Exercises — Boundary layer separation — adverse pressure gradient

Chain use karo: area change → speed change (continuity) → pressure change (Bernoulli).

• (a) Nozzle tung hota hai ⇒ $U\uparrow$ ⇒ $p\downarrow$ ⇒ favourable.
• (b) Front half: geometry $U\uparrow$ ko shoulder ki taraf force karti hai ⇒ $p\downarrow$ ⇒ favourable.
• (c) Diffuser chauda hota hai ⇒ $U\downarrow$ ⇒ $p\uparrow$ ⇒ adverse.
• (d) Rear half: $U$ ko rear stagnation point ki taraf girna padta hai ⇒ $p\uparrow$ ⇒ adverse.

Separation point: wall shear zero ho jaati hai, ${\tau }_w=\mu {\frac{\partial u}{\partial y}|}_{y=0}=0,$ jahan $\mu$ dynamic viscosity hai (fluid ki stickiness, upar define ki gayi). Bilkul downstream, ${\frac{\partial u}{\partial y}|}_{y=0}<0$ — near-wall flow reverse ho jaata hai (backflow).

Edge streamline par Bernoulli ($U_1,U_2$ free-stream speed $U$ hain jo do stations par padhi gayi hain): $p_1+\frac{1}{2}\rho U_1^2=p_2+\frac{1}{2}\rho U_2^2\Rightarrow \Delta p=\frac{1}{2}\rho \left(U_1^2-U_2^2\right).$ $\Delta p=\frac{1}{2}(1.2)\left(30^2-18^2\right)=0.6(900-324)=0.6\times 576=345.6\text{Pa}.$ $\Delta p>0$ ⇒ pressure downstream badhti hai ⇒ adverse gradient. Yeh positive $\Delta p$ exactly wahi "uphill" hai jo thaki hui wall fluid ko chadhna padta hai.

Pehle, yeh formula kyun sahi hai? Boundary-layer momentum equation lo aur use wall par bilkul $y=0$ par padho. No-slip along-wall speed $u=0$ aur across-layer speed $v=0$ wahan pin karta hai, toh dono convective terms $u\partial u/\partial x$ aur $v\partial u/\partial y$ vanish ho jaate hain. Jo bachta hai woh pressure push aur viscous curvature ke beech pure balance hai: $0=-\frac{1}{\rho }\frac{dp}{dx}+\nu {\frac{{\partial }^{2}u}{\partial y^2}|}_0\Rightarrow {\frac{{\partial }^{2}u}{\partial y^2}|}_0=\frac{1}{\mu }\frac{dp}{dx},$ $\mu =\rho \nu$ use karke. Toh wall curvature kuch nahi hai sirf pressure gradient ko $1/\mu$ se rescale karna. Ab plug in karo: ${\frac{{\partial }^{2}u}{\partial y^2}|}_0=\frac{1}{\mu }\frac{dp}{dx}=\frac{200}{1.8\times 10^{-5}}\approx 1.11\times 10^7\text{(m/s)}{\text{m}}^{-2}.$ Units ke baare mein: ${\partial }^{2}u/\partial y^2$ "velocity per length per length" hai — ek speed (m/s) ko length (m) se do baar differentiate karne par $\text{m/s}\cdot {\text{m}}^{-2}={\text{s}}^{-1}{\text{m}}^{-1}$ milta hai. Right side se check karo: $\frac{\text{Pa/m}}{\text{Pa⋅s}}=\frac{1}{\text{m⋅s}}$ — same. Dono taraf units match hona ek quiet confirmation hai ki formula dimensionally honest hai. Sign aur meaning: yeh positive hai. Wall se door curvature negative honi chahiye (taaki $u$ smoothly $U$ par bend kare). Sign change matlab inflection point zaroor hoga — adverse gradient aur reversal ki taraf prone profile ki pehchaan.

Differentiate karo. $\frac{\partial u}{\partial y}=U\left(\frac{3}{2\delta }-\frac{3}{2}\frac{y^2}{{\delta }^3}\right).$ $y=0$ par: $\frac{\partial u}{\partial y}{|}_0=\frac{3U}{2\delta }>0$ — positive slope ⇒ forward drag ⇒ attached. ✓ Second derivative: $\frac{{\partial }^{2}u}{\partial y^2}=U\left(-\frac{3y}{{\delta }^3}\right),\frac{{\partial }^{2}u}{\partial y^2}{|}_0=0.$ Wall par zero curvature matlab $\frac{1}{\mu }\frac{dp}{dx}=0$, yaani $dp/dx=0$. ✓ Yahi wajah hai ki yeh cubic standard zero-pressure-gradient profile hai.

