Viscous flow — Poiseuille flow, velocity profile in pipe: Worked examples

2.2.17 · D3 · Physics › Fluid Mechanics › Viscous flow — Poiseuille flow, velocity profile in pipe

Yahan sab kuch parent note par based hai: the topic note. Symbols jo use hue hain: $R$ = pipe radius, $L$ = pipe length, $Δ P$ = pressure drop across the length, $η$ = viscosity, $Q$ = volume flow rate (volume per second), $v (r)$ = speed at distance $r$ from the axis.

The scenario matrix

Koi bhi number se pehle, yahan poora map of case-classes hai jo ye topic tumpe throw kar sakta hai. Har worked example neeche us cell (C1–C12) ke saath tagged hai jo wo cover karta hai.

Cell	Case class	Kya cheez tricky banati hai
C1	Seedha $Q$ solve karo	Direct plug-in — baseline
C2	Koi hidden variable solve karo ( $Δ P$ , $η$ , $R$ , $L$ )	Formula rearrange karni padegi
C3	Velocity profile — ek point $r$ par speed	$v (r)$ use karo, $Q$ nahi
C4	Degenerate input: $r = 0$ (axis) aur $r = R$ (wall)	Parabola ke endpoints
C5	Limiting / scaling: $R$ double karo, $L$ half karo, etc.	Powers ( $R^{4}$ , $R^{2}$ ) dominate karte hain
C6	Zero / vanishing input: $Δ P = 0$ , ya $R \to 0$	Flow ruk jaati hai — formula ki sanity check
C7	Pipes series mein (same $Q$ , pressures add hote hain)	Continuity + additive $Δ P$
C8	Mean vs max speed	$\frac{1}{2}$ ka factor
C9	Real-world word problem (blood / oil)	Story ko symbols mein translate karo
C10	Exam twist: validity check (kya ye laminar bhi hai?)	Reynolds number gatekeeper
C11	Pipes parallel mein (same $Δ P$ , flows add hote hain)	Conductances add hote hain, resistances nahi
C12	Sign & singular limits: $Δ P < 0$ , $L \to 0$	Flow ulti ho jaati hai; $L \to 0$ par blow-up

Ab hum matrix ko, cell by cell, clear karte hain.

C1 — Seedha flow-rate calculation

Worked example Ek level pipe se oil

Oil ( $η = 0.10 Pa\cdots$ ) ek horizontal pipe se flow karti hai jiska radius $R = 1.0 cm$ aur length $L = 5.0 m$ hai, pressure drop $Δ P = 400 Pa$ ke saath. Volume flow rate $Q$ nikalo.

Forecast: Kya $Q$ bada hoga ya chhota? Padhne se pehle guess karo — $R^{4}$ mein $R$ metres mein bahut chhotha hai, toh expect karo bahut chhotha $Q$ .

Units convert karo. $R = 1.0 cm = 1.0 \times 1 0^{- 2} m$ , toh $R^{4} = (1 0^{- 2})^{4} = 1.0 \times 1 0^{- 8} m^{4}$ . Ye step kyun? Har SI formula ko metres chahiye; ek stray centimetre $R^{4}$ ko $1 0^{8}$ ke factor se bigaad deta hai.
Hagen–Poiseuille mein plug karo. $Q = \frac{π R ^{4} Δ P}{8 η L} = \frac{π ( 1.0 \times 1 0 ^{- 8} ) ( 400 )}{8 ( 0.10 ) ( 5.0 )} .$ Ye step kyun? Hume $Q$ chahiye aur sabhee chaar inputs diye hue hain — direct substitution.
Arithmetic. Numerator $= π \cdot 4.0 \times 1 0^{- 6} = 1.257 \times 1 0^{- 5}$ . Denominator $= 4.0$ . $Q = \frac{1.257 \times 1 0 ^{- 5}}{4.0} \approx 3.14 \times 1 0^{- 6} m^{3} / s .$

Verify: Units — $\frac{m ^{4} \cdot Pa}{Pa\cdots \cdot m} = \frac{m ^{4}}{s \cdot m} = m^{3} / s$ . ✓ Flow rate ke liye sahi dimensions. Chhotha answer ( $\sim 3 mL/s$ ) forecast se match karta hai: metres $R^{4}$ ko minuscule banaa dete hain.

