similarity

2. Deriving dynamic similarity from first principles

HOW to build the ratios — estimate each force by dimensions.

Take a characteristic length $L$ , velocity $V$ , density $\rho$ , viscosity $\mu$ .

Inertia force = mass × acceleration. Mass $\sim \rho L^3$ , acceleration $\sim \dfrac{V}{t}\sim \dfrac{V}{L/V}=\dfrac{V^2}{L}$ . $F_i \sim \rho L^3 \cdot \frac{V^2}{L} = \rho V^2 L^2$ Why this step? We replaced time by $L/V$ (the time to cross the body), the only timescale available.
Viscous force = shear stress × area. Stress $\tau = \mu\,\dfrac{dV}{dy}\sim \mu\dfrac{V}{L}$ , area $\sim L^2$ . $F_v \sim \mu \frac{V}{L}\cdot L^2 = \mu V L$

3. Reynolds similarity

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

Worked Example A — wind-tunnel speed

A car ( $L_p=4$ m) drives at $V_p=30$ m/s. A $1/5$ model ( $L_m=0.8$ m) is tested in air (same $\rho,\mu$ ). Find $V_m$ .

Why this step? No free surface, viscous drag matters → Reynolds similarity. $V_m = V_p\frac{L_p}{L_m}=30\times\frac{4}{0.8}=150\ \text{m/s}$
Why? Smaller body → must blow faster to keep $\rho VL/\mu$ the same. (This high speed is exactly why engineers often use water tunnels or pressurised tunnels instead.)

Worked Example B — scaling the force

With Re matched, the drag coefficient $C_D=\dfrac{F_D}{\frac12\rho V^2 A}$ is equal for both. Suppose model drag $F_{D,m}=12$ N. Find prototype drag. $\frac{F_{D,p}}{\rho_p V_p^2 A_p}=\frac{F_{D,m}}{\rho_m V_m^2 A_m}$ With same fluid and $V_m=V_p/L_r$ , $A=L^2$ : $F_{D,p}=F_{D,m}\frac{V_p^2 A_p}{V_m^2 A_m}=F_{D,m}\cdot L_r^2\cdot \frac{1}{L_r^2}=F_{D,m}=12\ \text{N}$

Why this surprising result? For same fluid with Re matched, the higher model speed exactly compensates the smaller area — forces are equal. (If fluids differ, plug actual $\rho,V,A$ — don't assume equality.)

Worked Example C — different fluid

A pipe ( $D_p=0.5$ m) carries water ( $\nu_p=1\times10^{-6}$ m²/s) at $V_p=2$ m/s. A model uses air ( $\nu_m=1.5\times10^{-5}$ m²/s), $D_m=0.1$ m. Find $V_m$ . $\frac{V_mD_m}{\nu_m}=\frac{V_pD_p}{\nu_p}\Rightarrow V_m=V_p\frac{D_p}{D_m}\frac{\nu_m}{\nu_p}=2\cdot\frac{0.5}{0.1}\cdot\frac{1.5\times10^{-5}}{1\times10^{-6}}=150\ \text{m/s}$

Why include $\nu$ ratio? General Re-matching keeps all three factors; only when fluids are identical does $\nu$ cancel.

4. Common mistakes (steel-manned)

5. Active recall

Recall Self-test (try before peeking)

Why does kinematic similarity require geometric similarity?
Derive Re from the inertia/viscous force ratio.
For same fluid, how does $V_m$ relate to $V_p$ under Re-matching?
When do you use Froude instead of Reynolds?
Why is matching Re enough to make $C_D$ equal?

Recall Feynman: explain to a 12-year-old

Imagine you build a tiny toy boat to learn about a giant ship. The toy only behaves like the real ship if it's the same shape (geometric), the water swirls in the same pattern (kinematic), and the push-and-drag forces balance the same way (dynamic). Scientists found a single "magic number," the Reynolds number, that mixes speed, size, and how thick the liquid is. If the toy's magic number equals the ship's magic number, then the toy is a perfect tiny copy of how the real one moves — so whatever you measure on the toy, you can scale up to the real ship. Cool trick: the tiny toy often has to move faster than the real ship to keep its magic number the same!

