Worked examples — Boundary layer thickness, displacement thickness, momentum thickness

2.2.21 · D3 · Physics › Fluid Mechanics › Boundary layer thickness, displacement thickness, momentum t

Yeh page ek drill floor hai. Parent note ne teen thicknesses $\delta$, ${\delta }^{\ast }$, $\theta$ ko first principles se build kiya tha. Yahan hum un formulas par har tarah ka input daalte hain taaki koi bhi exam question tumhe surprise na kar sake.

Shuru karne se pehle, teen tools ka ek reminder, seedhe alfazon mein:

Recall Teen definitions (inhe cold yaad rakho)

• $\delta$ ::: wall se woh distance jahan speed free-stream $U$ ka $99\%$ reach karti hai.
• ${\delta }^{\ast }={\int }_0^{\infty }\left(1-\frac{u}{U}\right)dy$ ::: mass flow ke "missing slab" ki thickness.
• $\theta ={\int }_0^{\infty }\frac{u}{U}\left(1-\frac{u}{U}\right)dy$ ::: momentum ke "missing slab" ki thickness.

Yahan $u=u(y)$ wall se $y$ height par fluid ki speed hai, $U$ free-stream speed hai, $\nu$ kinematic viscosity hai, aur sab kuch No-slip condition follow karta hai ($u=0$ at $y=0$).

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Scenario matrix

Inhe thicknesses ke baare mein har sawaal in boxes mein se kisi ek mein aata hai. Hum har box ko ek worked example se cover karenge.

Cell	Case class	Tricky kyun hai	Example
A	Linear profile	sabse simple polynomial, warm-up	Ex 1
B	Polynomial (cubic) profile	edge par slope + curvature match karna	Ex 2
C	Trigonometric (sine) profile	$\sin ,{\sin }^2$ integrate karna	Ex 3
D	Degenerate: plug/uniform flow ($u=U$ everywhere)	zero deficit — kya limits break hoti hain?	Ex 4
E	Degenerate: stagnant limit ($u\to 0$)	maximal deficit, sanity ceiling	Ex 4
F	Profile sirf $\delta$ tak defined (finite cutoff)	${\int }_0^{\infty }$ ka trap	Ex 5
G	Real-world word problem (numbers, units)	Reynolds number, mm-scale answer	Ex 6
H	Downstream growth / $x$ ke saath scaling	$\delta ,{\delta }^{\ast },\theta$ sab kaise scale karte hain	Ex 7
I	Exam twist: diya ${\delta }^{\ast }$ aur $H$, $\theta$ nikalo	Blasius solution shape factor reverse karo	Ex 8
J	Sign/limiting sanity: ordering $\theta <{\delta }^{\ast }<\delta$	inequality kyun kabhi flip nahi ho sakti	Ex 9

Ek baar-baar aane wala trick: aise profile ke liye jo sirf shape par depend kare, likho $\eta =y/\delta$ (ek pure number $0$ se $1$ tak). Tab $dy=\delta d\eta$ aur har thickness (ek pure number) × $\delta$ ban jaati hai. Woh pure number hi profiles ke beech badalta hai.

Neeche ki figure teen model profiles (linear, cubic, sine) ko same axes par plot karti hai, taaki tum dekh sako ki woh alag thicknesses kyun dete hain. Horizontal axis $u/U$ hai, vertical axis $\eta =y/\delta$ hai. Dekho ki har curve wall (bottom-left) se kaise nikalti hai aur top par $u=U$ (dashed navy line) tak pahunchti hai. Magenta shaded band linear curve aur $u=U$ ke beech exactly woh deficit $(1-u/U)$ hai jise ${\delta }^{\ast }$ integrate karta hai — jo curve dashed line ke zyada close rehti hai (jaise orange sine) woh thinner band chodti hai, isliye chhota ${\delta }^{\ast }$.

Yeh picture dimag mein rakho: jitna zyada koi profile dashed $u=U$ line se door hogi, utna fatter uska deficit, utna bada uska ${\delta }^{\ast }$.

