Exercises — Viscous flow — Poiseuille flow, velocity profile in pipe

2.2.17 · D4 · Physics › Fluid Mechanics › Viscous flow — Poiseuille flow, velocity profile in pipe

Poore page ke liye hamare do master tools:

Upar ki figure har problem ke peeche ka mental picture hai: speed ka ek red parabola, walls par zero, beech mein sabse fast. Ise dhyan mein rakho.

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Level 1 — Recognition

Kya tum pieces ko naam de sakte ho aur sahi formula choose kar sakte ho?

L1.1

Hagen–Poiseuille law mein woh quantity batao jo double hone par $Q$ ko sabse zyada change karti hai, aur kis factor se.

Recall Solution

$Q=\frac{\pi R^4\Delta P}{8\eta L}$. Exponents hain: $R^4$, $\Delta P^1$, ${\eta }^{-1}$, $L^{-1}$. Sabse bada exponent $R$ par hai. $R$ double karne se $Q$ ko $2^4=16$ se multiply karta hai. Kisi bhi doosre ko double karne se $Q$ sirf $2$ (ya $\frac{1}{2}$) factor se change hota hai.

L1.2

Ek pipe mein $v_{\max }=0.60\text{m/s}$ hai. Mean velocity $\bar{v}$ kya hai, aur wall par bilkul velocity kya hai?

Recall Solution

Parabolic profile ke liye average exactly peak ka aadha hota hai: $\bar{v}=\frac{1}{2}{v}_{\max }=\frac{1}{2}(0.60)=0.30\text{m/s}.$ Wall par no-slip condition force karti hai $v(R)=0\text{m/s}$ — wall ko touch karta fluid bilkul move nahi karta.

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Level 2 — Application

Ek formula mein carefully numbers daalo.

L2.1

Water ($\eta =1.0\times 10^{-3}\text{Pa⋅s}$) radius $R=2.0\text{mm}$, length $L=1.0\text{m}$ ki pipe mein se $\Delta P=500\text{Pa}$ se driven hokar flow karta hai. Flow rate $Q$ find karo.

Recall Solution

Step 1 — formula choose karo. Hum $R,L,\eta ,\Delta P$ se $Q$ chahte hain → Hagen–Poiseuille. Step 2 — $R^4$ compute karo. $R=2.0\times 10^{-3}\text{m}$, toh $R^4=(2.0\times 10^{-3}{)}^4=1.6\times 10^{-11}{\text{m}}^4$. Step 3 — substitute karo.

= \frac{\pi (8.0\times10^{-9})}{8.0\times10^{-3}}$$ $$Q = \frac{2.513\times10^{-8}}{8.0\times10^{-3}} \approx \mathbf{3.14\times10^{-6}\ \text{m}^3/\text{s}}.$$ Yeh lagbhag $3.1\ \text{mL/s}$ hai.

L2.2

L2.1 wali pipe ke liye, maximum (centre-line) velocity find karo.

Recall Solution

Step 1 — formula. $v_{\max }=\frac{\Delta P{R}^2}{4\eta L}$. Step 2 — numbers. $R^2=(2.0\times 10^{-3}{)}^2=4.0\times 10^{-6}{\text{m}}^2$.

= \frac{2.0\times10^{-3}}{4.0\times10^{-3}} = \mathbf{0.50\ \text{m/s}}.$$ **Cross-check:** $\bar v = \tfrac12 v_{\max}=0.25\ \text{m/s}$; aur $Q=\bar v\cdot\pi R^2 = 0.25\times\pi(4.0\times10^{-6}) = 3.14\times10^{-6}\ \text{m}^3/\text{s}$ ✓ L2.1 se match karta hai.

L2.3

Water $Q=2.0\times 10^{-5}{\text{m}}^3/\text{s}$ se ek pipe mein flow karta hai jisme $R=5.0\text{mm}$, $L=2.0\text{m}$, $\eta =1.0\times 10^{-3}\text{Pa⋅s}$ hai. $\Delta P$ find karo.

Recall Solution

Step 1 — rearrange karo. $\Delta P=\frac{8\eta LQ}{\pi R^4}$. Step 2 — $R^4$. $(5.0\times 10^{-3}{)}^4=6.25\times 10^{-10}{\text{m}}^4$. Step 3 — substitute karo.

= \frac{3.2\times10^{-7}}{1.963\times10^{-9}} \approx \mathbf{163\ \text{Pa}}.$$

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Level 3 — Analysis

Compare, ratio, aur "kya change hota hai" ke baare mein reason karo.

L3.1

Ek pipe ki radius uski original value ki $80\%$ tak reduce ho jaati hai (partly narrowed artery), baaki sab fixed. Original flow ka kitna fraction bacha rehta hai?

