Worked examples — Boundary layer separation — adverse pressure gradient

2.2.23 · D3 · Physics › Fluid Mechanics › Boundary layer separation — adverse pressure gradient

Ye page parent topic ka problem gym hai. Kuch bhi calculate karne se pehle, hum har tarah ki situation ko lay out karte hain jo separation ki physics aap par throw kar sakti hai, phir har cell ke liye ek example work karte hain. Yahan kuch bhi decoration nahi hai — har example ek specific hole fill karta hai taaki jab aap koi naya problem mile, toh aapne uski shape pehle se dekhi ho.

Jo kuch bhi hum use karte hain wo parent note mein build kiya gaya tha. Do tools jo hum constantly use karte hain:

• Wall-curvature law ${\frac{{\partial }^{2}u}{\partial y^2}|}_0=\frac{1}{\mu }\frac{dp}{dx}$ — pressure gradient set karta hai ki velocity profile bilkul wall par kaise bend karti hai.
• Separation condition ${\tau }_w=\mu {\frac{\partial u}{\partial y}|}_0=0$ — wall shear ($u$ ka near-wall slope) zero ho jaata hai.

Agar koi symbol unfamiliar lage, toh wo parent mein define hai — lekin hum important wale ko aage re-anchor karte jaayenge.

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The scenario matrix

Hume har sign, zero/degenerate cases, limits, ek word problem, aur ek exam twist cover karni hogi. Yahan pura grid hai; neeche har example us cell ke saath tagged hai jo wo fill karta hai.

Cell	Case class	Kya special hai	Example
A	$dp/dx<0$ (favourable)	Flow accelerate ho raha — separate HO NAHI SAKTA	Ex 1
B	$dp/dx=0$ (zero gradient, flat plate)	Degenerate boundary case	Ex 2
C	$dp/dx>0$ (adverse), find where ${\tau }_w=0$	Core computation	Ex 3
D	Sign $x$ ke saath badlta hai (favourable → adverse)	Switch locate karo, phir separation	Ex 4
E	Laminar vs turbulent, same geometry	Energy-rich wall fluid separation delay karta hai	Ex 5
F	Limiting input: gradient $\to 0^{+}$ / bahut bada	Separation infinity tak push / turant	Ex 6
G	Real-world word problem (diffuser design)	Geometry → $dp/dx$ → attached rakho	Ex 7
H	Exam twist: profile diya, inflection & separation test karo	Polynomial se curvature directly padho	Ex 8

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Ex 1 — Cell A: favourable gradient kabhi separate nahi karta

Forecast: Aage padhne se pehle curvature ka sign guess karo — upar ya neeche?

1. Bernoulli se $dp/dx$ nikalo. $p+\frac{1}{2}\rho U^2=\text{const}$ ko differentiate karo: $\frac{dp}{dx}=-\rho U\frac{dU}{dx}.$ Ye step kyun? Outer flow $p(x)$ impose karta hai; Bernoulli ek jaane-maane $U(x)$ ko us pressure gradient mein convert karta hai jo wall feel karti hai.

2. $x=0.5$ par numbers plug karo. Yahan $U=10+40(0.5)=30\text{m/s}$ aur $dU/dx=40{\text{s}}^{-1}$: $\frac{dp}{dx}=-(1.2)(30)(40)=-1440\text{Pa/m}.$ Ye step kyun? Negative $\Rightarrow$ favourable har jagah (kyunki $U$ sirf badhta hai), toh separation possible hi nahi.

3. Wall curvature. Curvature law se, ${\frac{{\partial }^{2}u}{\partial y^2}|}_0=\frac{1}{\mu }\frac{dp}{dx}=\frac{-1440}{1.8\times 10^{-5}}=-8.0\times 10^7{\text{s}}^{-1}{\text{m}}^{-1}.$ Ye step kyun? Negative curvature wall par matlab ek "fat" profile hai jisme strong forward slope hai — stable case, koi inflection point nahi.