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Example 1 — Linear profile (Cell A)

Forecast: compute karne se pehle guess karo — kya ${\delta }^{\ast }$, $\delta /2$ se bada hoga ya chhota? (Velocity puri layer mein $U$ se neeche hai, lekin deficit linearly zero tak shrink hoti hai...)

1. $\eta =y/\delta$ mein switch karo. Toh $u/U=\eta$, $dy=\delta d\eta$, limits $0\to 1$. Yeh step kyun? Yeh $\delta$ ko strip kar deta hai taaki integral sirf ek number ho jaye. Yeh figure mein red magenta line hai.
2. Displacement: ${\delta }^{\ast }={\int }_0^1(1-\eta )\delta d\eta =\delta {\left[\eta -\frac{{\eta }^2}{2}\right]}_0^1=\delta \left(1-\frac{1}{2}\right)=\frac{\delta }{2}.$ Yeh step kyun? $1-u/U=1-\eta$ fractional deficit hai — figure mein shaded band.
3. Momentum: $\theta ={\int }_0^1\eta (1-\eta )\delta d\eta =\delta {\left[\frac{{\eta }^2}{2}-\frac{{\eta }^3}{3}\right]}_0^1=\delta \left(\frac{1}{2}-\frac{1}{3}\right)=\frac{\delta }{6}.$ Yeh step kyun? Extra factor $\eta =u/U$ deficit ko weight karta hai ki actually kitna mass wahan hai.
4. Shape factor: $H=\frac{{\delta }^{\ast }}{\theta }=\frac{\delta /2}{\delta /6}=3.$ (Yaad karo $H$ upar box mein defined hai.)

Verify: $\theta =\delta /6<{\delta }^{\ast }=\delta /2<\delta$ ✓ (ordering hold karti hai). Units: har term $\times \delta$ hai, ek length ✓. Forecast check: ${\delta }^{\ast }=\delta /2$ exactly — linear deficit average hokar $1/2$ deta hai.

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Example 2 — Cubic profile (Cell B)

Forecast: yeh profile wall par steeper hai (bada slope at $\eta =0$) linear wali se — yeh figure mein violet curve hai. Kya uska $H$, $3$ se upar hoga ya neeche?

1. Displacement: $1-\frac{u}{U}=1-\frac{3}{2}\eta +\frac{1}{2}{\eta }^3$. ${\delta }^{\ast }=\delta {\int }_0^1\left(1-\frac{3}{2}\eta +\frac{1}{2}{\eta }^3\right)d\eta =\delta \left[1-\frac{3}{4}+\frac{1}{8}\right]=\frac{3\delta }{8}.$ Yeh step kyun? Bas term by term integrate karo; har power ${\eta }^n$ se $1/(n+1)$ milta hai.

2. Momentum — pehle integrand expand karo. $\frac{u}{U}\left(1-\frac{u}{U}\right)$ multiply out karo: $\left(\frac{3}{2}\eta -\frac{1}{2}{\eta }^3\right)\left(1-\frac{3}{2}\eta +\frac{1}{2}{\eta }^3\right).$ Term by term distribute karo: $=\underset{\frac{3}{2}\eta \cdot 1}{\underbrace{\frac{3}{2}\eta }}-\underset{\frac{3}{2}\eta \cdot (-\frac{3}{2}\eta )}{\underbrace{\frac{9}{4}{\eta }^2}}+\underset{\frac{3}{2}\eta \cdot \frac{1}{2}{\eta }^3}{\underbrace{\frac{3}{4}{\eta }^4}}-\underset{-\frac{1}{2}{\eta }^3\cdot 1}{\underbrace{\frac{1}{2}{\eta }^3}}+\underset{-\frac{1}{2}{\eta }^3\cdot (-\frac{3}{2}\eta )}{\underbrace{\frac{3}{4}{\eta }^4}}-\underset{-\frac{1}{2}{\eta }^3\cdot \frac{1}{2}{\eta }^3}{\underbrace{\frac{1}{4}{\eta }^6}}.$ Like powers collect karo (do $\frac{3}{4}{\eta }^4$ terms add hokar $\frac{3}{2}{\eta }^4$ bante hain): $=\frac{3}{2}\eta -\frac{9}{4}{\eta }^2-\frac{1}{2}{\eta }^3+\frac{3}{2}{\eta }^4-\frac{1}{4}{\eta }^6.$ Yeh step kyun? Momentum ko $u(1-u)$ chahiye; product ko pehle powers mein expand kiye bina integrate nahi kar sakte.

