Exercises — Viscosity — dynamic μ, kinematic ν = μ - ρ; Newtonian vs non-Newtonian

(a) $\nu =\frac{\mu }{\rho }=\frac{0.90}{900}=1.0\times 10^{-3}{\text{m}}^2/\text{s}$. Yeh operation kyun: $\nu$ ko dynamic viscosity divided by density ke roop mein define kiya gaya hai — woh division ek force-measure ko momentum-diffusion-measure (units m²/s) mein convert karta hai. (b) Reasoning galat hai. Newtonian hone ka matlab hai ki $\mu$ shear rate ke saath constant rehta hai, na ki $\mu$ bada hai. Honey bahut zyada viscous hai aur Newtonian bhi; ketchup kam viscous hai lekin non-Newtonian hai. $\mu$ ki magnitude Newtonian-ness ke baare mein kuch nahi kehti.

Rearrange karo: $\mu =\frac{\tau }{\dot{\gamma }}$.

• $\tau$ ek stress hai = force/area = $\frac{\text{N}}{{\text{m}}^2}=\frac{{\text{kg⋅m/s}}^2}{{\text{m}}^2}=\frac{\text{kg}}{{\text{m⋅s}}^2}$.
• $\dot{\gamma }$ ek gradient hai = (m/s)/m = ${\text{s}}^{-1}$. To $\mu =\frac{\text{kg}/({\text{m⋅s}}^2)}{1/\text{s}}=\frac{\text{kg}}{{\text{m⋅s}}^2}\cdot \text{s}=\frac{\text{kg}}{\text{m⋅s}}$. ✓ (Yeh Pa·s ke barabar hai.)

$\dot{\gamma }=du/dy=0$ kyunki layers ke beech speed mein koi difference nahi hai. Phir $\tau =\mu \times 0=0$ Pa. Viscous stress ke liye gradient chahiye, speed nahi — poora slab kisi bhi speed par zero internal friction ke saath move kar sakta hai.

Figure 1 (upar) dekho: fluid speed fixed floor par $0$ se moving lid par $U$ tak straight line mein badhti hai, to gradient (lal arrows ke tips ka tilt) har jagah constant hai. (a) $\dot{\gamma }=\frac{U}{h}=\frac{0.6}{0.0015}=400{\text{s}}^{-1}$. $U/h$ kyun: ek straight-line profile ka slope = (top speed − bottom speed)/(gap) = $U/h$ hota hai. (b) $\tau =\mu \dot{\gamma }=0.65\times 400=260\text{Pa}$. Kyun: Newton's law of viscosity. (c) $F=\tau A=260\times 0.4=104\mathbf{N}$. Kyun: stress force per area hai, to force = stress × area.

Pehle forward chain banao, ek physical link at a time, phir usse invert karo:

1. Force area par spread hoke stress banata hai: $\tau =F/A$. (Badi plate → same force se kam stress.)
2. Woh stress velocity gradient ke through Newton's law se carry hota hai: $\tau =\mu \dot{\gamma }$.
3. Straight Couette profile ke liye gradient hai $\dot{\gamma }=U/h$. 3 ko 2 mein substitute karo aur 1 ke barabar set karo: $\frac{F}{A}=\mu \frac{U}{h}$. Sirf $\mu$ unknown hai, to usse solve karo: $\mu =\frac{Fh}{UA}=\frac{30\times 0.002}{0.4\times 0.25}=\frac{0.06}{0.1}=0.6\text{Pa⋅s}.$ Algebra kyun legal hai: har physical link ek genuine equality hai, to unhe rearrange karna sirf us ek quantity ko isolate karna hai jo hume nahi pata — koi approximation chhupa nahi hai. Sanity check units: $\frac{\text{N}\cdot \text{m}}{(\text{m/s})\cdot {\text{m}}^2}=\frac{\text{N⋅s}}{{\text{m}}^2}=\text{Pa⋅s}$ ✓.