Verify: $dp/dx$ ki units: $[\rho ][U][dU/dx]=\frac{\text{kg}}{{\text{m}}^3}\cdot \frac{\text{m}}{\text{s}}\cdot \frac{1}{\text{s}}=\frac{\text{kg}}{{\text{m}}^2{\text{s}}^2}=\text{Pa/m}$ ✓. Sign negative = favourable ✓.

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Ex 2 — Cell B: zero-gradient flat plate (degenerate case)

Forecast: Agar wall par curvature zero hai, toh kya profile wahan seedha hai — ya sirf momentarily flat?

1. Pressure gradient. $U$ constant $\Rightarrow dU/dx=0\Rightarrow dp/dx=-\rho U\cdot 0=0$. Ye step kyun? Ye favourable aur adverse ke beech exact boundary hai — hamare matrix ki degenerate seam.

2. Wall curvature. ${\frac{{\partial }^{2}u}{\partial y^2}|}_0=\frac{1}{\mu }\cdot 0=0$. Ye step kyun? Profile mein wall par zero curvature hai — ye wall se seedhe line ke roop mein nikalta hai $U_{\infty }$ se milने से pehle upar jaake (curvature higher up negative hai).

3. Separation? $dp/dx=0$ ke saath wall fluid ko rokne ke liye koi rising pressure nahi, toh ${\tau }_w>0$ sabhi $x$ ke liye: koi separation nahi, kabhi bhi, flat plate par. Ye step kyun? Ye reference case hai (Blasius layer); separation ke liye kahin $dp/dx>0$ chahiye.

Verify: $dp/dx=0$, curvature $=0$, ${\tau }_w>0$ — consistent flat-plate Blasius behaviour ✓.

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Ex 3 — Cell C: adverse gradient, find where ${\tau }_w=0$

Yahan hum ek model velocity profile use karte hain taaki hum literally separation point solve kar sakein. Pehle hume do symbols anchor karne hain jo is example ko chahiye.

Ek standard textbook profile family (cubic Pohlhausen profile) hai

$\frac{u}{U}=\frac{3}{2}\eta -\frac{1}{2}{\eta }^3+\Lambda \frac{\eta }{4}(1-\eta {)}^2,\eta =\frac{y}{\delta },$

jahan $\Lambda =\frac{{\delta }^2}{\nu }\frac{dU}{dx}$ Pohlhausen pressure-gradient parameter hai — ek dimensionless dial: $\Lambda >0$ favourable, $\Lambda <0$ adverse.

Forecast: Separation matlab ek slope zero ho raha hai. Guess karo: critical $\Lambda$ ek chota negative number hai ya bada?

1. Profile ko wall par differentiate karo — term by term dikhaya. Hume $\frac{\partial u}{\partial y}$ chahiye $\eta =0$ par. Kyunki $\eta =y/\delta$, chain rule deta hai $\frac{\partial }{\partial y}=\frac{1}{\delta }\frac{\partial }{\partial \eta }$. $u/U$ ke har piece ko $\eta$ ke respect mein differentiate karo:

   \frac{d}{d\eta}\!\left(-\frac12\eta^3\right) = -\frac32\eta^2,\qquad \frac{d}{d\eta}\!\left(\Lambda\frac{\eta}{4}(1-\eta)^2\right) = \frac{\Lambda}{4}\big[(1-\eta)^2 - 2\eta(1-\eta)\big].$$ Ab $\eta=0$ set karo: middle term $-\tfrac32\eta^2 \to 0$, aur last bracket $(1-0)^2 - 0 = 1$, toh ye $\Lambda/4$ contribute karta hai. Hence $$\frac{1}{U}\frac{\partial u}{\partial y}\Big|_0 = \frac{1}{\delta}\left(\frac32 + \frac{\Lambda}{4}\right).$$ *Ye step kyun?* $\tau_w = \mu\,\partial u/\partial y|_0 \propto$ ye slope; separation exactly wahan hai jahan ye vanish ho. Har term dikhane se clear hota hai ki clean $\tfrac32 + \tfrac{\Lambda}{4}$ kahaan se aata hai — $\eta^3$ term wall par die kar jaata hai, toh sirf linear part aur $\Lambda$-term ka leading factor survive karta hai.

