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

2.2.3 · D2 · Physics › Fluid Mechanics › Viscosity — dynamic μ, kinematic ν = μ - ρ; Newtonian vs non

Hum ek sawaal ka jawaab denge: jab tum liquid ki ek layer ko kheenchte ho, to tumhe kitna sideways force per area feel hota hai, aur kyun?

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Step 1 — Woh picture jisse hum shuru kar sakte hain: do plates

KYA. Kuch fluid lo aur use do flat plates ke beech trap karo. Bottom plate bolted down hai. Top plate ko hum pakad ke sideways slide kar sakte hain. Plates ke beech ki vertical distance ek height hai jise hum $h$ bulayenge (bas "the gap").

KYUN. Hum chahte hain sabse simple situation jahan fluid layers ek doosre ke past slide karti hain. Koi pipe nahi, koi curve nahi, koi gravity trick nahi — bas ek flat sandwich. Mushkil cheezein baad mein aayengi; pehle humein ek clean picture chahiye jahan sirf ek cheez vary kare.

PICTURE. Neeche, bottom plate still hai. Hum top plate ko right ki taraf speed $U$ se push karte hain (ek speed, metres per second mein measure ki gayi). Beech ka neela material fluid hai.

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Step 2 — Fluid actually kya karta hai: speeds ka ek ramp

KYA. Fluid ek block ki tarah nahi chalta. Bottom plate se chipki layer bilkul nahi chalti. Top plate se chipki layer poori speed $U$ pe chalti hai. Beech ki har layer kisi beech wali speed pe chalti hai. Har layer se ek arrow draw karo: arrows neeche zero se upar $U$ tak badhte hain — ek seedha ramp.

KYUN bottom layer stuck hai. Real fluid us solid surface se chipak jaata hai jise wo touch karta hai. Ise no-slip condition kehte hain: wall ke bilkul paas wala fluid wall ki speed se match karta hai. Yeh ek experimental fact hai — dust kabhi poori tarah fan blade se nahi udhti isi wajah se. Iske bina humein koi drag hi nahi hoti.

KYUN ramp seedha hai. Is simple gap mein, koi cheez ek layer ko doosre se zyada speed up karne ko nahi kehti, isliye speed evenly badhti hai — bilkul perfectly even staircase ki steps ki tarah. (Hum abhi is seedhi line pe trust karenge; Step 4 exactly yeh batayega ki "seedhapan" humein kya deta hai.)

PICTURE. Chhote arrows ek triangle banate hain: zero-length neeche floor pe, full-length $U$ upar ceiling pe.

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Step 3 — Kyun layers ek doosre ko kheenchti hain: molecules momentum exchange karte hain

KYA. Itna zoom in karo jab tak tum molecules "dekh" sako. Fast (upper) layer ke molecules jiggling kar rahe hain aur kabhi kabhi neeche slow layer mein wander karte hain, apni fast sideways motion le jaate hain. Slow layer ke molecules upar wander karte hain apni sluggishness le jaate hain. Jab ek fast molecule slow waalon ke beech land karta hai, to unhe aage dhakelta hai; ek slow molecule fast waalon mein aata hai to unhe peeche kheenchta hai.

KYUN isse force banti hai. Sideways motion (momentum) ramp se neeche fast layers se slow layers tak hand ho raha hai. Momentum hand over karna hi force hai — yeh Newton's second law ulta padha gaya. Isliye har layer ko apne neighbours se ek sideways tug feel hota hai jo unke beech ki speed difference ko mitaane ki koshish karta hai. Yeh tug, ek area pe spread ho ke, woh hai jise hum shear stress bulayenge.

KEY consequence. Force neighbouring layers ke beech speed ke difference pe depend karta hai — na ki is baat pe ki wo kitni fast mil ke chalti hain. Agar layers ka poora deck ek speed pe chale, to koi molecule "out of step" arrive nahi karega, aur zero viscous force hogi. Speed akele kuch nahi karta; speed difference hi sab kuch hai.