${\nu }_{\text{gly}}=\frac{1.5}{1260}=1.19\times 10^{-3}{\text{m}}^2/\text{s}$. ${\nu }_{\text{water}}=\frac{1.0\times 10^{-3}}{1000}=1.0\times 10^{-6}{\text{m}}^2/\text{s}$. Ratio $\approx 1190$. Glycerin momentum zyada jaldi spread karta hai (larger $\nu$) — chahe woh kaafi zyada gooey lagta ho — kyunki $\nu$ momentum ke diffusion ko measure karta hai, aur glycerin ka huge $\mu$ uske thode bade $\rho$ par bhaari pad jaata hai.

Figure 2 is paint ko magenta curve ke roop mein dikhata hai, straight Newtonian line ke neeche bend karta hua — to jaise hum zyada shear karte hain origin se slope girta jaata hai. ${\mu }_{app}=\frac{K{\dot{\gamma }}^n}{\dot{\gamma }}=K{\dot{\gamma }}^{n-1}=3{\dot{\gamma }}^{-0.6}.$

• $\dot{\gamma }=1$ par: ${\mu }_{app}=3\times 1^{-0.6}=3.0\text{Pa⋅s}$.
• $\dot{\gamma }=100$ par: ${\mu }_{app}=3\times 100^{-0.6}=3\times 10^{-1.2}=3\times 0.0631=0.189\text{Pa⋅s}$. (Units check: ${\mu }_{app}=K{\dot{\gamma }}^{n-1}$ ke units ${\text{Pa⋅s}}^{0.4}\cdot ({\text{s}}^{-1}{)}^{-0.6}={\text{Pa⋅s}}^{0.4}\cdot {\text{s}}^{0.6}=\text{Pa⋅s}$ ✓.) Interpretation: wall par (near-zero shear) paint thick hota hai — $3\text{Pa⋅s}$ — to drip nahi karta. Brush par (high shear) yeh $\sim 0.19\text{Pa⋅s}$ tak thin ho jaata hai aur easily spread hota hai. Yahi shear-thinning ($n<1$) ka fayda hai.

Figure 2 par yeh orange curve hai, straight line ke upar bend karta hua — paint ka mirror image. ${\mu }_{app}=K{\dot{\gamma }}^{n-1}=0.8{\dot{\gamma }}^{0.5}.$

• $\dot{\gamma }=1$ par: ${\mu }_{app}=0.8\times 1^{0.5}=0.8\text{Pa⋅s}$.
• $\dot{\gamma }=100$ par: ${\mu }_{app}=0.8\times 100^{0.5}=0.8\times 10=8.0\text{Pa⋅s}$. Interpretation: jitna zyada shear karo ($n>1$), utna motha ho jaata hai. Dheere kado aur tum doob jaate ho (low shear, thin); tezi se stamping karo aur yeh tumhare pair ke neeche solid jam ho jaata hai (high shear, ~10× thicker) — to tum uske upar daaud sakte ho. Yeh L3.1 ke paint se power-law family ka poora ulta chor hai.

(a) Yield stress se neeche koi flow nahi hota: kyunki $150<200$, $\dot{\gamma }=0$ — paste tube mein wahi reh jaata hai. Isliye toothpaste brush par apni shape hold karta hai. (b) Yield ke upar, invert karo: $\dot{\gamma }=\frac{\tau -{\tau }_0}{{\mu }_p}=\frac{350-200}{5}=30{\text{s}}^{-1}$. Pehle ${\tau }_0$ kyun subtract karo: sirf yield value se zyada stress hi flow drive karta hai; pehle $200$ Pa "jammed structure unlock karne mein kharch" ho jaata hai kisi bhi sliding ke shuru hone se pehle.

(a) ${\nu }_{\text{air}}=\frac{1.8\times 10^{-5}}{1.2}=1.5\times 10^{-5}{\text{m}}^2/\text{s}$. ${\nu }_{\text{water}}=1.0\times 10^{-6}{\text{m}}^2/\text{s}$. Air 15× bada hai. Air ka tiny $\mu$ uske aur bhi tiny $\rho$ se haara jaata hai. (b) $\ell =\sqrt{\nu t}$.