2. Bracket ko zero set karo. $\frac{3}{2}+\frac{\Lambda }{4}=0\Rightarrow \Lambda =-6.$ Ye step kyun? ${\tau }_w=0$ ⇔ near-wall slope $=0$ ⇔ ye bracket $=0$. Toh separation $\Lambda =-6$ par predicted hai.

3. Interpret karo. $\Lambda =-6$ ek strong adverse gradient hai. $\Lambda <-6$ ke liye bracket negative hai ⇒ wall par backflow. Ye step kyun? Ye direction confirm karta hai: zyada negative $\Lambda$ (zyada adverse) $\Rightarrow$ reversed near-wall flow, exactly parent ki picture.

Neeche figure actual profile $u/U$ ko height $\eta$ ke against teen $\Lambda$ values ke liye plot karta hai, taaki aap dekh sako wall slope $\Lambda =-6$ par vertical collapse ho raha aur usse aage backflow mein tip kar raha.

Figure ko wall ($\eta =0$) par left-to-right padho: cyan curve ($\Lambda =-2$, attached) wall se daayein jhuk ke nikalti hai (positive slope, $u>0$); amber curve ($\Lambda =-6$) wall se vertically nikalti hai — wo vertical launch ${\tau }_w=0$ hai, separation point; white curve ($\Lambda =-9$) wall ke paas actually $u<0$ tak dip karti hai — reversed backflow.

Verify: $\Lambda =-6$ par: $\frac{3}{2}+(-6)/4=1.5-1.5=0$ ✓. $\Lambda =-4$ par: $1.5-1=0.5>0$ (attached); $\Lambda =-8$ par: $1.5-2=-0.5<0$ (reversed) ✓.

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Ex 4 — Cell D: gradient surface ke saath sign change karta hai

Forecast: Switch wahan hota hai jahan $U$ peak karta hai. Guess karo $x$ jahan flow sabse tez hai.

1. $dU/dx=0$ kahaan hai find karo (velocity peak). $\frac{dU}{dx}=12-12x=0\Rightarrow x=1\text{m}.$ Ye step kyun? $dp/dx=-\rho UdU/dx$, toh pressure gradient sign exactly wahan flip karta hai jahan $dU/dx$ sign flip karta hai — velocity maximum par.

2. Har side classify karo. $x<1$ ke liye: $dU/dx>0$ (accelerating) $\Rightarrow dp/dx<0$ favourable. $x>1$ ke liye: $dU/dx<0$ (decelerating) $\Rightarrow dp/dx>0$ adverse. Ye step kyun? Ye Flow over a cylinder and sphere ko mirror karta hai: front par attach, shoulder ke peeche adverse aur separation-prone.

3. Peak par velocity ka sanity check. $U(1)=20+12-6=26\text{m/s}$ — maximum, toh $p$ wahan minimum hai. Ye step kyun? Fastest point par lowest pressure exactly Bernoulli hai (Bernoulli's equation).

Verify: $dU/dx{|}_{x=1}=12-12=0$ ✓. $U(1)=26$, aur $U(0.5)=20+6-1.5=24.5$, $U(1.5)=20+18-13.5=24.5$ — peak $26$ ke baare mein symmetric, toh $x=1$ sach mein maximum hai ✓.

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Ex 5 — Cell E: same cylinder par laminar vs turbulent

Forecast: Turbulent baad mein separate hota hai (bada $\theta$). Kya bada angle yahan wider ya narrower wake matlab hai?

1. Transverse-extent proxy $w\propto \sin {\theta }_{\text{sep}}$. Ye measure karta hai ki flow centreline se kitna upar peel away karta hai. $\sin 82°=0.9903,\sin 120°=\mathrm{0.8660.}$ Ye step kyun? Baad mein separation aur peeche wrap karta hai, toh leaving point rear centreline ke closer hota hai — is measure se narrower wake. Drag — form vs skin friction dekho.

2. Fractional narrowing (proxy i). $\frac{w_L-w_T}{w_L}=\frac{0.9903-0.8660}{0.9903}=0.1256\approx 12.6\%.$ Ye step kyun? Narrower wake $\Rightarrow$ smaller low-pressure base $\Rightarrow$ less form drag — golf-ball dimple effect (Reynolds number & transition govern karta hai ki tripping kab hota hai).

