Worked examples — Boundary layer separation — adverse pressure gradient

This page is the problem gym for the parent topic. Before we compute anything, we lay out every kind of situation the physics of separation can throw at you, then work an example for each cell. Nothing here is decoration — each example plugs a specific hole so that when you meet a fresh problem, you have already seen its shape.

Everything we use was built in the parent note. Two tools we lean on constantly:

• The wall-curvature law ${\frac{{\partial }^{2}u}{\partial y^2}|}_0=\frac{1}{\mu }\frac{dp}{dx}$ — the pressure gradient sets how the velocity profile bends right at the wall.
• The separation condition ${\tau }_w=\mu {\frac{\partial u}{\partial y}|}_0=0$ — the wall shear (near-wall slope of $u$) hits zero.

If any symbol looks unfamiliar, it is defined in the parent — but we will re-anchor the important ones as we go.

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The scenario matrix

We must cover every sign, the zero/degenerate cases, the limits, a word problem, and an exam twist. Here is the full grid; each example below is tagged with the cell it fills.

Cell	Case class	What is special	Example
A	$dp/dx<0$ (favourable)	Flow accelerating — CANNOT separate	Ex 1
B	$dp/dx=0$ (zero gradient, flat plate)	The degenerate boundary case	Ex 2
C	$dp/dx>0$ (adverse), find where ${\tau }_w=0$	The core computation	Ex 3
D	Sign changes along $x$ (favourable → adverse)	Locate the switch, then separation	Ex 4
E	Laminar vs turbulent, same geometry	Energy-rich wall fluid delays separation	Ex 5
F	Limiting input: gradient $\to 0^{+}$ / very large	Separation pushed to infinity / immediately	Ex 6
G	Real-world word problem (diffuser design)	Translate geometry → $dp/dx$ → keep attached	Ex 7
H	Exam twist: profile given, test for inflection & separation	Read curvature straight off a polynomial	Ex 8

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Ex 1 — Cell A: favourable gradient never separates

Forecast: Guess the sign of the curvature before reading on — up or down?

1. Get $dp/dx$ from Bernoulli. Differentiate $p+\frac{1}{2}\rho U^2=\text{const}$: $\frac{dp}{dx}=-\rho U\frac{dU}{dx}.$ Why this step? The outer flow imposes $p(x)$; Bernoulli converts a known $U(x)$ into the pressure gradient the wall feels.

2. Plug numbers at $x=0.5$. Here $U=10+40(0.5)=30\text{m/s}$ and $dU/dx=40{\text{s}}^{-1}$: $\frac{dp}{dx}=-(1.2)(30)(40)=-1440\text{Pa/m}.$ Why this step? Negative $\Rightarrow$ favourable everywhere (since $U$ only rises), so no separation is even possible.

3. Wall curvature. By the curvature law, ${\frac{{\partial }^{2}u}{\partial y^2}|}_0=\frac{1}{\mu }\frac{dp}{dx}=\frac{-1440}{1.8\times 10^{-5}}=-8.0\times 10^7{\text{s}}^{-1}{\text{m}}^{-1}.$ Why this step? A negative curvature at the wall means a "fat" profile with strong forward slope — the stable case, no inflection point.

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

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

Forecast: If curvature at the wall is zero, is the profile straight there — or just momentarily flat?

1. Pressure gradient. $U$ constant $\Rightarrow dU/dx=0\Rightarrow dp/dx=-\rho U\cdot 0=0$. Why this step? This is the exact boundary between favourable and adverse — the degenerate seam of our matrix.

2. Wall curvature. ${\frac{{\partial }^{2}u}{\partial y^2}|}_0=\frac{1}{\mu }\cdot 0=0$. Why this step? The profile has zero curvature at the wall — it leaves the wall as a straight line before bending over to meet $U_{\infty }$ (curvature negative higher up).

3. Separation? With $dp/dx=0$ there is no rising pressure to stop the wall fluid, so ${\tau }_w>0$ for all $x$: no separation, ever, on a flat plate. Why this step? This is the reference case (Blasius layer); separation requires $dp/dx>0$ somewhere.

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 ${\tau }_w=0$

Here we use a model velocity profile so we can literally solve for the separation point. First we must anchor two symbols this example needs.

A standard textbook profile family (the cubic Pohlhausen profile) is

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

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

Forecast: Separation is a slope going to zero. Guess: is the critical $\Lambda$ a small negative number or a big one?

1. Differentiate the profile at the wall — shown term by term. We need $\frac{\partial u}{\partial y}$ at $\eta =0$. Since $\eta =y/\delta$, the chain rule gives $\frac{\partial }{\partial y}=\frac{1}{\delta }\frac{\partial }{\partial \eta }$. Differentiate each piece of $u/U$ with respect to $\eta$:

   \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].$$ Now set $\eta=0$: the middle term $-\tfrac32\eta^2 \to 0$, and the last bracket $(1-0)^2 - 0 = 1$, so it contributes $\Lambda/4$. Hence $$\frac{1}{U}\frac{\partial u}{\partial y}\Big|_0 = \frac{1}{\delta}\left(\frac32 + \frac{\Lambda}{4}\right).$$ *Why this step?* $\tau_w = \mu\,\partial u/\partial y|_0 \propto$ this slope; separation is exactly where it vanishes. Showing each term makes clear where the clean $\tfrac32 + \tfrac{\Lambda}{4}$ comes from — the $\eta^3$ term dies at the wall, so only the linear part and the $\Lambda$-term's leading factor survive.