• Air: $\sqrt{1.5\times 10^{-5}\times 10}=\sqrt{1.5\times 10^{-4}}\approx 0.0122\text{m}\approx 1.2\text{cm}$.
• Water: $\sqrt{1.0\times 10^{-6}\times 10}=\sqrt{10^{-5}}\approx 0.00316\text{m}\approx 3.2\text{mm}$. Momentum air mein per unit time zyada "door pahunchta" hai — Momentum Diffusion $\nu$ se govern hoti hai, is baat se nahi ki fluid kitna gooey lagta hai.

Pehle $\nu =\mu /\rho =10^{-3}/1000=10^{-6}{\text{m}}^2/\text{s}$. $Re=\frac{UD}{\nu }=\frac{2\times 0.05}{10^{-6}}=\frac{0.1}{10^{-6}}=1.0\times 10^5.$ Kyunki $10^5\gg 2300$, flow turbulent hai. Dekho Reynolds Number: $Re$ inertial push aur viscous smoothing ka ratio hai; yahan inertia $100,000$-to-1 dominate karta hai, to viscosity flow ko orderly nahi rakh sakti.

$R^4=(2\times 10^{-3}{)}^4=1.6\times 10^{-11}{\text{m}}^4$. $Q=\frac{\pi \times 1.6\times 10^{-11}\times 8000}{8\times 0.1\times 1}=\frac{\pi \times 1.28\times 10^{-7}}{0.8}=\pi \times 1.6\times 10^{-7}\approx 5.03\times 10^{-7}{\text{m}}^3/\text{s}.$ Agar $R$ double ho: $Q\propto R^4$, to $Q$ $2^4=16\times$ badhega. Radius par fourth-power punchline hai — thodi si pipe-widening flow ko massively boost karti hai, isliye arteries aur pipes radius ke itne sensitive hote hain.

Doobte sphere par teeno vertical forces set up karo (volume $V=\frac{4}{3}\pi a^3$):

• Weight (neeche): $W={\rho }_{s}Vg={\rho }_s\frac{4}{3}\pi a^{3}g$.
• Buoyancy (upar): $B={\rho }_{f}Vg={\rho }_f\frac{4}{3}\pi a^{3}g$ (displaced glycerin ka weight).
• Stokes drag (upar, sinking ka virodh karta hai): $F_d=6\pi \mu av$.

Terminal velocity par sphere accelerate karna band kar deta hai, to ups, down ko balance karte hain: $W=B+F_d\Rightarrow \frac{4}{3}\pi a^{3}g({\rho }_s-{\rho }_f)=6\pi \mu a{v}_t.$ $v_t$ ke liye solve karo — dono sides ko $6\pi \mu a$ se divide karo: $v_t=\frac{\frac{4}{3}\pi a^{3}g({\rho }_s-{\rho }_f)}{6\pi \mu a}=\frac{4\pi a^{3}g({\rho }_s-{\rho }_f)}{3\cdot 6\pi \mu a}=\frac{2a^2({\rho }_s-{\rho }_f)g}{9\mu }.$ ($\pi$ cancel hota hai, $a$ ka ek power cancel hoke $a^2$ bachta hai, aur $\frac{4}{18}=\frac{2}{9}$.) Evaluate karo: $a^2=10^{-6}{\text{m}}^2$, ${\rho }_s-{\rho }_f=6540{\text{kg/m}}^3$: $v_t=\frac{2\times 10^{-6}\times 6540\times 9.8}{9\times 1.5}=\frac{0.128184}{13.5}\approx 9.5\times 10^{-3}\text{m/s}(\approx 9.5\text{mm/s}).$ Slow, kyunki glycerin ka bada $\mu$ denominator mein hai. $a^2$ dikhata hai ki chote spheres dramatically slower girte hain.

