Exercises — Reynolds number Re = ρvL - μ — laminar vs turbulent criterion

WHAT: Hum fraction padhte hain. $\mu$ (viscosity) denominator mein hai; $\rho$, $v$, $L$ numerator mein hain. WHY it matters (physical): Viscosity neighbour-to-neighbour friction hai — ek fast layer literally ek slow layer ko saath kheenchti hai aur khud bhi kheenchi jaati hai. Yeh tugging velocity differences ko smooth out kar deta hai eddies mein curl hone se pehle. Toh badi $\mu$ physically kisi bhi budding swirl ko iron flat kar deti hai; mathematically yeh $\mu$ ke denominator mein dikhta hai, $Re$ ko laminar picture ki taraf drive karta hai. Viscosity peacemaker hai convention se nahi balki isliye kyunki shear friction physically us energy ko dissipate kar deta hai jo ek vortex ko chahiye hoti. Answer: $\mu$ denominator mein hai; yeh calming/smoothing (viscous) force ko represent karta hai.

Table se compare karo: $350<2000$, toh flow laminar hai (smooth, layered) — upar ki sabse left wali picture. Koi calculation nahi chahiye — bas number line par placement.

WHAT: SI units ko $\rho vL/\mu$ mein substitute karo. $\frac{[{\text{kg/m}}^3][\text{m/s}][\text{m}]}{[\text{Pa⋅s}]}=\frac{\text{kg}\cdot {\text{m}}^{-1}\cdot {\text{s}}^{-1}}{\text{Pa⋅s}}$ Ab $\text{Pa}={\text{N/m}}^2=\text{kg}{\text{m}}^{-1}{\text{s}}^{-2}$, toh $\text{Pa⋅s}=\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}$. $\frac{\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}}{\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}}=1\Rightarrow \text{dimensionless.}$ Answer: $Re$ ki koi units nahi hoti.

WHAT: SI numbers seedhe plug in karo. $Re=\frac{1000\times 0.30\times 0.010}{1.0\times 10^{-3}}=\frac{3.0}{1.0\times 10^{-3}}=3000$ WHY it comes out unitless — deeper point: yeh sirf lucky nahi hai ki units cancel hoti hain. $Re$ ek non-dimensional ratio hai inertial push $\rho v^2{L}^2$ aur viscous drag $\mu vL$ ka. Jab hum SI numbers daalaate hain toh units cancel honi hi chahiye kyunki hum literally ek force ko ek force se divide kar rahe hain — aur jo pure number bachta hai wahi physical answer hai: "$3000$ times zyada inertia friction se." SI mein kaam karna sirf humein us ratio ko directly padhne deta hai. Answer: $Re=3000$, jo $2000$–$4000$ mein aata hai ⇒ transitional (beech wali picture: intermittent wobbles aur pehle vortices).

$Re=\frac{1.2\times 15\times 2.0}{1.8\times 10^{-5}}=\frac{36}{1.8\times 10^{-5}}=2.0\times 10^6$ WHY: Lamba body + fast air + tiny viscosity sab milke $Re$ ko enormously zyada push karte hain. Itna bada number matlab inertia completely dominate karti hai ⇒ strongly turbulent boundary layer. Answer: $Re=2.0\times 10^6$.

WHAT: $\nu =\mu /\rho$. $\nu =\frac{1.0\times 10^{-3}}{1000}=1.0\times 10^{-6}{\text{m}}^2/\text{s}$ $Re=\frac{vL}{\nu }=\frac{0.30\times 0.010}{1.0\times 10^{-6}}=\frac{3.0\times 10^{-3}}{1.0\times 10^{-6}}=3000✓$ WHY this check matters (pedagogy): do alag tareekon se wahi $3000$ milna busywork nahi hai — yeh prove karta hai ki dono formulas same physics hain, alag kapde mein. Form $\rho vL/\mu$ density aur viscosity ko alag rakhta hai; form $vL/\nu$ ne unhe pehle se ek property $\nu$ mein fuse kar liya hai (momentum fluid mein kitni tezi se diffuse hota hai). Jab bhi koi problem tumhe $\rho$ aur $\mu$ ki jagah $\nu$ deti hai, ab tum jaante ho ki tum ise directly use kar sakte ho bina missing density dhoondhe. Aisi consistency checks se hi tum power of ten ki galti pakad sakte ho isse pehle ki wo poora answer kharab kar de. Answer: $Re=3000$, Exercise 2.1 se match karta hai.