Hum $U=1$ kyun set kar sakte hain. Separation condition $\partial u/\partial y{|}_0=0$ hai — ek equation jo zero ke barabar set ki gayi hai. Free-stream speed $U$ poori profile shape par sirf ek overall multiplier hai; poori profile ko $U$ se divide karna (yaani dimensionless $u/U$ ke saath kaam karna) yeh nahi badal sakta ki slope kahan zero hit karta hai. Toh hum safely $U=1$ lekr nondimensionalise kar sakte hain aur $\Lambda$ ko sirf shape se padh sakte hain; jawab kisi bhi $U>0$ ke liye hold karta hai.

$\eta =y/\delta$ lo. Tab $\frac{u}{U}=(2\eta -2{\eta }^3+{\eta }^4)+\frac{\Lambda }{U}\eta (1-\eta {)}^2.$ $\lambda =\Lambda /U$ (shape parameter) likhne par, $u/U$ ka wall slope hai ${\frac{d(u/U)}{d\eta }|}_0.$ Har piece ko $\eta =0$ par differentiate karo:

• $\frac{d}{d\eta }(2\eta -2{\eta }^3+{\eta }^4){|}_0=2.$
• $\eta (1-\eta {)}^2=\eta -2{\eta }^2+{\eta }^3$, derivative $=1-4\eta +3{\eta }^2$, $\eta =0$ par $1$ milta hai.

Toh $u/U$ ka wall slope $2+\lambda$ hai. Separation ke liye zero set karo: $2+\lambda =0\Rightarrow \lambda =\frac{\Lambda }{U}=-2\Rightarrow \boxed{\Lambda =-2U}.$ Negative $\Lambda$ = adverse gradient, exactly jaise expect kiya tha: wall tangent ko zero tak flatten karne ke liye kaafi strong uphill chahiye.

(a) $Re=\frac{\rho U_{\infty }D}{\mu }=\frac{1.2\times 45\times 0.1}{1.8\times 10^{-5}}=\frac{5.4}{1.8\times 10^{-5}}=3.0\times 10^5.$ (b) $Re=3\times 10^5$ transition threshold par bilkul baitha hai; isse neeche kuch bhi ho toh boundary layer separation par abhi bhi laminar hai ⇒ woh jaldi chhod deti hai, near $\theta \approx 82°$ par, ek wide low-pressure wake banati hai. (Is $Re$ se thoda aage push karo aur layer turbulent trip karti hai aur release $\approx 120°$ par jump karta hai — "drag crisis".) (c) Wide wake matlab cylinder ka rear low pressure mein baitha hai jabki front high pressure dekhta hai; woh front-back pressure difference (unbalanced kyunki wake kabhi pressure recover nahi karta) hi form (pressure) drag hai — dekho Drag — form vs skin friction.

Do competing drags (from Drag — form vs skin friction):

1. Skin friction — turbulent layer ke liye zyada (steeper wall slope, larger ${\tau }_w$).
2. Form (pressure) drag — wake width se set hoti hai.

Ek turbulent layer high-momentum outer fluid ko wall tak mix karti hai, toh near-wall fluid energy-rich hota hai aur ${\tau }_w\to 0$ se pehle zyada pressure hill chadh sakta hai. Separation $\approx 82°$ se $\approx 120°$ tak move karta hai, wake dramatically chhota ho jaata hai. Bluff body par (cylinder, sphere) form drag dominate karta hai, toh form drag mein badi drop skin friction mein choti rise se zyada hoti hai ⇒ net drag girta hai. Yahi wajah hai ki dimples golf ball ki madad karte hain — dekho Flow over a cylinder and sphere.