Figure 3 (upar) kinked profile dikhata hai: do straight segments alag slopes ke saath kyunki do films ka alag $\mu$ hai lekin same $\tau$ transmit karna zaroori hai. Har film Newton's law apne gradient ke saath obey karta hai. Film 1 height $h_1$ par speed $0\to u_i$ span karta hai; film 2 height $h_2$ par $u_i\to U$ span karta hai: $\tau ={\mu }_1\frac{u_i}{h_1}={\mu }_2\frac{U-u_i}{h_2}.$ Har film par velocity drop hai $\Delta u=\tau h/\mu$, aur dono drops ka total $U$ hona chahiye: $U=\tau \left(\frac{h_1}{{\mu }_1}+\frac{h_2}{{\mu }_2}\right)\Rightarrow \tau =\frac{U}{\frac{h_1}{{\mu }_1}+\frac{h_2}{{\mu }_2}}.$ Numbers: $\frac{h_1}{{\mu }_1}=\frac{0.001}{0.2}=5\times 10^{-3}$, $\frac{h_2}{{\mu }_2}=\frac{0.001}{0.5}=2\times 10^{-3}$, sum $=7\times 10^{-3}$. $\tau =\frac{0.3}{7\times 10^{-3}}=\frac{0.3}{0.007}\approx 42.857\text{Pa}.$ Interface speed, film 1 se: $u_i=\frac{\tau h_1}{{\mu }_1}=\frac{42.857\times 0.001}{0.2}=\frac{0.042857}{0.2}=0.2143\text{m/s}$. Film 2 se check karo: uska drop hai $U-u_i=0.3-0.2143=0.0857\text{m/s}$, aur $\frac{\tau h_2}{{\mu }_2}=\frac{42.857\times 0.001}{0.5}=0.0857$ ✓. Films series resistors ki tarah add hote hain ($h/\mu$ = resistance); zyada viscous film 2 ka gentler slope hai (same stress carry karne ke liye use kam gradient chahiye).

Real stack ki tarah same $\tau ={\mu }_{eff}\frac{U}{h}$, to ${\mu }_{eff}=\frac{\tau h}{U}=\frac{42.857\times 0.002}{0.3}=\frac{0.085714}{0.3}=0.2857\text{Pa⋅s}.$ Naive average hai $(0.2+0.5)/2=0.35$ — barabar nahi. Series films ek thickness-weighted harmonic mean ki tarah combine hote hain: $\frac{h}{{\mu }_{eff}}=\frac{h_1}{{\mu }_1}+\frac{h_2}{{\mu }_2}\Rightarrow {\mu }_{eff}=\frac{h_1+h_2}{\frac{h_1}{{\mu }_1}+\frac{h_2}{{\mu }_2}}=\frac{0.002}{7\times 10^{-3}}=\mathrm{0.2857.}$ Low-viscosity film dominate karta hai (yeh "easy" path hai), blend ko arithmetic mean se neeche kheenchta hai — exactly jaise series resistors mein current sabse choti slope wale element se set hoti hai jo sabse zyada deformation carry karta hai.

$Re=\frac{UD}{\nu }$ se $\nu$ ke liye solve karo: $\nu =\frac{UD}{Re}=\frac{1\times 0.02}{500}=4.0\times 10^{-5}{\text{m}}^2/\text{s}.$ Phir $\mu =\rho \nu =1200\times 4.0\times 10^{-5}=4.8\times 10^{-2}\text{Pa⋅s}$ ($=48$ cP, ek light oil). Reasoning: low $Re$ ka matlab hai viscosity ko inertia par dominate karna hai, to tum $\nu$ badhate ho (moota ya halka fluid). Yeh L4.1 ka ulta hai — wahan humne fluid se $Re$ compute kiya; yahan hum target $Re$ se fluid reverse-engineer karte hain.

L1 answers: $1.0\times 10^{-3}{\text{m}}^2/\text{s}$; unit derivation; $\tau =0$. L2 answers: $\dot{\gamma }=400,\tau =260,F=104\text{N}$; $\mu =0.6$; ${\nu }_{gly}=1.19\times 10^{-3},{\nu }_w=10^{-6}$. L3 answers: ${\mu }_{app}=3.0$ phir $0.189$ (thinning); $0.8$ phir $8.0$ (thickening); no flow phir $30{\text{s}}^{-1}$; ${\nu }_{air}=1.5\times 10^{-5}$, $1.2\text{cm}$ spread karta hai vs water $3.2\text{mm}$. L4 answers: $Re=10^5$ turbulent; $Q\approx 5.03\times 10^{-7}$; $v_t\approx 9.5\text{mm/s}$. L5 answers: $\tau \approx 42.857,u_i\approx 0.2143$; ${\mu }_{eff}=0.2857$; $\nu =4\times 10^{-5},\mu =0.048$.