C2 — Kisi hidden variable ke liye solve karo

Worked example Pipe kaunsi viscosity reveal karti hai?

Ek liquid ko $R = 2.0 mm$ , $L = 1.0 m$ wali pipe se $Δ P = 500 Pa$ ke saath push kiya jaata hai, aur $Q = 1.0 \times 1 0^{- 6} m^{3} / s$ produce hoti hai. Iska viscosity $η$ nikalo.

Forecast: Kyunki $Q \propto 1/ η$ , ek chhotha $Q$ ek bade $η$ ka hint deta hai. Guess karo kya ye water ( $1 0^{- 3} Pa\cdots$ ) se zyada ya kam viscous hai.

$η$ ke liye rearrange karo. $Q = \frac{π R ^{4} Δ P}{8 η L}$ se isolate karo: $η = \frac{π R ^{4} Δ P}{8 L Q} .$ Ye step kyun? $η$ denominator mein hai; dono taraf $η$ se multiply karo aur $Q$ se divide karo.
$R^{4}$ compute karo. $R = 2.0 \times 1 0^{- 3} m$ , toh $R^{4} = (2 \times 1 0^{- 3})^{4} = 1.6 \times 1 0^{- 11} m^{4}$ . Ye step kyun? Fourth power carefully evaluate karni padti hai — $2^{4} = 16$ aur $(1 0^{- 3})^{4} = 1 0^{- 12}$ .
Plug in karo. $η = \frac{π ( 1.6 \times 1 0 ^{- 11} ) ( 500 )}{8 ( 1.0 ) ( 1.0 \times 1 0 ^{- 6} )} = \frac{2.513 \times 1 0 ^{- 8}}{8.0 \times 1 0 ^{- 6}} \approx 3.14 \times 1 0^{- 3} Pa\cdots .$

Verify: $η \approx 3.1 \times 1 0^{- 3} Pa\cdots$ — water ki viscosity se kareeban teen guna, toh zyada viscous, jo "chhotha $Q$ → thick fluid" forecast se match karta hai. Units: $\frac{m ^{4} \cdot Pa}{m \cdot m ^{3} / s} = Pa\cdots$ . ✓

C3 & C4 — Velocity profile aur uske endpoints

Worked example Pipe ke across speed (ek picture problem)

Ek pipe ka $R = 6.0 mm$ hai, aur centre speed $v_{m a x} = 0.90 m/s$ hai. (a) axis $r = 0$ par, (b) $r = 3.0 mm$ par, (c) wall $r = R$ par speed nikalo. Ye C3 (ek point) aur C4 (dono degenerate endpoints) cover karta hai.

Forecast: Profile ek parabola hai. Compute karne se pehle apne dimaag mein sketch karo: beech mein sabse zyada, wall par zero, aur aadhe radius par abhi bhi kaafi fast hona chahiye (parent ka "75% at halfway" fact yaad karo).

