Visual walkthrough — Film cooling — effectiveness, coverage fraction

Here is the whole journey as a map before we start walking:

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Step 1 — Draw the scene: a hot wall and a thin cool film

WHAT. Picture the inside wall of a rocket combustion chamber as a flat floor. Far above it flows the hot core gas — the burning propellant at flame temperature. Right along the floor, injected from a slot, flows a thin ribbon of cool gas (the film). Gas flows left → right; distance along the wall is called $x$.

WHY. Before any algebra we must agree on what physical thing each symbol names. Three temperatures live in this picture, and everything else is built from them:

• $T_{hot}$ — the temperature the wall would feel from the raw hot gas with no film. (Precisely, it is the recovery temperature of the core stream.)
• $T_{cool}$ — the temperature of the fresh film gas right where it leaves the slot.
• $T_{aw}$ — the temperature the wall actually feels: the adiabatic wall temperature, meaning "the wall conducts no heat away, so it just floats to the temperature the gas next to it imposes."

PICTURE.

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Step 2 — Name the one quantity that matters: the coolness gap $\Delta T$

WHAT. Instead of tracking three temperatures we track a single number: how much cooler the wall is than the hot gas. Call it the coolness gap:

$\underset{\text{coolness gap}}{\underbrace{\Delta T}}\equiv \underset{\text{hot gas temp}}{\underbrace{T_{hot}}}-\underset{\text{wall temp now}}{\underbrace{T_{aw}}}$

• $\Delta T$ big → wall is much cooler than the fire → film is doing its job.
• $\Delta T=0$ → wall is as hot as the fire → film is gone.

WHY. The whole point of the film is the difference between wall and fire. A difference is one number, easier to track than two. And — crucially — the physics of mixing depends on this difference, as we will see next.

PICTURE. The vertical green arrow in the figure below is $\Delta T$. Watch it shrink as $x$ grows.

At the slot ($x=0$) the film is pure fresh coolant, so the wall sits at $T_{cool}$ and the gap is the biggest it will ever be:

$\Delta T_0\equiv T_{hot}-T_{cool}(\text{the gap at }x=0).$

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Step 3 — Why the gap shrinks: hot gas entrains into the film

WHAT. As the film flows downstream, the turbulent boundary between film and hot gas is ragged — it pulls (entrains) hot gas into the cool ribbon. (This is the entrainment process.) Each parcel of hot gas that mixes in warms the film, which warms the wall, which shrinks the gap $\Delta T$.

WHY. We need a rate rule: how fast does the gap shrink? Two competing effects set it:

• The driving force is the gap itself: the bigger the temperature difference, the more vigorously heat crosses into the film. This is exactly the convective heat transfer idea — heat flux $\propto h_g\Delta T$, where $h_g$ is the gas-side heat transfer coefficient (units ${\text{W/m}}^2\text{K}$: watts of heat per square metre per degree of gap).
• The resistance is the film's heat capacity: a heavier, higher-specific-heat film soaks up entrained heat with less temperature rise. That capacity per unit wall width is ${\dot{m}}_c{c}_p$, where ${\dot{m}}_c$ is the coolant mass flow per unit width ($\text{kg/(m⋅s)}$) and $c_p$ its specific heat ($\text{J/kgK}$).

PICTURE. Red arrows show hot gas being dragged into the blue film; the film gets paler (warmer) as more arrows land.

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Step 4 — Write the rate rule as an equation

WHAT. Turn the picture of Step 3 into symbols. Over a tiny distance $dx$, the gap changes by $d(\Delta T)$:

$\underset{\begin{array}{c}\text{how fast the gap}\\ \text{changes per metre}\end{array}}{\underbrace{\frac{d(\Delta T)}{dx}}}=-\underset{\begin{array}{c}\text{heating strength}\\ ÷\text{ film capacity}\end{array}}{\underbrace{\frac{h_g}{{\dot{m}}_c{c}_p}}}\underset{\begin{array}{c}\text{current}\\ \text{gap}\end{array}}{\underbrace{\Delta T}}$

WHY each piece.

• The minus sign ($-$): the gap only ever shrinks — hot gas can never make the film colder. Sign check passed.
• $\Delta T$ on the right: the driving force is the gap. Big gap → fast shrink; small gap → slow shrink. When $\Delta T=0$ the rate is $0$ and it stays there — the film cannot un-mix. Good.
• $h_g/({\dot{m}}_c{c}_p)$: the fixed "decay strength." More $h_g$ (fiercer hot gas) → faster decay. More ${\dot{m}}_c{c}_p$ (fatter, colder film) → slower decay. Both match intuition.

