Exercises — Boundary layer separation — adverse pressure gradient

Use the chain: area change → speed change (continuity) → pressure change (Bernoulli).

• (a) Nozzle narrows ⇒ $U\uparrow$ ⇒ $p\downarrow$ ⇒ favourable.
• (b) Front half: geometry forces $U\uparrow$ toward the shoulder ⇒ $p\downarrow$ ⇒ favourable.
• (c) Diffuser widens ⇒ $U\downarrow$ ⇒ $p\uparrow$ ⇒ adverse.
• (d) Rear half: $U$ must fall toward the rear stagnation point ⇒ $p\uparrow$ ⇒ adverse.

Separation point: wall shear vanishes, ${\tau }_w=\mu {\frac{\partial u}{\partial y}|}_{y=0}=0,$ where $\mu$ is the dynamic viscosity (the fluid's stickiness, defined above). Just downstream, ${\frac{\partial u}{\partial y}|}_{y=0}<0$ — the near-wall flow reverses (backflow).

Bernoulli along the edge streamline ($U_1,U_2$ are the free-stream speed $U$ read at the two stations): $p_1+\frac{1}{2}\rho U_1^2=p_2+\frac{1}{2}\rho U_2^2\Rightarrow \Delta p=\frac{1}{2}\rho \left(U_1^2-U_2^2\right).$ $\Delta p=\frac{1}{2}(1.2)\left(30^2-18^2\right)=0.6(900-324)=0.6\times 576=345.6\text{Pa}.$ $\Delta p>0$ ⇒ pressure rises downstream ⇒ adverse gradient. This positive $\Delta p$ is exactly the "uphill" the tired wall fluid must climb.

First, why does this formula hold? Take the boundary-layer momentum equation and read it right at the wall, $y=0$. No-slip pins the along-wall speed $u=0$ and the across-layer speed $v=0$ there, so both convective terms $u\partial u/\partial x$ and $v\partial u/\partial y$ vanish. What survives is a pure balance between the pressure push and the viscous curvature: $0=-\frac{1}{\rho }\frac{dp}{dx}+\nu {\frac{{\partial }^{2}u}{\partial y^2}|}_0\Rightarrow {\frac{{\partial }^{2}u}{\partial y^2}|}_0=\frac{1}{\mu }\frac{dp}{dx},$ using $\mu =\rho \nu$. So the wall curvature is nothing but the pressure gradient rescaled by $1/\mu$. Now plug in: ${\frac{{\partial }^{2}u}{\partial y^2}|}_0=\frac{1}{\mu }\frac{dp}{dx}=\frac{200}{1.8\times 10^{-5}}\approx 1.11\times 10^7\text{(m/s)}{\text{m}}^{-2}.$ On the units: ${\partial }^{2}u/\partial y^2$ is "velocity per length per length" — differentiating a speed (m/s) twice by a length (m) gives $\text{m/s}\cdot {\text{m}}^{-2}={\text{s}}^{-1}{\text{m}}^{-1}$. Check it from the right side: $\frac{\text{Pa/m}}{\text{Pa⋅s}}=\frac{1}{\text{m⋅s}}$ — the same. Units matching both ways is a quiet confirmation the formula is dimensionally honest. Sign & meaning: it is positive. Far from the wall the curvature must be negative (so $u$ bends over smoothly to $U$). A sign change means there must be an inflection point — the tell-tale of an adverse gradient and a profile prone to reversal.

Differentiate. $\frac{\partial u}{\partial y}=U\left(\frac{3}{2\delta }-\frac{3}{2}\frac{y^2}{{\delta }^3}\right).$ At $y=0$: $\frac{\partial u}{\partial y}{|}_0=\frac{3U}{2\delta }>0$ — positive slope ⇒ forward drag ⇒ attached. ✓ Second derivative: $\frac{{\partial }^{2}u}{\partial y^2}=U\left(-\frac{3y}{{\delta }^3}\right),\frac{{\partial }^{2}u}{\partial y^2}{|}_0=0.$ Zero curvature at the wall means $\frac{1}{\mu }\frac{dp}{dx}=0$, i.e. $dp/dx=0$. ✓ This is why this cubic is the standard zero-pressure-gradient profile.

Why we may set $U=1$. The separation condition is $\partial u/\partial y{|}_0=0$ — an equation set equal to zero. The free-stream speed $U$ is just an overall multiplier on the whole profile shape; dividing the entire profile by $U$ (i.e. working with the dimensionless $u/U$) cannot move where the slope hits zero. So we may safely nondimensionalise by taking $U=1$ and read $\Lambda$ off the shape alone; the answer holds for any $U>0$.

Let $\eta =y/\delta$. Then $\frac{u}{U}=(2\eta -2{\eta }^3+{\eta }^4)+\frac{\Lambda }{U}\eta (1-\eta {)}^2.$ Writing $\lambda =\Lambda /U$ (the shape parameter), the wall slope of $u/U$ is ${\frac{d(u/U)}{d\eta }|}_0.$ Differentiate each piece at $\eta =0$:

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

So the wall slope of $u/U$ is $2+\lambda$. Set to zero for separation: $2+\lambda =0\Rightarrow \lambda =\frac{\Lambda }{U}=-2\Rightarrow \boxed{\Lambda =-2U}.$ A negative $\Lambda$ = adverse gradient, exactly as expected: it takes a strong enough uphill to flatten the wall tangent to zero.