3. Har power integrate karo ${\int }_0^1{\eta }^{n}d\eta =\frac{1}{n+1}$ use karke, phir har term ko common denominator $280$ par laao:

   term	value	$\times 280$
   $+\frac{3}{2}\cdot \frac{1}{2}$	$+\frac{3}{4}$	$+210$
   $-\frac{9}{4}\cdot \frac{1}{3}$	$-\frac{3}{4}$	$-210$
   $-\frac{1}{2}\cdot \frac{1}{4}$	$-\frac{1}{8}$	$-35$
   $+\frac{3}{2}\cdot \frac{1}{5}$	$+\frac{3}{10}$	$+84$
   $-\frac{1}{4}\cdot \frac{1}{7}$	$-\frac{1}{28}$	$-10$

   Last column ka sum: $210-210-35+84-10=39$. Isliye $\theta =\frac{39}{280}\delta \approx 0.139\delta .$ Yeh step kyun? Table har fraction ko ek denominator ($280$) par force karta hai taaki koi term mis-scale na ho — kaafi fractions add karne ka safe tarika hai.

4. Shape factor: $H=\frac{3/8}{39/280}=\frac{3}{8}\cdot \frac{280}{39}=\frac{105}{39}=\frac{35}{13}\approx \mathrm{2.69.}$

Verify: $H\approx 2.69$ true Blasius solution value $2.59$ ke kaafi kareeb hai, linear profile ke $3$ se zyada — as expected, zyada realistic curved profile drag better predict karta hai. Ordering: $0.139\delta <0.375\delta <\delta$ ✓.

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Example 3 — Sine profile (Cell C)

Forecast: sine layer ke zyada hisse mein linear line se upar bulge karta hai (fast rise phir flatten) — figure mein orange curve, dashed $u=U$ line se chipki hui. Toh uska deficit chhota hai — expect karo ${\delta }^{\ast }<\delta /2$.

1. Displacement: ${\delta }^{\ast }=\delta {\int }_0^1\left(1-\sin \frac{\pi \eta }{2}\right)d\eta =\delta {\left[\eta +\frac{2}{\pi }\cos \frac{\pi \eta }{2}\right]}_0^1=\delta \left(1-\frac{2}{\pi }\right)\approx 0.363\delta .$ Yeh step kyun? ${\int }_0^1\sin \frac{\pi \eta }{2}d\eta =[-\frac{2}{\pi }\cos \frac{\pi \eta }{2}{]}_0^1=\frac{2}{\pi }$. Hum yahan ek trig tool use karte hain (polynomials nahi) kyunki profile khud trig hai — $\sin$ ka antiderivative ek jaana-pehchana clean function hai.
2. Momentum: $\frac{u}{U}\left(1-\frac{u}{U}\right)=\sin \frac{\pi \eta }{2}-{\sin }^2\frac{\pi \eta }{2}$. ${\int }_0^1\sin \frac{\pi \eta }{2}d\eta =\frac{2}{\pi }$ aur ${\int }_0^1{\sin }^2\frac{\pi \eta }{2}d\eta =\frac{1}{2}$ use karke: $\theta =\delta \left(\frac{2}{\pi }-\frac{1}{2}\right)\approx 0.137\delta .$ Yeh step kyun? Humne ${\sin }^2$ integrate karne ke liye half-angle identity ${\sin }^{2}x=\frac{1-\cos 2x}{2}$ use ki — isliye hum isko use karte hain, kyunki ${\sin }^2$ ko directly integrate nahi kar sakte.
3. Shape factor: $H=\frac{0.363}{0.137}\approx \mathrm{2.66.}$

Verify: $H\approx 2.66$, phir Blasius $2.59$ ke kareeb ✓. Forecast confirmed: ${\delta }^{\ast }\approx 0.363\delta <0.5\delta$ ✓.