Why a differential equation and not just algebra? Because the shrink rate depends on the current gap, which itself keeps changing — a chicken-and-egg loop. Algebra handles fixed numbers; a differential equation is the tool built precisely for "rate depends on the amount currently present." That signature ("rate $\propto$ amount") is the fingerprint of exponential decay.

PICTURE. The slope of the green curve at each point equals $-\frac{h_g}{{\dot{m}}_c{c}_p}$ times its height.

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Step 5 — Solve the ODE: the exponential appears

WHAT. Separate the variables (gap-stuff on the left, distance-stuff on the right) and integrate from the slot ($x=0$, gap $\Delta T_0$) to any station $x$ (gap $\Delta T$):

${\int }_{\Delta T_0}^{\Delta T}\frac{d(\Delta T)}{\Delta T}=-\frac{h_g}{{\dot{m}}_c{c}_p}{\int }_0^{x}dx$

The left integral is $\ln (\Delta T)-\ln (\Delta T_0)=\ln \frac{\Delta T}{\Delta T_0}$; the right is $-\frac{h_{g}x}{{\dot{m}}_c{c}_p}$. So

$\ln \frac{\Delta T}{\Delta T_0}=-\frac{h_{g}x}{{\dot{m}}_c{c}_p}\Rightarrow \boxed{\Delta T=\Delta T_0{e}^{-\frac{h_{g}x}{{\dot{m}}_c{c}_p}}}$

WHY. Integrating "$\frac{d(\text{stuff})}{\text{stuff}}$" always produces a $\ln$; undoing the $\ln$ produces $e$. That is why the exponential is unavoidable here — it is the only function whose rate of change is proportional to itself.

Reading the exponent. The group $\frac{h_{g}x}{{\dot{m}}_c{c}_p}$ is a pure number (a Stanton number-times-length parameter). When it equals $1$, the gap has fallen to $1/e\approx 37\%$; when it equals $4.5$, to about $1\%$.

PICTURE. A textbook exponential decay — steep at first, then a long tail.

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Step 6 — Rescale into effectiveness $\eta$

WHAT. The parent note's effectiveness is just the fraction of the original gap still surviving:

$\eta \equiv \frac{T_{hot}-T_{aw}}{T_{hot}-T_{cool}}=\frac{\Delta T}{\Delta T_0}.$

Divide the boxed result of Step 5 by $\Delta T_0$ and the $\Delta T_0$ cancels:

$\boxed{\eta (x)=\exp \left(-\frac{h_{g}x}{{\dot{m}}_c{c}_p}\right)}$

WHY this rescaling. $\Delta T_0$ changes from engine to engine, but $\eta$ is dimensionless and universal: it always starts at $1$ (full protection) at $x=0$ and always decays toward $0$ (no protection). Every curve collapses onto one shape.

Edge cases — check them all:

• $x=0$: exponent $=0$, $\eta =e^0=1$. Fresh film, wall = coolant. ✓
• $x\to \infty$: exponent $\to +\infty$, $\eta \to 0$. Film fully spent, wall = hot gas. ✓
• ${\dot{m}}_c\to 0$ (no coolant): exponent $\to \infty$ for any $x>0$, so $\eta \to 0$ everywhere — no film means no protection, instantly. ✓
• ${\dot{m}}_c\to \infty$ (infinite coolant): exponent $\to 0$, $\eta \to 1$ everywhere — an infinitely thick film never warms up. ✓
• $h_g\to 0$ (hot gas cannot transfer heat): exponent $\to 0$, $\eta \to 1$ — nothing to warm the film. ✓

Every limiting scenario behaves sensibly, so the formula is trustworthy.

PICTURE. Same curve as Step 5 but with the vertical axis relabelled from $\Delta T$ to $\eta$, running $1\to 0$, and the design threshold ${\eta }_{min}$ drawn as a horizontal line.