PICTURE. Do neighbouring layers; curvy arrows molecules ko unke beech cross karte dikhate hain, red arrows resulting tug dikhate hain jo unki speeds equalise karne ki koshish karta hai.

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Step 4 — "Speeds kitni different hain" measure karna: shear rate

KYA. Humein ek number chahiye "speed chadhte waqt kitni change hoti hai." Do nearby heights lo. Speed ne thodi si change ki, ise $du$ kaho, height mein thodi si rise $dy$ ke upar. Ratio $\frac{du}{dy}$ speed ramp ka slope hai — har metre chadhne pe kitne m/s gain hote hain.

KYUN ek ratio (aur yeh slope / derivative kyun hai). Step 3 ne bataya tha ki sirf difference matter karta hai. Lekin "difference kis distance ke upar?" — 1 m/s ka gap ek baal-jaisi thin layer mein kahin zyada violent hai bajaaye same gap ek metre ke upar. Isliye humein speed change ko us height se divide karna padega jis par wo hua. Woh "small change divided by small distance" exactly wahi hai jo ek derivative hoti hai: ek curve ki local steepness. Hum $\frac{du}{dy}$ ke liye reach karte hain kyunki yeh woh ek tool hai jo local steepness capture karta hai, aur Step 3 ne prove kiya ki local steepness hi force generate karta hai.

HAMARE SEEDHE RAMP KE LIYE slope har jagah same hai, isliye hum tiny quantities ki jagah whole quantities se compute kar sakte hain: $\underset{\text{shear rate}}{\underbrace{\dot{\gamma }}}=\frac{du}{dy}=\underset{\text{(top speed)}÷\text{(gap)}}{\underbrace{\frac{U}{h}}}$

Term by term: $\dot{\gamma }$ (padho "gamma-dot") slope ka humara naya naam hai; $U$ floor se ceiling tak poori speed change hai; $h$ woh height hai jis par yeh hua; $U/h$ woh change per metre hai. Units: $\frac{\text{m/s}}{\text{m}}={\text{s}}^{-1}$ — ek per-second, kyunki yeh shearing ki rate hai, koi distance nahi.

PICTURE. Ramp ke neeche ek right triangle draw kiya gaya: horizontal leg $=U$, vertical leg $=h$, aur slope $U/h$ ramp ki edge ki "steepness" hai.

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Step 5 — Newton ka experimental guess, knobs ki tarah draw kiya gaya

KYA. Ab sideways force per area, $\tau$ ("tau"), measure karo jo top plate ko glide karte rakhne ke liye chahiye. Newton ne do knobs ke baare mein socha jinhein tum turn kar sakte ho:

• Top plate ko faster push karo ($U$ up) → tumhe zyada force feel hoti hai. Isliye $\tau$ $U$ ke saath badhti hai.
• Gap ko wider karo ($h$ up) → speed difference zyada layers pe smear ho jaata hai, har ek barely out of step → tumhe kam force feel hoti hai. Isliye $\tau$ $h$ ke saath ghatati hai.

KYUN inhe $U/h$ ki tarah combine karo. Dono knobs ek hi buried quantity ki taraf point karte hain: force speed-per-gap, yaani $\tau \propto U/h=\dot{\gamma }$ track karta hai. Yeh coincidence nahi hai — Step 3 ne already bataya tha ki force local steepness se aati hai, aur Step 4 ne us steepness ko $\dot{\gamma }$ naam diya. Do knobs sirf ek dial ko turn karne ke do tarike hain jo matter karta hai.

PICTURE. Do side-by-side boards: left, faster plate ⇒ thicker force arrow; right, wider gap ⇒ thinner force arrow. Beech mein dial $\dot{\gamma }=U/h$ read karta hai.

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Step 6 — Constant ka naam: dynamic viscosity μ, aur law

KYA. Ek proportionality abhi equation nahi hai. $\propto$ ko $=$ mein convert karne ke liye hum ek number daaltе hain jo sirf kis fluid pe depend karta hai. Us number ko $\mu$ ("mu") bulao, dynamic viscosity.