3. Base-chord proxy $c\propto (1+\cos {\theta }_{\text{sep}})$, track karta hai ki rear surface kitna wetted rehta hai. Ek separation point aur peeche ($\theta$ bada, $180°$ ki taraf) ek shorter exposed base chhodta hai: $1+\cos 82°=1.1392,1+\cos 120°=0.5000,\frac{1.1392-0.5000}{1.1392}=\mathrm{0.561.}$ Ye step kyun? Ye doosra measure direction mein agree karta hai (turbulent → smaller base → less drag) lekin bahut zyada sensitive hai ($\sim 56\%$ drop). Golf balls par real drag reductions in crude proxies ke beech hoti hain — point ye hai ki dono geometric measures ek hi baat kehte hain: baad mein separation wake shrink karta hai.

Verify: $\sin 82°\approx 0.9903$, $\sin 120°=0.8660$, reduction $=0.1256$ ✓. $1+\cos 82°=1.1392$, $1+\cos 120°=0.5$, reduction $=0.561$ ✓.

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Ex 6 — Cell F: limiting inputs (gradient → 0 aur → bada)

Forecast: Gentle uphill — kya thaki hui fluid kabhi rukti hai, ya bas forever lag jaata hai?

1. Weak-gradient limit. Agar $dU/dx\to 0^{-}$, toh $\Lambda \to 0^{-}$, jo $-6$ ke paas kahin nahi. Layer ko $\delta$ enormously grow karna hoga (kyunki $\Lambda \propto {\delta }^2$) $\Lambda =-6$ tak pahunchne se pehle. Ye step kyun? Separation ke liye enough deceleration accumulate karna hota hai; adverse gradient ki ek whisper threshold sirf ek bahut lambe run ke baad reach karti hai — separation point infinity ki taraf jaata hai.

2. Quantify karo. $\Lambda =-6$ hit karne ke liye hume chahiye ${\delta }^2=\frac{-6\nu }{dU/dx}$. Jab $dU/dx\to 0^{-}$, $\delta \to \infty$: separation infinity tak push ho jaata hai (practically kabhi nahi). Ye step kyun? Intuition exact ban jaata hai — required thickness diverge hoti hai.

3. Strong-gradient limit. $\Lambda =-12<-6$: wall-slope bracket $\frac{3}{2}+\frac{\Lambda }{4}=1.5-3=-1.5<0$. Ye step kyun? Ek negative wall slope matlab flow already reversed hai — separation upstream occur ho chuki hai; $\Lambda =-6$ paar ho chuka tha.

Verify: Threshold thickness ${\delta }^2=-6\nu /(dU/dx)$, $dU/dx\to 0^{-}$ par diverge hoti hai ✓. $\Lambda =-12$ par bracket: $1.5-3=-1.5<0$ ✓.

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Ex 7 — Cell G: diffuser design word problem

Forecast: Flow slow karne se pressure badhta hai. Guess karo: construction se ye favourable hai ya adverse device?

1. Continuity se exit area ($U_1{A}_1=U_2{A}_2$): $A_2=\frac{U_1{A}_1}{U_2}=\frac{30\times 0.010}{12}=0.025{\text{m}}^2.$ Ye step kyun? Mass conservation (Diffusers and nozzles) fix karta hai ki duct kitna widen hona chahiye — aur widening hi deceleration force karta hai.

2. Bernoulli se pressure rise. $\Delta p=\frac{1}{2}\rho \left(U_1^2-U_2^2\right)=\frac{1}{2}(1.2)(30^2-12^2)=0.6(900-144)=453.6\text{Pa}.$ Ye step kyun? Ek diffuser kinetic energy ko pressure mein convert karta hai — ye pressure rise exactly wahi adverse gradient hai jo boundary layer ko survive karni hoti hai.