$v_{m a x}$ kahan se aata hai. Parent ke derived profile $v (r) = \frac{Δ P}{4 η L} (R^{2} - r^{2})$ se shuru karo. Axis par $r = 0$ par ye sabse bada hai: $v_{m a x} = v (0) = \frac{Δ P R ^{2}}{4 η L} .$ Profile ko is peak se divide karo aur messy $Δ P / (4 η L)$ cancel ho jaata hai: $v (r) = \frac{Δ P}{4 η L} (R^{2} - r^{2}) = v_{m a x} \frac{Δ P R ^{2}}{4 η L} (1 - \frac{r ^{2}}{R ^{2}}) = v_{m a x} (1 - \frac{r ^{2}}{R ^{2}}) .$ Ye step kyun? Ratio form koi naya law nahi hai — ye wahi parabola hai, rewrite ki gayi hai taaki $Δ P$ , $η$ , $L$ sab milkar ek single measured number $v_{m a x}$ ke andar pack ho jaayein. Kyunki problem humein $v_{m a x}$ directly deti hai, hume un teeno ki alag se zaroorat nahi padti.
(a) Axis, $r = 0$ : $v = v_{m a x} (1 - 0) = 0.90 m/s$ . Ye step kyun? Centre par kuch subtract nahi hota — ye peak ki definition hai. Figure mein ye red arrow ki tip hai.
(b) $r = 3.0 mm$ : $\frac{r ^{2}}{R ^{2}} = \frac{3 ^{2}}{6 ^{2}} = \frac{9}{36} = 0.25$ , toh $v = 0.90 (1 - 0.25) = 0.675 \approx 0.68 m/s .$ Ye step kyun? Aadhe raaste par ( $r = R /2$ ), speed ka sirf ek chauthai hissa kho jaata hai — parabola top ke paas flat hoti hai.
(c) Wall, $r = R$ : $\frac{R ^{2}}{R ^{2}} = 1$ , toh $v = 0.90 (1 - 1) = 0$ . Ye step kyun? Ye no-slip condition hai — wall se chipka fluid wahan stuck rehta hai. Figure mein arrow kinare par simat kar kuch nahi reh jaata.

Verify: Endpoints physics se match karte hain: centre par max, wall par zero. Half-radius value $0.68 m/s$ jo hai wo $0.90$ ka $75%$ hai — exactly parent ka fact. ✓

C5 — Scaling / limiting behaviour

Worked example Kai cheezein ek saath badlo

Ek pipe ka radius double kiya jaata hai, length triple ki jaati hai, aur pressure drop half kiya jaata hai, fluid same rahti hai. $Q$ kis factor se badalta hai?

Forecast: $R^{4}$ yahan giant hai — $R$ double karne se $Q$ $16$ guna ho jaata hai. Length aur pressure changes gentler hain. Guess karo: net increase, lekin kitna?

$Q$ ko proportionality ke roop mein likho. $Q \propto \frac{R ^{4} Δ P}{L} .$ Ye step kyun? $π$ , $8$ , $η$ unchanged hain, toh ratio mein cancel ho jaate hain — sirf varied quantities matter karti hain.
Har factor apply karo. New over old: $\frac{Q _{new}}{Q _{old}} = \frac{( 2 ) ^{4} \cdot \frac{1}{2}}{3} = \frac{16 \cdot 0.5}{3} = \frac{8}{3} \approx 2.67.$ Ye step kyun? Har variable apni power ke saath enter karta hai — $R$ 4th power mein, $Δ P$ 1st power mein, $L$ $- 1$ power mein.

Verify: $\frac{8}{3} \approx 2.67$ . Flow ~2.7× badhti hai. Dominant factor sach mein radius hai: akele ye ×16 deta, aur halved pressure (÷2) aur tripled length (÷3) isse sirf tame karte hain. ✓

C6 — Zero aur vanishing inputs (degenerate limits)

Worked example Kya ho agar push gayab ho jaaye — ya pipe band ho jaaye?

(a) Agar $Δ P = 0$ ho, toh $Q$ aur profile $v (r)$ kya hoga? (b) Jaise $R \to 0$ , $Q$ ka kya hoga?

Forecast: Koi push nahi matlab koi flow nahi. Ek pipe jo kuch bhi nahi reh jaati wo flow zero kar degi — lekin kitni tezi se?

(a) Dono formulas mein $Δ P = 0$ set karo. $Q = \frac{π R ^{4} ( 0 )}{8 η L} = 0, v (r) = \frac{0}{4 η L} (R^{2} - r^{2}) = 0 everywhere .$ Ye step kyun? Poiseuille flow pressure difference se driven hoti hai. Driver hatao aur chipchipaa fluid bas ruk jaata hai — koi residual motion nahi. Ye puri derivation ka ek crucial sanity check hai.
(b) $R \to 0$ lo. Kyunki $Q \propto R^{4}$ , $lim_{R \to 0} Q = lim_{R \to 0} \frac{π R ^{4} Δ P}{8 η L} = 0.$ Ye step kyun? $R^{4} \to 0$ kisi bhi lower power se tezi se. Ek pipe jo ek point tak simat jaati hai kuch bhi nahi jaane deti — aur ye quartically choke karti hai, cross-sectional area ( $\propto R^{2}$ ) ke sirf simatne se dramatically tezi se.