2. Set the bracket to zero. $\frac{3}{2}+\frac{\Lambda }{4}=0\Rightarrow \Lambda =-6.$ Why this step? ${\tau }_w=0$ ⇔ near-wall slope $=0$ ⇔ this bracket $=0$. So separation is predicted at $\Lambda =-6$.

3. Interpret. $\Lambda =-6$ is a strong adverse gradient. For $\Lambda <-6$ the bracket is negative ⇒ backflow at the wall. Why this step? It confirms the direction: more negative $\Lambda$ (more adverse) $\Rightarrow$ reversed near-wall flow, exactly the parent's picture.

The figure below plots the actual profile $u/U$ against height $\eta$ for three values of $\Lambda$, so you can see the wall slope collapse to vertical at $\Lambda =-6$ and tip into backflow beyond it.

Read the figure left-to-right at the wall ($\eta =0$): the cyan curve ($\Lambda =-2$, attached) leaves the wall leaning to the right (positive slope, $u>0$); the amber curve ($\Lambda =-6$) leaves the wall vertically — that vertical launch is ${\tau }_w=0$, the separation point; the white curve ($\Lambda =-9$) actually dips to $u<0$ near the wall — the reversed backflow.

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

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Ex 4 — Cell D: gradient changes sign along the surface

Forecast: The switch happens where $U$ peaks. Guess the $x$ where the flow is fastest.

1. Find where $dU/dx=0$ (the velocity peak). $\frac{dU}{dx}=12-12x=0\Rightarrow x=1\text{m}.$ Why this step? $dp/dx=-\rho UdU/dx$, so the pressure gradient flips sign exactly where $dU/dx$ flips sign — at the velocity maximum.

2. Classify each side. For $x<1$: $dU/dx>0$ (accelerating) $\Rightarrow dp/dx<0$ favourable. For $x>1$: $dU/dx<0$ (decelerating) $\Rightarrow dp/dx>0$ adverse. Why this step? This mirrors Flow over a cylinder and sphere: attach on the front, adverse and separation-prone behind the shoulder.

3. Sanity on velocity at the peak. $U(1)=20+12-6=26\text{m/s}$ — the maximum, so $p$ is minimum there. Why this step? Lowest pressure at the fastest point is exactly Bernoulli (Bernoulli's equation).

Verify: $dU/dx{|}_{x=1}=12-12=0$ ✓. $U(1)=26$, and $U(0.5)=20+6-1.5=24.5$, $U(1.5)=20+18-13.5=24.5$ — symmetric about the peak $26$, so $x=1$ is indeed the maximum ✓.

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Ex 5 — Cell E: laminar vs turbulent on the same cylinder

Forecast: Turbulent separates later (bigger $\theta$). Does a bigger angle mean a wider or narrower wake here?

1. Transverse-extent proxy $w\propto \sin {\theta }_{\text{sep}}$. This measures how high off the centreline the flow peels away. $\sin 82°=0.9903,\sin 120°=\mathrm{0.8660.}$ Why this step? Later separation wraps further around the back, so the leaving point is closer to the rear centreline — a narrower wake by this measure. See Drag — form vs skin friction.

2. Fractional narrowing (proxy i). $\frac{w_L-w_T}{w_L}=\frac{0.9903-0.8660}{0.9903}=0.1256\approx 12.6\%.$ Why this step? A narrower wake $\Rightarrow$ smaller low-pressure base $\Rightarrow$ less form drag — the golf-ball dimple effect (Reynolds number & transition governs when tripping happens).

3. Base-chord proxy $c\propto (1+\cos {\theta }_{\text{sep}})$, tracking how much of the rear surface stays wetted. A separation point further around the back ($\theta$ larger, toward $180°$) leaves a shorter exposed base: $1+\cos 82°=1.1392,1+\cos 120°=0.5000,\frac{1.1392-0.5000}{1.1392}=\mathrm{0.561.}$ Why this step? This second measure agrees in direction (turbulent → smaller base → less drag) but is far more sensitive (a $\sim 56\%$ drop). Real drag reductions on golf balls lie between these crude proxies — the point is that both geometric measures say the same thing: later separation shrinks the wake.

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 and → large)

Forecast: A gentle uphill — does the tired fluid ever stop, or does it just take forever?

1. Weak-gradient limit. If $dU/dx\to 0^{-}$, then $\Lambda \to 0^{-}$, which is nowhere near $-6$. The layer would have to grow $\delta$ enormously (since $\Lambda \propto {\delta }^2$) before reaching $\Lambda =-6$. Why this step? Separation requires accumulating enough deceleration; a whisper of adverse gradient only reaches the threshold after a very long run — the separation point recedes toward infinity.