Recall Solution

Step 1 — proportionality. Fixed $\Delta P,L,\eta$ par: $Q\propto R^4$. Step 2 — ratio. New radius $=0.80R$, toh $\frac{Q_{\text{new}}}{Q_{\text{old}}}=(0.80{)}^4=0.4096\approx 0.41.$ Sirf 41% flow hi bachta hai ek mere 20% narrowing mein — ek dramatic loss. Yahi wajah hai ki arterial narrowing itna serious hai (Blood flow and circulatory system).

L3.2

Do pipes same fluid ko same $\Delta P$ ke under carry karti hain. Pipe B twice as long aur pipe A ki $1.5\times$ radius wali hai. $Q_B/Q_A$ find karo.

Recall Solution

Step 1 — dependences collect karo. $Q\propto \frac{R^4}{L}$ ($\Delta P,\eta$ fixed ke saath). Step 2 — ratio.

= \frac{5.0625}{2} = \mathbf{2.53}.$$ $R^4$ boost (factor $5.06$) length penalty (factor $\tfrac12$) ko beat karta hai, toh pipe B phir bhi lagbhag $2.5\times$ zyada carry karti hai.

L3.3

Kis radial position $r$ par local velocity mean velocity $\bar{v}$ ke equal hai? $r$ ko $R$ ke fraction mein do.

Recall Solution

Step 1 — dono likho. $v(r)=v_{\max }\left(1-\frac{r^2}{R^2}\right)$ aur $\bar{v}=\frac{1}{2}{v}_{\max }$. Step 2 — equal set karo.

\;\Rightarrow\; 1-\frac{r^2}{R^2} = \tfrac12 \;\Rightarrow\; \frac{r^2}{R^2}=\tfrac12.$$ **Step 3 — solve karo.** $$r = \frac{R}{\sqrt2} \approx \mathbf{0.707\,R}.$$ Toh flow apni *average* speed par wall tak lagbhag 71\% jaane par move karta hai — neeche ki figure mein green line dekho.

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Level 4 — Synthesis

Poiseuille ko kisi aur idea (continuity, viscosity's law, power) ke saath combine karo.

L4.1

Ek fluid ek wide feeder pipe (radius $R_1=6\text{mm}$) mein mean speed ${\bar{v}}_1=0.20\text{m/s}$ se enter karta hai aur ek narrow pipe (radius $R_2=2\text{mm}$) mein chala jaata hai. Equation of continuity use karke, narrow pipe mein mean speed ${\bar{v}}_2$ find karo.

Recall Solution

Step 1 — continuity. Volume conserved hai: $Q={\bar{v}}_1\pi R_1^2={\bar{v}}_2\pi R_2^2$. Step 2 — solve karo.

= 0.20\times 9 = \mathbf{1.8\ \text{m/s}}.$$ **Note:** continuity area se *mean* speed fix karti hai; Poiseuille phir batata hai ki us mean speed ko *drive* karne ke liye pressure drop kitna chahiye. Woh alag-alag questions answer karte hain — pair karo, confuse mat karo.

L4.2

L2.1 wali pipe ki wall ($r=R$) par fluid jo shear stress $\tau$ exert karta hai woh find karo, Newton's law of viscosity $\tau =\eta |dv/dr|$ use karke (dekho Viscosity and Newton's law of viscosity). Yaad karo $R=2.0\text{mm}$, $\Delta P=500\text{Pa}$, $L=1.0\text{m}$.

Recall Solution

Step 1 — profile differentiate karo. $v(r)=\frac{\Delta P}{4\eta L}(R^2-r^2)\Rightarrow \frac{dv}{dr}=-\frac{\Delta P}{2\eta L}r$. Step 2 — wall $r=R$ par evaluate karo aur magnitude lo.

= \eta\cdot\frac{\Delta P}{2\eta L}R = \frac{\Delta P\,R}{2L}.$$ Notice karo $\eta$ cancel ho jaata hai! **Step 3 — numbers.** $$\tau_{\text{wall}} = \frac{(500)(2.0\times10^{-3})}{2(1.0)} = \mathbf{0.50\ \text{Pa}}.$$ **$\eta$ kyun cancel hota hai:** ek stickier fluid ka wall ke paas *steeper* gradient hota hai exactly utni amount se jo ek *given* pressure push ke liye uski extra stickiness ko offset karta hai.

L4.3

L2.1 mein flow sustain karne ke liye pump jo hydraulic power deliver karta hai woh find karo, $P_{\text{pump}}=\Delta P\cdot Q$ se diya gaya hai. $Q=3.14\times 10^{-6}{\text{m}}^3/\text{s}$ aur $\Delta P=500\text{Pa}$ use karo.

Recall Solution

Step 1 — yeh product kyun? Power = force × velocity = (pressure × area) × velocity = pressure × (area × velocity) = $\Delta P\cdot Q$. Units confirm karte hain: $\text{Pa}\cdot {\text{m}}^3/\text{s}=\frac{\text{N}}{{\text{m}}^2}\cdot \frac{{\text{m}}^3}{\text{s}}=\frac{\text{N⋅m}}{\text{s}}=\text{W}$. Step 2 — multiply karo.

