2.2.17
Real fluids chipchipe hote hain. Jab fluid ek pipe se bahta hai, to wall ko touch karne wali layer bilkul nahi hilti (no-slip condition ), jabki beech wala fluid sabse tez bhagta hai.
Is internal friction (viscosity ) ke against fluid ko chalate rehne ke liye aapko push karna padta hai —
isi liye pipe ke along pressure drop hota hai. Poiseuille flow hamen exactly batata hai ki speed pipe ke across kaise vary karti hai aur kitna fluid per second guzarta hai.
Definition Viscosity & shear stressJab adjacent fluid layers alag-alag speeds par ek doosre se slide karti hain, to ek frictional shear
stress τ τ = η d r d v η dynamic viscosity hai (units Pa⋅s d v / d r velocity gradient hai (speed flow ke across kitni tezi se change hoti hai).
WHY this form? Jitni tezi se ek layer apne neighbour ke relative move karti hai, utna hi zyada molecules ek doosre ko drag karte hain → stress gradient ke proportional hai, speed ke nahi.
Hum steady, laminar, incompressible flow consider karte hain ek horizontal cylindrical pipe mein jiska radius R L Δ P = P 1 − P 2
Intuition Profile ko padhna
Centre par maximum (r = 0 v m a x = 4 η L Δ P R 2
Wall par zero (r = R
R 2 − r 2 parabola hai → fast core, slow edges, bullet ki naak ki tarah.
R 4 fourth power ke saath scale karta hai. Pipe radius double karo → flow
2 4 = 16
Worked example Example 1 — Pressure drop nikaloPaani (η = 1.0 × 1 0 − 3 Pa⋅s Q = 2.0 × 1 0 − 5 m 3 / s R = 5.0 mm L = 2.0 m Δ P
Step 1: Hagen–Poiseuille rearrange karo: Δ P = π R 4 8 η L Q Kyun? Hamen Δ P
Step 2: Values daalo: R 4 = ( 5 × 1 0 − 3 ) 4 = 6.25 × 1 0 − 10 m 4
\approx \frac{3.2\times10^{-7}}{1.963\times10^{-9}} \approx 163\,\text{Pa}$$
*Yeh kyun matter karta hai:* ek modest flow ke liye sirf ek chota push chahiye kyunki $R^4$ denominator mein hai.
Worked example Example 2 — Centre vs. point speedEk pipe ke liye jisme v m a x = 0.40 m/s R = 4 mm r = 2 mm v
Step 1: Profile ratio form: v = v m a x ( 1 − R 2 r 2 ) Kyun? v m a x ( R 2 − r 2 ) / R 2 R 2
Step 2: R 2 r 2 = ( 4 ) 2 ( 2 ) 2 = 16 4 = 0.25 v = 0.40 ( 1 − 0.25 ) = 0.30 m/s Yeh kyun matter karta hai: aadhe raaste par, speed abhi bhi max ka 75% hai — slowdown wall ke paas concentrated hai.
Worked example Example 3 — Radius double karne ka effectAgar fixed Δ P , L , η Q
Step 1: Q ∝ R 4 Step 2: Q new / Q old = 2 4 = 16 16× zyada ho jaati hai. Kyun: zyada area aur flatter gradient dono help karte hain.
Common mistake "Velocity pipe ke across uniform hoti hai."Yeh sahi kyun lagta hai: idealized non-viscous flow mein hum speed ko constant treat karte hain; intro problems
ek single v Fix: real (viscous) flow mein parabolic profile hota hai — wall ka fluid
stationary hota hai. Jab ek single number chahiye tab v ˉ = 2 1 v m a x
Q ∝ R 2 Yeh sahi kyun lagta hai: area π R 2 R Fix:
velocity khud bhi R 2 ∝ R 2 Q ∝ R 2 ⋅ R 2 = R 4
Common mistake "Zyada viscosity → faster flow (yeh thick/heavy hoti hai)."Yeh sahi kyun lagta hai: "thick" ko "strong push" se confuse karna. Fix: Q ∝ 1/ η slower flow.
Common mistake "Poiseuille's law kisi bhi flow ke liye kaam karta hai."Yeh sahi kyun lagta hai: yeh ek clean formula hai. Fix: yeh laminar flow
(low Reynolds number), steady, incompressible, Newtonian fluid, long straight pipe assume karta hai. Turbulent
flow bilkul alag (non-linear) relation follow karta hai.
Recall Quick self-test (answers chhupa lo!)
Integration constant kis boundary condition se fix hota hai? → No-slip: v ( R ) = 0
Velocity profile ki shape kya hai? → Parabolic, v ∝ R 2 − r 2
Q R Q ∝ R 4 Mean speed vs max speed? → v ˉ = 2 1 v m a x
Kis tarah ka flow required hai? → Laminar, steady, incompressible, Newtonian.
Recall Feynman: ek 12-saal ke bachche ko explain karoSocho ek lambi line of people ko ek narrow hallway se push kar rahe ho. Jo log walls se brush kar rahe hain
woh phans jaate hain aur barely hilte hain, lekin beech wale freely daud sakte hain. Toh beech wala fastest jaata hai
aur edges rengti hain — yahi curved "parabola" shape hai. Sabko chalate rehne ke liye tumhe peeche se push karte rehna padta hai
(yahi pressure hai). Aur ek wider hallway itna zyada help karta hai ki width double karne par 16 guna zyada log guzar sakte hain —
kyunki zyada jagah aur kam wall-stickiness dono help karte hain.
Mnemonic Hagen–Poiseuille formula yaad karo"Pie R-Four Pushes Past Eighty Lazy" → Q = 8 η L π R 4 Δ P π R 4 P ressure upar; 8 η L neeche — "8 Lazy" eta.)
Aur profile: "R-squared minus r-squared" parabola hai.
Viscosity and Newton's law of viscosity — woh τ = η d v / d r Reynolds number and turbulence — batata hai ki Poiseuille's law kab break down hota hai.Equation of continuity — connected pipes mein Q Bernoulli's principle — non-viscous counterpart; Poiseuille friction losses add karta hai.Stokes' law and terminal velocity — viscous-drag ka ek aur application.Blood flow and circulatory system — R 4
Newton's law of viscosity (formula) τ = η d r d v
No-slip condition kaunsi boundary condition hai? Pipe wall par fluid velocity zero hoti hai, v ( R ) = 0
Poiseuille flow mein velocity profile v ( r ) = 4 η L Δ P ( R 2 − r 2 )
Pipe mein maximum (centre) velocity v m a x = 4 η L Δ P R 2
Hagen–Poiseuille flow rate Q = 8 η L π R 4 Δ P
Flow rate pipe radius par kaise depend karta hai? Q ∝ R 4
Mean aur max velocity ka relation v ˉ = 2 1 v m a x
Poiseuille's law ke assumptions kya hain? Steady, laminar, incompressible, Newtonian fluid ek long straight pipe mein.
Radius r, length L ke coaxial cylinder ke liye side-surface viscous force kaise likhte hain? F = η d r d v ( 2 π r L )
Zyada viscosity flow kyun reduce karti hai? Q ∝ 1/ η
fixes integration constant
steady flow net force zero
Newton law tau = eta dv/dr
Force balance on cylinder