KYUN ise is tarah define karo. Honey aur water identical plates mein identical $U$ aur $h$ ke saath baithte hain — same $\dot{\gamma }$ — phir bhi honey ko kaafi zyada force chahiye. Sirf difference fluid khud hai, isliye "yeh fluid kitna sticky hai" sab kuch ek single multiplier $\mu$ mein pack ho jaata hai. Yeh, by definition, wahi number hai jo force sahi nikalta hai: $\boxed{\tau =\mu \dot{\gamma }=\mu \frac{du}{dy}}$

Box ko left se right padhna:

• $\tau$ — woh sideways stress jo tumhe apply karna padega (Pa),
• $\mu$ — fluid ka stickiness dial (Pa·s); honey ke liye bada, air ke liye tiny,
• $\frac{du}{dy}$ — Step 4 se shear rate (s⁻¹); layers kitni out-of-step hain.

"Kitna sticky" ko "kitna out-of-step" se multiply karo aur milega "force per area kitna hai." Notice karo shape sabse simple honest wali hai: agar $\frac{du}{dy}=0$ (uniform flow) to $\tau =0$ automatically — exactly woh fact jo Step 3 ne demand kiya tha — aur gradient double karne se stress double hoti hai.

Units, memorise nahi ki, earn ki: $[\mu ]=\frac{[\tau ]}{[\dot{\gamma }]}=\frac{\text{Pa}}{{\text{s}}^{-1}}=\text{Pa⋅s}=\frac{\text{kg}}{\text{m⋅s}}$

PICTURE. Ek seedhi line ki tarah finished law: horizontal axis $\dot{\gamma }$, vertical axis $\tau$, origin se ek line jiska slope $\mu$ hai. Steeper line = stickier fluid.

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Step 7 — Edge cases jinhe line survive karna chahiye

Humne ek seedhi line draw ki. Ek law tabhi trustworthy hota hai jab extremes pe sahi behave kare. Har corner chalte hain:

Case A — zero shear ($\dot{\gamma }=0$). Sab layers alike chalti hain (ya kuch nahi chalta). $\tau =\mu \cdot 0=0$. Picture: line exactly origin se guzarti hai. Uniform motion mein koi viscous force nahi lagti — Step 3 ka misconception, ab formula se guaranteed.

Case B — drag reverse karo ($\dot{\gamma }<0$). Top plate ko right ki jagah left slide karo. Gradient sign flip karta hai, isliye $\tau$ sign flip karta hai: fluid ab doosri taraf kheenchta hai. Picture: line third quadrant mein seedhi continue karti hai. Viscous stress hamesha relative sliding ko oppose karti hai — sign apna khayal rakh leta hai.

Case C — dono plates bolt karo ($U=0$). Tab $\dot{\gamma }=0/h=0$, stress $0$. Ek still fluid mein koi viscous stress nahi feel hoti chahe kitni bhi thick ho. Stickiness dormant rehti hai jab tak kuch shear na kare.

Case D — gap zero ki taraf shrink ho ($h\to 0$). $U$ fixed ke saath, $\dot{\gamma }=U/h\to \infty$, isliye $\tau \to \infty$. Picture: plates ko saath squeeze karo aur sliding rakhne ki force blow up ho jaati hai — exactly yahi reason hai ki do nearly-touching surfaces (ek bearing) ke beech ek thin oil film enormous load carry kar sakti hai. Yeh Boundary Layer aur Poiseuille Flow ka seed hai, jahan thin high-gradient regions dominate karte hain.

Recall Edge cases pe khud ko check karo

Agar main dono plates ko same speed se rightward move karun, to viscous stress kya hogi? ::: Zero — same speed ka matlab $\dot{\gamma }=0$, isliye $\tau =\mu \cdot 0=0$. Plates ko closer push karna (smaller $h$) force kyun increase karta hai? ::: Kyunki $\dot{\gamma }=U/h$ badhta hai jab $h$ shrink hota hai; same speed change ek thinner gap mein pack ho jaati hai, isliye layers zyada out-of-step hoti hain.