2. Quantify. To hit $\Lambda =-6$ we need ${\delta }^2=\frac{-6\nu }{dU/dx}$. As $dU/dx\to 0^{-}$, $\delta \to \infty$: separation is pushed to infinity (never, practically). Why this step? Makes the intuition exact — the required thickness diverges.

3. Strong-gradient limit. $\Lambda =-12<-6$: the wall-slope bracket $\frac{3}{2}+\frac{\Lambda }{4}=1.5-3=-1.5<0$. Why this step? A negative wall slope means the flow is already reversed — separation has occurred upstream; $\Lambda =-6$ was passed.

Verify: Threshold thickness ${\delta }^2=-6\nu /(dU/dx)$ diverges as $dU/dx\to 0^{-}$ ✓. Bracket at $\Lambda =-12$: $1.5-3=-1.5<0$ ✓.

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Ex 7 — Cell G: diffuser design word problem

Forecast: Slowing the flow raises pressure. Guess: is this a favourable or adverse device by construction?

1. Exit area by continuity ($U_1{A}_1=U_2{A}_2$): $A_2=\frac{U_1{A}_1}{U_2}=\frac{30\times 0.010}{12}=0.025{\text{m}}^2.$ Why this step? Mass conservation (Diffusers and nozzles) fixes how much the duct must widen — and widening is what forces the deceleration.

2. Pressure rise by Bernoulli. $\Delta p=\frac{1}{2}\rho \left(U_1^2-U_2^2\right)=\frac{1}{2}(1.2)(30^2-12^2)=0.6(900-144)=453.6\text{Pa}.$ Why this step? A diffuser converts kinetic energy into pressure — that pressure rise is precisely the adverse gradient the boundary layer must survive.

3. Separation comment. Because $dp/dx>0$ throughout (adverse by design — the flow slows the whole way), the boundary layer is under threat everywhere in the diffuser. The $7°$ cone-angle cap keeps that adverse gradient gentle enough (the pressure rises slowly over a long length) that the wall shear ${\tau }_w$ stays positive and the flow remains attached. Open the cone wider than $\sim 7°$ and the same $453.6\text{Pa}$ rise is squeezed into a shorter distance ⇒ steeper $dp/dx$ ⇒ ${\tau }_w$ reaches zero ⇒ the diffuser stalls with a large separated region and heavy pressure-recovery losses. Why this step? This ties the number back to the switch of our matrix: an adverse gradient is inevitable in a diffuser, so the design lever is its steepness, not its sign.

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 — read curvature off a given profile

Forecast: You can read the pressure-gradient sign straight off the $y^2$ coefficient. Sign of $b$ → ?

1. Wall shear from the linear term. Differentiate: $\frac{\partial u}{\partial y}=a+2by+3c{y}^2$, so at $y=0$, $\frac{\partial u}{\partial y}{|}_0=a=200{\text{s}}^{-1}>0$. Then ${\tau }_w=\mu a=(1.8\times 10^{-5})(200)=3.6\times 10^{-3}\text{Pa}>0$. Why this step? The coefficient of $y$ is the wall slope (the higher terms all carry a factor of $y$ and vanish at the wall). Positive ⇒ not yet separated (separation needs $a=0$).

2. Wall curvature from the $y^2$ term. Differentiate again: $\frac{{\partial }^{2}u}{\partial y^2}=2b+6cy$, so at $y=0$, $\frac{{\partial }^{2}u}{\partial y^2}{|}_0=2b=-3.0\times 10^4{\text{s}}^{-1}{\text{m}}^{-1}<0$. Why this step? The curvature at the wall is $2b$ (the $y^3$ term contributes $6cy$, which is zero at $y=0$). Negative curvature ⇒ favourable profile, no inflection at the wall.

3. Pressure gradient. $\frac{dp}{dx}=\mu \frac{{\partial }^{2}u}{\partial y^2}{|}_0=(1.8\times 10^{-5})(-3.0\times 10^4)=-0.54\text{Pa/m}<0$. Why this step? The curvature law reads the gradient's sign directly: negative ⇒ favourable, confirming step 2. Attached and stable. (The $c$ term shapes the profile higher up but never affects wall shear or wall curvature — a 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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Recap of the matrix

Recall Which cells guarantee NO separation, and why?

A (favourable, $dp/dx<0$) and B (zero, $dp/dx=0$): no rising pressure hill, so ${\tau }_w>0$ always.

Recall In the cubic Pohlhausen model, at what

$\Lambda$ does separation occur? $\Lambda =-6$: the wall-slope bracket $\frac{3}{2}+\frac{\Lambda }{4}=0$.

Recall What is the wake-width narrowing (proxy) from laminar

$82°$ to turbulent $120°$ separation? About $12.6\%$ using $w\propto \sin {\theta }_{\text{sep}}$ (and even larger, $\sim 56\%$, using the base-chord proxy $1+\cos {\theta }_{\text{sep}}$ — both agree the wake shrinks).