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

• Reynolds number $\mathrm{Re}=\frac{\rho VL}{\mu }=\frac{VL}{\nu }$ — (inertia force)/(viscous force) ka ratio. Dimensionless hai.
• Froude number $\mathrm{Fr}=\frac{V}{\sqrt{gL}}$ — woh ratio jo free surfaces aur waves ko govern karta hai.
• Reynolds matching ${\mathrm{Re}}_m={\mathrm{Re}}_p\Rightarrow \frac{V_m{L}_m}{{\nu }_m}=\frac{V_p{L}_p}{{\nu }_p}$.
• Drag coefficient $C_D=\frac{F_D}{\frac{1}{2}\rho V^{2}A}$; jab Re match ho, toh $C_{D,m}=C_{D,p}$. Yahan subscript $m$ = model, $p$ = prototype (asli full-size cheez), $r$ = ratio $L_r=L_m/L_p$.

Reynolds. Ek fully submerged body mein no free surface hoti hai, isliye gravity waves ka koi role nahi; competition inertia aur viscosity ke beech hai, jo exactly $\mathrm{Re}=\rho VL/\mu$ measure karta hai. Froude tab hi aata hai jab liquid surface waves mein upar uth sakti ho.

Weakest → strongest: geometric → kinematic → dynamic.

• Kinematic (streamline patterns match karna) requires geometric (same shape).
• Dynamic (force ratios match karna) implies kinematic (matched forces matched motion produce karte hain). Toh dynamic similarity achieve karo aur baaki dono free mein aa jaate hain. Dynamic hi king hai.

$\frac{[\rho ][V][L]}{[\mu ]}=\frac{({\text{kg/m}}^3)(\text{m/s})(\text{m})}{\text{kg}/(\text{m}\cdot \text{s})}=\frac{\text{kg}\cdot {\text{m}}^{-1}\cdot {\text{s}}^{-1}}{\text{kg}\cdot {\text{m}}^{-1}\cdot {\text{s}}^{-1}}=1$ Haan — har unit cancel ho jaata hai, toh $\mathrm{Re}$ ek pure number hai. Isliye model aur prototype mein same value ka matlab same flow regime hai, size ki parwah kiye bina.

Same fluid ⇒ $\rho ,\mu$ cancel ho jaate hain ${\mathrm{Re}}_m={\mathrm{Re}}_p$ mein, baar $V_m{L}_m=V_p{L}_p$ bachta hai. $V_m=V_p\frac{L_p}{L_m}=25\times \frac{4.5}{0.75}=25\times 6=150\text{m/s}$ Chhote model ko $6\times$ tez blow karna padega taaki $\rho VL/\mu$ unchanged rahe.

Alag fluids ⇒ Re-matching mein teeno factors rakho: $\frac{V_m{D}_m}{{\nu }_m}=\frac{V_p{D}_p}{{\nu }_p}\Rightarrow V_m=V_p\frac{D_p}{D_m}\frac{{\nu }_m}{{\nu }_p}$ $V_m=1.5\times \frac{0.6}{0.15}\times \frac{1.5\times 10^{-5}}{1.0\times 10^{-6}}=1.5\times 4\times 15=90\text{m/s}$

Re matched ⇒ $C_{D,m}=C_{D,p}$, toh $\frac{F_{D,m}}{\rho V_m^2{A}_m}=\frac{F_{D,p}}{\rho V_p^2{A}_p}$. Same fluid ke saath, $A=L^2$, aur $V_m=V_p/L_r$ jahan $L_r=L_m/L_p=1/6$: $F_{D,p}=F_{D,m}\frac{V_p^2{A}_p}{V_m^2{A}_m}=F_{D,m}\cdot \frac{V_p^2}{(V_p/L_r{)}^2}\cdot \frac{L_p^2}{L_m^2}=F_{D,m}\cdot L_r^2\cdot \frac{1}{L_r^2}=F_{D,m}$ $F_{D,p}=18\text{N}$ Zyada model speed chhoti area ko exactly cancel kar deti hai — same fluid + Re matched ke liye, forces equal hoti hain.