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Example 4 — Degenerate limits: plug flow aur stagnant flow (Cells D, E)

Forecast: koi slowdown nahi toh deficit zero hoga — dono thicknesses vanish ho jayengi. Maximal slowdown ke saath deficit apni ceiling tak pahunchna chahiye: poori layer $\delta$.

1. (a) Plug flow: $1-u/U=0$ everywhere, isliye ${\delta }^{\ast }={\int }_0^{\infty }0dy=0,\theta ={\int }_0^{\infty }0dy=0.$ Yeh step kyun? Agar kuch bhi slow nahi hua, toh koi mass aur koi momentum "missing" nahi hai. Yeh sanity floor hai.
2. (b) Power-law $u/U={\eta }^n$: $\frac{{\delta }^{\ast }}{\delta }={\int }_0^1(1-{\eta }^n)d\eta =1-\frac{1}{n+1}=\frac{n}{n+1}.$ Yeh step kyun? Ek clean integral profiles ki poori family cover karta hai.
3. $n\to \infty$ lo: $\frac{{\delta }^{\ast }}{\delta }=\frac{n}{n+1}\to 1.$ Yeh step kyun? Jab $n\to \infty$, fluid almost sabhi $\eta <1$ ke liye near-zero hota hai phir $U$ par jump karta hai — deficit poori layer ko fill kar leta hai.

Verify: $n=1$ (linear) ke liye yeh $\frac{1}{2}$ deta hai, Example 1 se match karta hai ✓. Limit ${\delta }^{\ast }\to \delta$ theoretical ceiling hai: ${\delta }^{\ast }$ kabhi $\delta$ se exceed nahi kar sakta kyunki $1-u/U\le 1$. Plug flow $0$ deta hai, floor ✓.

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Example 5 — "${\int }_0^{\infty }$" trap (Cell F)

Forecast: $\delta$ se $\infty$ tak ka piece... kya contribute karta hai?

1. Integral ko $y=\delta$ (i.e. $\eta =1$) par split karo: ${\delta }^{\ast }={\int }_0^{\delta }\left(1-\frac{u}{U}\right)dy+{\int }_{\delta }^{\infty }\left(1-\frac{u}{U}\right)dy.$ Yeh step kyun? Profile piecewise defined hai, toh jahan pieces milte hain wahan split karo.
2. $\delta$ se aage, $u=U$ hai, toh $1-u/U=0$ aur doosra integral $0$ hai. Yeh step kyun? Yahi poora resolution hai — tail kuch contribute nahi karta. Isliye ${\int }_0^{\infty }$ aur ${\int }_0^{\delta }$ kisi bhi profile ke liye agree karte hain jo $\delta$ par $U$ reach karti hai.
3. Finite part compute karo $u/U=2\eta -{\eta }^2$ ke saath, toh $1-u/U=1-2\eta +{\eta }^2=(1-\eta {)}^2$: ${\delta }^{\ast }=\delta {\int }_0^1(1-\eta {)}^{2}d\eta =\delta {\left[-\frac{(1-\eta {)}^3}{3}\right]}_0^1=\frac{\delta }{3}.$

Verify: $(1-\eta {)}^2\ge 0$ toh ${\delta }^{\ast }>0$ ✓. Value $\delta /3\approx 0.333\delta <\delta$ ✓. Sanity: yeh parabola linear line se upar hai, $\delta /2$ se chhota deficit, aur indeed $\frac{1}{3}<\frac{1}{2}$ ✓.

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Example 6 — Real-world word problem (Cell G)

Forecast: everyday objects par boundary layers surprising tarike se patlii hoti hain — $\delta$ ko mm mein guess karo.

