2.2.3 · D3Viscosity — dynamic μ, kinematic ν = μ - ρ; Newtonian vs non
Intuition Yeh page kya haiParent note ne do viscosities build ki thi. Yahan hum cases drill karte hain — gradient ke har sign ka, zero-gradient trap ka, degenerate thin-gap limit ka, Newtonian aur non-Newtonian dono regimes ka, aur ek exam twist ka. Hum pehle forecast karte hain, phir solve karte hain, aur phir har number check karte hain.
Shuru karne se pehle, teen symbols jo parent ne kamaaye aur hum reuse karte hain. Agar koi unfamiliar lage, pehle parent dobara padh lo.
Recall Teen symbols jinpe hum takte hainτ μ τ = μ d u / d y γ ˙ d u / d y
γ γ ˙ woh rate hai jis par layers shear (slide) karti hain ek doosre se .
Is topic ke har problem in mein se kisi ek cell mein aati hai. Page ka baaki hissa har cell ko ek worked example se bharta hai.
Cell
Kya change hota hai
Example
A. Positive gradient speed height ke saath badhti hai, d u / d y > 0
Ex 1
B. Negative gradient speed height ke saath girti hai, d u / d y < 0 τ
Ex 2
C. Zero gradient (degenerate) uniform flow, d u / d y = 0
Ex 3
D. Thin-gap limit h → 0 Ex 4
E. μ ν kaun sa fluid momentum faster diffuse karta hai
Ex 5
F. Non-Newtonian, n < 1 shear-thinning apparent viscosity
Ex 6
G. Non-Newtonian, n > 1 shear-thickening apparent viscosity
Ex 7
H. Bingham yield (degenerate flow) τ > τ 0 Ex 8
I. Real-world word problem falling sphere, momentum-diffusion time
Ex 9
J. Exam twist do fluids stacked, stress matching
Ex 10
Figure mein charon gradient signs side by side dikhaye gaye hain — amber slope arrows dekho: upar-slope (A), flat (C), neeche-slope (B), aur near-vertical thin-gap case (D). Us slope ka sign aur steepness hi Newton's law ki poori kahaani hai .
Worked example Ex 1 — Standard Couette dragWater-jaisa oil, μ = 0.8 h = 2 mm = 0.002 y = h U = 0.3 A = 0.5 m 2
Forecast: andaza lagao — kya force 6 N ke kareebi hogi ya 600 N ke? Apna andaza likh lo.
Shear rate. γ ˙ = d y d u = h − 0 U − 0 = 0.002 0.3 = 150 s − 1 Yeh step kyun? Profile 0 U = U / h hi d u / d y Stress. τ = μ γ ˙ = 0.8 × 150 = 120 Yeh step kyun? Newton's law: stress = viscosity × gradient.Force. F = τ A = 120 × 0.5 = 60 Yeh step kyun? Stress force per area hoti hai; area se multiply karke force wapas milti hai.
Verify: units — ( Pa⋅s ) ( s − 1 ) = Pa ( Pa ) ( m 2 ) = N U τ wapas push karta hai, use drag karta hai, jaise expected tha.
Worked example Ex 2 — Speed height ke saath decrease ho rahi haiSame oil (μ = 0.8 girti hai: bottom par y = 0 u = 0.4 y = h = 0.002 u = 0.1 τ
Forecast: kya is baar τ
Shear rate. γ ˙ = h u top − u bottom = 0.002 0.1 − 0.4 = 0.002 − 0.3 = − 150 s − 1 Yeh step kyun? Gradient hai (top par value − bottom par value)/gap. Kyunki speed upar jaane par girti hai, numerator negative hai.Stress. τ = μ γ ˙ = 0.8 × ( − 150 ) = − 120 Yeh step kyun? Newton's law sign ko seedha carry karta hai.Interpret karo. Minus sign kehta hai upper layer par stress negative flow direction mein point karta hai — slow top fluid forward drag hota hai jabki woh fast bottom fluid ko wapas drag karta hai. Magnitude, 120 Pa, Ex 1 se bilkul same hai.
Yeh step kyun? Sign momentum transfer ki direction encode karta hai; magnitude encode karta hai kitna .
Verify: kaun si plate fast hai yeh flip karo to sign flip hota hai lekin size nahi — ek accha symmetry check. ∣ τ ∣ = 120
Worked example Ex 3 — Uniform streamSame oil ki puri nadi ek rigid slab ki tarah move karti hai: har layer u = 10
Forecast: 10 m/s fast hai — zaroor bada stress hoga? Padhne se pehle andaza lagao.