= 1.57\ \text{mW}.$$ Yeh saari power viscous friction se heat ke roop mein dissipate ho jaati hai — yeh ek sticky fluid ko push karne ki "cost" hai.

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Level 5 — Mastery

Multi-step, symbolic, ya edge-case reasoning — bina formula diye.

L5.1

Algebraically show karo ki Poiseuille flow ki mean velocity exactly maximum ki aadhi hai, yaani $\bar{v}=\frac{1}{2}{v}_{\max }$, profile ko cross-section par integrate karke.

Recall Solution

Step 1 — set up karo. $\bar{v}=\frac{Q}{\pi R^2}$ jahan $Q={\int }_0^{R}v(r)2\pi rdr$ aur $v(r)=v_{\max }\left(1-\frac{r^2}{R^2}\right)$. Step 2 — integrate karo.

= 2\pi v_{\max}\left[\frac{r^2}{2}-\frac{r^4}{4R^2}\right]_0^R = 2\pi v_{\max}\left(\frac{R^2}{2}-\frac{R^2}{4}\right)=2\pi v_{\max}\frac{R^2}{4}.$$ Toh $Q = \dfrac{\pi R^2 v_{\max}}{2}$. **Step 3 — divide karo.** $$\bar v = \frac{Q}{\pi R^2} = \frac{v_{\max}}{2}. \qquad\blacksquare$$

L5.2

Ek "power-law" pipe designer flow rate $Q$ ko triple karna chahta hai lekin $\Delta P$, $L$ aur $\eta$ fixed rakhna chahta hai. Radius kitne percentage se badhani chahiye?

Recall Solution

Step 1 — proportionality. $Q\propto R^4\Rightarrow \frac{R_{\text{new}}}{R_{\text{old}}}={\left(\frac{Q_{\text{new}}}{Q_{\text{old}}}\right)}^{1/4}=3^{1/4}$. Step 2 — evaluate karo. $3^{1/4}=1.3161$. Step 3 — percentage increase. $(1.3161-1)\times 100\%\approx 31.6\%.$ Ek chhota sa 32% widening flow triple kar deta hai — $R^4$ law wide pipes ke liye astonishingly generous hai.

L5.3

Centre-line fluid sabse fast move karta hai, phir bhi Newton's law kehta hai shear stress gradient $dv/dr$ ke proportional hoti hai. Explain karo ki shear stress exactly centre $r=0$ par kya hai, aur yeh kahan sabse badi hai. Apna claim algebra se support karo.

Recall Solution

Step 1 — gradient. $v(r)=\frac{\Delta P}{4\eta L}(R^2-r^2)$ se, $\frac{dv}{dr}=-\frac{\Delta P}{2\eta L}r.$ Step 2 — centre pad lo. $r=0$ par: $\frac{dv}{dr}=0$, toh $\tau (0)=\eta \left|\frac{dv}{dr}\right|=0.$ Step 3 — yeh kahan sabse badi hai? $|dv/dr|\propto r$, jo wall $r=R$ par sabse bada hai. Toh shear stress centre par zero se linearly grow karti hai wall par maximum tak — yahi case L4.2 mein numerically check kiya gaya tha (${\tau }_{\text{wall}}=0.50\text{Pa}$). Subtlety: "fastest" (high $v$) ka matlab "most sheared" (high $dv/dr$) nahi hota. Centre par parabola momentarily flat hoti hai (iska slope zero hai), toh wahan neighboring layers nearly same speed par move karti hain — koi rubbing nahi, koi shear nahi. Saari internal friction wall ke paas rehti hai.

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Active Recall

Recall One-line takeaways (answers cover karo!)

• $R$ double karne par $Q$ kitna multiply hota hai? ::: $2^4=16$.
• Radius $80\%$ tak cut hone par: flow kya ban jaata hai? ::: $(0.8{)}^4\approx 0.41$, yaani 41%.
• Local speed mean ke equal $r=$ par? ::: $R/\sqrt{2}\approx 0.71R$.
• Wall shear stress formula? ::: ${\tau }_{\text{wall}}=\frac{\Delta PR}{2L}$ ($\eta$ cancel hota hai).
• Pipe centre par shear stress? ::: Zero — profile wahan flat hai.
• Flow sustain karne ke liye pump power? ::: $P=\Delta P\cdot Q$.

Connections

• Parent: Poiseuille flow (Hinglish) — woh derivations jinhe yeh exercises drill karti hain.
• Viscosity and Newton's law of viscosity — L4.2 aur L5.3 mein wall shear ke liye use kiya gaya.
• Equation of continuity — L4.1 mein narrowing pipes mein speed-up ke liye use kiya gaya.
• Reynolds number and turbulence — har answer yahan laminar flow assume karta hai.
• Blood flow and circulatory system — L3.1 ka $R^4$ artery result.
• Bernoulli's principle — in viscous losses ka frictionless contrast.