1. Reynolds number: $R{e}_x=\frac{Ux}{\nu }=\frac{30\times 0.5}{1.5\times 10^{-5}}=10^6.$ Yeh step kyun? $R{e}_x$ humein batata hai flow laminar hai (transition band se kaafi neeche) aur yeh growth law drive karta hai.
2. Boundary layer thickness (Blasius solution growth law): $\delta =\frac{5x}{\sqrt{R{e}_x}}=\frac{5\times 0.5}{\sqrt{10^6}}=\frac{2.5}{1000}=2.5\times 10^{-3}\text{m}=2.5\text{mm}.$ Yeh step kyun? $5/\sqrt{R{e}_x}$ woh fraction hai $x$ ka jo slowed fluid occupy karta hai.
3. Ek profile model choose karo. Humne sirf $\delta$ measure kiya, poora velocity field $u(y)$ nahi. ${\delta }^{\ast }$ aur $\theta$ paane ke liye humein ek profile shape assume karni hogi; sabse simple Example 1 ki linear wali hai, jisne fixed ratios ${\delta }^{\ast }=\delta /2$ aur $\theta =\delta /6$ diye the. Yeh step kyun? ${\delta }^{\ast }$ aur $\theta$ $u(y)$ ke integrals hain — profile jaane bina woh undefined hain, toh ek model profile mandatory hai. Linear model sabse crude estimate hai lekin correct scale dikhata hai.
4. Displacement & momentum un ratios se: ${\delta }^{\ast }=\frac{\delta }{2}=1.25\text{mm},\theta =\frac{\delta }{6}=0.417\text{mm}.$
5. Shape factor — linear profile ke liye yeh ek pure number hai jo physical size par depend nahi karta: $H=\frac{{\delta }^{\ast }}{\theta }=\frac{1.25}{0.417}=3.$ Yeh step kyun? $H$ sirf profile shape par depend karta hai, $\delta$ par nahi, isliye yeh Example 1 ki value $3$ ke barabar hai.

Verify: Units: $\frac{(\text{m/s})(\text{m})}{{\text{m}}^2/\text{s}}=$ dimensionless ✓ $R{e}_x$ ke liye. $\delta =2.5\text{mm}$ — patla, as forecast ✓. Ordering $0.417<1.25<2.5\text{mm}$ ✓, aur $H=3$ Example 1 se match karta hai ✓.

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Example 7 — Downstream scaling (Cell H)

Forecast: zyada wall matlab zyada slowed fluid — teeno badhenge. Lekin kitna? $x$ ke saath linearly? Ya slower?

1. Growth law: $\delta =\frac{5x}{\sqrt{R{e}_x}}$ with $R{e}_x=\frac{Ux}{\nu }$, toh $\sqrt{R{e}_x}\propto \sqrt{x}$ aur $\delta \propto \frac{x}{\sqrt{x}}=\sqrt{x}.$ Yeh step kyun? $R{e}_x$ ko law mein substitute karo — numerator mein $x$, neeche $\sqrt{x}$ se partly cancel ho jaata hai.
2. Kyunki ${\delta }^{\ast }$ aur $\theta$ $\delta$ ke fixed fractions hain (self-similar laminar profile ke liye), woh bhi $\sqrt{x}$ ki tarah scale karte hain. Yeh step kyun? Self-similarity ka matlab hai profile shape $x$ ke saath unchanged rehti hai; sirf $\delta$ stretch karta hai, toh $H$ constant rehta hai jabki teeno lengths saath badhte hain.
3. 4× distance ke liye factor: $\sqrt{4}=2$. Teeno double ho jaate hain.

Verify: $\delta (2.0)=\frac{5\times 2.0}{\sqrt{30\times 2.0/1.5\times 10^{-5}}}=\frac{10}{\sqrt{4\times 10^6}}=\frac{10}{2000}=5\times 10^{-3}\text{m}=5\text{mm}$. Yeh exactly $2\times 2.5\text{mm}$ hai ✓. Layer $\sqrt{x}$ ki tarah thicken hoti hai, linearly nahi — Skin friction drag ke peeche ek key laminar fact.

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Example 8 — Exam twist: shape factor reverse karo (Cell I)

Forecast: badhta $H$ matlab ${\delta }^{\ast }$ $\theta$ ke relative badh raha hai — profile wall ke paas "emptier" ho rahi hai. Yeh physically kiske pehle aata hai?