(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) $Re=3\times 10^5$ sits right at the transition threshold; for anything below it the boundary layer is still laminar at separation ⇒ it lets go early, near $\theta \approx 82°$, forming a wide low-pressure wake. (Push a little past this $Re$ and the layer trips turbulent and the release jumps back to $\approx 120°$ — the "drag crisis".) (c) The wide wake means the rear of the cylinder sits in low pressure while the front sees high pressure; that front–back pressure difference (unbalanced because the wake never recovers pressure) is the form (pressure) drag — see Drag — form vs skin friction.

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

1. Skin friction — higher for a turbulent layer (steeper wall slope, larger ${\tau }_w$).
2. Form (pressure) drag — set by wake width.

A turbulent layer mixes high-momentum outer fluid down to the wall, so the near-wall fluid is energy-rich and can climb more of the pressure hill before ${\tau }_w\to 0$. Separation moves from $\approx 82°$ to $\approx 120°$, shrinking the wake dramatically. On a bluff body (cylinder, sphere) the form drag dominates, so the big drop in form drag outweighs the small rise in skin friction ⇒ net drag falls. This is exactly why dimples help a golf ball — see Flow over a cylinder and sphere.

Derivation. Start from the 2-D steady boundary-layer $x$-momentum equation, where $u$ is the along-wall speed and $v$ is the across-layer (wall-normal) speed defined at the top of this page: $u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=-\frac{1}{\rho }\frac{dp}{dx}+\nu \frac{{\partial }^{2}u}{\partial y^2}.$ Evaluate at the wall $y=0$. No-slip gives the along-wall speed $u=0$, and no-penetration (fluid cannot pass through the solid) gives the wall-normal speed $v=0$ there, so the left-hand convective terms both vanish: $0=-\frac{1}{\rho }\frac{dp}{dx}+\nu {\frac{{\partial }^{2}u}{\partial y^2}|}_0.$ Multiply by $\rho$ and use $\mu =\rho \nu$: $\mu {\frac{{\partial }^{2}u}{\partial y^2}|}_0=\frac{dp}{dx}.■$

No-separation argument. Suppose $dp/dx\le 0$ (favourable or zero) throughout the region.

• At the wall the curvature ${\frac{{\partial }^{2}u}{\partial y^2}|}_0=\frac{1}{\mu }\frac{dp}{dx}\le 0$.
• Since $u$ must rise from $0$ at the wall to $U>0$ at the layer's edge, it starts with a positive slope $\partial u/\partial y{|}_0>0$.
• Because there is no adverse stretch anywhere, the curvature stays $\le 0$ across the whole layer — nowhere does it become positive, so no inflection point can form. Without an inflection the profile stays "full": the slope $\partial u/\partial y$ decreases monotonically from its positive wall value out to zero at the edge, but never at the wall itself.
• Separation requires the wall slope to be driven all the way down to $\partial u/\partial y{|}_0=0$. The only thing that can bend the wall slope downward toward zero is a positive wall curvature (an adverse gradient), which we have excluded.
• Hence ${\frac{\partial u}{\partial y}|}_0$ stays strictly positive everywhere in the region, so ${\tau }_w=\mu \partial u/\partial y{|}_0>0$ and separation cannot occur. $■$

This is the rigorous form of the parent note's claim: viscosity creates the slow near-wall layer, but only an adverse pressure gradient ($dp/dx>0$) can act as the switch that drives the wall slope to zero and peels the flow off.

Area ratio $A_2/A_1=4\Rightarrow (r_2/r_1{)}^2=4\Rightarrow r_2=2r_1=40\text{mm}$. The radius grows by $\Delta r=r_2-r_1=20\text{mm}$ over length $L$. The cone half-angle satisfies $\tan \alpha =\frac{\Delta r}{L}\Rightarrow L=\frac{\Delta r}{\tan \alpha }.$ Why $\alpha =7°$ gives the shortest allowed length. As $\alpha$ decreases, $\tan \alpha$ decreases, so $L=\Delta r/\tan \alpha$ increases — gentler cones are longer. The constraint $\alpha \le 7°$ therefore caps the steepness, and the steepest allowed cone ($\alpha =7°$) is the shortest allowed diffuser. Setting $\alpha =7°$ gives that shortest length: $L_{\min }=\frac{20\text{mm}}{\tan 7°}=\frac{20}{0.12278}\approx 162.9\text{mm}\approx 0.163\text{m}.$ Any shorter than $\approx 163\text{mm}$ would force $\alpha >7°$, a stronger adverse gradient, and risk a "stalled" diffuser — see Diffusers and nozzles.

risk of separation into actual separation? An inflection point (from $dp/dx>0$) signals risk; separation actually occurs only when the wall slope reaches zero, $\partial u/\partial y{|}_0=0$.

$\tan \alpha$? On the cone's right triangle, the radius growth $\Delta r$ is the opposite side and length $L$ is the adjacent side, so $\tan \alpha =\text{opp}/\text{adj}=\Delta r/L$.

$U=1$ when finding the separation $\Lambda$? Because the separation condition is an equation set to zero and $U$ is only an overall multiplier on the profile; scaling by $U$ cannot move where the wall slope vanishes.

The sign of $dp/dx$: with $dp/dx\le 0$ everywhere the layer stays attached; separation needs an adverse stretch $dp/dx>0$ to drive the wall slope to zero.