Flashcards

Geometric similarity means

same shape — all lengths scale by one constant factor

L_r

, angles preserved.

Kinematic similarity means

same motion pattern — velocity/acceleration fields differ only by a constant scale; requires geometric similarity.

Dynamic similarity means

corresponding force ratios are equal in model and prototype; implies kinematic similarity.

Reynolds number physical meaning

ratio of inertia force to viscous force,

\mathrm{Re}=\rho VL/\mu = VL/\nu

Inertia force scaling estimate

F_i \sim \rho V^2 L^2

Viscous force scaling estimate

F_v \sim \mu V L

Reynolds similarity criterion

match

\mathrm{Re}_m=\mathrm{Re}_p

for viscous-dominated flows (no free surface).

Same-fluid Re-matching gives model speed

V_m = V_p (L_p/L_m) = V_p/L_r

(smaller model → faster).

When to use Froude not Reynolds

free-surface/gravity-wave flows (ships, spillways);

\mathrm{Fr}=V/\sqrt{gL}

Why matching Re makes

C_D

equal

non-dimensional Navier–Stokes has only

1/\mathrm{Re}

as coefficient, so equal Re → identical dimensionless solution.

Why can't you match Re and Fr together (same fluid)

Re needs

V\propto1/L

, Fr needs

V\propto\sqrt L

— contradictory.

Connections

Reynolds Number — laminar/turbulent transition uses the same number.
Buckingham Pi Theorem — formal source of dimensionless groups.
Navier-Stokes Equations — non-dimensionalisation justifies Re-similarity.
Drag Coefficient — the quantity preserved under dynamic similarity.
Froude Number — complementary criterion for free-surface flows.
Boundary Layer — Re controls its thickness and separation.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, idea simple hai: hum poora aeroplane ya ship har baar test nahi kar sakte, isliye chhota scale model banate hain aur wind tunnel ya water channel mein test karte hain. Lekin chhote model ka data tabhi kaam aata hai jab dono flow "similar" ho. Similarity ke teen level hote hain — pehla geometric (same shape, har length ek hi factor se scale), doosra kinematic (streamlines ka same pattern, motion same), teesra dynamic (forces ka ratio same). Yeh ek seedhi (ladder) hai: shape ke bina motion match nahi hoga, aur motion ke bina forces match nahi honge. Sabse powerful dynamic similarity hai — yeh mil gaya to baaki sab automatic aa jaata hai.

Ab forces ka ratio kaise match karein? Har force ko dimension se estimate karo. Inertia force $\sim \rho V^2 L^2$ aur viscous force $\sim \mu V L$ . Inka ratio nikaalo to milta hai Reynolds number $\mathrm{Re}=\rho VL/\mu = VL/\nu$ — bilkul dimensionless. Reynolds similarity ka matlab: jahan viscosity aur inertia main role play karte hain (pipe flow, submerged body, no free surface), wahan bas $\mathrm{Re}_m = \mathrm{Re}_p$ match kar do, to drag coefficient $C_D$ dono mein equal ho jaayega. Iska deep reason: Navier–Stokes equation ko non-dimensional karoge to sirf $1/\mathrm{Re}$ bachta hai — same Re matlab same dimensionless solution.

Ek mazedaar baat: agar same fluid use karo to Re match karne ke liye $V_m = V_p/L_r$ — yaani chhota model ko tej chalana padta hai (1/10 model ko 10 guna speed). Isiliye engineers kabhi water tunnel ya pressurised tunnel use karte hain. Common galti: log sochte hain "same speed = similarity" — galat! Similarity ratios ki baat hai, raw values ki nahi. Aur dono Re aur Froude ek saath match karna usually impossible hota hai, isliye dominant force pehchaano: submerged → Reynolds, free-surface/ship → Froude. Bas yahi 20% concept 80% problems solve kar dega.

Similarity — geometric, kinematic, dynamic; Reynolds similarity

1. The three levels of similarity

2. Deriving dynamic similarity from first principles

3. Reynolds similarity

Worked Example A — wind-tunnel speed

Worked Example B — scaling the force

Worked Example C — different fluid

4. Common mistakes (steel-manned)

5. Active recall

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Fluid Mechanics

Connections