Verify: Dono limits $0$ dete hain, exactly jaisa physical intuition demand karta hai. Formula apne degenerate corners par well-behaved hai — koi division-by-zero nahi, koi negative flow nahi. ✓

C7 — Do pipes series mein

Worked example Joined pipe mein pressure drop

Do horizontal pipes same $Q = 2.0 \times 1 0^{- 6} m^{3} / s$ ka fluid $η = 1.0 \times 1 0^{- 3} Pa\cdots$ ke saath carry karte hain, end to end joined. Pipe A: $R_{A} = 2.0 mm$ , $L_{A} = 1.0 m$ . Pipe B: $R_{B} = 1.0 mm$ , $L_{B} = 1.0 m$ . Total pressure drop nikalo.

Forecast: Same $Q$ dono se flow karta hai (continuity). Narrow pipe B, jiska $R^{4}$ sixteen guna chhotha hai, ko bahut zyada $Δ P$ chahiye. Guess karo kaun sa pipe dominate karega.

Pehle figure padho. Wide teal box pipe A hai, narrow orange box pipe B hai, end to end joined. Dono se seedha guzarta ek plum arrow flow $Q$ hai — same arrow, kyunki jo bhi volume A mein per second enter karta hai wo B se per second nikalna chahiye (continuity). Har pipe ke neeche figure uska apna pressure drop label karta hai; notice karo pipe B ka bahut zyada hai, even though dono pipes same length ki hain. Bottom line dikhata hai dono kaise add hote hain.

$Δ P$ ke liye Hagen–Poiseuille invert karo. $Q = \frac{π R ^{4} Δ P}{8 η L}$ se shuru karo. Hume $Δ P$ chahiye, toh dono taraf $8 η L$ multiply karo aur $π R^{4}$ se divide karo: $Q = \frac{π R ^{4} Δ P}{8 η L} ⟹ Q (8 η L) = π R^{4} Δ P ⟹ Δ P = \frac{8 η L Q}{π R ^{4}} .$ Ye step kyun? Har segment ek known $Q$ carry karta hai lekin ek unknown $Δ P$ , toh hum master formula ko pressure drop ke liye solve karte hain, flow ke liye nahi. Ye inverted form har "given the flow, find the push" problem ka tool hai.
Series rule — drops add hote hain. Har ek se same $Q$ ; A se push karne ke liye chahiye pressure plus B se push karne ke liye chahiye pressure hi total hai: $Δ P_{total} = Δ P_{A} + Δ P_{B} .$ Ye step kyun? Continuity (intuition box) ek common $Q$ force karta hai; joined pipe ke saath neeche jaate waqt tum A mein pressure kho te ho, phir B mein aur kho te ho, aur ek line mein losses simply sum hote hain.
Pipe A. $R_{A}^{4} = (2 \times 1 0^{- 3})^{4} = 1.6 \times 1 0^{- 11} m^{4}$ . $Δ P_{A} = \frac{8 ( 1 0 ^{- 3} ) ( 1.0 ) ( 2.0 \times 1 0 ^{- 6} )}{π ( 1.6 \times 1 0 ^{- 11} )} = \frac{1.6 \times 1 0 ^{- 8}}{5.027 \times 1 0 ^{- 11}} \approx 318 Pa .$ Ye step kyun? Segment A par inverted formula ka direct application.
Pipe B. $R_{B}^{4} = (1 \times 1 0^{- 3})^{4} = 1.0 \times 1 0^{- 12} m^{4}$ . $Δ P_{B} = \frac{1.6 \times 1 0 ^{- 8}}{π ( 1.0 \times 1 0 ^{- 12} )} = \frac{1.6 \times 1 0 ^{- 8}}{3.1416 \times 1 0 ^{- 12}} \approx 5093 Pa .$ Ye step kyun? Same numerator (same $η L Q$ ), lekin $R_{B}^{4}$ $16 \times$ chhotha hai, toh $Δ P_{B}$ $Δ P_{A}$ se $16 \times$ bada hai.
Add karo. $Δ P_{total} = 318 + 5093 \approx 5411 Pa$ .