WHAT: $Re=\rho vL/\mu$ ko $v$ ke liye rearrange karo. $v_c=\frac{Re\mu }{\rho L}=\frac{2000\times 1.0\times 10^{-3}}{1000\times 0.010}=\frac{2.0}{10}=0.20\text{m/s}$ WHY: $0.20\text{m/s}$ se neeche friction abhi bhi har disturbance ko smooth out kar sakta hai inertia ke badhne se pehle; exactly $v_c$ par dono rates balance ho jaate hain, aur uske upar inertia jeet jaati hai aur pehle vortices survive karte hain. Toh $v_c$ woh tipping speed hai jahan leftmost picture middle wali mein change hona shuru hoti hai. Answer: $v_c=0.20\text{m/s}$.

WHAT: Sirf $L$ change hota hai; $\rho ,v,\mu$ fixed hain. Kyunki $Re\propto L$, $\frac{R{e}_B}{R{e}_A}=\frac{L_B}{L_A}=3.$ WHY it looks like this (physical): pipe ko chauda karne se inertia ko ek bada cross-section milta hai momentum carry karne ke liye jabki shearing friction abhi bhi sirf wall region par act karti hai — toh inertia-to-friction balance upar ki taraf tip karta hai. Neeche ki figure $Re$ ko diameter ke against plot karti hai: yeh origin se seedhi line hai, toh do marked points A aur B exactly $1:3$ proportion ki heights par baithe hain, aur wahi speed jo narrow pipe mein laminar thi wide pipe mein turbulent ho sakti hai.

Figure: $Re$ versus pipe diameter fixed $\rho ,v,\mu$ par (water, $v=0.30\text{m/s}$). Line seedhi hai kyunki $Re\propto L$; pipe B (A ke diameter ka teen guna) A ki height ke teen gune par hai. Dashed lines laminar limit ($Re=2000$) aur turbulent onset ($Re=4000$) mark karti hain.

Answer: $Re$ $3\times$ bada ho jaata hai.

WHAT: $v,L$ equal hain, toh $Re\propto \rho /\mu$. $\frac{R{e}_{\text{water}}}{R{e}_{\text{honey}}}=\frac{{\rho }_w/{\mu }_w}{{\rho }_h/{\mu }_h}=\frac{1000/10^{-3}}{1400/10}=\frac{1.0\times 10^6}{140}\approx 7143$ WHY: Honey ke molecules ek doosre se bahut zyada chipke rehte hain, toh honey mein shuru hua koi bhi swirl almost instantly smothered ho jaata hai — uska momentum sideways diffuse ho jaata hai ek eddy mein roll hone se pehle. Woh physical stickiness hi denominator mein badi $\mu$ hai; yeh modest density difference ko swamp kar deti hai, water ka $Re$ hazaaron guna bada chhodke. Answer: $R{e}_{\text{water}}/R{e}_{\text{honey}}\approx 7.1\times 10^3$ (toh honey overwhelmingly zyada laminar hai).

WHAT (a): Pehle $Q$ calculate karo. $Q=\frac{\pi R^4\Delta P}{8\mu \ell }=\frac{\pi (5.0\times 10^{-3}{)}^4(8.0)}{8(1.0\times 10^{-3})(1.0)}$ Step — fourth power. $R^4=(5.0\times 10^{-3}{)}^4$. Mantissa aur exponent alag alag uthao: $5.0^4=625$ aur $(10^{-3}{)}^4=10^{-12}$, toh $R^4=625\times 10^{-12}=6.25\times 10^{-10}{\text{m}}^4$. (Pipe radius fourth power mein aata hai — isliye thodi si chaudi pipe dramatically zyada flow carry karti hai.) Step — top combine karo. Numerator $=\pi \times 6.25\times 10^{-10}\times 8.0=\pi \times 5.0\times 10^{-9}$. Step — divide karo. Denominator $=8\times 10^{-3}$, toh $Q=\frac{\pi \times 5.0\times 10^{-9}}{8.0\times 10^{-3}}=\pi \times 6.25\times 10^{-7}.$ Step — $\pi$ multiply karo. Pure number $6.25\times 10^{-7}$ ko $\pi \approx 3.1416$ se multiply karne par $6.25\times 3.1416\approx 19.63$ milta hai, matlab $Q\approx 19.63\times 10^{-7}=1.963\times 10^{-6}{\text{m}}^3/\text{s}$. (Hum $\pi$ ko symbolic rakhte hain last step tak taaki rounding kabhi pile up na ho.) WHAT (b): Average speed, phir $Re$ with $L=D=0.010\text{m}$: $v=\frac{Q}{\pi R^2}=\frac{1.963\times 10^{-6}}{\pi (5.0\times 10^{-3}{)}^2}=\frac{1.963\times 10^{-6}}{\pi \times 2.5\times 10^{-5}}=\frac{1.963\times 10^{-6}}{7.854\times 10^{-5}}\approx 0.025\text{m/s}$ $Re=\frac{\rho vD}{\mu }=\frac{1000\times 0.025\times 0.010}{1.0\times 10^{-3}}=\frac{0.25}{1.0\times 10^{-3}}=250$ WHY the check matters: Poiseuille's law sirf laminar flow mein valid hai (yeh assume karta hai left-hand picture ki woh tidy parallel layers). Yahan $Re=250<2000$ ⇒ laminar ⇒ law legitimately use ki gayi. Agar $Re$ $>2000$ aata, toh hamari saari $Q$ calculation ek broken assumption par tikti. Answer: $v\approx 0.025\text{m/s}$, $Re\approx 250$ — laminar, toh Poiseuille's law valid hai.