Froude matching: $\frac{V_m}{\sqrt{g{L}_m}}=\frac{V_p}{\sqrt{g{L}_p}}\Rightarrow V_m=V_p\sqrt{\frac{L_m}{L_p}}=V_p\sqrt{L_r}$ jahan $L_r=1/25$. $V_m=10\sqrt{1/25}=10\times \frac{1}{5}=2\text{m/s}$ Reynolds ratio (same $\nu$): $\frac{{\mathrm{Re}}_m}{{\mathrm{Re}}_p}=\frac{V_m{L}_m}{V_p{L}_p}=\sqrt{L_r}\cdot L_r=L_r^{3/2}=(1/25{)}^{3/2}=\frac{1}{125}$. Toh ${\mathrm{Re}}_m$ 125× bahut chhota hai — Re match nahi hota. Froude aur Reynolds ek hi fluid mein dono nahi ho sakte, isliye engineers Froude match karte hain (dominant, wave-making force) aur viscous drag separately correct karte hain.

Re-matching (same $\nu$): $V_m{L}_m=V_p{L}_p\Rightarrow V_r=1/L_r$. Fr-matching (same $g$): $V_m/\sqrt{L_m}=V_p/\sqrt{L_p}\Rightarrow V_r=\sqrt{L_r}$. Dono simultaneously true ⇒ $\frac{1}{L_r}=\sqrt{L_r}\Rightarrow L_r^{3/2}=1\Rightarrow L_r=1.$ $L_r=1$ matlab model prototype ke same size ka hai — koi scaling hi nahi. Isliye ek governing number choose karna padta hai. (Dono $L_r\ne 1$ ke saath satisfy karne ke liye ek alag fluid chahiye hoga specially chosen $\nu$ ke saath.)

$L$, $V$, $\rho V^2$ se non-dimensionalise karne par Navier–Stokes ek aisa equation ban jaata hai jiska sirf ek free coefficient $1/\mathrm{Re}$ hai. Agar do flows same shape share karte hain (geometry ⇒ dimensionless coordinates mein same boundary conditions) aur same Re, toh woh identical dimensionless equation ko identical boundary conditions ke saath solve karte hain ⇒ identical dimensionless velocity aur pressure fields. Kyunki $C_D$ un dimensionless fields ko (identical) dimensionless surface par integrate karke compute hota hai, $C_{D,m}=C_{D,p}$ automatically aa jaata hai. Yahi Drag Coefficient scaling ka poora engine hai.

(a) Froude: $V_r=\sqrt{L_r}=\sqrt{1/40}$. (b) $Q_r=V_r{A}_r=\sqrt{L_r}\cdot L_r^2=L_r^{5/2}=(1/40{)}^{5/2}$. $Q_r=\frac{1}{40^{2.5}}=\frac{1}{40^2\sqrt{40}}=\frac{1}{1600\times 6.3246}=\frac{1}{10119.3}=9.882\times 10^{-5}$ (c) $Q_m=Q_p{Q}_r=8000\times 9.882\times 10^{-5}=0.7906{\text{m}}^3/\text{s}$. Toh ek asli 8000 m³/s flood ko lab mein manageable ~0.79 m³/s se reproduce kiya jaata hai — yahi Froude scaling ki taakat hai.

$\frac{{\mathrm{Re}}_m}{{\mathrm{Re}}_p}=\frac{{\rho }_m{V}_m{L}_m/{\mu }_m}{{\rho }_p{V}_p{L}_p/{\mu }_p}=\frac{{\rho }_m}{{\rho }_p}\cdot \frac{V_m}{V_p}\cdot \frac{L_m}{L_p}\cdot \frac{{\mu }_p}{{\mu }_m}=4\times 1\times \frac{1}{8}\times 1=\frac{4}{8}=0.5$ Match nahi hua: ${\mathrm{Re}}_m$ prototype ka aadha hai. 4× density boost help karta hai lekin 1/8 size cut jeet jaata hai. $V_m=V_p$ par fully match karne ke liye ${\rho }_m=8{\rho }_p$ chahiye hoga (density utne hi factor se upar jitne se size neeche gayi). Yahi wajah hai ki real pressurised tunnels density bahut zyada badhaate hain.