Verify: $Δ P_{B} /Δ P_{A} = 5093/318 \approx 16.0 = 2^{4}$ ✓ — exactly $R^{4}$ s ka ratio. Narrow pipe total resistance ka ~94% carry karti hai, forecast se match karta hai. ✓

C8 — Mean speed vs maximum speed

C9 — Real-world word problem (blood)

Worked example Ek narrow artery

Blood ( $η = 3.5 \times 1 0^{- 3} Pa\cdots$ ) ek artery se flow karta hai jiska radius $R = 4.0 mm$ , length $L = 0.10 m$ hai, $Δ P = 300 Pa$ ke under. (a) Healthy flow rate $Q$ nikalo. (b) Agar plaque radius ko $20%$ narrow kar de (to $0.80 R$ ), original flow ka kaunsa fraction bachega?

Forecast: "Sirf" 20% narrowing minor lagta hai — lekin $R^{4}$ ise punish karta hai. Compute karne se pehle surviving fraction guess karo.

(a) Healthy $Q$ . Pehle $R^{4} = (4.0 \times 1 0^{- 3})^{4} = 2.56 \times 1 0^{- 10} m^{4}$ compute karo, phir saare chaar given quantities ko Hagen–Poiseuille mein substitute karo: $Q = \frac{π R ^{4} Δ P}{8 η L} = \frac{π ( 2.56 \times 1 0 ^{- 10} ) ( 300 )}{8 ( 3.5 \times 1 0 ^{- 3} ) ( 0.10 )} = \frac{2.413 \times 1 0 ^{- 7}}{2.8 \times 1 0 ^{- 3}} \approx 8.62 \times 1 0^{- 5} m^{3} / s .$ Ye step kyun? Problem directly $η$ , $R$ , $L$ aur $Δ P$ deti hai aur flow rate maangti hai, toh ye master formula ka seedha application hai — same move jaise C1, ab physiology ke roop mein dressed (dekhein Blood flow and circulatory system). Ek hi dhyaan rakhna hai: fourth power uthane se pehle radius ko metres mein convert karo.
(b) Narrowing ke baad fraction. Sirf $R$ badalta hai, toh $\frac{Q _{new}}{Q _{old}} = (\frac{0.80 R}{R})^{4} = (0.80)^{4} = 0.4096 \approx 0.41.$ Ye step kyun? $R$ ke alawa sab cancel ho jaata hai; ratio purely radius ratio ki fourth power hai.

Verify: 20% radius loss flow ko $41%$ tak cut kar deta hai — ek chhoti si pinch se lagbhag aadha ho jaata hai. Yahi reason hai ki arterial narrowing itna dangerous hai, aur $R^{4}$ law iska reason hai. $(0.8)^{4} = 0.4096$ ✓.

C10 — Exam twist: kya Poiseuille valid bhi hai?

Worked example Compute karne se pehle check karo

Water ( $η = 1.0 \times 1 0^{- 3} Pa\cdots$ , density $ρ = 1000 kg/m^{3}$ ) mean speed $\overset{v}{ˉ} = 2.0 m/s$ se radius $R = 5.0 mm$ ki pipe se flow karta hai. Kya Poiseuille's law applicable hai? Reynolds number hai $Re = \frac{ρ v ˉ ( 2 R )}{η}$ , aur flow tabhi laminar hai jab $Re ≲ 2000$ .

Forecast: Itni badi pipe mein high speed turbulent lagti hai. Guess karo kya $Re$ 2000 cross karega.