Shear rate. γ ˙ = h u top − u bottom = h 10 − 10 = 0 s − 1 Yeh step kyun? Viscosity layers ke beech ke farq ki parwah karta hai, unki common speed ki nahi. Koi farq nahi ⇒ zero gradient, chahe h Stress. τ = μ × 0 = 0 Yeh step kyun? Newton's law: koi gradient nahi, koi stress nahi — τ = μ d u / d y τ = μu
Verify: dimension-free sanity — uniformly moving fluid ko uss frame se dekha ja sakta hai jo uss ke saath move kar raha hai, jahan sab kuch rest mein hai aur obviously stress-free hai. Physics us frame choice par depend nahi kar sakti, to stress har frame mein 0
Worked example Ex 4 — Gap ko zero tak squeeze karnaEx 1 lo (μ = 0.8 U = 0.3 h = 0.2 = 2 × 1 0 − 4 τ h → 0
Forecast: das guna patla gap — stress kitne factor se badhega?
Naya shear rate. γ ˙ = 2 × 1 0 − 4 0.3 = 1500 s − 1 Yeh step kyun? Same speed spread tenth distance par ⇒ dus guna steep slope.Naya stress. τ = 0.8 × 1500 = 1200 10 × Yeh step kyun? Fixed U τ ∝ 1/ h h τ Limit. Jaise h → 0 γ ˙ = U / h → ∞ τ → ∞ Yeh step kyun? Isliye real machines ek finite oil film rakhte hain — infinitely thin film infinite force demand karti (aur fluid ko shred kar deti). Yahi Boundary Layer idea ka seed hai: wall ke paas gradient sabse steep hota hai.
Verify: ratio check τ new / τ old = 1200/120 = 10 = h old / h new
Worked example Ex 5 — Motion faster kaun spread karta hai, air ya water?Air: μ a = 1.8 × 1 0 − 5 ρ a = 1.2 kg/m 3 μ w = 1.0 × 1 0 − 3 ρ w = 1000 kg/m 3 ν
Forecast: water obviously air se "thicker" hai — to water ka ν
Air ka ν ν a = ρ a μ a = 1.2 1.8 × 1 0 − 5 = 1.5 × 1 0 − 5 m 2 / s Yeh step kyun? Definition ν = μ / ρ Water ka ν ν w = 1000 1.0 × 1 0 − 3 = 1.0 × 1 0 − 6 m 2 / s Rank. ν a / ν w = 1.0 × 1 0 − 6 1.5 × 1 0 − 5 = 15 15× faster diffuse karta hai per unit mass water se.
Yeh step kyun? Air ka μ ρ ν a > ν w Momentum Diffusion numbers mein hai.
Verify: units kg/m 3 Pa⋅s = kg/m 3 kg/(m⋅s) = m 2 / s
Worked example Ex 6 — Paint on a brushPower-law paint τ = K γ ˙ n K = 2 n = 0.5 γ ˙ = 4 s − 1 γ ˙ = 100 s − 1 apparent viscosity μ app = τ / γ ˙
Forecast: kya paint tez brush karne par aur thicki hogi ya patli?
Apparent-viscosity formula. μ app = γ ˙ τ = γ ˙ K γ ˙ n = K γ ˙ n − 1 = 2 γ ˙ − 0.5 Yeh step kyun? "Apparent viscosity" woh slope-to-origin hai jo tum us shear rate par maano Newtonian hai read karo. Hum ise use karte hain kyunki μ Slow shear. μ app = 2 × 4 − 0.5 = 2/ 4 = 2/2 = 1.0 Fast shear. μ app = 2 × 10 0 − 0.5 = 2/ 100 = 2/10 = 0.2 Yeh step kyun? n − 1 = − 0.5 < 0 γ ˙ μ app ghat jaati hai — yahi shear-thinning ki definition hai.
Verify: ratio 1.0/0.2 = 5 ( 100/4 ) 0.5 = ( 25 ) 0.5 = 5
Worked example Ex 7 — Oobleck jamsCornstarch suspension, τ = K γ ˙ n K = 0.5 n = 2 γ ˙ = 2 s − 1 γ ˙ = 50 s − 1 μ app
Forecast: ise tezi se punch karo — slow stir se softer hoga ya harder?