Re match ke liye speed: $\frac{{\rho }_m{V}_m{L}_m}{{\rho }_p{V}_p{L}_p}=1\Rightarrow 4\cdot \frac{V_m}{V_p}\cdot \frac{1}{8}=1\Rightarrow \frac{V_m}{V_p}=2$. Toh $V_m=2V_p$. Force ratio (koi bhi $C_D$-type force, $F\propto \rho V^{2}A$, $A\propto L^2$): $F_r=\frac{{\rho }_m{V}_m^2{A}_m}{{\rho }_p{V}_p^2{A}_p}=\frac{{\rho }_m}{{\rho }_p}{\left(\frac{V_m}{V_p}\right)}^2{\left(\frac{L_m}{L_p}\right)}^2=4\times 2^2\times {\left(\frac{1}{8}\right)}^2=4\times 4\times \frac{1}{64}=\frac{16}{64}=0.25$ Model force prototype force ka $1/4$ hai. (Dhyan do ki hum yahan equal forces assume nahi kar sakte kyunki fluid density alag hai — Drag Coefficient equal hai, lekin raw force nahi.)

(a) Variables $n=5$ ($F_D,\rho ,V,D,\mu$); fundamental dimensions $k=3$ (mass, length, time). Buckingham se, $\Pi$-groups ki sankhya $=n-k=5-3=2$. (b) Ek natural choice: ${\Pi }_1=\frac{F_D}{\rho V^2{D}^2}(\text{a drag coefficient}),{\Pi }_2=\frac{\rho VD}{\mu }=\mathrm{Re}.$ Theorem kehta hai ${\Pi }_1=\varphi ({\Pi }_2)$, yaani $C_D=\varphi (\mathrm{Re})$. Toh Re match karna $C_D$ fix karta hai — exactly wahi Reynolds-similarity claim, ab dimensional analysis se proven na ki assert kiya hua.

Error: ek ship water surface par chalti hai aur waves banati hai ⇒ gravity waves dominate ⇒ Froude, not Reynolds. Reynolds fully submerged / closed-pipe flow ke liye hai. Sahi scaling (Froude): $V_m=V_p\sqrt{L_r}=8\times \sqrt{1/10}=8\times 0.31623=2.53\text{m/s}$. Toh sahi channel speed lagbhag 2.53 m/s hai, 80 m/s nahi. Uska Reynolds answer model ko $\frac{80}{2.53}\approx 31.6=L_r^{-3/2}$ factor se over-speed karta hai — physically water channel mein impossible aur galat physics bhi.

Jab $C_D$ plateau par aa jaata hai (high-Re regime), toh woh fluid ki parwah kiye bina same number hota hai, isliye hum ise directly real conditions par apply karte hain: $F_{D,p}=\frac{1}{2}{C}_D{\rho }_p{V}_p^2{A}_p=\frac{1}{2}\times 1.2\times 1000\times 3^2\times 6$ $=0.6\times 1000\times 9\times 6=32400\text{N}=32.4\text{kN}.$ Model (air) ne sirf dimensionless $C_D$ measure kiya; prototype force prototype ke apne $\rho ,V,A$ use karta hai. Yahi roz ka workflow hai: model par $C_D$ measure karo, reality par apply karo.

Fully submerged submarine ke liye kaun sa number? ::: Reynolds (no free surface). Surface par ship ke liye kaun sa number? ::: Froude (gravity waves). Same fluid, Re matched — scale $L_r$ ke liye $V_m$ aur $V_p$ mein kya relation hai? ::: $V_m=V_p/L_r$. Same fluid, Re matched — model aur prototype forces compare kaise hoti hain? ::: equal ($F_m=F_p$). Kya Re aur Fr ek fluid mein saath match kar sakte ho? ::: sirf agar $L_r=1$ ho (no scaling).