WHAT: Particle diameter $L=2r=2.0\times 10^{-4}\text{m}$ use karo. $R{e}_p=\frac{1000\times 0.02\times 2.0\times 10^{-4}}{1.0\times 10^{-3}}=\frac{4.0\times 10^{-3}}{1.0\times 10^{-3}}=4.0$ WHY: Stokes assume karta hai ki viscous drag dominate karta hai; yeh sirf $R{e}_p\lesssim 1$ ke liye hold karta hai. Yahan $R{e}_p=4>1$, toh grain ke around inertial effects negligible nahi hain — Stokes' law drag ko over- ya mis-predict karega. Answer: $R{e}_p=4.0$; Stokes' law strictly valid nahi hai (yeh already safe $R{e}_p\lesssim 1$ window ke bahar hai).

WHAT: $Re=1.0\times 10^5$ par $v$ solve karo. $v=\frac{Re\mu }{\rho D}=\frac{1.0\times 10^5\times 1.0\times 10^{-3}}{1000\times 0.010}=\frac{100}{10}=10\text{m/s}$ WHY the rule bends: $R{e}_c\approx 2000$ nature ka law nahi hai; yeh woh point hai jahan jo disturbances actually exist karte hain unhe amplify kiya jaata hai. Ek ultra-smooth, vibration-free pipe mein amplify karne ke liye almost kuch nahi hota, toh flow bahut zyada $Re$ tak laminar rehti hai. Geometry aur disturbance level threshold set karte hain, fluid nahi. Answer: $v=10\text{m/s}$; $2000$ figure ordinary rough pipes ke liye ek practical guideline hai, hard ceiling nahi.

WHAT: $Re=vL/\nu$ use karke $R{e}_1=R{e}_2$ set karo. $\frac{v_1{L}_1}{{\nu }_{\text{air}}}=\frac{v_2{L}_2}{{\nu }_{\text{water}}}\Rightarrow v_2=v_1\frac{L_1}{L_2}\frac{{\nu }_{\text{water}}}{{\nu }_{\text{air}}}$ Ek ek factor plug karo: $\frac{L_1}{L_2}=\frac{2.0}{0.20}=10$ aur $\frac{{\nu }_{\text{water}}}{{\nu }_{\text{air}}}=\frac{1.0\times 10^{-6}}{1.5\times 10^{-5}}=0.0667$. Phir $v_2=50\times 10\times 0.0667=50\times 0.667=33.3\text{m/s}.$ WHY this works: agar do flows ka $Re$ same ho, toh unka inertia-to-viscosity balance identical hai, toh flow ki shape (jahan separate hota hai, jahan turbulent hota hai) same hai — isliye wind-tunnel aur water-tunnel models trustworthy hote hain. Answer: $v_2\approx 33.3\text{m/s}$.

WHAT: Stokes prediction: $C_d^{\text{Stokes}}=\frac{24}{R{e}_p}=\frac{24}{4.0}=6.0$ Measured $6.9$ ke against percent error: $\text{error}=\frac{|6.0-6.9|}{6.9}\times 100\%=\frac{0.9}{6.9}\times 100\%\approx 13\%$ WHY the gap: $R{e}_p=4$ par hum already pure-Stokes regime se aage hain; inertial effects extra drag add karte hain jo $24/Re$ formula ignore karta hai, toh yeh $C_d$ ko lagbhag $13\%$ underestimate karta hai. Yahi warning hai jo Exercise 4.2 mein thi, ab quantify ki gayi. Answer: Stokes $C_d=6.0$ predict karta hai; error $\approx 13\%$ low.