3. Separation comment. Kyunki $dp/dx>0$ poore time (adverse by design — flow poore raaste slow hoti hai), boundary layer diffuser mein har jagah threat mein hai. $7°$ cone-angle cap us adverse gradient ko gentle enough rakhta hai (pressure slowly ek lambi length mein badhta hai) ki wall shear ${\tau }_w$ positive rehta hai aur flow attached rehti hai. Cone ko $\sim 7°$ se wider karo aur wahi $453.6\text{Pa}$ rise ek shorter distance mein squeeze ho jaata hai ⇒ steeper $dp/dx$ ⇒ ${\tau }_w$ zero reach karta hai ⇒ diffuser stall kar jaata hai ek large separated region aur heavy pressure-recovery losses ke saath. Ye step kyun? Ye number ko hamare matrix ke switch se tie back karta hai: adverse gradient ek diffuser mein inevitable hai, toh design lever uski steepness hai, sign nahi.

Verify: $A_2=0.025{\text{m}}^2$; $\Delta p=0.6(900-144)=453.6\text{Pa}$ ✓. Units: $\rho U^2=\frac{\text{kg}}{{\text{m}}^3}\frac{{\text{m}}^2}{{\text{s}}^2}=\text{Pa}$ ✓.

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Ex 8 — Cell H: exam twist — diye gaye profile se curvature padho

Forecast: Aap pressure-gradient sign directly $y^2$ coefficient se padh sakte ho. $b$ ka sign → ?

1. Linear term se wall shear. Differentiate karo: $\frac{\partial u}{\partial y}=a+2by+3c{y}^2$, toh $y=0$ par, $\frac{\partial u}{\partial y}{|}_0=a=200{\text{s}}^{-1}>0$. Phir ${\tau }_w=\mu a=(1.8\times 10^{-5})(200)=3.6\times 10^{-3}\text{Pa}>0$. Ye step kyun? $y$ ka coefficient hai wall slope (higher terms sab $y$ ka factor carry karte hain aur wall par vanish ho jaate hain). Positive ⇒ abhi separate nahi hua (separation ko $a=0$ chahiye).

2. $y^2$ term se wall curvature. Phir differentiate karo: $\frac{{\partial }^{2}u}{\partial y^2}=2b+6cy$, toh $y=0$ par, $\frac{{\partial }^{2}u}{\partial y^2}{|}_0=2b=-3.0\times 10^4{\text{s}}^{-1}{\text{m}}^{-1}<0$. Ye step kyun? Wall par curvature $2b$ hai ($y^3$ term $6cy$ contribute karta hai, jo $y=0$ par zero hai). Negative curvature ⇒ favourable profile, wall par koi inflection nahi.

3. Pressure gradient. $\frac{dp}{dx}=\mu \frac{{\partial }^{2}u}{\partial y^2}{|}_0=(1.8\times 10^{-5})(-3.0\times 10^4)=-0.54\text{Pa/m}<0$. Ye step kyun? Curvature law sign directly read karta hai: negative ⇒ favourable, step 2 confirm karta hai. Attached aur stable. ($c$ term profile ko higher up shape karta hai lekin kabhi wall shear ya wall curvature affect nahi karta — ek classic exam trap.)

Verify: ${\tau }_w=\mu a=3.6\times 10^{-3}\text{Pa}>0$ ✓. $2b=-3.0\times 10^4$ ✓. $dp/dx=\mu (2b)=-0.54\text{Pa/m}$ ✓.

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Matrix ka Recap

Recall Kaun se cells NO separation guarantee karte hain, aur kyun?

A (favourable, $dp/dx<0$) aur B (zero, $dp/dx=0$): koi rising pressure hill nahi, toh ${\tau }_w>0$ hamesha.

Recall Cubic Pohlhausen model mein, kis

$\Lambda$ par separation hoti hai? $\Lambda =-6$: wall-slope bracket $\frac{3}{2}+\frac{\Lambda }{4}=0$.

Recall Laminar

$82°$ se turbulent $120°$ separation tak wake-width narrowing (proxy) kya hai? Lagbhag $12.6\%$ $w\propto \sin {\theta }_{\text{sep}}$ use karke (aur aur bhi bada, $\sim 56\%$, base-chord proxy $1+\cos {\theta }_{\text{sep}}$ use karke — dono agree karte hain ki wake shrink hoti hai).