Apparent viscosity. μ app = K γ ˙ n − 1 = 0.5 γ ˙ 1 = 0.5 γ ˙ Yeh step kyun? Same μ app = K γ ˙ n − 1 n − 1 = 1 > 0 Slow. μ app = 0.5 × 2 = 1.0 Fast. μ app = 0.5 × 50 = 25 Yeh step kyun? Positive exponent ka matlab hai μ app badhti hai — fast loading mein particles jam ho jaate hain. Isliye tum oobleck par daud sakte ho lekin khade raho to doob jaate ho.
Verify: ratio 25/1.0 = 25 = 50/2 γ ˙ n
Worked example Ex 8 — Toothpaste hilegi nahiEk Bingham plastic τ = τ 0 + μ p γ ˙ τ 0 = 60 μ p = 0.4 τ = 40 τ = 100
Forecast: 40 Pa par, kuch move hoga bhi?
Yield check karo. Flow tabhi hota hai jab τ > τ 0 40 < 60 γ ˙ = 0 Yeh step kyun? Yield se neeche structure ek soft solid ki tarah hold karta hai; koi gradient nahi kyunki koi motion nahi.Yield se upar (b). Rearrange karo: γ ˙ = μ p τ − τ 0 = 0.4 100 − 60 = 0.4 40 = 100 s − 1 Yeh step kyun? Sirf τ 0 zyada stress hi flow drive karta hai; us excess ko μ p
Verify: wapas plug karo — τ = 60 + 0.4 × 100 = 60 + 40 = 100
Worked example Ex 9 — Ek mug of water mein ek jolt reach karne mein kitna time lagegaEk spoon ek mug ki ek wall ko jolt karta hai. Momentum diffusion use karke estimate karo ki woh disturbance still water mein L = 3 = 0.03 ν w = 1.0 × 1 0 − 6 m 2 / s
Forecast: seconds, minutes, ya hours?
Diffusion length rule. Parent se: time t L ∼ ν t Yeh step kyun? ν ν t Momentum Diffusion .t L 2 = ν t ⇒ t = ν L 2 = 1.0 × 1 0 − 6 ( 0.03 ) 2 = 1 0 − 6 9 × 1 0 − 4 = 900 Yeh step kyun? Scaling law rearrange karne par woh timescale milta hai jis mein viscosity akele motion carry karti hai.
Verify: units m 2 / s m 2 = s 900 = 15 Boundary Layer mein chhupi rehti hain aur Reynolds Number inertia ko viscosity se kyun compare karta hai.
Worked example Ex 10 — Do stacked fluids, ek plateDo Newtonian layers ek fixed floor aur ek moving ceiling ke beech gap fill karti hain, ceiling U = 0.5 μ 1 = 0.2 h 1 = 1 mm μ 2 = 0.6 h 2 = 1 mm u i τ
Forecast: kya interface speed midpoint value 0.25
Stress continuity. Steady shear mein body force ke bina, τ μ 1 h 1 u i − 0 = μ 2 h 2 U − u i Yeh step kyun? Agar dono stresses alag hote, to thin interface layer par net force hota aur woh accelerate karta — steady state mein impossible hai.Numbers daalo (h 1 = h 2 = 0.001 0.001 0.2 u i = 0.001 0.6 ( 0.5 − u i ) ⇒ 0.2 u i = 0.6 ( 0.5 − u i ) Solve karo. 0.2 u i = 0.3 − 0.6 u i ⇒ 0.8 u i = 0.3 ⇒ u i = 0.375 Yeh step kyun? Algebra. Interface midpoint se upar hai kyunki kam resist karne wali bottom layer (chhota μ 1 Common stress. τ = μ 1 h 1 u i = 0.2 × 0.001 0.375 = 0.2 × 375 = 75 Yeh step kyun? Ab jab u i τ
Verify: top layer same stress deni chahiye: μ 2 h 2 U − u i = 0.6 × 0.001 0.5 − 0.375 = 0.6 × 125 = 75 u i = 0.375 > 0.25
Recall Kaun sa cell sabse mushkil tha?Zero-gradient trap (Cell C) ::: stress 0 gradients read karta hai, speeds nahi.
μ-vs-ν ranking (Cell E) ::: air, ν
Mnemonic Sign & case checklist"Slope upar → +τ, slope neeche → −τ, flat → 0; thin gap → huge τ; spread ki speed rank karne se pehle ρ se divide karo."