1. Shape factor ki definition: $H=\frac{{\delta }^{\ast }}{\theta }\Rightarrow \theta =\frac{{\delta }^{\ast }}{H}.$ Yeh step kyun? Single definition rearrange karna (upar definition box dekho) bas itna hi chahiye — koi integrals nahi.
2. (a) $\theta =\frac{1.8}{2.59}\approx 0.695\text{mm}.$
3. (b) Badhta $H$ aane wale Boundary layer separation ka signal deta hai — near-wall fluid itna slow ho raha hai ki reverse hone wala hai. Naya $\theta =\frac{1.8}{3.5}\approx 0.514\text{mm}.$ Yeh step kyun? Same formula; bada $H$ (${\delta }^{\ast }$ fixed ke saath) matlab chhota $\theta$ — wall ke paas momentum-carrying capacity collapse ho gayi hai.

Verify: $H\theta =2.59\times 0.695\approx 1.80\text{mm}={\delta }^{\ast }$ ✓. Aur $3.5\times 0.514\approx 1.80\text{mm}$ ✓. Dono measured ${\delta }^{\ast }$ reproduce karte hain.

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Example 9 — Kyun $\theta <{\delta }^{\ast }<\delta$ kabhi flip nahi ho sakti (Cell J)

Forecast: dono integrands ko term-by-term compare karo — ek hamesha doosre se neeche rehta hai. Kyun?

1. Dono integrands likho: ${\delta }^{\ast }$ mein $\left(1-\frac{u}{U}\right)$ hai; $\theta$ mein $\frac{u}{U}\left(1-\frac{u}{U}\right)$ hai. Yeh step kyun? Same factor $\left(1-\frac{u}{U}\right)\ge 0$ dono mein aata hai; sirf extra $\frac{u}{U}$ alag hai. Inhe compare karne se poori ordering ek extra factor ko $1$ se compare karne tak reduce ho jaati hai.
2. Extra multiplier $\le 1$ hai: kyunki $0\le u\le U$, hum jaante hain $0\le \frac{u}{U}\le 1$. Ek non-negative quantity ko $[0,1]$ mein kisi cheez se multiply karne par woh sirf shrink ho sakti hai (ya unchanged reh sakti hai): $\frac{u}{U}\left(1-\frac{u}{U}\right)\le \left(1-\frac{u}{U}\right).$ Yeh step kyun? Yeh poori ordering ka core hai — momentum deficit extra mass factor $u/U$ carry karta hai, aur mass fraction kabhi $1$ se exceed nahi karta.
3. Dono sides ko $y:0\to \infty$ par integrate karo. Integration $\le$ sign preserve karta hai, toh $\theta ={\int }_0^{\infty }\frac{u}{U}\left(1-\frac{u}{U}\right)dy\le {\int }_0^{\infty }\left(1-\frac{u}{U}\right)dy={\delta }^{\ast }.$ Yeh step kyun? Agar $f(y)\le g(y)$ everywhere aur dono integrable hain, tab $\int f\le \int g$.
4. Strictness. Step 2 mein equality ke liye $\frac{u}{U}=1$ (ya $\frac{u}{U}=0$) chahiye har height par. Ek real profile strictly $0$ se $U$ tak climb karti hai, toh yeh poori layer mein un extremes par nahi hoti — isliye inequality strict hai: $\theta <{\delta }^{\ast }$.
5. Top end. Isi tarah ${\delta }^{\ast }={\int }_0^{\delta }(1-u/U)dy<{\int }_0^{\delta }1dy=\delta$, kyunki $1-u/U<1$ jahan bhi $u>0$ hai. Chain karne par poori guarantee milti hai: $\boxed{\theta <{\delta }^{\ast }<\delta }$ Yeh single argument isliye hai ki hamare saare pehle examples ne ordering maani — yeh coincidence nahi hai, yeh sirf $0\le u\le U$ se follow karta hai.

Verify: Har pehle example check karo: linear $(\frac{1}{6}<\frac{1}{2}<1)$✓, cubic $(0.139<0.375<1)$✓, sine $(0.137<0.363<1)$✓, parabola ${\delta }^{\ast }=\frac{1}{3}<1$✓ — sab $\theta <{\delta }^{\ast }<\delta$ follow karte hain exactly jaise proof demand karta hai.

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