Derivation. 2-D steady boundary-layer $x$-momentum equation se shuru karo, jahan $u$ along-wall speed hai aur $v$ across-layer (wall-normal) speed hai jaise is page ke upar define ki gayi: $u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=-\frac{1}{\rho }\frac{dp}{dx}+\nu \frac{{\partial }^{2}u}{\partial y^2}.$ Wall $y=0$ par evaluate karo. No-slip along-wall speed $u=0$ deta hai, aur no-penetration (fluid solid ke through nahi ja sakta) wall-normal speed $v=0$ deta hai wahan, toh left-hand convective terms dono vanish ho jaate hain: $0=-\frac{1}{\rho }\frac{dp}{dx}+\nu {\frac{{\partial }^{2}u}{\partial y^2}|}_0.$ $\rho$ se multiply karo aur $\mu =\rho \nu$ use karo: $\mu {\frac{{\partial }^{2}u}{\partial y^2}|}_0=\frac{dp}{dx}.■$

No-separation argument. Maano $dp/dx\le 0$ (favourable ya zero) poore region mein hai.

• Wall par curvature ${\frac{{\partial }^{2}u}{\partial y^2}|}_0=\frac{1}{\mu }\frac{dp}{dx}\le 0$.
• Kyunki $u$ ko wall par $0$ se layer ke edge par $U>0$ tak badhna hai, yeh shuru hota hai positive slope $\partial u/\partial y{|}_0>0$ se.
• Kyunki kahin bhi adverse stretch nahi hai, curvature poori layer mein $\le 0$ rehti hai — kahin bhi positive nahi hoti, toh koi inflection point nahi ban sakta. Inflection ke bina profile "full" rehti hai: slope $\partial u/\partial y$ uski positive wall value se edge par zero tak monotonically decrease karta hai, lekin wall par khud kabhi nahi.
• Separation ke liye wall slope ko $\partial u/\partial y{|}_0=0$ tak drive hona chahiye. Jo cheez wall slope ko zero ki taraf neeche moda sakti hai woh sirf positive wall curvature hai (adverse gradient), jo hum ne exclude kar diya hai.
• Isliye ${\frac{\partial u}{\partial y}|}_0$ region mein har jagah strictly positive rehta hai, toh ${\tau }_w=\mu \partial u/\partial y{|}_0>0$ aur separation nahi ho sakti. $■$

Yeh parent note ke claim ka rigorous form hai: viscosity near-wall slow layer create karti hai, lekin sirf adverse pressure gradient ($dp/dx>0$) wall slope ko zero tak drive karne aur flow ko peel off karne ka switch ban sakta hai.

Area ratio $A_2/A_1=4\Rightarrow (r_2/r_1{)}^2=4\Rightarrow r_2=2r_1=40\text{mm}$. Radius $\Delta r=r_2-r_1=20\text{mm}$ se $L$ length par badhta hai. Cone half-angle satisfy karta hai $\tan \alpha =\frac{\Delta r}{L}\Rightarrow L=\frac{\Delta r}{\tan \alpha }.$ Kyun $\alpha =7°$ sabse choti allowed length deta hai. Jab $\alpha$ ghatta hai, $\tan \alpha$ ghatta hai, toh $L=\Delta r/\tan \alpha$ badhta hai — gentler cones longer hote hain. Constraint $\alpha \le 7°$ isliye steepness cap karti hai, aur sabse steep allowed cone ($\alpha =7°$) sabse chota allowed diffuser hai. $\alpha =7°$ set karne par woh shortest length milti hai: $L_{\min }=\frac{20\text{mm}}{\tan 7°}=\frac{20}{0.12278}\approx 162.9\text{mm}\approx 0.163\text{m}.$ $\approx 163\text{mm}$ se koi bhi chota $\alpha >7°$ force karega, stronger adverse gradient, aur ek "stalled" diffuser ka risk — dekho Diffusers and nozzles.

risk ko actual separation mein badal deti hai? Inflection point ($dp/dx>0$ se) risk signal karta hai; separation actually tab hoti hai jab wall slope zero ho jaata hai, $\partial u/\partial y{|}_0=0$.

$\tan \alpha$ kyun use karta hai? Cone ke right triangle par, radius growth $\Delta r$ opposite side hai aur length $L$ adjacent side hai, toh $\tan \alpha =\text{opp}/\text{adj}=\Delta r/L$.

$\Lambda$ nikaalte waqt hum $U=1$ set karne ki permission kyun rakhte hain? Kyunki separation condition ek equation hai jo zero par set ki gayi hai aur $U$ profile par sirf ek overall multiplier hai; $U$ se scale karna wahan nahi badal sakta jahan wall slope vanish karta hai.

$dp/dx$ ka sign: jab har jagah $dp/dx\le 0$ ho toh layer attached rehti hai; separation ke liye adverse stretch $dp/dx>0$ chahiye jo wall slope ko zero tak drive kare.