Diameter compute karo. $2 R = 2 (5.0 \times 1 0^{- 3}) = 1.0 \times 1 0^{- 2} m$ . Ye step kyun? Reynolds number characteristic length ke roop mein diameter use karta hai, radius nahi — ye ek classic exam trap hai.
$Re$ evaluate karo. $Re = \frac{ρ v ˉ ( 2 R )}{η} = \frac{( 1000 ) ( 2.0 ) ( 1.0 \times 1 0 ^{- 2} )}{1.0 \times 1 0 ^{- 3}} = \frac{20}{1.0 \times 1 0 ^{- 3}} = 20000.$ Ye step kyun? Chaar quantities definition mein plug karo; dekhein Reynolds number and turbulence.
Decide karo. $Re = 20000 ≫ 2000$ , toh flow turbulent hai — Poiseuille's law (jo laminar flow assume karta hai) apply nahi hota. Isse compute kiya hua koi bhi $Q$ physically galat hoga. Ye step kyun? Poiseuille ki poori derivation smooth, layered (laminar) flow par, bina mixing ke based hai. Jab turbulence aa jaata hai, neat parabolic profile destroy ho jaati hai aur flow ek alag, non-linear pressure law follow karta hai.

Verify: $Re = 20000$ , laminar ceiling $2000$ se das guna. Sahi exam move hai ye state karna ki Poiseuille yahan invalid hai, instead of blindly $Q = π R^{4} Δ P / (8 η L)$ mein numbers plug karne ke. ✓

C11 — Do pipes parallel mein

Worked example Do branches mein flow split karna

$Δ P = 200 Pa$ pressure drop do pipes mein apply hota hai jo side by side connected hain (dono ends shared), har ek ki length $L = 1.0 m$ , fluid $η = 1.0 \times 1 0^{- 3} Pa\cdots$ . Pipe A: $R_{A} = 2.0 mm$ . Pipe B: $R_{B} = 1.0 mm$ . Total flow rate nikalo.

Forecast: Parallel mein, dono pipes same push $Δ P$ feel karti hain (unke ends joined hain), aur unke flows add hote hain. Moti pipe dominate karegi — guess karo kitna.

Parallel rule. Har branch par same $Δ P$ ; total flow sum hai: $Q_{total} = Q_{A} + Q_{B}, Q = \frac{π R ^{4} Δ P}{8 η L} .$ Ye step kyun? Series ke ulta (jahan $Q$ shared tha aur $Δ P$ add hota tha), yahan pressure shared hai aur flows add hote hain — do branches independent paths offer karti hain, toh unki carrying capacities (conductances) add hoti hain, resistances nahi.
Pipe A. $R_{A}^{4} = (2 \times 1 0^{- 3})^{4} = 1.6 \times 1 0^{- 11} m^{4}$ . $Q_{A} = \frac{π ( 1.6 \times 1 0 ^{- 11} ) ( 200 )}{8 ( 1 0 ^{- 3} ) ( 1.0 )} = \frac{1.005 \times 1 0 ^{- 8}}{8.0 \times 1 0 ^{- 3}} \approx 1.257 \times 1 0^{- 6} m^{3} / s .$ Ye step kyun? Branch A ke liye direct plug-in.
Pipe B. $R_{B}^{4} = 1.0 \times 1 0^{- 12} m^{4}$ , jo $16 \times$ chhotha hai, toh $Q_{B} = \frac{Q _{A}}{16} \approx 7.85 \times 1 0^{- 8} m^{3} / s .$ Ye step kyun? Same $Δ P$ , $L$ , $η$ ; sirf $R^{4}$ factor $16$ se alag hai.
Add karo. $Q_{total} = 1.257 \times 1 0^{- 6} + 7.85 \times 1 0^{- 8} \approx 1.335 \times 1 0^{- 6} m^{3} / s$ .

Verify: $Q_{A} / Q_{B} = 16 = 2^{4}$ ✓. Moti pipe total flow ka lagbhag $94%$ carry karti hai — series case ka mirror image, jahan thin pipe $94%$ resistance hog leti thi. ✓

C12 — $Δ P$ ka sign aur singular short-pipe limit

Worked example Ulta push aur ek vanishing pipe

(a) Normally pipe mein $P_{1} > P_{2}$ hota hai toh $Δ P = P_{1} - P_{2} > 0$ . Agar hum $P_{1} < P_{2}$ kar dein, yaani $Δ P < 0$ , toh formula kya predict karta hai? (b) Jaise $L \to 0$ (infinitesimally short pipe), $Q$ ka kya hoga?