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Step 8 — Jab line seedhi rehne se mana kar deti hai (non-Newtonian)

KYA. Water, air, glycerin ke liye, $\mu$ ek fixed number hai — graph ek single seedhi line hai. Yeh Newtonian fluids hain. Lekin ketchup, cornstarch-water, aur toothpaste graph ko bend karte hain: unka slope harder shear karne pe change hota hai.

KYUN slope change ho sakta hai. Step 6 mein humne assume kiya tha ki fluid ki "stickiness" constant hai. Lekin kuch fluids mein internal structure hoti hai — tangled chains, packed grains — jo shear ke neeche rearrange hoti hai. Isliye effective slope ab fixed nahi rehta; hum ise apparent viscosity ${\mu }_{app}=\tau /\dot{\gamma }$ bolte hain, aur yeh $\dot{\gamma }$ pe depend karta hai.

Jaanne ke liye chaar boards:

• Newtonian — origin se seedha, slope $\mu$ fixed. (water, air)
• Shear-thinning — curve neeche bend karti hai; slope faster shear karne pe girta hai. (ketchup, blood, paint — chains align aur slip karti hain) → Reynolds Number contexts dekho jahan yeh matter karta hai.
• Shear-thickening — curve upar bend karti hai; slope shear ke saath badhta hai. (oobleck — grains jam ho jaate hain)
• Bingham plastic — yield stress ${\tau }_0$ tak flat refusal, phir ek line: $\tau ={\tau }_0+{\mu }_p\dot{\gamma }$. (toothpaste)

Bending waalon ke liye ek compact model: $\tau =K{\dot{\gamma }}^n$, $n=1$ Newtonian ke liye, $n<1$ thinning ke liye, $n>1$ thickening ke liye.

PICTURE. Ek $\tau$–$\dot{\gamma }$ board pe chaaon curves taaki shapes ek nazar mein contrast karein.

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Ek-picture summary

Upar sab kuch, ek board pe: plates (Step 1), no-slip ke saath speed ramp (Step 2), $\dot{\gamma }=U/h$ deta triangle (Step 4), aur $\tau$–$\dot{\gamma }$ graph jiska slope $\mu$ hai (Step 6), Newtonian line plus bending non-Newtonian curves ke saath (Step 8).

Recall Feynman retelling — walkthrough plain words mein

Kuch liquid ko do flat plates ke beech trap karo aur top wala slide karo. Bottom wala liquid bottom plate se chipka rehta hai aur nahi chalta; top wala liquid top plate ke saath saath ride karta hai. Beech mein, liquid speeds ke ek smooth ramp mein chalta hai — neeche slow, upar fast. Kisi ko bhi drag kyun feel hoti hai? Kyunki molecules layers ke beech hopping karte rehte hain, apni speed saath le jaate hain, aur ek fast molecule slow layer mein ghus ke use speed up karta hai jabki khud slow ho jaata hai. Woh motion ka trading hi ek force hai. Zaroori baat, yeh tabhi hota hai jab layers alag speeds pe chalti hain — agar poora slab saath drifts kare, koi out of step nahi aata aur zero drag hoti hai. Isliye hum measure karte hain "speeds per metre of height kitni different hain" — yeh ramp ka slope hai, $U$ gap $h$ se divide karke. Newton ne paya ki force per area bas woh slope times fluid ke liye ek number hai: sticky honey ke liye bada number, slippery water ke liye chhota. Woh number viscosity $\mu$ hai, aur law hai force-per-area = stickiness × steepness. Aakhir mein, zyaatar everyday liquids kitna bhi shear karo same stickiness rakhti hain — ek acchi straight-line graph. Lekin kuch (ketchup, oobleck, toothpaste) push karne pe andar se rearrange ho jaati hain, isliye unki line bend ho jaati hai: yeh thinner ya thicker ho jaati hain ya stubbornly flow karne se mana karti hain jab tak tum kaafi hard na dhakel do.

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