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Step 7 — Where does protection run out? The logarithm

WHAT. Engineers pick a minimum acceptable effectiveness ${\eta }_{min}$ (say $0.2$). The film "protects" the wall only up to the station $x_{prot}$ where $\eta$ first drops to ${\eta }_{min}$. Set $\eta (x_{prot})={\eta }_{min}$ and solve:

$\exp \left(-\frac{h_g{x}_{prot}}{{\dot{m}}_c{c}_p}\right)={\eta }_{min}\Rightarrow \frac{h_g{x}_{prot}}{{\dot{m}}_c{c}_p}=\ln \frac{1}{{\eta }_{min}}\Rightarrow x_{prot}=\frac{{\dot{m}}_c{c}_p}{h_g}\ln \frac{1}{{\eta }_{min}}.$

The coverage fraction $f$ is that protected length as a fraction of the chamber length $L$:

$\boxed{f=\frac{x_{prot}}{L}=\frac{{\dot{m}}_c{c}_p}{h_{g}L}\ln \frac{1}{{\eta }_{min}}}$

WHY the logarithm. To undo an exponential you take a logarithm — that is literally what $\ln$ is for. The exponential decay in Step 6 means the log appears the instant we ask "at what $x$ does $\eta$ reach a given value?" Physically: because protection dies exponentially, demanding a higher ${\eta }_{min}$ costs coolant with rapidly diminishing returns — the $\ln (1/{\eta }_{min})$ grows only slowly as ${\eta }_{min}\to 1$.

Degenerate check: if ${\eta }_{min}=1$ then $\ln (1/1)=\ln 1=0$, so $x_{prot}=0$ — demanding perfect protection means the film qualifies only exactly at the slot. Sensible.

PICTURE. The threshold line cuts the decay curve at $x_{prot}$; the shaded strip $[0,x_{prot}]$ is the covered wall.

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Step 8 — Plug in numbers (the parent's Example 3, seen)

WHAT. Take $h_g=1500{\text{W/m}}^2\text{K}$, ${\dot{m}}_c=0.05\text{kg/(m⋅s)}$, $c_p=2000\text{J/kgK}$, $L=0.50$ m, ${\eta }_{min}=0.2$.

$\frac{{\dot{m}}_c{c}_p}{h_g}=\frac{0.05\times 2000}{1500}=\frac{100}{1500}=0.0667\text{m},\ln \frac{1}{0.2}=\ln 5=1.609$ $x_{prot}=0.0667\times 1.609=0.1073\text{m},f=\frac{0.1073}{0.50}=0.215\approx 21\%.$

WHY it matters. One slot covers only ~21% of the wall at $\eta \ge 0.2$. To protect the whole chamber you distribute about $1/f\approx 5$ slots — this is how engineers literally count injection slots.

PICTURE. The bar of the full chamber, with the first slot's covered strip (~21%) shaded, and the four extra slots needed to fill the rest.

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The one-picture summary

Recall Feynman retelling — the whole walkthrough in plain words

You spray a cool gas along the burning-hot wall (Step 1). All that matters is how much cooler the wall is than the fire — call that the gap (Step 2). But the fire keeps sneaking into your cool ribbon and warming it, and it sneaks in faster when the gap is bigger (Step 3). "Change proportional to what's left" is the exact recipe for exponential fade (Steps 4–5). Rescale the gap so it always starts at $1$ and you get effectiveness $\eta =e^{-(\text{stuff})}$ — full at the slot, fading to nothing far downstream (Step 6). To find where the film "gives up," ask when $\eta$ falls to your minimum; undoing the exponential brings in a logarithm, and dividing by the chamber length gives the coverage fraction (Step 7). Numbers show one slot covers only about a fifth of the wall, so you need roughly five slots to protect it all (Step 8). More coolant or a fresh slot = more covered wall.

Recall Rebuild it yourself

Explain why the rate rule has a minus sign ::: The gap can only shrink — hot gas never cools the film — so $d(\Delta T)/dx<0$. Which mathematical tool turns "rate proportional to amount" into a solution, and why? ::: Integrating $d(\Delta T)/\Delta T$ gives a $\ln$, and undoing that gives $e^{-\dots }$ — the exponential is the only function equal to its own decay rate. Why does coverage fraction contain a logarithm? ::: Because $\eta$ decays exponentially; solving $\eta ={\eta }_{min}$ for $x$ requires the inverse of $\exp$, which is $\ln$. What is $\eta$ at $x=0$ and at $x\to \infty$? ::: $1$ (fresh film, wall = coolant) and $0$ (spent film, wall = hot gas).

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Connections

• Parent: Film cooling effectiveness & coverage
• Convective heat transfer coefficient — supplies $h_g$, the decay driver.
• Boundary layer & entrainment — the mixing that shrinks the gap.
• Adiabatic wall temperature & recovery factor — defines $T_{aw}$ and $T_{hot}$.
• Stanton number — the dimensionless group in the exponent.
• Regenerative cooling — the wall-side complement to gas-side film physics.
• Combustion chamber thermal design — where slot-counting lives.