Forecast: Jo end higher pressure par hai use flip karna sirf flow ki direction reverse karni chahiye — physics symmetric hai. Aur zero length ki pipe zero resistance offer karti hai, toh guess karo $Q$ kya karta hai.

(a) Flow ka sign. $Δ P < 0$ ke saath, formula deta hai $Q = \frac{π R ^{4} Δ P}{8 η L} < 0.$ Negative $Q$ ka matlab simply hai flow doosri taraf jaati hai — $P_{2}$ se $P_{1}$ ki taraf. Isi tarah har velocity $v (r) = \frac{Δ P}{4 η L} (R^{2} - r^{2})$ sign flip karti hai, toh poori parabola backward point karti hai. Ye step kyun? Formula linear aur odd in $Δ P$ hai: $Δ P \to - Δ P$ replace karo aur sab kuch negate ho jaata hai. Fluid hamesha pressure gradient down flow karta hai, high se low — $Δ P$ ka sign sirf ek bookkeeping choice hai ki tumne kaun sa end "1" kaha. Ye sign-convention check guarantee karta hai ki law physically consistent hai.
(b) Limit $L \to 0$ . Kyunki $L$ denominator mein hai, $lim_{L \to 0^{+}} Q = lim_{L \to 0^{+}} \frac{π R ^{4} Δ P}{8 η L} = + \infty.$ Mathematically flow rate blow up ho jaata hai — ek zero-length pipe mein zero viscous resistance hai, toh ek finite push infinite flow drive karega. Ye step kyun? Ye formula ki ek genuine singularity hai, aur ye flag karta hai ki model physical rehna band kar deta hai: bahut chhoti pipe ke liye "fully developed" parabolic flow ki assumption (fluid ko parabola mein settle hone ke liye pipe ki ek length chahiye) fail ho jaati hai, aur inertial/entry effects — jo Poiseuille capture nahi karta — takeover le lete hain. Blow-up formula ka honest tarika hai ye bolne ka ki ye apne domain se bahar hai.

Verify: Odd symmetry: $Q (- Δ P) = - Q (Δ P)$ ✓ (flow reverses, magnitude unchanged). Divergence: $Q \to \infty$ as $L \to 0^{+}$ ✓ — resistance $8 η L / (π R^{4}) \to 0$ . Dono behaviours exactly wahi hain jo formula ki algebra demand karti hai.

Active recall

Recall Kaunse sawaal ke liye kaunsi formula? (answers cover karo)

$v_{m a x}$ diya ho aur ek point par speed maangi ho, kaunsa form $η$ aur $Δ P$ se bachata hai? ::: Ratio form $v (r) = v_{m a x} (1 - r^{2} / R^{2})$ , jahan $v_{m a x} = Δ P R^{2} / (4 η L)$ . Series mein do pipes same kya carry karti hain, aur kya add hota hai? ::: Same $Q$ (continuity); pressure drops add hote hain. Parallel mein do pipes same kya share karti hain, aur kya add hota hai? ::: Same $Δ P$ ; flow rates $Q$ add hote hain. 20% radius reduction flow ka kaunsa fraction bachata hai? ::: $(0.8)^{4} \approx 0.41$ — lagbhag 41%. Negative $Δ P$ physically kya matlab hai? ::: Flow simply direction reverse karta hai; magnitude unchanged. Poiseuille apply karne se pehle, kaunsa number check karna zaroori hai? ::: Reynolds number ( $≲ 2000$ for laminar). Mean speed aur max speed mein kya relation hai? ::: $\overset{v}{ˉ} = \frac{1}{2} v_{m a x}$ .

Connections

Parent topic — wo derivations jinhe ye examples exercise karti hain.
Equation of continuity — "same $Q$ " rule jo series problem (C7) mein use hui.
Reynolds number and turbulence — C10 mein validity gate.
Blood flow and circulatory system — C9 ke peeche physiology.
Viscosity and Newton's law of viscosity — jahan $η$ aur shear stress aate hain.